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| author | kinitrupti | 2017-05-12 18:40:35 +0530 |
|---|---|---|
| committer | kinitrupti | 2017-05-12 18:40:35 +0530 |
| commit | d36fc3b8f88cc3108ffff6151e376b619b9abb01 (patch) | |
| tree | 9806b0d68a708d2cfc4efc8ae3751423c56b7721 /Engineering_Mechanics,_Schaum_Series_by_McLean | |
| parent | 1b1bb67e9ea912be5c8591523c8b328766e3680f (diff) | |
| download | Python-Textbook-Companions-d36fc3b8f88cc3108ffff6151e376b619b9abb01.tar.gz Python-Textbook-Companions-d36fc3b8f88cc3108ffff6151e376b619b9abb01.tar.bz2 Python-Textbook-Companions-d36fc3b8f88cc3108ffff6151e376b619b9abb01.zip | |
Revised list of TBCs
Diffstat (limited to 'Engineering_Mechanics,_Schaum_Series_by_McLean')
94 files changed, 0 insertions, 79092 deletions
diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter1.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter1.ipynb deleted file mode 100755 index c6f2d00b..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter1.ipynb +++ /dev/null @@ -1,928 +0,0 @@ -{
- "metadata": {
- "name": "chapter1.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 1:Vectors"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-1,Page no: 8"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initialisation of Variables\n",
- "\n",
- "f1=120 #lb\n",
- "f2=100 #lb\n",
- "theta=((60*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=sqrt(120**2+100**2-(2*120*100*cos(theta))) #Applying Thr rule of Cosines\n",
- "alpha1=(((arcsin(120*sin(theta)/111))*180)/pi) #Applying the Law of Sines\n",
- "alpha=alpha1+270 #As the vector lies in the fourth Quadrant by obsrevaton\n",
- "\n",
- "#Results\n",
- "\n",
- "print'The Resultant of The force system is equal to',round(R),\"lb\" #lb\n",
- "print'The Resultant is at',round(alpha),\"degrees\" #degrees\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Resultant of The force system is equal to 111.0 lb\n",
- "The Resultant is at 339.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-2,Page no: 9"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P=100 #lb\n",
- "Q=120 #lb\n",
- "theta=((30*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R_x=Q*cos(theta) #lb\n",
- "R_y=Q*sin(theta)-P #lb\n",
- "R=sqrt(R_x**2+R_y**2) #lb Triangle law\n",
- "Theta_1=((arctan(R_y/R_x))*180)/pi #degrees\n",
- "Theta_R=360+Theta_1 #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The resultant is at',round(Theta_R),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 111.0 lb\n",
- "The resultant is at 339.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-3,Page No: 9"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "# Initialization of variables\n",
- "\n",
- "R=400 # N\n",
- "F2=200 # N\n",
- "Theta1=((120*pi)/180) # radians\n",
- "Theta2=((20*pi)/180) # radians\n",
- "Theta=Theta1-Theta2 # radians\n",
- "\n",
- "# Calculation\n",
- "\n",
- "F=sqrt(R**2+F2**2-(2*R*F2*cos(Theta))) # N.Applying the Rule of Cosine\n",
- "Theta_r=arcsin((400*sin(Theta))/F) #radians Applying the rule of sines\n",
- "Theta_R=(Theta_r*180)/pi\n",
- "\n",
- "# Result\n",
- "\n",
- "print'The resultant of the force system is',round(F),\"N\"\n",
- "print'The Angle between F and 200N force is',round(Theta_R,1),\"degrees\"\n",
- "\n",
- "# Theta_R is off by 0.1 degrees"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 477.0 N\n",
- "The Angle between F and 200N force is 55.6 degrees\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-4, Page No: 9"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "# Initilization of variables\n",
- "\n",
- "F1=280 # N\n",
- "F2=130 # N\n",
- "Theta1=((320*pi)/180) # Radians\n",
- "Theta2=((60*pi)/180) # Radians\n",
- "\n",
- "# Calculations\n",
- "\n",
- "R_x=-F1*cos(Theta1)+F2*cos(Theta2) # N\n",
- "R_y=F1*sin(Theta1)-F2*sin(Theta2) # N\n",
- "R=sqrt(R_x**2+R_y**2) # N Applying Triangle Law\n",
- "ThetaR=arctan(R_y/R_x) # radians\n",
- "Theta_R=360-(ThetaR*180/pi) # degrees\n",
- "\n",
- "# Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"N\"\n",
- "print'The resultant is at',round(Theta_R),\"degrees\"\n",
- "\n",
- "# The answer for R waries from textbook. "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 329.0 N\n",
- "The resultant is at 297.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-5, Page No: 10"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "# Initialization of variables\n",
- "\n",
- "F1=26 #lb\n",
- "F2=39 #lb\n",
- "F3=63 #lb\n",
- "F4=57 #lb\n",
- "T1=((10*pi)/180) #Radians\n",
- "T2=((114*pi)/180) #Radians\n",
- "T3=((183*pi)/180) #radians\n",
- "T4=((261*pi)/180) #radians\n",
- "\n",
- "# Calculations\n",
- "\n",
- "R_x=F1*cos(T1)+F2*cos(T2)+F3*cos(T3)+F4*cos(T4) # lb Resolving vectors\n",
- "R_y=F1*sin(T1)+F2*sin(T2)+F3*sin(T3)+F4*sin(T4) # lb resolving vectors\n",
- "R=sqrt(R_x**2+R_y**2) # lb Applying Triangle Law\n",
- "theta=arctan(R_y/R_x)# radians\n",
- "Theta=(theta*180)/pi # degrees\n",
- "Theta_R=180+Theta\n",
- "\n",
- "# Results\n",
- "\n",
- "print'The Resultant of the force system is',round(R),\"lb\"\n",
- "print'The resultant is at',round(Theta_R),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Resultant of the force system is 65.0 lb\n",
- "The resultant is at 197.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 36
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-6, Page No: 11"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=10 #lb\n",
- "theta1=((60*pi)/180) #radians\n",
- "theta2=((45*pi)/180) #radians\n",
- "theta=theta1-theta2 #radians\n",
- "\n",
- "#Calculation\n",
- "\n",
- "F_OH=F/cos(theta) #lb resolving vectors\n",
- "\n",
- "# Result\n",
- "\n",
- "print'The component of F in the direction of OH is',round(F_OH,2),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The component of F in the direction of OH is 10.35 lb\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-7, Page No: 11"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "weight=80 #kg\n",
- "theta=((20*pi)/180) #radians\n",
- "theta_p=((70*pi)/180) # radians\n",
- "\n",
- "#Calcuations\n",
- "\n",
- "#Part (a)\n",
- "F=weight*9.81 # N\n",
- "R=F*cos(theta) #N\n",
- "#part (b)\n",
- "R_p=F*cos(theta_p) #N\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The normal component is',round(R),\"N\"\n",
- "print'The parallel component is',round(R_p),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The normal component is 737.0 N\n",
- "The parallel component is 268.0 N\n"
- ]
- }
- ],
- "prompt_number": 37
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-8, Page No: 11"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P=235 #N\n",
- "theta=((60*pi)/180) #radians\n",
- "bet=((22*pi)/180) #radians\n",
- "gam=((38*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Part (a)\n",
- "P_h=P*cos(theta) #N\n",
- "P_v=P*sin(theta) #N\n",
- "#Part (b)\n",
- "P_l=P*cos(theta-bet) #N\n",
- "P_p=P*sin(gam) #N\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The horizontal component is',round(P_h,1),\"N\"\n",
- "print'The vertical component is',round(P_v,1),\"N\"\n",
- "print'The component parallel to plane is',round(P_l),\"N\"\n",
- "print'The component perpendicular to the plane is',round(P_p,1),\"N\"\n",
- "\n",
- "#The decimal point accuracy might cause a small discrepancy in the answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The horizontal component is 117.5 N\n",
- "The vertical component is 203.5 N\n",
- "The component parallel to plane is 185.0 N\n",
- "The component perpendicular to the plane is 144.7 N\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-9, Page No: 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=90 #lb\n",
- "theta1=((40*pi)/180) #radians\n",
- "theta2=((30*pi)/180) #radians\n",
- "\n",
- "# Calculations\n",
- "\n",
- "R_x=0 #lb\n",
- "R_y=20 # lb\n",
- "#Taking the sum of forces in the X-Direction\n",
- "P=((F1*cos(theta1))/cos(theta2)) # lb\n",
- "# Taking the sum of the forces in the Y-Direction\n",
- "F=(P*sin(theta2))+(F1*sin(theta1))-20 #lb\n",
- "\n",
- "# Results\n",
- "\n",
- "print'The value of P is',round(P,1),\"lb\"\n",
- "print'The value of F is',round(F,1),\"lb\"\n",
- "\n",
- "# Decimal point error may cause a small discrepancy in the answers."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P is 79.6 lb\n",
- "The value of F is 77.7 lb\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-10, Page No: 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "x=4 # m\n",
- "y=3 # m\n",
- "z=2 # m\n",
- "F=50 #N\n",
- "\n",
- "# Calculations\n",
- "\n",
- "OP=sqrt(x**2+y**2+z**2) #m\n",
- "thetax=(x/OP) #radians\n",
- "thetay=(y/OP) #Radians\n",
- "thetaz=(z/OP) #radians\n",
- "P_x=F*(thetax) #N\n",
- "P_y=F*(thetay) #N\n",
- "P_z=F*(thetaz) #N\n",
- "\n",
- "# Result\n",
- "\n",
- "print'The vector P is',round(P_x,1),\"i +\",round(P_y,1),\"j +\",round(P_z,1),\"k\"\n",
- "\n",
- "# component of i is off by 0.1 units"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The vector P is 37.1 i + 27.9 j + 18.6 k\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-11, Page No: 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "x=2 \n",
- "y=-4\n",
- "z=1\n",
- "F=100 #N\n",
- "\n",
- "#Calculation\n",
- "\n",
- "thetax=x/sqrt(x**2+y**2+z**2) #radians\n",
- "thetay=y/sqrt(x**2+y**2+z**2) #radians\n",
- "thetaz=z/sqrt(x**2+y**2+z**2) #radians\n",
- "P_x=F*thetax #N\n",
- "P_y=F*thetay #N\n",
- "P_z=F*thetaz #N\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The vector P is',round(P_x,1),\"i\",round(P_y,1),\"j +\",round(P_z,1),\"k\"\n",
- "\n",
- "# component off i is off by 0.1 units"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The vector P is 43.6 i -87.3 j + 21.8 k\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-12, Page No: 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Solution\n",
- "\n",
- "print'P X Q=(Pxi+Pyj+Pzk)x(Qxi+Qyj+Qzk)'\n",
- "print' =(PxQx)i x i+(PxQy)i x j +(PxQz)i x k'\n",
- "print'But i x i = j x j = k x k =0; and i x j =k and j x i= -k, etc. Hence'\n",
- "print'P X Q= (PxQy)k-(PxQz)j-(PyQx)k+(PxQz)i+(PzQx)j-(PzQy)i'\n",
- "print'These terms can be grouped as'\n",
- "print'P X Q=(PyQz-PzQy)i+(PzQx-PxQz)j+(PxQy-PyQx)k'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "P X Q=(Pxi+Pyj+Pzk)x(Qxi+Qyj+Qzk)\n",
- " =(PxQx)i x i+(PxQy)i x j +(PxQz)i x k\n",
- "But i x i = j x j = k x k =0; and i x j =k and j x i= -k, etc. Hence\n",
- "P X Q= (PxQy)k-(PxQz)j-(PyQx)k+(PxQz)i+(PzQx)j-(PzQy)i\n",
- "These terms can be grouped as\n",
- "P X Q=(PyQz-PzQy)i+(PzQx-PxQz)j+(PxQy-PyQx)k\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-13,Page No: 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "Fx=2.63 #N\n",
- "Fy=4.28 #N\n",
- "Fz=-5.92 #N\n",
- "\n",
- "#Calculation\n",
- "\n",
- "F=sqrt(Fx**2+Fy**2+Fz**2) #N\n",
- "thetax=((arccos(Fx/F))*180)/pi #degrees\n",
- "thetay=((arccos(Fy/F))*180)/pi #degrees\n",
- "thetaz=((arccos(Fz/F))*180)/pi #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The magnitude of force is',round(F,2),\"N\"\n",
- "print'Thetax',round(thetax,1),\"degrees\"\n",
- "print'Thetay',round(thetay,1),\"degrees\"\n",
- "print'Thetaz',round(thetaz,1),\"degrees\"\n",
- "\n",
- "# Decimal point error may cause a small discrepancy in the answers."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The magnitude of force is 7.76 N\n",
- "Thetax 70.2 degrees\n",
- "Thetay 56.5 degrees\n",
- "Thetaz 139.7 degrees\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-14, Page No: 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P=[4.82, -2.33, 5.47] #N\n",
- "Q=[-2.81,-6.09,1.12 ] #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "M=P[0]*Q[0]+P[1]*Q[1]+P[2]*Q[2] #Nm\n",
- "\n",
- "#Results\n",
- "\n",
- "print'Result is',round(M,2),\"N.m\"\n",
- "\n",
- "# Decimal point error in calculation causes a small discrepancy in the answer."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Result is 6.77 N.m\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-15, Page No: 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "#Initilization of variables\n",
- "\n",
- "x1=2 #units\n",
- "x2=-2 #units\n",
- "y1=3 #units\n",
- "y2=4 #units\n",
- "z1=0 #units\n",
- "z2=6 #units\n",
- "P=np.array([2,3,-1]) #units\n",
- "\n",
- "#Calculations\n",
- "\n",
- "X=sqrt((x2-x1)**2+(y2-y1)**2+(z2-z1)**2) #units\n",
- "eLx=(x2-x1)/X #units\n",
- "eLy=(y2-y1)/X #units\n",
- "eLz=(z2-z1)/X #units\n",
- "Q=np.array([eLx,eLy,eLz]) #units\n",
- "Z=P[0]*Q[0]+P[1]*Q[1]+P[2]*Q[2] # units\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The unit vector is',round(eLx,3),\"i +\",round(eLy,3),\"j +\",round(eLz,3),\"k\"\n",
- "print'The projection of P is',round(Z,2),\"units\"\n",
- "\n",
- "#Note:The final answer for the projection of P is off by 0.1 units\n",
- "#The answer mentioned in the textbook is -1.41\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The unit vector is -0.549 i + 0.137 j + 0.824 k\n",
- "The projection of P is -1.51 units\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-16, Page No: 14"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "#Initilization of variables\n",
- "\n",
- "x1=2 #units\n",
- "x2=5 #units\n",
- "y1=-5 #units\n",
- "y2=2 #units\n",
- "z1=3 #units\n",
- "z2=-4 #units\n",
- "P=np.array([10,-8,14]) #units\n",
- "\n",
- "#Calculations\n",
- "\n",
- "X=sqrt((x2-x1)**2+(y2-y1)**2+(z2-z1)**2) #units\n",
- "eLx=(x2-x1)/X #units\n",
- "eLy=(y2-y1)/X #units\n",
- "eLz=(z2-z1)/X #units\n",
- "Q=np.array([eLx,eLy,eLz]) #units\n",
- "Z=P[0]*Q[0]+P[1]*Q[1]+P[2]*Q[2] #units\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The unit vector is',round(eLx,3),\"i +\",round(eLy,3),\"j\",round(eLz,3),\"k\" \n",
- "print'The projection of P is',round(Z),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The unit vector is 0.29 i + 0.677 j -0.677 k\n",
- "The projection of P is -12.0 lb\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-17, Page No: 14\n"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "Px=2.85 #ft\n",
- "Py=4.67 #ft\n",
- "Pz=-8.09 #ft\n",
- "Qx=28.3 #lb\n",
- "Qy=44.6 #lb\n",
- "Qz=53.3 #lb\n",
- "\n",
- "#Calculations\n",
- "\n",
- "X=(Py*Qz-Pz*Qy) #N.m \n",
- "Y=(Pz*Qx-Px*Qz) #N.m\n",
- "Z=(Px*Qy-Py*Qx) #N.m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The cross product is',round(X),\"i\",round(Y),\"j\",round(Z),\"k lb-ft\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The cross product is 610.0 i -381.0 j -5.0 k lb-ft\n"
- ]
- }
- ],
- "prompt_number": 45
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-18, Page No: 14"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Result\n",
- "#As this is symbolic solution directly print command is being used to give the required output\n",
- "\n",
- "print'The Time derivative is '\n",
- "print'dr/dt=(dx/dt)i+12*y(dy/dt)j-3*(dz/dt)k'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Time derivative is \n",
- "dr/dt=(dx/dt)i+12*y(dy/dt)j-3*(dz/dt)k\n"
- ]
- }
- ],
- "prompt_number": 47
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-19, Page No: 14"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return x**2\n",
- "a=1\n",
- "b=1\n",
- "I=quad(integrand, 1, 3, args=(a,b))\n",
- "\n",
- "def integrand(y, a, b):\n",
- " return 2*y\n",
- "a=1\n",
- "b=1\n",
- "J=quad(integrand, 1, 3, args=(a,b))\n",
- "\n",
- "def integrand(z, a, b):\n",
- " return 1\n",
- "a=1\n",
- "b=1\n",
- "K=quad(integrand, 1, 3, args=(a,b))\n",
- "\n",
- "# Results\n",
- "print'The answer is',round(I[0],2),\"i +\",round(J[0]),\"j -\",round(K[0]),\"k.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The answer is 8.67 i + 8.0 j - 2.0 k.\n"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter10.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter10.ipynb deleted file mode 100755 index 3ce0f857..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter10.ipynb +++ /dev/null @@ -1,426 +0,0 @@ -{
- "metadata": {
- "name": "chapter10.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 10: First Moments and Centroids"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-5, Page no 160"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "r=50 #mm\n",
- "L1=75 #mm\n",
- "L2=pi*r #mm\n",
- "L3=61.2 #mm\n",
- "# as theta1=45 degrees & theta2=60 degrees\n",
- "sintheta1=sqrt(2)**-1\n",
- "costheta1=sqrt(2)**-1\n",
- "sintheta2=sqrt(3)*2**-1\n",
- "costheta2=2**-1\n",
- "\n",
- "#Calculations\n",
- "x_bar=np.array([(L1/2)*costheta1,L1*costheta1+r,L1*costheta1+100+(L3/2)*costheta2]) #mm\n",
- "y_bar=np.array([(L1/2)*sintheta1,L1*sintheta1+(2*r)/pi,(L3/2)*sintheta2]) #mm\n",
- "#Centroid Calculations\n",
- "x=(L1*x_bar[0]+L2*x_bar[1]+L3*x_bar[2])/(L1+L2+L3) #mm\n",
- "y=(L1*y_bar[0]+L2*y_bar[1]+L3*y_bar[2])/(L1+L2+L3) #mm\n",
- "\n",
- "#Result\n",
- "print'The centroid is as follows:'\n",
- "print'x=',round(x,1),\"mm\"\n",
- "print'y=',round(y,1),\"mm\"\n",
- "\n",
- "# The answer may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid is as follows:\n",
- "x= 97.0 mm\n",
- "y= 57.7 mm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-6, Page no 160"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "theta=75 #degrees\n",
- "theta1=30 #degrees\n",
- "sintheta=0.96\n",
- "costheta=0.25\n",
- "sintheta1=2**-1\n",
- "costheta1=sqrt(3)*2**-1\n",
- "alpha=(150*pi)/180 #rad\n",
- "r=1\n",
- "lhor=14 #in\n",
- "\n",
- "#calculations\n",
- "a=((2*r)/alpha)*sintheta #in\n",
- "p=90-theta\n",
- "sinp=0.259\n",
- "y=-a*sinp #in\n",
- "#Length of arc\n",
- "l=r*alpha #in\n",
- "#Slope length calculations\n",
- "DF=7 #in\n",
- "AB=DF #in\n",
- "BC=1 #in\n",
- "BF=BC*costheta1 #in\n",
- "FC=BC*sintheta1 #in\n",
- "DC=DF+FC #in\n",
- "EC=DC/costheta1 #in\n",
- "#Centroid of EC is at G\n",
- "yslope=0.5*EC*sintheta1+BF #in\n",
- "#Y of composite figure\n",
- "Y=((2*l*y)+14*-1+(2*EC*yslope))/(2*l+lhor+2*EC) #in\n",
- "\n",
- "#Result\n",
- "print'The centroid is at Y=',round(Y,2),\"in\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid is at Y= 1.03 in\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-11, Page no 163"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "a=100 #mm\n",
- "b=150 #mm\n",
- "A1=2*10**4 #mm**2\n",
- "A2=5*10**3 #mm**2\n",
- "A3=(pi*(a/2)**2)/2 #mm**2\n",
- "\n",
- "#Calculations\n",
- "x=(A1*a+A2*(133.3)-A3*b)/(A1+A2-A3) #mm\n",
- "y=(A1*a*0.5+A2*(116.66)-A3*((4*a*0.5)/(3*pi)))/(A1+A2-A3) #mm\n",
- "\n",
- "#Result\n",
- "print'The centroidal distances are'\n",
- "print'x=',round(x,1),\"mm\"\n",
- "print'y=',round(y,1),\"mm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroidal distances are\n",
- "x= 98.6 mm\n",
- "y= 71.2 mm\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-16, Page no 166"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "V=np.array([1728*10**3,432*10**3,7.54*10**3])\n",
- "x_bar=np.array([60,140,60]) #mm\n",
- "y_bar=np.array([30,20,30]) #mm\n",
- "\n",
- "#Calculations\n",
- "x=(V[0]*x_bar[0]+V[1]*x_bar[1]+V[2]*x_bar[2])/(V[0]+V[1]+V[2]) #mm\n",
- "y=(V[0]*y_bar[0]+V[1]*y_bar[1]+V[2]*y_bar[2])/(V[0]+V[1]+V[2]) #mm\n",
- "z=120 #mm from symmetry\n",
- "\n",
- "#Result\n",
- "print'The centroid is at'\n",
- "print'x=',round(x,1),\"mm\"\n",
- "print'y=',round(y,1),\"mm\"\n",
- "print'z=',round(z,1),\"mm\"\n",
- "\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid is at\n",
- "x= 75.9 mm\n",
- "y= 28.0 mm\n",
- "z= 120.0 mm\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-17, Page no 166"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# Here tx=30 degrees,ty=45 degrees& tz=60 degrees,, thus\n",
- "sintx=2**-1\n",
- "costx=sqrt(3)*2**-1\n",
- "sinty=sqrt(2)**-1\n",
- "costy=sqrt(2)**-1\n",
- "sintz=sqrt(3)*2**-1\n",
- "costz=2**-1\n",
- "\n",
- "#Calculations\n",
- "V=np.array([10,15,25]) #in**3\n",
- "x_bar=np.array([4,12,24]) #in\n",
- "y_bar=np.array([4*costx,-6*costy,-4*costz])\n",
- "z_bar=np.array([-4*sintx,6*sinty,-4*sintz])\n",
- "#Centroid calculations\n",
- "x=(V[0]*x_bar[0]+V[1]*x_bar[1]+V[2]*x_bar[2])/(V[0]+V[1]+V[2]) #in\n",
- "y=(V[0]*y_bar[0]+V[1]*y_bar[1]+V[2]*y_bar[2])/(V[0]+V[1]+V[2]) #in\n",
- "z=(V[0]*z_bar[0]+V[1]*z_bar[1]+V[2]*z_bar[2])/(V[0]+V[1]+V[2]) #in\n",
- "\n",
- "#Result\n",
- "print'The centroid of three volumes is at'\n",
- "print'x=',round(x,1),\"in\"\n",
- "print'y=',round(y,2),\"in\"\n",
- "print'z=',round(z,2),\"in\"\n",
- "\n",
- "# The ans for x is off by 0.4 in"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid of three volumes is at\n",
- "x= 16.0 in\n",
- "y= -1.58 in\n",
- "z= -0.86 in\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-26, Page no 171"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Part a\n",
- "# Pefer textbook for part a.\n",
- "\n",
- "# Part b\n",
- "\n",
- "# Initilization of variables\n",
- "w=150 #lb/ft**2\n",
- "h=2 #ft height of the load\n",
- "s=8 #ft span\n",
- "b=2 #ft\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return x*(150*(x/4)*2)\n",
- "a=1\n",
- "b=1\n",
- "M=quad(integrand, 0, s, args=(a,b))\n",
- "Rr=M[0]/(2*s) #lb\n",
- "\n",
- "# Results\n",
- "print'The value of M is',round(M[0]),\"lb-ft\"\n",
- "print'The value of Rr is',round(Rr),\"lb\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of M is 12800.0 lb-ft\n",
- "The value of Rr is 800.0 lb\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-27, Page no 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "\n",
- "rho_m=1000 # kg/m**3\n",
- "h=0.3 # m height of hole\n",
- "b=0.6 # m width of hole\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(y, a, b):\n",
- " return y*9.8*rho_m*(1.2-y)*(0.6)\n",
- "a=1\n",
- "b=1\n",
- "I=quad(integrand, 0, h, args=(a,b))\n",
- "B=I[0]/(2*(0.3))\n",
- "\n",
- "# Results\n",
- "print'The value of B is',round(B),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of B is 441.0 N\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-28, Page no 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=62.4 #lb/ft**3\n",
- "h=12 #ft\n",
- "f=105 #lb/ft**3\n",
- "\n",
- "#Calculations\n",
- "p1=l*h #lb/ft**2\n",
- "#Total force on left side\n",
- "#Simplfying the equation we get a three degree equation in d\n",
- "#solving for d\n",
- "p=np.array([3**-1,0,-144,467])\n",
- "r=roots(p)\n",
- "d=r[2] #ft\n",
- "\n",
- "#Result\n",
- "print'The value of d is',round(d,2),\"feet\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of d is 3.33 feet\n"
- ]
- }
- ],
- "prompt_number": 24
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter10_1.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter10_1.ipynb deleted file mode 100755 index 3ce0f857..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter10_1.ipynb +++ /dev/null @@ -1,426 +0,0 @@ -{
- "metadata": {
- "name": "chapter10.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 10: First Moments and Centroids"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-5, Page no 160"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "r=50 #mm\n",
- "L1=75 #mm\n",
- "L2=pi*r #mm\n",
- "L3=61.2 #mm\n",
- "# as theta1=45 degrees & theta2=60 degrees\n",
- "sintheta1=sqrt(2)**-1\n",
- "costheta1=sqrt(2)**-1\n",
- "sintheta2=sqrt(3)*2**-1\n",
- "costheta2=2**-1\n",
- "\n",
- "#Calculations\n",
- "x_bar=np.array([(L1/2)*costheta1,L1*costheta1+r,L1*costheta1+100+(L3/2)*costheta2]) #mm\n",
- "y_bar=np.array([(L1/2)*sintheta1,L1*sintheta1+(2*r)/pi,(L3/2)*sintheta2]) #mm\n",
- "#Centroid Calculations\n",
- "x=(L1*x_bar[0]+L2*x_bar[1]+L3*x_bar[2])/(L1+L2+L3) #mm\n",
- "y=(L1*y_bar[0]+L2*y_bar[1]+L3*y_bar[2])/(L1+L2+L3) #mm\n",
- "\n",
- "#Result\n",
- "print'The centroid is as follows:'\n",
- "print'x=',round(x,1),\"mm\"\n",
- "print'y=',round(y,1),\"mm\"\n",
- "\n",
- "# The answer may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid is as follows:\n",
- "x= 97.0 mm\n",
- "y= 57.7 mm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-6, Page no 160"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "theta=75 #degrees\n",
- "theta1=30 #degrees\n",
- "sintheta=0.96\n",
- "costheta=0.25\n",
- "sintheta1=2**-1\n",
- "costheta1=sqrt(3)*2**-1\n",
- "alpha=(150*pi)/180 #rad\n",
- "r=1\n",
- "lhor=14 #in\n",
- "\n",
- "#calculations\n",
- "a=((2*r)/alpha)*sintheta #in\n",
- "p=90-theta\n",
- "sinp=0.259\n",
- "y=-a*sinp #in\n",
- "#Length of arc\n",
- "l=r*alpha #in\n",
- "#Slope length calculations\n",
- "DF=7 #in\n",
- "AB=DF #in\n",
- "BC=1 #in\n",
- "BF=BC*costheta1 #in\n",
- "FC=BC*sintheta1 #in\n",
- "DC=DF+FC #in\n",
- "EC=DC/costheta1 #in\n",
- "#Centroid of EC is at G\n",
- "yslope=0.5*EC*sintheta1+BF #in\n",
- "#Y of composite figure\n",
- "Y=((2*l*y)+14*-1+(2*EC*yslope))/(2*l+lhor+2*EC) #in\n",
- "\n",
- "#Result\n",
- "print'The centroid is at Y=',round(Y,2),\"in\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid is at Y= 1.03 in\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-11, Page no 163"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "a=100 #mm\n",
- "b=150 #mm\n",
- "A1=2*10**4 #mm**2\n",
- "A2=5*10**3 #mm**2\n",
- "A3=(pi*(a/2)**2)/2 #mm**2\n",
- "\n",
- "#Calculations\n",
- "x=(A1*a+A2*(133.3)-A3*b)/(A1+A2-A3) #mm\n",
- "y=(A1*a*0.5+A2*(116.66)-A3*((4*a*0.5)/(3*pi)))/(A1+A2-A3) #mm\n",
- "\n",
- "#Result\n",
- "print'The centroidal distances are'\n",
- "print'x=',round(x,1),\"mm\"\n",
- "print'y=',round(y,1),\"mm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroidal distances are\n",
- "x= 98.6 mm\n",
- "y= 71.2 mm\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-16, Page no 166"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "V=np.array([1728*10**3,432*10**3,7.54*10**3])\n",
- "x_bar=np.array([60,140,60]) #mm\n",
- "y_bar=np.array([30,20,30]) #mm\n",
- "\n",
- "#Calculations\n",
- "x=(V[0]*x_bar[0]+V[1]*x_bar[1]+V[2]*x_bar[2])/(V[0]+V[1]+V[2]) #mm\n",
- "y=(V[0]*y_bar[0]+V[1]*y_bar[1]+V[2]*y_bar[2])/(V[0]+V[1]+V[2]) #mm\n",
- "z=120 #mm from symmetry\n",
- "\n",
- "#Result\n",
- "print'The centroid is at'\n",
- "print'x=',round(x,1),\"mm\"\n",
- "print'y=',round(y,1),\"mm\"\n",
- "print'z=',round(z,1),\"mm\"\n",
- "\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid is at\n",
- "x= 75.9 mm\n",
- "y= 28.0 mm\n",
- "z= 120.0 mm\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-17, Page no 166"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# Here tx=30 degrees,ty=45 degrees& tz=60 degrees,, thus\n",
- "sintx=2**-1\n",
- "costx=sqrt(3)*2**-1\n",
- "sinty=sqrt(2)**-1\n",
- "costy=sqrt(2)**-1\n",
- "sintz=sqrt(3)*2**-1\n",
- "costz=2**-1\n",
- "\n",
- "#Calculations\n",
- "V=np.array([10,15,25]) #in**3\n",
- "x_bar=np.array([4,12,24]) #in\n",
- "y_bar=np.array([4*costx,-6*costy,-4*costz])\n",
- "z_bar=np.array([-4*sintx,6*sinty,-4*sintz])\n",
- "#Centroid calculations\n",
- "x=(V[0]*x_bar[0]+V[1]*x_bar[1]+V[2]*x_bar[2])/(V[0]+V[1]+V[2]) #in\n",
- "y=(V[0]*y_bar[0]+V[1]*y_bar[1]+V[2]*y_bar[2])/(V[0]+V[1]+V[2]) #in\n",
- "z=(V[0]*z_bar[0]+V[1]*z_bar[1]+V[2]*z_bar[2])/(V[0]+V[1]+V[2]) #in\n",
- "\n",
- "#Result\n",
- "print'The centroid of three volumes is at'\n",
- "print'x=',round(x,1),\"in\"\n",
- "print'y=',round(y,2),\"in\"\n",
- "print'z=',round(z,2),\"in\"\n",
- "\n",
- "# The ans for x is off by 0.4 in"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid of three volumes is at\n",
- "x= 16.0 in\n",
- "y= -1.58 in\n",
- "z= -0.86 in\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-26, Page no 171"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Part a\n",
- "# Pefer textbook for part a.\n",
- "\n",
- "# Part b\n",
- "\n",
- "# Initilization of variables\n",
- "w=150 #lb/ft**2\n",
- "h=2 #ft height of the load\n",
- "s=8 #ft span\n",
- "b=2 #ft\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return x*(150*(x/4)*2)\n",
- "a=1\n",
- "b=1\n",
- "M=quad(integrand, 0, s, args=(a,b))\n",
- "Rr=M[0]/(2*s) #lb\n",
- "\n",
- "# Results\n",
- "print'The value of M is',round(M[0]),\"lb-ft\"\n",
- "print'The value of Rr is',round(Rr),\"lb\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of M is 12800.0 lb-ft\n",
- "The value of Rr is 800.0 lb\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-27, Page no 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "\n",
- "rho_m=1000 # kg/m**3\n",
- "h=0.3 # m height of hole\n",
- "b=0.6 # m width of hole\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(y, a, b):\n",
- " return y*9.8*rho_m*(1.2-y)*(0.6)\n",
- "a=1\n",
- "b=1\n",
- "I=quad(integrand, 0, h, args=(a,b))\n",
- "B=I[0]/(2*(0.3))\n",
- "\n",
- "# Results\n",
- "print'The value of B is',round(B),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of B is 441.0 N\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-28, Page no 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=62.4 #lb/ft**3\n",
- "h=12 #ft\n",
- "f=105 #lb/ft**3\n",
- "\n",
- "#Calculations\n",
- "p1=l*h #lb/ft**2\n",
- "#Total force on left side\n",
- "#Simplfying the equation we get a three degree equation in d\n",
- "#solving for d\n",
- "p=np.array([3**-1,0,-144,467])\n",
- "r=roots(p)\n",
- "d=r[2] #ft\n",
- "\n",
- "#Result\n",
- "print'The value of d is',round(d,2),\"feet\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of d is 3.33 feet\n"
- ]
- }
- ],
- "prompt_number": 24
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter10_2.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter10_2.ipynb deleted file mode 100755 index 3ce0f857..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter10_2.ipynb +++ /dev/null @@ -1,426 +0,0 @@ -{
- "metadata": {
- "name": "chapter10.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 10: First Moments and Centroids"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-5, Page no 160"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "r=50 #mm\n",
- "L1=75 #mm\n",
- "L2=pi*r #mm\n",
- "L3=61.2 #mm\n",
- "# as theta1=45 degrees & theta2=60 degrees\n",
- "sintheta1=sqrt(2)**-1\n",
- "costheta1=sqrt(2)**-1\n",
- "sintheta2=sqrt(3)*2**-1\n",
- "costheta2=2**-1\n",
- "\n",
- "#Calculations\n",
- "x_bar=np.array([(L1/2)*costheta1,L1*costheta1+r,L1*costheta1+100+(L3/2)*costheta2]) #mm\n",
- "y_bar=np.array([(L1/2)*sintheta1,L1*sintheta1+(2*r)/pi,(L3/2)*sintheta2]) #mm\n",
- "#Centroid Calculations\n",
- "x=(L1*x_bar[0]+L2*x_bar[1]+L3*x_bar[2])/(L1+L2+L3) #mm\n",
- "y=(L1*y_bar[0]+L2*y_bar[1]+L3*y_bar[2])/(L1+L2+L3) #mm\n",
- "\n",
- "#Result\n",
- "print'The centroid is as follows:'\n",
- "print'x=',round(x,1),\"mm\"\n",
- "print'y=',round(y,1),\"mm\"\n",
- "\n",
- "# The answer may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid is as follows:\n",
- "x= 97.0 mm\n",
- "y= 57.7 mm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-6, Page no 160"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "theta=75 #degrees\n",
- "theta1=30 #degrees\n",
- "sintheta=0.96\n",
- "costheta=0.25\n",
- "sintheta1=2**-1\n",
- "costheta1=sqrt(3)*2**-1\n",
- "alpha=(150*pi)/180 #rad\n",
- "r=1\n",
- "lhor=14 #in\n",
- "\n",
- "#calculations\n",
- "a=((2*r)/alpha)*sintheta #in\n",
- "p=90-theta\n",
- "sinp=0.259\n",
- "y=-a*sinp #in\n",
- "#Length of arc\n",
- "l=r*alpha #in\n",
- "#Slope length calculations\n",
- "DF=7 #in\n",
- "AB=DF #in\n",
- "BC=1 #in\n",
- "BF=BC*costheta1 #in\n",
- "FC=BC*sintheta1 #in\n",
- "DC=DF+FC #in\n",
- "EC=DC/costheta1 #in\n",
- "#Centroid of EC is at G\n",
- "yslope=0.5*EC*sintheta1+BF #in\n",
- "#Y of composite figure\n",
- "Y=((2*l*y)+14*-1+(2*EC*yslope))/(2*l+lhor+2*EC) #in\n",
- "\n",
- "#Result\n",
- "print'The centroid is at Y=',round(Y,2),\"in\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid is at Y= 1.03 in\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-11, Page no 163"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "a=100 #mm\n",
- "b=150 #mm\n",
- "A1=2*10**4 #mm**2\n",
- "A2=5*10**3 #mm**2\n",
- "A3=(pi*(a/2)**2)/2 #mm**2\n",
- "\n",
- "#Calculations\n",
- "x=(A1*a+A2*(133.3)-A3*b)/(A1+A2-A3) #mm\n",
- "y=(A1*a*0.5+A2*(116.66)-A3*((4*a*0.5)/(3*pi)))/(A1+A2-A3) #mm\n",
- "\n",
- "#Result\n",
- "print'The centroidal distances are'\n",
- "print'x=',round(x,1),\"mm\"\n",
- "print'y=',round(y,1),\"mm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroidal distances are\n",
- "x= 98.6 mm\n",
- "y= 71.2 mm\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-16, Page no 166"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "V=np.array([1728*10**3,432*10**3,7.54*10**3])\n",
- "x_bar=np.array([60,140,60]) #mm\n",
- "y_bar=np.array([30,20,30]) #mm\n",
- "\n",
- "#Calculations\n",
- "x=(V[0]*x_bar[0]+V[1]*x_bar[1]+V[2]*x_bar[2])/(V[0]+V[1]+V[2]) #mm\n",
- "y=(V[0]*y_bar[0]+V[1]*y_bar[1]+V[2]*y_bar[2])/(V[0]+V[1]+V[2]) #mm\n",
- "z=120 #mm from symmetry\n",
- "\n",
- "#Result\n",
- "print'The centroid is at'\n",
- "print'x=',round(x,1),\"mm\"\n",
- "print'y=',round(y,1),\"mm\"\n",
- "print'z=',round(z,1),\"mm\"\n",
- "\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid is at\n",
- "x= 75.9 mm\n",
- "y= 28.0 mm\n",
- "z= 120.0 mm\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-17, Page no 166"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# Here tx=30 degrees,ty=45 degrees& tz=60 degrees,, thus\n",
- "sintx=2**-1\n",
- "costx=sqrt(3)*2**-1\n",
- "sinty=sqrt(2)**-1\n",
- "costy=sqrt(2)**-1\n",
- "sintz=sqrt(3)*2**-1\n",
- "costz=2**-1\n",
- "\n",
- "#Calculations\n",
- "V=np.array([10,15,25]) #in**3\n",
- "x_bar=np.array([4,12,24]) #in\n",
- "y_bar=np.array([4*costx,-6*costy,-4*costz])\n",
- "z_bar=np.array([-4*sintx,6*sinty,-4*sintz])\n",
- "#Centroid calculations\n",
- "x=(V[0]*x_bar[0]+V[1]*x_bar[1]+V[2]*x_bar[2])/(V[0]+V[1]+V[2]) #in\n",
- "y=(V[0]*y_bar[0]+V[1]*y_bar[1]+V[2]*y_bar[2])/(V[0]+V[1]+V[2]) #in\n",
- "z=(V[0]*z_bar[0]+V[1]*z_bar[1]+V[2]*z_bar[2])/(V[0]+V[1]+V[2]) #in\n",
- "\n",
- "#Result\n",
- "print'The centroid of three volumes is at'\n",
- "print'x=',round(x,1),\"in\"\n",
- "print'y=',round(y,2),\"in\"\n",
- "print'z=',round(z,2),\"in\"\n",
- "\n",
- "# The ans for x is off by 0.4 in"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid of three volumes is at\n",
- "x= 16.0 in\n",
- "y= -1.58 in\n",
- "z= -0.86 in\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-26, Page no 171"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Part a\n",
- "# Pefer textbook for part a.\n",
- "\n",
- "# Part b\n",
- "\n",
- "# Initilization of variables\n",
- "w=150 #lb/ft**2\n",
- "h=2 #ft height of the load\n",
- "s=8 #ft span\n",
- "b=2 #ft\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return x*(150*(x/4)*2)\n",
- "a=1\n",
- "b=1\n",
- "M=quad(integrand, 0, s, args=(a,b))\n",
- "Rr=M[0]/(2*s) #lb\n",
- "\n",
- "# Results\n",
- "print'The value of M is',round(M[0]),\"lb-ft\"\n",
- "print'The value of Rr is',round(Rr),\"lb\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of M is 12800.0 lb-ft\n",
- "The value of Rr is 800.0 lb\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-27, Page no 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "\n",
- "rho_m=1000 # kg/m**3\n",
- "h=0.3 # m height of hole\n",
- "b=0.6 # m width of hole\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(y, a, b):\n",
- " return y*9.8*rho_m*(1.2-y)*(0.6)\n",
- "a=1\n",
- "b=1\n",
- "I=quad(integrand, 0, h, args=(a,b))\n",
- "B=I[0]/(2*(0.3))\n",
- "\n",
- "# Results\n",
- "print'The value of B is',round(B),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of B is 441.0 N\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-28, Page no 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=62.4 #lb/ft**3\n",
- "h=12 #ft\n",
- "f=105 #lb/ft**3\n",
- "\n",
- "#Calculations\n",
- "p1=l*h #lb/ft**2\n",
- "#Total force on left side\n",
- "#Simplfying the equation we get a three degree equation in d\n",
- "#solving for d\n",
- "p=np.array([3**-1,0,-144,467])\n",
- "r=roots(p)\n",
- "d=r[2] #ft\n",
- "\n",
- "#Result\n",
- "print'The value of d is',round(d,2),\"feet\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of d is 3.33 feet\n"
- ]
- }
- ],
- "prompt_number": 24
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter10_3.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter10_3.ipynb deleted file mode 100755 index 3ce0f857..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter10_3.ipynb +++ /dev/null @@ -1,426 +0,0 @@ -{
- "metadata": {
- "name": "chapter10.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 10: First Moments and Centroids"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-5, Page no 160"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "r=50 #mm\n",
- "L1=75 #mm\n",
- "L2=pi*r #mm\n",
- "L3=61.2 #mm\n",
- "# as theta1=45 degrees & theta2=60 degrees\n",
- "sintheta1=sqrt(2)**-1\n",
- "costheta1=sqrt(2)**-1\n",
- "sintheta2=sqrt(3)*2**-1\n",
- "costheta2=2**-1\n",
- "\n",
- "#Calculations\n",
- "x_bar=np.array([(L1/2)*costheta1,L1*costheta1+r,L1*costheta1+100+(L3/2)*costheta2]) #mm\n",
- "y_bar=np.array([(L1/2)*sintheta1,L1*sintheta1+(2*r)/pi,(L3/2)*sintheta2]) #mm\n",
- "#Centroid Calculations\n",
- "x=(L1*x_bar[0]+L2*x_bar[1]+L3*x_bar[2])/(L1+L2+L3) #mm\n",
- "y=(L1*y_bar[0]+L2*y_bar[1]+L3*y_bar[2])/(L1+L2+L3) #mm\n",
- "\n",
- "#Result\n",
- "print'The centroid is as follows:'\n",
- "print'x=',round(x,1),\"mm\"\n",
- "print'y=',round(y,1),\"mm\"\n",
- "\n",
- "# The answer may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid is as follows:\n",
- "x= 97.0 mm\n",
- "y= 57.7 mm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-6, Page no 160"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "theta=75 #degrees\n",
- "theta1=30 #degrees\n",
- "sintheta=0.96\n",
- "costheta=0.25\n",
- "sintheta1=2**-1\n",
- "costheta1=sqrt(3)*2**-1\n",
- "alpha=(150*pi)/180 #rad\n",
- "r=1\n",
- "lhor=14 #in\n",
- "\n",
- "#calculations\n",
- "a=((2*r)/alpha)*sintheta #in\n",
- "p=90-theta\n",
- "sinp=0.259\n",
- "y=-a*sinp #in\n",
- "#Length of arc\n",
- "l=r*alpha #in\n",
- "#Slope length calculations\n",
- "DF=7 #in\n",
- "AB=DF #in\n",
- "BC=1 #in\n",
- "BF=BC*costheta1 #in\n",
- "FC=BC*sintheta1 #in\n",
- "DC=DF+FC #in\n",
- "EC=DC/costheta1 #in\n",
- "#Centroid of EC is at G\n",
- "yslope=0.5*EC*sintheta1+BF #in\n",
- "#Y of composite figure\n",
- "Y=((2*l*y)+14*-1+(2*EC*yslope))/(2*l+lhor+2*EC) #in\n",
- "\n",
- "#Result\n",
- "print'The centroid is at Y=',round(Y,2),\"in\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid is at Y= 1.03 in\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-11, Page no 163"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "a=100 #mm\n",
- "b=150 #mm\n",
- "A1=2*10**4 #mm**2\n",
- "A2=5*10**3 #mm**2\n",
- "A3=(pi*(a/2)**2)/2 #mm**2\n",
- "\n",
- "#Calculations\n",
- "x=(A1*a+A2*(133.3)-A3*b)/(A1+A2-A3) #mm\n",
- "y=(A1*a*0.5+A2*(116.66)-A3*((4*a*0.5)/(3*pi)))/(A1+A2-A3) #mm\n",
- "\n",
- "#Result\n",
- "print'The centroidal distances are'\n",
- "print'x=',round(x,1),\"mm\"\n",
- "print'y=',round(y,1),\"mm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroidal distances are\n",
- "x= 98.6 mm\n",
- "y= 71.2 mm\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-16, Page no 166"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "V=np.array([1728*10**3,432*10**3,7.54*10**3])\n",
- "x_bar=np.array([60,140,60]) #mm\n",
- "y_bar=np.array([30,20,30]) #mm\n",
- "\n",
- "#Calculations\n",
- "x=(V[0]*x_bar[0]+V[1]*x_bar[1]+V[2]*x_bar[2])/(V[0]+V[1]+V[2]) #mm\n",
- "y=(V[0]*y_bar[0]+V[1]*y_bar[1]+V[2]*y_bar[2])/(V[0]+V[1]+V[2]) #mm\n",
- "z=120 #mm from symmetry\n",
- "\n",
- "#Result\n",
- "print'The centroid is at'\n",
- "print'x=',round(x,1),\"mm\"\n",
- "print'y=',round(y,1),\"mm\"\n",
- "print'z=',round(z,1),\"mm\"\n",
- "\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid is at\n",
- "x= 75.9 mm\n",
- "y= 28.0 mm\n",
- "z= 120.0 mm\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-17, Page no 166"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# Here tx=30 degrees,ty=45 degrees& tz=60 degrees,, thus\n",
- "sintx=2**-1\n",
- "costx=sqrt(3)*2**-1\n",
- "sinty=sqrt(2)**-1\n",
- "costy=sqrt(2)**-1\n",
- "sintz=sqrt(3)*2**-1\n",
- "costz=2**-1\n",
- "\n",
- "#Calculations\n",
- "V=np.array([10,15,25]) #in**3\n",
- "x_bar=np.array([4,12,24]) #in\n",
- "y_bar=np.array([4*costx,-6*costy,-4*costz])\n",
- "z_bar=np.array([-4*sintx,6*sinty,-4*sintz])\n",
- "#Centroid calculations\n",
- "x=(V[0]*x_bar[0]+V[1]*x_bar[1]+V[2]*x_bar[2])/(V[0]+V[1]+V[2]) #in\n",
- "y=(V[0]*y_bar[0]+V[1]*y_bar[1]+V[2]*y_bar[2])/(V[0]+V[1]+V[2]) #in\n",
- "z=(V[0]*z_bar[0]+V[1]*z_bar[1]+V[2]*z_bar[2])/(V[0]+V[1]+V[2]) #in\n",
- "\n",
- "#Result\n",
- "print'The centroid of three volumes is at'\n",
- "print'x=',round(x,1),\"in\"\n",
- "print'y=',round(y,2),\"in\"\n",
- "print'z=',round(z,2),\"in\"\n",
- "\n",
- "# The ans for x is off by 0.4 in"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid of three volumes is at\n",
- "x= 16.0 in\n",
- "y= -1.58 in\n",
- "z= -0.86 in\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-26, Page no 171"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Part a\n",
- "# Pefer textbook for part a.\n",
- "\n",
- "# Part b\n",
- "\n",
- "# Initilization of variables\n",
- "w=150 #lb/ft**2\n",
- "h=2 #ft height of the load\n",
- "s=8 #ft span\n",
- "b=2 #ft\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return x*(150*(x/4)*2)\n",
- "a=1\n",
- "b=1\n",
- "M=quad(integrand, 0, s, args=(a,b))\n",
- "Rr=M[0]/(2*s) #lb\n",
- "\n",
- "# Results\n",
- "print'The value of M is',round(M[0]),\"lb-ft\"\n",
- "print'The value of Rr is',round(Rr),\"lb\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of M is 12800.0 lb-ft\n",
- "The value of Rr is 800.0 lb\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-27, Page no 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "\n",
- "rho_m=1000 # kg/m**3\n",
- "h=0.3 # m height of hole\n",
- "b=0.6 # m width of hole\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(y, a, b):\n",
- " return y*9.8*rho_m*(1.2-y)*(0.6)\n",
- "a=1\n",
- "b=1\n",
- "I=quad(integrand, 0, h, args=(a,b))\n",
- "B=I[0]/(2*(0.3))\n",
- "\n",
- "# Results\n",
- "print'The value of B is',round(B),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of B is 441.0 N\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-28, Page no 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=62.4 #lb/ft**3\n",
- "h=12 #ft\n",
- "f=105 #lb/ft**3\n",
- "\n",
- "#Calculations\n",
- "p1=l*h #lb/ft**2\n",
- "#Total force on left side\n",
- "#Simplfying the equation we get a three degree equation in d\n",
- "#solving for d\n",
- "p=np.array([3**-1,0,-144,467])\n",
- "r=roots(p)\n",
- "d=r[2] #ft\n",
- "\n",
- "#Result\n",
- "print'The value of d is',round(d,2),\"feet\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of d is 3.33 feet\n"
- ]
- }
- ],
- "prompt_number": 24
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter10_4.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter10_4.ipynb deleted file mode 100755 index 3ce0f857..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter10_4.ipynb +++ /dev/null @@ -1,426 +0,0 @@ -{
- "metadata": {
- "name": "chapter10.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 10: First Moments and Centroids"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-5, Page no 160"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "r=50 #mm\n",
- "L1=75 #mm\n",
- "L2=pi*r #mm\n",
- "L3=61.2 #mm\n",
- "# as theta1=45 degrees & theta2=60 degrees\n",
- "sintheta1=sqrt(2)**-1\n",
- "costheta1=sqrt(2)**-1\n",
- "sintheta2=sqrt(3)*2**-1\n",
- "costheta2=2**-1\n",
- "\n",
- "#Calculations\n",
- "x_bar=np.array([(L1/2)*costheta1,L1*costheta1+r,L1*costheta1+100+(L3/2)*costheta2]) #mm\n",
- "y_bar=np.array([(L1/2)*sintheta1,L1*sintheta1+(2*r)/pi,(L3/2)*sintheta2]) #mm\n",
- "#Centroid Calculations\n",
- "x=(L1*x_bar[0]+L2*x_bar[1]+L3*x_bar[2])/(L1+L2+L3) #mm\n",
- "y=(L1*y_bar[0]+L2*y_bar[1]+L3*y_bar[2])/(L1+L2+L3) #mm\n",
- "\n",
- "#Result\n",
- "print'The centroid is as follows:'\n",
- "print'x=',round(x,1),\"mm\"\n",
- "print'y=',round(y,1),\"mm\"\n",
- "\n",
- "# The answer may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid is as follows:\n",
- "x= 97.0 mm\n",
- "y= 57.7 mm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-6, Page no 160"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "theta=75 #degrees\n",
- "theta1=30 #degrees\n",
- "sintheta=0.96\n",
- "costheta=0.25\n",
- "sintheta1=2**-1\n",
- "costheta1=sqrt(3)*2**-1\n",
- "alpha=(150*pi)/180 #rad\n",
- "r=1\n",
- "lhor=14 #in\n",
- "\n",
- "#calculations\n",
- "a=((2*r)/alpha)*sintheta #in\n",
- "p=90-theta\n",
- "sinp=0.259\n",
- "y=-a*sinp #in\n",
- "#Length of arc\n",
- "l=r*alpha #in\n",
- "#Slope length calculations\n",
- "DF=7 #in\n",
- "AB=DF #in\n",
- "BC=1 #in\n",
- "BF=BC*costheta1 #in\n",
- "FC=BC*sintheta1 #in\n",
- "DC=DF+FC #in\n",
- "EC=DC/costheta1 #in\n",
- "#Centroid of EC is at G\n",
- "yslope=0.5*EC*sintheta1+BF #in\n",
- "#Y of composite figure\n",
- "Y=((2*l*y)+14*-1+(2*EC*yslope))/(2*l+lhor+2*EC) #in\n",
- "\n",
- "#Result\n",
- "print'The centroid is at Y=',round(Y,2),\"in\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid is at Y= 1.03 in\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-11, Page no 163"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "a=100 #mm\n",
- "b=150 #mm\n",
- "A1=2*10**4 #mm**2\n",
- "A2=5*10**3 #mm**2\n",
- "A3=(pi*(a/2)**2)/2 #mm**2\n",
- "\n",
- "#Calculations\n",
- "x=(A1*a+A2*(133.3)-A3*b)/(A1+A2-A3) #mm\n",
- "y=(A1*a*0.5+A2*(116.66)-A3*((4*a*0.5)/(3*pi)))/(A1+A2-A3) #mm\n",
- "\n",
- "#Result\n",
- "print'The centroidal distances are'\n",
- "print'x=',round(x,1),\"mm\"\n",
- "print'y=',round(y,1),\"mm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroidal distances are\n",
- "x= 98.6 mm\n",
- "y= 71.2 mm\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-16, Page no 166"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "V=np.array([1728*10**3,432*10**3,7.54*10**3])\n",
- "x_bar=np.array([60,140,60]) #mm\n",
- "y_bar=np.array([30,20,30]) #mm\n",
- "\n",
- "#Calculations\n",
- "x=(V[0]*x_bar[0]+V[1]*x_bar[1]+V[2]*x_bar[2])/(V[0]+V[1]+V[2]) #mm\n",
- "y=(V[0]*y_bar[0]+V[1]*y_bar[1]+V[2]*y_bar[2])/(V[0]+V[1]+V[2]) #mm\n",
- "z=120 #mm from symmetry\n",
- "\n",
- "#Result\n",
- "print'The centroid is at'\n",
- "print'x=',round(x,1),\"mm\"\n",
- "print'y=',round(y,1),\"mm\"\n",
- "print'z=',round(z,1),\"mm\"\n",
- "\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid is at\n",
- "x= 75.9 mm\n",
- "y= 28.0 mm\n",
- "z= 120.0 mm\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-17, Page no 166"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# Here tx=30 degrees,ty=45 degrees& tz=60 degrees,, thus\n",
- "sintx=2**-1\n",
- "costx=sqrt(3)*2**-1\n",
- "sinty=sqrt(2)**-1\n",
- "costy=sqrt(2)**-1\n",
- "sintz=sqrt(3)*2**-1\n",
- "costz=2**-1\n",
- "\n",
- "#Calculations\n",
- "V=np.array([10,15,25]) #in**3\n",
- "x_bar=np.array([4,12,24]) #in\n",
- "y_bar=np.array([4*costx,-6*costy,-4*costz])\n",
- "z_bar=np.array([-4*sintx,6*sinty,-4*sintz])\n",
- "#Centroid calculations\n",
- "x=(V[0]*x_bar[0]+V[1]*x_bar[1]+V[2]*x_bar[2])/(V[0]+V[1]+V[2]) #in\n",
- "y=(V[0]*y_bar[0]+V[1]*y_bar[1]+V[2]*y_bar[2])/(V[0]+V[1]+V[2]) #in\n",
- "z=(V[0]*z_bar[0]+V[1]*z_bar[1]+V[2]*z_bar[2])/(V[0]+V[1]+V[2]) #in\n",
- "\n",
- "#Result\n",
- "print'The centroid of three volumes is at'\n",
- "print'x=',round(x,1),\"in\"\n",
- "print'y=',round(y,2),\"in\"\n",
- "print'z=',round(z,2),\"in\"\n",
- "\n",
- "# The ans for x is off by 0.4 in"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid of three volumes is at\n",
- "x= 16.0 in\n",
- "y= -1.58 in\n",
- "z= -0.86 in\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-26, Page no 171"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Part a\n",
- "# Pefer textbook for part a.\n",
- "\n",
- "# Part b\n",
- "\n",
- "# Initilization of variables\n",
- "w=150 #lb/ft**2\n",
- "h=2 #ft height of the load\n",
- "s=8 #ft span\n",
- "b=2 #ft\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return x*(150*(x/4)*2)\n",
- "a=1\n",
- "b=1\n",
- "M=quad(integrand, 0, s, args=(a,b))\n",
- "Rr=M[0]/(2*s) #lb\n",
- "\n",
- "# Results\n",
- "print'The value of M is',round(M[0]),\"lb-ft\"\n",
- "print'The value of Rr is',round(Rr),\"lb\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of M is 12800.0 lb-ft\n",
- "The value of Rr is 800.0 lb\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-27, Page no 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "\n",
- "rho_m=1000 # kg/m**3\n",
- "h=0.3 # m height of hole\n",
- "b=0.6 # m width of hole\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(y, a, b):\n",
- " return y*9.8*rho_m*(1.2-y)*(0.6)\n",
- "a=1\n",
- "b=1\n",
- "I=quad(integrand, 0, h, args=(a,b))\n",
- "B=I[0]/(2*(0.3))\n",
- "\n",
- "# Results\n",
- "print'The value of B is',round(B),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of B is 441.0 N\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.10-28, Page no 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=62.4 #lb/ft**3\n",
- "h=12 #ft\n",
- "f=105 #lb/ft**3\n",
- "\n",
- "#Calculations\n",
- "p1=l*h #lb/ft**2\n",
- "#Total force on left side\n",
- "#Simplfying the equation we get a three degree equation in d\n",
- "#solving for d\n",
- "p=np.array([3**-1,0,-144,467])\n",
- "r=roots(p)\n",
- "d=r[2] #ft\n",
- "\n",
- "#Result\n",
- "print'The value of d is',round(d,2),\"feet\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of d is 3.33 feet\n"
- ]
- }
- ],
- "prompt_number": 24
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter11.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter11.ipynb deleted file mode 100755 index 5d465a0b..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter11.ipynb +++ /dev/null @@ -1,113 +0,0 @@ -{
- "metadata": {
- "name": "chapter11.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 11: Virtual Work"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11.11-6, Page no 188"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "N=100 #lb\n",
- "mu=0.3 #coefficient of friction\n",
- "l=5 #in compressed to length\n",
- "\n",
- "#Calculations\n",
- "#Simplfying the calculations we obtain\n",
- "M=8*(N+N*mu) #lb-in\n",
- "\n",
- "#Result\n",
- "print'The Moment is',round(M,3),\"lb-in\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Moment is 1040.0 lb-in\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11.11-7, Page no 189"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=10 #kg\n",
- "g=9.81 # m/s**2\n",
- "F=200 #N\n",
- "l=3 #m\n",
- "\n",
- "#Calculations\n",
- "#Applying Virtual work principle\n",
- "By=m*g*0.5 #N\n",
- "Bx=F*(2*3**-1) #N\n",
- "#By equations of equilibrium\n",
- "Ax=-Bx-F #N negative sign indictaes the LEFT orientation\n",
- "Ay=m*g-By #N\n",
- "\n",
- "#Result\n",
- "print'The values are'\n",
- "print'Ax=',round(Ax),\"N to left\"\n",
- "print'Ay=',round(Ay),\"N up\"\n",
- "print'Bx=',round(Bx),\"N to right\"\n",
- "print'By=',round(By),\"N up\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are\n",
- "Ax= -333.0 N to left\n",
- "Ay= 49.0 N up\n",
- "Bx= 133.0 N to right\n",
- "By= 49.0 N up\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter11_1.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter11_1.ipynb deleted file mode 100755 index 5d465a0b..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter11_1.ipynb +++ /dev/null @@ -1,113 +0,0 @@ -{
- "metadata": {
- "name": "chapter11.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 11: Virtual Work"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11.11-6, Page no 188"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "N=100 #lb\n",
- "mu=0.3 #coefficient of friction\n",
- "l=5 #in compressed to length\n",
- "\n",
- "#Calculations\n",
- "#Simplfying the calculations we obtain\n",
- "M=8*(N+N*mu) #lb-in\n",
- "\n",
- "#Result\n",
- "print'The Moment is',round(M,3),\"lb-in\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Moment is 1040.0 lb-in\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11.11-7, Page no 189"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=10 #kg\n",
- "g=9.81 # m/s**2\n",
- "F=200 #N\n",
- "l=3 #m\n",
- "\n",
- "#Calculations\n",
- "#Applying Virtual work principle\n",
- "By=m*g*0.5 #N\n",
- "Bx=F*(2*3**-1) #N\n",
- "#By equations of equilibrium\n",
- "Ax=-Bx-F #N negative sign indictaes the LEFT orientation\n",
- "Ay=m*g-By #N\n",
- "\n",
- "#Result\n",
- "print'The values are'\n",
- "print'Ax=',round(Ax),\"N to left\"\n",
- "print'Ay=',round(Ay),\"N up\"\n",
- "print'Bx=',round(Bx),\"N to right\"\n",
- "print'By=',round(By),\"N up\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are\n",
- "Ax= -333.0 N to left\n",
- "Ay= 49.0 N up\n",
- "Bx= 133.0 N to right\n",
- "By= 49.0 N up\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter11_2.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter11_2.ipynb deleted file mode 100755 index 5d465a0b..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter11_2.ipynb +++ /dev/null @@ -1,113 +0,0 @@ -{
- "metadata": {
- "name": "chapter11.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 11: Virtual Work"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11.11-6, Page no 188"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "N=100 #lb\n",
- "mu=0.3 #coefficient of friction\n",
- "l=5 #in compressed to length\n",
- "\n",
- "#Calculations\n",
- "#Simplfying the calculations we obtain\n",
- "M=8*(N+N*mu) #lb-in\n",
- "\n",
- "#Result\n",
- "print'The Moment is',round(M,3),\"lb-in\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Moment is 1040.0 lb-in\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11.11-7, Page no 189"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=10 #kg\n",
- "g=9.81 # m/s**2\n",
- "F=200 #N\n",
- "l=3 #m\n",
- "\n",
- "#Calculations\n",
- "#Applying Virtual work principle\n",
- "By=m*g*0.5 #N\n",
- "Bx=F*(2*3**-1) #N\n",
- "#By equations of equilibrium\n",
- "Ax=-Bx-F #N negative sign indictaes the LEFT orientation\n",
- "Ay=m*g-By #N\n",
- "\n",
- "#Result\n",
- "print'The values are'\n",
- "print'Ax=',round(Ax),\"N to left\"\n",
- "print'Ay=',round(Ay),\"N up\"\n",
- "print'Bx=',round(Bx),\"N to right\"\n",
- "print'By=',round(By),\"N up\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are\n",
- "Ax= -333.0 N to left\n",
- "Ay= 49.0 N up\n",
- "Bx= 133.0 N to right\n",
- "By= 49.0 N up\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter11_3.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter11_3.ipynb deleted file mode 100755 index 5d465a0b..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter11_3.ipynb +++ /dev/null @@ -1,113 +0,0 @@ -{
- "metadata": {
- "name": "chapter11.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 11: Virtual Work"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11.11-6, Page no 188"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "N=100 #lb\n",
- "mu=0.3 #coefficient of friction\n",
- "l=5 #in compressed to length\n",
- "\n",
- "#Calculations\n",
- "#Simplfying the calculations we obtain\n",
- "M=8*(N+N*mu) #lb-in\n",
- "\n",
- "#Result\n",
- "print'The Moment is',round(M,3),\"lb-in\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Moment is 1040.0 lb-in\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11.11-7, Page no 189"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=10 #kg\n",
- "g=9.81 # m/s**2\n",
- "F=200 #N\n",
- "l=3 #m\n",
- "\n",
- "#Calculations\n",
- "#Applying Virtual work principle\n",
- "By=m*g*0.5 #N\n",
- "Bx=F*(2*3**-1) #N\n",
- "#By equations of equilibrium\n",
- "Ax=-Bx-F #N negative sign indictaes the LEFT orientation\n",
- "Ay=m*g-By #N\n",
- "\n",
- "#Result\n",
- "print'The values are'\n",
- "print'Ax=',round(Ax),\"N to left\"\n",
- "print'Ay=',round(Ay),\"N up\"\n",
- "print'Bx=',round(Bx),\"N to right\"\n",
- "print'By=',round(By),\"N up\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are\n",
- "Ax= -333.0 N to left\n",
- "Ay= 49.0 N up\n",
- "Bx= 133.0 N to right\n",
- "By= 49.0 N up\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter11_4.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter11_4.ipynb deleted file mode 100755 index 5d465a0b..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter11_4.ipynb +++ /dev/null @@ -1,113 +0,0 @@ -{
- "metadata": {
- "name": "chapter11.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 11: Virtual Work"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11.11-6, Page no 188"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "N=100 #lb\n",
- "mu=0.3 #coefficient of friction\n",
- "l=5 #in compressed to length\n",
- "\n",
- "#Calculations\n",
- "#Simplfying the calculations we obtain\n",
- "M=8*(N+N*mu) #lb-in\n",
- "\n",
- "#Result\n",
- "print'The Moment is',round(M,3),\"lb-in\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Moment is 1040.0 lb-in\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11.11-7, Page no 189"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=10 #kg\n",
- "g=9.81 # m/s**2\n",
- "F=200 #N\n",
- "l=3 #m\n",
- "\n",
- "#Calculations\n",
- "#Applying Virtual work principle\n",
- "By=m*g*0.5 #N\n",
- "Bx=F*(2*3**-1) #N\n",
- "#By equations of equilibrium\n",
- "Ax=-Bx-F #N negative sign indictaes the LEFT orientation\n",
- "Ay=m*g-By #N\n",
- "\n",
- "#Result\n",
- "print'The values are'\n",
- "print'Ax=',round(Ax),\"N to left\"\n",
- "print'Ay=',round(Ay),\"N up\"\n",
- "print'Bx=',round(Bx),\"N to right\"\n",
- "print'By=',round(By),\"N up\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are\n",
- "Ax= -333.0 N to left\n",
- "Ay= 49.0 N up\n",
- "Bx= 133.0 N to right\n",
- "By= 49.0 N up\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter12.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter12.ipynb deleted file mode 100755 index 0f5cd8eb..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter12.ipynb +++ /dev/null @@ -1,1521 +0,0 @@ -{
- "metadata": {
- "name": "chapter12.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 12: Kinematics of a Particle"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex 12.12-1, Page No 200"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "# Initilization of variables\n",
- "t=4 #seconds\n",
- "\n",
- "# Calculations\n",
- "#Displacement \n",
- "x=3*t**3+t+2 #ft\n",
- "# Velocity\n",
- "v=9*t**2+1 # ft/s\n",
- "# Acceleration\n",
- "a=18*t # ft/s**2\n",
- "\n",
- "# Result\n",
- "print'The dipalacemnt is',round(x),\"ft\"\n",
- "print'The velocity is ',round(v),\"ft/s\"\n",
- "print'The acceleration is ',round(a),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The dipalacemnt is 198.0 ft\n",
- "The velocity is 145.0 ft/s\n",
- "The acceleration is 72.0 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-2, Page No 201"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "# Initilization of variables\n",
- "t1=4 #s\n",
- "t2=5 #s\n",
- "\n",
- "# Calculation\n",
- "v1=9*t1**2+1 # ft/s\n",
- "v2=9*t2**2+1 # ft/s\n",
- "a=(v2-v1)/(t2-t1) # m/s**2\n",
- "\n",
- "# Result\n",
- "print'The acceleration during fifth second is',round(a),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration during fifth second is 81.0 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-3, Page No 201"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Defining Matrices\n",
- "t=[0,1,2,3,4,5,10] #s\n",
- "# equation for s is s=8*t**2+2*t, Thus the different values of s corresponding to t is:\n",
- "#Displacement matrix\n",
- "s=[0,10,36,78,136,210,820]\n",
- "# Eqn for v is v=16*t+2,Thus the different values of v corresponding to t is:\n",
- "#Velocity Matrix\n",
- "v=[0,18,34,50,66,82,162]\n",
- "# Eqn for a is a=16, Thus the different values of a corresponding to t is:\n",
- "#Acceleration Matrix\n",
- "a=[16,16,16,16,16,16,16]\n",
- "#Plotting the curves\n",
- "#S-T curve\n",
- "plot(t,s)\n",
- "plot(t,v)\n",
- "plot(t,a)\n",
- "xlabel('t(s)')\n",
- "ylabel('s(m), v(m/s) & a(m/s**2)')\n",
- "\n",
- "#Result\n",
- "print'The graphs are the solutions'\n",
- "print'blue line is for \"s\" vs \"t\" '\n",
- "print'green line is for \"v\" vs \"t\" '\n",
- "print'red line is for \"a\" vs \"t\" '\n",
- "# All the 3 graphs have been combined into a single graph"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The graphs are the solutions\n",
- "blue line is for \"s\" vs \"t\" \n",
- "green line is for \"v\" vs \"t\" \n",
- "red line is for \"a\" vs \"t\" \n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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6dOlyx8E2hSQHIepWVaVWVI2PVxODwaB1RMIaNHty+OSTT5g2bRqdrWTOmyQH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\ndbbT+qMIjWk6x0un09XZ51j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- "text": [
- "<matplotlib.figure.Figure at 0x594d050>"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example No 12.12-4, Page No 202"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_o=0 #ft/s\n",
- "v_f=88#ft/s\n",
- "t=28 #s\n",
- "\n",
- "#Calculations\n",
- "k=(v_f-v_o)*t**-1 #ft/s**2\n",
- "s=((v_f-v_o)/2)*t #ft\n",
- "\n",
- "#Result\n",
- "print'The value of constant k is',round(k,2),\"ft/s**2\"\n",
- "print'The displacement is ',round(s),\"ft\"\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of constant k is 3.14 ft/s**2\n",
- "The displacement is 1232.0 ft\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-5, Page No 202"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_o=0 #ft/s\n",
- "v_f1=30 #ft/s\n",
- "v_f2=0 #ft/s\n",
- "t1=3 #s\n",
- "t2=2 #s\n",
- "\n",
- "#Calculations\n",
- "#Plotting the v-t curve\n",
- "#Velocity matrix \n",
- "v=[v_o,v_f1,v_f2]\n",
- "#Time matrix\n",
- "t=[0,3,5]\n",
- "plot(t,v)\n",
- "xlabel('t')\n",
- "ylabel('v')\n",
- "#Part \"b\"\n",
- "#Acceleration at 3s\n",
- "a1=(v_f1-v_o)/t1 #ft/s**2\n",
- "#Acceleration at 5s\n",
- "a2=(v_f2-v_f1)/t2 #ft/s**2\n",
- "#Part \"c\"\n",
- "s=(v_f1*t1*0.5)+(v_f1*t2*0.5) #ft\n",
- "#Part \"d\"\n",
- "#Simplfying the equation we get\n",
- "#7.5t**2-30t+5=0\n",
- "a=7.5\n",
- "b=-30\n",
- "c=5\n",
- "q=sqrt(b**2-4*a*c)\n",
- "x1=(-b+q)/(2*a)\n",
- "x2=(-b-q)/(2*a)\n",
- "#As x1 is greater than 2 it does not hold as a solution\n",
- "t=x2 #s\n",
- "#Hence total time is\n",
- "T=t1+t #s\n",
- "\n",
- "#Result\n",
- "print'The graph is the solution for part a'\n",
- "print'The acceleration at 3rd second is',round(a1),\"ft/s**2\"\n",
- "print'The acceleration at 5th second is',round(a2),\"ft/s**2\"\n",
- "print'The displacement is',round(s),\"ft\"\n",
- "print'The total time is',round(T,3),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The graph is the solution for part a\n",
- "The acceleration at 3rd second is 10.0 ft/s**2\n",
- "The acceleration at 5th second is -15.0 ft/s**2\n",
- "The displacement is 75.0 ft\n",
- "The total time is 3.174 s\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x594d2d0>"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example No 12.12-6, Page No 203"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_o=2 #m/s\n",
- "y_o=120 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Solve using ground as datum\n",
- "y=0\n",
- "#Simplfying the equation\n",
- "a=4.9\n",
- "b=-2\n",
- "c=-120\n",
- "q=sqrt(b**2-4*a*c)\n",
- "x1=(-b+q)/(2*a) #s\n",
- "x2=(-b-q)/(2*a) #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(x1,2),\"s\"\n",
- "#As x2 is negative and negative time does not make any physical sense\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 5.16 s\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-7, Page No 204"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Vo1=80 #ft/s\n",
- "Vo2=60 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Simplfying by equating the two times\n",
- "t=(-(Vo2*2)-(g*0.5*4))/(Vo1-Vo2-(g*0.5*4)) #s\n",
- "#Substituting this t in s we get\n",
- "s=(Vo1*t)-(0.5*g*t*t) #ft\n",
- "\n",
- "#Result\n",
- "print'The time obtained is',round(t,2),\"s\"\n",
- "print'and the ball meets at',round(s,1),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time obtained is 4.15 s\n",
- "and the ball meets at 54.5 ft\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-8, Page No 204"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=40 degrees\n",
- "costheta=0.77\n",
- "tantheta=0.83\n",
- "x=100 #ft\n",
- "ay=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Simplfying the equation\n",
- "t=((tantheta*x)*(ay/2)**-1)**0.5 #s\n",
- "#Velocity calculations\n",
- "Vo=100*(costheta*t)**-1 #ft/s\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The initial speed should be',round(Vo,1),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The initial speed should be 57.2 ft/s\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-9, Page No 204"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=[0,1,2,3,4,5,6] #s\n",
- "#Solving the Differential Equations we obtain\n",
- "# Eqn for s is s=(t+1)**3,Thus the different values of s corresponding to t is:\n",
- "#Displacement matrix\n",
- "s=[1,8,27,64,125,216,343]\n",
- "# Eqn for v is v=3*(t+1)**2, Thus the different values of v corresponding to t is:\n",
- "# Velocity matrix\n",
- "v=[3,12,27,48,75,108,147]\n",
- "# Eqn for a is a=6*(t+1),Thus the different values of a corresponding to t is:\n",
- "# Acceleration matrix\n",
- "a=[6,12,18,24,30,36,42]\n",
- "#Plotting\n",
- "plot(t,s)\n",
- "plot(t,v)\n",
- "plot(t,a)\n",
- "xlabel('t(s)')\n",
- "ylabel('s(ft) , v(ft/s) & a(ft/s**2)')\n",
- "\n",
- "#Result\n",
- "print'The result are the plots that have been generated'\n",
- "print'blue line is for \"s\" vs \"t\"'\n",
- "print'green line is for \"v\" vs \"t\"'\n",
- "print'red line is for \"a\" vs \"t\"'\n",
- "# All the graphs have been plotted on a single graph"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The result are the plots that have been generated\n",
- "blue line is for \"s\" vs \"t\"\n",
- "green line is for \"v\" vs \"t\"\n",
- "red line is for \"a\" vs \"t\"\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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w96effsqUKVOK3L5WrVr4+vry888/V0B0QpSfJAQhSmje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\n",
- "text": [
- "<matplotlib.figure.Figure at 0x58313f0>"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-10, Page No 205"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=3 #s\n",
- "\n",
- "#Calculations\n",
- "#After solving the differential equation\n",
- "s=(3**-1)*(t+2)**3 #ft\n",
- "v=(t+2)**2 #ft/s\n",
- "a=2*(t+2) #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The displacement at t=3s is',round(s,1),\"ft\"\n",
- "print'The velocity at t=3s is',round(v),\"ft/s\"\n",
- "print'The acceleration at t=3s is',round(a),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The displacement at t=3s is 41.7 ft\n",
- "The velocity at t=3s is 25.0 ft/s\n",
- "The acceleration at t=3s is 10.0 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-12, Page No 206"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#Calling upward direction positive\n",
- "xdot1=6 #ft/s\n",
- "xdot3=3 #ft/s\n",
- "xdoubledot=2 #ft/s**2\n",
- "xdoubledot3=-4 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "xdot=-xdot1 #ft/s\n",
- "xdot2=2*xdot-xdot3 #ft/s\n",
- "xdoubledot2=2*xdoubledot-xdoubledot3 #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of velocity is',round(xdot2,3),\"ft/s (down)\"\n",
- "print'The value of acceleration is',round(xdoubledot2,3),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of velocity is -15.0 ft/s (down)\n",
- "The value of acceleration is 8.0 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-16, Page No 207"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=4 #s\n",
- "\n",
- "#Calculations\n",
- "#Part (a)\n",
- "x=t**3 #in\n",
- "y=-2*t**2 #in\n",
- "z=2*t #in\n",
- "#Part (b)\n",
- "#Theory question\n",
- "#Part(c)\n",
- "#Unit vector calculation\n",
- "m=(4**2+1**1+(-3)**2)**0.5\n",
- "e_l=[4*m**-1,m**-1,-3*m**-1]\n",
- "v=[3*t**2,-4*t,2] #in/s\n",
- "#Projection of v on n at t=4s\n",
- "dot=[v[0]*e_l[0],v[1]*e_l[1],v[2]*e_l[2]]\n",
- "#dot=v.*e_l #in/s\n",
- "a=dot[0]+dot[1]+dot[2] #in/s\n",
- "\n",
- "#Result\n",
- "print'The co-ordinates of position are x=',round(x),\"in ,\",round(y),\"in and \",round(z),\"in respectively\"\n",
- "print'The projection of v on n at t=4s is',round(a,1),\"in/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The co-ordinates of position are x= 64.0 in , -32.0 in and 8.0 in respectively\n",
- "The projection of v on n at t=4s is 33.3 in/s\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-17, Page No 208"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "theta=pi/3 #rad\n",
- "\n",
- "#Calculations\n",
- "#Method (a)\n",
- "t=(theta)**0.5 #s\n",
- "r=2*theta\n",
- "rdot=4*t\n",
- "thetadot=2*t\n",
- "#Velocity calculations\n",
- "x=r*thetadot\n",
- "v=((rdot)**2+x**2)**0.5 #ft/s\n",
- "#Theta calculations\n",
- "thetax=30+arctan(rdot/x)*(180/pi) #degrees\n",
- "#Method (b)\n",
- "x=2*theta*cos(theta) #ft\n",
- "y=2*theta*sin(theta) #ft\n",
- "xdot=4*t*((cos(t**2)))+2*t**2*(-sin(t**2))*(2*t) #ft/s\n",
- "ydot=4*t**2*sin(t**2)+2*t**2*cos(t**2)*2*t #ft/s\n",
- "V=(xdot**2+ydot**2)**0.5 #ft/s\n",
- "Thetax=arctan(ydot/-xdot)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'By both the methods we obtain v and thetax as:'\n",
- "print'Method 1'\n",
- "print'v=',round(v,2),\"ft/s\",'and thetax=',round(thetax,1),\"degrees\"\n",
- "print'Method 2'\n",
- "print'V=',round(v,2),\"ft/s\",'and Thetax=',round(Thetax,1),\"degrees\"\n",
- "# The answer may wary due to decimal point accuracy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "By both the methods we obtain v and thetax as:\n",
- "Method 1\n",
- "v= 5.93 ft/s and thetax= 73.7 degrees\n",
- "Method 2\n",
- "V= 5.93 ft/s and Thetax= 73.9 degrees\n"
- ]
- }
- ],
- "prompt_number": 76
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-18, Page No 209"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "theta=pi/3 #rad\n",
- "\n",
- "#Calculations\n",
- "t=sqrt(theta) #s\n",
- "thetadot=2*t \n",
- "thetadoubledot=2\n",
- "r=2*t**2\n",
- "rdot=4*t\n",
- "rdoubledot=4\n",
- "ax=rdoubledot-(r*thetadoubledot*thetadoubledot) #ft/s**2\n",
- "ay=2*rdot*thetadot+r*thetadoubledot #ft/s**2\n",
- "a=sqrt(ax**2+ay**2) # fr/s**2\n",
- "thetax=30+arctan(ax/ay)*(180/pi) #degrees\n",
- "#Solving by cartesian co-ordinate system yields same solution\n",
- "\n",
- "#Result\n",
- "print'The value of acceleration is',round(a,1),\"ft/s**2\"\n",
- "print'The value of thetax is',round(thetax,1),\"degrees\"\n",
- "#Decimal accuracy causes discrepancy in answers\n",
- "# The ans for thetax is incorrcet in textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of acceleration is 21.4 ft/s**2\n",
- "The value of thetax is 18.2 degrees\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-21, Page No 211"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Va=5 #ft/s\n",
- "# as theta=70 degrees\n",
- "sintheta=0.94\n",
- "costheta=0.34\n",
- "l=6.24 #ft\n",
- "\n",
- "#Calculations\n",
- "Vb=(-costheta/sintheta)*Va #ft/s\n",
- "\n",
- "#Result\n",
- "print'The value of Vb is',round(Vb,2),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of Vb is -1.81 ft/s\n"
- ]
- }
- ],
- "prompt_number": 81
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.25-25, Page No 214"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "theta=linspace(0,360,13)\n",
- "\n",
- "#Calculations\n",
- "#Defining everything in terms of matrices \n",
- "t=(theta*pi)/(180) #s converting degrees to radians\n",
- "costheta=cos(t) \n",
- "sintheta=sin(t)\n",
- "x=2*costheta #ft\n",
- "v=-12*sintheta #ft/s\n",
- "a=-72*costheta #ft/s**2\n",
- "\n",
- "#Plotting\n",
- "# 1\n",
- "plot(t,x)\n",
- "# 2\n",
- "plot(t,v)\n",
- "# 3\n",
- "plot(t,a)\n",
- "xlabel('t(s)')\n",
- "ylabel('x(ft) , v(ft/s) ,a(ft/s**2)')\n",
- "\n",
- "#Result\n",
- "print'The results are the plots'\n",
- "print'The curve in blue colour represents t vs x'\n",
- "print'The curve in green represents t vs v'\n",
- "print'The curve in red represents t vs a'\n",
- "# All the 3 curves have been plotted in the same graph. "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The results are the plots\n",
- "The curve in blue colour represents t vs x\n",
- "The curve in green represents t vs v\n",
- "The curve in red represents t vs a\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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9jtmdU9i1axczZ87UDx+FhYVha2tb7GSzjY0NqTdSWX5sOZFxkexL2scQzyH6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\nhBBCT4qCEEIIvf8HaEMe+5zwRtYAAAAASUVORK5CYII=\n",
- "text": [
- "<matplotlib.figure.Figure at 0x5e5d030>"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-26, Page No 215"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=1.2 #m\n",
- "w0=0 #rpm\n",
- "w=2000 #rpm\n",
- "t=20 #s\n",
- "\n",
- "#Calculations\n",
- "alpha=(w-w0)/t \n",
- "alpha_rad=(alpha*2*pi)/60 #converting to radians/s**2\n",
- "\n",
- "#Result\n",
- "print'The angular acceleration is',round(alpha_rad,1),\"radians/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular acceleration is 10.5 radians/s**2\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-27, Page No 216"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w0=0 #rad/s\n",
- "w=209 #rad/s\n",
- "t=20 #s\n",
- "\n",
- "#Calculations\n",
- "theta=0.5*(w+w0)*t #rad\n",
- "theta_rev=round(theta/(2*pi)) #revolutions rounding off\n",
- "\n",
- "#Result\n",
- "print'The flywheel makes',round(theta_rev),\"revolutions\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The flywheel makes 333.0 revolutions\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-28, Page No 216"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w0=0 #rad/s\n",
- "alpha=10.5 #rad/s**2\n",
- "t=0.6 #s\n",
- "r=0.6 #m\n",
- "\n",
- "#Calculations\n",
- "w=w0+alpha*t #rad/s\n",
- "v=r*w #m/s\n",
- "a_t=r*alpha #m/s**2\n",
- "a_n=r*w*w #m/s**2\n",
- "a=sqrt(a_t**2+a_n**2) #m/s**2\n",
- "phi=arctan(a_t/a_n)*(180/pi) #degrees\n",
- "\n",
- "#result\n",
- "print'The tangential velocity is',round(v,2),\"m/s\"\n",
- "print'The acceleration is',round(a,1),\"m/s**2\"\n",
- "print'and the angle is',round(phi,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tangential velocity is 3.78 m/s\n",
- "The acceleration is 24.6 m/s**2\n",
- "and the angle is 14.8 degrees\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-29, Page No 216"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=4 #ft\n",
- "wb=40 #rpm\n",
- "we=60 #rpm\n",
- "\n",
- "#Calculations\n",
- "r=l/2 #ft\n",
- "vb=r*((wb*2*pi)/60) #ft/s\n",
- "ve=r*((we*2*pi)/60) # ft/s\n",
- "\n",
- "#Result\n",
- "print'The linear speeds are:'\n",
- "print'vb=',round(vb,2),\"ft/s\"\n",
- "print'and ve=',round(ve,1),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The linear speeds are:\n",
- "vb= 8.38 ft/s\n",
- "and ve= 12.6 ft/s\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-30, Page No 217"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "wb=40 #rpm\n",
- "we=60 #rpm\n",
- "t1=5 #s using different symbol to avoid conflict in decleration\n",
- "t=2 #s\n",
- "#Calculations\n",
- "\n",
- "alpha=(((we*2*pi)/60)-((wb*2*pi)/60))/t1 #rad/s**2\n",
- "w=((wb*2*pi)/60)+alpha*t #rad/s\n",
- "#Components of acceleration are\n",
- "a_t=r*alpha #ft/s**2\n",
- "a_n=r*w**2 #ft/s**2\n",
- "\n",
- "#result\n",
- "print'The tangential acceleration is',round(a_t,3),\"ft/s**2\"\n",
- "print'The normal acceleration is',round(a_n,1),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tangential acceleration is 0.838 ft/s**2\n",
- "The normal acceleration is 50.5 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-31, Page No 217"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=200 #mm\n",
- "w0=(800*2*pi)/60 #rpm\n",
- "w=0 #rpm\n",
- "t=600 #s\n",
- "\n",
- "#Calculations\n",
- "alpha=(w-w0)/t #rad/s**2 (deceleration)\n",
- "\n",
- "#result\n",
- "print'The angular acceleration is',round(alpha,2),\"radian/s**2\"\n",
- "# The negative sign indicates that the wheel decelerates\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular acceleration is -0.14 radian/s**2\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-32, Page No 217"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#The symbols used here differ from the textbook solution to avoid conflict \n",
- "t1=0 #s\n",
- "t2=0.5 #s\n",
- "t3=2.5 #s\n",
- "t4=3**-1 #s\n",
- "w=200 #rpm\n",
- "w0=0 #rpm\n",
- "\n",
- "#Calculations\n",
- "theta1=0.5*(w0+(w*60**-1))*t2 #rev\n",
- "theta2=(w*60**-1)*(t3-t2) #rev\n",
- "theta3=(2**-1)*((w*60**-1)+w0)*t4 #rev here the values of w and w0 are interchanged but essentially the value comes out to be the same hence the decleration has not been changed\n",
- "theta=theta1+theta2+theta3 #rev\n",
- "\n",
- "#Result\n",
- "print'The wheel undergoes',round(theta,2),\"revolutions\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wheel undergoes 8.06 revolutions\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-34, Page No 218"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=1 #s\n",
- "r=4 #m\n",
- "\n",
- "#Calculations\n",
- "s=t**3+3 #m\n",
- "theta=s/r #rad\n",
- "dtheta_dt=0.75*t**2 #rad/s\n",
- "Vx=-4*sin(theta)*dtheta_dt #m/s\n",
- "Vy=4*cos(theta)*dtheta_dt #m/s\n",
- "V=(Vx**2+Vy**2)**0.5 #m/s\n",
- "\n",
- "#Result\n",
- "print'The components of velocity are:'\n",
- "print'Vx=',round(Vx,2),\"m/s\"\n",
- "print'Vy=',round(Vy,2),\"m/s\"\n",
- "print'V=',round(V),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The components of velocity are:\n",
- "Vx= -2.52 m/s\n",
- "Vy= 1.62 m/s\n",
- "V= 3.0 m/s\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-35, Page No 218"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=1 #s\n",
- "theta=1 #rad\n",
- "\n",
- "#Calculations\n",
- "dtheta_dt=0.75*t**2 #rad/s\n",
- "acc=1.5*t #rad/s**2\n",
- "ax=-4*cos(theta)*dtheta_dt**2-(4*sin(theta)*acc) #m/s**2 (to left)\n",
- "ay=-4*sin(theta)*dtheta_dt**2+(4*cos(theta)*acc) #m/s**2 (up)\n",
- "a=sqrt(ax**2+ay**2) #m/s**2\n",
- "\n",
- "#result\n",
- "print'The acceleration is',round(a,2),\"m/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration is 6.41 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 38
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-36, Page No 218"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=2 #s\n",
- "\n",
- "#Calculations\n",
- "#Velocity\n",
- "vx=8*t-3 #ft/s\n",
- "vy=3*t**2 #ft/s\n",
- "v=sqrt(vx**2+vy**2) #ft/s\n",
- "theta_x=arctan(vy*vx**-1)*(180/pi) #degrees\n",
- "#Acceleration\n",
- "ax=8 #ft/s**2\n",
- "ay=6*t #ft/s**2\n",
- "a=sqrt(ax**2+ay**2) #ft/s**2\n",
- "phi_x=arctan(ay*ax**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The velocity is',round(v,1),\"ft/s\"\n",
- "print'and the angle is',round(theta_x,1),\"degrees\"\n",
- "print'The acceleration is',round(a,1),\"ft/s**2\"\n",
- "print'and the angle it makes is',round(phi_x,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity is 17.7 ft/s\n",
- "and the angle is 42.7 degrees\n",
- "The acceleration is 14.4 ft/s**2\n",
- "and the angle it makes is 56.3 degrees\n"
- ]
- }
- ],
- "prompt_number": 46
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-37, Page No 219"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "V_ao=29.3 #ft/s\n",
- "OA=50 #ft\n",
- "theta=45 #degrees\n",
- "OB=50*sqrt(2) #ft\n",
- "\n",
- "#Calculations\n",
- "w_ao=V_ao/OA #rad/s\n",
- "V_bo=V_ao*cos(theta) #ft/s\n",
- "w_bo=V_bo/OB #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular velocity with respect to the observer is',round(w_ao,3),\"rad/s\"\n",
- "print' The angular velocity after moving 50ft is',round(w_bo,3),\"rad/s\"\n",
- "# The answer for w_bo is incorrect in textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity with respect to the observer is 0.586 rad/s\n",
- " The angular velocity after moving 50ft is 0.218 rad/s\n"
- ]
- }
- ],
- "prompt_number": 47
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-38, Page No 219"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initiliztaion of variables\n",
- "# as theta=30 degrees\n",
- "costheta=sqrt(3)*2**-1\n",
- "tantheta=sqrt(3)**-1\n",
- "r=[100*tantheta*(180/pi),100] #ft\n",
- "v=17.6 #ft/s\n",
- "\n",
- "#Calculations\n",
- "v_1=100*costheta**-1*costheta**-1\n",
- "w=v/v_1 #rad/s (clockwise)\n",
- "\n",
- "#result\n",
- "print'The angular velocity is',round(w,3),\"rad/s clockwise\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity is 0.132 rad/s clockwise\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-39, Page No 220"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=2 #s\n",
- "\n",
- "#Calculations\n",
- "Vx=20*t+5 #m/s\n",
- "Vy=t**2-20 #m/s\n",
- "#As indefinite integral is not possible \n",
- "x=10*t**2+5*t+5 #m\n",
- "y=0.5*t**2-20*t-15 #m\n",
- "ax=20 #m/s**2\n",
- "ay=2*t #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The displacement components are x=',round(x),\"m\",'and y=',round(y),\"m.\"\n",
- "print'The velocity components are: Vx=',round(Vx),\"m/s\",'and Vy=',round(Vy),\"m/s\"\n",
- "print'The acceleration components are: ax=',round(ax),\"m/s**2\",'and ay=',round(ay),\"m/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The displacement components are x= 55.0 m and y= -53.0 m.\n",
- "The velocity components are: Vx= 45.0 m/s and Vy= -16.0 m/s\n",
- "The acceleration components are: ax= 20.0 m/s**2 and ay= 4.0 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 55
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-40, Page No 221"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=0.1 #m\n",
- "v=20 #m/s\n",
- "a_g=6 #m/s**2\n",
- "d2=0.150 #m\n",
- "\n",
- "#Calculations\n",
- "r=d/2 #m\n",
- "w=v/r #rad/s\n",
- "vb=d2*0.5*w #m/s\n",
- "alpha=a_g/r #rad/s**2\n",
- "a_t=d2*0.5*alpha #rad/s**2 tangential acceleration\n",
- "a_n=d2*0.5*w*w #m/s**2 normal acceleration\n",
- "a=sqrt(a_t**2+a_n**2) #m/s**2 linear acceleration\n",
- "\n",
- "#Result\n",
- "print'The linear velocity is',round(vb),\"m/s\"\n",
- "print'The acceleration is',round(a),\"m/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The linear velocity is 30.0 m/s\n",
- "The acceleration is 12000.0 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 57
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-41, Page No 221"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=40 degrees\n",
- "sintheta=0.64\n",
- "costheta=0.77\n",
- "tantheta=0.83\n",
- "x=100 #ft\n",
- "ax=0 #ft/s**2\n",
- "ay=-32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#vox=vocos40....(1)\n",
- "#voy=vox*t-1/2(32.2)t^2...(2)\n",
- "#Simplyfying eq (1) and eq(2)\n",
- "t_f=((x*tantheta)/(0.5*(-ay)))**0.5 #s time of flight\n",
- "Vo=x/(costheta*t_f) #ft/s\n",
- "#As the max height occurs at half wat through the flight\n",
- "t=t_f/2 #s\n",
- "ymax=Vo*sintheta*t+(0.5*ay*t*t) #ft the formula has positive sign as ay is defined negative\n",
- "\n",
- "#result\n",
- "print'The max height the ball will reach is',round(ymax,1),\"ft\"\n",
- "\n",
- "# The ans in textbook is incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The max height the ball will reach is 20.8 ft\n"
- ]
- }
- ],
- "prompt_number": 14
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter12_1.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter12_1.ipynb deleted file mode 100755 index 0f5cd8eb..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter12_1.ipynb +++ /dev/null @@ -1,1521 +0,0 @@ -{
- "metadata": {
- "name": "chapter12.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 12: Kinematics of a Particle"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex 12.12-1, Page No 200"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "# Initilization of variables\n",
- "t=4 #seconds\n",
- "\n",
- "# Calculations\n",
- "#Displacement \n",
- "x=3*t**3+t+2 #ft\n",
- "# Velocity\n",
- "v=9*t**2+1 # ft/s\n",
- "# Acceleration\n",
- "a=18*t # ft/s**2\n",
- "\n",
- "# Result\n",
- "print'The dipalacemnt is',round(x),\"ft\"\n",
- "print'The velocity is ',round(v),\"ft/s\"\n",
- "print'The acceleration is ',round(a),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The dipalacemnt is 198.0 ft\n",
- "The velocity is 145.0 ft/s\n",
- "The acceleration is 72.0 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-2, Page No 201"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "# Initilization of variables\n",
- "t1=4 #s\n",
- "t2=5 #s\n",
- "\n",
- "# Calculation\n",
- "v1=9*t1**2+1 # ft/s\n",
- "v2=9*t2**2+1 # ft/s\n",
- "a=(v2-v1)/(t2-t1) # m/s**2\n",
- "\n",
- "# Result\n",
- "print'The acceleration during fifth second is',round(a),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration during fifth second is 81.0 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-3, Page No 201"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Defining Matrices\n",
- "t=[0,1,2,3,4,5,10] #s\n",
- "# equation for s is s=8*t**2+2*t, Thus the different values of s corresponding to t is:\n",
- "#Displacement matrix\n",
- "s=[0,10,36,78,136,210,820]\n",
- "# Eqn for v is v=16*t+2,Thus the different values of v corresponding to t is:\n",
- "#Velocity Matrix\n",
- "v=[0,18,34,50,66,82,162]\n",
- "# Eqn for a is a=16, Thus the different values of a corresponding to t is:\n",
- "#Acceleration Matrix\n",
- "a=[16,16,16,16,16,16,16]\n",
- "#Plotting the curves\n",
- "#S-T curve\n",
- "plot(t,s)\n",
- "plot(t,v)\n",
- "plot(t,a)\n",
- "xlabel('t(s)')\n",
- "ylabel('s(m), v(m/s) & a(m/s**2)')\n",
- "\n",
- "#Result\n",
- "print'The graphs are the solutions'\n",
- "print'blue line is for \"s\" vs \"t\" '\n",
- "print'green line is for \"v\" vs \"t\" '\n",
- "print'red line is for \"a\" vs \"t\" '\n",
- "# All the 3 graphs have been combined into a single graph"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The graphs are the solutions\n",
- "blue line is for \"s\" vs \"t\" \n",
- "green line is for \"v\" vs \"t\" \n",
- "red line is for \"a\" vs \"t\" \n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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6dOlyx8E2hSQHIepWVaVWVI2PVxODwaB1RMIaNHty+OSTT5g2bRqdrWTOmyQH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\ndbbT+qMIjWk6x0un09XZ51j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- "text": [
- "<matplotlib.figure.Figure at 0x594d050>"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example No 12.12-4, Page No 202"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_o=0 #ft/s\n",
- "v_f=88#ft/s\n",
- "t=28 #s\n",
- "\n",
- "#Calculations\n",
- "k=(v_f-v_o)*t**-1 #ft/s**2\n",
- "s=((v_f-v_o)/2)*t #ft\n",
- "\n",
- "#Result\n",
- "print'The value of constant k is',round(k,2),\"ft/s**2\"\n",
- "print'The displacement is ',round(s),\"ft\"\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of constant k is 3.14 ft/s**2\n",
- "The displacement is 1232.0 ft\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-5, Page No 202"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_o=0 #ft/s\n",
- "v_f1=30 #ft/s\n",
- "v_f2=0 #ft/s\n",
- "t1=3 #s\n",
- "t2=2 #s\n",
- "\n",
- "#Calculations\n",
- "#Plotting the v-t curve\n",
- "#Velocity matrix \n",
- "v=[v_o,v_f1,v_f2]\n",
- "#Time matrix\n",
- "t=[0,3,5]\n",
- "plot(t,v)\n",
- "xlabel('t')\n",
- "ylabel('v')\n",
- "#Part \"b\"\n",
- "#Acceleration at 3s\n",
- "a1=(v_f1-v_o)/t1 #ft/s**2\n",
- "#Acceleration at 5s\n",
- "a2=(v_f2-v_f1)/t2 #ft/s**2\n",
- "#Part \"c\"\n",
- "s=(v_f1*t1*0.5)+(v_f1*t2*0.5) #ft\n",
- "#Part \"d\"\n",
- "#Simplfying the equation we get\n",
- "#7.5t**2-30t+5=0\n",
- "a=7.5\n",
- "b=-30\n",
- "c=5\n",
- "q=sqrt(b**2-4*a*c)\n",
- "x1=(-b+q)/(2*a)\n",
- "x2=(-b-q)/(2*a)\n",
- "#As x1 is greater than 2 it does not hold as a solution\n",
- "t=x2 #s\n",
- "#Hence total time is\n",
- "T=t1+t #s\n",
- "\n",
- "#Result\n",
- "print'The graph is the solution for part a'\n",
- "print'The acceleration at 3rd second is',round(a1),\"ft/s**2\"\n",
- "print'The acceleration at 5th second is',round(a2),\"ft/s**2\"\n",
- "print'The displacement is',round(s),\"ft\"\n",
- "print'The total time is',round(T,3),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The graph is the solution for part a\n",
- "The acceleration at 3rd second is 10.0 ft/s**2\n",
- "The acceleration at 5th second is -15.0 ft/s**2\n",
- "The displacement is 75.0 ft\n",
- "The total time is 3.174 s\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x594d2d0>"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example No 12.12-6, Page No 203"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_o=2 #m/s\n",
- "y_o=120 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Solve using ground as datum\n",
- "y=0\n",
- "#Simplfying the equation\n",
- "a=4.9\n",
- "b=-2\n",
- "c=-120\n",
- "q=sqrt(b**2-4*a*c)\n",
- "x1=(-b+q)/(2*a) #s\n",
- "x2=(-b-q)/(2*a) #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(x1,2),\"s\"\n",
- "#As x2 is negative and negative time does not make any physical sense\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 5.16 s\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-7, Page No 204"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Vo1=80 #ft/s\n",
- "Vo2=60 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Simplfying by equating the two times\n",
- "t=(-(Vo2*2)-(g*0.5*4))/(Vo1-Vo2-(g*0.5*4)) #s\n",
- "#Substituting this t in s we get\n",
- "s=(Vo1*t)-(0.5*g*t*t) #ft\n",
- "\n",
- "#Result\n",
- "print'The time obtained is',round(t,2),\"s\"\n",
- "print'and the ball meets at',round(s,1),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time obtained is 4.15 s\n",
- "and the ball meets at 54.5 ft\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-8, Page No 204"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=40 degrees\n",
- "costheta=0.77\n",
- "tantheta=0.83\n",
- "x=100 #ft\n",
- "ay=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Simplfying the equation\n",
- "t=((tantheta*x)*(ay/2)**-1)**0.5 #s\n",
- "#Velocity calculations\n",
- "Vo=100*(costheta*t)**-1 #ft/s\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The initial speed should be',round(Vo,1),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The initial speed should be 57.2 ft/s\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-9, Page No 204"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=[0,1,2,3,4,5,6] #s\n",
- "#Solving the Differential Equations we obtain\n",
- "# Eqn for s is s=(t+1)**3,Thus the different values of s corresponding to t is:\n",
- "#Displacement matrix\n",
- "s=[1,8,27,64,125,216,343]\n",
- "# Eqn for v is v=3*(t+1)**2, Thus the different values of v corresponding to t is:\n",
- "# Velocity matrix\n",
- "v=[3,12,27,48,75,108,147]\n",
- "# Eqn for a is a=6*(t+1),Thus the different values of a corresponding to t is:\n",
- "# Acceleration matrix\n",
- "a=[6,12,18,24,30,36,42]\n",
- "#Plotting\n",
- "plot(t,s)\n",
- "plot(t,v)\n",
- "plot(t,a)\n",
- "xlabel('t(s)')\n",
- "ylabel('s(ft) , v(ft/s) & a(ft/s**2)')\n",
- "\n",
- "#Result\n",
- "print'The result are the plots that have been generated'\n",
- "print'blue line is for \"s\" vs \"t\"'\n",
- "print'green line is for \"v\" vs \"t\"'\n",
- "print'red line is for \"a\" vs \"t\"'\n",
- "# All the graphs have been plotted on a single graph"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The result are the plots that have been generated\n",
- "blue line is for \"s\" vs \"t\"\n",
- "green line is for \"v\" vs \"t\"\n",
- "red line is for \"a\" vs \"t\"\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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w96effsqUKVOK3L5WrVr4+vry888/V0B0QpSfJAQhSmje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\n",
- "text": [
- "<matplotlib.figure.Figure at 0x58313f0>"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-10, Page No 205"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=3 #s\n",
- "\n",
- "#Calculations\n",
- "#After solving the differential equation\n",
- "s=(3**-1)*(t+2)**3 #ft\n",
- "v=(t+2)**2 #ft/s\n",
- "a=2*(t+2) #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The displacement at t=3s is',round(s,1),\"ft\"\n",
- "print'The velocity at t=3s is',round(v),\"ft/s\"\n",
- "print'The acceleration at t=3s is',round(a),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The displacement at t=3s is 41.7 ft\n",
- "The velocity at t=3s is 25.0 ft/s\n",
- "The acceleration at t=3s is 10.0 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-12, Page No 206"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#Calling upward direction positive\n",
- "xdot1=6 #ft/s\n",
- "xdot3=3 #ft/s\n",
- "xdoubledot=2 #ft/s**2\n",
- "xdoubledot3=-4 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "xdot=-xdot1 #ft/s\n",
- "xdot2=2*xdot-xdot3 #ft/s\n",
- "xdoubledot2=2*xdoubledot-xdoubledot3 #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of velocity is',round(xdot2,3),\"ft/s (down)\"\n",
- "print'The value of acceleration is',round(xdoubledot2,3),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of velocity is -15.0 ft/s (down)\n",
- "The value of acceleration is 8.0 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-16, Page No 207"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=4 #s\n",
- "\n",
- "#Calculations\n",
- "#Part (a)\n",
- "x=t**3 #in\n",
- "y=-2*t**2 #in\n",
- "z=2*t #in\n",
- "#Part (b)\n",
- "#Theory question\n",
- "#Part(c)\n",
- "#Unit vector calculation\n",
- "m=(4**2+1**1+(-3)**2)**0.5\n",
- "e_l=[4*m**-1,m**-1,-3*m**-1]\n",
- "v=[3*t**2,-4*t,2] #in/s\n",
- "#Projection of v on n at t=4s\n",
- "dot=[v[0]*e_l[0],v[1]*e_l[1],v[2]*e_l[2]]\n",
- "#dot=v.*e_l #in/s\n",
- "a=dot[0]+dot[1]+dot[2] #in/s\n",
- "\n",
- "#Result\n",
- "print'The co-ordinates of position are x=',round(x),\"in ,\",round(y),\"in and \",round(z),\"in respectively\"\n",
- "print'The projection of v on n at t=4s is',round(a,1),\"in/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The co-ordinates of position are x= 64.0 in , -32.0 in and 8.0 in respectively\n",
- "The projection of v on n at t=4s is 33.3 in/s\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-17, Page No 208"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "theta=pi/3 #rad\n",
- "\n",
- "#Calculations\n",
- "#Method (a)\n",
- "t=(theta)**0.5 #s\n",
- "r=2*theta\n",
- "rdot=4*t\n",
- "thetadot=2*t\n",
- "#Velocity calculations\n",
- "x=r*thetadot\n",
- "v=((rdot)**2+x**2)**0.5 #ft/s\n",
- "#Theta calculations\n",
- "thetax=30+arctan(rdot/x)*(180/pi) #degrees\n",
- "#Method (b)\n",
- "x=2*theta*cos(theta) #ft\n",
- "y=2*theta*sin(theta) #ft\n",
- "xdot=4*t*((cos(t**2)))+2*t**2*(-sin(t**2))*(2*t) #ft/s\n",
- "ydot=4*t**2*sin(t**2)+2*t**2*cos(t**2)*2*t #ft/s\n",
- "V=(xdot**2+ydot**2)**0.5 #ft/s\n",
- "Thetax=arctan(ydot/-xdot)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'By both the methods we obtain v and thetax as:'\n",
- "print'Method 1'\n",
- "print'v=',round(v,2),\"ft/s\",'and thetax=',round(thetax,1),\"degrees\"\n",
- "print'Method 2'\n",
- "print'V=',round(v,2),\"ft/s\",'and Thetax=',round(Thetax,1),\"degrees\"\n",
- "# The answer may wary due to decimal point accuracy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "By both the methods we obtain v and thetax as:\n",
- "Method 1\n",
- "v= 5.93 ft/s and thetax= 73.7 degrees\n",
- "Method 2\n",
- "V= 5.93 ft/s and Thetax= 73.9 degrees\n"
- ]
- }
- ],
- "prompt_number": 76
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-18, Page No 209"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "theta=pi/3 #rad\n",
- "\n",
- "#Calculations\n",
- "t=sqrt(theta) #s\n",
- "thetadot=2*t \n",
- "thetadoubledot=2\n",
- "r=2*t**2\n",
- "rdot=4*t\n",
- "rdoubledot=4\n",
- "ax=rdoubledot-(r*thetadoubledot*thetadoubledot) #ft/s**2\n",
- "ay=2*rdot*thetadot+r*thetadoubledot #ft/s**2\n",
- "a=sqrt(ax**2+ay**2) # fr/s**2\n",
- "thetax=30+arctan(ax/ay)*(180/pi) #degrees\n",
- "#Solving by cartesian co-ordinate system yields same solution\n",
- "\n",
- "#Result\n",
- "print'The value of acceleration is',round(a,1),\"ft/s**2\"\n",
- "print'The value of thetax is',round(thetax,1),\"degrees\"\n",
- "#Decimal accuracy causes discrepancy in answers\n",
- "# The ans for thetax is incorrcet in textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of acceleration is 21.4 ft/s**2\n",
- "The value of thetax is 18.2 degrees\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-21, Page No 211"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Va=5 #ft/s\n",
- "# as theta=70 degrees\n",
- "sintheta=0.94\n",
- "costheta=0.34\n",
- "l=6.24 #ft\n",
- "\n",
- "#Calculations\n",
- "Vb=(-costheta/sintheta)*Va #ft/s\n",
- "\n",
- "#Result\n",
- "print'The value of Vb is',round(Vb,2),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of Vb is -1.81 ft/s\n"
- ]
- }
- ],
- "prompt_number": 81
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.25-25, Page No 214"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "theta=linspace(0,360,13)\n",
- "\n",
- "#Calculations\n",
- "#Defining everything in terms of matrices \n",
- "t=(theta*pi)/(180) #s converting degrees to radians\n",
- "costheta=cos(t) \n",
- "sintheta=sin(t)\n",
- "x=2*costheta #ft\n",
- "v=-12*sintheta #ft/s\n",
- "a=-72*costheta #ft/s**2\n",
- "\n",
- "#Plotting\n",
- "# 1\n",
- "plot(t,x)\n",
- "# 2\n",
- "plot(t,v)\n",
- "# 3\n",
- "plot(t,a)\n",
- "xlabel('t(s)')\n",
- "ylabel('x(ft) , v(ft/s) ,a(ft/s**2)')\n",
- "\n",
- "#Result\n",
- "print'The results are the plots'\n",
- "print'The curve in blue colour represents t vs x'\n",
- "print'The curve in green represents t vs v'\n",
- "print'The curve in red represents t vs a'\n",
- "# All the 3 curves have been plotted in the same graph. "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The results are the plots\n",
- "The curve in blue colour represents t vs x\n",
- "The curve in green represents t vs v\n",
- "The curve in red represents t vs a\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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9jtmdU9i1axczZ87UDx+FhYVha2tb7GSzjY0NqTdSWX5sOZFxkexL2scQzyH6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\nhBBCT4qCEEIIvf8HaEMe+5zwRtYAAAAASUVORK5CYII=\n",
- "text": [
- "<matplotlib.figure.Figure at 0x5e5d030>"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-26, Page No 215"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=1.2 #m\n",
- "w0=0 #rpm\n",
- "w=2000 #rpm\n",
- "t=20 #s\n",
- "\n",
- "#Calculations\n",
- "alpha=(w-w0)/t \n",
- "alpha_rad=(alpha*2*pi)/60 #converting to radians/s**2\n",
- "\n",
- "#Result\n",
- "print'The angular acceleration is',round(alpha_rad,1),\"radians/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular acceleration is 10.5 radians/s**2\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-27, Page No 216"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w0=0 #rad/s\n",
- "w=209 #rad/s\n",
- "t=20 #s\n",
- "\n",
- "#Calculations\n",
- "theta=0.5*(w+w0)*t #rad\n",
- "theta_rev=round(theta/(2*pi)) #revolutions rounding off\n",
- "\n",
- "#Result\n",
- "print'The flywheel makes',round(theta_rev),\"revolutions\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The flywheel makes 333.0 revolutions\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-28, Page No 216"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w0=0 #rad/s\n",
- "alpha=10.5 #rad/s**2\n",
- "t=0.6 #s\n",
- "r=0.6 #m\n",
- "\n",
- "#Calculations\n",
- "w=w0+alpha*t #rad/s\n",
- "v=r*w #m/s\n",
- "a_t=r*alpha #m/s**2\n",
- "a_n=r*w*w #m/s**2\n",
- "a=sqrt(a_t**2+a_n**2) #m/s**2\n",
- "phi=arctan(a_t/a_n)*(180/pi) #degrees\n",
- "\n",
- "#result\n",
- "print'The tangential velocity is',round(v,2),\"m/s\"\n",
- "print'The acceleration is',round(a,1),\"m/s**2\"\n",
- "print'and the angle is',round(phi,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tangential velocity is 3.78 m/s\n",
- "The acceleration is 24.6 m/s**2\n",
- "and the angle is 14.8 degrees\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-29, Page No 216"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=4 #ft\n",
- "wb=40 #rpm\n",
- "we=60 #rpm\n",
- "\n",
- "#Calculations\n",
- "r=l/2 #ft\n",
- "vb=r*((wb*2*pi)/60) #ft/s\n",
- "ve=r*((we*2*pi)/60) # ft/s\n",
- "\n",
- "#Result\n",
- "print'The linear speeds are:'\n",
- "print'vb=',round(vb,2),\"ft/s\"\n",
- "print'and ve=',round(ve,1),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The linear speeds are:\n",
- "vb= 8.38 ft/s\n",
- "and ve= 12.6 ft/s\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-30, Page No 217"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "wb=40 #rpm\n",
- "we=60 #rpm\n",
- "t1=5 #s using different symbol to avoid conflict in decleration\n",
- "t=2 #s\n",
- "#Calculations\n",
- "\n",
- "alpha=(((we*2*pi)/60)-((wb*2*pi)/60))/t1 #rad/s**2\n",
- "w=((wb*2*pi)/60)+alpha*t #rad/s\n",
- "#Components of acceleration are\n",
- "a_t=r*alpha #ft/s**2\n",
- "a_n=r*w**2 #ft/s**2\n",
- "\n",
- "#result\n",
- "print'The tangential acceleration is',round(a_t,3),\"ft/s**2\"\n",
- "print'The normal acceleration is',round(a_n,1),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tangential acceleration is 0.838 ft/s**2\n",
- "The normal acceleration is 50.5 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-31, Page No 217"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=200 #mm\n",
- "w0=(800*2*pi)/60 #rpm\n",
- "w=0 #rpm\n",
- "t=600 #s\n",
- "\n",
- "#Calculations\n",
- "alpha=(w-w0)/t #rad/s**2 (deceleration)\n",
- "\n",
- "#result\n",
- "print'The angular acceleration is',round(alpha,2),\"radian/s**2\"\n",
- "# The negative sign indicates that the wheel decelerates\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular acceleration is -0.14 radian/s**2\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-32, Page No 217"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#The symbols used here differ from the textbook solution to avoid conflict \n",
- "t1=0 #s\n",
- "t2=0.5 #s\n",
- "t3=2.5 #s\n",
- "t4=3**-1 #s\n",
- "w=200 #rpm\n",
- "w0=0 #rpm\n",
- "\n",
- "#Calculations\n",
- "theta1=0.5*(w0+(w*60**-1))*t2 #rev\n",
- "theta2=(w*60**-1)*(t3-t2) #rev\n",
- "theta3=(2**-1)*((w*60**-1)+w0)*t4 #rev here the values of w and w0 are interchanged but essentially the value comes out to be the same hence the decleration has not been changed\n",
- "theta=theta1+theta2+theta3 #rev\n",
- "\n",
- "#Result\n",
- "print'The wheel undergoes',round(theta,2),\"revolutions\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wheel undergoes 8.06 revolutions\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-34, Page No 218"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=1 #s\n",
- "r=4 #m\n",
- "\n",
- "#Calculations\n",
- "s=t**3+3 #m\n",
- "theta=s/r #rad\n",
- "dtheta_dt=0.75*t**2 #rad/s\n",
- "Vx=-4*sin(theta)*dtheta_dt #m/s\n",
- "Vy=4*cos(theta)*dtheta_dt #m/s\n",
- "V=(Vx**2+Vy**2)**0.5 #m/s\n",
- "\n",
- "#Result\n",
- "print'The components of velocity are:'\n",
- "print'Vx=',round(Vx,2),\"m/s\"\n",
- "print'Vy=',round(Vy,2),\"m/s\"\n",
- "print'V=',round(V),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The components of velocity are:\n",
- "Vx= -2.52 m/s\n",
- "Vy= 1.62 m/s\n",
- "V= 3.0 m/s\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-35, Page No 218"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=1 #s\n",
- "theta=1 #rad\n",
- "\n",
- "#Calculations\n",
- "dtheta_dt=0.75*t**2 #rad/s\n",
- "acc=1.5*t #rad/s**2\n",
- "ax=-4*cos(theta)*dtheta_dt**2-(4*sin(theta)*acc) #m/s**2 (to left)\n",
- "ay=-4*sin(theta)*dtheta_dt**2+(4*cos(theta)*acc) #m/s**2 (up)\n",
- "a=sqrt(ax**2+ay**2) #m/s**2\n",
- "\n",
- "#result\n",
- "print'The acceleration is',round(a,2),\"m/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration is 6.41 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 38
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-36, Page No 218"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=2 #s\n",
- "\n",
- "#Calculations\n",
- "#Velocity\n",
- "vx=8*t-3 #ft/s\n",
- "vy=3*t**2 #ft/s\n",
- "v=sqrt(vx**2+vy**2) #ft/s\n",
- "theta_x=arctan(vy*vx**-1)*(180/pi) #degrees\n",
- "#Acceleration\n",
- "ax=8 #ft/s**2\n",
- "ay=6*t #ft/s**2\n",
- "a=sqrt(ax**2+ay**2) #ft/s**2\n",
- "phi_x=arctan(ay*ax**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The velocity is',round(v,1),\"ft/s\"\n",
- "print'and the angle is',round(theta_x,1),\"degrees\"\n",
- "print'The acceleration is',round(a,1),\"ft/s**2\"\n",
- "print'and the angle it makes is',round(phi_x,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity is 17.7 ft/s\n",
- "and the angle is 42.7 degrees\n",
- "The acceleration is 14.4 ft/s**2\n",
- "and the angle it makes is 56.3 degrees\n"
- ]
- }
- ],
- "prompt_number": 46
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-37, Page No 219"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "V_ao=29.3 #ft/s\n",
- "OA=50 #ft\n",
- "theta=45 #degrees\n",
- "OB=50*sqrt(2) #ft\n",
- "\n",
- "#Calculations\n",
- "w_ao=V_ao/OA #rad/s\n",
- "V_bo=V_ao*cos(theta) #ft/s\n",
- "w_bo=V_bo/OB #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular velocity with respect to the observer is',round(w_ao,3),\"rad/s\"\n",
- "print' The angular velocity after moving 50ft is',round(w_bo,3),\"rad/s\"\n",
- "# The answer for w_bo is incorrect in textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity with respect to the observer is 0.586 rad/s\n",
- " The angular velocity after moving 50ft is 0.218 rad/s\n"
- ]
- }
- ],
- "prompt_number": 47
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-38, Page No 219"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initiliztaion of variables\n",
- "# as theta=30 degrees\n",
- "costheta=sqrt(3)*2**-1\n",
- "tantheta=sqrt(3)**-1\n",
- "r=[100*tantheta*(180/pi),100] #ft\n",
- "v=17.6 #ft/s\n",
- "\n",
- "#Calculations\n",
- "v_1=100*costheta**-1*costheta**-1\n",
- "w=v/v_1 #rad/s (clockwise)\n",
- "\n",
- "#result\n",
- "print'The angular velocity is',round(w,3),\"rad/s clockwise\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity is 0.132 rad/s clockwise\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-39, Page No 220"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=2 #s\n",
- "\n",
- "#Calculations\n",
- "Vx=20*t+5 #m/s\n",
- "Vy=t**2-20 #m/s\n",
- "#As indefinite integral is not possible \n",
- "x=10*t**2+5*t+5 #m\n",
- "y=0.5*t**2-20*t-15 #m\n",
- "ax=20 #m/s**2\n",
- "ay=2*t #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The displacement components are x=',round(x),\"m\",'and y=',round(y),\"m.\"\n",
- "print'The velocity components are: Vx=',round(Vx),\"m/s\",'and Vy=',round(Vy),\"m/s\"\n",
- "print'The acceleration components are: ax=',round(ax),\"m/s**2\",'and ay=',round(ay),\"m/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The displacement components are x= 55.0 m and y= -53.0 m.\n",
- "The velocity components are: Vx= 45.0 m/s and Vy= -16.0 m/s\n",
- "The acceleration components are: ax= 20.0 m/s**2 and ay= 4.0 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 55
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-40, Page No 221"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=0.1 #m\n",
- "v=20 #m/s\n",
- "a_g=6 #m/s**2\n",
- "d2=0.150 #m\n",
- "\n",
- "#Calculations\n",
- "r=d/2 #m\n",
- "w=v/r #rad/s\n",
- "vb=d2*0.5*w #m/s\n",
- "alpha=a_g/r #rad/s**2\n",
- "a_t=d2*0.5*alpha #rad/s**2 tangential acceleration\n",
- "a_n=d2*0.5*w*w #m/s**2 normal acceleration\n",
- "a=sqrt(a_t**2+a_n**2) #m/s**2 linear acceleration\n",
- "\n",
- "#Result\n",
- "print'The linear velocity is',round(vb),\"m/s\"\n",
- "print'The acceleration is',round(a),\"m/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The linear velocity is 30.0 m/s\n",
- "The acceleration is 12000.0 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 57
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-41, Page No 221"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=40 degrees\n",
- "sintheta=0.64\n",
- "costheta=0.77\n",
- "tantheta=0.83\n",
- "x=100 #ft\n",
- "ax=0 #ft/s**2\n",
- "ay=-32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#vox=vocos40....(1)\n",
- "#voy=vox*t-1/2(32.2)t^2...(2)\n",
- "#Simplyfying eq (1) and eq(2)\n",
- "t_f=((x*tantheta)/(0.5*(-ay)))**0.5 #s time of flight\n",
- "Vo=x/(costheta*t_f) #ft/s\n",
- "#As the max height occurs at half wat through the flight\n",
- "t=t_f/2 #s\n",
- "ymax=Vo*sintheta*t+(0.5*ay*t*t) #ft the formula has positive sign as ay is defined negative\n",
- "\n",
- "#result\n",
- "print'The max height the ball will reach is',round(ymax,1),\"ft\"\n",
- "\n",
- "# The ans in textbook is incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The max height the ball will reach is 20.8 ft\n"
- ]
- }
- ],
- "prompt_number": 14
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter12_2.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter12_2.ipynb deleted file mode 100755 index 0f5cd8eb..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter12_2.ipynb +++ /dev/null @@ -1,1521 +0,0 @@ -{
- "metadata": {
- "name": "chapter12.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 12: Kinematics of a Particle"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex 12.12-1, Page No 200"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "# Initilization of variables\n",
- "t=4 #seconds\n",
- "\n",
- "# Calculations\n",
- "#Displacement \n",
- "x=3*t**3+t+2 #ft\n",
- "# Velocity\n",
- "v=9*t**2+1 # ft/s\n",
- "# Acceleration\n",
- "a=18*t # ft/s**2\n",
- "\n",
- "# Result\n",
- "print'The dipalacemnt is',round(x),\"ft\"\n",
- "print'The velocity is ',round(v),\"ft/s\"\n",
- "print'The acceleration is ',round(a),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The dipalacemnt is 198.0 ft\n",
- "The velocity is 145.0 ft/s\n",
- "The acceleration is 72.0 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-2, Page No 201"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "# Initilization of variables\n",
- "t1=4 #s\n",
- "t2=5 #s\n",
- "\n",
- "# Calculation\n",
- "v1=9*t1**2+1 # ft/s\n",
- "v2=9*t2**2+1 # ft/s\n",
- "a=(v2-v1)/(t2-t1) # m/s**2\n",
- "\n",
- "# Result\n",
- "print'The acceleration during fifth second is',round(a),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration during fifth second is 81.0 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-3, Page No 201"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Defining Matrices\n",
- "t=[0,1,2,3,4,5,10] #s\n",
- "# equation for s is s=8*t**2+2*t, Thus the different values of s corresponding to t is:\n",
- "#Displacement matrix\n",
- "s=[0,10,36,78,136,210,820]\n",
- "# Eqn for v is v=16*t+2,Thus the different values of v corresponding to t is:\n",
- "#Velocity Matrix\n",
- "v=[0,18,34,50,66,82,162]\n",
- "# Eqn for a is a=16, Thus the different values of a corresponding to t is:\n",
- "#Acceleration Matrix\n",
- "a=[16,16,16,16,16,16,16]\n",
- "#Plotting the curves\n",
- "#S-T curve\n",
- "plot(t,s)\n",
- "plot(t,v)\n",
- "plot(t,a)\n",
- "xlabel('t(s)')\n",
- "ylabel('s(m), v(m/s) & a(m/s**2)')\n",
- "\n",
- "#Result\n",
- "print'The graphs are the solutions'\n",
- "print'blue line is for \"s\" vs \"t\" '\n",
- "print'green line is for \"v\" vs \"t\" '\n",
- "print'red line is for \"a\" vs \"t\" '\n",
- "# All the 3 graphs have been combined into a single graph"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The graphs are the solutions\n",
- "blue line is for \"s\" vs \"t\" \n",
- "green line is for \"v\" vs \"t\" \n",
- "red line is for \"a\" vs \"t\" \n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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6dOlyx8E2hSQHIepWVaVWVI2PVxODwaB1RMIaNHty+OSTT5g2bRqdrWTOmyQH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\ndbbT+qMIjWk6x0un09XZ51j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- "text": [
- "<matplotlib.figure.Figure at 0x594d050>"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example No 12.12-4, Page No 202"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_o=0 #ft/s\n",
- "v_f=88#ft/s\n",
- "t=28 #s\n",
- "\n",
- "#Calculations\n",
- "k=(v_f-v_o)*t**-1 #ft/s**2\n",
- "s=((v_f-v_o)/2)*t #ft\n",
- "\n",
- "#Result\n",
- "print'The value of constant k is',round(k,2),\"ft/s**2\"\n",
- "print'The displacement is ',round(s),\"ft\"\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of constant k is 3.14 ft/s**2\n",
- "The displacement is 1232.0 ft\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-5, Page No 202"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_o=0 #ft/s\n",
- "v_f1=30 #ft/s\n",
- "v_f2=0 #ft/s\n",
- "t1=3 #s\n",
- "t2=2 #s\n",
- "\n",
- "#Calculations\n",
- "#Plotting the v-t curve\n",
- "#Velocity matrix \n",
- "v=[v_o,v_f1,v_f2]\n",
- "#Time matrix\n",
- "t=[0,3,5]\n",
- "plot(t,v)\n",
- "xlabel('t')\n",
- "ylabel('v')\n",
- "#Part \"b\"\n",
- "#Acceleration at 3s\n",
- "a1=(v_f1-v_o)/t1 #ft/s**2\n",
- "#Acceleration at 5s\n",
- "a2=(v_f2-v_f1)/t2 #ft/s**2\n",
- "#Part \"c\"\n",
- "s=(v_f1*t1*0.5)+(v_f1*t2*0.5) #ft\n",
- "#Part \"d\"\n",
- "#Simplfying the equation we get\n",
- "#7.5t**2-30t+5=0\n",
- "a=7.5\n",
- "b=-30\n",
- "c=5\n",
- "q=sqrt(b**2-4*a*c)\n",
- "x1=(-b+q)/(2*a)\n",
- "x2=(-b-q)/(2*a)\n",
- "#As x1 is greater than 2 it does not hold as a solution\n",
- "t=x2 #s\n",
- "#Hence total time is\n",
- "T=t1+t #s\n",
- "\n",
- "#Result\n",
- "print'The graph is the solution for part a'\n",
- "print'The acceleration at 3rd second is',round(a1),\"ft/s**2\"\n",
- "print'The acceleration at 5th second is',round(a2),\"ft/s**2\"\n",
- "print'The displacement is',round(s),\"ft\"\n",
- "print'The total time is',round(T,3),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The graph is the solution for part a\n",
- "The acceleration at 3rd second is 10.0 ft/s**2\n",
- "The acceleration at 5th second is -15.0 ft/s**2\n",
- "The displacement is 75.0 ft\n",
- "The total time is 3.174 s\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x594d2d0>"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example No 12.12-6, Page No 203"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_o=2 #m/s\n",
- "y_o=120 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Solve using ground as datum\n",
- "y=0\n",
- "#Simplfying the equation\n",
- "a=4.9\n",
- "b=-2\n",
- "c=-120\n",
- "q=sqrt(b**2-4*a*c)\n",
- "x1=(-b+q)/(2*a) #s\n",
- "x2=(-b-q)/(2*a) #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(x1,2),\"s\"\n",
- "#As x2 is negative and negative time does not make any physical sense\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 5.16 s\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-7, Page No 204"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Vo1=80 #ft/s\n",
- "Vo2=60 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Simplfying by equating the two times\n",
- "t=(-(Vo2*2)-(g*0.5*4))/(Vo1-Vo2-(g*0.5*4)) #s\n",
- "#Substituting this t in s we get\n",
- "s=(Vo1*t)-(0.5*g*t*t) #ft\n",
- "\n",
- "#Result\n",
- "print'The time obtained is',round(t,2),\"s\"\n",
- "print'and the ball meets at',round(s,1),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time obtained is 4.15 s\n",
- "and the ball meets at 54.5 ft\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-8, Page No 204"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=40 degrees\n",
- "costheta=0.77\n",
- "tantheta=0.83\n",
- "x=100 #ft\n",
- "ay=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Simplfying the equation\n",
- "t=((tantheta*x)*(ay/2)**-1)**0.5 #s\n",
- "#Velocity calculations\n",
- "Vo=100*(costheta*t)**-1 #ft/s\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The initial speed should be',round(Vo,1),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The initial speed should be 57.2 ft/s\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-9, Page No 204"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=[0,1,2,3,4,5,6] #s\n",
- "#Solving the Differential Equations we obtain\n",
- "# Eqn for s is s=(t+1)**3,Thus the different values of s corresponding to t is:\n",
- "#Displacement matrix\n",
- "s=[1,8,27,64,125,216,343]\n",
- "# Eqn for v is v=3*(t+1)**2, Thus the different values of v corresponding to t is:\n",
- "# Velocity matrix\n",
- "v=[3,12,27,48,75,108,147]\n",
- "# Eqn for a is a=6*(t+1),Thus the different values of a corresponding to t is:\n",
- "# Acceleration matrix\n",
- "a=[6,12,18,24,30,36,42]\n",
- "#Plotting\n",
- "plot(t,s)\n",
- "plot(t,v)\n",
- "plot(t,a)\n",
- "xlabel('t(s)')\n",
- "ylabel('s(ft) , v(ft/s) & a(ft/s**2)')\n",
- "\n",
- "#Result\n",
- "print'The result are the plots that have been generated'\n",
- "print'blue line is for \"s\" vs \"t\"'\n",
- "print'green line is for \"v\" vs \"t\"'\n",
- "print'red line is for \"a\" vs \"t\"'\n",
- "# All the graphs have been plotted on a single graph"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The result are the plots that have been generated\n",
- "blue line is for \"s\" vs \"t\"\n",
- "green line is for \"v\" vs \"t\"\n",
- "red line is for \"a\" vs \"t\"\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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w96effsqUKVOK3L5WrVr4+vry888/V0B0QpSfJAQhSmje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\n",
- "text": [
- "<matplotlib.figure.Figure at 0x58313f0>"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-10, Page No 205"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=3 #s\n",
- "\n",
- "#Calculations\n",
- "#After solving the differential equation\n",
- "s=(3**-1)*(t+2)**3 #ft\n",
- "v=(t+2)**2 #ft/s\n",
- "a=2*(t+2) #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The displacement at t=3s is',round(s,1),\"ft\"\n",
- "print'The velocity at t=3s is',round(v),\"ft/s\"\n",
- "print'The acceleration at t=3s is',round(a),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The displacement at t=3s is 41.7 ft\n",
- "The velocity at t=3s is 25.0 ft/s\n",
- "The acceleration at t=3s is 10.0 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-12, Page No 206"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#Calling upward direction positive\n",
- "xdot1=6 #ft/s\n",
- "xdot3=3 #ft/s\n",
- "xdoubledot=2 #ft/s**2\n",
- "xdoubledot3=-4 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "xdot=-xdot1 #ft/s\n",
- "xdot2=2*xdot-xdot3 #ft/s\n",
- "xdoubledot2=2*xdoubledot-xdoubledot3 #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of velocity is',round(xdot2,3),\"ft/s (down)\"\n",
- "print'The value of acceleration is',round(xdoubledot2,3),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of velocity is -15.0 ft/s (down)\n",
- "The value of acceleration is 8.0 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-16, Page No 207"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=4 #s\n",
- "\n",
- "#Calculations\n",
- "#Part (a)\n",
- "x=t**3 #in\n",
- "y=-2*t**2 #in\n",
- "z=2*t #in\n",
- "#Part (b)\n",
- "#Theory question\n",
- "#Part(c)\n",
- "#Unit vector calculation\n",
- "m=(4**2+1**1+(-3)**2)**0.5\n",
- "e_l=[4*m**-1,m**-1,-3*m**-1]\n",
- "v=[3*t**2,-4*t,2] #in/s\n",
- "#Projection of v on n at t=4s\n",
- "dot=[v[0]*e_l[0],v[1]*e_l[1],v[2]*e_l[2]]\n",
- "#dot=v.*e_l #in/s\n",
- "a=dot[0]+dot[1]+dot[2] #in/s\n",
- "\n",
- "#Result\n",
- "print'The co-ordinates of position are x=',round(x),\"in ,\",round(y),\"in and \",round(z),\"in respectively\"\n",
- "print'The projection of v on n at t=4s is',round(a,1),\"in/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The co-ordinates of position are x= 64.0 in , -32.0 in and 8.0 in respectively\n",
- "The projection of v on n at t=4s is 33.3 in/s\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-17, Page No 208"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "theta=pi/3 #rad\n",
- "\n",
- "#Calculations\n",
- "#Method (a)\n",
- "t=(theta)**0.5 #s\n",
- "r=2*theta\n",
- "rdot=4*t\n",
- "thetadot=2*t\n",
- "#Velocity calculations\n",
- "x=r*thetadot\n",
- "v=((rdot)**2+x**2)**0.5 #ft/s\n",
- "#Theta calculations\n",
- "thetax=30+arctan(rdot/x)*(180/pi) #degrees\n",
- "#Method (b)\n",
- "x=2*theta*cos(theta) #ft\n",
- "y=2*theta*sin(theta) #ft\n",
- "xdot=4*t*((cos(t**2)))+2*t**2*(-sin(t**2))*(2*t) #ft/s\n",
- "ydot=4*t**2*sin(t**2)+2*t**2*cos(t**2)*2*t #ft/s\n",
- "V=(xdot**2+ydot**2)**0.5 #ft/s\n",
- "Thetax=arctan(ydot/-xdot)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'By both the methods we obtain v and thetax as:'\n",
- "print'Method 1'\n",
- "print'v=',round(v,2),\"ft/s\",'and thetax=',round(thetax,1),\"degrees\"\n",
- "print'Method 2'\n",
- "print'V=',round(v,2),\"ft/s\",'and Thetax=',round(Thetax,1),\"degrees\"\n",
- "# The answer may wary due to decimal point accuracy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "By both the methods we obtain v and thetax as:\n",
- "Method 1\n",
- "v= 5.93 ft/s and thetax= 73.7 degrees\n",
- "Method 2\n",
- "V= 5.93 ft/s and Thetax= 73.9 degrees\n"
- ]
- }
- ],
- "prompt_number": 76
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-18, Page No 209"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "theta=pi/3 #rad\n",
- "\n",
- "#Calculations\n",
- "t=sqrt(theta) #s\n",
- "thetadot=2*t \n",
- "thetadoubledot=2\n",
- "r=2*t**2\n",
- "rdot=4*t\n",
- "rdoubledot=4\n",
- "ax=rdoubledot-(r*thetadoubledot*thetadoubledot) #ft/s**2\n",
- "ay=2*rdot*thetadot+r*thetadoubledot #ft/s**2\n",
- "a=sqrt(ax**2+ay**2) # fr/s**2\n",
- "thetax=30+arctan(ax/ay)*(180/pi) #degrees\n",
- "#Solving by cartesian co-ordinate system yields same solution\n",
- "\n",
- "#Result\n",
- "print'The value of acceleration is',round(a,1),\"ft/s**2\"\n",
- "print'The value of thetax is',round(thetax,1),\"degrees\"\n",
- "#Decimal accuracy causes discrepancy in answers\n",
- "# The ans for thetax is incorrcet in textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of acceleration is 21.4 ft/s**2\n",
- "The value of thetax is 18.2 degrees\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-21, Page No 211"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Va=5 #ft/s\n",
- "# as theta=70 degrees\n",
- "sintheta=0.94\n",
- "costheta=0.34\n",
- "l=6.24 #ft\n",
- "\n",
- "#Calculations\n",
- "Vb=(-costheta/sintheta)*Va #ft/s\n",
- "\n",
- "#Result\n",
- "print'The value of Vb is',round(Vb,2),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of Vb is -1.81 ft/s\n"
- ]
- }
- ],
- "prompt_number": 81
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.25-25, Page No 214"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "theta=linspace(0,360,13)\n",
- "\n",
- "#Calculations\n",
- "#Defining everything in terms of matrices \n",
- "t=(theta*pi)/(180) #s converting degrees to radians\n",
- "costheta=cos(t) \n",
- "sintheta=sin(t)\n",
- "x=2*costheta #ft\n",
- "v=-12*sintheta #ft/s\n",
- "a=-72*costheta #ft/s**2\n",
- "\n",
- "#Plotting\n",
- "# 1\n",
- "plot(t,x)\n",
- "# 2\n",
- "plot(t,v)\n",
- "# 3\n",
- "plot(t,a)\n",
- "xlabel('t(s)')\n",
- "ylabel('x(ft) , v(ft/s) ,a(ft/s**2)')\n",
- "\n",
- "#Result\n",
- "print'The results are the plots'\n",
- "print'The curve in blue colour represents t vs x'\n",
- "print'The curve in green represents t vs v'\n",
- "print'The curve in red represents t vs a'\n",
- "# All the 3 curves have been plotted in the same graph. "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The results are the plots\n",
- "The curve in blue colour represents t vs x\n",
- "The curve in green represents t vs v\n",
- "The curve in red represents t vs a\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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9jtmdU9i1axczZ87UDx+FhYVha2tb7GSzjY0NqTdSWX5sOZFxkexL2scQzyH6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\nhBBCT4qCEEIIvf8HaEMe+5zwRtYAAAAASUVORK5CYII=\n",
- "text": [
- "<matplotlib.figure.Figure at 0x5e5d030>"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-26, Page No 215"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=1.2 #m\n",
- "w0=0 #rpm\n",
- "w=2000 #rpm\n",
- "t=20 #s\n",
- "\n",
- "#Calculations\n",
- "alpha=(w-w0)/t \n",
- "alpha_rad=(alpha*2*pi)/60 #converting to radians/s**2\n",
- "\n",
- "#Result\n",
- "print'The angular acceleration is',round(alpha_rad,1),\"radians/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular acceleration is 10.5 radians/s**2\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-27, Page No 216"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w0=0 #rad/s\n",
- "w=209 #rad/s\n",
- "t=20 #s\n",
- "\n",
- "#Calculations\n",
- "theta=0.5*(w+w0)*t #rad\n",
- "theta_rev=round(theta/(2*pi)) #revolutions rounding off\n",
- "\n",
- "#Result\n",
- "print'The flywheel makes',round(theta_rev),\"revolutions\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The flywheel makes 333.0 revolutions\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-28, Page No 216"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w0=0 #rad/s\n",
- "alpha=10.5 #rad/s**2\n",
- "t=0.6 #s\n",
- "r=0.6 #m\n",
- "\n",
- "#Calculations\n",
- "w=w0+alpha*t #rad/s\n",
- "v=r*w #m/s\n",
- "a_t=r*alpha #m/s**2\n",
- "a_n=r*w*w #m/s**2\n",
- "a=sqrt(a_t**2+a_n**2) #m/s**2\n",
- "phi=arctan(a_t/a_n)*(180/pi) #degrees\n",
- "\n",
- "#result\n",
- "print'The tangential velocity is',round(v,2),\"m/s\"\n",
- "print'The acceleration is',round(a,1),\"m/s**2\"\n",
- "print'and the angle is',round(phi,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tangential velocity is 3.78 m/s\n",
- "The acceleration is 24.6 m/s**2\n",
- "and the angle is 14.8 degrees\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-29, Page No 216"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=4 #ft\n",
- "wb=40 #rpm\n",
- "we=60 #rpm\n",
- "\n",
- "#Calculations\n",
- "r=l/2 #ft\n",
- "vb=r*((wb*2*pi)/60) #ft/s\n",
- "ve=r*((we*2*pi)/60) # ft/s\n",
- "\n",
- "#Result\n",
- "print'The linear speeds are:'\n",
- "print'vb=',round(vb,2),\"ft/s\"\n",
- "print'and ve=',round(ve,1),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The linear speeds are:\n",
- "vb= 8.38 ft/s\n",
- "and ve= 12.6 ft/s\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-30, Page No 217"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "wb=40 #rpm\n",
- "we=60 #rpm\n",
- "t1=5 #s using different symbol to avoid conflict in decleration\n",
- "t=2 #s\n",
- "#Calculations\n",
- "\n",
- "alpha=(((we*2*pi)/60)-((wb*2*pi)/60))/t1 #rad/s**2\n",
- "w=((wb*2*pi)/60)+alpha*t #rad/s\n",
- "#Components of acceleration are\n",
- "a_t=r*alpha #ft/s**2\n",
- "a_n=r*w**2 #ft/s**2\n",
- "\n",
- "#result\n",
- "print'The tangential acceleration is',round(a_t,3),\"ft/s**2\"\n",
- "print'The normal acceleration is',round(a_n,1),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tangential acceleration is 0.838 ft/s**2\n",
- "The normal acceleration is 50.5 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-31, Page No 217"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=200 #mm\n",
- "w0=(800*2*pi)/60 #rpm\n",
- "w=0 #rpm\n",
- "t=600 #s\n",
- "\n",
- "#Calculations\n",
- "alpha=(w-w0)/t #rad/s**2 (deceleration)\n",
- "\n",
- "#result\n",
- "print'The angular acceleration is',round(alpha,2),\"radian/s**2\"\n",
- "# The negative sign indicates that the wheel decelerates\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular acceleration is -0.14 radian/s**2\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-32, Page No 217"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#The symbols used here differ from the textbook solution to avoid conflict \n",
- "t1=0 #s\n",
- "t2=0.5 #s\n",
- "t3=2.5 #s\n",
- "t4=3**-1 #s\n",
- "w=200 #rpm\n",
- "w0=0 #rpm\n",
- "\n",
- "#Calculations\n",
- "theta1=0.5*(w0+(w*60**-1))*t2 #rev\n",
- "theta2=(w*60**-1)*(t3-t2) #rev\n",
- "theta3=(2**-1)*((w*60**-1)+w0)*t4 #rev here the values of w and w0 are interchanged but essentially the value comes out to be the same hence the decleration has not been changed\n",
- "theta=theta1+theta2+theta3 #rev\n",
- "\n",
- "#Result\n",
- "print'The wheel undergoes',round(theta,2),\"revolutions\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wheel undergoes 8.06 revolutions\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-34, Page No 218"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=1 #s\n",
- "r=4 #m\n",
- "\n",
- "#Calculations\n",
- "s=t**3+3 #m\n",
- "theta=s/r #rad\n",
- "dtheta_dt=0.75*t**2 #rad/s\n",
- "Vx=-4*sin(theta)*dtheta_dt #m/s\n",
- "Vy=4*cos(theta)*dtheta_dt #m/s\n",
- "V=(Vx**2+Vy**2)**0.5 #m/s\n",
- "\n",
- "#Result\n",
- "print'The components of velocity are:'\n",
- "print'Vx=',round(Vx,2),\"m/s\"\n",
- "print'Vy=',round(Vy,2),\"m/s\"\n",
- "print'V=',round(V),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The components of velocity are:\n",
- "Vx= -2.52 m/s\n",
- "Vy= 1.62 m/s\n",
- "V= 3.0 m/s\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-35, Page No 218"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=1 #s\n",
- "theta=1 #rad\n",
- "\n",
- "#Calculations\n",
- "dtheta_dt=0.75*t**2 #rad/s\n",
- "acc=1.5*t #rad/s**2\n",
- "ax=-4*cos(theta)*dtheta_dt**2-(4*sin(theta)*acc) #m/s**2 (to left)\n",
- "ay=-4*sin(theta)*dtheta_dt**2+(4*cos(theta)*acc) #m/s**2 (up)\n",
- "a=sqrt(ax**2+ay**2) #m/s**2\n",
- "\n",
- "#result\n",
- "print'The acceleration is',round(a,2),\"m/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration is 6.41 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 38
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-36, Page No 218"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=2 #s\n",
- "\n",
- "#Calculations\n",
- "#Velocity\n",
- "vx=8*t-3 #ft/s\n",
- "vy=3*t**2 #ft/s\n",
- "v=sqrt(vx**2+vy**2) #ft/s\n",
- "theta_x=arctan(vy*vx**-1)*(180/pi) #degrees\n",
- "#Acceleration\n",
- "ax=8 #ft/s**2\n",
- "ay=6*t #ft/s**2\n",
- "a=sqrt(ax**2+ay**2) #ft/s**2\n",
- "phi_x=arctan(ay*ax**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The velocity is',round(v,1),\"ft/s\"\n",
- "print'and the angle is',round(theta_x,1),\"degrees\"\n",
- "print'The acceleration is',round(a,1),\"ft/s**2\"\n",
- "print'and the angle it makes is',round(phi_x,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity is 17.7 ft/s\n",
- "and the angle is 42.7 degrees\n",
- "The acceleration is 14.4 ft/s**2\n",
- "and the angle it makes is 56.3 degrees\n"
- ]
- }
- ],
- "prompt_number": 46
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-37, Page No 219"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "V_ao=29.3 #ft/s\n",
- "OA=50 #ft\n",
- "theta=45 #degrees\n",
- "OB=50*sqrt(2) #ft\n",
- "\n",
- "#Calculations\n",
- "w_ao=V_ao/OA #rad/s\n",
- "V_bo=V_ao*cos(theta) #ft/s\n",
- "w_bo=V_bo/OB #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular velocity with respect to the observer is',round(w_ao,3),\"rad/s\"\n",
- "print' The angular velocity after moving 50ft is',round(w_bo,3),\"rad/s\"\n",
- "# The answer for w_bo is incorrect in textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity with respect to the observer is 0.586 rad/s\n",
- " The angular velocity after moving 50ft is 0.218 rad/s\n"
- ]
- }
- ],
- "prompt_number": 47
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-38, Page No 219"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initiliztaion of variables\n",
- "# as theta=30 degrees\n",
- "costheta=sqrt(3)*2**-1\n",
- "tantheta=sqrt(3)**-1\n",
- "r=[100*tantheta*(180/pi),100] #ft\n",
- "v=17.6 #ft/s\n",
- "\n",
- "#Calculations\n",
- "v_1=100*costheta**-1*costheta**-1\n",
- "w=v/v_1 #rad/s (clockwise)\n",
- "\n",
- "#result\n",
- "print'The angular velocity is',round(w,3),\"rad/s clockwise\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity is 0.132 rad/s clockwise\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-39, Page No 220"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=2 #s\n",
- "\n",
- "#Calculations\n",
- "Vx=20*t+5 #m/s\n",
- "Vy=t**2-20 #m/s\n",
- "#As indefinite integral is not possible \n",
- "x=10*t**2+5*t+5 #m\n",
- "y=0.5*t**2-20*t-15 #m\n",
- "ax=20 #m/s**2\n",
- "ay=2*t #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The displacement components are x=',round(x),\"m\",'and y=',round(y),\"m.\"\n",
- "print'The velocity components are: Vx=',round(Vx),\"m/s\",'and Vy=',round(Vy),\"m/s\"\n",
- "print'The acceleration components are: ax=',round(ax),\"m/s**2\",'and ay=',round(ay),\"m/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The displacement components are x= 55.0 m and y= -53.0 m.\n",
- "The velocity components are: Vx= 45.0 m/s and Vy= -16.0 m/s\n",
- "The acceleration components are: ax= 20.0 m/s**2 and ay= 4.0 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 55
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-40, Page No 221"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=0.1 #m\n",
- "v=20 #m/s\n",
- "a_g=6 #m/s**2\n",
- "d2=0.150 #m\n",
- "\n",
- "#Calculations\n",
- "r=d/2 #m\n",
- "w=v/r #rad/s\n",
- "vb=d2*0.5*w #m/s\n",
- "alpha=a_g/r #rad/s**2\n",
- "a_t=d2*0.5*alpha #rad/s**2 tangential acceleration\n",
- "a_n=d2*0.5*w*w #m/s**2 normal acceleration\n",
- "a=sqrt(a_t**2+a_n**2) #m/s**2 linear acceleration\n",
- "\n",
- "#Result\n",
- "print'The linear velocity is',round(vb),\"m/s\"\n",
- "print'The acceleration is',round(a),\"m/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The linear velocity is 30.0 m/s\n",
- "The acceleration is 12000.0 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 57
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-41, Page No 221"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=40 degrees\n",
- "sintheta=0.64\n",
- "costheta=0.77\n",
- "tantheta=0.83\n",
- "x=100 #ft\n",
- "ax=0 #ft/s**2\n",
- "ay=-32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#vox=vocos40....(1)\n",
- "#voy=vox*t-1/2(32.2)t^2...(2)\n",
- "#Simplyfying eq (1) and eq(2)\n",
- "t_f=((x*tantheta)/(0.5*(-ay)))**0.5 #s time of flight\n",
- "Vo=x/(costheta*t_f) #ft/s\n",
- "#As the max height occurs at half wat through the flight\n",
- "t=t_f/2 #s\n",
- "ymax=Vo*sintheta*t+(0.5*ay*t*t) #ft the formula has positive sign as ay is defined negative\n",
- "\n",
- "#result\n",
- "print'The max height the ball will reach is',round(ymax,1),\"ft\"\n",
- "\n",
- "# The ans in textbook is incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The max height the ball will reach is 20.8 ft\n"
- ]
- }
- ],
- "prompt_number": 14
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter12_3.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter12_3.ipynb deleted file mode 100755 index ee679f70..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter12_3.ipynb +++ /dev/null @@ -1,1525 +0,0 @@ -{
- "metadata": {
- "name": "chapter12.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 12: Kinematics of a Particle"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex 12.12-1, Page No 200"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "# Initilization of variables\n",
- "t=4 #seconds\n",
- "\n",
- "# Calculations\n",
- "#Displacement \n",
- "x=3*t**3+t+2 #ft\n",
- "# Velocity\n",
- "v=9*t**2+1 # ft/s\n",
- "# Acceleration\n",
- "a=18*t # ft/s**2\n",
- "\n",
- "# Result\n",
- "print'The dipalacemnt is',round(x),\"ft\"\n",
- "print'The velocity is ',round(v),\"ft/s\"\n",
- "print'The acceleration is ',round(a),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The dipalacemnt is 198.0 ft\n",
- "The velocity is 145.0 ft/s\n",
- "The acceleration is 72.0 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-2, Page No 201"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "# Initilization of variables\n",
- "t1=4 #s\n",
- "t2=5 #s\n",
- "\n",
- "# Calculation\n",
- "v1=9*t1**2+1 # ft/s\n",
- "v2=9*t2**2+1 # ft/s\n",
- "a=(v2-v1)/(t2-t1) # m/s**2\n",
- "\n",
- "# Result\n",
- "print'The acceleration during fifth second is',round(a),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration during fifth second is 81.0 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-3, Page No 201"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Defining Matrices\n",
- "t=[0,1,2,3,4,5,10] #s\n",
- "# equation for s is s=8*t**2+2*t, Thus the different values of s corresponding to t is:\n",
- "#Displacement matrix\n",
- "s=[0,10,36,78,136,210,820]\n",
- "# Eqn for v is v=16*t+2,Thus the different values of v corresponding to t is:\n",
- "#Velocity Matrix\n",
- "v=[0,18,34,50,66,82,162]\n",
- "# Eqn for a is a=16, Thus the different values of a corresponding to t is:\n",
- "#Acceleration Matrix\n",
- "a=[16,16,16,16,16,16,16]\n",
- "#Plotting the curves\n",
- "#S-T curve\n",
- "plot(t,s)\n",
- "plot(t,v)\n",
- "plot(t,a)\n",
- "xlabel('t(s)')\n",
- "ylabel('s(m), v(m/s) & a(m/s**2)')\n",
- "\n",
- "#Result\n",
- "print'The graphs are the solutions'\n",
- "print'blue line is for \"s\" vs \"t\" '\n",
- "print'green line is for \"v\" vs \"t\" '\n",
- "print'red line is for \"a\" vs \"t\" '\n",
- "# All the 3 graphs have been combined into a single graph"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The graphs are the solutions\n",
- "blue line is for \"s\" vs \"t\" \n",
- "green line is for \"v\" vs \"t\" \n",
- "red line is for \"a\" vs \"t\" \n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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6dOlyx8E2hSQHIepWVaVWVI2PVxODwaB1RMIaNHty+OSTT5g2bRqdrWTOmyQH\nIWpXXq52I2VkQGws9O6tdUTCWjT7OofY2Fjc3d154okn2LlzJxUVFXcUoBCiZZSUqDu3Xbmithok\nMYg70aDZSuXl5ezevZvo6GgOHDiAv78/GzZssER8NUjLQYiazp9XN+gZNQrWrwebeiepi/am2buV\nbigvL2fPnj1s3LiR/fv38+OPPzY5yDshyUGI6s6ehcBACA2F8HBZ3CZur9m7lXbt2sVTTz2Fu7s7\nH3/8Mc888wz5+fl3FKQQonkkJcH48bB4sbofgyQG0VzqTQ6bNm1ixowZnDp1iqioKKZOnYpNA9qs\n165dY+zYsYwcORIvLy9eeeUVAAoKCvD398fDw4OAgACKiorMx0RERODu7o6npyfx8fF38LGEaPu+\n/FJtMaxdq5beFqI5tegK6dLSUrp160ZFRQXjxo3jjTfeIDY2ln79+rF06VJWr15NYWFhtT2kjxw5\nYt5DOiUlpUbBP+lWEgKio+HXv4aPPoKJE7WORrQGzd6tdCe6desGqOMVlZWV9O7dm9jYWMLCwgAI\nCwtj+/btAMTExBASEoKtrS1GoxE3NzcSExNbMjwhWqV169RupM8/l8QgWk6LJoeqqipGjhyJXq9n\n0qRJDB06lPz8fPR6PQB6vd48fpGTk4PhltU6BoOB7OzslgxPiFZFUeC119RupAMHYMQIrSMSbVmL\nTnjr0KED//nPfyguLiYwMJCvvvqq2us6nQ5dHSNotb0WHh5ufuzr64uvr29zhCuE1aqoUEttHzum\nbul5111aRySsXUJCAgkJCU0+vtHJISwsjG7duvHcc881uK6Svb0906ZN49///jd6vZ68vDwcHR3J\nzc3FwcEBACcnJzIzM83HZGVl4eTkdNvz3ZochGjrSkth7lx19fOXX0KPHlpHJFqDn984r1y5slHH\nN7pb6bnnnmPy5Ml88MEHdb7v4sWL5plIV69e5fPPP8fHx4egoCCioqIAiIqKYsaMGQAEBQWxdetW\nysvLOXfuHKdPn2bMmDGNDU+INqWgAPz9wd4eduyQxCAsp8VmKx07doywsDCqqqqoqqriiSeeYMmS\nJRQUFBAcHExGRgZGo5Ho6Gh69eoFwKpVq9i4cSM2NjZERkYSGBhYM2CZrSTaicxMdarqtGmwejXI\nTr3iTjTrCunz58/z0UcfsX//ftLS0sxVWSdMmMBjjz1m7hKyJEkOoj04cQKmTIEXX1RnJglxp5ot\nOTz99NMMqlziAAAXL0lEQVSkpqYyZcoUxowZY97PITc3l8TEROLi4nBzc+P9999vtuAbFLAkB9HG\nff01zJoFf/kLzJundTSirWi25PDDDz8wfPjwOg9uyHuamyQH0ZbFxsLTT8OHH6pdSkI0lxYrvGct\nJDmItmrDBvjd79QEMXq01tGItqbZV0jv2LEDHx8fevfujZ2dHXZ2dvTs2fOOghRC3KQo8Kc/qT/7\n9kliENah3paDq6srn376Kd7e3jXqHGlBWg6iLamshBdegIMHYfdu6N9f64hEW9XY7856F8EZDAaG\nDh1qFYlBiLakrAwefxwuXlRbDPb2WkckxE31JofVq1czZcoUJk2aRKdOnQA1Ay2W+XVCNFlxMcyc\nCX37QlwcWMkW7UKY1dsc+P3vf0+PHj24du0aly9f5vLly5SUlFgiNiHapNxctZqqlxds3SqJQVin\nesccvL29OX78uKXiqZeMOYjW7PRpdYrqggVqhVXZuU1YSrPPVpo6dSp79uy5o6CEEPDddzBhArz6\nqjplVRKDsGb1thx69OhBaWkpnTp1wtbWVj1Ip+PSpUsWCfDnpOUgWqP4eHXw+f33IShI62hEeySL\n4ISwMps3q/WRPvkEHnhA62hEe9Vs3Uqpqan1HtyQ9wjRnv3lL/DKK+o+DJIYRGtSa8thzpw5XLly\nhaCgIO65555qhfe+++47YmNjsbOzY+vWrZYNWFoOohWoqoLly2HnTtizBwYO1Doi0d41a7fSmTNn\n2Lp1K19//TXp6ekAuLi4MG7cOEJCQhg8ePCdR9xIkhyEtbt+XS2ed+aMmhz69NE6IiFkzEEITV2+\nDI89BjY2sG0bdOumdURCqJp9KutHH31knpn0hz/8gVmzZnH06NEGnTwzM5NJkyYxdOhQvL29Wbt2\nLQAFBQX4+/vj4eFBQECAeTtRgIiICNzd3fH09CQ+Pr7BH0QIrV28CJMnw4AB8OmnkhhEK6fUw9vb\nW1EURTlw4IAyceJEZceOHcro0aPrO0xRFEXJzc1VkpKSFEVRlJKSEsXDw0NJTk5WlixZoqxevVpR\nFEUxmUzKsmXLFEVRlBMnTigjRoxQysvLlXPnzimurq5KZWVltXM2IGQhLKqqSlF27VIUNzdFefVV\n9XchrE1jvzvrbTl07NgRgJ07d/LMM8/w8MMPc/369QYlHkdHR0aOHAmo6yWGDBlCdnY2sbGxhIWF\nARAWFsb27dsBiImJISQkBFtbW4xGI25ubiQmJjYh5QlhGUePgp8fvPQS/PnPatltWdwm2oJ6k4OT\nkxO//OUv2bZtG9OmTePatWtUVVU1+kJpaWkkJSUxduxY8vPz0ev1AOj1evLz8wHIycnBYDCYjzEY\nDGRnZzf6WkK0tLQ0dVHbtGnqGMPx47K4TbQt9VZljY6OJi4ujiVLltCrVy9yc3NZs2ZNoy5y+fJl\nHn30USIjI7Gzs6v2mk6nQ1fHrdbtXgsPDzc/9vX1xdfXt1HxCNFUhYWwahVs3Ai/+Q389a/ws/+k\nhbAKCQkJJCQkNPn4epND9+7defTRR82/9+/fn/6N2JHk+vXrPProozzxxBPMmDEDUFsLeXl5ODo6\nkpubi4ODA6C2UjIzM83HZmVl4eTkVOOctyYHISyhrAzWrQOTCWbNUlsKsjGPsGY/v3FeuXJlo45v\n0R18FEXh6aefxsvLixdffNH8fFBQEFFRUQBERUWZk0ZQUBBbt26lvLycc+fOcfr0acaMGdOSIQpR\np6oq+Oc/wdNT3ZBn3z54911JDKLta9F1DgcPHmTChAkMHz7c3D0UERHBmDFjCA4OJiMjA6PRSHR0\nNL169QJg1apVbNy4ERsbGyIjIwkMDKwesKxzEBby1VewZAl06ABr1qh7MAjRWskiOCHu0PHjsGwZ\nnDwJEREQHCwzkETr1+yL4IRoL7KzYeFCePBBCAhQk8OcOZIYRPskyUG0e5cuqZvvDB8O/fpBSgq8\n8IJs3ynaN0kOot26fh3eeQc8PCAzE5KS1NlIPw1/CdGu1TuVVYi2RlHU2kfLl4PRCHFx8NNCfiHE\nTyQ5iHblm2/g5ZfhyhV13UJAgNYRCdE8KqoqyL6UTUZxBhnFGaQXp5sfZxRnNPp8MltJtAspKeqO\nbEeOwB/+oJa++KlsmBCtQklZSbUv/PSidDIu/fRvcQZ5l/Nw6O6ASy8XnO2dce7pfPOxvTMjHEfI\nVFYhbjh/HlauVPdWWLIEnn8eunbVOiohqqtSqsgtyb3tXf+Nx+WV5TjbO+Nif/ML/9bHhp4GbDva\n1nqNxn53SreSaJOuXIE334S33lJbCf/9rzoTSQgtlF4vrX7HX5xR7a4/uySb3l16V7vr9+jrgd9g\nP3MS6NO1T5116JqbJAfRplRWwj/+AStWwAMPwOHD4OqqdVSiLVMUhfNXztd6x59RnMHl8ssM7Dmw\n2h2/r4svzsPVrh9DTwNdbLpo/VGqkeQg2gRFgV271JXNffrAv/4FY8dqHZVoC65VXCPrUlb1u/5b\nvvwzL2XSo1OPGl0945zHmR/f1f0uOuha18oBGXMQrZqiwMGDakshLw9Wr4aHH5ZVzaJhFEWh4GpB\nnXf9BVcLcLJzUr/4e7ng3NP55mN7Zwb2HEj3Tt21/ij1ktpKol24dg22boW1a9XxhZdfhvnzwUba\nwuIW1yuvk12Sfds7/hs/th1ta9z13zrQ69jDkY4dWv/UNkkOok3LzlY32HnvPfjFL9TZRwEBauVU\n0f4UXyu+7fTOG4/PXzmPYw/HWu/6ne2d6dm5p9YfwyJktpJocxRFXbz29tsQH6/OPjpwQC17Idqu\nyqpKci/n3vau/8a/lVWVuPRyqXan/7D+YfNjp55O2HSQr7mmkJaDsFrXrqnrE9auVYvj/eY38NRT\n0LN93Oi1eZfLL1fr3vn5XX9OSQ79uvWrdVGXi70Lvbr0suj0ztZMupVEq5eTo3Yd/d//gY+P2nX0\n0EPSddSaVClV5F/Or3NR19XrV81f9DUWd/VywcnOic42Uhq3uVhVt9KCBQv47LPPcHBw4NixYwAU\nFBQwZ84c0tPTa+wCFxERwcaNG+nYsSNr164lQArftBuKAocOqa2EPXsgNFTdktPTU+vIxO1cvX6V\nzEuZtS7qyrqURc/OPc13+i72LgzuPRhfo685CfTr1k/u+q1Yi7YcDhw4QI8ePXjyySfNyWHp0qX0\n69ePpUuXsnr1agoLCzGZTCQnJxMaGsqRI0fIzs7Gz8+PlJQUOvzsdlFaDm1LWRlER6tJoaBA7Tqa\nPx/s7bWOrP1SFIWLpRdr7efPKM6g+Foxhp6GOqd3drWVOiXWxKpaDuPHjyctLa3ac7Gxsezbtw+A\nsLAwfH19MZlMxMTEEBISgq2tLUajETc3NxITE7n33ntbMkShkdxc+Nvf4N131U12VqyAKVOkGJ4l\nlFeW17moK6M4g662XWt099xnuM/8nL6HvtUt6hKNY/Fh/Pz8fPR6PQB6vZ78/HwAcnJyqiUCg8FA\ndna2pcMTLezwYbWVsGsXhITAl1+Cl5fWUbUdiqJQdK2ozkVdF0sv0r9H/2p3/fcMuIdHvR413/Xb\ndbbT+qMIjWk6x0un09XZ51j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- "text": [
- "<matplotlib.figure.Figure at 0x53be4f0>"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example No 12.12-4, Page No 202"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_o=0 #ft/s\n",
- "v_f=88#ft/s\n",
- "t=28 #s\n",
- "\n",
- "#Calculations\n",
- "k=(v_f-v_o)*t**-1 #ft/s**2\n",
- "s=((v_f-v_o)/2)*t #ft\n",
- "\n",
- "#Result\n",
- "print'The value of constant k is',round(k,2),\"ft/s**2\"\n",
- "print'The displacement is ',round(s),\"ft\"\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of constant k is 3.14 ft/s**2\n",
- "The displacement is 1232.0 ft\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-5, Page No 202"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "v_o=0 #ft/s\n",
- "v_f1=30 #ft/s\n",
- "v_f2=0 #ft/s\n",
- "t1=3 #s\n",
- "t2=2 #s\n",
- "\n",
- "#Calculations\n",
- "#Plotting the v-t curve\n",
- "#Velocity matrix \n",
- "v=[v_o,v_f1,v_f2]\n",
- "#Time matrix\n",
- "t=[0,3,5]\n",
- "plot(t,v)\n",
- "xlabel('t')\n",
- "ylabel('v')\n",
- "#Part \"b\"\n",
- "#Acceleration at 3s\n",
- "a1=(v_f1-v_o)/t1 #ft/s**2\n",
- "#Acceleration at 5s\n",
- "a2=(v_f2-v_f1)/t2 #ft/s**2\n",
- "#Part \"c\"\n",
- "s=(v_f1*t1*0.5)+(v_f1*t2*0.5) #ft\n",
- "#Part \"d\"\n",
- "#Simplfying the equation we get\n",
- "#7.5t**2-30t+5=0\n",
- "a=7.5\n",
- "b=-30\n",
- "c=5\n",
- "q=sqrt(b**2-4*a*c)\n",
- "x1=(-b+q)/(2*a)\n",
- "x2=(-b-q)/(2*a)\n",
- "#As x1 is greater than 2 it does not hold as a solution\n",
- "t=x2 #s\n",
- "#Hence total time is\n",
- "T=t1+t #s\n",
- "\n",
- "#Result\n",
- "print'The graph is the solution for part a'\n",
- "print'The acceleration at 3rd second is',round(a1),\"ft/s**2\"\n",
- "print'The acceleration at 5th second is',round(a2),\"ft/s**2\"\n",
- "print'The displacement is',round(s),\"ft\"\n",
- "print'The total time is',round(T,3),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The graph is the solution for part a\n",
- "The acceleration at 3rd second is 10.0 ft/s**2\n",
- "The acceleration at 5th second is -15.0 ft/s**2\n",
- "The displacement is 75.0 ft\n",
- "The total time is 3.174 s\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x5bb16b0>"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example No 12.12-6, Page No 203"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_o=2 #m/s\n",
- "y_o=120 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Solve using ground as datum\n",
- "y=0\n",
- "#Simplfying the equation\n",
- "a=4.9\n",
- "b=-2\n",
- "c=-120\n",
- "q=sqrt(b**2-4*a*c)\n",
- "x1=(-b+q)/(2*a) #s\n",
- "x2=(-b-q)/(2*a) #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(x1,2),\"s\"\n",
- "#As x2 is negative and negative time does not make any physical sense\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 5.16 s\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-7, Page No 204"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Vo1=80 #ft/s\n",
- "Vo2=60 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Simplfying by equating the two times\n",
- "t=(-(Vo2*2)-(g*0.5*4))/(Vo1-Vo2-(g*0.5*4)) #s\n",
- "#Substituting this t in s we get\n",
- "s=(Vo1*t)-(0.5*g*t*t) #ft\n",
- "\n",
- "#Result\n",
- "print'The time obtained is',round(t,2),\"s\"\n",
- "print'and the ball meets at',round(s,1),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time obtained is 4.15 s\n",
- "and the ball meets at 54.5 ft\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-8, Page No 204"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=40 degrees\n",
- "costheta=0.77\n",
- "tantheta=0.83\n",
- "x=100 #ft\n",
- "ay=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Simplfying the equation\n",
- "t=((tantheta*x)*(ay/2)**-1)**0.5 #s\n",
- "#Velocity calculations\n",
- "Vo=100*(costheta*t)**-1 #ft/s\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The initial speed should be',round(Vo,1),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The initial speed should be 57.2 ft/s\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-9, Page No 204"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "t=[0,1,2,3,4,5,6] #s\n",
- "#Solving the Differential Equations we obtain\n",
- "# Eqn for s is s=(t+1)**3,Thus the different values of s corresponding to t is:\n",
- "#Displacement matrix\n",
- "s=[1,8,27,64,125,216,343]\n",
- "# Eqn for v is v=3*(t+1)**2, Thus the different values of v corresponding to t is:\n",
- "# Velocity matrix\n",
- "v=[3,12,27,48,75,108,147]\n",
- "# Eqn for a is a=6*(t+1),Thus the different values of a corresponding to t is:\n",
- "# Acceleration matrix\n",
- "a=[6,12,18,24,30,36,42]\n",
- "#Plotting\n",
- "plot(t,s)\n",
- "plot(t,v)\n",
- "plot(t,a)\n",
- "xlabel('t(s)')\n",
- "ylabel('s(ft) , v(ft/s) & a(ft/s**2)')\n",
- "\n",
- "#Result\n",
- "print'The result are the plots that have been generated'\n",
- "print'blue line is for \"s\" vs \"t\"'\n",
- "print'green line is for \"v\" vs \"t\"'\n",
- "print'red line is for \"a\" vs \"t\"'\n",
- "# All the graphs have been plotted on a single graph"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The result are the plots that have been generated\n",
- "blue line is for \"s\" vs \"t\"\n",
- "green line is for \"v\" vs \"t\"\n",
- "red line is for \"a\" vs \"t\"\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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1a9WFbkTpnE84z4rjK/jxrx8Z22Ess3rOop1DuwqNoSTXTqlOLoQoUkEBLF8O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\n0HKcbSp42rIQQFBQEEFBQaX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w96effsqUKVOK3L5WrVr4+vry888/V0B0QpSfJAQhSmje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\n",
- "text": [
- "<matplotlib.figure.Figure at 0x5ad4a10>"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-10, Page No 205"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=3 #s\n",
- "\n",
- "#Calculations\n",
- "#After solving the differential equation\n",
- "s=(3**-1)*(t+2)**3 #ft\n",
- "v=(t+2)**2 #ft/s\n",
- "a=2*(t+2) #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The displacement at t=3s is',round(s,1),\"ft\"\n",
- "print'The velocity at t=3s is',round(v),\"ft/s\"\n",
- "print'The acceleration at t=3s is',round(a),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The displacement at t=3s is 41.7 ft\n",
- "The velocity at t=3s is 25.0 ft/s\n",
- "The acceleration at t=3s is 10.0 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-12, Page No 206"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#Calling upward direction positive\n",
- "xdot1=6 #ft/s\n",
- "xdot3=3 #ft/s\n",
- "xdoubledot=2 #ft/s**2\n",
- "xdoubledot3=-4 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "xdot=-xdot1 #ft/s\n",
- "xdot2=2*xdot-xdot3 #ft/s\n",
- "xdoubledot2=2*xdoubledot-xdoubledot3 #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of velocity is',round(xdot2,3),\"ft/s (down)\"\n",
- "print'The value of acceleration is',round(xdoubledot2,3),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of velocity is -15.0 ft/s (down)\n",
- "The value of acceleration is 8.0 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-16, Page No 207"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=4 #s\n",
- "\n",
- "#Calculations\n",
- "#Part (a)\n",
- "x=t**3 #in\n",
- "y=-2*t**2 #in\n",
- "z=2*t #in\n",
- "#Part (b)\n",
- "#Theory question\n",
- "#Part(c)\n",
- "#Unit vector calculation\n",
- "m=(4**2+1**1+(-3)**2)**0.5\n",
- "e_l=[4*m**-1,m**-1,-3*m**-1]\n",
- "v=[3*t**2,-4*t,2] #in/s\n",
- "#Projection of v on n at t=4s\n",
- "dot=[v[0]*e_l[0],v[1]*e_l[1],v[2]*e_l[2]]\n",
- "#dot=v.*e_l #in/s\n",
- "a=dot[0]+dot[1]+dot[2] #in/s\n",
- "\n",
- "#Result\n",
- "print'The co-ordinates of position are x=',round(x),\"in ,\",round(y),\"in and \",round(z),\"in respectively\"\n",
- "print'The projection of v on n at t=4s is',round(a,1),\"in/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The co-ordinates of position are x= 64.0 in , -32.0 in and 8.0 in respectively\n",
- "The projection of v on n at t=4s is 33.3 in/s\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-17, Page No 208"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "theta=pi/3 #rad\n",
- "\n",
- "#Calculations\n",
- "#Method (a)\n",
- "t=(theta)**0.5 #s\n",
- "r=2*theta\n",
- "rdot=4*t\n",
- "thetadot=2*t\n",
- "#Velocity calculations\n",
- "x=r*thetadot\n",
- "v=((rdot)**2+x**2)**0.5 #ft/s\n",
- "#Theta calculations\n",
- "thetax=30+arctan(rdot/x)*(180/pi) #degrees\n",
- "#Method (b)\n",
- "x=2*theta*cos(theta) #ft\n",
- "y=2*theta*sin(theta) #ft\n",
- "xdot=4*t*((cos(t**2)))+2*t**2*(-sin(t**2))*(2*t) #ft/s\n",
- "ydot=4*t**2*sin(t**2)+2*t**2*cos(t**2)*2*t #ft/s\n",
- "V=(xdot**2+ydot**2)**0.5 #ft/s\n",
- "Thetax=arctan(ydot/-xdot)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'By both the methods we obtain v and thetax as:'\n",
- "print'Method 1'\n",
- "print'v=',round(v,2),\"ft/s\",'and thetax=',round(thetax,1),\"degrees\"\n",
- "print'Method 2'\n",
- "print'V=',round(v,2),\"ft/s\",'and Thetax=',round(Thetax,1),\"degrees\"\n",
- "# The answer may wary due to decimal point accuracy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "By both the methods we obtain v and thetax as:\n",
- "Method 1\n",
- "v= 5.93 ft/s and thetax= 73.7 degrees\n",
- "Method 2\n",
- "V= 5.93 ft/s and Thetax= 73.9 degrees\n"
- ]
- }
- ],
- "prompt_number": 76
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-18, Page No 209"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "theta=pi/3 #rad\n",
- "\n",
- "#Calculations\n",
- "t=sqrt(theta) #s\n",
- "thetadot=2*t \n",
- "thetadoubledot=2\n",
- "r=2*t**2\n",
- "rdot=4*t\n",
- "rdoubledot=4\n",
- "ax=rdoubledot-(r*thetadoubledot*thetadoubledot) #ft/s**2\n",
- "ay=2*rdot*thetadot+r*thetadoubledot #ft/s**2\n",
- "a=sqrt(ax**2+ay**2) # fr/s**2\n",
- "thetax=30+arctan(ax/ay)*(180/pi) #degrees\n",
- "#Solving by cartesian co-ordinate system yields same solution\n",
- "\n",
- "#Result\n",
- "print'The value of acceleration is',round(a,1),\"ft/s**2\"\n",
- "print'The value of thetax is',round(thetax,1),\"degrees\"\n",
- "#Decimal accuracy causes discrepancy in answers\n",
- "# The ans for thetax is incorrcet in textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of acceleration is 21.4 ft/s**2\n",
- "The value of thetax is 18.2 degrees\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-21, Page No 211"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Va=5 #ft/s\n",
- "# as theta=70 degrees\n",
- "sintheta=0.94\n",
- "costheta=0.34\n",
- "l=6.24 #ft\n",
- "\n",
- "#Calculations\n",
- "Vb=(-costheta/sintheta)*Va #ft/s\n",
- "\n",
- "#Result\n",
- "print'The value of Vb is',round(Vb,2),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of Vb is -1.81 ft/s\n"
- ]
- }
- ],
- "prompt_number": 81
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.25-25, Page No 214"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "theta=linspace(0,360,13)\n",
- "\n",
- "#Calculations\n",
- "#Defining everything in terms of matrices \n",
- "t=(theta*pi)/(180) #s converting degrees to radians\n",
- "costheta=cos(t) \n",
- "sintheta=sin(t)\n",
- "x=2*costheta #ft\n",
- "v=-12*sintheta #ft/s\n",
- "a=-72*costheta #ft/s**2\n",
- "\n",
- "#Plotting\n",
- "# 1\n",
- "plot(t,x)\n",
- "# 2\n",
- "plot(t,v)\n",
- "# 3\n",
- "plot(t,a)\n",
- "xlabel('t(s)')\n",
- "ylabel('x(ft) , v(ft/s) ,a(ft/s**2)')\n",
- "\n",
- "#Result\n",
- "print'The results are the plots'\n",
- "print'The curve in blue colour represents t vs x'\n",
- "print'The curve in green represents t vs v'\n",
- "print'The curve in red represents t vs a'\n",
- "# All the 3 curves have been plotted in the same graph. "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The results are the plots\n",
- "The curve in blue colour represents t vs x\n",
- "The curve in green represents t vs v\n",
- "The curve in red represents t vs a\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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9jtmdU9i1axczZ87UDx+FhYVha2tb7GSzjY0NqTdSWX5sOZFxkexL2scQzyH6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\nhBBCT4qCEEIIvf8HaEMe+5zwRtYAAAAASUVORK5CYII=\n",
- "text": [
- "<matplotlib.figure.Figure at 0x5bd53f0>"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-26, Page No 215"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=1.2 #m\n",
- "w0=0 #rpm\n",
- "w=2000 #rpm\n",
- "t=20 #s\n",
- "\n",
- "#Calculations\n",
- "alpha=(w-w0)/t \n",
- "alpha_rad=(alpha*2*pi)/60 #converting to radians/s**2\n",
- "\n",
- "#Result\n",
- "print'The angular acceleration is',round(alpha_rad,1),\"radians/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular acceleration is 10.5 radians/s**2\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-27, Page No 216"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w0=0 #rad/s\n",
- "w=209 #rad/s\n",
- "t=20 #s\n",
- "\n",
- "#Calculations\n",
- "theta=0.5*(w+w0)*t #rad\n",
- "theta_rev=round(theta/(2*pi)) #revolutions rounding off\n",
- "\n",
- "#Result\n",
- "print'The flywheel makes',round(theta_rev),\"revolutions\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The flywheel makes 333.0 revolutions\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-28, Page No 216"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w0=0 #rad/s\n",
- "alpha=10.5 #rad/s**2\n",
- "t=0.6 #s\n",
- "r=0.6 #m\n",
- "\n",
- "#Calculations\n",
- "w=w0+alpha*t #rad/s\n",
- "v=r*w #m/s\n",
- "a_t=r*alpha #m/s**2\n",
- "a_n=r*w*w #m/s**2\n",
- "a=sqrt(a_t**2+a_n**2) #m/s**2\n",
- "phi=arctan(a_t/a_n)*(180/pi) #degrees\n",
- "\n",
- "#result\n",
- "print'The tangential velocity is',round(v,2),\"m/s\"\n",
- "print'The acceleration is',round(a,1),\"m/s**2\"\n",
- "print'and the angle is',round(phi,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tangential velocity is 3.78 m/s\n",
- "The acceleration is 24.6 m/s**2\n",
- "and the angle is 14.8 degrees\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-29, Page No 216"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=4 #ft\n",
- "wb=40 #rpm\n",
- "we=60 #rpm\n",
- "\n",
- "#Calculations\n",
- "r=l/2 #ft\n",
- "vb=r*((wb*2*pi)/60) #ft/s\n",
- "ve=r*((we*2*pi)/60) # ft/s\n",
- "\n",
- "#Result\n",
- "print'The linear speeds are:'\n",
- "print'vb=',round(vb,2),\"ft/s\"\n",
- "print'and ve=',round(ve,1),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The linear speeds are:\n",
- "vb= 8.38 ft/s\n",
- "and ve= 12.6 ft/s\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-30, Page No 217"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "wb=40 #rpm\n",
- "we=60 #rpm\n",
- "t1=5 #s using different symbol to avoid conflict in decleration\n",
- "t=2 #s\n",
- "#Calculations\n",
- "\n",
- "alpha=(((we*2*pi)/60)-((wb*2*pi)/60))/t1 #rad/s**2\n",
- "w=((wb*2*pi)/60)+alpha*t #rad/s\n",
- "#Components of acceleration are\n",
- "a_t=r*alpha #ft/s**2\n",
- "a_n=r*w**2 #ft/s**2\n",
- "\n",
- "#result\n",
- "print'The tangential acceleration is',round(a_t,3),\"ft/s**2\"\n",
- "print'The normal acceleration is',round(a_n,1),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tangential acceleration is 0.838 ft/s**2\n",
- "The normal acceleration is 50.5 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-31, Page No 217"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=200 #mm\n",
- "w0=(800*2*pi)/60 #rpm\n",
- "w=0 #rpm\n",
- "t=600 #s\n",
- "\n",
- "#Calculations\n",
- "alpha=(w-w0)/t #rad/s**2 (deceleration)\n",
- "\n",
- "#result\n",
- "print'The angular acceleration is',round(alpha,2),\"radian/s**2\"\n",
- "# The negative sign indicates that the wheel decelerates\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular acceleration is -0.14 radian/s**2\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-32, Page No 217"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#The symbols used here differ from the textbook solution to avoid conflict \n",
- "t1=0 #s\n",
- "t2=0.5 #s\n",
- "t3=2.5 #s\n",
- "t4=3**-1 #s\n",
- "w=200 #rpm\n",
- "w0=0 #rpm\n",
- "\n",
- "#Calculations\n",
- "theta1=0.5*(w0+(w*60**-1))*t2 #rev\n",
- "theta2=(w*60**-1)*(t3-t2) #rev\n",
- "theta3=(2**-1)*((w*60**-1)+w0)*t4 #rev here the values of w and w0 are interchanged but essentially the value comes out to be the same hence the decleration has not been changed\n",
- "theta=theta1+theta2+theta3 #rev\n",
- "\n",
- "#Result\n",
- "print'The wheel undergoes',round(theta,2),\"revolutions\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wheel undergoes 8.06 revolutions\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-34, Page No 218"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=1 #s\n",
- "r=4 #m\n",
- "\n",
- "#Calculations\n",
- "s=t**3+3 #m\n",
- "theta=s/r #rad\n",
- "dtheta_dt=0.75*t**2 #rad/s\n",
- "Vx=-4*sin(theta)*dtheta_dt #m/s\n",
- "Vy=4*cos(theta)*dtheta_dt #m/s\n",
- "V=(Vx**2+Vy**2)**0.5 #m/s\n",
- "\n",
- "#Result\n",
- "print'The components of velocity are:'\n",
- "print'Vx=',round(Vx,2),\"m/s\"\n",
- "print'Vy=',round(Vy,2),\"m/s\"\n",
- "print'V=',round(V),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The components of velocity are:\n",
- "Vx= -2.52 m/s\n",
- "Vy= 1.62 m/s\n",
- "V= 3.0 m/s\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-35, Page No 218"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=1 #s\n",
- "theta=1 #rad\n",
- "\n",
- "#Calculations\n",
- "dtheta_dt=0.75*t**2 #rad/s\n",
- "acc=1.5*t #rad/s**2\n",
- "ax=-4*cos(theta)*dtheta_dt**2-(4*sin(theta)*acc) #m/s**2 (to left)\n",
- "ay=-4*sin(theta)*dtheta_dt**2+(4*cos(theta)*acc) #m/s**2 (up)\n",
- "a=sqrt(ax**2+ay**2) #m/s**2\n",
- "\n",
- "#result\n",
- "print'The acceleration is',round(a,2),\"m/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration is 6.41 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 38
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-36, Page No 218"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=2 #s\n",
- "\n",
- "#Calculations\n",
- "#Velocity\n",
- "vx=8*t-3 #ft/s\n",
- "vy=3*t**2 #ft/s\n",
- "v=sqrt(vx**2+vy**2) #ft/s\n",
- "theta_x=arctan(vy*vx**-1)*(180/pi) #degrees\n",
- "#Acceleration\n",
- "ax=8 #ft/s**2\n",
- "ay=6*t #ft/s**2\n",
- "a=sqrt(ax**2+ay**2) #ft/s**2\n",
- "phi_x=arctan(ay*ax**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The velocity is',round(v,1),\"ft/s\"\n",
- "print'and the angle is',round(theta_x,1),\"degrees\"\n",
- "print'The acceleration is',round(a,1),\"ft/s**2\"\n",
- "print'and the angle it makes is',round(phi_x,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity is 17.7 ft/s\n",
- "and the angle is 42.7 degrees\n",
- "The acceleration is 14.4 ft/s**2\n",
- "and the angle it makes is 56.3 degrees\n"
- ]
- }
- ],
- "prompt_number": 46
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-37, Page No 219"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "V_ao=29.3 #ft/s\n",
- "OA=50 #ft\n",
- "theta=45 #degrees\n",
- "OB=50*sqrt(2) #ft\n",
- "\n",
- "#Calculations\n",
- "w_ao=V_ao/OA #rad/s\n",
- "V_bo=V_ao*cos(theta) #ft/s\n",
- "w_bo=V_bo/OB #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular velocity with respect to the observer is',round(w_ao,3),\"rad/s\"\n",
- "print' The angular velocity after moving 50ft is',round(w_bo,3),\"rad/s\"\n",
- "# The answer for w_bo is incorrect in textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity with respect to the observer is 0.586 rad/s\n",
- " The angular velocity after moving 50ft is 0.218 rad/s\n"
- ]
- }
- ],
- "prompt_number": 47
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-38, Page No 219"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initiliztaion of variables\n",
- "# as theta=30 degrees\n",
- "costheta=sqrt(3)*2**-1\n",
- "tantheta=sqrt(3)**-1\n",
- "r=[100*tantheta*(180/pi),100] #ft\n",
- "v=17.6 #ft/s\n",
- "\n",
- "#Calculations\n",
- "v_1=100*costheta**-1*costheta**-1\n",
- "w=v/v_1 #rad/s (clockwise)\n",
- "\n",
- "#result\n",
- "print'The angular velocity is',round(w,3),\"rad/s clockwise\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity is 0.132 rad/s clockwise\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-39, Page No 220"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=2 #s\n",
- "\n",
- "#Calculations\n",
- "Vx=20*t+5 #m/s\n",
- "Vy=t**2-20 #m/s\n",
- "#As indefinite integral is not possible \n",
- "x=10*t**2+5*t+5 #m\n",
- "y=0.5*t**2-20*t-15 #m\n",
- "ax=20 #m/s**2\n",
- "ay=2*t #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The displacement components are x=',round(x),\"m\",'and y=',round(y),\"m.\"\n",
- "print'The velocity components are: Vx=',round(Vx),\"m/s\",'and Vy=',round(Vy),\"m/s\"\n",
- "print'The acceleration components are: ax=',round(ax),\"m/s**2\",'and ay=',round(ay),\"m/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The displacement components are x= 55.0 m and y= -53.0 m.\n",
- "The velocity components are: Vx= 45.0 m/s and Vy= -16.0 m/s\n",
- "The acceleration components are: ax= 20.0 m/s**2 and ay= 4.0 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 55
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-40, Page No 221"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=0.1 #m\n",
- "v=20 #m/s\n",
- "a_g=6 #m/s**2\n",
- "d2=0.150 #m\n",
- "\n",
- "#Calculations\n",
- "r=d/2 #m\n",
- "w=v/r #rad/s\n",
- "vb=d2*0.5*w #m/s\n",
- "alpha=a_g/r #rad/s**2\n",
- "a_t=d2*0.5*alpha #rad/s**2 tangential acceleration\n",
- "a_n=d2*0.5*w*w #m/s**2 normal acceleration\n",
- "a=sqrt(a_t**2+a_n**2) #m/s**2 linear acceleration\n",
- "\n",
- "#Result\n",
- "print'The linear velocity is',round(vb),\"m/s\"\n",
- "print'The acceleration is',round(a),\"m/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The linear velocity is 30.0 m/s\n",
- "The acceleration is 12000.0 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 57
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-41, Page No 221"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=40 degrees\n",
- "sintheta=0.64\n",
- "costheta=0.77\n",
- "tantheta=0.83\n",
- "x=100 #ft\n",
- "ax=0 #ft/s**2\n",
- "ay=-32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#vox=vocos40....(1)\n",
- "#voy=vox*t-1/2(32.2)t^2...(2)\n",
- "#Simplyfying eq (1) and eq(2)\n",
- "t_f=((x*tantheta)/(0.5*(-ay)))**0.5 #s time of flight\n",
- "Vo=x/(costheta*t_f) #ft/s\n",
- "#As the max height occurs at half wat through the flight\n",
- "t=t_f/2 #s\n",
- "ymax=Vo*sintheta*t+(0.5*ay*t*t) #ft the formula has positive sign as ay is defined negative\n",
- "\n",
- "#result\n",
- "print'The max height the ball will reach is',round(ymax,1),\"ft\"\n",
- "\n",
- "# The ans in textbook is incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The max height the ball will reach is 20.8 ft\n"
- ]
- }
- ],
- "prompt_number": 14
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter12_4.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter12_4.ipynb deleted file mode 100755 index ee679f70..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter12_4.ipynb +++ /dev/null @@ -1,1525 +0,0 @@ -{
- "metadata": {
- "name": "chapter12.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 12: Kinematics of a Particle"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex 12.12-1, Page No 200"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "# Initilization of variables\n",
- "t=4 #seconds\n",
- "\n",
- "# Calculations\n",
- "#Displacement \n",
- "x=3*t**3+t+2 #ft\n",
- "# Velocity\n",
- "v=9*t**2+1 # ft/s\n",
- "# Acceleration\n",
- "a=18*t # ft/s**2\n",
- "\n",
- "# Result\n",
- "print'The dipalacemnt is',round(x),\"ft\"\n",
- "print'The velocity is ',round(v),\"ft/s\"\n",
- "print'The acceleration is ',round(a),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The dipalacemnt is 198.0 ft\n",
- "The velocity is 145.0 ft/s\n",
- "The acceleration is 72.0 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-2, Page No 201"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "# Initilization of variables\n",
- "t1=4 #s\n",
- "t2=5 #s\n",
- "\n",
- "# Calculation\n",
- "v1=9*t1**2+1 # ft/s\n",
- "v2=9*t2**2+1 # ft/s\n",
- "a=(v2-v1)/(t2-t1) # m/s**2\n",
- "\n",
- "# Result\n",
- "print'The acceleration during fifth second is',round(a),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration during fifth second is 81.0 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-3, Page No 201"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Defining Matrices\n",
- "t=[0,1,2,3,4,5,10] #s\n",
- "# equation for s is s=8*t**2+2*t, Thus the different values of s corresponding to t is:\n",
- "#Displacement matrix\n",
- "s=[0,10,36,78,136,210,820]\n",
- "# Eqn for v is v=16*t+2,Thus the different values of v corresponding to t is:\n",
- "#Velocity Matrix\n",
- "v=[0,18,34,50,66,82,162]\n",
- "# Eqn for a is a=16, Thus the different values of a corresponding to t is:\n",
- "#Acceleration Matrix\n",
- "a=[16,16,16,16,16,16,16]\n",
- "#Plotting the curves\n",
- "#S-T curve\n",
- "plot(t,s)\n",
- "plot(t,v)\n",
- "plot(t,a)\n",
- "xlabel('t(s)')\n",
- "ylabel('s(m), v(m/s) & a(m/s**2)')\n",
- "\n",
- "#Result\n",
- "print'The graphs are the solutions'\n",
- "print'blue line is for \"s\" vs \"t\" '\n",
- "print'green line is for \"v\" vs \"t\" '\n",
- "print'red line is for \"a\" vs \"t\" '\n",
- "# All the 3 graphs have been combined into a single graph"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The graphs are the solutions\n",
- "blue line is for \"s\" vs \"t\" \n",
- "green line is for \"v\" vs \"t\" \n",
- "red line is for \"a\" vs \"t\" \n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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6dOlyx8E2hSQHIepWVaVWVI2PVxODwaB1RMIaNHty+OSTT5g2bRqdrWTOmyQH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\ndbbT+qMIjWk6x0un09XZ51j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- "text": [
- "<matplotlib.figure.Figure at 0x53be4f0>"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example No 12.12-4, Page No 202"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_o=0 #ft/s\n",
- "v_f=88#ft/s\n",
- "t=28 #s\n",
- "\n",
- "#Calculations\n",
- "k=(v_f-v_o)*t**-1 #ft/s**2\n",
- "s=((v_f-v_o)/2)*t #ft\n",
- "\n",
- "#Result\n",
- "print'The value of constant k is',round(k,2),\"ft/s**2\"\n",
- "print'The displacement is ',round(s),\"ft\"\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of constant k is 3.14 ft/s**2\n",
- "The displacement is 1232.0 ft\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-5, Page No 202"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "v_o=0 #ft/s\n",
- "v_f1=30 #ft/s\n",
- "v_f2=0 #ft/s\n",
- "t1=3 #s\n",
- "t2=2 #s\n",
- "\n",
- "#Calculations\n",
- "#Plotting the v-t curve\n",
- "#Velocity matrix \n",
- "v=[v_o,v_f1,v_f2]\n",
- "#Time matrix\n",
- "t=[0,3,5]\n",
- "plot(t,v)\n",
- "xlabel('t')\n",
- "ylabel('v')\n",
- "#Part \"b\"\n",
- "#Acceleration at 3s\n",
- "a1=(v_f1-v_o)/t1 #ft/s**2\n",
- "#Acceleration at 5s\n",
- "a2=(v_f2-v_f1)/t2 #ft/s**2\n",
- "#Part \"c\"\n",
- "s=(v_f1*t1*0.5)+(v_f1*t2*0.5) #ft\n",
- "#Part \"d\"\n",
- "#Simplfying the equation we get\n",
- "#7.5t**2-30t+5=0\n",
- "a=7.5\n",
- "b=-30\n",
- "c=5\n",
- "q=sqrt(b**2-4*a*c)\n",
- "x1=(-b+q)/(2*a)\n",
- "x2=(-b-q)/(2*a)\n",
- "#As x1 is greater than 2 it does not hold as a solution\n",
- "t=x2 #s\n",
- "#Hence total time is\n",
- "T=t1+t #s\n",
- "\n",
- "#Result\n",
- "print'The graph is the solution for part a'\n",
- "print'The acceleration at 3rd second is',round(a1),\"ft/s**2\"\n",
- "print'The acceleration at 5th second is',round(a2),\"ft/s**2\"\n",
- "print'The displacement is',round(s),\"ft\"\n",
- "print'The total time is',round(T,3),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The graph is the solution for part a\n",
- "The acceleration at 3rd second is 10.0 ft/s**2\n",
- "The acceleration at 5th second is -15.0 ft/s**2\n",
- "The displacement is 75.0 ft\n",
- "The total time is 3.174 s\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x5bb16b0>"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example No 12.12-6, Page No 203"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_o=2 #m/s\n",
- "y_o=120 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Solve using ground as datum\n",
- "y=0\n",
- "#Simplfying the equation\n",
- "a=4.9\n",
- "b=-2\n",
- "c=-120\n",
- "q=sqrt(b**2-4*a*c)\n",
- "x1=(-b+q)/(2*a) #s\n",
- "x2=(-b-q)/(2*a) #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(x1,2),\"s\"\n",
- "#As x2 is negative and negative time does not make any physical sense\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 5.16 s\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-7, Page No 204"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Vo1=80 #ft/s\n",
- "Vo2=60 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Simplfying by equating the two times\n",
- "t=(-(Vo2*2)-(g*0.5*4))/(Vo1-Vo2-(g*0.5*4)) #s\n",
- "#Substituting this t in s we get\n",
- "s=(Vo1*t)-(0.5*g*t*t) #ft\n",
- "\n",
- "#Result\n",
- "print'The time obtained is',round(t,2),\"s\"\n",
- "print'and the ball meets at',round(s,1),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time obtained is 4.15 s\n",
- "and the ball meets at 54.5 ft\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-8, Page No 204"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=40 degrees\n",
- "costheta=0.77\n",
- "tantheta=0.83\n",
- "x=100 #ft\n",
- "ay=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Simplfying the equation\n",
- "t=((tantheta*x)*(ay/2)**-1)**0.5 #s\n",
- "#Velocity calculations\n",
- "Vo=100*(costheta*t)**-1 #ft/s\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The initial speed should be',round(Vo,1),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The initial speed should be 57.2 ft/s\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-9, Page No 204"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "t=[0,1,2,3,4,5,6] #s\n",
- "#Solving the Differential Equations we obtain\n",
- "# Eqn for s is s=(t+1)**3,Thus the different values of s corresponding to t is:\n",
- "#Displacement matrix\n",
- "s=[1,8,27,64,125,216,343]\n",
- "# Eqn for v is v=3*(t+1)**2, Thus the different values of v corresponding to t is:\n",
- "# Velocity matrix\n",
- "v=[3,12,27,48,75,108,147]\n",
- "# Eqn for a is a=6*(t+1),Thus the different values of a corresponding to t is:\n",
- "# Acceleration matrix\n",
- "a=[6,12,18,24,30,36,42]\n",
- "#Plotting\n",
- "plot(t,s)\n",
- "plot(t,v)\n",
- "plot(t,a)\n",
- "xlabel('t(s)')\n",
- "ylabel('s(ft) , v(ft/s) & a(ft/s**2)')\n",
- "\n",
- "#Result\n",
- "print'The result are the plots that have been generated'\n",
- "print'blue line is for \"s\" vs \"t\"'\n",
- "print'green line is for \"v\" vs \"t\"'\n",
- "print'red line is for \"a\" vs \"t\"'\n",
- "# All the graphs have been plotted on a single graph"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The result are the plots that have been generated\n",
- "blue line is for \"s\" vs \"t\"\n",
- "green line is for \"v\" vs \"t\"\n",
- "red line is for \"a\" vs \"t\"\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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1a9WFbkTpnE84z4rjK/jxrx8Z22Ess3rOop1DuwqNoSTXTqlOLoQoUkEBLF8O\nH3ygTj6TZFByiqIQeDWQ5SeWczb+LC93f5nL/76MQx0HrUMrkiQEIcQDXb8OkyerSSEkBFq00Dqi\nyiE7P5vvzn7HihMrqGFRg9k9Z7Nj7A5qWdbSOrRiSUIQQhSiKPDttzB7NvznP+qjRg2tozJ/iZmJ\nrD25ljV/rKGLcxdWDlnJYy0eq1Rl/EudEN58801sbW2ZPn16uRfLEUKYl1u34IUX4MIFCAwEb2+t\nIzJ/FxMv8smJT/j+4vc84/EMByYfwKORh9ZhlUmpl7vu3r07NWrUYNasWaaIRwihkb17oXNncHWF\nU6ckGTyMoij8du03Hv/ucR7b9Biu9V259PIlvhj+RaVNBiCjjISo9u7cgblz1VIUGzZA//5aR2S+\ncvJz2HJ+C8uPL0ev6JndazbjO47H2tL81wiVUUZCiIc6dQomTFDvBs6ckRXOipJ0J4l1J9fx2R+f\n0dGxIx8O/JBBrQZVqv6BkpCEIEQ1lJ8PS5eqaxisXAnjxmkdkXm6lHSJT058wtYLW3nK/SkCJwbi\n2VjbAp+mJAlBiGrm6lW1MF3t2uodQtOmWkdkXhRFISgyiOUnlhMSHcLzXZ/nr5f+wrGeo9ahmVyx\nfQgXLlzg0KFDREZGotPpcHNzw9fXlw4dOlRUjIVIH4IQZaMosH49zJ8Pb70Fr7wCFqUeVlJ15Rbk\nsu38NpafWE52fjaze85mQqcJ1LaqrXVoRlGu0hWbN29m9erV2Nvb4+PjQ5MmTVAUhdjYWEJCQkhK\nSuLVV19lwoQJJgm+yIAlIQhRagkJMGMG3LihzjHQ6PecWYpKi2LTmU2sPbmW9g7tmd1rNkNaD8FC\nV7WyZbk6lVNSUti/fz82NjYP/Pz27dts3LixXAEKIUxv5051IZspU+D776Gm6QpqVhpZeVn89NdP\nbAzbyKnYU4z2GM2v43+ls1NnrUPTllKMI0eOlOi9B5k6darSuHFjxdPT0/DerVu3lAEDBiht2rRR\nBg4cqKSkpBg+W7x4sdK6dWulXbt2yt69ex94zBKELIRQFCU9XVFmzFAUNzdFOXRI62i0p9frleNR\nx5WZO2YqdkvtlEGbBylbzm1R7uTe0Tq0ClGSa2ex90Qvv/xyid57kKlTp7Jnz55C7y1dupSBAwcS\nHh5O//79Wbp0KaCuu7Bt2zYuXrzInj17ePHFF9Hr9SU6jxCisOPHwctLHU105gz4+modkXZi0mNY\ndmQZHmvwDOLpAAAajElEQVQ8mPTTJJo3aM7ZF86yd8JexnqOrTJ9BMZQZJPR8ePHOXbsGImJiSxf\nvtzQ9pSenl7iC7Wvry+RkZGF3tuxYwfBwcEATJ48GT8/P5YuXUpAQADjxo3DysoKNzc3WrduTUhI\nCD179izjVxOi+snLg0WL4MsvYc0aeOoprSPSRk5+Djsu7WBD2AaO3zzOqPaj+Gr4VzzS9JEqN3fA\nmIpMCLm5uaSnp1NQUEB6errh/fr16/PDDz+U+YTx8fE4OqrDtxwdHYmPjwcgJiam0MXf1dWV6Ojo\nMp9HiOrmr7/USWaNG0NYGDg5aR1RxVIUhVOxp9gYtpGt57fS2akzU72m8v0z31O3Zl2tw6sUikwI\nixYtYv/+/Vy8eJH33nvPJCfX6XQPzdZFfbZgwQLDaz8/P/z8/IwcmRCVh6LAZ5/BggXw//4f/Otf\nUJ1+BMdnxPPdue/YELaBzNxMpnhN4eTMk7g1cNM6NE0FBQURFBRUqn2KTAixsbEcO3aMs2fPEhoa\net/nXbp0KXWAoN4VxMXF4eTkRGxsrGFtZhcXF6Kiogzb3bx5ExcXlwce496EIER1FhMDzz0Hyclw\n7Bi0bat1RBUjtyCXX8J/YeOZjQRHBjPSfSSfDv0U3+a+VW64aFn988fywoULi92nyISwcOFCFi1a\nRHR0NHPmzLnv84MHD5YpSH9/fzZt2sTcuXPZtGkTI0eONLw/fvx4Zs+eTXR0NJcvX8bHx6dM5xCi\nOvjhB3jpJXjxRXjzTbCy0joi0zsTd4YNYRv477n/0r5Re6Z0nsK3T36LTa0HD48XpVTcMKSFCxeW\neZjT2LFjFWdnZ8XKykpxdXVVvv76a+XWrVtK//79Hzjs9P3331datWqltGvXTtmzZ88Dj1mCkIWo\n0lJTFWXiREVp00ZRTpzQOhrTS8xMVFaeWKl4fe6lNFvRTHnnwDvKlVtXtA6r0inJtbPImcrXrl2j\nZcuWD00mV69epVWrViZIU0WTmcqiOgsOVpe1HDoUPvoI6lbRvtJ8fT57ruxhQ9gG9l/bzxNtn2Cq\n11T6tegnTUJlVK7SFWPGjCEzMxN/f3+6deuGs7OzoXTFyZMn2bFjBzY2NmzdutUkwRcZsCQEUQ3l\n5MA776hlJ778EoYN0zoi07iQcIGNYRv59ty3tGjQgqleUxndYTS21rZah1bplSshAFy5coWtW7dy\n9OhRrl+/DkDz5s3p3bs348aNK/YOwhQkIYjq5tw5dThpy5bwxRfQqJHWERlXSlYKW85vYWPYRqLT\no5nUaRJTvKbQzqGd1qFVKeVOCOZIEoKoLvR6WLFCXbfggw/UWkRVZThpgb6Afdf2sSFsA3uv7GVI\n6yFM8ZrCwJYDqWFRQ+vwqiSjrJj2/fffM3jwYOrXr8///d//cfr0ad5+++0yDzsVQhTvxg21ryAv\nD0JCoEULrSMyjktJl9gYtpFvzn6Di40LU72m8vmwz7GrLUu1mYNie2cWLVpE/fr1OXLkCPv37+e5\n557j+eefr4jYhKh2FAW++w66dYPBg9VO5MqeDNKy0/jy1Jc8sv4R+m7sS74+n8AJgYTMCOGF7i9I\nMjAjxd4h1Kih3r7t2rWLGTNm8MQTT/DOO++YPDAhqpvkZHjhBTh/HvbuVdc5rqz0ip6DEQfZELaB\nXeG76N+yP2/6vsngVoOxqlENJkxUUsUmBBcXF2bOnMm+ffuYN28e2dnZUoVUCCPbt0+dcTxqFGzc\nqC5vWRldTb7KpjOb2HRmE/a17ZniNYVPhnyCQx0HrUMTJVBsp3JmZiZ79uyhU6dOtGnThtjYWM6d\nO8egQYMqKsZCpFNZVCVZWTB3Lvz0E2zYAAMGaB1R6WXkZvD9he/ZeGYjfyb+ybMdn2WK1xRZbMbM\nyCgjIcxYaCg8+yx07qyWqm7YUOuISk6v6Dl8/TAbwjbw818/09etL1O9pvJ4m8epWUOWZDNHkhCE\nMEMFBbBsGXzyCaxcCePGaR1RyV1PvW5oEqpjVYepXlN5tuOzONZz1Do0UQyjDDsVQhjPtWswcSJY\nW8OpU9C0qdYRFS8hM4Gdl3by3/P/5UzcGcZ6jmX7qO10ce4ii81UMXKHIEQFUBT4+muYN0+tTPrq\nq2BhxiV5LiVdIuBSAAGXAriQcIHBrQczqv0o/Nv5U8uyltbhiTKQJiMhzEBCAsycCZGRai0iT0+t\nI7pfgb6A36N/J+AvNQlk5Gbg386fEe1G4OfmJ0mgCpAmIyE0tmuXmgwmTYJt26CWGV1Xs/Ky+O3a\nbwRcCmBn+E4c6zoyot0Ivn3qW7o6d5XmoGpI7hCEMIGMDJgzBwID4ZtvwNdX64hUSXeS2BW+i4BL\nARyIOEAX5y6MaDcC/3b+tLSr+GKVouLIHYIQGjhxQu04fvRROHMG6tfXNp4ryVcMTUFn488yoOUA\nnnJ/iq+Gf4V9HXttgxNmRe4QhDCSvDx1kft169RF759+Wps49IqekOgQAv4KYEf4DpKzkvFv688I\n9xE81uIxrC2ttQlMaEruEISoAIoChw7B66+DgwOcPg3OzhUbQ3Z+NgciDvDzXz+zM3wnDWs3ZES7\nEXzt/zXdXbrLKmOiROQOQYgyysxURw19+qk62ew//4GpUytuzYJbd27xy+Vf2HFpB79d+41Ojp0Y\n0W4EI9xH0Lph64oJQlQaMuxUCBO4ckUtNXG3s/jll+GxxyomEVxLuWboDzgdd5rHWjzGiHYjGNZm\nGI3qVrGl1IRRSZOREEai16slqVevhpMnYdo0daZx8+YmPq+i51TMKcMksYTMBIa3Hc6cXnMY0HIA\nta0qaVlUYZbkDkGIh0hNVauQfvYZ2NrCv/8NY8aYtjx1Tn4OByMPGjqFbWraGJqCerj0kCUmRZnI\nHYIQZXTunNo3sH07PP44bN4MPXuarlkoJSuFXy//SsClAAKvBtKhcQdGtBvBgUkHZLF5UWE0Swhu\nbm7Ur1+fGjVqYGVlRUhICMnJyYwZM4br16/j5ubG9u3badCggVYhimomPx9+/llNBJcvw/PPw59/\ngpOTac53PfW6oSnoj+g/8HPzY0S7EaweulqqhwpNaNZk1KJFC06dOkXDe4rAv/HGGzg4OPDGG2+w\nbNkyUlJSWLp0aaH9pMlIGFtCAnz5JXz+ubp+8csvw5NPgpWRV3pUFIXTcacNncLR6dE80fYJRrQb\nwcCWA6lbs65xTyjEPcx6lFGLFi04efIk9vb/mynp7u5OcHAwjo6OxMXF4efnx19//VVoP0kIwlhC\nQtRO4l271KUrX3oJvLyMe47cglyCI4MJuBTAjks7qGVZS+0PaDeCR5o+Iv0BosKYdUJo2bIltra2\n1KhRg3/961/MmDEDOzs7UlJSAPXXVMOGDQ1/GwKWhCDKITtb7Rf49FNISlKTwNSpxl2tLC07jd1X\ndhNwKYA9V/bQzr6doVO4vUN7KRonNGHWncpHjx7F2dmZxMREBg4ciLu7e6HPdTpdkf/HWbBggeG1\nn58ffn5+JoxUVAVRUbB2LaxfD97e8O67MHQo1DDCD3RFUbicfJl9V/cRcCmAEzdP4NvclxHtRrB8\n0HKcbSp42rIQQFBQEEFBQaX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w96effsqUKVOK3L5WrVr4+vry888/V0B0QpSfJAQhSmje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\n",
- "text": [
- "<matplotlib.figure.Figure at 0x5ad4a10>"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-10, Page No 205"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=3 #s\n",
- "\n",
- "#Calculations\n",
- "#After solving the differential equation\n",
- "s=(3**-1)*(t+2)**3 #ft\n",
- "v=(t+2)**2 #ft/s\n",
- "a=2*(t+2) #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The displacement at t=3s is',round(s,1),\"ft\"\n",
- "print'The velocity at t=3s is',round(v),\"ft/s\"\n",
- "print'The acceleration at t=3s is',round(a),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The displacement at t=3s is 41.7 ft\n",
- "The velocity at t=3s is 25.0 ft/s\n",
- "The acceleration at t=3s is 10.0 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-12, Page No 206"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#Calling upward direction positive\n",
- "xdot1=6 #ft/s\n",
- "xdot3=3 #ft/s\n",
- "xdoubledot=2 #ft/s**2\n",
- "xdoubledot3=-4 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "xdot=-xdot1 #ft/s\n",
- "xdot2=2*xdot-xdot3 #ft/s\n",
- "xdoubledot2=2*xdoubledot-xdoubledot3 #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of velocity is',round(xdot2,3),\"ft/s (down)\"\n",
- "print'The value of acceleration is',round(xdoubledot2,3),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of velocity is -15.0 ft/s (down)\n",
- "The value of acceleration is 8.0 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-16, Page No 207"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=4 #s\n",
- "\n",
- "#Calculations\n",
- "#Part (a)\n",
- "x=t**3 #in\n",
- "y=-2*t**2 #in\n",
- "z=2*t #in\n",
- "#Part (b)\n",
- "#Theory question\n",
- "#Part(c)\n",
- "#Unit vector calculation\n",
- "m=(4**2+1**1+(-3)**2)**0.5\n",
- "e_l=[4*m**-1,m**-1,-3*m**-1]\n",
- "v=[3*t**2,-4*t,2] #in/s\n",
- "#Projection of v on n at t=4s\n",
- "dot=[v[0]*e_l[0],v[1]*e_l[1],v[2]*e_l[2]]\n",
- "#dot=v.*e_l #in/s\n",
- "a=dot[0]+dot[1]+dot[2] #in/s\n",
- "\n",
- "#Result\n",
- "print'The co-ordinates of position are x=',round(x),\"in ,\",round(y),\"in and \",round(z),\"in respectively\"\n",
- "print'The projection of v on n at t=4s is',round(a,1),\"in/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The co-ordinates of position are x= 64.0 in , -32.0 in and 8.0 in respectively\n",
- "The projection of v on n at t=4s is 33.3 in/s\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-17, Page No 208"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "theta=pi/3 #rad\n",
- "\n",
- "#Calculations\n",
- "#Method (a)\n",
- "t=(theta)**0.5 #s\n",
- "r=2*theta\n",
- "rdot=4*t\n",
- "thetadot=2*t\n",
- "#Velocity calculations\n",
- "x=r*thetadot\n",
- "v=((rdot)**2+x**2)**0.5 #ft/s\n",
- "#Theta calculations\n",
- "thetax=30+arctan(rdot/x)*(180/pi) #degrees\n",
- "#Method (b)\n",
- "x=2*theta*cos(theta) #ft\n",
- "y=2*theta*sin(theta) #ft\n",
- "xdot=4*t*((cos(t**2)))+2*t**2*(-sin(t**2))*(2*t) #ft/s\n",
- "ydot=4*t**2*sin(t**2)+2*t**2*cos(t**2)*2*t #ft/s\n",
- "V=(xdot**2+ydot**2)**0.5 #ft/s\n",
- "Thetax=arctan(ydot/-xdot)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'By both the methods we obtain v and thetax as:'\n",
- "print'Method 1'\n",
- "print'v=',round(v,2),\"ft/s\",'and thetax=',round(thetax,1),\"degrees\"\n",
- "print'Method 2'\n",
- "print'V=',round(v,2),\"ft/s\",'and Thetax=',round(Thetax,1),\"degrees\"\n",
- "# The answer may wary due to decimal point accuracy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "By both the methods we obtain v and thetax as:\n",
- "Method 1\n",
- "v= 5.93 ft/s and thetax= 73.7 degrees\n",
- "Method 2\n",
- "V= 5.93 ft/s and Thetax= 73.9 degrees\n"
- ]
- }
- ],
- "prompt_number": 76
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-18, Page No 209"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "theta=pi/3 #rad\n",
- "\n",
- "#Calculations\n",
- "t=sqrt(theta) #s\n",
- "thetadot=2*t \n",
- "thetadoubledot=2\n",
- "r=2*t**2\n",
- "rdot=4*t\n",
- "rdoubledot=4\n",
- "ax=rdoubledot-(r*thetadoubledot*thetadoubledot) #ft/s**2\n",
- "ay=2*rdot*thetadot+r*thetadoubledot #ft/s**2\n",
- "a=sqrt(ax**2+ay**2) # fr/s**2\n",
- "thetax=30+arctan(ax/ay)*(180/pi) #degrees\n",
- "#Solving by cartesian co-ordinate system yields same solution\n",
- "\n",
- "#Result\n",
- "print'The value of acceleration is',round(a,1),\"ft/s**2\"\n",
- "print'The value of thetax is',round(thetax,1),\"degrees\"\n",
- "#Decimal accuracy causes discrepancy in answers\n",
- "# The ans for thetax is incorrcet in textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of acceleration is 21.4 ft/s**2\n",
- "The value of thetax is 18.2 degrees\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-21, Page No 211"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Va=5 #ft/s\n",
- "# as theta=70 degrees\n",
- "sintheta=0.94\n",
- "costheta=0.34\n",
- "l=6.24 #ft\n",
- "\n",
- "#Calculations\n",
- "Vb=(-costheta/sintheta)*Va #ft/s\n",
- "\n",
- "#Result\n",
- "print'The value of Vb is',round(Vb,2),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of Vb is -1.81 ft/s\n"
- ]
- }
- ],
- "prompt_number": 81
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.25-25, Page No 214"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "theta=linspace(0,360,13)\n",
- "\n",
- "#Calculations\n",
- "#Defining everything in terms of matrices \n",
- "t=(theta*pi)/(180) #s converting degrees to radians\n",
- "costheta=cos(t) \n",
- "sintheta=sin(t)\n",
- "x=2*costheta #ft\n",
- "v=-12*sintheta #ft/s\n",
- "a=-72*costheta #ft/s**2\n",
- "\n",
- "#Plotting\n",
- "# 1\n",
- "plot(t,x)\n",
- "# 2\n",
- "plot(t,v)\n",
- "# 3\n",
- "plot(t,a)\n",
- "xlabel('t(s)')\n",
- "ylabel('x(ft) , v(ft/s) ,a(ft/s**2)')\n",
- "\n",
- "#Result\n",
- "print'The results are the plots'\n",
- "print'The curve in blue colour represents t vs x'\n",
- "print'The curve in green represents t vs v'\n",
- "print'The curve in red represents t vs a'\n",
- "# All the 3 curves have been plotted in the same graph. "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The results are the plots\n",
- "The curve in blue colour represents t vs x\n",
- "The curve in green represents t vs v\n",
- "The curve in red represents t vs a\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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9jtmdU9i1axczZ87UDx+FhYVha2tb7GSzjY0NqTdSWX5sOZFxkexL2scQzyH6\nC4NkEay/TJigrW45a9Ydd+XkGH6wuHZNe6ZW9A9fmhb+fq9vv83OyCs9FBQWsPX8ViKPaoWgae2m\n+qGhVg1aGffB7kOn054932v4sTxDlkUdaFaW9jMsT7G/9f1ate5yME9OBh8fOHoUmjSp0J+bKD+j\nnFPYtm0bHTt2pFatWvzwww/s37+fqVOn4uZmmgtx8vPzeeCBB9i4cSNNmzbFz8/vrieab419a4GI\nTYnlodYPEeQdxMBWA623QJw5A35+2tXL9eub5CF0ur+LS2kOVOU5uFWp8neRsLcv3Zjv7bdVsSsg\nrdY2LtSNIKHWcmoWOuORE0Tr3JE0tG1V7PMKC+9+jqk0t5X284oKQXa21mmVZRJDaYppzZr3OJgb\n08sva68/+cSEDyKMyWhLZx86dIhDhw7x9NNPM3HiRCIiIoiOjjZq2FutWbNGPyV1woQJvP7668VD\n3+MbKyoQEXERHEg5wJDWWgdhdQVi4kRtdsjs2aqTlJtOp51MLCoSeXmlP/jm5hdwOH0bW9Mi2XXt\nF+raOdO15kg6Vx+Jo03rEj+vNLPXSlOI7nVbjRp/H7wtejilqFuIiwNnZ9VpRCkYpSj4+voSGxvL\nrFmzcHFxYeLEiXTq1In9+/cbNWxZlPbitZQbKVqBOBrBwdSDPOz5MCO9RzLAY0DlLhBFXcKJE9Cg\ngeo0FaagsIDtCduJPBrJsmPLcHJw0s8Cae3YWnW8ymnqVK2yffyx6iSiFIxSFHr37s2gQYNYtGgR\nW7dupVGjRnTs2JHDhw8bNWxZlOeK5pQbKfwS9wuRcZH6AhHkHcQAjwFUs7Pgs6V3Uwm6hNIq1BWy\n/fx2Io5G8MuxX2jk0Ei/JIGno6fqeJWfdAsWxShFITk5mZ9++gk/Pz8efPBBzp8/T1RUFOPGjTNq\n2LIwdJmL5Ixk/RDTodRDDPUcqu8gLL5AnD0LXbtW6i6hqBBExkWyLG4ZjRwa6S8oe6DhA6rjWZ+X\nXtLGxj76SHUScR8GFYWBAwcyaNAgBg8ejJeXl0kClpcx1z5Kzkjml2O/EHE0giMXjzD0Aa1ABLYM\ntMwCMWmS9oxN4SZIplCoK2RHwg790JBjDUf90JAUAsWSkqBtW+kWLIBBRSE5OZm1a9fyxx9/8Oef\nf+Lv78/gwYPp37+/Sa9oLg1TLYiXlJGkH2I6fPEwPZv11O/S1KlJJ/PfFDw+Hjp31mYcWXiXUKgr\n5NilY0TFWvsLAAAcf0lEQVSfi9Ze4qNp7NBY6wh8RuLV0LyeqFg96RYsgtGWuSgoKGD37t2sWbOG\nTZs2Ub16dQYOHEhoaKjRwpZFRaySejHzYrGN089ePUs31276/V79XPzMr5N45hlo3BjeeUd1kjIr\nKCzgUOoh/c976/mt1KlWR789ZW+33ta51pClKOoWjh2D2y52FebDKEVh+/bt9OzZs9ht27Zt49y5\nc4wdO9bwlOWgYunsok3Bi/Z7PXbpGF1duuq3U+zm2o2a9jUrNFMxRV3CiRPg6KguRynlFeQRmxJL\ndLzWCWxP2I5zLWf9z7O3W29c67iqjinK4sUXtRUgP/xQdRJRAqNOSb2VpUxJNaXrOdfZfn67/pnt\nodRDdHDuoH9m26NZD2pXq11xgZ55Bho1grlzK+4xyyAnP4c9SXv0P6+dCTtxr+eu7wJ6u/XGqZY8\nw7RoiYnQrp10C2bMoKKwc+dOduzYwSeffMK0adP0XygjI4Nff/2VgwcPGj9xKZlDUbhdZm4muy7s\n0o+B70vah3cjb/1Br1fzXtSvYZorizl3Djp1MqsuISsvi90Xdut/HnsS9+DV0Et/juZBtwdpUMOy\nz3uIu3jhBW2lug8+UJ1E3IVBRSE6OprNmzfz1Vdf8dxzz+lvr127NkOHDqV1a3UXA5ljUbhd0abg\nRcNNuy7swqO+h9ZJuPfhweYP0sihkXEe7NlnoWFDpV1CRk4GOxJ26DuBAykHaOfUTl8UezbrSd3q\nRloVUJivom7h+HHt/JYwKwYVhX79+rFx40aCgoKIiIgwScDysoSicLvcglz2Je3THzS3J2ynSa0m\nuNVzw8nBSXup9ffrxg6NcXJwopFDI+xs77EqnIm7hEJdIWnZaaTeSCU1M/XO15mpJGUkcfLKSTo1\n6aQvet1du+NQVe0sNaHIlCnaWh7SLZgdg4qCt7c333zzDSEhIfz000933N+pUyfjpCwHSywKt8sv\nzCfuUhyJ1xOLHWQvZl4s9n5adhr1qtcrXjRuebvf+5HYN3Qm/51ZONVyomqVqqV67MtZl+95oC96\n+3LWZWpXrV2sYN2ewbmWM+2c2lXupUNE6V24AO3bS7dghgwqCpGRkXz77bds376dLl263HH/5s2b\njZOyHCpDUSitgsIC7QB+l4N3/tnTzJy+iuEzvfjT5gqXMi/hUNWh2AHbsYYj13Ku6T/vYuZFrt68\nSv3q9e95oC963cihUakKjRDFTJmirfj3/vuqk4hbGGX20ezZs3nrrbeMGsxQ1lQU7um557RlscPC\nAG2o52r21WKF40rWFepWr1u8UNR0vPeQlBCGkm7BLBlUFM6cOUPLlvfehvD06dN4eFT8bkJSFIDz\n58HXF/78UzvJLIS5ef55bVOH995TnUT8xaCiMGrUKDIzMxk2bBhdunShSZMm6HQ6kpOT2bt3LytX\nrqR27dosWbLEJOHvRYoC8M9/Qr16+i5BCLOTkAAdO2rdQiMjzbQTBjF4+OjUqVMsWbKE7du3c+7c\nOQDc3Nzo1asXo0ePvm8nYSpWXxTOn9f+2U6ckC5BmLfJk6F2bekWzITR1j4yN1ZfFCZP1vZbDA9X\nnUSIe5NuwayU5th5380A27dvz7vvvsvp06eNFkwYICEBli6F6dNVJxHi/po1g1GjZPVUC3LfTiE+\nPp6lS5cSERGBjY0NTzzxBEFBQTRv3ryiMt7BqjsFOXknLI1MijAbRukU3N3dee2119i3bx8///wz\nhw4dokULw5YwfvXVV2nTpg0dOnRgxIgRXLt2TX9fWFgYrVu3xsvLi3Xr1hn0OJVOQgIsWQKvvKI6\niRCl17w5BAVJt2AhSnVO4dZuoUqVKowaNYrpBgxfrF+/nn79+mFra8uMGTMACA8PJy4ujjFjxrBn\nzx4SExPp378/J06cwNa2eO2y2k5BugRhqaRbMAtG6RT8/f159NFHKSwsJDIykpiYGIMKAkBgYKD+\nQO/v78+FCxcAWLFiBaNHj8be3h53d3datWpFTEyMQY9VaVy4IF2CsFzNm8PIkfDxx6qTiPu472Wt\nixcvNukezd999x2jR48GICkpiW7duunvc3V1JTEx0WSPbVHCw2HCBJnBISzX669rizdOmybdghm7\nb1Eob0EIDAwkJSXljtvfffddhg4dCsDcuXOpWrUqY8aMKfHr2NjY3PX2mTNn6t8OCAggICCgXDkt\nwoUL8NNP2rQ+ISyVm9vf3cK776pOYxWioqKIiooq0+cou07hv//9LwsXLmTjxo1Ur66trhn+17z7\novMMgwYNYtasWfj7+xf7XKs7pyAbl4jKwgw3hLImZnvx2tq1a5k+fTrR0dE0vKWNLDrRHBMToz/R\nfOrUqTu6BasqCrJpiahszGBTKGtlkqKwZ88eXFxcaNq0abmDtW7dmtzcXBo00LZj7N69OwsWLAC0\n4aXvvvsOOzs75s2bx8CBA+8MbU1F4YUXoFo12QxdVB7x8dC5s3QLCpikKIwbN47Dhw/j6enJ0qVL\nDQpYXlZTFGQjdFFZPfOM1vm+847qJFbFpMNH169fp06dOuUKZiirKQovvghVq0qXICof6RaUMNtz\nCoayiqKQlARt20qXICqvSZO0v23pFiqMFAVL9tJLYGcnSwOIyuvsWejSBU6ehL/OLwrTkqJgqYq6\nhLg4cHZWnUYI05k0SfsbnzNHdRKrYLSikJmZSUJCAjY2Nri6uuLg4GC0kOVR6YvCSy9BlSqyJICo\n/KRbqFAGFYWMjAwWLlzIkiVLuHz5Mk5OTuh0OlJTU3F0dGTs2LFMmjSJWrVqmST8vVTqopCcDD4+\n0iUI6zFxIjRtCrNnq05S6RlUFPr168cTTzzB0KFDcb7t4JSSksLKlStZunQpGzduNF7iUqrURWHq\nVLC1lS5BWI8zZ8DPT5uJJN2CSck5BUsjXYKwVhMmgIuLdAsmZpSls/v161eq24QRvP8+BAdLQRDW\n5803YcECuHpVdRKrV+IqqdnZ2WRlZXHp0iXS0tL0t1+/fl2WszaF5GRYvBiOHlWdRIiK17IlPPII\nfPopzJqlOo1VK3H4aN68eXz66ackJSUVW+eodu3aPPPMM0yZMqXCQt6uUg4fTZsGhYXaP4UQ1qjo\n3MKpU1Cvnuo0lVJpjp0ldgo6nY6zZ88ye/Zs3nrrLaOHE7dISYH//le6BGHdWraEYcO0J0a37Jci\nKlaJnUKHDh04ePAgvr6+xMbGVnSue6p0nYJ0CUJoTp8Gf3/pFkzEoNlHo0ePZu/evSQmJuLh4XHH\nFz506JDxkpZRpSoKKSng7Q1HjmhztYWwduPHg7s7vP226iSVjsFTUlNSUhgwYACrVq264wu5u7sb\nJWR5VKqiMH065OfDvHmqkwhhHk6dgm7dpFswAblOwdylpkKbNtIlCHE76RZMwqDrFIYMGUJkZCRZ\nWVl33JeZmcnSpUt56KGHDE9pzT74AJ58UgqCELd7802YPx/S01UnsToldgoXL15k/vz5LFu2jCpV\nqtCkSRN0Oh0pKSnk5+czatQonn/+eRo1alTRmStHp3DiBPToAQcPaldyCiGKCwmBunXhk09UJ6k0\njDZ8lJKSwrlz5wBwc3O7Yy2kimbxRSE/Hx58EMaOBYXXewhh1i5fhvbt4eefoU8f1WkqBaMscxEX\nF4ezszP+/v74+/vj7OxMVFSUUQJ+9NFH2NraFrtiOiwsjNatW+Pl5cW6deuM8jhm58MPwcEBJk9W\nnUQI89WwIXz1lXZ+ISNDdRqrcd+iEBQUxHvvvYdOpyMrK4sXXniBGTNmGPzACQkJrF+/Hjc3N/1t\ncXFxLF26lLi4ONauXcvkyZMpLCw0+LHMyqFD2m5q332nrYYqhCjZ0KEQEACvvKI6idW471Fp9+7d\nJCQk0L17d/z8/GjSpAk7duww+IGnTZvG+++/X+y2FStWMHr0aOzt7XF3d6dVq1bExMQY/FhmIzcX\nxo3TFr5r3lx1GiEswyefwNq12oswufsWBTs7O2rUqEF2djY3b96kZcuW2Br4DHfFihW4urrSvn37\nYrcnJSXh6uqqf9/V1bVyLb43e7ZWDJ5+WnUSISxH3bpaZz1xoqyiWgFKXPuoiJ+fH8OGDWPv3r1c\nvnyZZ599ll9++YXIyMh7fl5gYCApKSl33D537lzCwsKKnS+414kPGxubu94+85a1UQICAggICLj3\nN6La7t3wzTdw4ACU8D0JIUrQrx88+ii88AL8+KPqNBYjKiqqzOeA7zv7aM+ePXTt2rXYbd9//z3j\nxo0rc0CAI0eO0K9fP2rWrAnAhQsXcHFxYffu3SxatAhAf85i0KBBzJo1C39//+KhLW32UXY2+Ppq\nm5OPHKk6jRCWKSsLOnaEsDB47DHVaSySRVzR3KJFC/bt20eDBg2Ii4tjzJgxxMTEkJiYSP/+/Tl1\n6tQd3YLFFYWXX9bWOPr5Z9VJhLBsO3bAiBHa9T1OTqrTWByDls6uKLce8L29vQkKCsLb2xs7OzsW\nLFhQ4vCRxYiKgogIbdaREMIwPXpo5+Seew6WL5ehWBNQ3imUh8V0ChkZ2sU38+fDkCGq0whROeTk\nQJcuEBoKTz2lOo1FsYjho/KwmKLwzDPaPgnffKM6iRCVS2wsDBwI+/ZBs2aq01gMKQoqrVkD//yn\nNmxUp47qNEJUPu+8A1u2wB9/yDBSKRllmQtRDmlpMGkSLFokBUEIU5kxQ1tF9csvVSepVKRTMIWx\nY7V1W2TjHCFM69gxbXHJ3bvhth0ixZ0sYvZRpbNsGezdq415CiFMq00bbe+Fp5/WZvpVqaI6kcWT\n4SNjSk3VlsJevBj+ujhPCGFiL72kLS4p+y4YhQwfGYtOB8OHQ9u2MHeu6jRCWJezZ6FrV4iOBh8f\n1WnMlpxorkjffw/x8fDWW6qTCGF9WrSAd9+F4GDIy1OdxqJJp2AMCQnQqRNs2AAdOqhOI4R10ung\noYegWzd4+23VacySXKdQEQoLtYto+vaFN95QnUYI65aYqC0+uWYNdO6sOo3ZkeGjivDll9pyFqGh\nqpMIIVxctBPO48bBzZuq01gk6RQMceoUdO8O27bBAw+oTiOEAG0YaeRIaNlS2+VQ6MnwkSkVFEDv\n3hAUpE2JE0KYj0uXtMUoly2Dnj1VpzEbMnxkSh9/DFWrajtBCSHMS6NG8MUX2mykGzdUp7Eo0imU\nx5Ej2onlPXvA3V1dDiHEvQUHQ61a8J//qE5iFmT4yBTy8sDfHyZP1jYSF0KYr/R0bRjp228hMFB1\nGuVk+MgU3nkHmjSBCRNUJxFC3E+9etp+JhMmaAVC3Jd0CmWxd692ccyBA9C0acU/vhCifCZPhqws\n+O9/VSdRyqw7hc8//5w2bdrQtm1bXnvtNf3tYWFhtG7dGi8vL9atW6cq3p1u3tTmPs+bJwVBCEvz\n/vva1PEVK1QnMXtKls7evHkzK1eu5NChQ9jb23Pp0iUA4uLiWLp0KXFxcSQmJtK/f39OnDiBra0Z\njHL961/aYndPPKE6iRCirGrV0rqEkSOhRw9tdpK4KyVH2y+++ILXX38de3t7ABr99QtasWIFo0eP\nxt7eHnd3d1q1akVMTIyKiMVt3Qo//QQLFsi2f0JYql694MkntW1yLW/UvMIoKQonT55ky5YtdOvW\njYCAAPbu3QtAUlISrq6u+o9zdXUlMTFRRcS/3bihbeDx5ZfabmpCCMs1Z462W9vPP6tOYrZMNnwU\nGBhISkrKHbfPnTuX/Px8rl69yq5du9izZw9BQUGcOXPmrl/HRvUz81df1a5cHjZMbQ4hhOGqV9eW\nuR88GAIC5PzgXZisKKxfv77E+7744gtGjBgBQNeuXbG1teXy5cu4uLiQkJCg/7gLFy7g4uJy168x\nc+ZM/dsBAQEEBAQYJXcxf/wBv/8Ohw4Z/2sLIdTo3Pnv64z+7/8q9ZBwVFQUUVFRZfocJVNSv/rq\nK5KSkpg1axYnTpygf//+nD9/nri4OMaMGUNMTIz+RPOpU6fu6BYqZErq1avaRS+LFkH//qZ9LCFE\nxcrL0/ZdeO45mDRJdZoKU5pjp5LZRyEhIYSEhNCuXTuqVq3K999/D4C3tzdBQUF4e3tjZ2fHggUL\n1Awf5edrey0/8ogUBCEqI3t7bRipTx9teFhWOdaTi9dupdPBypXw+uvg5ASrV4ODg/EfRwhhHr77\nTjtvOH689n/v6Kg6kUmZ9cVrZmfbNm3K2r//DR98AJs2SUEQorILCdEWuMzMBC8vCAvTrny2YlIU\njhzRZhY9+SQ8+yzExsKQIZX65JMQ4hZNmmjLbG/frv3/e3rC119rw8hWyHqLwvnzWsvYr5+2DPbx\n49oyFlWqqE4mhFDB0xMiImD5cliyRFvBYPlyq7vQzfqKwpUr8Mor2ubeLi5w4gS8/LI2f1kIIfz8\nYONGbZ2z2bO1LXfLOK3TkllPUcjK0sYLvby08cMjR7RlsOvWVZ1MCGFubGxg4EDYvx9efFE79/DQ\nQ3DwoOpkJlf5i0J+vjY+2Lq1Nl64fbs2ftikiepkQghzZ2sLY8Zow8uDB2uF4qmnID5edTKTqbxF\nQafTxgPbttXGB3/7TRsv9PRUnUwIYWmK9mM/eRI8PLSroqdOhb9WeK5MKmdRiIrSxgHnzNHGBTdu\nhK5dVacSQli62rVh5kyIi4OCAmjTRhuGzsxUncxoKldROHhQG/cLCdHGAfft09o9mV4qhDAmJyf4\n/HPYvVsrEK1ba8PSeXmqkxmschSF+HhtnG/gQG3c7/hxbRzQHDbnEUJUXh4e2l4rq1fDr7+Ct7c2\nTF1YqDpZuVn2UfPSJW1cr0sX7Zdz8qQ27le1qupkQghr0qkTrFundQvvv//3tFYLZLlF4Z13tPG8\nggI4elQb56tdW3UqIYQ1698fYmIgNFRbIWHAAG1aqwWx3KIQF6eN533+uTa+J4QQ5sDWFoKCtB3e\nhg/Xls356ivVqUpNVkkVQghTunEDbt40i+18S3PslKIghBBWQpbOFkIIUSZSFIQQQuhJURBCCKEn\nRUEIIYSekqIQExODn58fvr6+dO3alT179ujvCwsLo3Xr1nh5ebFu3ToV8YQQwmopKQqhoaHMmTOH\n2NhYZs+eTWhoKABxcXEsXbqUuLg41q5dy+TJkym04MvFSxJl4Rt2SH61JL86lpy9tJQUhSZNmnDt\n2jUA0tPTcXFxAWDFihWMHj0ae3t73N3dadWqFTExMSoimpSl/2FJfrUkvzqWnL207FQ8aHh4OL16\n9eKVV16hsLCQnTt3ApCUlES3bt30H+fq6kpiYqKKiEIIYZVMVhQCAwNJSUm54/a5c+fy2Wef8dln\nn/Hoo48SGRlJSEgI69evv+vXsZFlr4UQouLoFKhdu7b+7cLCQl2dOnV0Op1OFxYWpgsLC9PfN3Dg\nQN2uXbvu+HwPDw8dIC/yIi/yIi9lePHw8Ljv8VnJ8FGrVq2Ijo6mT58+bNq0Cc+/tsgcNmwYY8aM\nYdq0aSQmJnLy5En8/Pzu+PxTp05VdGQhhLAKSorC119/zfPPP09OTg41atTg66+/BsDb25ugoCC8\nvb2xs7NjwYIFMnwkhBAVyCIXxBNCCGEaFndF89q1a/Hy8qJ169a89957quOUSUhICE5OTrRr1051\nlHJJSEigb9+++Pj40LZtWz777DPVkcrk5s2b+Pv707FjR7y9vXn99ddVRyqzgoICfH19GTp0qOoo\nZebu7k779u3x9fW967CwuUtPT+fxxx+nTZs2eHt7s2vXLtWRSu3PP//E19dX/1K3bt2S/38NPmtc\ngfLz83UeHh66s2fP6nJzc3UdOnTQxcXFqY5Valu2bNHt379f17ZtW9VRyiU5OVkXGxur0+l0uoyM\nDJ2np6dF/fx1Op0uMzNTp9PpdHl5eTp/f3/d1q1bFScqm48++kg3ZswY3dChQ1VHKTN3d3fdlStX\nVMcot3Hjxum+/fZbnU6n/f2kp6crTlQ+BQUFOmdnZ9358+fver9FdQoxMTG0atUKd3d37O3teeKJ\nJ1ixYoXqWKX24IMPUr9+fdUxys3Z2ZmOHTsCUKtWLdq0aUNSUpLiVGVTs2ZNAHJzcykoKKBBgwaK\nE5XehQsX+P3335k4caLF7idiqbmvXbvG1q1bCQkJAcDOzo66desqTlU+GzZswMPDg2bNmt31fosq\nComJicW+Ebm4TZ34+HhiY2Px9/dXHaVMCgsL6dixI05OTvTt2xdvb2/VkUrt5Zdf5oMPPsDW1qL+\nbfVsbGzo378/Xbp0YeHCharjlMnZs2dp1KgR48ePp1OnTkyaNImsrCzVscplyZIljBkzpsT7Leqv\nS2YimYcbN27w+OOPM2/ePGrVqqU6TpnY2tpy4MABLly4wJYtWyxm2YLVq1fTuHFjfH19LfbZ9vbt\n24mNjWXNmjX85z//YevWraojlVp+fj779+9n8uTJ7N+/HwcHB8LDw1XHKrPc3FxWrVrFyJEjS/wY\niyoKLi4uJCQk6N9PSEjA1dVVYSLrk5eXx2OPPcaTTz7J8OHDVccpt7p16zJkyBD27t2rOkqp7Nix\ng5UrV9KiRQtGjx7Npk2bGDdunOpYZdKkSRMAGjVqxKOPPmpR65q5urri6upK165dAXj88cfZv3+/\n4lRlt2bNGjp37kyjRo1K/BiLKgpdunTh5MmTxMfHk5uby9KlSxk2bJjqWFZDp9MxYcIEvL29mTp1\nquo4ZXb58mXS09MByM7OZv369fj6+ipOVTrvvvsuCQkJnD17liVLlvCPf/yD77//XnWsUsvKyiIj\nIwOAzMxM1q1bZ1Gz8JydnWnWrBknTpwAtHF5Hx8fxanK7ueff2b06NH3/BglF6+Vl52dHfPnz2fg\nwIEUFBQwYcIE2rRpozpWqY0ePZro6GiuXLlCs2bNmD17NuPHj1cdq9S2b9/Ojz/+qJ9WCNr+F4MG\nDVKcrHSSk5MJDg6msLCQwsJCnnrqKfr166c6VrlY2lBqamoqjz76KKANxYwdO5YBAwYoTlU2n3/+\nOWPHjiU3NxcPDw8WLVqkOlKZZGZmsmHDhvuez5GL14QQQuhZ1PCREEII05KiIIQQQk+KghBCCD0p\nCkIIIfSkKAghhNCToiCEEEJPioIQZXTt2jW++OIL/fsXL15kyJAhJX58Tk4OvXv3prCwsCLiCWEQ\nKQpClNHVq1dZsGCB/v358+fz9NNPl/jx1apV48EHH+S3336rgHRCGEaKghBlNGPGDE6fPo2vry+h\noaEsW7ZM3ykcPXoUf39/fH196dChg34/8WHDhvHzzz+rjC1EqcgVzUKU0blz53j44Yc5fPgwKSkp\nBAYGcvjwYQBefPFFunXrxpgxY8jPzyc/P5/q1auTk5NDy5YtZal3YfYsau0jIczBrc+jzp07p1/9\nE6B79+7MnTuXCxcuMGLECFq1agVoQ0iFhYXcvHmT6tWrV3hmIUpLho+EMNCtRWL06NGsWrWKGjVq\n8NBDD7F58+ZiH2dpC9kJ6yNFQYgyql27tn4ZaDc3N1JSUvT3nT17lhYtWvDCCy/wyCOP6IeVcnJy\nqFKlCtWqVVOSWYjSkuEjIcrI0dGRnj170q5dOwYPHkx+fj6ZmZk4ODgQERHBDz/8gL29PU2aNOHN\nN98EIDY2lu7duytOLsT9yYlmIQw0c+ZM2rRpw6hRo0r8mDfeeIOuXbvq9xQQwlxJURDCQJcuXSI4\nOJjff//9rvfn5OQQGBhIdHS0nFMQZk+KghBCCD050SyEEEJPioIQQgg9KQpCCCH0pCgIIYTQk6Ig\nhBBCT4qCEEIIvf8HaEMe+5zwRtYAAAAASUVORK5CYII=\n",
- "text": [
- "<matplotlib.figure.Figure at 0x5bd53f0>"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-26, Page No 215"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=1.2 #m\n",
- "w0=0 #rpm\n",
- "w=2000 #rpm\n",
- "t=20 #s\n",
- "\n",
- "#Calculations\n",
- "alpha=(w-w0)/t \n",
- "alpha_rad=(alpha*2*pi)/60 #converting to radians/s**2\n",
- "\n",
- "#Result\n",
- "print'The angular acceleration is',round(alpha_rad,1),\"radians/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular acceleration is 10.5 radians/s**2\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-27, Page No 216"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w0=0 #rad/s\n",
- "w=209 #rad/s\n",
- "t=20 #s\n",
- "\n",
- "#Calculations\n",
- "theta=0.5*(w+w0)*t #rad\n",
- "theta_rev=round(theta/(2*pi)) #revolutions rounding off\n",
- "\n",
- "#Result\n",
- "print'The flywheel makes',round(theta_rev),\"revolutions\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The flywheel makes 333.0 revolutions\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-28, Page No 216"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w0=0 #rad/s\n",
- "alpha=10.5 #rad/s**2\n",
- "t=0.6 #s\n",
- "r=0.6 #m\n",
- "\n",
- "#Calculations\n",
- "w=w0+alpha*t #rad/s\n",
- "v=r*w #m/s\n",
- "a_t=r*alpha #m/s**2\n",
- "a_n=r*w*w #m/s**2\n",
- "a=sqrt(a_t**2+a_n**2) #m/s**2\n",
- "phi=arctan(a_t/a_n)*(180/pi) #degrees\n",
- "\n",
- "#result\n",
- "print'The tangential velocity is',round(v,2),\"m/s\"\n",
- "print'The acceleration is',round(a,1),\"m/s**2\"\n",
- "print'and the angle is',round(phi,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tangential velocity is 3.78 m/s\n",
- "The acceleration is 24.6 m/s**2\n",
- "and the angle is 14.8 degrees\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-29, Page No 216"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=4 #ft\n",
- "wb=40 #rpm\n",
- "we=60 #rpm\n",
- "\n",
- "#Calculations\n",
- "r=l/2 #ft\n",
- "vb=r*((wb*2*pi)/60) #ft/s\n",
- "ve=r*((we*2*pi)/60) # ft/s\n",
- "\n",
- "#Result\n",
- "print'The linear speeds are:'\n",
- "print'vb=',round(vb,2),\"ft/s\"\n",
- "print'and ve=',round(ve,1),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The linear speeds are:\n",
- "vb= 8.38 ft/s\n",
- "and ve= 12.6 ft/s\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-30, Page No 217"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "wb=40 #rpm\n",
- "we=60 #rpm\n",
- "t1=5 #s using different symbol to avoid conflict in decleration\n",
- "t=2 #s\n",
- "#Calculations\n",
- "\n",
- "alpha=(((we*2*pi)/60)-((wb*2*pi)/60))/t1 #rad/s**2\n",
- "w=((wb*2*pi)/60)+alpha*t #rad/s\n",
- "#Components of acceleration are\n",
- "a_t=r*alpha #ft/s**2\n",
- "a_n=r*w**2 #ft/s**2\n",
- "\n",
- "#result\n",
- "print'The tangential acceleration is',round(a_t,3),\"ft/s**2\"\n",
- "print'The normal acceleration is',round(a_n,1),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tangential acceleration is 0.838 ft/s**2\n",
- "The normal acceleration is 50.5 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-31, Page No 217"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=200 #mm\n",
- "w0=(800*2*pi)/60 #rpm\n",
- "w=0 #rpm\n",
- "t=600 #s\n",
- "\n",
- "#Calculations\n",
- "alpha=(w-w0)/t #rad/s**2 (deceleration)\n",
- "\n",
- "#result\n",
- "print'The angular acceleration is',round(alpha,2),\"radian/s**2\"\n",
- "# The negative sign indicates that the wheel decelerates\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular acceleration is -0.14 radian/s**2\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-32, Page No 217"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#The symbols used here differ from the textbook solution to avoid conflict \n",
- "t1=0 #s\n",
- "t2=0.5 #s\n",
- "t3=2.5 #s\n",
- "t4=3**-1 #s\n",
- "w=200 #rpm\n",
- "w0=0 #rpm\n",
- "\n",
- "#Calculations\n",
- "theta1=0.5*(w0+(w*60**-1))*t2 #rev\n",
- "theta2=(w*60**-1)*(t3-t2) #rev\n",
- "theta3=(2**-1)*((w*60**-1)+w0)*t4 #rev here the values of w and w0 are interchanged but essentially the value comes out to be the same hence the decleration has not been changed\n",
- "theta=theta1+theta2+theta3 #rev\n",
- "\n",
- "#Result\n",
- "print'The wheel undergoes',round(theta,2),\"revolutions\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wheel undergoes 8.06 revolutions\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-34, Page No 218"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=1 #s\n",
- "r=4 #m\n",
- "\n",
- "#Calculations\n",
- "s=t**3+3 #m\n",
- "theta=s/r #rad\n",
- "dtheta_dt=0.75*t**2 #rad/s\n",
- "Vx=-4*sin(theta)*dtheta_dt #m/s\n",
- "Vy=4*cos(theta)*dtheta_dt #m/s\n",
- "V=(Vx**2+Vy**2)**0.5 #m/s\n",
- "\n",
- "#Result\n",
- "print'The components of velocity are:'\n",
- "print'Vx=',round(Vx,2),\"m/s\"\n",
- "print'Vy=',round(Vy,2),\"m/s\"\n",
- "print'V=',round(V),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The components of velocity are:\n",
- "Vx= -2.52 m/s\n",
- "Vy= 1.62 m/s\n",
- "V= 3.0 m/s\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-35, Page No 218"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=1 #s\n",
- "theta=1 #rad\n",
- "\n",
- "#Calculations\n",
- "dtheta_dt=0.75*t**2 #rad/s\n",
- "acc=1.5*t #rad/s**2\n",
- "ax=-4*cos(theta)*dtheta_dt**2-(4*sin(theta)*acc) #m/s**2 (to left)\n",
- "ay=-4*sin(theta)*dtheta_dt**2+(4*cos(theta)*acc) #m/s**2 (up)\n",
- "a=sqrt(ax**2+ay**2) #m/s**2\n",
- "\n",
- "#result\n",
- "print'The acceleration is',round(a,2),\"m/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration is 6.41 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 38
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-36, Page No 218"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=2 #s\n",
- "\n",
- "#Calculations\n",
- "#Velocity\n",
- "vx=8*t-3 #ft/s\n",
- "vy=3*t**2 #ft/s\n",
- "v=sqrt(vx**2+vy**2) #ft/s\n",
- "theta_x=arctan(vy*vx**-1)*(180/pi) #degrees\n",
- "#Acceleration\n",
- "ax=8 #ft/s**2\n",
- "ay=6*t #ft/s**2\n",
- "a=sqrt(ax**2+ay**2) #ft/s**2\n",
- "phi_x=arctan(ay*ax**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The velocity is',round(v,1),\"ft/s\"\n",
- "print'and the angle is',round(theta_x,1),\"degrees\"\n",
- "print'The acceleration is',round(a,1),\"ft/s**2\"\n",
- "print'and the angle it makes is',round(phi_x,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity is 17.7 ft/s\n",
- "and the angle is 42.7 degrees\n",
- "The acceleration is 14.4 ft/s**2\n",
- "and the angle it makes is 56.3 degrees\n"
- ]
- }
- ],
- "prompt_number": 46
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-37, Page No 219"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "V_ao=29.3 #ft/s\n",
- "OA=50 #ft\n",
- "theta=45 #degrees\n",
- "OB=50*sqrt(2) #ft\n",
- "\n",
- "#Calculations\n",
- "w_ao=V_ao/OA #rad/s\n",
- "V_bo=V_ao*cos(theta) #ft/s\n",
- "w_bo=V_bo/OB #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular velocity with respect to the observer is',round(w_ao,3),\"rad/s\"\n",
- "print' The angular velocity after moving 50ft is',round(w_bo,3),\"rad/s\"\n",
- "# The answer for w_bo is incorrect in textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity with respect to the observer is 0.586 rad/s\n",
- " The angular velocity after moving 50ft is 0.218 rad/s\n"
- ]
- }
- ],
- "prompt_number": 47
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-38, Page No 219"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initiliztaion of variables\n",
- "# as theta=30 degrees\n",
- "costheta=sqrt(3)*2**-1\n",
- "tantheta=sqrt(3)**-1\n",
- "r=[100*tantheta*(180/pi),100] #ft\n",
- "v=17.6 #ft/s\n",
- "\n",
- "#Calculations\n",
- "v_1=100*costheta**-1*costheta**-1\n",
- "w=v/v_1 #rad/s (clockwise)\n",
- "\n",
- "#result\n",
- "print'The angular velocity is',round(w,3),\"rad/s clockwise\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity is 0.132 rad/s clockwise\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-39, Page No 220"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "t=2 #s\n",
- "\n",
- "#Calculations\n",
- "Vx=20*t+5 #m/s\n",
- "Vy=t**2-20 #m/s\n",
- "#As indefinite integral is not possible \n",
- "x=10*t**2+5*t+5 #m\n",
- "y=0.5*t**2-20*t-15 #m\n",
- "ax=20 #m/s**2\n",
- "ay=2*t #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The displacement components are x=',round(x),\"m\",'and y=',round(y),\"m.\"\n",
- "print'The velocity components are: Vx=',round(Vx),\"m/s\",'and Vy=',round(Vy),\"m/s\"\n",
- "print'The acceleration components are: ax=',round(ax),\"m/s**2\",'and ay=',round(ay),\"m/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The displacement components are x= 55.0 m and y= -53.0 m.\n",
- "The velocity components are: Vx= 45.0 m/s and Vy= -16.0 m/s\n",
- "The acceleration components are: ax= 20.0 m/s**2 and ay= 4.0 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 55
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-40, Page No 221"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=0.1 #m\n",
- "v=20 #m/s\n",
- "a_g=6 #m/s**2\n",
- "d2=0.150 #m\n",
- "\n",
- "#Calculations\n",
- "r=d/2 #m\n",
- "w=v/r #rad/s\n",
- "vb=d2*0.5*w #m/s\n",
- "alpha=a_g/r #rad/s**2\n",
- "a_t=d2*0.5*alpha #rad/s**2 tangential acceleration\n",
- "a_n=d2*0.5*w*w #m/s**2 normal acceleration\n",
- "a=sqrt(a_t**2+a_n**2) #m/s**2 linear acceleration\n",
- "\n",
- "#Result\n",
- "print'The linear velocity is',round(vb),\"m/s\"\n",
- "print'The acceleration is',round(a),\"m/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The linear velocity is 30.0 m/s\n",
- "The acceleration is 12000.0 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 57
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12-41, Page No 221"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=40 degrees\n",
- "sintheta=0.64\n",
- "costheta=0.77\n",
- "tantheta=0.83\n",
- "x=100 #ft\n",
- "ax=0 #ft/s**2\n",
- "ay=-32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#vox=vocos40....(1)\n",
- "#voy=vox*t-1/2(32.2)t^2...(2)\n",
- "#Simplyfying eq (1) and eq(2)\n",
- "t_f=((x*tantheta)/(0.5*(-ay)))**0.5 #s time of flight\n",
- "Vo=x/(costheta*t_f) #ft/s\n",
- "#As the max height occurs at half wat through the flight\n",
- "t=t_f/2 #s\n",
- "ymax=Vo*sintheta*t+(0.5*ay*t*t) #ft the formula has positive sign as ay is defined negative\n",
- "\n",
- "#result\n",
- "print'The max height the ball will reach is',round(ymax,1),\"ft\"\n",
- "\n",
- "# The ans in textbook is incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The max height the ball will reach is 20.8 ft\n"
- ]
- }
- ],
- "prompt_number": 14
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter13.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter13.ipynb deleted file mode 100755 index a77abe8e..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter13.ipynb +++ /dev/null @@ -1,899 +0,0 @@ -{
- "metadata": {
- "name": "chapter13.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 13: Dynamics of a Particle"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-1, Page No 230"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=2 #lb\n",
- "F=1.5 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Angles are with respect to the plane are,\n",
- "# theta1=10 degrees & theta2=30 degrees\n",
- "sintheta1=0.17\n",
- "costheta1=0.99\n",
- "sintheta2=0.5\n",
- "costheta2=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Now here the forces are considered as parallel and perpendicular to the plane \n",
- "#Applying Newtond Principle\n",
- "ax=(g/2)*(F*costheta1-(W*sintheta2)) #ft/s**2\n",
- "N1=(2*costheta2-(F*sintheta1)) #lb\n",
- "\n",
- "#result\n",
- "print'The force on the particle is',round(N1,2),\"lb\"\n",
- "print'The acceleration is',round(ax,2),\"ft/s**2\"\n",
- "\n",
- "# The answer may wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force on the particle is 1.48 lb\n",
- "The acceleration is 7.81 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-2, Page No 231"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=5 #kg\n",
- "s=12 #m\n",
- "v=4 #m/s\n",
- "vo=0 #m/s\n",
- "g=9.8 #m/s**2\n",
- "mu=0.25\n",
- "\n",
- "#Calculations\n",
- "#Using the kinematic equations of motion\n",
- "a=(v**2-vo**2)*(2*s)**-1 #m/s**2\n",
- "#Using Newtons Principle\n",
- "N1=g*m #N\n",
- "P=m*a+mu*N1 #N\n",
- "\n",
- "#Result\n",
- "print'The value of P is',round(P,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P is 15.6 N\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-3, Page No 232"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=2 #kg\n",
- "vo=0 #m/s\n",
- "v=3 #m/s\n",
- "s=0.8 #m\n",
- "# as theta=20 degrees,\n",
- "sintheta=0.34\n",
- "costheta=0.94\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "N=m*g*costheta #N\n",
- "a=(vo**2-v**2)*(2*s)**-1 #m/s**2\n",
- "u=-((2*a)+(m*g*sintheta))/N \n",
- "#Solving for return speed\n",
- "#Symbol convention is different from textbook\n",
- "a_ret=((m*g*sintheta)-(u*N))/2 #m/s**2\n",
- "vf=sqrt((2*a_ret*s)) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed is',round(vf,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed is 1.3 m/s\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-4, Page No 232"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=1800 #lb\n",
- "r=2000 #ft\n",
- "v=58.7 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "F=(W*v*v)/(g*r) #lb\n",
- "\n",
- "#Result\n",
- "print'The frictional force to be exerted is',round(F,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frictional force to be exerted is 96.3 lb\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-7, Page No 234"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "# Initilization of variables\n",
- "W=10 #lb\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=sqrt(3)*2**-1\n",
- "l=2 #ft\n",
- "w=10 #rev/min\n",
- "g=32.2 # ft/s**2\n",
- "\n",
- "# Calculations\n",
- "r=l*costheta # ft\n",
- "a_n=r*(((w*2*pi)/60)**2) #ft/s**2\n",
- "#Applying Newtons Principle\n",
- "#Solving by matrix method\n",
- "A=np.array([[costheta,-sintheta],[sintheta,costheta]]) \n",
- "B=np.array([[(W*a_n)/g],[W]]) \n",
- "C=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The value of T is',round(C[0],2),\"lb\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of T is 5.51 lb\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-8, Page No 235"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=4 #lb\n",
- "v=6 #ft/s\n",
- "r=2 #ft\n",
- "# as theta1=40 degrees & theta2=20 degrees\n",
- "sintheta1=0.64\n",
- "costheta1=0.77\n",
- "sintheta2=0.34\n",
- "costheta2=0.94\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "a_n=v**2/r #ft/s**2\n",
- "#Applying Newtons Principle\n",
- "Fi=(m*a_n)/g #lb\n",
- "#Solving by matrix method\n",
- "A=np.array([[costheta1,costheta2],[sintheta1,-sintheta2]])\n",
- "B=np.array([[m],[Fi]]) \n",
- "C=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The values are: T=',round(C[0],2),\"lb\",'and C=',round(C[1],2),\"lb\"\n",
- "\n",
- "# The ans for C waries due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are: T= 4.01 lb and C= 0.97 lb\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-10, Page No 237"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=2 #kg\n",
- "# as theta=20 degrees,\n",
- "sintheta=0.34\n",
- "m2=4 #kg\n",
- "t=4 #s\n",
- "g=9.8 #m/s**2\n",
- "vo=0 #m/s\n",
- "\n",
- "#Calculations\n",
- "#Applying Newtons Principle\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,-2],[1,4]])\n",
- "B=np.array([[m1*g*sintheta],[m2*g]])\n",
- "C=np.linalg.solve(A,B)\n",
- "a=C[1] #m/s**2\n",
- "v=vo+a*t #m/s\n",
- "\n",
- "#Result\n",
- "print'The velocity of 4kg mass is',round(v,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity of 4kg mass is 21.7 m/s\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-11, Page No 237"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m_A=20 #lb\n",
- "m_B=60 #lb\n",
- "u=0.3 #coefficient of friction\n",
- "t=4 #s\n",
- "# as theta1=30 degrees & theta2=60 degrees,\n",
- "sintheta1=2**-1\n",
- "costheta1=sqrt(3)*2**-1\n",
- "sintheta2=sqrt(3)*2**-1\n",
- "costheta2=2**-1\n",
- "g=32.2 #ft/s^2\n",
- "vo=0 #ft/s\n",
- "\n",
- "#Calculations\n",
- "N1=m_A*costheta1 #lb\n",
- "N2=m_B*costheta2 #lb\n",
- "#Solving for T and a using matrix method\n",
- "A=np.array([[1,-m_A/g],[-1,-m_B/g]])\n",
- "B=np.array([[(m_A*sintheta1+u*N1)],[(-m_B*sintheta2+u*N2)]]) \n",
- "C=np.linalg.solve(A,B)\n",
- "a=C[1] #ft/s**2\n",
- "v=vo+a*t #ft/s\n",
- "\n",
- "#Result\n",
- "print'The velocity is',round(v,1),\"ft/s\"\n",
- "\n",
- "# The ans in the textbook is incorrect."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity is 44.7 ft/s\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-12, Page No 238"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m_A=40 #kg\n",
- "m_B=15 #kg\n",
- "F=500 #N\n",
- "g=9.8 #m/s**2\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "m=m_A+m_B #kg\n",
- "a=(F-m*g*sintheta)/(m) #m/s**2\n",
- "#Summing forces parallel and perpendicular to the plane\n",
- "#Simplfying equation (1) and (2)\n",
- "Nb=m_B*g+(m_B*a*sintheta) #N\n",
- "#Substituting this in eq(1)\n",
- "u=-(m_B*g*costheta-(Nb*costheta))/(Nb*sintheta)\n",
- "\n",
- "#Result\n",
- "print'The value of u is',round(u,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of u is 0.31\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-13, Page No 239"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "P=70 #N\n",
- "m_A=16 #kg\n",
- "u_AH=0.25 #coefficient of friction between Block A and Horizontal Plane\n",
- "m_B=4 #kg\n",
- "u_BH=0.5 #coefficient of friction between Block B and Horizontal Plane\n",
- "# as theta=10 degrees,\n",
- "sintheta=0.17\n",
- "costheta=0.98\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying sum of forces to both the FBD's\n",
- "#Solving by matrix method \n",
- "A=np.array([[-costheta,-u_AH,-m_A,0],[-sintheta,1,0,0],[costheta,0,-m_B,-u_BH],[sintheta,0,0,1]]) \n",
- "B=np.array([[-P],[m_A*g],[0],[m_B*g]])\n",
- "C=np.linalg.solve(A,B) \n",
- "\n",
- "#Result\n",
- "print'The Value of T is',round(C[0],1),\"N\"\n",
- "\n",
- "# The ans waries due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Value of T is 20.7 N\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-14, Page No 239"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=10 degrees\n",
- "sintheta=0.1736\n",
- "costheta=0.9848\n",
- "v=10 #ft/s\n",
- "v0=0 #ft/s\n",
- "u=3**-1 #coefficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Equations of motion for box are\n",
- "#Simplfying the equations by sybstitution\n",
- "a=((u*costheta)-(sintheta))*g #ft/s**2\n",
- "#Time calculations\n",
- "t=(v-v0)/a #s\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,2),\"ft/s**2\",'and the time required is',round(t),\"seconds\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is 4.98 ft/s**2 and the time required is 2.0 seconds\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-15, Page No 240"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Simplfying the equations we can solve for T2 and aA first to obtain the solution\n",
- "#Solving by matrix method\n",
- "A=np.array([[-1.5,-4],[-3.5,24]])\n",
- "B=np.array([[-4*g],[-24*g]])\n",
- "C=np.linalg.solve(A,B) \n",
- "T2=C[0] #N\n",
- "T1=T2/2 #N\n",
- "T3=T2/2 #N\n",
- "#Acceleration calculations\n",
- "a1=1*g-T1 #m/s**2\n",
- "a2=(2*g-T1)/2 #m/s**2\n",
- "a3=(3*g-T3)/3 #m/s**2\n",
- "a4=(4*g-T3)/4 #m/s**2\n",
- "#Tension in fixed cord\n",
- "T_f=2*T2 #N\n",
- "\n",
- "#Result\n",
- "print'The acceleration values are: a1=',round(a1),\"m/s**2 (up)\",',',round(a2,1),\"m/s**2 (down)\",',',round(a3,2),\"m/s**2 (down)\",',',round(a4,1),\"m/s**2 (down) respectively.\"\n",
- "print'The tension in the fixed cord is',round(T_f,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration values are: a1= -9.0 m/s**2 (up) , 0.4 m/s**2 (down) , 3.53 m/s**2 (down) , 5.1 m/s**2 (down) respectively.\n",
- "The tension in the fixed cord is 75.3 N\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-16, Page No 241"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=14 #kg\n",
- "m2=7 #kg\n",
- "# as theta=45 degrees,\n",
- "sintheta=sqrt(2)**-1\n",
- "costheta=sqrt(2)**-1\n",
- "u_1=4**-1 #coefficient of friction between mass 1 and plane\n",
- "u_2=3*8**-1 #coefficient of friction between mass 2 and plane\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#The equations of motion for m1 are\n",
- "N1=m1*g*costheta #N\n",
- "F1=u_1*N1 #N\n",
- "#The equations of motion for m2 are\n",
- "N2=m2*g*costheta #N\n",
- "F2=u_2*N2 #N\n",
- "#Now to get T and a we solve using matrix method\n",
- "A=np.array([[-1,-m1],[1,-m2]])\n",
- "B=np.array([[-(m1*g*sintheta-F1)],[-(m2*g*sintheta-F2)]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The Value of T is',round(C[0]),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Value of T is 4.0 N\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-19, Page No 244"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=12 #oz\n",
- "k=2 #oz/in\n",
- "M=0.34 #kg\n",
- "K=22 #N/m\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Part(a)\n",
- "a=(k*W*g)/16\n",
- "b=W*16**-1\n",
- "f=((2*pi)**-1)*((a/b)**0.5) #Hz for simplicity the numerator and denominator have been computed seperately as a and b\n",
- "#Part(b)\n",
- "F=((2*pi)**-1)*((K/M)**0.5) #Hz\n",
- "\n",
- "#Result\n",
- "print'The frequency in part (a) is',round(f,2),\"Hz\",'and in part(b) is',round(F,2),\"Hz\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frequency in part (a) is 1.28 Hz and in part(b) is 1.28 Hz\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-20, Page No 244"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#As the entire question is theoritical\n",
- "#theta is directly computed \n",
- "theta=arccos(2*3**-1)*(180/pi) #degrees\n",
- "\n",
- "#result\n",
- "print'The value of theta is',round(theta,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of theta is 48.2 degrees\n"
- ]
- }
- ],
- "prompt_number": 43
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-28, Page No 254"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "G=6.658*(10**-8)**-1 #cm**3/g.s**2\n",
- "#Calculations\n",
- "G1=G*((3.281*10**2)/((2.205*32.2**-1)*10**4)) #ft**3/slug-s**2\n",
- "G2=G1 #ft**4/lb-s**4\n",
- "\n",
- "#Result\n",
- "print'The ans is',round(G2,2),\"ft**4/lb-s**4\"\n",
- "\n",
- "# The ans waries slightly due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ans is 319004859.68 ft**4/lb-s**4\n"
- ]
- }
- ],
- "prompt_number": 51
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-29, Page No 254"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#Modifying the value of C without vo**2 in it\n",
- "C=5000*5280\n",
- "G=3.43*10**-8 #Gravatational Constant\n",
- "M=4.09*10**23 #Mass of the Earth\n",
- "a=5.31*10**8\n",
- "#When the orbit is circular e=0\n",
- "vo1=(a)**0.5 #ft/s\n",
- "#When the orbit is parabolic e=1\n",
- "vo2=((C*a+G*M)/C)**0.5 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The value of vo1=',round(vo1),\"ft/s\",'is smaller than vo2=',round(vo2),\"ft/s, hence the Satellite will enter a hyperbolic path and never return to earth.\"\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of vo1= 23043.0 ft/s is smaller than vo2= 32594.0 ft/s, hence the Satellite will enter a hyperbolic path and never return to earth.\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-30, Page No 255"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=3940+500 #mi\n",
- "phi=0 #degrees\n",
- "vo=36000 #ft/s\n",
- "C=4440*5280*vo\n",
- "G=3.43*10**-8\n",
- "M=4.09*10**23 #kg\n",
- "\n",
- "#Calculations\n",
- "e=((C*vo)/(G*M))-1\n",
- "\n",
- "#Result\n",
- "print'The value of e=',round(e,2),\",hence the path is Hyperbolic\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of e= 1.17 ,hence the path is Hyperbolic\n"
- ]
- }
- ],
- "prompt_number": 54
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-31, Page No 255"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "a=92.9*10**6 #mi\n",
- "G=3.43*10**-8\n",
- "T=365*24*3600 #s\n",
- "c=5280\n",
- "\n",
- "#Calculations\n",
- "M=(4*pi**2*a**3*c**3)/(G*T**2) #slugs\n",
- "\n",
- "#Result\n",
- "print'The mass of the sun is',round(M,1),\"slugs\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The mass of the sun is 1.36584467048e+29 slugs\n"
- ]
- }
- ],
- "prompt_number": 56
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter13_1.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter13_1.ipynb deleted file mode 100755 index a77abe8e..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter13_1.ipynb +++ /dev/null @@ -1,899 +0,0 @@ -{
- "metadata": {
- "name": "chapter13.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 13: Dynamics of a Particle"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-1, Page No 230"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=2 #lb\n",
- "F=1.5 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Angles are with respect to the plane are,\n",
- "# theta1=10 degrees & theta2=30 degrees\n",
- "sintheta1=0.17\n",
- "costheta1=0.99\n",
- "sintheta2=0.5\n",
- "costheta2=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Now here the forces are considered as parallel and perpendicular to the plane \n",
- "#Applying Newtond Principle\n",
- "ax=(g/2)*(F*costheta1-(W*sintheta2)) #ft/s**2\n",
- "N1=(2*costheta2-(F*sintheta1)) #lb\n",
- "\n",
- "#result\n",
- "print'The force on the particle is',round(N1,2),\"lb\"\n",
- "print'The acceleration is',round(ax,2),\"ft/s**2\"\n",
- "\n",
- "# The answer may wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force on the particle is 1.48 lb\n",
- "The acceleration is 7.81 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-2, Page No 231"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=5 #kg\n",
- "s=12 #m\n",
- "v=4 #m/s\n",
- "vo=0 #m/s\n",
- "g=9.8 #m/s**2\n",
- "mu=0.25\n",
- "\n",
- "#Calculations\n",
- "#Using the kinematic equations of motion\n",
- "a=(v**2-vo**2)*(2*s)**-1 #m/s**2\n",
- "#Using Newtons Principle\n",
- "N1=g*m #N\n",
- "P=m*a+mu*N1 #N\n",
- "\n",
- "#Result\n",
- "print'The value of P is',round(P,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P is 15.6 N\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-3, Page No 232"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=2 #kg\n",
- "vo=0 #m/s\n",
- "v=3 #m/s\n",
- "s=0.8 #m\n",
- "# as theta=20 degrees,\n",
- "sintheta=0.34\n",
- "costheta=0.94\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "N=m*g*costheta #N\n",
- "a=(vo**2-v**2)*(2*s)**-1 #m/s**2\n",
- "u=-((2*a)+(m*g*sintheta))/N \n",
- "#Solving for return speed\n",
- "#Symbol convention is different from textbook\n",
- "a_ret=((m*g*sintheta)-(u*N))/2 #m/s**2\n",
- "vf=sqrt((2*a_ret*s)) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed is',round(vf,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed is 1.3 m/s\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-4, Page No 232"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=1800 #lb\n",
- "r=2000 #ft\n",
- "v=58.7 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "F=(W*v*v)/(g*r) #lb\n",
- "\n",
- "#Result\n",
- "print'The frictional force to be exerted is',round(F,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frictional force to be exerted is 96.3 lb\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-7, Page No 234"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "# Initilization of variables\n",
- "W=10 #lb\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=sqrt(3)*2**-1\n",
- "l=2 #ft\n",
- "w=10 #rev/min\n",
- "g=32.2 # ft/s**2\n",
- "\n",
- "# Calculations\n",
- "r=l*costheta # ft\n",
- "a_n=r*(((w*2*pi)/60)**2) #ft/s**2\n",
- "#Applying Newtons Principle\n",
- "#Solving by matrix method\n",
- "A=np.array([[costheta,-sintheta],[sintheta,costheta]]) \n",
- "B=np.array([[(W*a_n)/g],[W]]) \n",
- "C=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The value of T is',round(C[0],2),\"lb\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of T is 5.51 lb\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-8, Page No 235"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=4 #lb\n",
- "v=6 #ft/s\n",
- "r=2 #ft\n",
- "# as theta1=40 degrees & theta2=20 degrees\n",
- "sintheta1=0.64\n",
- "costheta1=0.77\n",
- "sintheta2=0.34\n",
- "costheta2=0.94\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "a_n=v**2/r #ft/s**2\n",
- "#Applying Newtons Principle\n",
- "Fi=(m*a_n)/g #lb\n",
- "#Solving by matrix method\n",
- "A=np.array([[costheta1,costheta2],[sintheta1,-sintheta2]])\n",
- "B=np.array([[m],[Fi]]) \n",
- "C=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The values are: T=',round(C[0],2),\"lb\",'and C=',round(C[1],2),\"lb\"\n",
- "\n",
- "# The ans for C waries due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are: T= 4.01 lb and C= 0.97 lb\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-10, Page No 237"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=2 #kg\n",
- "# as theta=20 degrees,\n",
- "sintheta=0.34\n",
- "m2=4 #kg\n",
- "t=4 #s\n",
- "g=9.8 #m/s**2\n",
- "vo=0 #m/s\n",
- "\n",
- "#Calculations\n",
- "#Applying Newtons Principle\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,-2],[1,4]])\n",
- "B=np.array([[m1*g*sintheta],[m2*g]])\n",
- "C=np.linalg.solve(A,B)\n",
- "a=C[1] #m/s**2\n",
- "v=vo+a*t #m/s\n",
- "\n",
- "#Result\n",
- "print'The velocity of 4kg mass is',round(v,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity of 4kg mass is 21.7 m/s\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-11, Page No 237"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m_A=20 #lb\n",
- "m_B=60 #lb\n",
- "u=0.3 #coefficient of friction\n",
- "t=4 #s\n",
- "# as theta1=30 degrees & theta2=60 degrees,\n",
- "sintheta1=2**-1\n",
- "costheta1=sqrt(3)*2**-1\n",
- "sintheta2=sqrt(3)*2**-1\n",
- "costheta2=2**-1\n",
- "g=32.2 #ft/s^2\n",
- "vo=0 #ft/s\n",
- "\n",
- "#Calculations\n",
- "N1=m_A*costheta1 #lb\n",
- "N2=m_B*costheta2 #lb\n",
- "#Solving for T and a using matrix method\n",
- "A=np.array([[1,-m_A/g],[-1,-m_B/g]])\n",
- "B=np.array([[(m_A*sintheta1+u*N1)],[(-m_B*sintheta2+u*N2)]]) \n",
- "C=np.linalg.solve(A,B)\n",
- "a=C[1] #ft/s**2\n",
- "v=vo+a*t #ft/s\n",
- "\n",
- "#Result\n",
- "print'The velocity is',round(v,1),\"ft/s\"\n",
- "\n",
- "# The ans in the textbook is incorrect."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity is 44.7 ft/s\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-12, Page No 238"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m_A=40 #kg\n",
- "m_B=15 #kg\n",
- "F=500 #N\n",
- "g=9.8 #m/s**2\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "m=m_A+m_B #kg\n",
- "a=(F-m*g*sintheta)/(m) #m/s**2\n",
- "#Summing forces parallel and perpendicular to the plane\n",
- "#Simplfying equation (1) and (2)\n",
- "Nb=m_B*g+(m_B*a*sintheta) #N\n",
- "#Substituting this in eq(1)\n",
- "u=-(m_B*g*costheta-(Nb*costheta))/(Nb*sintheta)\n",
- "\n",
- "#Result\n",
- "print'The value of u is',round(u,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of u is 0.31\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-13, Page No 239"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "P=70 #N\n",
- "m_A=16 #kg\n",
- "u_AH=0.25 #coefficient of friction between Block A and Horizontal Plane\n",
- "m_B=4 #kg\n",
- "u_BH=0.5 #coefficient of friction between Block B and Horizontal Plane\n",
- "# as theta=10 degrees,\n",
- "sintheta=0.17\n",
- "costheta=0.98\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying sum of forces to both the FBD's\n",
- "#Solving by matrix method \n",
- "A=np.array([[-costheta,-u_AH,-m_A,0],[-sintheta,1,0,0],[costheta,0,-m_B,-u_BH],[sintheta,0,0,1]]) \n",
- "B=np.array([[-P],[m_A*g],[0],[m_B*g]])\n",
- "C=np.linalg.solve(A,B) \n",
- "\n",
- "#Result\n",
- "print'The Value of T is',round(C[0],1),\"N\"\n",
- "\n",
- "# The ans waries due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Value of T is 20.7 N\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-14, Page No 239"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=10 degrees\n",
- "sintheta=0.1736\n",
- "costheta=0.9848\n",
- "v=10 #ft/s\n",
- "v0=0 #ft/s\n",
- "u=3**-1 #coefficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Equations of motion for box are\n",
- "#Simplfying the equations by sybstitution\n",
- "a=((u*costheta)-(sintheta))*g #ft/s**2\n",
- "#Time calculations\n",
- "t=(v-v0)/a #s\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,2),\"ft/s**2\",'and the time required is',round(t),\"seconds\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is 4.98 ft/s**2 and the time required is 2.0 seconds\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-15, Page No 240"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Simplfying the equations we can solve for T2 and aA first to obtain the solution\n",
- "#Solving by matrix method\n",
- "A=np.array([[-1.5,-4],[-3.5,24]])\n",
- "B=np.array([[-4*g],[-24*g]])\n",
- "C=np.linalg.solve(A,B) \n",
- "T2=C[0] #N\n",
- "T1=T2/2 #N\n",
- "T3=T2/2 #N\n",
- "#Acceleration calculations\n",
- "a1=1*g-T1 #m/s**2\n",
- "a2=(2*g-T1)/2 #m/s**2\n",
- "a3=(3*g-T3)/3 #m/s**2\n",
- "a4=(4*g-T3)/4 #m/s**2\n",
- "#Tension in fixed cord\n",
- "T_f=2*T2 #N\n",
- "\n",
- "#Result\n",
- "print'The acceleration values are: a1=',round(a1),\"m/s**2 (up)\",',',round(a2,1),\"m/s**2 (down)\",',',round(a3,2),\"m/s**2 (down)\",',',round(a4,1),\"m/s**2 (down) respectively.\"\n",
- "print'The tension in the fixed cord is',round(T_f,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration values are: a1= -9.0 m/s**2 (up) , 0.4 m/s**2 (down) , 3.53 m/s**2 (down) , 5.1 m/s**2 (down) respectively.\n",
- "The tension in the fixed cord is 75.3 N\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-16, Page No 241"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=14 #kg\n",
- "m2=7 #kg\n",
- "# as theta=45 degrees,\n",
- "sintheta=sqrt(2)**-1\n",
- "costheta=sqrt(2)**-1\n",
- "u_1=4**-1 #coefficient of friction between mass 1 and plane\n",
- "u_2=3*8**-1 #coefficient of friction between mass 2 and plane\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#The equations of motion for m1 are\n",
- "N1=m1*g*costheta #N\n",
- "F1=u_1*N1 #N\n",
- "#The equations of motion for m2 are\n",
- "N2=m2*g*costheta #N\n",
- "F2=u_2*N2 #N\n",
- "#Now to get T and a we solve using matrix method\n",
- "A=np.array([[-1,-m1],[1,-m2]])\n",
- "B=np.array([[-(m1*g*sintheta-F1)],[-(m2*g*sintheta-F2)]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The Value of T is',round(C[0]),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Value of T is 4.0 N\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-19, Page No 244"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=12 #oz\n",
- "k=2 #oz/in\n",
- "M=0.34 #kg\n",
- "K=22 #N/m\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Part(a)\n",
- "a=(k*W*g)/16\n",
- "b=W*16**-1\n",
- "f=((2*pi)**-1)*((a/b)**0.5) #Hz for simplicity the numerator and denominator have been computed seperately as a and b\n",
- "#Part(b)\n",
- "F=((2*pi)**-1)*((K/M)**0.5) #Hz\n",
- "\n",
- "#Result\n",
- "print'The frequency in part (a) is',round(f,2),\"Hz\",'and in part(b) is',round(F,2),\"Hz\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frequency in part (a) is 1.28 Hz and in part(b) is 1.28 Hz\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-20, Page No 244"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#As the entire question is theoritical\n",
- "#theta is directly computed \n",
- "theta=arccos(2*3**-1)*(180/pi) #degrees\n",
- "\n",
- "#result\n",
- "print'The value of theta is',round(theta,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of theta is 48.2 degrees\n"
- ]
- }
- ],
- "prompt_number": 43
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-28, Page No 254"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "G=6.658*(10**-8)**-1 #cm**3/g.s**2\n",
- "#Calculations\n",
- "G1=G*((3.281*10**2)/((2.205*32.2**-1)*10**4)) #ft**3/slug-s**2\n",
- "G2=G1 #ft**4/lb-s**4\n",
- "\n",
- "#Result\n",
- "print'The ans is',round(G2,2),\"ft**4/lb-s**4\"\n",
- "\n",
- "# The ans waries slightly due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ans is 319004859.68 ft**4/lb-s**4\n"
- ]
- }
- ],
- "prompt_number": 51
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-29, Page No 254"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#Modifying the value of C without vo**2 in it\n",
- "C=5000*5280\n",
- "G=3.43*10**-8 #Gravatational Constant\n",
- "M=4.09*10**23 #Mass of the Earth\n",
- "a=5.31*10**8\n",
- "#When the orbit is circular e=0\n",
- "vo1=(a)**0.5 #ft/s\n",
- "#When the orbit is parabolic e=1\n",
- "vo2=((C*a+G*M)/C)**0.5 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The value of vo1=',round(vo1),\"ft/s\",'is smaller than vo2=',round(vo2),\"ft/s, hence the Satellite will enter a hyperbolic path and never return to earth.\"\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of vo1= 23043.0 ft/s is smaller than vo2= 32594.0 ft/s, hence the Satellite will enter a hyperbolic path and never return to earth.\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-30, Page No 255"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=3940+500 #mi\n",
- "phi=0 #degrees\n",
- "vo=36000 #ft/s\n",
- "C=4440*5280*vo\n",
- "G=3.43*10**-8\n",
- "M=4.09*10**23 #kg\n",
- "\n",
- "#Calculations\n",
- "e=((C*vo)/(G*M))-1\n",
- "\n",
- "#Result\n",
- "print'The value of e=',round(e,2),\",hence the path is Hyperbolic\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of e= 1.17 ,hence the path is Hyperbolic\n"
- ]
- }
- ],
- "prompt_number": 54
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-31, Page No 255"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "a=92.9*10**6 #mi\n",
- "G=3.43*10**-8\n",
- "T=365*24*3600 #s\n",
- "c=5280\n",
- "\n",
- "#Calculations\n",
- "M=(4*pi**2*a**3*c**3)/(G*T**2) #slugs\n",
- "\n",
- "#Result\n",
- "print'The mass of the sun is',round(M,1),\"slugs\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The mass of the sun is 1.36584467048e+29 slugs\n"
- ]
- }
- ],
- "prompt_number": 56
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter13_2.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter13_2.ipynb deleted file mode 100755 index a77abe8e..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter13_2.ipynb +++ /dev/null @@ -1,899 +0,0 @@ -{
- "metadata": {
- "name": "chapter13.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 13: Dynamics of a Particle"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-1, Page No 230"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=2 #lb\n",
- "F=1.5 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Angles are with respect to the plane are,\n",
- "# theta1=10 degrees & theta2=30 degrees\n",
- "sintheta1=0.17\n",
- "costheta1=0.99\n",
- "sintheta2=0.5\n",
- "costheta2=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Now here the forces are considered as parallel and perpendicular to the plane \n",
- "#Applying Newtond Principle\n",
- "ax=(g/2)*(F*costheta1-(W*sintheta2)) #ft/s**2\n",
- "N1=(2*costheta2-(F*sintheta1)) #lb\n",
- "\n",
- "#result\n",
- "print'The force on the particle is',round(N1,2),\"lb\"\n",
- "print'The acceleration is',round(ax,2),\"ft/s**2\"\n",
- "\n",
- "# The answer may wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force on the particle is 1.48 lb\n",
- "The acceleration is 7.81 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-2, Page No 231"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=5 #kg\n",
- "s=12 #m\n",
- "v=4 #m/s\n",
- "vo=0 #m/s\n",
- "g=9.8 #m/s**2\n",
- "mu=0.25\n",
- "\n",
- "#Calculations\n",
- "#Using the kinematic equations of motion\n",
- "a=(v**2-vo**2)*(2*s)**-1 #m/s**2\n",
- "#Using Newtons Principle\n",
- "N1=g*m #N\n",
- "P=m*a+mu*N1 #N\n",
- "\n",
- "#Result\n",
- "print'The value of P is',round(P,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P is 15.6 N\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-3, Page No 232"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=2 #kg\n",
- "vo=0 #m/s\n",
- "v=3 #m/s\n",
- "s=0.8 #m\n",
- "# as theta=20 degrees,\n",
- "sintheta=0.34\n",
- "costheta=0.94\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "N=m*g*costheta #N\n",
- "a=(vo**2-v**2)*(2*s)**-1 #m/s**2\n",
- "u=-((2*a)+(m*g*sintheta))/N \n",
- "#Solving for return speed\n",
- "#Symbol convention is different from textbook\n",
- "a_ret=((m*g*sintheta)-(u*N))/2 #m/s**2\n",
- "vf=sqrt((2*a_ret*s)) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed is',round(vf,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed is 1.3 m/s\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-4, Page No 232"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=1800 #lb\n",
- "r=2000 #ft\n",
- "v=58.7 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "F=(W*v*v)/(g*r) #lb\n",
- "\n",
- "#Result\n",
- "print'The frictional force to be exerted is',round(F,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frictional force to be exerted is 96.3 lb\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-7, Page No 234"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "# Initilization of variables\n",
- "W=10 #lb\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=sqrt(3)*2**-1\n",
- "l=2 #ft\n",
- "w=10 #rev/min\n",
- "g=32.2 # ft/s**2\n",
- "\n",
- "# Calculations\n",
- "r=l*costheta # ft\n",
- "a_n=r*(((w*2*pi)/60)**2) #ft/s**2\n",
- "#Applying Newtons Principle\n",
- "#Solving by matrix method\n",
- "A=np.array([[costheta,-sintheta],[sintheta,costheta]]) \n",
- "B=np.array([[(W*a_n)/g],[W]]) \n",
- "C=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The value of T is',round(C[0],2),\"lb\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of T is 5.51 lb\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-8, Page No 235"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=4 #lb\n",
- "v=6 #ft/s\n",
- "r=2 #ft\n",
- "# as theta1=40 degrees & theta2=20 degrees\n",
- "sintheta1=0.64\n",
- "costheta1=0.77\n",
- "sintheta2=0.34\n",
- "costheta2=0.94\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "a_n=v**2/r #ft/s**2\n",
- "#Applying Newtons Principle\n",
- "Fi=(m*a_n)/g #lb\n",
- "#Solving by matrix method\n",
- "A=np.array([[costheta1,costheta2],[sintheta1,-sintheta2]])\n",
- "B=np.array([[m],[Fi]]) \n",
- "C=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The values are: T=',round(C[0],2),\"lb\",'and C=',round(C[1],2),\"lb\"\n",
- "\n",
- "# The ans for C waries due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are: T= 4.01 lb and C= 0.97 lb\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-10, Page No 237"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=2 #kg\n",
- "# as theta=20 degrees,\n",
- "sintheta=0.34\n",
- "m2=4 #kg\n",
- "t=4 #s\n",
- "g=9.8 #m/s**2\n",
- "vo=0 #m/s\n",
- "\n",
- "#Calculations\n",
- "#Applying Newtons Principle\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,-2],[1,4]])\n",
- "B=np.array([[m1*g*sintheta],[m2*g]])\n",
- "C=np.linalg.solve(A,B)\n",
- "a=C[1] #m/s**2\n",
- "v=vo+a*t #m/s\n",
- "\n",
- "#Result\n",
- "print'The velocity of 4kg mass is',round(v,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity of 4kg mass is 21.7 m/s\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-11, Page No 237"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m_A=20 #lb\n",
- "m_B=60 #lb\n",
- "u=0.3 #coefficient of friction\n",
- "t=4 #s\n",
- "# as theta1=30 degrees & theta2=60 degrees,\n",
- "sintheta1=2**-1\n",
- "costheta1=sqrt(3)*2**-1\n",
- "sintheta2=sqrt(3)*2**-1\n",
- "costheta2=2**-1\n",
- "g=32.2 #ft/s^2\n",
- "vo=0 #ft/s\n",
- "\n",
- "#Calculations\n",
- "N1=m_A*costheta1 #lb\n",
- "N2=m_B*costheta2 #lb\n",
- "#Solving for T and a using matrix method\n",
- "A=np.array([[1,-m_A/g],[-1,-m_B/g]])\n",
- "B=np.array([[(m_A*sintheta1+u*N1)],[(-m_B*sintheta2+u*N2)]]) \n",
- "C=np.linalg.solve(A,B)\n",
- "a=C[1] #ft/s**2\n",
- "v=vo+a*t #ft/s\n",
- "\n",
- "#Result\n",
- "print'The velocity is',round(v,1),\"ft/s\"\n",
- "\n",
- "# The ans in the textbook is incorrect."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity is 44.7 ft/s\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-12, Page No 238"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m_A=40 #kg\n",
- "m_B=15 #kg\n",
- "F=500 #N\n",
- "g=9.8 #m/s**2\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "m=m_A+m_B #kg\n",
- "a=(F-m*g*sintheta)/(m) #m/s**2\n",
- "#Summing forces parallel and perpendicular to the plane\n",
- "#Simplfying equation (1) and (2)\n",
- "Nb=m_B*g+(m_B*a*sintheta) #N\n",
- "#Substituting this in eq(1)\n",
- "u=-(m_B*g*costheta-(Nb*costheta))/(Nb*sintheta)\n",
- "\n",
- "#Result\n",
- "print'The value of u is',round(u,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of u is 0.31\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-13, Page No 239"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "P=70 #N\n",
- "m_A=16 #kg\n",
- "u_AH=0.25 #coefficient of friction between Block A and Horizontal Plane\n",
- "m_B=4 #kg\n",
- "u_BH=0.5 #coefficient of friction between Block B and Horizontal Plane\n",
- "# as theta=10 degrees,\n",
- "sintheta=0.17\n",
- "costheta=0.98\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying sum of forces to both the FBD's\n",
- "#Solving by matrix method \n",
- "A=np.array([[-costheta,-u_AH,-m_A,0],[-sintheta,1,0,0],[costheta,0,-m_B,-u_BH],[sintheta,0,0,1]]) \n",
- "B=np.array([[-P],[m_A*g],[0],[m_B*g]])\n",
- "C=np.linalg.solve(A,B) \n",
- "\n",
- "#Result\n",
- "print'The Value of T is',round(C[0],1),\"N\"\n",
- "\n",
- "# The ans waries due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Value of T is 20.7 N\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-14, Page No 239"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=10 degrees\n",
- "sintheta=0.1736\n",
- "costheta=0.9848\n",
- "v=10 #ft/s\n",
- "v0=0 #ft/s\n",
- "u=3**-1 #coefficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Equations of motion for box are\n",
- "#Simplfying the equations by sybstitution\n",
- "a=((u*costheta)-(sintheta))*g #ft/s**2\n",
- "#Time calculations\n",
- "t=(v-v0)/a #s\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,2),\"ft/s**2\",'and the time required is',round(t),\"seconds\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is 4.98 ft/s**2 and the time required is 2.0 seconds\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-15, Page No 240"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Simplfying the equations we can solve for T2 and aA first to obtain the solution\n",
- "#Solving by matrix method\n",
- "A=np.array([[-1.5,-4],[-3.5,24]])\n",
- "B=np.array([[-4*g],[-24*g]])\n",
- "C=np.linalg.solve(A,B) \n",
- "T2=C[0] #N\n",
- "T1=T2/2 #N\n",
- "T3=T2/2 #N\n",
- "#Acceleration calculations\n",
- "a1=1*g-T1 #m/s**2\n",
- "a2=(2*g-T1)/2 #m/s**2\n",
- "a3=(3*g-T3)/3 #m/s**2\n",
- "a4=(4*g-T3)/4 #m/s**2\n",
- "#Tension in fixed cord\n",
- "T_f=2*T2 #N\n",
- "\n",
- "#Result\n",
- "print'The acceleration values are: a1=',round(a1),\"m/s**2 (up)\",',',round(a2,1),\"m/s**2 (down)\",',',round(a3,2),\"m/s**2 (down)\",',',round(a4,1),\"m/s**2 (down) respectively.\"\n",
- "print'The tension in the fixed cord is',round(T_f,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration values are: a1= -9.0 m/s**2 (up) , 0.4 m/s**2 (down) , 3.53 m/s**2 (down) , 5.1 m/s**2 (down) respectively.\n",
- "The tension in the fixed cord is 75.3 N\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-16, Page No 241"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=14 #kg\n",
- "m2=7 #kg\n",
- "# as theta=45 degrees,\n",
- "sintheta=sqrt(2)**-1\n",
- "costheta=sqrt(2)**-1\n",
- "u_1=4**-1 #coefficient of friction between mass 1 and plane\n",
- "u_2=3*8**-1 #coefficient of friction between mass 2 and plane\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#The equations of motion for m1 are\n",
- "N1=m1*g*costheta #N\n",
- "F1=u_1*N1 #N\n",
- "#The equations of motion for m2 are\n",
- "N2=m2*g*costheta #N\n",
- "F2=u_2*N2 #N\n",
- "#Now to get T and a we solve using matrix method\n",
- "A=np.array([[-1,-m1],[1,-m2]])\n",
- "B=np.array([[-(m1*g*sintheta-F1)],[-(m2*g*sintheta-F2)]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The Value of T is',round(C[0]),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Value of T is 4.0 N\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-19, Page No 244"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=12 #oz\n",
- "k=2 #oz/in\n",
- "M=0.34 #kg\n",
- "K=22 #N/m\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Part(a)\n",
- "a=(k*W*g)/16\n",
- "b=W*16**-1\n",
- "f=((2*pi)**-1)*((a/b)**0.5) #Hz for simplicity the numerator and denominator have been computed seperately as a and b\n",
- "#Part(b)\n",
- "F=((2*pi)**-1)*((K/M)**0.5) #Hz\n",
- "\n",
- "#Result\n",
- "print'The frequency in part (a) is',round(f,2),\"Hz\",'and in part(b) is',round(F,2),\"Hz\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frequency in part (a) is 1.28 Hz and in part(b) is 1.28 Hz\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-20, Page No 244"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#As the entire question is theoritical\n",
- "#theta is directly computed \n",
- "theta=arccos(2*3**-1)*(180/pi) #degrees\n",
- "\n",
- "#result\n",
- "print'The value of theta is',round(theta,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of theta is 48.2 degrees\n"
- ]
- }
- ],
- "prompt_number": 43
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-28, Page No 254"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "G=6.658*(10**-8)**-1 #cm**3/g.s**2\n",
- "#Calculations\n",
- "G1=G*((3.281*10**2)/((2.205*32.2**-1)*10**4)) #ft**3/slug-s**2\n",
- "G2=G1 #ft**4/lb-s**4\n",
- "\n",
- "#Result\n",
- "print'The ans is',round(G2,2),\"ft**4/lb-s**4\"\n",
- "\n",
- "# The ans waries slightly due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ans is 319004859.68 ft**4/lb-s**4\n"
- ]
- }
- ],
- "prompt_number": 51
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-29, Page No 254"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#Modifying the value of C without vo**2 in it\n",
- "C=5000*5280\n",
- "G=3.43*10**-8 #Gravatational Constant\n",
- "M=4.09*10**23 #Mass of the Earth\n",
- "a=5.31*10**8\n",
- "#When the orbit is circular e=0\n",
- "vo1=(a)**0.5 #ft/s\n",
- "#When the orbit is parabolic e=1\n",
- "vo2=((C*a+G*M)/C)**0.5 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The value of vo1=',round(vo1),\"ft/s\",'is smaller than vo2=',round(vo2),\"ft/s, hence the Satellite will enter a hyperbolic path and never return to earth.\"\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of vo1= 23043.0 ft/s is smaller than vo2= 32594.0 ft/s, hence the Satellite will enter a hyperbolic path and never return to earth.\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-30, Page No 255"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=3940+500 #mi\n",
- "phi=0 #degrees\n",
- "vo=36000 #ft/s\n",
- "C=4440*5280*vo\n",
- "G=3.43*10**-8\n",
- "M=4.09*10**23 #kg\n",
- "\n",
- "#Calculations\n",
- "e=((C*vo)/(G*M))-1\n",
- "\n",
- "#Result\n",
- "print'The value of e=',round(e,2),\",hence the path is Hyperbolic\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of e= 1.17 ,hence the path is Hyperbolic\n"
- ]
- }
- ],
- "prompt_number": 54
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-31, Page No 255"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "a=92.9*10**6 #mi\n",
- "G=3.43*10**-8\n",
- "T=365*24*3600 #s\n",
- "c=5280\n",
- "\n",
- "#Calculations\n",
- "M=(4*pi**2*a**3*c**3)/(G*T**2) #slugs\n",
- "\n",
- "#Result\n",
- "print'The mass of the sun is',round(M,1),\"slugs\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The mass of the sun is 1.36584467048e+29 slugs\n"
- ]
- }
- ],
- "prompt_number": 56
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter13_3.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter13_3.ipynb deleted file mode 100755 index a77abe8e..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter13_3.ipynb +++ /dev/null @@ -1,899 +0,0 @@ -{
- "metadata": {
- "name": "chapter13.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 13: Dynamics of a Particle"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-1, Page No 230"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=2 #lb\n",
- "F=1.5 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Angles are with respect to the plane are,\n",
- "# theta1=10 degrees & theta2=30 degrees\n",
- "sintheta1=0.17\n",
- "costheta1=0.99\n",
- "sintheta2=0.5\n",
- "costheta2=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Now here the forces are considered as parallel and perpendicular to the plane \n",
- "#Applying Newtond Principle\n",
- "ax=(g/2)*(F*costheta1-(W*sintheta2)) #ft/s**2\n",
- "N1=(2*costheta2-(F*sintheta1)) #lb\n",
- "\n",
- "#result\n",
- "print'The force on the particle is',round(N1,2),\"lb\"\n",
- "print'The acceleration is',round(ax,2),\"ft/s**2\"\n",
- "\n",
- "# The answer may wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force on the particle is 1.48 lb\n",
- "The acceleration is 7.81 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-2, Page No 231"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=5 #kg\n",
- "s=12 #m\n",
- "v=4 #m/s\n",
- "vo=0 #m/s\n",
- "g=9.8 #m/s**2\n",
- "mu=0.25\n",
- "\n",
- "#Calculations\n",
- "#Using the kinematic equations of motion\n",
- "a=(v**2-vo**2)*(2*s)**-1 #m/s**2\n",
- "#Using Newtons Principle\n",
- "N1=g*m #N\n",
- "P=m*a+mu*N1 #N\n",
- "\n",
- "#Result\n",
- "print'The value of P is',round(P,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P is 15.6 N\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-3, Page No 232"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=2 #kg\n",
- "vo=0 #m/s\n",
- "v=3 #m/s\n",
- "s=0.8 #m\n",
- "# as theta=20 degrees,\n",
- "sintheta=0.34\n",
- "costheta=0.94\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "N=m*g*costheta #N\n",
- "a=(vo**2-v**2)*(2*s)**-1 #m/s**2\n",
- "u=-((2*a)+(m*g*sintheta))/N \n",
- "#Solving for return speed\n",
- "#Symbol convention is different from textbook\n",
- "a_ret=((m*g*sintheta)-(u*N))/2 #m/s**2\n",
- "vf=sqrt((2*a_ret*s)) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed is',round(vf,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed is 1.3 m/s\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-4, Page No 232"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=1800 #lb\n",
- "r=2000 #ft\n",
- "v=58.7 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "F=(W*v*v)/(g*r) #lb\n",
- "\n",
- "#Result\n",
- "print'The frictional force to be exerted is',round(F,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frictional force to be exerted is 96.3 lb\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-7, Page No 234"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "# Initilization of variables\n",
- "W=10 #lb\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=sqrt(3)*2**-1\n",
- "l=2 #ft\n",
- "w=10 #rev/min\n",
- "g=32.2 # ft/s**2\n",
- "\n",
- "# Calculations\n",
- "r=l*costheta # ft\n",
- "a_n=r*(((w*2*pi)/60)**2) #ft/s**2\n",
- "#Applying Newtons Principle\n",
- "#Solving by matrix method\n",
- "A=np.array([[costheta,-sintheta],[sintheta,costheta]]) \n",
- "B=np.array([[(W*a_n)/g],[W]]) \n",
- "C=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The value of T is',round(C[0],2),\"lb\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of T is 5.51 lb\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-8, Page No 235"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=4 #lb\n",
- "v=6 #ft/s\n",
- "r=2 #ft\n",
- "# as theta1=40 degrees & theta2=20 degrees\n",
- "sintheta1=0.64\n",
- "costheta1=0.77\n",
- "sintheta2=0.34\n",
- "costheta2=0.94\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "a_n=v**2/r #ft/s**2\n",
- "#Applying Newtons Principle\n",
- "Fi=(m*a_n)/g #lb\n",
- "#Solving by matrix method\n",
- "A=np.array([[costheta1,costheta2],[sintheta1,-sintheta2]])\n",
- "B=np.array([[m],[Fi]]) \n",
- "C=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The values are: T=',round(C[0],2),\"lb\",'and C=',round(C[1],2),\"lb\"\n",
- "\n",
- "# The ans for C waries due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are: T= 4.01 lb and C= 0.97 lb\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-10, Page No 237"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=2 #kg\n",
- "# as theta=20 degrees,\n",
- "sintheta=0.34\n",
- "m2=4 #kg\n",
- "t=4 #s\n",
- "g=9.8 #m/s**2\n",
- "vo=0 #m/s\n",
- "\n",
- "#Calculations\n",
- "#Applying Newtons Principle\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,-2],[1,4]])\n",
- "B=np.array([[m1*g*sintheta],[m2*g]])\n",
- "C=np.linalg.solve(A,B)\n",
- "a=C[1] #m/s**2\n",
- "v=vo+a*t #m/s\n",
- "\n",
- "#Result\n",
- "print'The velocity of 4kg mass is',round(v,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity of 4kg mass is 21.7 m/s\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-11, Page No 237"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m_A=20 #lb\n",
- "m_B=60 #lb\n",
- "u=0.3 #coefficient of friction\n",
- "t=4 #s\n",
- "# as theta1=30 degrees & theta2=60 degrees,\n",
- "sintheta1=2**-1\n",
- "costheta1=sqrt(3)*2**-1\n",
- "sintheta2=sqrt(3)*2**-1\n",
- "costheta2=2**-1\n",
- "g=32.2 #ft/s^2\n",
- "vo=0 #ft/s\n",
- "\n",
- "#Calculations\n",
- "N1=m_A*costheta1 #lb\n",
- "N2=m_B*costheta2 #lb\n",
- "#Solving for T and a using matrix method\n",
- "A=np.array([[1,-m_A/g],[-1,-m_B/g]])\n",
- "B=np.array([[(m_A*sintheta1+u*N1)],[(-m_B*sintheta2+u*N2)]]) \n",
- "C=np.linalg.solve(A,B)\n",
- "a=C[1] #ft/s**2\n",
- "v=vo+a*t #ft/s\n",
- "\n",
- "#Result\n",
- "print'The velocity is',round(v,1),\"ft/s\"\n",
- "\n",
- "# The ans in the textbook is incorrect."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity is 44.7 ft/s\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-12, Page No 238"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m_A=40 #kg\n",
- "m_B=15 #kg\n",
- "F=500 #N\n",
- "g=9.8 #m/s**2\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "m=m_A+m_B #kg\n",
- "a=(F-m*g*sintheta)/(m) #m/s**2\n",
- "#Summing forces parallel and perpendicular to the plane\n",
- "#Simplfying equation (1) and (2)\n",
- "Nb=m_B*g+(m_B*a*sintheta) #N\n",
- "#Substituting this in eq(1)\n",
- "u=-(m_B*g*costheta-(Nb*costheta))/(Nb*sintheta)\n",
- "\n",
- "#Result\n",
- "print'The value of u is',round(u,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of u is 0.31\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-13, Page No 239"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "P=70 #N\n",
- "m_A=16 #kg\n",
- "u_AH=0.25 #coefficient of friction between Block A and Horizontal Plane\n",
- "m_B=4 #kg\n",
- "u_BH=0.5 #coefficient of friction between Block B and Horizontal Plane\n",
- "# as theta=10 degrees,\n",
- "sintheta=0.17\n",
- "costheta=0.98\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying sum of forces to both the FBD's\n",
- "#Solving by matrix method \n",
- "A=np.array([[-costheta,-u_AH,-m_A,0],[-sintheta,1,0,0],[costheta,0,-m_B,-u_BH],[sintheta,0,0,1]]) \n",
- "B=np.array([[-P],[m_A*g],[0],[m_B*g]])\n",
- "C=np.linalg.solve(A,B) \n",
- "\n",
- "#Result\n",
- "print'The Value of T is',round(C[0],1),\"N\"\n",
- "\n",
- "# The ans waries due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Value of T is 20.7 N\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-14, Page No 239"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=10 degrees\n",
- "sintheta=0.1736\n",
- "costheta=0.9848\n",
- "v=10 #ft/s\n",
- "v0=0 #ft/s\n",
- "u=3**-1 #coefficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Equations of motion for box are\n",
- "#Simplfying the equations by sybstitution\n",
- "a=((u*costheta)-(sintheta))*g #ft/s**2\n",
- "#Time calculations\n",
- "t=(v-v0)/a #s\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,2),\"ft/s**2\",'and the time required is',round(t),\"seconds\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is 4.98 ft/s**2 and the time required is 2.0 seconds\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-15, Page No 240"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Simplfying the equations we can solve for T2 and aA first to obtain the solution\n",
- "#Solving by matrix method\n",
- "A=np.array([[-1.5,-4],[-3.5,24]])\n",
- "B=np.array([[-4*g],[-24*g]])\n",
- "C=np.linalg.solve(A,B) \n",
- "T2=C[0] #N\n",
- "T1=T2/2 #N\n",
- "T3=T2/2 #N\n",
- "#Acceleration calculations\n",
- "a1=1*g-T1 #m/s**2\n",
- "a2=(2*g-T1)/2 #m/s**2\n",
- "a3=(3*g-T3)/3 #m/s**2\n",
- "a4=(4*g-T3)/4 #m/s**2\n",
- "#Tension in fixed cord\n",
- "T_f=2*T2 #N\n",
- "\n",
- "#Result\n",
- "print'The acceleration values are: a1=',round(a1),\"m/s**2 (up)\",',',round(a2,1),\"m/s**2 (down)\",',',round(a3,2),\"m/s**2 (down)\",',',round(a4,1),\"m/s**2 (down) respectively.\"\n",
- "print'The tension in the fixed cord is',round(T_f,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration values are: a1= -9.0 m/s**2 (up) , 0.4 m/s**2 (down) , 3.53 m/s**2 (down) , 5.1 m/s**2 (down) respectively.\n",
- "The tension in the fixed cord is 75.3 N\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-16, Page No 241"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=14 #kg\n",
- "m2=7 #kg\n",
- "# as theta=45 degrees,\n",
- "sintheta=sqrt(2)**-1\n",
- "costheta=sqrt(2)**-1\n",
- "u_1=4**-1 #coefficient of friction between mass 1 and plane\n",
- "u_2=3*8**-1 #coefficient of friction between mass 2 and plane\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#The equations of motion for m1 are\n",
- "N1=m1*g*costheta #N\n",
- "F1=u_1*N1 #N\n",
- "#The equations of motion for m2 are\n",
- "N2=m2*g*costheta #N\n",
- "F2=u_2*N2 #N\n",
- "#Now to get T and a we solve using matrix method\n",
- "A=np.array([[-1,-m1],[1,-m2]])\n",
- "B=np.array([[-(m1*g*sintheta-F1)],[-(m2*g*sintheta-F2)]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The Value of T is',round(C[0]),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Value of T is 4.0 N\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-19, Page No 244"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=12 #oz\n",
- "k=2 #oz/in\n",
- "M=0.34 #kg\n",
- "K=22 #N/m\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Part(a)\n",
- "a=(k*W*g)/16\n",
- "b=W*16**-1\n",
- "f=((2*pi)**-1)*((a/b)**0.5) #Hz for simplicity the numerator and denominator have been computed seperately as a and b\n",
- "#Part(b)\n",
- "F=((2*pi)**-1)*((K/M)**0.5) #Hz\n",
- "\n",
- "#Result\n",
- "print'The frequency in part (a) is',round(f,2),\"Hz\",'and in part(b) is',round(F,2),\"Hz\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frequency in part (a) is 1.28 Hz and in part(b) is 1.28 Hz\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-20, Page No 244"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#As the entire question is theoritical\n",
- "#theta is directly computed \n",
- "theta=arccos(2*3**-1)*(180/pi) #degrees\n",
- "\n",
- "#result\n",
- "print'The value of theta is',round(theta,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of theta is 48.2 degrees\n"
- ]
- }
- ],
- "prompt_number": 43
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-28, Page No 254"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "G=6.658*(10**-8)**-1 #cm**3/g.s**2\n",
- "#Calculations\n",
- "G1=G*((3.281*10**2)/((2.205*32.2**-1)*10**4)) #ft**3/slug-s**2\n",
- "G2=G1 #ft**4/lb-s**4\n",
- "\n",
- "#Result\n",
- "print'The ans is',round(G2,2),\"ft**4/lb-s**4\"\n",
- "\n",
- "# The ans waries slightly due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ans is 319004859.68 ft**4/lb-s**4\n"
- ]
- }
- ],
- "prompt_number": 51
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-29, Page No 254"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#Modifying the value of C without vo**2 in it\n",
- "C=5000*5280\n",
- "G=3.43*10**-8 #Gravatational Constant\n",
- "M=4.09*10**23 #Mass of the Earth\n",
- "a=5.31*10**8\n",
- "#When the orbit is circular e=0\n",
- "vo1=(a)**0.5 #ft/s\n",
- "#When the orbit is parabolic e=1\n",
- "vo2=((C*a+G*M)/C)**0.5 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The value of vo1=',round(vo1),\"ft/s\",'is smaller than vo2=',round(vo2),\"ft/s, hence the Satellite will enter a hyperbolic path and never return to earth.\"\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of vo1= 23043.0 ft/s is smaller than vo2= 32594.0 ft/s, hence the Satellite will enter a hyperbolic path and never return to earth.\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-30, Page No 255"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=3940+500 #mi\n",
- "phi=0 #degrees\n",
- "vo=36000 #ft/s\n",
- "C=4440*5280*vo\n",
- "G=3.43*10**-8\n",
- "M=4.09*10**23 #kg\n",
- "\n",
- "#Calculations\n",
- "e=((C*vo)/(G*M))-1\n",
- "\n",
- "#Result\n",
- "print'The value of e=',round(e,2),\",hence the path is Hyperbolic\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of e= 1.17 ,hence the path is Hyperbolic\n"
- ]
- }
- ],
- "prompt_number": 54
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-31, Page No 255"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "a=92.9*10**6 #mi\n",
- "G=3.43*10**-8\n",
- "T=365*24*3600 #s\n",
- "c=5280\n",
- "\n",
- "#Calculations\n",
- "M=(4*pi**2*a**3*c**3)/(G*T**2) #slugs\n",
- "\n",
- "#Result\n",
- "print'The mass of the sun is',round(M,1),\"slugs\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The mass of the sun is 1.36584467048e+29 slugs\n"
- ]
- }
- ],
- "prompt_number": 56
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter13_4.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter13_4.ipynb deleted file mode 100755 index a77abe8e..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter13_4.ipynb +++ /dev/null @@ -1,899 +0,0 @@ -{
- "metadata": {
- "name": "chapter13.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 13: Dynamics of a Particle"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-1, Page No 230"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=2 #lb\n",
- "F=1.5 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Angles are with respect to the plane are,\n",
- "# theta1=10 degrees & theta2=30 degrees\n",
- "sintheta1=0.17\n",
- "costheta1=0.99\n",
- "sintheta2=0.5\n",
- "costheta2=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Now here the forces are considered as parallel and perpendicular to the plane \n",
- "#Applying Newtond Principle\n",
- "ax=(g/2)*(F*costheta1-(W*sintheta2)) #ft/s**2\n",
- "N1=(2*costheta2-(F*sintheta1)) #lb\n",
- "\n",
- "#result\n",
- "print'The force on the particle is',round(N1,2),\"lb\"\n",
- "print'The acceleration is',round(ax,2),\"ft/s**2\"\n",
- "\n",
- "# The answer may wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force on the particle is 1.48 lb\n",
- "The acceleration is 7.81 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-2, Page No 231"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=5 #kg\n",
- "s=12 #m\n",
- "v=4 #m/s\n",
- "vo=0 #m/s\n",
- "g=9.8 #m/s**2\n",
- "mu=0.25\n",
- "\n",
- "#Calculations\n",
- "#Using the kinematic equations of motion\n",
- "a=(v**2-vo**2)*(2*s)**-1 #m/s**2\n",
- "#Using Newtons Principle\n",
- "N1=g*m #N\n",
- "P=m*a+mu*N1 #N\n",
- "\n",
- "#Result\n",
- "print'The value of P is',round(P,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P is 15.6 N\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-3, Page No 232"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=2 #kg\n",
- "vo=0 #m/s\n",
- "v=3 #m/s\n",
- "s=0.8 #m\n",
- "# as theta=20 degrees,\n",
- "sintheta=0.34\n",
- "costheta=0.94\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "N=m*g*costheta #N\n",
- "a=(vo**2-v**2)*(2*s)**-1 #m/s**2\n",
- "u=-((2*a)+(m*g*sintheta))/N \n",
- "#Solving for return speed\n",
- "#Symbol convention is different from textbook\n",
- "a_ret=((m*g*sintheta)-(u*N))/2 #m/s**2\n",
- "vf=sqrt((2*a_ret*s)) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed is',round(vf,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed is 1.3 m/s\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-4, Page No 232"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=1800 #lb\n",
- "r=2000 #ft\n",
- "v=58.7 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "F=(W*v*v)/(g*r) #lb\n",
- "\n",
- "#Result\n",
- "print'The frictional force to be exerted is',round(F,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frictional force to be exerted is 96.3 lb\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-7, Page No 234"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "# Initilization of variables\n",
- "W=10 #lb\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=sqrt(3)*2**-1\n",
- "l=2 #ft\n",
- "w=10 #rev/min\n",
- "g=32.2 # ft/s**2\n",
- "\n",
- "# Calculations\n",
- "r=l*costheta # ft\n",
- "a_n=r*(((w*2*pi)/60)**2) #ft/s**2\n",
- "#Applying Newtons Principle\n",
- "#Solving by matrix method\n",
- "A=np.array([[costheta,-sintheta],[sintheta,costheta]]) \n",
- "B=np.array([[(W*a_n)/g],[W]]) \n",
- "C=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The value of T is',round(C[0],2),\"lb\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of T is 5.51 lb\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-8, Page No 235"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=4 #lb\n",
- "v=6 #ft/s\n",
- "r=2 #ft\n",
- "# as theta1=40 degrees & theta2=20 degrees\n",
- "sintheta1=0.64\n",
- "costheta1=0.77\n",
- "sintheta2=0.34\n",
- "costheta2=0.94\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "a_n=v**2/r #ft/s**2\n",
- "#Applying Newtons Principle\n",
- "Fi=(m*a_n)/g #lb\n",
- "#Solving by matrix method\n",
- "A=np.array([[costheta1,costheta2],[sintheta1,-sintheta2]])\n",
- "B=np.array([[m],[Fi]]) \n",
- "C=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The values are: T=',round(C[0],2),\"lb\",'and C=',round(C[1],2),\"lb\"\n",
- "\n",
- "# The ans for C waries due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are: T= 4.01 lb and C= 0.97 lb\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-10, Page No 237"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=2 #kg\n",
- "# as theta=20 degrees,\n",
- "sintheta=0.34\n",
- "m2=4 #kg\n",
- "t=4 #s\n",
- "g=9.8 #m/s**2\n",
- "vo=0 #m/s\n",
- "\n",
- "#Calculations\n",
- "#Applying Newtons Principle\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,-2],[1,4]])\n",
- "B=np.array([[m1*g*sintheta],[m2*g]])\n",
- "C=np.linalg.solve(A,B)\n",
- "a=C[1] #m/s**2\n",
- "v=vo+a*t #m/s\n",
- "\n",
- "#Result\n",
- "print'The velocity of 4kg mass is',round(v,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity of 4kg mass is 21.7 m/s\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-11, Page No 237"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m_A=20 #lb\n",
- "m_B=60 #lb\n",
- "u=0.3 #coefficient of friction\n",
- "t=4 #s\n",
- "# as theta1=30 degrees & theta2=60 degrees,\n",
- "sintheta1=2**-1\n",
- "costheta1=sqrt(3)*2**-1\n",
- "sintheta2=sqrt(3)*2**-1\n",
- "costheta2=2**-1\n",
- "g=32.2 #ft/s^2\n",
- "vo=0 #ft/s\n",
- "\n",
- "#Calculations\n",
- "N1=m_A*costheta1 #lb\n",
- "N2=m_B*costheta2 #lb\n",
- "#Solving for T and a using matrix method\n",
- "A=np.array([[1,-m_A/g],[-1,-m_B/g]])\n",
- "B=np.array([[(m_A*sintheta1+u*N1)],[(-m_B*sintheta2+u*N2)]]) \n",
- "C=np.linalg.solve(A,B)\n",
- "a=C[1] #ft/s**2\n",
- "v=vo+a*t #ft/s\n",
- "\n",
- "#Result\n",
- "print'The velocity is',round(v,1),\"ft/s\"\n",
- "\n",
- "# The ans in the textbook is incorrect."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity is 44.7 ft/s\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-12, Page No 238"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m_A=40 #kg\n",
- "m_B=15 #kg\n",
- "F=500 #N\n",
- "g=9.8 #m/s**2\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "m=m_A+m_B #kg\n",
- "a=(F-m*g*sintheta)/(m) #m/s**2\n",
- "#Summing forces parallel and perpendicular to the plane\n",
- "#Simplfying equation (1) and (2)\n",
- "Nb=m_B*g+(m_B*a*sintheta) #N\n",
- "#Substituting this in eq(1)\n",
- "u=-(m_B*g*costheta-(Nb*costheta))/(Nb*sintheta)\n",
- "\n",
- "#Result\n",
- "print'The value of u is',round(u,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of u is 0.31\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-13, Page No 239"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "P=70 #N\n",
- "m_A=16 #kg\n",
- "u_AH=0.25 #coefficient of friction between Block A and Horizontal Plane\n",
- "m_B=4 #kg\n",
- "u_BH=0.5 #coefficient of friction between Block B and Horizontal Plane\n",
- "# as theta=10 degrees,\n",
- "sintheta=0.17\n",
- "costheta=0.98\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying sum of forces to both the FBD's\n",
- "#Solving by matrix method \n",
- "A=np.array([[-costheta,-u_AH,-m_A,0],[-sintheta,1,0,0],[costheta,0,-m_B,-u_BH],[sintheta,0,0,1]]) \n",
- "B=np.array([[-P],[m_A*g],[0],[m_B*g]])\n",
- "C=np.linalg.solve(A,B) \n",
- "\n",
- "#Result\n",
- "print'The Value of T is',round(C[0],1),\"N\"\n",
- "\n",
- "# The ans waries due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Value of T is 20.7 N\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-14, Page No 239"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=10 degrees\n",
- "sintheta=0.1736\n",
- "costheta=0.9848\n",
- "v=10 #ft/s\n",
- "v0=0 #ft/s\n",
- "u=3**-1 #coefficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Equations of motion for box are\n",
- "#Simplfying the equations by sybstitution\n",
- "a=((u*costheta)-(sintheta))*g #ft/s**2\n",
- "#Time calculations\n",
- "t=(v-v0)/a #s\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,2),\"ft/s**2\",'and the time required is',round(t),\"seconds\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is 4.98 ft/s**2 and the time required is 2.0 seconds\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-15, Page No 240"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Simplfying the equations we can solve for T2 and aA first to obtain the solution\n",
- "#Solving by matrix method\n",
- "A=np.array([[-1.5,-4],[-3.5,24]])\n",
- "B=np.array([[-4*g],[-24*g]])\n",
- "C=np.linalg.solve(A,B) \n",
- "T2=C[0] #N\n",
- "T1=T2/2 #N\n",
- "T3=T2/2 #N\n",
- "#Acceleration calculations\n",
- "a1=1*g-T1 #m/s**2\n",
- "a2=(2*g-T1)/2 #m/s**2\n",
- "a3=(3*g-T3)/3 #m/s**2\n",
- "a4=(4*g-T3)/4 #m/s**2\n",
- "#Tension in fixed cord\n",
- "T_f=2*T2 #N\n",
- "\n",
- "#Result\n",
- "print'The acceleration values are: a1=',round(a1),\"m/s**2 (up)\",',',round(a2,1),\"m/s**2 (down)\",',',round(a3,2),\"m/s**2 (down)\",',',round(a4,1),\"m/s**2 (down) respectively.\"\n",
- "print'The tension in the fixed cord is',round(T_f,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration values are: a1= -9.0 m/s**2 (up) , 0.4 m/s**2 (down) , 3.53 m/s**2 (down) , 5.1 m/s**2 (down) respectively.\n",
- "The tension in the fixed cord is 75.3 N\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-16, Page No 241"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=14 #kg\n",
- "m2=7 #kg\n",
- "# as theta=45 degrees,\n",
- "sintheta=sqrt(2)**-1\n",
- "costheta=sqrt(2)**-1\n",
- "u_1=4**-1 #coefficient of friction between mass 1 and plane\n",
- "u_2=3*8**-1 #coefficient of friction between mass 2 and plane\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#The equations of motion for m1 are\n",
- "N1=m1*g*costheta #N\n",
- "F1=u_1*N1 #N\n",
- "#The equations of motion for m2 are\n",
- "N2=m2*g*costheta #N\n",
- "F2=u_2*N2 #N\n",
- "#Now to get T and a we solve using matrix method\n",
- "A=np.array([[-1,-m1],[1,-m2]])\n",
- "B=np.array([[-(m1*g*sintheta-F1)],[-(m2*g*sintheta-F2)]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The Value of T is',round(C[0]),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Value of T is 4.0 N\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-19, Page No 244"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=12 #oz\n",
- "k=2 #oz/in\n",
- "M=0.34 #kg\n",
- "K=22 #N/m\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Part(a)\n",
- "a=(k*W*g)/16\n",
- "b=W*16**-1\n",
- "f=((2*pi)**-1)*((a/b)**0.5) #Hz for simplicity the numerator and denominator have been computed seperately as a and b\n",
- "#Part(b)\n",
- "F=((2*pi)**-1)*((K/M)**0.5) #Hz\n",
- "\n",
- "#Result\n",
- "print'The frequency in part (a) is',round(f,2),\"Hz\",'and in part(b) is',round(F,2),\"Hz\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frequency in part (a) is 1.28 Hz and in part(b) is 1.28 Hz\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-20, Page No 244"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#As the entire question is theoritical\n",
- "#theta is directly computed \n",
- "theta=arccos(2*3**-1)*(180/pi) #degrees\n",
- "\n",
- "#result\n",
- "print'The value of theta is',round(theta,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of theta is 48.2 degrees\n"
- ]
- }
- ],
- "prompt_number": 43
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-28, Page No 254"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "G=6.658*(10**-8)**-1 #cm**3/g.s**2\n",
- "#Calculations\n",
- "G1=G*((3.281*10**2)/((2.205*32.2**-1)*10**4)) #ft**3/slug-s**2\n",
- "G2=G1 #ft**4/lb-s**4\n",
- "\n",
- "#Result\n",
- "print'The ans is',round(G2,2),\"ft**4/lb-s**4\"\n",
- "\n",
- "# The ans waries slightly due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ans is 319004859.68 ft**4/lb-s**4\n"
- ]
- }
- ],
- "prompt_number": 51
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-29, Page No 254"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#Modifying the value of C without vo**2 in it\n",
- "C=5000*5280\n",
- "G=3.43*10**-8 #Gravatational Constant\n",
- "M=4.09*10**23 #Mass of the Earth\n",
- "a=5.31*10**8\n",
- "#When the orbit is circular e=0\n",
- "vo1=(a)**0.5 #ft/s\n",
- "#When the orbit is parabolic e=1\n",
- "vo2=((C*a+G*M)/C)**0.5 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The value of vo1=',round(vo1),\"ft/s\",'is smaller than vo2=',round(vo2),\"ft/s, hence the Satellite will enter a hyperbolic path and never return to earth.\"\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of vo1= 23043.0 ft/s is smaller than vo2= 32594.0 ft/s, hence the Satellite will enter a hyperbolic path and never return to earth.\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-30, Page No 255"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=3940+500 #mi\n",
- "phi=0 #degrees\n",
- "vo=36000 #ft/s\n",
- "C=4440*5280*vo\n",
- "G=3.43*10**-8\n",
- "M=4.09*10**23 #kg\n",
- "\n",
- "#Calculations\n",
- "e=((C*vo)/(G*M))-1\n",
- "\n",
- "#Result\n",
- "print'The value of e=',round(e,2),\",hence the path is Hyperbolic\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of e= 1.17 ,hence the path is Hyperbolic\n"
- ]
- }
- ],
- "prompt_number": 54
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13-31, Page No 255"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "a=92.9*10**6 #mi\n",
- "G=3.43*10**-8\n",
- "T=365*24*3600 #s\n",
- "c=5280\n",
- "\n",
- "#Calculations\n",
- "M=(4*pi**2*a**3*c**3)/(G*T**2) #slugs\n",
- "\n",
- "#Result\n",
- "print'The mass of the sun is',round(M,1),\"slugs\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The mass of the sun is 1.36584467048e+29 slugs\n"
- ]
- }
- ],
- "prompt_number": 56
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter14.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter14.ipynb deleted file mode 100755 index c04236af..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter14.ipynb +++ /dev/null @@ -1,772 +0,0 @@ -{
- "metadata": {
- "name": "chapter14.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 14: Kinematics of a Rigid Body in Plane Motion"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-2, Page no 265"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=500 #mm\n",
- "wo=0 #rpm\n",
- "w=300 #rpm\n",
- "t=20 #s\n",
- "t1=2 #s\n",
- "\n",
- "#Calculations\n",
- "alpha=(2*pi*(60**-1)*(w-wo))/t #rad/s**2\n",
- "w1=wo+alpha*t1 #rad/s\n",
- "v=(d*(2*1000)**-1)*w1 #m/s\n",
- "a_n=(d*(2*1000)**-1)*w1**2 #m/s**2\n",
- "a_t=(d*(2*1000)**-1)*alpha #m/s**2\n",
- "a=(a_n**2+a_t**2)**0.5 #m/s**2\n",
- "theta=arccos(a_n/a)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The computed values are:'\n",
- "print'alpha=',round(alpha,2),\"rad/s**2\"\n",
- "print'w1=',round(w1,2),\"rad/s\"\n",
- "print'v=',round(v,3),\"m/s\"\n",
- "print'a=',round(a,2),\"m/s**2\"\n",
- "print'theta=',round(theta,1),\"degrees\"\n",
- "\n",
- "# The answers may wary in decimal points."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The computed values are:\n",
- "alpha= 1.57 rad/s**2\n",
- "w1= 3.14 rad/s\n",
- "v= 0.785 m/s\n",
- "a= 2.5 m/s**2\n",
- "theta= 9.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-3, Page no 266"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "s_BC=2 #m\n",
- "s_C=2.5 #m\n",
- "\n",
- "#Calculations\n",
- "s_B=(s_BC**2+s_C**2)**0.5 #m\n",
- "theta=arctan(s_BC*s_C**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The absolute displacement is',round(s_B,1),\"m\",'and the angle made by the vector is',round(theta,1),\"degrees.\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The absolute displacement is 3.2 m and the angle made by the vector is 38.7 degrees.\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-4, Page no 266"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "V_A=20 #mi/h\n",
- "V_B=70 #mi/h\n",
- "# as theta1=60 degrees,\n",
- "sintheta1=sqrt(3)*2**-1\n",
- "costheta1=2**-1\n",
- "# also phi=45 degrees, thus\n",
- "sinphi=sqrt(2)**-1\n",
- "cosphi=sqrt(2)**-1\n",
- "\n",
- "#Result\n",
- "#Vector's in matrix form\n",
- "v_A=np.array([-V_A*cosphi,V_A*sinphi]) #mi/h\n",
- "v_B=np.array([V_B*costheta1,V_B*sintheta1]) #mi/h\n",
- "a=v_A[0]+v_B[0] #mi/h\n",
- "b=v_A[1]+v_B[1] #mi/h\n",
- "v_ab=(a**2+b**2)**0.5 #mi/h\n",
- "theta=arctan(b/a)*(180/pi) #degrees\n",
- "#The relative velocity v_ba is just different in sign while the magnitude stays the same\n",
- "\n",
- "#Resul\n",
- "print'The relative velocity is',round(v_ab,1),\"mi/h\",'making an angle',round(theta,1),\"degrees\"\n",
- "\n",
- "# The ans may wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The relative velocity is 77.6 mi/h making an angle 74.4 degrees\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-9, Page no 271"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=2.5 #m\n",
- "v_A=4 #m/s\n",
- "a_A=5 #m/s**2\n",
- "theta=30 #degrees\n",
- "\n",
- "#Calculations\n",
- "#Vector triangle yields v_a.b=2.93 m/s\n",
- "v_ab=2.93 #m/s\n",
- "w=v_ab*l**-1 #rad/s (clockwise)\n",
- "#Ploygon yields alpha_a/b=2.75 m/s**2\n",
- "alpha_ab=2.75 #m/s**2\n",
- "alpha=alpha_ab*l**-1 #rad/s**2 (counterclockwise)\n",
- "\n",
- "#Result\n",
- "print'The value of angular velocity is',round(w,2),\"rad/s\"\n",
- "print'The value of angular acceleration is',round(alpha,1),\"rad/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of angular velocity is 1.17 rad/s\n",
- "The value of angular acceleration is 1.1 rad/s**2\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-10, Page no 272"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=(2*pi*120)/60 #rad/s\n",
- "l=24 #in\n",
- "l_c=4 #in\n",
- "# as th=30 degrees,\n",
- "sinth=2**-1\n",
- "\n",
- "#Calculations\n",
- "v=(l_c*12**-1)*w #ft/s\n",
- "betaa=arcsin((l_c*sinth)/l)*(180/pi) #degrees\n",
- "# betaa yeilds 4.8 degrees, thus value of cosbetaa is,\n",
- "cosbetaa=0.996\n",
- "theta=60-betaa #degrees\n",
- "# here theta yeilds 55.2 degrees, thus value of costheta is,\n",
- "costheta=0.57\n",
- "#Component of velocity along connecting rod is \n",
- "v1=v*costheta #ft/s\n",
- "v_p=v1/cosbetaa #ft/s\n",
- "\n",
- "#Result\n",
- "print'The absoulte velocity is',round(v_p,2),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The absoulte velocity is 2.4 ft/s\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-13, Page no 274"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_pc=3.68 #ft/s\n",
- "l=2 #ft\n",
- "\n",
- "#Calculations\n",
- "w=v_pc/l #rad/s counterclockwise\n",
- "\n",
- "#Result\n",
- "print'The angular velocity is',round(w,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity is 1.84 rad/s\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-14, Page no 274"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#This problem is a combination of numerical and graphical solution\n",
- "#The program only deals with the numerical solution parts the rest can be verified by graphical solution\n",
- "#Initilization of variables\n",
- "r=4*12**-1 #ft\n",
- "w=4*pi #rad/s\n",
- "l=2 #ft\n",
- "w2=1.84 #rad/s\n",
- "\n",
- "#Calculations\n",
- "ac_n=r*w**2 #ft/s**2\n",
- "a_pc_n=l*w2**2 #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of ac_n is',round(ac_n,1),\"ft/s**2\"\n",
- "print'The value of a_pc_n is',round(a_pc_n,2),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of ac_n is 52.6 ft/s**2\n",
- "The value of a_pc_n is 6.77 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-15, Page no 275"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w_bc=10 #rad/s\n",
- "AB=250 #mm\n",
- "BC=150 #mm\n",
- "AC=179 #mm\n",
- "AD=200 #mm\n",
- "# as theta1=45 degrees,\n",
- "sintheta1=(2**0.5)**-1\n",
- "costheta1=(2**0.5)**-1\n",
- "\n",
- "#Calculations\n",
- "v_c=(BC*1000**-1)*w_bc #m/s\n",
- "AC=((AB**2+BC**2)-(2*AB*BC*costheta1))**0.5 #m\n",
- "betaa=arcsin((BC*sintheta1)/AC)*(180/pi) #degrees\n",
- "gammaa=arcsin((AB*sintheta1)/AC)*(180/pi)#degrees answer in the textbook is incorrect\n",
- "ang=60-betaa #degrees\n",
- "# ang yeilds 23.7 degrees, thus\n",
- "sinang=0.40056\n",
- "cosang=0.916\n",
- "CD=sqrt(AD**2+AC**2-(2*AD*AC*cosang)) #mm\n",
- "D=arcsin((AC*sinang)/CD)*(180/pi) #degrees\n",
- "# D yeilds 63.2 degrees,thus\n",
- "sinD=0.8925\n",
- "theta=arcsin((AD*sinD)/AC)*(180/pi) #degrees\n",
- "n=360-(theta+gammaa+90) #degrees\n",
- "# n yeilds 101.8 degrees, thus \n",
- "cosn=-0.2045\n",
- "v_cd=v_c*cosn #m/s\n",
- "delta=180-(90+D) #degrees\n",
- "# Delts yeilds 26.8 degrees, thus\n",
- "cosdelta=0.8925\n",
- "v_D=v_cd/cosdelta #m/s\n",
- "w_AD=v_D/(AD*1000**-1) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular Velocity of AD is',round(w_AD,2),\"rad/s clockwise.\" #Negative sign indicates clockwise orientation \n",
- "#Answer in the textbook is incorrect\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular Velocity of AD is -1.72 rad/s clockwise.\n"
- ]
- }
- ],
- "prompt_number": 47
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-18, Page No 277"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta1=73.9 degrees,theta2=60 degrees and theta3=46.1 degrees\n",
- "sintheta1=0.96\n",
- "sintheta2=sqrt(3)*2**-1\n",
- "sintheta3=0.72\n",
- "V=900 #mm/s\n",
- "\n",
- "#Calculations\n",
- "BC=((350*350)+(86.6*86.6))**0.5 #mm\n",
- "CD=400 #mm\n",
- "v_cb=(V*sintheta2)/(sintheta1) #mm/s\n",
- "v_c=((V*sintheta3))/(sintheta1) #mm/s\n",
- "w_dc=v_c/CD #rad/s\n",
- "w_bc=v_cb/BC #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular velocities are: w_dc=',round(w_dc,2),\"rad/s\",'and w_bc=',round(w_bc,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocities are: w_dc= 1.69 rad/s and w_bc= 2.25 rad/s\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-19, Page No 278"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Calculations\n",
- "#After equating the i and j terms we obtain simplified equations\n",
- "#Solving by matrix method\n",
- "A=np.array([[346,86.7],[200,-350]])\n",
- "B=np.array([[-3700],[-1790]]) \n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The angular accelerations are alpha_DC=',round(C[0],3),\"rad/s**2\",'and alpha_BC=',round(C[1],2),\"rad/s**2\" \n",
- "#The signs only indicate that the originally assumed orientations are incorrect and are opposite to those assumed\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular accelerations are alpha_DC= -10.475 rad/s**2 and alpha_BC= -0.87 rad/s**2\n"
- ]
- }
- ],
- "prompt_number": 80
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-20, Page No 279"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=3 #m\n",
- "w=8 #rad/s (clockwise)\n",
- "alpha=4 #rad/s**2 (counterclockwise)\n",
- "r=d*2**-1 #m\n",
- "\n",
- "#Calculations\n",
- "vo=r*w #m/s\n",
- "ao=r*alpha #m/s**2\n",
- "#Here OB is r\n",
- "OB=r #m\n",
- "v_bo=OB*w #m/s\n",
- "v_B=v_bo+vo #m/s\n",
- "#Also\n",
- "a_bo=r*alpha #m/s**2 (directed left)\n",
- "a_bo_n=r*w**2 #m/s**2\n",
- "a_h=ao+a_bo #m/s**2\n",
- "a_v=a_bo_n #m/s**2\n",
- "a_B=((a_h**2)+(a_v**2))**0.5 #m/s**2\n",
- "phi=arctan(a_h/a_v)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The linear velocity at B is',round(v_B),\"m/s\",'and the acceleration is',round(a_B,1),\"m/s**2\",'making an angle of',round(phi,2),\"degrees with horizontal\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The linear velocity at B is 24.0 m/s and the acceleration is 96.7 m/s**2 making an angle of 7.13 degrees with horizontal\n"
- ]
- }
- ],
- "prompt_number": 96
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-21, Page No 281"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "OA=0.6 #m\n",
- "w=8 #rad/s\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "v_O=12 #m/s\n",
- "alpha=4 #rad/s**2\n",
- "a_O=6 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Velocity Calculations\n",
- "v_AO=OA*w #m/s\n",
- "v_Ah=v_AO*sintheta+v_O #m/s horizontal component\n",
- "v_Av=v_AO*costheta #m/s\n",
- "v_A=((v_Ah**2)+(v_Av**2))**0.5 #m/s\n",
- "phi=arctan(v_Av/v_Ah)*(180/pi) #degrees\n",
- "#Acceleration Calculations\n",
- "a_AOt=OA*alpha #m/s**2\n",
- "a_AOn=OA*w**2 #m/s**2\n",
- "a_Ah=-a_O-a_AOn*costheta-a_AOt*sintheta #m/s**2\n",
- "a_Av=-a_AOn*sintheta+a_AOt*costheta #m/s**2\n",
- "a_A=((a_Ah**2)+(a_Av**2))**0.5 #m/s**2\n",
- "phi2=arctan(a_Av*a_Ah**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The velocity is',round(v_A),\"m/s\",'making an angle of',round(phi,1),\"degrees with horizontal.\"\n",
- "print'The acceleration is',round(a_A),\"m/s**2\",'making an angle of',round(phi2,1),\"degrees with horizontal\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity is 15.0 m/s making an angle of 16.1 degrees with horizontal.\n",
- "The acceleration is 44.0 m/s**2 making an angle of 22.9 degrees with horizontal\n"
- ]
- }
- ],
- "prompt_number": 100
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-22, Page No 282"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "AL=5 #ft\n",
- "d=10 #ft displacement\n",
- "\n",
- "#Calculations\n",
- "theta=d/AL #radians\n",
- "s_o=3*theta#ft\n",
- "\n",
- "#Result\n",
- "print'The displacement So is',round(s_o),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The displacement So is 6.0 ft\n"
- ]
- }
- ],
- "prompt_number": 101
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-23, Page No 283"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#Speed and acceleration at the center\n",
- "v=12 #in/s\n",
- "a=18 #in/s**2\n",
- "\n",
- "#Calculations\n",
- "v_D=((a+v*0.5)*a**-1)*v #in/s\n",
- "#Speed at point F\n",
- "v_F=((v/2)*v**-1)*v_D #in/s\n",
- "#Acceleration at D\n",
- "a_D=(24/a)*a #in/s**2\n",
- "#Acceleration at F\n",
- "a_F=((v/2)*v**-1)*24 #in/s**2\n",
- "\n",
- "#Result\n",
- "print'The velocity and acceleration of weight A are',round(v_F),\"in/s\",'and',round(a_F),\"in/s**2 respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity and acceleration of weight A are 8.0 in/s and 12.0 in/s**2 respectively.\n"
- ]
- }
- ],
- "prompt_number": 103
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-24, Page No 283"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Calculations\n",
- "#Speed and acceleration of D\n",
- "sD=((18-6)*18**-1)*12 #in/s\n",
- "aD=(12*18**-1)*18 #in/s**2\n",
- "#Speed and acceleration of F\n",
- "sF=(6*12**-1)*8 #in/s\n",
- "aF=(6*12**-1)*12 #in/s^2\n",
- "\n",
- "#Result\n",
- "print'The velocity and acceleration of weight A are',round(sF),\"in/s\",'and',round(aF),\"in/s**2 respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity and acceleration of weight A are 4.0 in/s and 6.0 in/s**2 respectively.\n"
- ]
- }
- ],
- "prompt_number": 104
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-26, Page No 284"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_BG=300 #mm/s\n",
- "v_G=300 #mm/s\n",
- "a_BGt=500 #mm/s**2\n",
- "a_BGn=3600#mm/s**2\n",
- "a_Gh=500 #mm/s**2\n",
- "a_Bv=1800 #mm/s**2\n",
- "\n",
- "#Calculations\n",
- "w=((75-25)/25)*6 #rad/s\n",
- "alpha=((75-25)/25)*10 #rad/s**2\n",
- "v_B=(v_BG**2+v_G**2)**0.5 #mm/s\n",
- "a_v=a_Bv-a_BGt #mm/s**2\n",
- "a_h=a_BGn-a_Gh #mm/s**2\n",
- "a_B=(a_v**2+a_h**2)**0.5 #mm/s**2\n",
- "\n",
- "#Result \n",
- "print'The velocity and acceleration of point B are',round(v_B),\"mm/s\",'and',round(a_B),\"mm/s**2 respectively.\"\n",
- "\n",
- "# The ans for a_B is incorrectin textbook."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity and acceleration of point B are 424.0 mm/s and 3362.0 mm/s**2 respectively.\n"
- ]
- }
- ],
- "prompt_number": 106
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter14_1.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter14_1.ipynb deleted file mode 100755 index c04236af..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter14_1.ipynb +++ /dev/null @@ -1,772 +0,0 @@ -{
- "metadata": {
- "name": "chapter14.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 14: Kinematics of a Rigid Body in Plane Motion"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-2, Page no 265"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=500 #mm\n",
- "wo=0 #rpm\n",
- "w=300 #rpm\n",
- "t=20 #s\n",
- "t1=2 #s\n",
- "\n",
- "#Calculations\n",
- "alpha=(2*pi*(60**-1)*(w-wo))/t #rad/s**2\n",
- "w1=wo+alpha*t1 #rad/s\n",
- "v=(d*(2*1000)**-1)*w1 #m/s\n",
- "a_n=(d*(2*1000)**-1)*w1**2 #m/s**2\n",
- "a_t=(d*(2*1000)**-1)*alpha #m/s**2\n",
- "a=(a_n**2+a_t**2)**0.5 #m/s**2\n",
- "theta=arccos(a_n/a)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The computed values are:'\n",
- "print'alpha=',round(alpha,2),\"rad/s**2\"\n",
- "print'w1=',round(w1,2),\"rad/s\"\n",
- "print'v=',round(v,3),\"m/s\"\n",
- "print'a=',round(a,2),\"m/s**2\"\n",
- "print'theta=',round(theta,1),\"degrees\"\n",
- "\n",
- "# The answers may wary in decimal points."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The computed values are:\n",
- "alpha= 1.57 rad/s**2\n",
- "w1= 3.14 rad/s\n",
- "v= 0.785 m/s\n",
- "a= 2.5 m/s**2\n",
- "theta= 9.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-3, Page no 266"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "s_BC=2 #m\n",
- "s_C=2.5 #m\n",
- "\n",
- "#Calculations\n",
- "s_B=(s_BC**2+s_C**2)**0.5 #m\n",
- "theta=arctan(s_BC*s_C**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The absolute displacement is',round(s_B,1),\"m\",'and the angle made by the vector is',round(theta,1),\"degrees.\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The absolute displacement is 3.2 m and the angle made by the vector is 38.7 degrees.\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-4, Page no 266"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "V_A=20 #mi/h\n",
- "V_B=70 #mi/h\n",
- "# as theta1=60 degrees,\n",
- "sintheta1=sqrt(3)*2**-1\n",
- "costheta1=2**-1\n",
- "# also phi=45 degrees, thus\n",
- "sinphi=sqrt(2)**-1\n",
- "cosphi=sqrt(2)**-1\n",
- "\n",
- "#Result\n",
- "#Vector's in matrix form\n",
- "v_A=np.array([-V_A*cosphi,V_A*sinphi]) #mi/h\n",
- "v_B=np.array([V_B*costheta1,V_B*sintheta1]) #mi/h\n",
- "a=v_A[0]+v_B[0] #mi/h\n",
- "b=v_A[1]+v_B[1] #mi/h\n",
- "v_ab=(a**2+b**2)**0.5 #mi/h\n",
- "theta=arctan(b/a)*(180/pi) #degrees\n",
- "#The relative velocity v_ba is just different in sign while the magnitude stays the same\n",
- "\n",
- "#Resul\n",
- "print'The relative velocity is',round(v_ab,1),\"mi/h\",'making an angle',round(theta,1),\"degrees\"\n",
- "\n",
- "# The ans may wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The relative velocity is 77.6 mi/h making an angle 74.4 degrees\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-9, Page no 271"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=2.5 #m\n",
- "v_A=4 #m/s\n",
- "a_A=5 #m/s**2\n",
- "theta=30 #degrees\n",
- "\n",
- "#Calculations\n",
- "#Vector triangle yields v_a.b=2.93 m/s\n",
- "v_ab=2.93 #m/s\n",
- "w=v_ab*l**-1 #rad/s (clockwise)\n",
- "#Ploygon yields alpha_a/b=2.75 m/s**2\n",
- "alpha_ab=2.75 #m/s**2\n",
- "alpha=alpha_ab*l**-1 #rad/s**2 (counterclockwise)\n",
- "\n",
- "#Result\n",
- "print'The value of angular velocity is',round(w,2),\"rad/s\"\n",
- "print'The value of angular acceleration is',round(alpha,1),\"rad/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of angular velocity is 1.17 rad/s\n",
- "The value of angular acceleration is 1.1 rad/s**2\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-10, Page no 272"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=(2*pi*120)/60 #rad/s\n",
- "l=24 #in\n",
- "l_c=4 #in\n",
- "# as th=30 degrees,\n",
- "sinth=2**-1\n",
- "\n",
- "#Calculations\n",
- "v=(l_c*12**-1)*w #ft/s\n",
- "betaa=arcsin((l_c*sinth)/l)*(180/pi) #degrees\n",
- "# betaa yeilds 4.8 degrees, thus value of cosbetaa is,\n",
- "cosbetaa=0.996\n",
- "theta=60-betaa #degrees\n",
- "# here theta yeilds 55.2 degrees, thus value of costheta is,\n",
- "costheta=0.57\n",
- "#Component of velocity along connecting rod is \n",
- "v1=v*costheta #ft/s\n",
- "v_p=v1/cosbetaa #ft/s\n",
- "\n",
- "#Result\n",
- "print'The absoulte velocity is',round(v_p,2),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The absoulte velocity is 2.4 ft/s\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-13, Page no 274"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_pc=3.68 #ft/s\n",
- "l=2 #ft\n",
- "\n",
- "#Calculations\n",
- "w=v_pc/l #rad/s counterclockwise\n",
- "\n",
- "#Result\n",
- "print'The angular velocity is',round(w,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity is 1.84 rad/s\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-14, Page no 274"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#This problem is a combination of numerical and graphical solution\n",
- "#The program only deals with the numerical solution parts the rest can be verified by graphical solution\n",
- "#Initilization of variables\n",
- "r=4*12**-1 #ft\n",
- "w=4*pi #rad/s\n",
- "l=2 #ft\n",
- "w2=1.84 #rad/s\n",
- "\n",
- "#Calculations\n",
- "ac_n=r*w**2 #ft/s**2\n",
- "a_pc_n=l*w2**2 #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of ac_n is',round(ac_n,1),\"ft/s**2\"\n",
- "print'The value of a_pc_n is',round(a_pc_n,2),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of ac_n is 52.6 ft/s**2\n",
- "The value of a_pc_n is 6.77 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-15, Page no 275"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w_bc=10 #rad/s\n",
- "AB=250 #mm\n",
- "BC=150 #mm\n",
- "AC=179 #mm\n",
- "AD=200 #mm\n",
- "# as theta1=45 degrees,\n",
- "sintheta1=(2**0.5)**-1\n",
- "costheta1=(2**0.5)**-1\n",
- "\n",
- "#Calculations\n",
- "v_c=(BC*1000**-1)*w_bc #m/s\n",
- "AC=((AB**2+BC**2)-(2*AB*BC*costheta1))**0.5 #m\n",
- "betaa=arcsin((BC*sintheta1)/AC)*(180/pi) #degrees\n",
- "gammaa=arcsin((AB*sintheta1)/AC)*(180/pi)#degrees answer in the textbook is incorrect\n",
- "ang=60-betaa #degrees\n",
- "# ang yeilds 23.7 degrees, thus\n",
- "sinang=0.40056\n",
- "cosang=0.916\n",
- "CD=sqrt(AD**2+AC**2-(2*AD*AC*cosang)) #mm\n",
- "D=arcsin((AC*sinang)/CD)*(180/pi) #degrees\n",
- "# D yeilds 63.2 degrees,thus\n",
- "sinD=0.8925\n",
- "theta=arcsin((AD*sinD)/AC)*(180/pi) #degrees\n",
- "n=360-(theta+gammaa+90) #degrees\n",
- "# n yeilds 101.8 degrees, thus \n",
- "cosn=-0.2045\n",
- "v_cd=v_c*cosn #m/s\n",
- "delta=180-(90+D) #degrees\n",
- "# Delts yeilds 26.8 degrees, thus\n",
- "cosdelta=0.8925\n",
- "v_D=v_cd/cosdelta #m/s\n",
- "w_AD=v_D/(AD*1000**-1) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular Velocity of AD is',round(w_AD,2),\"rad/s clockwise.\" #Negative sign indicates clockwise orientation \n",
- "#Answer in the textbook is incorrect\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular Velocity of AD is -1.72 rad/s clockwise.\n"
- ]
- }
- ],
- "prompt_number": 47
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-18, Page No 277"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta1=73.9 degrees,theta2=60 degrees and theta3=46.1 degrees\n",
- "sintheta1=0.96\n",
- "sintheta2=sqrt(3)*2**-1\n",
- "sintheta3=0.72\n",
- "V=900 #mm/s\n",
- "\n",
- "#Calculations\n",
- "BC=((350*350)+(86.6*86.6))**0.5 #mm\n",
- "CD=400 #mm\n",
- "v_cb=(V*sintheta2)/(sintheta1) #mm/s\n",
- "v_c=((V*sintheta3))/(sintheta1) #mm/s\n",
- "w_dc=v_c/CD #rad/s\n",
- "w_bc=v_cb/BC #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular velocities are: w_dc=',round(w_dc,2),\"rad/s\",'and w_bc=',round(w_bc,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocities are: w_dc= 1.69 rad/s and w_bc= 2.25 rad/s\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-19, Page No 278"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Calculations\n",
- "#After equating the i and j terms we obtain simplified equations\n",
- "#Solving by matrix method\n",
- "A=np.array([[346,86.7],[200,-350]])\n",
- "B=np.array([[-3700],[-1790]]) \n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The angular accelerations are alpha_DC=',round(C[0],3),\"rad/s**2\",'and alpha_BC=',round(C[1],2),\"rad/s**2\" \n",
- "#The signs only indicate that the originally assumed orientations are incorrect and are opposite to those assumed\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular accelerations are alpha_DC= -10.475 rad/s**2 and alpha_BC= -0.87 rad/s**2\n"
- ]
- }
- ],
- "prompt_number": 80
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-20, Page No 279"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=3 #m\n",
- "w=8 #rad/s (clockwise)\n",
- "alpha=4 #rad/s**2 (counterclockwise)\n",
- "r=d*2**-1 #m\n",
- "\n",
- "#Calculations\n",
- "vo=r*w #m/s\n",
- "ao=r*alpha #m/s**2\n",
- "#Here OB is r\n",
- "OB=r #m\n",
- "v_bo=OB*w #m/s\n",
- "v_B=v_bo+vo #m/s\n",
- "#Also\n",
- "a_bo=r*alpha #m/s**2 (directed left)\n",
- "a_bo_n=r*w**2 #m/s**2\n",
- "a_h=ao+a_bo #m/s**2\n",
- "a_v=a_bo_n #m/s**2\n",
- "a_B=((a_h**2)+(a_v**2))**0.5 #m/s**2\n",
- "phi=arctan(a_h/a_v)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The linear velocity at B is',round(v_B),\"m/s\",'and the acceleration is',round(a_B,1),\"m/s**2\",'making an angle of',round(phi,2),\"degrees with horizontal\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The linear velocity at B is 24.0 m/s and the acceleration is 96.7 m/s**2 making an angle of 7.13 degrees with horizontal\n"
- ]
- }
- ],
- "prompt_number": 96
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-21, Page No 281"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "OA=0.6 #m\n",
- "w=8 #rad/s\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "v_O=12 #m/s\n",
- "alpha=4 #rad/s**2\n",
- "a_O=6 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Velocity Calculations\n",
- "v_AO=OA*w #m/s\n",
- "v_Ah=v_AO*sintheta+v_O #m/s horizontal component\n",
- "v_Av=v_AO*costheta #m/s\n",
- "v_A=((v_Ah**2)+(v_Av**2))**0.5 #m/s\n",
- "phi=arctan(v_Av/v_Ah)*(180/pi) #degrees\n",
- "#Acceleration Calculations\n",
- "a_AOt=OA*alpha #m/s**2\n",
- "a_AOn=OA*w**2 #m/s**2\n",
- "a_Ah=-a_O-a_AOn*costheta-a_AOt*sintheta #m/s**2\n",
- "a_Av=-a_AOn*sintheta+a_AOt*costheta #m/s**2\n",
- "a_A=((a_Ah**2)+(a_Av**2))**0.5 #m/s**2\n",
- "phi2=arctan(a_Av*a_Ah**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The velocity is',round(v_A),\"m/s\",'making an angle of',round(phi,1),\"degrees with horizontal.\"\n",
- "print'The acceleration is',round(a_A),\"m/s**2\",'making an angle of',round(phi2,1),\"degrees with horizontal\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity is 15.0 m/s making an angle of 16.1 degrees with horizontal.\n",
- "The acceleration is 44.0 m/s**2 making an angle of 22.9 degrees with horizontal\n"
- ]
- }
- ],
- "prompt_number": 100
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-22, Page No 282"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "AL=5 #ft\n",
- "d=10 #ft displacement\n",
- "\n",
- "#Calculations\n",
- "theta=d/AL #radians\n",
- "s_o=3*theta#ft\n",
- "\n",
- "#Result\n",
- "print'The displacement So is',round(s_o),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The displacement So is 6.0 ft\n"
- ]
- }
- ],
- "prompt_number": 101
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-23, Page No 283"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#Speed and acceleration at the center\n",
- "v=12 #in/s\n",
- "a=18 #in/s**2\n",
- "\n",
- "#Calculations\n",
- "v_D=((a+v*0.5)*a**-1)*v #in/s\n",
- "#Speed at point F\n",
- "v_F=((v/2)*v**-1)*v_D #in/s\n",
- "#Acceleration at D\n",
- "a_D=(24/a)*a #in/s**2\n",
- "#Acceleration at F\n",
- "a_F=((v/2)*v**-1)*24 #in/s**2\n",
- "\n",
- "#Result\n",
- "print'The velocity and acceleration of weight A are',round(v_F),\"in/s\",'and',round(a_F),\"in/s**2 respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity and acceleration of weight A are 8.0 in/s and 12.0 in/s**2 respectively.\n"
- ]
- }
- ],
- "prompt_number": 103
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-24, Page No 283"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Calculations\n",
- "#Speed and acceleration of D\n",
- "sD=((18-6)*18**-1)*12 #in/s\n",
- "aD=(12*18**-1)*18 #in/s**2\n",
- "#Speed and acceleration of F\n",
- "sF=(6*12**-1)*8 #in/s\n",
- "aF=(6*12**-1)*12 #in/s^2\n",
- "\n",
- "#Result\n",
- "print'The velocity and acceleration of weight A are',round(sF),\"in/s\",'and',round(aF),\"in/s**2 respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity and acceleration of weight A are 4.0 in/s and 6.0 in/s**2 respectively.\n"
- ]
- }
- ],
- "prompt_number": 104
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-26, Page No 284"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_BG=300 #mm/s\n",
- "v_G=300 #mm/s\n",
- "a_BGt=500 #mm/s**2\n",
- "a_BGn=3600#mm/s**2\n",
- "a_Gh=500 #mm/s**2\n",
- "a_Bv=1800 #mm/s**2\n",
- "\n",
- "#Calculations\n",
- "w=((75-25)/25)*6 #rad/s\n",
- "alpha=((75-25)/25)*10 #rad/s**2\n",
- "v_B=(v_BG**2+v_G**2)**0.5 #mm/s\n",
- "a_v=a_Bv-a_BGt #mm/s**2\n",
- "a_h=a_BGn-a_Gh #mm/s**2\n",
- "a_B=(a_v**2+a_h**2)**0.5 #mm/s**2\n",
- "\n",
- "#Result \n",
- "print'The velocity and acceleration of point B are',round(v_B),\"mm/s\",'and',round(a_B),\"mm/s**2 respectively.\"\n",
- "\n",
- "# The ans for a_B is incorrectin textbook."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity and acceleration of point B are 424.0 mm/s and 3362.0 mm/s**2 respectively.\n"
- ]
- }
- ],
- "prompt_number": 106
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter14_2.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter14_2.ipynb deleted file mode 100755 index c04236af..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter14_2.ipynb +++ /dev/null @@ -1,772 +0,0 @@ -{
- "metadata": {
- "name": "chapter14.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 14: Kinematics of a Rigid Body in Plane Motion"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-2, Page no 265"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=500 #mm\n",
- "wo=0 #rpm\n",
- "w=300 #rpm\n",
- "t=20 #s\n",
- "t1=2 #s\n",
- "\n",
- "#Calculations\n",
- "alpha=(2*pi*(60**-1)*(w-wo))/t #rad/s**2\n",
- "w1=wo+alpha*t1 #rad/s\n",
- "v=(d*(2*1000)**-1)*w1 #m/s\n",
- "a_n=(d*(2*1000)**-1)*w1**2 #m/s**2\n",
- "a_t=(d*(2*1000)**-1)*alpha #m/s**2\n",
- "a=(a_n**2+a_t**2)**0.5 #m/s**2\n",
- "theta=arccos(a_n/a)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The computed values are:'\n",
- "print'alpha=',round(alpha,2),\"rad/s**2\"\n",
- "print'w1=',round(w1,2),\"rad/s\"\n",
- "print'v=',round(v,3),\"m/s\"\n",
- "print'a=',round(a,2),\"m/s**2\"\n",
- "print'theta=',round(theta,1),\"degrees\"\n",
- "\n",
- "# The answers may wary in decimal points."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The computed values are:\n",
- "alpha= 1.57 rad/s**2\n",
- "w1= 3.14 rad/s\n",
- "v= 0.785 m/s\n",
- "a= 2.5 m/s**2\n",
- "theta= 9.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-3, Page no 266"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "s_BC=2 #m\n",
- "s_C=2.5 #m\n",
- "\n",
- "#Calculations\n",
- "s_B=(s_BC**2+s_C**2)**0.5 #m\n",
- "theta=arctan(s_BC*s_C**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The absolute displacement is',round(s_B,1),\"m\",'and the angle made by the vector is',round(theta,1),\"degrees.\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The absolute displacement is 3.2 m and the angle made by the vector is 38.7 degrees.\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-4, Page no 266"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "V_A=20 #mi/h\n",
- "V_B=70 #mi/h\n",
- "# as theta1=60 degrees,\n",
- "sintheta1=sqrt(3)*2**-1\n",
- "costheta1=2**-1\n",
- "# also phi=45 degrees, thus\n",
- "sinphi=sqrt(2)**-1\n",
- "cosphi=sqrt(2)**-1\n",
- "\n",
- "#Result\n",
- "#Vector's in matrix form\n",
- "v_A=np.array([-V_A*cosphi,V_A*sinphi]) #mi/h\n",
- "v_B=np.array([V_B*costheta1,V_B*sintheta1]) #mi/h\n",
- "a=v_A[0]+v_B[0] #mi/h\n",
- "b=v_A[1]+v_B[1] #mi/h\n",
- "v_ab=(a**2+b**2)**0.5 #mi/h\n",
- "theta=arctan(b/a)*(180/pi) #degrees\n",
- "#The relative velocity v_ba is just different in sign while the magnitude stays the same\n",
- "\n",
- "#Resul\n",
- "print'The relative velocity is',round(v_ab,1),\"mi/h\",'making an angle',round(theta,1),\"degrees\"\n",
- "\n",
- "# The ans may wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The relative velocity is 77.6 mi/h making an angle 74.4 degrees\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-9, Page no 271"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=2.5 #m\n",
- "v_A=4 #m/s\n",
- "a_A=5 #m/s**2\n",
- "theta=30 #degrees\n",
- "\n",
- "#Calculations\n",
- "#Vector triangle yields v_a.b=2.93 m/s\n",
- "v_ab=2.93 #m/s\n",
- "w=v_ab*l**-1 #rad/s (clockwise)\n",
- "#Ploygon yields alpha_a/b=2.75 m/s**2\n",
- "alpha_ab=2.75 #m/s**2\n",
- "alpha=alpha_ab*l**-1 #rad/s**2 (counterclockwise)\n",
- "\n",
- "#Result\n",
- "print'The value of angular velocity is',round(w,2),\"rad/s\"\n",
- "print'The value of angular acceleration is',round(alpha,1),\"rad/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of angular velocity is 1.17 rad/s\n",
- "The value of angular acceleration is 1.1 rad/s**2\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-10, Page no 272"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=(2*pi*120)/60 #rad/s\n",
- "l=24 #in\n",
- "l_c=4 #in\n",
- "# as th=30 degrees,\n",
- "sinth=2**-1\n",
- "\n",
- "#Calculations\n",
- "v=(l_c*12**-1)*w #ft/s\n",
- "betaa=arcsin((l_c*sinth)/l)*(180/pi) #degrees\n",
- "# betaa yeilds 4.8 degrees, thus value of cosbetaa is,\n",
- "cosbetaa=0.996\n",
- "theta=60-betaa #degrees\n",
- "# here theta yeilds 55.2 degrees, thus value of costheta is,\n",
- "costheta=0.57\n",
- "#Component of velocity along connecting rod is \n",
- "v1=v*costheta #ft/s\n",
- "v_p=v1/cosbetaa #ft/s\n",
- "\n",
- "#Result\n",
- "print'The absoulte velocity is',round(v_p,2),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The absoulte velocity is 2.4 ft/s\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-13, Page no 274"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_pc=3.68 #ft/s\n",
- "l=2 #ft\n",
- "\n",
- "#Calculations\n",
- "w=v_pc/l #rad/s counterclockwise\n",
- "\n",
- "#Result\n",
- "print'The angular velocity is',round(w,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity is 1.84 rad/s\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-14, Page no 274"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#This problem is a combination of numerical and graphical solution\n",
- "#The program only deals with the numerical solution parts the rest can be verified by graphical solution\n",
- "#Initilization of variables\n",
- "r=4*12**-1 #ft\n",
- "w=4*pi #rad/s\n",
- "l=2 #ft\n",
- "w2=1.84 #rad/s\n",
- "\n",
- "#Calculations\n",
- "ac_n=r*w**2 #ft/s**2\n",
- "a_pc_n=l*w2**2 #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of ac_n is',round(ac_n,1),\"ft/s**2\"\n",
- "print'The value of a_pc_n is',round(a_pc_n,2),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of ac_n is 52.6 ft/s**2\n",
- "The value of a_pc_n is 6.77 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-15, Page no 275"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w_bc=10 #rad/s\n",
- "AB=250 #mm\n",
- "BC=150 #mm\n",
- "AC=179 #mm\n",
- "AD=200 #mm\n",
- "# as theta1=45 degrees,\n",
- "sintheta1=(2**0.5)**-1\n",
- "costheta1=(2**0.5)**-1\n",
- "\n",
- "#Calculations\n",
- "v_c=(BC*1000**-1)*w_bc #m/s\n",
- "AC=((AB**2+BC**2)-(2*AB*BC*costheta1))**0.5 #m\n",
- "betaa=arcsin((BC*sintheta1)/AC)*(180/pi) #degrees\n",
- "gammaa=arcsin((AB*sintheta1)/AC)*(180/pi)#degrees answer in the textbook is incorrect\n",
- "ang=60-betaa #degrees\n",
- "# ang yeilds 23.7 degrees, thus\n",
- "sinang=0.40056\n",
- "cosang=0.916\n",
- "CD=sqrt(AD**2+AC**2-(2*AD*AC*cosang)) #mm\n",
- "D=arcsin((AC*sinang)/CD)*(180/pi) #degrees\n",
- "# D yeilds 63.2 degrees,thus\n",
- "sinD=0.8925\n",
- "theta=arcsin((AD*sinD)/AC)*(180/pi) #degrees\n",
- "n=360-(theta+gammaa+90) #degrees\n",
- "# n yeilds 101.8 degrees, thus \n",
- "cosn=-0.2045\n",
- "v_cd=v_c*cosn #m/s\n",
- "delta=180-(90+D) #degrees\n",
- "# Delts yeilds 26.8 degrees, thus\n",
- "cosdelta=0.8925\n",
- "v_D=v_cd/cosdelta #m/s\n",
- "w_AD=v_D/(AD*1000**-1) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular Velocity of AD is',round(w_AD,2),\"rad/s clockwise.\" #Negative sign indicates clockwise orientation \n",
- "#Answer in the textbook is incorrect\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular Velocity of AD is -1.72 rad/s clockwise.\n"
- ]
- }
- ],
- "prompt_number": 47
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-18, Page No 277"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta1=73.9 degrees,theta2=60 degrees and theta3=46.1 degrees\n",
- "sintheta1=0.96\n",
- "sintheta2=sqrt(3)*2**-1\n",
- "sintheta3=0.72\n",
- "V=900 #mm/s\n",
- "\n",
- "#Calculations\n",
- "BC=((350*350)+(86.6*86.6))**0.5 #mm\n",
- "CD=400 #mm\n",
- "v_cb=(V*sintheta2)/(sintheta1) #mm/s\n",
- "v_c=((V*sintheta3))/(sintheta1) #mm/s\n",
- "w_dc=v_c/CD #rad/s\n",
- "w_bc=v_cb/BC #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular velocities are: w_dc=',round(w_dc,2),\"rad/s\",'and w_bc=',round(w_bc,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocities are: w_dc= 1.69 rad/s and w_bc= 2.25 rad/s\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-19, Page No 278"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Calculations\n",
- "#After equating the i and j terms we obtain simplified equations\n",
- "#Solving by matrix method\n",
- "A=np.array([[346,86.7],[200,-350]])\n",
- "B=np.array([[-3700],[-1790]]) \n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The angular accelerations are alpha_DC=',round(C[0],3),\"rad/s**2\",'and alpha_BC=',round(C[1],2),\"rad/s**2\" \n",
- "#The signs only indicate that the originally assumed orientations are incorrect and are opposite to those assumed\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular accelerations are alpha_DC= -10.475 rad/s**2 and alpha_BC= -0.87 rad/s**2\n"
- ]
- }
- ],
- "prompt_number": 80
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-20, Page No 279"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=3 #m\n",
- "w=8 #rad/s (clockwise)\n",
- "alpha=4 #rad/s**2 (counterclockwise)\n",
- "r=d*2**-1 #m\n",
- "\n",
- "#Calculations\n",
- "vo=r*w #m/s\n",
- "ao=r*alpha #m/s**2\n",
- "#Here OB is r\n",
- "OB=r #m\n",
- "v_bo=OB*w #m/s\n",
- "v_B=v_bo+vo #m/s\n",
- "#Also\n",
- "a_bo=r*alpha #m/s**2 (directed left)\n",
- "a_bo_n=r*w**2 #m/s**2\n",
- "a_h=ao+a_bo #m/s**2\n",
- "a_v=a_bo_n #m/s**2\n",
- "a_B=((a_h**2)+(a_v**2))**0.5 #m/s**2\n",
- "phi=arctan(a_h/a_v)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The linear velocity at B is',round(v_B),\"m/s\",'and the acceleration is',round(a_B,1),\"m/s**2\",'making an angle of',round(phi,2),\"degrees with horizontal\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The linear velocity at B is 24.0 m/s and the acceleration is 96.7 m/s**2 making an angle of 7.13 degrees with horizontal\n"
- ]
- }
- ],
- "prompt_number": 96
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-21, Page No 281"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "OA=0.6 #m\n",
- "w=8 #rad/s\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "v_O=12 #m/s\n",
- "alpha=4 #rad/s**2\n",
- "a_O=6 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Velocity Calculations\n",
- "v_AO=OA*w #m/s\n",
- "v_Ah=v_AO*sintheta+v_O #m/s horizontal component\n",
- "v_Av=v_AO*costheta #m/s\n",
- "v_A=((v_Ah**2)+(v_Av**2))**0.5 #m/s\n",
- "phi=arctan(v_Av/v_Ah)*(180/pi) #degrees\n",
- "#Acceleration Calculations\n",
- "a_AOt=OA*alpha #m/s**2\n",
- "a_AOn=OA*w**2 #m/s**2\n",
- "a_Ah=-a_O-a_AOn*costheta-a_AOt*sintheta #m/s**2\n",
- "a_Av=-a_AOn*sintheta+a_AOt*costheta #m/s**2\n",
- "a_A=((a_Ah**2)+(a_Av**2))**0.5 #m/s**2\n",
- "phi2=arctan(a_Av*a_Ah**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The velocity is',round(v_A),\"m/s\",'making an angle of',round(phi,1),\"degrees with horizontal.\"\n",
- "print'The acceleration is',round(a_A),\"m/s**2\",'making an angle of',round(phi2,1),\"degrees with horizontal\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity is 15.0 m/s making an angle of 16.1 degrees with horizontal.\n",
- "The acceleration is 44.0 m/s**2 making an angle of 22.9 degrees with horizontal\n"
- ]
- }
- ],
- "prompt_number": 100
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-22, Page No 282"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "AL=5 #ft\n",
- "d=10 #ft displacement\n",
- "\n",
- "#Calculations\n",
- "theta=d/AL #radians\n",
- "s_o=3*theta#ft\n",
- "\n",
- "#Result\n",
- "print'The displacement So is',round(s_o),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The displacement So is 6.0 ft\n"
- ]
- }
- ],
- "prompt_number": 101
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-23, Page No 283"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#Speed and acceleration at the center\n",
- "v=12 #in/s\n",
- "a=18 #in/s**2\n",
- "\n",
- "#Calculations\n",
- "v_D=((a+v*0.5)*a**-1)*v #in/s\n",
- "#Speed at point F\n",
- "v_F=((v/2)*v**-1)*v_D #in/s\n",
- "#Acceleration at D\n",
- "a_D=(24/a)*a #in/s**2\n",
- "#Acceleration at F\n",
- "a_F=((v/2)*v**-1)*24 #in/s**2\n",
- "\n",
- "#Result\n",
- "print'The velocity and acceleration of weight A are',round(v_F),\"in/s\",'and',round(a_F),\"in/s**2 respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity and acceleration of weight A are 8.0 in/s and 12.0 in/s**2 respectively.\n"
- ]
- }
- ],
- "prompt_number": 103
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-24, Page No 283"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Calculations\n",
- "#Speed and acceleration of D\n",
- "sD=((18-6)*18**-1)*12 #in/s\n",
- "aD=(12*18**-1)*18 #in/s**2\n",
- "#Speed and acceleration of F\n",
- "sF=(6*12**-1)*8 #in/s\n",
- "aF=(6*12**-1)*12 #in/s^2\n",
- "\n",
- "#Result\n",
- "print'The velocity and acceleration of weight A are',round(sF),\"in/s\",'and',round(aF),\"in/s**2 respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity and acceleration of weight A are 4.0 in/s and 6.0 in/s**2 respectively.\n"
- ]
- }
- ],
- "prompt_number": 104
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-26, Page No 284"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_BG=300 #mm/s\n",
- "v_G=300 #mm/s\n",
- "a_BGt=500 #mm/s**2\n",
- "a_BGn=3600#mm/s**2\n",
- "a_Gh=500 #mm/s**2\n",
- "a_Bv=1800 #mm/s**2\n",
- "\n",
- "#Calculations\n",
- "w=((75-25)/25)*6 #rad/s\n",
- "alpha=((75-25)/25)*10 #rad/s**2\n",
- "v_B=(v_BG**2+v_G**2)**0.5 #mm/s\n",
- "a_v=a_Bv-a_BGt #mm/s**2\n",
- "a_h=a_BGn-a_Gh #mm/s**2\n",
- "a_B=(a_v**2+a_h**2)**0.5 #mm/s**2\n",
- "\n",
- "#Result \n",
- "print'The velocity and acceleration of point B are',round(v_B),\"mm/s\",'and',round(a_B),\"mm/s**2 respectively.\"\n",
- "\n",
- "# The ans for a_B is incorrectin textbook."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity and acceleration of point B are 424.0 mm/s and 3362.0 mm/s**2 respectively.\n"
- ]
- }
- ],
- "prompt_number": 106
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter14_3.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter14_3.ipynb deleted file mode 100755 index c04236af..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter14_3.ipynb +++ /dev/null @@ -1,772 +0,0 @@ -{
- "metadata": {
- "name": "chapter14.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 14: Kinematics of a Rigid Body in Plane Motion"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-2, Page no 265"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=500 #mm\n",
- "wo=0 #rpm\n",
- "w=300 #rpm\n",
- "t=20 #s\n",
- "t1=2 #s\n",
- "\n",
- "#Calculations\n",
- "alpha=(2*pi*(60**-1)*(w-wo))/t #rad/s**2\n",
- "w1=wo+alpha*t1 #rad/s\n",
- "v=(d*(2*1000)**-1)*w1 #m/s\n",
- "a_n=(d*(2*1000)**-1)*w1**2 #m/s**2\n",
- "a_t=(d*(2*1000)**-1)*alpha #m/s**2\n",
- "a=(a_n**2+a_t**2)**0.5 #m/s**2\n",
- "theta=arccos(a_n/a)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The computed values are:'\n",
- "print'alpha=',round(alpha,2),\"rad/s**2\"\n",
- "print'w1=',round(w1,2),\"rad/s\"\n",
- "print'v=',round(v,3),\"m/s\"\n",
- "print'a=',round(a,2),\"m/s**2\"\n",
- "print'theta=',round(theta,1),\"degrees\"\n",
- "\n",
- "# The answers may wary in decimal points."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The computed values are:\n",
- "alpha= 1.57 rad/s**2\n",
- "w1= 3.14 rad/s\n",
- "v= 0.785 m/s\n",
- "a= 2.5 m/s**2\n",
- "theta= 9.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-3, Page no 266"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "s_BC=2 #m\n",
- "s_C=2.5 #m\n",
- "\n",
- "#Calculations\n",
- "s_B=(s_BC**2+s_C**2)**0.5 #m\n",
- "theta=arctan(s_BC*s_C**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The absolute displacement is',round(s_B,1),\"m\",'and the angle made by the vector is',round(theta,1),\"degrees.\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The absolute displacement is 3.2 m and the angle made by the vector is 38.7 degrees.\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-4, Page no 266"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "V_A=20 #mi/h\n",
- "V_B=70 #mi/h\n",
- "# as theta1=60 degrees,\n",
- "sintheta1=sqrt(3)*2**-1\n",
- "costheta1=2**-1\n",
- "# also phi=45 degrees, thus\n",
- "sinphi=sqrt(2)**-1\n",
- "cosphi=sqrt(2)**-1\n",
- "\n",
- "#Result\n",
- "#Vector's in matrix form\n",
- "v_A=np.array([-V_A*cosphi,V_A*sinphi]) #mi/h\n",
- "v_B=np.array([V_B*costheta1,V_B*sintheta1]) #mi/h\n",
- "a=v_A[0]+v_B[0] #mi/h\n",
- "b=v_A[1]+v_B[1] #mi/h\n",
- "v_ab=(a**2+b**2)**0.5 #mi/h\n",
- "theta=arctan(b/a)*(180/pi) #degrees\n",
- "#The relative velocity v_ba is just different in sign while the magnitude stays the same\n",
- "\n",
- "#Resul\n",
- "print'The relative velocity is',round(v_ab,1),\"mi/h\",'making an angle',round(theta,1),\"degrees\"\n",
- "\n",
- "# The ans may wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The relative velocity is 77.6 mi/h making an angle 74.4 degrees\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-9, Page no 271"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=2.5 #m\n",
- "v_A=4 #m/s\n",
- "a_A=5 #m/s**2\n",
- "theta=30 #degrees\n",
- "\n",
- "#Calculations\n",
- "#Vector triangle yields v_a.b=2.93 m/s\n",
- "v_ab=2.93 #m/s\n",
- "w=v_ab*l**-1 #rad/s (clockwise)\n",
- "#Ploygon yields alpha_a/b=2.75 m/s**2\n",
- "alpha_ab=2.75 #m/s**2\n",
- "alpha=alpha_ab*l**-1 #rad/s**2 (counterclockwise)\n",
- "\n",
- "#Result\n",
- "print'The value of angular velocity is',round(w,2),\"rad/s\"\n",
- "print'The value of angular acceleration is',round(alpha,1),\"rad/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of angular velocity is 1.17 rad/s\n",
- "The value of angular acceleration is 1.1 rad/s**2\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-10, Page no 272"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=(2*pi*120)/60 #rad/s\n",
- "l=24 #in\n",
- "l_c=4 #in\n",
- "# as th=30 degrees,\n",
- "sinth=2**-1\n",
- "\n",
- "#Calculations\n",
- "v=(l_c*12**-1)*w #ft/s\n",
- "betaa=arcsin((l_c*sinth)/l)*(180/pi) #degrees\n",
- "# betaa yeilds 4.8 degrees, thus value of cosbetaa is,\n",
- "cosbetaa=0.996\n",
- "theta=60-betaa #degrees\n",
- "# here theta yeilds 55.2 degrees, thus value of costheta is,\n",
- "costheta=0.57\n",
- "#Component of velocity along connecting rod is \n",
- "v1=v*costheta #ft/s\n",
- "v_p=v1/cosbetaa #ft/s\n",
- "\n",
- "#Result\n",
- "print'The absoulte velocity is',round(v_p,2),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The absoulte velocity is 2.4 ft/s\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-13, Page no 274"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_pc=3.68 #ft/s\n",
- "l=2 #ft\n",
- "\n",
- "#Calculations\n",
- "w=v_pc/l #rad/s counterclockwise\n",
- "\n",
- "#Result\n",
- "print'The angular velocity is',round(w,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity is 1.84 rad/s\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-14, Page no 274"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#This problem is a combination of numerical and graphical solution\n",
- "#The program only deals with the numerical solution parts the rest can be verified by graphical solution\n",
- "#Initilization of variables\n",
- "r=4*12**-1 #ft\n",
- "w=4*pi #rad/s\n",
- "l=2 #ft\n",
- "w2=1.84 #rad/s\n",
- "\n",
- "#Calculations\n",
- "ac_n=r*w**2 #ft/s**2\n",
- "a_pc_n=l*w2**2 #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of ac_n is',round(ac_n,1),\"ft/s**2\"\n",
- "print'The value of a_pc_n is',round(a_pc_n,2),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of ac_n is 52.6 ft/s**2\n",
- "The value of a_pc_n is 6.77 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-15, Page no 275"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w_bc=10 #rad/s\n",
- "AB=250 #mm\n",
- "BC=150 #mm\n",
- "AC=179 #mm\n",
- "AD=200 #mm\n",
- "# as theta1=45 degrees,\n",
- "sintheta1=(2**0.5)**-1\n",
- "costheta1=(2**0.5)**-1\n",
- "\n",
- "#Calculations\n",
- "v_c=(BC*1000**-1)*w_bc #m/s\n",
- "AC=((AB**2+BC**2)-(2*AB*BC*costheta1))**0.5 #m\n",
- "betaa=arcsin((BC*sintheta1)/AC)*(180/pi) #degrees\n",
- "gammaa=arcsin((AB*sintheta1)/AC)*(180/pi)#degrees answer in the textbook is incorrect\n",
- "ang=60-betaa #degrees\n",
- "# ang yeilds 23.7 degrees, thus\n",
- "sinang=0.40056\n",
- "cosang=0.916\n",
- "CD=sqrt(AD**2+AC**2-(2*AD*AC*cosang)) #mm\n",
- "D=arcsin((AC*sinang)/CD)*(180/pi) #degrees\n",
- "# D yeilds 63.2 degrees,thus\n",
- "sinD=0.8925\n",
- "theta=arcsin((AD*sinD)/AC)*(180/pi) #degrees\n",
- "n=360-(theta+gammaa+90) #degrees\n",
- "# n yeilds 101.8 degrees, thus \n",
- "cosn=-0.2045\n",
- "v_cd=v_c*cosn #m/s\n",
- "delta=180-(90+D) #degrees\n",
- "# Delts yeilds 26.8 degrees, thus\n",
- "cosdelta=0.8925\n",
- "v_D=v_cd/cosdelta #m/s\n",
- "w_AD=v_D/(AD*1000**-1) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular Velocity of AD is',round(w_AD,2),\"rad/s clockwise.\" #Negative sign indicates clockwise orientation \n",
- "#Answer in the textbook is incorrect\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular Velocity of AD is -1.72 rad/s clockwise.\n"
- ]
- }
- ],
- "prompt_number": 47
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-18, Page No 277"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta1=73.9 degrees,theta2=60 degrees and theta3=46.1 degrees\n",
- "sintheta1=0.96\n",
- "sintheta2=sqrt(3)*2**-1\n",
- "sintheta3=0.72\n",
- "V=900 #mm/s\n",
- "\n",
- "#Calculations\n",
- "BC=((350*350)+(86.6*86.6))**0.5 #mm\n",
- "CD=400 #mm\n",
- "v_cb=(V*sintheta2)/(sintheta1) #mm/s\n",
- "v_c=((V*sintheta3))/(sintheta1) #mm/s\n",
- "w_dc=v_c/CD #rad/s\n",
- "w_bc=v_cb/BC #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular velocities are: w_dc=',round(w_dc,2),\"rad/s\",'and w_bc=',round(w_bc,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocities are: w_dc= 1.69 rad/s and w_bc= 2.25 rad/s\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-19, Page No 278"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Calculations\n",
- "#After equating the i and j terms we obtain simplified equations\n",
- "#Solving by matrix method\n",
- "A=np.array([[346,86.7],[200,-350]])\n",
- "B=np.array([[-3700],[-1790]]) \n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The angular accelerations are alpha_DC=',round(C[0],3),\"rad/s**2\",'and alpha_BC=',round(C[1],2),\"rad/s**2\" \n",
- "#The signs only indicate that the originally assumed orientations are incorrect and are opposite to those assumed\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular accelerations are alpha_DC= -10.475 rad/s**2 and alpha_BC= -0.87 rad/s**2\n"
- ]
- }
- ],
- "prompt_number": 80
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-20, Page No 279"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=3 #m\n",
- "w=8 #rad/s (clockwise)\n",
- "alpha=4 #rad/s**2 (counterclockwise)\n",
- "r=d*2**-1 #m\n",
- "\n",
- "#Calculations\n",
- "vo=r*w #m/s\n",
- "ao=r*alpha #m/s**2\n",
- "#Here OB is r\n",
- "OB=r #m\n",
- "v_bo=OB*w #m/s\n",
- "v_B=v_bo+vo #m/s\n",
- "#Also\n",
- "a_bo=r*alpha #m/s**2 (directed left)\n",
- "a_bo_n=r*w**2 #m/s**2\n",
- "a_h=ao+a_bo #m/s**2\n",
- "a_v=a_bo_n #m/s**2\n",
- "a_B=((a_h**2)+(a_v**2))**0.5 #m/s**2\n",
- "phi=arctan(a_h/a_v)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The linear velocity at B is',round(v_B),\"m/s\",'and the acceleration is',round(a_B,1),\"m/s**2\",'making an angle of',round(phi,2),\"degrees with horizontal\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The linear velocity at B is 24.0 m/s and the acceleration is 96.7 m/s**2 making an angle of 7.13 degrees with horizontal\n"
- ]
- }
- ],
- "prompt_number": 96
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-21, Page No 281"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "OA=0.6 #m\n",
- "w=8 #rad/s\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "v_O=12 #m/s\n",
- "alpha=4 #rad/s**2\n",
- "a_O=6 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Velocity Calculations\n",
- "v_AO=OA*w #m/s\n",
- "v_Ah=v_AO*sintheta+v_O #m/s horizontal component\n",
- "v_Av=v_AO*costheta #m/s\n",
- "v_A=((v_Ah**2)+(v_Av**2))**0.5 #m/s\n",
- "phi=arctan(v_Av/v_Ah)*(180/pi) #degrees\n",
- "#Acceleration Calculations\n",
- "a_AOt=OA*alpha #m/s**2\n",
- "a_AOn=OA*w**2 #m/s**2\n",
- "a_Ah=-a_O-a_AOn*costheta-a_AOt*sintheta #m/s**2\n",
- "a_Av=-a_AOn*sintheta+a_AOt*costheta #m/s**2\n",
- "a_A=((a_Ah**2)+(a_Av**2))**0.5 #m/s**2\n",
- "phi2=arctan(a_Av*a_Ah**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The velocity is',round(v_A),\"m/s\",'making an angle of',round(phi,1),\"degrees with horizontal.\"\n",
- "print'The acceleration is',round(a_A),\"m/s**2\",'making an angle of',round(phi2,1),\"degrees with horizontal\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity is 15.0 m/s making an angle of 16.1 degrees with horizontal.\n",
- "The acceleration is 44.0 m/s**2 making an angle of 22.9 degrees with horizontal\n"
- ]
- }
- ],
- "prompt_number": 100
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-22, Page No 282"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "AL=5 #ft\n",
- "d=10 #ft displacement\n",
- "\n",
- "#Calculations\n",
- "theta=d/AL #radians\n",
- "s_o=3*theta#ft\n",
- "\n",
- "#Result\n",
- "print'The displacement So is',round(s_o),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The displacement So is 6.0 ft\n"
- ]
- }
- ],
- "prompt_number": 101
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-23, Page No 283"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#Speed and acceleration at the center\n",
- "v=12 #in/s\n",
- "a=18 #in/s**2\n",
- "\n",
- "#Calculations\n",
- "v_D=((a+v*0.5)*a**-1)*v #in/s\n",
- "#Speed at point F\n",
- "v_F=((v/2)*v**-1)*v_D #in/s\n",
- "#Acceleration at D\n",
- "a_D=(24/a)*a #in/s**2\n",
- "#Acceleration at F\n",
- "a_F=((v/2)*v**-1)*24 #in/s**2\n",
- "\n",
- "#Result\n",
- "print'The velocity and acceleration of weight A are',round(v_F),\"in/s\",'and',round(a_F),\"in/s**2 respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity and acceleration of weight A are 8.0 in/s and 12.0 in/s**2 respectively.\n"
- ]
- }
- ],
- "prompt_number": 103
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-24, Page No 283"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Calculations\n",
- "#Speed and acceleration of D\n",
- "sD=((18-6)*18**-1)*12 #in/s\n",
- "aD=(12*18**-1)*18 #in/s**2\n",
- "#Speed and acceleration of F\n",
- "sF=(6*12**-1)*8 #in/s\n",
- "aF=(6*12**-1)*12 #in/s^2\n",
- "\n",
- "#Result\n",
- "print'The velocity and acceleration of weight A are',round(sF),\"in/s\",'and',round(aF),\"in/s**2 respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity and acceleration of weight A are 4.0 in/s and 6.0 in/s**2 respectively.\n"
- ]
- }
- ],
- "prompt_number": 104
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-26, Page No 284"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_BG=300 #mm/s\n",
- "v_G=300 #mm/s\n",
- "a_BGt=500 #mm/s**2\n",
- "a_BGn=3600#mm/s**2\n",
- "a_Gh=500 #mm/s**2\n",
- "a_Bv=1800 #mm/s**2\n",
- "\n",
- "#Calculations\n",
- "w=((75-25)/25)*6 #rad/s\n",
- "alpha=((75-25)/25)*10 #rad/s**2\n",
- "v_B=(v_BG**2+v_G**2)**0.5 #mm/s\n",
- "a_v=a_Bv-a_BGt #mm/s**2\n",
- "a_h=a_BGn-a_Gh #mm/s**2\n",
- "a_B=(a_v**2+a_h**2)**0.5 #mm/s**2\n",
- "\n",
- "#Result \n",
- "print'The velocity and acceleration of point B are',round(v_B),\"mm/s\",'and',round(a_B),\"mm/s**2 respectively.\"\n",
- "\n",
- "# The ans for a_B is incorrectin textbook."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity and acceleration of point B are 424.0 mm/s and 3362.0 mm/s**2 respectively.\n"
- ]
- }
- ],
- "prompt_number": 106
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter14_4.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter14_4.ipynb deleted file mode 100755 index c04236af..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter14_4.ipynb +++ /dev/null @@ -1,772 +0,0 @@ -{
- "metadata": {
- "name": "chapter14.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 14: Kinematics of a Rigid Body in Plane Motion"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-2, Page no 265"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=500 #mm\n",
- "wo=0 #rpm\n",
- "w=300 #rpm\n",
- "t=20 #s\n",
- "t1=2 #s\n",
- "\n",
- "#Calculations\n",
- "alpha=(2*pi*(60**-1)*(w-wo))/t #rad/s**2\n",
- "w1=wo+alpha*t1 #rad/s\n",
- "v=(d*(2*1000)**-1)*w1 #m/s\n",
- "a_n=(d*(2*1000)**-1)*w1**2 #m/s**2\n",
- "a_t=(d*(2*1000)**-1)*alpha #m/s**2\n",
- "a=(a_n**2+a_t**2)**0.5 #m/s**2\n",
- "theta=arccos(a_n/a)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The computed values are:'\n",
- "print'alpha=',round(alpha,2),\"rad/s**2\"\n",
- "print'w1=',round(w1,2),\"rad/s\"\n",
- "print'v=',round(v,3),\"m/s\"\n",
- "print'a=',round(a,2),\"m/s**2\"\n",
- "print'theta=',round(theta,1),\"degrees\"\n",
- "\n",
- "# The answers may wary in decimal points."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The computed values are:\n",
- "alpha= 1.57 rad/s**2\n",
- "w1= 3.14 rad/s\n",
- "v= 0.785 m/s\n",
- "a= 2.5 m/s**2\n",
- "theta= 9.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-3, Page no 266"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "s_BC=2 #m\n",
- "s_C=2.5 #m\n",
- "\n",
- "#Calculations\n",
- "s_B=(s_BC**2+s_C**2)**0.5 #m\n",
- "theta=arctan(s_BC*s_C**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The absolute displacement is',round(s_B,1),\"m\",'and the angle made by the vector is',round(theta,1),\"degrees.\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The absolute displacement is 3.2 m and the angle made by the vector is 38.7 degrees.\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-4, Page no 266"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "V_A=20 #mi/h\n",
- "V_B=70 #mi/h\n",
- "# as theta1=60 degrees,\n",
- "sintheta1=sqrt(3)*2**-1\n",
- "costheta1=2**-1\n",
- "# also phi=45 degrees, thus\n",
- "sinphi=sqrt(2)**-1\n",
- "cosphi=sqrt(2)**-1\n",
- "\n",
- "#Result\n",
- "#Vector's in matrix form\n",
- "v_A=np.array([-V_A*cosphi,V_A*sinphi]) #mi/h\n",
- "v_B=np.array([V_B*costheta1,V_B*sintheta1]) #mi/h\n",
- "a=v_A[0]+v_B[0] #mi/h\n",
- "b=v_A[1]+v_B[1] #mi/h\n",
- "v_ab=(a**2+b**2)**0.5 #mi/h\n",
- "theta=arctan(b/a)*(180/pi) #degrees\n",
- "#The relative velocity v_ba is just different in sign while the magnitude stays the same\n",
- "\n",
- "#Resul\n",
- "print'The relative velocity is',round(v_ab,1),\"mi/h\",'making an angle',round(theta,1),\"degrees\"\n",
- "\n",
- "# The ans may wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The relative velocity is 77.6 mi/h making an angle 74.4 degrees\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-9, Page no 271"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=2.5 #m\n",
- "v_A=4 #m/s\n",
- "a_A=5 #m/s**2\n",
- "theta=30 #degrees\n",
- "\n",
- "#Calculations\n",
- "#Vector triangle yields v_a.b=2.93 m/s\n",
- "v_ab=2.93 #m/s\n",
- "w=v_ab*l**-1 #rad/s (clockwise)\n",
- "#Ploygon yields alpha_a/b=2.75 m/s**2\n",
- "alpha_ab=2.75 #m/s**2\n",
- "alpha=alpha_ab*l**-1 #rad/s**2 (counterclockwise)\n",
- "\n",
- "#Result\n",
- "print'The value of angular velocity is',round(w,2),\"rad/s\"\n",
- "print'The value of angular acceleration is',round(alpha,1),\"rad/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of angular velocity is 1.17 rad/s\n",
- "The value of angular acceleration is 1.1 rad/s**2\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-10, Page no 272"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=(2*pi*120)/60 #rad/s\n",
- "l=24 #in\n",
- "l_c=4 #in\n",
- "# as th=30 degrees,\n",
- "sinth=2**-1\n",
- "\n",
- "#Calculations\n",
- "v=(l_c*12**-1)*w #ft/s\n",
- "betaa=arcsin((l_c*sinth)/l)*(180/pi) #degrees\n",
- "# betaa yeilds 4.8 degrees, thus value of cosbetaa is,\n",
- "cosbetaa=0.996\n",
- "theta=60-betaa #degrees\n",
- "# here theta yeilds 55.2 degrees, thus value of costheta is,\n",
- "costheta=0.57\n",
- "#Component of velocity along connecting rod is \n",
- "v1=v*costheta #ft/s\n",
- "v_p=v1/cosbetaa #ft/s\n",
- "\n",
- "#Result\n",
- "print'The absoulte velocity is',round(v_p,2),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The absoulte velocity is 2.4 ft/s\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-13, Page no 274"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_pc=3.68 #ft/s\n",
- "l=2 #ft\n",
- "\n",
- "#Calculations\n",
- "w=v_pc/l #rad/s counterclockwise\n",
- "\n",
- "#Result\n",
- "print'The angular velocity is',round(w,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity is 1.84 rad/s\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-14, Page no 274"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#This problem is a combination of numerical and graphical solution\n",
- "#The program only deals with the numerical solution parts the rest can be verified by graphical solution\n",
- "#Initilization of variables\n",
- "r=4*12**-1 #ft\n",
- "w=4*pi #rad/s\n",
- "l=2 #ft\n",
- "w2=1.84 #rad/s\n",
- "\n",
- "#Calculations\n",
- "ac_n=r*w**2 #ft/s**2\n",
- "a_pc_n=l*w2**2 #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of ac_n is',round(ac_n,1),\"ft/s**2\"\n",
- "print'The value of a_pc_n is',round(a_pc_n,2),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of ac_n is 52.6 ft/s**2\n",
- "The value of a_pc_n is 6.77 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-15, Page no 275"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w_bc=10 #rad/s\n",
- "AB=250 #mm\n",
- "BC=150 #mm\n",
- "AC=179 #mm\n",
- "AD=200 #mm\n",
- "# as theta1=45 degrees,\n",
- "sintheta1=(2**0.5)**-1\n",
- "costheta1=(2**0.5)**-1\n",
- "\n",
- "#Calculations\n",
- "v_c=(BC*1000**-1)*w_bc #m/s\n",
- "AC=((AB**2+BC**2)-(2*AB*BC*costheta1))**0.5 #m\n",
- "betaa=arcsin((BC*sintheta1)/AC)*(180/pi) #degrees\n",
- "gammaa=arcsin((AB*sintheta1)/AC)*(180/pi)#degrees answer in the textbook is incorrect\n",
- "ang=60-betaa #degrees\n",
- "# ang yeilds 23.7 degrees, thus\n",
- "sinang=0.40056\n",
- "cosang=0.916\n",
- "CD=sqrt(AD**2+AC**2-(2*AD*AC*cosang)) #mm\n",
- "D=arcsin((AC*sinang)/CD)*(180/pi) #degrees\n",
- "# D yeilds 63.2 degrees,thus\n",
- "sinD=0.8925\n",
- "theta=arcsin((AD*sinD)/AC)*(180/pi) #degrees\n",
- "n=360-(theta+gammaa+90) #degrees\n",
- "# n yeilds 101.8 degrees, thus \n",
- "cosn=-0.2045\n",
- "v_cd=v_c*cosn #m/s\n",
- "delta=180-(90+D) #degrees\n",
- "# Delts yeilds 26.8 degrees, thus\n",
- "cosdelta=0.8925\n",
- "v_D=v_cd/cosdelta #m/s\n",
- "w_AD=v_D/(AD*1000**-1) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular Velocity of AD is',round(w_AD,2),\"rad/s clockwise.\" #Negative sign indicates clockwise orientation \n",
- "#Answer in the textbook is incorrect\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular Velocity of AD is -1.72 rad/s clockwise.\n"
- ]
- }
- ],
- "prompt_number": 47
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-18, Page No 277"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta1=73.9 degrees,theta2=60 degrees and theta3=46.1 degrees\n",
- "sintheta1=0.96\n",
- "sintheta2=sqrt(3)*2**-1\n",
- "sintheta3=0.72\n",
- "V=900 #mm/s\n",
- "\n",
- "#Calculations\n",
- "BC=((350*350)+(86.6*86.6))**0.5 #mm\n",
- "CD=400 #mm\n",
- "v_cb=(V*sintheta2)/(sintheta1) #mm/s\n",
- "v_c=((V*sintheta3))/(sintheta1) #mm/s\n",
- "w_dc=v_c/CD #rad/s\n",
- "w_bc=v_cb/BC #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular velocities are: w_dc=',round(w_dc,2),\"rad/s\",'and w_bc=',round(w_bc,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocities are: w_dc= 1.69 rad/s and w_bc= 2.25 rad/s\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-19, Page No 278"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Calculations\n",
- "#After equating the i and j terms we obtain simplified equations\n",
- "#Solving by matrix method\n",
- "A=np.array([[346,86.7],[200,-350]])\n",
- "B=np.array([[-3700],[-1790]]) \n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The angular accelerations are alpha_DC=',round(C[0],3),\"rad/s**2\",'and alpha_BC=',round(C[1],2),\"rad/s**2\" \n",
- "#The signs only indicate that the originally assumed orientations are incorrect and are opposite to those assumed\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular accelerations are alpha_DC= -10.475 rad/s**2 and alpha_BC= -0.87 rad/s**2\n"
- ]
- }
- ],
- "prompt_number": 80
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-20, Page No 279"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=3 #m\n",
- "w=8 #rad/s (clockwise)\n",
- "alpha=4 #rad/s**2 (counterclockwise)\n",
- "r=d*2**-1 #m\n",
- "\n",
- "#Calculations\n",
- "vo=r*w #m/s\n",
- "ao=r*alpha #m/s**2\n",
- "#Here OB is r\n",
- "OB=r #m\n",
- "v_bo=OB*w #m/s\n",
- "v_B=v_bo+vo #m/s\n",
- "#Also\n",
- "a_bo=r*alpha #m/s**2 (directed left)\n",
- "a_bo_n=r*w**2 #m/s**2\n",
- "a_h=ao+a_bo #m/s**2\n",
- "a_v=a_bo_n #m/s**2\n",
- "a_B=((a_h**2)+(a_v**2))**0.5 #m/s**2\n",
- "phi=arctan(a_h/a_v)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The linear velocity at B is',round(v_B),\"m/s\",'and the acceleration is',round(a_B,1),\"m/s**2\",'making an angle of',round(phi,2),\"degrees with horizontal\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The linear velocity at B is 24.0 m/s and the acceleration is 96.7 m/s**2 making an angle of 7.13 degrees with horizontal\n"
- ]
- }
- ],
- "prompt_number": 96
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-21, Page No 281"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "OA=0.6 #m\n",
- "w=8 #rad/s\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "v_O=12 #m/s\n",
- "alpha=4 #rad/s**2\n",
- "a_O=6 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Velocity Calculations\n",
- "v_AO=OA*w #m/s\n",
- "v_Ah=v_AO*sintheta+v_O #m/s horizontal component\n",
- "v_Av=v_AO*costheta #m/s\n",
- "v_A=((v_Ah**2)+(v_Av**2))**0.5 #m/s\n",
- "phi=arctan(v_Av/v_Ah)*(180/pi) #degrees\n",
- "#Acceleration Calculations\n",
- "a_AOt=OA*alpha #m/s**2\n",
- "a_AOn=OA*w**2 #m/s**2\n",
- "a_Ah=-a_O-a_AOn*costheta-a_AOt*sintheta #m/s**2\n",
- "a_Av=-a_AOn*sintheta+a_AOt*costheta #m/s**2\n",
- "a_A=((a_Ah**2)+(a_Av**2))**0.5 #m/s**2\n",
- "phi2=arctan(a_Av*a_Ah**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The velocity is',round(v_A),\"m/s\",'making an angle of',round(phi,1),\"degrees with horizontal.\"\n",
- "print'The acceleration is',round(a_A),\"m/s**2\",'making an angle of',round(phi2,1),\"degrees with horizontal\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity is 15.0 m/s making an angle of 16.1 degrees with horizontal.\n",
- "The acceleration is 44.0 m/s**2 making an angle of 22.9 degrees with horizontal\n"
- ]
- }
- ],
- "prompt_number": 100
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-22, Page No 282"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "AL=5 #ft\n",
- "d=10 #ft displacement\n",
- "\n",
- "#Calculations\n",
- "theta=d/AL #radians\n",
- "s_o=3*theta#ft\n",
- "\n",
- "#Result\n",
- "print'The displacement So is',round(s_o),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The displacement So is 6.0 ft\n"
- ]
- }
- ],
- "prompt_number": 101
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-23, Page No 283"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#Speed and acceleration at the center\n",
- "v=12 #in/s\n",
- "a=18 #in/s**2\n",
- "\n",
- "#Calculations\n",
- "v_D=((a+v*0.5)*a**-1)*v #in/s\n",
- "#Speed at point F\n",
- "v_F=((v/2)*v**-1)*v_D #in/s\n",
- "#Acceleration at D\n",
- "a_D=(24/a)*a #in/s**2\n",
- "#Acceleration at F\n",
- "a_F=((v/2)*v**-1)*24 #in/s**2\n",
- "\n",
- "#Result\n",
- "print'The velocity and acceleration of weight A are',round(v_F),\"in/s\",'and',round(a_F),\"in/s**2 respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity and acceleration of weight A are 8.0 in/s and 12.0 in/s**2 respectively.\n"
- ]
- }
- ],
- "prompt_number": 103
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-24, Page No 283"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Calculations\n",
- "#Speed and acceleration of D\n",
- "sD=((18-6)*18**-1)*12 #in/s\n",
- "aD=(12*18**-1)*18 #in/s**2\n",
- "#Speed and acceleration of F\n",
- "sF=(6*12**-1)*8 #in/s\n",
- "aF=(6*12**-1)*12 #in/s^2\n",
- "\n",
- "#Result\n",
- "print'The velocity and acceleration of weight A are',round(sF),\"in/s\",'and',round(aF),\"in/s**2 respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity and acceleration of weight A are 4.0 in/s and 6.0 in/s**2 respectively.\n"
- ]
- }
- ],
- "prompt_number": 104
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14-26, Page No 284"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v_BG=300 #mm/s\n",
- "v_G=300 #mm/s\n",
- "a_BGt=500 #mm/s**2\n",
- "a_BGn=3600#mm/s**2\n",
- "a_Gh=500 #mm/s**2\n",
- "a_Bv=1800 #mm/s**2\n",
- "\n",
- "#Calculations\n",
- "w=((75-25)/25)*6 #rad/s\n",
- "alpha=((75-25)/25)*10 #rad/s**2\n",
- "v_B=(v_BG**2+v_G**2)**0.5 #mm/s\n",
- "a_v=a_Bv-a_BGt #mm/s**2\n",
- "a_h=a_BGn-a_Gh #mm/s**2\n",
- "a_B=(a_v**2+a_h**2)**0.5 #mm/s**2\n",
- "\n",
- "#Result \n",
- "print'The velocity and acceleration of point B are',round(v_B),\"mm/s\",'and',round(a_B),\"mm/s**2 respectively.\"\n",
- "\n",
- "# The ans for a_B is incorrectin textbook."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity and acceleration of point B are 424.0 mm/s and 3362.0 mm/s**2 respectively.\n"
- ]
- }
- ],
- "prompt_number": 106
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter15.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter15.ipynb deleted file mode 100755 index da6b1b6a..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter15.ipynb +++ /dev/null @@ -1,510 +0,0 @@ -{
- "metadata": {
- "name": "chapter15.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 15: Moments of Inertia"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-11, Page no 305"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "y1=1 #in\n",
- "y2=4 #in\n",
- "d1=2.2-1 #in\n",
- "d2=4-2.2 #in\n",
- "A1=12 #in**2\n",
- "A2=8 #in**2\n",
- "b1=6 #in\n",
- "b2=2 #in\n",
- "h1=2 #in\n",
- "h2=4 #in\n",
- "\n",
- "#Calculations\n",
- "y_bar=(A1*y1+A2*y2)/(A1+A2) #in\n",
- "I1=(12**-1)*(b1)*(h1**3) #in**4\n",
- "I2=(12**-1)*(b2)*(h2**3) #in**4\n",
- "#Using Parallel Axes Theorem\n",
- "I=(I1+(A1*d1**2))+(I2+(A2*d2**2)) #in**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(I,1),\"in**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 57.9 in**4\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-12, Page no 306"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=60 #mm diameter of the hole\n",
- "#Areas\n",
- "At=100*100 #mm**2\n",
- "Ab=200*100 #mm**2\n",
- "Ac=((pi/4)*d**2) #mm**2\n",
- "bt=100 #mm\n",
- "ht=100 #mm\n",
- "bb=200 #mm\n",
- "hb=100 #mm\n",
- "#Distance of centroids of each area\n",
- "yt=150 #mm\n",
- "yb=50 #mm\n",
- "yc=150 #mm\n",
- "\n",
- "#Calculations\n",
- "y_bar=((At*yt)+(Ab*yb)-(Ac*yc))/(At+Ab-Ac) #mm\n",
- "#Distances\n",
- "dt=yt-y_bar #mm\n",
- "db=y_bar-yb #mm\n",
- "dc=yc-y_bar #mm\n",
- "#Values of Inertia\n",
- "It=(12**-1)*(bt)*(ht**3) #mm**4\n",
- "Ib=(12**-1)*(bb)*(hb**3) #mm**4\n",
- "Ic=(4**-1)*(pi)*((d/2)**4) #mm**4\n",
- "#Moment of inertia\n",
- "I=((It+At*dt**2)+(Ib+Ab*db**2)-(Ic+Ac*dc**2)) #mm**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(I,1),\"mm**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 77156533.6 mm**4\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-14, Page no 308"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "b1=2 #in\n",
- "b2=4 #in\n",
- "h1=8 #in\n",
- "h2=2 #in\n",
- "bo=8 #in\n",
- "ho=8 #in\n",
- "bi=4 #in\n",
- "hi=4 #in\n",
- "\n",
- "#Calculations\n",
- "I1=(12**-1)*(b1)*(h1**3) #in**4\n",
- "I2=(12**-1)*(b2)*(h2**3) #in**4\n",
- "I=2*(I1+I2) #in**4\n",
- "Io=(12**-1)*(bo)*(ho**3) #in**4\n",
- "Ii=(12**-1)*(bi)*(hi**3) #in**4\n",
- "I_bar=Io-Ii #in**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(I_bar),\"in**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 320.0 in**4\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-15, Page no 308"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "b1=75 #mm\n",
- "b2=12 #mm\n",
- "h1=12 #mm\n",
- "h2=162 #mm\n",
- "d1=75 #mm\n",
- "\n",
- "#Calculations\n",
- "A=(h2*b2)+(2*b1*h1) #mm**2\n",
- "I1=(12**-1)*(b1)*(h1**3)+(b1*h1*d1**2) #mm**4\n",
- "I2=(12**-1)*(b2)*(h2**3) #mm**4\n",
- "I_bar=2*I1+I2 #mm**4\n",
- "k=sqrt(I_bar/A) #mm\n",
- "\n",
- "#Result\n",
- "print'The axial moment of inertia is',round(I_bar,1),\"mm**4\"\n",
- "print'The radius of gyration is',round(k,1),\"mm\"\n",
- "\n",
- "# Here value of k is off by 0.1 mm"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The axial moment of inertia is 14398128.0 mm**4\n",
- "The radius of gyration is 62.0 mm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-20, Page no 311"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=50 #mm\n",
- "\n",
- "#Calculations\n",
- "Ixy=(8**-1)*(50**4) #mm**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(Ixy,2),\"mm**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 781250.0 mm**4\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-24, Page no 314"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#The notation has been changed for ease\n",
- "\n",
- "#Calculations\n",
- "x=(5*1*3.5+8*1*0.5)/(5*1+8*1) #in\n",
- "y=(5*1*0.5+8*1*4)/13 #in\n",
- "#Moment of inertia \n",
- "Ix=(12**-1)*(5)*(1**3)+(5*2.15*2.15)+(12**-1)*(1*8**3)+(8*1.35**2) #in**4\n",
- "Iy=(12**-1)*(1)*(5**3)+(5*1.85*1.85)+(12**-1)*(8)*(1**3)+(8*1.15**2) #in**4\n",
- "Ixy=(8*1*(-1.15)*1.35)+(5*1*1.85*(-2.15)) #in**4\n",
- "#Mohr circle calculations\n",
- "d=0.5*(Ix+Iy) #distance to center of the cirlce \n",
- "r=sqrt((21**2)+(32.3**2)) \n",
- "maxI=d+r #in**4\n",
- "theta=arctan(32.3/21)*(180/pi) #degrees maxI occurs at this angle\n",
- "minI=d-r #in**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertias are as follows:'\n",
- "print'Ix=',round(Ix,1),\"in**4\"\n",
- "print'Iy=',round(Iy,1),\"in**4\"\n",
- "print'Ixy=',round(Ixy,1),\"in**4\"\n",
- "print'maxI=',round(maxI,1),\"in**4\"\n",
- "print'minI=',round(minI,1),\"in**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertias are as follows:\n",
- "Ix= 80.8 in**4\n",
- "Iy= 38.8 in**4\n",
- "Ixy= -32.3 in**4\n",
- "maxI= 98.3 in**4\n",
- "minI= 21.2 in**4\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-25, Page no 315"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Notations have been changed\n",
- "\n",
- "#Calculations\n",
- "x=-(25*125*0.5*125+25*100*0.5*25)/(25*125+25*100) #mm\n",
- "y=(25*125*0.5*25+25*100*75)/5625 #mm \n",
- "Iy=(12**-1)*25*125**3+25*125*(62.5-40.3)**2+(12**-1)*100*25**3+100*25*(40.3-12.5)**2 #mm**4\n",
- "Ix=Iy #mm**4 for L-section\n",
- "#The second computation checks the first\n",
- "Ixy=(125*25*22.2*27.8)+(100*25*(-27.8)*(-34.7)) #mm**4\n",
- "#Mohr Circle analysis\n",
- "Imax=Ix+Ixy #mm**4\n",
- "Imin=Ix-Ixy #mm**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertias are as follows:'\n",
- "print'Ix=',round(Ix,2),\"mm**4\"\n",
- "print'Iy=',round(Iy,2),\"mm**4\"\n",
- "print'Ixy=',round(Ixy,2),\"mm**4\"\n",
- "print'Imax=',round(Imax),\"mm**4\"\n",
- "print'Imin=',round(Imin,2),\"mm**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertias are as follows:\n",
- "Ix= 7671443.75 mm**4\n",
- "Iy= 7671443.75 mm**4\n",
- "Ixy= 4340275.0 mm**4\n",
- "Imax= 12011719.0 mm**4\n",
- "Imin= 3331168.75 mm**4\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-30, Page no 320"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "rho=490 #lb/ft**3\n",
- "t=0.02 #in\n",
- "d=4 #in\n",
- "r=d/2 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "W=(pi*r**2*t*rho)*1728**-1 #lb\n",
- "#Mass\n",
- "m=W*g**-1 #slugs\n",
- "#Momemt of inertia\n",
- "I=(4**-1)*m*(r*12**-1)**2 #slug-ft**2\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(I,6),\"slug-ft**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 1.5e-05 slug-ft**2\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-36, Page no 322"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#The integration involves variables hence the direct formula is being used in this coding\n",
- "m=500 #kg\n",
- "R=0.25 #m\n",
- "h=0.5 #m\n",
- "\n",
- "#Calculations\n",
- "Ix=(3*10**-1)*m*R**2 #kg.m**2\n",
- "Iy=(3*5**-1)*m*((4**-1)*R**2+h**2) #kg.m**2\n",
- "\n",
- "#Result\n",
- "print'Hence proved that Ix=',round(Ix,2),\"kg.m**2\" \n",
- "print'and Iy=',round(Iy,1),\"kg.m**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Hence proved that Ix= 9.38 kg.m**2\n",
- "and Iy= 79.7 kg.m**2\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-37, Page no 323"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "delta=450 #lb/ft**3\n",
- "h1=9*12**-1 #ft\n",
- "h2=10*12**-1 #ft\n",
- "ro1=4*12**-1 #ft\n",
- "ri1=2*12**-1 #ft\n",
- "ro2=18*12**-1 #ft\n",
- "ri2=16*12**-1 #ft\n",
- "a=2.5*24**-1 #ft\n",
- "b=3.5*24**-1 #ft\n",
- "l=1 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "Whub=(pi*ro1**2-pi*ri1**2)*h1*delta #lb\n",
- "Wrim=(pi*ro2**2-pi*ri2**2)*h2*delta #lb\n",
- "#For one spoke\n",
- "Wspoke=(pi*a*b*l*delta) #lb\n",
- "#Moment of inertia calculations\n",
- "Ihub=0.5*(Whub*g**-1)*(ro1**2+ri1**2) #lb-s**2-ft\n",
- "Irim=0.5*(Wrim*g**-1)*(ro2**2+ri2**2) #lb-s**2-ft\n",
- "Ispoke=6*((12**-1)*(Wspoke*g**-1)*l**2+(Wspoke*g**-1)*h2**2) #lb-s**2-ft\n",
- "Iwheel=Ihub+Irim+Ispoke #lb-s**2-ft\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia of the wheel is',round(Iwheel,1),\"lb-s**s-ft\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia of the wheel is 38.1 lb-s**s-ft\n"
- ]
- }
- ],
- "prompt_number": 20
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter15_1.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter15_1.ipynb deleted file mode 100755 index da6b1b6a..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter15_1.ipynb +++ /dev/null @@ -1,510 +0,0 @@ -{
- "metadata": {
- "name": "chapter15.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 15: Moments of Inertia"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-11, Page no 305"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "y1=1 #in\n",
- "y2=4 #in\n",
- "d1=2.2-1 #in\n",
- "d2=4-2.2 #in\n",
- "A1=12 #in**2\n",
- "A2=8 #in**2\n",
- "b1=6 #in\n",
- "b2=2 #in\n",
- "h1=2 #in\n",
- "h2=4 #in\n",
- "\n",
- "#Calculations\n",
- "y_bar=(A1*y1+A2*y2)/(A1+A2) #in\n",
- "I1=(12**-1)*(b1)*(h1**3) #in**4\n",
- "I2=(12**-1)*(b2)*(h2**3) #in**4\n",
- "#Using Parallel Axes Theorem\n",
- "I=(I1+(A1*d1**2))+(I2+(A2*d2**2)) #in**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(I,1),\"in**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 57.9 in**4\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-12, Page no 306"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=60 #mm diameter of the hole\n",
- "#Areas\n",
- "At=100*100 #mm**2\n",
- "Ab=200*100 #mm**2\n",
- "Ac=((pi/4)*d**2) #mm**2\n",
- "bt=100 #mm\n",
- "ht=100 #mm\n",
- "bb=200 #mm\n",
- "hb=100 #mm\n",
- "#Distance of centroids of each area\n",
- "yt=150 #mm\n",
- "yb=50 #mm\n",
- "yc=150 #mm\n",
- "\n",
- "#Calculations\n",
- "y_bar=((At*yt)+(Ab*yb)-(Ac*yc))/(At+Ab-Ac) #mm\n",
- "#Distances\n",
- "dt=yt-y_bar #mm\n",
- "db=y_bar-yb #mm\n",
- "dc=yc-y_bar #mm\n",
- "#Values of Inertia\n",
- "It=(12**-1)*(bt)*(ht**3) #mm**4\n",
- "Ib=(12**-1)*(bb)*(hb**3) #mm**4\n",
- "Ic=(4**-1)*(pi)*((d/2)**4) #mm**4\n",
- "#Moment of inertia\n",
- "I=((It+At*dt**2)+(Ib+Ab*db**2)-(Ic+Ac*dc**2)) #mm**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(I,1),\"mm**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 77156533.6 mm**4\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-14, Page no 308"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "b1=2 #in\n",
- "b2=4 #in\n",
- "h1=8 #in\n",
- "h2=2 #in\n",
- "bo=8 #in\n",
- "ho=8 #in\n",
- "bi=4 #in\n",
- "hi=4 #in\n",
- "\n",
- "#Calculations\n",
- "I1=(12**-1)*(b1)*(h1**3) #in**4\n",
- "I2=(12**-1)*(b2)*(h2**3) #in**4\n",
- "I=2*(I1+I2) #in**4\n",
- "Io=(12**-1)*(bo)*(ho**3) #in**4\n",
- "Ii=(12**-1)*(bi)*(hi**3) #in**4\n",
- "I_bar=Io-Ii #in**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(I_bar),\"in**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 320.0 in**4\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-15, Page no 308"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "b1=75 #mm\n",
- "b2=12 #mm\n",
- "h1=12 #mm\n",
- "h2=162 #mm\n",
- "d1=75 #mm\n",
- "\n",
- "#Calculations\n",
- "A=(h2*b2)+(2*b1*h1) #mm**2\n",
- "I1=(12**-1)*(b1)*(h1**3)+(b1*h1*d1**2) #mm**4\n",
- "I2=(12**-1)*(b2)*(h2**3) #mm**4\n",
- "I_bar=2*I1+I2 #mm**4\n",
- "k=sqrt(I_bar/A) #mm\n",
- "\n",
- "#Result\n",
- "print'The axial moment of inertia is',round(I_bar,1),\"mm**4\"\n",
- "print'The radius of gyration is',round(k,1),\"mm\"\n",
- "\n",
- "# Here value of k is off by 0.1 mm"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The axial moment of inertia is 14398128.0 mm**4\n",
- "The radius of gyration is 62.0 mm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-20, Page no 311"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=50 #mm\n",
- "\n",
- "#Calculations\n",
- "Ixy=(8**-1)*(50**4) #mm**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(Ixy,2),\"mm**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 781250.0 mm**4\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-24, Page no 314"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#The notation has been changed for ease\n",
- "\n",
- "#Calculations\n",
- "x=(5*1*3.5+8*1*0.5)/(5*1+8*1) #in\n",
- "y=(5*1*0.5+8*1*4)/13 #in\n",
- "#Moment of inertia \n",
- "Ix=(12**-1)*(5)*(1**3)+(5*2.15*2.15)+(12**-1)*(1*8**3)+(8*1.35**2) #in**4\n",
- "Iy=(12**-1)*(1)*(5**3)+(5*1.85*1.85)+(12**-1)*(8)*(1**3)+(8*1.15**2) #in**4\n",
- "Ixy=(8*1*(-1.15)*1.35)+(5*1*1.85*(-2.15)) #in**4\n",
- "#Mohr circle calculations\n",
- "d=0.5*(Ix+Iy) #distance to center of the cirlce \n",
- "r=sqrt((21**2)+(32.3**2)) \n",
- "maxI=d+r #in**4\n",
- "theta=arctan(32.3/21)*(180/pi) #degrees maxI occurs at this angle\n",
- "minI=d-r #in**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertias are as follows:'\n",
- "print'Ix=',round(Ix,1),\"in**4\"\n",
- "print'Iy=',round(Iy,1),\"in**4\"\n",
- "print'Ixy=',round(Ixy,1),\"in**4\"\n",
- "print'maxI=',round(maxI,1),\"in**4\"\n",
- "print'minI=',round(minI,1),\"in**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertias are as follows:\n",
- "Ix= 80.8 in**4\n",
- "Iy= 38.8 in**4\n",
- "Ixy= -32.3 in**4\n",
- "maxI= 98.3 in**4\n",
- "minI= 21.2 in**4\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-25, Page no 315"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Notations have been changed\n",
- "\n",
- "#Calculations\n",
- "x=-(25*125*0.5*125+25*100*0.5*25)/(25*125+25*100) #mm\n",
- "y=(25*125*0.5*25+25*100*75)/5625 #mm \n",
- "Iy=(12**-1)*25*125**3+25*125*(62.5-40.3)**2+(12**-1)*100*25**3+100*25*(40.3-12.5)**2 #mm**4\n",
- "Ix=Iy #mm**4 for L-section\n",
- "#The second computation checks the first\n",
- "Ixy=(125*25*22.2*27.8)+(100*25*(-27.8)*(-34.7)) #mm**4\n",
- "#Mohr Circle analysis\n",
- "Imax=Ix+Ixy #mm**4\n",
- "Imin=Ix-Ixy #mm**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertias are as follows:'\n",
- "print'Ix=',round(Ix,2),\"mm**4\"\n",
- "print'Iy=',round(Iy,2),\"mm**4\"\n",
- "print'Ixy=',round(Ixy,2),\"mm**4\"\n",
- "print'Imax=',round(Imax),\"mm**4\"\n",
- "print'Imin=',round(Imin,2),\"mm**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertias are as follows:\n",
- "Ix= 7671443.75 mm**4\n",
- "Iy= 7671443.75 mm**4\n",
- "Ixy= 4340275.0 mm**4\n",
- "Imax= 12011719.0 mm**4\n",
- "Imin= 3331168.75 mm**4\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-30, Page no 320"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "rho=490 #lb/ft**3\n",
- "t=0.02 #in\n",
- "d=4 #in\n",
- "r=d/2 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "W=(pi*r**2*t*rho)*1728**-1 #lb\n",
- "#Mass\n",
- "m=W*g**-1 #slugs\n",
- "#Momemt of inertia\n",
- "I=(4**-1)*m*(r*12**-1)**2 #slug-ft**2\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(I,6),\"slug-ft**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 1.5e-05 slug-ft**2\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-36, Page no 322"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#The integration involves variables hence the direct formula is being used in this coding\n",
- "m=500 #kg\n",
- "R=0.25 #m\n",
- "h=0.5 #m\n",
- "\n",
- "#Calculations\n",
- "Ix=(3*10**-1)*m*R**2 #kg.m**2\n",
- "Iy=(3*5**-1)*m*((4**-1)*R**2+h**2) #kg.m**2\n",
- "\n",
- "#Result\n",
- "print'Hence proved that Ix=',round(Ix,2),\"kg.m**2\" \n",
- "print'and Iy=',round(Iy,1),\"kg.m**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Hence proved that Ix= 9.38 kg.m**2\n",
- "and Iy= 79.7 kg.m**2\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-37, Page no 323"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "delta=450 #lb/ft**3\n",
- "h1=9*12**-1 #ft\n",
- "h2=10*12**-1 #ft\n",
- "ro1=4*12**-1 #ft\n",
- "ri1=2*12**-1 #ft\n",
- "ro2=18*12**-1 #ft\n",
- "ri2=16*12**-1 #ft\n",
- "a=2.5*24**-1 #ft\n",
- "b=3.5*24**-1 #ft\n",
- "l=1 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "Whub=(pi*ro1**2-pi*ri1**2)*h1*delta #lb\n",
- "Wrim=(pi*ro2**2-pi*ri2**2)*h2*delta #lb\n",
- "#For one spoke\n",
- "Wspoke=(pi*a*b*l*delta) #lb\n",
- "#Moment of inertia calculations\n",
- "Ihub=0.5*(Whub*g**-1)*(ro1**2+ri1**2) #lb-s**2-ft\n",
- "Irim=0.5*(Wrim*g**-1)*(ro2**2+ri2**2) #lb-s**2-ft\n",
- "Ispoke=6*((12**-1)*(Wspoke*g**-1)*l**2+(Wspoke*g**-1)*h2**2) #lb-s**2-ft\n",
- "Iwheel=Ihub+Irim+Ispoke #lb-s**2-ft\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia of the wheel is',round(Iwheel,1),\"lb-s**s-ft\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia of the wheel is 38.1 lb-s**s-ft\n"
- ]
- }
- ],
- "prompt_number": 20
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter15_2.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter15_2.ipynb deleted file mode 100755 index da6b1b6a..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter15_2.ipynb +++ /dev/null @@ -1,510 +0,0 @@ -{
- "metadata": {
- "name": "chapter15.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 15: Moments of Inertia"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-11, Page no 305"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "y1=1 #in\n",
- "y2=4 #in\n",
- "d1=2.2-1 #in\n",
- "d2=4-2.2 #in\n",
- "A1=12 #in**2\n",
- "A2=8 #in**2\n",
- "b1=6 #in\n",
- "b2=2 #in\n",
- "h1=2 #in\n",
- "h2=4 #in\n",
- "\n",
- "#Calculations\n",
- "y_bar=(A1*y1+A2*y2)/(A1+A2) #in\n",
- "I1=(12**-1)*(b1)*(h1**3) #in**4\n",
- "I2=(12**-1)*(b2)*(h2**3) #in**4\n",
- "#Using Parallel Axes Theorem\n",
- "I=(I1+(A1*d1**2))+(I2+(A2*d2**2)) #in**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(I,1),\"in**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 57.9 in**4\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-12, Page no 306"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=60 #mm diameter of the hole\n",
- "#Areas\n",
- "At=100*100 #mm**2\n",
- "Ab=200*100 #mm**2\n",
- "Ac=((pi/4)*d**2) #mm**2\n",
- "bt=100 #mm\n",
- "ht=100 #mm\n",
- "bb=200 #mm\n",
- "hb=100 #mm\n",
- "#Distance of centroids of each area\n",
- "yt=150 #mm\n",
- "yb=50 #mm\n",
- "yc=150 #mm\n",
- "\n",
- "#Calculations\n",
- "y_bar=((At*yt)+(Ab*yb)-(Ac*yc))/(At+Ab-Ac) #mm\n",
- "#Distances\n",
- "dt=yt-y_bar #mm\n",
- "db=y_bar-yb #mm\n",
- "dc=yc-y_bar #mm\n",
- "#Values of Inertia\n",
- "It=(12**-1)*(bt)*(ht**3) #mm**4\n",
- "Ib=(12**-1)*(bb)*(hb**3) #mm**4\n",
- "Ic=(4**-1)*(pi)*((d/2)**4) #mm**4\n",
- "#Moment of inertia\n",
- "I=((It+At*dt**2)+(Ib+Ab*db**2)-(Ic+Ac*dc**2)) #mm**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(I,1),\"mm**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 77156533.6 mm**4\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-14, Page no 308"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "b1=2 #in\n",
- "b2=4 #in\n",
- "h1=8 #in\n",
- "h2=2 #in\n",
- "bo=8 #in\n",
- "ho=8 #in\n",
- "bi=4 #in\n",
- "hi=4 #in\n",
- "\n",
- "#Calculations\n",
- "I1=(12**-1)*(b1)*(h1**3) #in**4\n",
- "I2=(12**-1)*(b2)*(h2**3) #in**4\n",
- "I=2*(I1+I2) #in**4\n",
- "Io=(12**-1)*(bo)*(ho**3) #in**4\n",
- "Ii=(12**-1)*(bi)*(hi**3) #in**4\n",
- "I_bar=Io-Ii #in**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(I_bar),\"in**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 320.0 in**4\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-15, Page no 308"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "b1=75 #mm\n",
- "b2=12 #mm\n",
- "h1=12 #mm\n",
- "h2=162 #mm\n",
- "d1=75 #mm\n",
- "\n",
- "#Calculations\n",
- "A=(h2*b2)+(2*b1*h1) #mm**2\n",
- "I1=(12**-1)*(b1)*(h1**3)+(b1*h1*d1**2) #mm**4\n",
- "I2=(12**-1)*(b2)*(h2**3) #mm**4\n",
- "I_bar=2*I1+I2 #mm**4\n",
- "k=sqrt(I_bar/A) #mm\n",
- "\n",
- "#Result\n",
- "print'The axial moment of inertia is',round(I_bar,1),\"mm**4\"\n",
- "print'The radius of gyration is',round(k,1),\"mm\"\n",
- "\n",
- "# Here value of k is off by 0.1 mm"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The axial moment of inertia is 14398128.0 mm**4\n",
- "The radius of gyration is 62.0 mm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-20, Page no 311"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=50 #mm\n",
- "\n",
- "#Calculations\n",
- "Ixy=(8**-1)*(50**4) #mm**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(Ixy,2),\"mm**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 781250.0 mm**4\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-24, Page no 314"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#The notation has been changed for ease\n",
- "\n",
- "#Calculations\n",
- "x=(5*1*3.5+8*1*0.5)/(5*1+8*1) #in\n",
- "y=(5*1*0.5+8*1*4)/13 #in\n",
- "#Moment of inertia \n",
- "Ix=(12**-1)*(5)*(1**3)+(5*2.15*2.15)+(12**-1)*(1*8**3)+(8*1.35**2) #in**4\n",
- "Iy=(12**-1)*(1)*(5**3)+(5*1.85*1.85)+(12**-1)*(8)*(1**3)+(8*1.15**2) #in**4\n",
- "Ixy=(8*1*(-1.15)*1.35)+(5*1*1.85*(-2.15)) #in**4\n",
- "#Mohr circle calculations\n",
- "d=0.5*(Ix+Iy) #distance to center of the cirlce \n",
- "r=sqrt((21**2)+(32.3**2)) \n",
- "maxI=d+r #in**4\n",
- "theta=arctan(32.3/21)*(180/pi) #degrees maxI occurs at this angle\n",
- "minI=d-r #in**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertias are as follows:'\n",
- "print'Ix=',round(Ix,1),\"in**4\"\n",
- "print'Iy=',round(Iy,1),\"in**4\"\n",
- "print'Ixy=',round(Ixy,1),\"in**4\"\n",
- "print'maxI=',round(maxI,1),\"in**4\"\n",
- "print'minI=',round(minI,1),\"in**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertias are as follows:\n",
- "Ix= 80.8 in**4\n",
- "Iy= 38.8 in**4\n",
- "Ixy= -32.3 in**4\n",
- "maxI= 98.3 in**4\n",
- "minI= 21.2 in**4\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-25, Page no 315"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Notations have been changed\n",
- "\n",
- "#Calculations\n",
- "x=-(25*125*0.5*125+25*100*0.5*25)/(25*125+25*100) #mm\n",
- "y=(25*125*0.5*25+25*100*75)/5625 #mm \n",
- "Iy=(12**-1)*25*125**3+25*125*(62.5-40.3)**2+(12**-1)*100*25**3+100*25*(40.3-12.5)**2 #mm**4\n",
- "Ix=Iy #mm**4 for L-section\n",
- "#The second computation checks the first\n",
- "Ixy=(125*25*22.2*27.8)+(100*25*(-27.8)*(-34.7)) #mm**4\n",
- "#Mohr Circle analysis\n",
- "Imax=Ix+Ixy #mm**4\n",
- "Imin=Ix-Ixy #mm**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertias are as follows:'\n",
- "print'Ix=',round(Ix,2),\"mm**4\"\n",
- "print'Iy=',round(Iy,2),\"mm**4\"\n",
- "print'Ixy=',round(Ixy,2),\"mm**4\"\n",
- "print'Imax=',round(Imax),\"mm**4\"\n",
- "print'Imin=',round(Imin,2),\"mm**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertias are as follows:\n",
- "Ix= 7671443.75 mm**4\n",
- "Iy= 7671443.75 mm**4\n",
- "Ixy= 4340275.0 mm**4\n",
- "Imax= 12011719.0 mm**4\n",
- "Imin= 3331168.75 mm**4\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-30, Page no 320"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "rho=490 #lb/ft**3\n",
- "t=0.02 #in\n",
- "d=4 #in\n",
- "r=d/2 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "W=(pi*r**2*t*rho)*1728**-1 #lb\n",
- "#Mass\n",
- "m=W*g**-1 #slugs\n",
- "#Momemt of inertia\n",
- "I=(4**-1)*m*(r*12**-1)**2 #slug-ft**2\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(I,6),\"slug-ft**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 1.5e-05 slug-ft**2\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-36, Page no 322"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#The integration involves variables hence the direct formula is being used in this coding\n",
- "m=500 #kg\n",
- "R=0.25 #m\n",
- "h=0.5 #m\n",
- "\n",
- "#Calculations\n",
- "Ix=(3*10**-1)*m*R**2 #kg.m**2\n",
- "Iy=(3*5**-1)*m*((4**-1)*R**2+h**2) #kg.m**2\n",
- "\n",
- "#Result\n",
- "print'Hence proved that Ix=',round(Ix,2),\"kg.m**2\" \n",
- "print'and Iy=',round(Iy,1),\"kg.m**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Hence proved that Ix= 9.38 kg.m**2\n",
- "and Iy= 79.7 kg.m**2\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-37, Page no 323"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "delta=450 #lb/ft**3\n",
- "h1=9*12**-1 #ft\n",
- "h2=10*12**-1 #ft\n",
- "ro1=4*12**-1 #ft\n",
- "ri1=2*12**-1 #ft\n",
- "ro2=18*12**-1 #ft\n",
- "ri2=16*12**-1 #ft\n",
- "a=2.5*24**-1 #ft\n",
- "b=3.5*24**-1 #ft\n",
- "l=1 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "Whub=(pi*ro1**2-pi*ri1**2)*h1*delta #lb\n",
- "Wrim=(pi*ro2**2-pi*ri2**2)*h2*delta #lb\n",
- "#For one spoke\n",
- "Wspoke=(pi*a*b*l*delta) #lb\n",
- "#Moment of inertia calculations\n",
- "Ihub=0.5*(Whub*g**-1)*(ro1**2+ri1**2) #lb-s**2-ft\n",
- "Irim=0.5*(Wrim*g**-1)*(ro2**2+ri2**2) #lb-s**2-ft\n",
- "Ispoke=6*((12**-1)*(Wspoke*g**-1)*l**2+(Wspoke*g**-1)*h2**2) #lb-s**2-ft\n",
- "Iwheel=Ihub+Irim+Ispoke #lb-s**2-ft\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia of the wheel is',round(Iwheel,1),\"lb-s**s-ft\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia of the wheel is 38.1 lb-s**s-ft\n"
- ]
- }
- ],
- "prompt_number": 20
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter15_3.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter15_3.ipynb deleted file mode 100755 index da6b1b6a..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter15_3.ipynb +++ /dev/null @@ -1,510 +0,0 @@ -{
- "metadata": {
- "name": "chapter15.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 15: Moments of Inertia"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-11, Page no 305"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "y1=1 #in\n",
- "y2=4 #in\n",
- "d1=2.2-1 #in\n",
- "d2=4-2.2 #in\n",
- "A1=12 #in**2\n",
- "A2=8 #in**2\n",
- "b1=6 #in\n",
- "b2=2 #in\n",
- "h1=2 #in\n",
- "h2=4 #in\n",
- "\n",
- "#Calculations\n",
- "y_bar=(A1*y1+A2*y2)/(A1+A2) #in\n",
- "I1=(12**-1)*(b1)*(h1**3) #in**4\n",
- "I2=(12**-1)*(b2)*(h2**3) #in**4\n",
- "#Using Parallel Axes Theorem\n",
- "I=(I1+(A1*d1**2))+(I2+(A2*d2**2)) #in**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(I,1),\"in**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 57.9 in**4\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-12, Page no 306"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=60 #mm diameter of the hole\n",
- "#Areas\n",
- "At=100*100 #mm**2\n",
- "Ab=200*100 #mm**2\n",
- "Ac=((pi/4)*d**2) #mm**2\n",
- "bt=100 #mm\n",
- "ht=100 #mm\n",
- "bb=200 #mm\n",
- "hb=100 #mm\n",
- "#Distance of centroids of each area\n",
- "yt=150 #mm\n",
- "yb=50 #mm\n",
- "yc=150 #mm\n",
- "\n",
- "#Calculations\n",
- "y_bar=((At*yt)+(Ab*yb)-(Ac*yc))/(At+Ab-Ac) #mm\n",
- "#Distances\n",
- "dt=yt-y_bar #mm\n",
- "db=y_bar-yb #mm\n",
- "dc=yc-y_bar #mm\n",
- "#Values of Inertia\n",
- "It=(12**-1)*(bt)*(ht**3) #mm**4\n",
- "Ib=(12**-1)*(bb)*(hb**3) #mm**4\n",
- "Ic=(4**-1)*(pi)*((d/2)**4) #mm**4\n",
- "#Moment of inertia\n",
- "I=((It+At*dt**2)+(Ib+Ab*db**2)-(Ic+Ac*dc**2)) #mm**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(I,1),\"mm**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 77156533.6 mm**4\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-14, Page no 308"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "b1=2 #in\n",
- "b2=4 #in\n",
- "h1=8 #in\n",
- "h2=2 #in\n",
- "bo=8 #in\n",
- "ho=8 #in\n",
- "bi=4 #in\n",
- "hi=4 #in\n",
- "\n",
- "#Calculations\n",
- "I1=(12**-1)*(b1)*(h1**3) #in**4\n",
- "I2=(12**-1)*(b2)*(h2**3) #in**4\n",
- "I=2*(I1+I2) #in**4\n",
- "Io=(12**-1)*(bo)*(ho**3) #in**4\n",
- "Ii=(12**-1)*(bi)*(hi**3) #in**4\n",
- "I_bar=Io-Ii #in**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(I_bar),\"in**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 320.0 in**4\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-15, Page no 308"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "b1=75 #mm\n",
- "b2=12 #mm\n",
- "h1=12 #mm\n",
- "h2=162 #mm\n",
- "d1=75 #mm\n",
- "\n",
- "#Calculations\n",
- "A=(h2*b2)+(2*b1*h1) #mm**2\n",
- "I1=(12**-1)*(b1)*(h1**3)+(b1*h1*d1**2) #mm**4\n",
- "I2=(12**-1)*(b2)*(h2**3) #mm**4\n",
- "I_bar=2*I1+I2 #mm**4\n",
- "k=sqrt(I_bar/A) #mm\n",
- "\n",
- "#Result\n",
- "print'The axial moment of inertia is',round(I_bar,1),\"mm**4\"\n",
- "print'The radius of gyration is',round(k,1),\"mm\"\n",
- "\n",
- "# Here value of k is off by 0.1 mm"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The axial moment of inertia is 14398128.0 mm**4\n",
- "The radius of gyration is 62.0 mm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-20, Page no 311"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=50 #mm\n",
- "\n",
- "#Calculations\n",
- "Ixy=(8**-1)*(50**4) #mm**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(Ixy,2),\"mm**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 781250.0 mm**4\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-24, Page no 314"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#The notation has been changed for ease\n",
- "\n",
- "#Calculations\n",
- "x=(5*1*3.5+8*1*0.5)/(5*1+8*1) #in\n",
- "y=(5*1*0.5+8*1*4)/13 #in\n",
- "#Moment of inertia \n",
- "Ix=(12**-1)*(5)*(1**3)+(5*2.15*2.15)+(12**-1)*(1*8**3)+(8*1.35**2) #in**4\n",
- "Iy=(12**-1)*(1)*(5**3)+(5*1.85*1.85)+(12**-1)*(8)*(1**3)+(8*1.15**2) #in**4\n",
- "Ixy=(8*1*(-1.15)*1.35)+(5*1*1.85*(-2.15)) #in**4\n",
- "#Mohr circle calculations\n",
- "d=0.5*(Ix+Iy) #distance to center of the cirlce \n",
- "r=sqrt((21**2)+(32.3**2)) \n",
- "maxI=d+r #in**4\n",
- "theta=arctan(32.3/21)*(180/pi) #degrees maxI occurs at this angle\n",
- "minI=d-r #in**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertias are as follows:'\n",
- "print'Ix=',round(Ix,1),\"in**4\"\n",
- "print'Iy=',round(Iy,1),\"in**4\"\n",
- "print'Ixy=',round(Ixy,1),\"in**4\"\n",
- "print'maxI=',round(maxI,1),\"in**4\"\n",
- "print'minI=',round(minI,1),\"in**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertias are as follows:\n",
- "Ix= 80.8 in**4\n",
- "Iy= 38.8 in**4\n",
- "Ixy= -32.3 in**4\n",
- "maxI= 98.3 in**4\n",
- "minI= 21.2 in**4\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-25, Page no 315"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Notations have been changed\n",
- "\n",
- "#Calculations\n",
- "x=-(25*125*0.5*125+25*100*0.5*25)/(25*125+25*100) #mm\n",
- "y=(25*125*0.5*25+25*100*75)/5625 #mm \n",
- "Iy=(12**-1)*25*125**3+25*125*(62.5-40.3)**2+(12**-1)*100*25**3+100*25*(40.3-12.5)**2 #mm**4\n",
- "Ix=Iy #mm**4 for L-section\n",
- "#The second computation checks the first\n",
- "Ixy=(125*25*22.2*27.8)+(100*25*(-27.8)*(-34.7)) #mm**4\n",
- "#Mohr Circle analysis\n",
- "Imax=Ix+Ixy #mm**4\n",
- "Imin=Ix-Ixy #mm**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertias are as follows:'\n",
- "print'Ix=',round(Ix,2),\"mm**4\"\n",
- "print'Iy=',round(Iy,2),\"mm**4\"\n",
- "print'Ixy=',round(Ixy,2),\"mm**4\"\n",
- "print'Imax=',round(Imax),\"mm**4\"\n",
- "print'Imin=',round(Imin,2),\"mm**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertias are as follows:\n",
- "Ix= 7671443.75 mm**4\n",
- "Iy= 7671443.75 mm**4\n",
- "Ixy= 4340275.0 mm**4\n",
- "Imax= 12011719.0 mm**4\n",
- "Imin= 3331168.75 mm**4\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-30, Page no 320"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "rho=490 #lb/ft**3\n",
- "t=0.02 #in\n",
- "d=4 #in\n",
- "r=d/2 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "W=(pi*r**2*t*rho)*1728**-1 #lb\n",
- "#Mass\n",
- "m=W*g**-1 #slugs\n",
- "#Momemt of inertia\n",
- "I=(4**-1)*m*(r*12**-1)**2 #slug-ft**2\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(I,6),\"slug-ft**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 1.5e-05 slug-ft**2\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-36, Page no 322"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#The integration involves variables hence the direct formula is being used in this coding\n",
- "m=500 #kg\n",
- "R=0.25 #m\n",
- "h=0.5 #m\n",
- "\n",
- "#Calculations\n",
- "Ix=(3*10**-1)*m*R**2 #kg.m**2\n",
- "Iy=(3*5**-1)*m*((4**-1)*R**2+h**2) #kg.m**2\n",
- "\n",
- "#Result\n",
- "print'Hence proved that Ix=',round(Ix,2),\"kg.m**2\" \n",
- "print'and Iy=',round(Iy,1),\"kg.m**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Hence proved that Ix= 9.38 kg.m**2\n",
- "and Iy= 79.7 kg.m**2\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-37, Page no 323"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "delta=450 #lb/ft**3\n",
- "h1=9*12**-1 #ft\n",
- "h2=10*12**-1 #ft\n",
- "ro1=4*12**-1 #ft\n",
- "ri1=2*12**-1 #ft\n",
- "ro2=18*12**-1 #ft\n",
- "ri2=16*12**-1 #ft\n",
- "a=2.5*24**-1 #ft\n",
- "b=3.5*24**-1 #ft\n",
- "l=1 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "Whub=(pi*ro1**2-pi*ri1**2)*h1*delta #lb\n",
- "Wrim=(pi*ro2**2-pi*ri2**2)*h2*delta #lb\n",
- "#For one spoke\n",
- "Wspoke=(pi*a*b*l*delta) #lb\n",
- "#Moment of inertia calculations\n",
- "Ihub=0.5*(Whub*g**-1)*(ro1**2+ri1**2) #lb-s**2-ft\n",
- "Irim=0.5*(Wrim*g**-1)*(ro2**2+ri2**2) #lb-s**2-ft\n",
- "Ispoke=6*((12**-1)*(Wspoke*g**-1)*l**2+(Wspoke*g**-1)*h2**2) #lb-s**2-ft\n",
- "Iwheel=Ihub+Irim+Ispoke #lb-s**2-ft\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia of the wheel is',round(Iwheel,1),\"lb-s**s-ft\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia of the wheel is 38.1 lb-s**s-ft\n"
- ]
- }
- ],
- "prompt_number": 20
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter15_4.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter15_4.ipynb deleted file mode 100755 index da6b1b6a..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter15_4.ipynb +++ /dev/null @@ -1,510 +0,0 @@ -{
- "metadata": {
- "name": "chapter15.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 15: Moments of Inertia"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-11, Page no 305"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "y1=1 #in\n",
- "y2=4 #in\n",
- "d1=2.2-1 #in\n",
- "d2=4-2.2 #in\n",
- "A1=12 #in**2\n",
- "A2=8 #in**2\n",
- "b1=6 #in\n",
- "b2=2 #in\n",
- "h1=2 #in\n",
- "h2=4 #in\n",
- "\n",
- "#Calculations\n",
- "y_bar=(A1*y1+A2*y2)/(A1+A2) #in\n",
- "I1=(12**-1)*(b1)*(h1**3) #in**4\n",
- "I2=(12**-1)*(b2)*(h2**3) #in**4\n",
- "#Using Parallel Axes Theorem\n",
- "I=(I1+(A1*d1**2))+(I2+(A2*d2**2)) #in**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(I,1),\"in**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 57.9 in**4\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-12, Page no 306"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=60 #mm diameter of the hole\n",
- "#Areas\n",
- "At=100*100 #mm**2\n",
- "Ab=200*100 #mm**2\n",
- "Ac=((pi/4)*d**2) #mm**2\n",
- "bt=100 #mm\n",
- "ht=100 #mm\n",
- "bb=200 #mm\n",
- "hb=100 #mm\n",
- "#Distance of centroids of each area\n",
- "yt=150 #mm\n",
- "yb=50 #mm\n",
- "yc=150 #mm\n",
- "\n",
- "#Calculations\n",
- "y_bar=((At*yt)+(Ab*yb)-(Ac*yc))/(At+Ab-Ac) #mm\n",
- "#Distances\n",
- "dt=yt-y_bar #mm\n",
- "db=y_bar-yb #mm\n",
- "dc=yc-y_bar #mm\n",
- "#Values of Inertia\n",
- "It=(12**-1)*(bt)*(ht**3) #mm**4\n",
- "Ib=(12**-1)*(bb)*(hb**3) #mm**4\n",
- "Ic=(4**-1)*(pi)*((d/2)**4) #mm**4\n",
- "#Moment of inertia\n",
- "I=((It+At*dt**2)+(Ib+Ab*db**2)-(Ic+Ac*dc**2)) #mm**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(I,1),\"mm**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 77156533.6 mm**4\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-14, Page no 308"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "b1=2 #in\n",
- "b2=4 #in\n",
- "h1=8 #in\n",
- "h2=2 #in\n",
- "bo=8 #in\n",
- "ho=8 #in\n",
- "bi=4 #in\n",
- "hi=4 #in\n",
- "\n",
- "#Calculations\n",
- "I1=(12**-1)*(b1)*(h1**3) #in**4\n",
- "I2=(12**-1)*(b2)*(h2**3) #in**4\n",
- "I=2*(I1+I2) #in**4\n",
- "Io=(12**-1)*(bo)*(ho**3) #in**4\n",
- "Ii=(12**-1)*(bi)*(hi**3) #in**4\n",
- "I_bar=Io-Ii #in**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(I_bar),\"in**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 320.0 in**4\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-15, Page no 308"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "b1=75 #mm\n",
- "b2=12 #mm\n",
- "h1=12 #mm\n",
- "h2=162 #mm\n",
- "d1=75 #mm\n",
- "\n",
- "#Calculations\n",
- "A=(h2*b2)+(2*b1*h1) #mm**2\n",
- "I1=(12**-1)*(b1)*(h1**3)+(b1*h1*d1**2) #mm**4\n",
- "I2=(12**-1)*(b2)*(h2**3) #mm**4\n",
- "I_bar=2*I1+I2 #mm**4\n",
- "k=sqrt(I_bar/A) #mm\n",
- "\n",
- "#Result\n",
- "print'The axial moment of inertia is',round(I_bar,1),\"mm**4\"\n",
- "print'The radius of gyration is',round(k,1),\"mm\"\n",
- "\n",
- "# Here value of k is off by 0.1 mm"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The axial moment of inertia is 14398128.0 mm**4\n",
- "The radius of gyration is 62.0 mm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-20, Page no 311"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=50 #mm\n",
- "\n",
- "#Calculations\n",
- "Ixy=(8**-1)*(50**4) #mm**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(Ixy,2),\"mm**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 781250.0 mm**4\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-24, Page no 314"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#The notation has been changed for ease\n",
- "\n",
- "#Calculations\n",
- "x=(5*1*3.5+8*1*0.5)/(5*1+8*1) #in\n",
- "y=(5*1*0.5+8*1*4)/13 #in\n",
- "#Moment of inertia \n",
- "Ix=(12**-1)*(5)*(1**3)+(5*2.15*2.15)+(12**-1)*(1*8**3)+(8*1.35**2) #in**4\n",
- "Iy=(12**-1)*(1)*(5**3)+(5*1.85*1.85)+(12**-1)*(8)*(1**3)+(8*1.15**2) #in**4\n",
- "Ixy=(8*1*(-1.15)*1.35)+(5*1*1.85*(-2.15)) #in**4\n",
- "#Mohr circle calculations\n",
- "d=0.5*(Ix+Iy) #distance to center of the cirlce \n",
- "r=sqrt((21**2)+(32.3**2)) \n",
- "maxI=d+r #in**4\n",
- "theta=arctan(32.3/21)*(180/pi) #degrees maxI occurs at this angle\n",
- "minI=d-r #in**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertias are as follows:'\n",
- "print'Ix=',round(Ix,1),\"in**4\"\n",
- "print'Iy=',round(Iy,1),\"in**4\"\n",
- "print'Ixy=',round(Ixy,1),\"in**4\"\n",
- "print'maxI=',round(maxI,1),\"in**4\"\n",
- "print'minI=',round(minI,1),\"in**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertias are as follows:\n",
- "Ix= 80.8 in**4\n",
- "Iy= 38.8 in**4\n",
- "Ixy= -32.3 in**4\n",
- "maxI= 98.3 in**4\n",
- "minI= 21.2 in**4\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-25, Page no 315"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Notations have been changed\n",
- "\n",
- "#Calculations\n",
- "x=-(25*125*0.5*125+25*100*0.5*25)/(25*125+25*100) #mm\n",
- "y=(25*125*0.5*25+25*100*75)/5625 #mm \n",
- "Iy=(12**-1)*25*125**3+25*125*(62.5-40.3)**2+(12**-1)*100*25**3+100*25*(40.3-12.5)**2 #mm**4\n",
- "Ix=Iy #mm**4 for L-section\n",
- "#The second computation checks the first\n",
- "Ixy=(125*25*22.2*27.8)+(100*25*(-27.8)*(-34.7)) #mm**4\n",
- "#Mohr Circle analysis\n",
- "Imax=Ix+Ixy #mm**4\n",
- "Imin=Ix-Ixy #mm**4\n",
- "\n",
- "#Result\n",
- "print'The moment of inertias are as follows:'\n",
- "print'Ix=',round(Ix,2),\"mm**4\"\n",
- "print'Iy=',round(Iy,2),\"mm**4\"\n",
- "print'Ixy=',round(Ixy,2),\"mm**4\"\n",
- "print'Imax=',round(Imax),\"mm**4\"\n",
- "print'Imin=',round(Imin,2),\"mm**4\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertias are as follows:\n",
- "Ix= 7671443.75 mm**4\n",
- "Iy= 7671443.75 mm**4\n",
- "Ixy= 4340275.0 mm**4\n",
- "Imax= 12011719.0 mm**4\n",
- "Imin= 3331168.75 mm**4\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-30, Page no 320"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "rho=490 #lb/ft**3\n",
- "t=0.02 #in\n",
- "d=4 #in\n",
- "r=d/2 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "W=(pi*r**2*t*rho)*1728**-1 #lb\n",
- "#Mass\n",
- "m=W*g**-1 #slugs\n",
- "#Momemt of inertia\n",
- "I=(4**-1)*m*(r*12**-1)**2 #slug-ft**2\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia is',round(I,6),\"slug-ft**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia is 1.5e-05 slug-ft**2\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-36, Page no 322"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#The integration involves variables hence the direct formula is being used in this coding\n",
- "m=500 #kg\n",
- "R=0.25 #m\n",
- "h=0.5 #m\n",
- "\n",
- "#Calculations\n",
- "Ix=(3*10**-1)*m*R**2 #kg.m**2\n",
- "Iy=(3*5**-1)*m*((4**-1)*R**2+h**2) #kg.m**2\n",
- "\n",
- "#Result\n",
- "print'Hence proved that Ix=',round(Ix,2),\"kg.m**2\" \n",
- "print'and Iy=',round(Iy,1),\"kg.m**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Hence proved that Ix= 9.38 kg.m**2\n",
- "and Iy= 79.7 kg.m**2\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.15-37, Page no 323"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "delta=450 #lb/ft**3\n",
- "h1=9*12**-1 #ft\n",
- "h2=10*12**-1 #ft\n",
- "ro1=4*12**-1 #ft\n",
- "ri1=2*12**-1 #ft\n",
- "ro2=18*12**-1 #ft\n",
- "ri2=16*12**-1 #ft\n",
- "a=2.5*24**-1 #ft\n",
- "b=3.5*24**-1 #ft\n",
- "l=1 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "Whub=(pi*ro1**2-pi*ri1**2)*h1*delta #lb\n",
- "Wrim=(pi*ro2**2-pi*ri2**2)*h2*delta #lb\n",
- "#For one spoke\n",
- "Wspoke=(pi*a*b*l*delta) #lb\n",
- "#Moment of inertia calculations\n",
- "Ihub=0.5*(Whub*g**-1)*(ro1**2+ri1**2) #lb-s**2-ft\n",
- "Irim=0.5*(Wrim*g**-1)*(ro2**2+ri2**2) #lb-s**2-ft\n",
- "Ispoke=6*((12**-1)*(Wspoke*g**-1)*l**2+(Wspoke*g**-1)*h2**2) #lb-s**2-ft\n",
- "Iwheel=Ihub+Irim+Ispoke #lb-s**2-ft\n",
- "\n",
- "#Result\n",
- "print'The moment of inertia of the wheel is',round(Iwheel,1),\"lb-s**s-ft\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of inertia of the wheel is 38.1 lb-s**s-ft\n"
- ]
- }
- ],
- "prompt_number": 20
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter16.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter16.ipynb deleted file mode 100755 index 22430a7a..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter16.ipynb +++ /dev/null @@ -1,1911 +0,0 @@ -{
- "metadata": {
- "name": "chapter16.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 16: Dynamics of a Rigid Body in Plane Motion"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-2, Page No 335"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Intilization of variables\n",
- "W=600 #lb\n",
- "d=30 #in\n",
- "# as theta=25 degrees,\n",
- "sintheta=0.422\n",
- "costheta=0.906\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "m=W/g #lb-s**2/ft\n",
- "#Moment of inertia\n",
- "I=0.5*m*((d/2)/12)**2 #lb-s**2-ft\n",
- "#Applying Newtons law and coservation of angular momentum and rolling\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,m,0,0],[0,0,0,1],[((d/2)/12),0,-I,0],[0,1,-((d/2)/12),0]])\n",
- "B=np.array([[W*sintheta],[W*costheta],[0],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The Frictional Force is',round(C[0],1),\"lb\",'and the acceleration is',round(C[1],2),\"ft/s**2\"\n",
- "\n",
- "# The answers wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Frictional Force is 84.4 lb and the acceleration is 9.06 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-3, Page No 336"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=18 #kg\n",
- "d=0.6 #m\n",
- "vo=3 #m/s\n",
- "# as theta=20 degrees,\n",
- "sintheta=0.342\n",
- "costheta=0.939\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of Inertia\n",
- "I=0.5*m*(d/2)**2 \n",
- "#Applying Newtons second Law a\n",
- "A=np.array([[1,m,0,0],[0,0,1,0],[d/2,0,0,-I],[0,1,0,(-d/2)]])\n",
- "B=np.array([[g*m*sintheta],[g*m*costheta],[0],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing the answers in variables\n",
- "F=C[0] #N\n",
- "ax=C[1] #m/s**2\n",
- "Na=C[2] #N\n",
- "alpha=C[3] #rad/s**2\n",
- "#Time Calculations\n",
- "v=0 #m/s**2\n",
- "t=(vo)/ax #s\n",
- "\n",
- "#Result\n",
- "print'It takes',round(t,2),\"s to reach the highest point of travel\"\n",
- "\n",
- "# The ans is off by 0.01 sec."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "It takes 1.34 s to reach the highest point of travel\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-5, Page No 337"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=20 #kg\n",
- "F1=40 #N\n",
- "ro=0.6 #m\n",
- "ri=0.45 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of inertia\n",
- "I=(2*5**-1)*m*ro**2 #kg-m**2\n",
- "#Applying Newtons Law and conservation of angular Momentum\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,m],[ro,-I/ro]])\n",
- "B=np.array([[F1],[F1*ri]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing answers in variables\n",
- "F=C[0] #N\n",
- "a=C[1] #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The acceleration is',round(a,2),\"m/s**2\"\n",
- "print'The force is',round(F,2),\"N\"\n",
- "#The solution in the textbook is incorrect\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration is 0.36 m/s**2\n",
- "The force is 32.86 N\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-6, Page No 338"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=16.1 #lb\n",
- "u=0.10 #co-efficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "F=1.39 #lb\n",
- "\n",
- "#Calculations\n",
- "#Applying Newtons Second Law\n",
- "#Using F=1.39 lb\n",
- "a=((W*sintheta)-F)/(W*g**-1) #ft/s**2\n",
- "alpha=(F*0.5*5/2)/((W*g**-1)*(0.5**2)) #rad/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,1),\"ft/s**2\",'and alpha is ',round(alpha,1),\"rad/s**2.\"\n",
- "print'Hence the sphere will both,roll and slip'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is 13.3 ft/s**2 and alpha is 13.9 rad/s**2.\n",
- "Hence the sphere will both,roll and slip\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-8, Page No 339"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=30 degrees\n",
- "sintheta=2**-1\n",
- "W=80 #lb\n",
- "Ww=100 #lb\n",
- "I=4 #slug-ft**2\n",
- "r=0.5 #ft\n",
- "v= 20 #ft/s\n",
- "vo=0 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using Equations of motion\n",
- "#Solving the system of linear equatinons by matrix method\n",
- "A=np.array([[-1,0,-W/g],[1,-1,-Ww/g],[0,r,-2*I]])\n",
- "B=np.array([[-W],[Ww*sintheta],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing values in variables\n",
- "T=C[0] #lb\n",
- "F=C[1] #lb\n",
- "a=C[2] #ft/s**2\n",
- "#Time calculations\n",
- "t=(v-vo)/a #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t,1),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 14.4 s\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-9, Page No 340"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "M=70 #kg\n",
- "ko=0.4 #m\n",
- "ri=0.45 #m\n",
- "ro=0.6 #m\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "m=35 #kg\n",
- "g= 9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "I=M*ko**2 #kg-m**2\n",
- "#Using Equations of motion\n",
- "#Solving the equations by matrix method\n",
- "A=np.array([[-1,-m*0.15,0],[1,-M*ro,-1],[-ri,-I,ro]])\n",
- "B=np.array([[-m*g],[M*g*sintheta],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "F=C[2] #N\n",
- "Na=M*g*costheta #N\n",
- "#Required coefficient of friction\n",
- "u=F/Na #coefficient of friction\n",
- "\n",
- "#Result\n",
- "print'The value of alpha is',round(C[1],2),\"rad/s**2\",'and tension is',round(C[0]),\"N\"\n",
- "print'F=',round(F),\"N,\",'Na=',round(Na),\"N\",'and u=',round(u,2),\"(coefficient of friction)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of alpha is -4.15 rad/s**2 and tension is 365.0 N\n",
- "F= 196.0 N, Na= 594.0 N and u= 0.33 (coefficient of friction)\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Exampe 16.16-10, Page No 341"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=200 #kg\n",
- "g=9.8 #m/s**2\n",
- "r=1.2 #m\n",
- "F1=1000 #N\n",
- "F2=1400 #N\n",
- "\n",
- "#Calculations\n",
- "N=m*g #N\n",
- "I=(2*5**-1)*(m)*r**2 #kg-m**2\n",
- "#Using equations of motion\n",
- "#Solving for F and alpha using matrix method\n",
- "#Applying equations of motion\n",
- "A=np.array([[1,-m],[-r,-I/r]])\n",
- "B=np.array([[F1-F2],[F1*r]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing values\n",
- "F=C[0] #N\n",
- "alpha=C[1] #rad/s**2\n",
- "a=r*alpha #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,2),\"m/s**2\",'and F is',round(F),\"N\"\n",
- "#The negative signs indicate that the direction is opposite to what was origninally assumed\n",
- "# The answers wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is -2.57 m/s**2 and F is -829.0 N\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-11, Page No 341"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "Wa=161 #lb\n",
- "Wb=193.2 #lb\n",
- "Wc=300 #lb\n",
- "ka=3 #ft\n",
- "kb=2.5 #ft\n",
- "# as theta1=30 degrees & theta2=45 degrees,\n",
- "sintheta1=2**-1\n",
- "sintheta2=(2**0.5)**-1\n",
- "costheta2=(2**0.5)**-1\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of inertia Calculations\n",
- "Ia=(Wa/g)*ka**2 #lb-s**2-ft\n",
- "Ib=(Wb/g)*kb**2 #lb-s**2-ft\n",
- "#Using equations of motion for A and B and C\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,1,-Wa/g,0,0],[1,-4,-Ia*(4**-1),0,0],[-2,0,-Ib*(5*8**-1),4,0],[0,0,-(Wc/g)*(5*2**-1),-1,-0.25],[0,0,0,0,1]])\n",
- "B=np.array([[Wa*sintheta1],[0],[0],[-Wc*costheta2],[Wc*sintheta2]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing values in the variables\n",
- "T1=C[0] #lb\n",
- "T2=C[3] #lb\n",
- "a=C[2] #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The values are:'\n",
- "print'a=',round(a,2),\"ft/s**2 ,\",'T1=',round(T1),\"lb\",'and T2=',round(T2,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are:\n",
- "a= 3.93 ft/s**2 , T1= 89.0 lb and T2= 67.5 lb\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-12, Page No 342"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=644 #lb\n",
- "F=30 #lb\n",
- "# as theta=30 degrees,\n",
- "sintheta=(2)**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "r=1.5 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,-W/g],[-r,-(2**-1)*(W/g)*(2*2)*(r**-1)]])\n",
- "B=np.array([[(W*sintheta)-(F*costheta)],[-F*2]])\n",
- "C=np.linalg.solve(A,B)\n",
- "a=C[1] #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,2),\"ft/s**2\"\n",
- "# The negative sign indicates that the cylinder will roll down the plane."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is -6.78 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-14, Page No 344"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=20 #lb\n",
- "g=32.2 #ft/s**2\n",
- "vb=0.5 #rad/s\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "#Solving the three equations simultaneously by matrix method\n",
- "X=np.array([[0,1,-(W*g**-1)*5.2],[-1,0,-(W*g**-1)*3],[3,-3,-(12**-1)*(W*g**-1)*12**2]])\n",
- "Y=np.array([[-0.75*(W*g**-1)],[(W*g**-1)*1.3-W],[0]])\n",
- "C=np.linalg.solve(X,Y)\n",
- "A=C[0] #lb\n",
- "B=C[1] #lb\n",
- "alpha=C[2] #rad/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of alpha is',round(alpha,1),\"rad/s**2 countercockwise\",'and of A and B are',round(A,1),\"lb up and\",round(B,2),\"lb to the right\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of alpha is 2.6 rad/s**2 countercockwise and of A and B are 14.4 lb up and 7.91 lb to the right\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-16, Page No 345"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mc=7.25 #kg\n",
- "d=0.9 #m\n",
- "la=0.2 #m\n",
- "ma=9 #kg\n",
- "F=45 #N\n",
- "ay=0 #m/s**2\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "I=2*(0.5*mc*(d*2**-1)**2)+0.5*ma*(la*2**-1)**2 #kg-m**2\n",
- "#Using the equations of motion\n",
- "Na=(2*mc+ma)*g #N\n",
- "#Simplfying using radial velocity formula\n",
- "#Solving the two equations using matrix method\n",
- "A=np.array([[-1,-(2*mc+ma)],[(d*2**-1),-I/(d*2**-1)]])\n",
- "B=np.array([[-F],[F*(la*2**-1)]])\n",
- "C=np.linalg.solve(A,B)\n",
- "F=C[0] #N\n",
- "ax=C[1] #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The computation yields ax=',round(ax,2),\"m/s**2 to the right.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The computation yields ax= 1.13 m/s**2 to the right.\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-18, Page No 347"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=0.05 #m cylinder radius\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Here the equation has been solved in terms of the veriables\n",
- "#Hence we directly consider the final result\n",
- "av=(2*g)/3 #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of av is',round(av,2),\"m/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of av is 6.53 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-21, Page No 349"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#initilization of variables\n",
- "W=16.1 #lb\n",
- "v=9 #ft/s\n",
- "# as phi=30 degrees,\n",
- "sinphi=(2)**-1\n",
- "cosphi=(3**0.5)*2**-1\n",
- "r=0.5 #ft\n",
- "g=32.2 #ft/s**2\n",
- "OG=4.5 #ft\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "an=v**2/OG #ft/s**2\n",
- "#Solving for alpha we get\n",
- "N=(W*g**-1)*an+W*cosphi #lb\n",
- "#Using equations of motion\n",
- "A=np.array([[1,-r],[-1,-r*r]])\n",
- "B=np.array([[W*sinphi],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "F=C[0] #lb\n",
- "at=C[1] #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of N and F are',round(N,1),\"lb and\",round(F,2),\"lb respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of N and F are 22.9 lb and 2.68 lb respectively.\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-25, Page No 353"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=50 #lb\n",
- "P=10 #lb\n",
- "t=5 #s\n",
- "vo=0 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "ax=(P*g)/W #ft/s**2\n",
- "#Solving by matrix method for A and B\n",
- "F=np.array([[1,1],[-4,4]])\n",
- "Q=np.array([[W],[P]])\n",
- "R=np.linalg.solve(F,Q)\n",
- "#Velocity calculations\n",
- "v=vo+ax*t #ft/s\n",
- "A=R[0] #lb\n",
- "B=R[1] #lb\n",
- "\n",
- "#Result\n",
- "print'The velocity of the door after 5s is',round(v,1),\"ft/s\",'and A=',round(A,1),\"lb\",'and B=',round(B,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity of the door after 5s is 32.2 ft/s and A= 23.8 lb and B= 26.3 lb\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-26, Page No 354"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "#Initilization of variables\n",
- "AB=2 #m\n",
- "m=2 #kg\n",
- "F=20 #N\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equation of motion\n",
- "a=F/m #m/s**2\n",
- "#Solving by matrix method for Na and Nb\n",
- "A=np.array([[1,-1],[4*5**-1,4*5**-1]])\n",
- "B=np.array([[m*g],[F*(3*5**-1)]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a),\"m/s**2\",'and the reactions are: Na=',round(C[0],1),\"N up\",'and Nb=',round(C[1],1),\"N up\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is 10.0 m/s**2 and the reactions are: Na= 17.3 N up and Nb= -2.3 N up\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-27, Page No 354"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "vo=0 #ft/s\n",
- "\n",
- "#Calculations\n",
- "s=(0.011*5280*2)/(2*0.004)\n",
- "\n",
- "#Result\n",
- "print'It travels',round(s),\"ft along the level before coming to rest\"\n",
- "#Answer in the textbook is incorrect by 20ft\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "It travels 14520.0 ft along the level before coming to rest\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-28, Page No 355"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "u=0.3 #coefficient of friction\n",
- "m=70 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#CASE 1\n",
- "#Using equations of motion\n",
- "Na=m*g #N\n",
- "ah=(u*Na)/m #m/s**2\n",
- "#CASE 2\n",
- "#Applying sum of moments equal to zero\n",
- "F=(Na*0.3)/1.2 #N\n",
- "a_h=F/m #m/s**2\n",
- "\n",
- "#Result\n",
- "#Intutive insights can be attained after we get these results\n",
- "print'The value of Na is',round(Na),\"N\"\n",
- "print'and that of acceleration are:'\n",
- "print'1st value=',round(ah,2),\"m/s**2\",'and 2nd value is',round(a_h,2),\"m/s**2\"\n",
- "print'and the value of F is',round(F),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of Na is 686.0 N\n",
- "and that of acceleration are:\n",
- "1st value= 2.94 m/s**2 and 2nd value is 2.45 m/s**2\n",
- "and the value of F is 172.0 N\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-29, Page No 356"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=60 #kg\n",
- "me=660 #kg\n",
- "a=6 #m/s**2\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "P=m*a+m*g #N\n",
- "#Scale reading\n",
- "R=P/g #kg\n",
- "#Increase in mass\n",
- "I=R-m #kg\n",
- "#Tension\n",
- "T=me*a+me*g #N\n",
- "\n",
- "#Result\n",
- "print'The value of P is',round(P),\"n\"\n",
- "print'The apparent icrease in weight is',round(I,1),\"kg\",'and the tension in the cable is',round(T),\"N\"\n",
- "#Answer in the textbook is off by 28 #N in Tension\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P is 948.0 n\n",
- "The apparent icrease in weight is 36.7 kg and the tension in the cable is 10428.0 N\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-30, Page No 356"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "u=0.2 #coefficient of friction\n",
- "ma=1.2 #kg\n",
- "mb=2 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "Nb=mb*g #N\n",
- "F=u*Nb #N\n",
- "#Using equations of motion\n",
- "#Solving for T and a\n",
- "A=np.array([[-1,-ma],[1,-mb]])\n",
- "B=np.array([[-ma*g],[F]])\n",
- "C=np.linalg.solve(A,B)\n",
- "T=C[0] #N\n",
- "a=C[1] #m/s**2\n",
- "#Taking the sum of the moments\n",
- "x_m=-(F*0.15+T*0.15)/Nb #m\n",
- "x=x_m*1000 #mm\n",
- "\n",
- "#Result\n",
- "print'The acceleration of block A is',round(a,2),\"m/s**2\" \n",
- "print'and Nb acts at a distance of',round(x,1),\"mm.(Negative sign indictaes that the side assumed is incorrect)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration of block A is 2.45 m/s**2\n",
- "and Nb acts at a distance of -97.5 mm.(Negative sign indictaes that the side assumed is incorrect)\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-31, Page No 357"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "a=2.5 #m/s**2\n",
- "mA=3 #kg\n",
- "mB=7 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "F=(mA+mB)*a #N\n",
- "#Using equations of motion\n",
- "Py=mB*g #N\n",
- "#Solving for Px and H\n",
- "A=np.array([[1,1],[-0.0375,0.0375]])\n",
- "B=np.array([[mB*a],[Py*0.05]])\n",
- "C=np.linalg.solve(A,B)\n",
- "Px=C[0] #N\n",
- "H=C[1] #N\n",
- "\n",
- "#Result\n",
- "print'The value of H is',round(H,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of H is 54.5 N\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-32, Page No 358"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=20 #kg\n",
- "g=9.8 #m/s**2\n",
- "vo=3 #m/s\n",
- "v=0 #m/s\n",
- "s=4 #m\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "Na=m*g #N\n",
- "F=(Na*0.075)/0.125 #N\n",
- "a=F/m #m/s**2\n",
- "#Displacement \n",
- "d=-(v**2-vo**2)/(2*a) #m\n",
- "displ=s-d #m\n",
- "v_f=sqrt(2*a*displ) #m/s\n",
- "\n",
- "#Result\n",
- "print'The final velocity is',round(v_f,2),\"m/s to the left.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The final velocity is 6.17 m/s to the left.\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-33, Page No 358"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mA=30 #kg\n",
- "mB=45 #kg\n",
- "u_ab=3**-1 #coefficient of friction between two blocks\n",
- "u_bp=10**-1 #coefficient of friction between block and horizontal plane\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#By inspection\n",
- "Na=mA*g #N\n",
- "Nb=Na+mB*g #N\n",
- "a=(u_ab*Na-u_bp*Nb)/mB #m/s**2\n",
- "P=(mA*a+u_ab*Na) #N\n",
- "#For block A\n",
- "#Solving for P,F and a\n",
- "A=np.array([[1,-1,-mA],[-0.05,-0.075,0],[0,1,-mB]])\n",
- "B=np.array([[0],[-Na*0.050],[Nb*u_bp]])\n",
- "C=np.linalg.solve(A,B)\n",
- "P_new=C[0] #N\n",
- "\n",
- "#Result\n",
- "#As p < p_new\n",
- "print'The maximum value of P is',round(P),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum value of P is 114.0 N\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-34, Page No 359"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Vo=1.5 #m/s\n",
- "V=0 #m/s\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "a=(g*0.2)/0.75 #m/s**2\n",
- "t=-(V-Vo)/a #s\n",
- "\n",
- "#Result\n",
- "print'The maximum acceleration is',round(a,2),\"m/s**2\",'and minimum time is',round(t,2),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum acceleration is 2.61 m/s**2 and minimum time is 0.57 s\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-36, Page No 361"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "vo=0 #mi/h\n",
- "v=60 #mi/h\n",
- "t=13.8 #s\n",
- "W=3385 #lb\n",
- "xb=46 #in\n",
- "xf=66 #in\n",
- "xv=31 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "a=(((v*88*60)/3600)-vo)/t #ft/s**2\n",
- "#Summing horizontal forces\n",
- "F=(W/g)*a #lb\n",
- "#Solving for Rf and Rr\n",
- "A=np.array([[1,1],[-xf,xb]])\n",
- "B=np.array([[W],[-F*xv]])\n",
- "C=np.linalg.solve(A,B)\n",
- "Rr=C[0] #lb\n",
- "Rf=C[1] #lb\n",
- "\n",
- "#Result\n",
- "print'The value of reactions are Rf=',round(Rf),\"lb\",'and Rr=',round(Rr),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of reactions are Rf= 1809.0 lb and Rr= 1576.0 lb\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-43, Page No 366"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=161 #lb\n",
- "F=16.1 #lb\n",
- "r=18 #ft radius\n",
- "t=2 #s\n",
- "g=32.2 #ft/s**2\n",
- "wo=0 #rad/s\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "#Solving for T and alpha\n",
- "A=np.array([[r*12**-1,-0.5*(W*g**-1)*(r*12**-1)**2],[-1,-F*g**-1]])\n",
- "B=np.array([[0],[-F]])\n",
- "C=np.linalg.solve(A,B)\n",
- "alpha=C[1] #rad/s**2\n",
- "w=wo+(alpha*t) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular speed is',round(w,2),\"rad/s\"\n",
- "\n",
- "#The ans is incorrect in textbook."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular speed is 7.58 rad/s\n"
- ]
- }
- ],
- "prompt_number": 43
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-47, Page No 369"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization fo variables\n",
- "r=2000 #ft\n",
- "g=32.2 #ft/s**2\n",
- "d=4.71 #ft\n",
- "v=176 #ft/s\n",
- "\n",
- "#Calculations\n",
- "e=(d*v**2)/(g*r) #ft\n",
- "\n",
- "#Result\n",
- "print'The superelevation is',round(e,2),\"ft\"\n",
- "#Watch the unit in the final answer\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The superelevation is 2.27 ft\n"
- ]
- }
- ],
- "prompt_number": 45
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-48, Page No 369"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "a=5 #ft/s**2\n",
- "C=50 #lb-ft\n",
- "W=161 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "T=0.5*(W/g)*1**2*a+C #lb\n",
- "Ox=-T*(2/sqrt(a)) #lb\n",
- "Oy=T*(1/sqrt(a))+W #lb\n",
- "Wa=T/(1-(a/g)) #lb\n",
- "\n",
- "#Result\n",
- "print'The values are: T=',round(T,1),\"lb\",', Wa=',round(Wa),\"b\",',Ox=',round(Ox,1),\"lb\",'and Oy=+',round(Oy),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are: T= 62.5 lb , Wa= 74.0 b ,Ox= -55.9 lb and Oy=+ 189.0 lb\n"
- ]
- }
- ],
- "prompt_number": 49
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-49, Page No 370"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=100 #kg\n",
- "mr=20 #kg\n",
- "w=8 #rad/s\n",
- "l1=300 #mm\n",
- "l2=600 #mm\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "r_bar=(mr*l1+m*750)/120 #mm\n",
- "I=(3**-1)*mr*(l2*1000**-1)**2+(2*5**-1)*m*(l1*2000**-1)**2+m*(0.75)**2 #kg.m**2\n",
- "alpha=(m+mr)*g*(r_bar*1000**-1)/I #rad/s**2\n",
- "On=(m+mr)*(r_bar*1000**-1)*w**2 #N\n",
- "Ot=((m+mr)*(r_bar*1000**-1)*alpha)-(m+mr)*g #N\n",
- "\n",
- "#Result\n",
- "print'The angular acceleration is',round(alpha,1),\"rad/s**2\",'and On=',round(On),\"N\",'and Ot=',round(Ot,1),\"N\"\n",
- "\n",
- "#Due to decimal accuracy there is discrepancy in answers with the textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular acceleration is 13.3 rad/s**2 and On= 5184.0 N and Ot= -96.3 N\n"
- ]
- }
- ],
- "prompt_number": 48
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-50, Page No 370"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=40 #lb\n",
- "w=10 #rad/s\n",
- "alpha=2 #rad/s**2\n",
- "r=2 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "On=(W*g**-1)*(1*6**-1)*w**2 #lb\n",
- "Ot=(W*g**-1)*(1*6**-1)*alpha\n",
- "Io=(0.5*(W*g**-1)*0.5**2)*2+((W*g**-1)*(1*6**-1)**2)*2\n",
- "\n",
- "#Result\n",
- "print'The reaction components are On=',round(On,1),\"lb to the right\",'and Ot=',round(Ot,2),\"lb up\"\n",
- "print'The value of Io is',round(Io,2),\"lb-ft\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction components are On= 20.7 lb to the right and Ot= 0.41 lb up\n",
- "The value of Io is 0.38 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 59
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-51, Page No 371"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilizatin of variables\n",
- "W=6 #lb\n",
- "l=8 #ft\n",
- "v=10 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "# as theta1=60 degrees & theta2=30 degrees\n",
- "costheta1=2**-1\n",
- "costheta2=(3**0.5)*2**-1\n",
- "\n",
- "#Calculations\n",
- "Fe=(W*v**2)*(g*l*0.5)**-1 #lb\n",
- "#Using equations of motion\n",
- "#Solving for C and T\n",
- "A=np.array([[costheta1,-costheta2],[costheta2,costheta1]])\n",
- "B=np.array([[-Fe],[W]])\n",
- "P=np.linalg.solve(A,B) #lb\n",
- "C=P[0] #lb\n",
- "T=P[1] #lb\n",
- "\n",
- "#Result\n",
- "print'The value of C is',round(C,2),\"lb\",'and T is',round(T,2),\"lb\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of C is 2.87 lb and T is 7.03 lb\n"
- ]
- }
- ],
- "prompt_number": 49
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-52, Page No 372"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=32.2 #lb\n",
- "T=120 #lb\n",
- "m=1 #slug\n",
- "r=6*12**-1 #ft\n",
- "\n",
- "#Calculations\n",
- "w=sqrt((T*(3*5**-1)*4)*(m*r*3)**-1) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular speed permissible is',round(w,1),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular speed permissible is 13.9 rad/s\n"
- ]
- }
- ],
- "prompt_number": 50
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-53, Page No 372"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=30 #kg\n",
- "k=0.45 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Using equations of motion\n",
- "#Solving for T1,T2 and alpha\n",
- "A=np.array([[1,0,-m],[0,-1,-45],[-0.6,0.3,-m*k**2]])\n",
- "B=np.array([[50*g],[-150*g],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The values are: T1=',round(C[0]),\"N\",',T2=',round(C[1]),\"N\",'and alpha=',round(C[2],1),\"rad/s**2\"\n",
- "\n",
- "# The answer for T2 and alpha is off by 4 & 0.1 units"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are: T1= 607.0 N ,T2= 1294.0 N and alpha= 3.9 rad/s**2\n"
- ]
- }
- ],
- "prompt_number": 51
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-54, Page No 373"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wc=28 #lb\n",
- "v=16 #ft/s\n",
- "Ib=12 #ft-lb-s**2\n",
- "u=0.4 #coefficient of friction\n",
- "t=2 #s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "T=Wc+(Wc*g**-1)*8 #lb\n",
- "alpha=(8*12)*15**-1 #rad/s**2\n",
- "F=((Ib*alpha)+(T*1.25))/t #lb\n",
- "N=F/u #lb\n",
- "#Summing moments about D\n",
- "P=(N*8+F*3)/40 #lb\n",
- "#Summing forces horizontally and vertically\n",
- "Dx=151-P #lb\n",
- "Dy=-F #lb\n",
- "\n",
- "#Result\n",
- "print'The reactions at D are: Dx=',round(Dx),\"lb to right\",'and Dy=',round(Dy,1),\"lb down\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reactions at D are: Dx= 116.0 lb to right and Dy= -60.2 lb down\n"
- ]
- }
- ],
- "prompt_number": 65
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-55, Page No 374"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=8 #kg\n",
- "n=90 #rpm\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "Fg=m*g #N\n",
- "w=2*pi*n/60 #rad/s\n",
- "#using equations of motion\n",
- "By=m*g #N\n",
- "#Solving for Bx and C\n",
- "A=np.array([[1,1],[-0.3,0.9]])\n",
- "B=np.array([[m*0.3*w**2],[By*0.3]])\n",
- "C=np.linalg.solve(A,B) #N\n",
- "\n",
- "#Result\n",
- "print'The solution is: Bx=',round(C[0]),\"N\",',By=',round(By,1),\"N\",'and C=',round(C[1],1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is: Bx= 140.0 N ,By= 78.4 N and C= 72.9 N\n"
- ]
- }
- ],
- "prompt_number": 79
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-56, Page No 375"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=8 #kg\n",
- "n=90 #rpm\n",
- "g=9.8 #m/s**2\n",
- "r=0.3 #m\n",
- "\n",
- "#calculations\n",
- "w=2*pi*n/60 #rad/s\n",
- "#Using equations of motion\n",
- "C=(m*g*0.3+m*r*w**2*r)/1.2 #N\n",
- "Bx=-C+m*r*w**2 #N\n",
- "By=m*g #N\n",
- "\n",
- "#Result\n",
- "print'The solution is: Bx=',round(Bx),\"N\",',By=',round(By,1),\"N\",'and C=',round(C,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is: Bx= 140.0 N ,By= 78.4 N and C= 72.9 N\n"
- ]
- }
- ],
- "prompt_number": 66
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-57, Page No 375"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Na=294 #N\n",
- "Nb=735 #N\n",
- "\n",
- "#Calculations\n",
- "a=(10**-1*Nb-3**-1*Na)/45 #m/s**2\n",
- "P=(3**-1*Na)-30*a #N\n",
- "\n",
- "#Result\n",
- "print'The solution is P=',round(P),\"N\",'and a=',round(a,3),\"m/s**2\"\n",
- "# The negative sign indicates the assumed direction is incorrect."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is P= 114.0 N and a= -0.544 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 68
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-58, Page No 376"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=50 #lb\n",
- "g=32.2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "a=(10/(W/g)) #ft/s**2\n",
- "B=((2.5*(W/g)*a)+4*W-1.5*10)/8 #lb\n",
- "A=50-B #lb\n",
- "\n",
- "#Result\n",
- "print'The solution is A=',round(A,1),\"lb\",',B=',round(B,1),\"lb\",'and a=',round(a,2),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is A= 23.8 lb ,B= 26.3 lb and a= 6.44 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 69
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-60, Page No 377"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "g=9.8 #m/s**2\n",
- "r1=0.3 #m\n",
- "m1=20 #kg\n",
- "m2=100 #kg\n",
- "r2=0.75 #m\n",
- "\n",
- "#Calculations\n",
- "alpha=(m1*g*r1+m2*g*r2)*(m1*r1**2+(m1/12)*0.6**2+m2*r2**2+(2*5**-1)*m2*0.15**2)**-1 #rad/s**2\n",
- "\n",
- "#Result\n",
- "print'The angular acceleration is',round(alpha,1),\"rad/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular acceleration is 13.4 rad/s**2\n"
- ]
- }
- ],
- "prompt_number": 71
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-61, Page No 378"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=15*12**-1 #ft\n",
- "W=600 #lb\n",
- "# as theta=25 degrees,\n",
- "sintheta=0.422\n",
- "\n",
- "#calculations\n",
- "ax=(r*W*sintheta)*((r**-1)*14.5+r*18.6)**-1 #ft/s**2\n",
- "F=(W*sintheta)-(18.6*ax) #lb\n",
- "\n",
- "#Result\n",
- "print'The solution is F=',round(F,1),\"lb\",'and ax=',round(ax,2),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is F= 84.3 lb and ax= 9.08 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-62, Page No 378"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=7 #kg\n",
- "g=9.8 #m/s**2\n",
- "r=0.5 #m\n",
- "I=0.875 #kg.m**2\n",
- "\n",
- "#Calculations\n",
- "#Solving for alpha and T\n",
- "alpha=(m*g*r)/(I+m*r*0.5) #rad/s**2\n",
- "T=(I*alpha)/r #N\n",
- "\n",
- "#Result\n",
- "print'The soultion is alpha=',round(alpha,1),\"rad/s**2\",'and T=',round(T,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The soultion is alpha= 13.1 rad/s**2 and T= 22.9 N\n"
- ]
- }
- ],
- "prompt_number": 79
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter16_1.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter16_1.ipynb deleted file mode 100755 index 22430a7a..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter16_1.ipynb +++ /dev/null @@ -1,1911 +0,0 @@ -{
- "metadata": {
- "name": "chapter16.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 16: Dynamics of a Rigid Body in Plane Motion"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-2, Page No 335"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Intilization of variables\n",
- "W=600 #lb\n",
- "d=30 #in\n",
- "# as theta=25 degrees,\n",
- "sintheta=0.422\n",
- "costheta=0.906\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "m=W/g #lb-s**2/ft\n",
- "#Moment of inertia\n",
- "I=0.5*m*((d/2)/12)**2 #lb-s**2-ft\n",
- "#Applying Newtons law and coservation of angular momentum and rolling\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,m,0,0],[0,0,0,1],[((d/2)/12),0,-I,0],[0,1,-((d/2)/12),0]])\n",
- "B=np.array([[W*sintheta],[W*costheta],[0],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The Frictional Force is',round(C[0],1),\"lb\",'and the acceleration is',round(C[1],2),\"ft/s**2\"\n",
- "\n",
- "# The answers wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Frictional Force is 84.4 lb and the acceleration is 9.06 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-3, Page No 336"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=18 #kg\n",
- "d=0.6 #m\n",
- "vo=3 #m/s\n",
- "# as theta=20 degrees,\n",
- "sintheta=0.342\n",
- "costheta=0.939\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of Inertia\n",
- "I=0.5*m*(d/2)**2 \n",
- "#Applying Newtons second Law a\n",
- "A=np.array([[1,m,0,0],[0,0,1,0],[d/2,0,0,-I],[0,1,0,(-d/2)]])\n",
- "B=np.array([[g*m*sintheta],[g*m*costheta],[0],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing the answers in variables\n",
- "F=C[0] #N\n",
- "ax=C[1] #m/s**2\n",
- "Na=C[2] #N\n",
- "alpha=C[3] #rad/s**2\n",
- "#Time Calculations\n",
- "v=0 #m/s**2\n",
- "t=(vo)/ax #s\n",
- "\n",
- "#Result\n",
- "print'It takes',round(t,2),\"s to reach the highest point of travel\"\n",
- "\n",
- "# The ans is off by 0.01 sec."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "It takes 1.34 s to reach the highest point of travel\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-5, Page No 337"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=20 #kg\n",
- "F1=40 #N\n",
- "ro=0.6 #m\n",
- "ri=0.45 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of inertia\n",
- "I=(2*5**-1)*m*ro**2 #kg-m**2\n",
- "#Applying Newtons Law and conservation of angular Momentum\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,m],[ro,-I/ro]])\n",
- "B=np.array([[F1],[F1*ri]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing answers in variables\n",
- "F=C[0] #N\n",
- "a=C[1] #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The acceleration is',round(a,2),\"m/s**2\"\n",
- "print'The force is',round(F,2),\"N\"\n",
- "#The solution in the textbook is incorrect\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration is 0.36 m/s**2\n",
- "The force is 32.86 N\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-6, Page No 338"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=16.1 #lb\n",
- "u=0.10 #co-efficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "F=1.39 #lb\n",
- "\n",
- "#Calculations\n",
- "#Applying Newtons Second Law\n",
- "#Using F=1.39 lb\n",
- "a=((W*sintheta)-F)/(W*g**-1) #ft/s**2\n",
- "alpha=(F*0.5*5/2)/((W*g**-1)*(0.5**2)) #rad/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,1),\"ft/s**2\",'and alpha is ',round(alpha,1),\"rad/s**2.\"\n",
- "print'Hence the sphere will both,roll and slip'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is 13.3 ft/s**2 and alpha is 13.9 rad/s**2.\n",
- "Hence the sphere will both,roll and slip\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-8, Page No 339"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=30 degrees\n",
- "sintheta=2**-1\n",
- "W=80 #lb\n",
- "Ww=100 #lb\n",
- "I=4 #slug-ft**2\n",
- "r=0.5 #ft\n",
- "v= 20 #ft/s\n",
- "vo=0 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using Equations of motion\n",
- "#Solving the system of linear equatinons by matrix method\n",
- "A=np.array([[-1,0,-W/g],[1,-1,-Ww/g],[0,r,-2*I]])\n",
- "B=np.array([[-W],[Ww*sintheta],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing values in variables\n",
- "T=C[0] #lb\n",
- "F=C[1] #lb\n",
- "a=C[2] #ft/s**2\n",
- "#Time calculations\n",
- "t=(v-vo)/a #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t,1),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 14.4 s\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-9, Page No 340"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "M=70 #kg\n",
- "ko=0.4 #m\n",
- "ri=0.45 #m\n",
- "ro=0.6 #m\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "m=35 #kg\n",
- "g= 9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "I=M*ko**2 #kg-m**2\n",
- "#Using Equations of motion\n",
- "#Solving the equations by matrix method\n",
- "A=np.array([[-1,-m*0.15,0],[1,-M*ro,-1],[-ri,-I,ro]])\n",
- "B=np.array([[-m*g],[M*g*sintheta],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "F=C[2] #N\n",
- "Na=M*g*costheta #N\n",
- "#Required coefficient of friction\n",
- "u=F/Na #coefficient of friction\n",
- "\n",
- "#Result\n",
- "print'The value of alpha is',round(C[1],2),\"rad/s**2\",'and tension is',round(C[0]),\"N\"\n",
- "print'F=',round(F),\"N,\",'Na=',round(Na),\"N\",'and u=',round(u,2),\"(coefficient of friction)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of alpha is -4.15 rad/s**2 and tension is 365.0 N\n",
- "F= 196.0 N, Na= 594.0 N and u= 0.33 (coefficient of friction)\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Exampe 16.16-10, Page No 341"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=200 #kg\n",
- "g=9.8 #m/s**2\n",
- "r=1.2 #m\n",
- "F1=1000 #N\n",
- "F2=1400 #N\n",
- "\n",
- "#Calculations\n",
- "N=m*g #N\n",
- "I=(2*5**-1)*(m)*r**2 #kg-m**2\n",
- "#Using equations of motion\n",
- "#Solving for F and alpha using matrix method\n",
- "#Applying equations of motion\n",
- "A=np.array([[1,-m],[-r,-I/r]])\n",
- "B=np.array([[F1-F2],[F1*r]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing values\n",
- "F=C[0] #N\n",
- "alpha=C[1] #rad/s**2\n",
- "a=r*alpha #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,2),\"m/s**2\",'and F is',round(F),\"N\"\n",
- "#The negative signs indicate that the direction is opposite to what was origninally assumed\n",
- "# The answers wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is -2.57 m/s**2 and F is -829.0 N\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-11, Page No 341"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "Wa=161 #lb\n",
- "Wb=193.2 #lb\n",
- "Wc=300 #lb\n",
- "ka=3 #ft\n",
- "kb=2.5 #ft\n",
- "# as theta1=30 degrees & theta2=45 degrees,\n",
- "sintheta1=2**-1\n",
- "sintheta2=(2**0.5)**-1\n",
- "costheta2=(2**0.5)**-1\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of inertia Calculations\n",
- "Ia=(Wa/g)*ka**2 #lb-s**2-ft\n",
- "Ib=(Wb/g)*kb**2 #lb-s**2-ft\n",
- "#Using equations of motion for A and B and C\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,1,-Wa/g,0,0],[1,-4,-Ia*(4**-1),0,0],[-2,0,-Ib*(5*8**-1),4,0],[0,0,-(Wc/g)*(5*2**-1),-1,-0.25],[0,0,0,0,1]])\n",
- "B=np.array([[Wa*sintheta1],[0],[0],[-Wc*costheta2],[Wc*sintheta2]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing values in the variables\n",
- "T1=C[0] #lb\n",
- "T2=C[3] #lb\n",
- "a=C[2] #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The values are:'\n",
- "print'a=',round(a,2),\"ft/s**2 ,\",'T1=',round(T1),\"lb\",'and T2=',round(T2,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are:\n",
- "a= 3.93 ft/s**2 , T1= 89.0 lb and T2= 67.5 lb\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-12, Page No 342"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=644 #lb\n",
- "F=30 #lb\n",
- "# as theta=30 degrees,\n",
- "sintheta=(2)**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "r=1.5 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,-W/g],[-r,-(2**-1)*(W/g)*(2*2)*(r**-1)]])\n",
- "B=np.array([[(W*sintheta)-(F*costheta)],[-F*2]])\n",
- "C=np.linalg.solve(A,B)\n",
- "a=C[1] #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,2),\"ft/s**2\"\n",
- "# The negative sign indicates that the cylinder will roll down the plane."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is -6.78 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-14, Page No 344"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=20 #lb\n",
- "g=32.2 #ft/s**2\n",
- "vb=0.5 #rad/s\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "#Solving the three equations simultaneously by matrix method\n",
- "X=np.array([[0,1,-(W*g**-1)*5.2],[-1,0,-(W*g**-1)*3],[3,-3,-(12**-1)*(W*g**-1)*12**2]])\n",
- "Y=np.array([[-0.75*(W*g**-1)],[(W*g**-1)*1.3-W],[0]])\n",
- "C=np.linalg.solve(X,Y)\n",
- "A=C[0] #lb\n",
- "B=C[1] #lb\n",
- "alpha=C[2] #rad/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of alpha is',round(alpha,1),\"rad/s**2 countercockwise\",'and of A and B are',round(A,1),\"lb up and\",round(B,2),\"lb to the right\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of alpha is 2.6 rad/s**2 countercockwise and of A and B are 14.4 lb up and 7.91 lb to the right\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-16, Page No 345"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mc=7.25 #kg\n",
- "d=0.9 #m\n",
- "la=0.2 #m\n",
- "ma=9 #kg\n",
- "F=45 #N\n",
- "ay=0 #m/s**2\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "I=2*(0.5*mc*(d*2**-1)**2)+0.5*ma*(la*2**-1)**2 #kg-m**2\n",
- "#Using the equations of motion\n",
- "Na=(2*mc+ma)*g #N\n",
- "#Simplfying using radial velocity formula\n",
- "#Solving the two equations using matrix method\n",
- "A=np.array([[-1,-(2*mc+ma)],[(d*2**-1),-I/(d*2**-1)]])\n",
- "B=np.array([[-F],[F*(la*2**-1)]])\n",
- "C=np.linalg.solve(A,B)\n",
- "F=C[0] #N\n",
- "ax=C[1] #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The computation yields ax=',round(ax,2),\"m/s**2 to the right.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The computation yields ax= 1.13 m/s**2 to the right.\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-18, Page No 347"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=0.05 #m cylinder radius\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Here the equation has been solved in terms of the veriables\n",
- "#Hence we directly consider the final result\n",
- "av=(2*g)/3 #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of av is',round(av,2),\"m/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of av is 6.53 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-21, Page No 349"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#initilization of variables\n",
- "W=16.1 #lb\n",
- "v=9 #ft/s\n",
- "# as phi=30 degrees,\n",
- "sinphi=(2)**-1\n",
- "cosphi=(3**0.5)*2**-1\n",
- "r=0.5 #ft\n",
- "g=32.2 #ft/s**2\n",
- "OG=4.5 #ft\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "an=v**2/OG #ft/s**2\n",
- "#Solving for alpha we get\n",
- "N=(W*g**-1)*an+W*cosphi #lb\n",
- "#Using equations of motion\n",
- "A=np.array([[1,-r],[-1,-r*r]])\n",
- "B=np.array([[W*sinphi],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "F=C[0] #lb\n",
- "at=C[1] #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of N and F are',round(N,1),\"lb and\",round(F,2),\"lb respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of N and F are 22.9 lb and 2.68 lb respectively.\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-25, Page No 353"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=50 #lb\n",
- "P=10 #lb\n",
- "t=5 #s\n",
- "vo=0 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "ax=(P*g)/W #ft/s**2\n",
- "#Solving by matrix method for A and B\n",
- "F=np.array([[1,1],[-4,4]])\n",
- "Q=np.array([[W],[P]])\n",
- "R=np.linalg.solve(F,Q)\n",
- "#Velocity calculations\n",
- "v=vo+ax*t #ft/s\n",
- "A=R[0] #lb\n",
- "B=R[1] #lb\n",
- "\n",
- "#Result\n",
- "print'The velocity of the door after 5s is',round(v,1),\"ft/s\",'and A=',round(A,1),\"lb\",'and B=',round(B,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity of the door after 5s is 32.2 ft/s and A= 23.8 lb and B= 26.3 lb\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-26, Page No 354"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "#Initilization of variables\n",
- "AB=2 #m\n",
- "m=2 #kg\n",
- "F=20 #N\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equation of motion\n",
- "a=F/m #m/s**2\n",
- "#Solving by matrix method for Na and Nb\n",
- "A=np.array([[1,-1],[4*5**-1,4*5**-1]])\n",
- "B=np.array([[m*g],[F*(3*5**-1)]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a),\"m/s**2\",'and the reactions are: Na=',round(C[0],1),\"N up\",'and Nb=',round(C[1],1),\"N up\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is 10.0 m/s**2 and the reactions are: Na= 17.3 N up and Nb= -2.3 N up\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-27, Page No 354"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "vo=0 #ft/s\n",
- "\n",
- "#Calculations\n",
- "s=(0.011*5280*2)/(2*0.004)\n",
- "\n",
- "#Result\n",
- "print'It travels',round(s),\"ft along the level before coming to rest\"\n",
- "#Answer in the textbook is incorrect by 20ft\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "It travels 14520.0 ft along the level before coming to rest\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-28, Page No 355"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "u=0.3 #coefficient of friction\n",
- "m=70 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#CASE 1\n",
- "#Using equations of motion\n",
- "Na=m*g #N\n",
- "ah=(u*Na)/m #m/s**2\n",
- "#CASE 2\n",
- "#Applying sum of moments equal to zero\n",
- "F=(Na*0.3)/1.2 #N\n",
- "a_h=F/m #m/s**2\n",
- "\n",
- "#Result\n",
- "#Intutive insights can be attained after we get these results\n",
- "print'The value of Na is',round(Na),\"N\"\n",
- "print'and that of acceleration are:'\n",
- "print'1st value=',round(ah,2),\"m/s**2\",'and 2nd value is',round(a_h,2),\"m/s**2\"\n",
- "print'and the value of F is',round(F),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of Na is 686.0 N\n",
- "and that of acceleration are:\n",
- "1st value= 2.94 m/s**2 and 2nd value is 2.45 m/s**2\n",
- "and the value of F is 172.0 N\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-29, Page No 356"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=60 #kg\n",
- "me=660 #kg\n",
- "a=6 #m/s**2\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "P=m*a+m*g #N\n",
- "#Scale reading\n",
- "R=P/g #kg\n",
- "#Increase in mass\n",
- "I=R-m #kg\n",
- "#Tension\n",
- "T=me*a+me*g #N\n",
- "\n",
- "#Result\n",
- "print'The value of P is',round(P),\"n\"\n",
- "print'The apparent icrease in weight is',round(I,1),\"kg\",'and the tension in the cable is',round(T),\"N\"\n",
- "#Answer in the textbook is off by 28 #N in Tension\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P is 948.0 n\n",
- "The apparent icrease in weight is 36.7 kg and the tension in the cable is 10428.0 N\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-30, Page No 356"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "u=0.2 #coefficient of friction\n",
- "ma=1.2 #kg\n",
- "mb=2 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "Nb=mb*g #N\n",
- "F=u*Nb #N\n",
- "#Using equations of motion\n",
- "#Solving for T and a\n",
- "A=np.array([[-1,-ma],[1,-mb]])\n",
- "B=np.array([[-ma*g],[F]])\n",
- "C=np.linalg.solve(A,B)\n",
- "T=C[0] #N\n",
- "a=C[1] #m/s**2\n",
- "#Taking the sum of the moments\n",
- "x_m=-(F*0.15+T*0.15)/Nb #m\n",
- "x=x_m*1000 #mm\n",
- "\n",
- "#Result\n",
- "print'The acceleration of block A is',round(a,2),\"m/s**2\" \n",
- "print'and Nb acts at a distance of',round(x,1),\"mm.(Negative sign indictaes that the side assumed is incorrect)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration of block A is 2.45 m/s**2\n",
- "and Nb acts at a distance of -97.5 mm.(Negative sign indictaes that the side assumed is incorrect)\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-31, Page No 357"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "a=2.5 #m/s**2\n",
- "mA=3 #kg\n",
- "mB=7 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "F=(mA+mB)*a #N\n",
- "#Using equations of motion\n",
- "Py=mB*g #N\n",
- "#Solving for Px and H\n",
- "A=np.array([[1,1],[-0.0375,0.0375]])\n",
- "B=np.array([[mB*a],[Py*0.05]])\n",
- "C=np.linalg.solve(A,B)\n",
- "Px=C[0] #N\n",
- "H=C[1] #N\n",
- "\n",
- "#Result\n",
- "print'The value of H is',round(H,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of H is 54.5 N\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-32, Page No 358"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=20 #kg\n",
- "g=9.8 #m/s**2\n",
- "vo=3 #m/s\n",
- "v=0 #m/s\n",
- "s=4 #m\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "Na=m*g #N\n",
- "F=(Na*0.075)/0.125 #N\n",
- "a=F/m #m/s**2\n",
- "#Displacement \n",
- "d=-(v**2-vo**2)/(2*a) #m\n",
- "displ=s-d #m\n",
- "v_f=sqrt(2*a*displ) #m/s\n",
- "\n",
- "#Result\n",
- "print'The final velocity is',round(v_f,2),\"m/s to the left.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The final velocity is 6.17 m/s to the left.\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-33, Page No 358"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mA=30 #kg\n",
- "mB=45 #kg\n",
- "u_ab=3**-1 #coefficient of friction between two blocks\n",
- "u_bp=10**-1 #coefficient of friction between block and horizontal plane\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#By inspection\n",
- "Na=mA*g #N\n",
- "Nb=Na+mB*g #N\n",
- "a=(u_ab*Na-u_bp*Nb)/mB #m/s**2\n",
- "P=(mA*a+u_ab*Na) #N\n",
- "#For block A\n",
- "#Solving for P,F and a\n",
- "A=np.array([[1,-1,-mA],[-0.05,-0.075,0],[0,1,-mB]])\n",
- "B=np.array([[0],[-Na*0.050],[Nb*u_bp]])\n",
- "C=np.linalg.solve(A,B)\n",
- "P_new=C[0] #N\n",
- "\n",
- "#Result\n",
- "#As p < p_new\n",
- "print'The maximum value of P is',round(P),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum value of P is 114.0 N\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-34, Page No 359"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Vo=1.5 #m/s\n",
- "V=0 #m/s\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "a=(g*0.2)/0.75 #m/s**2\n",
- "t=-(V-Vo)/a #s\n",
- "\n",
- "#Result\n",
- "print'The maximum acceleration is',round(a,2),\"m/s**2\",'and minimum time is',round(t,2),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum acceleration is 2.61 m/s**2 and minimum time is 0.57 s\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-36, Page No 361"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "vo=0 #mi/h\n",
- "v=60 #mi/h\n",
- "t=13.8 #s\n",
- "W=3385 #lb\n",
- "xb=46 #in\n",
- "xf=66 #in\n",
- "xv=31 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "a=(((v*88*60)/3600)-vo)/t #ft/s**2\n",
- "#Summing horizontal forces\n",
- "F=(W/g)*a #lb\n",
- "#Solving for Rf and Rr\n",
- "A=np.array([[1,1],[-xf,xb]])\n",
- "B=np.array([[W],[-F*xv]])\n",
- "C=np.linalg.solve(A,B)\n",
- "Rr=C[0] #lb\n",
- "Rf=C[1] #lb\n",
- "\n",
- "#Result\n",
- "print'The value of reactions are Rf=',round(Rf),\"lb\",'and Rr=',round(Rr),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of reactions are Rf= 1809.0 lb and Rr= 1576.0 lb\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-43, Page No 366"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=161 #lb\n",
- "F=16.1 #lb\n",
- "r=18 #ft radius\n",
- "t=2 #s\n",
- "g=32.2 #ft/s**2\n",
- "wo=0 #rad/s\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "#Solving for T and alpha\n",
- "A=np.array([[r*12**-1,-0.5*(W*g**-1)*(r*12**-1)**2],[-1,-F*g**-1]])\n",
- "B=np.array([[0],[-F]])\n",
- "C=np.linalg.solve(A,B)\n",
- "alpha=C[1] #rad/s**2\n",
- "w=wo+(alpha*t) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular speed is',round(w,2),\"rad/s\"\n",
- "\n",
- "#The ans is incorrect in textbook."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular speed is 7.58 rad/s\n"
- ]
- }
- ],
- "prompt_number": 43
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-47, Page No 369"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization fo variables\n",
- "r=2000 #ft\n",
- "g=32.2 #ft/s**2\n",
- "d=4.71 #ft\n",
- "v=176 #ft/s\n",
- "\n",
- "#Calculations\n",
- "e=(d*v**2)/(g*r) #ft\n",
- "\n",
- "#Result\n",
- "print'The superelevation is',round(e,2),\"ft\"\n",
- "#Watch the unit in the final answer\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The superelevation is 2.27 ft\n"
- ]
- }
- ],
- "prompt_number": 45
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-48, Page No 369"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "a=5 #ft/s**2\n",
- "C=50 #lb-ft\n",
- "W=161 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "T=0.5*(W/g)*1**2*a+C #lb\n",
- "Ox=-T*(2/sqrt(a)) #lb\n",
- "Oy=T*(1/sqrt(a))+W #lb\n",
- "Wa=T/(1-(a/g)) #lb\n",
- "\n",
- "#Result\n",
- "print'The values are: T=',round(T,1),\"lb\",', Wa=',round(Wa),\"b\",',Ox=',round(Ox,1),\"lb\",'and Oy=+',round(Oy),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are: T= 62.5 lb , Wa= 74.0 b ,Ox= -55.9 lb and Oy=+ 189.0 lb\n"
- ]
- }
- ],
- "prompt_number": 49
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-49, Page No 370"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=100 #kg\n",
- "mr=20 #kg\n",
- "w=8 #rad/s\n",
- "l1=300 #mm\n",
- "l2=600 #mm\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "r_bar=(mr*l1+m*750)/120 #mm\n",
- "I=(3**-1)*mr*(l2*1000**-1)**2+(2*5**-1)*m*(l1*2000**-1)**2+m*(0.75)**2 #kg.m**2\n",
- "alpha=(m+mr)*g*(r_bar*1000**-1)/I #rad/s**2\n",
- "On=(m+mr)*(r_bar*1000**-1)*w**2 #N\n",
- "Ot=((m+mr)*(r_bar*1000**-1)*alpha)-(m+mr)*g #N\n",
- "\n",
- "#Result\n",
- "print'The angular acceleration is',round(alpha,1),\"rad/s**2\",'and On=',round(On),\"N\",'and Ot=',round(Ot,1),\"N\"\n",
- "\n",
- "#Due to decimal accuracy there is discrepancy in answers with the textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular acceleration is 13.3 rad/s**2 and On= 5184.0 N and Ot= -96.3 N\n"
- ]
- }
- ],
- "prompt_number": 48
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-50, Page No 370"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=40 #lb\n",
- "w=10 #rad/s\n",
- "alpha=2 #rad/s**2\n",
- "r=2 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "On=(W*g**-1)*(1*6**-1)*w**2 #lb\n",
- "Ot=(W*g**-1)*(1*6**-1)*alpha\n",
- "Io=(0.5*(W*g**-1)*0.5**2)*2+((W*g**-1)*(1*6**-1)**2)*2\n",
- "\n",
- "#Result\n",
- "print'The reaction components are On=',round(On,1),\"lb to the right\",'and Ot=',round(Ot,2),\"lb up\"\n",
- "print'The value of Io is',round(Io,2),\"lb-ft\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction components are On= 20.7 lb to the right and Ot= 0.41 lb up\n",
- "The value of Io is 0.38 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 59
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-51, Page No 371"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilizatin of variables\n",
- "W=6 #lb\n",
- "l=8 #ft\n",
- "v=10 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "# as theta1=60 degrees & theta2=30 degrees\n",
- "costheta1=2**-1\n",
- "costheta2=(3**0.5)*2**-1\n",
- "\n",
- "#Calculations\n",
- "Fe=(W*v**2)*(g*l*0.5)**-1 #lb\n",
- "#Using equations of motion\n",
- "#Solving for C and T\n",
- "A=np.array([[costheta1,-costheta2],[costheta2,costheta1]])\n",
- "B=np.array([[-Fe],[W]])\n",
- "P=np.linalg.solve(A,B) #lb\n",
- "C=P[0] #lb\n",
- "T=P[1] #lb\n",
- "\n",
- "#Result\n",
- "print'The value of C is',round(C,2),\"lb\",'and T is',round(T,2),\"lb\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of C is 2.87 lb and T is 7.03 lb\n"
- ]
- }
- ],
- "prompt_number": 49
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-52, Page No 372"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=32.2 #lb\n",
- "T=120 #lb\n",
- "m=1 #slug\n",
- "r=6*12**-1 #ft\n",
- "\n",
- "#Calculations\n",
- "w=sqrt((T*(3*5**-1)*4)*(m*r*3)**-1) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular speed permissible is',round(w,1),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular speed permissible is 13.9 rad/s\n"
- ]
- }
- ],
- "prompt_number": 50
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-53, Page No 372"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=30 #kg\n",
- "k=0.45 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Using equations of motion\n",
- "#Solving for T1,T2 and alpha\n",
- "A=np.array([[1,0,-m],[0,-1,-45],[-0.6,0.3,-m*k**2]])\n",
- "B=np.array([[50*g],[-150*g],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The values are: T1=',round(C[0]),\"N\",',T2=',round(C[1]),\"N\",'and alpha=',round(C[2],1),\"rad/s**2\"\n",
- "\n",
- "# The answer for T2 and alpha is off by 4 & 0.1 units"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are: T1= 607.0 N ,T2= 1294.0 N and alpha= 3.9 rad/s**2\n"
- ]
- }
- ],
- "prompt_number": 51
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-54, Page No 373"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wc=28 #lb\n",
- "v=16 #ft/s\n",
- "Ib=12 #ft-lb-s**2\n",
- "u=0.4 #coefficient of friction\n",
- "t=2 #s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "T=Wc+(Wc*g**-1)*8 #lb\n",
- "alpha=(8*12)*15**-1 #rad/s**2\n",
- "F=((Ib*alpha)+(T*1.25))/t #lb\n",
- "N=F/u #lb\n",
- "#Summing moments about D\n",
- "P=(N*8+F*3)/40 #lb\n",
- "#Summing forces horizontally and vertically\n",
- "Dx=151-P #lb\n",
- "Dy=-F #lb\n",
- "\n",
- "#Result\n",
- "print'The reactions at D are: Dx=',round(Dx),\"lb to right\",'and Dy=',round(Dy,1),\"lb down\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reactions at D are: Dx= 116.0 lb to right and Dy= -60.2 lb down\n"
- ]
- }
- ],
- "prompt_number": 65
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-55, Page No 374"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=8 #kg\n",
- "n=90 #rpm\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "Fg=m*g #N\n",
- "w=2*pi*n/60 #rad/s\n",
- "#using equations of motion\n",
- "By=m*g #N\n",
- "#Solving for Bx and C\n",
- "A=np.array([[1,1],[-0.3,0.9]])\n",
- "B=np.array([[m*0.3*w**2],[By*0.3]])\n",
- "C=np.linalg.solve(A,B) #N\n",
- "\n",
- "#Result\n",
- "print'The solution is: Bx=',round(C[0]),\"N\",',By=',round(By,1),\"N\",'and C=',round(C[1],1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is: Bx= 140.0 N ,By= 78.4 N and C= 72.9 N\n"
- ]
- }
- ],
- "prompt_number": 79
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-56, Page No 375"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=8 #kg\n",
- "n=90 #rpm\n",
- "g=9.8 #m/s**2\n",
- "r=0.3 #m\n",
- "\n",
- "#calculations\n",
- "w=2*pi*n/60 #rad/s\n",
- "#Using equations of motion\n",
- "C=(m*g*0.3+m*r*w**2*r)/1.2 #N\n",
- "Bx=-C+m*r*w**2 #N\n",
- "By=m*g #N\n",
- "\n",
- "#Result\n",
- "print'The solution is: Bx=',round(Bx),\"N\",',By=',round(By,1),\"N\",'and C=',round(C,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is: Bx= 140.0 N ,By= 78.4 N and C= 72.9 N\n"
- ]
- }
- ],
- "prompt_number": 66
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-57, Page No 375"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Na=294 #N\n",
- "Nb=735 #N\n",
- "\n",
- "#Calculations\n",
- "a=(10**-1*Nb-3**-1*Na)/45 #m/s**2\n",
- "P=(3**-1*Na)-30*a #N\n",
- "\n",
- "#Result\n",
- "print'The solution is P=',round(P),\"N\",'and a=',round(a,3),\"m/s**2\"\n",
- "# The negative sign indicates the assumed direction is incorrect."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is P= 114.0 N and a= -0.544 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 68
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-58, Page No 376"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=50 #lb\n",
- "g=32.2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "a=(10/(W/g)) #ft/s**2\n",
- "B=((2.5*(W/g)*a)+4*W-1.5*10)/8 #lb\n",
- "A=50-B #lb\n",
- "\n",
- "#Result\n",
- "print'The solution is A=',round(A,1),\"lb\",',B=',round(B,1),\"lb\",'and a=',round(a,2),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is A= 23.8 lb ,B= 26.3 lb and a= 6.44 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 69
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-60, Page No 377"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "g=9.8 #m/s**2\n",
- "r1=0.3 #m\n",
- "m1=20 #kg\n",
- "m2=100 #kg\n",
- "r2=0.75 #m\n",
- "\n",
- "#Calculations\n",
- "alpha=(m1*g*r1+m2*g*r2)*(m1*r1**2+(m1/12)*0.6**2+m2*r2**2+(2*5**-1)*m2*0.15**2)**-1 #rad/s**2\n",
- "\n",
- "#Result\n",
- "print'The angular acceleration is',round(alpha,1),\"rad/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular acceleration is 13.4 rad/s**2\n"
- ]
- }
- ],
- "prompt_number": 71
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-61, Page No 378"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=15*12**-1 #ft\n",
- "W=600 #lb\n",
- "# as theta=25 degrees,\n",
- "sintheta=0.422\n",
- "\n",
- "#calculations\n",
- "ax=(r*W*sintheta)*((r**-1)*14.5+r*18.6)**-1 #ft/s**2\n",
- "F=(W*sintheta)-(18.6*ax) #lb\n",
- "\n",
- "#Result\n",
- "print'The solution is F=',round(F,1),\"lb\",'and ax=',round(ax,2),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is F= 84.3 lb and ax= 9.08 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-62, Page No 378"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=7 #kg\n",
- "g=9.8 #m/s**2\n",
- "r=0.5 #m\n",
- "I=0.875 #kg.m**2\n",
- "\n",
- "#Calculations\n",
- "#Solving for alpha and T\n",
- "alpha=(m*g*r)/(I+m*r*0.5) #rad/s**2\n",
- "T=(I*alpha)/r #N\n",
- "\n",
- "#Result\n",
- "print'The soultion is alpha=',round(alpha,1),\"rad/s**2\",'and T=',round(T,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The soultion is alpha= 13.1 rad/s**2 and T= 22.9 N\n"
- ]
- }
- ],
- "prompt_number": 79
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter16_2.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter16_2.ipynb deleted file mode 100755 index 22430a7a..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter16_2.ipynb +++ /dev/null @@ -1,1911 +0,0 @@ -{
- "metadata": {
- "name": "chapter16.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 16: Dynamics of a Rigid Body in Plane Motion"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-2, Page No 335"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Intilization of variables\n",
- "W=600 #lb\n",
- "d=30 #in\n",
- "# as theta=25 degrees,\n",
- "sintheta=0.422\n",
- "costheta=0.906\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "m=W/g #lb-s**2/ft\n",
- "#Moment of inertia\n",
- "I=0.5*m*((d/2)/12)**2 #lb-s**2-ft\n",
- "#Applying Newtons law and coservation of angular momentum and rolling\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,m,0,0],[0,0,0,1],[((d/2)/12),0,-I,0],[0,1,-((d/2)/12),0]])\n",
- "B=np.array([[W*sintheta],[W*costheta],[0],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The Frictional Force is',round(C[0],1),\"lb\",'and the acceleration is',round(C[1],2),\"ft/s**2\"\n",
- "\n",
- "# The answers wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Frictional Force is 84.4 lb and the acceleration is 9.06 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-3, Page No 336"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=18 #kg\n",
- "d=0.6 #m\n",
- "vo=3 #m/s\n",
- "# as theta=20 degrees,\n",
- "sintheta=0.342\n",
- "costheta=0.939\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of Inertia\n",
- "I=0.5*m*(d/2)**2 \n",
- "#Applying Newtons second Law a\n",
- "A=np.array([[1,m,0,0],[0,0,1,0],[d/2,0,0,-I],[0,1,0,(-d/2)]])\n",
- "B=np.array([[g*m*sintheta],[g*m*costheta],[0],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing the answers in variables\n",
- "F=C[0] #N\n",
- "ax=C[1] #m/s**2\n",
- "Na=C[2] #N\n",
- "alpha=C[3] #rad/s**2\n",
- "#Time Calculations\n",
- "v=0 #m/s**2\n",
- "t=(vo)/ax #s\n",
- "\n",
- "#Result\n",
- "print'It takes',round(t,2),\"s to reach the highest point of travel\"\n",
- "\n",
- "# The ans is off by 0.01 sec."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "It takes 1.34 s to reach the highest point of travel\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-5, Page No 337"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=20 #kg\n",
- "F1=40 #N\n",
- "ro=0.6 #m\n",
- "ri=0.45 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of inertia\n",
- "I=(2*5**-1)*m*ro**2 #kg-m**2\n",
- "#Applying Newtons Law and conservation of angular Momentum\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,m],[ro,-I/ro]])\n",
- "B=np.array([[F1],[F1*ri]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing answers in variables\n",
- "F=C[0] #N\n",
- "a=C[1] #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The acceleration is',round(a,2),\"m/s**2\"\n",
- "print'The force is',round(F,2),\"N\"\n",
- "#The solution in the textbook is incorrect\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration is 0.36 m/s**2\n",
- "The force is 32.86 N\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-6, Page No 338"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=16.1 #lb\n",
- "u=0.10 #co-efficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "F=1.39 #lb\n",
- "\n",
- "#Calculations\n",
- "#Applying Newtons Second Law\n",
- "#Using F=1.39 lb\n",
- "a=((W*sintheta)-F)/(W*g**-1) #ft/s**2\n",
- "alpha=(F*0.5*5/2)/((W*g**-1)*(0.5**2)) #rad/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,1),\"ft/s**2\",'and alpha is ',round(alpha,1),\"rad/s**2.\"\n",
- "print'Hence the sphere will both,roll and slip'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is 13.3 ft/s**2 and alpha is 13.9 rad/s**2.\n",
- "Hence the sphere will both,roll and slip\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-8, Page No 339"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=30 degrees\n",
- "sintheta=2**-1\n",
- "W=80 #lb\n",
- "Ww=100 #lb\n",
- "I=4 #slug-ft**2\n",
- "r=0.5 #ft\n",
- "v= 20 #ft/s\n",
- "vo=0 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using Equations of motion\n",
- "#Solving the system of linear equatinons by matrix method\n",
- "A=np.array([[-1,0,-W/g],[1,-1,-Ww/g],[0,r,-2*I]])\n",
- "B=np.array([[-W],[Ww*sintheta],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing values in variables\n",
- "T=C[0] #lb\n",
- "F=C[1] #lb\n",
- "a=C[2] #ft/s**2\n",
- "#Time calculations\n",
- "t=(v-vo)/a #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t,1),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 14.4 s\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-9, Page No 340"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "M=70 #kg\n",
- "ko=0.4 #m\n",
- "ri=0.45 #m\n",
- "ro=0.6 #m\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "m=35 #kg\n",
- "g= 9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "I=M*ko**2 #kg-m**2\n",
- "#Using Equations of motion\n",
- "#Solving the equations by matrix method\n",
- "A=np.array([[-1,-m*0.15,0],[1,-M*ro,-1],[-ri,-I,ro]])\n",
- "B=np.array([[-m*g],[M*g*sintheta],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "F=C[2] #N\n",
- "Na=M*g*costheta #N\n",
- "#Required coefficient of friction\n",
- "u=F/Na #coefficient of friction\n",
- "\n",
- "#Result\n",
- "print'The value of alpha is',round(C[1],2),\"rad/s**2\",'and tension is',round(C[0]),\"N\"\n",
- "print'F=',round(F),\"N,\",'Na=',round(Na),\"N\",'and u=',round(u,2),\"(coefficient of friction)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of alpha is -4.15 rad/s**2 and tension is 365.0 N\n",
- "F= 196.0 N, Na= 594.0 N and u= 0.33 (coefficient of friction)\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Exampe 16.16-10, Page No 341"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=200 #kg\n",
- "g=9.8 #m/s**2\n",
- "r=1.2 #m\n",
- "F1=1000 #N\n",
- "F2=1400 #N\n",
- "\n",
- "#Calculations\n",
- "N=m*g #N\n",
- "I=(2*5**-1)*(m)*r**2 #kg-m**2\n",
- "#Using equations of motion\n",
- "#Solving for F and alpha using matrix method\n",
- "#Applying equations of motion\n",
- "A=np.array([[1,-m],[-r,-I/r]])\n",
- "B=np.array([[F1-F2],[F1*r]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing values\n",
- "F=C[0] #N\n",
- "alpha=C[1] #rad/s**2\n",
- "a=r*alpha #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,2),\"m/s**2\",'and F is',round(F),\"N\"\n",
- "#The negative signs indicate that the direction is opposite to what was origninally assumed\n",
- "# The answers wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is -2.57 m/s**2 and F is -829.0 N\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-11, Page No 341"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "Wa=161 #lb\n",
- "Wb=193.2 #lb\n",
- "Wc=300 #lb\n",
- "ka=3 #ft\n",
- "kb=2.5 #ft\n",
- "# as theta1=30 degrees & theta2=45 degrees,\n",
- "sintheta1=2**-1\n",
- "sintheta2=(2**0.5)**-1\n",
- "costheta2=(2**0.5)**-1\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of inertia Calculations\n",
- "Ia=(Wa/g)*ka**2 #lb-s**2-ft\n",
- "Ib=(Wb/g)*kb**2 #lb-s**2-ft\n",
- "#Using equations of motion for A and B and C\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,1,-Wa/g,0,0],[1,-4,-Ia*(4**-1),0,0],[-2,0,-Ib*(5*8**-1),4,0],[0,0,-(Wc/g)*(5*2**-1),-1,-0.25],[0,0,0,0,1]])\n",
- "B=np.array([[Wa*sintheta1],[0],[0],[-Wc*costheta2],[Wc*sintheta2]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing values in the variables\n",
- "T1=C[0] #lb\n",
- "T2=C[3] #lb\n",
- "a=C[2] #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The values are:'\n",
- "print'a=',round(a,2),\"ft/s**2 ,\",'T1=',round(T1),\"lb\",'and T2=',round(T2,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are:\n",
- "a= 3.93 ft/s**2 , T1= 89.0 lb and T2= 67.5 lb\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-12, Page No 342"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=644 #lb\n",
- "F=30 #lb\n",
- "# as theta=30 degrees,\n",
- "sintheta=(2)**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "r=1.5 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,-W/g],[-r,-(2**-1)*(W/g)*(2*2)*(r**-1)]])\n",
- "B=np.array([[(W*sintheta)-(F*costheta)],[-F*2]])\n",
- "C=np.linalg.solve(A,B)\n",
- "a=C[1] #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,2),\"ft/s**2\"\n",
- "# The negative sign indicates that the cylinder will roll down the plane."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is -6.78 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-14, Page No 344"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=20 #lb\n",
- "g=32.2 #ft/s**2\n",
- "vb=0.5 #rad/s\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "#Solving the three equations simultaneously by matrix method\n",
- "X=np.array([[0,1,-(W*g**-1)*5.2],[-1,0,-(W*g**-1)*3],[3,-3,-(12**-1)*(W*g**-1)*12**2]])\n",
- "Y=np.array([[-0.75*(W*g**-1)],[(W*g**-1)*1.3-W],[0]])\n",
- "C=np.linalg.solve(X,Y)\n",
- "A=C[0] #lb\n",
- "B=C[1] #lb\n",
- "alpha=C[2] #rad/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of alpha is',round(alpha,1),\"rad/s**2 countercockwise\",'and of A and B are',round(A,1),\"lb up and\",round(B,2),\"lb to the right\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of alpha is 2.6 rad/s**2 countercockwise and of A and B are 14.4 lb up and 7.91 lb to the right\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-16, Page No 345"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mc=7.25 #kg\n",
- "d=0.9 #m\n",
- "la=0.2 #m\n",
- "ma=9 #kg\n",
- "F=45 #N\n",
- "ay=0 #m/s**2\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "I=2*(0.5*mc*(d*2**-1)**2)+0.5*ma*(la*2**-1)**2 #kg-m**2\n",
- "#Using the equations of motion\n",
- "Na=(2*mc+ma)*g #N\n",
- "#Simplfying using radial velocity formula\n",
- "#Solving the two equations using matrix method\n",
- "A=np.array([[-1,-(2*mc+ma)],[(d*2**-1),-I/(d*2**-1)]])\n",
- "B=np.array([[-F],[F*(la*2**-1)]])\n",
- "C=np.linalg.solve(A,B)\n",
- "F=C[0] #N\n",
- "ax=C[1] #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The computation yields ax=',round(ax,2),\"m/s**2 to the right.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The computation yields ax= 1.13 m/s**2 to the right.\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-18, Page No 347"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=0.05 #m cylinder radius\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Here the equation has been solved in terms of the veriables\n",
- "#Hence we directly consider the final result\n",
- "av=(2*g)/3 #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of av is',round(av,2),\"m/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of av is 6.53 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-21, Page No 349"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#initilization of variables\n",
- "W=16.1 #lb\n",
- "v=9 #ft/s\n",
- "# as phi=30 degrees,\n",
- "sinphi=(2)**-1\n",
- "cosphi=(3**0.5)*2**-1\n",
- "r=0.5 #ft\n",
- "g=32.2 #ft/s**2\n",
- "OG=4.5 #ft\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "an=v**2/OG #ft/s**2\n",
- "#Solving for alpha we get\n",
- "N=(W*g**-1)*an+W*cosphi #lb\n",
- "#Using equations of motion\n",
- "A=np.array([[1,-r],[-1,-r*r]])\n",
- "B=np.array([[W*sinphi],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "F=C[0] #lb\n",
- "at=C[1] #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of N and F are',round(N,1),\"lb and\",round(F,2),\"lb respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of N and F are 22.9 lb and 2.68 lb respectively.\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-25, Page No 353"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=50 #lb\n",
- "P=10 #lb\n",
- "t=5 #s\n",
- "vo=0 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "ax=(P*g)/W #ft/s**2\n",
- "#Solving by matrix method for A and B\n",
- "F=np.array([[1,1],[-4,4]])\n",
- "Q=np.array([[W],[P]])\n",
- "R=np.linalg.solve(F,Q)\n",
- "#Velocity calculations\n",
- "v=vo+ax*t #ft/s\n",
- "A=R[0] #lb\n",
- "B=R[1] #lb\n",
- "\n",
- "#Result\n",
- "print'The velocity of the door after 5s is',round(v,1),\"ft/s\",'and A=',round(A,1),\"lb\",'and B=',round(B,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity of the door after 5s is 32.2 ft/s and A= 23.8 lb and B= 26.3 lb\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-26, Page No 354"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "#Initilization of variables\n",
- "AB=2 #m\n",
- "m=2 #kg\n",
- "F=20 #N\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equation of motion\n",
- "a=F/m #m/s**2\n",
- "#Solving by matrix method for Na and Nb\n",
- "A=np.array([[1,-1],[4*5**-1,4*5**-1]])\n",
- "B=np.array([[m*g],[F*(3*5**-1)]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a),\"m/s**2\",'and the reactions are: Na=',round(C[0],1),\"N up\",'and Nb=',round(C[1],1),\"N up\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is 10.0 m/s**2 and the reactions are: Na= 17.3 N up and Nb= -2.3 N up\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-27, Page No 354"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "vo=0 #ft/s\n",
- "\n",
- "#Calculations\n",
- "s=(0.011*5280*2)/(2*0.004)\n",
- "\n",
- "#Result\n",
- "print'It travels',round(s),\"ft along the level before coming to rest\"\n",
- "#Answer in the textbook is incorrect by 20ft\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "It travels 14520.0 ft along the level before coming to rest\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-28, Page No 355"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "u=0.3 #coefficient of friction\n",
- "m=70 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#CASE 1\n",
- "#Using equations of motion\n",
- "Na=m*g #N\n",
- "ah=(u*Na)/m #m/s**2\n",
- "#CASE 2\n",
- "#Applying sum of moments equal to zero\n",
- "F=(Na*0.3)/1.2 #N\n",
- "a_h=F/m #m/s**2\n",
- "\n",
- "#Result\n",
- "#Intutive insights can be attained after we get these results\n",
- "print'The value of Na is',round(Na),\"N\"\n",
- "print'and that of acceleration are:'\n",
- "print'1st value=',round(ah,2),\"m/s**2\",'and 2nd value is',round(a_h,2),\"m/s**2\"\n",
- "print'and the value of F is',round(F),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of Na is 686.0 N\n",
- "and that of acceleration are:\n",
- "1st value= 2.94 m/s**2 and 2nd value is 2.45 m/s**2\n",
- "and the value of F is 172.0 N\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-29, Page No 356"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=60 #kg\n",
- "me=660 #kg\n",
- "a=6 #m/s**2\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "P=m*a+m*g #N\n",
- "#Scale reading\n",
- "R=P/g #kg\n",
- "#Increase in mass\n",
- "I=R-m #kg\n",
- "#Tension\n",
- "T=me*a+me*g #N\n",
- "\n",
- "#Result\n",
- "print'The value of P is',round(P),\"n\"\n",
- "print'The apparent icrease in weight is',round(I,1),\"kg\",'and the tension in the cable is',round(T),\"N\"\n",
- "#Answer in the textbook is off by 28 #N in Tension\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P is 948.0 n\n",
- "The apparent icrease in weight is 36.7 kg and the tension in the cable is 10428.0 N\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-30, Page No 356"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "u=0.2 #coefficient of friction\n",
- "ma=1.2 #kg\n",
- "mb=2 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "Nb=mb*g #N\n",
- "F=u*Nb #N\n",
- "#Using equations of motion\n",
- "#Solving for T and a\n",
- "A=np.array([[-1,-ma],[1,-mb]])\n",
- "B=np.array([[-ma*g],[F]])\n",
- "C=np.linalg.solve(A,B)\n",
- "T=C[0] #N\n",
- "a=C[1] #m/s**2\n",
- "#Taking the sum of the moments\n",
- "x_m=-(F*0.15+T*0.15)/Nb #m\n",
- "x=x_m*1000 #mm\n",
- "\n",
- "#Result\n",
- "print'The acceleration of block A is',round(a,2),\"m/s**2\" \n",
- "print'and Nb acts at a distance of',round(x,1),\"mm.(Negative sign indictaes that the side assumed is incorrect)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration of block A is 2.45 m/s**2\n",
- "and Nb acts at a distance of -97.5 mm.(Negative sign indictaes that the side assumed is incorrect)\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-31, Page No 357"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "a=2.5 #m/s**2\n",
- "mA=3 #kg\n",
- "mB=7 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "F=(mA+mB)*a #N\n",
- "#Using equations of motion\n",
- "Py=mB*g #N\n",
- "#Solving for Px and H\n",
- "A=np.array([[1,1],[-0.0375,0.0375]])\n",
- "B=np.array([[mB*a],[Py*0.05]])\n",
- "C=np.linalg.solve(A,B)\n",
- "Px=C[0] #N\n",
- "H=C[1] #N\n",
- "\n",
- "#Result\n",
- "print'The value of H is',round(H,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of H is 54.5 N\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-32, Page No 358"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=20 #kg\n",
- "g=9.8 #m/s**2\n",
- "vo=3 #m/s\n",
- "v=0 #m/s\n",
- "s=4 #m\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "Na=m*g #N\n",
- "F=(Na*0.075)/0.125 #N\n",
- "a=F/m #m/s**2\n",
- "#Displacement \n",
- "d=-(v**2-vo**2)/(2*a) #m\n",
- "displ=s-d #m\n",
- "v_f=sqrt(2*a*displ) #m/s\n",
- "\n",
- "#Result\n",
- "print'The final velocity is',round(v_f,2),\"m/s to the left.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The final velocity is 6.17 m/s to the left.\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-33, Page No 358"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mA=30 #kg\n",
- "mB=45 #kg\n",
- "u_ab=3**-1 #coefficient of friction between two blocks\n",
- "u_bp=10**-1 #coefficient of friction between block and horizontal plane\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#By inspection\n",
- "Na=mA*g #N\n",
- "Nb=Na+mB*g #N\n",
- "a=(u_ab*Na-u_bp*Nb)/mB #m/s**2\n",
- "P=(mA*a+u_ab*Na) #N\n",
- "#For block A\n",
- "#Solving for P,F and a\n",
- "A=np.array([[1,-1,-mA],[-0.05,-0.075,0],[0,1,-mB]])\n",
- "B=np.array([[0],[-Na*0.050],[Nb*u_bp]])\n",
- "C=np.linalg.solve(A,B)\n",
- "P_new=C[0] #N\n",
- "\n",
- "#Result\n",
- "#As p < p_new\n",
- "print'The maximum value of P is',round(P),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum value of P is 114.0 N\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-34, Page No 359"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Vo=1.5 #m/s\n",
- "V=0 #m/s\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "a=(g*0.2)/0.75 #m/s**2\n",
- "t=-(V-Vo)/a #s\n",
- "\n",
- "#Result\n",
- "print'The maximum acceleration is',round(a,2),\"m/s**2\",'and minimum time is',round(t,2),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum acceleration is 2.61 m/s**2 and minimum time is 0.57 s\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-36, Page No 361"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "vo=0 #mi/h\n",
- "v=60 #mi/h\n",
- "t=13.8 #s\n",
- "W=3385 #lb\n",
- "xb=46 #in\n",
- "xf=66 #in\n",
- "xv=31 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "a=(((v*88*60)/3600)-vo)/t #ft/s**2\n",
- "#Summing horizontal forces\n",
- "F=(W/g)*a #lb\n",
- "#Solving for Rf and Rr\n",
- "A=np.array([[1,1],[-xf,xb]])\n",
- "B=np.array([[W],[-F*xv]])\n",
- "C=np.linalg.solve(A,B)\n",
- "Rr=C[0] #lb\n",
- "Rf=C[1] #lb\n",
- "\n",
- "#Result\n",
- "print'The value of reactions are Rf=',round(Rf),\"lb\",'and Rr=',round(Rr),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of reactions are Rf= 1809.0 lb and Rr= 1576.0 lb\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-43, Page No 366"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=161 #lb\n",
- "F=16.1 #lb\n",
- "r=18 #ft radius\n",
- "t=2 #s\n",
- "g=32.2 #ft/s**2\n",
- "wo=0 #rad/s\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "#Solving for T and alpha\n",
- "A=np.array([[r*12**-1,-0.5*(W*g**-1)*(r*12**-1)**2],[-1,-F*g**-1]])\n",
- "B=np.array([[0],[-F]])\n",
- "C=np.linalg.solve(A,B)\n",
- "alpha=C[1] #rad/s**2\n",
- "w=wo+(alpha*t) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular speed is',round(w,2),\"rad/s\"\n",
- "\n",
- "#The ans is incorrect in textbook."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular speed is 7.58 rad/s\n"
- ]
- }
- ],
- "prompt_number": 43
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-47, Page No 369"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization fo variables\n",
- "r=2000 #ft\n",
- "g=32.2 #ft/s**2\n",
- "d=4.71 #ft\n",
- "v=176 #ft/s\n",
- "\n",
- "#Calculations\n",
- "e=(d*v**2)/(g*r) #ft\n",
- "\n",
- "#Result\n",
- "print'The superelevation is',round(e,2),\"ft\"\n",
- "#Watch the unit in the final answer\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The superelevation is 2.27 ft\n"
- ]
- }
- ],
- "prompt_number": 45
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-48, Page No 369"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "a=5 #ft/s**2\n",
- "C=50 #lb-ft\n",
- "W=161 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "T=0.5*(W/g)*1**2*a+C #lb\n",
- "Ox=-T*(2/sqrt(a)) #lb\n",
- "Oy=T*(1/sqrt(a))+W #lb\n",
- "Wa=T/(1-(a/g)) #lb\n",
- "\n",
- "#Result\n",
- "print'The values are: T=',round(T,1),\"lb\",', Wa=',round(Wa),\"b\",',Ox=',round(Ox,1),\"lb\",'and Oy=+',round(Oy),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are: T= 62.5 lb , Wa= 74.0 b ,Ox= -55.9 lb and Oy=+ 189.0 lb\n"
- ]
- }
- ],
- "prompt_number": 49
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-49, Page No 370"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=100 #kg\n",
- "mr=20 #kg\n",
- "w=8 #rad/s\n",
- "l1=300 #mm\n",
- "l2=600 #mm\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "r_bar=(mr*l1+m*750)/120 #mm\n",
- "I=(3**-1)*mr*(l2*1000**-1)**2+(2*5**-1)*m*(l1*2000**-1)**2+m*(0.75)**2 #kg.m**2\n",
- "alpha=(m+mr)*g*(r_bar*1000**-1)/I #rad/s**2\n",
- "On=(m+mr)*(r_bar*1000**-1)*w**2 #N\n",
- "Ot=((m+mr)*(r_bar*1000**-1)*alpha)-(m+mr)*g #N\n",
- "\n",
- "#Result\n",
- "print'The angular acceleration is',round(alpha,1),\"rad/s**2\",'and On=',round(On),\"N\",'and Ot=',round(Ot,1),\"N\"\n",
- "\n",
- "#Due to decimal accuracy there is discrepancy in answers with the textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular acceleration is 13.3 rad/s**2 and On= 5184.0 N and Ot= -96.3 N\n"
- ]
- }
- ],
- "prompt_number": 48
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-50, Page No 370"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=40 #lb\n",
- "w=10 #rad/s\n",
- "alpha=2 #rad/s**2\n",
- "r=2 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "On=(W*g**-1)*(1*6**-1)*w**2 #lb\n",
- "Ot=(W*g**-1)*(1*6**-1)*alpha\n",
- "Io=(0.5*(W*g**-1)*0.5**2)*2+((W*g**-1)*(1*6**-1)**2)*2\n",
- "\n",
- "#Result\n",
- "print'The reaction components are On=',round(On,1),\"lb to the right\",'and Ot=',round(Ot,2),\"lb up\"\n",
- "print'The value of Io is',round(Io,2),\"lb-ft\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction components are On= 20.7 lb to the right and Ot= 0.41 lb up\n",
- "The value of Io is 0.38 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 59
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-51, Page No 371"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilizatin of variables\n",
- "W=6 #lb\n",
- "l=8 #ft\n",
- "v=10 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "# as theta1=60 degrees & theta2=30 degrees\n",
- "costheta1=2**-1\n",
- "costheta2=(3**0.5)*2**-1\n",
- "\n",
- "#Calculations\n",
- "Fe=(W*v**2)*(g*l*0.5)**-1 #lb\n",
- "#Using equations of motion\n",
- "#Solving for C and T\n",
- "A=np.array([[costheta1,-costheta2],[costheta2,costheta1]])\n",
- "B=np.array([[-Fe],[W]])\n",
- "P=np.linalg.solve(A,B) #lb\n",
- "C=P[0] #lb\n",
- "T=P[1] #lb\n",
- "\n",
- "#Result\n",
- "print'The value of C is',round(C,2),\"lb\",'and T is',round(T,2),\"lb\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of C is 2.87 lb and T is 7.03 lb\n"
- ]
- }
- ],
- "prompt_number": 49
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-52, Page No 372"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=32.2 #lb\n",
- "T=120 #lb\n",
- "m=1 #slug\n",
- "r=6*12**-1 #ft\n",
- "\n",
- "#Calculations\n",
- "w=sqrt((T*(3*5**-1)*4)*(m*r*3)**-1) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular speed permissible is',round(w,1),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular speed permissible is 13.9 rad/s\n"
- ]
- }
- ],
- "prompt_number": 50
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-53, Page No 372"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=30 #kg\n",
- "k=0.45 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Using equations of motion\n",
- "#Solving for T1,T2 and alpha\n",
- "A=np.array([[1,0,-m],[0,-1,-45],[-0.6,0.3,-m*k**2]])\n",
- "B=np.array([[50*g],[-150*g],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The values are: T1=',round(C[0]),\"N\",',T2=',round(C[1]),\"N\",'and alpha=',round(C[2],1),\"rad/s**2\"\n",
- "\n",
- "# The answer for T2 and alpha is off by 4 & 0.1 units"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are: T1= 607.0 N ,T2= 1294.0 N and alpha= 3.9 rad/s**2\n"
- ]
- }
- ],
- "prompt_number": 51
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-54, Page No 373"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wc=28 #lb\n",
- "v=16 #ft/s\n",
- "Ib=12 #ft-lb-s**2\n",
- "u=0.4 #coefficient of friction\n",
- "t=2 #s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "T=Wc+(Wc*g**-1)*8 #lb\n",
- "alpha=(8*12)*15**-1 #rad/s**2\n",
- "F=((Ib*alpha)+(T*1.25))/t #lb\n",
- "N=F/u #lb\n",
- "#Summing moments about D\n",
- "P=(N*8+F*3)/40 #lb\n",
- "#Summing forces horizontally and vertically\n",
- "Dx=151-P #lb\n",
- "Dy=-F #lb\n",
- "\n",
- "#Result\n",
- "print'The reactions at D are: Dx=',round(Dx),\"lb to right\",'and Dy=',round(Dy,1),\"lb down\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reactions at D are: Dx= 116.0 lb to right and Dy= -60.2 lb down\n"
- ]
- }
- ],
- "prompt_number": 65
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-55, Page No 374"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=8 #kg\n",
- "n=90 #rpm\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "Fg=m*g #N\n",
- "w=2*pi*n/60 #rad/s\n",
- "#using equations of motion\n",
- "By=m*g #N\n",
- "#Solving for Bx and C\n",
- "A=np.array([[1,1],[-0.3,0.9]])\n",
- "B=np.array([[m*0.3*w**2],[By*0.3]])\n",
- "C=np.linalg.solve(A,B) #N\n",
- "\n",
- "#Result\n",
- "print'The solution is: Bx=',round(C[0]),\"N\",',By=',round(By,1),\"N\",'and C=',round(C[1],1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is: Bx= 140.0 N ,By= 78.4 N and C= 72.9 N\n"
- ]
- }
- ],
- "prompt_number": 79
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-56, Page No 375"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=8 #kg\n",
- "n=90 #rpm\n",
- "g=9.8 #m/s**2\n",
- "r=0.3 #m\n",
- "\n",
- "#calculations\n",
- "w=2*pi*n/60 #rad/s\n",
- "#Using equations of motion\n",
- "C=(m*g*0.3+m*r*w**2*r)/1.2 #N\n",
- "Bx=-C+m*r*w**2 #N\n",
- "By=m*g #N\n",
- "\n",
- "#Result\n",
- "print'The solution is: Bx=',round(Bx),\"N\",',By=',round(By,1),\"N\",'and C=',round(C,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is: Bx= 140.0 N ,By= 78.4 N and C= 72.9 N\n"
- ]
- }
- ],
- "prompt_number": 66
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-57, Page No 375"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Na=294 #N\n",
- "Nb=735 #N\n",
- "\n",
- "#Calculations\n",
- "a=(10**-1*Nb-3**-1*Na)/45 #m/s**2\n",
- "P=(3**-1*Na)-30*a #N\n",
- "\n",
- "#Result\n",
- "print'The solution is P=',round(P),\"N\",'and a=',round(a,3),\"m/s**2\"\n",
- "# The negative sign indicates the assumed direction is incorrect."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is P= 114.0 N and a= -0.544 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 68
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-58, Page No 376"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=50 #lb\n",
- "g=32.2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "a=(10/(W/g)) #ft/s**2\n",
- "B=((2.5*(W/g)*a)+4*W-1.5*10)/8 #lb\n",
- "A=50-B #lb\n",
- "\n",
- "#Result\n",
- "print'The solution is A=',round(A,1),\"lb\",',B=',round(B,1),\"lb\",'and a=',round(a,2),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is A= 23.8 lb ,B= 26.3 lb and a= 6.44 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 69
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-60, Page No 377"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "g=9.8 #m/s**2\n",
- "r1=0.3 #m\n",
- "m1=20 #kg\n",
- "m2=100 #kg\n",
- "r2=0.75 #m\n",
- "\n",
- "#Calculations\n",
- "alpha=(m1*g*r1+m2*g*r2)*(m1*r1**2+(m1/12)*0.6**2+m2*r2**2+(2*5**-1)*m2*0.15**2)**-1 #rad/s**2\n",
- "\n",
- "#Result\n",
- "print'The angular acceleration is',round(alpha,1),\"rad/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular acceleration is 13.4 rad/s**2\n"
- ]
- }
- ],
- "prompt_number": 71
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-61, Page No 378"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=15*12**-1 #ft\n",
- "W=600 #lb\n",
- "# as theta=25 degrees,\n",
- "sintheta=0.422\n",
- "\n",
- "#calculations\n",
- "ax=(r*W*sintheta)*((r**-1)*14.5+r*18.6)**-1 #ft/s**2\n",
- "F=(W*sintheta)-(18.6*ax) #lb\n",
- "\n",
- "#Result\n",
- "print'The solution is F=',round(F,1),\"lb\",'and ax=',round(ax,2),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is F= 84.3 lb and ax= 9.08 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-62, Page No 378"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=7 #kg\n",
- "g=9.8 #m/s**2\n",
- "r=0.5 #m\n",
- "I=0.875 #kg.m**2\n",
- "\n",
- "#Calculations\n",
- "#Solving for alpha and T\n",
- "alpha=(m*g*r)/(I+m*r*0.5) #rad/s**2\n",
- "T=(I*alpha)/r #N\n",
- "\n",
- "#Result\n",
- "print'The soultion is alpha=',round(alpha,1),\"rad/s**2\",'and T=',round(T,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The soultion is alpha= 13.1 rad/s**2 and T= 22.9 N\n"
- ]
- }
- ],
- "prompt_number": 79
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter16_3.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter16_3.ipynb deleted file mode 100755 index 22430a7a..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter16_3.ipynb +++ /dev/null @@ -1,1911 +0,0 @@ -{
- "metadata": {
- "name": "chapter16.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 16: Dynamics of a Rigid Body in Plane Motion"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-2, Page No 335"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Intilization of variables\n",
- "W=600 #lb\n",
- "d=30 #in\n",
- "# as theta=25 degrees,\n",
- "sintheta=0.422\n",
- "costheta=0.906\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "m=W/g #lb-s**2/ft\n",
- "#Moment of inertia\n",
- "I=0.5*m*((d/2)/12)**2 #lb-s**2-ft\n",
- "#Applying Newtons law and coservation of angular momentum and rolling\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,m,0,0],[0,0,0,1],[((d/2)/12),0,-I,0],[0,1,-((d/2)/12),0]])\n",
- "B=np.array([[W*sintheta],[W*costheta],[0],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The Frictional Force is',round(C[0],1),\"lb\",'and the acceleration is',round(C[1],2),\"ft/s**2\"\n",
- "\n",
- "# The answers wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Frictional Force is 84.4 lb and the acceleration is 9.06 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-3, Page No 336"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=18 #kg\n",
- "d=0.6 #m\n",
- "vo=3 #m/s\n",
- "# as theta=20 degrees,\n",
- "sintheta=0.342\n",
- "costheta=0.939\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of Inertia\n",
- "I=0.5*m*(d/2)**2 \n",
- "#Applying Newtons second Law a\n",
- "A=np.array([[1,m,0,0],[0,0,1,0],[d/2,0,0,-I],[0,1,0,(-d/2)]])\n",
- "B=np.array([[g*m*sintheta],[g*m*costheta],[0],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing the answers in variables\n",
- "F=C[0] #N\n",
- "ax=C[1] #m/s**2\n",
- "Na=C[2] #N\n",
- "alpha=C[3] #rad/s**2\n",
- "#Time Calculations\n",
- "v=0 #m/s**2\n",
- "t=(vo)/ax #s\n",
- "\n",
- "#Result\n",
- "print'It takes',round(t,2),\"s to reach the highest point of travel\"\n",
- "\n",
- "# The ans is off by 0.01 sec."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "It takes 1.34 s to reach the highest point of travel\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-5, Page No 337"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=20 #kg\n",
- "F1=40 #N\n",
- "ro=0.6 #m\n",
- "ri=0.45 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of inertia\n",
- "I=(2*5**-1)*m*ro**2 #kg-m**2\n",
- "#Applying Newtons Law and conservation of angular Momentum\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,m],[ro,-I/ro]])\n",
- "B=np.array([[F1],[F1*ri]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing answers in variables\n",
- "F=C[0] #N\n",
- "a=C[1] #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The acceleration is',round(a,2),\"m/s**2\"\n",
- "print'The force is',round(F,2),\"N\"\n",
- "#The solution in the textbook is incorrect\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration is 0.36 m/s**2\n",
- "The force is 32.86 N\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-6, Page No 338"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=16.1 #lb\n",
- "u=0.10 #co-efficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "F=1.39 #lb\n",
- "\n",
- "#Calculations\n",
- "#Applying Newtons Second Law\n",
- "#Using F=1.39 lb\n",
- "a=((W*sintheta)-F)/(W*g**-1) #ft/s**2\n",
- "alpha=(F*0.5*5/2)/((W*g**-1)*(0.5**2)) #rad/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,1),\"ft/s**2\",'and alpha is ',round(alpha,1),\"rad/s**2.\"\n",
- "print'Hence the sphere will both,roll and slip'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is 13.3 ft/s**2 and alpha is 13.9 rad/s**2.\n",
- "Hence the sphere will both,roll and slip\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-8, Page No 339"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=30 degrees\n",
- "sintheta=2**-1\n",
- "W=80 #lb\n",
- "Ww=100 #lb\n",
- "I=4 #slug-ft**2\n",
- "r=0.5 #ft\n",
- "v= 20 #ft/s\n",
- "vo=0 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using Equations of motion\n",
- "#Solving the system of linear equatinons by matrix method\n",
- "A=np.array([[-1,0,-W/g],[1,-1,-Ww/g],[0,r,-2*I]])\n",
- "B=np.array([[-W],[Ww*sintheta],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing values in variables\n",
- "T=C[0] #lb\n",
- "F=C[1] #lb\n",
- "a=C[2] #ft/s**2\n",
- "#Time calculations\n",
- "t=(v-vo)/a #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t,1),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 14.4 s\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-9, Page No 340"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "M=70 #kg\n",
- "ko=0.4 #m\n",
- "ri=0.45 #m\n",
- "ro=0.6 #m\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "m=35 #kg\n",
- "g= 9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "I=M*ko**2 #kg-m**2\n",
- "#Using Equations of motion\n",
- "#Solving the equations by matrix method\n",
- "A=np.array([[-1,-m*0.15,0],[1,-M*ro,-1],[-ri,-I,ro]])\n",
- "B=np.array([[-m*g],[M*g*sintheta],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "F=C[2] #N\n",
- "Na=M*g*costheta #N\n",
- "#Required coefficient of friction\n",
- "u=F/Na #coefficient of friction\n",
- "\n",
- "#Result\n",
- "print'The value of alpha is',round(C[1],2),\"rad/s**2\",'and tension is',round(C[0]),\"N\"\n",
- "print'F=',round(F),\"N,\",'Na=',round(Na),\"N\",'and u=',round(u,2),\"(coefficient of friction)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of alpha is -4.15 rad/s**2 and tension is 365.0 N\n",
- "F= 196.0 N, Na= 594.0 N and u= 0.33 (coefficient of friction)\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Exampe 16.16-10, Page No 341"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=200 #kg\n",
- "g=9.8 #m/s**2\n",
- "r=1.2 #m\n",
- "F1=1000 #N\n",
- "F2=1400 #N\n",
- "\n",
- "#Calculations\n",
- "N=m*g #N\n",
- "I=(2*5**-1)*(m)*r**2 #kg-m**2\n",
- "#Using equations of motion\n",
- "#Solving for F and alpha using matrix method\n",
- "#Applying equations of motion\n",
- "A=np.array([[1,-m],[-r,-I/r]])\n",
- "B=np.array([[F1-F2],[F1*r]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing values\n",
- "F=C[0] #N\n",
- "alpha=C[1] #rad/s**2\n",
- "a=r*alpha #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,2),\"m/s**2\",'and F is',round(F),\"N\"\n",
- "#The negative signs indicate that the direction is opposite to what was origninally assumed\n",
- "# The answers wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is -2.57 m/s**2 and F is -829.0 N\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-11, Page No 341"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "Wa=161 #lb\n",
- "Wb=193.2 #lb\n",
- "Wc=300 #lb\n",
- "ka=3 #ft\n",
- "kb=2.5 #ft\n",
- "# as theta1=30 degrees & theta2=45 degrees,\n",
- "sintheta1=2**-1\n",
- "sintheta2=(2**0.5)**-1\n",
- "costheta2=(2**0.5)**-1\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of inertia Calculations\n",
- "Ia=(Wa/g)*ka**2 #lb-s**2-ft\n",
- "Ib=(Wb/g)*kb**2 #lb-s**2-ft\n",
- "#Using equations of motion for A and B and C\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,1,-Wa/g,0,0],[1,-4,-Ia*(4**-1),0,0],[-2,0,-Ib*(5*8**-1),4,0],[0,0,-(Wc/g)*(5*2**-1),-1,-0.25],[0,0,0,0,1]])\n",
- "B=np.array([[Wa*sintheta1],[0],[0],[-Wc*costheta2],[Wc*sintheta2]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing values in the variables\n",
- "T1=C[0] #lb\n",
- "T2=C[3] #lb\n",
- "a=C[2] #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The values are:'\n",
- "print'a=',round(a,2),\"ft/s**2 ,\",'T1=',round(T1),\"lb\",'and T2=',round(T2,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are:\n",
- "a= 3.93 ft/s**2 , T1= 89.0 lb and T2= 67.5 lb\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-12, Page No 342"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=644 #lb\n",
- "F=30 #lb\n",
- "# as theta=30 degrees,\n",
- "sintheta=(2)**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "r=1.5 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,-W/g],[-r,-(2**-1)*(W/g)*(2*2)*(r**-1)]])\n",
- "B=np.array([[(W*sintheta)-(F*costheta)],[-F*2]])\n",
- "C=np.linalg.solve(A,B)\n",
- "a=C[1] #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,2),\"ft/s**2\"\n",
- "# The negative sign indicates that the cylinder will roll down the plane."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is -6.78 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-14, Page No 344"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=20 #lb\n",
- "g=32.2 #ft/s**2\n",
- "vb=0.5 #rad/s\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "#Solving the three equations simultaneously by matrix method\n",
- "X=np.array([[0,1,-(W*g**-1)*5.2],[-1,0,-(W*g**-1)*3],[3,-3,-(12**-1)*(W*g**-1)*12**2]])\n",
- "Y=np.array([[-0.75*(W*g**-1)],[(W*g**-1)*1.3-W],[0]])\n",
- "C=np.linalg.solve(X,Y)\n",
- "A=C[0] #lb\n",
- "B=C[1] #lb\n",
- "alpha=C[2] #rad/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of alpha is',round(alpha,1),\"rad/s**2 countercockwise\",'and of A and B are',round(A,1),\"lb up and\",round(B,2),\"lb to the right\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of alpha is 2.6 rad/s**2 countercockwise and of A and B are 14.4 lb up and 7.91 lb to the right\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-16, Page No 345"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mc=7.25 #kg\n",
- "d=0.9 #m\n",
- "la=0.2 #m\n",
- "ma=9 #kg\n",
- "F=45 #N\n",
- "ay=0 #m/s**2\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "I=2*(0.5*mc*(d*2**-1)**2)+0.5*ma*(la*2**-1)**2 #kg-m**2\n",
- "#Using the equations of motion\n",
- "Na=(2*mc+ma)*g #N\n",
- "#Simplfying using radial velocity formula\n",
- "#Solving the two equations using matrix method\n",
- "A=np.array([[-1,-(2*mc+ma)],[(d*2**-1),-I/(d*2**-1)]])\n",
- "B=np.array([[-F],[F*(la*2**-1)]])\n",
- "C=np.linalg.solve(A,B)\n",
- "F=C[0] #N\n",
- "ax=C[1] #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The computation yields ax=',round(ax,2),\"m/s**2 to the right.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The computation yields ax= 1.13 m/s**2 to the right.\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-18, Page No 347"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=0.05 #m cylinder radius\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Here the equation has been solved in terms of the veriables\n",
- "#Hence we directly consider the final result\n",
- "av=(2*g)/3 #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of av is',round(av,2),\"m/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of av is 6.53 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-21, Page No 349"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#initilization of variables\n",
- "W=16.1 #lb\n",
- "v=9 #ft/s\n",
- "# as phi=30 degrees,\n",
- "sinphi=(2)**-1\n",
- "cosphi=(3**0.5)*2**-1\n",
- "r=0.5 #ft\n",
- "g=32.2 #ft/s**2\n",
- "OG=4.5 #ft\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "an=v**2/OG #ft/s**2\n",
- "#Solving for alpha we get\n",
- "N=(W*g**-1)*an+W*cosphi #lb\n",
- "#Using equations of motion\n",
- "A=np.array([[1,-r],[-1,-r*r]])\n",
- "B=np.array([[W*sinphi],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "F=C[0] #lb\n",
- "at=C[1] #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of N and F are',round(N,1),\"lb and\",round(F,2),\"lb respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of N and F are 22.9 lb and 2.68 lb respectively.\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-25, Page No 353"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=50 #lb\n",
- "P=10 #lb\n",
- "t=5 #s\n",
- "vo=0 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "ax=(P*g)/W #ft/s**2\n",
- "#Solving by matrix method for A and B\n",
- "F=np.array([[1,1],[-4,4]])\n",
- "Q=np.array([[W],[P]])\n",
- "R=np.linalg.solve(F,Q)\n",
- "#Velocity calculations\n",
- "v=vo+ax*t #ft/s\n",
- "A=R[0] #lb\n",
- "B=R[1] #lb\n",
- "\n",
- "#Result\n",
- "print'The velocity of the door after 5s is',round(v,1),\"ft/s\",'and A=',round(A,1),\"lb\",'and B=',round(B,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity of the door after 5s is 32.2 ft/s and A= 23.8 lb and B= 26.3 lb\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-26, Page No 354"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "#Initilization of variables\n",
- "AB=2 #m\n",
- "m=2 #kg\n",
- "F=20 #N\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equation of motion\n",
- "a=F/m #m/s**2\n",
- "#Solving by matrix method for Na and Nb\n",
- "A=np.array([[1,-1],[4*5**-1,4*5**-1]])\n",
- "B=np.array([[m*g],[F*(3*5**-1)]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a),\"m/s**2\",'and the reactions are: Na=',round(C[0],1),\"N up\",'and Nb=',round(C[1],1),\"N up\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is 10.0 m/s**2 and the reactions are: Na= 17.3 N up and Nb= -2.3 N up\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-27, Page No 354"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "vo=0 #ft/s\n",
- "\n",
- "#Calculations\n",
- "s=(0.011*5280*2)/(2*0.004)\n",
- "\n",
- "#Result\n",
- "print'It travels',round(s),\"ft along the level before coming to rest\"\n",
- "#Answer in the textbook is incorrect by 20ft\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "It travels 14520.0 ft along the level before coming to rest\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-28, Page No 355"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "u=0.3 #coefficient of friction\n",
- "m=70 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#CASE 1\n",
- "#Using equations of motion\n",
- "Na=m*g #N\n",
- "ah=(u*Na)/m #m/s**2\n",
- "#CASE 2\n",
- "#Applying sum of moments equal to zero\n",
- "F=(Na*0.3)/1.2 #N\n",
- "a_h=F/m #m/s**2\n",
- "\n",
- "#Result\n",
- "#Intutive insights can be attained after we get these results\n",
- "print'The value of Na is',round(Na),\"N\"\n",
- "print'and that of acceleration are:'\n",
- "print'1st value=',round(ah,2),\"m/s**2\",'and 2nd value is',round(a_h,2),\"m/s**2\"\n",
- "print'and the value of F is',round(F),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of Na is 686.0 N\n",
- "and that of acceleration are:\n",
- "1st value= 2.94 m/s**2 and 2nd value is 2.45 m/s**2\n",
- "and the value of F is 172.0 N\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-29, Page No 356"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=60 #kg\n",
- "me=660 #kg\n",
- "a=6 #m/s**2\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "P=m*a+m*g #N\n",
- "#Scale reading\n",
- "R=P/g #kg\n",
- "#Increase in mass\n",
- "I=R-m #kg\n",
- "#Tension\n",
- "T=me*a+me*g #N\n",
- "\n",
- "#Result\n",
- "print'The value of P is',round(P),\"n\"\n",
- "print'The apparent icrease in weight is',round(I,1),\"kg\",'and the tension in the cable is',round(T),\"N\"\n",
- "#Answer in the textbook is off by 28 #N in Tension\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P is 948.0 n\n",
- "The apparent icrease in weight is 36.7 kg and the tension in the cable is 10428.0 N\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-30, Page No 356"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "u=0.2 #coefficient of friction\n",
- "ma=1.2 #kg\n",
- "mb=2 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "Nb=mb*g #N\n",
- "F=u*Nb #N\n",
- "#Using equations of motion\n",
- "#Solving for T and a\n",
- "A=np.array([[-1,-ma],[1,-mb]])\n",
- "B=np.array([[-ma*g],[F]])\n",
- "C=np.linalg.solve(A,B)\n",
- "T=C[0] #N\n",
- "a=C[1] #m/s**2\n",
- "#Taking the sum of the moments\n",
- "x_m=-(F*0.15+T*0.15)/Nb #m\n",
- "x=x_m*1000 #mm\n",
- "\n",
- "#Result\n",
- "print'The acceleration of block A is',round(a,2),\"m/s**2\" \n",
- "print'and Nb acts at a distance of',round(x,1),\"mm.(Negative sign indictaes that the side assumed is incorrect)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration of block A is 2.45 m/s**2\n",
- "and Nb acts at a distance of -97.5 mm.(Negative sign indictaes that the side assumed is incorrect)\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-31, Page No 357"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "a=2.5 #m/s**2\n",
- "mA=3 #kg\n",
- "mB=7 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "F=(mA+mB)*a #N\n",
- "#Using equations of motion\n",
- "Py=mB*g #N\n",
- "#Solving for Px and H\n",
- "A=np.array([[1,1],[-0.0375,0.0375]])\n",
- "B=np.array([[mB*a],[Py*0.05]])\n",
- "C=np.linalg.solve(A,B)\n",
- "Px=C[0] #N\n",
- "H=C[1] #N\n",
- "\n",
- "#Result\n",
- "print'The value of H is',round(H,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of H is 54.5 N\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-32, Page No 358"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=20 #kg\n",
- "g=9.8 #m/s**2\n",
- "vo=3 #m/s\n",
- "v=0 #m/s\n",
- "s=4 #m\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "Na=m*g #N\n",
- "F=(Na*0.075)/0.125 #N\n",
- "a=F/m #m/s**2\n",
- "#Displacement \n",
- "d=-(v**2-vo**2)/(2*a) #m\n",
- "displ=s-d #m\n",
- "v_f=sqrt(2*a*displ) #m/s\n",
- "\n",
- "#Result\n",
- "print'The final velocity is',round(v_f,2),\"m/s to the left.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The final velocity is 6.17 m/s to the left.\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-33, Page No 358"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mA=30 #kg\n",
- "mB=45 #kg\n",
- "u_ab=3**-1 #coefficient of friction between two blocks\n",
- "u_bp=10**-1 #coefficient of friction between block and horizontal plane\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#By inspection\n",
- "Na=mA*g #N\n",
- "Nb=Na+mB*g #N\n",
- "a=(u_ab*Na-u_bp*Nb)/mB #m/s**2\n",
- "P=(mA*a+u_ab*Na) #N\n",
- "#For block A\n",
- "#Solving for P,F and a\n",
- "A=np.array([[1,-1,-mA],[-0.05,-0.075,0],[0,1,-mB]])\n",
- "B=np.array([[0],[-Na*0.050],[Nb*u_bp]])\n",
- "C=np.linalg.solve(A,B)\n",
- "P_new=C[0] #N\n",
- "\n",
- "#Result\n",
- "#As p < p_new\n",
- "print'The maximum value of P is',round(P),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum value of P is 114.0 N\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-34, Page No 359"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Vo=1.5 #m/s\n",
- "V=0 #m/s\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "a=(g*0.2)/0.75 #m/s**2\n",
- "t=-(V-Vo)/a #s\n",
- "\n",
- "#Result\n",
- "print'The maximum acceleration is',round(a,2),\"m/s**2\",'and minimum time is',round(t,2),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum acceleration is 2.61 m/s**2 and minimum time is 0.57 s\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-36, Page No 361"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "vo=0 #mi/h\n",
- "v=60 #mi/h\n",
- "t=13.8 #s\n",
- "W=3385 #lb\n",
- "xb=46 #in\n",
- "xf=66 #in\n",
- "xv=31 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "a=(((v*88*60)/3600)-vo)/t #ft/s**2\n",
- "#Summing horizontal forces\n",
- "F=(W/g)*a #lb\n",
- "#Solving for Rf and Rr\n",
- "A=np.array([[1,1],[-xf,xb]])\n",
- "B=np.array([[W],[-F*xv]])\n",
- "C=np.linalg.solve(A,B)\n",
- "Rr=C[0] #lb\n",
- "Rf=C[1] #lb\n",
- "\n",
- "#Result\n",
- "print'The value of reactions are Rf=',round(Rf),\"lb\",'and Rr=',round(Rr),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of reactions are Rf= 1809.0 lb and Rr= 1576.0 lb\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-43, Page No 366"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=161 #lb\n",
- "F=16.1 #lb\n",
- "r=18 #ft radius\n",
- "t=2 #s\n",
- "g=32.2 #ft/s**2\n",
- "wo=0 #rad/s\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "#Solving for T and alpha\n",
- "A=np.array([[r*12**-1,-0.5*(W*g**-1)*(r*12**-1)**2],[-1,-F*g**-1]])\n",
- "B=np.array([[0],[-F]])\n",
- "C=np.linalg.solve(A,B)\n",
- "alpha=C[1] #rad/s**2\n",
- "w=wo+(alpha*t) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular speed is',round(w,2),\"rad/s\"\n",
- "\n",
- "#The ans is incorrect in textbook."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular speed is 7.58 rad/s\n"
- ]
- }
- ],
- "prompt_number": 43
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-47, Page No 369"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization fo variables\n",
- "r=2000 #ft\n",
- "g=32.2 #ft/s**2\n",
- "d=4.71 #ft\n",
- "v=176 #ft/s\n",
- "\n",
- "#Calculations\n",
- "e=(d*v**2)/(g*r) #ft\n",
- "\n",
- "#Result\n",
- "print'The superelevation is',round(e,2),\"ft\"\n",
- "#Watch the unit in the final answer\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The superelevation is 2.27 ft\n"
- ]
- }
- ],
- "prompt_number": 45
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-48, Page No 369"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "a=5 #ft/s**2\n",
- "C=50 #lb-ft\n",
- "W=161 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "T=0.5*(W/g)*1**2*a+C #lb\n",
- "Ox=-T*(2/sqrt(a)) #lb\n",
- "Oy=T*(1/sqrt(a))+W #lb\n",
- "Wa=T/(1-(a/g)) #lb\n",
- "\n",
- "#Result\n",
- "print'The values are: T=',round(T,1),\"lb\",', Wa=',round(Wa),\"b\",',Ox=',round(Ox,1),\"lb\",'and Oy=+',round(Oy),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are: T= 62.5 lb , Wa= 74.0 b ,Ox= -55.9 lb and Oy=+ 189.0 lb\n"
- ]
- }
- ],
- "prompt_number": 49
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-49, Page No 370"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=100 #kg\n",
- "mr=20 #kg\n",
- "w=8 #rad/s\n",
- "l1=300 #mm\n",
- "l2=600 #mm\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "r_bar=(mr*l1+m*750)/120 #mm\n",
- "I=(3**-1)*mr*(l2*1000**-1)**2+(2*5**-1)*m*(l1*2000**-1)**2+m*(0.75)**2 #kg.m**2\n",
- "alpha=(m+mr)*g*(r_bar*1000**-1)/I #rad/s**2\n",
- "On=(m+mr)*(r_bar*1000**-1)*w**2 #N\n",
- "Ot=((m+mr)*(r_bar*1000**-1)*alpha)-(m+mr)*g #N\n",
- "\n",
- "#Result\n",
- "print'The angular acceleration is',round(alpha,1),\"rad/s**2\",'and On=',round(On),\"N\",'and Ot=',round(Ot,1),\"N\"\n",
- "\n",
- "#Due to decimal accuracy there is discrepancy in answers with the textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular acceleration is 13.3 rad/s**2 and On= 5184.0 N and Ot= -96.3 N\n"
- ]
- }
- ],
- "prompt_number": 48
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-50, Page No 370"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=40 #lb\n",
- "w=10 #rad/s\n",
- "alpha=2 #rad/s**2\n",
- "r=2 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "On=(W*g**-1)*(1*6**-1)*w**2 #lb\n",
- "Ot=(W*g**-1)*(1*6**-1)*alpha\n",
- "Io=(0.5*(W*g**-1)*0.5**2)*2+((W*g**-1)*(1*6**-1)**2)*2\n",
- "\n",
- "#Result\n",
- "print'The reaction components are On=',round(On,1),\"lb to the right\",'and Ot=',round(Ot,2),\"lb up\"\n",
- "print'The value of Io is',round(Io,2),\"lb-ft\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction components are On= 20.7 lb to the right and Ot= 0.41 lb up\n",
- "The value of Io is 0.38 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 59
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-51, Page No 371"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilizatin of variables\n",
- "W=6 #lb\n",
- "l=8 #ft\n",
- "v=10 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "# as theta1=60 degrees & theta2=30 degrees\n",
- "costheta1=2**-1\n",
- "costheta2=(3**0.5)*2**-1\n",
- "\n",
- "#Calculations\n",
- "Fe=(W*v**2)*(g*l*0.5)**-1 #lb\n",
- "#Using equations of motion\n",
- "#Solving for C and T\n",
- "A=np.array([[costheta1,-costheta2],[costheta2,costheta1]])\n",
- "B=np.array([[-Fe],[W]])\n",
- "P=np.linalg.solve(A,B) #lb\n",
- "C=P[0] #lb\n",
- "T=P[1] #lb\n",
- "\n",
- "#Result\n",
- "print'The value of C is',round(C,2),\"lb\",'and T is',round(T,2),\"lb\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of C is 2.87 lb and T is 7.03 lb\n"
- ]
- }
- ],
- "prompt_number": 49
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-52, Page No 372"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=32.2 #lb\n",
- "T=120 #lb\n",
- "m=1 #slug\n",
- "r=6*12**-1 #ft\n",
- "\n",
- "#Calculations\n",
- "w=sqrt((T*(3*5**-1)*4)*(m*r*3)**-1) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular speed permissible is',round(w,1),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular speed permissible is 13.9 rad/s\n"
- ]
- }
- ],
- "prompt_number": 50
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-53, Page No 372"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=30 #kg\n",
- "k=0.45 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Using equations of motion\n",
- "#Solving for T1,T2 and alpha\n",
- "A=np.array([[1,0,-m],[0,-1,-45],[-0.6,0.3,-m*k**2]])\n",
- "B=np.array([[50*g],[-150*g],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The values are: T1=',round(C[0]),\"N\",',T2=',round(C[1]),\"N\",'and alpha=',round(C[2],1),\"rad/s**2\"\n",
- "\n",
- "# The answer for T2 and alpha is off by 4 & 0.1 units"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are: T1= 607.0 N ,T2= 1294.0 N and alpha= 3.9 rad/s**2\n"
- ]
- }
- ],
- "prompt_number": 51
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-54, Page No 373"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wc=28 #lb\n",
- "v=16 #ft/s\n",
- "Ib=12 #ft-lb-s**2\n",
- "u=0.4 #coefficient of friction\n",
- "t=2 #s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "T=Wc+(Wc*g**-1)*8 #lb\n",
- "alpha=(8*12)*15**-1 #rad/s**2\n",
- "F=((Ib*alpha)+(T*1.25))/t #lb\n",
- "N=F/u #lb\n",
- "#Summing moments about D\n",
- "P=(N*8+F*3)/40 #lb\n",
- "#Summing forces horizontally and vertically\n",
- "Dx=151-P #lb\n",
- "Dy=-F #lb\n",
- "\n",
- "#Result\n",
- "print'The reactions at D are: Dx=',round(Dx),\"lb to right\",'and Dy=',round(Dy,1),\"lb down\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reactions at D are: Dx= 116.0 lb to right and Dy= -60.2 lb down\n"
- ]
- }
- ],
- "prompt_number": 65
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-55, Page No 374"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=8 #kg\n",
- "n=90 #rpm\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "Fg=m*g #N\n",
- "w=2*pi*n/60 #rad/s\n",
- "#using equations of motion\n",
- "By=m*g #N\n",
- "#Solving for Bx and C\n",
- "A=np.array([[1,1],[-0.3,0.9]])\n",
- "B=np.array([[m*0.3*w**2],[By*0.3]])\n",
- "C=np.linalg.solve(A,B) #N\n",
- "\n",
- "#Result\n",
- "print'The solution is: Bx=',round(C[0]),\"N\",',By=',round(By,1),\"N\",'and C=',round(C[1],1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is: Bx= 140.0 N ,By= 78.4 N and C= 72.9 N\n"
- ]
- }
- ],
- "prompt_number": 79
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-56, Page No 375"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=8 #kg\n",
- "n=90 #rpm\n",
- "g=9.8 #m/s**2\n",
- "r=0.3 #m\n",
- "\n",
- "#calculations\n",
- "w=2*pi*n/60 #rad/s\n",
- "#Using equations of motion\n",
- "C=(m*g*0.3+m*r*w**2*r)/1.2 #N\n",
- "Bx=-C+m*r*w**2 #N\n",
- "By=m*g #N\n",
- "\n",
- "#Result\n",
- "print'The solution is: Bx=',round(Bx),\"N\",',By=',round(By,1),\"N\",'and C=',round(C,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is: Bx= 140.0 N ,By= 78.4 N and C= 72.9 N\n"
- ]
- }
- ],
- "prompt_number": 66
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-57, Page No 375"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Na=294 #N\n",
- "Nb=735 #N\n",
- "\n",
- "#Calculations\n",
- "a=(10**-1*Nb-3**-1*Na)/45 #m/s**2\n",
- "P=(3**-1*Na)-30*a #N\n",
- "\n",
- "#Result\n",
- "print'The solution is P=',round(P),\"N\",'and a=',round(a,3),\"m/s**2\"\n",
- "# The negative sign indicates the assumed direction is incorrect."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is P= 114.0 N and a= -0.544 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 68
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-58, Page No 376"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=50 #lb\n",
- "g=32.2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "a=(10/(W/g)) #ft/s**2\n",
- "B=((2.5*(W/g)*a)+4*W-1.5*10)/8 #lb\n",
- "A=50-B #lb\n",
- "\n",
- "#Result\n",
- "print'The solution is A=',round(A,1),\"lb\",',B=',round(B,1),\"lb\",'and a=',round(a,2),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is A= 23.8 lb ,B= 26.3 lb and a= 6.44 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 69
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-60, Page No 377"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "g=9.8 #m/s**2\n",
- "r1=0.3 #m\n",
- "m1=20 #kg\n",
- "m2=100 #kg\n",
- "r2=0.75 #m\n",
- "\n",
- "#Calculations\n",
- "alpha=(m1*g*r1+m2*g*r2)*(m1*r1**2+(m1/12)*0.6**2+m2*r2**2+(2*5**-1)*m2*0.15**2)**-1 #rad/s**2\n",
- "\n",
- "#Result\n",
- "print'The angular acceleration is',round(alpha,1),\"rad/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular acceleration is 13.4 rad/s**2\n"
- ]
- }
- ],
- "prompt_number": 71
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-61, Page No 378"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=15*12**-1 #ft\n",
- "W=600 #lb\n",
- "# as theta=25 degrees,\n",
- "sintheta=0.422\n",
- "\n",
- "#calculations\n",
- "ax=(r*W*sintheta)*((r**-1)*14.5+r*18.6)**-1 #ft/s**2\n",
- "F=(W*sintheta)-(18.6*ax) #lb\n",
- "\n",
- "#Result\n",
- "print'The solution is F=',round(F,1),\"lb\",'and ax=',round(ax,2),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is F= 84.3 lb and ax= 9.08 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-62, Page No 378"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=7 #kg\n",
- "g=9.8 #m/s**2\n",
- "r=0.5 #m\n",
- "I=0.875 #kg.m**2\n",
- "\n",
- "#Calculations\n",
- "#Solving for alpha and T\n",
- "alpha=(m*g*r)/(I+m*r*0.5) #rad/s**2\n",
- "T=(I*alpha)/r #N\n",
- "\n",
- "#Result\n",
- "print'The soultion is alpha=',round(alpha,1),\"rad/s**2\",'and T=',round(T,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The soultion is alpha= 13.1 rad/s**2 and T= 22.9 N\n"
- ]
- }
- ],
- "prompt_number": 79
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter16_4.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter16_4.ipynb deleted file mode 100755 index 22430a7a..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter16_4.ipynb +++ /dev/null @@ -1,1911 +0,0 @@ -{
- "metadata": {
- "name": "chapter16.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 16: Dynamics of a Rigid Body in Plane Motion"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-2, Page No 335"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Intilization of variables\n",
- "W=600 #lb\n",
- "d=30 #in\n",
- "# as theta=25 degrees,\n",
- "sintheta=0.422\n",
- "costheta=0.906\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "m=W/g #lb-s**2/ft\n",
- "#Moment of inertia\n",
- "I=0.5*m*((d/2)/12)**2 #lb-s**2-ft\n",
- "#Applying Newtons law and coservation of angular momentum and rolling\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,m,0,0],[0,0,0,1],[((d/2)/12),0,-I,0],[0,1,-((d/2)/12),0]])\n",
- "B=np.array([[W*sintheta],[W*costheta],[0],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The Frictional Force is',round(C[0],1),\"lb\",'and the acceleration is',round(C[1],2),\"ft/s**2\"\n",
- "\n",
- "# The answers wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Frictional Force is 84.4 lb and the acceleration is 9.06 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-3, Page No 336"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=18 #kg\n",
- "d=0.6 #m\n",
- "vo=3 #m/s\n",
- "# as theta=20 degrees,\n",
- "sintheta=0.342\n",
- "costheta=0.939\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of Inertia\n",
- "I=0.5*m*(d/2)**2 \n",
- "#Applying Newtons second Law a\n",
- "A=np.array([[1,m,0,0],[0,0,1,0],[d/2,0,0,-I],[0,1,0,(-d/2)]])\n",
- "B=np.array([[g*m*sintheta],[g*m*costheta],[0],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing the answers in variables\n",
- "F=C[0] #N\n",
- "ax=C[1] #m/s**2\n",
- "Na=C[2] #N\n",
- "alpha=C[3] #rad/s**2\n",
- "#Time Calculations\n",
- "v=0 #m/s**2\n",
- "t=(vo)/ax #s\n",
- "\n",
- "#Result\n",
- "print'It takes',round(t,2),\"s to reach the highest point of travel\"\n",
- "\n",
- "# The ans is off by 0.01 sec."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "It takes 1.34 s to reach the highest point of travel\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-5, Page No 337"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=20 #kg\n",
- "F1=40 #N\n",
- "ro=0.6 #m\n",
- "ri=0.45 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of inertia\n",
- "I=(2*5**-1)*m*ro**2 #kg-m**2\n",
- "#Applying Newtons Law and conservation of angular Momentum\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,m],[ro,-I/ro]])\n",
- "B=np.array([[F1],[F1*ri]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing answers in variables\n",
- "F=C[0] #N\n",
- "a=C[1] #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The acceleration is',round(a,2),\"m/s**2\"\n",
- "print'The force is',round(F,2),\"N\"\n",
- "#The solution in the textbook is incorrect\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration is 0.36 m/s**2\n",
- "The force is 32.86 N\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-6, Page No 338"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=16.1 #lb\n",
- "u=0.10 #co-efficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "F=1.39 #lb\n",
- "\n",
- "#Calculations\n",
- "#Applying Newtons Second Law\n",
- "#Using F=1.39 lb\n",
- "a=((W*sintheta)-F)/(W*g**-1) #ft/s**2\n",
- "alpha=(F*0.5*5/2)/((W*g**-1)*(0.5**2)) #rad/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,1),\"ft/s**2\",'and alpha is ',round(alpha,1),\"rad/s**2.\"\n",
- "print'Hence the sphere will both,roll and slip'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is 13.3 ft/s**2 and alpha is 13.9 rad/s**2.\n",
- "Hence the sphere will both,roll and slip\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-8, Page No 339"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=30 degrees\n",
- "sintheta=2**-1\n",
- "W=80 #lb\n",
- "Ww=100 #lb\n",
- "I=4 #slug-ft**2\n",
- "r=0.5 #ft\n",
- "v= 20 #ft/s\n",
- "vo=0 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using Equations of motion\n",
- "#Solving the system of linear equatinons by matrix method\n",
- "A=np.array([[-1,0,-W/g],[1,-1,-Ww/g],[0,r,-2*I]])\n",
- "B=np.array([[-W],[Ww*sintheta],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing values in variables\n",
- "T=C[0] #lb\n",
- "F=C[1] #lb\n",
- "a=C[2] #ft/s**2\n",
- "#Time calculations\n",
- "t=(v-vo)/a #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t,1),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 14.4 s\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-9, Page No 340"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "M=70 #kg\n",
- "ko=0.4 #m\n",
- "ri=0.45 #m\n",
- "ro=0.6 #m\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "m=35 #kg\n",
- "g= 9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "I=M*ko**2 #kg-m**2\n",
- "#Using Equations of motion\n",
- "#Solving the equations by matrix method\n",
- "A=np.array([[-1,-m*0.15,0],[1,-M*ro,-1],[-ri,-I,ro]])\n",
- "B=np.array([[-m*g],[M*g*sintheta],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "F=C[2] #N\n",
- "Na=M*g*costheta #N\n",
- "#Required coefficient of friction\n",
- "u=F/Na #coefficient of friction\n",
- "\n",
- "#Result\n",
- "print'The value of alpha is',round(C[1],2),\"rad/s**2\",'and tension is',round(C[0]),\"N\"\n",
- "print'F=',round(F),\"N,\",'Na=',round(Na),\"N\",'and u=',round(u,2),\"(coefficient of friction)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of alpha is -4.15 rad/s**2 and tension is 365.0 N\n",
- "F= 196.0 N, Na= 594.0 N and u= 0.33 (coefficient of friction)\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Exampe 16.16-10, Page No 341"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=200 #kg\n",
- "g=9.8 #m/s**2\n",
- "r=1.2 #m\n",
- "F1=1000 #N\n",
- "F2=1400 #N\n",
- "\n",
- "#Calculations\n",
- "N=m*g #N\n",
- "I=(2*5**-1)*(m)*r**2 #kg-m**2\n",
- "#Using equations of motion\n",
- "#Solving for F and alpha using matrix method\n",
- "#Applying equations of motion\n",
- "A=np.array([[1,-m],[-r,-I/r]])\n",
- "B=np.array([[F1-F2],[F1*r]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing values\n",
- "F=C[0] #N\n",
- "alpha=C[1] #rad/s**2\n",
- "a=r*alpha #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,2),\"m/s**2\",'and F is',round(F),\"N\"\n",
- "#The negative signs indicate that the direction is opposite to what was origninally assumed\n",
- "# The answers wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is -2.57 m/s**2 and F is -829.0 N\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-11, Page No 341"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "Wa=161 #lb\n",
- "Wb=193.2 #lb\n",
- "Wc=300 #lb\n",
- "ka=3 #ft\n",
- "kb=2.5 #ft\n",
- "# as theta1=30 degrees & theta2=45 degrees,\n",
- "sintheta1=2**-1\n",
- "sintheta2=(2**0.5)**-1\n",
- "costheta2=(2**0.5)**-1\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of inertia Calculations\n",
- "Ia=(Wa/g)*ka**2 #lb-s**2-ft\n",
- "Ib=(Wb/g)*kb**2 #lb-s**2-ft\n",
- "#Using equations of motion for A and B and C\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,1,-Wa/g,0,0],[1,-4,-Ia*(4**-1),0,0],[-2,0,-Ib*(5*8**-1),4,0],[0,0,-(Wc/g)*(5*2**-1),-1,-0.25],[0,0,0,0,1]])\n",
- "B=np.array([[Wa*sintheta1],[0],[0],[-Wc*costheta2],[Wc*sintheta2]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing values in the variables\n",
- "T1=C[0] #lb\n",
- "T2=C[3] #lb\n",
- "a=C[2] #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The values are:'\n",
- "print'a=',round(a,2),\"ft/s**2 ,\",'T1=',round(T1),\"lb\",'and T2=',round(T2,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are:\n",
- "a= 3.93 ft/s**2 , T1= 89.0 lb and T2= 67.5 lb\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-12, Page No 342"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=644 #lb\n",
- "F=30 #lb\n",
- "# as theta=30 degrees,\n",
- "sintheta=(2)**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "r=1.5 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,-W/g],[-r,-(2**-1)*(W/g)*(2*2)*(r**-1)]])\n",
- "B=np.array([[(W*sintheta)-(F*costheta)],[-F*2]])\n",
- "C=np.linalg.solve(A,B)\n",
- "a=C[1] #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,2),\"ft/s**2\"\n",
- "# The negative sign indicates that the cylinder will roll down the plane."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is -6.78 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-14, Page No 344"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=20 #lb\n",
- "g=32.2 #ft/s**2\n",
- "vb=0.5 #rad/s\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "#Solving the three equations simultaneously by matrix method\n",
- "X=np.array([[0,1,-(W*g**-1)*5.2],[-1,0,-(W*g**-1)*3],[3,-3,-(12**-1)*(W*g**-1)*12**2]])\n",
- "Y=np.array([[-0.75*(W*g**-1)],[(W*g**-1)*1.3-W],[0]])\n",
- "C=np.linalg.solve(X,Y)\n",
- "A=C[0] #lb\n",
- "B=C[1] #lb\n",
- "alpha=C[2] #rad/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of alpha is',round(alpha,1),\"rad/s**2 countercockwise\",'and of A and B are',round(A,1),\"lb up and\",round(B,2),\"lb to the right\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of alpha is 2.6 rad/s**2 countercockwise and of A and B are 14.4 lb up and 7.91 lb to the right\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-16, Page No 345"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mc=7.25 #kg\n",
- "d=0.9 #m\n",
- "la=0.2 #m\n",
- "ma=9 #kg\n",
- "F=45 #N\n",
- "ay=0 #m/s**2\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "I=2*(0.5*mc*(d*2**-1)**2)+0.5*ma*(la*2**-1)**2 #kg-m**2\n",
- "#Using the equations of motion\n",
- "Na=(2*mc+ma)*g #N\n",
- "#Simplfying using radial velocity formula\n",
- "#Solving the two equations using matrix method\n",
- "A=np.array([[-1,-(2*mc+ma)],[(d*2**-1),-I/(d*2**-1)]])\n",
- "B=np.array([[-F],[F*(la*2**-1)]])\n",
- "C=np.linalg.solve(A,B)\n",
- "F=C[0] #N\n",
- "ax=C[1] #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The computation yields ax=',round(ax,2),\"m/s**2 to the right.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The computation yields ax= 1.13 m/s**2 to the right.\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-18, Page No 347"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=0.05 #m cylinder radius\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Here the equation has been solved in terms of the veriables\n",
- "#Hence we directly consider the final result\n",
- "av=(2*g)/3 #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of av is',round(av,2),\"m/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of av is 6.53 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-21, Page No 349"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#initilization of variables\n",
- "W=16.1 #lb\n",
- "v=9 #ft/s\n",
- "# as phi=30 degrees,\n",
- "sinphi=(2)**-1\n",
- "cosphi=(3**0.5)*2**-1\n",
- "r=0.5 #ft\n",
- "g=32.2 #ft/s**2\n",
- "OG=4.5 #ft\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "an=v**2/OG #ft/s**2\n",
- "#Solving for alpha we get\n",
- "N=(W*g**-1)*an+W*cosphi #lb\n",
- "#Using equations of motion\n",
- "A=np.array([[1,-r],[-1,-r*r]])\n",
- "B=np.array([[W*sinphi],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "F=C[0] #lb\n",
- "at=C[1] #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of N and F are',round(N,1),\"lb and\",round(F,2),\"lb respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of N and F are 22.9 lb and 2.68 lb respectively.\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-25, Page No 353"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=50 #lb\n",
- "P=10 #lb\n",
- "t=5 #s\n",
- "vo=0 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "ax=(P*g)/W #ft/s**2\n",
- "#Solving by matrix method for A and B\n",
- "F=np.array([[1,1],[-4,4]])\n",
- "Q=np.array([[W],[P]])\n",
- "R=np.linalg.solve(F,Q)\n",
- "#Velocity calculations\n",
- "v=vo+ax*t #ft/s\n",
- "A=R[0] #lb\n",
- "B=R[1] #lb\n",
- "\n",
- "#Result\n",
- "print'The velocity of the door after 5s is',round(v,1),\"ft/s\",'and A=',round(A,1),\"lb\",'and B=',round(B,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity of the door after 5s is 32.2 ft/s and A= 23.8 lb and B= 26.3 lb\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-26, Page No 354"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "#Initilization of variables\n",
- "AB=2 #m\n",
- "m=2 #kg\n",
- "F=20 #N\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equation of motion\n",
- "a=F/m #m/s**2\n",
- "#Solving by matrix method for Na and Nb\n",
- "A=np.array([[1,-1],[4*5**-1,4*5**-1]])\n",
- "B=np.array([[m*g],[F*(3*5**-1)]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a),\"m/s**2\",'and the reactions are: Na=',round(C[0],1),\"N up\",'and Nb=',round(C[1],1),\"N up\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is 10.0 m/s**2 and the reactions are: Na= 17.3 N up and Nb= -2.3 N up\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-27, Page No 354"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "vo=0 #ft/s\n",
- "\n",
- "#Calculations\n",
- "s=(0.011*5280*2)/(2*0.004)\n",
- "\n",
- "#Result\n",
- "print'It travels',round(s),\"ft along the level before coming to rest\"\n",
- "#Answer in the textbook is incorrect by 20ft\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "It travels 14520.0 ft along the level before coming to rest\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-28, Page No 355"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "u=0.3 #coefficient of friction\n",
- "m=70 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#CASE 1\n",
- "#Using equations of motion\n",
- "Na=m*g #N\n",
- "ah=(u*Na)/m #m/s**2\n",
- "#CASE 2\n",
- "#Applying sum of moments equal to zero\n",
- "F=(Na*0.3)/1.2 #N\n",
- "a_h=F/m #m/s**2\n",
- "\n",
- "#Result\n",
- "#Intutive insights can be attained after we get these results\n",
- "print'The value of Na is',round(Na),\"N\"\n",
- "print'and that of acceleration are:'\n",
- "print'1st value=',round(ah,2),\"m/s**2\",'and 2nd value is',round(a_h,2),\"m/s**2\"\n",
- "print'and the value of F is',round(F),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of Na is 686.0 N\n",
- "and that of acceleration are:\n",
- "1st value= 2.94 m/s**2 and 2nd value is 2.45 m/s**2\n",
- "and the value of F is 172.0 N\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-29, Page No 356"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=60 #kg\n",
- "me=660 #kg\n",
- "a=6 #m/s**2\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "P=m*a+m*g #N\n",
- "#Scale reading\n",
- "R=P/g #kg\n",
- "#Increase in mass\n",
- "I=R-m #kg\n",
- "#Tension\n",
- "T=me*a+me*g #N\n",
- "\n",
- "#Result\n",
- "print'The value of P is',round(P),\"n\"\n",
- "print'The apparent icrease in weight is',round(I,1),\"kg\",'and the tension in the cable is',round(T),\"N\"\n",
- "#Answer in the textbook is off by 28 #N in Tension\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P is 948.0 n\n",
- "The apparent icrease in weight is 36.7 kg and the tension in the cable is 10428.0 N\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-30, Page No 356"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "u=0.2 #coefficient of friction\n",
- "ma=1.2 #kg\n",
- "mb=2 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "Nb=mb*g #N\n",
- "F=u*Nb #N\n",
- "#Using equations of motion\n",
- "#Solving for T and a\n",
- "A=np.array([[-1,-ma],[1,-mb]])\n",
- "B=np.array([[-ma*g],[F]])\n",
- "C=np.linalg.solve(A,B)\n",
- "T=C[0] #N\n",
- "a=C[1] #m/s**2\n",
- "#Taking the sum of the moments\n",
- "x_m=-(F*0.15+T*0.15)/Nb #m\n",
- "x=x_m*1000 #mm\n",
- "\n",
- "#Result\n",
- "print'The acceleration of block A is',round(a,2),\"m/s**2\" \n",
- "print'and Nb acts at a distance of',round(x,1),\"mm.(Negative sign indictaes that the side assumed is incorrect)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration of block A is 2.45 m/s**2\n",
- "and Nb acts at a distance of -97.5 mm.(Negative sign indictaes that the side assumed is incorrect)\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-31, Page No 357"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "a=2.5 #m/s**2\n",
- "mA=3 #kg\n",
- "mB=7 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "F=(mA+mB)*a #N\n",
- "#Using equations of motion\n",
- "Py=mB*g #N\n",
- "#Solving for Px and H\n",
- "A=np.array([[1,1],[-0.0375,0.0375]])\n",
- "B=np.array([[mB*a],[Py*0.05]])\n",
- "C=np.linalg.solve(A,B)\n",
- "Px=C[0] #N\n",
- "H=C[1] #N\n",
- "\n",
- "#Result\n",
- "print'The value of H is',round(H,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of H is 54.5 N\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-32, Page No 358"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=20 #kg\n",
- "g=9.8 #m/s**2\n",
- "vo=3 #m/s\n",
- "v=0 #m/s\n",
- "s=4 #m\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "Na=m*g #N\n",
- "F=(Na*0.075)/0.125 #N\n",
- "a=F/m #m/s**2\n",
- "#Displacement \n",
- "d=-(v**2-vo**2)/(2*a) #m\n",
- "displ=s-d #m\n",
- "v_f=sqrt(2*a*displ) #m/s\n",
- "\n",
- "#Result\n",
- "print'The final velocity is',round(v_f,2),\"m/s to the left.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The final velocity is 6.17 m/s to the left.\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-33, Page No 358"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mA=30 #kg\n",
- "mB=45 #kg\n",
- "u_ab=3**-1 #coefficient of friction between two blocks\n",
- "u_bp=10**-1 #coefficient of friction between block and horizontal plane\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#By inspection\n",
- "Na=mA*g #N\n",
- "Nb=Na+mB*g #N\n",
- "a=(u_ab*Na-u_bp*Nb)/mB #m/s**2\n",
- "P=(mA*a+u_ab*Na) #N\n",
- "#For block A\n",
- "#Solving for P,F and a\n",
- "A=np.array([[1,-1,-mA],[-0.05,-0.075,0],[0,1,-mB]])\n",
- "B=np.array([[0],[-Na*0.050],[Nb*u_bp]])\n",
- "C=np.linalg.solve(A,B)\n",
- "P_new=C[0] #N\n",
- "\n",
- "#Result\n",
- "#As p < p_new\n",
- "print'The maximum value of P is',round(P),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum value of P is 114.0 N\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-34, Page No 359"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Vo=1.5 #m/s\n",
- "V=0 #m/s\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "a=(g*0.2)/0.75 #m/s**2\n",
- "t=-(V-Vo)/a #s\n",
- "\n",
- "#Result\n",
- "print'The maximum acceleration is',round(a,2),\"m/s**2\",'and minimum time is',round(t,2),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum acceleration is 2.61 m/s**2 and minimum time is 0.57 s\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-36, Page No 361"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "vo=0 #mi/h\n",
- "v=60 #mi/h\n",
- "t=13.8 #s\n",
- "W=3385 #lb\n",
- "xb=46 #in\n",
- "xf=66 #in\n",
- "xv=31 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "a=(((v*88*60)/3600)-vo)/t #ft/s**2\n",
- "#Summing horizontal forces\n",
- "F=(W/g)*a #lb\n",
- "#Solving for Rf and Rr\n",
- "A=np.array([[1,1],[-xf,xb]])\n",
- "B=np.array([[W],[-F*xv]])\n",
- "C=np.linalg.solve(A,B)\n",
- "Rr=C[0] #lb\n",
- "Rf=C[1] #lb\n",
- "\n",
- "#Result\n",
- "print'The value of reactions are Rf=',round(Rf),\"lb\",'and Rr=',round(Rr),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of reactions are Rf= 1809.0 lb and Rr= 1576.0 lb\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-43, Page No 366"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=161 #lb\n",
- "F=16.1 #lb\n",
- "r=18 #ft radius\n",
- "t=2 #s\n",
- "g=32.2 #ft/s**2\n",
- "wo=0 #rad/s\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "#Solving for T and alpha\n",
- "A=np.array([[r*12**-1,-0.5*(W*g**-1)*(r*12**-1)**2],[-1,-F*g**-1]])\n",
- "B=np.array([[0],[-F]])\n",
- "C=np.linalg.solve(A,B)\n",
- "alpha=C[1] #rad/s**2\n",
- "w=wo+(alpha*t) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular speed is',round(w,2),\"rad/s\"\n",
- "\n",
- "#The ans is incorrect in textbook."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular speed is 7.58 rad/s\n"
- ]
- }
- ],
- "prompt_number": 43
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-47, Page No 369"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization fo variables\n",
- "r=2000 #ft\n",
- "g=32.2 #ft/s**2\n",
- "d=4.71 #ft\n",
- "v=176 #ft/s\n",
- "\n",
- "#Calculations\n",
- "e=(d*v**2)/(g*r) #ft\n",
- "\n",
- "#Result\n",
- "print'The superelevation is',round(e,2),\"ft\"\n",
- "#Watch the unit in the final answer\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The superelevation is 2.27 ft\n"
- ]
- }
- ],
- "prompt_number": 45
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-48, Page No 369"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "a=5 #ft/s**2\n",
- "C=50 #lb-ft\n",
- "W=161 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "T=0.5*(W/g)*1**2*a+C #lb\n",
- "Ox=-T*(2/sqrt(a)) #lb\n",
- "Oy=T*(1/sqrt(a))+W #lb\n",
- "Wa=T/(1-(a/g)) #lb\n",
- "\n",
- "#Result\n",
- "print'The values are: T=',round(T,1),\"lb\",', Wa=',round(Wa),\"b\",',Ox=',round(Ox,1),\"lb\",'and Oy=+',round(Oy),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are: T= 62.5 lb , Wa= 74.0 b ,Ox= -55.9 lb and Oy=+ 189.0 lb\n"
- ]
- }
- ],
- "prompt_number": 49
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-49, Page No 370"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=100 #kg\n",
- "mr=20 #kg\n",
- "w=8 #rad/s\n",
- "l1=300 #mm\n",
- "l2=600 #mm\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "r_bar=(mr*l1+m*750)/120 #mm\n",
- "I=(3**-1)*mr*(l2*1000**-1)**2+(2*5**-1)*m*(l1*2000**-1)**2+m*(0.75)**2 #kg.m**2\n",
- "alpha=(m+mr)*g*(r_bar*1000**-1)/I #rad/s**2\n",
- "On=(m+mr)*(r_bar*1000**-1)*w**2 #N\n",
- "Ot=((m+mr)*(r_bar*1000**-1)*alpha)-(m+mr)*g #N\n",
- "\n",
- "#Result\n",
- "print'The angular acceleration is',round(alpha,1),\"rad/s**2\",'and On=',round(On),\"N\",'and Ot=',round(Ot,1),\"N\"\n",
- "\n",
- "#Due to decimal accuracy there is discrepancy in answers with the textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular acceleration is 13.3 rad/s**2 and On= 5184.0 N and Ot= -96.3 N\n"
- ]
- }
- ],
- "prompt_number": 48
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-50, Page No 370"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=40 #lb\n",
- "w=10 #rad/s\n",
- "alpha=2 #rad/s**2\n",
- "r=2 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "On=(W*g**-1)*(1*6**-1)*w**2 #lb\n",
- "Ot=(W*g**-1)*(1*6**-1)*alpha\n",
- "Io=(0.5*(W*g**-1)*0.5**2)*2+((W*g**-1)*(1*6**-1)**2)*2\n",
- "\n",
- "#Result\n",
- "print'The reaction components are On=',round(On,1),\"lb to the right\",'and Ot=',round(Ot,2),\"lb up\"\n",
- "print'The value of Io is',round(Io,2),\"lb-ft\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction components are On= 20.7 lb to the right and Ot= 0.41 lb up\n",
- "The value of Io is 0.38 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 59
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-51, Page No 371"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilizatin of variables\n",
- "W=6 #lb\n",
- "l=8 #ft\n",
- "v=10 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "# as theta1=60 degrees & theta2=30 degrees\n",
- "costheta1=2**-1\n",
- "costheta2=(3**0.5)*2**-1\n",
- "\n",
- "#Calculations\n",
- "Fe=(W*v**2)*(g*l*0.5)**-1 #lb\n",
- "#Using equations of motion\n",
- "#Solving for C and T\n",
- "A=np.array([[costheta1,-costheta2],[costheta2,costheta1]])\n",
- "B=np.array([[-Fe],[W]])\n",
- "P=np.linalg.solve(A,B) #lb\n",
- "C=P[0] #lb\n",
- "T=P[1] #lb\n",
- "\n",
- "#Result\n",
- "print'The value of C is',round(C,2),\"lb\",'and T is',round(T,2),\"lb\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of C is 2.87 lb and T is 7.03 lb\n"
- ]
- }
- ],
- "prompt_number": 49
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-52, Page No 372"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=32.2 #lb\n",
- "T=120 #lb\n",
- "m=1 #slug\n",
- "r=6*12**-1 #ft\n",
- "\n",
- "#Calculations\n",
- "w=sqrt((T*(3*5**-1)*4)*(m*r*3)**-1) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular speed permissible is',round(w,1),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular speed permissible is 13.9 rad/s\n"
- ]
- }
- ],
- "prompt_number": 50
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-53, Page No 372"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=30 #kg\n",
- "k=0.45 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Using equations of motion\n",
- "#Solving for T1,T2 and alpha\n",
- "A=np.array([[1,0,-m],[0,-1,-45],[-0.6,0.3,-m*k**2]])\n",
- "B=np.array([[50*g],[-150*g],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The values are: T1=',round(C[0]),\"N\",',T2=',round(C[1]),\"N\",'and alpha=',round(C[2],1),\"rad/s**2\"\n",
- "\n",
- "# The answer for T2 and alpha is off by 4 & 0.1 units"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are: T1= 607.0 N ,T2= 1294.0 N and alpha= 3.9 rad/s**2\n"
- ]
- }
- ],
- "prompt_number": 51
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-54, Page No 373"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wc=28 #lb\n",
- "v=16 #ft/s\n",
- "Ib=12 #ft-lb-s**2\n",
- "u=0.4 #coefficient of friction\n",
- "t=2 #s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "T=Wc+(Wc*g**-1)*8 #lb\n",
- "alpha=(8*12)*15**-1 #rad/s**2\n",
- "F=((Ib*alpha)+(T*1.25))/t #lb\n",
- "N=F/u #lb\n",
- "#Summing moments about D\n",
- "P=(N*8+F*3)/40 #lb\n",
- "#Summing forces horizontally and vertically\n",
- "Dx=151-P #lb\n",
- "Dy=-F #lb\n",
- "\n",
- "#Result\n",
- "print'The reactions at D are: Dx=',round(Dx),\"lb to right\",'and Dy=',round(Dy,1),\"lb down\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reactions at D are: Dx= 116.0 lb to right and Dy= -60.2 lb down\n"
- ]
- }
- ],
- "prompt_number": 65
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-55, Page No 374"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=8 #kg\n",
- "n=90 #rpm\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "Fg=m*g #N\n",
- "w=2*pi*n/60 #rad/s\n",
- "#using equations of motion\n",
- "By=m*g #N\n",
- "#Solving for Bx and C\n",
- "A=np.array([[1,1],[-0.3,0.9]])\n",
- "B=np.array([[m*0.3*w**2],[By*0.3]])\n",
- "C=np.linalg.solve(A,B) #N\n",
- "\n",
- "#Result\n",
- "print'The solution is: Bx=',round(C[0]),\"N\",',By=',round(By,1),\"N\",'and C=',round(C[1],1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is: Bx= 140.0 N ,By= 78.4 N and C= 72.9 N\n"
- ]
- }
- ],
- "prompt_number": 79
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-56, Page No 375"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=8 #kg\n",
- "n=90 #rpm\n",
- "g=9.8 #m/s**2\n",
- "r=0.3 #m\n",
- "\n",
- "#calculations\n",
- "w=2*pi*n/60 #rad/s\n",
- "#Using equations of motion\n",
- "C=(m*g*0.3+m*r*w**2*r)/1.2 #N\n",
- "Bx=-C+m*r*w**2 #N\n",
- "By=m*g #N\n",
- "\n",
- "#Result\n",
- "print'The solution is: Bx=',round(Bx),\"N\",',By=',round(By,1),\"N\",'and C=',round(C,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is: Bx= 140.0 N ,By= 78.4 N and C= 72.9 N\n"
- ]
- }
- ],
- "prompt_number": 66
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-57, Page No 375"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Na=294 #N\n",
- "Nb=735 #N\n",
- "\n",
- "#Calculations\n",
- "a=(10**-1*Nb-3**-1*Na)/45 #m/s**2\n",
- "P=(3**-1*Na)-30*a #N\n",
- "\n",
- "#Result\n",
- "print'The solution is P=',round(P),\"N\",'and a=',round(a,3),\"m/s**2\"\n",
- "# The negative sign indicates the assumed direction is incorrect."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is P= 114.0 N and a= -0.544 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 68
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-58, Page No 376"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=50 #lb\n",
- "g=32.2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "a=(10/(W/g)) #ft/s**2\n",
- "B=((2.5*(W/g)*a)+4*W-1.5*10)/8 #lb\n",
- "A=50-B #lb\n",
- "\n",
- "#Result\n",
- "print'The solution is A=',round(A,1),\"lb\",',B=',round(B,1),\"lb\",'and a=',round(a,2),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is A= 23.8 lb ,B= 26.3 lb and a= 6.44 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 69
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-60, Page No 377"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "g=9.8 #m/s**2\n",
- "r1=0.3 #m\n",
- "m1=20 #kg\n",
- "m2=100 #kg\n",
- "r2=0.75 #m\n",
- "\n",
- "#Calculations\n",
- "alpha=(m1*g*r1+m2*g*r2)*(m1*r1**2+(m1/12)*0.6**2+m2*r2**2+(2*5**-1)*m2*0.15**2)**-1 #rad/s**2\n",
- "\n",
- "#Result\n",
- "print'The angular acceleration is',round(alpha,1),\"rad/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular acceleration is 13.4 rad/s**2\n"
- ]
- }
- ],
- "prompt_number": 71
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-61, Page No 378"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=15*12**-1 #ft\n",
- "W=600 #lb\n",
- "# as theta=25 degrees,\n",
- "sintheta=0.422\n",
- "\n",
- "#calculations\n",
- "ax=(r*W*sintheta)*((r**-1)*14.5+r*18.6)**-1 #ft/s**2\n",
- "F=(W*sintheta)-(18.6*ax) #lb\n",
- "\n",
- "#Result\n",
- "print'The solution is F=',round(F,1),\"lb\",'and ax=',round(ax,2),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is F= 84.3 lb and ax= 9.08 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-62, Page No 378"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=7 #kg\n",
- "g=9.8 #m/s**2\n",
- "r=0.5 #m\n",
- "I=0.875 #kg.m**2\n",
- "\n",
- "#Calculations\n",
- "#Solving for alpha and T\n",
- "alpha=(m*g*r)/(I+m*r*0.5) #rad/s**2\n",
- "T=(I*alpha)/r #N\n",
- "\n",
- "#Result\n",
- "print'The soultion is alpha=',round(alpha,1),\"rad/s**2\",'and T=',round(T,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The soultion is alpha= 13.1 rad/s**2 and T= 22.9 N\n"
- ]
- }
- ],
- "prompt_number": 79
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter16_5.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter16_5.ipynb deleted file mode 100755 index 22430a7a..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter16_5.ipynb +++ /dev/null @@ -1,1911 +0,0 @@ -{
- "metadata": {
- "name": "chapter16.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 16: Dynamics of a Rigid Body in Plane Motion"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-2, Page No 335"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Intilization of variables\n",
- "W=600 #lb\n",
- "d=30 #in\n",
- "# as theta=25 degrees,\n",
- "sintheta=0.422\n",
- "costheta=0.906\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "m=W/g #lb-s**2/ft\n",
- "#Moment of inertia\n",
- "I=0.5*m*((d/2)/12)**2 #lb-s**2-ft\n",
- "#Applying Newtons law and coservation of angular momentum and rolling\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,m,0,0],[0,0,0,1],[((d/2)/12),0,-I,0],[0,1,-((d/2)/12),0]])\n",
- "B=np.array([[W*sintheta],[W*costheta],[0],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The Frictional Force is',round(C[0],1),\"lb\",'and the acceleration is',round(C[1],2),\"ft/s**2\"\n",
- "\n",
- "# The answers wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Frictional Force is 84.4 lb and the acceleration is 9.06 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-3, Page No 336"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=18 #kg\n",
- "d=0.6 #m\n",
- "vo=3 #m/s\n",
- "# as theta=20 degrees,\n",
- "sintheta=0.342\n",
- "costheta=0.939\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of Inertia\n",
- "I=0.5*m*(d/2)**2 \n",
- "#Applying Newtons second Law a\n",
- "A=np.array([[1,m,0,0],[0,0,1,0],[d/2,0,0,-I],[0,1,0,(-d/2)]])\n",
- "B=np.array([[g*m*sintheta],[g*m*costheta],[0],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing the answers in variables\n",
- "F=C[0] #N\n",
- "ax=C[1] #m/s**2\n",
- "Na=C[2] #N\n",
- "alpha=C[3] #rad/s**2\n",
- "#Time Calculations\n",
- "v=0 #m/s**2\n",
- "t=(vo)/ax #s\n",
- "\n",
- "#Result\n",
- "print'It takes',round(t,2),\"s to reach the highest point of travel\"\n",
- "\n",
- "# The ans is off by 0.01 sec."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "It takes 1.34 s to reach the highest point of travel\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-5, Page No 337"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=20 #kg\n",
- "F1=40 #N\n",
- "ro=0.6 #m\n",
- "ri=0.45 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of inertia\n",
- "I=(2*5**-1)*m*ro**2 #kg-m**2\n",
- "#Applying Newtons Law and conservation of angular Momentum\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,m],[ro,-I/ro]])\n",
- "B=np.array([[F1],[F1*ri]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing answers in variables\n",
- "F=C[0] #N\n",
- "a=C[1] #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The acceleration is',round(a,2),\"m/s**2\"\n",
- "print'The force is',round(F,2),\"N\"\n",
- "#The solution in the textbook is incorrect\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration is 0.36 m/s**2\n",
- "The force is 32.86 N\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-6, Page No 338"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=16.1 #lb\n",
- "u=0.10 #co-efficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "F=1.39 #lb\n",
- "\n",
- "#Calculations\n",
- "#Applying Newtons Second Law\n",
- "#Using F=1.39 lb\n",
- "a=((W*sintheta)-F)/(W*g**-1) #ft/s**2\n",
- "alpha=(F*0.5*5/2)/((W*g**-1)*(0.5**2)) #rad/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,1),\"ft/s**2\",'and alpha is ',round(alpha,1),\"rad/s**2.\"\n",
- "print'Hence the sphere will both,roll and slip'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is 13.3 ft/s**2 and alpha is 13.9 rad/s**2.\n",
- "Hence the sphere will both,roll and slip\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-8, Page No 339"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=30 degrees\n",
- "sintheta=2**-1\n",
- "W=80 #lb\n",
- "Ww=100 #lb\n",
- "I=4 #slug-ft**2\n",
- "r=0.5 #ft\n",
- "v= 20 #ft/s\n",
- "vo=0 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using Equations of motion\n",
- "#Solving the system of linear equatinons by matrix method\n",
- "A=np.array([[-1,0,-W/g],[1,-1,-Ww/g],[0,r,-2*I]])\n",
- "B=np.array([[-W],[Ww*sintheta],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing values in variables\n",
- "T=C[0] #lb\n",
- "F=C[1] #lb\n",
- "a=C[2] #ft/s**2\n",
- "#Time calculations\n",
- "t=(v-vo)/a #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t,1),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 14.4 s\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-9, Page No 340"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "M=70 #kg\n",
- "ko=0.4 #m\n",
- "ri=0.45 #m\n",
- "ro=0.6 #m\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "m=35 #kg\n",
- "g= 9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "I=M*ko**2 #kg-m**2\n",
- "#Using Equations of motion\n",
- "#Solving the equations by matrix method\n",
- "A=np.array([[-1,-m*0.15,0],[1,-M*ro,-1],[-ri,-I,ro]])\n",
- "B=np.array([[-m*g],[M*g*sintheta],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "F=C[2] #N\n",
- "Na=M*g*costheta #N\n",
- "#Required coefficient of friction\n",
- "u=F/Na #coefficient of friction\n",
- "\n",
- "#Result\n",
- "print'The value of alpha is',round(C[1],2),\"rad/s**2\",'and tension is',round(C[0]),\"N\"\n",
- "print'F=',round(F),\"N,\",'Na=',round(Na),\"N\",'and u=',round(u,2),\"(coefficient of friction)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of alpha is -4.15 rad/s**2 and tension is 365.0 N\n",
- "F= 196.0 N, Na= 594.0 N and u= 0.33 (coefficient of friction)\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Exampe 16.16-10, Page No 341"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=200 #kg\n",
- "g=9.8 #m/s**2\n",
- "r=1.2 #m\n",
- "F1=1000 #N\n",
- "F2=1400 #N\n",
- "\n",
- "#Calculations\n",
- "N=m*g #N\n",
- "I=(2*5**-1)*(m)*r**2 #kg-m**2\n",
- "#Using equations of motion\n",
- "#Solving for F and alpha using matrix method\n",
- "#Applying equations of motion\n",
- "A=np.array([[1,-m],[-r,-I/r]])\n",
- "B=np.array([[F1-F2],[F1*r]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing values\n",
- "F=C[0] #N\n",
- "alpha=C[1] #rad/s**2\n",
- "a=r*alpha #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,2),\"m/s**2\",'and F is',round(F),\"N\"\n",
- "#The negative signs indicate that the direction is opposite to what was origninally assumed\n",
- "# The answers wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is -2.57 m/s**2 and F is -829.0 N\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-11, Page No 341"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "Wa=161 #lb\n",
- "Wb=193.2 #lb\n",
- "Wc=300 #lb\n",
- "ka=3 #ft\n",
- "kb=2.5 #ft\n",
- "# as theta1=30 degrees & theta2=45 degrees,\n",
- "sintheta1=2**-1\n",
- "sintheta2=(2**0.5)**-1\n",
- "costheta2=(2**0.5)**-1\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of inertia Calculations\n",
- "Ia=(Wa/g)*ka**2 #lb-s**2-ft\n",
- "Ib=(Wb/g)*kb**2 #lb-s**2-ft\n",
- "#Using equations of motion for A and B and C\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,1,-Wa/g,0,0],[1,-4,-Ia*(4**-1),0,0],[-2,0,-Ib*(5*8**-1),4,0],[0,0,-(Wc/g)*(5*2**-1),-1,-0.25],[0,0,0,0,1]])\n",
- "B=np.array([[Wa*sintheta1],[0],[0],[-Wc*costheta2],[Wc*sintheta2]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Storing values in the variables\n",
- "T1=C[0] #lb\n",
- "T2=C[3] #lb\n",
- "a=C[2] #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The values are:'\n",
- "print'a=',round(a,2),\"ft/s**2 ,\",'T1=',round(T1),\"lb\",'and T2=',round(T2,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are:\n",
- "a= 3.93 ft/s**2 , T1= 89.0 lb and T2= 67.5 lb\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-12, Page No 342"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=644 #lb\n",
- "F=30 #lb\n",
- "# as theta=30 degrees,\n",
- "sintheta=(2)**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "r=1.5 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,-W/g],[-r,-(2**-1)*(W/g)*(2*2)*(r**-1)]])\n",
- "B=np.array([[(W*sintheta)-(F*costheta)],[-F*2]])\n",
- "C=np.linalg.solve(A,B)\n",
- "a=C[1] #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a,2),\"ft/s**2\"\n",
- "# The negative sign indicates that the cylinder will roll down the plane."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is -6.78 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-14, Page No 344"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=20 #lb\n",
- "g=32.2 #ft/s**2\n",
- "vb=0.5 #rad/s\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "#Solving the three equations simultaneously by matrix method\n",
- "X=np.array([[0,1,-(W*g**-1)*5.2],[-1,0,-(W*g**-1)*3],[3,-3,-(12**-1)*(W*g**-1)*12**2]])\n",
- "Y=np.array([[-0.75*(W*g**-1)],[(W*g**-1)*1.3-W],[0]])\n",
- "C=np.linalg.solve(X,Y)\n",
- "A=C[0] #lb\n",
- "B=C[1] #lb\n",
- "alpha=C[2] #rad/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of alpha is',round(alpha,1),\"rad/s**2 countercockwise\",'and of A and B are',round(A,1),\"lb up and\",round(B,2),\"lb to the right\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of alpha is 2.6 rad/s**2 countercockwise and of A and B are 14.4 lb up and 7.91 lb to the right\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-16, Page No 345"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mc=7.25 #kg\n",
- "d=0.9 #m\n",
- "la=0.2 #m\n",
- "ma=9 #kg\n",
- "F=45 #N\n",
- "ay=0 #m/s**2\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "I=2*(0.5*mc*(d*2**-1)**2)+0.5*ma*(la*2**-1)**2 #kg-m**2\n",
- "#Using the equations of motion\n",
- "Na=(2*mc+ma)*g #N\n",
- "#Simplfying using radial velocity formula\n",
- "#Solving the two equations using matrix method\n",
- "A=np.array([[-1,-(2*mc+ma)],[(d*2**-1),-I/(d*2**-1)]])\n",
- "B=np.array([[-F],[F*(la*2**-1)]])\n",
- "C=np.linalg.solve(A,B)\n",
- "F=C[0] #N\n",
- "ax=C[1] #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The computation yields ax=',round(ax,2),\"m/s**2 to the right.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The computation yields ax= 1.13 m/s**2 to the right.\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-18, Page No 347"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=0.05 #m cylinder radius\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Here the equation has been solved in terms of the veriables\n",
- "#Hence we directly consider the final result\n",
- "av=(2*g)/3 #m/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of av is',round(av,2),\"m/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of av is 6.53 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-21, Page No 349"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#initilization of variables\n",
- "W=16.1 #lb\n",
- "v=9 #ft/s\n",
- "# as phi=30 degrees,\n",
- "sinphi=(2)**-1\n",
- "cosphi=(3**0.5)*2**-1\n",
- "r=0.5 #ft\n",
- "g=32.2 #ft/s**2\n",
- "OG=4.5 #ft\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "an=v**2/OG #ft/s**2\n",
- "#Solving for alpha we get\n",
- "N=(W*g**-1)*an+W*cosphi #lb\n",
- "#Using equations of motion\n",
- "A=np.array([[1,-r],[-1,-r*r]])\n",
- "B=np.array([[W*sinphi],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "F=C[0] #lb\n",
- "at=C[1] #ft/s**2\n",
- "\n",
- "#Result\n",
- "print'The value of N and F are',round(N,1),\"lb and\",round(F,2),\"lb respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of N and F are 22.9 lb and 2.68 lb respectively.\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-25, Page No 353"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=50 #lb\n",
- "P=10 #lb\n",
- "t=5 #s\n",
- "vo=0 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "ax=(P*g)/W #ft/s**2\n",
- "#Solving by matrix method for A and B\n",
- "F=np.array([[1,1],[-4,4]])\n",
- "Q=np.array([[W],[P]])\n",
- "R=np.linalg.solve(F,Q)\n",
- "#Velocity calculations\n",
- "v=vo+ax*t #ft/s\n",
- "A=R[0] #lb\n",
- "B=R[1] #lb\n",
- "\n",
- "#Result\n",
- "print'The velocity of the door after 5s is',round(v,1),\"ft/s\",'and A=',round(A,1),\"lb\",'and B=',round(B,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity of the door after 5s is 32.2 ft/s and A= 23.8 lb and B= 26.3 lb\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-26, Page No 354"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "#Initilization of variables\n",
- "AB=2 #m\n",
- "m=2 #kg\n",
- "F=20 #N\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equation of motion\n",
- "a=F/m #m/s**2\n",
- "#Solving by matrix method for Na and Nb\n",
- "A=np.array([[1,-1],[4*5**-1,4*5**-1]])\n",
- "B=np.array([[m*g],[F*(3*5**-1)]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The value of a is',round(a),\"m/s**2\",'and the reactions are: Na=',round(C[0],1),\"N up\",'and Nb=',round(C[1],1),\"N up\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of a is 10.0 m/s**2 and the reactions are: Na= 17.3 N up and Nb= -2.3 N up\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-27, Page No 354"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "vo=0 #ft/s\n",
- "\n",
- "#Calculations\n",
- "s=(0.011*5280*2)/(2*0.004)\n",
- "\n",
- "#Result\n",
- "print'It travels',round(s),\"ft along the level before coming to rest\"\n",
- "#Answer in the textbook is incorrect by 20ft\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "It travels 14520.0 ft along the level before coming to rest\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-28, Page No 355"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "u=0.3 #coefficient of friction\n",
- "m=70 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#CASE 1\n",
- "#Using equations of motion\n",
- "Na=m*g #N\n",
- "ah=(u*Na)/m #m/s**2\n",
- "#CASE 2\n",
- "#Applying sum of moments equal to zero\n",
- "F=(Na*0.3)/1.2 #N\n",
- "a_h=F/m #m/s**2\n",
- "\n",
- "#Result\n",
- "#Intutive insights can be attained after we get these results\n",
- "print'The value of Na is',round(Na),\"N\"\n",
- "print'and that of acceleration are:'\n",
- "print'1st value=',round(ah,2),\"m/s**2\",'and 2nd value is',round(a_h,2),\"m/s**2\"\n",
- "print'and the value of F is',round(F),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of Na is 686.0 N\n",
- "and that of acceleration are:\n",
- "1st value= 2.94 m/s**2 and 2nd value is 2.45 m/s**2\n",
- "and the value of F is 172.0 N\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-29, Page No 356"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=60 #kg\n",
- "me=660 #kg\n",
- "a=6 #m/s**2\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "P=m*a+m*g #N\n",
- "#Scale reading\n",
- "R=P/g #kg\n",
- "#Increase in mass\n",
- "I=R-m #kg\n",
- "#Tension\n",
- "T=me*a+me*g #N\n",
- "\n",
- "#Result\n",
- "print'The value of P is',round(P),\"n\"\n",
- "print'The apparent icrease in weight is',round(I,1),\"kg\",'and the tension in the cable is',round(T),\"N\"\n",
- "#Answer in the textbook is off by 28 #N in Tension\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P is 948.0 n\n",
- "The apparent icrease in weight is 36.7 kg and the tension in the cable is 10428.0 N\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-30, Page No 356"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "u=0.2 #coefficient of friction\n",
- "ma=1.2 #kg\n",
- "mb=2 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "Nb=mb*g #N\n",
- "F=u*Nb #N\n",
- "#Using equations of motion\n",
- "#Solving for T and a\n",
- "A=np.array([[-1,-ma],[1,-mb]])\n",
- "B=np.array([[-ma*g],[F]])\n",
- "C=np.linalg.solve(A,B)\n",
- "T=C[0] #N\n",
- "a=C[1] #m/s**2\n",
- "#Taking the sum of the moments\n",
- "x_m=-(F*0.15+T*0.15)/Nb #m\n",
- "x=x_m*1000 #mm\n",
- "\n",
- "#Result\n",
- "print'The acceleration of block A is',round(a,2),\"m/s**2\" \n",
- "print'and Nb acts at a distance of',round(x,1),\"mm.(Negative sign indictaes that the side assumed is incorrect)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The acceleration of block A is 2.45 m/s**2\n",
- "and Nb acts at a distance of -97.5 mm.(Negative sign indictaes that the side assumed is incorrect)\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-31, Page No 357"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "a=2.5 #m/s**2\n",
- "mA=3 #kg\n",
- "mB=7 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "F=(mA+mB)*a #N\n",
- "#Using equations of motion\n",
- "Py=mB*g #N\n",
- "#Solving for Px and H\n",
- "A=np.array([[1,1],[-0.0375,0.0375]])\n",
- "B=np.array([[mB*a],[Py*0.05]])\n",
- "C=np.linalg.solve(A,B)\n",
- "Px=C[0] #N\n",
- "H=C[1] #N\n",
- "\n",
- "#Result\n",
- "print'The value of H is',round(H,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of H is 54.5 N\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-32, Page No 358"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=20 #kg\n",
- "g=9.8 #m/s**2\n",
- "vo=3 #m/s\n",
- "v=0 #m/s\n",
- "s=4 #m\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "Na=m*g #N\n",
- "F=(Na*0.075)/0.125 #N\n",
- "a=F/m #m/s**2\n",
- "#Displacement \n",
- "d=-(v**2-vo**2)/(2*a) #m\n",
- "displ=s-d #m\n",
- "v_f=sqrt(2*a*displ) #m/s\n",
- "\n",
- "#Result\n",
- "print'The final velocity is',round(v_f,2),\"m/s to the left.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The final velocity is 6.17 m/s to the left.\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-33, Page No 358"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mA=30 #kg\n",
- "mB=45 #kg\n",
- "u_ab=3**-1 #coefficient of friction between two blocks\n",
- "u_bp=10**-1 #coefficient of friction between block and horizontal plane\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#By inspection\n",
- "Na=mA*g #N\n",
- "Nb=Na+mB*g #N\n",
- "a=(u_ab*Na-u_bp*Nb)/mB #m/s**2\n",
- "P=(mA*a+u_ab*Na) #N\n",
- "#For block A\n",
- "#Solving for P,F and a\n",
- "A=np.array([[1,-1,-mA],[-0.05,-0.075,0],[0,1,-mB]])\n",
- "B=np.array([[0],[-Na*0.050],[Nb*u_bp]])\n",
- "C=np.linalg.solve(A,B)\n",
- "P_new=C[0] #N\n",
- "\n",
- "#Result\n",
- "#As p < p_new\n",
- "print'The maximum value of P is',round(P),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum value of P is 114.0 N\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-34, Page No 359"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Vo=1.5 #m/s\n",
- "V=0 #m/s\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "a=(g*0.2)/0.75 #m/s**2\n",
- "t=-(V-Vo)/a #s\n",
- "\n",
- "#Result\n",
- "print'The maximum acceleration is',round(a,2),\"m/s**2\",'and minimum time is',round(t,2),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum acceleration is 2.61 m/s**2 and minimum time is 0.57 s\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-36, Page No 361"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "vo=0 #mi/h\n",
- "v=60 #mi/h\n",
- "t=13.8 #s\n",
- "W=3385 #lb\n",
- "xb=46 #in\n",
- "xf=66 #in\n",
- "xv=31 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "a=(((v*88*60)/3600)-vo)/t #ft/s**2\n",
- "#Summing horizontal forces\n",
- "F=(W/g)*a #lb\n",
- "#Solving for Rf and Rr\n",
- "A=np.array([[1,1],[-xf,xb]])\n",
- "B=np.array([[W],[-F*xv]])\n",
- "C=np.linalg.solve(A,B)\n",
- "Rr=C[0] #lb\n",
- "Rf=C[1] #lb\n",
- "\n",
- "#Result\n",
- "print'The value of reactions are Rf=',round(Rf),\"lb\",'and Rr=',round(Rr),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of reactions are Rf= 1809.0 lb and Rr= 1576.0 lb\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-43, Page No 366"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=161 #lb\n",
- "F=16.1 #lb\n",
- "r=18 #ft radius\n",
- "t=2 #s\n",
- "g=32.2 #ft/s**2\n",
- "wo=0 #rad/s\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "#Solving for T and alpha\n",
- "A=np.array([[r*12**-1,-0.5*(W*g**-1)*(r*12**-1)**2],[-1,-F*g**-1]])\n",
- "B=np.array([[0],[-F]])\n",
- "C=np.linalg.solve(A,B)\n",
- "alpha=C[1] #rad/s**2\n",
- "w=wo+(alpha*t) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular speed is',round(w,2),\"rad/s\"\n",
- "\n",
- "#The ans is incorrect in textbook."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular speed is 7.58 rad/s\n"
- ]
- }
- ],
- "prompt_number": 43
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-47, Page No 369"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization fo variables\n",
- "r=2000 #ft\n",
- "g=32.2 #ft/s**2\n",
- "d=4.71 #ft\n",
- "v=176 #ft/s\n",
- "\n",
- "#Calculations\n",
- "e=(d*v**2)/(g*r) #ft\n",
- "\n",
- "#Result\n",
- "print'The superelevation is',round(e,2),\"ft\"\n",
- "#Watch the unit in the final answer\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The superelevation is 2.27 ft\n"
- ]
- }
- ],
- "prompt_number": 45
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-48, Page No 369"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "a=5 #ft/s**2\n",
- "C=50 #lb-ft\n",
- "W=161 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "T=0.5*(W/g)*1**2*a+C #lb\n",
- "Ox=-T*(2/sqrt(a)) #lb\n",
- "Oy=T*(1/sqrt(a))+W #lb\n",
- "Wa=T/(1-(a/g)) #lb\n",
- "\n",
- "#Result\n",
- "print'The values are: T=',round(T,1),\"lb\",', Wa=',round(Wa),\"b\",',Ox=',round(Ox,1),\"lb\",'and Oy=+',round(Oy),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are: T= 62.5 lb , Wa= 74.0 b ,Ox= -55.9 lb and Oy=+ 189.0 lb\n"
- ]
- }
- ],
- "prompt_number": 49
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-49, Page No 370"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=100 #kg\n",
- "mr=20 #kg\n",
- "w=8 #rad/s\n",
- "l1=300 #mm\n",
- "l2=600 #mm\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "r_bar=(mr*l1+m*750)/120 #mm\n",
- "I=(3**-1)*mr*(l2*1000**-1)**2+(2*5**-1)*m*(l1*2000**-1)**2+m*(0.75)**2 #kg.m**2\n",
- "alpha=(m+mr)*g*(r_bar*1000**-1)/I #rad/s**2\n",
- "On=(m+mr)*(r_bar*1000**-1)*w**2 #N\n",
- "Ot=((m+mr)*(r_bar*1000**-1)*alpha)-(m+mr)*g #N\n",
- "\n",
- "#Result\n",
- "print'The angular acceleration is',round(alpha,1),\"rad/s**2\",'and On=',round(On),\"N\",'and Ot=',round(Ot,1),\"N\"\n",
- "\n",
- "#Due to decimal accuracy there is discrepancy in answers with the textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular acceleration is 13.3 rad/s**2 and On= 5184.0 N and Ot= -96.3 N\n"
- ]
- }
- ],
- "prompt_number": 48
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-50, Page No 370"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=40 #lb\n",
- "w=10 #rad/s\n",
- "alpha=2 #rad/s**2\n",
- "r=2 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "On=(W*g**-1)*(1*6**-1)*w**2 #lb\n",
- "Ot=(W*g**-1)*(1*6**-1)*alpha\n",
- "Io=(0.5*(W*g**-1)*0.5**2)*2+((W*g**-1)*(1*6**-1)**2)*2\n",
- "\n",
- "#Result\n",
- "print'The reaction components are On=',round(On,1),\"lb to the right\",'and Ot=',round(Ot,2),\"lb up\"\n",
- "print'The value of Io is',round(Io,2),\"lb-ft\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction components are On= 20.7 lb to the right and Ot= 0.41 lb up\n",
- "The value of Io is 0.38 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 59
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-51, Page No 371"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilizatin of variables\n",
- "W=6 #lb\n",
- "l=8 #ft\n",
- "v=10 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "# as theta1=60 degrees & theta2=30 degrees\n",
- "costheta1=2**-1\n",
- "costheta2=(3**0.5)*2**-1\n",
- "\n",
- "#Calculations\n",
- "Fe=(W*v**2)*(g*l*0.5)**-1 #lb\n",
- "#Using equations of motion\n",
- "#Solving for C and T\n",
- "A=np.array([[costheta1,-costheta2],[costheta2,costheta1]])\n",
- "B=np.array([[-Fe],[W]])\n",
- "P=np.linalg.solve(A,B) #lb\n",
- "C=P[0] #lb\n",
- "T=P[1] #lb\n",
- "\n",
- "#Result\n",
- "print'The value of C is',round(C,2),\"lb\",'and T is',round(T,2),\"lb\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of C is 2.87 lb and T is 7.03 lb\n"
- ]
- }
- ],
- "prompt_number": 49
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-52, Page No 372"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=32.2 #lb\n",
- "T=120 #lb\n",
- "m=1 #slug\n",
- "r=6*12**-1 #ft\n",
- "\n",
- "#Calculations\n",
- "w=sqrt((T*(3*5**-1)*4)*(m*r*3)**-1) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular speed permissible is',round(w,1),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular speed permissible is 13.9 rad/s\n"
- ]
- }
- ],
- "prompt_number": 50
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-53, Page No 372"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=30 #kg\n",
- "k=0.45 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Using equations of motion\n",
- "#Solving for T1,T2 and alpha\n",
- "A=np.array([[1,0,-m],[0,-1,-45],[-0.6,0.3,-m*k**2]])\n",
- "B=np.array([[50*g],[-150*g],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The values are: T1=',round(C[0]),\"N\",',T2=',round(C[1]),\"N\",'and alpha=',round(C[2],1),\"rad/s**2\"\n",
- "\n",
- "# The answer for T2 and alpha is off by 4 & 0.1 units"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are: T1= 607.0 N ,T2= 1294.0 N and alpha= 3.9 rad/s**2\n"
- ]
- }
- ],
- "prompt_number": 51
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-54, Page No 373"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wc=28 #lb\n",
- "v=16 #ft/s\n",
- "Ib=12 #ft-lb-s**2\n",
- "u=0.4 #coefficient of friction\n",
- "t=2 #s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "T=Wc+(Wc*g**-1)*8 #lb\n",
- "alpha=(8*12)*15**-1 #rad/s**2\n",
- "F=((Ib*alpha)+(T*1.25))/t #lb\n",
- "N=F/u #lb\n",
- "#Summing moments about D\n",
- "P=(N*8+F*3)/40 #lb\n",
- "#Summing forces horizontally and vertically\n",
- "Dx=151-P #lb\n",
- "Dy=-F #lb\n",
- "\n",
- "#Result\n",
- "print'The reactions at D are: Dx=',round(Dx),\"lb to right\",'and Dy=',round(Dy,1),\"lb down\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reactions at D are: Dx= 116.0 lb to right and Dy= -60.2 lb down\n"
- ]
- }
- ],
- "prompt_number": 65
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-55, Page No 374"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=8 #kg\n",
- "n=90 #rpm\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "Fg=m*g #N\n",
- "w=2*pi*n/60 #rad/s\n",
- "#using equations of motion\n",
- "By=m*g #N\n",
- "#Solving for Bx and C\n",
- "A=np.array([[1,1],[-0.3,0.9]])\n",
- "B=np.array([[m*0.3*w**2],[By*0.3]])\n",
- "C=np.linalg.solve(A,B) #N\n",
- "\n",
- "#Result\n",
- "print'The solution is: Bx=',round(C[0]),\"N\",',By=',round(By,1),\"N\",'and C=',round(C[1],1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is: Bx= 140.0 N ,By= 78.4 N and C= 72.9 N\n"
- ]
- }
- ],
- "prompt_number": 79
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-56, Page No 375"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=8 #kg\n",
- "n=90 #rpm\n",
- "g=9.8 #m/s**2\n",
- "r=0.3 #m\n",
- "\n",
- "#calculations\n",
- "w=2*pi*n/60 #rad/s\n",
- "#Using equations of motion\n",
- "C=(m*g*0.3+m*r*w**2*r)/1.2 #N\n",
- "Bx=-C+m*r*w**2 #N\n",
- "By=m*g #N\n",
- "\n",
- "#Result\n",
- "print'The solution is: Bx=',round(Bx),\"N\",',By=',round(By,1),\"N\",'and C=',round(C,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is: Bx= 140.0 N ,By= 78.4 N and C= 72.9 N\n"
- ]
- }
- ],
- "prompt_number": 66
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-57, Page No 375"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Na=294 #N\n",
- "Nb=735 #N\n",
- "\n",
- "#Calculations\n",
- "a=(10**-1*Nb-3**-1*Na)/45 #m/s**2\n",
- "P=(3**-1*Na)-30*a #N\n",
- "\n",
- "#Result\n",
- "print'The solution is P=',round(P),\"N\",'and a=',round(a,3),\"m/s**2\"\n",
- "# The negative sign indicates the assumed direction is incorrect."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is P= 114.0 N and a= -0.544 m/s**2\n"
- ]
- }
- ],
- "prompt_number": 68
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-58, Page No 376"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=50 #lb\n",
- "g=32.2\n",
- "\n",
- "#Calculations\n",
- "#Using equations of motion\n",
- "a=(10/(W/g)) #ft/s**2\n",
- "B=((2.5*(W/g)*a)+4*W-1.5*10)/8 #lb\n",
- "A=50-B #lb\n",
- "\n",
- "#Result\n",
- "print'The solution is A=',round(A,1),\"lb\",',B=',round(B,1),\"lb\",'and a=',round(a,2),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is A= 23.8 lb ,B= 26.3 lb and a= 6.44 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 69
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-60, Page No 377"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "g=9.8 #m/s**2\n",
- "r1=0.3 #m\n",
- "m1=20 #kg\n",
- "m2=100 #kg\n",
- "r2=0.75 #m\n",
- "\n",
- "#Calculations\n",
- "alpha=(m1*g*r1+m2*g*r2)*(m1*r1**2+(m1/12)*0.6**2+m2*r2**2+(2*5**-1)*m2*0.15**2)**-1 #rad/s**2\n",
- "\n",
- "#Result\n",
- "print'The angular acceleration is',round(alpha,1),\"rad/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular acceleration is 13.4 rad/s**2\n"
- ]
- }
- ],
- "prompt_number": 71
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-61, Page No 378"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=15*12**-1 #ft\n",
- "W=600 #lb\n",
- "# as theta=25 degrees,\n",
- "sintheta=0.422\n",
- "\n",
- "#calculations\n",
- "ax=(r*W*sintheta)*((r**-1)*14.5+r*18.6)**-1 #ft/s**2\n",
- "F=(W*sintheta)-(18.6*ax) #lb\n",
- "\n",
- "#Result\n",
- "print'The solution is F=',round(F,1),\"lb\",'and ax=',round(ax,2),\"ft/s**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solution is F= 84.3 lb and ax= 9.08 ft/s**2\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16-62, Page No 378"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=7 #kg\n",
- "g=9.8 #m/s**2\n",
- "r=0.5 #m\n",
- "I=0.875 #kg.m**2\n",
- "\n",
- "#Calculations\n",
- "#Solving for alpha and T\n",
- "alpha=(m*g*r)/(I+m*r*0.5) #rad/s**2\n",
- "T=(I*alpha)/r #N\n",
- "\n",
- "#Result\n",
- "print'The soultion is alpha=',round(alpha,1),\"rad/s**2\",'and T=',round(T,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The soultion is alpha= 13.1 rad/s**2 and T= 22.9 N\n"
- ]
- }
- ],
- "prompt_number": 79
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter17.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter17.ipynb deleted file mode 100755 index 7a620362..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter17.ipynb +++ /dev/null @@ -1,1273 +0,0 @@ -{
- "metadata": {
- "name": "chapter17.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 17: Work and Energy"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-4, Page No 400"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "s1=2 # compression of the spring- initial\n",
- "s2=5 # compression of the spring- final\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(s, a, b):\n",
- " return 20*s\n",
- "a=1\n",
- "b=1\n",
- "U=quad(integrand, s1, s2, args=(a,b))\n",
- "\n",
- "# Results\n",
- "print'The word done in compressing the spring is',round(U[0]),\"in-lb\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The word done in compressing the spring is 210.0 in-lb\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-6, Page No 400"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=5 #kg\n",
- "d=6 #m\n",
- "# as theta1=30 degrees & theta2=10 degrees,\n",
- "sintheta1=2**-1\n",
- "sintheta2=0.1736\n",
- "costheta1=(3**0.5)*2**-1\n",
- "costheta2=0.9848\n",
- "u=0.2 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "F=70 #N\n",
- "\n",
- "#Calculations\n",
- "#Using free body diagram\n",
- "Na=(m*g*costheta1)-(F*sintheta2) #N\n",
- "#work done by each force\n",
- "W=[F*costheta2,-m*g*sintheta1,0,-u*Na*d] #N.m\n",
- "#Total Work Done\n",
- "W_tot=W[0]+W[1]+W[2]+W[3] #N.m\n",
- "#Using resultant\n",
- "R=F*costheta2-(u*Na)-(m*g*sintheta1) #N\n",
- "W_d=R*d #N.m (Work Done)\n",
- "\n",
- "#Result\n",
- "print'The work done is',round(W_d),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The work done is 230.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-7, Page No 401"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=20 #kg\n",
- "d=1.5 #m\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "u=0.25 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "F=130 #N\n",
- "\n",
- "#Calculations\n",
- "W=F*d-(m*g*sintheta*d) #N.m\n",
- "\n",
- "#Result\n",
- "print'The work done is',round(W),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The work done is 48.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-9, Page No 402"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=6*12**-1 #ft\n",
- "l=8*12**-1 #ft\n",
- "l_c=3.2 #in\n",
- "y=1.82 #in**2\n",
- "\n",
- "#Calculations\n",
- "V=1*4**-1*pi*d**2*l #ft**3\n",
- "#One horizontal inch \n",
- "h_i=V/l_c #ft**3\n",
- "#One vertical inch\n",
- "v_i=100*144 #lb/ft**2\n",
- "#Then 1.82 in**2 represents\n",
- "x=y*v_i*h_i #ft-lb\n",
- "\n",
- "#Result\n",
- "print'The work capacity is',round(x),\"ft-lb\"\n",
- "\n",
- "# The ans in the textbook is incorrect."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The work capacity is 1072.0 ft-lb\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-10, Page No 402"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "speed=90000 #m/h\n",
- "P=100*1000 #N\n",
- "\n",
- "#Calculations\n",
- "Power=P*((speed)/3600) #J/s\n",
- "\n",
- "#Result\n",
- "print'The power developed is',round(Power*10**-6,1),\"MJ/s\"\n",
- "# Note the unit used."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power developed is 2.5 MJ/s\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-11, Page No 403"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=0.6 #m\n",
- "T_t=800 #N\n",
- "T_s=180 #N\n",
- "w=200 #rpm\n",
- "\n",
- "#Calculations\n",
- "r=d/2 #m radius\n",
- "#Torque\n",
- "M=(T_t-T_s)*r #N.m\n",
- "#Power\n",
- "w_new=(2*pi*w)/60 #rad/s\n",
- "Power=M*(w_new) #W\n",
- "\n",
- "#Result\n",
- "print'The power transmitted is',round(Power*10**-3,1),\"kW\"\n",
- "\n",
- "# The answer in the book is incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power transmitted is 3.9 kW\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-12, Page No 403"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "P=25.6 #lb\n",
- "w=600 #rpm\n",
- "a=36 #in\n",
- "b=12 #in\n",
- "\n",
- "#Calculations\n",
- "M=P*(((b*2**-1)+a)/12) #lb-ft\n",
- "w_new=(2*pi*w)/60 #rad/s\n",
- "Hp=(M*w_new)/550 #hp\n",
- "\n",
- "#Result\n",
- "print'The power being transmitted is',round(Hp,1),\"hp\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power being transmitted is 10.2 hp\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-13, Page No 403"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Pout=3.8 #bhp\n",
- "Pin=4.1 #ihp\n",
- "\n",
- "#Calculations\n",
- "Efficiency=round((Pout/Pin)*100) #Percent\n",
- "\n",
- "#Result\n",
- "print'The efficiency of the engine is',round(Efficiency),\"%\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The efficiency of the engine is 93.0 %\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-15, Page No 404"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return -(3/x)\n",
- "a=1\n",
- "b=1\n",
- "U=quad(integrand, 6, 3, args=(a,b))\n",
- "g=32.2 # ft/s**2\n",
- "deltaT=((U[0]*32.2*2)/(4*16**-1))**0.5\n",
- "# Results\n",
- "print'The speed of the disc will be',round(deltaT,1),\"ft/s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the disc will be 23.1 ft/s\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-16, Page No 404"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=2 #m\n",
- "m=4 #kg\n",
- "w_1=20 #rpm\n",
- "w_2=50 #rpm\n",
- "rev=10 #no of revolution\n",
- "\n",
- "#Calculations\n",
- "Io=(3**-1)*(m)*l**2 #kg.m**2\n",
- "w1=(2*pi*w_1)/60 #rad/s\n",
- "w2=(2*pi*w_2)/60 #rad/s\n",
- "theta=2*pi*rev #rad\n",
- "M=(0.5*Io*(w2**2-w1**2))/theta #N.m\n",
- "\n",
- "#Result\n",
- "print'The constant moment required is',round(M,3),\"N.m\"\n",
- "# The ans waries in decimal places."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The constant moment required is 0.977 N.m\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-18, Page No 405"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=1000 #lb\n",
- "w_w=200 #lb weight of the individual wheel\n",
- "d_w=2.5 #ft diameter of the wheel\n",
- "v=22 #ft/s\n",
- "t=2 #minutes\n",
- "\n",
- "#Calculations\n",
- "#T1=Initial Kinetic Energy and T2=Final Kinetic Energy\n",
- "F=(-0.5*W*32.2**-1*v**2-4*0.5*w_w*32.2**-1*(v**2+0.5*v**2))/(10560) #lb\n",
- "#Negative sign in the answer tells it oposses the motion\n",
- "\n",
- "#Result\n",
- "print'The rolling resistance is',round(F,2),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The rolling resistance is -1.57 lb\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-19, Page No 405"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=100 #lb\n",
- "lo=4 #ft\n",
- "# as theta=45 degrees\n",
- "costheta=(2**0.5)**-1\n",
- "g=32.2 #ft/s**2\n",
- "l=8/3 #ft\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point O and equating it to zero\n",
- "alpha=(W*(lo*0.5)*costheta)/((W/g)*(l)*2) #rad/s**2\n",
- "#Summing forces in the t direction\n",
- "Ot=(W*costheta)-((W/g)*lo*0.5*alpha) #lb\n",
- "#Work Done\n",
- "Work=W*(lo*0.5*costheta) #ft/lb\n",
- "#Moment of inertia\n",
- "Io=(3**-1)*(W/g)*(lo**2) #kg-ft**2\n",
- "#Using the concept for work done=chane in K.E\n",
- "w=(Work/(0.5*Io))**0.5 #rad/s\n",
- "#Summing forces along the bar\n",
- "On=-(-((W/g)*lo*0.5*w**2)-(W*costheta)) #lb\n",
- "\n",
- "#Result\n",
- "print'The bearing reaction at O on the rod is',round(On),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The bearing reaction at O on the rod is 177.0 lb\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-21, Page No 406"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "vo=9 #m/s\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "x=((7*10**-1)*vo**2)*(g*sintheta)**-1 #m\n",
- "\n",
- "#Result\n",
- "print'The ball will roll',round(x,1),\"m up the plane\"\n",
- "\n",
- "#The textbook wrongly mentions the unit of displacement as in\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ball will roll 11.6 m up the plane\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-22, Page No 406"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=322 #lb\n",
- "F=12 #lb\n",
- "a=0 #lower limit (where the cyliner starts rolling)\n",
- "b=pi/2 #Upper Limit (where the cyliner stops rolling)\n",
- "d=3.2 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "dR=1.6 #Differential Radius\n",
- "d_U=2*dR*F #differential work done\n",
- "#Integration Calculations\n",
- "#As it is a simple integration we can resort to this\n",
- "U=d_U*(b-a) #ft-lb\n",
- "#Determination of K.E\n",
- "w=sqrt(U/((0.5*(W/g)*(1/(d/2)**2))+((0.5*0.5)*(W/g)*(d/2)**2))) #rad/s\n",
- "\n",
- "#Result \n",
- "print'The angular velocity of the cylinder is',round(w,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity of the cylinder is 2.69 rad/s\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-24, Page No 407"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wa=161 #lb\n",
- "Wb=193.2 #lb\n",
- "Wc=322 #lb\n",
- "v1=5 #ft/s\n",
- "lc=6 #in\n",
- "k=6 #lb/ft\n",
- "l=4 #ft\n",
- "u=0.2 #coefficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "Ib=(2**-1)*(Wb/g)*(2**-1)**2 #Moment of inertia\n",
- "w1=v1*0.5**-1 #rad/s\n",
- "T1=(0.5*(Wc/g)*v1**2)+(0.5*Ib*w1**2)+(0.5*(Wa/g)*v1**2) #ft-lb\n",
- "#Work Done on the system\n",
- "#The textbook is ambigious on the calculations hence the result is dispalyed directly\n",
- "U=26.4 #ft-lb\n",
- "#Velocity Calculations\n",
- "v=((T1+U)*9**-1)**0.5 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The velocity of the block is',round(v,2),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity of the block is 5.29 ft/s\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-25, Page No 408"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Mm=70 #kg\n",
- "Mc=45 #kg\n",
- "R=0.6 #m\n",
- "g=9.8 #m/s**2\n",
- "l=5 #m\n",
- "# as theta=50 degrees,\n",
- "sintheta=0.77\n",
- "\n",
- "#Calculations\n",
- "#T2 calculations except for v term in it as it cannot be declared as a number\n",
- "T2=68.7 #without the v term in it\n",
- "v=((g*Mm*l-g*Mc*l*sintheta)/T2)**0.5 #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed is',round(v,2),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed is 5.02 m/s\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-26, Page No 408"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#The textbook has a typo in printing the question number\n",
- "#Initilization of variables\n",
- "W1=96.6 #lb\n",
- "W2=128.8 #lb\n",
- "v=8 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "\n",
- "#Calculations\n",
- "#Initial KE of the system is T1=0\n",
- "T2=(0.5*(W1/g)*v**2)+(0.5*(W2/g)*(v/2)**2) #ft-lb\n",
- "#Work Done without s term\n",
- "U=-(W1*sintheta)+W2*0.5\n",
- "#S calculations\n",
- "s=T2*U**-1 #ft\n",
- "\n",
- "#Result\n",
- "print'The block attains a speed of 8 ft/s in',round(s,2),\"ft (up the plane)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The block attains a speed of 8 ft/s in 7.95 ft (up the plane)\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-28, Page No 409"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "l=6 #m length of the cable\n",
- "m=50 #kg mass of the cable\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return 81.7*(6-x)\n",
- "a=1\n",
- "b=1\n",
- "Work=quad(integrand, 0, 6, args=(a,b))\n",
- "\n",
- "# Results\n",
- "print'The word done is',round(Work[0]),\"N.m\"\n",
- "# The answer in textbook is off by 1 N.m"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The word done is 1471.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-31, Page No 410"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "k=0.044 # spring constant\n",
- "#x=0.300 #m length of compression from 450 to 150\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return 0.5*0.044*x**2\n",
- "a=1\n",
- "b=1\n",
- "W=quad(integrand, 0, 300, args=(a,b))\n",
- "\n",
- "# Results\n",
- "print'The word done on the balls is',round(W[0]*10**-5,2),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The word done on the balls is 1.98 N.m\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-32, Page No 410"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=10 #kg\n",
- "d=1.2 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Initilial KE is zero\n",
- "#Final KE is(without v^2 term in it)\n",
- "KE2=(3*4**-1)*10\n",
- "#Work Done\n",
- "U=m*g*d #N.m\n",
- "#Velocity calculations\n",
- "v=sqrt(U*KE2**-1) #m/s\n",
- "\n",
- "#Result\n",
- "print'The velocity is',round(v,2),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity is 3.96 m/s\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-33, Page No 411"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=161 #lb\n",
- "wa=150 #lb\n",
- "wb=100 #lb\n",
- "la=2 #ft\n",
- "lb=4 #ft\n",
- "\n",
- "#Calculations\n",
- "#Work Done\n",
- "T1=wb*lb-wa*la #ft-lb\n",
- "#Final KE=zero\n",
- "T2=0 #ft-lb\n",
- "#Work Done on the system=T2-T1\n",
- "#Hence the equation becomes\n",
- "#50x-50x^2+100=0\n",
- "#where\n",
- "a=-50\n",
- "b=50\n",
- "c=100\n",
- "#Solution\n",
- "d=sqrt(b**2-4*a*c) \n",
- "x1=(-b+d)/(2*a) #ft\n",
- "x2=(-b-d)/(2*a) #ft\n",
- "\n",
- "#Result\n",
- "print'The stretch of the spring is',round(x2),\"ft\"\n",
- "#Here even x1 could have been the solution,but the stretch in the string is elongation not compression hence x2 is the valid answer\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The stretch of the spring is 2.0 ft\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-34, Page No 411"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "I=100 #slug-ft**2\n",
- "w=4 #rad/s\n",
- "theta=6 #rad\n",
- "Mc=64.4 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "vb=2*w #ft/s\n",
- "vc=0.5*w #ft/s\n",
- "Mb=(0.5*I*w**2+0.5*(Mc/g)*vc**2+0.5*Mc*theta)/(2*theta-(0.5*vb**2*(1/g))) #lb\n",
- "\n",
- "#Result\n",
- "print'The weight of the block B is',round(Mb,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The weight of the block B is 90.6 lb\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-35, Page No 411"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wa=96.6 #lb\n",
- "Wb=128.8 #lb\n",
- "g=32.2 #ft/s**2\n",
- "I=12 #slug-ft**2\n",
- "v=16 #ft/s\n",
- "ratio=3**-1 #ratio of Sb/Sa\n",
- "r=3#ft\n",
- "va=6 #ft/s\n",
- "vb=2 #ft/s\n",
- "\n",
- "#Calculations\n",
- "#Work Done without S in it\n",
- "W=Wa-(ratio*Wb)\n",
- "#System has zero KE initially and final KE is given by\n",
- "w=va/r #rad/s\n",
- "T2=(0.5*(Wa/g)*va**2+0.5*I*w**2+0.5*(Wb/g)*vb**2) #ft-lb\n",
- "#Distance Calculations\n",
- "S=T2*W**-1 #ft\n",
- "\n",
- "#Result\n",
- "print'The distance through which A falls is',round(S,2),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The distance through which A falls is 1.6 ft\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-36, Page No 412"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#initilization of variables\n",
- "u=0.25 #coefficient of friction\n",
- "k=2800 #N/m\n",
- "x=0.075 #m\n",
- "g=9.8 #m/s**2\n",
- "m=7 #kg\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Normal Reaction\n",
- "N=g*m*costheta #N\n",
- "#Frictional Force\n",
- "Fr=u*N #N\n",
- "#Component of force along the plane\n",
- "F=g*m*sintheta #N\n",
- "#Spring work is\n",
- "W=0.5*k*x*x #N.m\n",
- "s=(W+Fr*x-F*x)/(F-Fr) #m\n",
- "S=(s*1000) #mm\n",
- "\n",
- "#Result\n",
- "print'The value of S is',round(S),\"mm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of S is 330.0 mm\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-37, Page No 412"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=5 #kg\n",
- "l=2 #m\n",
- "k=10000 #N/m\n",
- "x=0.1 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "drop=l+x #m mass drop length\n",
- "#Work Done by Gravity\n",
- "Wg=g*m*drop #N.m\n",
- "#Work Done by Spring\n",
- "Ws=0.5*k*x**2 #N.m\n",
- "#Increase in KE is without v**2\n",
- "KE=0.5*m #kg\n",
- "#Velocity Calculations\n",
- "v=sqrt((Wg-Ws)/KE) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed is',round(v,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed is 4.6 m/s\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-38, Page No 412"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=6 #ft\n",
- "k=20 #lb/in\n",
- "x=8 #in\n",
- "\n",
- "#Calculations\n",
- "#Work Done by Gravity\n",
- "Wg=(l*12+x) #in without W\n",
- "#Work Done by Spring\n",
- "Ws=0.5*k*x**2 #in-lb\n",
- "#Change in the kinetic energy is zero\n",
- "W=Ws/Wg #lb\n",
- "\n",
- "#Result\n",
- "print'The weight is',round(W),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The weight is 8.0 lb\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-40, Page No 413"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=8 #lb\n",
- "\n",
- "#Calculations\n",
- "#work done by the spring woithout k\n",
- "Ws=0.5*((9*12**-1)**2-(12**-1)**2) \n",
- "#Work done by gravity\n",
- "Wg=W*(10.5*12**-1) #ft-lb\n",
- "#Change in KE is zero\n",
- "k=Wg/Ws #lb/ft\n",
- "\n",
- "#Result\n",
- "print'The value of k is',round(k,1),\"lb/ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of k is 25.2 lb/ft\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-41, Page No 414"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wc=100 #lb\n",
- "r= 1 #ft\n",
- "F=80 #lb\n",
- "k=50 #lb/ft\n",
- "s=6 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Work done on the system\n",
- "U=-0.5*k*(1)+F*(s*12**-1) #ft-lb\n",
- "#Initial KE is zero\n",
- "Vo=(U/(0.5*(Wc/g+0.5*(Wc/g)*r)))**0.5 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The initial speed is',round(Vo,2),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The initial speed is 2.54 ft/s\n"
- ]
- }
- ],
- "prompt_number": 27
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter17_1.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter17_1.ipynb deleted file mode 100755 index 7a620362..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter17_1.ipynb +++ /dev/null @@ -1,1273 +0,0 @@ -{
- "metadata": {
- "name": "chapter17.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 17: Work and Energy"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-4, Page No 400"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "s1=2 # compression of the spring- initial\n",
- "s2=5 # compression of the spring- final\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(s, a, b):\n",
- " return 20*s\n",
- "a=1\n",
- "b=1\n",
- "U=quad(integrand, s1, s2, args=(a,b))\n",
- "\n",
- "# Results\n",
- "print'The word done in compressing the spring is',round(U[0]),\"in-lb\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The word done in compressing the spring is 210.0 in-lb\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-6, Page No 400"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=5 #kg\n",
- "d=6 #m\n",
- "# as theta1=30 degrees & theta2=10 degrees,\n",
- "sintheta1=2**-1\n",
- "sintheta2=0.1736\n",
- "costheta1=(3**0.5)*2**-1\n",
- "costheta2=0.9848\n",
- "u=0.2 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "F=70 #N\n",
- "\n",
- "#Calculations\n",
- "#Using free body diagram\n",
- "Na=(m*g*costheta1)-(F*sintheta2) #N\n",
- "#work done by each force\n",
- "W=[F*costheta2,-m*g*sintheta1,0,-u*Na*d] #N.m\n",
- "#Total Work Done\n",
- "W_tot=W[0]+W[1]+W[2]+W[3] #N.m\n",
- "#Using resultant\n",
- "R=F*costheta2-(u*Na)-(m*g*sintheta1) #N\n",
- "W_d=R*d #N.m (Work Done)\n",
- "\n",
- "#Result\n",
- "print'The work done is',round(W_d),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The work done is 230.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-7, Page No 401"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=20 #kg\n",
- "d=1.5 #m\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "u=0.25 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "F=130 #N\n",
- "\n",
- "#Calculations\n",
- "W=F*d-(m*g*sintheta*d) #N.m\n",
- "\n",
- "#Result\n",
- "print'The work done is',round(W),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The work done is 48.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-9, Page No 402"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=6*12**-1 #ft\n",
- "l=8*12**-1 #ft\n",
- "l_c=3.2 #in\n",
- "y=1.82 #in**2\n",
- "\n",
- "#Calculations\n",
- "V=1*4**-1*pi*d**2*l #ft**3\n",
- "#One horizontal inch \n",
- "h_i=V/l_c #ft**3\n",
- "#One vertical inch\n",
- "v_i=100*144 #lb/ft**2\n",
- "#Then 1.82 in**2 represents\n",
- "x=y*v_i*h_i #ft-lb\n",
- "\n",
- "#Result\n",
- "print'The work capacity is',round(x),\"ft-lb\"\n",
- "\n",
- "# The ans in the textbook is incorrect."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The work capacity is 1072.0 ft-lb\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-10, Page No 402"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "speed=90000 #m/h\n",
- "P=100*1000 #N\n",
- "\n",
- "#Calculations\n",
- "Power=P*((speed)/3600) #J/s\n",
- "\n",
- "#Result\n",
- "print'The power developed is',round(Power*10**-6,1),\"MJ/s\"\n",
- "# Note the unit used."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power developed is 2.5 MJ/s\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-11, Page No 403"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=0.6 #m\n",
- "T_t=800 #N\n",
- "T_s=180 #N\n",
- "w=200 #rpm\n",
- "\n",
- "#Calculations\n",
- "r=d/2 #m radius\n",
- "#Torque\n",
- "M=(T_t-T_s)*r #N.m\n",
- "#Power\n",
- "w_new=(2*pi*w)/60 #rad/s\n",
- "Power=M*(w_new) #W\n",
- "\n",
- "#Result\n",
- "print'The power transmitted is',round(Power*10**-3,1),\"kW\"\n",
- "\n",
- "# The answer in the book is incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power transmitted is 3.9 kW\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-12, Page No 403"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "P=25.6 #lb\n",
- "w=600 #rpm\n",
- "a=36 #in\n",
- "b=12 #in\n",
- "\n",
- "#Calculations\n",
- "M=P*(((b*2**-1)+a)/12) #lb-ft\n",
- "w_new=(2*pi*w)/60 #rad/s\n",
- "Hp=(M*w_new)/550 #hp\n",
- "\n",
- "#Result\n",
- "print'The power being transmitted is',round(Hp,1),\"hp\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power being transmitted is 10.2 hp\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-13, Page No 403"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Pout=3.8 #bhp\n",
- "Pin=4.1 #ihp\n",
- "\n",
- "#Calculations\n",
- "Efficiency=round((Pout/Pin)*100) #Percent\n",
- "\n",
- "#Result\n",
- "print'The efficiency of the engine is',round(Efficiency),\"%\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The efficiency of the engine is 93.0 %\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-15, Page No 404"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return -(3/x)\n",
- "a=1\n",
- "b=1\n",
- "U=quad(integrand, 6, 3, args=(a,b))\n",
- "g=32.2 # ft/s**2\n",
- "deltaT=((U[0]*32.2*2)/(4*16**-1))**0.5\n",
- "# Results\n",
- "print'The speed of the disc will be',round(deltaT,1),\"ft/s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the disc will be 23.1 ft/s\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-16, Page No 404"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=2 #m\n",
- "m=4 #kg\n",
- "w_1=20 #rpm\n",
- "w_2=50 #rpm\n",
- "rev=10 #no of revolution\n",
- "\n",
- "#Calculations\n",
- "Io=(3**-1)*(m)*l**2 #kg.m**2\n",
- "w1=(2*pi*w_1)/60 #rad/s\n",
- "w2=(2*pi*w_2)/60 #rad/s\n",
- "theta=2*pi*rev #rad\n",
- "M=(0.5*Io*(w2**2-w1**2))/theta #N.m\n",
- "\n",
- "#Result\n",
- "print'The constant moment required is',round(M,3),\"N.m\"\n",
- "# The ans waries in decimal places."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The constant moment required is 0.977 N.m\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-18, Page No 405"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=1000 #lb\n",
- "w_w=200 #lb weight of the individual wheel\n",
- "d_w=2.5 #ft diameter of the wheel\n",
- "v=22 #ft/s\n",
- "t=2 #minutes\n",
- "\n",
- "#Calculations\n",
- "#T1=Initial Kinetic Energy and T2=Final Kinetic Energy\n",
- "F=(-0.5*W*32.2**-1*v**2-4*0.5*w_w*32.2**-1*(v**2+0.5*v**2))/(10560) #lb\n",
- "#Negative sign in the answer tells it oposses the motion\n",
- "\n",
- "#Result\n",
- "print'The rolling resistance is',round(F,2),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The rolling resistance is -1.57 lb\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-19, Page No 405"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=100 #lb\n",
- "lo=4 #ft\n",
- "# as theta=45 degrees\n",
- "costheta=(2**0.5)**-1\n",
- "g=32.2 #ft/s**2\n",
- "l=8/3 #ft\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point O and equating it to zero\n",
- "alpha=(W*(lo*0.5)*costheta)/((W/g)*(l)*2) #rad/s**2\n",
- "#Summing forces in the t direction\n",
- "Ot=(W*costheta)-((W/g)*lo*0.5*alpha) #lb\n",
- "#Work Done\n",
- "Work=W*(lo*0.5*costheta) #ft/lb\n",
- "#Moment of inertia\n",
- "Io=(3**-1)*(W/g)*(lo**2) #kg-ft**2\n",
- "#Using the concept for work done=chane in K.E\n",
- "w=(Work/(0.5*Io))**0.5 #rad/s\n",
- "#Summing forces along the bar\n",
- "On=-(-((W/g)*lo*0.5*w**2)-(W*costheta)) #lb\n",
- "\n",
- "#Result\n",
- "print'The bearing reaction at O on the rod is',round(On),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The bearing reaction at O on the rod is 177.0 lb\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-21, Page No 406"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "vo=9 #m/s\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "x=((7*10**-1)*vo**2)*(g*sintheta)**-1 #m\n",
- "\n",
- "#Result\n",
- "print'The ball will roll',round(x,1),\"m up the plane\"\n",
- "\n",
- "#The textbook wrongly mentions the unit of displacement as in\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ball will roll 11.6 m up the plane\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-22, Page No 406"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=322 #lb\n",
- "F=12 #lb\n",
- "a=0 #lower limit (where the cyliner starts rolling)\n",
- "b=pi/2 #Upper Limit (where the cyliner stops rolling)\n",
- "d=3.2 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "dR=1.6 #Differential Radius\n",
- "d_U=2*dR*F #differential work done\n",
- "#Integration Calculations\n",
- "#As it is a simple integration we can resort to this\n",
- "U=d_U*(b-a) #ft-lb\n",
- "#Determination of K.E\n",
- "w=sqrt(U/((0.5*(W/g)*(1/(d/2)**2))+((0.5*0.5)*(W/g)*(d/2)**2))) #rad/s\n",
- "\n",
- "#Result \n",
- "print'The angular velocity of the cylinder is',round(w,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity of the cylinder is 2.69 rad/s\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-24, Page No 407"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wa=161 #lb\n",
- "Wb=193.2 #lb\n",
- "Wc=322 #lb\n",
- "v1=5 #ft/s\n",
- "lc=6 #in\n",
- "k=6 #lb/ft\n",
- "l=4 #ft\n",
- "u=0.2 #coefficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "Ib=(2**-1)*(Wb/g)*(2**-1)**2 #Moment of inertia\n",
- "w1=v1*0.5**-1 #rad/s\n",
- "T1=(0.5*(Wc/g)*v1**2)+(0.5*Ib*w1**2)+(0.5*(Wa/g)*v1**2) #ft-lb\n",
- "#Work Done on the system\n",
- "#The textbook is ambigious on the calculations hence the result is dispalyed directly\n",
- "U=26.4 #ft-lb\n",
- "#Velocity Calculations\n",
- "v=((T1+U)*9**-1)**0.5 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The velocity of the block is',round(v,2),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity of the block is 5.29 ft/s\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-25, Page No 408"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Mm=70 #kg\n",
- "Mc=45 #kg\n",
- "R=0.6 #m\n",
- "g=9.8 #m/s**2\n",
- "l=5 #m\n",
- "# as theta=50 degrees,\n",
- "sintheta=0.77\n",
- "\n",
- "#Calculations\n",
- "#T2 calculations except for v term in it as it cannot be declared as a number\n",
- "T2=68.7 #without the v term in it\n",
- "v=((g*Mm*l-g*Mc*l*sintheta)/T2)**0.5 #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed is',round(v,2),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed is 5.02 m/s\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-26, Page No 408"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#The textbook has a typo in printing the question number\n",
- "#Initilization of variables\n",
- "W1=96.6 #lb\n",
- "W2=128.8 #lb\n",
- "v=8 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "\n",
- "#Calculations\n",
- "#Initial KE of the system is T1=0\n",
- "T2=(0.5*(W1/g)*v**2)+(0.5*(W2/g)*(v/2)**2) #ft-lb\n",
- "#Work Done without s term\n",
- "U=-(W1*sintheta)+W2*0.5\n",
- "#S calculations\n",
- "s=T2*U**-1 #ft\n",
- "\n",
- "#Result\n",
- "print'The block attains a speed of 8 ft/s in',round(s,2),\"ft (up the plane)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The block attains a speed of 8 ft/s in 7.95 ft (up the plane)\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-28, Page No 409"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "l=6 #m length of the cable\n",
- "m=50 #kg mass of the cable\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return 81.7*(6-x)\n",
- "a=1\n",
- "b=1\n",
- "Work=quad(integrand, 0, 6, args=(a,b))\n",
- "\n",
- "# Results\n",
- "print'The word done is',round(Work[0]),\"N.m\"\n",
- "# The answer in textbook is off by 1 N.m"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The word done is 1471.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-31, Page No 410"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "k=0.044 # spring constant\n",
- "#x=0.300 #m length of compression from 450 to 150\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return 0.5*0.044*x**2\n",
- "a=1\n",
- "b=1\n",
- "W=quad(integrand, 0, 300, args=(a,b))\n",
- "\n",
- "# Results\n",
- "print'The word done on the balls is',round(W[0]*10**-5,2),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The word done on the balls is 1.98 N.m\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-32, Page No 410"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=10 #kg\n",
- "d=1.2 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Initilial KE is zero\n",
- "#Final KE is(without v^2 term in it)\n",
- "KE2=(3*4**-1)*10\n",
- "#Work Done\n",
- "U=m*g*d #N.m\n",
- "#Velocity calculations\n",
- "v=sqrt(U*KE2**-1) #m/s\n",
- "\n",
- "#Result\n",
- "print'The velocity is',round(v,2),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity is 3.96 m/s\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-33, Page No 411"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=161 #lb\n",
- "wa=150 #lb\n",
- "wb=100 #lb\n",
- "la=2 #ft\n",
- "lb=4 #ft\n",
- "\n",
- "#Calculations\n",
- "#Work Done\n",
- "T1=wb*lb-wa*la #ft-lb\n",
- "#Final KE=zero\n",
- "T2=0 #ft-lb\n",
- "#Work Done on the system=T2-T1\n",
- "#Hence the equation becomes\n",
- "#50x-50x^2+100=0\n",
- "#where\n",
- "a=-50\n",
- "b=50\n",
- "c=100\n",
- "#Solution\n",
- "d=sqrt(b**2-4*a*c) \n",
- "x1=(-b+d)/(2*a) #ft\n",
- "x2=(-b-d)/(2*a) #ft\n",
- "\n",
- "#Result\n",
- "print'The stretch of the spring is',round(x2),\"ft\"\n",
- "#Here even x1 could have been the solution,but the stretch in the string is elongation not compression hence x2 is the valid answer\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The stretch of the spring is 2.0 ft\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-34, Page No 411"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "I=100 #slug-ft**2\n",
- "w=4 #rad/s\n",
- "theta=6 #rad\n",
- "Mc=64.4 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "vb=2*w #ft/s\n",
- "vc=0.5*w #ft/s\n",
- "Mb=(0.5*I*w**2+0.5*(Mc/g)*vc**2+0.5*Mc*theta)/(2*theta-(0.5*vb**2*(1/g))) #lb\n",
- "\n",
- "#Result\n",
- "print'The weight of the block B is',round(Mb,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The weight of the block B is 90.6 lb\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-35, Page No 411"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wa=96.6 #lb\n",
- "Wb=128.8 #lb\n",
- "g=32.2 #ft/s**2\n",
- "I=12 #slug-ft**2\n",
- "v=16 #ft/s\n",
- "ratio=3**-1 #ratio of Sb/Sa\n",
- "r=3#ft\n",
- "va=6 #ft/s\n",
- "vb=2 #ft/s\n",
- "\n",
- "#Calculations\n",
- "#Work Done without S in it\n",
- "W=Wa-(ratio*Wb)\n",
- "#System has zero KE initially and final KE is given by\n",
- "w=va/r #rad/s\n",
- "T2=(0.5*(Wa/g)*va**2+0.5*I*w**2+0.5*(Wb/g)*vb**2) #ft-lb\n",
- "#Distance Calculations\n",
- "S=T2*W**-1 #ft\n",
- "\n",
- "#Result\n",
- "print'The distance through which A falls is',round(S,2),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The distance through which A falls is 1.6 ft\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-36, Page No 412"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#initilization of variables\n",
- "u=0.25 #coefficient of friction\n",
- "k=2800 #N/m\n",
- "x=0.075 #m\n",
- "g=9.8 #m/s**2\n",
- "m=7 #kg\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Normal Reaction\n",
- "N=g*m*costheta #N\n",
- "#Frictional Force\n",
- "Fr=u*N #N\n",
- "#Component of force along the plane\n",
- "F=g*m*sintheta #N\n",
- "#Spring work is\n",
- "W=0.5*k*x*x #N.m\n",
- "s=(W+Fr*x-F*x)/(F-Fr) #m\n",
- "S=(s*1000) #mm\n",
- "\n",
- "#Result\n",
- "print'The value of S is',round(S),\"mm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of S is 330.0 mm\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-37, Page No 412"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=5 #kg\n",
- "l=2 #m\n",
- "k=10000 #N/m\n",
- "x=0.1 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "drop=l+x #m mass drop length\n",
- "#Work Done by Gravity\n",
- "Wg=g*m*drop #N.m\n",
- "#Work Done by Spring\n",
- "Ws=0.5*k*x**2 #N.m\n",
- "#Increase in KE is without v**2\n",
- "KE=0.5*m #kg\n",
- "#Velocity Calculations\n",
- "v=sqrt((Wg-Ws)/KE) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed is',round(v,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed is 4.6 m/s\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-38, Page No 412"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=6 #ft\n",
- "k=20 #lb/in\n",
- "x=8 #in\n",
- "\n",
- "#Calculations\n",
- "#Work Done by Gravity\n",
- "Wg=(l*12+x) #in without W\n",
- "#Work Done by Spring\n",
- "Ws=0.5*k*x**2 #in-lb\n",
- "#Change in the kinetic energy is zero\n",
- "W=Ws/Wg #lb\n",
- "\n",
- "#Result\n",
- "print'The weight is',round(W),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The weight is 8.0 lb\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-40, Page No 413"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=8 #lb\n",
- "\n",
- "#Calculations\n",
- "#work done by the spring woithout k\n",
- "Ws=0.5*((9*12**-1)**2-(12**-1)**2) \n",
- "#Work done by gravity\n",
- "Wg=W*(10.5*12**-1) #ft-lb\n",
- "#Change in KE is zero\n",
- "k=Wg/Ws #lb/ft\n",
- "\n",
- "#Result\n",
- "print'The value of k is',round(k,1),\"lb/ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of k is 25.2 lb/ft\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-41, Page No 414"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wc=100 #lb\n",
- "r= 1 #ft\n",
- "F=80 #lb\n",
- "k=50 #lb/ft\n",
- "s=6 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Work done on the system\n",
- "U=-0.5*k*(1)+F*(s*12**-1) #ft-lb\n",
- "#Initial KE is zero\n",
- "Vo=(U/(0.5*(Wc/g+0.5*(Wc/g)*r)))**0.5 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The initial speed is',round(Vo,2),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The initial speed is 2.54 ft/s\n"
- ]
- }
- ],
- "prompt_number": 27
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter17_2.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter17_2.ipynb deleted file mode 100755 index 7a620362..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter17_2.ipynb +++ /dev/null @@ -1,1273 +0,0 @@ -{
- "metadata": {
- "name": "chapter17.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 17: Work and Energy"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-4, Page No 400"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "s1=2 # compression of the spring- initial\n",
- "s2=5 # compression of the spring- final\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(s, a, b):\n",
- " return 20*s\n",
- "a=1\n",
- "b=1\n",
- "U=quad(integrand, s1, s2, args=(a,b))\n",
- "\n",
- "# Results\n",
- "print'The word done in compressing the spring is',round(U[0]),\"in-lb\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The word done in compressing the spring is 210.0 in-lb\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-6, Page No 400"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=5 #kg\n",
- "d=6 #m\n",
- "# as theta1=30 degrees & theta2=10 degrees,\n",
- "sintheta1=2**-1\n",
- "sintheta2=0.1736\n",
- "costheta1=(3**0.5)*2**-1\n",
- "costheta2=0.9848\n",
- "u=0.2 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "F=70 #N\n",
- "\n",
- "#Calculations\n",
- "#Using free body diagram\n",
- "Na=(m*g*costheta1)-(F*sintheta2) #N\n",
- "#work done by each force\n",
- "W=[F*costheta2,-m*g*sintheta1,0,-u*Na*d] #N.m\n",
- "#Total Work Done\n",
- "W_tot=W[0]+W[1]+W[2]+W[3] #N.m\n",
- "#Using resultant\n",
- "R=F*costheta2-(u*Na)-(m*g*sintheta1) #N\n",
- "W_d=R*d #N.m (Work Done)\n",
- "\n",
- "#Result\n",
- "print'The work done is',round(W_d),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The work done is 230.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-7, Page No 401"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=20 #kg\n",
- "d=1.5 #m\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "u=0.25 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "F=130 #N\n",
- "\n",
- "#Calculations\n",
- "W=F*d-(m*g*sintheta*d) #N.m\n",
- "\n",
- "#Result\n",
- "print'The work done is',round(W),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The work done is 48.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-9, Page No 402"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=6*12**-1 #ft\n",
- "l=8*12**-1 #ft\n",
- "l_c=3.2 #in\n",
- "y=1.82 #in**2\n",
- "\n",
- "#Calculations\n",
- "V=1*4**-1*pi*d**2*l #ft**3\n",
- "#One horizontal inch \n",
- "h_i=V/l_c #ft**3\n",
- "#One vertical inch\n",
- "v_i=100*144 #lb/ft**2\n",
- "#Then 1.82 in**2 represents\n",
- "x=y*v_i*h_i #ft-lb\n",
- "\n",
- "#Result\n",
- "print'The work capacity is',round(x),\"ft-lb\"\n",
- "\n",
- "# The ans in the textbook is incorrect."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The work capacity is 1072.0 ft-lb\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-10, Page No 402"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "speed=90000 #m/h\n",
- "P=100*1000 #N\n",
- "\n",
- "#Calculations\n",
- "Power=P*((speed)/3600) #J/s\n",
- "\n",
- "#Result\n",
- "print'The power developed is',round(Power*10**-6,1),\"MJ/s\"\n",
- "# Note the unit used."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power developed is 2.5 MJ/s\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-11, Page No 403"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=0.6 #m\n",
- "T_t=800 #N\n",
- "T_s=180 #N\n",
- "w=200 #rpm\n",
- "\n",
- "#Calculations\n",
- "r=d/2 #m radius\n",
- "#Torque\n",
- "M=(T_t-T_s)*r #N.m\n",
- "#Power\n",
- "w_new=(2*pi*w)/60 #rad/s\n",
- "Power=M*(w_new) #W\n",
- "\n",
- "#Result\n",
- "print'The power transmitted is',round(Power*10**-3,1),\"kW\"\n",
- "\n",
- "# The answer in the book is incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power transmitted is 3.9 kW\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-12, Page No 403"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "P=25.6 #lb\n",
- "w=600 #rpm\n",
- "a=36 #in\n",
- "b=12 #in\n",
- "\n",
- "#Calculations\n",
- "M=P*(((b*2**-1)+a)/12) #lb-ft\n",
- "w_new=(2*pi*w)/60 #rad/s\n",
- "Hp=(M*w_new)/550 #hp\n",
- "\n",
- "#Result\n",
- "print'The power being transmitted is',round(Hp,1),\"hp\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power being transmitted is 10.2 hp\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-13, Page No 403"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Pout=3.8 #bhp\n",
- "Pin=4.1 #ihp\n",
- "\n",
- "#Calculations\n",
- "Efficiency=round((Pout/Pin)*100) #Percent\n",
- "\n",
- "#Result\n",
- "print'The efficiency of the engine is',round(Efficiency),\"%\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The efficiency of the engine is 93.0 %\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-15, Page No 404"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return -(3/x)\n",
- "a=1\n",
- "b=1\n",
- "U=quad(integrand, 6, 3, args=(a,b))\n",
- "g=32.2 # ft/s**2\n",
- "deltaT=((U[0]*32.2*2)/(4*16**-1))**0.5\n",
- "# Results\n",
- "print'The speed of the disc will be',round(deltaT,1),\"ft/s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the disc will be 23.1 ft/s\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-16, Page No 404"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=2 #m\n",
- "m=4 #kg\n",
- "w_1=20 #rpm\n",
- "w_2=50 #rpm\n",
- "rev=10 #no of revolution\n",
- "\n",
- "#Calculations\n",
- "Io=(3**-1)*(m)*l**2 #kg.m**2\n",
- "w1=(2*pi*w_1)/60 #rad/s\n",
- "w2=(2*pi*w_2)/60 #rad/s\n",
- "theta=2*pi*rev #rad\n",
- "M=(0.5*Io*(w2**2-w1**2))/theta #N.m\n",
- "\n",
- "#Result\n",
- "print'The constant moment required is',round(M,3),\"N.m\"\n",
- "# The ans waries in decimal places."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The constant moment required is 0.977 N.m\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-18, Page No 405"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=1000 #lb\n",
- "w_w=200 #lb weight of the individual wheel\n",
- "d_w=2.5 #ft diameter of the wheel\n",
- "v=22 #ft/s\n",
- "t=2 #minutes\n",
- "\n",
- "#Calculations\n",
- "#T1=Initial Kinetic Energy and T2=Final Kinetic Energy\n",
- "F=(-0.5*W*32.2**-1*v**2-4*0.5*w_w*32.2**-1*(v**2+0.5*v**2))/(10560) #lb\n",
- "#Negative sign in the answer tells it oposses the motion\n",
- "\n",
- "#Result\n",
- "print'The rolling resistance is',round(F,2),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The rolling resistance is -1.57 lb\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-19, Page No 405"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=100 #lb\n",
- "lo=4 #ft\n",
- "# as theta=45 degrees\n",
- "costheta=(2**0.5)**-1\n",
- "g=32.2 #ft/s**2\n",
- "l=8/3 #ft\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point O and equating it to zero\n",
- "alpha=(W*(lo*0.5)*costheta)/((W/g)*(l)*2) #rad/s**2\n",
- "#Summing forces in the t direction\n",
- "Ot=(W*costheta)-((W/g)*lo*0.5*alpha) #lb\n",
- "#Work Done\n",
- "Work=W*(lo*0.5*costheta) #ft/lb\n",
- "#Moment of inertia\n",
- "Io=(3**-1)*(W/g)*(lo**2) #kg-ft**2\n",
- "#Using the concept for work done=chane in K.E\n",
- "w=(Work/(0.5*Io))**0.5 #rad/s\n",
- "#Summing forces along the bar\n",
- "On=-(-((W/g)*lo*0.5*w**2)-(W*costheta)) #lb\n",
- "\n",
- "#Result\n",
- "print'The bearing reaction at O on the rod is',round(On),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The bearing reaction at O on the rod is 177.0 lb\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-21, Page No 406"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "vo=9 #m/s\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "x=((7*10**-1)*vo**2)*(g*sintheta)**-1 #m\n",
- "\n",
- "#Result\n",
- "print'The ball will roll',round(x,1),\"m up the plane\"\n",
- "\n",
- "#The textbook wrongly mentions the unit of displacement as in\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ball will roll 11.6 m up the plane\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-22, Page No 406"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=322 #lb\n",
- "F=12 #lb\n",
- "a=0 #lower limit (where the cyliner starts rolling)\n",
- "b=pi/2 #Upper Limit (where the cyliner stops rolling)\n",
- "d=3.2 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "dR=1.6 #Differential Radius\n",
- "d_U=2*dR*F #differential work done\n",
- "#Integration Calculations\n",
- "#As it is a simple integration we can resort to this\n",
- "U=d_U*(b-a) #ft-lb\n",
- "#Determination of K.E\n",
- "w=sqrt(U/((0.5*(W/g)*(1/(d/2)**2))+((0.5*0.5)*(W/g)*(d/2)**2))) #rad/s\n",
- "\n",
- "#Result \n",
- "print'The angular velocity of the cylinder is',round(w,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity of the cylinder is 2.69 rad/s\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-24, Page No 407"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wa=161 #lb\n",
- "Wb=193.2 #lb\n",
- "Wc=322 #lb\n",
- "v1=5 #ft/s\n",
- "lc=6 #in\n",
- "k=6 #lb/ft\n",
- "l=4 #ft\n",
- "u=0.2 #coefficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "Ib=(2**-1)*(Wb/g)*(2**-1)**2 #Moment of inertia\n",
- "w1=v1*0.5**-1 #rad/s\n",
- "T1=(0.5*(Wc/g)*v1**2)+(0.5*Ib*w1**2)+(0.5*(Wa/g)*v1**2) #ft-lb\n",
- "#Work Done on the system\n",
- "#The textbook is ambigious on the calculations hence the result is dispalyed directly\n",
- "U=26.4 #ft-lb\n",
- "#Velocity Calculations\n",
- "v=((T1+U)*9**-1)**0.5 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The velocity of the block is',round(v,2),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity of the block is 5.29 ft/s\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-25, Page No 408"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Mm=70 #kg\n",
- "Mc=45 #kg\n",
- "R=0.6 #m\n",
- "g=9.8 #m/s**2\n",
- "l=5 #m\n",
- "# as theta=50 degrees,\n",
- "sintheta=0.77\n",
- "\n",
- "#Calculations\n",
- "#T2 calculations except for v term in it as it cannot be declared as a number\n",
- "T2=68.7 #without the v term in it\n",
- "v=((g*Mm*l-g*Mc*l*sintheta)/T2)**0.5 #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed is',round(v,2),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed is 5.02 m/s\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-26, Page No 408"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#The textbook has a typo in printing the question number\n",
- "#Initilization of variables\n",
- "W1=96.6 #lb\n",
- "W2=128.8 #lb\n",
- "v=8 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "\n",
- "#Calculations\n",
- "#Initial KE of the system is T1=0\n",
- "T2=(0.5*(W1/g)*v**2)+(0.5*(W2/g)*(v/2)**2) #ft-lb\n",
- "#Work Done without s term\n",
- "U=-(W1*sintheta)+W2*0.5\n",
- "#S calculations\n",
- "s=T2*U**-1 #ft\n",
- "\n",
- "#Result\n",
- "print'The block attains a speed of 8 ft/s in',round(s,2),\"ft (up the plane)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The block attains a speed of 8 ft/s in 7.95 ft (up the plane)\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-28, Page No 409"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "l=6 #m length of the cable\n",
- "m=50 #kg mass of the cable\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return 81.7*(6-x)\n",
- "a=1\n",
- "b=1\n",
- "Work=quad(integrand, 0, 6, args=(a,b))\n",
- "\n",
- "# Results\n",
- "print'The word done is',round(Work[0]),\"N.m\"\n",
- "# The answer in textbook is off by 1 N.m"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The word done is 1471.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-31, Page No 410"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "k=0.044 # spring constant\n",
- "#x=0.300 #m length of compression from 450 to 150\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return 0.5*0.044*x**2\n",
- "a=1\n",
- "b=1\n",
- "W=quad(integrand, 0, 300, args=(a,b))\n",
- "\n",
- "# Results\n",
- "print'The word done on the balls is',round(W[0]*10**-5,2),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The word done on the balls is 1.98 N.m\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-32, Page No 410"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=10 #kg\n",
- "d=1.2 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Initilial KE is zero\n",
- "#Final KE is(without v^2 term in it)\n",
- "KE2=(3*4**-1)*10\n",
- "#Work Done\n",
- "U=m*g*d #N.m\n",
- "#Velocity calculations\n",
- "v=sqrt(U*KE2**-1) #m/s\n",
- "\n",
- "#Result\n",
- "print'The velocity is',round(v,2),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity is 3.96 m/s\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-33, Page No 411"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=161 #lb\n",
- "wa=150 #lb\n",
- "wb=100 #lb\n",
- "la=2 #ft\n",
- "lb=4 #ft\n",
- "\n",
- "#Calculations\n",
- "#Work Done\n",
- "T1=wb*lb-wa*la #ft-lb\n",
- "#Final KE=zero\n",
- "T2=0 #ft-lb\n",
- "#Work Done on the system=T2-T1\n",
- "#Hence the equation becomes\n",
- "#50x-50x^2+100=0\n",
- "#where\n",
- "a=-50\n",
- "b=50\n",
- "c=100\n",
- "#Solution\n",
- "d=sqrt(b**2-4*a*c) \n",
- "x1=(-b+d)/(2*a) #ft\n",
- "x2=(-b-d)/(2*a) #ft\n",
- "\n",
- "#Result\n",
- "print'The stretch of the spring is',round(x2),\"ft\"\n",
- "#Here even x1 could have been the solution,but the stretch in the string is elongation not compression hence x2 is the valid answer\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The stretch of the spring is 2.0 ft\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-34, Page No 411"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "I=100 #slug-ft**2\n",
- "w=4 #rad/s\n",
- "theta=6 #rad\n",
- "Mc=64.4 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "vb=2*w #ft/s\n",
- "vc=0.5*w #ft/s\n",
- "Mb=(0.5*I*w**2+0.5*(Mc/g)*vc**2+0.5*Mc*theta)/(2*theta-(0.5*vb**2*(1/g))) #lb\n",
- "\n",
- "#Result\n",
- "print'The weight of the block B is',round(Mb,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The weight of the block B is 90.6 lb\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-35, Page No 411"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wa=96.6 #lb\n",
- "Wb=128.8 #lb\n",
- "g=32.2 #ft/s**2\n",
- "I=12 #slug-ft**2\n",
- "v=16 #ft/s\n",
- "ratio=3**-1 #ratio of Sb/Sa\n",
- "r=3#ft\n",
- "va=6 #ft/s\n",
- "vb=2 #ft/s\n",
- "\n",
- "#Calculations\n",
- "#Work Done without S in it\n",
- "W=Wa-(ratio*Wb)\n",
- "#System has zero KE initially and final KE is given by\n",
- "w=va/r #rad/s\n",
- "T2=(0.5*(Wa/g)*va**2+0.5*I*w**2+0.5*(Wb/g)*vb**2) #ft-lb\n",
- "#Distance Calculations\n",
- "S=T2*W**-1 #ft\n",
- "\n",
- "#Result\n",
- "print'The distance through which A falls is',round(S,2),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The distance through which A falls is 1.6 ft\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-36, Page No 412"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#initilization of variables\n",
- "u=0.25 #coefficient of friction\n",
- "k=2800 #N/m\n",
- "x=0.075 #m\n",
- "g=9.8 #m/s**2\n",
- "m=7 #kg\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Normal Reaction\n",
- "N=g*m*costheta #N\n",
- "#Frictional Force\n",
- "Fr=u*N #N\n",
- "#Component of force along the plane\n",
- "F=g*m*sintheta #N\n",
- "#Spring work is\n",
- "W=0.5*k*x*x #N.m\n",
- "s=(W+Fr*x-F*x)/(F-Fr) #m\n",
- "S=(s*1000) #mm\n",
- "\n",
- "#Result\n",
- "print'The value of S is',round(S),\"mm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of S is 330.0 mm\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-37, Page No 412"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=5 #kg\n",
- "l=2 #m\n",
- "k=10000 #N/m\n",
- "x=0.1 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "drop=l+x #m mass drop length\n",
- "#Work Done by Gravity\n",
- "Wg=g*m*drop #N.m\n",
- "#Work Done by Spring\n",
- "Ws=0.5*k*x**2 #N.m\n",
- "#Increase in KE is without v**2\n",
- "KE=0.5*m #kg\n",
- "#Velocity Calculations\n",
- "v=sqrt((Wg-Ws)/KE) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed is',round(v,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed is 4.6 m/s\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-38, Page No 412"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=6 #ft\n",
- "k=20 #lb/in\n",
- "x=8 #in\n",
- "\n",
- "#Calculations\n",
- "#Work Done by Gravity\n",
- "Wg=(l*12+x) #in without W\n",
- "#Work Done by Spring\n",
- "Ws=0.5*k*x**2 #in-lb\n",
- "#Change in the kinetic energy is zero\n",
- "W=Ws/Wg #lb\n",
- "\n",
- "#Result\n",
- "print'The weight is',round(W),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The weight is 8.0 lb\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-40, Page No 413"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=8 #lb\n",
- "\n",
- "#Calculations\n",
- "#work done by the spring woithout k\n",
- "Ws=0.5*((9*12**-1)**2-(12**-1)**2) \n",
- "#Work done by gravity\n",
- "Wg=W*(10.5*12**-1) #ft-lb\n",
- "#Change in KE is zero\n",
- "k=Wg/Ws #lb/ft\n",
- "\n",
- "#Result\n",
- "print'The value of k is',round(k,1),\"lb/ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of k is 25.2 lb/ft\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-41, Page No 414"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wc=100 #lb\n",
- "r= 1 #ft\n",
- "F=80 #lb\n",
- "k=50 #lb/ft\n",
- "s=6 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Work done on the system\n",
- "U=-0.5*k*(1)+F*(s*12**-1) #ft-lb\n",
- "#Initial KE is zero\n",
- "Vo=(U/(0.5*(Wc/g+0.5*(Wc/g)*r)))**0.5 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The initial speed is',round(Vo,2),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The initial speed is 2.54 ft/s\n"
- ]
- }
- ],
- "prompt_number": 27
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter17_3.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter17_3.ipynb deleted file mode 100755 index 7a620362..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter17_3.ipynb +++ /dev/null @@ -1,1273 +0,0 @@ -{
- "metadata": {
- "name": "chapter17.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 17: Work and Energy"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-4, Page No 400"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "s1=2 # compression of the spring- initial\n",
- "s2=5 # compression of the spring- final\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(s, a, b):\n",
- " return 20*s\n",
- "a=1\n",
- "b=1\n",
- "U=quad(integrand, s1, s2, args=(a,b))\n",
- "\n",
- "# Results\n",
- "print'The word done in compressing the spring is',round(U[0]),\"in-lb\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The word done in compressing the spring is 210.0 in-lb\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-6, Page No 400"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=5 #kg\n",
- "d=6 #m\n",
- "# as theta1=30 degrees & theta2=10 degrees,\n",
- "sintheta1=2**-1\n",
- "sintheta2=0.1736\n",
- "costheta1=(3**0.5)*2**-1\n",
- "costheta2=0.9848\n",
- "u=0.2 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "F=70 #N\n",
- "\n",
- "#Calculations\n",
- "#Using free body diagram\n",
- "Na=(m*g*costheta1)-(F*sintheta2) #N\n",
- "#work done by each force\n",
- "W=[F*costheta2,-m*g*sintheta1,0,-u*Na*d] #N.m\n",
- "#Total Work Done\n",
- "W_tot=W[0]+W[1]+W[2]+W[3] #N.m\n",
- "#Using resultant\n",
- "R=F*costheta2-(u*Na)-(m*g*sintheta1) #N\n",
- "W_d=R*d #N.m (Work Done)\n",
- "\n",
- "#Result\n",
- "print'The work done is',round(W_d),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The work done is 230.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-7, Page No 401"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=20 #kg\n",
- "d=1.5 #m\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "u=0.25 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "F=130 #N\n",
- "\n",
- "#Calculations\n",
- "W=F*d-(m*g*sintheta*d) #N.m\n",
- "\n",
- "#Result\n",
- "print'The work done is',round(W),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The work done is 48.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-9, Page No 402"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=6*12**-1 #ft\n",
- "l=8*12**-1 #ft\n",
- "l_c=3.2 #in\n",
- "y=1.82 #in**2\n",
- "\n",
- "#Calculations\n",
- "V=1*4**-1*pi*d**2*l #ft**3\n",
- "#One horizontal inch \n",
- "h_i=V/l_c #ft**3\n",
- "#One vertical inch\n",
- "v_i=100*144 #lb/ft**2\n",
- "#Then 1.82 in**2 represents\n",
- "x=y*v_i*h_i #ft-lb\n",
- "\n",
- "#Result\n",
- "print'The work capacity is',round(x),\"ft-lb\"\n",
- "\n",
- "# The ans in the textbook is incorrect."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The work capacity is 1072.0 ft-lb\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-10, Page No 402"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "speed=90000 #m/h\n",
- "P=100*1000 #N\n",
- "\n",
- "#Calculations\n",
- "Power=P*((speed)/3600) #J/s\n",
- "\n",
- "#Result\n",
- "print'The power developed is',round(Power*10**-6,1),\"MJ/s\"\n",
- "# Note the unit used."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power developed is 2.5 MJ/s\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-11, Page No 403"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=0.6 #m\n",
- "T_t=800 #N\n",
- "T_s=180 #N\n",
- "w=200 #rpm\n",
- "\n",
- "#Calculations\n",
- "r=d/2 #m radius\n",
- "#Torque\n",
- "M=(T_t-T_s)*r #N.m\n",
- "#Power\n",
- "w_new=(2*pi*w)/60 #rad/s\n",
- "Power=M*(w_new) #W\n",
- "\n",
- "#Result\n",
- "print'The power transmitted is',round(Power*10**-3,1),\"kW\"\n",
- "\n",
- "# The answer in the book is incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power transmitted is 3.9 kW\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-12, Page No 403"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "P=25.6 #lb\n",
- "w=600 #rpm\n",
- "a=36 #in\n",
- "b=12 #in\n",
- "\n",
- "#Calculations\n",
- "M=P*(((b*2**-1)+a)/12) #lb-ft\n",
- "w_new=(2*pi*w)/60 #rad/s\n",
- "Hp=(M*w_new)/550 #hp\n",
- "\n",
- "#Result\n",
- "print'The power being transmitted is',round(Hp,1),\"hp\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power being transmitted is 10.2 hp\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-13, Page No 403"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Pout=3.8 #bhp\n",
- "Pin=4.1 #ihp\n",
- "\n",
- "#Calculations\n",
- "Efficiency=round((Pout/Pin)*100) #Percent\n",
- "\n",
- "#Result\n",
- "print'The efficiency of the engine is',round(Efficiency),\"%\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The efficiency of the engine is 93.0 %\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-15, Page No 404"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return -(3/x)\n",
- "a=1\n",
- "b=1\n",
- "U=quad(integrand, 6, 3, args=(a,b))\n",
- "g=32.2 # ft/s**2\n",
- "deltaT=((U[0]*32.2*2)/(4*16**-1))**0.5\n",
- "# Results\n",
- "print'The speed of the disc will be',round(deltaT,1),\"ft/s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the disc will be 23.1 ft/s\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-16, Page No 404"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=2 #m\n",
- "m=4 #kg\n",
- "w_1=20 #rpm\n",
- "w_2=50 #rpm\n",
- "rev=10 #no of revolution\n",
- "\n",
- "#Calculations\n",
- "Io=(3**-1)*(m)*l**2 #kg.m**2\n",
- "w1=(2*pi*w_1)/60 #rad/s\n",
- "w2=(2*pi*w_2)/60 #rad/s\n",
- "theta=2*pi*rev #rad\n",
- "M=(0.5*Io*(w2**2-w1**2))/theta #N.m\n",
- "\n",
- "#Result\n",
- "print'The constant moment required is',round(M,3),\"N.m\"\n",
- "# The ans waries in decimal places."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The constant moment required is 0.977 N.m\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-18, Page No 405"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=1000 #lb\n",
- "w_w=200 #lb weight of the individual wheel\n",
- "d_w=2.5 #ft diameter of the wheel\n",
- "v=22 #ft/s\n",
- "t=2 #minutes\n",
- "\n",
- "#Calculations\n",
- "#T1=Initial Kinetic Energy and T2=Final Kinetic Energy\n",
- "F=(-0.5*W*32.2**-1*v**2-4*0.5*w_w*32.2**-1*(v**2+0.5*v**2))/(10560) #lb\n",
- "#Negative sign in the answer tells it oposses the motion\n",
- "\n",
- "#Result\n",
- "print'The rolling resistance is',round(F,2),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The rolling resistance is -1.57 lb\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-19, Page No 405"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=100 #lb\n",
- "lo=4 #ft\n",
- "# as theta=45 degrees\n",
- "costheta=(2**0.5)**-1\n",
- "g=32.2 #ft/s**2\n",
- "l=8/3 #ft\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point O and equating it to zero\n",
- "alpha=(W*(lo*0.5)*costheta)/((W/g)*(l)*2) #rad/s**2\n",
- "#Summing forces in the t direction\n",
- "Ot=(W*costheta)-((W/g)*lo*0.5*alpha) #lb\n",
- "#Work Done\n",
- "Work=W*(lo*0.5*costheta) #ft/lb\n",
- "#Moment of inertia\n",
- "Io=(3**-1)*(W/g)*(lo**2) #kg-ft**2\n",
- "#Using the concept for work done=chane in K.E\n",
- "w=(Work/(0.5*Io))**0.5 #rad/s\n",
- "#Summing forces along the bar\n",
- "On=-(-((W/g)*lo*0.5*w**2)-(W*costheta)) #lb\n",
- "\n",
- "#Result\n",
- "print'The bearing reaction at O on the rod is',round(On),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The bearing reaction at O on the rod is 177.0 lb\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-21, Page No 406"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "vo=9 #m/s\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "x=((7*10**-1)*vo**2)*(g*sintheta)**-1 #m\n",
- "\n",
- "#Result\n",
- "print'The ball will roll',round(x,1),\"m up the plane\"\n",
- "\n",
- "#The textbook wrongly mentions the unit of displacement as in\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ball will roll 11.6 m up the plane\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-22, Page No 406"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=322 #lb\n",
- "F=12 #lb\n",
- "a=0 #lower limit (where the cyliner starts rolling)\n",
- "b=pi/2 #Upper Limit (where the cyliner stops rolling)\n",
- "d=3.2 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "dR=1.6 #Differential Radius\n",
- "d_U=2*dR*F #differential work done\n",
- "#Integration Calculations\n",
- "#As it is a simple integration we can resort to this\n",
- "U=d_U*(b-a) #ft-lb\n",
- "#Determination of K.E\n",
- "w=sqrt(U/((0.5*(W/g)*(1/(d/2)**2))+((0.5*0.5)*(W/g)*(d/2)**2))) #rad/s\n",
- "\n",
- "#Result \n",
- "print'The angular velocity of the cylinder is',round(w,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity of the cylinder is 2.69 rad/s\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-24, Page No 407"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wa=161 #lb\n",
- "Wb=193.2 #lb\n",
- "Wc=322 #lb\n",
- "v1=5 #ft/s\n",
- "lc=6 #in\n",
- "k=6 #lb/ft\n",
- "l=4 #ft\n",
- "u=0.2 #coefficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "Ib=(2**-1)*(Wb/g)*(2**-1)**2 #Moment of inertia\n",
- "w1=v1*0.5**-1 #rad/s\n",
- "T1=(0.5*(Wc/g)*v1**2)+(0.5*Ib*w1**2)+(0.5*(Wa/g)*v1**2) #ft-lb\n",
- "#Work Done on the system\n",
- "#The textbook is ambigious on the calculations hence the result is dispalyed directly\n",
- "U=26.4 #ft-lb\n",
- "#Velocity Calculations\n",
- "v=((T1+U)*9**-1)**0.5 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The velocity of the block is',round(v,2),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity of the block is 5.29 ft/s\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-25, Page No 408"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Mm=70 #kg\n",
- "Mc=45 #kg\n",
- "R=0.6 #m\n",
- "g=9.8 #m/s**2\n",
- "l=5 #m\n",
- "# as theta=50 degrees,\n",
- "sintheta=0.77\n",
- "\n",
- "#Calculations\n",
- "#T2 calculations except for v term in it as it cannot be declared as a number\n",
- "T2=68.7 #without the v term in it\n",
- "v=((g*Mm*l-g*Mc*l*sintheta)/T2)**0.5 #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed is',round(v,2),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed is 5.02 m/s\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-26, Page No 408"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#The textbook has a typo in printing the question number\n",
- "#Initilization of variables\n",
- "W1=96.6 #lb\n",
- "W2=128.8 #lb\n",
- "v=8 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "\n",
- "#Calculations\n",
- "#Initial KE of the system is T1=0\n",
- "T2=(0.5*(W1/g)*v**2)+(0.5*(W2/g)*(v/2)**2) #ft-lb\n",
- "#Work Done without s term\n",
- "U=-(W1*sintheta)+W2*0.5\n",
- "#S calculations\n",
- "s=T2*U**-1 #ft\n",
- "\n",
- "#Result\n",
- "print'The block attains a speed of 8 ft/s in',round(s,2),\"ft (up the plane)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The block attains a speed of 8 ft/s in 7.95 ft (up the plane)\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-28, Page No 409"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "l=6 #m length of the cable\n",
- "m=50 #kg mass of the cable\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return 81.7*(6-x)\n",
- "a=1\n",
- "b=1\n",
- "Work=quad(integrand, 0, 6, args=(a,b))\n",
- "\n",
- "# Results\n",
- "print'The word done is',round(Work[0]),\"N.m\"\n",
- "# The answer in textbook is off by 1 N.m"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The word done is 1471.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-31, Page No 410"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "k=0.044 # spring constant\n",
- "#x=0.300 #m length of compression from 450 to 150\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return 0.5*0.044*x**2\n",
- "a=1\n",
- "b=1\n",
- "W=quad(integrand, 0, 300, args=(a,b))\n",
- "\n",
- "# Results\n",
- "print'The word done on the balls is',round(W[0]*10**-5,2),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The word done on the balls is 1.98 N.m\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-32, Page No 410"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=10 #kg\n",
- "d=1.2 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Initilial KE is zero\n",
- "#Final KE is(without v^2 term in it)\n",
- "KE2=(3*4**-1)*10\n",
- "#Work Done\n",
- "U=m*g*d #N.m\n",
- "#Velocity calculations\n",
- "v=sqrt(U*KE2**-1) #m/s\n",
- "\n",
- "#Result\n",
- "print'The velocity is',round(v,2),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity is 3.96 m/s\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-33, Page No 411"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=161 #lb\n",
- "wa=150 #lb\n",
- "wb=100 #lb\n",
- "la=2 #ft\n",
- "lb=4 #ft\n",
- "\n",
- "#Calculations\n",
- "#Work Done\n",
- "T1=wb*lb-wa*la #ft-lb\n",
- "#Final KE=zero\n",
- "T2=0 #ft-lb\n",
- "#Work Done on the system=T2-T1\n",
- "#Hence the equation becomes\n",
- "#50x-50x^2+100=0\n",
- "#where\n",
- "a=-50\n",
- "b=50\n",
- "c=100\n",
- "#Solution\n",
- "d=sqrt(b**2-4*a*c) \n",
- "x1=(-b+d)/(2*a) #ft\n",
- "x2=(-b-d)/(2*a) #ft\n",
- "\n",
- "#Result\n",
- "print'The stretch of the spring is',round(x2),\"ft\"\n",
- "#Here even x1 could have been the solution,but the stretch in the string is elongation not compression hence x2 is the valid answer\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The stretch of the spring is 2.0 ft\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-34, Page No 411"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "I=100 #slug-ft**2\n",
- "w=4 #rad/s\n",
- "theta=6 #rad\n",
- "Mc=64.4 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "vb=2*w #ft/s\n",
- "vc=0.5*w #ft/s\n",
- "Mb=(0.5*I*w**2+0.5*(Mc/g)*vc**2+0.5*Mc*theta)/(2*theta-(0.5*vb**2*(1/g))) #lb\n",
- "\n",
- "#Result\n",
- "print'The weight of the block B is',round(Mb,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The weight of the block B is 90.6 lb\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-35, Page No 411"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wa=96.6 #lb\n",
- "Wb=128.8 #lb\n",
- "g=32.2 #ft/s**2\n",
- "I=12 #slug-ft**2\n",
- "v=16 #ft/s\n",
- "ratio=3**-1 #ratio of Sb/Sa\n",
- "r=3#ft\n",
- "va=6 #ft/s\n",
- "vb=2 #ft/s\n",
- "\n",
- "#Calculations\n",
- "#Work Done without S in it\n",
- "W=Wa-(ratio*Wb)\n",
- "#System has zero KE initially and final KE is given by\n",
- "w=va/r #rad/s\n",
- "T2=(0.5*(Wa/g)*va**2+0.5*I*w**2+0.5*(Wb/g)*vb**2) #ft-lb\n",
- "#Distance Calculations\n",
- "S=T2*W**-1 #ft\n",
- "\n",
- "#Result\n",
- "print'The distance through which A falls is',round(S,2),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The distance through which A falls is 1.6 ft\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-36, Page No 412"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#initilization of variables\n",
- "u=0.25 #coefficient of friction\n",
- "k=2800 #N/m\n",
- "x=0.075 #m\n",
- "g=9.8 #m/s**2\n",
- "m=7 #kg\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Normal Reaction\n",
- "N=g*m*costheta #N\n",
- "#Frictional Force\n",
- "Fr=u*N #N\n",
- "#Component of force along the plane\n",
- "F=g*m*sintheta #N\n",
- "#Spring work is\n",
- "W=0.5*k*x*x #N.m\n",
- "s=(W+Fr*x-F*x)/(F-Fr) #m\n",
- "S=(s*1000) #mm\n",
- "\n",
- "#Result\n",
- "print'The value of S is',round(S),\"mm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of S is 330.0 mm\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-37, Page No 412"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=5 #kg\n",
- "l=2 #m\n",
- "k=10000 #N/m\n",
- "x=0.1 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "drop=l+x #m mass drop length\n",
- "#Work Done by Gravity\n",
- "Wg=g*m*drop #N.m\n",
- "#Work Done by Spring\n",
- "Ws=0.5*k*x**2 #N.m\n",
- "#Increase in KE is without v**2\n",
- "KE=0.5*m #kg\n",
- "#Velocity Calculations\n",
- "v=sqrt((Wg-Ws)/KE) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed is',round(v,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed is 4.6 m/s\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-38, Page No 412"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=6 #ft\n",
- "k=20 #lb/in\n",
- "x=8 #in\n",
- "\n",
- "#Calculations\n",
- "#Work Done by Gravity\n",
- "Wg=(l*12+x) #in without W\n",
- "#Work Done by Spring\n",
- "Ws=0.5*k*x**2 #in-lb\n",
- "#Change in the kinetic energy is zero\n",
- "W=Ws/Wg #lb\n",
- "\n",
- "#Result\n",
- "print'The weight is',round(W),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The weight is 8.0 lb\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-40, Page No 413"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=8 #lb\n",
- "\n",
- "#Calculations\n",
- "#work done by the spring woithout k\n",
- "Ws=0.5*((9*12**-1)**2-(12**-1)**2) \n",
- "#Work done by gravity\n",
- "Wg=W*(10.5*12**-1) #ft-lb\n",
- "#Change in KE is zero\n",
- "k=Wg/Ws #lb/ft\n",
- "\n",
- "#Result\n",
- "print'The value of k is',round(k,1),\"lb/ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of k is 25.2 lb/ft\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-41, Page No 414"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wc=100 #lb\n",
- "r= 1 #ft\n",
- "F=80 #lb\n",
- "k=50 #lb/ft\n",
- "s=6 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Work done on the system\n",
- "U=-0.5*k*(1)+F*(s*12**-1) #ft-lb\n",
- "#Initial KE is zero\n",
- "Vo=(U/(0.5*(Wc/g+0.5*(Wc/g)*r)))**0.5 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The initial speed is',round(Vo,2),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The initial speed is 2.54 ft/s\n"
- ]
- }
- ],
- "prompt_number": 27
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter17_4.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter17_4.ipynb deleted file mode 100755 index 7a620362..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter17_4.ipynb +++ /dev/null @@ -1,1273 +0,0 @@ -{
- "metadata": {
- "name": "chapter17.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 17: Work and Energy"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-4, Page No 400"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "s1=2 # compression of the spring- initial\n",
- "s2=5 # compression of the spring- final\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(s, a, b):\n",
- " return 20*s\n",
- "a=1\n",
- "b=1\n",
- "U=quad(integrand, s1, s2, args=(a,b))\n",
- "\n",
- "# Results\n",
- "print'The word done in compressing the spring is',round(U[0]),\"in-lb\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The word done in compressing the spring is 210.0 in-lb\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-6, Page No 400"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=5 #kg\n",
- "d=6 #m\n",
- "# as theta1=30 degrees & theta2=10 degrees,\n",
- "sintheta1=2**-1\n",
- "sintheta2=0.1736\n",
- "costheta1=(3**0.5)*2**-1\n",
- "costheta2=0.9848\n",
- "u=0.2 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "F=70 #N\n",
- "\n",
- "#Calculations\n",
- "#Using free body diagram\n",
- "Na=(m*g*costheta1)-(F*sintheta2) #N\n",
- "#work done by each force\n",
- "W=[F*costheta2,-m*g*sintheta1,0,-u*Na*d] #N.m\n",
- "#Total Work Done\n",
- "W_tot=W[0]+W[1]+W[2]+W[3] #N.m\n",
- "#Using resultant\n",
- "R=F*costheta2-(u*Na)-(m*g*sintheta1) #N\n",
- "W_d=R*d #N.m (Work Done)\n",
- "\n",
- "#Result\n",
- "print'The work done is',round(W_d),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The work done is 230.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-7, Page No 401"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=20 #kg\n",
- "d=1.5 #m\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "u=0.25 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "F=130 #N\n",
- "\n",
- "#Calculations\n",
- "W=F*d-(m*g*sintheta*d) #N.m\n",
- "\n",
- "#Result\n",
- "print'The work done is',round(W),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The work done is 48.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-9, Page No 402"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=6*12**-1 #ft\n",
- "l=8*12**-1 #ft\n",
- "l_c=3.2 #in\n",
- "y=1.82 #in**2\n",
- "\n",
- "#Calculations\n",
- "V=1*4**-1*pi*d**2*l #ft**3\n",
- "#One horizontal inch \n",
- "h_i=V/l_c #ft**3\n",
- "#One vertical inch\n",
- "v_i=100*144 #lb/ft**2\n",
- "#Then 1.82 in**2 represents\n",
- "x=y*v_i*h_i #ft-lb\n",
- "\n",
- "#Result\n",
- "print'The work capacity is',round(x),\"ft-lb\"\n",
- "\n",
- "# The ans in the textbook is incorrect."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The work capacity is 1072.0 ft-lb\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-10, Page No 402"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "speed=90000 #m/h\n",
- "P=100*1000 #N\n",
- "\n",
- "#Calculations\n",
- "Power=P*((speed)/3600) #J/s\n",
- "\n",
- "#Result\n",
- "print'The power developed is',round(Power*10**-6,1),\"MJ/s\"\n",
- "# Note the unit used."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power developed is 2.5 MJ/s\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-11, Page No 403"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=0.6 #m\n",
- "T_t=800 #N\n",
- "T_s=180 #N\n",
- "w=200 #rpm\n",
- "\n",
- "#Calculations\n",
- "r=d/2 #m radius\n",
- "#Torque\n",
- "M=(T_t-T_s)*r #N.m\n",
- "#Power\n",
- "w_new=(2*pi*w)/60 #rad/s\n",
- "Power=M*(w_new) #W\n",
- "\n",
- "#Result\n",
- "print'The power transmitted is',round(Power*10**-3,1),\"kW\"\n",
- "\n",
- "# The answer in the book is incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power transmitted is 3.9 kW\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-12, Page No 403"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "P=25.6 #lb\n",
- "w=600 #rpm\n",
- "a=36 #in\n",
- "b=12 #in\n",
- "\n",
- "#Calculations\n",
- "M=P*(((b*2**-1)+a)/12) #lb-ft\n",
- "w_new=(2*pi*w)/60 #rad/s\n",
- "Hp=(M*w_new)/550 #hp\n",
- "\n",
- "#Result\n",
- "print'The power being transmitted is',round(Hp,1),\"hp\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power being transmitted is 10.2 hp\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-13, Page No 403"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Pout=3.8 #bhp\n",
- "Pin=4.1 #ihp\n",
- "\n",
- "#Calculations\n",
- "Efficiency=round((Pout/Pin)*100) #Percent\n",
- "\n",
- "#Result\n",
- "print'The efficiency of the engine is',round(Efficiency),\"%\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The efficiency of the engine is 93.0 %\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-15, Page No 404"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return -(3/x)\n",
- "a=1\n",
- "b=1\n",
- "U=quad(integrand, 6, 3, args=(a,b))\n",
- "g=32.2 # ft/s**2\n",
- "deltaT=((U[0]*32.2*2)/(4*16**-1))**0.5\n",
- "# Results\n",
- "print'The speed of the disc will be',round(deltaT,1),\"ft/s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the disc will be 23.1 ft/s\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-16, Page No 404"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=2 #m\n",
- "m=4 #kg\n",
- "w_1=20 #rpm\n",
- "w_2=50 #rpm\n",
- "rev=10 #no of revolution\n",
- "\n",
- "#Calculations\n",
- "Io=(3**-1)*(m)*l**2 #kg.m**2\n",
- "w1=(2*pi*w_1)/60 #rad/s\n",
- "w2=(2*pi*w_2)/60 #rad/s\n",
- "theta=2*pi*rev #rad\n",
- "M=(0.5*Io*(w2**2-w1**2))/theta #N.m\n",
- "\n",
- "#Result\n",
- "print'The constant moment required is',round(M,3),\"N.m\"\n",
- "# The ans waries in decimal places."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The constant moment required is 0.977 N.m\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-18, Page No 405"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=1000 #lb\n",
- "w_w=200 #lb weight of the individual wheel\n",
- "d_w=2.5 #ft diameter of the wheel\n",
- "v=22 #ft/s\n",
- "t=2 #minutes\n",
- "\n",
- "#Calculations\n",
- "#T1=Initial Kinetic Energy and T2=Final Kinetic Energy\n",
- "F=(-0.5*W*32.2**-1*v**2-4*0.5*w_w*32.2**-1*(v**2+0.5*v**2))/(10560) #lb\n",
- "#Negative sign in the answer tells it oposses the motion\n",
- "\n",
- "#Result\n",
- "print'The rolling resistance is',round(F,2),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The rolling resistance is -1.57 lb\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-19, Page No 405"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=100 #lb\n",
- "lo=4 #ft\n",
- "# as theta=45 degrees\n",
- "costheta=(2**0.5)**-1\n",
- "g=32.2 #ft/s**2\n",
- "l=8/3 #ft\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point O and equating it to zero\n",
- "alpha=(W*(lo*0.5)*costheta)/((W/g)*(l)*2) #rad/s**2\n",
- "#Summing forces in the t direction\n",
- "Ot=(W*costheta)-((W/g)*lo*0.5*alpha) #lb\n",
- "#Work Done\n",
- "Work=W*(lo*0.5*costheta) #ft/lb\n",
- "#Moment of inertia\n",
- "Io=(3**-1)*(W/g)*(lo**2) #kg-ft**2\n",
- "#Using the concept for work done=chane in K.E\n",
- "w=(Work/(0.5*Io))**0.5 #rad/s\n",
- "#Summing forces along the bar\n",
- "On=-(-((W/g)*lo*0.5*w**2)-(W*costheta)) #lb\n",
- "\n",
- "#Result\n",
- "print'The bearing reaction at O on the rod is',round(On),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The bearing reaction at O on the rod is 177.0 lb\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-21, Page No 406"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "vo=9 #m/s\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "x=((7*10**-1)*vo**2)*(g*sintheta)**-1 #m\n",
- "\n",
- "#Result\n",
- "print'The ball will roll',round(x,1),\"m up the plane\"\n",
- "\n",
- "#The textbook wrongly mentions the unit of displacement as in\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ball will roll 11.6 m up the plane\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-22, Page No 406"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=322 #lb\n",
- "F=12 #lb\n",
- "a=0 #lower limit (where the cyliner starts rolling)\n",
- "b=pi/2 #Upper Limit (where the cyliner stops rolling)\n",
- "d=3.2 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "dR=1.6 #Differential Radius\n",
- "d_U=2*dR*F #differential work done\n",
- "#Integration Calculations\n",
- "#As it is a simple integration we can resort to this\n",
- "U=d_U*(b-a) #ft-lb\n",
- "#Determination of K.E\n",
- "w=sqrt(U/((0.5*(W/g)*(1/(d/2)**2))+((0.5*0.5)*(W/g)*(d/2)**2))) #rad/s\n",
- "\n",
- "#Result \n",
- "print'The angular velocity of the cylinder is',round(w,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity of the cylinder is 2.69 rad/s\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-24, Page No 407"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wa=161 #lb\n",
- "Wb=193.2 #lb\n",
- "Wc=322 #lb\n",
- "v1=5 #ft/s\n",
- "lc=6 #in\n",
- "k=6 #lb/ft\n",
- "l=4 #ft\n",
- "u=0.2 #coefficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "Ib=(2**-1)*(Wb/g)*(2**-1)**2 #Moment of inertia\n",
- "w1=v1*0.5**-1 #rad/s\n",
- "T1=(0.5*(Wc/g)*v1**2)+(0.5*Ib*w1**2)+(0.5*(Wa/g)*v1**2) #ft-lb\n",
- "#Work Done on the system\n",
- "#The textbook is ambigious on the calculations hence the result is dispalyed directly\n",
- "U=26.4 #ft-lb\n",
- "#Velocity Calculations\n",
- "v=((T1+U)*9**-1)**0.5 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The velocity of the block is',round(v,2),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity of the block is 5.29 ft/s\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-25, Page No 408"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Mm=70 #kg\n",
- "Mc=45 #kg\n",
- "R=0.6 #m\n",
- "g=9.8 #m/s**2\n",
- "l=5 #m\n",
- "# as theta=50 degrees,\n",
- "sintheta=0.77\n",
- "\n",
- "#Calculations\n",
- "#T2 calculations except for v term in it as it cannot be declared as a number\n",
- "T2=68.7 #without the v term in it\n",
- "v=((g*Mm*l-g*Mc*l*sintheta)/T2)**0.5 #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed is',round(v,2),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed is 5.02 m/s\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-26, Page No 408"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#The textbook has a typo in printing the question number\n",
- "#Initilization of variables\n",
- "W1=96.6 #lb\n",
- "W2=128.8 #lb\n",
- "v=8 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "\n",
- "#Calculations\n",
- "#Initial KE of the system is T1=0\n",
- "T2=(0.5*(W1/g)*v**2)+(0.5*(W2/g)*(v/2)**2) #ft-lb\n",
- "#Work Done without s term\n",
- "U=-(W1*sintheta)+W2*0.5\n",
- "#S calculations\n",
- "s=T2*U**-1 #ft\n",
- "\n",
- "#Result\n",
- "print'The block attains a speed of 8 ft/s in',round(s,2),\"ft (up the plane)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The block attains a speed of 8 ft/s in 7.95 ft (up the plane)\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-28, Page No 409"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "l=6 #m length of the cable\n",
- "m=50 #kg mass of the cable\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return 81.7*(6-x)\n",
- "a=1\n",
- "b=1\n",
- "Work=quad(integrand, 0, 6, args=(a,b))\n",
- "\n",
- "# Results\n",
- "print'The word done is',round(Work[0]),\"N.m\"\n",
- "# The answer in textbook is off by 1 N.m"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The word done is 1471.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-31, Page No 410"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "k=0.044 # spring constant\n",
- "#x=0.300 #m length of compression from 450 to 150\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return 0.5*0.044*x**2\n",
- "a=1\n",
- "b=1\n",
- "W=quad(integrand, 0, 300, args=(a,b))\n",
- "\n",
- "# Results\n",
- "print'The word done on the balls is',round(W[0]*10**-5,2),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The word done on the balls is 1.98 N.m\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-32, Page No 410"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=10 #kg\n",
- "d=1.2 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Initilial KE is zero\n",
- "#Final KE is(without v^2 term in it)\n",
- "KE2=(3*4**-1)*10\n",
- "#Work Done\n",
- "U=m*g*d #N.m\n",
- "#Velocity calculations\n",
- "v=sqrt(U*KE2**-1) #m/s\n",
- "\n",
- "#Result\n",
- "print'The velocity is',round(v,2),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity is 3.96 m/s\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-33, Page No 411"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=161 #lb\n",
- "wa=150 #lb\n",
- "wb=100 #lb\n",
- "la=2 #ft\n",
- "lb=4 #ft\n",
- "\n",
- "#Calculations\n",
- "#Work Done\n",
- "T1=wb*lb-wa*la #ft-lb\n",
- "#Final KE=zero\n",
- "T2=0 #ft-lb\n",
- "#Work Done on the system=T2-T1\n",
- "#Hence the equation becomes\n",
- "#50x-50x^2+100=0\n",
- "#where\n",
- "a=-50\n",
- "b=50\n",
- "c=100\n",
- "#Solution\n",
- "d=sqrt(b**2-4*a*c) \n",
- "x1=(-b+d)/(2*a) #ft\n",
- "x2=(-b-d)/(2*a) #ft\n",
- "\n",
- "#Result\n",
- "print'The stretch of the spring is',round(x2),\"ft\"\n",
- "#Here even x1 could have been the solution,but the stretch in the string is elongation not compression hence x2 is the valid answer\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The stretch of the spring is 2.0 ft\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-34, Page No 411"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "I=100 #slug-ft**2\n",
- "w=4 #rad/s\n",
- "theta=6 #rad\n",
- "Mc=64.4 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "vb=2*w #ft/s\n",
- "vc=0.5*w #ft/s\n",
- "Mb=(0.5*I*w**2+0.5*(Mc/g)*vc**2+0.5*Mc*theta)/(2*theta-(0.5*vb**2*(1/g))) #lb\n",
- "\n",
- "#Result\n",
- "print'The weight of the block B is',round(Mb,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The weight of the block B is 90.6 lb\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-35, Page No 411"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wa=96.6 #lb\n",
- "Wb=128.8 #lb\n",
- "g=32.2 #ft/s**2\n",
- "I=12 #slug-ft**2\n",
- "v=16 #ft/s\n",
- "ratio=3**-1 #ratio of Sb/Sa\n",
- "r=3#ft\n",
- "va=6 #ft/s\n",
- "vb=2 #ft/s\n",
- "\n",
- "#Calculations\n",
- "#Work Done without S in it\n",
- "W=Wa-(ratio*Wb)\n",
- "#System has zero KE initially and final KE is given by\n",
- "w=va/r #rad/s\n",
- "T2=(0.5*(Wa/g)*va**2+0.5*I*w**2+0.5*(Wb/g)*vb**2) #ft-lb\n",
- "#Distance Calculations\n",
- "S=T2*W**-1 #ft\n",
- "\n",
- "#Result\n",
- "print'The distance through which A falls is',round(S,2),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The distance through which A falls is 1.6 ft\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-36, Page No 412"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#initilization of variables\n",
- "u=0.25 #coefficient of friction\n",
- "k=2800 #N/m\n",
- "x=0.075 #m\n",
- "g=9.8 #m/s**2\n",
- "m=7 #kg\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Normal Reaction\n",
- "N=g*m*costheta #N\n",
- "#Frictional Force\n",
- "Fr=u*N #N\n",
- "#Component of force along the plane\n",
- "F=g*m*sintheta #N\n",
- "#Spring work is\n",
- "W=0.5*k*x*x #N.m\n",
- "s=(W+Fr*x-F*x)/(F-Fr) #m\n",
- "S=(s*1000) #mm\n",
- "\n",
- "#Result\n",
- "print'The value of S is',round(S),\"mm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of S is 330.0 mm\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-37, Page No 412"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=5 #kg\n",
- "l=2 #m\n",
- "k=10000 #N/m\n",
- "x=0.1 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "drop=l+x #m mass drop length\n",
- "#Work Done by Gravity\n",
- "Wg=g*m*drop #N.m\n",
- "#Work Done by Spring\n",
- "Ws=0.5*k*x**2 #N.m\n",
- "#Increase in KE is without v**2\n",
- "KE=0.5*m #kg\n",
- "#Velocity Calculations\n",
- "v=sqrt((Wg-Ws)/KE) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed is',round(v,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed is 4.6 m/s\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-38, Page No 412"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=6 #ft\n",
- "k=20 #lb/in\n",
- "x=8 #in\n",
- "\n",
- "#Calculations\n",
- "#Work Done by Gravity\n",
- "Wg=(l*12+x) #in without W\n",
- "#Work Done by Spring\n",
- "Ws=0.5*k*x**2 #in-lb\n",
- "#Change in the kinetic energy is zero\n",
- "W=Ws/Wg #lb\n",
- "\n",
- "#Result\n",
- "print'The weight is',round(W),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The weight is 8.0 lb\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-40, Page No 413"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=8 #lb\n",
- "\n",
- "#Calculations\n",
- "#work done by the spring woithout k\n",
- "Ws=0.5*((9*12**-1)**2-(12**-1)**2) \n",
- "#Work done by gravity\n",
- "Wg=W*(10.5*12**-1) #ft-lb\n",
- "#Change in KE is zero\n",
- "k=Wg/Ws #lb/ft\n",
- "\n",
- "#Result\n",
- "print'The value of k is',round(k,1),\"lb/ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of k is 25.2 lb/ft\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-41, Page No 414"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wc=100 #lb\n",
- "r= 1 #ft\n",
- "F=80 #lb\n",
- "k=50 #lb/ft\n",
- "s=6 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Work done on the system\n",
- "U=-0.5*k*(1)+F*(s*12**-1) #ft-lb\n",
- "#Initial KE is zero\n",
- "Vo=(U/(0.5*(Wc/g+0.5*(Wc/g)*r)))**0.5 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The initial speed is',round(Vo,2),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The initial speed is 2.54 ft/s\n"
- ]
- }
- ],
- "prompt_number": 27
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter17_5.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter17_5.ipynb deleted file mode 100755 index 7a620362..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter17_5.ipynb +++ /dev/null @@ -1,1273 +0,0 @@ -{
- "metadata": {
- "name": "chapter17.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 17: Work and Energy"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-4, Page No 400"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "s1=2 # compression of the spring- initial\n",
- "s2=5 # compression of the spring- final\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(s, a, b):\n",
- " return 20*s\n",
- "a=1\n",
- "b=1\n",
- "U=quad(integrand, s1, s2, args=(a,b))\n",
- "\n",
- "# Results\n",
- "print'The word done in compressing the spring is',round(U[0]),\"in-lb\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The word done in compressing the spring is 210.0 in-lb\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-6, Page No 400"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=5 #kg\n",
- "d=6 #m\n",
- "# as theta1=30 degrees & theta2=10 degrees,\n",
- "sintheta1=2**-1\n",
- "sintheta2=0.1736\n",
- "costheta1=(3**0.5)*2**-1\n",
- "costheta2=0.9848\n",
- "u=0.2 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "F=70 #N\n",
- "\n",
- "#Calculations\n",
- "#Using free body diagram\n",
- "Na=(m*g*costheta1)-(F*sintheta2) #N\n",
- "#work done by each force\n",
- "W=[F*costheta2,-m*g*sintheta1,0,-u*Na*d] #N.m\n",
- "#Total Work Done\n",
- "W_tot=W[0]+W[1]+W[2]+W[3] #N.m\n",
- "#Using resultant\n",
- "R=F*costheta2-(u*Na)-(m*g*sintheta1) #N\n",
- "W_d=R*d #N.m (Work Done)\n",
- "\n",
- "#Result\n",
- "print'The work done is',round(W_d),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The work done is 230.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-7, Page No 401"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=20 #kg\n",
- "d=1.5 #m\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "u=0.25 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "F=130 #N\n",
- "\n",
- "#Calculations\n",
- "W=F*d-(m*g*sintheta*d) #N.m\n",
- "\n",
- "#Result\n",
- "print'The work done is',round(W),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The work done is 48.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-9, Page No 402"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=6*12**-1 #ft\n",
- "l=8*12**-1 #ft\n",
- "l_c=3.2 #in\n",
- "y=1.82 #in**2\n",
- "\n",
- "#Calculations\n",
- "V=1*4**-1*pi*d**2*l #ft**3\n",
- "#One horizontal inch \n",
- "h_i=V/l_c #ft**3\n",
- "#One vertical inch\n",
- "v_i=100*144 #lb/ft**2\n",
- "#Then 1.82 in**2 represents\n",
- "x=y*v_i*h_i #ft-lb\n",
- "\n",
- "#Result\n",
- "print'The work capacity is',round(x),\"ft-lb\"\n",
- "\n",
- "# The ans in the textbook is incorrect."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The work capacity is 1072.0 ft-lb\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-10, Page No 402"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "speed=90000 #m/h\n",
- "P=100*1000 #N\n",
- "\n",
- "#Calculations\n",
- "Power=P*((speed)/3600) #J/s\n",
- "\n",
- "#Result\n",
- "print'The power developed is',round(Power*10**-6,1),\"MJ/s\"\n",
- "# Note the unit used."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power developed is 2.5 MJ/s\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-11, Page No 403"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=0.6 #m\n",
- "T_t=800 #N\n",
- "T_s=180 #N\n",
- "w=200 #rpm\n",
- "\n",
- "#Calculations\n",
- "r=d/2 #m radius\n",
- "#Torque\n",
- "M=(T_t-T_s)*r #N.m\n",
- "#Power\n",
- "w_new=(2*pi*w)/60 #rad/s\n",
- "Power=M*(w_new) #W\n",
- "\n",
- "#Result\n",
- "print'The power transmitted is',round(Power*10**-3,1),\"kW\"\n",
- "\n",
- "# The answer in the book is incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power transmitted is 3.9 kW\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-12, Page No 403"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "P=25.6 #lb\n",
- "w=600 #rpm\n",
- "a=36 #in\n",
- "b=12 #in\n",
- "\n",
- "#Calculations\n",
- "M=P*(((b*2**-1)+a)/12) #lb-ft\n",
- "w_new=(2*pi*w)/60 #rad/s\n",
- "Hp=(M*w_new)/550 #hp\n",
- "\n",
- "#Result\n",
- "print'The power being transmitted is',round(Hp,1),\"hp\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power being transmitted is 10.2 hp\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-13, Page No 403"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Pout=3.8 #bhp\n",
- "Pin=4.1 #ihp\n",
- "\n",
- "#Calculations\n",
- "Efficiency=round((Pout/Pin)*100) #Percent\n",
- "\n",
- "#Result\n",
- "print'The efficiency of the engine is',round(Efficiency),\"%\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The efficiency of the engine is 93.0 %\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-15, Page No 404"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return -(3/x)\n",
- "a=1\n",
- "b=1\n",
- "U=quad(integrand, 6, 3, args=(a,b))\n",
- "g=32.2 # ft/s**2\n",
- "deltaT=((U[0]*32.2*2)/(4*16**-1))**0.5\n",
- "# Results\n",
- "print'The speed of the disc will be',round(deltaT,1),\"ft/s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the disc will be 23.1 ft/s\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-16, Page No 404"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=2 #m\n",
- "m=4 #kg\n",
- "w_1=20 #rpm\n",
- "w_2=50 #rpm\n",
- "rev=10 #no of revolution\n",
- "\n",
- "#Calculations\n",
- "Io=(3**-1)*(m)*l**2 #kg.m**2\n",
- "w1=(2*pi*w_1)/60 #rad/s\n",
- "w2=(2*pi*w_2)/60 #rad/s\n",
- "theta=2*pi*rev #rad\n",
- "M=(0.5*Io*(w2**2-w1**2))/theta #N.m\n",
- "\n",
- "#Result\n",
- "print'The constant moment required is',round(M,3),\"N.m\"\n",
- "# The ans waries in decimal places."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The constant moment required is 0.977 N.m\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-18, Page No 405"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=1000 #lb\n",
- "w_w=200 #lb weight of the individual wheel\n",
- "d_w=2.5 #ft diameter of the wheel\n",
- "v=22 #ft/s\n",
- "t=2 #minutes\n",
- "\n",
- "#Calculations\n",
- "#T1=Initial Kinetic Energy and T2=Final Kinetic Energy\n",
- "F=(-0.5*W*32.2**-1*v**2-4*0.5*w_w*32.2**-1*(v**2+0.5*v**2))/(10560) #lb\n",
- "#Negative sign in the answer tells it oposses the motion\n",
- "\n",
- "#Result\n",
- "print'The rolling resistance is',round(F,2),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The rolling resistance is -1.57 lb\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-19, Page No 405"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=100 #lb\n",
- "lo=4 #ft\n",
- "# as theta=45 degrees\n",
- "costheta=(2**0.5)**-1\n",
- "g=32.2 #ft/s**2\n",
- "l=8/3 #ft\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point O and equating it to zero\n",
- "alpha=(W*(lo*0.5)*costheta)/((W/g)*(l)*2) #rad/s**2\n",
- "#Summing forces in the t direction\n",
- "Ot=(W*costheta)-((W/g)*lo*0.5*alpha) #lb\n",
- "#Work Done\n",
- "Work=W*(lo*0.5*costheta) #ft/lb\n",
- "#Moment of inertia\n",
- "Io=(3**-1)*(W/g)*(lo**2) #kg-ft**2\n",
- "#Using the concept for work done=chane in K.E\n",
- "w=(Work/(0.5*Io))**0.5 #rad/s\n",
- "#Summing forces along the bar\n",
- "On=-(-((W/g)*lo*0.5*w**2)-(W*costheta)) #lb\n",
- "\n",
- "#Result\n",
- "print'The bearing reaction at O on the rod is',round(On),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The bearing reaction at O on the rod is 177.0 lb\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-21, Page No 406"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "vo=9 #m/s\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "x=((7*10**-1)*vo**2)*(g*sintheta)**-1 #m\n",
- "\n",
- "#Result\n",
- "print'The ball will roll',round(x,1),\"m up the plane\"\n",
- "\n",
- "#The textbook wrongly mentions the unit of displacement as in\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ball will roll 11.6 m up the plane\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-22, Page No 406"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=322 #lb\n",
- "F=12 #lb\n",
- "a=0 #lower limit (where the cyliner starts rolling)\n",
- "b=pi/2 #Upper Limit (where the cyliner stops rolling)\n",
- "d=3.2 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "dR=1.6 #Differential Radius\n",
- "d_U=2*dR*F #differential work done\n",
- "#Integration Calculations\n",
- "#As it is a simple integration we can resort to this\n",
- "U=d_U*(b-a) #ft-lb\n",
- "#Determination of K.E\n",
- "w=sqrt(U/((0.5*(W/g)*(1/(d/2)**2))+((0.5*0.5)*(W/g)*(d/2)**2))) #rad/s\n",
- "\n",
- "#Result \n",
- "print'The angular velocity of the cylinder is',round(w,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity of the cylinder is 2.69 rad/s\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-24, Page No 407"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wa=161 #lb\n",
- "Wb=193.2 #lb\n",
- "Wc=322 #lb\n",
- "v1=5 #ft/s\n",
- "lc=6 #in\n",
- "k=6 #lb/ft\n",
- "l=4 #ft\n",
- "u=0.2 #coefficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "Ib=(2**-1)*(Wb/g)*(2**-1)**2 #Moment of inertia\n",
- "w1=v1*0.5**-1 #rad/s\n",
- "T1=(0.5*(Wc/g)*v1**2)+(0.5*Ib*w1**2)+(0.5*(Wa/g)*v1**2) #ft-lb\n",
- "#Work Done on the system\n",
- "#The textbook is ambigious on the calculations hence the result is dispalyed directly\n",
- "U=26.4 #ft-lb\n",
- "#Velocity Calculations\n",
- "v=((T1+U)*9**-1)**0.5 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The velocity of the block is',round(v,2),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity of the block is 5.29 ft/s\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-25, Page No 408"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Mm=70 #kg\n",
- "Mc=45 #kg\n",
- "R=0.6 #m\n",
- "g=9.8 #m/s**2\n",
- "l=5 #m\n",
- "# as theta=50 degrees,\n",
- "sintheta=0.77\n",
- "\n",
- "#Calculations\n",
- "#T2 calculations except for v term in it as it cannot be declared as a number\n",
- "T2=68.7 #without the v term in it\n",
- "v=((g*Mm*l-g*Mc*l*sintheta)/T2)**0.5 #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed is',round(v,2),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed is 5.02 m/s\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-26, Page No 408"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#The textbook has a typo in printing the question number\n",
- "#Initilization of variables\n",
- "W1=96.6 #lb\n",
- "W2=128.8 #lb\n",
- "v=8 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "\n",
- "#Calculations\n",
- "#Initial KE of the system is T1=0\n",
- "T2=(0.5*(W1/g)*v**2)+(0.5*(W2/g)*(v/2)**2) #ft-lb\n",
- "#Work Done without s term\n",
- "U=-(W1*sintheta)+W2*0.5\n",
- "#S calculations\n",
- "s=T2*U**-1 #ft\n",
- "\n",
- "#Result\n",
- "print'The block attains a speed of 8 ft/s in',round(s,2),\"ft (up the plane)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The block attains a speed of 8 ft/s in 7.95 ft (up the plane)\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-28, Page No 409"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "l=6 #m length of the cable\n",
- "m=50 #kg mass of the cable\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return 81.7*(6-x)\n",
- "a=1\n",
- "b=1\n",
- "Work=quad(integrand, 0, 6, args=(a,b))\n",
- "\n",
- "# Results\n",
- "print'The word done is',round(Work[0]),\"N.m\"\n",
- "# The answer in textbook is off by 1 N.m"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The word done is 1471.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-31, Page No 410"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "k=0.044 # spring constant\n",
- "#x=0.300 #m length of compression from 450 to 150\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return 0.5*0.044*x**2\n",
- "a=1\n",
- "b=1\n",
- "W=quad(integrand, 0, 300, args=(a,b))\n",
- "\n",
- "# Results\n",
- "print'The word done on the balls is',round(W[0]*10**-5,2),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The word done on the balls is 1.98 N.m\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-32, Page No 410"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=10 #kg\n",
- "d=1.2 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Initilial KE is zero\n",
- "#Final KE is(without v^2 term in it)\n",
- "KE2=(3*4**-1)*10\n",
- "#Work Done\n",
- "U=m*g*d #N.m\n",
- "#Velocity calculations\n",
- "v=sqrt(U*KE2**-1) #m/s\n",
- "\n",
- "#Result\n",
- "print'The velocity is',round(v,2),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocity is 3.96 m/s\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-33, Page No 411"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=161 #lb\n",
- "wa=150 #lb\n",
- "wb=100 #lb\n",
- "la=2 #ft\n",
- "lb=4 #ft\n",
- "\n",
- "#Calculations\n",
- "#Work Done\n",
- "T1=wb*lb-wa*la #ft-lb\n",
- "#Final KE=zero\n",
- "T2=0 #ft-lb\n",
- "#Work Done on the system=T2-T1\n",
- "#Hence the equation becomes\n",
- "#50x-50x^2+100=0\n",
- "#where\n",
- "a=-50\n",
- "b=50\n",
- "c=100\n",
- "#Solution\n",
- "d=sqrt(b**2-4*a*c) \n",
- "x1=(-b+d)/(2*a) #ft\n",
- "x2=(-b-d)/(2*a) #ft\n",
- "\n",
- "#Result\n",
- "print'The stretch of the spring is',round(x2),\"ft\"\n",
- "#Here even x1 could have been the solution,but the stretch in the string is elongation not compression hence x2 is the valid answer\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The stretch of the spring is 2.0 ft\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-34, Page No 411"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "I=100 #slug-ft**2\n",
- "w=4 #rad/s\n",
- "theta=6 #rad\n",
- "Mc=64.4 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "vb=2*w #ft/s\n",
- "vc=0.5*w #ft/s\n",
- "Mb=(0.5*I*w**2+0.5*(Mc/g)*vc**2+0.5*Mc*theta)/(2*theta-(0.5*vb**2*(1/g))) #lb\n",
- "\n",
- "#Result\n",
- "print'The weight of the block B is',round(Mb,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The weight of the block B is 90.6 lb\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-35, Page No 411"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wa=96.6 #lb\n",
- "Wb=128.8 #lb\n",
- "g=32.2 #ft/s**2\n",
- "I=12 #slug-ft**2\n",
- "v=16 #ft/s\n",
- "ratio=3**-1 #ratio of Sb/Sa\n",
- "r=3#ft\n",
- "va=6 #ft/s\n",
- "vb=2 #ft/s\n",
- "\n",
- "#Calculations\n",
- "#Work Done without S in it\n",
- "W=Wa-(ratio*Wb)\n",
- "#System has zero KE initially and final KE is given by\n",
- "w=va/r #rad/s\n",
- "T2=(0.5*(Wa/g)*va**2+0.5*I*w**2+0.5*(Wb/g)*vb**2) #ft-lb\n",
- "#Distance Calculations\n",
- "S=T2*W**-1 #ft\n",
- "\n",
- "#Result\n",
- "print'The distance through which A falls is',round(S,2),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The distance through which A falls is 1.6 ft\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-36, Page No 412"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#initilization of variables\n",
- "u=0.25 #coefficient of friction\n",
- "k=2800 #N/m\n",
- "x=0.075 #m\n",
- "g=9.8 #m/s**2\n",
- "m=7 #kg\n",
- "# as theta=30 degrees,\n",
- "sintheta=2**-1\n",
- "costheta=(3**0.5)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Normal Reaction\n",
- "N=g*m*costheta #N\n",
- "#Frictional Force\n",
- "Fr=u*N #N\n",
- "#Component of force along the plane\n",
- "F=g*m*sintheta #N\n",
- "#Spring work is\n",
- "W=0.5*k*x*x #N.m\n",
- "s=(W+Fr*x-F*x)/(F-Fr) #m\n",
- "S=(s*1000) #mm\n",
- "\n",
- "#Result\n",
- "print'The value of S is',round(S),\"mm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of S is 330.0 mm\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-37, Page No 412"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=5 #kg\n",
- "l=2 #m\n",
- "k=10000 #N/m\n",
- "x=0.1 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "drop=l+x #m mass drop length\n",
- "#Work Done by Gravity\n",
- "Wg=g*m*drop #N.m\n",
- "#Work Done by Spring\n",
- "Ws=0.5*k*x**2 #N.m\n",
- "#Increase in KE is without v**2\n",
- "KE=0.5*m #kg\n",
- "#Velocity Calculations\n",
- "v=sqrt((Wg-Ws)/KE) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed is',round(v,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed is 4.6 m/s\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-38, Page No 412"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=6 #ft\n",
- "k=20 #lb/in\n",
- "x=8 #in\n",
- "\n",
- "#Calculations\n",
- "#Work Done by Gravity\n",
- "Wg=(l*12+x) #in without W\n",
- "#Work Done by Spring\n",
- "Ws=0.5*k*x**2 #in-lb\n",
- "#Change in the kinetic energy is zero\n",
- "W=Ws/Wg #lb\n",
- "\n",
- "#Result\n",
- "print'The weight is',round(W),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The weight is 8.0 lb\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-40, Page No 413"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=8 #lb\n",
- "\n",
- "#Calculations\n",
- "#work done by the spring woithout k\n",
- "Ws=0.5*((9*12**-1)**2-(12**-1)**2) \n",
- "#Work done by gravity\n",
- "Wg=W*(10.5*12**-1) #ft-lb\n",
- "#Change in KE is zero\n",
- "k=Wg/Ws #lb/ft\n",
- "\n",
- "#Result\n",
- "print'The value of k is',round(k,1),\"lb/ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of k is 25.2 lb/ft\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.17-41, Page No 414"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wc=100 #lb\n",
- "r= 1 #ft\n",
- "F=80 #lb\n",
- "k=50 #lb/ft\n",
- "s=6 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Work done on the system\n",
- "U=-0.5*k*(1)+F*(s*12**-1) #ft-lb\n",
- "#Initial KE is zero\n",
- "Vo=(U/(0.5*(Wc/g+0.5*(Wc/g)*r)))**0.5 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The initial speed is',round(Vo,2),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The initial speed is 2.54 ft/s\n"
- ]
- }
- ],
- "prompt_number": 27
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter18.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter18.ipynb deleted file mode 100755 index bbd27129..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter18.ipynb +++ /dev/null @@ -1,1503 +0,0 @@ -{
- "metadata": {
- "name": "chapter18.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 18: Impulse and Momentum"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-8, Page No 435"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=100 #lb\n",
- "u=0.2 #coefficient of friction\n",
- "t=5 #s\n",
- "v1=5 #ft/s\n",
- "v2=10 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "ll=0 #lower limit of integration\n",
- "ul=5 #upper limit of integration\n",
- "\n",
- "#Calculations\n",
- "Fr=u*W #lb\n",
- "#Using The impulse momentum theorem\n",
- "#Since the integration is just subtraction of limits we can skip that\n",
- "F=((W/g)*v2-(W/g)*v1+Fr*ul)/ul #lb\n",
- "\n",
- "#Result\n",
- "print'The Force is',round(F,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Force is 23.1 lb\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-9, Page No 435"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=10 #kg\n",
- "#theta=30,thus\n",
- "cos30=sqrt(3)*2**-1\n",
- "sin30=2**-1\n",
- "u=0.3 #coefficient of kinetic friction\n",
- "t=5 #s\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Asthere is no motion in the vertical direction \n",
- "#Summing forces along vertical direction\n",
- "Na=m*g*cos30 #N\n",
- "#Using impulse momentum theorem\n",
- "vx=(m*g*sin30-u*Na)*(t*m**-1) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed after 5s is',round(vx,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed after 5s is 11.8 m/s\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-10, Page No 436"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(t, a, b):\n",
- " return 20*t-16\n",
- "a=1\n",
- "b=1\n",
- "F=quad(integrand, 1, 5, args=(a,b))\n",
- "g=32.2 # ft/s**2\n",
- "v=(F[0]*32.2)/80\n",
- "\n",
- "# Results\n",
- "print'The speed of the block is',round(v,1),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the block is 70.8 ft/s\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-12, Page No 436"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=40 #kg\n",
- "m2=10 #kg\n",
- "m3=15 #kg\n",
- "v0=2.5 #m/s\n",
- "vf=5 #m/s\n",
- "t=12 #s\n",
- "u=0.1 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "#theta=45 degrees,thus\n",
- "sin45=(sqrt(2))**-1\n",
- "cos45=(sqrt(2))**-1\n",
- "\n",
- "#Calculations\n",
- "#Applying Impulse-Momentum Theoroem\n",
- "P=(((m1+m2+m3)*(vf-v0))+(t*(-m2*g*sin45+u*g*m2*cos45+u*g*m3+g*m1)))*t**-1 #N\n",
- "\n",
- "#Result\n",
- "print'The value of P is',round(P),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P is 358.0 N\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-13, Page Noo 437"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W1=4 #lb\n",
- "W2=2 #lb\n",
- "t2=0.04 #s\n",
- "W3=-2 #lb\n",
- "t3=0.02 #s\n",
- "t=3 #s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Algebraic sum of two areas\n",
- "A=t2*W2+t3*W3 #lb-s\n",
- "#Using Impulse Momentum Theorem\n",
- "v=(A*g)/W1 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The spped after 3s is',round(v,3),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The spped after 3s is 0.322 ft/s\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-14, Page No 437"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "f_r=1 #in/s rate of fall of mercury\n",
- "ll=18 #in length of left column\n",
- "lr=22 #in length of right column\n",
- "rho=850 #lb/ft**3\n",
- "d=4**-1 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying Impulse momentum theorem\n",
- "M=((d*pi*d**2*4)/12**3)*(rho/g)*(12**-1) #lb-s\n",
- "\n",
- "#Result\n",
- "print'The upward momentum of mercury is +',round(M,5),\"lb-s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The upward momentum of mercury is + 0.00025 lb-s\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-15, Page No 437"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=2000 #kg\n",
- "k=1.200 #m\n",
- "w=120 #rpm\n",
- "t=200 #s\n",
- "\n",
- "#Calculations\n",
- "#Applying Angular Momentum theorem\n",
- "M=((m*k**2*(w*2*pi))/60)/t #N.m\n",
- "\n",
- "#Result\n",
- "print'The Momentum necessary is',round(M),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Momentum necessary is 181.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-17, Page No 438"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=5 #kg\n",
- "m2=7 #kg\n",
- "mp=5 #kg\n",
- "r=0.6 #m\n",
- "k=0.45 #m\n",
- "vi=3 #m/s\n",
- "vf=6 #m/s\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "I=m1*k**2 #kg.m**2\n",
- "wnet=(vf/r)-(vi/r) #rad/s\n",
- "#Solving the system of linear equations\n",
- "#Simplfying the equation we get\n",
- "t=((I*wnet)+m1*(vf-vi)+m2*(vf-vi))*r/(r*(m2-m1)*g) #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t,2),\"s\"\n",
- "# The ans in the textbook is incorrect.\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 2.1 s\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-18, Page No 438"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=50 #kg\n",
- "vo=4 #m/s\n",
- "vf=8 #m/s\n",
- "t=6 #s\n",
- "g=9.8 #m/s**2\n",
- "r=0.8 #m\n",
- "u=0.25 #coefficient of friction\n",
- "I=30 #kg-m**2\n",
- "\n",
- "#Calculations\n",
- "Na=m*g #N\n",
- "F=u*Na #N\n",
- "#Angular Speeds\n",
- "wo=vo/r #rad/s\n",
- "wf=vf/r #rad/s\n",
- "#Applying impulse momentum theorem\n",
- "mb=(I*wo+m*vo*r-F*r*t-I*wf-m*vf*r)/(vf*r-vo*r-g*r*t) #kg\n",
- "\n",
- "#Result\n",
- "print'The mass of block B is',round(mb,1),\"kg\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The mass of block B is 20.5 kg\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-19, Page No 439"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Ws=250 #lb\n",
- "Wl=500 #lb\n",
- "W3=161 #lb\n",
- "W4=64.4 #lb\n",
- "wo=100 #rpm\n",
- "wf=300 #rpm\n",
- "rl=3 #ft\n",
- "rs=2 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment Of Inertia\n",
- "I=(0.5*(Wl/g)*rl**2+0.5*(Ws/g)*rs**2) #slug-ft**2\n",
- "#Change in angular Momentum\n",
- "change1=I*((wf-wo)*2*(pi/60)) #lb-s-ft\n",
- "#Change in angular momentum about G for 161lb\n",
- "change2=2*((W3*g**-1)*(wf-wo)*(4*60**-1)*pi) #lb-s-ft\n",
- "#Similarly change in 64lb is\n",
- "change3=3*((W4*g**-1)*(wf-wo)*(6*60**-1)*pi) #lb-s-ft\n",
- "#Change in linear impulse\n",
- "#Without t term in it\n",
- "m1=2*W3\n",
- "m2=-3*W4\n",
- "#Total angular impulse\n",
- "t=(change1+change2+change3)/(m1+m2) #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 20.0 s\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-21, Page No 440"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "d=3 #ft\n",
- "W=300 #lb\n",
- "# as theta=20 degrees\n",
- "sintheta=0.342\n",
- "F=250 #lb\n",
- "t=6 #s\n",
- "vo=0 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying linear impulse momentum theorem\n",
- "#Solving by matrix method\n",
- "A=np.array([[-W*g**-1,1*t],[-((0.5)*(W*g**-1)*(d*2**-1)**2*(d*2**-1)**-1),(-t)*d*2**-1]])\n",
- "B=np.array([[-F*t+W*sintheta*t],[-F*(d*2**-1)*6]]) \n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The speed after 6s is',round(C[0]),\"ft/s,parallel to the plane\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed after 6s is 171.0 ft/s,parallel to the plane\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-22, Page No 440"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Initilization of variables\n",
- "#theta=30 degrees\n",
- "sin30=2**-1\n",
- "vo=20 #ft/s\n",
- "r=4 #ft\n",
- "vf=0 #ft/s\n",
- "W=300 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "wo=vo*r**-1 #rad/s\n",
- "wf=vf*r**-1 #rad/s\n",
- "#Applying impulse momentum theorem\n",
- "#Solving simultaneous equations\n",
- "t=-((W*g**-1)*(vf-vo)+((0.5*(W*g**-1)*r**2*(wf-wo))*r**-1))/(W*sin30) #s\n",
- "\n",
- "#Result\n",
- "print'The time t is',round(t,2),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time t is 1.86 s\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-23, Page No 441"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mw=75 #kg\n",
- "k=0.9 #m\n",
- "wi=10 #rad/s\n",
- "wf=6 #rad/s\n",
- "r=1.2 #m\n",
- "m=30 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Initial speed\n",
- "vi=-r*wi #m/s\n",
- "vf=r*wf #m/s\n",
- "#Initial speed of B is\n",
- "vib=-0.8*wi+vi #m/s\n",
- "#Similarly\n",
- "vfb=12 #m/s\n",
- "#Applying impulse momentum theorem\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,-1,-(mw*(vf-vi))],[0.8,1.2,-(mw*(k**2)*(wf+wi))],[-1,0,-(m*(vfb-vi))]])\n",
- "B=np.array([[0],[0],[-g*m]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Here t is calculated as 1/t for simplicity\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(C[2]**-1,2),\"s\" \n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 7.04 s\n"
- ]
- }
- ],
- "prompt_number": 59
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-24, Page No 441"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=8 #in\n",
- "W=96.6 #lb\n",
- "w=36 #rad/s\n",
- "u=0.15 #coefficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "r=(d/2)*12**-1 #m\n",
- "N=W #lb\n",
- "F=u*N #lb\n",
- "m=W/g #slugs\n",
- "I=0.5*m*(r**2) #slug-ft**2\n",
- "#Applying the impulse momentum theorem\n",
- "#Solving the two equations simultaneously\n",
- "A=np.array([[F,-m],[F*r,I*(1*r**-1)]])\n",
- "B=np.array([[0],[w*I]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Distance travelled\n",
- "s=0.5*C[1]*C[0] #ft\n",
- "t=C[0] #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t,2),\"s\",'and it travels',round(s,2),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 0.83 s and it travels 1.66 ft\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-25, Page No 442"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilisation of variables\n",
- "d=2*12**-1 #ft\n",
- "v=80 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Mass flow reate without time\n",
- "m=(1*4**-1)*pi*d**2*v*(62.4/g) \n",
- "#Let P=force of plate on mass m of water\n",
- "P=m*(0-v) #lb\n",
- "\n",
- "#Result\n",
- "print'The force water exerts on the plate is +',round(-P),\"lb,that is,to the right\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force water exerts on the plate is + 271.0 lb,that is,to the right\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-26, Page No 442"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v1=20 #ft/s\n",
- "vw=80 #ft/s\n",
- "d=2*12**-1 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "v=vw-v1 #ft/s\n",
- "#mass flow rate without t\n",
- "m=(4**-1)*(pi*d**2)*(62.4/g)*v\n",
- "#Applying impulse momentum theorem\n",
- "P=m*v #lb\n",
- "\n",
- "#Result\n",
- "print'The force exerted by water on the plate is +',round(P),\"lb,that is,to the right\"\n",
- "\n",
- "# Decimal poinat accuracy causes a small discrepancy in the answer"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force exerted by water on the plate is + 152.0 lb,that is,to the right\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-27, Page No 443"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "A=2000*10**-6 #m**2\n",
- "v=10 #m/s\n",
- "rho=1000 #kg/m**3\n",
- "#theta=45 degrees,thus\n",
- "cos45=(2**0.5)**-1\n",
- "sin45=(2**0.5)**-1\n",
- "\n",
- "#Calculations\n",
- "#Mass flow \n",
- "m=A*v*rho\n",
- "#Applying impulse momentum theorem\n",
- "Px=m*(-v*cos45-v) #N\n",
- "Py=m*(v*sin45-0) #N\n",
- "\n",
- "#Result\n",
- "print'The x component of force is',round(Px),\"N (to left on water)\",'and y component is +',round(Py),\"N (up on the water)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The x component of force is -341.0 N (to left on water) and y component is + 141.0 N (up on the water)\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-28, Page No 443"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilisation of variables\n",
- "W=150 #lb\n",
- "v=20 #ft/s\n",
- "A=0.2 #in**2\n",
- "t=60 #s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Mass flow\n",
- "m=(A/12**2)*v*(62.4/g)\n",
- "#Force\n",
- "F=m*(0-v) #lb\n",
- "#At t=60s the tank holds\n",
- "Ww=(A/12**2)*v*t*62.4 #lb\n",
- "#Total reading on scale\n",
- "S=-F+W+Ww #lb\n",
- "\n",
- "#Result\n",
- "print'The scale reading is',round(S),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The scale reading is 255.0 lb\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-29, Page No 444"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wp=130 #lb\n",
- "Wb=150 #lb\n",
- "Wbullet=2*16**-1 #lb\n",
- "g=32.2 #ft/s**2\n",
- "vbullet=1200 #ft/s\n",
- "\n",
- "#Calculations\n",
- "v=((-Wbullet*g**-1)*vbullet)/((Wb+Wp)/g) #ft/s\n",
- "\n",
- "#Result\n",
- "print'The speed of the boat is',round(v,2),\"ft/s\"\n",
- "#Negative sign indicates direction opposite to that of bullet\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the boat is -0.54 ft/s\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-30, Page No 444"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "mb=0.06 #kg\n",
- "ms=50 #kg\n",
- "h=0.03 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Speed of bag+bullet\n",
- "v2=sqrt(2*g*h) #m/s\n",
- "#Applying conservation of momentum \n",
- "v1=((mb+ms)*v2)/mb #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed of bullet as it entered the bag was',round(v1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of bullet as it entered the bag was 640.0 m/s\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-32, Page No 445"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "mb=0.06 #kg\n",
- "vb=500 #m/s\n",
- "mblock=5 #kg\n",
- "vblock=30 #m/s\n",
- "\n",
- "#Calculations\n",
- "#Applying conservation of momentum\n",
- "v=(mb*vb+mblock*vblock)/(mb+mblock) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed of the system is',round(v,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the system is 35.6 m/s\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-33, Page No 445"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "W1=2 #lb\n",
- "W2=3 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return 12*(2-x)\n",
- "a=1\n",
- "b=1\n",
- "W=quad(integrand, 0, 2, args=(a,b))\n",
- "Work=W[0]/12 # ft-lb\n",
- "\n",
- "# Solving the simultaneousequations\n",
- "v3=(Work*(0.5*(W2*g**-1)+0.5*(W1*g**-1)*(-W2*W1**-1)**2)**-1)**0.5 #ft/s\n",
- "v2=-(W2*W1**-1)*v3 #ft/s\n",
- "\n",
- "# Results\n",
- "print'The speed of 2lb block is',round(v2,2),\"ft/s (to the left)\",'and that of 3lb block is',round(v3,2),\"ft/s (to the right) respectively.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of 2lb block is -6.22 ft/s (to the left) and that of 3lb block is 4.14 ft/s (to the right) respectively.\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-34, Page No 445"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#Here the integration is indefinite hence it will be computed manually and entered\n",
- "W=10 #lb\n",
- "l=4 #ft\n",
- "w=2 #rad/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Part (a)\n",
- "wf=1.5 #rad/s\n",
- "t=(((W/g)*(l*w*l))-((W/g)*(l*wf*l)))**0.5 #s\n",
- "#Part (b)\n",
- "#Applying conservation of angular momentum\n",
- "r=(l*wf*l)*(l*w)**-1 #ft\n",
- "\n",
- "#Result\n",
- "print'The answer for part (a) is',round(t,2),\"s\",'and the answer for part (b) is',round(r,2),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The answer for part (a) is 1.58 s and the answer for part (b) is 3.0 ft\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-35, Page No 446"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=2.5 #lb\n",
- "w=36 #rad/s\n",
- "Idisk=0.4 #slug-ft**2\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "Ii=Idisk+(2*(W*g**-1)*(3*12**-1)**2) #slug-ft**2\n",
- "If=Idisk+(2*(W*g**-1)*(11*12**-1)**2) #slug-ft**2\n",
- "#Since no external moments act\n",
- "#Applying conservation of momentum\n",
- "wf=(Ii*w)*If**-1 #rad/s\n",
- "\n",
- "#Result\n",
- "print'The final angular speed is',round(wf,1),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The final angular speed is 27.8 rad/s\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-39, Page No 448"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "u1=6 #ft/s\n",
- "u2=-8 #ft/s\n",
- "e=0.8 #coefficient of restitution\n",
- "\n",
- "#Calculations\n",
- "#Solving both simultaneous equations\n",
- "A=np.array([[1,-1],[1,1]])\n",
- "B=np.array([[11.2],[-2]])\n",
- "C=np.linalg.solve(A,B) #ft/s\n",
- "\n",
- "#Result\n",
- "print'The velocities are v1=',round(C[1],1),\"ft/s\",'and v2=',round(C[0],1),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocities are v1= -6.6 ft/s and v2= 4.6 ft/s\n"
- ]
- }
- ],
- "prompt_number": 52
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-40, Page No 448"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "h1=20 #m\n",
- "h2=14 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "u1=sqrt(2*g*h1) #m/s\n",
- "u2=0 #m/s\n",
- "v1=-sqrt(2*g*h2) #m/s\n",
- "v2=0 #m/s\n",
- "e=(v2-v1)/(u1-u2) #coefficient of restitution\n",
- "\n",
- "#Result\n",
- "print'The value of coefficient of restitution is',round(e,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of coefficient of restitution is 0.84\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-41, Page No 448"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "u=6.55 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "L=6 #ft\n",
- "W=5 #lb\n",
- "c=0.7 #fraction of impulse acting in second phase\n",
- "\n",
- "#Calculations\n",
- "#Impulse\n",
- "I=(W*g**-1)*(u*3**-1) #N.s\n",
- "#Second Phase\n",
- "v=((-c*10.9)/2)+u #ft/s\n",
- "wprime=(1.09*60+c*(W*u*(3**-1)*6))/60 #rad/s\n",
- "\n",
- "#Result\n",
- "print'The value is v=',round(v,2),\"ft/s to the right\",' and that of w is',round(wprime,2),\"rad/s counterclockwise.\"\n",
- "\n",
- "#The value of w is incorrect in the textbook\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value is v= 2.73 ft/s to the right and that of w is 1.85 rad/s counterclockwise.\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-42, Page No 449"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=9 #kg\n",
- "m2=5.5 #kg\n",
- "u1=-3 #m/s\n",
- "u2=1.77 #m/s\n",
- "e=0.8 #coefficient of restitution\n",
- "\n",
- "#Calculations\n",
- "#Solving by matrix method after we get the two equations\n",
- "A=np.array([[-1,1],[m1,m2]])\n",
- "B=np.array([[(e*u1-e*u2)],[m1*u1+m2*u2]])\n",
- "C=np.linalg.solve(A,B) #m/s\n",
- "\n",
- "#Result\n",
- "print'The 9kg ball will rebound up the speed of',round(C[0],2),\"m/s\",'and the 5.5 kg ball will move to the right and down with components of',round(u2,2),\"m/s\",'and',round(-C[1],2),\"m/s respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The 9kg ball will rebound up the speed of 0.26 m/s and the 5.5 kg ball will move to the right and down with components of 1.77 m/s and 3.56 m/s respectively.\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-46, Page No 451"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v=4 #m/s\n",
- "m=9 #kg\n",
- "s=1.5 #m\n",
- "\n",
- "#Calculations\n",
- "Io=(2*3**-1)*(m*s**2) #kg.m\n",
- "w=(m*v*s*0.5)/Io #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular velocity of the box is',round(w),\"rad/s clockwise.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity of the box is 2.0 rad/s clockwise.\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-48, Page No 452"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=2000 #kg\n",
- "mf=8500 #kg\n",
- "vr=2000 #m/s\n",
- "a=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying Newtons Second Law\n",
- "dm_dt=-(-((m+mf)*a)-(m+mf)*a)/(-vr) #kg/s\n",
- "\n",
- "#Result\n",
- "print'dm/dt=',round(dm_dt),\"kg/s\"\n",
- "#The negative sign indicates the loss in the mass of the system\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "dm/dt= -103.0 kg/s\n"
- ]
- }
- ],
- "prompt_number": 72
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-50, Page No 453"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=4000 #lb\n",
- "k=3 #ft\n",
- "wp=(60**-1)*2*pi #rad/s\n",
- "ws=(300/60)*2*pi #rad/s\n",
- "d=3.5 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "I=(W/g)*k**2 #slug-ft**2\n",
- "M=I*ws*wp #lb-ft\n",
- "#Now equating M to Rf-Rr gives one equations and vertical sum yields other\n",
- "#solving them by matrix method\n",
- "A=np.array([[1,-1],[1,1]])\n",
- "B=np.array([[M*(2/d)],[W]])\n",
- "C=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The weight of Rf and Rr are',round(C[0]),\"lb\",'and',round(C[1]),\"lb respectively.\"\n",
- "\n",
- "# The answers in the textbook are incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The weight of Rf and Rr are 3051.0 lb and 949.0 lb respectively.\n"
- ]
- }
- ],
- "prompt_number": 74
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-51, Page No 454"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#As the integration is indefinite we will directly consider the equation with R\n",
- "#Initillization of variables\n",
- "GM=1.41*10**16 #ft**3/s**2\n",
- "r=2640000 #ft\n",
- "theta=60 #degrees\n",
- "R=21120000 #ft\n",
- "\n",
- "#Calculations\n",
- "v1=sqrt((GM*((R**-1)-((R+r)**-1)))/2.031) #ft/s\n",
- "\n",
- "#Result\n",
- "print'The speed required will be',round(v1),\"ft/s\"\n",
- "\n",
- "# Answer may wary due to decimal point discrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed required will be 6043.0 ft/s\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-52, Page No 454"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "k=4 #lb/ft\n",
- "so=1 #ft\n",
- "W=2**-1 #lb\n",
- "g=32.2 #ft/s**2\n",
- "vo=5 #ft/s\n",
- "\n",
- "#Calculations\n",
- "m=W/g #kg\n",
- "#Angular momentum is conserved\n",
- "v=sqrt((0.5*k*so**2*2*2*g)+vo**2) #ft/s\n",
- "#Using vd=15\n",
- "d=15/v #ft\n",
- "\n",
- "#Result\n",
- "print'The ball passes',round(d,2),\"ft close to the fixed pin\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ball passes 0.89 ft close to the fixed pin\n"
- ]
- }
- ],
- "prompt_number": 26
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter18_1.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter18_1.ipynb deleted file mode 100755 index bbd27129..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter18_1.ipynb +++ /dev/null @@ -1,1503 +0,0 @@ -{
- "metadata": {
- "name": "chapter18.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 18: Impulse and Momentum"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-8, Page No 435"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=100 #lb\n",
- "u=0.2 #coefficient of friction\n",
- "t=5 #s\n",
- "v1=5 #ft/s\n",
- "v2=10 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "ll=0 #lower limit of integration\n",
- "ul=5 #upper limit of integration\n",
- "\n",
- "#Calculations\n",
- "Fr=u*W #lb\n",
- "#Using The impulse momentum theorem\n",
- "#Since the integration is just subtraction of limits we can skip that\n",
- "F=((W/g)*v2-(W/g)*v1+Fr*ul)/ul #lb\n",
- "\n",
- "#Result\n",
- "print'The Force is',round(F,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Force is 23.1 lb\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-9, Page No 435"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=10 #kg\n",
- "#theta=30,thus\n",
- "cos30=sqrt(3)*2**-1\n",
- "sin30=2**-1\n",
- "u=0.3 #coefficient of kinetic friction\n",
- "t=5 #s\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Asthere is no motion in the vertical direction \n",
- "#Summing forces along vertical direction\n",
- "Na=m*g*cos30 #N\n",
- "#Using impulse momentum theorem\n",
- "vx=(m*g*sin30-u*Na)*(t*m**-1) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed after 5s is',round(vx,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed after 5s is 11.8 m/s\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-10, Page No 436"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(t, a, b):\n",
- " return 20*t-16\n",
- "a=1\n",
- "b=1\n",
- "F=quad(integrand, 1, 5, args=(a,b))\n",
- "g=32.2 # ft/s**2\n",
- "v=(F[0]*32.2)/80\n",
- "\n",
- "# Results\n",
- "print'The speed of the block is',round(v,1),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the block is 70.8 ft/s\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-12, Page No 436"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=40 #kg\n",
- "m2=10 #kg\n",
- "m3=15 #kg\n",
- "v0=2.5 #m/s\n",
- "vf=5 #m/s\n",
- "t=12 #s\n",
- "u=0.1 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "#theta=45 degrees,thus\n",
- "sin45=(sqrt(2))**-1\n",
- "cos45=(sqrt(2))**-1\n",
- "\n",
- "#Calculations\n",
- "#Applying Impulse-Momentum Theoroem\n",
- "P=(((m1+m2+m3)*(vf-v0))+(t*(-m2*g*sin45+u*g*m2*cos45+u*g*m3+g*m1)))*t**-1 #N\n",
- "\n",
- "#Result\n",
- "print'The value of P is',round(P),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P is 358.0 N\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-13, Page Noo 437"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W1=4 #lb\n",
- "W2=2 #lb\n",
- "t2=0.04 #s\n",
- "W3=-2 #lb\n",
- "t3=0.02 #s\n",
- "t=3 #s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Algebraic sum of two areas\n",
- "A=t2*W2+t3*W3 #lb-s\n",
- "#Using Impulse Momentum Theorem\n",
- "v=(A*g)/W1 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The spped after 3s is',round(v,3),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The spped after 3s is 0.322 ft/s\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-14, Page No 437"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "f_r=1 #in/s rate of fall of mercury\n",
- "ll=18 #in length of left column\n",
- "lr=22 #in length of right column\n",
- "rho=850 #lb/ft**3\n",
- "d=4**-1 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying Impulse momentum theorem\n",
- "M=((d*pi*d**2*4)/12**3)*(rho/g)*(12**-1) #lb-s\n",
- "\n",
- "#Result\n",
- "print'The upward momentum of mercury is +',round(M,5),\"lb-s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The upward momentum of mercury is + 0.00025 lb-s\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-15, Page No 437"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=2000 #kg\n",
- "k=1.200 #m\n",
- "w=120 #rpm\n",
- "t=200 #s\n",
- "\n",
- "#Calculations\n",
- "#Applying Angular Momentum theorem\n",
- "M=((m*k**2*(w*2*pi))/60)/t #N.m\n",
- "\n",
- "#Result\n",
- "print'The Momentum necessary is',round(M),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Momentum necessary is 181.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-17, Page No 438"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=5 #kg\n",
- "m2=7 #kg\n",
- "mp=5 #kg\n",
- "r=0.6 #m\n",
- "k=0.45 #m\n",
- "vi=3 #m/s\n",
- "vf=6 #m/s\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "I=m1*k**2 #kg.m**2\n",
- "wnet=(vf/r)-(vi/r) #rad/s\n",
- "#Solving the system of linear equations\n",
- "#Simplfying the equation we get\n",
- "t=((I*wnet)+m1*(vf-vi)+m2*(vf-vi))*r/(r*(m2-m1)*g) #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t,2),\"s\"\n",
- "# The ans in the textbook is incorrect.\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 2.1 s\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-18, Page No 438"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=50 #kg\n",
- "vo=4 #m/s\n",
- "vf=8 #m/s\n",
- "t=6 #s\n",
- "g=9.8 #m/s**2\n",
- "r=0.8 #m\n",
- "u=0.25 #coefficient of friction\n",
- "I=30 #kg-m**2\n",
- "\n",
- "#Calculations\n",
- "Na=m*g #N\n",
- "F=u*Na #N\n",
- "#Angular Speeds\n",
- "wo=vo/r #rad/s\n",
- "wf=vf/r #rad/s\n",
- "#Applying impulse momentum theorem\n",
- "mb=(I*wo+m*vo*r-F*r*t-I*wf-m*vf*r)/(vf*r-vo*r-g*r*t) #kg\n",
- "\n",
- "#Result\n",
- "print'The mass of block B is',round(mb,1),\"kg\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The mass of block B is 20.5 kg\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-19, Page No 439"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Ws=250 #lb\n",
- "Wl=500 #lb\n",
- "W3=161 #lb\n",
- "W4=64.4 #lb\n",
- "wo=100 #rpm\n",
- "wf=300 #rpm\n",
- "rl=3 #ft\n",
- "rs=2 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment Of Inertia\n",
- "I=(0.5*(Wl/g)*rl**2+0.5*(Ws/g)*rs**2) #slug-ft**2\n",
- "#Change in angular Momentum\n",
- "change1=I*((wf-wo)*2*(pi/60)) #lb-s-ft\n",
- "#Change in angular momentum about G for 161lb\n",
- "change2=2*((W3*g**-1)*(wf-wo)*(4*60**-1)*pi) #lb-s-ft\n",
- "#Similarly change in 64lb is\n",
- "change3=3*((W4*g**-1)*(wf-wo)*(6*60**-1)*pi) #lb-s-ft\n",
- "#Change in linear impulse\n",
- "#Without t term in it\n",
- "m1=2*W3\n",
- "m2=-3*W4\n",
- "#Total angular impulse\n",
- "t=(change1+change2+change3)/(m1+m2) #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 20.0 s\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-21, Page No 440"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "d=3 #ft\n",
- "W=300 #lb\n",
- "# as theta=20 degrees\n",
- "sintheta=0.342\n",
- "F=250 #lb\n",
- "t=6 #s\n",
- "vo=0 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying linear impulse momentum theorem\n",
- "#Solving by matrix method\n",
- "A=np.array([[-W*g**-1,1*t],[-((0.5)*(W*g**-1)*(d*2**-1)**2*(d*2**-1)**-1),(-t)*d*2**-1]])\n",
- "B=np.array([[-F*t+W*sintheta*t],[-F*(d*2**-1)*6]]) \n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The speed after 6s is',round(C[0]),\"ft/s,parallel to the plane\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed after 6s is 171.0 ft/s,parallel to the plane\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-22, Page No 440"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Initilization of variables\n",
- "#theta=30 degrees\n",
- "sin30=2**-1\n",
- "vo=20 #ft/s\n",
- "r=4 #ft\n",
- "vf=0 #ft/s\n",
- "W=300 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "wo=vo*r**-1 #rad/s\n",
- "wf=vf*r**-1 #rad/s\n",
- "#Applying impulse momentum theorem\n",
- "#Solving simultaneous equations\n",
- "t=-((W*g**-1)*(vf-vo)+((0.5*(W*g**-1)*r**2*(wf-wo))*r**-1))/(W*sin30) #s\n",
- "\n",
- "#Result\n",
- "print'The time t is',round(t,2),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time t is 1.86 s\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-23, Page No 441"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mw=75 #kg\n",
- "k=0.9 #m\n",
- "wi=10 #rad/s\n",
- "wf=6 #rad/s\n",
- "r=1.2 #m\n",
- "m=30 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Initial speed\n",
- "vi=-r*wi #m/s\n",
- "vf=r*wf #m/s\n",
- "#Initial speed of B is\n",
- "vib=-0.8*wi+vi #m/s\n",
- "#Similarly\n",
- "vfb=12 #m/s\n",
- "#Applying impulse momentum theorem\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,-1,-(mw*(vf-vi))],[0.8,1.2,-(mw*(k**2)*(wf+wi))],[-1,0,-(m*(vfb-vi))]])\n",
- "B=np.array([[0],[0],[-g*m]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Here t is calculated as 1/t for simplicity\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(C[2]**-1,2),\"s\" \n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 7.04 s\n"
- ]
- }
- ],
- "prompt_number": 59
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-24, Page No 441"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=8 #in\n",
- "W=96.6 #lb\n",
- "w=36 #rad/s\n",
- "u=0.15 #coefficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "r=(d/2)*12**-1 #m\n",
- "N=W #lb\n",
- "F=u*N #lb\n",
- "m=W/g #slugs\n",
- "I=0.5*m*(r**2) #slug-ft**2\n",
- "#Applying the impulse momentum theorem\n",
- "#Solving the two equations simultaneously\n",
- "A=np.array([[F,-m],[F*r,I*(1*r**-1)]])\n",
- "B=np.array([[0],[w*I]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Distance travelled\n",
- "s=0.5*C[1]*C[0] #ft\n",
- "t=C[0] #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t,2),\"s\",'and it travels',round(s,2),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 0.83 s and it travels 1.66 ft\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-25, Page No 442"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilisation of variables\n",
- "d=2*12**-1 #ft\n",
- "v=80 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Mass flow reate without time\n",
- "m=(1*4**-1)*pi*d**2*v*(62.4/g) \n",
- "#Let P=force of plate on mass m of water\n",
- "P=m*(0-v) #lb\n",
- "\n",
- "#Result\n",
- "print'The force water exerts on the plate is +',round(-P),\"lb,that is,to the right\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force water exerts on the plate is + 271.0 lb,that is,to the right\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-26, Page No 442"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v1=20 #ft/s\n",
- "vw=80 #ft/s\n",
- "d=2*12**-1 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "v=vw-v1 #ft/s\n",
- "#mass flow rate without t\n",
- "m=(4**-1)*(pi*d**2)*(62.4/g)*v\n",
- "#Applying impulse momentum theorem\n",
- "P=m*v #lb\n",
- "\n",
- "#Result\n",
- "print'The force exerted by water on the plate is +',round(P),\"lb,that is,to the right\"\n",
- "\n",
- "# Decimal poinat accuracy causes a small discrepancy in the answer"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force exerted by water on the plate is + 152.0 lb,that is,to the right\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-27, Page No 443"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "A=2000*10**-6 #m**2\n",
- "v=10 #m/s\n",
- "rho=1000 #kg/m**3\n",
- "#theta=45 degrees,thus\n",
- "cos45=(2**0.5)**-1\n",
- "sin45=(2**0.5)**-1\n",
- "\n",
- "#Calculations\n",
- "#Mass flow \n",
- "m=A*v*rho\n",
- "#Applying impulse momentum theorem\n",
- "Px=m*(-v*cos45-v) #N\n",
- "Py=m*(v*sin45-0) #N\n",
- "\n",
- "#Result\n",
- "print'The x component of force is',round(Px),\"N (to left on water)\",'and y component is +',round(Py),\"N (up on the water)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The x component of force is -341.0 N (to left on water) and y component is + 141.0 N (up on the water)\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-28, Page No 443"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilisation of variables\n",
- "W=150 #lb\n",
- "v=20 #ft/s\n",
- "A=0.2 #in**2\n",
- "t=60 #s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Mass flow\n",
- "m=(A/12**2)*v*(62.4/g)\n",
- "#Force\n",
- "F=m*(0-v) #lb\n",
- "#At t=60s the tank holds\n",
- "Ww=(A/12**2)*v*t*62.4 #lb\n",
- "#Total reading on scale\n",
- "S=-F+W+Ww #lb\n",
- "\n",
- "#Result\n",
- "print'The scale reading is',round(S),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The scale reading is 255.0 lb\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-29, Page No 444"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wp=130 #lb\n",
- "Wb=150 #lb\n",
- "Wbullet=2*16**-1 #lb\n",
- "g=32.2 #ft/s**2\n",
- "vbullet=1200 #ft/s\n",
- "\n",
- "#Calculations\n",
- "v=((-Wbullet*g**-1)*vbullet)/((Wb+Wp)/g) #ft/s\n",
- "\n",
- "#Result\n",
- "print'The speed of the boat is',round(v,2),\"ft/s\"\n",
- "#Negative sign indicates direction opposite to that of bullet\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the boat is -0.54 ft/s\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-30, Page No 444"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "mb=0.06 #kg\n",
- "ms=50 #kg\n",
- "h=0.03 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Speed of bag+bullet\n",
- "v2=sqrt(2*g*h) #m/s\n",
- "#Applying conservation of momentum \n",
- "v1=((mb+ms)*v2)/mb #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed of bullet as it entered the bag was',round(v1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of bullet as it entered the bag was 640.0 m/s\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-32, Page No 445"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "mb=0.06 #kg\n",
- "vb=500 #m/s\n",
- "mblock=5 #kg\n",
- "vblock=30 #m/s\n",
- "\n",
- "#Calculations\n",
- "#Applying conservation of momentum\n",
- "v=(mb*vb+mblock*vblock)/(mb+mblock) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed of the system is',round(v,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the system is 35.6 m/s\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-33, Page No 445"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "W1=2 #lb\n",
- "W2=3 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return 12*(2-x)\n",
- "a=1\n",
- "b=1\n",
- "W=quad(integrand, 0, 2, args=(a,b))\n",
- "Work=W[0]/12 # ft-lb\n",
- "\n",
- "# Solving the simultaneousequations\n",
- "v3=(Work*(0.5*(W2*g**-1)+0.5*(W1*g**-1)*(-W2*W1**-1)**2)**-1)**0.5 #ft/s\n",
- "v2=-(W2*W1**-1)*v3 #ft/s\n",
- "\n",
- "# Results\n",
- "print'The speed of 2lb block is',round(v2,2),\"ft/s (to the left)\",'and that of 3lb block is',round(v3,2),\"ft/s (to the right) respectively.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of 2lb block is -6.22 ft/s (to the left) and that of 3lb block is 4.14 ft/s (to the right) respectively.\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-34, Page No 445"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#Here the integration is indefinite hence it will be computed manually and entered\n",
- "W=10 #lb\n",
- "l=4 #ft\n",
- "w=2 #rad/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Part (a)\n",
- "wf=1.5 #rad/s\n",
- "t=(((W/g)*(l*w*l))-((W/g)*(l*wf*l)))**0.5 #s\n",
- "#Part (b)\n",
- "#Applying conservation of angular momentum\n",
- "r=(l*wf*l)*(l*w)**-1 #ft\n",
- "\n",
- "#Result\n",
- "print'The answer for part (a) is',round(t,2),\"s\",'and the answer for part (b) is',round(r,2),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The answer for part (a) is 1.58 s and the answer for part (b) is 3.0 ft\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-35, Page No 446"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=2.5 #lb\n",
- "w=36 #rad/s\n",
- "Idisk=0.4 #slug-ft**2\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "Ii=Idisk+(2*(W*g**-1)*(3*12**-1)**2) #slug-ft**2\n",
- "If=Idisk+(2*(W*g**-1)*(11*12**-1)**2) #slug-ft**2\n",
- "#Since no external moments act\n",
- "#Applying conservation of momentum\n",
- "wf=(Ii*w)*If**-1 #rad/s\n",
- "\n",
- "#Result\n",
- "print'The final angular speed is',round(wf,1),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The final angular speed is 27.8 rad/s\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-39, Page No 448"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "u1=6 #ft/s\n",
- "u2=-8 #ft/s\n",
- "e=0.8 #coefficient of restitution\n",
- "\n",
- "#Calculations\n",
- "#Solving both simultaneous equations\n",
- "A=np.array([[1,-1],[1,1]])\n",
- "B=np.array([[11.2],[-2]])\n",
- "C=np.linalg.solve(A,B) #ft/s\n",
- "\n",
- "#Result\n",
- "print'The velocities are v1=',round(C[1],1),\"ft/s\",'and v2=',round(C[0],1),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocities are v1= -6.6 ft/s and v2= 4.6 ft/s\n"
- ]
- }
- ],
- "prompt_number": 52
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-40, Page No 448"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "h1=20 #m\n",
- "h2=14 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "u1=sqrt(2*g*h1) #m/s\n",
- "u2=0 #m/s\n",
- "v1=-sqrt(2*g*h2) #m/s\n",
- "v2=0 #m/s\n",
- "e=(v2-v1)/(u1-u2) #coefficient of restitution\n",
- "\n",
- "#Result\n",
- "print'The value of coefficient of restitution is',round(e,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of coefficient of restitution is 0.84\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-41, Page No 448"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "u=6.55 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "L=6 #ft\n",
- "W=5 #lb\n",
- "c=0.7 #fraction of impulse acting in second phase\n",
- "\n",
- "#Calculations\n",
- "#Impulse\n",
- "I=(W*g**-1)*(u*3**-1) #N.s\n",
- "#Second Phase\n",
- "v=((-c*10.9)/2)+u #ft/s\n",
- "wprime=(1.09*60+c*(W*u*(3**-1)*6))/60 #rad/s\n",
- "\n",
- "#Result\n",
- "print'The value is v=',round(v,2),\"ft/s to the right\",' and that of w is',round(wprime,2),\"rad/s counterclockwise.\"\n",
- "\n",
- "#The value of w is incorrect in the textbook\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value is v= 2.73 ft/s to the right and that of w is 1.85 rad/s counterclockwise.\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-42, Page No 449"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=9 #kg\n",
- "m2=5.5 #kg\n",
- "u1=-3 #m/s\n",
- "u2=1.77 #m/s\n",
- "e=0.8 #coefficient of restitution\n",
- "\n",
- "#Calculations\n",
- "#Solving by matrix method after we get the two equations\n",
- "A=np.array([[-1,1],[m1,m2]])\n",
- "B=np.array([[(e*u1-e*u2)],[m1*u1+m2*u2]])\n",
- "C=np.linalg.solve(A,B) #m/s\n",
- "\n",
- "#Result\n",
- "print'The 9kg ball will rebound up the speed of',round(C[0],2),\"m/s\",'and the 5.5 kg ball will move to the right and down with components of',round(u2,2),\"m/s\",'and',round(-C[1],2),\"m/s respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The 9kg ball will rebound up the speed of 0.26 m/s and the 5.5 kg ball will move to the right and down with components of 1.77 m/s and 3.56 m/s respectively.\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-46, Page No 451"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v=4 #m/s\n",
- "m=9 #kg\n",
- "s=1.5 #m\n",
- "\n",
- "#Calculations\n",
- "Io=(2*3**-1)*(m*s**2) #kg.m\n",
- "w=(m*v*s*0.5)/Io #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular velocity of the box is',round(w),\"rad/s clockwise.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity of the box is 2.0 rad/s clockwise.\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-48, Page No 452"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=2000 #kg\n",
- "mf=8500 #kg\n",
- "vr=2000 #m/s\n",
- "a=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying Newtons Second Law\n",
- "dm_dt=-(-((m+mf)*a)-(m+mf)*a)/(-vr) #kg/s\n",
- "\n",
- "#Result\n",
- "print'dm/dt=',round(dm_dt),\"kg/s\"\n",
- "#The negative sign indicates the loss in the mass of the system\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "dm/dt= -103.0 kg/s\n"
- ]
- }
- ],
- "prompt_number": 72
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-50, Page No 453"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=4000 #lb\n",
- "k=3 #ft\n",
- "wp=(60**-1)*2*pi #rad/s\n",
- "ws=(300/60)*2*pi #rad/s\n",
- "d=3.5 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "I=(W/g)*k**2 #slug-ft**2\n",
- "M=I*ws*wp #lb-ft\n",
- "#Now equating M to Rf-Rr gives one equations and vertical sum yields other\n",
- "#solving them by matrix method\n",
- "A=np.array([[1,-1],[1,1]])\n",
- "B=np.array([[M*(2/d)],[W]])\n",
- "C=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The weight of Rf and Rr are',round(C[0]),\"lb\",'and',round(C[1]),\"lb respectively.\"\n",
- "\n",
- "# The answers in the textbook are incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The weight of Rf and Rr are 3051.0 lb and 949.0 lb respectively.\n"
- ]
- }
- ],
- "prompt_number": 74
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-51, Page No 454"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#As the integration is indefinite we will directly consider the equation with R\n",
- "#Initillization of variables\n",
- "GM=1.41*10**16 #ft**3/s**2\n",
- "r=2640000 #ft\n",
- "theta=60 #degrees\n",
- "R=21120000 #ft\n",
- "\n",
- "#Calculations\n",
- "v1=sqrt((GM*((R**-1)-((R+r)**-1)))/2.031) #ft/s\n",
- "\n",
- "#Result\n",
- "print'The speed required will be',round(v1),\"ft/s\"\n",
- "\n",
- "# Answer may wary due to decimal point discrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed required will be 6043.0 ft/s\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-52, Page No 454"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "k=4 #lb/ft\n",
- "so=1 #ft\n",
- "W=2**-1 #lb\n",
- "g=32.2 #ft/s**2\n",
- "vo=5 #ft/s\n",
- "\n",
- "#Calculations\n",
- "m=W/g #kg\n",
- "#Angular momentum is conserved\n",
- "v=sqrt((0.5*k*so**2*2*2*g)+vo**2) #ft/s\n",
- "#Using vd=15\n",
- "d=15/v #ft\n",
- "\n",
- "#Result\n",
- "print'The ball passes',round(d,2),\"ft close to the fixed pin\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ball passes 0.89 ft close to the fixed pin\n"
- ]
- }
- ],
- "prompt_number": 26
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter18_2.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter18_2.ipynb deleted file mode 100755 index bbd27129..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter18_2.ipynb +++ /dev/null @@ -1,1503 +0,0 @@ -{
- "metadata": {
- "name": "chapter18.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 18: Impulse and Momentum"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-8, Page No 435"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=100 #lb\n",
- "u=0.2 #coefficient of friction\n",
- "t=5 #s\n",
- "v1=5 #ft/s\n",
- "v2=10 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "ll=0 #lower limit of integration\n",
- "ul=5 #upper limit of integration\n",
- "\n",
- "#Calculations\n",
- "Fr=u*W #lb\n",
- "#Using The impulse momentum theorem\n",
- "#Since the integration is just subtraction of limits we can skip that\n",
- "F=((W/g)*v2-(W/g)*v1+Fr*ul)/ul #lb\n",
- "\n",
- "#Result\n",
- "print'The Force is',round(F,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Force is 23.1 lb\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-9, Page No 435"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=10 #kg\n",
- "#theta=30,thus\n",
- "cos30=sqrt(3)*2**-1\n",
- "sin30=2**-1\n",
- "u=0.3 #coefficient of kinetic friction\n",
- "t=5 #s\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Asthere is no motion in the vertical direction \n",
- "#Summing forces along vertical direction\n",
- "Na=m*g*cos30 #N\n",
- "#Using impulse momentum theorem\n",
- "vx=(m*g*sin30-u*Na)*(t*m**-1) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed after 5s is',round(vx,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed after 5s is 11.8 m/s\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-10, Page No 436"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(t, a, b):\n",
- " return 20*t-16\n",
- "a=1\n",
- "b=1\n",
- "F=quad(integrand, 1, 5, args=(a,b))\n",
- "g=32.2 # ft/s**2\n",
- "v=(F[0]*32.2)/80\n",
- "\n",
- "# Results\n",
- "print'The speed of the block is',round(v,1),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the block is 70.8 ft/s\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-12, Page No 436"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=40 #kg\n",
- "m2=10 #kg\n",
- "m3=15 #kg\n",
- "v0=2.5 #m/s\n",
- "vf=5 #m/s\n",
- "t=12 #s\n",
- "u=0.1 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "#theta=45 degrees,thus\n",
- "sin45=(sqrt(2))**-1\n",
- "cos45=(sqrt(2))**-1\n",
- "\n",
- "#Calculations\n",
- "#Applying Impulse-Momentum Theoroem\n",
- "P=(((m1+m2+m3)*(vf-v0))+(t*(-m2*g*sin45+u*g*m2*cos45+u*g*m3+g*m1)))*t**-1 #N\n",
- "\n",
- "#Result\n",
- "print'The value of P is',round(P),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P is 358.0 N\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-13, Page Noo 437"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W1=4 #lb\n",
- "W2=2 #lb\n",
- "t2=0.04 #s\n",
- "W3=-2 #lb\n",
- "t3=0.02 #s\n",
- "t=3 #s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Algebraic sum of two areas\n",
- "A=t2*W2+t3*W3 #lb-s\n",
- "#Using Impulse Momentum Theorem\n",
- "v=(A*g)/W1 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The spped after 3s is',round(v,3),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The spped after 3s is 0.322 ft/s\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-14, Page No 437"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "f_r=1 #in/s rate of fall of mercury\n",
- "ll=18 #in length of left column\n",
- "lr=22 #in length of right column\n",
- "rho=850 #lb/ft**3\n",
- "d=4**-1 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying Impulse momentum theorem\n",
- "M=((d*pi*d**2*4)/12**3)*(rho/g)*(12**-1) #lb-s\n",
- "\n",
- "#Result\n",
- "print'The upward momentum of mercury is +',round(M,5),\"lb-s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The upward momentum of mercury is + 0.00025 lb-s\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-15, Page No 437"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=2000 #kg\n",
- "k=1.200 #m\n",
- "w=120 #rpm\n",
- "t=200 #s\n",
- "\n",
- "#Calculations\n",
- "#Applying Angular Momentum theorem\n",
- "M=((m*k**2*(w*2*pi))/60)/t #N.m\n",
- "\n",
- "#Result\n",
- "print'The Momentum necessary is',round(M),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Momentum necessary is 181.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-17, Page No 438"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=5 #kg\n",
- "m2=7 #kg\n",
- "mp=5 #kg\n",
- "r=0.6 #m\n",
- "k=0.45 #m\n",
- "vi=3 #m/s\n",
- "vf=6 #m/s\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "I=m1*k**2 #kg.m**2\n",
- "wnet=(vf/r)-(vi/r) #rad/s\n",
- "#Solving the system of linear equations\n",
- "#Simplfying the equation we get\n",
- "t=((I*wnet)+m1*(vf-vi)+m2*(vf-vi))*r/(r*(m2-m1)*g) #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t,2),\"s\"\n",
- "# The ans in the textbook is incorrect.\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 2.1 s\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-18, Page No 438"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=50 #kg\n",
- "vo=4 #m/s\n",
- "vf=8 #m/s\n",
- "t=6 #s\n",
- "g=9.8 #m/s**2\n",
- "r=0.8 #m\n",
- "u=0.25 #coefficient of friction\n",
- "I=30 #kg-m**2\n",
- "\n",
- "#Calculations\n",
- "Na=m*g #N\n",
- "F=u*Na #N\n",
- "#Angular Speeds\n",
- "wo=vo/r #rad/s\n",
- "wf=vf/r #rad/s\n",
- "#Applying impulse momentum theorem\n",
- "mb=(I*wo+m*vo*r-F*r*t-I*wf-m*vf*r)/(vf*r-vo*r-g*r*t) #kg\n",
- "\n",
- "#Result\n",
- "print'The mass of block B is',round(mb,1),\"kg\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The mass of block B is 20.5 kg\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-19, Page No 439"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Ws=250 #lb\n",
- "Wl=500 #lb\n",
- "W3=161 #lb\n",
- "W4=64.4 #lb\n",
- "wo=100 #rpm\n",
- "wf=300 #rpm\n",
- "rl=3 #ft\n",
- "rs=2 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment Of Inertia\n",
- "I=(0.5*(Wl/g)*rl**2+0.5*(Ws/g)*rs**2) #slug-ft**2\n",
- "#Change in angular Momentum\n",
- "change1=I*((wf-wo)*2*(pi/60)) #lb-s-ft\n",
- "#Change in angular momentum about G for 161lb\n",
- "change2=2*((W3*g**-1)*(wf-wo)*(4*60**-1)*pi) #lb-s-ft\n",
- "#Similarly change in 64lb is\n",
- "change3=3*((W4*g**-1)*(wf-wo)*(6*60**-1)*pi) #lb-s-ft\n",
- "#Change in linear impulse\n",
- "#Without t term in it\n",
- "m1=2*W3\n",
- "m2=-3*W4\n",
- "#Total angular impulse\n",
- "t=(change1+change2+change3)/(m1+m2) #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 20.0 s\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-21, Page No 440"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "d=3 #ft\n",
- "W=300 #lb\n",
- "# as theta=20 degrees\n",
- "sintheta=0.342\n",
- "F=250 #lb\n",
- "t=6 #s\n",
- "vo=0 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying linear impulse momentum theorem\n",
- "#Solving by matrix method\n",
- "A=np.array([[-W*g**-1,1*t],[-((0.5)*(W*g**-1)*(d*2**-1)**2*(d*2**-1)**-1),(-t)*d*2**-1]])\n",
- "B=np.array([[-F*t+W*sintheta*t],[-F*(d*2**-1)*6]]) \n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The speed after 6s is',round(C[0]),\"ft/s,parallel to the plane\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed after 6s is 171.0 ft/s,parallel to the plane\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-22, Page No 440"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Initilization of variables\n",
- "#theta=30 degrees\n",
- "sin30=2**-1\n",
- "vo=20 #ft/s\n",
- "r=4 #ft\n",
- "vf=0 #ft/s\n",
- "W=300 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "wo=vo*r**-1 #rad/s\n",
- "wf=vf*r**-1 #rad/s\n",
- "#Applying impulse momentum theorem\n",
- "#Solving simultaneous equations\n",
- "t=-((W*g**-1)*(vf-vo)+((0.5*(W*g**-1)*r**2*(wf-wo))*r**-1))/(W*sin30) #s\n",
- "\n",
- "#Result\n",
- "print'The time t is',round(t,2),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time t is 1.86 s\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-23, Page No 441"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mw=75 #kg\n",
- "k=0.9 #m\n",
- "wi=10 #rad/s\n",
- "wf=6 #rad/s\n",
- "r=1.2 #m\n",
- "m=30 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Initial speed\n",
- "vi=-r*wi #m/s\n",
- "vf=r*wf #m/s\n",
- "#Initial speed of B is\n",
- "vib=-0.8*wi+vi #m/s\n",
- "#Similarly\n",
- "vfb=12 #m/s\n",
- "#Applying impulse momentum theorem\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,-1,-(mw*(vf-vi))],[0.8,1.2,-(mw*(k**2)*(wf+wi))],[-1,0,-(m*(vfb-vi))]])\n",
- "B=np.array([[0],[0],[-g*m]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Here t is calculated as 1/t for simplicity\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(C[2]**-1,2),\"s\" \n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 7.04 s\n"
- ]
- }
- ],
- "prompt_number": 59
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-24, Page No 441"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=8 #in\n",
- "W=96.6 #lb\n",
- "w=36 #rad/s\n",
- "u=0.15 #coefficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "r=(d/2)*12**-1 #m\n",
- "N=W #lb\n",
- "F=u*N #lb\n",
- "m=W/g #slugs\n",
- "I=0.5*m*(r**2) #slug-ft**2\n",
- "#Applying the impulse momentum theorem\n",
- "#Solving the two equations simultaneously\n",
- "A=np.array([[F,-m],[F*r,I*(1*r**-1)]])\n",
- "B=np.array([[0],[w*I]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Distance travelled\n",
- "s=0.5*C[1]*C[0] #ft\n",
- "t=C[0] #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t,2),\"s\",'and it travels',round(s,2),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 0.83 s and it travels 1.66 ft\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-25, Page No 442"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilisation of variables\n",
- "d=2*12**-1 #ft\n",
- "v=80 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Mass flow reate without time\n",
- "m=(1*4**-1)*pi*d**2*v*(62.4/g) \n",
- "#Let P=force of plate on mass m of water\n",
- "P=m*(0-v) #lb\n",
- "\n",
- "#Result\n",
- "print'The force water exerts on the plate is +',round(-P),\"lb,that is,to the right\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force water exerts on the plate is + 271.0 lb,that is,to the right\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-26, Page No 442"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v1=20 #ft/s\n",
- "vw=80 #ft/s\n",
- "d=2*12**-1 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "v=vw-v1 #ft/s\n",
- "#mass flow rate without t\n",
- "m=(4**-1)*(pi*d**2)*(62.4/g)*v\n",
- "#Applying impulse momentum theorem\n",
- "P=m*v #lb\n",
- "\n",
- "#Result\n",
- "print'The force exerted by water on the plate is +',round(P),\"lb,that is,to the right\"\n",
- "\n",
- "# Decimal poinat accuracy causes a small discrepancy in the answer"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force exerted by water on the plate is + 152.0 lb,that is,to the right\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-27, Page No 443"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "A=2000*10**-6 #m**2\n",
- "v=10 #m/s\n",
- "rho=1000 #kg/m**3\n",
- "#theta=45 degrees,thus\n",
- "cos45=(2**0.5)**-1\n",
- "sin45=(2**0.5)**-1\n",
- "\n",
- "#Calculations\n",
- "#Mass flow \n",
- "m=A*v*rho\n",
- "#Applying impulse momentum theorem\n",
- "Px=m*(-v*cos45-v) #N\n",
- "Py=m*(v*sin45-0) #N\n",
- "\n",
- "#Result\n",
- "print'The x component of force is',round(Px),\"N (to left on water)\",'and y component is +',round(Py),\"N (up on the water)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The x component of force is -341.0 N (to left on water) and y component is + 141.0 N (up on the water)\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-28, Page No 443"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilisation of variables\n",
- "W=150 #lb\n",
- "v=20 #ft/s\n",
- "A=0.2 #in**2\n",
- "t=60 #s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Mass flow\n",
- "m=(A/12**2)*v*(62.4/g)\n",
- "#Force\n",
- "F=m*(0-v) #lb\n",
- "#At t=60s the tank holds\n",
- "Ww=(A/12**2)*v*t*62.4 #lb\n",
- "#Total reading on scale\n",
- "S=-F+W+Ww #lb\n",
- "\n",
- "#Result\n",
- "print'The scale reading is',round(S),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The scale reading is 255.0 lb\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-29, Page No 444"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wp=130 #lb\n",
- "Wb=150 #lb\n",
- "Wbullet=2*16**-1 #lb\n",
- "g=32.2 #ft/s**2\n",
- "vbullet=1200 #ft/s\n",
- "\n",
- "#Calculations\n",
- "v=((-Wbullet*g**-1)*vbullet)/((Wb+Wp)/g) #ft/s\n",
- "\n",
- "#Result\n",
- "print'The speed of the boat is',round(v,2),\"ft/s\"\n",
- "#Negative sign indicates direction opposite to that of bullet\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the boat is -0.54 ft/s\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-30, Page No 444"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "mb=0.06 #kg\n",
- "ms=50 #kg\n",
- "h=0.03 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Speed of bag+bullet\n",
- "v2=sqrt(2*g*h) #m/s\n",
- "#Applying conservation of momentum \n",
- "v1=((mb+ms)*v2)/mb #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed of bullet as it entered the bag was',round(v1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of bullet as it entered the bag was 640.0 m/s\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-32, Page No 445"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "mb=0.06 #kg\n",
- "vb=500 #m/s\n",
- "mblock=5 #kg\n",
- "vblock=30 #m/s\n",
- "\n",
- "#Calculations\n",
- "#Applying conservation of momentum\n",
- "v=(mb*vb+mblock*vblock)/(mb+mblock) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed of the system is',round(v,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the system is 35.6 m/s\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-33, Page No 445"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "W1=2 #lb\n",
- "W2=3 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return 12*(2-x)\n",
- "a=1\n",
- "b=1\n",
- "W=quad(integrand, 0, 2, args=(a,b))\n",
- "Work=W[0]/12 # ft-lb\n",
- "\n",
- "# Solving the simultaneousequations\n",
- "v3=(Work*(0.5*(W2*g**-1)+0.5*(W1*g**-1)*(-W2*W1**-1)**2)**-1)**0.5 #ft/s\n",
- "v2=-(W2*W1**-1)*v3 #ft/s\n",
- "\n",
- "# Results\n",
- "print'The speed of 2lb block is',round(v2,2),\"ft/s (to the left)\",'and that of 3lb block is',round(v3,2),\"ft/s (to the right) respectively.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of 2lb block is -6.22 ft/s (to the left) and that of 3lb block is 4.14 ft/s (to the right) respectively.\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-34, Page No 445"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#Here the integration is indefinite hence it will be computed manually and entered\n",
- "W=10 #lb\n",
- "l=4 #ft\n",
- "w=2 #rad/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Part (a)\n",
- "wf=1.5 #rad/s\n",
- "t=(((W/g)*(l*w*l))-((W/g)*(l*wf*l)))**0.5 #s\n",
- "#Part (b)\n",
- "#Applying conservation of angular momentum\n",
- "r=(l*wf*l)*(l*w)**-1 #ft\n",
- "\n",
- "#Result\n",
- "print'The answer for part (a) is',round(t,2),\"s\",'and the answer for part (b) is',round(r,2),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The answer for part (a) is 1.58 s and the answer for part (b) is 3.0 ft\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-35, Page No 446"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=2.5 #lb\n",
- "w=36 #rad/s\n",
- "Idisk=0.4 #slug-ft**2\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "Ii=Idisk+(2*(W*g**-1)*(3*12**-1)**2) #slug-ft**2\n",
- "If=Idisk+(2*(W*g**-1)*(11*12**-1)**2) #slug-ft**2\n",
- "#Since no external moments act\n",
- "#Applying conservation of momentum\n",
- "wf=(Ii*w)*If**-1 #rad/s\n",
- "\n",
- "#Result\n",
- "print'The final angular speed is',round(wf,1),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The final angular speed is 27.8 rad/s\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-39, Page No 448"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "u1=6 #ft/s\n",
- "u2=-8 #ft/s\n",
- "e=0.8 #coefficient of restitution\n",
- "\n",
- "#Calculations\n",
- "#Solving both simultaneous equations\n",
- "A=np.array([[1,-1],[1,1]])\n",
- "B=np.array([[11.2],[-2]])\n",
- "C=np.linalg.solve(A,B) #ft/s\n",
- "\n",
- "#Result\n",
- "print'The velocities are v1=',round(C[1],1),\"ft/s\",'and v2=',round(C[0],1),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocities are v1= -6.6 ft/s and v2= 4.6 ft/s\n"
- ]
- }
- ],
- "prompt_number": 52
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-40, Page No 448"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "h1=20 #m\n",
- "h2=14 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "u1=sqrt(2*g*h1) #m/s\n",
- "u2=0 #m/s\n",
- "v1=-sqrt(2*g*h2) #m/s\n",
- "v2=0 #m/s\n",
- "e=(v2-v1)/(u1-u2) #coefficient of restitution\n",
- "\n",
- "#Result\n",
- "print'The value of coefficient of restitution is',round(e,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of coefficient of restitution is 0.84\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-41, Page No 448"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "u=6.55 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "L=6 #ft\n",
- "W=5 #lb\n",
- "c=0.7 #fraction of impulse acting in second phase\n",
- "\n",
- "#Calculations\n",
- "#Impulse\n",
- "I=(W*g**-1)*(u*3**-1) #N.s\n",
- "#Second Phase\n",
- "v=((-c*10.9)/2)+u #ft/s\n",
- "wprime=(1.09*60+c*(W*u*(3**-1)*6))/60 #rad/s\n",
- "\n",
- "#Result\n",
- "print'The value is v=',round(v,2),\"ft/s to the right\",' and that of w is',round(wprime,2),\"rad/s counterclockwise.\"\n",
- "\n",
- "#The value of w is incorrect in the textbook\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value is v= 2.73 ft/s to the right and that of w is 1.85 rad/s counterclockwise.\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-42, Page No 449"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=9 #kg\n",
- "m2=5.5 #kg\n",
- "u1=-3 #m/s\n",
- "u2=1.77 #m/s\n",
- "e=0.8 #coefficient of restitution\n",
- "\n",
- "#Calculations\n",
- "#Solving by matrix method after we get the two equations\n",
- "A=np.array([[-1,1],[m1,m2]])\n",
- "B=np.array([[(e*u1-e*u2)],[m1*u1+m2*u2]])\n",
- "C=np.linalg.solve(A,B) #m/s\n",
- "\n",
- "#Result\n",
- "print'The 9kg ball will rebound up the speed of',round(C[0],2),\"m/s\",'and the 5.5 kg ball will move to the right and down with components of',round(u2,2),\"m/s\",'and',round(-C[1],2),\"m/s respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The 9kg ball will rebound up the speed of 0.26 m/s and the 5.5 kg ball will move to the right and down with components of 1.77 m/s and 3.56 m/s respectively.\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-46, Page No 451"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v=4 #m/s\n",
- "m=9 #kg\n",
- "s=1.5 #m\n",
- "\n",
- "#Calculations\n",
- "Io=(2*3**-1)*(m*s**2) #kg.m\n",
- "w=(m*v*s*0.5)/Io #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular velocity of the box is',round(w),\"rad/s clockwise.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity of the box is 2.0 rad/s clockwise.\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-48, Page No 452"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=2000 #kg\n",
- "mf=8500 #kg\n",
- "vr=2000 #m/s\n",
- "a=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying Newtons Second Law\n",
- "dm_dt=-(-((m+mf)*a)-(m+mf)*a)/(-vr) #kg/s\n",
- "\n",
- "#Result\n",
- "print'dm/dt=',round(dm_dt),\"kg/s\"\n",
- "#The negative sign indicates the loss in the mass of the system\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "dm/dt= -103.0 kg/s\n"
- ]
- }
- ],
- "prompt_number": 72
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-50, Page No 453"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=4000 #lb\n",
- "k=3 #ft\n",
- "wp=(60**-1)*2*pi #rad/s\n",
- "ws=(300/60)*2*pi #rad/s\n",
- "d=3.5 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "I=(W/g)*k**2 #slug-ft**2\n",
- "M=I*ws*wp #lb-ft\n",
- "#Now equating M to Rf-Rr gives one equations and vertical sum yields other\n",
- "#solving them by matrix method\n",
- "A=np.array([[1,-1],[1,1]])\n",
- "B=np.array([[M*(2/d)],[W]])\n",
- "C=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The weight of Rf and Rr are',round(C[0]),\"lb\",'and',round(C[1]),\"lb respectively.\"\n",
- "\n",
- "# The answers in the textbook are incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The weight of Rf and Rr are 3051.0 lb and 949.0 lb respectively.\n"
- ]
- }
- ],
- "prompt_number": 74
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-51, Page No 454"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#As the integration is indefinite we will directly consider the equation with R\n",
- "#Initillization of variables\n",
- "GM=1.41*10**16 #ft**3/s**2\n",
- "r=2640000 #ft\n",
- "theta=60 #degrees\n",
- "R=21120000 #ft\n",
- "\n",
- "#Calculations\n",
- "v1=sqrt((GM*((R**-1)-((R+r)**-1)))/2.031) #ft/s\n",
- "\n",
- "#Result\n",
- "print'The speed required will be',round(v1),\"ft/s\"\n",
- "\n",
- "# Answer may wary due to decimal point discrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed required will be 6043.0 ft/s\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-52, Page No 454"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "k=4 #lb/ft\n",
- "so=1 #ft\n",
- "W=2**-1 #lb\n",
- "g=32.2 #ft/s**2\n",
- "vo=5 #ft/s\n",
- "\n",
- "#Calculations\n",
- "m=W/g #kg\n",
- "#Angular momentum is conserved\n",
- "v=sqrt((0.5*k*so**2*2*2*g)+vo**2) #ft/s\n",
- "#Using vd=15\n",
- "d=15/v #ft\n",
- "\n",
- "#Result\n",
- "print'The ball passes',round(d,2),\"ft close to the fixed pin\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ball passes 0.89 ft close to the fixed pin\n"
- ]
- }
- ],
- "prompt_number": 26
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter18_3.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter18_3.ipynb deleted file mode 100755 index bbd27129..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter18_3.ipynb +++ /dev/null @@ -1,1503 +0,0 @@ -{
- "metadata": {
- "name": "chapter18.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 18: Impulse and Momentum"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-8, Page No 435"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=100 #lb\n",
- "u=0.2 #coefficient of friction\n",
- "t=5 #s\n",
- "v1=5 #ft/s\n",
- "v2=10 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "ll=0 #lower limit of integration\n",
- "ul=5 #upper limit of integration\n",
- "\n",
- "#Calculations\n",
- "Fr=u*W #lb\n",
- "#Using The impulse momentum theorem\n",
- "#Since the integration is just subtraction of limits we can skip that\n",
- "F=((W/g)*v2-(W/g)*v1+Fr*ul)/ul #lb\n",
- "\n",
- "#Result\n",
- "print'The Force is',round(F,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Force is 23.1 lb\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-9, Page No 435"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=10 #kg\n",
- "#theta=30,thus\n",
- "cos30=sqrt(3)*2**-1\n",
- "sin30=2**-1\n",
- "u=0.3 #coefficient of kinetic friction\n",
- "t=5 #s\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Asthere is no motion in the vertical direction \n",
- "#Summing forces along vertical direction\n",
- "Na=m*g*cos30 #N\n",
- "#Using impulse momentum theorem\n",
- "vx=(m*g*sin30-u*Na)*(t*m**-1) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed after 5s is',round(vx,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed after 5s is 11.8 m/s\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-10, Page No 436"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(t, a, b):\n",
- " return 20*t-16\n",
- "a=1\n",
- "b=1\n",
- "F=quad(integrand, 1, 5, args=(a,b))\n",
- "g=32.2 # ft/s**2\n",
- "v=(F[0]*32.2)/80\n",
- "\n",
- "# Results\n",
- "print'The speed of the block is',round(v,1),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the block is 70.8 ft/s\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-12, Page No 436"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=40 #kg\n",
- "m2=10 #kg\n",
- "m3=15 #kg\n",
- "v0=2.5 #m/s\n",
- "vf=5 #m/s\n",
- "t=12 #s\n",
- "u=0.1 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "#theta=45 degrees,thus\n",
- "sin45=(sqrt(2))**-1\n",
- "cos45=(sqrt(2))**-1\n",
- "\n",
- "#Calculations\n",
- "#Applying Impulse-Momentum Theoroem\n",
- "P=(((m1+m2+m3)*(vf-v0))+(t*(-m2*g*sin45+u*g*m2*cos45+u*g*m3+g*m1)))*t**-1 #N\n",
- "\n",
- "#Result\n",
- "print'The value of P is',round(P),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P is 358.0 N\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-13, Page Noo 437"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W1=4 #lb\n",
- "W2=2 #lb\n",
- "t2=0.04 #s\n",
- "W3=-2 #lb\n",
- "t3=0.02 #s\n",
- "t=3 #s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Algebraic sum of two areas\n",
- "A=t2*W2+t3*W3 #lb-s\n",
- "#Using Impulse Momentum Theorem\n",
- "v=(A*g)/W1 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The spped after 3s is',round(v,3),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The spped after 3s is 0.322 ft/s\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-14, Page No 437"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "f_r=1 #in/s rate of fall of mercury\n",
- "ll=18 #in length of left column\n",
- "lr=22 #in length of right column\n",
- "rho=850 #lb/ft**3\n",
- "d=4**-1 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying Impulse momentum theorem\n",
- "M=((d*pi*d**2*4)/12**3)*(rho/g)*(12**-1) #lb-s\n",
- "\n",
- "#Result\n",
- "print'The upward momentum of mercury is +',round(M,5),\"lb-s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The upward momentum of mercury is + 0.00025 lb-s\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-15, Page No 437"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=2000 #kg\n",
- "k=1.200 #m\n",
- "w=120 #rpm\n",
- "t=200 #s\n",
- "\n",
- "#Calculations\n",
- "#Applying Angular Momentum theorem\n",
- "M=((m*k**2*(w*2*pi))/60)/t #N.m\n",
- "\n",
- "#Result\n",
- "print'The Momentum necessary is',round(M),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Momentum necessary is 181.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-17, Page No 438"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=5 #kg\n",
- "m2=7 #kg\n",
- "mp=5 #kg\n",
- "r=0.6 #m\n",
- "k=0.45 #m\n",
- "vi=3 #m/s\n",
- "vf=6 #m/s\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "I=m1*k**2 #kg.m**2\n",
- "wnet=(vf/r)-(vi/r) #rad/s\n",
- "#Solving the system of linear equations\n",
- "#Simplfying the equation we get\n",
- "t=((I*wnet)+m1*(vf-vi)+m2*(vf-vi))*r/(r*(m2-m1)*g) #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t,2),\"s\"\n",
- "# The ans in the textbook is incorrect.\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 2.1 s\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-18, Page No 438"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=50 #kg\n",
- "vo=4 #m/s\n",
- "vf=8 #m/s\n",
- "t=6 #s\n",
- "g=9.8 #m/s**2\n",
- "r=0.8 #m\n",
- "u=0.25 #coefficient of friction\n",
- "I=30 #kg-m**2\n",
- "\n",
- "#Calculations\n",
- "Na=m*g #N\n",
- "F=u*Na #N\n",
- "#Angular Speeds\n",
- "wo=vo/r #rad/s\n",
- "wf=vf/r #rad/s\n",
- "#Applying impulse momentum theorem\n",
- "mb=(I*wo+m*vo*r-F*r*t-I*wf-m*vf*r)/(vf*r-vo*r-g*r*t) #kg\n",
- "\n",
- "#Result\n",
- "print'The mass of block B is',round(mb,1),\"kg\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The mass of block B is 20.5 kg\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-19, Page No 439"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Ws=250 #lb\n",
- "Wl=500 #lb\n",
- "W3=161 #lb\n",
- "W4=64.4 #lb\n",
- "wo=100 #rpm\n",
- "wf=300 #rpm\n",
- "rl=3 #ft\n",
- "rs=2 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment Of Inertia\n",
- "I=(0.5*(Wl/g)*rl**2+0.5*(Ws/g)*rs**2) #slug-ft**2\n",
- "#Change in angular Momentum\n",
- "change1=I*((wf-wo)*2*(pi/60)) #lb-s-ft\n",
- "#Change in angular momentum about G for 161lb\n",
- "change2=2*((W3*g**-1)*(wf-wo)*(4*60**-1)*pi) #lb-s-ft\n",
- "#Similarly change in 64lb is\n",
- "change3=3*((W4*g**-1)*(wf-wo)*(6*60**-1)*pi) #lb-s-ft\n",
- "#Change in linear impulse\n",
- "#Without t term in it\n",
- "m1=2*W3\n",
- "m2=-3*W4\n",
- "#Total angular impulse\n",
- "t=(change1+change2+change3)/(m1+m2) #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 20.0 s\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-21, Page No 440"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "d=3 #ft\n",
- "W=300 #lb\n",
- "# as theta=20 degrees\n",
- "sintheta=0.342\n",
- "F=250 #lb\n",
- "t=6 #s\n",
- "vo=0 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying linear impulse momentum theorem\n",
- "#Solving by matrix method\n",
- "A=np.array([[-W*g**-1,1*t],[-((0.5)*(W*g**-1)*(d*2**-1)**2*(d*2**-1)**-1),(-t)*d*2**-1]])\n",
- "B=np.array([[-F*t+W*sintheta*t],[-F*(d*2**-1)*6]]) \n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The speed after 6s is',round(C[0]),\"ft/s,parallel to the plane\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed after 6s is 171.0 ft/s,parallel to the plane\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-22, Page No 440"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Initilization of variables\n",
- "#theta=30 degrees\n",
- "sin30=2**-1\n",
- "vo=20 #ft/s\n",
- "r=4 #ft\n",
- "vf=0 #ft/s\n",
- "W=300 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "wo=vo*r**-1 #rad/s\n",
- "wf=vf*r**-1 #rad/s\n",
- "#Applying impulse momentum theorem\n",
- "#Solving simultaneous equations\n",
- "t=-((W*g**-1)*(vf-vo)+((0.5*(W*g**-1)*r**2*(wf-wo))*r**-1))/(W*sin30) #s\n",
- "\n",
- "#Result\n",
- "print'The time t is',round(t,2),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time t is 1.86 s\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-23, Page No 441"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mw=75 #kg\n",
- "k=0.9 #m\n",
- "wi=10 #rad/s\n",
- "wf=6 #rad/s\n",
- "r=1.2 #m\n",
- "m=30 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Initial speed\n",
- "vi=-r*wi #m/s\n",
- "vf=r*wf #m/s\n",
- "#Initial speed of B is\n",
- "vib=-0.8*wi+vi #m/s\n",
- "#Similarly\n",
- "vfb=12 #m/s\n",
- "#Applying impulse momentum theorem\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,-1,-(mw*(vf-vi))],[0.8,1.2,-(mw*(k**2)*(wf+wi))],[-1,0,-(m*(vfb-vi))]])\n",
- "B=np.array([[0],[0],[-g*m]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Here t is calculated as 1/t for simplicity\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(C[2]**-1,2),\"s\" \n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 7.04 s\n"
- ]
- }
- ],
- "prompt_number": 59
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-24, Page No 441"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=8 #in\n",
- "W=96.6 #lb\n",
- "w=36 #rad/s\n",
- "u=0.15 #coefficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "r=(d/2)*12**-1 #m\n",
- "N=W #lb\n",
- "F=u*N #lb\n",
- "m=W/g #slugs\n",
- "I=0.5*m*(r**2) #slug-ft**2\n",
- "#Applying the impulse momentum theorem\n",
- "#Solving the two equations simultaneously\n",
- "A=np.array([[F,-m],[F*r,I*(1*r**-1)]])\n",
- "B=np.array([[0],[w*I]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Distance travelled\n",
- "s=0.5*C[1]*C[0] #ft\n",
- "t=C[0] #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t,2),\"s\",'and it travels',round(s,2),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 0.83 s and it travels 1.66 ft\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-25, Page No 442"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilisation of variables\n",
- "d=2*12**-1 #ft\n",
- "v=80 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Mass flow reate without time\n",
- "m=(1*4**-1)*pi*d**2*v*(62.4/g) \n",
- "#Let P=force of plate on mass m of water\n",
- "P=m*(0-v) #lb\n",
- "\n",
- "#Result\n",
- "print'The force water exerts on the plate is +',round(-P),\"lb,that is,to the right\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force water exerts on the plate is + 271.0 lb,that is,to the right\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-26, Page No 442"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v1=20 #ft/s\n",
- "vw=80 #ft/s\n",
- "d=2*12**-1 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "v=vw-v1 #ft/s\n",
- "#mass flow rate without t\n",
- "m=(4**-1)*(pi*d**2)*(62.4/g)*v\n",
- "#Applying impulse momentum theorem\n",
- "P=m*v #lb\n",
- "\n",
- "#Result\n",
- "print'The force exerted by water on the plate is +',round(P),\"lb,that is,to the right\"\n",
- "\n",
- "# Decimal poinat accuracy causes a small discrepancy in the answer"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force exerted by water on the plate is + 152.0 lb,that is,to the right\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-27, Page No 443"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "A=2000*10**-6 #m**2\n",
- "v=10 #m/s\n",
- "rho=1000 #kg/m**3\n",
- "#theta=45 degrees,thus\n",
- "cos45=(2**0.5)**-1\n",
- "sin45=(2**0.5)**-1\n",
- "\n",
- "#Calculations\n",
- "#Mass flow \n",
- "m=A*v*rho\n",
- "#Applying impulse momentum theorem\n",
- "Px=m*(-v*cos45-v) #N\n",
- "Py=m*(v*sin45-0) #N\n",
- "\n",
- "#Result\n",
- "print'The x component of force is',round(Px),\"N (to left on water)\",'and y component is +',round(Py),\"N (up on the water)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The x component of force is -341.0 N (to left on water) and y component is + 141.0 N (up on the water)\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-28, Page No 443"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilisation of variables\n",
- "W=150 #lb\n",
- "v=20 #ft/s\n",
- "A=0.2 #in**2\n",
- "t=60 #s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Mass flow\n",
- "m=(A/12**2)*v*(62.4/g)\n",
- "#Force\n",
- "F=m*(0-v) #lb\n",
- "#At t=60s the tank holds\n",
- "Ww=(A/12**2)*v*t*62.4 #lb\n",
- "#Total reading on scale\n",
- "S=-F+W+Ww #lb\n",
- "\n",
- "#Result\n",
- "print'The scale reading is',round(S),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The scale reading is 255.0 lb\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-29, Page No 444"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wp=130 #lb\n",
- "Wb=150 #lb\n",
- "Wbullet=2*16**-1 #lb\n",
- "g=32.2 #ft/s**2\n",
- "vbullet=1200 #ft/s\n",
- "\n",
- "#Calculations\n",
- "v=((-Wbullet*g**-1)*vbullet)/((Wb+Wp)/g) #ft/s\n",
- "\n",
- "#Result\n",
- "print'The speed of the boat is',round(v,2),\"ft/s\"\n",
- "#Negative sign indicates direction opposite to that of bullet\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the boat is -0.54 ft/s\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-30, Page No 444"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "mb=0.06 #kg\n",
- "ms=50 #kg\n",
- "h=0.03 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Speed of bag+bullet\n",
- "v2=sqrt(2*g*h) #m/s\n",
- "#Applying conservation of momentum \n",
- "v1=((mb+ms)*v2)/mb #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed of bullet as it entered the bag was',round(v1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of bullet as it entered the bag was 640.0 m/s\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-32, Page No 445"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "mb=0.06 #kg\n",
- "vb=500 #m/s\n",
- "mblock=5 #kg\n",
- "vblock=30 #m/s\n",
- "\n",
- "#Calculations\n",
- "#Applying conservation of momentum\n",
- "v=(mb*vb+mblock*vblock)/(mb+mblock) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed of the system is',round(v,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the system is 35.6 m/s\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-33, Page No 445"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "W1=2 #lb\n",
- "W2=3 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return 12*(2-x)\n",
- "a=1\n",
- "b=1\n",
- "W=quad(integrand, 0, 2, args=(a,b))\n",
- "Work=W[0]/12 # ft-lb\n",
- "\n",
- "# Solving the simultaneousequations\n",
- "v3=(Work*(0.5*(W2*g**-1)+0.5*(W1*g**-1)*(-W2*W1**-1)**2)**-1)**0.5 #ft/s\n",
- "v2=-(W2*W1**-1)*v3 #ft/s\n",
- "\n",
- "# Results\n",
- "print'The speed of 2lb block is',round(v2,2),\"ft/s (to the left)\",'and that of 3lb block is',round(v3,2),\"ft/s (to the right) respectively.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of 2lb block is -6.22 ft/s (to the left) and that of 3lb block is 4.14 ft/s (to the right) respectively.\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-34, Page No 445"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#Here the integration is indefinite hence it will be computed manually and entered\n",
- "W=10 #lb\n",
- "l=4 #ft\n",
- "w=2 #rad/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Part (a)\n",
- "wf=1.5 #rad/s\n",
- "t=(((W/g)*(l*w*l))-((W/g)*(l*wf*l)))**0.5 #s\n",
- "#Part (b)\n",
- "#Applying conservation of angular momentum\n",
- "r=(l*wf*l)*(l*w)**-1 #ft\n",
- "\n",
- "#Result\n",
- "print'The answer for part (a) is',round(t,2),\"s\",'and the answer for part (b) is',round(r,2),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The answer for part (a) is 1.58 s and the answer for part (b) is 3.0 ft\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-35, Page No 446"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=2.5 #lb\n",
- "w=36 #rad/s\n",
- "Idisk=0.4 #slug-ft**2\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "Ii=Idisk+(2*(W*g**-1)*(3*12**-1)**2) #slug-ft**2\n",
- "If=Idisk+(2*(W*g**-1)*(11*12**-1)**2) #slug-ft**2\n",
- "#Since no external moments act\n",
- "#Applying conservation of momentum\n",
- "wf=(Ii*w)*If**-1 #rad/s\n",
- "\n",
- "#Result\n",
- "print'The final angular speed is',round(wf,1),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The final angular speed is 27.8 rad/s\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-39, Page No 448"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "u1=6 #ft/s\n",
- "u2=-8 #ft/s\n",
- "e=0.8 #coefficient of restitution\n",
- "\n",
- "#Calculations\n",
- "#Solving both simultaneous equations\n",
- "A=np.array([[1,-1],[1,1]])\n",
- "B=np.array([[11.2],[-2]])\n",
- "C=np.linalg.solve(A,B) #ft/s\n",
- "\n",
- "#Result\n",
- "print'The velocities are v1=',round(C[1],1),\"ft/s\",'and v2=',round(C[0],1),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocities are v1= -6.6 ft/s and v2= 4.6 ft/s\n"
- ]
- }
- ],
- "prompt_number": 52
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-40, Page No 448"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "h1=20 #m\n",
- "h2=14 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "u1=sqrt(2*g*h1) #m/s\n",
- "u2=0 #m/s\n",
- "v1=-sqrt(2*g*h2) #m/s\n",
- "v2=0 #m/s\n",
- "e=(v2-v1)/(u1-u2) #coefficient of restitution\n",
- "\n",
- "#Result\n",
- "print'The value of coefficient of restitution is',round(e,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of coefficient of restitution is 0.84\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-41, Page No 448"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "u=6.55 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "L=6 #ft\n",
- "W=5 #lb\n",
- "c=0.7 #fraction of impulse acting in second phase\n",
- "\n",
- "#Calculations\n",
- "#Impulse\n",
- "I=(W*g**-1)*(u*3**-1) #N.s\n",
- "#Second Phase\n",
- "v=((-c*10.9)/2)+u #ft/s\n",
- "wprime=(1.09*60+c*(W*u*(3**-1)*6))/60 #rad/s\n",
- "\n",
- "#Result\n",
- "print'The value is v=',round(v,2),\"ft/s to the right\",' and that of w is',round(wprime,2),\"rad/s counterclockwise.\"\n",
- "\n",
- "#The value of w is incorrect in the textbook\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value is v= 2.73 ft/s to the right and that of w is 1.85 rad/s counterclockwise.\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-42, Page No 449"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=9 #kg\n",
- "m2=5.5 #kg\n",
- "u1=-3 #m/s\n",
- "u2=1.77 #m/s\n",
- "e=0.8 #coefficient of restitution\n",
- "\n",
- "#Calculations\n",
- "#Solving by matrix method after we get the two equations\n",
- "A=np.array([[-1,1],[m1,m2]])\n",
- "B=np.array([[(e*u1-e*u2)],[m1*u1+m2*u2]])\n",
- "C=np.linalg.solve(A,B) #m/s\n",
- "\n",
- "#Result\n",
- "print'The 9kg ball will rebound up the speed of',round(C[0],2),\"m/s\",'and the 5.5 kg ball will move to the right and down with components of',round(u2,2),\"m/s\",'and',round(-C[1],2),\"m/s respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The 9kg ball will rebound up the speed of 0.26 m/s and the 5.5 kg ball will move to the right and down with components of 1.77 m/s and 3.56 m/s respectively.\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-46, Page No 451"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v=4 #m/s\n",
- "m=9 #kg\n",
- "s=1.5 #m\n",
- "\n",
- "#Calculations\n",
- "Io=(2*3**-1)*(m*s**2) #kg.m\n",
- "w=(m*v*s*0.5)/Io #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular velocity of the box is',round(w),\"rad/s clockwise.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity of the box is 2.0 rad/s clockwise.\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-48, Page No 452"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=2000 #kg\n",
- "mf=8500 #kg\n",
- "vr=2000 #m/s\n",
- "a=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying Newtons Second Law\n",
- "dm_dt=-(-((m+mf)*a)-(m+mf)*a)/(-vr) #kg/s\n",
- "\n",
- "#Result\n",
- "print'dm/dt=',round(dm_dt),\"kg/s\"\n",
- "#The negative sign indicates the loss in the mass of the system\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "dm/dt= -103.0 kg/s\n"
- ]
- }
- ],
- "prompt_number": 72
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-50, Page No 453"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=4000 #lb\n",
- "k=3 #ft\n",
- "wp=(60**-1)*2*pi #rad/s\n",
- "ws=(300/60)*2*pi #rad/s\n",
- "d=3.5 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "I=(W/g)*k**2 #slug-ft**2\n",
- "M=I*ws*wp #lb-ft\n",
- "#Now equating M to Rf-Rr gives one equations and vertical sum yields other\n",
- "#solving them by matrix method\n",
- "A=np.array([[1,-1],[1,1]])\n",
- "B=np.array([[M*(2/d)],[W]])\n",
- "C=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The weight of Rf and Rr are',round(C[0]),\"lb\",'and',round(C[1]),\"lb respectively.\"\n",
- "\n",
- "# The answers in the textbook are incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The weight of Rf and Rr are 3051.0 lb and 949.0 lb respectively.\n"
- ]
- }
- ],
- "prompt_number": 74
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-51, Page No 454"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#As the integration is indefinite we will directly consider the equation with R\n",
- "#Initillization of variables\n",
- "GM=1.41*10**16 #ft**3/s**2\n",
- "r=2640000 #ft\n",
- "theta=60 #degrees\n",
- "R=21120000 #ft\n",
- "\n",
- "#Calculations\n",
- "v1=sqrt((GM*((R**-1)-((R+r)**-1)))/2.031) #ft/s\n",
- "\n",
- "#Result\n",
- "print'The speed required will be',round(v1),\"ft/s\"\n",
- "\n",
- "# Answer may wary due to decimal point discrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed required will be 6043.0 ft/s\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-52, Page No 454"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "k=4 #lb/ft\n",
- "so=1 #ft\n",
- "W=2**-1 #lb\n",
- "g=32.2 #ft/s**2\n",
- "vo=5 #ft/s\n",
- "\n",
- "#Calculations\n",
- "m=W/g #kg\n",
- "#Angular momentum is conserved\n",
- "v=sqrt((0.5*k*so**2*2*2*g)+vo**2) #ft/s\n",
- "#Using vd=15\n",
- "d=15/v #ft\n",
- "\n",
- "#Result\n",
- "print'The ball passes',round(d,2),\"ft close to the fixed pin\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ball passes 0.89 ft close to the fixed pin\n"
- ]
- }
- ],
- "prompt_number": 26
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter18_4.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter18_4.ipynb deleted file mode 100755 index bbd27129..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter18_4.ipynb +++ /dev/null @@ -1,1503 +0,0 @@ -{
- "metadata": {
- "name": "chapter18.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 18: Impulse and Momentum"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-8, Page No 435"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=100 #lb\n",
- "u=0.2 #coefficient of friction\n",
- "t=5 #s\n",
- "v1=5 #ft/s\n",
- "v2=10 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "ll=0 #lower limit of integration\n",
- "ul=5 #upper limit of integration\n",
- "\n",
- "#Calculations\n",
- "Fr=u*W #lb\n",
- "#Using The impulse momentum theorem\n",
- "#Since the integration is just subtraction of limits we can skip that\n",
- "F=((W/g)*v2-(W/g)*v1+Fr*ul)/ul #lb\n",
- "\n",
- "#Result\n",
- "print'The Force is',round(F,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Force is 23.1 lb\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-9, Page No 435"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=10 #kg\n",
- "#theta=30,thus\n",
- "cos30=sqrt(3)*2**-1\n",
- "sin30=2**-1\n",
- "u=0.3 #coefficient of kinetic friction\n",
- "t=5 #s\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Asthere is no motion in the vertical direction \n",
- "#Summing forces along vertical direction\n",
- "Na=m*g*cos30 #N\n",
- "#Using impulse momentum theorem\n",
- "vx=(m*g*sin30-u*Na)*(t*m**-1) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed after 5s is',round(vx,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed after 5s is 11.8 m/s\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-10, Page No 436"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(t, a, b):\n",
- " return 20*t-16\n",
- "a=1\n",
- "b=1\n",
- "F=quad(integrand, 1, 5, args=(a,b))\n",
- "g=32.2 # ft/s**2\n",
- "v=(F[0]*32.2)/80\n",
- "\n",
- "# Results\n",
- "print'The speed of the block is',round(v,1),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the block is 70.8 ft/s\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-12, Page No 436"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=40 #kg\n",
- "m2=10 #kg\n",
- "m3=15 #kg\n",
- "v0=2.5 #m/s\n",
- "vf=5 #m/s\n",
- "t=12 #s\n",
- "u=0.1 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "#theta=45 degrees,thus\n",
- "sin45=(sqrt(2))**-1\n",
- "cos45=(sqrt(2))**-1\n",
- "\n",
- "#Calculations\n",
- "#Applying Impulse-Momentum Theoroem\n",
- "P=(((m1+m2+m3)*(vf-v0))+(t*(-m2*g*sin45+u*g*m2*cos45+u*g*m3+g*m1)))*t**-1 #N\n",
- "\n",
- "#Result\n",
- "print'The value of P is',round(P),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P is 358.0 N\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-13, Page Noo 437"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W1=4 #lb\n",
- "W2=2 #lb\n",
- "t2=0.04 #s\n",
- "W3=-2 #lb\n",
- "t3=0.02 #s\n",
- "t=3 #s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Algebraic sum of two areas\n",
- "A=t2*W2+t3*W3 #lb-s\n",
- "#Using Impulse Momentum Theorem\n",
- "v=(A*g)/W1 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The spped after 3s is',round(v,3),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The spped after 3s is 0.322 ft/s\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-14, Page No 437"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "f_r=1 #in/s rate of fall of mercury\n",
- "ll=18 #in length of left column\n",
- "lr=22 #in length of right column\n",
- "rho=850 #lb/ft**3\n",
- "d=4**-1 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying Impulse momentum theorem\n",
- "M=((d*pi*d**2*4)/12**3)*(rho/g)*(12**-1) #lb-s\n",
- "\n",
- "#Result\n",
- "print'The upward momentum of mercury is +',round(M,5),\"lb-s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The upward momentum of mercury is + 0.00025 lb-s\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-15, Page No 437"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=2000 #kg\n",
- "k=1.200 #m\n",
- "w=120 #rpm\n",
- "t=200 #s\n",
- "\n",
- "#Calculations\n",
- "#Applying Angular Momentum theorem\n",
- "M=((m*k**2*(w*2*pi))/60)/t #N.m\n",
- "\n",
- "#Result\n",
- "print'The Momentum necessary is',round(M),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Momentum necessary is 181.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-17, Page No 438"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=5 #kg\n",
- "m2=7 #kg\n",
- "mp=5 #kg\n",
- "r=0.6 #m\n",
- "k=0.45 #m\n",
- "vi=3 #m/s\n",
- "vf=6 #m/s\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "I=m1*k**2 #kg.m**2\n",
- "wnet=(vf/r)-(vi/r) #rad/s\n",
- "#Solving the system of linear equations\n",
- "#Simplfying the equation we get\n",
- "t=((I*wnet)+m1*(vf-vi)+m2*(vf-vi))*r/(r*(m2-m1)*g) #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t,2),\"s\"\n",
- "# The ans in the textbook is incorrect.\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 2.1 s\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-18, Page No 438"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=50 #kg\n",
- "vo=4 #m/s\n",
- "vf=8 #m/s\n",
- "t=6 #s\n",
- "g=9.8 #m/s**2\n",
- "r=0.8 #m\n",
- "u=0.25 #coefficient of friction\n",
- "I=30 #kg-m**2\n",
- "\n",
- "#Calculations\n",
- "Na=m*g #N\n",
- "F=u*Na #N\n",
- "#Angular Speeds\n",
- "wo=vo/r #rad/s\n",
- "wf=vf/r #rad/s\n",
- "#Applying impulse momentum theorem\n",
- "mb=(I*wo+m*vo*r-F*r*t-I*wf-m*vf*r)/(vf*r-vo*r-g*r*t) #kg\n",
- "\n",
- "#Result\n",
- "print'The mass of block B is',round(mb,1),\"kg\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The mass of block B is 20.5 kg\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-19, Page No 439"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Ws=250 #lb\n",
- "Wl=500 #lb\n",
- "W3=161 #lb\n",
- "W4=64.4 #lb\n",
- "wo=100 #rpm\n",
- "wf=300 #rpm\n",
- "rl=3 #ft\n",
- "rs=2 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment Of Inertia\n",
- "I=(0.5*(Wl/g)*rl**2+0.5*(Ws/g)*rs**2) #slug-ft**2\n",
- "#Change in angular Momentum\n",
- "change1=I*((wf-wo)*2*(pi/60)) #lb-s-ft\n",
- "#Change in angular momentum about G for 161lb\n",
- "change2=2*((W3*g**-1)*(wf-wo)*(4*60**-1)*pi) #lb-s-ft\n",
- "#Similarly change in 64lb is\n",
- "change3=3*((W4*g**-1)*(wf-wo)*(6*60**-1)*pi) #lb-s-ft\n",
- "#Change in linear impulse\n",
- "#Without t term in it\n",
- "m1=2*W3\n",
- "m2=-3*W4\n",
- "#Total angular impulse\n",
- "t=(change1+change2+change3)/(m1+m2) #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 20.0 s\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-21, Page No 440"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "d=3 #ft\n",
- "W=300 #lb\n",
- "# as theta=20 degrees\n",
- "sintheta=0.342\n",
- "F=250 #lb\n",
- "t=6 #s\n",
- "vo=0 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying linear impulse momentum theorem\n",
- "#Solving by matrix method\n",
- "A=np.array([[-W*g**-1,1*t],[-((0.5)*(W*g**-1)*(d*2**-1)**2*(d*2**-1)**-1),(-t)*d*2**-1]])\n",
- "B=np.array([[-F*t+W*sintheta*t],[-F*(d*2**-1)*6]]) \n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The speed after 6s is',round(C[0]),\"ft/s,parallel to the plane\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed after 6s is 171.0 ft/s,parallel to the plane\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-22, Page No 440"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Initilization of variables\n",
- "#theta=30 degrees\n",
- "sin30=2**-1\n",
- "vo=20 #ft/s\n",
- "r=4 #ft\n",
- "vf=0 #ft/s\n",
- "W=300 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "wo=vo*r**-1 #rad/s\n",
- "wf=vf*r**-1 #rad/s\n",
- "#Applying impulse momentum theorem\n",
- "#Solving simultaneous equations\n",
- "t=-((W*g**-1)*(vf-vo)+((0.5*(W*g**-1)*r**2*(wf-wo))*r**-1))/(W*sin30) #s\n",
- "\n",
- "#Result\n",
- "print'The time t is',round(t,2),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time t is 1.86 s\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-23, Page No 441"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mw=75 #kg\n",
- "k=0.9 #m\n",
- "wi=10 #rad/s\n",
- "wf=6 #rad/s\n",
- "r=1.2 #m\n",
- "m=30 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Initial speed\n",
- "vi=-r*wi #m/s\n",
- "vf=r*wf #m/s\n",
- "#Initial speed of B is\n",
- "vib=-0.8*wi+vi #m/s\n",
- "#Similarly\n",
- "vfb=12 #m/s\n",
- "#Applying impulse momentum theorem\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,-1,-(mw*(vf-vi))],[0.8,1.2,-(mw*(k**2)*(wf+wi))],[-1,0,-(m*(vfb-vi))]])\n",
- "B=np.array([[0],[0],[-g*m]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Here t is calculated as 1/t for simplicity\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(C[2]**-1,2),\"s\" \n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 7.04 s\n"
- ]
- }
- ],
- "prompt_number": 59
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-24, Page No 441"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=8 #in\n",
- "W=96.6 #lb\n",
- "w=36 #rad/s\n",
- "u=0.15 #coefficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "r=(d/2)*12**-1 #m\n",
- "N=W #lb\n",
- "F=u*N #lb\n",
- "m=W/g #slugs\n",
- "I=0.5*m*(r**2) #slug-ft**2\n",
- "#Applying the impulse momentum theorem\n",
- "#Solving the two equations simultaneously\n",
- "A=np.array([[F,-m],[F*r,I*(1*r**-1)]])\n",
- "B=np.array([[0],[w*I]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Distance travelled\n",
- "s=0.5*C[1]*C[0] #ft\n",
- "t=C[0] #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t,2),\"s\",'and it travels',round(s,2),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 0.83 s and it travels 1.66 ft\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-25, Page No 442"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilisation of variables\n",
- "d=2*12**-1 #ft\n",
- "v=80 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Mass flow reate without time\n",
- "m=(1*4**-1)*pi*d**2*v*(62.4/g) \n",
- "#Let P=force of plate on mass m of water\n",
- "P=m*(0-v) #lb\n",
- "\n",
- "#Result\n",
- "print'The force water exerts on the plate is +',round(-P),\"lb,that is,to the right\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force water exerts on the plate is + 271.0 lb,that is,to the right\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-26, Page No 442"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v1=20 #ft/s\n",
- "vw=80 #ft/s\n",
- "d=2*12**-1 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "v=vw-v1 #ft/s\n",
- "#mass flow rate without t\n",
- "m=(4**-1)*(pi*d**2)*(62.4/g)*v\n",
- "#Applying impulse momentum theorem\n",
- "P=m*v #lb\n",
- "\n",
- "#Result\n",
- "print'The force exerted by water on the plate is +',round(P),\"lb,that is,to the right\"\n",
- "\n",
- "# Decimal poinat accuracy causes a small discrepancy in the answer"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force exerted by water on the plate is + 152.0 lb,that is,to the right\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-27, Page No 443"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "A=2000*10**-6 #m**2\n",
- "v=10 #m/s\n",
- "rho=1000 #kg/m**3\n",
- "#theta=45 degrees,thus\n",
- "cos45=(2**0.5)**-1\n",
- "sin45=(2**0.5)**-1\n",
- "\n",
- "#Calculations\n",
- "#Mass flow \n",
- "m=A*v*rho\n",
- "#Applying impulse momentum theorem\n",
- "Px=m*(-v*cos45-v) #N\n",
- "Py=m*(v*sin45-0) #N\n",
- "\n",
- "#Result\n",
- "print'The x component of force is',round(Px),\"N (to left on water)\",'and y component is +',round(Py),\"N (up on the water)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The x component of force is -341.0 N (to left on water) and y component is + 141.0 N (up on the water)\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-28, Page No 443"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilisation of variables\n",
- "W=150 #lb\n",
- "v=20 #ft/s\n",
- "A=0.2 #in**2\n",
- "t=60 #s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Mass flow\n",
- "m=(A/12**2)*v*(62.4/g)\n",
- "#Force\n",
- "F=m*(0-v) #lb\n",
- "#At t=60s the tank holds\n",
- "Ww=(A/12**2)*v*t*62.4 #lb\n",
- "#Total reading on scale\n",
- "S=-F+W+Ww #lb\n",
- "\n",
- "#Result\n",
- "print'The scale reading is',round(S),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The scale reading is 255.0 lb\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-29, Page No 444"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wp=130 #lb\n",
- "Wb=150 #lb\n",
- "Wbullet=2*16**-1 #lb\n",
- "g=32.2 #ft/s**2\n",
- "vbullet=1200 #ft/s\n",
- "\n",
- "#Calculations\n",
- "v=((-Wbullet*g**-1)*vbullet)/((Wb+Wp)/g) #ft/s\n",
- "\n",
- "#Result\n",
- "print'The speed of the boat is',round(v,2),\"ft/s\"\n",
- "#Negative sign indicates direction opposite to that of bullet\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the boat is -0.54 ft/s\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-30, Page No 444"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "mb=0.06 #kg\n",
- "ms=50 #kg\n",
- "h=0.03 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Speed of bag+bullet\n",
- "v2=sqrt(2*g*h) #m/s\n",
- "#Applying conservation of momentum \n",
- "v1=((mb+ms)*v2)/mb #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed of bullet as it entered the bag was',round(v1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of bullet as it entered the bag was 640.0 m/s\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-32, Page No 445"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "mb=0.06 #kg\n",
- "vb=500 #m/s\n",
- "mblock=5 #kg\n",
- "vblock=30 #m/s\n",
- "\n",
- "#Calculations\n",
- "#Applying conservation of momentum\n",
- "v=(mb*vb+mblock*vblock)/(mb+mblock) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed of the system is',round(v,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the system is 35.6 m/s\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-33, Page No 445"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "W1=2 #lb\n",
- "W2=3 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return 12*(2-x)\n",
- "a=1\n",
- "b=1\n",
- "W=quad(integrand, 0, 2, args=(a,b))\n",
- "Work=W[0]/12 # ft-lb\n",
- "\n",
- "# Solving the simultaneousequations\n",
- "v3=(Work*(0.5*(W2*g**-1)+0.5*(W1*g**-1)*(-W2*W1**-1)**2)**-1)**0.5 #ft/s\n",
- "v2=-(W2*W1**-1)*v3 #ft/s\n",
- "\n",
- "# Results\n",
- "print'The speed of 2lb block is',round(v2,2),\"ft/s (to the left)\",'and that of 3lb block is',round(v3,2),\"ft/s (to the right) respectively.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of 2lb block is -6.22 ft/s (to the left) and that of 3lb block is 4.14 ft/s (to the right) respectively.\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-34, Page No 445"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#Here the integration is indefinite hence it will be computed manually and entered\n",
- "W=10 #lb\n",
- "l=4 #ft\n",
- "w=2 #rad/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Part (a)\n",
- "wf=1.5 #rad/s\n",
- "t=(((W/g)*(l*w*l))-((W/g)*(l*wf*l)))**0.5 #s\n",
- "#Part (b)\n",
- "#Applying conservation of angular momentum\n",
- "r=(l*wf*l)*(l*w)**-1 #ft\n",
- "\n",
- "#Result\n",
- "print'The answer for part (a) is',round(t,2),\"s\",'and the answer for part (b) is',round(r,2),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The answer for part (a) is 1.58 s and the answer for part (b) is 3.0 ft\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-35, Page No 446"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=2.5 #lb\n",
- "w=36 #rad/s\n",
- "Idisk=0.4 #slug-ft**2\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "Ii=Idisk+(2*(W*g**-1)*(3*12**-1)**2) #slug-ft**2\n",
- "If=Idisk+(2*(W*g**-1)*(11*12**-1)**2) #slug-ft**2\n",
- "#Since no external moments act\n",
- "#Applying conservation of momentum\n",
- "wf=(Ii*w)*If**-1 #rad/s\n",
- "\n",
- "#Result\n",
- "print'The final angular speed is',round(wf,1),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The final angular speed is 27.8 rad/s\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-39, Page No 448"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "u1=6 #ft/s\n",
- "u2=-8 #ft/s\n",
- "e=0.8 #coefficient of restitution\n",
- "\n",
- "#Calculations\n",
- "#Solving both simultaneous equations\n",
- "A=np.array([[1,-1],[1,1]])\n",
- "B=np.array([[11.2],[-2]])\n",
- "C=np.linalg.solve(A,B) #ft/s\n",
- "\n",
- "#Result\n",
- "print'The velocities are v1=',round(C[1],1),\"ft/s\",'and v2=',round(C[0],1),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocities are v1= -6.6 ft/s and v2= 4.6 ft/s\n"
- ]
- }
- ],
- "prompt_number": 52
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-40, Page No 448"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "h1=20 #m\n",
- "h2=14 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "u1=sqrt(2*g*h1) #m/s\n",
- "u2=0 #m/s\n",
- "v1=-sqrt(2*g*h2) #m/s\n",
- "v2=0 #m/s\n",
- "e=(v2-v1)/(u1-u2) #coefficient of restitution\n",
- "\n",
- "#Result\n",
- "print'The value of coefficient of restitution is',round(e,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of coefficient of restitution is 0.84\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-41, Page No 448"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "u=6.55 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "L=6 #ft\n",
- "W=5 #lb\n",
- "c=0.7 #fraction of impulse acting in second phase\n",
- "\n",
- "#Calculations\n",
- "#Impulse\n",
- "I=(W*g**-1)*(u*3**-1) #N.s\n",
- "#Second Phase\n",
- "v=((-c*10.9)/2)+u #ft/s\n",
- "wprime=(1.09*60+c*(W*u*(3**-1)*6))/60 #rad/s\n",
- "\n",
- "#Result\n",
- "print'The value is v=',round(v,2),\"ft/s to the right\",' and that of w is',round(wprime,2),\"rad/s counterclockwise.\"\n",
- "\n",
- "#The value of w is incorrect in the textbook\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value is v= 2.73 ft/s to the right and that of w is 1.85 rad/s counterclockwise.\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-42, Page No 449"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=9 #kg\n",
- "m2=5.5 #kg\n",
- "u1=-3 #m/s\n",
- "u2=1.77 #m/s\n",
- "e=0.8 #coefficient of restitution\n",
- "\n",
- "#Calculations\n",
- "#Solving by matrix method after we get the two equations\n",
- "A=np.array([[-1,1],[m1,m2]])\n",
- "B=np.array([[(e*u1-e*u2)],[m1*u1+m2*u2]])\n",
- "C=np.linalg.solve(A,B) #m/s\n",
- "\n",
- "#Result\n",
- "print'The 9kg ball will rebound up the speed of',round(C[0],2),\"m/s\",'and the 5.5 kg ball will move to the right and down with components of',round(u2,2),\"m/s\",'and',round(-C[1],2),\"m/s respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The 9kg ball will rebound up the speed of 0.26 m/s and the 5.5 kg ball will move to the right and down with components of 1.77 m/s and 3.56 m/s respectively.\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-46, Page No 451"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v=4 #m/s\n",
- "m=9 #kg\n",
- "s=1.5 #m\n",
- "\n",
- "#Calculations\n",
- "Io=(2*3**-1)*(m*s**2) #kg.m\n",
- "w=(m*v*s*0.5)/Io #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular velocity of the box is',round(w),\"rad/s clockwise.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity of the box is 2.0 rad/s clockwise.\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-48, Page No 452"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=2000 #kg\n",
- "mf=8500 #kg\n",
- "vr=2000 #m/s\n",
- "a=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying Newtons Second Law\n",
- "dm_dt=-(-((m+mf)*a)-(m+mf)*a)/(-vr) #kg/s\n",
- "\n",
- "#Result\n",
- "print'dm/dt=',round(dm_dt),\"kg/s\"\n",
- "#The negative sign indicates the loss in the mass of the system\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "dm/dt= -103.0 kg/s\n"
- ]
- }
- ],
- "prompt_number": 72
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-50, Page No 453"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=4000 #lb\n",
- "k=3 #ft\n",
- "wp=(60**-1)*2*pi #rad/s\n",
- "ws=(300/60)*2*pi #rad/s\n",
- "d=3.5 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "I=(W/g)*k**2 #slug-ft**2\n",
- "M=I*ws*wp #lb-ft\n",
- "#Now equating M to Rf-Rr gives one equations and vertical sum yields other\n",
- "#solving them by matrix method\n",
- "A=np.array([[1,-1],[1,1]])\n",
- "B=np.array([[M*(2/d)],[W]])\n",
- "C=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The weight of Rf and Rr are',round(C[0]),\"lb\",'and',round(C[1]),\"lb respectively.\"\n",
- "\n",
- "# The answers in the textbook are incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The weight of Rf and Rr are 3051.0 lb and 949.0 lb respectively.\n"
- ]
- }
- ],
- "prompt_number": 74
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-51, Page No 454"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#As the integration is indefinite we will directly consider the equation with R\n",
- "#Initillization of variables\n",
- "GM=1.41*10**16 #ft**3/s**2\n",
- "r=2640000 #ft\n",
- "theta=60 #degrees\n",
- "R=21120000 #ft\n",
- "\n",
- "#Calculations\n",
- "v1=sqrt((GM*((R**-1)-((R+r)**-1)))/2.031) #ft/s\n",
- "\n",
- "#Result\n",
- "print'The speed required will be',round(v1),\"ft/s\"\n",
- "\n",
- "# Answer may wary due to decimal point discrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed required will be 6043.0 ft/s\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-52, Page No 454"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "k=4 #lb/ft\n",
- "so=1 #ft\n",
- "W=2**-1 #lb\n",
- "g=32.2 #ft/s**2\n",
- "vo=5 #ft/s\n",
- "\n",
- "#Calculations\n",
- "m=W/g #kg\n",
- "#Angular momentum is conserved\n",
- "v=sqrt((0.5*k*so**2*2*2*g)+vo**2) #ft/s\n",
- "#Using vd=15\n",
- "d=15/v #ft\n",
- "\n",
- "#Result\n",
- "print'The ball passes',round(d,2),\"ft close to the fixed pin\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ball passes 0.89 ft close to the fixed pin\n"
- ]
- }
- ],
- "prompt_number": 26
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter18_5.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter18_5.ipynb deleted file mode 100755 index bbd27129..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter18_5.ipynb +++ /dev/null @@ -1,1503 +0,0 @@ -{
- "metadata": {
- "name": "chapter18.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 18: Impulse and Momentum"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-8, Page No 435"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=100 #lb\n",
- "u=0.2 #coefficient of friction\n",
- "t=5 #s\n",
- "v1=5 #ft/s\n",
- "v2=10 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "ll=0 #lower limit of integration\n",
- "ul=5 #upper limit of integration\n",
- "\n",
- "#Calculations\n",
- "Fr=u*W #lb\n",
- "#Using The impulse momentum theorem\n",
- "#Since the integration is just subtraction of limits we can skip that\n",
- "F=((W/g)*v2-(W/g)*v1+Fr*ul)/ul #lb\n",
- "\n",
- "#Result\n",
- "print'The Force is',round(F,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Force is 23.1 lb\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-9, Page No 435"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=10 #kg\n",
- "#theta=30,thus\n",
- "cos30=sqrt(3)*2**-1\n",
- "sin30=2**-1\n",
- "u=0.3 #coefficient of kinetic friction\n",
- "t=5 #s\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Asthere is no motion in the vertical direction \n",
- "#Summing forces along vertical direction\n",
- "Na=m*g*cos30 #N\n",
- "#Using impulse momentum theorem\n",
- "vx=(m*g*sin30-u*Na)*(t*m**-1) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed after 5s is',round(vx,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed after 5s is 11.8 m/s\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-10, Page No 436"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(t, a, b):\n",
- " return 20*t-16\n",
- "a=1\n",
- "b=1\n",
- "F=quad(integrand, 1, 5, args=(a,b))\n",
- "g=32.2 # ft/s**2\n",
- "v=(F[0]*32.2)/80\n",
- "\n",
- "# Results\n",
- "print'The speed of the block is',round(v,1),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the block is 70.8 ft/s\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-12, Page No 436"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=40 #kg\n",
- "m2=10 #kg\n",
- "m3=15 #kg\n",
- "v0=2.5 #m/s\n",
- "vf=5 #m/s\n",
- "t=12 #s\n",
- "u=0.1 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "#theta=45 degrees,thus\n",
- "sin45=(sqrt(2))**-1\n",
- "cos45=(sqrt(2))**-1\n",
- "\n",
- "#Calculations\n",
- "#Applying Impulse-Momentum Theoroem\n",
- "P=(((m1+m2+m3)*(vf-v0))+(t*(-m2*g*sin45+u*g*m2*cos45+u*g*m3+g*m1)))*t**-1 #N\n",
- "\n",
- "#Result\n",
- "print'The value of P is',round(P),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P is 358.0 N\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-13, Page Noo 437"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W1=4 #lb\n",
- "W2=2 #lb\n",
- "t2=0.04 #s\n",
- "W3=-2 #lb\n",
- "t3=0.02 #s\n",
- "t=3 #s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Algebraic sum of two areas\n",
- "A=t2*W2+t3*W3 #lb-s\n",
- "#Using Impulse Momentum Theorem\n",
- "v=(A*g)/W1 #ft/s\n",
- "\n",
- "#Result\n",
- "print'The spped after 3s is',round(v,3),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The spped after 3s is 0.322 ft/s\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-14, Page No 437"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "f_r=1 #in/s rate of fall of mercury\n",
- "ll=18 #in length of left column\n",
- "lr=22 #in length of right column\n",
- "rho=850 #lb/ft**3\n",
- "d=4**-1 #in\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying Impulse momentum theorem\n",
- "M=((d*pi*d**2*4)/12**3)*(rho/g)*(12**-1) #lb-s\n",
- "\n",
- "#Result\n",
- "print'The upward momentum of mercury is +',round(M,5),\"lb-s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The upward momentum of mercury is + 0.00025 lb-s\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-15, Page No 437"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=2000 #kg\n",
- "k=1.200 #m\n",
- "w=120 #rpm\n",
- "t=200 #s\n",
- "\n",
- "#Calculations\n",
- "#Applying Angular Momentum theorem\n",
- "M=((m*k**2*(w*2*pi))/60)/t #N.m\n",
- "\n",
- "#Result\n",
- "print'The Momentum necessary is',round(M),\"N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Momentum necessary is 181.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-17, Page No 438"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=5 #kg\n",
- "m2=7 #kg\n",
- "mp=5 #kg\n",
- "r=0.6 #m\n",
- "k=0.45 #m\n",
- "vi=3 #m/s\n",
- "vf=6 #m/s\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "I=m1*k**2 #kg.m**2\n",
- "wnet=(vf/r)-(vi/r) #rad/s\n",
- "#Solving the system of linear equations\n",
- "#Simplfying the equation we get\n",
- "t=((I*wnet)+m1*(vf-vi)+m2*(vf-vi))*r/(r*(m2-m1)*g) #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t,2),\"s\"\n",
- "# The ans in the textbook is incorrect.\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 2.1 s\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-18, Page No 438"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=50 #kg\n",
- "vo=4 #m/s\n",
- "vf=8 #m/s\n",
- "t=6 #s\n",
- "g=9.8 #m/s**2\n",
- "r=0.8 #m\n",
- "u=0.25 #coefficient of friction\n",
- "I=30 #kg-m**2\n",
- "\n",
- "#Calculations\n",
- "Na=m*g #N\n",
- "F=u*Na #N\n",
- "#Angular Speeds\n",
- "wo=vo/r #rad/s\n",
- "wf=vf/r #rad/s\n",
- "#Applying impulse momentum theorem\n",
- "mb=(I*wo+m*vo*r-F*r*t-I*wf-m*vf*r)/(vf*r-vo*r-g*r*t) #kg\n",
- "\n",
- "#Result\n",
- "print'The mass of block B is',round(mb,1),\"kg\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The mass of block B is 20.5 kg\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-19, Page No 439"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Ws=250 #lb\n",
- "Wl=500 #lb\n",
- "W3=161 #lb\n",
- "W4=64.4 #lb\n",
- "wo=100 #rpm\n",
- "wf=300 #rpm\n",
- "rl=3 #ft\n",
- "rs=2 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment Of Inertia\n",
- "I=(0.5*(Wl/g)*rl**2+0.5*(Ws/g)*rs**2) #slug-ft**2\n",
- "#Change in angular Momentum\n",
- "change1=I*((wf-wo)*2*(pi/60)) #lb-s-ft\n",
- "#Change in angular momentum about G for 161lb\n",
- "change2=2*((W3*g**-1)*(wf-wo)*(4*60**-1)*pi) #lb-s-ft\n",
- "#Similarly change in 64lb is\n",
- "change3=3*((W4*g**-1)*(wf-wo)*(6*60**-1)*pi) #lb-s-ft\n",
- "#Change in linear impulse\n",
- "#Without t term in it\n",
- "m1=2*W3\n",
- "m2=-3*W4\n",
- "#Total angular impulse\n",
- "t=(change1+change2+change3)/(m1+m2) #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 20.0 s\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-21, Page No 440"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "d=3 #ft\n",
- "W=300 #lb\n",
- "# as theta=20 degrees\n",
- "sintheta=0.342\n",
- "F=250 #lb\n",
- "t=6 #s\n",
- "vo=0 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying linear impulse momentum theorem\n",
- "#Solving by matrix method\n",
- "A=np.array([[-W*g**-1,1*t],[-((0.5)*(W*g**-1)*(d*2**-1)**2*(d*2**-1)**-1),(-t)*d*2**-1]])\n",
- "B=np.array([[-F*t+W*sintheta*t],[-F*(d*2**-1)*6]]) \n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The speed after 6s is',round(C[0]),\"ft/s,parallel to the plane\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed after 6s is 171.0 ft/s,parallel to the plane\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-22, Page No 440"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Initilization of variables\n",
- "#theta=30 degrees\n",
- "sin30=2**-1\n",
- "vo=20 #ft/s\n",
- "r=4 #ft\n",
- "vf=0 #ft/s\n",
- "W=300 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "wo=vo*r**-1 #rad/s\n",
- "wf=vf*r**-1 #rad/s\n",
- "#Applying impulse momentum theorem\n",
- "#Solving simultaneous equations\n",
- "t=-((W*g**-1)*(vf-vo)+((0.5*(W*g**-1)*r**2*(wf-wo))*r**-1))/(W*sin30) #s\n",
- "\n",
- "#Result\n",
- "print'The time t is',round(t,2),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time t is 1.86 s\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-23, Page No 441"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mw=75 #kg\n",
- "k=0.9 #m\n",
- "wi=10 #rad/s\n",
- "wf=6 #rad/s\n",
- "r=1.2 #m\n",
- "m=30 #kg\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Initial speed\n",
- "vi=-r*wi #m/s\n",
- "vf=r*wf #m/s\n",
- "#Initial speed of B is\n",
- "vib=-0.8*wi+vi #m/s\n",
- "#Similarly\n",
- "vfb=12 #m/s\n",
- "#Applying impulse momentum theorem\n",
- "#Solving by matrix method\n",
- "A=np.array([[1,-1,-(mw*(vf-vi))],[0.8,1.2,-(mw*(k**2)*(wf+wi))],[-1,0,-(m*(vfb-vi))]])\n",
- "B=np.array([[0],[0],[-g*m]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Here t is calculated as 1/t for simplicity\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(C[2]**-1,2),\"s\" \n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 7.04 s\n"
- ]
- }
- ],
- "prompt_number": 59
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-24, Page No 441"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=8 #in\n",
- "W=96.6 #lb\n",
- "w=36 #rad/s\n",
- "u=0.15 #coefficient of friction\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "r=(d/2)*12**-1 #m\n",
- "N=W #lb\n",
- "F=u*N #lb\n",
- "m=W/g #slugs\n",
- "I=0.5*m*(r**2) #slug-ft**2\n",
- "#Applying the impulse momentum theorem\n",
- "#Solving the two equations simultaneously\n",
- "A=np.array([[F,-m],[F*r,I*(1*r**-1)]])\n",
- "B=np.array([[0],[w*I]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Distance travelled\n",
- "s=0.5*C[1]*C[0] #ft\n",
- "t=C[0] #s\n",
- "\n",
- "#Result\n",
- "print'The time required is',round(t,2),\"s\",'and it travels',round(s,2),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time required is 0.83 s and it travels 1.66 ft\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-25, Page No 442"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilisation of variables\n",
- "d=2*12**-1 #ft\n",
- "v=80 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Mass flow reate without time\n",
- "m=(1*4**-1)*pi*d**2*v*(62.4/g) \n",
- "#Let P=force of plate on mass m of water\n",
- "P=m*(0-v) #lb\n",
- "\n",
- "#Result\n",
- "print'The force water exerts on the plate is +',round(-P),\"lb,that is,to the right\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force water exerts on the plate is + 271.0 lb,that is,to the right\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-26, Page No 442"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v1=20 #ft/s\n",
- "vw=80 #ft/s\n",
- "d=2*12**-1 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "v=vw-v1 #ft/s\n",
- "#mass flow rate without t\n",
- "m=(4**-1)*(pi*d**2)*(62.4/g)*v\n",
- "#Applying impulse momentum theorem\n",
- "P=m*v #lb\n",
- "\n",
- "#Result\n",
- "print'The force exerted by water on the plate is +',round(P),\"lb,that is,to the right\"\n",
- "\n",
- "# Decimal poinat accuracy causes a small discrepancy in the answer"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force exerted by water on the plate is + 152.0 lb,that is,to the right\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-27, Page No 443"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "A=2000*10**-6 #m**2\n",
- "v=10 #m/s\n",
- "rho=1000 #kg/m**3\n",
- "#theta=45 degrees,thus\n",
- "cos45=(2**0.5)**-1\n",
- "sin45=(2**0.5)**-1\n",
- "\n",
- "#Calculations\n",
- "#Mass flow \n",
- "m=A*v*rho\n",
- "#Applying impulse momentum theorem\n",
- "Px=m*(-v*cos45-v) #N\n",
- "Py=m*(v*sin45-0) #N\n",
- "\n",
- "#Result\n",
- "print'The x component of force is',round(Px),\"N (to left on water)\",'and y component is +',round(Py),\"N (up on the water)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The x component of force is -341.0 N (to left on water) and y component is + 141.0 N (up on the water)\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-28, Page No 443"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilisation of variables\n",
- "W=150 #lb\n",
- "v=20 #ft/s\n",
- "A=0.2 #in**2\n",
- "t=60 #s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Mass flow\n",
- "m=(A/12**2)*v*(62.4/g)\n",
- "#Force\n",
- "F=m*(0-v) #lb\n",
- "#At t=60s the tank holds\n",
- "Ww=(A/12**2)*v*t*62.4 #lb\n",
- "#Total reading on scale\n",
- "S=-F+W+Ww #lb\n",
- "\n",
- "#Result\n",
- "print'The scale reading is',round(S),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The scale reading is 255.0 lb\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-29, Page No 444"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wp=130 #lb\n",
- "Wb=150 #lb\n",
- "Wbullet=2*16**-1 #lb\n",
- "g=32.2 #ft/s**2\n",
- "vbullet=1200 #ft/s\n",
- "\n",
- "#Calculations\n",
- "v=((-Wbullet*g**-1)*vbullet)/((Wb+Wp)/g) #ft/s\n",
- "\n",
- "#Result\n",
- "print'The speed of the boat is',round(v,2),\"ft/s\"\n",
- "#Negative sign indicates direction opposite to that of bullet\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the boat is -0.54 ft/s\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-30, Page No 444"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "mb=0.06 #kg\n",
- "ms=50 #kg\n",
- "h=0.03 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Speed of bag+bullet\n",
- "v2=sqrt(2*g*h) #m/s\n",
- "#Applying conservation of momentum \n",
- "v1=((mb+ms)*v2)/mb #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed of bullet as it entered the bag was',round(v1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of bullet as it entered the bag was 640.0 m/s\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-32, Page No 445"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "mb=0.06 #kg\n",
- "vb=500 #m/s\n",
- "mblock=5 #kg\n",
- "vblock=30 #m/s\n",
- "\n",
- "#Calculations\n",
- "#Applying conservation of momentum\n",
- "v=(mb*vb+mblock*vblock)/(mb+mblock) #m/s\n",
- "\n",
- "#Result\n",
- "print'The speed of the system is',round(v,1),\"m/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of the system is 35.6 m/s\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-33, Page No 445"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Initilization of variables\n",
- "W1=2 #lb\n",
- "W2=3 #lb\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return 12*(2-x)\n",
- "a=1\n",
- "b=1\n",
- "W=quad(integrand, 0, 2, args=(a,b))\n",
- "Work=W[0]/12 # ft-lb\n",
- "\n",
- "# Solving the simultaneousequations\n",
- "v3=(Work*(0.5*(W2*g**-1)+0.5*(W1*g**-1)*(-W2*W1**-1)**2)**-1)**0.5 #ft/s\n",
- "v2=-(W2*W1**-1)*v3 #ft/s\n",
- "\n",
- "# Results\n",
- "print'The speed of 2lb block is',round(v2,2),\"ft/s (to the left)\",'and that of 3lb block is',round(v3,2),\"ft/s (to the right) respectively.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed of 2lb block is -6.22 ft/s (to the left) and that of 3lb block is 4.14 ft/s (to the right) respectively.\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-34, Page No 445"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#Here the integration is indefinite hence it will be computed manually and entered\n",
- "W=10 #lb\n",
- "l=4 #ft\n",
- "w=2 #rad/s\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "#Part (a)\n",
- "wf=1.5 #rad/s\n",
- "t=(((W/g)*(l*w*l))-((W/g)*(l*wf*l)))**0.5 #s\n",
- "#Part (b)\n",
- "#Applying conservation of angular momentum\n",
- "r=(l*wf*l)*(l*w)**-1 #ft\n",
- "\n",
- "#Result\n",
- "print'The answer for part (a) is',round(t,2),\"s\",'and the answer for part (b) is',round(r,2),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The answer for part (a) is 1.58 s and the answer for part (b) is 3.0 ft\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-35, Page No 446"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=2.5 #lb\n",
- "w=36 #rad/s\n",
- "Idisk=0.4 #slug-ft**2\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "Ii=Idisk+(2*(W*g**-1)*(3*12**-1)**2) #slug-ft**2\n",
- "If=Idisk+(2*(W*g**-1)*(11*12**-1)**2) #slug-ft**2\n",
- "#Since no external moments act\n",
- "#Applying conservation of momentum\n",
- "wf=(Ii*w)*If**-1 #rad/s\n",
- "\n",
- "#Result\n",
- "print'The final angular speed is',round(wf,1),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The final angular speed is 27.8 rad/s\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-39, Page No 448"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "u1=6 #ft/s\n",
- "u2=-8 #ft/s\n",
- "e=0.8 #coefficient of restitution\n",
- "\n",
- "#Calculations\n",
- "#Solving both simultaneous equations\n",
- "A=np.array([[1,-1],[1,1]])\n",
- "B=np.array([[11.2],[-2]])\n",
- "C=np.linalg.solve(A,B) #ft/s\n",
- "\n",
- "#Result\n",
- "print'The velocities are v1=',round(C[1],1),\"ft/s\",'and v2=',round(C[0],1),\"ft/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The velocities are v1= -6.6 ft/s and v2= 4.6 ft/s\n"
- ]
- }
- ],
- "prompt_number": 52
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-40, Page No 448"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "h1=20 #m\n",
- "h2=14 #m\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "u1=sqrt(2*g*h1) #m/s\n",
- "u2=0 #m/s\n",
- "v1=-sqrt(2*g*h2) #m/s\n",
- "v2=0 #m/s\n",
- "e=(v2-v1)/(u1-u2) #coefficient of restitution\n",
- "\n",
- "#Result\n",
- "print'The value of coefficient of restitution is',round(e,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of coefficient of restitution is 0.84\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-41, Page No 448"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "u=6.55 #ft/s\n",
- "g=32.2 #ft/s**2\n",
- "L=6 #ft\n",
- "W=5 #lb\n",
- "c=0.7 #fraction of impulse acting in second phase\n",
- "\n",
- "#Calculations\n",
- "#Impulse\n",
- "I=(W*g**-1)*(u*3**-1) #N.s\n",
- "#Second Phase\n",
- "v=((-c*10.9)/2)+u #ft/s\n",
- "wprime=(1.09*60+c*(W*u*(3**-1)*6))/60 #rad/s\n",
- "\n",
- "#Result\n",
- "print'The value is v=',round(v,2),\"ft/s to the right\",' and that of w is',round(wprime,2),\"rad/s counterclockwise.\"\n",
- "\n",
- "#The value of w is incorrect in the textbook\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value is v= 2.73 ft/s to the right and that of w is 1.85 rad/s counterclockwise.\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-42, Page No 449"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=9 #kg\n",
- "m2=5.5 #kg\n",
- "u1=-3 #m/s\n",
- "u2=1.77 #m/s\n",
- "e=0.8 #coefficient of restitution\n",
- "\n",
- "#Calculations\n",
- "#Solving by matrix method after we get the two equations\n",
- "A=np.array([[-1,1],[m1,m2]])\n",
- "B=np.array([[(e*u1-e*u2)],[m1*u1+m2*u2]])\n",
- "C=np.linalg.solve(A,B) #m/s\n",
- "\n",
- "#Result\n",
- "print'The 9kg ball will rebound up the speed of',round(C[0],2),\"m/s\",'and the 5.5 kg ball will move to the right and down with components of',round(u2,2),\"m/s\",'and',round(-C[1],2),\"m/s respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The 9kg ball will rebound up the speed of 0.26 m/s and the 5.5 kg ball will move to the right and down with components of 1.77 m/s and 3.56 m/s respectively.\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-46, Page No 451"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "v=4 #m/s\n",
- "m=9 #kg\n",
- "s=1.5 #m\n",
- "\n",
- "#Calculations\n",
- "Io=(2*3**-1)*(m*s**2) #kg.m\n",
- "w=(m*v*s*0.5)/Io #rad/s\n",
- "\n",
- "#Result\n",
- "print'The angular velocity of the box is',round(w),\"rad/s clockwise.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angular velocity of the box is 2.0 rad/s clockwise.\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-48, Page No 452"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=2000 #kg\n",
- "mf=8500 #kg\n",
- "vr=2000 #m/s\n",
- "a=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Applying Newtons Second Law\n",
- "dm_dt=-(-((m+mf)*a)-(m+mf)*a)/(-vr) #kg/s\n",
- "\n",
- "#Result\n",
- "print'dm/dt=',round(dm_dt),\"kg/s\"\n",
- "#The negative sign indicates the loss in the mass of the system\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "dm/dt= -103.0 kg/s\n"
- ]
- }
- ],
- "prompt_number": 72
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-50, Page No 453"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=4000 #lb\n",
- "k=3 #ft\n",
- "wp=(60**-1)*2*pi #rad/s\n",
- "ws=(300/60)*2*pi #rad/s\n",
- "d=3.5 #ft\n",
- "g=32.2 #ft/s**2\n",
- "\n",
- "#Calculations\n",
- "I=(W/g)*k**2 #slug-ft**2\n",
- "M=I*ws*wp #lb-ft\n",
- "#Now equating M to Rf-Rr gives one equations and vertical sum yields other\n",
- "#solving them by matrix method\n",
- "A=np.array([[1,-1],[1,1]])\n",
- "B=np.array([[M*(2/d)],[W]])\n",
- "C=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The weight of Rf and Rr are',round(C[0]),\"lb\",'and',round(C[1]),\"lb respectively.\"\n",
- "\n",
- "# The answers in the textbook are incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The weight of Rf and Rr are 3051.0 lb and 949.0 lb respectively.\n"
- ]
- }
- ],
- "prompt_number": 74
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-51, Page No 454"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#As the integration is indefinite we will directly consider the equation with R\n",
- "#Initillization of variables\n",
- "GM=1.41*10**16 #ft**3/s**2\n",
- "r=2640000 #ft\n",
- "theta=60 #degrees\n",
- "R=21120000 #ft\n",
- "\n",
- "#Calculations\n",
- "v1=sqrt((GM*((R**-1)-((R+r)**-1)))/2.031) #ft/s\n",
- "\n",
- "#Result\n",
- "print'The speed required will be',round(v1),\"ft/s\"\n",
- "\n",
- "# Answer may wary due to decimal point discrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed required will be 6043.0 ft/s\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18-52, Page No 454"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "k=4 #lb/ft\n",
- "so=1 #ft\n",
- "W=2**-1 #lb\n",
- "g=32.2 #ft/s**2\n",
- "vo=5 #ft/s\n",
- "\n",
- "#Calculations\n",
- "m=W/g #kg\n",
- "#Angular momentum is conserved\n",
- "v=sqrt((0.5*k*so**2*2*2*g)+vo**2) #ft/s\n",
- "#Using vd=15\n",
- "d=15/v #ft\n",
- "\n",
- "#Result\n",
- "print'The ball passes',round(d,2),\"ft close to the fixed pin\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ball passes 0.89 ft close to the fixed pin\n"
- ]
- }
- ],
- "prompt_number": 26
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter19.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter19.ipynb deleted file mode 100755 index d723a0d2..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter19.ipynb +++ /dev/null @@ -1,503 +0,0 @@ -{
- "metadata": {
- "name": "chapter19.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 19: Mechanical Vibrations"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-2, Page No:465"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#Initilization of variables\n",
- "k=18 #lb/in\n",
- "g=386 #in/s**2\n",
- "W=35 #lb\n",
- "\n",
- "#Calculations\n",
- "f=(1/(2*pi))*sqrt((k*g/W)) #cps\n",
- "period=1/f #s\n",
- "\n",
- "#Result\n",
- "print'The period of vibration is',round(period,2),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The period of vibration is 0.45 s\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-11, Page No:471"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "ds=0.2 #m\n",
- "ts=0.05 #m\n",
- "rhos=7850 #kg/m**3 density of steel\n",
- "dw=0.002 #m\n",
- "lw=0.9 #m\n",
- "G=80*10**9 #Pa\n",
- "\n",
- "#Calculations\n",
- "#Torsional Constant\n",
- "K=(pi*dw**4*G)/(32*lw) #m/rad\n",
- "#Mass Calculations\n",
- "m=(4**-1)*pi*(ds**2)*ts*rhos #kg\n",
- "#Moment of Inertia\n",
- "Io=(0.5)*m*(ds*2**-1)**2 #kg.m**2\n",
- "#Frequency\n",
- "f=(1*(2*pi)**-1)*(sqrt(K*Io**-1)) #Hz\n",
- "\n",
- "#Result\n",
- "print'The natural frequency of the system is',round(f,2),\"Hz\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The natural frequency of the system is 0.24 Hz\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-13, Page No 472"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=120 #kg\n",
- "k=0.3 #m\n",
- "ls=0.6 #m\n",
- "ds=0.05 #m\n",
- "G=80*10**9 #Pa\n",
- "\n",
- "#Calculations\n",
- "#Polar Moment of Inertia\n",
- "J1=m*k**2 #kg.m**2\n",
- "J2=J1 #kg.m**2\n",
- "J=(32**-1)*pi*(ds**4) #m**4\n",
- "#Frequency\n",
- "f=(1/(2*pi))*(sqrt((J*G*(J1+J2))/(ls*J1*J2))) #Hz\n",
- "\n",
- "#Result\n",
- "print'The natural frequency of the torsional oscillation is',round(f,1),\"Hz\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The natural frequency of the torsional oscillation is 19.6 Hz\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-14, Page No: 473"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "ds=2 #in\n",
- "L=15 #in\n",
- "Wf1=300 #lb\n",
- "k1=6 #in\n",
- "Wf2=100 #lb\n",
- "k2=4 #in\n",
- "G=12*10**6 #Pa\n",
- "g=386 #in/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of inertia of flywheel\n",
- "Jf=(Wf1*g**-1)*k1**2 #lb-s**2-in\n",
- "#Moment of inertia of the rotor\n",
- "Jr=(Wf2*g**-1)*k2**2 #lb-s**2-in\n",
- "#Moment of inertia of the shaft cross section\n",
- "J=(32**-1)*pi*ds**4 #in**4\n",
- "#Frequency\n",
- "f=((pi*2)**-1)*(sqrt((J*G*(Jf+Jr))*(L*Jf*Jr)**-1)) #cps\n",
- "\n",
- "#Result\n",
- "print'The natural frequency of the system is',round(f,1),\"cps\"\n",
- "\n",
- "#The answer may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The natural frequency of the system is 93.9 cps\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-15, Page No: 473"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=10 #lb\n",
- "A=2 #in**2\n",
- "#Calculations\n",
- "\n",
- "wn=sqrt(((A*144**-1)*5*62.4*5)/2.59) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The frequency of oscillation is',round(wn,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frequency of oscillation is 2.89 rad/s\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-16, Page No:474"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "f=50 #cps\n",
- "g=386 #in/s**2\n",
- "E=30*10**6 #lb/in**2\n",
- "l=4 #in\n",
- "I=2.08*10**-6 #in**4\n",
- "\n",
- "#Calculations\n",
- "W=(3*E*I*g)/(((f*2*pi)**2)*l**3) #lb\n",
- "\n",
- "#Result\n",
- "print'The value of W is',round(W,3),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of W is 0.011 lb\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-19, Page No:478"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=10 #lb\n",
- "v=20 #in/s\n",
- "g=386 #in/s\n",
- "W=12 #lb\n",
- "k=20 #lb/in\n",
- "\n",
- "#Calculations\n",
- "#Coefficient of damping\n",
- "c=F*(v**-1) #lb-s/in\n",
- "#Natural Frequency\n",
- "wn=sqrt((k*g)/W) #rad/s\n",
- "#Critical Damping coefficient\n",
- "cr=(2*W*(g**-1))*wn #lb-s/in\n",
- "#Damping Coefficient\n",
- "d=c*(cr**-1)\n",
- "#Frequency of damped vibrations\n",
- "wd=sqrt(1-d**2)*wn #rad/s\n",
- "\n",
- "#Result\n",
- "print'The frequency of damped vibrations is',round(wd,1),\"rad/s\"\n",
- "\n",
- "# The answer is off by 0.1 units"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frequency of damped vibrations is 24.0 rad/s\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-20, Page No 478"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "wn=25.4 #rad/s\n",
- "t=0.261 #s\n",
- "d=0.316\n",
- "\n",
- "#Calculations\n",
- "delta=d*t*wn #logarithmic decay\n",
- "\n",
- "#Result\n",
- "print'The rate of decay is',round(delta,3)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The rate of decay is 2.095\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-24, Page No 483"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=9 #N\n",
- "m=5 #kg\n",
- "k=6000 #N/m\n",
- "f1=1 #Hz\n",
- "f2=5.4 #Hz\n",
- "f3=50 #Hz\n",
- "\n",
- "#Calculations\n",
- "#Natural Frequency\n",
- "fn=((pi*2)**-1)*(sqrt(k/m)) #Hz\n",
- "deltaf=F*(k/1000)**-1 #mm\n",
- "#Part(a)\n",
- "r1=f1*fn**-1\n",
- "amp1=deltaf*(1-r1**2)**-1 #mm amplitude\n",
- "#Part (b)\n",
- "r2=f2*fn**-1\n",
- "amp2=deltaf/(1-r2**2) #mm amplitude\n",
- "#Part (c)\n",
- "r3=f3*fn**-1\n",
- "amp3=deltaf/(1-r3**2) #mm amplitude\n",
- "\n",
- "#Result\n",
- "print'The amplitudes in part (a),(b) and (c) respectively are',round(amp1,3),\"mm ,\",round(amp2,1),\"mm and\",round(amp3,3),\"mm\"\n",
- "\n",
- "# The answer for amp2 is incorrect in textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The amplitudes in part (a),(b) and (c) respectively are 1.551 mm , 36.9 mm and -0.018 mm\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-25, Page No 483"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of vraiables\n",
- "g=386 #in/s**2\n",
- "W=20 #lb\n",
- "w=600 #rpm\n",
- "ratio=12**-1\n",
- "\n",
- "#Calculations\n",
- "r=sqrt((1*ratio**-1)+1) \n",
- "fn=((w/60)/r) #cps\n",
- "k=((fn*2*pi)**2*W)/(g) #lb/in\n",
- "\n",
- "#Result\n",
- "print'The value of k is',round(k,1),\"lb/in\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of k is 15.7 lb/in\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-28, Page No 487"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "X=12 #mm\n",
- "me_M=1.3 #mm\n",
- "\n",
- "#Calculations\n",
- "d=(me_M)/(2*X)\n",
- "\n",
- "#Result\n",
- "print'The damping ratio is',round(d,3)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The damping ratio is 0.054\n"
- ]
- }
- ],
- "prompt_number": 29
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter19_1.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter19_1.ipynb deleted file mode 100755 index d723a0d2..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter19_1.ipynb +++ /dev/null @@ -1,503 +0,0 @@ -{
- "metadata": {
- "name": "chapter19.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 19: Mechanical Vibrations"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-2, Page No:465"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#Initilization of variables\n",
- "k=18 #lb/in\n",
- "g=386 #in/s**2\n",
- "W=35 #lb\n",
- "\n",
- "#Calculations\n",
- "f=(1/(2*pi))*sqrt((k*g/W)) #cps\n",
- "period=1/f #s\n",
- "\n",
- "#Result\n",
- "print'The period of vibration is',round(period,2),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The period of vibration is 0.45 s\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-11, Page No:471"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "ds=0.2 #m\n",
- "ts=0.05 #m\n",
- "rhos=7850 #kg/m**3 density of steel\n",
- "dw=0.002 #m\n",
- "lw=0.9 #m\n",
- "G=80*10**9 #Pa\n",
- "\n",
- "#Calculations\n",
- "#Torsional Constant\n",
- "K=(pi*dw**4*G)/(32*lw) #m/rad\n",
- "#Mass Calculations\n",
- "m=(4**-1)*pi*(ds**2)*ts*rhos #kg\n",
- "#Moment of Inertia\n",
- "Io=(0.5)*m*(ds*2**-1)**2 #kg.m**2\n",
- "#Frequency\n",
- "f=(1*(2*pi)**-1)*(sqrt(K*Io**-1)) #Hz\n",
- "\n",
- "#Result\n",
- "print'The natural frequency of the system is',round(f,2),\"Hz\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The natural frequency of the system is 0.24 Hz\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-13, Page No 472"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=120 #kg\n",
- "k=0.3 #m\n",
- "ls=0.6 #m\n",
- "ds=0.05 #m\n",
- "G=80*10**9 #Pa\n",
- "\n",
- "#Calculations\n",
- "#Polar Moment of Inertia\n",
- "J1=m*k**2 #kg.m**2\n",
- "J2=J1 #kg.m**2\n",
- "J=(32**-1)*pi*(ds**4) #m**4\n",
- "#Frequency\n",
- "f=(1/(2*pi))*(sqrt((J*G*(J1+J2))/(ls*J1*J2))) #Hz\n",
- "\n",
- "#Result\n",
- "print'The natural frequency of the torsional oscillation is',round(f,1),\"Hz\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The natural frequency of the torsional oscillation is 19.6 Hz\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-14, Page No: 473"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "ds=2 #in\n",
- "L=15 #in\n",
- "Wf1=300 #lb\n",
- "k1=6 #in\n",
- "Wf2=100 #lb\n",
- "k2=4 #in\n",
- "G=12*10**6 #Pa\n",
- "g=386 #in/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of inertia of flywheel\n",
- "Jf=(Wf1*g**-1)*k1**2 #lb-s**2-in\n",
- "#Moment of inertia of the rotor\n",
- "Jr=(Wf2*g**-1)*k2**2 #lb-s**2-in\n",
- "#Moment of inertia of the shaft cross section\n",
- "J=(32**-1)*pi*ds**4 #in**4\n",
- "#Frequency\n",
- "f=((pi*2)**-1)*(sqrt((J*G*(Jf+Jr))*(L*Jf*Jr)**-1)) #cps\n",
- "\n",
- "#Result\n",
- "print'The natural frequency of the system is',round(f,1),\"cps\"\n",
- "\n",
- "#The answer may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The natural frequency of the system is 93.9 cps\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-15, Page No: 473"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=10 #lb\n",
- "A=2 #in**2\n",
- "#Calculations\n",
- "\n",
- "wn=sqrt(((A*144**-1)*5*62.4*5)/2.59) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The frequency of oscillation is',round(wn,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frequency of oscillation is 2.89 rad/s\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-16, Page No:474"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "f=50 #cps\n",
- "g=386 #in/s**2\n",
- "E=30*10**6 #lb/in**2\n",
- "l=4 #in\n",
- "I=2.08*10**-6 #in**4\n",
- "\n",
- "#Calculations\n",
- "W=(3*E*I*g)/(((f*2*pi)**2)*l**3) #lb\n",
- "\n",
- "#Result\n",
- "print'The value of W is',round(W,3),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of W is 0.011 lb\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-19, Page No:478"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=10 #lb\n",
- "v=20 #in/s\n",
- "g=386 #in/s\n",
- "W=12 #lb\n",
- "k=20 #lb/in\n",
- "\n",
- "#Calculations\n",
- "#Coefficient of damping\n",
- "c=F*(v**-1) #lb-s/in\n",
- "#Natural Frequency\n",
- "wn=sqrt((k*g)/W) #rad/s\n",
- "#Critical Damping coefficient\n",
- "cr=(2*W*(g**-1))*wn #lb-s/in\n",
- "#Damping Coefficient\n",
- "d=c*(cr**-1)\n",
- "#Frequency of damped vibrations\n",
- "wd=sqrt(1-d**2)*wn #rad/s\n",
- "\n",
- "#Result\n",
- "print'The frequency of damped vibrations is',round(wd,1),\"rad/s\"\n",
- "\n",
- "# The answer is off by 0.1 units"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frequency of damped vibrations is 24.0 rad/s\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-20, Page No 478"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "wn=25.4 #rad/s\n",
- "t=0.261 #s\n",
- "d=0.316\n",
- "\n",
- "#Calculations\n",
- "delta=d*t*wn #logarithmic decay\n",
- "\n",
- "#Result\n",
- "print'The rate of decay is',round(delta,3)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The rate of decay is 2.095\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-24, Page No 483"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=9 #N\n",
- "m=5 #kg\n",
- "k=6000 #N/m\n",
- "f1=1 #Hz\n",
- "f2=5.4 #Hz\n",
- "f3=50 #Hz\n",
- "\n",
- "#Calculations\n",
- "#Natural Frequency\n",
- "fn=((pi*2)**-1)*(sqrt(k/m)) #Hz\n",
- "deltaf=F*(k/1000)**-1 #mm\n",
- "#Part(a)\n",
- "r1=f1*fn**-1\n",
- "amp1=deltaf*(1-r1**2)**-1 #mm amplitude\n",
- "#Part (b)\n",
- "r2=f2*fn**-1\n",
- "amp2=deltaf/(1-r2**2) #mm amplitude\n",
- "#Part (c)\n",
- "r3=f3*fn**-1\n",
- "amp3=deltaf/(1-r3**2) #mm amplitude\n",
- "\n",
- "#Result\n",
- "print'The amplitudes in part (a),(b) and (c) respectively are',round(amp1,3),\"mm ,\",round(amp2,1),\"mm and\",round(amp3,3),\"mm\"\n",
- "\n",
- "# The answer for amp2 is incorrect in textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The amplitudes in part (a),(b) and (c) respectively are 1.551 mm , 36.9 mm and -0.018 mm\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-25, Page No 483"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of vraiables\n",
- "g=386 #in/s**2\n",
- "W=20 #lb\n",
- "w=600 #rpm\n",
- "ratio=12**-1\n",
- "\n",
- "#Calculations\n",
- "r=sqrt((1*ratio**-1)+1) \n",
- "fn=((w/60)/r) #cps\n",
- "k=((fn*2*pi)**2*W)/(g) #lb/in\n",
- "\n",
- "#Result\n",
- "print'The value of k is',round(k,1),\"lb/in\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of k is 15.7 lb/in\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-28, Page No 487"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "X=12 #mm\n",
- "me_M=1.3 #mm\n",
- "\n",
- "#Calculations\n",
- "d=(me_M)/(2*X)\n",
- "\n",
- "#Result\n",
- "print'The damping ratio is',round(d,3)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The damping ratio is 0.054\n"
- ]
- }
- ],
- "prompt_number": 29
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter19_2.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter19_2.ipynb deleted file mode 100755 index d723a0d2..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter19_2.ipynb +++ /dev/null @@ -1,503 +0,0 @@ -{
- "metadata": {
- "name": "chapter19.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 19: Mechanical Vibrations"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-2, Page No:465"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#Initilization of variables\n",
- "k=18 #lb/in\n",
- "g=386 #in/s**2\n",
- "W=35 #lb\n",
- "\n",
- "#Calculations\n",
- "f=(1/(2*pi))*sqrt((k*g/W)) #cps\n",
- "period=1/f #s\n",
- "\n",
- "#Result\n",
- "print'The period of vibration is',round(period,2),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The period of vibration is 0.45 s\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-11, Page No:471"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "ds=0.2 #m\n",
- "ts=0.05 #m\n",
- "rhos=7850 #kg/m**3 density of steel\n",
- "dw=0.002 #m\n",
- "lw=0.9 #m\n",
- "G=80*10**9 #Pa\n",
- "\n",
- "#Calculations\n",
- "#Torsional Constant\n",
- "K=(pi*dw**4*G)/(32*lw) #m/rad\n",
- "#Mass Calculations\n",
- "m=(4**-1)*pi*(ds**2)*ts*rhos #kg\n",
- "#Moment of Inertia\n",
- "Io=(0.5)*m*(ds*2**-1)**2 #kg.m**2\n",
- "#Frequency\n",
- "f=(1*(2*pi)**-1)*(sqrt(K*Io**-1)) #Hz\n",
- "\n",
- "#Result\n",
- "print'The natural frequency of the system is',round(f,2),\"Hz\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The natural frequency of the system is 0.24 Hz\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-13, Page No 472"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=120 #kg\n",
- "k=0.3 #m\n",
- "ls=0.6 #m\n",
- "ds=0.05 #m\n",
- "G=80*10**9 #Pa\n",
- "\n",
- "#Calculations\n",
- "#Polar Moment of Inertia\n",
- "J1=m*k**2 #kg.m**2\n",
- "J2=J1 #kg.m**2\n",
- "J=(32**-1)*pi*(ds**4) #m**4\n",
- "#Frequency\n",
- "f=(1/(2*pi))*(sqrt((J*G*(J1+J2))/(ls*J1*J2))) #Hz\n",
- "\n",
- "#Result\n",
- "print'The natural frequency of the torsional oscillation is',round(f,1),\"Hz\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The natural frequency of the torsional oscillation is 19.6 Hz\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-14, Page No: 473"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "ds=2 #in\n",
- "L=15 #in\n",
- "Wf1=300 #lb\n",
- "k1=6 #in\n",
- "Wf2=100 #lb\n",
- "k2=4 #in\n",
- "G=12*10**6 #Pa\n",
- "g=386 #in/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of inertia of flywheel\n",
- "Jf=(Wf1*g**-1)*k1**2 #lb-s**2-in\n",
- "#Moment of inertia of the rotor\n",
- "Jr=(Wf2*g**-1)*k2**2 #lb-s**2-in\n",
- "#Moment of inertia of the shaft cross section\n",
- "J=(32**-1)*pi*ds**4 #in**4\n",
- "#Frequency\n",
- "f=((pi*2)**-1)*(sqrt((J*G*(Jf+Jr))*(L*Jf*Jr)**-1)) #cps\n",
- "\n",
- "#Result\n",
- "print'The natural frequency of the system is',round(f,1),\"cps\"\n",
- "\n",
- "#The answer may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The natural frequency of the system is 93.9 cps\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-15, Page No: 473"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=10 #lb\n",
- "A=2 #in**2\n",
- "#Calculations\n",
- "\n",
- "wn=sqrt(((A*144**-1)*5*62.4*5)/2.59) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The frequency of oscillation is',round(wn,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frequency of oscillation is 2.89 rad/s\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-16, Page No:474"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "f=50 #cps\n",
- "g=386 #in/s**2\n",
- "E=30*10**6 #lb/in**2\n",
- "l=4 #in\n",
- "I=2.08*10**-6 #in**4\n",
- "\n",
- "#Calculations\n",
- "W=(3*E*I*g)/(((f*2*pi)**2)*l**3) #lb\n",
- "\n",
- "#Result\n",
- "print'The value of W is',round(W,3),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of W is 0.011 lb\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-19, Page No:478"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=10 #lb\n",
- "v=20 #in/s\n",
- "g=386 #in/s\n",
- "W=12 #lb\n",
- "k=20 #lb/in\n",
- "\n",
- "#Calculations\n",
- "#Coefficient of damping\n",
- "c=F*(v**-1) #lb-s/in\n",
- "#Natural Frequency\n",
- "wn=sqrt((k*g)/W) #rad/s\n",
- "#Critical Damping coefficient\n",
- "cr=(2*W*(g**-1))*wn #lb-s/in\n",
- "#Damping Coefficient\n",
- "d=c*(cr**-1)\n",
- "#Frequency of damped vibrations\n",
- "wd=sqrt(1-d**2)*wn #rad/s\n",
- "\n",
- "#Result\n",
- "print'The frequency of damped vibrations is',round(wd,1),\"rad/s\"\n",
- "\n",
- "# The answer is off by 0.1 units"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frequency of damped vibrations is 24.0 rad/s\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-20, Page No 478"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "wn=25.4 #rad/s\n",
- "t=0.261 #s\n",
- "d=0.316\n",
- "\n",
- "#Calculations\n",
- "delta=d*t*wn #logarithmic decay\n",
- "\n",
- "#Result\n",
- "print'The rate of decay is',round(delta,3)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The rate of decay is 2.095\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-24, Page No 483"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=9 #N\n",
- "m=5 #kg\n",
- "k=6000 #N/m\n",
- "f1=1 #Hz\n",
- "f2=5.4 #Hz\n",
- "f3=50 #Hz\n",
- "\n",
- "#Calculations\n",
- "#Natural Frequency\n",
- "fn=((pi*2)**-1)*(sqrt(k/m)) #Hz\n",
- "deltaf=F*(k/1000)**-1 #mm\n",
- "#Part(a)\n",
- "r1=f1*fn**-1\n",
- "amp1=deltaf*(1-r1**2)**-1 #mm amplitude\n",
- "#Part (b)\n",
- "r2=f2*fn**-1\n",
- "amp2=deltaf/(1-r2**2) #mm amplitude\n",
- "#Part (c)\n",
- "r3=f3*fn**-1\n",
- "amp3=deltaf/(1-r3**2) #mm amplitude\n",
- "\n",
- "#Result\n",
- "print'The amplitudes in part (a),(b) and (c) respectively are',round(amp1,3),\"mm ,\",round(amp2,1),\"mm and\",round(amp3,3),\"mm\"\n",
- "\n",
- "# The answer for amp2 is incorrect in textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The amplitudes in part (a),(b) and (c) respectively are 1.551 mm , 36.9 mm and -0.018 mm\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-25, Page No 483"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of vraiables\n",
- "g=386 #in/s**2\n",
- "W=20 #lb\n",
- "w=600 #rpm\n",
- "ratio=12**-1\n",
- "\n",
- "#Calculations\n",
- "r=sqrt((1*ratio**-1)+1) \n",
- "fn=((w/60)/r) #cps\n",
- "k=((fn*2*pi)**2*W)/(g) #lb/in\n",
- "\n",
- "#Result\n",
- "print'The value of k is',round(k,1),\"lb/in\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of k is 15.7 lb/in\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-28, Page No 487"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "X=12 #mm\n",
- "me_M=1.3 #mm\n",
- "\n",
- "#Calculations\n",
- "d=(me_M)/(2*X)\n",
- "\n",
- "#Result\n",
- "print'The damping ratio is',round(d,3)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The damping ratio is 0.054\n"
- ]
- }
- ],
- "prompt_number": 29
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter19_3.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter19_3.ipynb deleted file mode 100755 index d723a0d2..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter19_3.ipynb +++ /dev/null @@ -1,503 +0,0 @@ -{
- "metadata": {
- "name": "chapter19.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 19: Mechanical Vibrations"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-2, Page No:465"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#Initilization of variables\n",
- "k=18 #lb/in\n",
- "g=386 #in/s**2\n",
- "W=35 #lb\n",
- "\n",
- "#Calculations\n",
- "f=(1/(2*pi))*sqrt((k*g/W)) #cps\n",
- "period=1/f #s\n",
- "\n",
- "#Result\n",
- "print'The period of vibration is',round(period,2),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The period of vibration is 0.45 s\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-11, Page No:471"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "ds=0.2 #m\n",
- "ts=0.05 #m\n",
- "rhos=7850 #kg/m**3 density of steel\n",
- "dw=0.002 #m\n",
- "lw=0.9 #m\n",
- "G=80*10**9 #Pa\n",
- "\n",
- "#Calculations\n",
- "#Torsional Constant\n",
- "K=(pi*dw**4*G)/(32*lw) #m/rad\n",
- "#Mass Calculations\n",
- "m=(4**-1)*pi*(ds**2)*ts*rhos #kg\n",
- "#Moment of Inertia\n",
- "Io=(0.5)*m*(ds*2**-1)**2 #kg.m**2\n",
- "#Frequency\n",
- "f=(1*(2*pi)**-1)*(sqrt(K*Io**-1)) #Hz\n",
- "\n",
- "#Result\n",
- "print'The natural frequency of the system is',round(f,2),\"Hz\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The natural frequency of the system is 0.24 Hz\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-13, Page No 472"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=120 #kg\n",
- "k=0.3 #m\n",
- "ls=0.6 #m\n",
- "ds=0.05 #m\n",
- "G=80*10**9 #Pa\n",
- "\n",
- "#Calculations\n",
- "#Polar Moment of Inertia\n",
- "J1=m*k**2 #kg.m**2\n",
- "J2=J1 #kg.m**2\n",
- "J=(32**-1)*pi*(ds**4) #m**4\n",
- "#Frequency\n",
- "f=(1/(2*pi))*(sqrt((J*G*(J1+J2))/(ls*J1*J2))) #Hz\n",
- "\n",
- "#Result\n",
- "print'The natural frequency of the torsional oscillation is',round(f,1),\"Hz\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The natural frequency of the torsional oscillation is 19.6 Hz\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-14, Page No: 473"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "ds=2 #in\n",
- "L=15 #in\n",
- "Wf1=300 #lb\n",
- "k1=6 #in\n",
- "Wf2=100 #lb\n",
- "k2=4 #in\n",
- "G=12*10**6 #Pa\n",
- "g=386 #in/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of inertia of flywheel\n",
- "Jf=(Wf1*g**-1)*k1**2 #lb-s**2-in\n",
- "#Moment of inertia of the rotor\n",
- "Jr=(Wf2*g**-1)*k2**2 #lb-s**2-in\n",
- "#Moment of inertia of the shaft cross section\n",
- "J=(32**-1)*pi*ds**4 #in**4\n",
- "#Frequency\n",
- "f=((pi*2)**-1)*(sqrt((J*G*(Jf+Jr))*(L*Jf*Jr)**-1)) #cps\n",
- "\n",
- "#Result\n",
- "print'The natural frequency of the system is',round(f,1),\"cps\"\n",
- "\n",
- "#The answer may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The natural frequency of the system is 93.9 cps\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-15, Page No: 473"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=10 #lb\n",
- "A=2 #in**2\n",
- "#Calculations\n",
- "\n",
- "wn=sqrt(((A*144**-1)*5*62.4*5)/2.59) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The frequency of oscillation is',round(wn,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frequency of oscillation is 2.89 rad/s\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-16, Page No:474"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "f=50 #cps\n",
- "g=386 #in/s**2\n",
- "E=30*10**6 #lb/in**2\n",
- "l=4 #in\n",
- "I=2.08*10**-6 #in**4\n",
- "\n",
- "#Calculations\n",
- "W=(3*E*I*g)/(((f*2*pi)**2)*l**3) #lb\n",
- "\n",
- "#Result\n",
- "print'The value of W is',round(W,3),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of W is 0.011 lb\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-19, Page No:478"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=10 #lb\n",
- "v=20 #in/s\n",
- "g=386 #in/s\n",
- "W=12 #lb\n",
- "k=20 #lb/in\n",
- "\n",
- "#Calculations\n",
- "#Coefficient of damping\n",
- "c=F*(v**-1) #lb-s/in\n",
- "#Natural Frequency\n",
- "wn=sqrt((k*g)/W) #rad/s\n",
- "#Critical Damping coefficient\n",
- "cr=(2*W*(g**-1))*wn #lb-s/in\n",
- "#Damping Coefficient\n",
- "d=c*(cr**-1)\n",
- "#Frequency of damped vibrations\n",
- "wd=sqrt(1-d**2)*wn #rad/s\n",
- "\n",
- "#Result\n",
- "print'The frequency of damped vibrations is',round(wd,1),\"rad/s\"\n",
- "\n",
- "# The answer is off by 0.1 units"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frequency of damped vibrations is 24.0 rad/s\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-20, Page No 478"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "wn=25.4 #rad/s\n",
- "t=0.261 #s\n",
- "d=0.316\n",
- "\n",
- "#Calculations\n",
- "delta=d*t*wn #logarithmic decay\n",
- "\n",
- "#Result\n",
- "print'The rate of decay is',round(delta,3)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The rate of decay is 2.095\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-24, Page No 483"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=9 #N\n",
- "m=5 #kg\n",
- "k=6000 #N/m\n",
- "f1=1 #Hz\n",
- "f2=5.4 #Hz\n",
- "f3=50 #Hz\n",
- "\n",
- "#Calculations\n",
- "#Natural Frequency\n",
- "fn=((pi*2)**-1)*(sqrt(k/m)) #Hz\n",
- "deltaf=F*(k/1000)**-1 #mm\n",
- "#Part(a)\n",
- "r1=f1*fn**-1\n",
- "amp1=deltaf*(1-r1**2)**-1 #mm amplitude\n",
- "#Part (b)\n",
- "r2=f2*fn**-1\n",
- "amp2=deltaf/(1-r2**2) #mm amplitude\n",
- "#Part (c)\n",
- "r3=f3*fn**-1\n",
- "amp3=deltaf/(1-r3**2) #mm amplitude\n",
- "\n",
- "#Result\n",
- "print'The amplitudes in part (a),(b) and (c) respectively are',round(amp1,3),\"mm ,\",round(amp2,1),\"mm and\",round(amp3,3),\"mm\"\n",
- "\n",
- "# The answer for amp2 is incorrect in textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The amplitudes in part (a),(b) and (c) respectively are 1.551 mm , 36.9 mm and -0.018 mm\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-25, Page No 483"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of vraiables\n",
- "g=386 #in/s**2\n",
- "W=20 #lb\n",
- "w=600 #rpm\n",
- "ratio=12**-1\n",
- "\n",
- "#Calculations\n",
- "r=sqrt((1*ratio**-1)+1) \n",
- "fn=((w/60)/r) #cps\n",
- "k=((fn*2*pi)**2*W)/(g) #lb/in\n",
- "\n",
- "#Result\n",
- "print'The value of k is',round(k,1),\"lb/in\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of k is 15.7 lb/in\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-28, Page No 487"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "X=12 #mm\n",
- "me_M=1.3 #mm\n",
- "\n",
- "#Calculations\n",
- "d=(me_M)/(2*X)\n",
- "\n",
- "#Result\n",
- "print'The damping ratio is',round(d,3)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The damping ratio is 0.054\n"
- ]
- }
- ],
- "prompt_number": 29
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter19_4.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter19_4.ipynb deleted file mode 100755 index d723a0d2..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter19_4.ipynb +++ /dev/null @@ -1,503 +0,0 @@ -{
- "metadata": {
- "name": "chapter19.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 19: Mechanical Vibrations"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-2, Page No:465"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#Initilization of variables\n",
- "k=18 #lb/in\n",
- "g=386 #in/s**2\n",
- "W=35 #lb\n",
- "\n",
- "#Calculations\n",
- "f=(1/(2*pi))*sqrt((k*g/W)) #cps\n",
- "period=1/f #s\n",
- "\n",
- "#Result\n",
- "print'The period of vibration is',round(period,2),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The period of vibration is 0.45 s\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-11, Page No:471"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "ds=0.2 #m\n",
- "ts=0.05 #m\n",
- "rhos=7850 #kg/m**3 density of steel\n",
- "dw=0.002 #m\n",
- "lw=0.9 #m\n",
- "G=80*10**9 #Pa\n",
- "\n",
- "#Calculations\n",
- "#Torsional Constant\n",
- "K=(pi*dw**4*G)/(32*lw) #m/rad\n",
- "#Mass Calculations\n",
- "m=(4**-1)*pi*(ds**2)*ts*rhos #kg\n",
- "#Moment of Inertia\n",
- "Io=(0.5)*m*(ds*2**-1)**2 #kg.m**2\n",
- "#Frequency\n",
- "f=(1*(2*pi)**-1)*(sqrt(K*Io**-1)) #Hz\n",
- "\n",
- "#Result\n",
- "print'The natural frequency of the system is',round(f,2),\"Hz\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The natural frequency of the system is 0.24 Hz\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-13, Page No 472"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=120 #kg\n",
- "k=0.3 #m\n",
- "ls=0.6 #m\n",
- "ds=0.05 #m\n",
- "G=80*10**9 #Pa\n",
- "\n",
- "#Calculations\n",
- "#Polar Moment of Inertia\n",
- "J1=m*k**2 #kg.m**2\n",
- "J2=J1 #kg.m**2\n",
- "J=(32**-1)*pi*(ds**4) #m**4\n",
- "#Frequency\n",
- "f=(1/(2*pi))*(sqrt((J*G*(J1+J2))/(ls*J1*J2))) #Hz\n",
- "\n",
- "#Result\n",
- "print'The natural frequency of the torsional oscillation is',round(f,1),\"Hz\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The natural frequency of the torsional oscillation is 19.6 Hz\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-14, Page No: 473"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "ds=2 #in\n",
- "L=15 #in\n",
- "Wf1=300 #lb\n",
- "k1=6 #in\n",
- "Wf2=100 #lb\n",
- "k2=4 #in\n",
- "G=12*10**6 #Pa\n",
- "g=386 #in/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of inertia of flywheel\n",
- "Jf=(Wf1*g**-1)*k1**2 #lb-s**2-in\n",
- "#Moment of inertia of the rotor\n",
- "Jr=(Wf2*g**-1)*k2**2 #lb-s**2-in\n",
- "#Moment of inertia of the shaft cross section\n",
- "J=(32**-1)*pi*ds**4 #in**4\n",
- "#Frequency\n",
- "f=((pi*2)**-1)*(sqrt((J*G*(Jf+Jr))*(L*Jf*Jr)**-1)) #cps\n",
- "\n",
- "#Result\n",
- "print'The natural frequency of the system is',round(f,1),\"cps\"\n",
- "\n",
- "#The answer may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The natural frequency of the system is 93.9 cps\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-15, Page No: 473"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=10 #lb\n",
- "A=2 #in**2\n",
- "#Calculations\n",
- "\n",
- "wn=sqrt(((A*144**-1)*5*62.4*5)/2.59) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The frequency of oscillation is',round(wn,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frequency of oscillation is 2.89 rad/s\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-16, Page No:474"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "f=50 #cps\n",
- "g=386 #in/s**2\n",
- "E=30*10**6 #lb/in**2\n",
- "l=4 #in\n",
- "I=2.08*10**-6 #in**4\n",
- "\n",
- "#Calculations\n",
- "W=(3*E*I*g)/(((f*2*pi)**2)*l**3) #lb\n",
- "\n",
- "#Result\n",
- "print'The value of W is',round(W,3),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of W is 0.011 lb\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-19, Page No:478"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=10 #lb\n",
- "v=20 #in/s\n",
- "g=386 #in/s\n",
- "W=12 #lb\n",
- "k=20 #lb/in\n",
- "\n",
- "#Calculations\n",
- "#Coefficient of damping\n",
- "c=F*(v**-1) #lb-s/in\n",
- "#Natural Frequency\n",
- "wn=sqrt((k*g)/W) #rad/s\n",
- "#Critical Damping coefficient\n",
- "cr=(2*W*(g**-1))*wn #lb-s/in\n",
- "#Damping Coefficient\n",
- "d=c*(cr**-1)\n",
- "#Frequency of damped vibrations\n",
- "wd=sqrt(1-d**2)*wn #rad/s\n",
- "\n",
- "#Result\n",
- "print'The frequency of damped vibrations is',round(wd,1),\"rad/s\"\n",
- "\n",
- "# The answer is off by 0.1 units"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frequency of damped vibrations is 24.0 rad/s\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-20, Page No 478"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "wn=25.4 #rad/s\n",
- "t=0.261 #s\n",
- "d=0.316\n",
- "\n",
- "#Calculations\n",
- "delta=d*t*wn #logarithmic decay\n",
- "\n",
- "#Result\n",
- "print'The rate of decay is',round(delta,3)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The rate of decay is 2.095\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-24, Page No 483"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=9 #N\n",
- "m=5 #kg\n",
- "k=6000 #N/m\n",
- "f1=1 #Hz\n",
- "f2=5.4 #Hz\n",
- "f3=50 #Hz\n",
- "\n",
- "#Calculations\n",
- "#Natural Frequency\n",
- "fn=((pi*2)**-1)*(sqrt(k/m)) #Hz\n",
- "deltaf=F*(k/1000)**-1 #mm\n",
- "#Part(a)\n",
- "r1=f1*fn**-1\n",
- "amp1=deltaf*(1-r1**2)**-1 #mm amplitude\n",
- "#Part (b)\n",
- "r2=f2*fn**-1\n",
- "amp2=deltaf/(1-r2**2) #mm amplitude\n",
- "#Part (c)\n",
- "r3=f3*fn**-1\n",
- "amp3=deltaf/(1-r3**2) #mm amplitude\n",
- "\n",
- "#Result\n",
- "print'The amplitudes in part (a),(b) and (c) respectively are',round(amp1,3),\"mm ,\",round(amp2,1),\"mm and\",round(amp3,3),\"mm\"\n",
- "\n",
- "# The answer for amp2 is incorrect in textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The amplitudes in part (a),(b) and (c) respectively are 1.551 mm , 36.9 mm and -0.018 mm\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-25, Page No 483"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of vraiables\n",
- "g=386 #in/s**2\n",
- "W=20 #lb\n",
- "w=600 #rpm\n",
- "ratio=12**-1\n",
- "\n",
- "#Calculations\n",
- "r=sqrt((1*ratio**-1)+1) \n",
- "fn=((w/60)/r) #cps\n",
- "k=((fn*2*pi)**2*W)/(g) #lb/in\n",
- "\n",
- "#Result\n",
- "print'The value of k is',round(k,1),\"lb/in\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of k is 15.7 lb/in\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-28, Page No 487"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "X=12 #mm\n",
- "me_M=1.3 #mm\n",
- "\n",
- "#Calculations\n",
- "d=(me_M)/(2*X)\n",
- "\n",
- "#Result\n",
- "print'The damping ratio is',round(d,3)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The damping ratio is 0.054\n"
- ]
- }
- ],
- "prompt_number": 29
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter19_5.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter19_5.ipynb deleted file mode 100755 index d723a0d2..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter19_5.ipynb +++ /dev/null @@ -1,503 +0,0 @@ -{
- "metadata": {
- "name": "chapter19.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 19: Mechanical Vibrations"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-2, Page No:465"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#Initilization of variables\n",
- "k=18 #lb/in\n",
- "g=386 #in/s**2\n",
- "W=35 #lb\n",
- "\n",
- "#Calculations\n",
- "f=(1/(2*pi))*sqrt((k*g/W)) #cps\n",
- "period=1/f #s\n",
- "\n",
- "#Result\n",
- "print'The period of vibration is',round(period,2),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The period of vibration is 0.45 s\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-11, Page No:471"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "ds=0.2 #m\n",
- "ts=0.05 #m\n",
- "rhos=7850 #kg/m**3 density of steel\n",
- "dw=0.002 #m\n",
- "lw=0.9 #m\n",
- "G=80*10**9 #Pa\n",
- "\n",
- "#Calculations\n",
- "#Torsional Constant\n",
- "K=(pi*dw**4*G)/(32*lw) #m/rad\n",
- "#Mass Calculations\n",
- "m=(4**-1)*pi*(ds**2)*ts*rhos #kg\n",
- "#Moment of Inertia\n",
- "Io=(0.5)*m*(ds*2**-1)**2 #kg.m**2\n",
- "#Frequency\n",
- "f=(1*(2*pi)**-1)*(sqrt(K*Io**-1)) #Hz\n",
- "\n",
- "#Result\n",
- "print'The natural frequency of the system is',round(f,2),\"Hz\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The natural frequency of the system is 0.24 Hz\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-13, Page No 472"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=120 #kg\n",
- "k=0.3 #m\n",
- "ls=0.6 #m\n",
- "ds=0.05 #m\n",
- "G=80*10**9 #Pa\n",
- "\n",
- "#Calculations\n",
- "#Polar Moment of Inertia\n",
- "J1=m*k**2 #kg.m**2\n",
- "J2=J1 #kg.m**2\n",
- "J=(32**-1)*pi*(ds**4) #m**4\n",
- "#Frequency\n",
- "f=(1/(2*pi))*(sqrt((J*G*(J1+J2))/(ls*J1*J2))) #Hz\n",
- "\n",
- "#Result\n",
- "print'The natural frequency of the torsional oscillation is',round(f,1),\"Hz\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The natural frequency of the torsional oscillation is 19.6 Hz\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-14, Page No: 473"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "ds=2 #in\n",
- "L=15 #in\n",
- "Wf1=300 #lb\n",
- "k1=6 #in\n",
- "Wf2=100 #lb\n",
- "k2=4 #in\n",
- "G=12*10**6 #Pa\n",
- "g=386 #in/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of inertia of flywheel\n",
- "Jf=(Wf1*g**-1)*k1**2 #lb-s**2-in\n",
- "#Moment of inertia of the rotor\n",
- "Jr=(Wf2*g**-1)*k2**2 #lb-s**2-in\n",
- "#Moment of inertia of the shaft cross section\n",
- "J=(32**-1)*pi*ds**4 #in**4\n",
- "#Frequency\n",
- "f=((pi*2)**-1)*(sqrt((J*G*(Jf+Jr))*(L*Jf*Jr)**-1)) #cps\n",
- "\n",
- "#Result\n",
- "print'The natural frequency of the system is',round(f,1),\"cps\"\n",
- "\n",
- "#The answer may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The natural frequency of the system is 93.9 cps\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-15, Page No: 473"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=10 #lb\n",
- "A=2 #in**2\n",
- "#Calculations\n",
- "\n",
- "wn=sqrt(((A*144**-1)*5*62.4*5)/2.59) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The frequency of oscillation is',round(wn,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frequency of oscillation is 2.89 rad/s\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-16, Page No:474"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "f=50 #cps\n",
- "g=386 #in/s**2\n",
- "E=30*10**6 #lb/in**2\n",
- "l=4 #in\n",
- "I=2.08*10**-6 #in**4\n",
- "\n",
- "#Calculations\n",
- "W=(3*E*I*g)/(((f*2*pi)**2)*l**3) #lb\n",
- "\n",
- "#Result\n",
- "print'The value of W is',round(W,3),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of W is 0.011 lb\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-19, Page No:478"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=10 #lb\n",
- "v=20 #in/s\n",
- "g=386 #in/s\n",
- "W=12 #lb\n",
- "k=20 #lb/in\n",
- "\n",
- "#Calculations\n",
- "#Coefficient of damping\n",
- "c=F*(v**-1) #lb-s/in\n",
- "#Natural Frequency\n",
- "wn=sqrt((k*g)/W) #rad/s\n",
- "#Critical Damping coefficient\n",
- "cr=(2*W*(g**-1))*wn #lb-s/in\n",
- "#Damping Coefficient\n",
- "d=c*(cr**-1)\n",
- "#Frequency of damped vibrations\n",
- "wd=sqrt(1-d**2)*wn #rad/s\n",
- "\n",
- "#Result\n",
- "print'The frequency of damped vibrations is',round(wd,1),\"rad/s\"\n",
- "\n",
- "# The answer is off by 0.1 units"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frequency of damped vibrations is 24.0 rad/s\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-20, Page No 478"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "wn=25.4 #rad/s\n",
- "t=0.261 #s\n",
- "d=0.316\n",
- "\n",
- "#Calculations\n",
- "delta=d*t*wn #logarithmic decay\n",
- "\n",
- "#Result\n",
- "print'The rate of decay is',round(delta,3)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The rate of decay is 2.095\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-24, Page No 483"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=9 #N\n",
- "m=5 #kg\n",
- "k=6000 #N/m\n",
- "f1=1 #Hz\n",
- "f2=5.4 #Hz\n",
- "f3=50 #Hz\n",
- "\n",
- "#Calculations\n",
- "#Natural Frequency\n",
- "fn=((pi*2)**-1)*(sqrt(k/m)) #Hz\n",
- "deltaf=F*(k/1000)**-1 #mm\n",
- "#Part(a)\n",
- "r1=f1*fn**-1\n",
- "amp1=deltaf*(1-r1**2)**-1 #mm amplitude\n",
- "#Part (b)\n",
- "r2=f2*fn**-1\n",
- "amp2=deltaf/(1-r2**2) #mm amplitude\n",
- "#Part (c)\n",
- "r3=f3*fn**-1\n",
- "amp3=deltaf/(1-r3**2) #mm amplitude\n",
- "\n",
- "#Result\n",
- "print'The amplitudes in part (a),(b) and (c) respectively are',round(amp1,3),\"mm ,\",round(amp2,1),\"mm and\",round(amp3,3),\"mm\"\n",
- "\n",
- "# The answer for amp2 is incorrect in textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The amplitudes in part (a),(b) and (c) respectively are 1.551 mm , 36.9 mm and -0.018 mm\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-25, Page No 483"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of vraiables\n",
- "g=386 #in/s**2\n",
- "W=20 #lb\n",
- "w=600 #rpm\n",
- "ratio=12**-1\n",
- "\n",
- "#Calculations\n",
- "r=sqrt((1*ratio**-1)+1) \n",
- "fn=((w/60)/r) #cps\n",
- "k=((fn*2*pi)**2*W)/(g) #lb/in\n",
- "\n",
- "#Result\n",
- "print'The value of k is',round(k,1),\"lb/in\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of k is 15.7 lb/in\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-28, Page No 487"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "X=12 #mm\n",
- "me_M=1.3 #mm\n",
- "\n",
- "#Calculations\n",
- "d=(me_M)/(2*X)\n",
- "\n",
- "#Result\n",
- "print'The damping ratio is',round(d,3)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The damping ratio is 0.054\n"
- ]
- }
- ],
- "prompt_number": 29
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter19_6.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter19_6.ipynb deleted file mode 100755 index d723a0d2..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter19_6.ipynb +++ /dev/null @@ -1,503 +0,0 @@ -{
- "metadata": {
- "name": "chapter19.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 19: Mechanical Vibrations"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-2, Page No:465"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#Initilization of variables\n",
- "k=18 #lb/in\n",
- "g=386 #in/s**2\n",
- "W=35 #lb\n",
- "\n",
- "#Calculations\n",
- "f=(1/(2*pi))*sqrt((k*g/W)) #cps\n",
- "period=1/f #s\n",
- "\n",
- "#Result\n",
- "print'The period of vibration is',round(period,2),\"s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The period of vibration is 0.45 s\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-11, Page No:471"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "ds=0.2 #m\n",
- "ts=0.05 #m\n",
- "rhos=7850 #kg/m**3 density of steel\n",
- "dw=0.002 #m\n",
- "lw=0.9 #m\n",
- "G=80*10**9 #Pa\n",
- "\n",
- "#Calculations\n",
- "#Torsional Constant\n",
- "K=(pi*dw**4*G)/(32*lw) #m/rad\n",
- "#Mass Calculations\n",
- "m=(4**-1)*pi*(ds**2)*ts*rhos #kg\n",
- "#Moment of Inertia\n",
- "Io=(0.5)*m*(ds*2**-1)**2 #kg.m**2\n",
- "#Frequency\n",
- "f=(1*(2*pi)**-1)*(sqrt(K*Io**-1)) #Hz\n",
- "\n",
- "#Result\n",
- "print'The natural frequency of the system is',round(f,2),\"Hz\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The natural frequency of the system is 0.24 Hz\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-13, Page No 472"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=120 #kg\n",
- "k=0.3 #m\n",
- "ls=0.6 #m\n",
- "ds=0.05 #m\n",
- "G=80*10**9 #Pa\n",
- "\n",
- "#Calculations\n",
- "#Polar Moment of Inertia\n",
- "J1=m*k**2 #kg.m**2\n",
- "J2=J1 #kg.m**2\n",
- "J=(32**-1)*pi*(ds**4) #m**4\n",
- "#Frequency\n",
- "f=(1/(2*pi))*(sqrt((J*G*(J1+J2))/(ls*J1*J2))) #Hz\n",
- "\n",
- "#Result\n",
- "print'The natural frequency of the torsional oscillation is',round(f,1),\"Hz\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The natural frequency of the torsional oscillation is 19.6 Hz\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-14, Page No: 473"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "ds=2 #in\n",
- "L=15 #in\n",
- "Wf1=300 #lb\n",
- "k1=6 #in\n",
- "Wf2=100 #lb\n",
- "k2=4 #in\n",
- "G=12*10**6 #Pa\n",
- "g=386 #in/s**2\n",
- "\n",
- "#Calculations\n",
- "#Moment of inertia of flywheel\n",
- "Jf=(Wf1*g**-1)*k1**2 #lb-s**2-in\n",
- "#Moment of inertia of the rotor\n",
- "Jr=(Wf2*g**-1)*k2**2 #lb-s**2-in\n",
- "#Moment of inertia of the shaft cross section\n",
- "J=(32**-1)*pi*ds**4 #in**4\n",
- "#Frequency\n",
- "f=((pi*2)**-1)*(sqrt((J*G*(Jf+Jr))*(L*Jf*Jr)**-1)) #cps\n",
- "\n",
- "#Result\n",
- "print'The natural frequency of the system is',round(f,1),\"cps\"\n",
- "\n",
- "#The answer may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The natural frequency of the system is 93.9 cps\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-15, Page No: 473"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=10 #lb\n",
- "A=2 #in**2\n",
- "#Calculations\n",
- "\n",
- "wn=sqrt(((A*144**-1)*5*62.4*5)/2.59) #rad/s\n",
- "\n",
- "#Result\n",
- "print'The frequency of oscillation is',round(wn,2),\"rad/s\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frequency of oscillation is 2.89 rad/s\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-16, Page No:474"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "f=50 #cps\n",
- "g=386 #in/s**2\n",
- "E=30*10**6 #lb/in**2\n",
- "l=4 #in\n",
- "I=2.08*10**-6 #in**4\n",
- "\n",
- "#Calculations\n",
- "W=(3*E*I*g)/(((f*2*pi)**2)*l**3) #lb\n",
- "\n",
- "#Result\n",
- "print'The value of W is',round(W,3),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of W is 0.011 lb\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-19, Page No:478"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=10 #lb\n",
- "v=20 #in/s\n",
- "g=386 #in/s\n",
- "W=12 #lb\n",
- "k=20 #lb/in\n",
- "\n",
- "#Calculations\n",
- "#Coefficient of damping\n",
- "c=F*(v**-1) #lb-s/in\n",
- "#Natural Frequency\n",
- "wn=sqrt((k*g)/W) #rad/s\n",
- "#Critical Damping coefficient\n",
- "cr=(2*W*(g**-1))*wn #lb-s/in\n",
- "#Damping Coefficient\n",
- "d=c*(cr**-1)\n",
- "#Frequency of damped vibrations\n",
- "wd=sqrt(1-d**2)*wn #rad/s\n",
- "\n",
- "#Result\n",
- "print'The frequency of damped vibrations is',round(wd,1),\"rad/s\"\n",
- "\n",
- "# The answer is off by 0.1 units"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The frequency of damped vibrations is 24.0 rad/s\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-20, Page No 478"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "wn=25.4 #rad/s\n",
- "t=0.261 #s\n",
- "d=0.316\n",
- "\n",
- "#Calculations\n",
- "delta=d*t*wn #logarithmic decay\n",
- "\n",
- "#Result\n",
- "print'The rate of decay is',round(delta,3)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The rate of decay is 2.095\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-24, Page No 483"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=9 #N\n",
- "m=5 #kg\n",
- "k=6000 #N/m\n",
- "f1=1 #Hz\n",
- "f2=5.4 #Hz\n",
- "f3=50 #Hz\n",
- "\n",
- "#Calculations\n",
- "#Natural Frequency\n",
- "fn=((pi*2)**-1)*(sqrt(k/m)) #Hz\n",
- "deltaf=F*(k/1000)**-1 #mm\n",
- "#Part(a)\n",
- "r1=f1*fn**-1\n",
- "amp1=deltaf*(1-r1**2)**-1 #mm amplitude\n",
- "#Part (b)\n",
- "r2=f2*fn**-1\n",
- "amp2=deltaf/(1-r2**2) #mm amplitude\n",
- "#Part (c)\n",
- "r3=f3*fn**-1\n",
- "amp3=deltaf/(1-r3**2) #mm amplitude\n",
- "\n",
- "#Result\n",
- "print'The amplitudes in part (a),(b) and (c) respectively are',round(amp1,3),\"mm ,\",round(amp2,1),\"mm and\",round(amp3,3),\"mm\"\n",
- "\n",
- "# The answer for amp2 is incorrect in textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The amplitudes in part (a),(b) and (c) respectively are 1.551 mm , 36.9 mm and -0.018 mm\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-25, Page No 483"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of vraiables\n",
- "g=386 #in/s**2\n",
- "W=20 #lb\n",
- "w=600 #rpm\n",
- "ratio=12**-1\n",
- "\n",
- "#Calculations\n",
- "r=sqrt((1*ratio**-1)+1) \n",
- "fn=((w/60)/r) #cps\n",
- "k=((fn*2*pi)**2*W)/(g) #lb/in\n",
- "\n",
- "#Result\n",
- "print'The value of k is',round(k,1),\"lb/in\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of k is 15.7 lb/in\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.19-28, Page No 487"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "X=12 #mm\n",
- "me_M=1.3 #mm\n",
- "\n",
- "#Calculations\n",
- "d=(me_M)/(2*X)\n",
- "\n",
- "#Result\n",
- "print'The damping ratio is',round(d,3)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The damping ratio is 0.054\n"
- ]
- }
- ],
- "prompt_number": 29
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter1_1.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter1_1.ipynb deleted file mode 100755 index c6f2d00b..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter1_1.ipynb +++ /dev/null @@ -1,928 +0,0 @@ -{
- "metadata": {
- "name": "chapter1.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 1:Vectors"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-1,Page no: 8"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initialisation of Variables\n",
- "\n",
- "f1=120 #lb\n",
- "f2=100 #lb\n",
- "theta=((60*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=sqrt(120**2+100**2-(2*120*100*cos(theta))) #Applying Thr rule of Cosines\n",
- "alpha1=(((arcsin(120*sin(theta)/111))*180)/pi) #Applying the Law of Sines\n",
- "alpha=alpha1+270 #As the vector lies in the fourth Quadrant by obsrevaton\n",
- "\n",
- "#Results\n",
- "\n",
- "print'The Resultant of The force system is equal to',round(R),\"lb\" #lb\n",
- "print'The Resultant is at',round(alpha),\"degrees\" #degrees\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Resultant of The force system is equal to 111.0 lb\n",
- "The Resultant is at 339.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-2,Page no: 9"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P=100 #lb\n",
- "Q=120 #lb\n",
- "theta=((30*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R_x=Q*cos(theta) #lb\n",
- "R_y=Q*sin(theta)-P #lb\n",
- "R=sqrt(R_x**2+R_y**2) #lb Triangle law\n",
- "Theta_1=((arctan(R_y/R_x))*180)/pi #degrees\n",
- "Theta_R=360+Theta_1 #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The resultant is at',round(Theta_R),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 111.0 lb\n",
- "The resultant is at 339.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-3,Page No: 9"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "# Initialization of variables\n",
- "\n",
- "R=400 # N\n",
- "F2=200 # N\n",
- "Theta1=((120*pi)/180) # radians\n",
- "Theta2=((20*pi)/180) # radians\n",
- "Theta=Theta1-Theta2 # radians\n",
- "\n",
- "# Calculation\n",
- "\n",
- "F=sqrt(R**2+F2**2-(2*R*F2*cos(Theta))) # N.Applying the Rule of Cosine\n",
- "Theta_r=arcsin((400*sin(Theta))/F) #radians Applying the rule of sines\n",
- "Theta_R=(Theta_r*180)/pi\n",
- "\n",
- "# Result\n",
- "\n",
- "print'The resultant of the force system is',round(F),\"N\"\n",
- "print'The Angle between F and 200N force is',round(Theta_R,1),\"degrees\"\n",
- "\n",
- "# Theta_R is off by 0.1 degrees"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 477.0 N\n",
- "The Angle between F and 200N force is 55.6 degrees\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-4, Page No: 9"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "# Initilization of variables\n",
- "\n",
- "F1=280 # N\n",
- "F2=130 # N\n",
- "Theta1=((320*pi)/180) # Radians\n",
- "Theta2=((60*pi)/180) # Radians\n",
- "\n",
- "# Calculations\n",
- "\n",
- "R_x=-F1*cos(Theta1)+F2*cos(Theta2) # N\n",
- "R_y=F1*sin(Theta1)-F2*sin(Theta2) # N\n",
- "R=sqrt(R_x**2+R_y**2) # N Applying Triangle Law\n",
- "ThetaR=arctan(R_y/R_x) # radians\n",
- "Theta_R=360-(ThetaR*180/pi) # degrees\n",
- "\n",
- "# Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"N\"\n",
- "print'The resultant is at',round(Theta_R),\"degrees\"\n",
- "\n",
- "# The answer for R waries from textbook. "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 329.0 N\n",
- "The resultant is at 297.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-5, Page No: 10"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "# Initialization of variables\n",
- "\n",
- "F1=26 #lb\n",
- "F2=39 #lb\n",
- "F3=63 #lb\n",
- "F4=57 #lb\n",
- "T1=((10*pi)/180) #Radians\n",
- "T2=((114*pi)/180) #Radians\n",
- "T3=((183*pi)/180) #radians\n",
- "T4=((261*pi)/180) #radians\n",
- "\n",
- "# Calculations\n",
- "\n",
- "R_x=F1*cos(T1)+F2*cos(T2)+F3*cos(T3)+F4*cos(T4) # lb Resolving vectors\n",
- "R_y=F1*sin(T1)+F2*sin(T2)+F3*sin(T3)+F4*sin(T4) # lb resolving vectors\n",
- "R=sqrt(R_x**2+R_y**2) # lb Applying Triangle Law\n",
- "theta=arctan(R_y/R_x)# radians\n",
- "Theta=(theta*180)/pi # degrees\n",
- "Theta_R=180+Theta\n",
- "\n",
- "# Results\n",
- "\n",
- "print'The Resultant of the force system is',round(R),\"lb\"\n",
- "print'The resultant is at',round(Theta_R),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Resultant of the force system is 65.0 lb\n",
- "The resultant is at 197.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 36
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-6, Page No: 11"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=10 #lb\n",
- "theta1=((60*pi)/180) #radians\n",
- "theta2=((45*pi)/180) #radians\n",
- "theta=theta1-theta2 #radians\n",
- "\n",
- "#Calculation\n",
- "\n",
- "F_OH=F/cos(theta) #lb resolving vectors\n",
- "\n",
- "# Result\n",
- "\n",
- "print'The component of F in the direction of OH is',round(F_OH,2),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The component of F in the direction of OH is 10.35 lb\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-7, Page No: 11"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "weight=80 #kg\n",
- "theta=((20*pi)/180) #radians\n",
- "theta_p=((70*pi)/180) # radians\n",
- "\n",
- "#Calcuations\n",
- "\n",
- "#Part (a)\n",
- "F=weight*9.81 # N\n",
- "R=F*cos(theta) #N\n",
- "#part (b)\n",
- "R_p=F*cos(theta_p) #N\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The normal component is',round(R),\"N\"\n",
- "print'The parallel component is',round(R_p),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The normal component is 737.0 N\n",
- "The parallel component is 268.0 N\n"
- ]
- }
- ],
- "prompt_number": 37
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-8, Page No: 11"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P=235 #N\n",
- "theta=((60*pi)/180) #radians\n",
- "bet=((22*pi)/180) #radians\n",
- "gam=((38*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Part (a)\n",
- "P_h=P*cos(theta) #N\n",
- "P_v=P*sin(theta) #N\n",
- "#Part (b)\n",
- "P_l=P*cos(theta-bet) #N\n",
- "P_p=P*sin(gam) #N\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The horizontal component is',round(P_h,1),\"N\"\n",
- "print'The vertical component is',round(P_v,1),\"N\"\n",
- "print'The component parallel to plane is',round(P_l),\"N\"\n",
- "print'The component perpendicular to the plane is',round(P_p,1),\"N\"\n",
- "\n",
- "#The decimal point accuracy might cause a small discrepancy in the answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The horizontal component is 117.5 N\n",
- "The vertical component is 203.5 N\n",
- "The component parallel to plane is 185.0 N\n",
- "The component perpendicular to the plane is 144.7 N\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-9, Page No: 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=90 #lb\n",
- "theta1=((40*pi)/180) #radians\n",
- "theta2=((30*pi)/180) #radians\n",
- "\n",
- "# Calculations\n",
- "\n",
- "R_x=0 #lb\n",
- "R_y=20 # lb\n",
- "#Taking the sum of forces in the X-Direction\n",
- "P=((F1*cos(theta1))/cos(theta2)) # lb\n",
- "# Taking the sum of the forces in the Y-Direction\n",
- "F=(P*sin(theta2))+(F1*sin(theta1))-20 #lb\n",
- "\n",
- "# Results\n",
- "\n",
- "print'The value of P is',round(P,1),\"lb\"\n",
- "print'The value of F is',round(F,1),\"lb\"\n",
- "\n",
- "# Decimal point error may cause a small discrepancy in the answers."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P is 79.6 lb\n",
- "The value of F is 77.7 lb\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-10, Page No: 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "x=4 # m\n",
- "y=3 # m\n",
- "z=2 # m\n",
- "F=50 #N\n",
- "\n",
- "# Calculations\n",
- "\n",
- "OP=sqrt(x**2+y**2+z**2) #m\n",
- "thetax=(x/OP) #radians\n",
- "thetay=(y/OP) #Radians\n",
- "thetaz=(z/OP) #radians\n",
- "P_x=F*(thetax) #N\n",
- "P_y=F*(thetay) #N\n",
- "P_z=F*(thetaz) #N\n",
- "\n",
- "# Result\n",
- "\n",
- "print'The vector P is',round(P_x,1),\"i +\",round(P_y,1),\"j +\",round(P_z,1),\"k\"\n",
- "\n",
- "# component of i is off by 0.1 units"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The vector P is 37.1 i + 27.9 j + 18.6 k\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-11, Page No: 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "x=2 \n",
- "y=-4\n",
- "z=1\n",
- "F=100 #N\n",
- "\n",
- "#Calculation\n",
- "\n",
- "thetax=x/sqrt(x**2+y**2+z**2) #radians\n",
- "thetay=y/sqrt(x**2+y**2+z**2) #radians\n",
- "thetaz=z/sqrt(x**2+y**2+z**2) #radians\n",
- "P_x=F*thetax #N\n",
- "P_y=F*thetay #N\n",
- "P_z=F*thetaz #N\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The vector P is',round(P_x,1),\"i\",round(P_y,1),\"j +\",round(P_z,1),\"k\"\n",
- "\n",
- "# component off i is off by 0.1 units"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The vector P is 43.6 i -87.3 j + 21.8 k\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-12, Page No: 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Solution\n",
- "\n",
- "print'P X Q=(Pxi+Pyj+Pzk)x(Qxi+Qyj+Qzk)'\n",
- "print' =(PxQx)i x i+(PxQy)i x j +(PxQz)i x k'\n",
- "print'But i x i = j x j = k x k =0; and i x j =k and j x i= -k, etc. Hence'\n",
- "print'P X Q= (PxQy)k-(PxQz)j-(PyQx)k+(PxQz)i+(PzQx)j-(PzQy)i'\n",
- "print'These terms can be grouped as'\n",
- "print'P X Q=(PyQz-PzQy)i+(PzQx-PxQz)j+(PxQy-PyQx)k'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "P X Q=(Pxi+Pyj+Pzk)x(Qxi+Qyj+Qzk)\n",
- " =(PxQx)i x i+(PxQy)i x j +(PxQz)i x k\n",
- "But i x i = j x j = k x k =0; and i x j =k and j x i= -k, etc. Hence\n",
- "P X Q= (PxQy)k-(PxQz)j-(PyQx)k+(PxQz)i+(PzQx)j-(PzQy)i\n",
- "These terms can be grouped as\n",
- "P X Q=(PyQz-PzQy)i+(PzQx-PxQz)j+(PxQy-PyQx)k\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-13,Page No: 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "Fx=2.63 #N\n",
- "Fy=4.28 #N\n",
- "Fz=-5.92 #N\n",
- "\n",
- "#Calculation\n",
- "\n",
- "F=sqrt(Fx**2+Fy**2+Fz**2) #N\n",
- "thetax=((arccos(Fx/F))*180)/pi #degrees\n",
- "thetay=((arccos(Fy/F))*180)/pi #degrees\n",
- "thetaz=((arccos(Fz/F))*180)/pi #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The magnitude of force is',round(F,2),\"N\"\n",
- "print'Thetax',round(thetax,1),\"degrees\"\n",
- "print'Thetay',round(thetay,1),\"degrees\"\n",
- "print'Thetaz',round(thetaz,1),\"degrees\"\n",
- "\n",
- "# Decimal point error may cause a small discrepancy in the answers."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The magnitude of force is 7.76 N\n",
- "Thetax 70.2 degrees\n",
- "Thetay 56.5 degrees\n",
- "Thetaz 139.7 degrees\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-14, Page No: 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P=[4.82, -2.33, 5.47] #N\n",
- "Q=[-2.81,-6.09,1.12 ] #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "M=P[0]*Q[0]+P[1]*Q[1]+P[2]*Q[2] #Nm\n",
- "\n",
- "#Results\n",
- "\n",
- "print'Result is',round(M,2),\"N.m\"\n",
- "\n",
- "# Decimal point error in calculation causes a small discrepancy in the answer."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Result is 6.77 N.m\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-15, Page No: 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "#Initilization of variables\n",
- "\n",
- "x1=2 #units\n",
- "x2=-2 #units\n",
- "y1=3 #units\n",
- "y2=4 #units\n",
- "z1=0 #units\n",
- "z2=6 #units\n",
- "P=np.array([2,3,-1]) #units\n",
- "\n",
- "#Calculations\n",
- "\n",
- "X=sqrt((x2-x1)**2+(y2-y1)**2+(z2-z1)**2) #units\n",
- "eLx=(x2-x1)/X #units\n",
- "eLy=(y2-y1)/X #units\n",
- "eLz=(z2-z1)/X #units\n",
- "Q=np.array([eLx,eLy,eLz]) #units\n",
- "Z=P[0]*Q[0]+P[1]*Q[1]+P[2]*Q[2] # units\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The unit vector is',round(eLx,3),\"i +\",round(eLy,3),\"j +\",round(eLz,3),\"k\"\n",
- "print'The projection of P is',round(Z,2),\"units\"\n",
- "\n",
- "#Note:The final answer for the projection of P is off by 0.1 units\n",
- "#The answer mentioned in the textbook is -1.41\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The unit vector is -0.549 i + 0.137 j + 0.824 k\n",
- "The projection of P is -1.51 units\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-16, Page No: 14"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "#Initilization of variables\n",
- "\n",
- "x1=2 #units\n",
- "x2=5 #units\n",
- "y1=-5 #units\n",
- "y2=2 #units\n",
- "z1=3 #units\n",
- "z2=-4 #units\n",
- "P=np.array([10,-8,14]) #units\n",
- "\n",
- "#Calculations\n",
- "\n",
- "X=sqrt((x2-x1)**2+(y2-y1)**2+(z2-z1)**2) #units\n",
- "eLx=(x2-x1)/X #units\n",
- "eLy=(y2-y1)/X #units\n",
- "eLz=(z2-z1)/X #units\n",
- "Q=np.array([eLx,eLy,eLz]) #units\n",
- "Z=P[0]*Q[0]+P[1]*Q[1]+P[2]*Q[2] #units\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The unit vector is',round(eLx,3),\"i +\",round(eLy,3),\"j\",round(eLz,3),\"k\" \n",
- "print'The projection of P is',round(Z),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The unit vector is 0.29 i + 0.677 j -0.677 k\n",
- "The projection of P is -12.0 lb\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-17, Page No: 14\n"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "Px=2.85 #ft\n",
- "Py=4.67 #ft\n",
- "Pz=-8.09 #ft\n",
- "Qx=28.3 #lb\n",
- "Qy=44.6 #lb\n",
- "Qz=53.3 #lb\n",
- "\n",
- "#Calculations\n",
- "\n",
- "X=(Py*Qz-Pz*Qy) #N.m \n",
- "Y=(Pz*Qx-Px*Qz) #N.m\n",
- "Z=(Px*Qy-Py*Qx) #N.m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The cross product is',round(X),\"i\",round(Y),\"j\",round(Z),\"k lb-ft\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The cross product is 610.0 i -381.0 j -5.0 k lb-ft\n"
- ]
- }
- ],
- "prompt_number": 45
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-18, Page No: 14"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Result\n",
- "#As this is symbolic solution directly print command is being used to give the required output\n",
- "\n",
- "print'The Time derivative is '\n",
- "print'dr/dt=(dx/dt)i+12*y(dy/dt)j-3*(dz/dt)k'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Time derivative is \n",
- "dr/dt=(dx/dt)i+12*y(dy/dt)j-3*(dz/dt)k\n"
- ]
- }
- ],
- "prompt_number": 47
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-19, Page No: 14"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return x**2\n",
- "a=1\n",
- "b=1\n",
- "I=quad(integrand, 1, 3, args=(a,b))\n",
- "\n",
- "def integrand(y, a, b):\n",
- " return 2*y\n",
- "a=1\n",
- "b=1\n",
- "J=quad(integrand, 1, 3, args=(a,b))\n",
- "\n",
- "def integrand(z, a, b):\n",
- " return 1\n",
- "a=1\n",
- "b=1\n",
- "K=quad(integrand, 1, 3, args=(a,b))\n",
- "\n",
- "# Results\n",
- "print'The answer is',round(I[0],2),\"i +\",round(J[0]),\"j -\",round(K[0]),\"k.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The answer is 8.67 i + 8.0 j - 2.0 k.\n"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter1_2.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter1_2.ipynb deleted file mode 100755 index c6f2d00b..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter1_2.ipynb +++ /dev/null @@ -1,928 +0,0 @@ -{
- "metadata": {
- "name": "chapter1.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 1:Vectors"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-1,Page no: 8"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initialisation of Variables\n",
- "\n",
- "f1=120 #lb\n",
- "f2=100 #lb\n",
- "theta=((60*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=sqrt(120**2+100**2-(2*120*100*cos(theta))) #Applying Thr rule of Cosines\n",
- "alpha1=(((arcsin(120*sin(theta)/111))*180)/pi) #Applying the Law of Sines\n",
- "alpha=alpha1+270 #As the vector lies in the fourth Quadrant by obsrevaton\n",
- "\n",
- "#Results\n",
- "\n",
- "print'The Resultant of The force system is equal to',round(R),\"lb\" #lb\n",
- "print'The Resultant is at',round(alpha),\"degrees\" #degrees\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Resultant of The force system is equal to 111.0 lb\n",
- "The Resultant is at 339.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-2,Page no: 9"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P=100 #lb\n",
- "Q=120 #lb\n",
- "theta=((30*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R_x=Q*cos(theta) #lb\n",
- "R_y=Q*sin(theta)-P #lb\n",
- "R=sqrt(R_x**2+R_y**2) #lb Triangle law\n",
- "Theta_1=((arctan(R_y/R_x))*180)/pi #degrees\n",
- "Theta_R=360+Theta_1 #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The resultant is at',round(Theta_R),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 111.0 lb\n",
- "The resultant is at 339.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-3,Page No: 9"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "# Initialization of variables\n",
- "\n",
- "R=400 # N\n",
- "F2=200 # N\n",
- "Theta1=((120*pi)/180) # radians\n",
- "Theta2=((20*pi)/180) # radians\n",
- "Theta=Theta1-Theta2 # radians\n",
- "\n",
- "# Calculation\n",
- "\n",
- "F=sqrt(R**2+F2**2-(2*R*F2*cos(Theta))) # N.Applying the Rule of Cosine\n",
- "Theta_r=arcsin((400*sin(Theta))/F) #radians Applying the rule of sines\n",
- "Theta_R=(Theta_r*180)/pi\n",
- "\n",
- "# Result\n",
- "\n",
- "print'The resultant of the force system is',round(F),\"N\"\n",
- "print'The Angle between F and 200N force is',round(Theta_R,1),\"degrees\"\n",
- "\n",
- "# Theta_R is off by 0.1 degrees"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 477.0 N\n",
- "The Angle between F and 200N force is 55.6 degrees\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-4, Page No: 9"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "# Initilization of variables\n",
- "\n",
- "F1=280 # N\n",
- "F2=130 # N\n",
- "Theta1=((320*pi)/180) # Radians\n",
- "Theta2=((60*pi)/180) # Radians\n",
- "\n",
- "# Calculations\n",
- "\n",
- "R_x=-F1*cos(Theta1)+F2*cos(Theta2) # N\n",
- "R_y=F1*sin(Theta1)-F2*sin(Theta2) # N\n",
- "R=sqrt(R_x**2+R_y**2) # N Applying Triangle Law\n",
- "ThetaR=arctan(R_y/R_x) # radians\n",
- "Theta_R=360-(ThetaR*180/pi) # degrees\n",
- "\n",
- "# Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"N\"\n",
- "print'The resultant is at',round(Theta_R),\"degrees\"\n",
- "\n",
- "# The answer for R waries from textbook. "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 329.0 N\n",
- "The resultant is at 297.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-5, Page No: 10"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "# Initialization of variables\n",
- "\n",
- "F1=26 #lb\n",
- "F2=39 #lb\n",
- "F3=63 #lb\n",
- "F4=57 #lb\n",
- "T1=((10*pi)/180) #Radians\n",
- "T2=((114*pi)/180) #Radians\n",
- "T3=((183*pi)/180) #radians\n",
- "T4=((261*pi)/180) #radians\n",
- "\n",
- "# Calculations\n",
- "\n",
- "R_x=F1*cos(T1)+F2*cos(T2)+F3*cos(T3)+F4*cos(T4) # lb Resolving vectors\n",
- "R_y=F1*sin(T1)+F2*sin(T2)+F3*sin(T3)+F4*sin(T4) # lb resolving vectors\n",
- "R=sqrt(R_x**2+R_y**2) # lb Applying Triangle Law\n",
- "theta=arctan(R_y/R_x)# radians\n",
- "Theta=(theta*180)/pi # degrees\n",
- "Theta_R=180+Theta\n",
- "\n",
- "# Results\n",
- "\n",
- "print'The Resultant of the force system is',round(R),\"lb\"\n",
- "print'The resultant is at',round(Theta_R),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Resultant of the force system is 65.0 lb\n",
- "The resultant is at 197.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 36
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-6, Page No: 11"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=10 #lb\n",
- "theta1=((60*pi)/180) #radians\n",
- "theta2=((45*pi)/180) #radians\n",
- "theta=theta1-theta2 #radians\n",
- "\n",
- "#Calculation\n",
- "\n",
- "F_OH=F/cos(theta) #lb resolving vectors\n",
- "\n",
- "# Result\n",
- "\n",
- "print'The component of F in the direction of OH is',round(F_OH,2),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The component of F in the direction of OH is 10.35 lb\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-7, Page No: 11"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "weight=80 #kg\n",
- "theta=((20*pi)/180) #radians\n",
- "theta_p=((70*pi)/180) # radians\n",
- "\n",
- "#Calcuations\n",
- "\n",
- "#Part (a)\n",
- "F=weight*9.81 # N\n",
- "R=F*cos(theta) #N\n",
- "#part (b)\n",
- "R_p=F*cos(theta_p) #N\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The normal component is',round(R),\"N\"\n",
- "print'The parallel component is',round(R_p),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The normal component is 737.0 N\n",
- "The parallel component is 268.0 N\n"
- ]
- }
- ],
- "prompt_number": 37
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-8, Page No: 11"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P=235 #N\n",
- "theta=((60*pi)/180) #radians\n",
- "bet=((22*pi)/180) #radians\n",
- "gam=((38*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Part (a)\n",
- "P_h=P*cos(theta) #N\n",
- "P_v=P*sin(theta) #N\n",
- "#Part (b)\n",
- "P_l=P*cos(theta-bet) #N\n",
- "P_p=P*sin(gam) #N\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The horizontal component is',round(P_h,1),\"N\"\n",
- "print'The vertical component is',round(P_v,1),\"N\"\n",
- "print'The component parallel to plane is',round(P_l),\"N\"\n",
- "print'The component perpendicular to the plane is',round(P_p,1),\"N\"\n",
- "\n",
- "#The decimal point accuracy might cause a small discrepancy in the answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The horizontal component is 117.5 N\n",
- "The vertical component is 203.5 N\n",
- "The component parallel to plane is 185.0 N\n",
- "The component perpendicular to the plane is 144.7 N\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-9, Page No: 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=90 #lb\n",
- "theta1=((40*pi)/180) #radians\n",
- "theta2=((30*pi)/180) #radians\n",
- "\n",
- "# Calculations\n",
- "\n",
- "R_x=0 #lb\n",
- "R_y=20 # lb\n",
- "#Taking the sum of forces in the X-Direction\n",
- "P=((F1*cos(theta1))/cos(theta2)) # lb\n",
- "# Taking the sum of the forces in the Y-Direction\n",
- "F=(P*sin(theta2))+(F1*sin(theta1))-20 #lb\n",
- "\n",
- "# Results\n",
- "\n",
- "print'The value of P is',round(P,1),\"lb\"\n",
- "print'The value of F is',round(F,1),\"lb\"\n",
- "\n",
- "# Decimal point error may cause a small discrepancy in the answers."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P is 79.6 lb\n",
- "The value of F is 77.7 lb\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-10, Page No: 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "x=4 # m\n",
- "y=3 # m\n",
- "z=2 # m\n",
- "F=50 #N\n",
- "\n",
- "# Calculations\n",
- "\n",
- "OP=sqrt(x**2+y**2+z**2) #m\n",
- "thetax=(x/OP) #radians\n",
- "thetay=(y/OP) #Radians\n",
- "thetaz=(z/OP) #radians\n",
- "P_x=F*(thetax) #N\n",
- "P_y=F*(thetay) #N\n",
- "P_z=F*(thetaz) #N\n",
- "\n",
- "# Result\n",
- "\n",
- "print'The vector P is',round(P_x,1),\"i +\",round(P_y,1),\"j +\",round(P_z,1),\"k\"\n",
- "\n",
- "# component of i is off by 0.1 units"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The vector P is 37.1 i + 27.9 j + 18.6 k\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-11, Page No: 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "x=2 \n",
- "y=-4\n",
- "z=1\n",
- "F=100 #N\n",
- "\n",
- "#Calculation\n",
- "\n",
- "thetax=x/sqrt(x**2+y**2+z**2) #radians\n",
- "thetay=y/sqrt(x**2+y**2+z**2) #radians\n",
- "thetaz=z/sqrt(x**2+y**2+z**2) #radians\n",
- "P_x=F*thetax #N\n",
- "P_y=F*thetay #N\n",
- "P_z=F*thetaz #N\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The vector P is',round(P_x,1),\"i\",round(P_y,1),\"j +\",round(P_z,1),\"k\"\n",
- "\n",
- "# component off i is off by 0.1 units"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The vector P is 43.6 i -87.3 j + 21.8 k\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-12, Page No: 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Solution\n",
- "\n",
- "print'P X Q=(Pxi+Pyj+Pzk)x(Qxi+Qyj+Qzk)'\n",
- "print' =(PxQx)i x i+(PxQy)i x j +(PxQz)i x k'\n",
- "print'But i x i = j x j = k x k =0; and i x j =k and j x i= -k, etc. Hence'\n",
- "print'P X Q= (PxQy)k-(PxQz)j-(PyQx)k+(PxQz)i+(PzQx)j-(PzQy)i'\n",
- "print'These terms can be grouped as'\n",
- "print'P X Q=(PyQz-PzQy)i+(PzQx-PxQz)j+(PxQy-PyQx)k'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "P X Q=(Pxi+Pyj+Pzk)x(Qxi+Qyj+Qzk)\n",
- " =(PxQx)i x i+(PxQy)i x j +(PxQz)i x k\n",
- "But i x i = j x j = k x k =0; and i x j =k and j x i= -k, etc. Hence\n",
- "P X Q= (PxQy)k-(PxQz)j-(PyQx)k+(PxQz)i+(PzQx)j-(PzQy)i\n",
- "These terms can be grouped as\n",
- "P X Q=(PyQz-PzQy)i+(PzQx-PxQz)j+(PxQy-PyQx)k\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-13,Page No: 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "Fx=2.63 #N\n",
- "Fy=4.28 #N\n",
- "Fz=-5.92 #N\n",
- "\n",
- "#Calculation\n",
- "\n",
- "F=sqrt(Fx**2+Fy**2+Fz**2) #N\n",
- "thetax=((arccos(Fx/F))*180)/pi #degrees\n",
- "thetay=((arccos(Fy/F))*180)/pi #degrees\n",
- "thetaz=((arccos(Fz/F))*180)/pi #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The magnitude of force is',round(F,2),\"N\"\n",
- "print'Thetax',round(thetax,1),\"degrees\"\n",
- "print'Thetay',round(thetay,1),\"degrees\"\n",
- "print'Thetaz',round(thetaz,1),\"degrees\"\n",
- "\n",
- "# Decimal point error may cause a small discrepancy in the answers."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The magnitude of force is 7.76 N\n",
- "Thetax 70.2 degrees\n",
- "Thetay 56.5 degrees\n",
- "Thetaz 139.7 degrees\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-14, Page No: 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P=[4.82, -2.33, 5.47] #N\n",
- "Q=[-2.81,-6.09,1.12 ] #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "M=P[0]*Q[0]+P[1]*Q[1]+P[2]*Q[2] #Nm\n",
- "\n",
- "#Results\n",
- "\n",
- "print'Result is',round(M,2),\"N.m\"\n",
- "\n",
- "# Decimal point error in calculation causes a small discrepancy in the answer."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Result is 6.77 N.m\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-15, Page No: 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "#Initilization of variables\n",
- "\n",
- "x1=2 #units\n",
- "x2=-2 #units\n",
- "y1=3 #units\n",
- "y2=4 #units\n",
- "z1=0 #units\n",
- "z2=6 #units\n",
- "P=np.array([2,3,-1]) #units\n",
- "\n",
- "#Calculations\n",
- "\n",
- "X=sqrt((x2-x1)**2+(y2-y1)**2+(z2-z1)**2) #units\n",
- "eLx=(x2-x1)/X #units\n",
- "eLy=(y2-y1)/X #units\n",
- "eLz=(z2-z1)/X #units\n",
- "Q=np.array([eLx,eLy,eLz]) #units\n",
- "Z=P[0]*Q[0]+P[1]*Q[1]+P[2]*Q[2] # units\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The unit vector is',round(eLx,3),\"i +\",round(eLy,3),\"j +\",round(eLz,3),\"k\"\n",
- "print'The projection of P is',round(Z,2),\"units\"\n",
- "\n",
- "#Note:The final answer for the projection of P is off by 0.1 units\n",
- "#The answer mentioned in the textbook is -1.41\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The unit vector is -0.549 i + 0.137 j + 0.824 k\n",
- "The projection of P is -1.51 units\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-16, Page No: 14"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "#Initilization of variables\n",
- "\n",
- "x1=2 #units\n",
- "x2=5 #units\n",
- "y1=-5 #units\n",
- "y2=2 #units\n",
- "z1=3 #units\n",
- "z2=-4 #units\n",
- "P=np.array([10,-8,14]) #units\n",
- "\n",
- "#Calculations\n",
- "\n",
- "X=sqrt((x2-x1)**2+(y2-y1)**2+(z2-z1)**2) #units\n",
- "eLx=(x2-x1)/X #units\n",
- "eLy=(y2-y1)/X #units\n",
- "eLz=(z2-z1)/X #units\n",
- "Q=np.array([eLx,eLy,eLz]) #units\n",
- "Z=P[0]*Q[0]+P[1]*Q[1]+P[2]*Q[2] #units\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The unit vector is',round(eLx,3),\"i +\",round(eLy,3),\"j\",round(eLz,3),\"k\" \n",
- "print'The projection of P is',round(Z),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The unit vector is 0.29 i + 0.677 j -0.677 k\n",
- "The projection of P is -12.0 lb\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-17, Page No: 14\n"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "Px=2.85 #ft\n",
- "Py=4.67 #ft\n",
- "Pz=-8.09 #ft\n",
- "Qx=28.3 #lb\n",
- "Qy=44.6 #lb\n",
- "Qz=53.3 #lb\n",
- "\n",
- "#Calculations\n",
- "\n",
- "X=(Py*Qz-Pz*Qy) #N.m \n",
- "Y=(Pz*Qx-Px*Qz) #N.m\n",
- "Z=(Px*Qy-Py*Qx) #N.m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The cross product is',round(X),\"i\",round(Y),\"j\",round(Z),\"k lb-ft\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The cross product is 610.0 i -381.0 j -5.0 k lb-ft\n"
- ]
- }
- ],
- "prompt_number": 45
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-18, Page No: 14"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Result\n",
- "#As this is symbolic solution directly print command is being used to give the required output\n",
- "\n",
- "print'The Time derivative is '\n",
- "print'dr/dt=(dx/dt)i+12*y(dy/dt)j-3*(dz/dt)k'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Time derivative is \n",
- "dr/dt=(dx/dt)i+12*y(dy/dt)j-3*(dz/dt)k\n"
- ]
- }
- ],
- "prompt_number": 47
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-19, Page No: 14"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return x**2\n",
- "a=1\n",
- "b=1\n",
- "I=quad(integrand, 1, 3, args=(a,b))\n",
- "\n",
- "def integrand(y, a, b):\n",
- " return 2*y\n",
- "a=1\n",
- "b=1\n",
- "J=quad(integrand, 1, 3, args=(a,b))\n",
- "\n",
- "def integrand(z, a, b):\n",
- " return 1\n",
- "a=1\n",
- "b=1\n",
- "K=quad(integrand, 1, 3, args=(a,b))\n",
- "\n",
- "# Results\n",
- "print'The answer is',round(I[0],2),\"i +\",round(J[0]),\"j -\",round(K[0]),\"k.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The answer is 8.67 i + 8.0 j - 2.0 k.\n"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter1_3.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter1_3.ipynb deleted file mode 100755 index c6f2d00b..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter1_3.ipynb +++ /dev/null @@ -1,928 +0,0 @@ -{
- "metadata": {
- "name": "chapter1.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 1:Vectors"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-1,Page no: 8"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initialisation of Variables\n",
- "\n",
- "f1=120 #lb\n",
- "f2=100 #lb\n",
- "theta=((60*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=sqrt(120**2+100**2-(2*120*100*cos(theta))) #Applying Thr rule of Cosines\n",
- "alpha1=(((arcsin(120*sin(theta)/111))*180)/pi) #Applying the Law of Sines\n",
- "alpha=alpha1+270 #As the vector lies in the fourth Quadrant by obsrevaton\n",
- "\n",
- "#Results\n",
- "\n",
- "print'The Resultant of The force system is equal to',round(R),\"lb\" #lb\n",
- "print'The Resultant is at',round(alpha),\"degrees\" #degrees\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Resultant of The force system is equal to 111.0 lb\n",
- "The Resultant is at 339.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-2,Page no: 9"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P=100 #lb\n",
- "Q=120 #lb\n",
- "theta=((30*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R_x=Q*cos(theta) #lb\n",
- "R_y=Q*sin(theta)-P #lb\n",
- "R=sqrt(R_x**2+R_y**2) #lb Triangle law\n",
- "Theta_1=((arctan(R_y/R_x))*180)/pi #degrees\n",
- "Theta_R=360+Theta_1 #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The resultant is at',round(Theta_R),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 111.0 lb\n",
- "The resultant is at 339.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-3,Page No: 9"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "# Initialization of variables\n",
- "\n",
- "R=400 # N\n",
- "F2=200 # N\n",
- "Theta1=((120*pi)/180) # radians\n",
- "Theta2=((20*pi)/180) # radians\n",
- "Theta=Theta1-Theta2 # radians\n",
- "\n",
- "# Calculation\n",
- "\n",
- "F=sqrt(R**2+F2**2-(2*R*F2*cos(Theta))) # N.Applying the Rule of Cosine\n",
- "Theta_r=arcsin((400*sin(Theta))/F) #radians Applying the rule of sines\n",
- "Theta_R=(Theta_r*180)/pi\n",
- "\n",
- "# Result\n",
- "\n",
- "print'The resultant of the force system is',round(F),\"N\"\n",
- "print'The Angle between F and 200N force is',round(Theta_R,1),\"degrees\"\n",
- "\n",
- "# Theta_R is off by 0.1 degrees"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 477.0 N\n",
- "The Angle between F and 200N force is 55.6 degrees\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-4, Page No: 9"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "# Initilization of variables\n",
- "\n",
- "F1=280 # N\n",
- "F2=130 # N\n",
- "Theta1=((320*pi)/180) # Radians\n",
- "Theta2=((60*pi)/180) # Radians\n",
- "\n",
- "# Calculations\n",
- "\n",
- "R_x=-F1*cos(Theta1)+F2*cos(Theta2) # N\n",
- "R_y=F1*sin(Theta1)-F2*sin(Theta2) # N\n",
- "R=sqrt(R_x**2+R_y**2) # N Applying Triangle Law\n",
- "ThetaR=arctan(R_y/R_x) # radians\n",
- "Theta_R=360-(ThetaR*180/pi) # degrees\n",
- "\n",
- "# Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"N\"\n",
- "print'The resultant is at',round(Theta_R),\"degrees\"\n",
- "\n",
- "# The answer for R waries from textbook. "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 329.0 N\n",
- "The resultant is at 297.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-5, Page No: 10"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "# Initialization of variables\n",
- "\n",
- "F1=26 #lb\n",
- "F2=39 #lb\n",
- "F3=63 #lb\n",
- "F4=57 #lb\n",
- "T1=((10*pi)/180) #Radians\n",
- "T2=((114*pi)/180) #Radians\n",
- "T3=((183*pi)/180) #radians\n",
- "T4=((261*pi)/180) #radians\n",
- "\n",
- "# Calculations\n",
- "\n",
- "R_x=F1*cos(T1)+F2*cos(T2)+F3*cos(T3)+F4*cos(T4) # lb Resolving vectors\n",
- "R_y=F1*sin(T1)+F2*sin(T2)+F3*sin(T3)+F4*sin(T4) # lb resolving vectors\n",
- "R=sqrt(R_x**2+R_y**2) # lb Applying Triangle Law\n",
- "theta=arctan(R_y/R_x)# radians\n",
- "Theta=(theta*180)/pi # degrees\n",
- "Theta_R=180+Theta\n",
- "\n",
- "# Results\n",
- "\n",
- "print'The Resultant of the force system is',round(R),\"lb\"\n",
- "print'The resultant is at',round(Theta_R),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Resultant of the force system is 65.0 lb\n",
- "The resultant is at 197.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 36
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-6, Page No: 11"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=10 #lb\n",
- "theta1=((60*pi)/180) #radians\n",
- "theta2=((45*pi)/180) #radians\n",
- "theta=theta1-theta2 #radians\n",
- "\n",
- "#Calculation\n",
- "\n",
- "F_OH=F/cos(theta) #lb resolving vectors\n",
- "\n",
- "# Result\n",
- "\n",
- "print'The component of F in the direction of OH is',round(F_OH,2),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The component of F in the direction of OH is 10.35 lb\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-7, Page No: 11"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "weight=80 #kg\n",
- "theta=((20*pi)/180) #radians\n",
- "theta_p=((70*pi)/180) # radians\n",
- "\n",
- "#Calcuations\n",
- "\n",
- "#Part (a)\n",
- "F=weight*9.81 # N\n",
- "R=F*cos(theta) #N\n",
- "#part (b)\n",
- "R_p=F*cos(theta_p) #N\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The normal component is',round(R),\"N\"\n",
- "print'The parallel component is',round(R_p),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The normal component is 737.0 N\n",
- "The parallel component is 268.0 N\n"
- ]
- }
- ],
- "prompt_number": 37
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-8, Page No: 11"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P=235 #N\n",
- "theta=((60*pi)/180) #radians\n",
- "bet=((22*pi)/180) #radians\n",
- "gam=((38*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Part (a)\n",
- "P_h=P*cos(theta) #N\n",
- "P_v=P*sin(theta) #N\n",
- "#Part (b)\n",
- "P_l=P*cos(theta-bet) #N\n",
- "P_p=P*sin(gam) #N\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The horizontal component is',round(P_h,1),\"N\"\n",
- "print'The vertical component is',round(P_v,1),\"N\"\n",
- "print'The component parallel to plane is',round(P_l),\"N\"\n",
- "print'The component perpendicular to the plane is',round(P_p,1),\"N\"\n",
- "\n",
- "#The decimal point accuracy might cause a small discrepancy in the answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The horizontal component is 117.5 N\n",
- "The vertical component is 203.5 N\n",
- "The component parallel to plane is 185.0 N\n",
- "The component perpendicular to the plane is 144.7 N\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-9, Page No: 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=90 #lb\n",
- "theta1=((40*pi)/180) #radians\n",
- "theta2=((30*pi)/180) #radians\n",
- "\n",
- "# Calculations\n",
- "\n",
- "R_x=0 #lb\n",
- "R_y=20 # lb\n",
- "#Taking the sum of forces in the X-Direction\n",
- "P=((F1*cos(theta1))/cos(theta2)) # lb\n",
- "# Taking the sum of the forces in the Y-Direction\n",
- "F=(P*sin(theta2))+(F1*sin(theta1))-20 #lb\n",
- "\n",
- "# Results\n",
- "\n",
- "print'The value of P is',round(P,1),\"lb\"\n",
- "print'The value of F is',round(F,1),\"lb\"\n",
- "\n",
- "# Decimal point error may cause a small discrepancy in the answers."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P is 79.6 lb\n",
- "The value of F is 77.7 lb\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-10, Page No: 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "x=4 # m\n",
- "y=3 # m\n",
- "z=2 # m\n",
- "F=50 #N\n",
- "\n",
- "# Calculations\n",
- "\n",
- "OP=sqrt(x**2+y**2+z**2) #m\n",
- "thetax=(x/OP) #radians\n",
- "thetay=(y/OP) #Radians\n",
- "thetaz=(z/OP) #radians\n",
- "P_x=F*(thetax) #N\n",
- "P_y=F*(thetay) #N\n",
- "P_z=F*(thetaz) #N\n",
- "\n",
- "# Result\n",
- "\n",
- "print'The vector P is',round(P_x,1),\"i +\",round(P_y,1),\"j +\",round(P_z,1),\"k\"\n",
- "\n",
- "# component of i is off by 0.1 units"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The vector P is 37.1 i + 27.9 j + 18.6 k\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-11, Page No: 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "x=2 \n",
- "y=-4\n",
- "z=1\n",
- "F=100 #N\n",
- "\n",
- "#Calculation\n",
- "\n",
- "thetax=x/sqrt(x**2+y**2+z**2) #radians\n",
- "thetay=y/sqrt(x**2+y**2+z**2) #radians\n",
- "thetaz=z/sqrt(x**2+y**2+z**2) #radians\n",
- "P_x=F*thetax #N\n",
- "P_y=F*thetay #N\n",
- "P_z=F*thetaz #N\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The vector P is',round(P_x,1),\"i\",round(P_y,1),\"j +\",round(P_z,1),\"k\"\n",
- "\n",
- "# component off i is off by 0.1 units"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The vector P is 43.6 i -87.3 j + 21.8 k\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-12, Page No: 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Solution\n",
- "\n",
- "print'P X Q=(Pxi+Pyj+Pzk)x(Qxi+Qyj+Qzk)'\n",
- "print' =(PxQx)i x i+(PxQy)i x j +(PxQz)i x k'\n",
- "print'But i x i = j x j = k x k =0; and i x j =k and j x i= -k, etc. Hence'\n",
- "print'P X Q= (PxQy)k-(PxQz)j-(PyQx)k+(PxQz)i+(PzQx)j-(PzQy)i'\n",
- "print'These terms can be grouped as'\n",
- "print'P X Q=(PyQz-PzQy)i+(PzQx-PxQz)j+(PxQy-PyQx)k'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "P X Q=(Pxi+Pyj+Pzk)x(Qxi+Qyj+Qzk)\n",
- " =(PxQx)i x i+(PxQy)i x j +(PxQz)i x k\n",
- "But i x i = j x j = k x k =0; and i x j =k and j x i= -k, etc. Hence\n",
- "P X Q= (PxQy)k-(PxQz)j-(PyQx)k+(PxQz)i+(PzQx)j-(PzQy)i\n",
- "These terms can be grouped as\n",
- "P X Q=(PyQz-PzQy)i+(PzQx-PxQz)j+(PxQy-PyQx)k\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-13,Page No: 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "Fx=2.63 #N\n",
- "Fy=4.28 #N\n",
- "Fz=-5.92 #N\n",
- "\n",
- "#Calculation\n",
- "\n",
- "F=sqrt(Fx**2+Fy**2+Fz**2) #N\n",
- "thetax=((arccos(Fx/F))*180)/pi #degrees\n",
- "thetay=((arccos(Fy/F))*180)/pi #degrees\n",
- "thetaz=((arccos(Fz/F))*180)/pi #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The magnitude of force is',round(F,2),\"N\"\n",
- "print'Thetax',round(thetax,1),\"degrees\"\n",
- "print'Thetay',round(thetay,1),\"degrees\"\n",
- "print'Thetaz',round(thetaz,1),\"degrees\"\n",
- "\n",
- "# Decimal point error may cause a small discrepancy in the answers."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The magnitude of force is 7.76 N\n",
- "Thetax 70.2 degrees\n",
- "Thetay 56.5 degrees\n",
- "Thetaz 139.7 degrees\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-14, Page No: 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P=[4.82, -2.33, 5.47] #N\n",
- "Q=[-2.81,-6.09,1.12 ] #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "M=P[0]*Q[0]+P[1]*Q[1]+P[2]*Q[2] #Nm\n",
- "\n",
- "#Results\n",
- "\n",
- "print'Result is',round(M,2),\"N.m\"\n",
- "\n",
- "# Decimal point error in calculation causes a small discrepancy in the answer."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Result is 6.77 N.m\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-15, Page No: 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "#Initilization of variables\n",
- "\n",
- "x1=2 #units\n",
- "x2=-2 #units\n",
- "y1=3 #units\n",
- "y2=4 #units\n",
- "z1=0 #units\n",
- "z2=6 #units\n",
- "P=np.array([2,3,-1]) #units\n",
- "\n",
- "#Calculations\n",
- "\n",
- "X=sqrt((x2-x1)**2+(y2-y1)**2+(z2-z1)**2) #units\n",
- "eLx=(x2-x1)/X #units\n",
- "eLy=(y2-y1)/X #units\n",
- "eLz=(z2-z1)/X #units\n",
- "Q=np.array([eLx,eLy,eLz]) #units\n",
- "Z=P[0]*Q[0]+P[1]*Q[1]+P[2]*Q[2] # units\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The unit vector is',round(eLx,3),\"i +\",round(eLy,3),\"j +\",round(eLz,3),\"k\"\n",
- "print'The projection of P is',round(Z,2),\"units\"\n",
- "\n",
- "#Note:The final answer for the projection of P is off by 0.1 units\n",
- "#The answer mentioned in the textbook is -1.41\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The unit vector is -0.549 i + 0.137 j + 0.824 k\n",
- "The projection of P is -1.51 units\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-16, Page No: 14"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "#Initilization of variables\n",
- "\n",
- "x1=2 #units\n",
- "x2=5 #units\n",
- "y1=-5 #units\n",
- "y2=2 #units\n",
- "z1=3 #units\n",
- "z2=-4 #units\n",
- "P=np.array([10,-8,14]) #units\n",
- "\n",
- "#Calculations\n",
- "\n",
- "X=sqrt((x2-x1)**2+(y2-y1)**2+(z2-z1)**2) #units\n",
- "eLx=(x2-x1)/X #units\n",
- "eLy=(y2-y1)/X #units\n",
- "eLz=(z2-z1)/X #units\n",
- "Q=np.array([eLx,eLy,eLz]) #units\n",
- "Z=P[0]*Q[0]+P[1]*Q[1]+P[2]*Q[2] #units\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The unit vector is',round(eLx,3),\"i +\",round(eLy,3),\"j\",round(eLz,3),\"k\" \n",
- "print'The projection of P is',round(Z),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The unit vector is 0.29 i + 0.677 j -0.677 k\n",
- "The projection of P is -12.0 lb\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-17, Page No: 14\n"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "Px=2.85 #ft\n",
- "Py=4.67 #ft\n",
- "Pz=-8.09 #ft\n",
- "Qx=28.3 #lb\n",
- "Qy=44.6 #lb\n",
- "Qz=53.3 #lb\n",
- "\n",
- "#Calculations\n",
- "\n",
- "X=(Py*Qz-Pz*Qy) #N.m \n",
- "Y=(Pz*Qx-Px*Qz) #N.m\n",
- "Z=(Px*Qy-Py*Qx) #N.m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The cross product is',round(X),\"i\",round(Y),\"j\",round(Z),\"k lb-ft\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The cross product is 610.0 i -381.0 j -5.0 k lb-ft\n"
- ]
- }
- ],
- "prompt_number": 45
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-18, Page No: 14"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Result\n",
- "#As this is symbolic solution directly print command is being used to give the required output\n",
- "\n",
- "print'The Time derivative is '\n",
- "print'dr/dt=(dx/dt)i+12*y(dy/dt)j-3*(dz/dt)k'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Time derivative is \n",
- "dr/dt=(dx/dt)i+12*y(dy/dt)j-3*(dz/dt)k\n"
- ]
- }
- ],
- "prompt_number": 47
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1-19, Page No: 14"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return x**2\n",
- "a=1\n",
- "b=1\n",
- "I=quad(integrand, 1, 3, args=(a,b))\n",
- "\n",
- "def integrand(y, a, b):\n",
- " return 2*y\n",
- "a=1\n",
- "b=1\n",
- "J=quad(integrand, 1, 3, args=(a,b))\n",
- "\n",
- "def integrand(z, a, b):\n",
- " return 1\n",
- "a=1\n",
- "b=1\n",
- "K=quad(integrand, 1, 3, args=(a,b))\n",
- "\n",
- "# Results\n",
- "print'The answer is',round(I[0],2),\"i +\",round(J[0]),\"j -\",round(K[0]),\"k.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The answer is 8.67 i + 8.0 j - 2.0 k.\n"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter2.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter2.ipynb deleted file mode 100755 index 9a6ee7f9..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter2.ipynb +++ /dev/null @@ -1,562 +0,0 @@ -{
- "metadata": {
- "name": "chapter 2.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 2: Operations with Forces"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-1, Page No: 21"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=20 #lb\n",
- "L=4.33 #ft\n",
- "#Calculation\n",
- "M=-F*L #lb-ft\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The moment of force F about O is',round(M,1),\"lb-ft\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of force F about O is -86.6 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-2, Page no: 22"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=20 #lb\n",
- "theta=((60*pi)/180) #radians\n",
- "L=5 #ft\n",
- "\n",
- "#Calculations\n",
- "\n",
- "F_x=F*cos(theta) #Resloving the vector\n",
- "F_y=F*sin(theta) #Resloving the vector\n",
- "M=-F_y*L #Appling Varignon's theorem\n",
- "#Negative sign tells that moment is clockwise\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The moment of the force about O is',round(M,1),\"lb-ft\"\n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of the force about O is -86.6 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-3,Page No: 22"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=100 #N\n",
- "x1=2 #m\n",
- "x2=5 #m\n",
- "y1=0 #m\n",
- "y2=1 #m\n",
- "z1=4 #m\n",
- "z2=1 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "xside=(x2-x1) #m\n",
- "yside=(y2-y1) #m\n",
- "zside=(z2-z1) #m\n",
- "LD=sqrt(xside**2+yside**2+zside**2)\n",
- "Fx=(xside/LD)*F #N\n",
- "Fy=(yside/LD)*F #N\n",
- "Fz=(zside/LD)*F #N\n",
- "Mx=-Fy*z1 #N-m\n",
- "My=Fx*x1-Fz*z1 #N-m\n",
- "Mz=Fy*x1 #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'Fx is',round(Fx,1),\"N\"\n",
- "print'Fy is',round(Fy,1),\"N\"\n",
- "print'Fz is',round(Fz,1),\"N\"\n",
- "print'Moment about X-Axis is',round(Mx,1),\"N.m\"\n",
- "print'Moment about Y-Axis is +',round(My),\"N.m\"\n",
- "print'Moment about Z-Axis is +',round(Mz,1),\"N.m\"\n",
- "\n",
- "# Decimal point error in calculation causes a small discrepancy in the resulting solutions."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Fx is 68.8 N\n",
- "Fy is 22.9 N\n",
- "Fz is -68.8 N\n",
- "Moment about X-Axis is -91.8 N.m\n",
- "Moment about Y-Axis is + 413.0 N.m\n",
- "Moment about Z-Axis is + 45.9 N.m\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-4, Page No: 23"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "Fx=68.7 #N\n",
- "Fy=22.9 #N\n",
- "Fz=-68.7 #N\n",
- "rx=2 #m\n",
- "ry=0 #m\n",
- "rz=4 #m\n",
- "rx1=5 #m\n",
- "ry1=1 #m\n",
- "rz1=1 #m\n",
- "\n",
- "#Calculation\n",
- "\n",
- "Mx=Fz*ry-Fy*rz #N-m\n",
- "My=-(Fz*rx-Fx*rz) #N-m\n",
- "Mz=Fy*rx-Fx*ry #N-m\n",
- "Mx1=Fz*ry1-Fy*rz1 #N-m\n",
- "My1=-(Fz*rx1-Fx*rz1) #N-m\n",
- "Mz1=Fy*rx1-Fx*ry1 #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'Moment with respect to origin using point(2,0,4)is',round(Mx,1),\"i +\",round(My),\"j +\",round(Mz,1),\"k N.m\"\n",
- "print'Moment with respect to origin using point (5,1,1) is',round(Mx1,1),\"i +\",round(My1),\"j +\",round(Mz1,1),\"k N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Moment with respect to origin using point(2,0,4)is -91.6 i + 412.0 j + 45.8 k N.m\n",
- "Moment with respect to origin using point (5,1,1) is -91.6 i + 412.0 j + 45.8 k N.m\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-5, Page no: 23"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "Fx=2 #lb\n",
- "Fy=3 #lb\n",
- "Fz=-1 #lb\n",
- "rx=1 #ft\n",
- "ry=-4 #ft\n",
- "rz=3 #ft\n",
- "#Coordinates of points\n",
- "ax=3 #ft\n",
- "ay=1 #ft\n",
- "az=1 #ft\n",
- "bx=3 #ft\n",
- "by=-1 #ft\n",
- "bz=1 #ft\n",
- "cx=2 #ft\n",
- "cy=5 #ft\n",
- "cz=-2 #ft\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Rx=ax-cx #ft\n",
- "Ry=ay-cy #ft\n",
- "Rz=az-cz #ft\n",
- "Mx=(Ry*Fz)-(Rz*Fy) #lb-ft\n",
- "My=-((Rx*Fz)-(Rz*Fx)) #lb-ft\n",
- "Mz=(Rx*Fy)-(Ry*Fx) #lb-ft\n",
- "E_u=sqrt((bx-cx)**2+(by-cy)**2+(bz-cz)**2) #ft\n",
- "ex=(bx-cx)/E_u #ft\n",
- "ey=(by-cy)/E_u #ft\n",
- "ez=(bz-cz)/E_u #ft\n",
- "M_lx=Mx*ex #lb-ft\n",
- "M_ly=My*ey #lb-ft\n",
- "M_lz=Mz*ez #lb-ft\n",
- "M_l=M_lx+M_ly+M_lz #lb-ft\n",
- "\n",
- "#Result\n",
- "\n",
- "print'Hence the moment about line is',round(M_l,2),\"lb-ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Hence the moment about line is -2.06 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-6, Page No: 24"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P_x=22 #N\n",
- "P_y=23 #N\n",
- "P_z=7 #N\n",
- "p1=1 #m\n",
- "p2=-1 #m\n",
- "p3=-2 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Mx=(p2*P_z)-(p3*P_y) #N-m\n",
- "My=-((p1*P_z)-(p3*P_x)) #N-m\n",
- "Mz=(p1*P_y)-(p2*P_x) #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The moment about the line from the origin is',round(Mx),\"i\",round(My),\"j +\",round(Mz),\"k N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment about the line from the origin is 39.0 i -51.0 j + 45.0 k N.m\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-8, Page No: 24"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=10 #N Force couple\n",
- "a=3 #m Moment arm\n",
- "#Calculations\n",
- "C=-F*a #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant couple is',round(C),\"N-m\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant couple is -30.0 N-m\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-11, Page No: 26"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "C1=20 #N-m\n",
- "C2=40 #N-m\n",
- "C3=-55 #N-m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "C=sqrt(C1**2+C2**2+C3**2) #N-m\n",
- "thetax=C2/C \n",
- "thetay=C3/C\n",
- "thetaz=C1/C\n",
- "Cx=C*thetax #N-m\n",
- "Cy=C*thetay #N-m\n",
- "Cz=C*thetaz #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'Couple in vector notation is',round(Cx),\"i\",round(Cy),\"j +\",round(Cz),\"k N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Couple in vector notation is 40.0 i -55.0 j + 20.0 k N.m\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-12,Page No: 26"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=25 #lb\n",
- "F2=25 #lb\n",
- "L1=14 #in\n",
- "L2=20 #in\n",
- "\n",
- "#Calculations\n",
- "\n",
- "C=F1*L1 #lb-in\n",
- "M=-F2*L2 #lb-in\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The twisting couple is',round(C),\"lb-in\"\n",
- "print'The bending moment is',round(M),\"lb-in\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The twisting couple is 350.0 lb-in\n",
- "The bending moment is -500.0 lb-in\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-13, Page No: 27"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "rx=20 #in\n",
- "ry=0 #in\n",
- "rz=14 #in\n",
- "Fx=0 #lb\n",
- "Fy=-25 #lb\n",
- "Fz=0 #lb\n",
- "\n",
- "#Calculation\n",
- "\n",
- "Mx=ry*Fz-rz*Fy #lb-in\n",
- "My=rx*Fz-rz*Fx #lb-in\n",
- "Mz=rx*Fy-ry*Fx #lb-in\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The moment of the 25-lb force is',round(Mx),\"i +\",round(My),\"j\",round(Mz),\"k lb-in\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of the 25-lb force is 350.0 i + 0.0 j -500.0 k lb-in\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-14,Page No: 27"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "#Co-ordinates with respect to point O\n",
- "x=17.9 #ft\n",
- "y=6.91 #ft\n",
- "z=46.3 #ft\n",
- "Fz=-4000 #lb\n",
- "Fy=0 #lb\n",
- "\n",
- "#Calculation\n",
- "\n",
- "Mx=y*Fz-z*Fy #lb-ft\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The scalar coefficient of the i term is the moment about the X-Axis is',round(Mx,3),\"lb-ft\"\n",
- "\n",
- "# The answer given in the textbook is incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The scalar coefficient of the i term is the moment about the X-Axis is -27640.0 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 16
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter2_1.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter2_1.ipynb deleted file mode 100755 index 9a6ee7f9..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter2_1.ipynb +++ /dev/null @@ -1,562 +0,0 @@ -{
- "metadata": {
- "name": "chapter 2.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 2: Operations with Forces"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-1, Page No: 21"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=20 #lb\n",
- "L=4.33 #ft\n",
- "#Calculation\n",
- "M=-F*L #lb-ft\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The moment of force F about O is',round(M,1),\"lb-ft\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of force F about O is -86.6 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-2, Page no: 22"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=20 #lb\n",
- "theta=((60*pi)/180) #radians\n",
- "L=5 #ft\n",
- "\n",
- "#Calculations\n",
- "\n",
- "F_x=F*cos(theta) #Resloving the vector\n",
- "F_y=F*sin(theta) #Resloving the vector\n",
- "M=-F_y*L #Appling Varignon's theorem\n",
- "#Negative sign tells that moment is clockwise\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The moment of the force about O is',round(M,1),\"lb-ft\"\n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of the force about O is -86.6 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-3,Page No: 22"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=100 #N\n",
- "x1=2 #m\n",
- "x2=5 #m\n",
- "y1=0 #m\n",
- "y2=1 #m\n",
- "z1=4 #m\n",
- "z2=1 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "xside=(x2-x1) #m\n",
- "yside=(y2-y1) #m\n",
- "zside=(z2-z1) #m\n",
- "LD=sqrt(xside**2+yside**2+zside**2)\n",
- "Fx=(xside/LD)*F #N\n",
- "Fy=(yside/LD)*F #N\n",
- "Fz=(zside/LD)*F #N\n",
- "Mx=-Fy*z1 #N-m\n",
- "My=Fx*x1-Fz*z1 #N-m\n",
- "Mz=Fy*x1 #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'Fx is',round(Fx,1),\"N\"\n",
- "print'Fy is',round(Fy,1),\"N\"\n",
- "print'Fz is',round(Fz,1),\"N\"\n",
- "print'Moment about X-Axis is',round(Mx,1),\"N.m\"\n",
- "print'Moment about Y-Axis is +',round(My),\"N.m\"\n",
- "print'Moment about Z-Axis is +',round(Mz,1),\"N.m\"\n",
- "\n",
- "# Decimal point error in calculation causes a small discrepancy in the resulting solutions."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Fx is 68.8 N\n",
- "Fy is 22.9 N\n",
- "Fz is -68.8 N\n",
- "Moment about X-Axis is -91.8 N.m\n",
- "Moment about Y-Axis is + 413.0 N.m\n",
- "Moment about Z-Axis is + 45.9 N.m\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-4, Page No: 23"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "Fx=68.7 #N\n",
- "Fy=22.9 #N\n",
- "Fz=-68.7 #N\n",
- "rx=2 #m\n",
- "ry=0 #m\n",
- "rz=4 #m\n",
- "rx1=5 #m\n",
- "ry1=1 #m\n",
- "rz1=1 #m\n",
- "\n",
- "#Calculation\n",
- "\n",
- "Mx=Fz*ry-Fy*rz #N-m\n",
- "My=-(Fz*rx-Fx*rz) #N-m\n",
- "Mz=Fy*rx-Fx*ry #N-m\n",
- "Mx1=Fz*ry1-Fy*rz1 #N-m\n",
- "My1=-(Fz*rx1-Fx*rz1) #N-m\n",
- "Mz1=Fy*rx1-Fx*ry1 #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'Moment with respect to origin using point(2,0,4)is',round(Mx,1),\"i +\",round(My),\"j +\",round(Mz,1),\"k N.m\"\n",
- "print'Moment with respect to origin using point (5,1,1) is',round(Mx1,1),\"i +\",round(My1),\"j +\",round(Mz1,1),\"k N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Moment with respect to origin using point(2,0,4)is -91.6 i + 412.0 j + 45.8 k N.m\n",
- "Moment with respect to origin using point (5,1,1) is -91.6 i + 412.0 j + 45.8 k N.m\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-5, Page no: 23"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "Fx=2 #lb\n",
- "Fy=3 #lb\n",
- "Fz=-1 #lb\n",
- "rx=1 #ft\n",
- "ry=-4 #ft\n",
- "rz=3 #ft\n",
- "#Coordinates of points\n",
- "ax=3 #ft\n",
- "ay=1 #ft\n",
- "az=1 #ft\n",
- "bx=3 #ft\n",
- "by=-1 #ft\n",
- "bz=1 #ft\n",
- "cx=2 #ft\n",
- "cy=5 #ft\n",
- "cz=-2 #ft\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Rx=ax-cx #ft\n",
- "Ry=ay-cy #ft\n",
- "Rz=az-cz #ft\n",
- "Mx=(Ry*Fz)-(Rz*Fy) #lb-ft\n",
- "My=-((Rx*Fz)-(Rz*Fx)) #lb-ft\n",
- "Mz=(Rx*Fy)-(Ry*Fx) #lb-ft\n",
- "E_u=sqrt((bx-cx)**2+(by-cy)**2+(bz-cz)**2) #ft\n",
- "ex=(bx-cx)/E_u #ft\n",
- "ey=(by-cy)/E_u #ft\n",
- "ez=(bz-cz)/E_u #ft\n",
- "M_lx=Mx*ex #lb-ft\n",
- "M_ly=My*ey #lb-ft\n",
- "M_lz=Mz*ez #lb-ft\n",
- "M_l=M_lx+M_ly+M_lz #lb-ft\n",
- "\n",
- "#Result\n",
- "\n",
- "print'Hence the moment about line is',round(M_l,2),\"lb-ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Hence the moment about line is -2.06 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-6, Page No: 24"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P_x=22 #N\n",
- "P_y=23 #N\n",
- "P_z=7 #N\n",
- "p1=1 #m\n",
- "p2=-1 #m\n",
- "p3=-2 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Mx=(p2*P_z)-(p3*P_y) #N-m\n",
- "My=-((p1*P_z)-(p3*P_x)) #N-m\n",
- "Mz=(p1*P_y)-(p2*P_x) #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The moment about the line from the origin is',round(Mx),\"i\",round(My),\"j +\",round(Mz),\"k N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment about the line from the origin is 39.0 i -51.0 j + 45.0 k N.m\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-8, Page No: 24"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=10 #N Force couple\n",
- "a=3 #m Moment arm\n",
- "#Calculations\n",
- "C=-F*a #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant couple is',round(C),\"N-m\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant couple is -30.0 N-m\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-11, Page No: 26"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "C1=20 #N-m\n",
- "C2=40 #N-m\n",
- "C3=-55 #N-m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "C=sqrt(C1**2+C2**2+C3**2) #N-m\n",
- "thetax=C2/C \n",
- "thetay=C3/C\n",
- "thetaz=C1/C\n",
- "Cx=C*thetax #N-m\n",
- "Cy=C*thetay #N-m\n",
- "Cz=C*thetaz #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'Couple in vector notation is',round(Cx),\"i\",round(Cy),\"j +\",round(Cz),\"k N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Couple in vector notation is 40.0 i -55.0 j + 20.0 k N.m\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-12,Page No: 26"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=25 #lb\n",
- "F2=25 #lb\n",
- "L1=14 #in\n",
- "L2=20 #in\n",
- "\n",
- "#Calculations\n",
- "\n",
- "C=F1*L1 #lb-in\n",
- "M=-F2*L2 #lb-in\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The twisting couple is',round(C),\"lb-in\"\n",
- "print'The bending moment is',round(M),\"lb-in\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The twisting couple is 350.0 lb-in\n",
- "The bending moment is -500.0 lb-in\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-13, Page No: 27"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "rx=20 #in\n",
- "ry=0 #in\n",
- "rz=14 #in\n",
- "Fx=0 #lb\n",
- "Fy=-25 #lb\n",
- "Fz=0 #lb\n",
- "\n",
- "#Calculation\n",
- "\n",
- "Mx=ry*Fz-rz*Fy #lb-in\n",
- "My=rx*Fz-rz*Fx #lb-in\n",
- "Mz=rx*Fy-ry*Fx #lb-in\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The moment of the 25-lb force is',round(Mx),\"i +\",round(My),\"j\",round(Mz),\"k lb-in\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of the 25-lb force is 350.0 i + 0.0 j -500.0 k lb-in\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-14,Page No: 27"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "#Co-ordinates with respect to point O\n",
- "x=17.9 #ft\n",
- "y=6.91 #ft\n",
- "z=46.3 #ft\n",
- "Fz=-4000 #lb\n",
- "Fy=0 #lb\n",
- "\n",
- "#Calculation\n",
- "\n",
- "Mx=y*Fz-z*Fy #lb-ft\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The scalar coefficient of the i term is the moment about the X-Axis is',round(Mx,3),\"lb-ft\"\n",
- "\n",
- "# The answer given in the textbook is incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The scalar coefficient of the i term is the moment about the X-Axis is -27640.0 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 16
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter2_2.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter2_2.ipynb deleted file mode 100755 index 9a6ee7f9..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter2_2.ipynb +++ /dev/null @@ -1,562 +0,0 @@ -{
- "metadata": {
- "name": "chapter 2.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 2: Operations with Forces"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-1, Page No: 21"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=20 #lb\n",
- "L=4.33 #ft\n",
- "#Calculation\n",
- "M=-F*L #lb-ft\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The moment of force F about O is',round(M,1),\"lb-ft\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of force F about O is -86.6 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-2, Page no: 22"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=20 #lb\n",
- "theta=((60*pi)/180) #radians\n",
- "L=5 #ft\n",
- "\n",
- "#Calculations\n",
- "\n",
- "F_x=F*cos(theta) #Resloving the vector\n",
- "F_y=F*sin(theta) #Resloving the vector\n",
- "M=-F_y*L #Appling Varignon's theorem\n",
- "#Negative sign tells that moment is clockwise\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The moment of the force about O is',round(M,1),\"lb-ft\"\n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of the force about O is -86.6 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-3,Page No: 22"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=100 #N\n",
- "x1=2 #m\n",
- "x2=5 #m\n",
- "y1=0 #m\n",
- "y2=1 #m\n",
- "z1=4 #m\n",
- "z2=1 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "xside=(x2-x1) #m\n",
- "yside=(y2-y1) #m\n",
- "zside=(z2-z1) #m\n",
- "LD=sqrt(xside**2+yside**2+zside**2)\n",
- "Fx=(xside/LD)*F #N\n",
- "Fy=(yside/LD)*F #N\n",
- "Fz=(zside/LD)*F #N\n",
- "Mx=-Fy*z1 #N-m\n",
- "My=Fx*x1-Fz*z1 #N-m\n",
- "Mz=Fy*x1 #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'Fx is',round(Fx,1),\"N\"\n",
- "print'Fy is',round(Fy,1),\"N\"\n",
- "print'Fz is',round(Fz,1),\"N\"\n",
- "print'Moment about X-Axis is',round(Mx,1),\"N.m\"\n",
- "print'Moment about Y-Axis is +',round(My),\"N.m\"\n",
- "print'Moment about Z-Axis is +',round(Mz,1),\"N.m\"\n",
- "\n",
- "# Decimal point error in calculation causes a small discrepancy in the resulting solutions."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Fx is 68.8 N\n",
- "Fy is 22.9 N\n",
- "Fz is -68.8 N\n",
- "Moment about X-Axis is -91.8 N.m\n",
- "Moment about Y-Axis is + 413.0 N.m\n",
- "Moment about Z-Axis is + 45.9 N.m\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-4, Page No: 23"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "Fx=68.7 #N\n",
- "Fy=22.9 #N\n",
- "Fz=-68.7 #N\n",
- "rx=2 #m\n",
- "ry=0 #m\n",
- "rz=4 #m\n",
- "rx1=5 #m\n",
- "ry1=1 #m\n",
- "rz1=1 #m\n",
- "\n",
- "#Calculation\n",
- "\n",
- "Mx=Fz*ry-Fy*rz #N-m\n",
- "My=-(Fz*rx-Fx*rz) #N-m\n",
- "Mz=Fy*rx-Fx*ry #N-m\n",
- "Mx1=Fz*ry1-Fy*rz1 #N-m\n",
- "My1=-(Fz*rx1-Fx*rz1) #N-m\n",
- "Mz1=Fy*rx1-Fx*ry1 #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'Moment with respect to origin using point(2,0,4)is',round(Mx,1),\"i +\",round(My),\"j +\",round(Mz,1),\"k N.m\"\n",
- "print'Moment with respect to origin using point (5,1,1) is',round(Mx1,1),\"i +\",round(My1),\"j +\",round(Mz1,1),\"k N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Moment with respect to origin using point(2,0,4)is -91.6 i + 412.0 j + 45.8 k N.m\n",
- "Moment with respect to origin using point (5,1,1) is -91.6 i + 412.0 j + 45.8 k N.m\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-5, Page no: 23"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "Fx=2 #lb\n",
- "Fy=3 #lb\n",
- "Fz=-1 #lb\n",
- "rx=1 #ft\n",
- "ry=-4 #ft\n",
- "rz=3 #ft\n",
- "#Coordinates of points\n",
- "ax=3 #ft\n",
- "ay=1 #ft\n",
- "az=1 #ft\n",
- "bx=3 #ft\n",
- "by=-1 #ft\n",
- "bz=1 #ft\n",
- "cx=2 #ft\n",
- "cy=5 #ft\n",
- "cz=-2 #ft\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Rx=ax-cx #ft\n",
- "Ry=ay-cy #ft\n",
- "Rz=az-cz #ft\n",
- "Mx=(Ry*Fz)-(Rz*Fy) #lb-ft\n",
- "My=-((Rx*Fz)-(Rz*Fx)) #lb-ft\n",
- "Mz=(Rx*Fy)-(Ry*Fx) #lb-ft\n",
- "E_u=sqrt((bx-cx)**2+(by-cy)**2+(bz-cz)**2) #ft\n",
- "ex=(bx-cx)/E_u #ft\n",
- "ey=(by-cy)/E_u #ft\n",
- "ez=(bz-cz)/E_u #ft\n",
- "M_lx=Mx*ex #lb-ft\n",
- "M_ly=My*ey #lb-ft\n",
- "M_lz=Mz*ez #lb-ft\n",
- "M_l=M_lx+M_ly+M_lz #lb-ft\n",
- "\n",
- "#Result\n",
- "\n",
- "print'Hence the moment about line is',round(M_l,2),\"lb-ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Hence the moment about line is -2.06 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-6, Page No: 24"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P_x=22 #N\n",
- "P_y=23 #N\n",
- "P_z=7 #N\n",
- "p1=1 #m\n",
- "p2=-1 #m\n",
- "p3=-2 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Mx=(p2*P_z)-(p3*P_y) #N-m\n",
- "My=-((p1*P_z)-(p3*P_x)) #N-m\n",
- "Mz=(p1*P_y)-(p2*P_x) #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The moment about the line from the origin is',round(Mx),\"i\",round(My),\"j +\",round(Mz),\"k N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment about the line from the origin is 39.0 i -51.0 j + 45.0 k N.m\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-8, Page No: 24"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=10 #N Force couple\n",
- "a=3 #m Moment arm\n",
- "#Calculations\n",
- "C=-F*a #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant couple is',round(C),\"N-m\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant couple is -30.0 N-m\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-11, Page No: 26"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "C1=20 #N-m\n",
- "C2=40 #N-m\n",
- "C3=-55 #N-m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "C=sqrt(C1**2+C2**2+C3**2) #N-m\n",
- "thetax=C2/C \n",
- "thetay=C3/C\n",
- "thetaz=C1/C\n",
- "Cx=C*thetax #N-m\n",
- "Cy=C*thetay #N-m\n",
- "Cz=C*thetaz #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'Couple in vector notation is',round(Cx),\"i\",round(Cy),\"j +\",round(Cz),\"k N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Couple in vector notation is 40.0 i -55.0 j + 20.0 k N.m\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-12,Page No: 26"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=25 #lb\n",
- "F2=25 #lb\n",
- "L1=14 #in\n",
- "L2=20 #in\n",
- "\n",
- "#Calculations\n",
- "\n",
- "C=F1*L1 #lb-in\n",
- "M=-F2*L2 #lb-in\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The twisting couple is',round(C),\"lb-in\"\n",
- "print'The bending moment is',round(M),\"lb-in\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The twisting couple is 350.0 lb-in\n",
- "The bending moment is -500.0 lb-in\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-13, Page No: 27"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "rx=20 #in\n",
- "ry=0 #in\n",
- "rz=14 #in\n",
- "Fx=0 #lb\n",
- "Fy=-25 #lb\n",
- "Fz=0 #lb\n",
- "\n",
- "#Calculation\n",
- "\n",
- "Mx=ry*Fz-rz*Fy #lb-in\n",
- "My=rx*Fz-rz*Fx #lb-in\n",
- "Mz=rx*Fy-ry*Fx #lb-in\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The moment of the 25-lb force is',round(Mx),\"i +\",round(My),\"j\",round(Mz),\"k lb-in\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of the 25-lb force is 350.0 i + 0.0 j -500.0 k lb-in\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-14,Page No: 27"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "#Co-ordinates with respect to point O\n",
- "x=17.9 #ft\n",
- "y=6.91 #ft\n",
- "z=46.3 #ft\n",
- "Fz=-4000 #lb\n",
- "Fy=0 #lb\n",
- "\n",
- "#Calculation\n",
- "\n",
- "Mx=y*Fz-z*Fy #lb-ft\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The scalar coefficient of the i term is the moment about the X-Axis is',round(Mx,3),\"lb-ft\"\n",
- "\n",
- "# The answer given in the textbook is incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The scalar coefficient of the i term is the moment about the X-Axis is -27640.0 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 16
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter2_3.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter2_3.ipynb deleted file mode 100755 index 9a6ee7f9..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter2_3.ipynb +++ /dev/null @@ -1,562 +0,0 @@ -{
- "metadata": {
- "name": "chapter 2.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 2: Operations with Forces"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-1, Page No: 21"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=20 #lb\n",
- "L=4.33 #ft\n",
- "#Calculation\n",
- "M=-F*L #lb-ft\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The moment of force F about O is',round(M,1),\"lb-ft\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of force F about O is -86.6 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-2, Page no: 22"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=20 #lb\n",
- "theta=((60*pi)/180) #radians\n",
- "L=5 #ft\n",
- "\n",
- "#Calculations\n",
- "\n",
- "F_x=F*cos(theta) #Resloving the vector\n",
- "F_y=F*sin(theta) #Resloving the vector\n",
- "M=-F_y*L #Appling Varignon's theorem\n",
- "#Negative sign tells that moment is clockwise\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The moment of the force about O is',round(M,1),\"lb-ft\"\n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of the force about O is -86.6 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-3,Page No: 22"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=100 #N\n",
- "x1=2 #m\n",
- "x2=5 #m\n",
- "y1=0 #m\n",
- "y2=1 #m\n",
- "z1=4 #m\n",
- "z2=1 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "xside=(x2-x1) #m\n",
- "yside=(y2-y1) #m\n",
- "zside=(z2-z1) #m\n",
- "LD=sqrt(xside**2+yside**2+zside**2)\n",
- "Fx=(xside/LD)*F #N\n",
- "Fy=(yside/LD)*F #N\n",
- "Fz=(zside/LD)*F #N\n",
- "Mx=-Fy*z1 #N-m\n",
- "My=Fx*x1-Fz*z1 #N-m\n",
- "Mz=Fy*x1 #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'Fx is',round(Fx,1),\"N\"\n",
- "print'Fy is',round(Fy,1),\"N\"\n",
- "print'Fz is',round(Fz,1),\"N\"\n",
- "print'Moment about X-Axis is',round(Mx,1),\"N.m\"\n",
- "print'Moment about Y-Axis is +',round(My),\"N.m\"\n",
- "print'Moment about Z-Axis is +',round(Mz,1),\"N.m\"\n",
- "\n",
- "# Decimal point error in calculation causes a small discrepancy in the resulting solutions."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Fx is 68.8 N\n",
- "Fy is 22.9 N\n",
- "Fz is -68.8 N\n",
- "Moment about X-Axis is -91.8 N.m\n",
- "Moment about Y-Axis is + 413.0 N.m\n",
- "Moment about Z-Axis is + 45.9 N.m\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-4, Page No: 23"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "Fx=68.7 #N\n",
- "Fy=22.9 #N\n",
- "Fz=-68.7 #N\n",
- "rx=2 #m\n",
- "ry=0 #m\n",
- "rz=4 #m\n",
- "rx1=5 #m\n",
- "ry1=1 #m\n",
- "rz1=1 #m\n",
- "\n",
- "#Calculation\n",
- "\n",
- "Mx=Fz*ry-Fy*rz #N-m\n",
- "My=-(Fz*rx-Fx*rz) #N-m\n",
- "Mz=Fy*rx-Fx*ry #N-m\n",
- "Mx1=Fz*ry1-Fy*rz1 #N-m\n",
- "My1=-(Fz*rx1-Fx*rz1) #N-m\n",
- "Mz1=Fy*rx1-Fx*ry1 #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'Moment with respect to origin using point(2,0,4)is',round(Mx,1),\"i +\",round(My),\"j +\",round(Mz,1),\"k N.m\"\n",
- "print'Moment with respect to origin using point (5,1,1) is',round(Mx1,1),\"i +\",round(My1),\"j +\",round(Mz1,1),\"k N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Moment with respect to origin using point(2,0,4)is -91.6 i + 412.0 j + 45.8 k N.m\n",
- "Moment with respect to origin using point (5,1,1) is -91.6 i + 412.0 j + 45.8 k N.m\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-5, Page no: 23"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "Fx=2 #lb\n",
- "Fy=3 #lb\n",
- "Fz=-1 #lb\n",
- "rx=1 #ft\n",
- "ry=-4 #ft\n",
- "rz=3 #ft\n",
- "#Coordinates of points\n",
- "ax=3 #ft\n",
- "ay=1 #ft\n",
- "az=1 #ft\n",
- "bx=3 #ft\n",
- "by=-1 #ft\n",
- "bz=1 #ft\n",
- "cx=2 #ft\n",
- "cy=5 #ft\n",
- "cz=-2 #ft\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Rx=ax-cx #ft\n",
- "Ry=ay-cy #ft\n",
- "Rz=az-cz #ft\n",
- "Mx=(Ry*Fz)-(Rz*Fy) #lb-ft\n",
- "My=-((Rx*Fz)-(Rz*Fx)) #lb-ft\n",
- "Mz=(Rx*Fy)-(Ry*Fx) #lb-ft\n",
- "E_u=sqrt((bx-cx)**2+(by-cy)**2+(bz-cz)**2) #ft\n",
- "ex=(bx-cx)/E_u #ft\n",
- "ey=(by-cy)/E_u #ft\n",
- "ez=(bz-cz)/E_u #ft\n",
- "M_lx=Mx*ex #lb-ft\n",
- "M_ly=My*ey #lb-ft\n",
- "M_lz=Mz*ez #lb-ft\n",
- "M_l=M_lx+M_ly+M_lz #lb-ft\n",
- "\n",
- "#Result\n",
- "\n",
- "print'Hence the moment about line is',round(M_l,2),\"lb-ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Hence the moment about line is -2.06 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-6, Page No: 24"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P_x=22 #N\n",
- "P_y=23 #N\n",
- "P_z=7 #N\n",
- "p1=1 #m\n",
- "p2=-1 #m\n",
- "p3=-2 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Mx=(p2*P_z)-(p3*P_y) #N-m\n",
- "My=-((p1*P_z)-(p3*P_x)) #N-m\n",
- "Mz=(p1*P_y)-(p2*P_x) #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The moment about the line from the origin is',round(Mx),\"i\",round(My),\"j +\",round(Mz),\"k N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment about the line from the origin is 39.0 i -51.0 j + 45.0 k N.m\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-8, Page No: 24"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=10 #N Force couple\n",
- "a=3 #m Moment arm\n",
- "#Calculations\n",
- "C=-F*a #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant couple is',round(C),\"N-m\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant couple is -30.0 N-m\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-11, Page No: 26"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "C1=20 #N-m\n",
- "C2=40 #N-m\n",
- "C3=-55 #N-m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "C=sqrt(C1**2+C2**2+C3**2) #N-m\n",
- "thetax=C2/C \n",
- "thetay=C3/C\n",
- "thetaz=C1/C\n",
- "Cx=C*thetax #N-m\n",
- "Cy=C*thetay #N-m\n",
- "Cz=C*thetaz #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'Couple in vector notation is',round(Cx),\"i\",round(Cy),\"j +\",round(Cz),\"k N.m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Couple in vector notation is 40.0 i -55.0 j + 20.0 k N.m\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-12,Page No: 26"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=25 #lb\n",
- "F2=25 #lb\n",
- "L1=14 #in\n",
- "L2=20 #in\n",
- "\n",
- "#Calculations\n",
- "\n",
- "C=F1*L1 #lb-in\n",
- "M=-F2*L2 #lb-in\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The twisting couple is',round(C),\"lb-in\"\n",
- "print'The bending moment is',round(M),\"lb-in\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The twisting couple is 350.0 lb-in\n",
- "The bending moment is -500.0 lb-in\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-13, Page No: 27"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "rx=20 #in\n",
- "ry=0 #in\n",
- "rz=14 #in\n",
- "Fx=0 #lb\n",
- "Fy=-25 #lb\n",
- "Fz=0 #lb\n",
- "\n",
- "#Calculation\n",
- "\n",
- "Mx=ry*Fz-rz*Fy #lb-in\n",
- "My=rx*Fz-rz*Fx #lb-in\n",
- "Mz=rx*Fy-ry*Fx #lb-in\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The moment of the 25-lb force is',round(Mx),\"i +\",round(My),\"j\",round(Mz),\"k lb-in\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment of the 25-lb force is 350.0 i + 0.0 j -500.0 k lb-in\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2-14,Page No: 27"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "#Co-ordinates with respect to point O\n",
- "x=17.9 #ft\n",
- "y=6.91 #ft\n",
- "z=46.3 #ft\n",
- "Fz=-4000 #lb\n",
- "Fy=0 #lb\n",
- "\n",
- "#Calculation\n",
- "\n",
- "Mx=y*Fz-z*Fy #lb-ft\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The scalar coefficient of the i term is the moment about the X-Axis is',round(Mx,3),\"lb-ft\"\n",
- "\n",
- "# The answer given in the textbook is incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The scalar coefficient of the i term is the moment about the X-Axis is -27640.0 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 16
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter3.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter3.ipynb deleted file mode 100755 index 62463123..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter3.ipynb +++ /dev/null @@ -1,797 +0,0 @@ -{
- "metadata": {
- "name": "chapter3.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 3: Resultants of Coplanar Force Systems"
- ]
- },
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Example 3.3-1, Page no: 32"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=150 #lb\n",
- "F2=200 #lb\n",
- "F3=80 #lb\n",
- "F4=180 #lb\n",
- "theta1=((30*pi)/180) #radians\n",
- "theta2=((150*pi)/180) #radians\n",
- "theta3=((240*pi)/180) #radians\n",
- "theta4=((315*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "F1x=F1*cos(theta1) #lb\n",
- "F1y=F1*sin(theta1) #lb\n",
- "F2x=F2*cos(theta2) #lb\n",
- "F2y=F2*sin(theta2) #lb\n",
- "F3x=F3*cos(theta3) #lb\n",
- "F3y=F3*sin(theta3) #lb\n",
- "F4x=F4*cos(theta4) #lb\n",
- "F4y=F4*sin(theta4) #lb\n",
- "Fx=F1x+F2x+F3x+F4x #lb\n",
- "Fy=F1y+F2y+F3y+F4y #lb\n",
- "R=sqrt(Fx**2+Fy**2) #lb\n",
- "theta=((arctan(Fy/Fx))*180)/pi #degrees\n",
- "theta_R=360+theta #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The resultant is at',round(theta_R),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 49.0 lb\n",
- "The resultant is at 334.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-2, Page no: 33"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=50 #N\n",
- "F2=100 #N\n",
- "F3=30 #N\n",
- "\n",
- "#Calculation\n",
- "\n",
- "#The book has a misprint for squareroot of 1**2\n",
- "F1x=F1/sqrt(2) #N \n",
- "F1y=F1/sqrt(2) #N\n",
- "F2x=-(F2*3)/sqrt(10) #N\n",
- "F2y=(-F2)/sqrt(10) #N\n",
- "F3x=F3/sqrt(5) #N\n",
- "F3y=(-F3*2)/sqrt(5) #N\n",
- "Fx=F1x+F2x+F3x #N\n",
- "Fy=F1y+F2y+F3y #N\n",
- "R=sqrt(Fx**2+Fy**2) #N\n",
- "theta=arctan(Fy/Fx) #radians\n",
- "theta_x=180+(theta*180)/pi #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant is',round(R,1),\"N\"\n",
- "print'The resultant makes an angle of',round(theta_x),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant is 51.6 N\n",
- "The resultant makes an angle of 207.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-3, Page no: 33"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=70 #lb\n",
- "F2=100 #lb\n",
- "F3=125 #lb\n",
- "theta1=0 #radians\n",
- "theta2=((10*pi)/180) #radians\n",
- "theta3=((30*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Fx=F1-(F2*cos(theta3))-(125*sin(theta2)) #lb\n",
- "Fy=125*cos(theta2)-(100*sin(theta3)) #lb\n",
- "R=sqrt(Fx**2+Fy**2) #lb\n",
- "theta=arctan(Fy/Fx) #radians\n",
- "theta_x=180+(theta*180)/pi #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R,1),\"lb\"\n",
- "print'The resultant with respect to X axis is at',round(theta_x),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 82.5 lb\n",
- "The resultant with respect to X axis is at 118.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-4, Page No: 34"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=-20 #N\n",
- "F2=30 #N\n",
- "F3=5 #N\n",
- "F4=-40 #N\n",
- "#Distances with respect to point O\n",
- "x1=6 #m\n",
- "x2=0 #m\n",
- "x3=8 #m\n",
- "x4=13 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=F1+F2+F3+F4 #N\n",
- "# Applying moment about point O equal to zero\n",
- "M_O=-(F1*x1)+(F2*x2)+(F3*x3)+(F4*x4) #N-m\n",
- "#Applying moment about point O equal to R*x\n",
- "x=M_O/R #m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of force system is',round(R),\"N\"\n",
- "print'The moment about point O is',round(M_O),\"N.m\"\n",
- "print'The resultant of moment acts at',round(x,1),\"m (to the right of O)\"\n",
- "\n",
- "# The answer for M_O & R is correct but x waries due to some discrepancy in python."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of force system is -25.0 N\n",
- "The moment about point O is -360.0 N.m\n",
- "The resultant of moment acts at 14.0 m (to the right of O)\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-5, Page No: 34"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=-100 #lb\n",
- "F2=200 #lb\n",
- "F3=-200 #lb\n",
- "F4=400 #lb\n",
- "F5=-300 #lb\n",
- "#Distance with respect to point O\n",
- "x1=0 #ft\n",
- "x2=2 #ft\n",
- "x3=5 #ft\n",
- "x4=9 #ft\n",
- "x5=11 #ft\n",
- "\n",
- "#Calculation\n",
- "\n",
- "R=F1+F2+F3+F4+F5 #lb\n",
- "M_O=(F1*x1)+(F2*x2)+(F3*x3)+(F4*x4)+(F5*x5) #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The moment about point O is',round(M_O),\"lb-ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 0.0 lb\n",
- "The moment about point O is -300.0 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-6, Page no: 35"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=20 #lb\n",
- "F2=20 #lb\n",
- "F3=-40 #lb\n",
- "#Distance from point O\n",
- "x1=3 #ft\n",
- "x2=3 #ft\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=F1+F2+F3 #lb\n",
- "M_O=-(F1*x1)+(F2*x2) #lb-ft\n",
- "\n",
- "#Results\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The Moment about point O is',round(M_O),\"lb-ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 0.0 lb\n",
- "The Moment about point O is 0.0 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-7, Page No: 35"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=500 #N\n",
- "F2=-400 #N\n",
- "F3=-200 #N\n",
- "C=1500 #N-m\n",
- "#Distance from point O\n",
- "x1=2 #m\n",
- "x2=4 #m\n",
- "x3=6 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=F1+F2+F3 #N\n",
- "M_O=(F1*x1)+(F2*x2)+(F3*x3)+C #N-m\n",
- "#Applying Varignons theorem\n",
- "x=M_O/R #m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"N\"\n",
- "print'The moment about point O is',round(M_O),\"N.m\"\n",
- "print'The resultant acts at',round(x),\"m (from point O)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is -100.0 N\n",
- "The moment about point O is -300.0 N.m\n",
- "The resultant acts at 3.0 m (from point O)\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-8,Page No: 35"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=50 #lb\n",
- "F2=100 #lb\n",
- "theta1=((45*pi)/180) #radians\n",
- "#Distance from point O\n",
- "x1=5 #ft\n",
- "x2=4 #ft\n",
- "\n",
- "#Calculation\n",
- "\n",
- "F_x=F1-(F2*cos(theta1)) #lb\n",
- "F_y=F1-(F2*sin(theta1)) #lb\n",
- "R=sqrt(F_x**2+F_y**2) #lb\n",
- "M_O=F1*x1-(x2*F1) #lb-ft\n",
- "#Applying Varignons Theorem\n",
- "x=M_O/R #ft\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R,1),\"lb\"\n",
- "print'The Moment about point O is',round(M_O),\"lb-ft\"\n",
- "print'The Resultant acts at',round(x,2),\"ft (from point O)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 29.3 lb\n",
- "The Moment about point O is 50.0 lb-ft\n",
- "The Resultant acts at 1.71 ft (from point O)\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-9, Page no: 36"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "A=80 #N\n",
- "B=120 #N\n",
- "C=100 #N\n",
- "D=50 #N\n",
- "thetaA=((90*pi)/180) #radians\n",
- "thetaB=((150*pi)/180) #radians\n",
- "thetaC=((45*pi)/180) #radians\n",
- "thetaD=((340*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Ax=A*cos(thetaA) #N\n",
- "Ay=A*sin(thetaA) #N\n",
- "Bx=B*cos(thetaB) #N\n",
- "By=B*sin(thetaB) #N\n",
- "Cx=C*cos(thetaC) #N\n",
- "Cy=C*sin(thetaC) #N\n",
- "Dx=D*cos(thetaD) #N\n",
- "Dy=D*sin(thetaD) #N\n",
- "M_Ax=0 #N-m\n",
- "M_Ay=0 #N-m\n",
- "M_Bx=-Bx*5 #N-m\n",
- "M_By=By*8 #N-m\n",
- "M_Cx=-Cx*1 #N-m\n",
- "M_Cy=Cy*1 #N-m\n",
- "M_Dx=-Dx*-1 #N-m\n",
- "M_Dy=Dy*8 #N-m\n",
- "Fx=Ax+Bx+Cx+Dx #N\n",
- "Fy=Ay+By+Cy+Dy #N\n",
- "R=sqrt(Fx**2+Fy**2) #N\n",
- "M_O=M_Dx+M_Dy+M_Cx+M_Cy+M_Bx+M_By+M_Ax+M_Ay #N-m\n",
- "theta=arctan(Fy/Fx) #radians\n",
- "theta_x=(theta*180)/pi #degrees\n",
- "#Appliying Varignons theorem\n",
- "x=M_O/R #m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"N\"\n",
- "print'The moment about point O is +',round(M_O),\"N.m\"\n",
- "print'The resultant acts at and angle of',round(theta_x),\"degrees (with respect to X axis)\"\n",
- "print'The resultant of the force system acts at',round(x,1),\"m (from point O)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 194.0 N\n",
- "The moment about point O is + 910.0 N.m\n",
- "The resultant acts at and angle of 86.0 degrees (with respect to X axis)\n",
- "The resultant of the force system acts at 4.7 m (from point O)\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-10, Page No: 37"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=100 #lb\n",
- "F2=80 #lb\n",
- "F3=120 #lb\n",
- "F4=150 #lb\n",
- "theta1=((60*pi)/180) #radians\n",
- "theta2=((45*pi)/180) #radians\n",
- "theta3=((90*pi)/180) #radians\n",
- "theta4=((75*pi)/180) #radians\n",
- "#Distance from point O\n",
- "x1=-5 #ft\n",
- "y1=20 #ft\n",
- "x2=10 #ft\n",
- "y2=10 #ft\n",
- "x3=25 #ft\n",
- "y3=25 #ft\n",
- "x4=35 #ft\n",
- "y4=15 #ft\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Fx=F1*cos(theta1)+F2*cos(theta2)+F4*cos(theta4) #lb\n",
- "Fy=-F1*sin(theta1)+F2*sin(theta2)-F3-F4*sin(theta4) #lb\n",
- "R=sqrt(Fx**2+Fy**2) #lb\n",
- "theta=arctan(Fy/Fx) #radians\n",
- "theta_x=(theta*180)/pi #degrees\n",
- "M_O=-(F1*cos(theta1)*y1)+(-x1)*(F1*sin(theta1))-(x2)*(F2*cos(theta2))+(y2)*(F2*sin(theta2))-(x3*F3)-(y4*F4*cos(theta4))-(x4*F4*sin(theta4)) #lb-ft\n",
- "#Applying varignons theorem\n",
- "x=M_O/Fy #ft\n",
- "y=-M_O/Fx #ft\n",
- "\n",
- "#Results\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The resultant acts at',round(theta_x,1),\"degrees(with respect to X axis)\"\n",
- "print'The moment about point O is',round(M_O),\"lb-ft\"\n",
- "print'The x intercept of resultant is',round(x,1),\"ft\"\n",
- "print'The y intercept of resultant is',round(y,1),\"ft\"\n",
- "#Answer for angle should be negative which has not been mentioned in the tectbook but a schematic shows the angle in fourth quadrant to clarify the doubt \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 329.0 lb\n",
- "The resultant acts at -63.8 degrees(with respect to X axis)\n",
- "The moment about point O is -9220.0 lb-ft\n",
- "The x intercept of resultant is 31.3 ft\n",
- "The y intercept of resultant is 63.4 ft\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-11, Page No: 38"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=150 #lb\n",
- "F2=80 #lb\n",
- "F3=100 #lb\n",
- "F4=50 #lb\n",
- "theta1=((45*pi)/180) #radians\n",
- "r=3 #units\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Fh=F1-F3*cos(theta1) #lb\n",
- "Fv=F4-F2-F3*sin(theta1) #lb\n",
- "R=sqrt(Fh**2+Fv**2) #lb\n",
- "#Applying the Varignons Theorem\n",
- "a=(F4*r-F1*r+F2*r-F3*r)/R # ft\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The resultant acts at',round(a,2),\"ft (from point O)\"\n",
- "#Negative sign indicates a negative moment caused by the resultant\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 128.0 lb\n",
- "The resultant acts at -2.81 ft (from point O)\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-12, Page no: 39"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=150 #lb\n",
- "F2=200 #lb\n",
- "F3=200 #lb\n",
- "F4=225 #lb\n",
- "M=900 #lb-ft\n",
- "Theta1=(45*pi)/180 #radians\n",
- "Theta2=(30*pi)/180 #radians\n",
- "x1=3 #ft\n",
- "x2=15 #ft\n",
- "x3=12 #ft\n",
- "x4=6 #ft\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Fx=F1*cos(Theta1)+F2-F4*cos(Theta2) #Applying sum of all forces equal to zero in X direction\n",
- "Fy=F1*sin(Theta1)-F4*sin(Theta2)+F2 #Applying sum of all forces equal to zero in Y direction\n",
- "R=sqrt(Fx**2+Fy**2) #lb\n",
- "theta=(arctan(Fy/Fx))*(180/pi) #degrees\n",
- "M_o=x1*F2-x2*F1*cos(Theta1)+x3*F1*sin(Theta1)-x4*F2+M+x4*F4*cos(Theta2)-x1*F4*sin(Theta2) #Moment about point O\n",
- "x=M_o/Fy # in -Varignons Theorem \n",
- "\n",
- "#Result\n",
- "\n",
- "print'The x intercept of resultant position is',round(x,1),\"in\"\n",
- "print'The Resultant is',round(R),\"lb\"\n",
- "print'The resultant acts at an angle of',round(theta),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The x intercept of resultant position is 4.2 in\n",
- "The Resultant is 223.0 lb\n",
- "The resultant acts at an angle of 60.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-13, Page No: 39"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return 20\n",
- "a=1\n",
- "b=1\n",
- "I=quad(integrand, 0, 6, args=(a,b))\n",
- "\n",
- "def integrand(x, a, b):\n",
- " return 20*x\n",
- "a=1\n",
- "b=1\n",
- "J=quad(integrand, 0, 6, args=(a,b))\n",
- "d=J[0]/I[0]\n",
- "\n",
- "# Results\n",
- "print'The value of R is',round(I[0]),\"lb\"\n",
- "print'The value of d is',round(d),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of R is 120.0 lb\n",
- "The value of d is 3.0 ft\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-14, Page no: 39"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return (x/9)*30\n",
- "a=1\n",
- "b=1\n",
- "I=quad(integrand, 0, 9, args=(a,b))\n",
- "\n",
- "def integrand(y, a, b):\n",
- " return y*(y/9)*30\n",
- "a=1\n",
- "b=1\n",
- "J=quad(integrand, 0, 9, args=(a,b))\n",
- "d=J[0]/I[0]\n",
- "\n",
- "# Results\n",
- "print'The value of R is',round(I[0]),\"N\"\n",
- "print'The value of d is',round(d),\"m\"\n",
- "\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of R is 135.0 N\n",
- "The value of d is 6.0 m\n"
- ]
- }
- ],
- "prompt_number": 4
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter3_1.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter3_1.ipynb deleted file mode 100755 index 62463123..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter3_1.ipynb +++ /dev/null @@ -1,797 +0,0 @@ -{
- "metadata": {
- "name": "chapter3.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 3: Resultants of Coplanar Force Systems"
- ]
- },
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Example 3.3-1, Page no: 32"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=150 #lb\n",
- "F2=200 #lb\n",
- "F3=80 #lb\n",
- "F4=180 #lb\n",
- "theta1=((30*pi)/180) #radians\n",
- "theta2=((150*pi)/180) #radians\n",
- "theta3=((240*pi)/180) #radians\n",
- "theta4=((315*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "F1x=F1*cos(theta1) #lb\n",
- "F1y=F1*sin(theta1) #lb\n",
- "F2x=F2*cos(theta2) #lb\n",
- "F2y=F2*sin(theta2) #lb\n",
- "F3x=F3*cos(theta3) #lb\n",
- "F3y=F3*sin(theta3) #lb\n",
- "F4x=F4*cos(theta4) #lb\n",
- "F4y=F4*sin(theta4) #lb\n",
- "Fx=F1x+F2x+F3x+F4x #lb\n",
- "Fy=F1y+F2y+F3y+F4y #lb\n",
- "R=sqrt(Fx**2+Fy**2) #lb\n",
- "theta=((arctan(Fy/Fx))*180)/pi #degrees\n",
- "theta_R=360+theta #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The resultant is at',round(theta_R),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 49.0 lb\n",
- "The resultant is at 334.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-2, Page no: 33"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=50 #N\n",
- "F2=100 #N\n",
- "F3=30 #N\n",
- "\n",
- "#Calculation\n",
- "\n",
- "#The book has a misprint for squareroot of 1**2\n",
- "F1x=F1/sqrt(2) #N \n",
- "F1y=F1/sqrt(2) #N\n",
- "F2x=-(F2*3)/sqrt(10) #N\n",
- "F2y=(-F2)/sqrt(10) #N\n",
- "F3x=F3/sqrt(5) #N\n",
- "F3y=(-F3*2)/sqrt(5) #N\n",
- "Fx=F1x+F2x+F3x #N\n",
- "Fy=F1y+F2y+F3y #N\n",
- "R=sqrt(Fx**2+Fy**2) #N\n",
- "theta=arctan(Fy/Fx) #radians\n",
- "theta_x=180+(theta*180)/pi #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant is',round(R,1),\"N\"\n",
- "print'The resultant makes an angle of',round(theta_x),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant is 51.6 N\n",
- "The resultant makes an angle of 207.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-3, Page no: 33"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=70 #lb\n",
- "F2=100 #lb\n",
- "F3=125 #lb\n",
- "theta1=0 #radians\n",
- "theta2=((10*pi)/180) #radians\n",
- "theta3=((30*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Fx=F1-(F2*cos(theta3))-(125*sin(theta2)) #lb\n",
- "Fy=125*cos(theta2)-(100*sin(theta3)) #lb\n",
- "R=sqrt(Fx**2+Fy**2) #lb\n",
- "theta=arctan(Fy/Fx) #radians\n",
- "theta_x=180+(theta*180)/pi #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R,1),\"lb\"\n",
- "print'The resultant with respect to X axis is at',round(theta_x),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 82.5 lb\n",
- "The resultant with respect to X axis is at 118.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-4, Page No: 34"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=-20 #N\n",
- "F2=30 #N\n",
- "F3=5 #N\n",
- "F4=-40 #N\n",
- "#Distances with respect to point O\n",
- "x1=6 #m\n",
- "x2=0 #m\n",
- "x3=8 #m\n",
- "x4=13 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=F1+F2+F3+F4 #N\n",
- "# Applying moment about point O equal to zero\n",
- "M_O=-(F1*x1)+(F2*x2)+(F3*x3)+(F4*x4) #N-m\n",
- "#Applying moment about point O equal to R*x\n",
- "x=M_O/R #m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of force system is',round(R),\"N\"\n",
- "print'The moment about point O is',round(M_O),\"N.m\"\n",
- "print'The resultant of moment acts at',round(x,1),\"m (to the right of O)\"\n",
- "\n",
- "# The answer for M_O & R is correct but x waries due to some discrepancy in python."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of force system is -25.0 N\n",
- "The moment about point O is -360.0 N.m\n",
- "The resultant of moment acts at 14.0 m (to the right of O)\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-5, Page No: 34"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=-100 #lb\n",
- "F2=200 #lb\n",
- "F3=-200 #lb\n",
- "F4=400 #lb\n",
- "F5=-300 #lb\n",
- "#Distance with respect to point O\n",
- "x1=0 #ft\n",
- "x2=2 #ft\n",
- "x3=5 #ft\n",
- "x4=9 #ft\n",
- "x5=11 #ft\n",
- "\n",
- "#Calculation\n",
- "\n",
- "R=F1+F2+F3+F4+F5 #lb\n",
- "M_O=(F1*x1)+(F2*x2)+(F3*x3)+(F4*x4)+(F5*x5) #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The moment about point O is',round(M_O),\"lb-ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 0.0 lb\n",
- "The moment about point O is -300.0 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-6, Page no: 35"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=20 #lb\n",
- "F2=20 #lb\n",
- "F3=-40 #lb\n",
- "#Distance from point O\n",
- "x1=3 #ft\n",
- "x2=3 #ft\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=F1+F2+F3 #lb\n",
- "M_O=-(F1*x1)+(F2*x2) #lb-ft\n",
- "\n",
- "#Results\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The Moment about point O is',round(M_O),\"lb-ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 0.0 lb\n",
- "The Moment about point O is 0.0 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-7, Page No: 35"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=500 #N\n",
- "F2=-400 #N\n",
- "F3=-200 #N\n",
- "C=1500 #N-m\n",
- "#Distance from point O\n",
- "x1=2 #m\n",
- "x2=4 #m\n",
- "x3=6 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=F1+F2+F3 #N\n",
- "M_O=(F1*x1)+(F2*x2)+(F3*x3)+C #N-m\n",
- "#Applying Varignons theorem\n",
- "x=M_O/R #m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"N\"\n",
- "print'The moment about point O is',round(M_O),\"N.m\"\n",
- "print'The resultant acts at',round(x),\"m (from point O)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is -100.0 N\n",
- "The moment about point O is -300.0 N.m\n",
- "The resultant acts at 3.0 m (from point O)\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-8,Page No: 35"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=50 #lb\n",
- "F2=100 #lb\n",
- "theta1=((45*pi)/180) #radians\n",
- "#Distance from point O\n",
- "x1=5 #ft\n",
- "x2=4 #ft\n",
- "\n",
- "#Calculation\n",
- "\n",
- "F_x=F1-(F2*cos(theta1)) #lb\n",
- "F_y=F1-(F2*sin(theta1)) #lb\n",
- "R=sqrt(F_x**2+F_y**2) #lb\n",
- "M_O=F1*x1-(x2*F1) #lb-ft\n",
- "#Applying Varignons Theorem\n",
- "x=M_O/R #ft\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R,1),\"lb\"\n",
- "print'The Moment about point O is',round(M_O),\"lb-ft\"\n",
- "print'The Resultant acts at',round(x,2),\"ft (from point O)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 29.3 lb\n",
- "The Moment about point O is 50.0 lb-ft\n",
- "The Resultant acts at 1.71 ft (from point O)\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-9, Page no: 36"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "A=80 #N\n",
- "B=120 #N\n",
- "C=100 #N\n",
- "D=50 #N\n",
- "thetaA=((90*pi)/180) #radians\n",
- "thetaB=((150*pi)/180) #radians\n",
- "thetaC=((45*pi)/180) #radians\n",
- "thetaD=((340*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Ax=A*cos(thetaA) #N\n",
- "Ay=A*sin(thetaA) #N\n",
- "Bx=B*cos(thetaB) #N\n",
- "By=B*sin(thetaB) #N\n",
- "Cx=C*cos(thetaC) #N\n",
- "Cy=C*sin(thetaC) #N\n",
- "Dx=D*cos(thetaD) #N\n",
- "Dy=D*sin(thetaD) #N\n",
- "M_Ax=0 #N-m\n",
- "M_Ay=0 #N-m\n",
- "M_Bx=-Bx*5 #N-m\n",
- "M_By=By*8 #N-m\n",
- "M_Cx=-Cx*1 #N-m\n",
- "M_Cy=Cy*1 #N-m\n",
- "M_Dx=-Dx*-1 #N-m\n",
- "M_Dy=Dy*8 #N-m\n",
- "Fx=Ax+Bx+Cx+Dx #N\n",
- "Fy=Ay+By+Cy+Dy #N\n",
- "R=sqrt(Fx**2+Fy**2) #N\n",
- "M_O=M_Dx+M_Dy+M_Cx+M_Cy+M_Bx+M_By+M_Ax+M_Ay #N-m\n",
- "theta=arctan(Fy/Fx) #radians\n",
- "theta_x=(theta*180)/pi #degrees\n",
- "#Appliying Varignons theorem\n",
- "x=M_O/R #m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"N\"\n",
- "print'The moment about point O is +',round(M_O),\"N.m\"\n",
- "print'The resultant acts at and angle of',round(theta_x),\"degrees (with respect to X axis)\"\n",
- "print'The resultant of the force system acts at',round(x,1),\"m (from point O)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 194.0 N\n",
- "The moment about point O is + 910.0 N.m\n",
- "The resultant acts at and angle of 86.0 degrees (with respect to X axis)\n",
- "The resultant of the force system acts at 4.7 m (from point O)\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-10, Page No: 37"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=100 #lb\n",
- "F2=80 #lb\n",
- "F3=120 #lb\n",
- "F4=150 #lb\n",
- "theta1=((60*pi)/180) #radians\n",
- "theta2=((45*pi)/180) #radians\n",
- "theta3=((90*pi)/180) #radians\n",
- "theta4=((75*pi)/180) #radians\n",
- "#Distance from point O\n",
- "x1=-5 #ft\n",
- "y1=20 #ft\n",
- "x2=10 #ft\n",
- "y2=10 #ft\n",
- "x3=25 #ft\n",
- "y3=25 #ft\n",
- "x4=35 #ft\n",
- "y4=15 #ft\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Fx=F1*cos(theta1)+F2*cos(theta2)+F4*cos(theta4) #lb\n",
- "Fy=-F1*sin(theta1)+F2*sin(theta2)-F3-F4*sin(theta4) #lb\n",
- "R=sqrt(Fx**2+Fy**2) #lb\n",
- "theta=arctan(Fy/Fx) #radians\n",
- "theta_x=(theta*180)/pi #degrees\n",
- "M_O=-(F1*cos(theta1)*y1)+(-x1)*(F1*sin(theta1))-(x2)*(F2*cos(theta2))+(y2)*(F2*sin(theta2))-(x3*F3)-(y4*F4*cos(theta4))-(x4*F4*sin(theta4)) #lb-ft\n",
- "#Applying varignons theorem\n",
- "x=M_O/Fy #ft\n",
- "y=-M_O/Fx #ft\n",
- "\n",
- "#Results\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The resultant acts at',round(theta_x,1),\"degrees(with respect to X axis)\"\n",
- "print'The moment about point O is',round(M_O),\"lb-ft\"\n",
- "print'The x intercept of resultant is',round(x,1),\"ft\"\n",
- "print'The y intercept of resultant is',round(y,1),\"ft\"\n",
- "#Answer for angle should be negative which has not been mentioned in the tectbook but a schematic shows the angle in fourth quadrant to clarify the doubt \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 329.0 lb\n",
- "The resultant acts at -63.8 degrees(with respect to X axis)\n",
- "The moment about point O is -9220.0 lb-ft\n",
- "The x intercept of resultant is 31.3 ft\n",
- "The y intercept of resultant is 63.4 ft\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-11, Page No: 38"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=150 #lb\n",
- "F2=80 #lb\n",
- "F3=100 #lb\n",
- "F4=50 #lb\n",
- "theta1=((45*pi)/180) #radians\n",
- "r=3 #units\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Fh=F1-F3*cos(theta1) #lb\n",
- "Fv=F4-F2-F3*sin(theta1) #lb\n",
- "R=sqrt(Fh**2+Fv**2) #lb\n",
- "#Applying the Varignons Theorem\n",
- "a=(F4*r-F1*r+F2*r-F3*r)/R # ft\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The resultant acts at',round(a,2),\"ft (from point O)\"\n",
- "#Negative sign indicates a negative moment caused by the resultant\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 128.0 lb\n",
- "The resultant acts at -2.81 ft (from point O)\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-12, Page no: 39"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=150 #lb\n",
- "F2=200 #lb\n",
- "F3=200 #lb\n",
- "F4=225 #lb\n",
- "M=900 #lb-ft\n",
- "Theta1=(45*pi)/180 #radians\n",
- "Theta2=(30*pi)/180 #radians\n",
- "x1=3 #ft\n",
- "x2=15 #ft\n",
- "x3=12 #ft\n",
- "x4=6 #ft\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Fx=F1*cos(Theta1)+F2-F4*cos(Theta2) #Applying sum of all forces equal to zero in X direction\n",
- "Fy=F1*sin(Theta1)-F4*sin(Theta2)+F2 #Applying sum of all forces equal to zero in Y direction\n",
- "R=sqrt(Fx**2+Fy**2) #lb\n",
- "theta=(arctan(Fy/Fx))*(180/pi) #degrees\n",
- "M_o=x1*F2-x2*F1*cos(Theta1)+x3*F1*sin(Theta1)-x4*F2+M+x4*F4*cos(Theta2)-x1*F4*sin(Theta2) #Moment about point O\n",
- "x=M_o/Fy # in -Varignons Theorem \n",
- "\n",
- "#Result\n",
- "\n",
- "print'The x intercept of resultant position is',round(x,1),\"in\"\n",
- "print'The Resultant is',round(R),\"lb\"\n",
- "print'The resultant acts at an angle of',round(theta),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The x intercept of resultant position is 4.2 in\n",
- "The Resultant is 223.0 lb\n",
- "The resultant acts at an angle of 60.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-13, Page No: 39"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return 20\n",
- "a=1\n",
- "b=1\n",
- "I=quad(integrand, 0, 6, args=(a,b))\n",
- "\n",
- "def integrand(x, a, b):\n",
- " return 20*x\n",
- "a=1\n",
- "b=1\n",
- "J=quad(integrand, 0, 6, args=(a,b))\n",
- "d=J[0]/I[0]\n",
- "\n",
- "# Results\n",
- "print'The value of R is',round(I[0]),\"lb\"\n",
- "print'The value of d is',round(d),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of R is 120.0 lb\n",
- "The value of d is 3.0 ft\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-14, Page no: 39"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return (x/9)*30\n",
- "a=1\n",
- "b=1\n",
- "I=quad(integrand, 0, 9, args=(a,b))\n",
- "\n",
- "def integrand(y, a, b):\n",
- " return y*(y/9)*30\n",
- "a=1\n",
- "b=1\n",
- "J=quad(integrand, 0, 9, args=(a,b))\n",
- "d=J[0]/I[0]\n",
- "\n",
- "# Results\n",
- "print'The value of R is',round(I[0]),\"N\"\n",
- "print'The value of d is',round(d),\"m\"\n",
- "\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of R is 135.0 N\n",
- "The value of d is 6.0 m\n"
- ]
- }
- ],
- "prompt_number": 4
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter3_2.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter3_2.ipynb deleted file mode 100755 index 62463123..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter3_2.ipynb +++ /dev/null @@ -1,797 +0,0 @@ -{
- "metadata": {
- "name": "chapter3.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 3: Resultants of Coplanar Force Systems"
- ]
- },
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Example 3.3-1, Page no: 32"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=150 #lb\n",
- "F2=200 #lb\n",
- "F3=80 #lb\n",
- "F4=180 #lb\n",
- "theta1=((30*pi)/180) #radians\n",
- "theta2=((150*pi)/180) #radians\n",
- "theta3=((240*pi)/180) #radians\n",
- "theta4=((315*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "F1x=F1*cos(theta1) #lb\n",
- "F1y=F1*sin(theta1) #lb\n",
- "F2x=F2*cos(theta2) #lb\n",
- "F2y=F2*sin(theta2) #lb\n",
- "F3x=F3*cos(theta3) #lb\n",
- "F3y=F3*sin(theta3) #lb\n",
- "F4x=F4*cos(theta4) #lb\n",
- "F4y=F4*sin(theta4) #lb\n",
- "Fx=F1x+F2x+F3x+F4x #lb\n",
- "Fy=F1y+F2y+F3y+F4y #lb\n",
- "R=sqrt(Fx**2+Fy**2) #lb\n",
- "theta=((arctan(Fy/Fx))*180)/pi #degrees\n",
- "theta_R=360+theta #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The resultant is at',round(theta_R),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 49.0 lb\n",
- "The resultant is at 334.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-2, Page no: 33"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=50 #N\n",
- "F2=100 #N\n",
- "F3=30 #N\n",
- "\n",
- "#Calculation\n",
- "\n",
- "#The book has a misprint for squareroot of 1**2\n",
- "F1x=F1/sqrt(2) #N \n",
- "F1y=F1/sqrt(2) #N\n",
- "F2x=-(F2*3)/sqrt(10) #N\n",
- "F2y=(-F2)/sqrt(10) #N\n",
- "F3x=F3/sqrt(5) #N\n",
- "F3y=(-F3*2)/sqrt(5) #N\n",
- "Fx=F1x+F2x+F3x #N\n",
- "Fy=F1y+F2y+F3y #N\n",
- "R=sqrt(Fx**2+Fy**2) #N\n",
- "theta=arctan(Fy/Fx) #radians\n",
- "theta_x=180+(theta*180)/pi #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant is',round(R,1),\"N\"\n",
- "print'The resultant makes an angle of',round(theta_x),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant is 51.6 N\n",
- "The resultant makes an angle of 207.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-3, Page no: 33"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=70 #lb\n",
- "F2=100 #lb\n",
- "F3=125 #lb\n",
- "theta1=0 #radians\n",
- "theta2=((10*pi)/180) #radians\n",
- "theta3=((30*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Fx=F1-(F2*cos(theta3))-(125*sin(theta2)) #lb\n",
- "Fy=125*cos(theta2)-(100*sin(theta3)) #lb\n",
- "R=sqrt(Fx**2+Fy**2) #lb\n",
- "theta=arctan(Fy/Fx) #radians\n",
- "theta_x=180+(theta*180)/pi #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R,1),\"lb\"\n",
- "print'The resultant with respect to X axis is at',round(theta_x),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 82.5 lb\n",
- "The resultant with respect to X axis is at 118.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-4, Page No: 34"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=-20 #N\n",
- "F2=30 #N\n",
- "F3=5 #N\n",
- "F4=-40 #N\n",
- "#Distances with respect to point O\n",
- "x1=6 #m\n",
- "x2=0 #m\n",
- "x3=8 #m\n",
- "x4=13 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=F1+F2+F3+F4 #N\n",
- "# Applying moment about point O equal to zero\n",
- "M_O=-(F1*x1)+(F2*x2)+(F3*x3)+(F4*x4) #N-m\n",
- "#Applying moment about point O equal to R*x\n",
- "x=M_O/R #m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of force system is',round(R),\"N\"\n",
- "print'The moment about point O is',round(M_O),\"N.m\"\n",
- "print'The resultant of moment acts at',round(x,1),\"m (to the right of O)\"\n",
- "\n",
- "# The answer for M_O & R is correct but x waries due to some discrepancy in python."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of force system is -25.0 N\n",
- "The moment about point O is -360.0 N.m\n",
- "The resultant of moment acts at 14.0 m (to the right of O)\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-5, Page No: 34"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=-100 #lb\n",
- "F2=200 #lb\n",
- "F3=-200 #lb\n",
- "F4=400 #lb\n",
- "F5=-300 #lb\n",
- "#Distance with respect to point O\n",
- "x1=0 #ft\n",
- "x2=2 #ft\n",
- "x3=5 #ft\n",
- "x4=9 #ft\n",
- "x5=11 #ft\n",
- "\n",
- "#Calculation\n",
- "\n",
- "R=F1+F2+F3+F4+F5 #lb\n",
- "M_O=(F1*x1)+(F2*x2)+(F3*x3)+(F4*x4)+(F5*x5) #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The moment about point O is',round(M_O),\"lb-ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 0.0 lb\n",
- "The moment about point O is -300.0 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-6, Page no: 35"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=20 #lb\n",
- "F2=20 #lb\n",
- "F3=-40 #lb\n",
- "#Distance from point O\n",
- "x1=3 #ft\n",
- "x2=3 #ft\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=F1+F2+F3 #lb\n",
- "M_O=-(F1*x1)+(F2*x2) #lb-ft\n",
- "\n",
- "#Results\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The Moment about point O is',round(M_O),\"lb-ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 0.0 lb\n",
- "The Moment about point O is 0.0 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-7, Page No: 35"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=500 #N\n",
- "F2=-400 #N\n",
- "F3=-200 #N\n",
- "C=1500 #N-m\n",
- "#Distance from point O\n",
- "x1=2 #m\n",
- "x2=4 #m\n",
- "x3=6 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=F1+F2+F3 #N\n",
- "M_O=(F1*x1)+(F2*x2)+(F3*x3)+C #N-m\n",
- "#Applying Varignons theorem\n",
- "x=M_O/R #m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"N\"\n",
- "print'The moment about point O is',round(M_O),\"N.m\"\n",
- "print'The resultant acts at',round(x),\"m (from point O)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is -100.0 N\n",
- "The moment about point O is -300.0 N.m\n",
- "The resultant acts at 3.0 m (from point O)\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-8,Page No: 35"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=50 #lb\n",
- "F2=100 #lb\n",
- "theta1=((45*pi)/180) #radians\n",
- "#Distance from point O\n",
- "x1=5 #ft\n",
- "x2=4 #ft\n",
- "\n",
- "#Calculation\n",
- "\n",
- "F_x=F1-(F2*cos(theta1)) #lb\n",
- "F_y=F1-(F2*sin(theta1)) #lb\n",
- "R=sqrt(F_x**2+F_y**2) #lb\n",
- "M_O=F1*x1-(x2*F1) #lb-ft\n",
- "#Applying Varignons Theorem\n",
- "x=M_O/R #ft\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R,1),\"lb\"\n",
- "print'The Moment about point O is',round(M_O),\"lb-ft\"\n",
- "print'The Resultant acts at',round(x,2),\"ft (from point O)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 29.3 lb\n",
- "The Moment about point O is 50.0 lb-ft\n",
- "The Resultant acts at 1.71 ft (from point O)\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-9, Page no: 36"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "A=80 #N\n",
- "B=120 #N\n",
- "C=100 #N\n",
- "D=50 #N\n",
- "thetaA=((90*pi)/180) #radians\n",
- "thetaB=((150*pi)/180) #radians\n",
- "thetaC=((45*pi)/180) #radians\n",
- "thetaD=((340*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Ax=A*cos(thetaA) #N\n",
- "Ay=A*sin(thetaA) #N\n",
- "Bx=B*cos(thetaB) #N\n",
- "By=B*sin(thetaB) #N\n",
- "Cx=C*cos(thetaC) #N\n",
- "Cy=C*sin(thetaC) #N\n",
- "Dx=D*cos(thetaD) #N\n",
- "Dy=D*sin(thetaD) #N\n",
- "M_Ax=0 #N-m\n",
- "M_Ay=0 #N-m\n",
- "M_Bx=-Bx*5 #N-m\n",
- "M_By=By*8 #N-m\n",
- "M_Cx=-Cx*1 #N-m\n",
- "M_Cy=Cy*1 #N-m\n",
- "M_Dx=-Dx*-1 #N-m\n",
- "M_Dy=Dy*8 #N-m\n",
- "Fx=Ax+Bx+Cx+Dx #N\n",
- "Fy=Ay+By+Cy+Dy #N\n",
- "R=sqrt(Fx**2+Fy**2) #N\n",
- "M_O=M_Dx+M_Dy+M_Cx+M_Cy+M_Bx+M_By+M_Ax+M_Ay #N-m\n",
- "theta=arctan(Fy/Fx) #radians\n",
- "theta_x=(theta*180)/pi #degrees\n",
- "#Appliying Varignons theorem\n",
- "x=M_O/R #m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"N\"\n",
- "print'The moment about point O is +',round(M_O),\"N.m\"\n",
- "print'The resultant acts at and angle of',round(theta_x),\"degrees (with respect to X axis)\"\n",
- "print'The resultant of the force system acts at',round(x,1),\"m (from point O)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 194.0 N\n",
- "The moment about point O is + 910.0 N.m\n",
- "The resultant acts at and angle of 86.0 degrees (with respect to X axis)\n",
- "The resultant of the force system acts at 4.7 m (from point O)\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-10, Page No: 37"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=100 #lb\n",
- "F2=80 #lb\n",
- "F3=120 #lb\n",
- "F4=150 #lb\n",
- "theta1=((60*pi)/180) #radians\n",
- "theta2=((45*pi)/180) #radians\n",
- "theta3=((90*pi)/180) #radians\n",
- "theta4=((75*pi)/180) #radians\n",
- "#Distance from point O\n",
- "x1=-5 #ft\n",
- "y1=20 #ft\n",
- "x2=10 #ft\n",
- "y2=10 #ft\n",
- "x3=25 #ft\n",
- "y3=25 #ft\n",
- "x4=35 #ft\n",
- "y4=15 #ft\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Fx=F1*cos(theta1)+F2*cos(theta2)+F4*cos(theta4) #lb\n",
- "Fy=-F1*sin(theta1)+F2*sin(theta2)-F3-F4*sin(theta4) #lb\n",
- "R=sqrt(Fx**2+Fy**2) #lb\n",
- "theta=arctan(Fy/Fx) #radians\n",
- "theta_x=(theta*180)/pi #degrees\n",
- "M_O=-(F1*cos(theta1)*y1)+(-x1)*(F1*sin(theta1))-(x2)*(F2*cos(theta2))+(y2)*(F2*sin(theta2))-(x3*F3)-(y4*F4*cos(theta4))-(x4*F4*sin(theta4)) #lb-ft\n",
- "#Applying varignons theorem\n",
- "x=M_O/Fy #ft\n",
- "y=-M_O/Fx #ft\n",
- "\n",
- "#Results\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The resultant acts at',round(theta_x,1),\"degrees(with respect to X axis)\"\n",
- "print'The moment about point O is',round(M_O),\"lb-ft\"\n",
- "print'The x intercept of resultant is',round(x,1),\"ft\"\n",
- "print'The y intercept of resultant is',round(y,1),\"ft\"\n",
- "#Answer for angle should be negative which has not been mentioned in the tectbook but a schematic shows the angle in fourth quadrant to clarify the doubt \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 329.0 lb\n",
- "The resultant acts at -63.8 degrees(with respect to X axis)\n",
- "The moment about point O is -9220.0 lb-ft\n",
- "The x intercept of resultant is 31.3 ft\n",
- "The y intercept of resultant is 63.4 ft\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-11, Page No: 38"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=150 #lb\n",
- "F2=80 #lb\n",
- "F3=100 #lb\n",
- "F4=50 #lb\n",
- "theta1=((45*pi)/180) #radians\n",
- "r=3 #units\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Fh=F1-F3*cos(theta1) #lb\n",
- "Fv=F4-F2-F3*sin(theta1) #lb\n",
- "R=sqrt(Fh**2+Fv**2) #lb\n",
- "#Applying the Varignons Theorem\n",
- "a=(F4*r-F1*r+F2*r-F3*r)/R # ft\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The resultant acts at',round(a,2),\"ft (from point O)\"\n",
- "#Negative sign indicates a negative moment caused by the resultant\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 128.0 lb\n",
- "The resultant acts at -2.81 ft (from point O)\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-12, Page no: 39"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=150 #lb\n",
- "F2=200 #lb\n",
- "F3=200 #lb\n",
- "F4=225 #lb\n",
- "M=900 #lb-ft\n",
- "Theta1=(45*pi)/180 #radians\n",
- "Theta2=(30*pi)/180 #radians\n",
- "x1=3 #ft\n",
- "x2=15 #ft\n",
- "x3=12 #ft\n",
- "x4=6 #ft\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Fx=F1*cos(Theta1)+F2-F4*cos(Theta2) #Applying sum of all forces equal to zero in X direction\n",
- "Fy=F1*sin(Theta1)-F4*sin(Theta2)+F2 #Applying sum of all forces equal to zero in Y direction\n",
- "R=sqrt(Fx**2+Fy**2) #lb\n",
- "theta=(arctan(Fy/Fx))*(180/pi) #degrees\n",
- "M_o=x1*F2-x2*F1*cos(Theta1)+x3*F1*sin(Theta1)-x4*F2+M+x4*F4*cos(Theta2)-x1*F4*sin(Theta2) #Moment about point O\n",
- "x=M_o/Fy # in -Varignons Theorem \n",
- "\n",
- "#Result\n",
- "\n",
- "print'The x intercept of resultant position is',round(x,1),\"in\"\n",
- "print'The Resultant is',round(R),\"lb\"\n",
- "print'The resultant acts at an angle of',round(theta),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The x intercept of resultant position is 4.2 in\n",
- "The Resultant is 223.0 lb\n",
- "The resultant acts at an angle of 60.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-13, Page No: 39"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return 20\n",
- "a=1\n",
- "b=1\n",
- "I=quad(integrand, 0, 6, args=(a,b))\n",
- "\n",
- "def integrand(x, a, b):\n",
- " return 20*x\n",
- "a=1\n",
- "b=1\n",
- "J=quad(integrand, 0, 6, args=(a,b))\n",
- "d=J[0]/I[0]\n",
- "\n",
- "# Results\n",
- "print'The value of R is',round(I[0]),\"lb\"\n",
- "print'The value of d is',round(d),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of R is 120.0 lb\n",
- "The value of d is 3.0 ft\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-14, Page no: 39"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return (x/9)*30\n",
- "a=1\n",
- "b=1\n",
- "I=quad(integrand, 0, 9, args=(a,b))\n",
- "\n",
- "def integrand(y, a, b):\n",
- " return y*(y/9)*30\n",
- "a=1\n",
- "b=1\n",
- "J=quad(integrand, 0, 9, args=(a,b))\n",
- "d=J[0]/I[0]\n",
- "\n",
- "# Results\n",
- "print'The value of R is',round(I[0]),\"N\"\n",
- "print'The value of d is',round(d),\"m\"\n",
- "\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of R is 135.0 N\n",
- "The value of d is 6.0 m\n"
- ]
- }
- ],
- "prompt_number": 4
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter3_3.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter3_3.ipynb deleted file mode 100755 index 62463123..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter3_3.ipynb +++ /dev/null @@ -1,797 +0,0 @@ -{
- "metadata": {
- "name": "chapter3.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 3: Resultants of Coplanar Force Systems"
- ]
- },
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Example 3.3-1, Page no: 32"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=150 #lb\n",
- "F2=200 #lb\n",
- "F3=80 #lb\n",
- "F4=180 #lb\n",
- "theta1=((30*pi)/180) #radians\n",
- "theta2=((150*pi)/180) #radians\n",
- "theta3=((240*pi)/180) #radians\n",
- "theta4=((315*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "F1x=F1*cos(theta1) #lb\n",
- "F1y=F1*sin(theta1) #lb\n",
- "F2x=F2*cos(theta2) #lb\n",
- "F2y=F2*sin(theta2) #lb\n",
- "F3x=F3*cos(theta3) #lb\n",
- "F3y=F3*sin(theta3) #lb\n",
- "F4x=F4*cos(theta4) #lb\n",
- "F4y=F4*sin(theta4) #lb\n",
- "Fx=F1x+F2x+F3x+F4x #lb\n",
- "Fy=F1y+F2y+F3y+F4y #lb\n",
- "R=sqrt(Fx**2+Fy**2) #lb\n",
- "theta=((arctan(Fy/Fx))*180)/pi #degrees\n",
- "theta_R=360+theta #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The resultant is at',round(theta_R),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 49.0 lb\n",
- "The resultant is at 334.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-2, Page no: 33"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=50 #N\n",
- "F2=100 #N\n",
- "F3=30 #N\n",
- "\n",
- "#Calculation\n",
- "\n",
- "#The book has a misprint for squareroot of 1**2\n",
- "F1x=F1/sqrt(2) #N \n",
- "F1y=F1/sqrt(2) #N\n",
- "F2x=-(F2*3)/sqrt(10) #N\n",
- "F2y=(-F2)/sqrt(10) #N\n",
- "F3x=F3/sqrt(5) #N\n",
- "F3y=(-F3*2)/sqrt(5) #N\n",
- "Fx=F1x+F2x+F3x #N\n",
- "Fy=F1y+F2y+F3y #N\n",
- "R=sqrt(Fx**2+Fy**2) #N\n",
- "theta=arctan(Fy/Fx) #radians\n",
- "theta_x=180+(theta*180)/pi #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant is',round(R,1),\"N\"\n",
- "print'The resultant makes an angle of',round(theta_x),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant is 51.6 N\n",
- "The resultant makes an angle of 207.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-3, Page no: 33"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=70 #lb\n",
- "F2=100 #lb\n",
- "F3=125 #lb\n",
- "theta1=0 #radians\n",
- "theta2=((10*pi)/180) #radians\n",
- "theta3=((30*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Fx=F1-(F2*cos(theta3))-(125*sin(theta2)) #lb\n",
- "Fy=125*cos(theta2)-(100*sin(theta3)) #lb\n",
- "R=sqrt(Fx**2+Fy**2) #lb\n",
- "theta=arctan(Fy/Fx) #radians\n",
- "theta_x=180+(theta*180)/pi #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R,1),\"lb\"\n",
- "print'The resultant with respect to X axis is at',round(theta_x),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 82.5 lb\n",
- "The resultant with respect to X axis is at 118.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-4, Page No: 34"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=-20 #N\n",
- "F2=30 #N\n",
- "F3=5 #N\n",
- "F4=-40 #N\n",
- "#Distances with respect to point O\n",
- "x1=6 #m\n",
- "x2=0 #m\n",
- "x3=8 #m\n",
- "x4=13 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=F1+F2+F3+F4 #N\n",
- "# Applying moment about point O equal to zero\n",
- "M_O=-(F1*x1)+(F2*x2)+(F3*x3)+(F4*x4) #N-m\n",
- "#Applying moment about point O equal to R*x\n",
- "x=M_O/R #m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of force system is',round(R),\"N\"\n",
- "print'The moment about point O is',round(M_O),\"N.m\"\n",
- "print'The resultant of moment acts at',round(x,1),\"m (to the right of O)\"\n",
- "\n",
- "# The answer for M_O & R is correct but x waries due to some discrepancy in python."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of force system is -25.0 N\n",
- "The moment about point O is -360.0 N.m\n",
- "The resultant of moment acts at 14.0 m (to the right of O)\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-5, Page No: 34"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=-100 #lb\n",
- "F2=200 #lb\n",
- "F3=-200 #lb\n",
- "F4=400 #lb\n",
- "F5=-300 #lb\n",
- "#Distance with respect to point O\n",
- "x1=0 #ft\n",
- "x2=2 #ft\n",
- "x3=5 #ft\n",
- "x4=9 #ft\n",
- "x5=11 #ft\n",
- "\n",
- "#Calculation\n",
- "\n",
- "R=F1+F2+F3+F4+F5 #lb\n",
- "M_O=(F1*x1)+(F2*x2)+(F3*x3)+(F4*x4)+(F5*x5) #N-m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The moment about point O is',round(M_O),\"lb-ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 0.0 lb\n",
- "The moment about point O is -300.0 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-6, Page no: 35"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=20 #lb\n",
- "F2=20 #lb\n",
- "F3=-40 #lb\n",
- "#Distance from point O\n",
- "x1=3 #ft\n",
- "x2=3 #ft\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=F1+F2+F3 #lb\n",
- "M_O=-(F1*x1)+(F2*x2) #lb-ft\n",
- "\n",
- "#Results\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The Moment about point O is',round(M_O),\"lb-ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 0.0 lb\n",
- "The Moment about point O is 0.0 lb-ft\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-7, Page No: 35"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=500 #N\n",
- "F2=-400 #N\n",
- "F3=-200 #N\n",
- "C=1500 #N-m\n",
- "#Distance from point O\n",
- "x1=2 #m\n",
- "x2=4 #m\n",
- "x3=6 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=F1+F2+F3 #N\n",
- "M_O=(F1*x1)+(F2*x2)+(F3*x3)+C #N-m\n",
- "#Applying Varignons theorem\n",
- "x=M_O/R #m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"N\"\n",
- "print'The moment about point O is',round(M_O),\"N.m\"\n",
- "print'The resultant acts at',round(x),\"m (from point O)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is -100.0 N\n",
- "The moment about point O is -300.0 N.m\n",
- "The resultant acts at 3.0 m (from point O)\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-8,Page No: 35"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=50 #lb\n",
- "F2=100 #lb\n",
- "theta1=((45*pi)/180) #radians\n",
- "#Distance from point O\n",
- "x1=5 #ft\n",
- "x2=4 #ft\n",
- "\n",
- "#Calculation\n",
- "\n",
- "F_x=F1-(F2*cos(theta1)) #lb\n",
- "F_y=F1-(F2*sin(theta1)) #lb\n",
- "R=sqrt(F_x**2+F_y**2) #lb\n",
- "M_O=F1*x1-(x2*F1) #lb-ft\n",
- "#Applying Varignons Theorem\n",
- "x=M_O/R #ft\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R,1),\"lb\"\n",
- "print'The Moment about point O is',round(M_O),\"lb-ft\"\n",
- "print'The Resultant acts at',round(x,2),\"ft (from point O)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 29.3 lb\n",
- "The Moment about point O is 50.0 lb-ft\n",
- "The Resultant acts at 1.71 ft (from point O)\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-9, Page no: 36"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "A=80 #N\n",
- "B=120 #N\n",
- "C=100 #N\n",
- "D=50 #N\n",
- "thetaA=((90*pi)/180) #radians\n",
- "thetaB=((150*pi)/180) #radians\n",
- "thetaC=((45*pi)/180) #radians\n",
- "thetaD=((340*pi)/180) #radians\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Ax=A*cos(thetaA) #N\n",
- "Ay=A*sin(thetaA) #N\n",
- "Bx=B*cos(thetaB) #N\n",
- "By=B*sin(thetaB) #N\n",
- "Cx=C*cos(thetaC) #N\n",
- "Cy=C*sin(thetaC) #N\n",
- "Dx=D*cos(thetaD) #N\n",
- "Dy=D*sin(thetaD) #N\n",
- "M_Ax=0 #N-m\n",
- "M_Ay=0 #N-m\n",
- "M_Bx=-Bx*5 #N-m\n",
- "M_By=By*8 #N-m\n",
- "M_Cx=-Cx*1 #N-m\n",
- "M_Cy=Cy*1 #N-m\n",
- "M_Dx=-Dx*-1 #N-m\n",
- "M_Dy=Dy*8 #N-m\n",
- "Fx=Ax+Bx+Cx+Dx #N\n",
- "Fy=Ay+By+Cy+Dy #N\n",
- "R=sqrt(Fx**2+Fy**2) #N\n",
- "M_O=M_Dx+M_Dy+M_Cx+M_Cy+M_Bx+M_By+M_Ax+M_Ay #N-m\n",
- "theta=arctan(Fy/Fx) #radians\n",
- "theta_x=(theta*180)/pi #degrees\n",
- "#Appliying Varignons theorem\n",
- "x=M_O/R #m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"N\"\n",
- "print'The moment about point O is +',round(M_O),\"N.m\"\n",
- "print'The resultant acts at and angle of',round(theta_x),\"degrees (with respect to X axis)\"\n",
- "print'The resultant of the force system acts at',round(x,1),\"m (from point O)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 194.0 N\n",
- "The moment about point O is + 910.0 N.m\n",
- "The resultant acts at and angle of 86.0 degrees (with respect to X axis)\n",
- "The resultant of the force system acts at 4.7 m (from point O)\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-10, Page No: 37"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=100 #lb\n",
- "F2=80 #lb\n",
- "F3=120 #lb\n",
- "F4=150 #lb\n",
- "theta1=((60*pi)/180) #radians\n",
- "theta2=((45*pi)/180) #radians\n",
- "theta3=((90*pi)/180) #radians\n",
- "theta4=((75*pi)/180) #radians\n",
- "#Distance from point O\n",
- "x1=-5 #ft\n",
- "y1=20 #ft\n",
- "x2=10 #ft\n",
- "y2=10 #ft\n",
- "x3=25 #ft\n",
- "y3=25 #ft\n",
- "x4=35 #ft\n",
- "y4=15 #ft\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Fx=F1*cos(theta1)+F2*cos(theta2)+F4*cos(theta4) #lb\n",
- "Fy=-F1*sin(theta1)+F2*sin(theta2)-F3-F4*sin(theta4) #lb\n",
- "R=sqrt(Fx**2+Fy**2) #lb\n",
- "theta=arctan(Fy/Fx) #radians\n",
- "theta_x=(theta*180)/pi #degrees\n",
- "M_O=-(F1*cos(theta1)*y1)+(-x1)*(F1*sin(theta1))-(x2)*(F2*cos(theta2))+(y2)*(F2*sin(theta2))-(x3*F3)-(y4*F4*cos(theta4))-(x4*F4*sin(theta4)) #lb-ft\n",
- "#Applying varignons theorem\n",
- "x=M_O/Fy #ft\n",
- "y=-M_O/Fx #ft\n",
- "\n",
- "#Results\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The resultant acts at',round(theta_x,1),\"degrees(with respect to X axis)\"\n",
- "print'The moment about point O is',round(M_O),\"lb-ft\"\n",
- "print'The x intercept of resultant is',round(x,1),\"ft\"\n",
- "print'The y intercept of resultant is',round(y,1),\"ft\"\n",
- "#Answer for angle should be negative which has not been mentioned in the tectbook but a schematic shows the angle in fourth quadrant to clarify the doubt \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 329.0 lb\n",
- "The resultant acts at -63.8 degrees(with respect to X axis)\n",
- "The moment about point O is -9220.0 lb-ft\n",
- "The x intercept of resultant is 31.3 ft\n",
- "The y intercept of resultant is 63.4 ft\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-11, Page No: 38"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=150 #lb\n",
- "F2=80 #lb\n",
- "F3=100 #lb\n",
- "F4=50 #lb\n",
- "theta1=((45*pi)/180) #radians\n",
- "r=3 #units\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Fh=F1-F3*cos(theta1) #lb\n",
- "Fv=F4-F2-F3*sin(theta1) #lb\n",
- "R=sqrt(Fh**2+Fv**2) #lb\n",
- "#Applying the Varignons Theorem\n",
- "a=(F4*r-F1*r+F2*r-F3*r)/R # ft\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R),\"lb\"\n",
- "print'The resultant acts at',round(a,2),\"ft (from point O)\"\n",
- "#Negative sign indicates a negative moment caused by the resultant\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 128.0 lb\n",
- "The resultant acts at -2.81 ft (from point O)\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-12, Page no: 39"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F1=150 #lb\n",
- "F2=200 #lb\n",
- "F3=200 #lb\n",
- "F4=225 #lb\n",
- "M=900 #lb-ft\n",
- "Theta1=(45*pi)/180 #radians\n",
- "Theta2=(30*pi)/180 #radians\n",
- "x1=3 #ft\n",
- "x2=15 #ft\n",
- "x3=12 #ft\n",
- "x4=6 #ft\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Fx=F1*cos(Theta1)+F2-F4*cos(Theta2) #Applying sum of all forces equal to zero in X direction\n",
- "Fy=F1*sin(Theta1)-F4*sin(Theta2)+F2 #Applying sum of all forces equal to zero in Y direction\n",
- "R=sqrt(Fx**2+Fy**2) #lb\n",
- "theta=(arctan(Fy/Fx))*(180/pi) #degrees\n",
- "M_o=x1*F2-x2*F1*cos(Theta1)+x3*F1*sin(Theta1)-x4*F2+M+x4*F4*cos(Theta2)-x1*F4*sin(Theta2) #Moment about point O\n",
- "x=M_o/Fy # in -Varignons Theorem \n",
- "\n",
- "#Result\n",
- "\n",
- "print'The x intercept of resultant position is',round(x,1),\"in\"\n",
- "print'The Resultant is',round(R),\"lb\"\n",
- "print'The resultant acts at an angle of',round(theta),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The x intercept of resultant position is 4.2 in\n",
- "The Resultant is 223.0 lb\n",
- "The resultant acts at an angle of 60.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-13, Page No: 39"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return 20\n",
- "a=1\n",
- "b=1\n",
- "I=quad(integrand, 0, 6, args=(a,b))\n",
- "\n",
- "def integrand(x, a, b):\n",
- " return 20*x\n",
- "a=1\n",
- "b=1\n",
- "J=quad(integrand, 0, 6, args=(a,b))\n",
- "d=J[0]/I[0]\n",
- "\n",
- "# Results\n",
- "print'The value of R is',round(I[0]),\"lb\"\n",
- "print'The value of d is',round(d),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of R is 120.0 lb\n",
- "The value of d is 3.0 ft\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3-14, Page no: 39"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "from scipy.integrate import quad\n",
- "def integrand(x, a, b):\n",
- " return (x/9)*30\n",
- "a=1\n",
- "b=1\n",
- "I=quad(integrand, 0, 9, args=(a,b))\n",
- "\n",
- "def integrand(y, a, b):\n",
- " return y*(y/9)*30\n",
- "a=1\n",
- "b=1\n",
- "J=quad(integrand, 0, 9, args=(a,b))\n",
- "d=J[0]/I[0]\n",
- "\n",
- "# Results\n",
- "print'The value of R is',round(I[0]),\"N\"\n",
- "print'The value of d is',round(d),\"m\"\n",
- "\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of R is 135.0 N\n",
- "The value of d is 6.0 m\n"
- ]
- }
- ],
- "prompt_number": 4
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter4.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter4.ipynb deleted file mode 100755 index 3133c796..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter4.ipynb +++ /dev/null @@ -1,365 +0,0 @@ -{
- "metadata": {
- "name": "chapter4.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 4: Resultants of Non-coplanar Force Systems"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4-1, Page no: 48"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "#Initilization Of Variables\n",
- "\n",
- "F1=20 #lb\n",
- "F2=15 #lb\n",
- "F3=30 #lb\n",
- "F4=50 #lb\n",
- "#Co-ordinates of Forces\n",
- "C1=np.array([2,1,6])\n",
- "C2=np.array([4,-2,5])\n",
- "C3=np.array([-3,-2,1])\n",
- "C4=np.array([5,1,-2])\n",
- "\n",
- "#Calculations\n",
- "\n",
- "A=(C1[0]**2+C1[1]**2+C1[2]**2)**0.5\n",
- "B=(C2[0]**2+C2[1]**2+C2[2]**2)**0.5\n",
- "C=(C3[0]**2+C3[1]**2+C3[2]**2)**0.5\n",
- "D=(C4[0]**2+C4[1]**2+C4[2]**2)**0.5\n",
- "#Calculations for cos(thetax),cos(thetay) and cos(thetaz)\n",
- "theta1=(A**-1)*np.array([C1[0],C1[1],C1[2]])\n",
- "theta2=(B**-1)*np.array([C2[0],C2[1],C2[2]])\n",
- "theta3=(C**-1)*np.array([C3[0],C3[1],C3[2]])\n",
- "theta4=(D**-1)*np.array([C4[0],C4[1],C4[2]])\n",
- "#Calculations for forces (in form of force vectors)\n",
- "Fa=F1*np.array([theta1[0],theta1[1],theta1[2]]) #lb\n",
- "Fb=F2*np.array([theta2[0],theta2[1],theta2[2]]) #lb\n",
- "Fc=F3*np.array([theta3[0],theta3[1],theta3[2]]) #lb\n",
- "Fd=F4*np.array([theta4[0],theta4[1],theta4[2]]) #lb\n",
- "Fx=Fa[0]+Fb[0]+Fc[0]+Fd[0] #lb\n",
- "Fy=Fa[1]+Fb[1]+Fc[1]+Fd[1] #lb\n",
- "Fz=Fa[2]+Fb[2]+Fc[2]+Fd[2] #lb\n",
- "R=(Fx**2+Fy**2+Fz**2)**0.5 #lb\n",
- "thetax=arccos(Fx*R**-1)*(180/pi) #degrees\n",
- "thetay=180-((180*arccos(Fy*R**-1))/pi) #degrees\n",
- "thetaz=(180*arccos(Fz*R**-1))/pi #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R,1),\"lb\"\n",
- "print'The angle of the resultant with respect to x axis is',round(thetax,1),\"degree\"\n",
- "print'The angle of the resultant with respect to y axis is',round(thetay),\"degree\"\n",
- "print'The angle of the resultant with respect to a axis is',round(thetaz,1),\"degree\"\n",
- "\n",
- "# Thetax is off by 0.1 degrees"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 42.5 lb\n",
- "The angle of the resultant with respect to x axis is 30.1 degree\n",
- "The angle of the resultant with respect to y axis is 79.0 degree\n",
- "The angle of the resultant with respect to a axis is 62.4 degree\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4-2, Page no: 49"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=[20,-10,30] #N\n",
- "#co-ordinates in meters\n",
- "a=2 #m\n",
- "b=4 #m\n",
- "c=7 #m\n",
- "d=3 #m\n",
- "e=2 #m\n",
- "f=4 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=F[0]+F[1]+F[2] #N\n",
- "M_o=F[0]*a+F[1]*b+F[2]*c #N.m\n",
- "x=M_o*R**-1 #m\n",
- "M_x=-F[2]*f-F[0]*d-F[1]*e #N.m\n",
- "z=-M_x/R #m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant is +',round(R),\"N\"\n",
- "print'The moment about point O is +',round(M_o),\"N.m\"\n",
- "print'The position of R is at',round(x,2),\"m (from origin)\"\n",
- "print'The moment is',round(M_x),\"N.m\"\n",
- "print'The z co-ordinate is +',round(z),\"m\"\n",
- "\n",
- "# Here the value of R & M_o is correct which should yeild the vaue of x (M_o/R) correctly. However dueto some error in the software the correct value is not being printed.\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant is + 40.0 N\n",
- "The moment about point O is + 210.0 N.m\n",
- "The position of R is at 5.25 m (from origin)\n",
- "The moment is -160.0 N.m\n",
- "The z co-ordinate is + 4.0 m\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4-3, Page No: 49"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=[100,50,-150] #Force vector N\n",
- "a=2 #m\n",
- "b=2 #m \n",
- "c=3 #m\n",
- "d=2 #m\n",
- "e=4 #m\n",
- "f=8 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=F[0]+F[1]+F[2] #N\n",
- "M_x=-F[0]*a+F[1]*b-F[2]*c #N.m\n",
- "M_z=F[0]*d+F[1]*e+F[2]*f #N.m\n",
- "C=sqrt(M_x**2+M_z**2) #N-m\n",
- "thetax=arctan(M_x*-M_z**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant is',round(R),\"N\"\n",
- "print'The moment about x axis is +',round(M_x),\"N.m\"\n",
- "print'The moment about z axis is',round(M_z),\"N.m\"\n",
- "print'The couple acting is',round(C),\"N.m\"\n",
- "print'The trace makes an angle with x axis of',round(thetax,1),\"degrees\"\n",
- "\n",
- "# The answer for C waries by 1 N.m as compared to the book."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant is 0.0 N\n",
- "The moment about x axis is + 350.0 N.m\n",
- "The moment about z axis is -800.0 N.m\n",
- "The couple acting is 873.0 N.m\n",
- "The trace makes an angle with x axis of 23.6 degrees\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4-4, Page No: 50"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "x1=-2\n",
- "y1=2\n",
- "z1=-2\n",
- "x2=3\n",
- "y2=0\n",
- "z2=-4\n",
- "x3=3\n",
- "y3=2\n",
- "z3=2\n",
- "F1=40 #lb\n",
- "F2=30 #lb\n",
- "F3=20 #lb\n",
- "Mxm=np.array([-92.4,-48,-19.4])\n",
- "Mym=np.array([-46.2,72,9.8])\n",
- "Mzm=np.array([46.2,-36,19.4])\n",
- "\n",
- "#Calculations\n",
- "mag1=(x1**2+y1**2+z1**2)**0.5\n",
- "mag2=(x2**2+y2**2+z2**2)**0.5\n",
- "mag3=(x3**2+y3**2+z3**2)**0.5\n",
- "thetax1=(x1*mag1**-1) #degrees\n",
- "thetay1=(y1*mag1**-1) #degrees\n",
- "thetaz1=(z1*mag1**-1) #degrees\n",
- "thetax2=(x2*mag2**-1) #degrees\n",
- "thetay2=(y2*mag2**-1) #degrees\n",
- "thetaz2=(z2*mag2**-1) #degrees\n",
- "thetax3=(x3*mag3**-1) #degrees\n",
- "thetay3=(y3*mag3**-1) #degrees\n",
- "thetaz3=(z3*mag3**-1) #degrees\n",
- "#Now we will define all the components in terms of matrices for simplicity of computation\n",
- "F=np.array([F1,F2,F3]) #lb\n",
- "Fx1=F[0]*thetax1\n",
- "Fy1=F[0]*thetay1\n",
- "Fz1=F[0]*thetaz1\n",
- "Fx2=F[1]*thetax2\n",
- "Fy2=F[1]*thetay2\n",
- "Fz2=F[1]*thetaz2\n",
- "Fx3=F[2]*thetax3\n",
- "Fy3=F[2]*thetay3\n",
- "Fz3=F[2]*thetaz3\n",
- "Fx=Fx1+Fx2+Fx3 #lb\n",
- "Fy=Fy1+Fy2+Fy3 #lb\n",
- "Fz=Fz1+Fz2+Fz3 #lb\n",
- "R=(Fx**2+Fy**2+Fz**2)**0.5 #lb\n",
- "thetax=arccos(Fx*R**-1)*(180/pi) #degrees\n",
- "thetay=arccos(Fy*R**-1)*(180/pi) #degrees\n",
- "thetaz=arccos(Fz*R**-1)*(180/pi) #degrees\n",
- "#Moment calculations\n",
- "Mx=Mxm[0]+Mxm[1]+Mxm[2] #lb-ft\n",
- "My=Mym[0]+Mym[1]+Mym[2] #lb-ft\n",
- "Mz=Mzm[0]+Mzm[1]+Mzm[2] #lb-ft\n",
- "C=(Mx**2+My**2+Mz**2)**0.5 #lb-ft\n",
- "#Direction cosines\n",
- "PHIx=arccos(Mx*C**-1)*(180/pi) #degrees\n",
- "PHIy=arccos(My*C**-1)*(180/pi) #degrees\n",
- "PHIz=arccos(Mz*C**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The result of the force is',round(R,1),\"lb\"\n",
- "print'The angles with respect to X-Axis,Y-Axis and Z-axis are:',round(thetax,1),\"degrees\",',',round(thetay,1),\"degrees\",'and',round(thetaz,1),\"degrees respectively.\"\n",
- "print'The magnitude of resultant couple is',round(C),\"lb-ft\"\n",
- "print'The angles are as follows: Cosphix=',round(PHIx,1),\"degrees\",',Cosphiy=',round(PHIy,1),\"degrees\",'and Cosphiz=',round(PHIz,1),\"degrees\"\n",
- "\n",
- "# The answers may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The result of the force is 50.6 lb\n",
- "The angles with respect to X-Axis,Y-Axis and Z-axis are: 79.2 degrees , 49.6 degrees and 137.6 degrees respectively.\n",
- "The magnitude of resultant couple is 166.0 lb-ft\n",
- "The angles are as follows: Cosphix= 163.8 degrees ,Cosphiy= 77.6 degrees and Cosphiz= 79.8 degrees\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4-5, Page no 52"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "F=np.array([150,90,160]) #lb force vector kind of decleration\n",
- "# Co-ordinates defined as [x,y,z] all the co-ordinates are in feet\n",
- "C_1=np.array([2,0,0]) \n",
- "C_2=np.array([0,0,1])\n",
- "C_3=np.array([0,-2,-1])\n",
- "C_4=np.array([-1,0,-1])\n",
- "\n",
- "#Calculations\n",
- "A=C_2-C_1\n",
- "B=C_4-C_3\n",
- "F1=(F[0]*A)/(A[0]**2+A[1]**2+A[2]**2)**0.5\n",
- "F2=(F[1]*B)/(B[0]**2+B[1]**2+B[2]**2)**0.5\n",
- "R=F1+F2\n",
- "# Determine C1 & C2\n",
- "# Calculating the cross products of C1 & C2 as,\n",
- "C1=np.array([[C_1[1]*F1[2]-C_1[2]*F1[1]],[-(C_1[0]*F1[2]-C_1[2]*F1[0])],[C_1[0]*F1[1]-C_1[1]*F1[0]]]) # lb-ft\n",
- "C2=np.array([[C_3[1]*F2[2]-C_3[2]*F2[1]],[-(C_3[0]*F2[2]-C_3[2]*F2[0])],[C_3[0]*F2[1]-C_3[1]*F2[0]]]) # lb-ft\n",
- "C3=np.array([[0],[0],[160]]) # lb-ft\n",
- "C=C1+C2+C3\n",
- "\n",
- "#Result\n",
- "print'The resultant force couple is',round(C[0],1),\"i\",round(C[1],1),\"j +\",round(C[2],1),\"k lb-ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant force couple is 80.5 i -93.9 j + 79.5 k lb-ft\n"
- ]
- }
- ],
- "prompt_number": 48
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter4_1.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter4_1.ipynb deleted file mode 100755 index 3133c796..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter4_1.ipynb +++ /dev/null @@ -1,365 +0,0 @@ -{
- "metadata": {
- "name": "chapter4.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 4: Resultants of Non-coplanar Force Systems"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4-1, Page no: 48"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "#Initilization Of Variables\n",
- "\n",
- "F1=20 #lb\n",
- "F2=15 #lb\n",
- "F3=30 #lb\n",
- "F4=50 #lb\n",
- "#Co-ordinates of Forces\n",
- "C1=np.array([2,1,6])\n",
- "C2=np.array([4,-2,5])\n",
- "C3=np.array([-3,-2,1])\n",
- "C4=np.array([5,1,-2])\n",
- "\n",
- "#Calculations\n",
- "\n",
- "A=(C1[0]**2+C1[1]**2+C1[2]**2)**0.5\n",
- "B=(C2[0]**2+C2[1]**2+C2[2]**2)**0.5\n",
- "C=(C3[0]**2+C3[1]**2+C3[2]**2)**0.5\n",
- "D=(C4[0]**2+C4[1]**2+C4[2]**2)**0.5\n",
- "#Calculations for cos(thetax),cos(thetay) and cos(thetaz)\n",
- "theta1=(A**-1)*np.array([C1[0],C1[1],C1[2]])\n",
- "theta2=(B**-1)*np.array([C2[0],C2[1],C2[2]])\n",
- "theta3=(C**-1)*np.array([C3[0],C3[1],C3[2]])\n",
- "theta4=(D**-1)*np.array([C4[0],C4[1],C4[2]])\n",
- "#Calculations for forces (in form of force vectors)\n",
- "Fa=F1*np.array([theta1[0],theta1[1],theta1[2]]) #lb\n",
- "Fb=F2*np.array([theta2[0],theta2[1],theta2[2]]) #lb\n",
- "Fc=F3*np.array([theta3[0],theta3[1],theta3[2]]) #lb\n",
- "Fd=F4*np.array([theta4[0],theta4[1],theta4[2]]) #lb\n",
- "Fx=Fa[0]+Fb[0]+Fc[0]+Fd[0] #lb\n",
- "Fy=Fa[1]+Fb[1]+Fc[1]+Fd[1] #lb\n",
- "Fz=Fa[2]+Fb[2]+Fc[2]+Fd[2] #lb\n",
- "R=(Fx**2+Fy**2+Fz**2)**0.5 #lb\n",
- "thetax=arccos(Fx*R**-1)*(180/pi) #degrees\n",
- "thetay=180-((180*arccos(Fy*R**-1))/pi) #degrees\n",
- "thetaz=(180*arccos(Fz*R**-1))/pi #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R,1),\"lb\"\n",
- "print'The angle of the resultant with respect to x axis is',round(thetax,1),\"degree\"\n",
- "print'The angle of the resultant with respect to y axis is',round(thetay),\"degree\"\n",
- "print'The angle of the resultant with respect to a axis is',round(thetaz,1),\"degree\"\n",
- "\n",
- "# Thetax is off by 0.1 degrees"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 42.5 lb\n",
- "The angle of the resultant with respect to x axis is 30.1 degree\n",
- "The angle of the resultant with respect to y axis is 79.0 degree\n",
- "The angle of the resultant with respect to a axis is 62.4 degree\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4-2, Page no: 49"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=[20,-10,30] #N\n",
- "#co-ordinates in meters\n",
- "a=2 #m\n",
- "b=4 #m\n",
- "c=7 #m\n",
- "d=3 #m\n",
- "e=2 #m\n",
- "f=4 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=F[0]+F[1]+F[2] #N\n",
- "M_o=F[0]*a+F[1]*b+F[2]*c #N.m\n",
- "x=M_o*R**-1 #m\n",
- "M_x=-F[2]*f-F[0]*d-F[1]*e #N.m\n",
- "z=-M_x/R #m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant is +',round(R),\"N\"\n",
- "print'The moment about point O is +',round(M_o),\"N.m\"\n",
- "print'The position of R is at',round(x,2),\"m (from origin)\"\n",
- "print'The moment is',round(M_x),\"N.m\"\n",
- "print'The z co-ordinate is +',round(z),\"m\"\n",
- "\n",
- "# Here the value of R & M_o is correct which should yeild the vaue of x (M_o/R) correctly. However dueto some error in the software the correct value is not being printed.\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant is + 40.0 N\n",
- "The moment about point O is + 210.0 N.m\n",
- "The position of R is at 5.25 m (from origin)\n",
- "The moment is -160.0 N.m\n",
- "The z co-ordinate is + 4.0 m\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4-3, Page No: 49"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=[100,50,-150] #Force vector N\n",
- "a=2 #m\n",
- "b=2 #m \n",
- "c=3 #m\n",
- "d=2 #m\n",
- "e=4 #m\n",
- "f=8 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=F[0]+F[1]+F[2] #N\n",
- "M_x=-F[0]*a+F[1]*b-F[2]*c #N.m\n",
- "M_z=F[0]*d+F[1]*e+F[2]*f #N.m\n",
- "C=sqrt(M_x**2+M_z**2) #N-m\n",
- "thetax=arctan(M_x*-M_z**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant is',round(R),\"N\"\n",
- "print'The moment about x axis is +',round(M_x),\"N.m\"\n",
- "print'The moment about z axis is',round(M_z),\"N.m\"\n",
- "print'The couple acting is',round(C),\"N.m\"\n",
- "print'The trace makes an angle with x axis of',round(thetax,1),\"degrees\"\n",
- "\n",
- "# The answer for C waries by 1 N.m as compared to the book."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant is 0.0 N\n",
- "The moment about x axis is + 350.0 N.m\n",
- "The moment about z axis is -800.0 N.m\n",
- "The couple acting is 873.0 N.m\n",
- "The trace makes an angle with x axis of 23.6 degrees\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4-4, Page No: 50"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "x1=-2\n",
- "y1=2\n",
- "z1=-2\n",
- "x2=3\n",
- "y2=0\n",
- "z2=-4\n",
- "x3=3\n",
- "y3=2\n",
- "z3=2\n",
- "F1=40 #lb\n",
- "F2=30 #lb\n",
- "F3=20 #lb\n",
- "Mxm=np.array([-92.4,-48,-19.4])\n",
- "Mym=np.array([-46.2,72,9.8])\n",
- "Mzm=np.array([46.2,-36,19.4])\n",
- "\n",
- "#Calculations\n",
- "mag1=(x1**2+y1**2+z1**2)**0.5\n",
- "mag2=(x2**2+y2**2+z2**2)**0.5\n",
- "mag3=(x3**2+y3**2+z3**2)**0.5\n",
- "thetax1=(x1*mag1**-1) #degrees\n",
- "thetay1=(y1*mag1**-1) #degrees\n",
- "thetaz1=(z1*mag1**-1) #degrees\n",
- "thetax2=(x2*mag2**-1) #degrees\n",
- "thetay2=(y2*mag2**-1) #degrees\n",
- "thetaz2=(z2*mag2**-1) #degrees\n",
- "thetax3=(x3*mag3**-1) #degrees\n",
- "thetay3=(y3*mag3**-1) #degrees\n",
- "thetaz3=(z3*mag3**-1) #degrees\n",
- "#Now we will define all the components in terms of matrices for simplicity of computation\n",
- "F=np.array([F1,F2,F3]) #lb\n",
- "Fx1=F[0]*thetax1\n",
- "Fy1=F[0]*thetay1\n",
- "Fz1=F[0]*thetaz1\n",
- "Fx2=F[1]*thetax2\n",
- "Fy2=F[1]*thetay2\n",
- "Fz2=F[1]*thetaz2\n",
- "Fx3=F[2]*thetax3\n",
- "Fy3=F[2]*thetay3\n",
- "Fz3=F[2]*thetaz3\n",
- "Fx=Fx1+Fx2+Fx3 #lb\n",
- "Fy=Fy1+Fy2+Fy3 #lb\n",
- "Fz=Fz1+Fz2+Fz3 #lb\n",
- "R=(Fx**2+Fy**2+Fz**2)**0.5 #lb\n",
- "thetax=arccos(Fx*R**-1)*(180/pi) #degrees\n",
- "thetay=arccos(Fy*R**-1)*(180/pi) #degrees\n",
- "thetaz=arccos(Fz*R**-1)*(180/pi) #degrees\n",
- "#Moment calculations\n",
- "Mx=Mxm[0]+Mxm[1]+Mxm[2] #lb-ft\n",
- "My=Mym[0]+Mym[1]+Mym[2] #lb-ft\n",
- "Mz=Mzm[0]+Mzm[1]+Mzm[2] #lb-ft\n",
- "C=(Mx**2+My**2+Mz**2)**0.5 #lb-ft\n",
- "#Direction cosines\n",
- "PHIx=arccos(Mx*C**-1)*(180/pi) #degrees\n",
- "PHIy=arccos(My*C**-1)*(180/pi) #degrees\n",
- "PHIz=arccos(Mz*C**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The result of the force is',round(R,1),\"lb\"\n",
- "print'The angles with respect to X-Axis,Y-Axis and Z-axis are:',round(thetax,1),\"degrees\",',',round(thetay,1),\"degrees\",'and',round(thetaz,1),\"degrees respectively.\"\n",
- "print'The magnitude of resultant couple is',round(C),\"lb-ft\"\n",
- "print'The angles are as follows: Cosphix=',round(PHIx,1),\"degrees\",',Cosphiy=',round(PHIy,1),\"degrees\",'and Cosphiz=',round(PHIz,1),\"degrees\"\n",
- "\n",
- "# The answers may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The result of the force is 50.6 lb\n",
- "The angles with respect to X-Axis,Y-Axis and Z-axis are: 79.2 degrees , 49.6 degrees and 137.6 degrees respectively.\n",
- "The magnitude of resultant couple is 166.0 lb-ft\n",
- "The angles are as follows: Cosphix= 163.8 degrees ,Cosphiy= 77.6 degrees and Cosphiz= 79.8 degrees\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4-5, Page no 52"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "F=np.array([150,90,160]) #lb force vector kind of decleration\n",
- "# Co-ordinates defined as [x,y,z] all the co-ordinates are in feet\n",
- "C_1=np.array([2,0,0]) \n",
- "C_2=np.array([0,0,1])\n",
- "C_3=np.array([0,-2,-1])\n",
- "C_4=np.array([-1,0,-1])\n",
- "\n",
- "#Calculations\n",
- "A=C_2-C_1\n",
- "B=C_4-C_3\n",
- "F1=(F[0]*A)/(A[0]**2+A[1]**2+A[2]**2)**0.5\n",
- "F2=(F[1]*B)/(B[0]**2+B[1]**2+B[2]**2)**0.5\n",
- "R=F1+F2\n",
- "# Determine C1 & C2\n",
- "# Calculating the cross products of C1 & C2 as,\n",
- "C1=np.array([[C_1[1]*F1[2]-C_1[2]*F1[1]],[-(C_1[0]*F1[2]-C_1[2]*F1[0])],[C_1[0]*F1[1]-C_1[1]*F1[0]]]) # lb-ft\n",
- "C2=np.array([[C_3[1]*F2[2]-C_3[2]*F2[1]],[-(C_3[0]*F2[2]-C_3[2]*F2[0])],[C_3[0]*F2[1]-C_3[1]*F2[0]]]) # lb-ft\n",
- "C3=np.array([[0],[0],[160]]) # lb-ft\n",
- "C=C1+C2+C3\n",
- "\n",
- "#Result\n",
- "print'The resultant force couple is',round(C[0],1),\"i\",round(C[1],1),\"j +\",round(C[2],1),\"k lb-ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant force couple is 80.5 i -93.9 j + 79.5 k lb-ft\n"
- ]
- }
- ],
- "prompt_number": 48
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter4_2.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter4_2.ipynb deleted file mode 100755 index 3133c796..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter4_2.ipynb +++ /dev/null @@ -1,365 +0,0 @@ -{
- "metadata": {
- "name": "chapter4.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 4: Resultants of Non-coplanar Force Systems"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4-1, Page no: 48"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "#Initilization Of Variables\n",
- "\n",
- "F1=20 #lb\n",
- "F2=15 #lb\n",
- "F3=30 #lb\n",
- "F4=50 #lb\n",
- "#Co-ordinates of Forces\n",
- "C1=np.array([2,1,6])\n",
- "C2=np.array([4,-2,5])\n",
- "C3=np.array([-3,-2,1])\n",
- "C4=np.array([5,1,-2])\n",
- "\n",
- "#Calculations\n",
- "\n",
- "A=(C1[0]**2+C1[1]**2+C1[2]**2)**0.5\n",
- "B=(C2[0]**2+C2[1]**2+C2[2]**2)**0.5\n",
- "C=(C3[0]**2+C3[1]**2+C3[2]**2)**0.5\n",
- "D=(C4[0]**2+C4[1]**2+C4[2]**2)**0.5\n",
- "#Calculations for cos(thetax),cos(thetay) and cos(thetaz)\n",
- "theta1=(A**-1)*np.array([C1[0],C1[1],C1[2]])\n",
- "theta2=(B**-1)*np.array([C2[0],C2[1],C2[2]])\n",
- "theta3=(C**-1)*np.array([C3[0],C3[1],C3[2]])\n",
- "theta4=(D**-1)*np.array([C4[0],C4[1],C4[2]])\n",
- "#Calculations for forces (in form of force vectors)\n",
- "Fa=F1*np.array([theta1[0],theta1[1],theta1[2]]) #lb\n",
- "Fb=F2*np.array([theta2[0],theta2[1],theta2[2]]) #lb\n",
- "Fc=F3*np.array([theta3[0],theta3[1],theta3[2]]) #lb\n",
- "Fd=F4*np.array([theta4[0],theta4[1],theta4[2]]) #lb\n",
- "Fx=Fa[0]+Fb[0]+Fc[0]+Fd[0] #lb\n",
- "Fy=Fa[1]+Fb[1]+Fc[1]+Fd[1] #lb\n",
- "Fz=Fa[2]+Fb[2]+Fc[2]+Fd[2] #lb\n",
- "R=(Fx**2+Fy**2+Fz**2)**0.5 #lb\n",
- "thetax=arccos(Fx*R**-1)*(180/pi) #degrees\n",
- "thetay=180-((180*arccos(Fy*R**-1))/pi) #degrees\n",
- "thetaz=(180*arccos(Fz*R**-1))/pi #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R,1),\"lb\"\n",
- "print'The angle of the resultant with respect to x axis is',round(thetax,1),\"degree\"\n",
- "print'The angle of the resultant with respect to y axis is',round(thetay),\"degree\"\n",
- "print'The angle of the resultant with respect to a axis is',round(thetaz,1),\"degree\"\n",
- "\n",
- "# Thetax is off by 0.1 degrees"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 42.5 lb\n",
- "The angle of the resultant with respect to x axis is 30.1 degree\n",
- "The angle of the resultant with respect to y axis is 79.0 degree\n",
- "The angle of the resultant with respect to a axis is 62.4 degree\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4-2, Page no: 49"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=[20,-10,30] #N\n",
- "#co-ordinates in meters\n",
- "a=2 #m\n",
- "b=4 #m\n",
- "c=7 #m\n",
- "d=3 #m\n",
- "e=2 #m\n",
- "f=4 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=F[0]+F[1]+F[2] #N\n",
- "M_o=F[0]*a+F[1]*b+F[2]*c #N.m\n",
- "x=M_o*R**-1 #m\n",
- "M_x=-F[2]*f-F[0]*d-F[1]*e #N.m\n",
- "z=-M_x/R #m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant is +',round(R),\"N\"\n",
- "print'The moment about point O is +',round(M_o),\"N.m\"\n",
- "print'The position of R is at',round(x,2),\"m (from origin)\"\n",
- "print'The moment is',round(M_x),\"N.m\"\n",
- "print'The z co-ordinate is +',round(z),\"m\"\n",
- "\n",
- "# Here the value of R & M_o is correct which should yeild the vaue of x (M_o/R) correctly. However dueto some error in the software the correct value is not being printed.\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant is + 40.0 N\n",
- "The moment about point O is + 210.0 N.m\n",
- "The position of R is at 5.25 m (from origin)\n",
- "The moment is -160.0 N.m\n",
- "The z co-ordinate is + 4.0 m\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4-3, Page No: 49"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=[100,50,-150] #Force vector N\n",
- "a=2 #m\n",
- "b=2 #m \n",
- "c=3 #m\n",
- "d=2 #m\n",
- "e=4 #m\n",
- "f=8 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=F[0]+F[1]+F[2] #N\n",
- "M_x=-F[0]*a+F[1]*b-F[2]*c #N.m\n",
- "M_z=F[0]*d+F[1]*e+F[2]*f #N.m\n",
- "C=sqrt(M_x**2+M_z**2) #N-m\n",
- "thetax=arctan(M_x*-M_z**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant is',round(R),\"N\"\n",
- "print'The moment about x axis is +',round(M_x),\"N.m\"\n",
- "print'The moment about z axis is',round(M_z),\"N.m\"\n",
- "print'The couple acting is',round(C),\"N.m\"\n",
- "print'The trace makes an angle with x axis of',round(thetax,1),\"degrees\"\n",
- "\n",
- "# The answer for C waries by 1 N.m as compared to the book."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant is 0.0 N\n",
- "The moment about x axis is + 350.0 N.m\n",
- "The moment about z axis is -800.0 N.m\n",
- "The couple acting is 873.0 N.m\n",
- "The trace makes an angle with x axis of 23.6 degrees\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4-4, Page No: 50"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "x1=-2\n",
- "y1=2\n",
- "z1=-2\n",
- "x2=3\n",
- "y2=0\n",
- "z2=-4\n",
- "x3=3\n",
- "y3=2\n",
- "z3=2\n",
- "F1=40 #lb\n",
- "F2=30 #lb\n",
- "F3=20 #lb\n",
- "Mxm=np.array([-92.4,-48,-19.4])\n",
- "Mym=np.array([-46.2,72,9.8])\n",
- "Mzm=np.array([46.2,-36,19.4])\n",
- "\n",
- "#Calculations\n",
- "mag1=(x1**2+y1**2+z1**2)**0.5\n",
- "mag2=(x2**2+y2**2+z2**2)**0.5\n",
- "mag3=(x3**2+y3**2+z3**2)**0.5\n",
- "thetax1=(x1*mag1**-1) #degrees\n",
- "thetay1=(y1*mag1**-1) #degrees\n",
- "thetaz1=(z1*mag1**-1) #degrees\n",
- "thetax2=(x2*mag2**-1) #degrees\n",
- "thetay2=(y2*mag2**-1) #degrees\n",
- "thetaz2=(z2*mag2**-1) #degrees\n",
- "thetax3=(x3*mag3**-1) #degrees\n",
- "thetay3=(y3*mag3**-1) #degrees\n",
- "thetaz3=(z3*mag3**-1) #degrees\n",
- "#Now we will define all the components in terms of matrices for simplicity of computation\n",
- "F=np.array([F1,F2,F3]) #lb\n",
- "Fx1=F[0]*thetax1\n",
- "Fy1=F[0]*thetay1\n",
- "Fz1=F[0]*thetaz1\n",
- "Fx2=F[1]*thetax2\n",
- "Fy2=F[1]*thetay2\n",
- "Fz2=F[1]*thetaz2\n",
- "Fx3=F[2]*thetax3\n",
- "Fy3=F[2]*thetay3\n",
- "Fz3=F[2]*thetaz3\n",
- "Fx=Fx1+Fx2+Fx3 #lb\n",
- "Fy=Fy1+Fy2+Fy3 #lb\n",
- "Fz=Fz1+Fz2+Fz3 #lb\n",
- "R=(Fx**2+Fy**2+Fz**2)**0.5 #lb\n",
- "thetax=arccos(Fx*R**-1)*(180/pi) #degrees\n",
- "thetay=arccos(Fy*R**-1)*(180/pi) #degrees\n",
- "thetaz=arccos(Fz*R**-1)*(180/pi) #degrees\n",
- "#Moment calculations\n",
- "Mx=Mxm[0]+Mxm[1]+Mxm[2] #lb-ft\n",
- "My=Mym[0]+Mym[1]+Mym[2] #lb-ft\n",
- "Mz=Mzm[0]+Mzm[1]+Mzm[2] #lb-ft\n",
- "C=(Mx**2+My**2+Mz**2)**0.5 #lb-ft\n",
- "#Direction cosines\n",
- "PHIx=arccos(Mx*C**-1)*(180/pi) #degrees\n",
- "PHIy=arccos(My*C**-1)*(180/pi) #degrees\n",
- "PHIz=arccos(Mz*C**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The result of the force is',round(R,1),\"lb\"\n",
- "print'The angles with respect to X-Axis,Y-Axis and Z-axis are:',round(thetax,1),\"degrees\",',',round(thetay,1),\"degrees\",'and',round(thetaz,1),\"degrees respectively.\"\n",
- "print'The magnitude of resultant couple is',round(C),\"lb-ft\"\n",
- "print'The angles are as follows: Cosphix=',round(PHIx,1),\"degrees\",',Cosphiy=',round(PHIy,1),\"degrees\",'and Cosphiz=',round(PHIz,1),\"degrees\"\n",
- "\n",
- "# The answers may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The result of the force is 50.6 lb\n",
- "The angles with respect to X-Axis,Y-Axis and Z-axis are: 79.2 degrees , 49.6 degrees and 137.6 degrees respectively.\n",
- "The magnitude of resultant couple is 166.0 lb-ft\n",
- "The angles are as follows: Cosphix= 163.8 degrees ,Cosphiy= 77.6 degrees and Cosphiz= 79.8 degrees\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4-5, Page no 52"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "F=np.array([150,90,160]) #lb force vector kind of decleration\n",
- "# Co-ordinates defined as [x,y,z] all the co-ordinates are in feet\n",
- "C_1=np.array([2,0,0]) \n",
- "C_2=np.array([0,0,1])\n",
- "C_3=np.array([0,-2,-1])\n",
- "C_4=np.array([-1,0,-1])\n",
- "\n",
- "#Calculations\n",
- "A=C_2-C_1\n",
- "B=C_4-C_3\n",
- "F1=(F[0]*A)/(A[0]**2+A[1]**2+A[2]**2)**0.5\n",
- "F2=(F[1]*B)/(B[0]**2+B[1]**2+B[2]**2)**0.5\n",
- "R=F1+F2\n",
- "# Determine C1 & C2\n",
- "# Calculating the cross products of C1 & C2 as,\n",
- "C1=np.array([[C_1[1]*F1[2]-C_1[2]*F1[1]],[-(C_1[0]*F1[2]-C_1[2]*F1[0])],[C_1[0]*F1[1]-C_1[1]*F1[0]]]) # lb-ft\n",
- "C2=np.array([[C_3[1]*F2[2]-C_3[2]*F2[1]],[-(C_3[0]*F2[2]-C_3[2]*F2[0])],[C_3[0]*F2[1]-C_3[1]*F2[0]]]) # lb-ft\n",
- "C3=np.array([[0],[0],[160]]) # lb-ft\n",
- "C=C1+C2+C3\n",
- "\n",
- "#Result\n",
- "print'The resultant force couple is',round(C[0],1),\"i\",round(C[1],1),\"j +\",round(C[2],1),\"k lb-ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant force couple is 80.5 i -93.9 j + 79.5 k lb-ft\n"
- ]
- }
- ],
- "prompt_number": 48
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter4_3.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter4_3.ipynb deleted file mode 100755 index 3133c796..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter4_3.ipynb +++ /dev/null @@ -1,365 +0,0 @@ -{
- "metadata": {
- "name": "chapter4.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 4: Resultants of Non-coplanar Force Systems"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4-1, Page no: 48"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "#Initilization Of Variables\n",
- "\n",
- "F1=20 #lb\n",
- "F2=15 #lb\n",
- "F3=30 #lb\n",
- "F4=50 #lb\n",
- "#Co-ordinates of Forces\n",
- "C1=np.array([2,1,6])\n",
- "C2=np.array([4,-2,5])\n",
- "C3=np.array([-3,-2,1])\n",
- "C4=np.array([5,1,-2])\n",
- "\n",
- "#Calculations\n",
- "\n",
- "A=(C1[0]**2+C1[1]**2+C1[2]**2)**0.5\n",
- "B=(C2[0]**2+C2[1]**2+C2[2]**2)**0.5\n",
- "C=(C3[0]**2+C3[1]**2+C3[2]**2)**0.5\n",
- "D=(C4[0]**2+C4[1]**2+C4[2]**2)**0.5\n",
- "#Calculations for cos(thetax),cos(thetay) and cos(thetaz)\n",
- "theta1=(A**-1)*np.array([C1[0],C1[1],C1[2]])\n",
- "theta2=(B**-1)*np.array([C2[0],C2[1],C2[2]])\n",
- "theta3=(C**-1)*np.array([C3[0],C3[1],C3[2]])\n",
- "theta4=(D**-1)*np.array([C4[0],C4[1],C4[2]])\n",
- "#Calculations for forces (in form of force vectors)\n",
- "Fa=F1*np.array([theta1[0],theta1[1],theta1[2]]) #lb\n",
- "Fb=F2*np.array([theta2[0],theta2[1],theta2[2]]) #lb\n",
- "Fc=F3*np.array([theta3[0],theta3[1],theta3[2]]) #lb\n",
- "Fd=F4*np.array([theta4[0],theta4[1],theta4[2]]) #lb\n",
- "Fx=Fa[0]+Fb[0]+Fc[0]+Fd[0] #lb\n",
- "Fy=Fa[1]+Fb[1]+Fc[1]+Fd[1] #lb\n",
- "Fz=Fa[2]+Fb[2]+Fc[2]+Fd[2] #lb\n",
- "R=(Fx**2+Fy**2+Fz**2)**0.5 #lb\n",
- "thetax=arccos(Fx*R**-1)*(180/pi) #degrees\n",
- "thetay=180-((180*arccos(Fy*R**-1))/pi) #degrees\n",
- "thetaz=(180*arccos(Fz*R**-1))/pi #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant of the force system is',round(R,1),\"lb\"\n",
- "print'The angle of the resultant with respect to x axis is',round(thetax,1),\"degree\"\n",
- "print'The angle of the resultant with respect to y axis is',round(thetay),\"degree\"\n",
- "print'The angle of the resultant with respect to a axis is',round(thetaz,1),\"degree\"\n",
- "\n",
- "# Thetax is off by 0.1 degrees"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant of the force system is 42.5 lb\n",
- "The angle of the resultant with respect to x axis is 30.1 degree\n",
- "The angle of the resultant with respect to y axis is 79.0 degree\n",
- "The angle of the resultant with respect to a axis is 62.4 degree\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4-2, Page no: 49"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=[20,-10,30] #N\n",
- "#co-ordinates in meters\n",
- "a=2 #m\n",
- "b=4 #m\n",
- "c=7 #m\n",
- "d=3 #m\n",
- "e=2 #m\n",
- "f=4 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=F[0]+F[1]+F[2] #N\n",
- "M_o=F[0]*a+F[1]*b+F[2]*c #N.m\n",
- "x=M_o*R**-1 #m\n",
- "M_x=-F[2]*f-F[0]*d-F[1]*e #N.m\n",
- "z=-M_x/R #m\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant is +',round(R),\"N\"\n",
- "print'The moment about point O is +',round(M_o),\"N.m\"\n",
- "print'The position of R is at',round(x,2),\"m (from origin)\"\n",
- "print'The moment is',round(M_x),\"N.m\"\n",
- "print'The z co-ordinate is +',round(z),\"m\"\n",
- "\n",
- "# Here the value of R & M_o is correct which should yeild the vaue of x (M_o/R) correctly. However dueto some error in the software the correct value is not being printed.\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant is + 40.0 N\n",
- "The moment about point O is + 210.0 N.m\n",
- "The position of R is at 5.25 m (from origin)\n",
- "The moment is -160.0 N.m\n",
- "The z co-ordinate is + 4.0 m\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4-3, Page No: 49"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=[100,50,-150] #Force vector N\n",
- "a=2 #m\n",
- "b=2 #m \n",
- "c=3 #m\n",
- "d=2 #m\n",
- "e=4 #m\n",
- "f=8 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R=F[0]+F[1]+F[2] #N\n",
- "M_x=-F[0]*a+F[1]*b-F[2]*c #N.m\n",
- "M_z=F[0]*d+F[1]*e+F[2]*f #N.m\n",
- "C=sqrt(M_x**2+M_z**2) #N-m\n",
- "thetax=arctan(M_x*-M_z**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The resultant is',round(R),\"N\"\n",
- "print'The moment about x axis is +',round(M_x),\"N.m\"\n",
- "print'The moment about z axis is',round(M_z),\"N.m\"\n",
- "print'The couple acting is',round(C),\"N.m\"\n",
- "print'The trace makes an angle with x axis of',round(thetax,1),\"degrees\"\n",
- "\n",
- "# The answer for C waries by 1 N.m as compared to the book."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant is 0.0 N\n",
- "The moment about x axis is + 350.0 N.m\n",
- "The moment about z axis is -800.0 N.m\n",
- "The couple acting is 873.0 N.m\n",
- "The trace makes an angle with x axis of 23.6 degrees\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4-4, Page No: 50"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "x1=-2\n",
- "y1=2\n",
- "z1=-2\n",
- "x2=3\n",
- "y2=0\n",
- "z2=-4\n",
- "x3=3\n",
- "y3=2\n",
- "z3=2\n",
- "F1=40 #lb\n",
- "F2=30 #lb\n",
- "F3=20 #lb\n",
- "Mxm=np.array([-92.4,-48,-19.4])\n",
- "Mym=np.array([-46.2,72,9.8])\n",
- "Mzm=np.array([46.2,-36,19.4])\n",
- "\n",
- "#Calculations\n",
- "mag1=(x1**2+y1**2+z1**2)**0.5\n",
- "mag2=(x2**2+y2**2+z2**2)**0.5\n",
- "mag3=(x3**2+y3**2+z3**2)**0.5\n",
- "thetax1=(x1*mag1**-1) #degrees\n",
- "thetay1=(y1*mag1**-1) #degrees\n",
- "thetaz1=(z1*mag1**-1) #degrees\n",
- "thetax2=(x2*mag2**-1) #degrees\n",
- "thetay2=(y2*mag2**-1) #degrees\n",
- "thetaz2=(z2*mag2**-1) #degrees\n",
- "thetax3=(x3*mag3**-1) #degrees\n",
- "thetay3=(y3*mag3**-1) #degrees\n",
- "thetaz3=(z3*mag3**-1) #degrees\n",
- "#Now we will define all the components in terms of matrices for simplicity of computation\n",
- "F=np.array([F1,F2,F3]) #lb\n",
- "Fx1=F[0]*thetax1\n",
- "Fy1=F[0]*thetay1\n",
- "Fz1=F[0]*thetaz1\n",
- "Fx2=F[1]*thetax2\n",
- "Fy2=F[1]*thetay2\n",
- "Fz2=F[1]*thetaz2\n",
- "Fx3=F[2]*thetax3\n",
- "Fy3=F[2]*thetay3\n",
- "Fz3=F[2]*thetaz3\n",
- "Fx=Fx1+Fx2+Fx3 #lb\n",
- "Fy=Fy1+Fy2+Fy3 #lb\n",
- "Fz=Fz1+Fz2+Fz3 #lb\n",
- "R=(Fx**2+Fy**2+Fz**2)**0.5 #lb\n",
- "thetax=arccos(Fx*R**-1)*(180/pi) #degrees\n",
- "thetay=arccos(Fy*R**-1)*(180/pi) #degrees\n",
- "thetaz=arccos(Fz*R**-1)*(180/pi) #degrees\n",
- "#Moment calculations\n",
- "Mx=Mxm[0]+Mxm[1]+Mxm[2] #lb-ft\n",
- "My=Mym[0]+Mym[1]+Mym[2] #lb-ft\n",
- "Mz=Mzm[0]+Mzm[1]+Mzm[2] #lb-ft\n",
- "C=(Mx**2+My**2+Mz**2)**0.5 #lb-ft\n",
- "#Direction cosines\n",
- "PHIx=arccos(Mx*C**-1)*(180/pi) #degrees\n",
- "PHIy=arccos(My*C**-1)*(180/pi) #degrees\n",
- "PHIz=arccos(Mz*C**-1)*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The result of the force is',round(R,1),\"lb\"\n",
- "print'The angles with respect to X-Axis,Y-Axis and Z-axis are:',round(thetax,1),\"degrees\",',',round(thetay,1),\"degrees\",'and',round(thetaz,1),\"degrees respectively.\"\n",
- "print'The magnitude of resultant couple is',round(C),\"lb-ft\"\n",
- "print'The angles are as follows: Cosphix=',round(PHIx,1),\"degrees\",',Cosphiy=',round(PHIy,1),\"degrees\",'and Cosphiz=',round(PHIz,1),\"degrees\"\n",
- "\n",
- "# The answers may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The result of the force is 50.6 lb\n",
- "The angles with respect to X-Axis,Y-Axis and Z-axis are: 79.2 degrees , 49.6 degrees and 137.6 degrees respectively.\n",
- "The magnitude of resultant couple is 166.0 lb-ft\n",
- "The angles are as follows: Cosphix= 163.8 degrees ,Cosphiy= 77.6 degrees and Cosphiz= 79.8 degrees\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4-5, Page no 52"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "F=np.array([150,90,160]) #lb force vector kind of decleration\n",
- "# Co-ordinates defined as [x,y,z] all the co-ordinates are in feet\n",
- "C_1=np.array([2,0,0]) \n",
- "C_2=np.array([0,0,1])\n",
- "C_3=np.array([0,-2,-1])\n",
- "C_4=np.array([-1,0,-1])\n",
- "\n",
- "#Calculations\n",
- "A=C_2-C_1\n",
- "B=C_4-C_3\n",
- "F1=(F[0]*A)/(A[0]**2+A[1]**2+A[2]**2)**0.5\n",
- "F2=(F[1]*B)/(B[0]**2+B[1]**2+B[2]**2)**0.5\n",
- "R=F1+F2\n",
- "# Determine C1 & C2\n",
- "# Calculating the cross products of C1 & C2 as,\n",
- "C1=np.array([[C_1[1]*F1[2]-C_1[2]*F1[1]],[-(C_1[0]*F1[2]-C_1[2]*F1[0])],[C_1[0]*F1[1]-C_1[1]*F1[0]]]) # lb-ft\n",
- "C2=np.array([[C_3[1]*F2[2]-C_3[2]*F2[1]],[-(C_3[0]*F2[2]-C_3[2]*F2[0])],[C_3[0]*F2[1]-C_3[1]*F2[0]]]) # lb-ft\n",
- "C3=np.array([[0],[0],[160]]) # lb-ft\n",
- "C=C1+C2+C3\n",
- "\n",
- "#Result\n",
- "print'The resultant force couple is',round(C[0],1),\"i\",round(C[1],1),\"j +\",round(C[2],1),\"k lb-ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resultant force couple is 80.5 i -93.9 j + 79.5 k lb-ft\n"
- ]
- }
- ],
- "prompt_number": 48
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter5.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter5.ipynb deleted file mode 100755 index 8951e1db..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter5.ipynb +++ /dev/null @@ -1,799 +0,0 @@ -{
- "metadata": {
- "name": "chapter5.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 5: Equilibrium of Coplanar Force Systems."
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-1, Page no 58"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "# From eqn's 1&2\n",
- "D=np.array([[6/sqrt(40),-4/sqrt(20)],[2/sqrt(40),2/sqrt(20)]])\n",
- "B=np.array([0,25]) #lb\n",
- "\n",
- "#Calculations\n",
- "\n",
- "X=np.linalg.solve(D,B)\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The tension in cable AB is',round(X[1],1),\"lb\"\n",
- "print'The tension in cable AC is',round(X[0],1),\"lb\"\n",
- "\n",
- "# The tensions in the cable AB & AC is off by 0.1 lb"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in cable AB is 33.5 lb\n",
- "The tension in cable AC is 31.6 lb\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-2, Page no 59"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F1=100 #lb\n",
- "R=16 #in\n",
- "\n",
- "#Calculations\n",
- "theta=arcsin(14*R**-1)*(180/pi) #degrees\n",
- "# since theta=61 degrees,\n",
- "sin61=0.8746\n",
- "cos61=0.4848\n",
- "N=F1/sin61 #lb\n",
- "P=N*cos61 #lb\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The value of normal reaction offered is',round(N,1),\"lb\"\n",
- "print'The push required is',round(P,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of normal reaction offered is 114.3 lb\n",
- "The push required is 55.4 lb\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-3,Page no 59"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "L=20 #m\n",
- "M=1200 #kg\n",
- "g=9.81 #m/s**2\n",
- "H=10 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "AB=sqrt(L**2-H**2) #Applying Pythagoras Theorem\n",
- "costheta=17.3/20\n",
- "F1=M*g*H/AB #N\n",
- "F2=M*g/costheta #N\n",
- "\n",
- "#Result\n",
- "\n",
- "print'Force F1 is',round(F1),\"N\"\n",
- "print'Force F2 is',round(F2),\"N\"\n",
- "\n",
- "#Decimal accuracy causes discrepancy in answers compared to the textbook answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force F1 is 6797.0 N\n",
- "Force F2 is 13609.0 N\n"
- ]
- }
- ],
- "prompt_number": 45
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-4, Page No 60"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "Fx=1000 #lb\n",
- "Fy=1000 #lb\n",
- "costheta=9*15**-1\n",
- "cosbeta=12*15**-1\n",
- "sintheta=4*5**-1\n",
- "sinbeta=3*5**-1\n",
- "\n",
- "#Calculations\n",
- "#Matrix solution\n",
- "A=np.array([[costheta,-cosbeta],[sintheta,sinbeta]]) \n",
- "B=np.array([-1000,1000])\n",
- "X=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The force in AB is',round(X[0]),\"lb compression\"\n",
- "print'The force in BC is',round(X[1]),\"lb compression\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force in AB is 200.0 lb compression\n",
- "The force in BC is 1400.0 lb compression\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-5, Page no 61"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "w=10 #lb/ft\n",
- "L=12 #ft\n",
- "# as theta=30 degrees,\n",
- "sin30=2**-1\n",
- "cos30=sqrt(3)*2**-1\n",
- "\n",
- "#Calculation\n",
- "#Matrix Calculations\n",
- "A=np.array([[cos30,-cos30],[sin30,sin30]]) \n",
- "B=np.array([0,120]) \n",
- "X=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The tension in the cable is,T=',round(X[0]),\"lb\"\n",
- "print'The reaction at B is,R',round(X[1]),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in the cable is,T= 120.0 lb\n",
- "The reaction at B is,R 120.0 lb\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-6,Page no 61"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W1=40 #lb\n",
- "W2=30 #lb\n",
- "# as theta1=30 degrees,\n",
- "sin30=2**-1\n",
- "\n",
- "#Calculations\n",
- "#Summing the forces parallel to 30 degree plane\n",
- "T=W1*sin30\n",
- "theta=arcsin(T/W2)*(180/pi)\n",
- "\n",
- "#Result\n",
- "print'The tension in the cable is',round(T),\"lb\"\n",
- "print'The angle is',round(theta,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in the cable is 20.0 lb\n",
- "The angle is 41.8 degrees\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-8,Page no 62"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F1=125 #N\n",
- "F2=200 #N\n",
- "F3=340 #N\n",
- "F4=180 #N\n",
- "x1=4 #m\n",
- "x2=3 #m\n",
- "x3=10 #m\n",
- "x4=15 #m\n",
- "x5=17 #m\n",
- "\n",
- "#Calculations\n",
- "Rb=(-F1*x1+F2*x2+F3*x3+F4*x4)/x5 #moment about point A\n",
- "Ra=(F1*(x1+x5)+F3*(x5-x3)+F2*(x5-x2)+F4*(x5-x4))/x5 #moment about point B\n",
- "\n",
- "#Result\n",
- "print'The reaction at A is',round(Ra),\"N\"\n",
- "print'The reaction at B is',round(Rb),\"N\"\n",
- "\n",
- "# The ans for B is off by 1 N"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction at A is 480.0 N\n",
- "The reaction at B is 364.0 N\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-9, Page no 63"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F1=1000 #lb\n",
- "F2=1200 #lb\n",
- "F3=2000 #lb\n",
- "x1=1 #ft\n",
- "x2=7 #ft\n",
- "x4=2 #ft\n",
- "x3=6 #ft\n",
- "\n",
- "#Calculation\n",
- "#Equilibrium equations\n",
- "Rn=(F3*(x1+x2+x3)+F2*(x1+x2)+F1*x1)/(x1+x3+x2+x4) #Moment about point M\n",
- "Rm=(F1*(x2+x3+x4)+F2*(x3+x4)+F3*x4)/(x1+x2+x3+x4) #Moment about point N\n",
- "\n",
- "#Result\n",
- "print'The reaction at M is',round(Rm),\"lb\"\n",
- "print'The reaction at N is',round(Rn),\"lb\"\n",
- "\n",
- "#Decimal Accuracy causes discrepancy in answers between computation and textbook\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction at M is 1787.0 lb\n",
- "The reaction at N is 2412.0 lb\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-10, Page no 64"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "P=10 #kg\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "# equilibrium at fig b\n",
- "T1=P*g/2 #N\n",
- "# equilibrium at fig c\n",
- "T2=T1/2 #N\n",
- "#equilibrium at fig d\n",
- "P=T2\n",
- "\n",
- "#Result\n",
- "print'The force P is',round(P,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force P is 24.5 N\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-11, Page no 64"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "k=20 # lb/in\n",
- "w=20 # lb/ft\n",
- "x1=4 #ft\n",
- "x2=10 # ft\n",
- "x3=8 #ft\n",
- "x4=6 #ft\n",
- "x5=9 #ft\n",
- "F1=1920 #lb.rad\n",
- "F2=3360 #lb.rad\n",
- "\n",
- "#calculations\n",
- "theta=(w*x2*x5)*(F1*x3+F2*(x3+x4))**-1 #radians\n",
- "FB=F1*theta\n",
- "FC=F2*theta\n",
- "A=(w*x2)-FB-FC\n",
- "\n",
- "#Result\n",
- "print'The force in spring B is',round(FB,1),\"lb\"\n",
- "print'The force in spring C is',round(FC,1),\"lb\"\n",
- "print'The reaction at A is',round(A,1),\"lb up\"\n",
- "\n",
- " # The answer waries slightly due to decimal point discrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force in spring B is 55.4 lb\n",
- "The force in spring C is 96.9 lb\n",
- "The reaction at A is 47.7 lb up\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-12, Page no 65"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "L=3.8 #m\n",
- "w=10 # kg/m\n",
- "P=1000 #N\n",
- "t=0.8 #m\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "Gf=L*w*g #N\n",
- "A=(P*L+Gf*L*0.5)/t #N Taking moment about point B\n",
- "B=(P*(L-t)+Gf*(0.5*L-t))/t #N Taking moment about point A\n",
- "\n",
- "#Result\n",
- "print'The reaction at point A is',round(A),\"N\"\n",
- "print'The reaction at point B is',round(B),\"N\"\n",
- "\n",
- "# The answers in the textbook are incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction at point A is 5635.0 N\n",
- "The reaction at point B is 4263.0 N\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-13, Page no 65"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wa=400 #lb\n",
- "Wb=200 #lb\n",
- "# as theta=30 degrees,\n",
- "sin30=2**-1\n",
- "\n",
- "#Calculations\n",
- "Ta=Wa*sin30 #lb\n",
- "Tb=Wb*sin30 #lb\n",
- "#Taking moment about point O\n",
- "P=(Tb*12+Ta*6)/24 #lb\n",
- "\n",
- "#Result\n",
- "print'The value of Ta is',round(Ta,3),\"lb\"\n",
- "print'The value of Tb is',round(Tb,3),\"lb\"\n",
- "print'The value of P is',round(P,3),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of Ta is 200.0 lb\n",
- "The value of Tb is 100.0 lb\n",
- "The value of P is 100.0 lb\n"
- ]
- }
- ],
- "prompt_number": 37
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-15, Page no 66"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "F=np.array([5,2,3,1.5]) #kN Forces are defined as a cloumn matrix\n",
- "theta=(pi*np.array([90,60,45,80]))/180 #degrees angles are also defined as a column matrix\n",
- "d=np.array([2,6,13,17]) #distances from point C of each force\n",
- "c=np.array([17,15,11,4]) #distance form point D of each force\n",
- "#Calculations\n",
- "\n",
- "#Summing horizontal forces\n",
- "Ch=F[1]*cos(theta[1])-F[2]*cos(theta[2])+F[3]*cos(theta[3]) #kN \"which indidcates that Ch acts to the left instead of the assumed\"\n",
- "#Taking moment about point C\n",
- "D=(F[0]*d[0]+F[1]*sin(theta[1])*d[1]+F[2]*sin(theta[2])*d[2]+F[3]*sin(theta[3])*d[3])/d[3] #kN\n",
- "#Taking moment about point D\n",
- "Cv=(F[0]*c[1]+F[1]*sin(theta[1])*c[2]+F[2]*sin(theta[2])*c[3])/c[1]\n",
- "#Result\n",
- "\n",
- "print'The values of Ch,D and Cv are:',round(Ch,2),\"kN ,\",round(D,1),\"kN\",'and',round(Cv,2),\"kN\"\n",
- "\n",
- "# The ans of Cv is incorrect in textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values of Ch,D and Cv are: -0.86 kN , 4.3 kN and 6.84 kN\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-16, Page no 67"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=100 #N/m\n",
- "F1=200 #N\n",
- "M=500 #N.m\n",
- "Lw=2 #m\n",
- "#Distance from point A\n",
- "d=np.array([1,2,3,4,5]) #m\n",
- "#Distance from point B\n",
- "b=np.array([5,4,3,2,1]) #m\n",
- "\n",
- "#Calculations\n",
- "#Taking moment aboout point A\n",
- "Rb=(w*Lw*d[0]+F1*d[2]-M)/d[3] #N\n",
- "#Taking moment about point B\n",
- "Ra=(w*Lw*b[2]+F1*b[4]+M)/b[1] #N\n",
- "\n",
- "#Result\n",
- "print'The value of reaction at A is',round(Ra),\"N\"\n",
- "print'The value of reaction at B is',round(Rb),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of reaction at A is 325.0 N\n",
- "The value of reaction at B is 75.0 N\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-18, Page no 68"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "# The values of theta are=[60,60,45] degrees, therefore its values are as,\n",
- "costheta2=sqrt(2)**-1\n",
- "sintheta2=sqrt(2)**-1\n",
- "d=np.array([4.46,3.54,2]) #feet defined as a matrix\n",
- "F=400 #lb\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point A\n",
- "Re=(F*(8-d[1]))/8 #lb\n",
- "Ra=400-Re #lb here i have used the summation of forces in the vertical direction\n",
- "#Taking moment about point B\n",
- "Dv=(-F*3.644)*5.77**-1 #lb\n",
- "#Taking moment about point D\n",
- "Bv=(F*2.126)/5.77 #lb\n",
- "#Taking summation of forces in the vertical direction\n",
- "Cv=-223-Dv #lb\n",
- "#Taking moment about point D\n",
- "Ch=((223*d[2]*costheta2)-(Cv*5.173*costheta2))*(5.173*sintheta2)**-1 #lb\n",
- "#Taking summation of forces in the horizontal direction\n",
- "Dh=-Ch #lb\n",
- "#Taking sum of forces in horizontal direction\n",
- "Bh=-Dh #lb\n",
- "\n",
- "#Result\n",
- "print'The Floor reactions are'\n",
- "print'Ra=',round(Ra),\"lb up\"\n",
- "print'Re=',round(Re),\"lb up\"\n",
- "\n",
- "print'Pin reaction at C on CE are'\n",
- "print'Ch=',round(Ch,1),\"lb to right\"\n",
- "print'Cv=',round(Cv,1),\"lb up\"\n",
- "\n",
- "print'The pin reactions at B on AC are:'\n",
- "print'Bh=',round(Bh,1),\"lb to right\"\n",
- "print'Bv=',round(Bv,1),\"lb down\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Floor reactions are\n",
- "Ra= 177.0 lb up\n",
- "Re= 223.0 lb up\n",
- "Pin reaction at C on CE are\n",
- "Ch= 56.6 lb to right\n",
- "Cv= 29.6 lb up\n",
- "The pin reactions at B on AC are:\n",
- "Bh= 56.6 lb to right\n",
- "Bv= 147.4 lb down\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-19, Page no 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=0.5 #m\n",
- "m=10 #kg\n",
- "g=9.81 #m/s**2\n",
- "# since theta=60 degrees,\n",
- "sin30=2**-1\n",
- "cos30=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Due to symmetry the reaction will be shared by the structure\n",
- "A=m*g*r #N\n",
- "B=A #N\n",
- "#Vertical forces summed\n",
- "N1=m*g/(2*sin30) #N\n",
- "#Taking moment about point C\n",
- "T=(N1*0.866+B*sin30)*(1.5*cos30)**-1\n",
- " \n",
- "#Result\n",
- "print'The value of N1 is',round(N1),\"N\"\n",
- "print'The value of T is',round(T,1),\"N\"\n",
- "\n",
- "# The ans for T is off by 0.1 N"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of N1 is 98.0 N\n",
- "The value of T is 84.3 N\n"
- ]
- }
- ],
- "prompt_number": 52
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter5_1.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter5_1.ipynb deleted file mode 100755 index 8951e1db..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter5_1.ipynb +++ /dev/null @@ -1,799 +0,0 @@ -{
- "metadata": {
- "name": "chapter5.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 5: Equilibrium of Coplanar Force Systems."
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-1, Page no 58"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "# From eqn's 1&2\n",
- "D=np.array([[6/sqrt(40),-4/sqrt(20)],[2/sqrt(40),2/sqrt(20)]])\n",
- "B=np.array([0,25]) #lb\n",
- "\n",
- "#Calculations\n",
- "\n",
- "X=np.linalg.solve(D,B)\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The tension in cable AB is',round(X[1],1),\"lb\"\n",
- "print'The tension in cable AC is',round(X[0],1),\"lb\"\n",
- "\n",
- "# The tensions in the cable AB & AC is off by 0.1 lb"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in cable AB is 33.5 lb\n",
- "The tension in cable AC is 31.6 lb\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-2, Page no 59"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F1=100 #lb\n",
- "R=16 #in\n",
- "\n",
- "#Calculations\n",
- "theta=arcsin(14*R**-1)*(180/pi) #degrees\n",
- "# since theta=61 degrees,\n",
- "sin61=0.8746\n",
- "cos61=0.4848\n",
- "N=F1/sin61 #lb\n",
- "P=N*cos61 #lb\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The value of normal reaction offered is',round(N,1),\"lb\"\n",
- "print'The push required is',round(P,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of normal reaction offered is 114.3 lb\n",
- "The push required is 55.4 lb\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-3,Page no 59"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "L=20 #m\n",
- "M=1200 #kg\n",
- "g=9.81 #m/s**2\n",
- "H=10 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "AB=sqrt(L**2-H**2) #Applying Pythagoras Theorem\n",
- "costheta=17.3/20\n",
- "F1=M*g*H/AB #N\n",
- "F2=M*g/costheta #N\n",
- "\n",
- "#Result\n",
- "\n",
- "print'Force F1 is',round(F1),\"N\"\n",
- "print'Force F2 is',round(F2),\"N\"\n",
- "\n",
- "#Decimal accuracy causes discrepancy in answers compared to the textbook answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force F1 is 6797.0 N\n",
- "Force F2 is 13609.0 N\n"
- ]
- }
- ],
- "prompt_number": 45
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-4, Page No 60"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "Fx=1000 #lb\n",
- "Fy=1000 #lb\n",
- "costheta=9*15**-1\n",
- "cosbeta=12*15**-1\n",
- "sintheta=4*5**-1\n",
- "sinbeta=3*5**-1\n",
- "\n",
- "#Calculations\n",
- "#Matrix solution\n",
- "A=np.array([[costheta,-cosbeta],[sintheta,sinbeta]]) \n",
- "B=np.array([-1000,1000])\n",
- "X=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The force in AB is',round(X[0]),\"lb compression\"\n",
- "print'The force in BC is',round(X[1]),\"lb compression\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force in AB is 200.0 lb compression\n",
- "The force in BC is 1400.0 lb compression\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-5, Page no 61"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "w=10 #lb/ft\n",
- "L=12 #ft\n",
- "# as theta=30 degrees,\n",
- "sin30=2**-1\n",
- "cos30=sqrt(3)*2**-1\n",
- "\n",
- "#Calculation\n",
- "#Matrix Calculations\n",
- "A=np.array([[cos30,-cos30],[sin30,sin30]]) \n",
- "B=np.array([0,120]) \n",
- "X=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The tension in the cable is,T=',round(X[0]),\"lb\"\n",
- "print'The reaction at B is,R',round(X[1]),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in the cable is,T= 120.0 lb\n",
- "The reaction at B is,R 120.0 lb\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-6,Page no 61"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W1=40 #lb\n",
- "W2=30 #lb\n",
- "# as theta1=30 degrees,\n",
- "sin30=2**-1\n",
- "\n",
- "#Calculations\n",
- "#Summing the forces parallel to 30 degree plane\n",
- "T=W1*sin30\n",
- "theta=arcsin(T/W2)*(180/pi)\n",
- "\n",
- "#Result\n",
- "print'The tension in the cable is',round(T),\"lb\"\n",
- "print'The angle is',round(theta,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in the cable is 20.0 lb\n",
- "The angle is 41.8 degrees\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-8,Page no 62"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F1=125 #N\n",
- "F2=200 #N\n",
- "F3=340 #N\n",
- "F4=180 #N\n",
- "x1=4 #m\n",
- "x2=3 #m\n",
- "x3=10 #m\n",
- "x4=15 #m\n",
- "x5=17 #m\n",
- "\n",
- "#Calculations\n",
- "Rb=(-F1*x1+F2*x2+F3*x3+F4*x4)/x5 #moment about point A\n",
- "Ra=(F1*(x1+x5)+F3*(x5-x3)+F2*(x5-x2)+F4*(x5-x4))/x5 #moment about point B\n",
- "\n",
- "#Result\n",
- "print'The reaction at A is',round(Ra),\"N\"\n",
- "print'The reaction at B is',round(Rb),\"N\"\n",
- "\n",
- "# The ans for B is off by 1 N"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction at A is 480.0 N\n",
- "The reaction at B is 364.0 N\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-9, Page no 63"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F1=1000 #lb\n",
- "F2=1200 #lb\n",
- "F3=2000 #lb\n",
- "x1=1 #ft\n",
- "x2=7 #ft\n",
- "x4=2 #ft\n",
- "x3=6 #ft\n",
- "\n",
- "#Calculation\n",
- "#Equilibrium equations\n",
- "Rn=(F3*(x1+x2+x3)+F2*(x1+x2)+F1*x1)/(x1+x3+x2+x4) #Moment about point M\n",
- "Rm=(F1*(x2+x3+x4)+F2*(x3+x4)+F3*x4)/(x1+x2+x3+x4) #Moment about point N\n",
- "\n",
- "#Result\n",
- "print'The reaction at M is',round(Rm),\"lb\"\n",
- "print'The reaction at N is',round(Rn),\"lb\"\n",
- "\n",
- "#Decimal Accuracy causes discrepancy in answers between computation and textbook\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction at M is 1787.0 lb\n",
- "The reaction at N is 2412.0 lb\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-10, Page no 64"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "P=10 #kg\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "# equilibrium at fig b\n",
- "T1=P*g/2 #N\n",
- "# equilibrium at fig c\n",
- "T2=T1/2 #N\n",
- "#equilibrium at fig d\n",
- "P=T2\n",
- "\n",
- "#Result\n",
- "print'The force P is',round(P,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force P is 24.5 N\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-11, Page no 64"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "k=20 # lb/in\n",
- "w=20 # lb/ft\n",
- "x1=4 #ft\n",
- "x2=10 # ft\n",
- "x3=8 #ft\n",
- "x4=6 #ft\n",
- "x5=9 #ft\n",
- "F1=1920 #lb.rad\n",
- "F2=3360 #lb.rad\n",
- "\n",
- "#calculations\n",
- "theta=(w*x2*x5)*(F1*x3+F2*(x3+x4))**-1 #radians\n",
- "FB=F1*theta\n",
- "FC=F2*theta\n",
- "A=(w*x2)-FB-FC\n",
- "\n",
- "#Result\n",
- "print'The force in spring B is',round(FB,1),\"lb\"\n",
- "print'The force in spring C is',round(FC,1),\"lb\"\n",
- "print'The reaction at A is',round(A,1),\"lb up\"\n",
- "\n",
- " # The answer waries slightly due to decimal point discrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force in spring B is 55.4 lb\n",
- "The force in spring C is 96.9 lb\n",
- "The reaction at A is 47.7 lb up\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-12, Page no 65"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "L=3.8 #m\n",
- "w=10 # kg/m\n",
- "P=1000 #N\n",
- "t=0.8 #m\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "Gf=L*w*g #N\n",
- "A=(P*L+Gf*L*0.5)/t #N Taking moment about point B\n",
- "B=(P*(L-t)+Gf*(0.5*L-t))/t #N Taking moment about point A\n",
- "\n",
- "#Result\n",
- "print'The reaction at point A is',round(A),\"N\"\n",
- "print'The reaction at point B is',round(B),\"N\"\n",
- "\n",
- "# The answers in the textbook are incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction at point A is 5635.0 N\n",
- "The reaction at point B is 4263.0 N\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-13, Page no 65"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wa=400 #lb\n",
- "Wb=200 #lb\n",
- "# as theta=30 degrees,\n",
- "sin30=2**-1\n",
- "\n",
- "#Calculations\n",
- "Ta=Wa*sin30 #lb\n",
- "Tb=Wb*sin30 #lb\n",
- "#Taking moment about point O\n",
- "P=(Tb*12+Ta*6)/24 #lb\n",
- "\n",
- "#Result\n",
- "print'The value of Ta is',round(Ta,3),\"lb\"\n",
- "print'The value of Tb is',round(Tb,3),\"lb\"\n",
- "print'The value of P is',round(P,3),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of Ta is 200.0 lb\n",
- "The value of Tb is 100.0 lb\n",
- "The value of P is 100.0 lb\n"
- ]
- }
- ],
- "prompt_number": 37
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-15, Page no 66"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "F=np.array([5,2,3,1.5]) #kN Forces are defined as a cloumn matrix\n",
- "theta=(pi*np.array([90,60,45,80]))/180 #degrees angles are also defined as a column matrix\n",
- "d=np.array([2,6,13,17]) #distances from point C of each force\n",
- "c=np.array([17,15,11,4]) #distance form point D of each force\n",
- "#Calculations\n",
- "\n",
- "#Summing horizontal forces\n",
- "Ch=F[1]*cos(theta[1])-F[2]*cos(theta[2])+F[3]*cos(theta[3]) #kN \"which indidcates that Ch acts to the left instead of the assumed\"\n",
- "#Taking moment about point C\n",
- "D=(F[0]*d[0]+F[1]*sin(theta[1])*d[1]+F[2]*sin(theta[2])*d[2]+F[3]*sin(theta[3])*d[3])/d[3] #kN\n",
- "#Taking moment about point D\n",
- "Cv=(F[0]*c[1]+F[1]*sin(theta[1])*c[2]+F[2]*sin(theta[2])*c[3])/c[1]\n",
- "#Result\n",
- "\n",
- "print'The values of Ch,D and Cv are:',round(Ch,2),\"kN ,\",round(D,1),\"kN\",'and',round(Cv,2),\"kN\"\n",
- "\n",
- "# The ans of Cv is incorrect in textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values of Ch,D and Cv are: -0.86 kN , 4.3 kN and 6.84 kN\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-16, Page no 67"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=100 #N/m\n",
- "F1=200 #N\n",
- "M=500 #N.m\n",
- "Lw=2 #m\n",
- "#Distance from point A\n",
- "d=np.array([1,2,3,4,5]) #m\n",
- "#Distance from point B\n",
- "b=np.array([5,4,3,2,1]) #m\n",
- "\n",
- "#Calculations\n",
- "#Taking moment aboout point A\n",
- "Rb=(w*Lw*d[0]+F1*d[2]-M)/d[3] #N\n",
- "#Taking moment about point B\n",
- "Ra=(w*Lw*b[2]+F1*b[4]+M)/b[1] #N\n",
- "\n",
- "#Result\n",
- "print'The value of reaction at A is',round(Ra),\"N\"\n",
- "print'The value of reaction at B is',round(Rb),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of reaction at A is 325.0 N\n",
- "The value of reaction at B is 75.0 N\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-18, Page no 68"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "# The values of theta are=[60,60,45] degrees, therefore its values are as,\n",
- "costheta2=sqrt(2)**-1\n",
- "sintheta2=sqrt(2)**-1\n",
- "d=np.array([4.46,3.54,2]) #feet defined as a matrix\n",
- "F=400 #lb\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point A\n",
- "Re=(F*(8-d[1]))/8 #lb\n",
- "Ra=400-Re #lb here i have used the summation of forces in the vertical direction\n",
- "#Taking moment about point B\n",
- "Dv=(-F*3.644)*5.77**-1 #lb\n",
- "#Taking moment about point D\n",
- "Bv=(F*2.126)/5.77 #lb\n",
- "#Taking summation of forces in the vertical direction\n",
- "Cv=-223-Dv #lb\n",
- "#Taking moment about point D\n",
- "Ch=((223*d[2]*costheta2)-(Cv*5.173*costheta2))*(5.173*sintheta2)**-1 #lb\n",
- "#Taking summation of forces in the horizontal direction\n",
- "Dh=-Ch #lb\n",
- "#Taking sum of forces in horizontal direction\n",
- "Bh=-Dh #lb\n",
- "\n",
- "#Result\n",
- "print'The Floor reactions are'\n",
- "print'Ra=',round(Ra),\"lb up\"\n",
- "print'Re=',round(Re),\"lb up\"\n",
- "\n",
- "print'Pin reaction at C on CE are'\n",
- "print'Ch=',round(Ch,1),\"lb to right\"\n",
- "print'Cv=',round(Cv,1),\"lb up\"\n",
- "\n",
- "print'The pin reactions at B on AC are:'\n",
- "print'Bh=',round(Bh,1),\"lb to right\"\n",
- "print'Bv=',round(Bv,1),\"lb down\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Floor reactions are\n",
- "Ra= 177.0 lb up\n",
- "Re= 223.0 lb up\n",
- "Pin reaction at C on CE are\n",
- "Ch= 56.6 lb to right\n",
- "Cv= 29.6 lb up\n",
- "The pin reactions at B on AC are:\n",
- "Bh= 56.6 lb to right\n",
- "Bv= 147.4 lb down\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-19, Page no 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=0.5 #m\n",
- "m=10 #kg\n",
- "g=9.81 #m/s**2\n",
- "# since theta=60 degrees,\n",
- "sin30=2**-1\n",
- "cos30=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Due to symmetry the reaction will be shared by the structure\n",
- "A=m*g*r #N\n",
- "B=A #N\n",
- "#Vertical forces summed\n",
- "N1=m*g/(2*sin30) #N\n",
- "#Taking moment about point C\n",
- "T=(N1*0.866+B*sin30)*(1.5*cos30)**-1\n",
- " \n",
- "#Result\n",
- "print'The value of N1 is',round(N1),\"N\"\n",
- "print'The value of T is',round(T,1),\"N\"\n",
- "\n",
- "# The ans for T is off by 0.1 N"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of N1 is 98.0 N\n",
- "The value of T is 84.3 N\n"
- ]
- }
- ],
- "prompt_number": 52
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter5_2.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter5_2.ipynb deleted file mode 100755 index 8951e1db..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter5_2.ipynb +++ /dev/null @@ -1,799 +0,0 @@ -{
- "metadata": {
- "name": "chapter5.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 5: Equilibrium of Coplanar Force Systems."
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-1, Page no 58"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "# From eqn's 1&2\n",
- "D=np.array([[6/sqrt(40),-4/sqrt(20)],[2/sqrt(40),2/sqrt(20)]])\n",
- "B=np.array([0,25]) #lb\n",
- "\n",
- "#Calculations\n",
- "\n",
- "X=np.linalg.solve(D,B)\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The tension in cable AB is',round(X[1],1),\"lb\"\n",
- "print'The tension in cable AC is',round(X[0],1),\"lb\"\n",
- "\n",
- "# The tensions in the cable AB & AC is off by 0.1 lb"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in cable AB is 33.5 lb\n",
- "The tension in cable AC is 31.6 lb\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-2, Page no 59"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F1=100 #lb\n",
- "R=16 #in\n",
- "\n",
- "#Calculations\n",
- "theta=arcsin(14*R**-1)*(180/pi) #degrees\n",
- "# since theta=61 degrees,\n",
- "sin61=0.8746\n",
- "cos61=0.4848\n",
- "N=F1/sin61 #lb\n",
- "P=N*cos61 #lb\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The value of normal reaction offered is',round(N,1),\"lb\"\n",
- "print'The push required is',round(P,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of normal reaction offered is 114.3 lb\n",
- "The push required is 55.4 lb\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-3,Page no 59"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "L=20 #m\n",
- "M=1200 #kg\n",
- "g=9.81 #m/s**2\n",
- "H=10 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "AB=sqrt(L**2-H**2) #Applying Pythagoras Theorem\n",
- "costheta=17.3/20\n",
- "F1=M*g*H/AB #N\n",
- "F2=M*g/costheta #N\n",
- "\n",
- "#Result\n",
- "\n",
- "print'Force F1 is',round(F1),\"N\"\n",
- "print'Force F2 is',round(F2),\"N\"\n",
- "\n",
- "#Decimal accuracy causes discrepancy in answers compared to the textbook answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force F1 is 6797.0 N\n",
- "Force F2 is 13609.0 N\n"
- ]
- }
- ],
- "prompt_number": 45
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-4, Page No 60"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "Fx=1000 #lb\n",
- "Fy=1000 #lb\n",
- "costheta=9*15**-1\n",
- "cosbeta=12*15**-1\n",
- "sintheta=4*5**-1\n",
- "sinbeta=3*5**-1\n",
- "\n",
- "#Calculations\n",
- "#Matrix solution\n",
- "A=np.array([[costheta,-cosbeta],[sintheta,sinbeta]]) \n",
- "B=np.array([-1000,1000])\n",
- "X=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The force in AB is',round(X[0]),\"lb compression\"\n",
- "print'The force in BC is',round(X[1]),\"lb compression\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force in AB is 200.0 lb compression\n",
- "The force in BC is 1400.0 lb compression\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-5, Page no 61"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "w=10 #lb/ft\n",
- "L=12 #ft\n",
- "# as theta=30 degrees,\n",
- "sin30=2**-1\n",
- "cos30=sqrt(3)*2**-1\n",
- "\n",
- "#Calculation\n",
- "#Matrix Calculations\n",
- "A=np.array([[cos30,-cos30],[sin30,sin30]]) \n",
- "B=np.array([0,120]) \n",
- "X=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The tension in the cable is,T=',round(X[0]),\"lb\"\n",
- "print'The reaction at B is,R',round(X[1]),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in the cable is,T= 120.0 lb\n",
- "The reaction at B is,R 120.0 lb\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-6,Page no 61"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W1=40 #lb\n",
- "W2=30 #lb\n",
- "# as theta1=30 degrees,\n",
- "sin30=2**-1\n",
- "\n",
- "#Calculations\n",
- "#Summing the forces parallel to 30 degree plane\n",
- "T=W1*sin30\n",
- "theta=arcsin(T/W2)*(180/pi)\n",
- "\n",
- "#Result\n",
- "print'The tension in the cable is',round(T),\"lb\"\n",
- "print'The angle is',round(theta,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in the cable is 20.0 lb\n",
- "The angle is 41.8 degrees\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-8,Page no 62"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F1=125 #N\n",
- "F2=200 #N\n",
- "F3=340 #N\n",
- "F4=180 #N\n",
- "x1=4 #m\n",
- "x2=3 #m\n",
- "x3=10 #m\n",
- "x4=15 #m\n",
- "x5=17 #m\n",
- "\n",
- "#Calculations\n",
- "Rb=(-F1*x1+F2*x2+F3*x3+F4*x4)/x5 #moment about point A\n",
- "Ra=(F1*(x1+x5)+F3*(x5-x3)+F2*(x5-x2)+F4*(x5-x4))/x5 #moment about point B\n",
- "\n",
- "#Result\n",
- "print'The reaction at A is',round(Ra),\"N\"\n",
- "print'The reaction at B is',round(Rb),\"N\"\n",
- "\n",
- "# The ans for B is off by 1 N"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction at A is 480.0 N\n",
- "The reaction at B is 364.0 N\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-9, Page no 63"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F1=1000 #lb\n",
- "F2=1200 #lb\n",
- "F3=2000 #lb\n",
- "x1=1 #ft\n",
- "x2=7 #ft\n",
- "x4=2 #ft\n",
- "x3=6 #ft\n",
- "\n",
- "#Calculation\n",
- "#Equilibrium equations\n",
- "Rn=(F3*(x1+x2+x3)+F2*(x1+x2)+F1*x1)/(x1+x3+x2+x4) #Moment about point M\n",
- "Rm=(F1*(x2+x3+x4)+F2*(x3+x4)+F3*x4)/(x1+x2+x3+x4) #Moment about point N\n",
- "\n",
- "#Result\n",
- "print'The reaction at M is',round(Rm),\"lb\"\n",
- "print'The reaction at N is',round(Rn),\"lb\"\n",
- "\n",
- "#Decimal Accuracy causes discrepancy in answers between computation and textbook\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction at M is 1787.0 lb\n",
- "The reaction at N is 2412.0 lb\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-10, Page no 64"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "P=10 #kg\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "# equilibrium at fig b\n",
- "T1=P*g/2 #N\n",
- "# equilibrium at fig c\n",
- "T2=T1/2 #N\n",
- "#equilibrium at fig d\n",
- "P=T2\n",
- "\n",
- "#Result\n",
- "print'The force P is',round(P,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force P is 24.5 N\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-11, Page no 64"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "k=20 # lb/in\n",
- "w=20 # lb/ft\n",
- "x1=4 #ft\n",
- "x2=10 # ft\n",
- "x3=8 #ft\n",
- "x4=6 #ft\n",
- "x5=9 #ft\n",
- "F1=1920 #lb.rad\n",
- "F2=3360 #lb.rad\n",
- "\n",
- "#calculations\n",
- "theta=(w*x2*x5)*(F1*x3+F2*(x3+x4))**-1 #radians\n",
- "FB=F1*theta\n",
- "FC=F2*theta\n",
- "A=(w*x2)-FB-FC\n",
- "\n",
- "#Result\n",
- "print'The force in spring B is',round(FB,1),\"lb\"\n",
- "print'The force in spring C is',round(FC,1),\"lb\"\n",
- "print'The reaction at A is',round(A,1),\"lb up\"\n",
- "\n",
- " # The answer waries slightly due to decimal point discrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force in spring B is 55.4 lb\n",
- "The force in spring C is 96.9 lb\n",
- "The reaction at A is 47.7 lb up\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-12, Page no 65"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "L=3.8 #m\n",
- "w=10 # kg/m\n",
- "P=1000 #N\n",
- "t=0.8 #m\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "Gf=L*w*g #N\n",
- "A=(P*L+Gf*L*0.5)/t #N Taking moment about point B\n",
- "B=(P*(L-t)+Gf*(0.5*L-t))/t #N Taking moment about point A\n",
- "\n",
- "#Result\n",
- "print'The reaction at point A is',round(A),\"N\"\n",
- "print'The reaction at point B is',round(B),\"N\"\n",
- "\n",
- "# The answers in the textbook are incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction at point A is 5635.0 N\n",
- "The reaction at point B is 4263.0 N\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-13, Page no 65"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wa=400 #lb\n",
- "Wb=200 #lb\n",
- "# as theta=30 degrees,\n",
- "sin30=2**-1\n",
- "\n",
- "#Calculations\n",
- "Ta=Wa*sin30 #lb\n",
- "Tb=Wb*sin30 #lb\n",
- "#Taking moment about point O\n",
- "P=(Tb*12+Ta*6)/24 #lb\n",
- "\n",
- "#Result\n",
- "print'The value of Ta is',round(Ta,3),\"lb\"\n",
- "print'The value of Tb is',round(Tb,3),\"lb\"\n",
- "print'The value of P is',round(P,3),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of Ta is 200.0 lb\n",
- "The value of Tb is 100.0 lb\n",
- "The value of P is 100.0 lb\n"
- ]
- }
- ],
- "prompt_number": 37
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-15, Page no 66"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "F=np.array([5,2,3,1.5]) #kN Forces are defined as a cloumn matrix\n",
- "theta=(pi*np.array([90,60,45,80]))/180 #degrees angles are also defined as a column matrix\n",
- "d=np.array([2,6,13,17]) #distances from point C of each force\n",
- "c=np.array([17,15,11,4]) #distance form point D of each force\n",
- "#Calculations\n",
- "\n",
- "#Summing horizontal forces\n",
- "Ch=F[1]*cos(theta[1])-F[2]*cos(theta[2])+F[3]*cos(theta[3]) #kN \"which indidcates that Ch acts to the left instead of the assumed\"\n",
- "#Taking moment about point C\n",
- "D=(F[0]*d[0]+F[1]*sin(theta[1])*d[1]+F[2]*sin(theta[2])*d[2]+F[3]*sin(theta[3])*d[3])/d[3] #kN\n",
- "#Taking moment about point D\n",
- "Cv=(F[0]*c[1]+F[1]*sin(theta[1])*c[2]+F[2]*sin(theta[2])*c[3])/c[1]\n",
- "#Result\n",
- "\n",
- "print'The values of Ch,D and Cv are:',round(Ch,2),\"kN ,\",round(D,1),\"kN\",'and',round(Cv,2),\"kN\"\n",
- "\n",
- "# The ans of Cv is incorrect in textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values of Ch,D and Cv are: -0.86 kN , 4.3 kN and 6.84 kN\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-16, Page no 67"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=100 #N/m\n",
- "F1=200 #N\n",
- "M=500 #N.m\n",
- "Lw=2 #m\n",
- "#Distance from point A\n",
- "d=np.array([1,2,3,4,5]) #m\n",
- "#Distance from point B\n",
- "b=np.array([5,4,3,2,1]) #m\n",
- "\n",
- "#Calculations\n",
- "#Taking moment aboout point A\n",
- "Rb=(w*Lw*d[0]+F1*d[2]-M)/d[3] #N\n",
- "#Taking moment about point B\n",
- "Ra=(w*Lw*b[2]+F1*b[4]+M)/b[1] #N\n",
- "\n",
- "#Result\n",
- "print'The value of reaction at A is',round(Ra),\"N\"\n",
- "print'The value of reaction at B is',round(Rb),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of reaction at A is 325.0 N\n",
- "The value of reaction at B is 75.0 N\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-18, Page no 68"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "# The values of theta are=[60,60,45] degrees, therefore its values are as,\n",
- "costheta2=sqrt(2)**-1\n",
- "sintheta2=sqrt(2)**-1\n",
- "d=np.array([4.46,3.54,2]) #feet defined as a matrix\n",
- "F=400 #lb\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point A\n",
- "Re=(F*(8-d[1]))/8 #lb\n",
- "Ra=400-Re #lb here i have used the summation of forces in the vertical direction\n",
- "#Taking moment about point B\n",
- "Dv=(-F*3.644)*5.77**-1 #lb\n",
- "#Taking moment about point D\n",
- "Bv=(F*2.126)/5.77 #lb\n",
- "#Taking summation of forces in the vertical direction\n",
- "Cv=-223-Dv #lb\n",
- "#Taking moment about point D\n",
- "Ch=((223*d[2]*costheta2)-(Cv*5.173*costheta2))*(5.173*sintheta2)**-1 #lb\n",
- "#Taking summation of forces in the horizontal direction\n",
- "Dh=-Ch #lb\n",
- "#Taking sum of forces in horizontal direction\n",
- "Bh=-Dh #lb\n",
- "\n",
- "#Result\n",
- "print'The Floor reactions are'\n",
- "print'Ra=',round(Ra),\"lb up\"\n",
- "print'Re=',round(Re),\"lb up\"\n",
- "\n",
- "print'Pin reaction at C on CE are'\n",
- "print'Ch=',round(Ch,1),\"lb to right\"\n",
- "print'Cv=',round(Cv,1),\"lb up\"\n",
- "\n",
- "print'The pin reactions at B on AC are:'\n",
- "print'Bh=',round(Bh,1),\"lb to right\"\n",
- "print'Bv=',round(Bv,1),\"lb down\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Floor reactions are\n",
- "Ra= 177.0 lb up\n",
- "Re= 223.0 lb up\n",
- "Pin reaction at C on CE are\n",
- "Ch= 56.6 lb to right\n",
- "Cv= 29.6 lb up\n",
- "The pin reactions at B on AC are:\n",
- "Bh= 56.6 lb to right\n",
- "Bv= 147.4 lb down\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-19, Page no 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=0.5 #m\n",
- "m=10 #kg\n",
- "g=9.81 #m/s**2\n",
- "# since theta=60 degrees,\n",
- "sin30=2**-1\n",
- "cos30=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Due to symmetry the reaction will be shared by the structure\n",
- "A=m*g*r #N\n",
- "B=A #N\n",
- "#Vertical forces summed\n",
- "N1=m*g/(2*sin30) #N\n",
- "#Taking moment about point C\n",
- "T=(N1*0.866+B*sin30)*(1.5*cos30)**-1\n",
- " \n",
- "#Result\n",
- "print'The value of N1 is',round(N1),\"N\"\n",
- "print'The value of T is',round(T,1),\"N\"\n",
- "\n",
- "# The ans for T is off by 0.1 N"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of N1 is 98.0 N\n",
- "The value of T is 84.3 N\n"
- ]
- }
- ],
- "prompt_number": 52
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter5_3.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter5_3.ipynb deleted file mode 100755 index 8951e1db..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter5_3.ipynb +++ /dev/null @@ -1,799 +0,0 @@ -{
- "metadata": {
- "name": "chapter5.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 5: Equilibrium of Coplanar Force Systems."
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-1, Page no 58"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "# From eqn's 1&2\n",
- "D=np.array([[6/sqrt(40),-4/sqrt(20)],[2/sqrt(40),2/sqrt(20)]])\n",
- "B=np.array([0,25]) #lb\n",
- "\n",
- "#Calculations\n",
- "\n",
- "X=np.linalg.solve(D,B)\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The tension in cable AB is',round(X[1],1),\"lb\"\n",
- "print'The tension in cable AC is',round(X[0],1),\"lb\"\n",
- "\n",
- "# The tensions in the cable AB & AC is off by 0.1 lb"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in cable AB is 33.5 lb\n",
- "The tension in cable AC is 31.6 lb\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-2, Page no 59"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F1=100 #lb\n",
- "R=16 #in\n",
- "\n",
- "#Calculations\n",
- "theta=arcsin(14*R**-1)*(180/pi) #degrees\n",
- "# since theta=61 degrees,\n",
- "sin61=0.8746\n",
- "cos61=0.4848\n",
- "N=F1/sin61 #lb\n",
- "P=N*cos61 #lb\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The value of normal reaction offered is',round(N,1),\"lb\"\n",
- "print'The push required is',round(P,1),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of normal reaction offered is 114.3 lb\n",
- "The push required is 55.4 lb\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-3,Page no 59"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "L=20 #m\n",
- "M=1200 #kg\n",
- "g=9.81 #m/s**2\n",
- "H=10 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "AB=sqrt(L**2-H**2) #Applying Pythagoras Theorem\n",
- "costheta=17.3/20\n",
- "F1=M*g*H/AB #N\n",
- "F2=M*g/costheta #N\n",
- "\n",
- "#Result\n",
- "\n",
- "print'Force F1 is',round(F1),\"N\"\n",
- "print'Force F2 is',round(F2),\"N\"\n",
- "\n",
- "#Decimal accuracy causes discrepancy in answers compared to the textbook answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force F1 is 6797.0 N\n",
- "Force F2 is 13609.0 N\n"
- ]
- }
- ],
- "prompt_number": 45
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-4, Page No 60"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "Fx=1000 #lb\n",
- "Fy=1000 #lb\n",
- "costheta=9*15**-1\n",
- "cosbeta=12*15**-1\n",
- "sintheta=4*5**-1\n",
- "sinbeta=3*5**-1\n",
- "\n",
- "#Calculations\n",
- "#Matrix solution\n",
- "A=np.array([[costheta,-cosbeta],[sintheta,sinbeta]]) \n",
- "B=np.array([-1000,1000])\n",
- "X=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The force in AB is',round(X[0]),\"lb compression\"\n",
- "print'The force in BC is',round(X[1]),\"lb compression\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force in AB is 200.0 lb compression\n",
- "The force in BC is 1400.0 lb compression\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-5, Page no 61"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "w=10 #lb/ft\n",
- "L=12 #ft\n",
- "# as theta=30 degrees,\n",
- "sin30=2**-1\n",
- "cos30=sqrt(3)*2**-1\n",
- "\n",
- "#Calculation\n",
- "#Matrix Calculations\n",
- "A=np.array([[cos30,-cos30],[sin30,sin30]]) \n",
- "B=np.array([0,120]) \n",
- "X=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The tension in the cable is,T=',round(X[0]),\"lb\"\n",
- "print'The reaction at B is,R',round(X[1]),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in the cable is,T= 120.0 lb\n",
- "The reaction at B is,R 120.0 lb\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-6,Page no 61"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W1=40 #lb\n",
- "W2=30 #lb\n",
- "# as theta1=30 degrees,\n",
- "sin30=2**-1\n",
- "\n",
- "#Calculations\n",
- "#Summing the forces parallel to 30 degree plane\n",
- "T=W1*sin30\n",
- "theta=arcsin(T/W2)*(180/pi)\n",
- "\n",
- "#Result\n",
- "print'The tension in the cable is',round(T),\"lb\"\n",
- "print'The angle is',round(theta,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in the cable is 20.0 lb\n",
- "The angle is 41.8 degrees\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-8,Page no 62"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F1=125 #N\n",
- "F2=200 #N\n",
- "F3=340 #N\n",
- "F4=180 #N\n",
- "x1=4 #m\n",
- "x2=3 #m\n",
- "x3=10 #m\n",
- "x4=15 #m\n",
- "x5=17 #m\n",
- "\n",
- "#Calculations\n",
- "Rb=(-F1*x1+F2*x2+F3*x3+F4*x4)/x5 #moment about point A\n",
- "Ra=(F1*(x1+x5)+F3*(x5-x3)+F2*(x5-x2)+F4*(x5-x4))/x5 #moment about point B\n",
- "\n",
- "#Result\n",
- "print'The reaction at A is',round(Ra),\"N\"\n",
- "print'The reaction at B is',round(Rb),\"N\"\n",
- "\n",
- "# The ans for B is off by 1 N"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction at A is 480.0 N\n",
- "The reaction at B is 364.0 N\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-9, Page no 63"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F1=1000 #lb\n",
- "F2=1200 #lb\n",
- "F3=2000 #lb\n",
- "x1=1 #ft\n",
- "x2=7 #ft\n",
- "x4=2 #ft\n",
- "x3=6 #ft\n",
- "\n",
- "#Calculation\n",
- "#Equilibrium equations\n",
- "Rn=(F3*(x1+x2+x3)+F2*(x1+x2)+F1*x1)/(x1+x3+x2+x4) #Moment about point M\n",
- "Rm=(F1*(x2+x3+x4)+F2*(x3+x4)+F3*x4)/(x1+x2+x3+x4) #Moment about point N\n",
- "\n",
- "#Result\n",
- "print'The reaction at M is',round(Rm),\"lb\"\n",
- "print'The reaction at N is',round(Rn),\"lb\"\n",
- "\n",
- "#Decimal Accuracy causes discrepancy in answers between computation and textbook\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction at M is 1787.0 lb\n",
- "The reaction at N is 2412.0 lb\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-10, Page no 64"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "P=10 #kg\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "# equilibrium at fig b\n",
- "T1=P*g/2 #N\n",
- "# equilibrium at fig c\n",
- "T2=T1/2 #N\n",
- "#equilibrium at fig d\n",
- "P=T2\n",
- "\n",
- "#Result\n",
- "print'The force P is',round(P,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force P is 24.5 N\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-11, Page no 64"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "k=20 # lb/in\n",
- "w=20 # lb/ft\n",
- "x1=4 #ft\n",
- "x2=10 # ft\n",
- "x3=8 #ft\n",
- "x4=6 #ft\n",
- "x5=9 #ft\n",
- "F1=1920 #lb.rad\n",
- "F2=3360 #lb.rad\n",
- "\n",
- "#calculations\n",
- "theta=(w*x2*x5)*(F1*x3+F2*(x3+x4))**-1 #radians\n",
- "FB=F1*theta\n",
- "FC=F2*theta\n",
- "A=(w*x2)-FB-FC\n",
- "\n",
- "#Result\n",
- "print'The force in spring B is',round(FB,1),\"lb\"\n",
- "print'The force in spring C is',round(FC,1),\"lb\"\n",
- "print'The reaction at A is',round(A,1),\"lb up\"\n",
- "\n",
- " # The answer waries slightly due to decimal point discrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force in spring B is 55.4 lb\n",
- "The force in spring C is 96.9 lb\n",
- "The reaction at A is 47.7 lb up\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-12, Page no 65"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "L=3.8 #m\n",
- "w=10 # kg/m\n",
- "P=1000 #N\n",
- "t=0.8 #m\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "Gf=L*w*g #N\n",
- "A=(P*L+Gf*L*0.5)/t #N Taking moment about point B\n",
- "B=(P*(L-t)+Gf*(0.5*L-t))/t #N Taking moment about point A\n",
- "\n",
- "#Result\n",
- "print'The reaction at point A is',round(A),\"N\"\n",
- "print'The reaction at point B is',round(B),\"N\"\n",
- "\n",
- "# The answers in the textbook are incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction at point A is 5635.0 N\n",
- "The reaction at point B is 4263.0 N\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-13, Page no 65"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "Wa=400 #lb\n",
- "Wb=200 #lb\n",
- "# as theta=30 degrees,\n",
- "sin30=2**-1\n",
- "\n",
- "#Calculations\n",
- "Ta=Wa*sin30 #lb\n",
- "Tb=Wb*sin30 #lb\n",
- "#Taking moment about point O\n",
- "P=(Tb*12+Ta*6)/24 #lb\n",
- "\n",
- "#Result\n",
- "print'The value of Ta is',round(Ta,3),\"lb\"\n",
- "print'The value of Tb is',round(Tb,3),\"lb\"\n",
- "print'The value of P is',round(P,3),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of Ta is 200.0 lb\n",
- "The value of Tb is 100.0 lb\n",
- "The value of P is 100.0 lb\n"
- ]
- }
- ],
- "prompt_number": 37
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-15, Page no 66"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "F=np.array([5,2,3,1.5]) #kN Forces are defined as a cloumn matrix\n",
- "theta=(pi*np.array([90,60,45,80]))/180 #degrees angles are also defined as a column matrix\n",
- "d=np.array([2,6,13,17]) #distances from point C of each force\n",
- "c=np.array([17,15,11,4]) #distance form point D of each force\n",
- "#Calculations\n",
- "\n",
- "#Summing horizontal forces\n",
- "Ch=F[1]*cos(theta[1])-F[2]*cos(theta[2])+F[3]*cos(theta[3]) #kN \"which indidcates that Ch acts to the left instead of the assumed\"\n",
- "#Taking moment about point C\n",
- "D=(F[0]*d[0]+F[1]*sin(theta[1])*d[1]+F[2]*sin(theta[2])*d[2]+F[3]*sin(theta[3])*d[3])/d[3] #kN\n",
- "#Taking moment about point D\n",
- "Cv=(F[0]*c[1]+F[1]*sin(theta[1])*c[2]+F[2]*sin(theta[2])*c[3])/c[1]\n",
- "#Result\n",
- "\n",
- "print'The values of Ch,D and Cv are:',round(Ch,2),\"kN ,\",round(D,1),\"kN\",'and',round(Cv,2),\"kN\"\n",
- "\n",
- "# The ans of Cv is incorrect in textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values of Ch,D and Cv are: -0.86 kN , 4.3 kN and 6.84 kN\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-16, Page no 67"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=100 #N/m\n",
- "F1=200 #N\n",
- "M=500 #N.m\n",
- "Lw=2 #m\n",
- "#Distance from point A\n",
- "d=np.array([1,2,3,4,5]) #m\n",
- "#Distance from point B\n",
- "b=np.array([5,4,3,2,1]) #m\n",
- "\n",
- "#Calculations\n",
- "#Taking moment aboout point A\n",
- "Rb=(w*Lw*d[0]+F1*d[2]-M)/d[3] #N\n",
- "#Taking moment about point B\n",
- "Ra=(w*Lw*b[2]+F1*b[4]+M)/b[1] #N\n",
- "\n",
- "#Result\n",
- "print'The value of reaction at A is',round(Ra),\"N\"\n",
- "print'The value of reaction at B is',round(Rb),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of reaction at A is 325.0 N\n",
- "The value of reaction at B is 75.0 N\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-18, Page no 68"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "# The values of theta are=[60,60,45] degrees, therefore its values are as,\n",
- "costheta2=sqrt(2)**-1\n",
- "sintheta2=sqrt(2)**-1\n",
- "d=np.array([4.46,3.54,2]) #feet defined as a matrix\n",
- "F=400 #lb\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point A\n",
- "Re=(F*(8-d[1]))/8 #lb\n",
- "Ra=400-Re #lb here i have used the summation of forces in the vertical direction\n",
- "#Taking moment about point B\n",
- "Dv=(-F*3.644)*5.77**-1 #lb\n",
- "#Taking moment about point D\n",
- "Bv=(F*2.126)/5.77 #lb\n",
- "#Taking summation of forces in the vertical direction\n",
- "Cv=-223-Dv #lb\n",
- "#Taking moment about point D\n",
- "Ch=((223*d[2]*costheta2)-(Cv*5.173*costheta2))*(5.173*sintheta2)**-1 #lb\n",
- "#Taking summation of forces in the horizontal direction\n",
- "Dh=-Ch #lb\n",
- "#Taking sum of forces in horizontal direction\n",
- "Bh=-Dh #lb\n",
- "\n",
- "#Result\n",
- "print'The Floor reactions are'\n",
- "print'Ra=',round(Ra),\"lb up\"\n",
- "print'Re=',round(Re),\"lb up\"\n",
- "\n",
- "print'Pin reaction at C on CE are'\n",
- "print'Ch=',round(Ch,1),\"lb to right\"\n",
- "print'Cv=',round(Cv,1),\"lb up\"\n",
- "\n",
- "print'The pin reactions at B on AC are:'\n",
- "print'Bh=',round(Bh,1),\"lb to right\"\n",
- "print'Bv=',round(Bv,1),\"lb down\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Floor reactions are\n",
- "Ra= 177.0 lb up\n",
- "Re= 223.0 lb up\n",
- "Pin reaction at C on CE are\n",
- "Ch= 56.6 lb to right\n",
- "Cv= 29.6 lb up\n",
- "The pin reactions at B on AC are:\n",
- "Bh= 56.6 lb to right\n",
- "Bv= 147.4 lb down\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5-19, Page no 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r=0.5 #m\n",
- "m=10 #kg\n",
- "g=9.81 #m/s**2\n",
- "# since theta=60 degrees,\n",
- "sin30=2**-1\n",
- "cos30=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Due to symmetry the reaction will be shared by the structure\n",
- "A=m*g*r #N\n",
- "B=A #N\n",
- "#Vertical forces summed\n",
- "N1=m*g/(2*sin30) #N\n",
- "#Taking moment about point C\n",
- "T=(N1*0.866+B*sin30)*(1.5*cos30)**-1\n",
- " \n",
- "#Result\n",
- "print'The value of N1 is',round(N1),\"N\"\n",
- "print'The value of T is',round(T,1),\"N\"\n",
- "\n",
- "# The ans for T is off by 0.1 N"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of N1 is 98.0 N\n",
- "The value of T is 84.3 N\n"
- ]
- }
- ],
- "prompt_number": 52
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter6.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter6.ipynb deleted file mode 100755 index be27eeca..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter6.ipynb +++ /dev/null @@ -1,770 +0,0 @@ -{
- "metadata": {
- "name": "chapter6.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 6: Equilibrium of Non Coplanar Force Systems "
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-1, Page no 81"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "H=30 #ft\n",
- "F=150 #lb\n",
- "# Here theta1=10, theta2=30 & theta3=60 degrees. Thus the values are,\n",
- "sin10=0.1736\n",
- "cos10=0.9848\n",
- "sin30=2**-1\n",
- "cos30=sqrt(3)*2**-1\n",
- "sin60=sqrt(3)*2**-1\n",
- "cos60=2**-1\n",
- "\n",
- "#Calculations\n",
- "#Matrix solution of simultaneous equations\n",
- "X=np.array([[cos60*sin30, -cos60*sin30],[cos60*cos30, cos60*cos30]]) \n",
- "Y=np.array([[0],[F*cos10]]) \n",
- "R=np.linalg.solve(X,Y)\n",
- "#To find P,sum the forces vertically along the y-axis\n",
- "P=F*sin10+2*R[0]*sin60 #lb Compression\n",
- "\n",
- "#Result\n",
- "print'The value of A and B is',round(R[0]),\"lb (T)\"\n",
- "print'The value of P is',round(P),\"lb (C)\"\n",
- "\n",
- "# The value of P is off by 1 lb"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of A and B is 171.0 lb (T)\n",
- "The value of P is 321.0 lb (C)\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-2, Page no 82"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "#Initilization of variables\n",
- "F=150 #lb\n",
- "# Here theta1=10, theta2=30 & theta3=60 degrees. Thus the values are,\n",
- "sin10=0.1736\n",
- "cos10=0.9848\n",
- "sin30=2**-1\n",
- "cos30=sqrt(3)*2**-1\n",
- "sin60=sqrt(3)*2**-1\n",
- "cos60=2**-1\n",
- "\n",
- "#Calculations\n",
- "A=[-cos60*cos30,-sin60,cos60*sin30] \n",
- "B=[-cos60*cos60,-sin60,cos60*sin30]\n",
- "# 150lb force is actually a vector\n",
- "F_v=[F*cos10,F*sin10,0] #lb\n",
- "#Postion vector relative to C \n",
- "r=[0,30,0]\n",
- "# Moment about point C is zero\n",
- "#solution by matrix\n",
- "X=np.array([[7.5,-7.5],[13,13]]) \n",
- "Y=np.array([[0],[4470]]) \n",
- "R=np.linalg.solve(X,Y)\n",
- "A=R[0] #lb\n",
- "B=R[1] #lb\n",
- "#Summing forces in y direction\n",
- "Cy=0.866*A+0.866*B+25.9 #lb\n",
- "\n",
- "#Result\n",
- "print'The value of A and B is',round(A),\"lb\"\n",
- "print'The value of Cy is',round(Cy),\"lb\"\n",
- "\n",
- "# The answer may wary due to decimal point descripancy in computation"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of A and B is 172.0 lb\n",
- "The value of Cy is 324.0 lb\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-3, Page no 82"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initililization of variables\n",
- "m=6.12 #kg\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "AD=sqrt(3**2+2**2+6**2) \n",
- "AC=sqrt(4**2+2**2)\n",
- "AB=5\n",
- "#Sum of forces in the y direction\n",
- "T1=(m*g*AD)/6 #N\n",
- "#sum of forces in the x and z direction\n",
- "#Matrix solution of the folllowing simultaneous equations\n",
- "X=np.array([[4*4.47**-1,-3*5**-1],[-2*4.47**-1,4*5**-1]])\n",
- "Y=np.array([[T1*(3*7**-1)],[T1*(2*7**-1)]]) \n",
- "R=np.linalg.solve(X,Y)\n",
- "T2=R[0] #N\n",
- "T3=R[1] #N\n",
- "\n",
- "#Result\n",
- "print'The value of T1 is',round(T1),\"N\"\n",
- "print'The value of T2 is',round(T2,1),\"N\"\n",
- "print'The vaue of T3 is',round(T3),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of T1 is 70.0 N\n",
- "The value of T2 is 80.5 N\n",
- "The vaue of T3 is 70.0 N\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-4, Page no 83"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Intilization of variables\n",
- "\n",
- "F=np.array([0,60,0]) #Force defined as a matrix\n",
- "t1=np.array([-3*7**-1,6*7**-1,2*7**-1]) #Tension defined as a matrix\n",
- "t2=np.array([4*4.47**-1,0,-2*4.47**-1]) #tension defined as a mtrix\n",
- "t3=np.array([-3*5**-1,0,4*5**-1]) #Tension defined as a matrix\n",
- "\n",
- "#Calculations\n",
- "#Summation of forces in the y-direction\n",
- "T1=F[1]*(t1[1]**-1) #N\n",
- "#Summation of forces in the x-direction and z direction\n",
- "M1=np.array([[t2[0],t3[0]],[t2[2],t3[2]]])\n",
- "M2=np.array([[-1*t1[0]*T1],[t1[2]*T1]]) \n",
- "R=np.linalg.solve(M1,M2)\n",
- "\n",
- "#Result\n",
- "print'The tension in the strings are: T1=',round(T1),\"N\",',T2=',round(R[0],1),\"N\",'and T3=',round(R[1]),\"N respectively\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in the strings are: T1= 70.0 N ,T2= 80.5 N and T3= 70.0 N respectively\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-5, Page no 84"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=80 #kg\n",
- "g=9.81 # m/s**2\n",
- "#Co-ordinates of points in Meters\n",
- "A=np.array([1,3,0])\n",
- "B=np.array([3,3,-4])\n",
- "C=np.array([4,3,0])\n",
- "D=np.array([2,0,-1])\n",
- "\n",
- "#Calculations\n",
- "#Tension in DC will be\n",
- "a=np.array([[C[0]-D[0]],[C[1]-D[1]],[C[2]-D[2]]])\n",
- "h=((C[0]-D[0])**2+(C[1]-D[1])**2+(C[2]-D[2])**2)**0.5\n",
- "c=a*h**-1\n",
- "#Unit vector calculations\n",
- "e=np.array([[B[0]-A[0]],[B[1]-A[1]],[B[2]-A[2]]])\n",
- "v=((B[0]-A[0])**2+(B[1]-A[1])**2+(B[2]-A[2])**2)**0.5\n",
- "e_ab=e*v**-1\n",
- "#Position vector AD\n",
- "r_ad=np.array([[D[0]-A[0]],[D[1]-A[1]],[D[2]-A[2]]])\n",
- "#Moment Calculations\n",
- "O=np.array([[1,0,0],[1,-3,-1],[0,-m*g,0]])\n",
- "P=np.array([[0,1,0],[1,-3,-1],[0,-m*g,0]])\n",
- "Q=np.array([[0,0,1],[1,-3,-1],[0,-m*g,0]])\n",
- "C1=np.array([[1,0,0],[1,-3,-1],[2,3,1]])\n",
- "C2=np.array([[0,1,0],[1,-3,-1],[2,3,1]])\n",
- "C3=np.array([[0,0,1],[1,-3,-1],[2,3,1]])\n",
- "rxF1=np.array([[det(O),det(P),det(Q)]])\n",
- "rxF2=np.array([(det(C1)*h**-1),(det(C2)*h**-1),(det(C3)*h**-1)])\n",
- "#Final Moment calculations\n",
- "rxF=rxF1+rxF2 \n",
- "#Taking dot product\n",
- "dot1=e_ab*rxF\n",
- "dot2=e_ab*rxF2\n",
- "#equating dot product to zero to obtain C\n",
- "C=-(dot1[0,0]+dot1[2,2])/dot2[2,2]\n",
- "\n",
- "#Result \n",
- "print'The tension in CD is',round(C),\"N\"\n",
- "\n",
- "# The ans is off by 1 N due to Decimal point descrepancy. "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in CD is 162.0 N\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-6, Page no 85"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=200 #lb\n",
- "Dh=4 #ft\n",
- "\n",
- "#Calculation\n",
- "theta=arctan(2*Dh**-1)*(180/pi) #degrees\n",
- "T=w/(3*cos(theta)) #lb\n",
- "\n",
- "#Result\n",
- "print'The Tension in each rope is',round(T,1),\"lb\"\n",
- "print'The angle is',round(theta,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Tension in each rope is 482.9 lb\n",
- "The angle is 26.6 degrees\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-7, Page no 85"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "F=np.array([100,0,0]) #N\n",
- "CE=5 #m\n",
- "BC=(34)**0.5 #m\n",
- "AC=(41)**0.5 #m\n",
- "\n",
- "#Calculations\n",
- "#solving as a matrix for system of linear equations\n",
- "A=np.array([[3*BC**-1,-4*AC**-1,0],[0,0,(6*4)*CE**-1],[-3*BC**-1,-3*AC**-1,-3*CE**-1]])\n",
- "B=np.array([[0],[F[0]*4],[-F[0]]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The forces F1,F2 and F3 are as',round(C[0],1),\"N\",',',round(C[1],1),\"N\",'and',round(C[2],1),\"N\"\n",
- "print'Here F3 is compression assumed and rest are Tension'"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces F1,F2 and F3 are as 55.5 N , 45.7 N and 83.3 N\n",
- "Here F3 is compression assumed and rest are Tension\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-8, Page no 85"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "F=np.array([100,0,0]) #N\n",
- "CE=5 #m\n",
- "BC=(34)**0.5 #m\n",
- "AC=(41)**0.5 #m\n",
- "\n",
- "#Calculations\n",
- "#solving as a matrix for system of linear equations\n",
- "A=np.array([[3/BC,-4/AC,0],[-4/BC,-4/AC,4*CE**-1],[-3/BC,-3/AC,-3*CE**-1]])\n",
- "B=np.array([[0],[0],[-F[0]]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The forces are: F1=',round(C[0],1),\"N\",',F2=',round(C[1],1),\"N\",'and F3=',round(C[2],1),\"N\"\n",
- "print'Here F3 is compression assumed and rest are Tension'"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces are: F1= 55.5 N ,F2= 45.7 N and F3= 83.3 N\n",
- "Here F3 is compression assumed and rest are Tension\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-9, Page no 86"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "#here forces will be defines as matrices along with their co-ordinates\n",
- "#Force in N and co-ordinates in mm\n",
- "F1=[30,200,300] \n",
- "F2=[10,400,200]\n",
- "F3=[20,200,500]\n",
- "F4=[50,400,500]\n",
- "#Calculations\n",
- "#solving as system of linear equations\n",
- "A=np.array([[1,1,1],[-600,-600,0],[0,600,600]])\n",
- "B=np.array([[F1[0]+F2[0]+F3[0]+F4[0]],[-(F3[0]*F3[2]+F1[0]*F1[2]+F4[0]*F4[2]+F2[0]*F2[2])],[-(-F3[0]*F3[1]-F1[0]*F1[1]-F4[0]*F4[1]-F2[0]*F2[1])]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The reactions are as R1=',round(C[0],1),\"N\",',R2=',round(C[1],1),\"N\",'and R3=',round(C[2],1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reactions are as R1= 53.3 N ,R2= 23.3 N and R3= 33.3 N\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-10, Page no87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=1 #kg\n",
- "g=9.81 #m/s**2\n",
- "# As t1=45 degrees and t2=30 degrees\n",
- "cost1=sqrt(2)**-1\n",
- "cost2=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Solving as system of linear equations\n",
- "A=np.array([[1,0,-cost1,0],[0,1,0,3*5**-1],[-5,g*m*cost1*cost2,0,0],[-1,0,0,4*5**-1]])\n",
- "B=np.array([[0],[g*m],[g*m*5*cost1*cost2],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The forces are: Nb=',round(C[0],1),',Nc=',round(C[1],1),',Tc=',round(C[2],1),'and Tb=',round(C[3],1),\"respectively\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces are: Nb= 3.0 ,Nc= 7.5 ,Tc= 4.3 and Tb= 3.8 respectively\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-11, Page no 87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "w=50 #lb wind load\n",
- "W=60 #lb weight of door\n",
- "\n",
- "#Calculations\n",
- "#Calculation as system of linear equations\n",
- "A=np.array([[0,0,33],[1,1,-1],[28,10,-28]])\n",
- "B=np.array([[50*18],[-50],[-50*24]])\n",
- "C=np.linalg.solve(A,B)\n",
- "P=C[2]*(cos(20*pi*180**-1))**-1\n",
- "D=np.array([[-28,-10],[1,1]])\n",
- "E=np.array([1080-(28*(P*sin((20*pi)*180**-1))),P*sin((20*pi)*180**-1)])\n",
- "F=np.linalg.solve(D,E)\n",
- "By=60\n",
- "\n",
- "#Result\n",
- "print'The forces are as follows:'\n",
- "print'Az=',round(C[0],1),\"lb\",',Bz=',round(C[1],1),\"lb\",',Pz=',round(C[2],1),\"lb\",',Ax=',round(F[0]),\"lb\",',Bx=',round(F[1]),\"lb\",'and By=',round(By),\"lb respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces are as follows:\n",
- "Az= -11.6 lb ,Bz= -11.1 lb ,Pz= 27.3 lb ,Ax= -50.0 lb ,Bx= 60.0 lb and By= 60.0 lb respectively.\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-12, Page no 89"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=1 #kg\n",
- "g=9.81 #m/s**2\n",
- "# Since t1=45 degrees and t2=30 degrees\n",
- "cost1=sqrt(2)**-1\n",
- "cost2=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Solving as system of linear equations\n",
- "A=np.array([[1,0,-cost1,0],[0,1,0,3*5**-1],[-5,g*m*cost1*cost2,0,0],[-1,0,0,4*5**-1]])\n",
- "B=np.array([[0],[g*m],[g*m*5*cost1*cost2],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The forces are: Nb=',round(C[0],2),\"N\",',Nc=',round(C[1],2),\"N\",',Tc=',round(C[2],2),\"N\",'and Tb=',round(C[3],2),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " The forces are: Nb= 3.04 N ,Nc= 7.53 N ,Tc= 4.3 N and Tb= 3.8 N\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-13, Page no 89"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=200 #lb\n",
- "cos25=0.9063\n",
- "sin25=0.4226\n",
- "\n",
- "#Calculations\n",
- "P=l*5*12**-1 #lb\n",
- "#Solving as system of linear equations\n",
- "A=np.array([[0,-36,0,0],[0,0,0,36],[0,0,1,1],[1,1,0,0]])\n",
- "B=np.array([[-P*cos25*48],[l*20+P*sin25*48],[P*sin25+200],[P*cos25]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The forces are: Az=',round(C[0],1),\"lb\",',Bz=',round(C[1]),\"lb\",',Ay=',round(C[2],1),\"lb\",'and By=',round(C[3]),\"lb\"\n",
- "\n",
- "# The answer for Az waries due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces are: Az= -25.2 lb ,Bz= 101.0 lb ,Ay= 77.2 lb and By= 158.0 lb\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-14, Page No 90"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "A=80 #lb\n",
- "B=40 #lb\n",
- "C=60 #lb\n",
- "l1=2 #in\n",
- "l2=4 #in\n",
- "l3=6 #in\n",
- "l4=9 #in\n",
- "l5=3 #in\n",
- "l6=7 #in\n",
- "\n",
- "#Calculations\n",
- "P=-(-A*l1+B*l2-C*l2)/l1\n",
- "By=-(A*l3+P*l3)/l4\n",
- "Ay=(-A*l5-P*l5)/l4\n",
- "Bz=-(-C*l1-B*l1)/l4\n",
- "Az=(C*l6+B*l6)/l4\n",
- "\n",
- "#Result\n",
- "print'The forces are:Ay=',round(Ay,1),\"lb\",',By=',round(By,1),\"lb\",',Az=',round(Az,1),\"lb forward\",'and Bz=',round(Bz,1),\"lb forward\"\n",
- "\n",
- "# The answers may wary due to rounding of the values"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces are:Ay= -67.0 lb ,By= -134.0 lb ,Az= 77.0 lb forward and Bz= 22.0 lb forward\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-15, Page no 91"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=138 #lb\n",
- "w=80 #lb\n",
- "\n",
- "#Calculations\n",
- "u=(3*3+4*4+6*6)**0.5\n",
- "a=np.array([-3*u**-1,4*u**-1,-6*u**-1])\n",
- "v=(3*3+3*3+3*3)**0.5\n",
- "c=np.array([3*v**-1,3*v**-1,-3*v**-1])\n",
- "P=np.array([[1,0,0],[0,0,8],[0,-W,0]])\n",
- "Q=np.array([[0,0,1],[0,0,8],[0,-W,0]])\n",
- "R=np.array([[1,0,0],[0,0,4],[0,-w,0]])\n",
- "S=np.array([[0,0,1],[0,0,4],[0,-w,0]])\n",
- "T=np.array([[1,0,0],[0,0,6],[a[0],a[1],a[2]]])\n",
- "U=np.array([[0,1,0],[0,0,6],[a[0],a[1],a[2]]])\n",
- "V=np.array([[1,0,0],[0,0,3],[c[0],c[1],c[2]]])\n",
- "Y=np.array([[0,1,0],[0,0,3],[c[0],c[1],c[2]]])\n",
- "#Solving for A and C\n",
- "MAT1=np.array([[det(T),det(V)],[det(U),det(Y)]])\n",
- "MAT2=np.array([[det(P)+det(R)],[0]])\n",
- "res=np.linalg.solve(-MAT1,MAT2)\n",
- "A=np.array([a[0]*res[0],a[1]*res[0],a[2]*res[0]])\n",
- "C=np.array([c[0]*res[1],c[1]*res[1],c[2]*res[1]])\n",
- "E=np.array([-(A[0]+C[0]),-(-w-W+A[1]+C[1]),-(A[2]+C[2])])\n",
- "\n",
- "#Result\n",
- "print'The force vectors are as follows:'\n",
- "print'A=',round(A[0]),\"i +\",round(A[1]),\"j \",round(A[2]),\"k\",\n",
- "print'and C=',round(C[0]),\"i +\",round(C[1]),\"j \",round(C[2]),\"k\"\n",
- "print'also,Ex=',round(E[0]),\"lb\",',Ey=',round(E[1]),\"lb\",'and Ez=',round(E[2]),\"lb\"\n",
- "#Decimal accuracy causes discrepancy in the answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force vectors are as follows:\n",
- "A= -102.0 i + 136.0 j -203.0 k and C= 203.0 i + 203.0 j -203.0 k\n",
- "also,Ex= -102.0 lb ,Ey= -121.0 lb and Ez= 407.0 lb\n"
- ]
- }
- ],
- "prompt_number": 9
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter6_1.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter6_1.ipynb deleted file mode 100755 index be27eeca..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter6_1.ipynb +++ /dev/null @@ -1,770 +0,0 @@ -{
- "metadata": {
- "name": "chapter6.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 6: Equilibrium of Non Coplanar Force Systems "
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-1, Page no 81"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "H=30 #ft\n",
- "F=150 #lb\n",
- "# Here theta1=10, theta2=30 & theta3=60 degrees. Thus the values are,\n",
- "sin10=0.1736\n",
- "cos10=0.9848\n",
- "sin30=2**-1\n",
- "cos30=sqrt(3)*2**-1\n",
- "sin60=sqrt(3)*2**-1\n",
- "cos60=2**-1\n",
- "\n",
- "#Calculations\n",
- "#Matrix solution of simultaneous equations\n",
- "X=np.array([[cos60*sin30, -cos60*sin30],[cos60*cos30, cos60*cos30]]) \n",
- "Y=np.array([[0],[F*cos10]]) \n",
- "R=np.linalg.solve(X,Y)\n",
- "#To find P,sum the forces vertically along the y-axis\n",
- "P=F*sin10+2*R[0]*sin60 #lb Compression\n",
- "\n",
- "#Result\n",
- "print'The value of A and B is',round(R[0]),\"lb (T)\"\n",
- "print'The value of P is',round(P),\"lb (C)\"\n",
- "\n",
- "# The value of P is off by 1 lb"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of A and B is 171.0 lb (T)\n",
- "The value of P is 321.0 lb (C)\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-2, Page no 82"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "#Initilization of variables\n",
- "F=150 #lb\n",
- "# Here theta1=10, theta2=30 & theta3=60 degrees. Thus the values are,\n",
- "sin10=0.1736\n",
- "cos10=0.9848\n",
- "sin30=2**-1\n",
- "cos30=sqrt(3)*2**-1\n",
- "sin60=sqrt(3)*2**-1\n",
- "cos60=2**-1\n",
- "\n",
- "#Calculations\n",
- "A=[-cos60*cos30,-sin60,cos60*sin30] \n",
- "B=[-cos60*cos60,-sin60,cos60*sin30]\n",
- "# 150lb force is actually a vector\n",
- "F_v=[F*cos10,F*sin10,0] #lb\n",
- "#Postion vector relative to C \n",
- "r=[0,30,0]\n",
- "# Moment about point C is zero\n",
- "#solution by matrix\n",
- "X=np.array([[7.5,-7.5],[13,13]]) \n",
- "Y=np.array([[0],[4470]]) \n",
- "R=np.linalg.solve(X,Y)\n",
- "A=R[0] #lb\n",
- "B=R[1] #lb\n",
- "#Summing forces in y direction\n",
- "Cy=0.866*A+0.866*B+25.9 #lb\n",
- "\n",
- "#Result\n",
- "print'The value of A and B is',round(A),\"lb\"\n",
- "print'The value of Cy is',round(Cy),\"lb\"\n",
- "\n",
- "# The answer may wary due to decimal point descripancy in computation"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of A and B is 172.0 lb\n",
- "The value of Cy is 324.0 lb\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-3, Page no 82"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initililization of variables\n",
- "m=6.12 #kg\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "AD=sqrt(3**2+2**2+6**2) \n",
- "AC=sqrt(4**2+2**2)\n",
- "AB=5\n",
- "#Sum of forces in the y direction\n",
- "T1=(m*g*AD)/6 #N\n",
- "#sum of forces in the x and z direction\n",
- "#Matrix solution of the folllowing simultaneous equations\n",
- "X=np.array([[4*4.47**-1,-3*5**-1],[-2*4.47**-1,4*5**-1]])\n",
- "Y=np.array([[T1*(3*7**-1)],[T1*(2*7**-1)]]) \n",
- "R=np.linalg.solve(X,Y)\n",
- "T2=R[0] #N\n",
- "T3=R[1] #N\n",
- "\n",
- "#Result\n",
- "print'The value of T1 is',round(T1),\"N\"\n",
- "print'The value of T2 is',round(T2,1),\"N\"\n",
- "print'The vaue of T3 is',round(T3),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of T1 is 70.0 N\n",
- "The value of T2 is 80.5 N\n",
- "The vaue of T3 is 70.0 N\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-4, Page no 83"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Intilization of variables\n",
- "\n",
- "F=np.array([0,60,0]) #Force defined as a matrix\n",
- "t1=np.array([-3*7**-1,6*7**-1,2*7**-1]) #Tension defined as a matrix\n",
- "t2=np.array([4*4.47**-1,0,-2*4.47**-1]) #tension defined as a mtrix\n",
- "t3=np.array([-3*5**-1,0,4*5**-1]) #Tension defined as a matrix\n",
- "\n",
- "#Calculations\n",
- "#Summation of forces in the y-direction\n",
- "T1=F[1]*(t1[1]**-1) #N\n",
- "#Summation of forces in the x-direction and z direction\n",
- "M1=np.array([[t2[0],t3[0]],[t2[2],t3[2]]])\n",
- "M2=np.array([[-1*t1[0]*T1],[t1[2]*T1]]) \n",
- "R=np.linalg.solve(M1,M2)\n",
- "\n",
- "#Result\n",
- "print'The tension in the strings are: T1=',round(T1),\"N\",',T2=',round(R[0],1),\"N\",'and T3=',round(R[1]),\"N respectively\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in the strings are: T1= 70.0 N ,T2= 80.5 N and T3= 70.0 N respectively\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-5, Page no 84"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=80 #kg\n",
- "g=9.81 # m/s**2\n",
- "#Co-ordinates of points in Meters\n",
- "A=np.array([1,3,0])\n",
- "B=np.array([3,3,-4])\n",
- "C=np.array([4,3,0])\n",
- "D=np.array([2,0,-1])\n",
- "\n",
- "#Calculations\n",
- "#Tension in DC will be\n",
- "a=np.array([[C[0]-D[0]],[C[1]-D[1]],[C[2]-D[2]]])\n",
- "h=((C[0]-D[0])**2+(C[1]-D[1])**2+(C[2]-D[2])**2)**0.5\n",
- "c=a*h**-1\n",
- "#Unit vector calculations\n",
- "e=np.array([[B[0]-A[0]],[B[1]-A[1]],[B[2]-A[2]]])\n",
- "v=((B[0]-A[0])**2+(B[1]-A[1])**2+(B[2]-A[2])**2)**0.5\n",
- "e_ab=e*v**-1\n",
- "#Position vector AD\n",
- "r_ad=np.array([[D[0]-A[0]],[D[1]-A[1]],[D[2]-A[2]]])\n",
- "#Moment Calculations\n",
- "O=np.array([[1,0,0],[1,-3,-1],[0,-m*g,0]])\n",
- "P=np.array([[0,1,0],[1,-3,-1],[0,-m*g,0]])\n",
- "Q=np.array([[0,0,1],[1,-3,-1],[0,-m*g,0]])\n",
- "C1=np.array([[1,0,0],[1,-3,-1],[2,3,1]])\n",
- "C2=np.array([[0,1,0],[1,-3,-1],[2,3,1]])\n",
- "C3=np.array([[0,0,1],[1,-3,-1],[2,3,1]])\n",
- "rxF1=np.array([[det(O),det(P),det(Q)]])\n",
- "rxF2=np.array([(det(C1)*h**-1),(det(C2)*h**-1),(det(C3)*h**-1)])\n",
- "#Final Moment calculations\n",
- "rxF=rxF1+rxF2 \n",
- "#Taking dot product\n",
- "dot1=e_ab*rxF\n",
- "dot2=e_ab*rxF2\n",
- "#equating dot product to zero to obtain C\n",
- "C=-(dot1[0,0]+dot1[2,2])/dot2[2,2]\n",
- "\n",
- "#Result \n",
- "print'The tension in CD is',round(C),\"N\"\n",
- "\n",
- "# The ans is off by 1 N due to Decimal point descrepancy. "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in CD is 162.0 N\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-6, Page no 85"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=200 #lb\n",
- "Dh=4 #ft\n",
- "\n",
- "#Calculation\n",
- "theta=arctan(2*Dh**-1)*(180/pi) #degrees\n",
- "T=w/(3*cos(theta)) #lb\n",
- "\n",
- "#Result\n",
- "print'The Tension in each rope is',round(T,1),\"lb\"\n",
- "print'The angle is',round(theta,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Tension in each rope is 482.9 lb\n",
- "The angle is 26.6 degrees\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-7, Page no 85"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "F=np.array([100,0,0]) #N\n",
- "CE=5 #m\n",
- "BC=(34)**0.5 #m\n",
- "AC=(41)**0.5 #m\n",
- "\n",
- "#Calculations\n",
- "#solving as a matrix for system of linear equations\n",
- "A=np.array([[3*BC**-1,-4*AC**-1,0],[0,0,(6*4)*CE**-1],[-3*BC**-1,-3*AC**-1,-3*CE**-1]])\n",
- "B=np.array([[0],[F[0]*4],[-F[0]]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The forces F1,F2 and F3 are as',round(C[0],1),\"N\",',',round(C[1],1),\"N\",'and',round(C[2],1),\"N\"\n",
- "print'Here F3 is compression assumed and rest are Tension'"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces F1,F2 and F3 are as 55.5 N , 45.7 N and 83.3 N\n",
- "Here F3 is compression assumed and rest are Tension\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-8, Page no 85"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "F=np.array([100,0,0]) #N\n",
- "CE=5 #m\n",
- "BC=(34)**0.5 #m\n",
- "AC=(41)**0.5 #m\n",
- "\n",
- "#Calculations\n",
- "#solving as a matrix for system of linear equations\n",
- "A=np.array([[3/BC,-4/AC,0],[-4/BC,-4/AC,4*CE**-1],[-3/BC,-3/AC,-3*CE**-1]])\n",
- "B=np.array([[0],[0],[-F[0]]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The forces are: F1=',round(C[0],1),\"N\",',F2=',round(C[1],1),\"N\",'and F3=',round(C[2],1),\"N\"\n",
- "print'Here F3 is compression assumed and rest are Tension'"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces are: F1= 55.5 N ,F2= 45.7 N and F3= 83.3 N\n",
- "Here F3 is compression assumed and rest are Tension\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-9, Page no 86"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "#here forces will be defines as matrices along with their co-ordinates\n",
- "#Force in N and co-ordinates in mm\n",
- "F1=[30,200,300] \n",
- "F2=[10,400,200]\n",
- "F3=[20,200,500]\n",
- "F4=[50,400,500]\n",
- "#Calculations\n",
- "#solving as system of linear equations\n",
- "A=np.array([[1,1,1],[-600,-600,0],[0,600,600]])\n",
- "B=np.array([[F1[0]+F2[0]+F3[0]+F4[0]],[-(F3[0]*F3[2]+F1[0]*F1[2]+F4[0]*F4[2]+F2[0]*F2[2])],[-(-F3[0]*F3[1]-F1[0]*F1[1]-F4[0]*F4[1]-F2[0]*F2[1])]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The reactions are as R1=',round(C[0],1),\"N\",',R2=',round(C[1],1),\"N\",'and R3=',round(C[2],1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reactions are as R1= 53.3 N ,R2= 23.3 N and R3= 33.3 N\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-10, Page no87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=1 #kg\n",
- "g=9.81 #m/s**2\n",
- "# As t1=45 degrees and t2=30 degrees\n",
- "cost1=sqrt(2)**-1\n",
- "cost2=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Solving as system of linear equations\n",
- "A=np.array([[1,0,-cost1,0],[0,1,0,3*5**-1],[-5,g*m*cost1*cost2,0,0],[-1,0,0,4*5**-1]])\n",
- "B=np.array([[0],[g*m],[g*m*5*cost1*cost2],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The forces are: Nb=',round(C[0],1),',Nc=',round(C[1],1),',Tc=',round(C[2],1),'and Tb=',round(C[3],1),\"respectively\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces are: Nb= 3.0 ,Nc= 7.5 ,Tc= 4.3 and Tb= 3.8 respectively\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-11, Page no 87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "w=50 #lb wind load\n",
- "W=60 #lb weight of door\n",
- "\n",
- "#Calculations\n",
- "#Calculation as system of linear equations\n",
- "A=np.array([[0,0,33],[1,1,-1],[28,10,-28]])\n",
- "B=np.array([[50*18],[-50],[-50*24]])\n",
- "C=np.linalg.solve(A,B)\n",
- "P=C[2]*(cos(20*pi*180**-1))**-1\n",
- "D=np.array([[-28,-10],[1,1]])\n",
- "E=np.array([1080-(28*(P*sin((20*pi)*180**-1))),P*sin((20*pi)*180**-1)])\n",
- "F=np.linalg.solve(D,E)\n",
- "By=60\n",
- "\n",
- "#Result\n",
- "print'The forces are as follows:'\n",
- "print'Az=',round(C[0],1),\"lb\",',Bz=',round(C[1],1),\"lb\",',Pz=',round(C[2],1),\"lb\",',Ax=',round(F[0]),\"lb\",',Bx=',round(F[1]),\"lb\",'and By=',round(By),\"lb respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces are as follows:\n",
- "Az= -11.6 lb ,Bz= -11.1 lb ,Pz= 27.3 lb ,Ax= -50.0 lb ,Bx= 60.0 lb and By= 60.0 lb respectively.\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-12, Page no 89"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=1 #kg\n",
- "g=9.81 #m/s**2\n",
- "# Since t1=45 degrees and t2=30 degrees\n",
- "cost1=sqrt(2)**-1\n",
- "cost2=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Solving as system of linear equations\n",
- "A=np.array([[1,0,-cost1,0],[0,1,0,3*5**-1],[-5,g*m*cost1*cost2,0,0],[-1,0,0,4*5**-1]])\n",
- "B=np.array([[0],[g*m],[g*m*5*cost1*cost2],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The forces are: Nb=',round(C[0],2),\"N\",',Nc=',round(C[1],2),\"N\",',Tc=',round(C[2],2),\"N\",'and Tb=',round(C[3],2),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " The forces are: Nb= 3.04 N ,Nc= 7.53 N ,Tc= 4.3 N and Tb= 3.8 N\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-13, Page no 89"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=200 #lb\n",
- "cos25=0.9063\n",
- "sin25=0.4226\n",
- "\n",
- "#Calculations\n",
- "P=l*5*12**-1 #lb\n",
- "#Solving as system of linear equations\n",
- "A=np.array([[0,-36,0,0],[0,0,0,36],[0,0,1,1],[1,1,0,0]])\n",
- "B=np.array([[-P*cos25*48],[l*20+P*sin25*48],[P*sin25+200],[P*cos25]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The forces are: Az=',round(C[0],1),\"lb\",',Bz=',round(C[1]),\"lb\",',Ay=',round(C[2],1),\"lb\",'and By=',round(C[3]),\"lb\"\n",
- "\n",
- "# The answer for Az waries due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces are: Az= -25.2 lb ,Bz= 101.0 lb ,Ay= 77.2 lb and By= 158.0 lb\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-14, Page No 90"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "A=80 #lb\n",
- "B=40 #lb\n",
- "C=60 #lb\n",
- "l1=2 #in\n",
- "l2=4 #in\n",
- "l3=6 #in\n",
- "l4=9 #in\n",
- "l5=3 #in\n",
- "l6=7 #in\n",
- "\n",
- "#Calculations\n",
- "P=-(-A*l1+B*l2-C*l2)/l1\n",
- "By=-(A*l3+P*l3)/l4\n",
- "Ay=(-A*l5-P*l5)/l4\n",
- "Bz=-(-C*l1-B*l1)/l4\n",
- "Az=(C*l6+B*l6)/l4\n",
- "\n",
- "#Result\n",
- "print'The forces are:Ay=',round(Ay,1),\"lb\",',By=',round(By,1),\"lb\",',Az=',round(Az,1),\"lb forward\",'and Bz=',round(Bz,1),\"lb forward\"\n",
- "\n",
- "# The answers may wary due to rounding of the values"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces are:Ay= -67.0 lb ,By= -134.0 lb ,Az= 77.0 lb forward and Bz= 22.0 lb forward\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-15, Page no 91"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=138 #lb\n",
- "w=80 #lb\n",
- "\n",
- "#Calculations\n",
- "u=(3*3+4*4+6*6)**0.5\n",
- "a=np.array([-3*u**-1,4*u**-1,-6*u**-1])\n",
- "v=(3*3+3*3+3*3)**0.5\n",
- "c=np.array([3*v**-1,3*v**-1,-3*v**-1])\n",
- "P=np.array([[1,0,0],[0,0,8],[0,-W,0]])\n",
- "Q=np.array([[0,0,1],[0,0,8],[0,-W,0]])\n",
- "R=np.array([[1,0,0],[0,0,4],[0,-w,0]])\n",
- "S=np.array([[0,0,1],[0,0,4],[0,-w,0]])\n",
- "T=np.array([[1,0,0],[0,0,6],[a[0],a[1],a[2]]])\n",
- "U=np.array([[0,1,0],[0,0,6],[a[0],a[1],a[2]]])\n",
- "V=np.array([[1,0,0],[0,0,3],[c[0],c[1],c[2]]])\n",
- "Y=np.array([[0,1,0],[0,0,3],[c[0],c[1],c[2]]])\n",
- "#Solving for A and C\n",
- "MAT1=np.array([[det(T),det(V)],[det(U),det(Y)]])\n",
- "MAT2=np.array([[det(P)+det(R)],[0]])\n",
- "res=np.linalg.solve(-MAT1,MAT2)\n",
- "A=np.array([a[0]*res[0],a[1]*res[0],a[2]*res[0]])\n",
- "C=np.array([c[0]*res[1],c[1]*res[1],c[2]*res[1]])\n",
- "E=np.array([-(A[0]+C[0]),-(-w-W+A[1]+C[1]),-(A[2]+C[2])])\n",
- "\n",
- "#Result\n",
- "print'The force vectors are as follows:'\n",
- "print'A=',round(A[0]),\"i +\",round(A[1]),\"j \",round(A[2]),\"k\",\n",
- "print'and C=',round(C[0]),\"i +\",round(C[1]),\"j \",round(C[2]),\"k\"\n",
- "print'also,Ex=',round(E[0]),\"lb\",',Ey=',round(E[1]),\"lb\",'and Ez=',round(E[2]),\"lb\"\n",
- "#Decimal accuracy causes discrepancy in the answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force vectors are as follows:\n",
- "A= -102.0 i + 136.0 j -203.0 k and C= 203.0 i + 203.0 j -203.0 k\n",
- "also,Ex= -102.0 lb ,Ey= -121.0 lb and Ez= 407.0 lb\n"
- ]
- }
- ],
- "prompt_number": 9
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter6_2.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter6_2.ipynb deleted file mode 100755 index be27eeca..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter6_2.ipynb +++ /dev/null @@ -1,770 +0,0 @@ -{
- "metadata": {
- "name": "chapter6.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 6: Equilibrium of Non Coplanar Force Systems "
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-1, Page no 81"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "H=30 #ft\n",
- "F=150 #lb\n",
- "# Here theta1=10, theta2=30 & theta3=60 degrees. Thus the values are,\n",
- "sin10=0.1736\n",
- "cos10=0.9848\n",
- "sin30=2**-1\n",
- "cos30=sqrt(3)*2**-1\n",
- "sin60=sqrt(3)*2**-1\n",
- "cos60=2**-1\n",
- "\n",
- "#Calculations\n",
- "#Matrix solution of simultaneous equations\n",
- "X=np.array([[cos60*sin30, -cos60*sin30],[cos60*cos30, cos60*cos30]]) \n",
- "Y=np.array([[0],[F*cos10]]) \n",
- "R=np.linalg.solve(X,Y)\n",
- "#To find P,sum the forces vertically along the y-axis\n",
- "P=F*sin10+2*R[0]*sin60 #lb Compression\n",
- "\n",
- "#Result\n",
- "print'The value of A and B is',round(R[0]),\"lb (T)\"\n",
- "print'The value of P is',round(P),\"lb (C)\"\n",
- "\n",
- "# The value of P is off by 1 lb"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of A and B is 171.0 lb (T)\n",
- "The value of P is 321.0 lb (C)\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-2, Page no 82"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "#Initilization of variables\n",
- "F=150 #lb\n",
- "# Here theta1=10, theta2=30 & theta3=60 degrees. Thus the values are,\n",
- "sin10=0.1736\n",
- "cos10=0.9848\n",
- "sin30=2**-1\n",
- "cos30=sqrt(3)*2**-1\n",
- "sin60=sqrt(3)*2**-1\n",
- "cos60=2**-1\n",
- "\n",
- "#Calculations\n",
- "A=[-cos60*cos30,-sin60,cos60*sin30] \n",
- "B=[-cos60*cos60,-sin60,cos60*sin30]\n",
- "# 150lb force is actually a vector\n",
- "F_v=[F*cos10,F*sin10,0] #lb\n",
- "#Postion vector relative to C \n",
- "r=[0,30,0]\n",
- "# Moment about point C is zero\n",
- "#solution by matrix\n",
- "X=np.array([[7.5,-7.5],[13,13]]) \n",
- "Y=np.array([[0],[4470]]) \n",
- "R=np.linalg.solve(X,Y)\n",
- "A=R[0] #lb\n",
- "B=R[1] #lb\n",
- "#Summing forces in y direction\n",
- "Cy=0.866*A+0.866*B+25.9 #lb\n",
- "\n",
- "#Result\n",
- "print'The value of A and B is',round(A),\"lb\"\n",
- "print'The value of Cy is',round(Cy),\"lb\"\n",
- "\n",
- "# The answer may wary due to decimal point descripancy in computation"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of A and B is 172.0 lb\n",
- "The value of Cy is 324.0 lb\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-3, Page no 82"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initililization of variables\n",
- "m=6.12 #kg\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "AD=sqrt(3**2+2**2+6**2) \n",
- "AC=sqrt(4**2+2**2)\n",
- "AB=5\n",
- "#Sum of forces in the y direction\n",
- "T1=(m*g*AD)/6 #N\n",
- "#sum of forces in the x and z direction\n",
- "#Matrix solution of the folllowing simultaneous equations\n",
- "X=np.array([[4*4.47**-1,-3*5**-1],[-2*4.47**-1,4*5**-1]])\n",
- "Y=np.array([[T1*(3*7**-1)],[T1*(2*7**-1)]]) \n",
- "R=np.linalg.solve(X,Y)\n",
- "T2=R[0] #N\n",
- "T3=R[1] #N\n",
- "\n",
- "#Result\n",
- "print'The value of T1 is',round(T1),\"N\"\n",
- "print'The value of T2 is',round(T2,1),\"N\"\n",
- "print'The vaue of T3 is',round(T3),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of T1 is 70.0 N\n",
- "The value of T2 is 80.5 N\n",
- "The vaue of T3 is 70.0 N\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-4, Page no 83"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Intilization of variables\n",
- "\n",
- "F=np.array([0,60,0]) #Force defined as a matrix\n",
- "t1=np.array([-3*7**-1,6*7**-1,2*7**-1]) #Tension defined as a matrix\n",
- "t2=np.array([4*4.47**-1,0,-2*4.47**-1]) #tension defined as a mtrix\n",
- "t3=np.array([-3*5**-1,0,4*5**-1]) #Tension defined as a matrix\n",
- "\n",
- "#Calculations\n",
- "#Summation of forces in the y-direction\n",
- "T1=F[1]*(t1[1]**-1) #N\n",
- "#Summation of forces in the x-direction and z direction\n",
- "M1=np.array([[t2[0],t3[0]],[t2[2],t3[2]]])\n",
- "M2=np.array([[-1*t1[0]*T1],[t1[2]*T1]]) \n",
- "R=np.linalg.solve(M1,M2)\n",
- "\n",
- "#Result\n",
- "print'The tension in the strings are: T1=',round(T1),\"N\",',T2=',round(R[0],1),\"N\",'and T3=',round(R[1]),\"N respectively\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in the strings are: T1= 70.0 N ,T2= 80.5 N and T3= 70.0 N respectively\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-5, Page no 84"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=80 #kg\n",
- "g=9.81 # m/s**2\n",
- "#Co-ordinates of points in Meters\n",
- "A=np.array([1,3,0])\n",
- "B=np.array([3,3,-4])\n",
- "C=np.array([4,3,0])\n",
- "D=np.array([2,0,-1])\n",
- "\n",
- "#Calculations\n",
- "#Tension in DC will be\n",
- "a=np.array([[C[0]-D[0]],[C[1]-D[1]],[C[2]-D[2]]])\n",
- "h=((C[0]-D[0])**2+(C[1]-D[1])**2+(C[2]-D[2])**2)**0.5\n",
- "c=a*h**-1\n",
- "#Unit vector calculations\n",
- "e=np.array([[B[0]-A[0]],[B[1]-A[1]],[B[2]-A[2]]])\n",
- "v=((B[0]-A[0])**2+(B[1]-A[1])**2+(B[2]-A[2])**2)**0.5\n",
- "e_ab=e*v**-1\n",
- "#Position vector AD\n",
- "r_ad=np.array([[D[0]-A[0]],[D[1]-A[1]],[D[2]-A[2]]])\n",
- "#Moment Calculations\n",
- "O=np.array([[1,0,0],[1,-3,-1],[0,-m*g,0]])\n",
- "P=np.array([[0,1,0],[1,-3,-1],[0,-m*g,0]])\n",
- "Q=np.array([[0,0,1],[1,-3,-1],[0,-m*g,0]])\n",
- "C1=np.array([[1,0,0],[1,-3,-1],[2,3,1]])\n",
- "C2=np.array([[0,1,0],[1,-3,-1],[2,3,1]])\n",
- "C3=np.array([[0,0,1],[1,-3,-1],[2,3,1]])\n",
- "rxF1=np.array([[det(O),det(P),det(Q)]])\n",
- "rxF2=np.array([(det(C1)*h**-1),(det(C2)*h**-1),(det(C3)*h**-1)])\n",
- "#Final Moment calculations\n",
- "rxF=rxF1+rxF2 \n",
- "#Taking dot product\n",
- "dot1=e_ab*rxF\n",
- "dot2=e_ab*rxF2\n",
- "#equating dot product to zero to obtain C\n",
- "C=-(dot1[0,0]+dot1[2,2])/dot2[2,2]\n",
- "\n",
- "#Result \n",
- "print'The tension in CD is',round(C),\"N\"\n",
- "\n",
- "# The ans is off by 1 N due to Decimal point descrepancy. "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in CD is 162.0 N\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-6, Page no 85"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=200 #lb\n",
- "Dh=4 #ft\n",
- "\n",
- "#Calculation\n",
- "theta=arctan(2*Dh**-1)*(180/pi) #degrees\n",
- "T=w/(3*cos(theta)) #lb\n",
- "\n",
- "#Result\n",
- "print'The Tension in each rope is',round(T,1),\"lb\"\n",
- "print'The angle is',round(theta,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Tension in each rope is 482.9 lb\n",
- "The angle is 26.6 degrees\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-7, Page no 85"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "F=np.array([100,0,0]) #N\n",
- "CE=5 #m\n",
- "BC=(34)**0.5 #m\n",
- "AC=(41)**0.5 #m\n",
- "\n",
- "#Calculations\n",
- "#solving as a matrix for system of linear equations\n",
- "A=np.array([[3*BC**-1,-4*AC**-1,0],[0,0,(6*4)*CE**-1],[-3*BC**-1,-3*AC**-1,-3*CE**-1]])\n",
- "B=np.array([[0],[F[0]*4],[-F[0]]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The forces F1,F2 and F3 are as',round(C[0],1),\"N\",',',round(C[1],1),\"N\",'and',round(C[2],1),\"N\"\n",
- "print'Here F3 is compression assumed and rest are Tension'"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces F1,F2 and F3 are as 55.5 N , 45.7 N and 83.3 N\n",
- "Here F3 is compression assumed and rest are Tension\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-8, Page no 85"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "F=np.array([100,0,0]) #N\n",
- "CE=5 #m\n",
- "BC=(34)**0.5 #m\n",
- "AC=(41)**0.5 #m\n",
- "\n",
- "#Calculations\n",
- "#solving as a matrix for system of linear equations\n",
- "A=np.array([[3/BC,-4/AC,0],[-4/BC,-4/AC,4*CE**-1],[-3/BC,-3/AC,-3*CE**-1]])\n",
- "B=np.array([[0],[0],[-F[0]]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The forces are: F1=',round(C[0],1),\"N\",',F2=',round(C[1],1),\"N\",'and F3=',round(C[2],1),\"N\"\n",
- "print'Here F3 is compression assumed and rest are Tension'"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces are: F1= 55.5 N ,F2= 45.7 N and F3= 83.3 N\n",
- "Here F3 is compression assumed and rest are Tension\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-9, Page no 86"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "#here forces will be defines as matrices along with their co-ordinates\n",
- "#Force in N and co-ordinates in mm\n",
- "F1=[30,200,300] \n",
- "F2=[10,400,200]\n",
- "F3=[20,200,500]\n",
- "F4=[50,400,500]\n",
- "#Calculations\n",
- "#solving as system of linear equations\n",
- "A=np.array([[1,1,1],[-600,-600,0],[0,600,600]])\n",
- "B=np.array([[F1[0]+F2[0]+F3[0]+F4[0]],[-(F3[0]*F3[2]+F1[0]*F1[2]+F4[0]*F4[2]+F2[0]*F2[2])],[-(-F3[0]*F3[1]-F1[0]*F1[1]-F4[0]*F4[1]-F2[0]*F2[1])]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The reactions are as R1=',round(C[0],1),\"N\",',R2=',round(C[1],1),\"N\",'and R3=',round(C[2],1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reactions are as R1= 53.3 N ,R2= 23.3 N and R3= 33.3 N\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-10, Page no87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=1 #kg\n",
- "g=9.81 #m/s**2\n",
- "# As t1=45 degrees and t2=30 degrees\n",
- "cost1=sqrt(2)**-1\n",
- "cost2=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Solving as system of linear equations\n",
- "A=np.array([[1,0,-cost1,0],[0,1,0,3*5**-1],[-5,g*m*cost1*cost2,0,0],[-1,0,0,4*5**-1]])\n",
- "B=np.array([[0],[g*m],[g*m*5*cost1*cost2],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The forces are: Nb=',round(C[0],1),',Nc=',round(C[1],1),',Tc=',round(C[2],1),'and Tb=',round(C[3],1),\"respectively\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces are: Nb= 3.0 ,Nc= 7.5 ,Tc= 4.3 and Tb= 3.8 respectively\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-11, Page no 87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "w=50 #lb wind load\n",
- "W=60 #lb weight of door\n",
- "\n",
- "#Calculations\n",
- "#Calculation as system of linear equations\n",
- "A=np.array([[0,0,33],[1,1,-1],[28,10,-28]])\n",
- "B=np.array([[50*18],[-50],[-50*24]])\n",
- "C=np.linalg.solve(A,B)\n",
- "P=C[2]*(cos(20*pi*180**-1))**-1\n",
- "D=np.array([[-28,-10],[1,1]])\n",
- "E=np.array([1080-(28*(P*sin((20*pi)*180**-1))),P*sin((20*pi)*180**-1)])\n",
- "F=np.linalg.solve(D,E)\n",
- "By=60\n",
- "\n",
- "#Result\n",
- "print'The forces are as follows:'\n",
- "print'Az=',round(C[0],1),\"lb\",',Bz=',round(C[1],1),\"lb\",',Pz=',round(C[2],1),\"lb\",',Ax=',round(F[0]),\"lb\",',Bx=',round(F[1]),\"lb\",'and By=',round(By),\"lb respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces are as follows:\n",
- "Az= -11.6 lb ,Bz= -11.1 lb ,Pz= 27.3 lb ,Ax= -50.0 lb ,Bx= 60.0 lb and By= 60.0 lb respectively.\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-12, Page no 89"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=1 #kg\n",
- "g=9.81 #m/s**2\n",
- "# Since t1=45 degrees and t2=30 degrees\n",
- "cost1=sqrt(2)**-1\n",
- "cost2=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Solving as system of linear equations\n",
- "A=np.array([[1,0,-cost1,0],[0,1,0,3*5**-1],[-5,g*m*cost1*cost2,0,0],[-1,0,0,4*5**-1]])\n",
- "B=np.array([[0],[g*m],[g*m*5*cost1*cost2],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The forces are: Nb=',round(C[0],2),\"N\",',Nc=',round(C[1],2),\"N\",',Tc=',round(C[2],2),\"N\",'and Tb=',round(C[3],2),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " The forces are: Nb= 3.04 N ,Nc= 7.53 N ,Tc= 4.3 N and Tb= 3.8 N\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-13, Page no 89"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=200 #lb\n",
- "cos25=0.9063\n",
- "sin25=0.4226\n",
- "\n",
- "#Calculations\n",
- "P=l*5*12**-1 #lb\n",
- "#Solving as system of linear equations\n",
- "A=np.array([[0,-36,0,0],[0,0,0,36],[0,0,1,1],[1,1,0,0]])\n",
- "B=np.array([[-P*cos25*48],[l*20+P*sin25*48],[P*sin25+200],[P*cos25]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The forces are: Az=',round(C[0],1),\"lb\",',Bz=',round(C[1]),\"lb\",',Ay=',round(C[2],1),\"lb\",'and By=',round(C[3]),\"lb\"\n",
- "\n",
- "# The answer for Az waries due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces are: Az= -25.2 lb ,Bz= 101.0 lb ,Ay= 77.2 lb and By= 158.0 lb\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-14, Page No 90"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "A=80 #lb\n",
- "B=40 #lb\n",
- "C=60 #lb\n",
- "l1=2 #in\n",
- "l2=4 #in\n",
- "l3=6 #in\n",
- "l4=9 #in\n",
- "l5=3 #in\n",
- "l6=7 #in\n",
- "\n",
- "#Calculations\n",
- "P=-(-A*l1+B*l2-C*l2)/l1\n",
- "By=-(A*l3+P*l3)/l4\n",
- "Ay=(-A*l5-P*l5)/l4\n",
- "Bz=-(-C*l1-B*l1)/l4\n",
- "Az=(C*l6+B*l6)/l4\n",
- "\n",
- "#Result\n",
- "print'The forces are:Ay=',round(Ay,1),\"lb\",',By=',round(By,1),\"lb\",',Az=',round(Az,1),\"lb forward\",'and Bz=',round(Bz,1),\"lb forward\"\n",
- "\n",
- "# The answers may wary due to rounding of the values"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces are:Ay= -67.0 lb ,By= -134.0 lb ,Az= 77.0 lb forward and Bz= 22.0 lb forward\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-15, Page no 91"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=138 #lb\n",
- "w=80 #lb\n",
- "\n",
- "#Calculations\n",
- "u=(3*3+4*4+6*6)**0.5\n",
- "a=np.array([-3*u**-1,4*u**-1,-6*u**-1])\n",
- "v=(3*3+3*3+3*3)**0.5\n",
- "c=np.array([3*v**-1,3*v**-1,-3*v**-1])\n",
- "P=np.array([[1,0,0],[0,0,8],[0,-W,0]])\n",
- "Q=np.array([[0,0,1],[0,0,8],[0,-W,0]])\n",
- "R=np.array([[1,0,0],[0,0,4],[0,-w,0]])\n",
- "S=np.array([[0,0,1],[0,0,4],[0,-w,0]])\n",
- "T=np.array([[1,0,0],[0,0,6],[a[0],a[1],a[2]]])\n",
- "U=np.array([[0,1,0],[0,0,6],[a[0],a[1],a[2]]])\n",
- "V=np.array([[1,0,0],[0,0,3],[c[0],c[1],c[2]]])\n",
- "Y=np.array([[0,1,0],[0,0,3],[c[0],c[1],c[2]]])\n",
- "#Solving for A and C\n",
- "MAT1=np.array([[det(T),det(V)],[det(U),det(Y)]])\n",
- "MAT2=np.array([[det(P)+det(R)],[0]])\n",
- "res=np.linalg.solve(-MAT1,MAT2)\n",
- "A=np.array([a[0]*res[0],a[1]*res[0],a[2]*res[0]])\n",
- "C=np.array([c[0]*res[1],c[1]*res[1],c[2]*res[1]])\n",
- "E=np.array([-(A[0]+C[0]),-(-w-W+A[1]+C[1]),-(A[2]+C[2])])\n",
- "\n",
- "#Result\n",
- "print'The force vectors are as follows:'\n",
- "print'A=',round(A[0]),\"i +\",round(A[1]),\"j \",round(A[2]),\"k\",\n",
- "print'and C=',round(C[0]),\"i +\",round(C[1]),\"j \",round(C[2]),\"k\"\n",
- "print'also,Ex=',round(E[0]),\"lb\",',Ey=',round(E[1]),\"lb\",'and Ez=',round(E[2]),\"lb\"\n",
- "#Decimal accuracy causes discrepancy in the answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force vectors are as follows:\n",
- "A= -102.0 i + 136.0 j -203.0 k and C= 203.0 i + 203.0 j -203.0 k\n",
- "also,Ex= -102.0 lb ,Ey= -121.0 lb and Ez= 407.0 lb\n"
- ]
- }
- ],
- "prompt_number": 9
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter6_3.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter6_3.ipynb deleted file mode 100755 index be27eeca..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter6_3.ipynb +++ /dev/null @@ -1,770 +0,0 @@ -{
- "metadata": {
- "name": "chapter6.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 6: Equilibrium of Non Coplanar Force Systems "
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-1, Page no 81"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "H=30 #ft\n",
- "F=150 #lb\n",
- "# Here theta1=10, theta2=30 & theta3=60 degrees. Thus the values are,\n",
- "sin10=0.1736\n",
- "cos10=0.9848\n",
- "sin30=2**-1\n",
- "cos30=sqrt(3)*2**-1\n",
- "sin60=sqrt(3)*2**-1\n",
- "cos60=2**-1\n",
- "\n",
- "#Calculations\n",
- "#Matrix solution of simultaneous equations\n",
- "X=np.array([[cos60*sin30, -cos60*sin30],[cos60*cos30, cos60*cos30]]) \n",
- "Y=np.array([[0],[F*cos10]]) \n",
- "R=np.linalg.solve(X,Y)\n",
- "#To find P,sum the forces vertically along the y-axis\n",
- "P=F*sin10+2*R[0]*sin60 #lb Compression\n",
- "\n",
- "#Result\n",
- "print'The value of A and B is',round(R[0]),\"lb (T)\"\n",
- "print'The value of P is',round(P),\"lb (C)\"\n",
- "\n",
- "# The value of P is off by 1 lb"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of A and B is 171.0 lb (T)\n",
- "The value of P is 321.0 lb (C)\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-2, Page no 82"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "#Initilization of variables\n",
- "F=150 #lb\n",
- "# Here theta1=10, theta2=30 & theta3=60 degrees. Thus the values are,\n",
- "sin10=0.1736\n",
- "cos10=0.9848\n",
- "sin30=2**-1\n",
- "cos30=sqrt(3)*2**-1\n",
- "sin60=sqrt(3)*2**-1\n",
- "cos60=2**-1\n",
- "\n",
- "#Calculations\n",
- "A=[-cos60*cos30,-sin60,cos60*sin30] \n",
- "B=[-cos60*cos60,-sin60,cos60*sin30]\n",
- "# 150lb force is actually a vector\n",
- "F_v=[F*cos10,F*sin10,0] #lb\n",
- "#Postion vector relative to C \n",
- "r=[0,30,0]\n",
- "# Moment about point C is zero\n",
- "#solution by matrix\n",
- "X=np.array([[7.5,-7.5],[13,13]]) \n",
- "Y=np.array([[0],[4470]]) \n",
- "R=np.linalg.solve(X,Y)\n",
- "A=R[0] #lb\n",
- "B=R[1] #lb\n",
- "#Summing forces in y direction\n",
- "Cy=0.866*A+0.866*B+25.9 #lb\n",
- "\n",
- "#Result\n",
- "print'The value of A and B is',round(A),\"lb\"\n",
- "print'The value of Cy is',round(Cy),\"lb\"\n",
- "\n",
- "# The answer may wary due to decimal point descripancy in computation"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of A and B is 172.0 lb\n",
- "The value of Cy is 324.0 lb\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-3, Page no 82"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initililization of variables\n",
- "m=6.12 #kg\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "AD=sqrt(3**2+2**2+6**2) \n",
- "AC=sqrt(4**2+2**2)\n",
- "AB=5\n",
- "#Sum of forces in the y direction\n",
- "T1=(m*g*AD)/6 #N\n",
- "#sum of forces in the x and z direction\n",
- "#Matrix solution of the folllowing simultaneous equations\n",
- "X=np.array([[4*4.47**-1,-3*5**-1],[-2*4.47**-1,4*5**-1]])\n",
- "Y=np.array([[T1*(3*7**-1)],[T1*(2*7**-1)]]) \n",
- "R=np.linalg.solve(X,Y)\n",
- "T2=R[0] #N\n",
- "T3=R[1] #N\n",
- "\n",
- "#Result\n",
- "print'The value of T1 is',round(T1),\"N\"\n",
- "print'The value of T2 is',round(T2,1),\"N\"\n",
- "print'The vaue of T3 is',round(T3),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of T1 is 70.0 N\n",
- "The value of T2 is 80.5 N\n",
- "The vaue of T3 is 70.0 N\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-4, Page no 83"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Intilization of variables\n",
- "\n",
- "F=np.array([0,60,0]) #Force defined as a matrix\n",
- "t1=np.array([-3*7**-1,6*7**-1,2*7**-1]) #Tension defined as a matrix\n",
- "t2=np.array([4*4.47**-1,0,-2*4.47**-1]) #tension defined as a mtrix\n",
- "t3=np.array([-3*5**-1,0,4*5**-1]) #Tension defined as a matrix\n",
- "\n",
- "#Calculations\n",
- "#Summation of forces in the y-direction\n",
- "T1=F[1]*(t1[1]**-1) #N\n",
- "#Summation of forces in the x-direction and z direction\n",
- "M1=np.array([[t2[0],t3[0]],[t2[2],t3[2]]])\n",
- "M2=np.array([[-1*t1[0]*T1],[t1[2]*T1]]) \n",
- "R=np.linalg.solve(M1,M2)\n",
- "\n",
- "#Result\n",
- "print'The tension in the strings are: T1=',round(T1),\"N\",',T2=',round(R[0],1),\"N\",'and T3=',round(R[1]),\"N respectively\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in the strings are: T1= 70.0 N ,T2= 80.5 N and T3= 70.0 N respectively\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-5, Page no 84"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=80 #kg\n",
- "g=9.81 # m/s**2\n",
- "#Co-ordinates of points in Meters\n",
- "A=np.array([1,3,0])\n",
- "B=np.array([3,3,-4])\n",
- "C=np.array([4,3,0])\n",
- "D=np.array([2,0,-1])\n",
- "\n",
- "#Calculations\n",
- "#Tension in DC will be\n",
- "a=np.array([[C[0]-D[0]],[C[1]-D[1]],[C[2]-D[2]]])\n",
- "h=((C[0]-D[0])**2+(C[1]-D[1])**2+(C[2]-D[2])**2)**0.5\n",
- "c=a*h**-1\n",
- "#Unit vector calculations\n",
- "e=np.array([[B[0]-A[0]],[B[1]-A[1]],[B[2]-A[2]]])\n",
- "v=((B[0]-A[0])**2+(B[1]-A[1])**2+(B[2]-A[2])**2)**0.5\n",
- "e_ab=e*v**-1\n",
- "#Position vector AD\n",
- "r_ad=np.array([[D[0]-A[0]],[D[1]-A[1]],[D[2]-A[2]]])\n",
- "#Moment Calculations\n",
- "O=np.array([[1,0,0],[1,-3,-1],[0,-m*g,0]])\n",
- "P=np.array([[0,1,0],[1,-3,-1],[0,-m*g,0]])\n",
- "Q=np.array([[0,0,1],[1,-3,-1],[0,-m*g,0]])\n",
- "C1=np.array([[1,0,0],[1,-3,-1],[2,3,1]])\n",
- "C2=np.array([[0,1,0],[1,-3,-1],[2,3,1]])\n",
- "C3=np.array([[0,0,1],[1,-3,-1],[2,3,1]])\n",
- "rxF1=np.array([[det(O),det(P),det(Q)]])\n",
- "rxF2=np.array([(det(C1)*h**-1),(det(C2)*h**-1),(det(C3)*h**-1)])\n",
- "#Final Moment calculations\n",
- "rxF=rxF1+rxF2 \n",
- "#Taking dot product\n",
- "dot1=e_ab*rxF\n",
- "dot2=e_ab*rxF2\n",
- "#equating dot product to zero to obtain C\n",
- "C=-(dot1[0,0]+dot1[2,2])/dot2[2,2]\n",
- "\n",
- "#Result \n",
- "print'The tension in CD is',round(C),\"N\"\n",
- "\n",
- "# The ans is off by 1 N due to Decimal point descrepancy. "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in CD is 162.0 N\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-6, Page no 85"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=200 #lb\n",
- "Dh=4 #ft\n",
- "\n",
- "#Calculation\n",
- "theta=arctan(2*Dh**-1)*(180/pi) #degrees\n",
- "T=w/(3*cos(theta)) #lb\n",
- "\n",
- "#Result\n",
- "print'The Tension in each rope is',round(T,1),\"lb\"\n",
- "print'The angle is',round(theta,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Tension in each rope is 482.9 lb\n",
- "The angle is 26.6 degrees\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-7, Page no 85"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "F=np.array([100,0,0]) #N\n",
- "CE=5 #m\n",
- "BC=(34)**0.5 #m\n",
- "AC=(41)**0.5 #m\n",
- "\n",
- "#Calculations\n",
- "#solving as a matrix for system of linear equations\n",
- "A=np.array([[3*BC**-1,-4*AC**-1,0],[0,0,(6*4)*CE**-1],[-3*BC**-1,-3*AC**-1,-3*CE**-1]])\n",
- "B=np.array([[0],[F[0]*4],[-F[0]]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The forces F1,F2 and F3 are as',round(C[0],1),\"N\",',',round(C[1],1),\"N\",'and',round(C[2],1),\"N\"\n",
- "print'Here F3 is compression assumed and rest are Tension'"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces F1,F2 and F3 are as 55.5 N , 45.7 N and 83.3 N\n",
- "Here F3 is compression assumed and rest are Tension\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-8, Page no 85"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "F=np.array([100,0,0]) #N\n",
- "CE=5 #m\n",
- "BC=(34)**0.5 #m\n",
- "AC=(41)**0.5 #m\n",
- "\n",
- "#Calculations\n",
- "#solving as a matrix for system of linear equations\n",
- "A=np.array([[3/BC,-4/AC,0],[-4/BC,-4/AC,4*CE**-1],[-3/BC,-3/AC,-3*CE**-1]])\n",
- "B=np.array([[0],[0],[-F[0]]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The forces are: F1=',round(C[0],1),\"N\",',F2=',round(C[1],1),\"N\",'and F3=',round(C[2],1),\"N\"\n",
- "print'Here F3 is compression assumed and rest are Tension'"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces are: F1= 55.5 N ,F2= 45.7 N and F3= 83.3 N\n",
- "Here F3 is compression assumed and rest are Tension\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-9, Page no 86"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "#here forces will be defines as matrices along with their co-ordinates\n",
- "#Force in N and co-ordinates in mm\n",
- "F1=[30,200,300] \n",
- "F2=[10,400,200]\n",
- "F3=[20,200,500]\n",
- "F4=[50,400,500]\n",
- "#Calculations\n",
- "#solving as system of linear equations\n",
- "A=np.array([[1,1,1],[-600,-600,0],[0,600,600]])\n",
- "B=np.array([[F1[0]+F2[0]+F3[0]+F4[0]],[-(F3[0]*F3[2]+F1[0]*F1[2]+F4[0]*F4[2]+F2[0]*F2[2])],[-(-F3[0]*F3[1]-F1[0]*F1[1]-F4[0]*F4[1]-F2[0]*F2[1])]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The reactions are as R1=',round(C[0],1),\"N\",',R2=',round(C[1],1),\"N\",'and R3=',round(C[2],1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reactions are as R1= 53.3 N ,R2= 23.3 N and R3= 33.3 N\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-10, Page no87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=1 #kg\n",
- "g=9.81 #m/s**2\n",
- "# As t1=45 degrees and t2=30 degrees\n",
- "cost1=sqrt(2)**-1\n",
- "cost2=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Solving as system of linear equations\n",
- "A=np.array([[1,0,-cost1,0],[0,1,0,3*5**-1],[-5,g*m*cost1*cost2,0,0],[-1,0,0,4*5**-1]])\n",
- "B=np.array([[0],[g*m],[g*m*5*cost1*cost2],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The forces are: Nb=',round(C[0],1),',Nc=',round(C[1],1),',Tc=',round(C[2],1),'and Tb=',round(C[3],1),\"respectively\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces are: Nb= 3.0 ,Nc= 7.5 ,Tc= 4.3 and Tb= 3.8 respectively\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-11, Page no 87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "w=50 #lb wind load\n",
- "W=60 #lb weight of door\n",
- "\n",
- "#Calculations\n",
- "#Calculation as system of linear equations\n",
- "A=np.array([[0,0,33],[1,1,-1],[28,10,-28]])\n",
- "B=np.array([[50*18],[-50],[-50*24]])\n",
- "C=np.linalg.solve(A,B)\n",
- "P=C[2]*(cos(20*pi*180**-1))**-1\n",
- "D=np.array([[-28,-10],[1,1]])\n",
- "E=np.array([1080-(28*(P*sin((20*pi)*180**-1))),P*sin((20*pi)*180**-1)])\n",
- "F=np.linalg.solve(D,E)\n",
- "By=60\n",
- "\n",
- "#Result\n",
- "print'The forces are as follows:'\n",
- "print'Az=',round(C[0],1),\"lb\",',Bz=',round(C[1],1),\"lb\",',Pz=',round(C[2],1),\"lb\",',Ax=',round(F[0]),\"lb\",',Bx=',round(F[1]),\"lb\",'and By=',round(By),\"lb respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces are as follows:\n",
- "Az= -11.6 lb ,Bz= -11.1 lb ,Pz= 27.3 lb ,Ax= -50.0 lb ,Bx= 60.0 lb and By= 60.0 lb respectively.\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-12, Page no 89"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=1 #kg\n",
- "g=9.81 #m/s**2\n",
- "# Since t1=45 degrees and t2=30 degrees\n",
- "cost1=sqrt(2)**-1\n",
- "cost2=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Solving as system of linear equations\n",
- "A=np.array([[1,0,-cost1,0],[0,1,0,3*5**-1],[-5,g*m*cost1*cost2,0,0],[-1,0,0,4*5**-1]])\n",
- "B=np.array([[0],[g*m],[g*m*5*cost1*cost2],[0]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The forces are: Nb=',round(C[0],2),\"N\",',Nc=',round(C[1],2),\"N\",',Tc=',round(C[2],2),\"N\",'and Tb=',round(C[3],2),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " The forces are: Nb= 3.04 N ,Nc= 7.53 N ,Tc= 4.3 N and Tb= 3.8 N\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-13, Page no 89"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=200 #lb\n",
- "cos25=0.9063\n",
- "sin25=0.4226\n",
- "\n",
- "#Calculations\n",
- "P=l*5*12**-1 #lb\n",
- "#Solving as system of linear equations\n",
- "A=np.array([[0,-36,0,0],[0,0,0,36],[0,0,1,1],[1,1,0,0]])\n",
- "B=np.array([[-P*cos25*48],[l*20+P*sin25*48],[P*sin25+200],[P*cos25]])\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#Result\n",
- "print'The forces are: Az=',round(C[0],1),\"lb\",',Bz=',round(C[1]),\"lb\",',Ay=',round(C[2],1),\"lb\",'and By=',round(C[3]),\"lb\"\n",
- "\n",
- "# The answer for Az waries due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces are: Az= -25.2 lb ,Bz= 101.0 lb ,Ay= 77.2 lb and By= 158.0 lb\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-14, Page No 90"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "A=80 #lb\n",
- "B=40 #lb\n",
- "C=60 #lb\n",
- "l1=2 #in\n",
- "l2=4 #in\n",
- "l3=6 #in\n",
- "l4=9 #in\n",
- "l5=3 #in\n",
- "l6=7 #in\n",
- "\n",
- "#Calculations\n",
- "P=-(-A*l1+B*l2-C*l2)/l1\n",
- "By=-(A*l3+P*l3)/l4\n",
- "Ay=(-A*l5-P*l5)/l4\n",
- "Bz=-(-C*l1-B*l1)/l4\n",
- "Az=(C*l6+B*l6)/l4\n",
- "\n",
- "#Result\n",
- "print'The forces are:Ay=',round(Ay,1),\"lb\",',By=',round(By,1),\"lb\",',Az=',round(Az,1),\"lb forward\",'and Bz=',round(Bz,1),\"lb forward\"\n",
- "\n",
- "# The answers may wary due to rounding of the values"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces are:Ay= -67.0 lb ,By= -134.0 lb ,Az= 77.0 lb forward and Bz= 22.0 lb forward\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6-15, Page no 91"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=138 #lb\n",
- "w=80 #lb\n",
- "\n",
- "#Calculations\n",
- "u=(3*3+4*4+6*6)**0.5\n",
- "a=np.array([-3*u**-1,4*u**-1,-6*u**-1])\n",
- "v=(3*3+3*3+3*3)**0.5\n",
- "c=np.array([3*v**-1,3*v**-1,-3*v**-1])\n",
- "P=np.array([[1,0,0],[0,0,8],[0,-W,0]])\n",
- "Q=np.array([[0,0,1],[0,0,8],[0,-W,0]])\n",
- "R=np.array([[1,0,0],[0,0,4],[0,-w,0]])\n",
- "S=np.array([[0,0,1],[0,0,4],[0,-w,0]])\n",
- "T=np.array([[1,0,0],[0,0,6],[a[0],a[1],a[2]]])\n",
- "U=np.array([[0,1,0],[0,0,6],[a[0],a[1],a[2]]])\n",
- "V=np.array([[1,0,0],[0,0,3],[c[0],c[1],c[2]]])\n",
- "Y=np.array([[0,1,0],[0,0,3],[c[0],c[1],c[2]]])\n",
- "#Solving for A and C\n",
- "MAT1=np.array([[det(T),det(V)],[det(U),det(Y)]])\n",
- "MAT2=np.array([[det(P)+det(R)],[0]])\n",
- "res=np.linalg.solve(-MAT1,MAT2)\n",
- "A=np.array([a[0]*res[0],a[1]*res[0],a[2]*res[0]])\n",
- "C=np.array([c[0]*res[1],c[1]*res[1],c[2]*res[1]])\n",
- "E=np.array([-(A[0]+C[0]),-(-w-W+A[1]+C[1]),-(A[2]+C[2])])\n",
- "\n",
- "#Result\n",
- "print'The force vectors are as follows:'\n",
- "print'A=',round(A[0]),\"i +\",round(A[1]),\"j \",round(A[2]),\"k\",\n",
- "print'and C=',round(C[0]),\"i +\",round(C[1]),\"j \",round(C[2]),\"k\"\n",
- "print'also,Ex=',round(E[0]),\"lb\",',Ey=',round(E[1]),\"lb\",'and Ez=',round(E[2]),\"lb\"\n",
- "#Decimal accuracy causes discrepancy in the answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force vectors are as follows:\n",
- "A= -102.0 i + 136.0 j -203.0 k and C= 203.0 i + 203.0 j -203.0 k\n",
- "also,Ex= -102.0 lb ,Ey= -121.0 lb and Ez= 407.0 lb\n"
- ]
- }
- ],
- "prompt_number": 9
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter7.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter7.ipynb deleted file mode 100755 index 1343d7ec..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter7.ipynb +++ /dev/null @@ -1,668 +0,0 @@ -{
- "metadata": {
- "name": "chapter7.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 7: Trusses And Cables"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-1, Page no 99"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "F1=2000 #lb\n",
- "F2=4000 #lb\n",
- "l1=10 #ft\n",
- "l2=30 #ft\n",
- "l3=20 #ft\n",
- "l4=40 #ft\n",
- "# as t=60 degrees\n",
- "sint=sqrt(3)*2**-1\n",
- "cost=2**-1\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point B and A\n",
- "Ra=(F1*l2+F2*l1)/l4 \n",
- "Rb=(F2*l2+F1*l1)/l4\n",
- "#Consider fig 7-4(c)\n",
- "A=np.array([[1,-cost],[0,-sint]])\n",
- "B=np.array([[0],[-2500]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Consider figure 7-4(d)\n",
- "A1=np.array([[1,cost],[0,-sint]])\n",
- "B1=np.array([-C[1]*cost,-C[1]*sint+F1])\n",
- "C1=np.linalg.solve(A1,B1)\n",
- "#Consider figure 7-4(e)\n",
- "CD=577\n",
- "CE=C[0]+C1[1]*cost+CD*cost\n",
- "#Consider figure 7-4(f)\n",
- "DE=Rb/sint\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The reactions are:Ra=',round(Ra),\"lb\",'and Rb=',round(Rb),\"lb\"\n",
- "print'Force in member AB=',round(C[1]),\"lb (C)\",'and AC=',round(C[0]),\"lb (T)\"\n",
- "print'Force in member BC=',round(C1[1]),\"lb (T)\",'and BD=',round(C1[0]),\"lb (-ve sign indicates compression)\"\n",
- "print'Force in member CD=',round(CD),\"lb (C)\",'and CE=',round(CE),\"lb (T)\"\n",
- "print'Force in member DE=',round(DE),\"lb (C)\"\n",
- "\n",
- "#Decimal Accuracy causes discrepancy in answers.Thus answers wary as compared to the textbook.\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reactions are:Ra= 2500.0 lb and Rb= 3500.0 lb\n",
- "Force in member AB= 2887.0 lb (C) and AC= 1443.0 lb (T)\n",
- "Force in member BC= 577.0 lb (T) and BD= -1732.0 lb (-ve sign indicates compression)\n",
- "Force in member CD= 577.0 lb (C) and CE= 2021.0 lb (T)\n",
- "Force in member DE= 4041.0 lb (C)\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-2, Page no 101"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "s=4 #m length of sides\n",
- "l=2 #kN load acting on each node\n",
- "r=7 #kN by inspection reaction at A\n",
- "sin60=sqrt(3)*2**-1\n",
- "tan30=sqrt(3)**-1\n",
- "tan60=sqrt(3)\n",
- "\n",
- "#Calculation\n",
- "#Taking Moment about point G\n",
- "FH=(-r*12+2*10+2*6+2*2)/(2*tan60) #kN Compressive\n",
- "#Taking moment about point H\n",
- "GI=(r*14-2*12-2*8-2*4)/(2*tan30) #kN Tension\n",
- "#Summing forces in the vertical direction\n",
- "HG=-(r-(l*3))/sin60 #kN Compression\n",
- "#Taking moment about point J yields\n",
- "IK=(-2*4-2*8+r*10)/(2*tan60) #kN\n",
- "\n",
- "#Result\n",
- "print'The value of the forces in the components are as follows'\n",
- "print'FH=',round(FH,1),\"kN (C)\",',GI=',round(GI,1),\"kN (T)\",',HG=',round(HG,2),\"kN (C)\",'and IK=',round(IK,1),\"kN (T)\"\n",
- "print'The answer in the text book for GI is wrong'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of the forces in the components are as follows\n",
- "FH= -13.9 kN (C) ,GI= 43.3 kN (T) ,HG= -1.15 kN (C) and IK= 13.3 kN (T)\n",
- "The answer in the text book for GI is wrong\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-3, Page no 101"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=30 degrees,\n",
- "sin30=2**-1\n",
- "EF=40000 #lb\n",
- "l=36 #feet\n",
- "\n",
- "#Calculation\n",
- "#Taking moment about point D and setting EF=40000lbs\n",
- "P=-(EF*sin30*l)/l #lb\n",
- "\n",
- "#Result\n",
- "print'The maximum value of P is',round(P),\"lb\"\n",
- "print'The negative sign indicates the downward direction'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum value of P is -20000.0 lb\n",
- "The negative sign indicates the downward direction\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-4, Page no 102"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=12 #m\n",
- "# as theta1=30 degrees\n",
- "cos30=sqrt(3)*2**-1\n",
- "cos60=2**-1\n",
- "sin30=2**-1\n",
- "\n",
- "F1=1000 #N\n",
- "F2=2000 #N\n",
- "\n",
- "#Calculation\n",
- "FG=l*cos30 #m\n",
- "DG=(l+(l/2))/cos30 #m\n",
- "#Taking moment about point G\n",
- "A=(F1*l+F2*FG+F1*DG)/(l*3) #N\n",
- "#Summing forces in horizontal direction\n",
- "G_x=(2*F1+F2)*sin30 #N\n",
- "#Summing forces in the vertical direction\n",
- "G_y=(2*F1+F2)*cos30+F1-A #N\n",
- "#Taking moment about point C\n",
- "BD=-(A*l)/(l/2) #N\n",
- "#Taking moment about point D\n",
- "CE=(A*(l+(l/2)))/FG #N\n",
- "theta=arctan((l/2)/FG) #degrees \n",
- "#Summing forces in the vertical direction\n",
- "CD=(A+(BD*cos60))/cos(theta) #N\n",
- "\n",
- "#Result\n",
- "print'The values of the forces are as follows'\n",
- "print'A=',round(A),\"N\",',G_x=',round(G_x),\"N\",',G_y=',round(G_y),\"N\",',BD=',round(BD),\"N (C)\",',CE=',round(CE),\"N (T)\",'and CD=',round(CD),\"N(T)\"\n",
- "#Decimal Accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values of the forces are as follows\n",
- "A= 1488.0 N ,G_x= 2000.0 N ,G_y= 2976.0 N ,BD= -2976.0 N (C) ,CE= 2577.0 N (T) and CD= 0.0 N(T)\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-5, Page no 103"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "A=2000 #lb\n",
- "E=2000 #lb\n",
- "# as theta=60 degrees and theta1=30 degrees, which means:\n",
- "sin60=sqrt(3)*2**-1\n",
- "cos60=2**-1\n",
- "sin30=2**-1\n",
- "cos30=sqrt(3)*2**-1\n",
- "\n",
- "#Sign convention positive means Tension and negative means Compression\n",
- "#Taking sum of forces along x and y direction in fig7-13\n",
- "AB=-A/sin60 #lb\n",
- "AG=-AB*cos60 #lb\n",
- "#Taking sum of forces along x and y direction in fig7-14\n",
- "BG=((-AB*cos30)-1000)/(cos30) #lb\n",
- "BC=((AB*sin30)-(BG*sin30)) #lb\n",
- "#Taking sum of forces along x and y direction in fig7-15\n",
- "GC=-(BG*sin60)/sin60 #lb\n",
- "GF=AG+BG*cos60-GC*(cos60) #lb\n",
- "#By symmetry of structure\n",
- "DE=AB #lb\n",
- "FE=AG #lb\n",
- "DF=BG #lb\n",
- "CD=BC #lb\n",
- "\n",
- "#Result\n",
- "print'The forces in the truess are'\n",
- "print'AB=DE=',round(AB),\"lb (C)\",',AG=FE=',round(AG),\"lb (T)\",',BG=DF=',round(BG),\"lb (T)\",',BC=CD=',round(BC),\"lb (C)\",'and CG=CF=',round(GC),\"lb (C)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces in the truess are\n",
- "AB=DE= -2309.0 lb (C) ,AG=FE= 1155.0 lb (T) ,BG=DF= 1155.0 lb (T) ,BC=CD= -1732.0 lb (C) and CG=CF= -1155.0 lb (C)\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-6, Page no 104"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=500 #N\n",
- "A=1000 #N\n",
- "# as theta=60 degrees,\n",
- "sin60=sqrt(3)*2**-1\n",
- "l=20 #m\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point G\n",
- "R_c=(20*3*A+50*F+30*F+10*F)/40 #N\n",
- "#Returning to fig7-17\n",
- "#Taking moment about point C\n",
- "BD=(l*A+(l/2)*F)/(l*sin60) #N\n",
- "#Taking sum of forces in vertical direction\n",
- "CD=(A+F-R_c)/sin60 #N\n",
- "\n",
- "#Result\n",
- "print'The forces in the members are as follows'\n",
- "print'BD=',round(BD),\"N (T)\",'and CD=',round(CD),\"N (C).\",'Also the reaction at C is',round(R_c),\"N\"\n",
- "#Decimal accuracy causes discrepancey in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces in the members are as follows\n",
- "BD= 1443.0 N (T) and CD= -1299.0 N (C). Also the reaction at C is 2625.0 N\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-7, Page no 104"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=800 #lb/ft\n",
- "a=600 #ft\n",
- "d=40 #ft\n",
- "\n",
- "#Calculations\n",
- "T=0.5*w*a*(sqrt(1+(a**2/(16*d**2)))) #lb\n",
- "H=(w*a**2)/(8*d) #lb\n",
- "#Taking the first two terms of the series\n",
- "l=a*(1+(8/3)*(d*a**-1)**2-(32/5)*0.00002) #ft\n",
- "\n",
- "#Result\n",
- "print'The value of T=',round(T),\"lb\",'and that of H=',round(H),\"lb.\",'Also l=',round(l),\"ft\"\n",
- "#Deciaml accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of T= 929516.0 lb and that of H= 900000.0 lb. Also l= 605.0 ft\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-8,Page no 105"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=800*300 #lb\n",
- "\n",
- "#Calculations\n",
- "#Summing forces in horizontal and vertical direction\n",
- "theta=arctan(40*150**-1) #degrees\n",
- "H=l/tan(theta) #lb\n",
- "T_max=sqrt(l**2+H**2) #lb\n",
- "\n",
- "#Result\n",
- "print'The maximun tension is',round(T_max),\"lb\",'and H=',round(H),\"lb\"\n",
- "#Decimal accuracy causes discrepancy in answers"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximun tension is 931450.0 lb and H= 900000.0 lb\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-9,Page no 105"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#For simplicity a1 and a2 values are being considered as constant free of H\n",
- "a_1=sqrt(10*14.7**-1)\n",
- "a_2=sqrt(30/14.7)\n",
- "y=10 #m\n",
- "\n",
- "#Calculations\n",
- "H=(300/(a_1+a_2))**2 #N\n",
- "#Now reconsidering a1 and a2 actual values\n",
- "a1=a_1*sqrt(H) #m\n",
- "a2=a_2*sqrt(H) #m\n",
- "#Theta calculations\n",
- "x=a1\n",
- "theta=arctan(2*y/x)\n",
- "#T calculations\n",
- "T=sqrt((864*a2**2)+H**2) #N\n",
- "\n",
- "#Result\n",
- "print'The tension in the cable is',round(T,2),\"*10**-3 kN\"\n",
- "# The answer may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in the cable is 18585.57 *10**-3 kN\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-10, Page no 106"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "T=140000 #N\n",
- "w=2000 #N/m\n",
- "a=20 #m\n",
- "\n",
- "#Calculations\n",
- "#Calculation step by step\n",
- "lhs=(140000*2)*(2000*20)**-1 \n",
- "d=sqrt(1/((((lhs**2)-1)*16)/(20**2))) #m\n",
- "l=a*(1+(8/3)*(d/a)**2) #m\n",
- "\n",
- "#Result\n",
- "print'The sag in the cable is',round(d,2),\"m\"\n",
- "print'The required length is',round(l,2),\"m\"\n",
- "\n",
- "# Value of l is off by 0.2 m"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The sag in the cable is 0.72 m\n",
- "The required length is 20.05 m\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-11, Page no 106"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=10*16**-1 #lb/ft\n",
- "a=80 #ft\n",
- "P=500 #lb\n",
- "\n",
- "#Calculations\n",
- "lhs=(P*2)/(w*a)\n",
- "d=sqrt(1*((((lhs**2)-1)*16)/(80**2))**-1) #ft\n",
- "\n",
- "#Result\n",
- "print'The sag in the cable is',round(d),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The sag in the cable is 1.0 ft\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-12, Page no 107"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=0.518 #lb/ft\n",
- "d=50 #ft\n",
- "l=500 #ft\n",
- "#Plot coding\n",
- "A=linspace(0,800,9) #defined x axis\n",
- "B=A+50\n",
- "C=[50000,500*(2*100)**-1,500*(2*200)**-1,500*(2*300)**-1,500*(2*400)**-1,500*(2*500)**-1,500*(2*600)**-1,500*(2*700)**-1,500*(2*800)**-1]\n",
- "D=cosh(C)\n",
- "E=([D[0]*A[0],D[1]*A[1],D[2]*A[2],D[3]*A[3],D[4]*A[4],D[5]*A[5],D[6]*A[6],D[7]*A[7],D[8]*A[8]])\n",
- "plot(A,B,A,E) #plotting two lines on the same plot\n",
- "\n",
- "#Calculations\n",
- "#By close observation of plot taking c around 650\n",
- "#consider c=635\n",
- "c=635\n",
- "T_max=w*(c+d) #lb\n",
- "a=c+d\n",
- "l=(4*(a*a-c*c))**0.5\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The maximum tension is',round(T_max),\"lb\",'and length required is',round(l),\"ft respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum tension is 355.0 lb and length required is 514.0 ft respectively.\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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Zv4RPkz6lk10nJveYzEDXgTd1Ebas4mL49FOYORMef9zUo765LJIS9ZyM8C1Q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\n",
- "text": [
- "<matplotlib.figure.Figure at 0x5cb46f0>"
- ]
- }
- ],
- "prompt_number": 70
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-13, Page no 108"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=0.6 #kg/m\n",
- "l=240 #m\n",
- "d=24 #m\n",
- "\n",
- "#Calculations\n",
- "c=((((1*4**-1)*(l**2))-(24*24)))/(2*d)\n",
- "T_max=9.8*m*(c+d) #N\n",
- "a=arcsinh((l)/(2*c))*576\n",
- "\n",
- "#Result\n",
- "print'The maximun tension is',round(T_max),\"N\"\n",
- "print'The value of a is',round(a)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximun tension is 1835.0 N\n",
- "The value of a is 234.0\n"
- ]
- }
- ],
- "prompt_number": 25
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter7_1.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter7_1.ipynb deleted file mode 100755 index 1343d7ec..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter7_1.ipynb +++ /dev/null @@ -1,668 +0,0 @@ -{
- "metadata": {
- "name": "chapter7.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 7: Trusses And Cables"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-1, Page no 99"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "F1=2000 #lb\n",
- "F2=4000 #lb\n",
- "l1=10 #ft\n",
- "l2=30 #ft\n",
- "l3=20 #ft\n",
- "l4=40 #ft\n",
- "# as t=60 degrees\n",
- "sint=sqrt(3)*2**-1\n",
- "cost=2**-1\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point B and A\n",
- "Ra=(F1*l2+F2*l1)/l4 \n",
- "Rb=(F2*l2+F1*l1)/l4\n",
- "#Consider fig 7-4(c)\n",
- "A=np.array([[1,-cost],[0,-sint]])\n",
- "B=np.array([[0],[-2500]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Consider figure 7-4(d)\n",
- "A1=np.array([[1,cost],[0,-sint]])\n",
- "B1=np.array([-C[1]*cost,-C[1]*sint+F1])\n",
- "C1=np.linalg.solve(A1,B1)\n",
- "#Consider figure 7-4(e)\n",
- "CD=577\n",
- "CE=C[0]+C1[1]*cost+CD*cost\n",
- "#Consider figure 7-4(f)\n",
- "DE=Rb/sint\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The reactions are:Ra=',round(Ra),\"lb\",'and Rb=',round(Rb),\"lb\"\n",
- "print'Force in member AB=',round(C[1]),\"lb (C)\",'and AC=',round(C[0]),\"lb (T)\"\n",
- "print'Force in member BC=',round(C1[1]),\"lb (T)\",'and BD=',round(C1[0]),\"lb (-ve sign indicates compression)\"\n",
- "print'Force in member CD=',round(CD),\"lb (C)\",'and CE=',round(CE),\"lb (T)\"\n",
- "print'Force in member DE=',round(DE),\"lb (C)\"\n",
- "\n",
- "#Decimal Accuracy causes discrepancy in answers.Thus answers wary as compared to the textbook.\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reactions are:Ra= 2500.0 lb and Rb= 3500.0 lb\n",
- "Force in member AB= 2887.0 lb (C) and AC= 1443.0 lb (T)\n",
- "Force in member BC= 577.0 lb (T) and BD= -1732.0 lb (-ve sign indicates compression)\n",
- "Force in member CD= 577.0 lb (C) and CE= 2021.0 lb (T)\n",
- "Force in member DE= 4041.0 lb (C)\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-2, Page no 101"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "s=4 #m length of sides\n",
- "l=2 #kN load acting on each node\n",
- "r=7 #kN by inspection reaction at A\n",
- "sin60=sqrt(3)*2**-1\n",
- "tan30=sqrt(3)**-1\n",
- "tan60=sqrt(3)\n",
- "\n",
- "#Calculation\n",
- "#Taking Moment about point G\n",
- "FH=(-r*12+2*10+2*6+2*2)/(2*tan60) #kN Compressive\n",
- "#Taking moment about point H\n",
- "GI=(r*14-2*12-2*8-2*4)/(2*tan30) #kN Tension\n",
- "#Summing forces in the vertical direction\n",
- "HG=-(r-(l*3))/sin60 #kN Compression\n",
- "#Taking moment about point J yields\n",
- "IK=(-2*4-2*8+r*10)/(2*tan60) #kN\n",
- "\n",
- "#Result\n",
- "print'The value of the forces in the components are as follows'\n",
- "print'FH=',round(FH,1),\"kN (C)\",',GI=',round(GI,1),\"kN (T)\",',HG=',round(HG,2),\"kN (C)\",'and IK=',round(IK,1),\"kN (T)\"\n",
- "print'The answer in the text book for GI is wrong'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of the forces in the components are as follows\n",
- "FH= -13.9 kN (C) ,GI= 43.3 kN (T) ,HG= -1.15 kN (C) and IK= 13.3 kN (T)\n",
- "The answer in the text book for GI is wrong\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-3, Page no 101"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=30 degrees,\n",
- "sin30=2**-1\n",
- "EF=40000 #lb\n",
- "l=36 #feet\n",
- "\n",
- "#Calculation\n",
- "#Taking moment about point D and setting EF=40000lbs\n",
- "P=-(EF*sin30*l)/l #lb\n",
- "\n",
- "#Result\n",
- "print'The maximum value of P is',round(P),\"lb\"\n",
- "print'The negative sign indicates the downward direction'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum value of P is -20000.0 lb\n",
- "The negative sign indicates the downward direction\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-4, Page no 102"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=12 #m\n",
- "# as theta1=30 degrees\n",
- "cos30=sqrt(3)*2**-1\n",
- "cos60=2**-1\n",
- "sin30=2**-1\n",
- "\n",
- "F1=1000 #N\n",
- "F2=2000 #N\n",
- "\n",
- "#Calculation\n",
- "FG=l*cos30 #m\n",
- "DG=(l+(l/2))/cos30 #m\n",
- "#Taking moment about point G\n",
- "A=(F1*l+F2*FG+F1*DG)/(l*3) #N\n",
- "#Summing forces in horizontal direction\n",
- "G_x=(2*F1+F2)*sin30 #N\n",
- "#Summing forces in the vertical direction\n",
- "G_y=(2*F1+F2)*cos30+F1-A #N\n",
- "#Taking moment about point C\n",
- "BD=-(A*l)/(l/2) #N\n",
- "#Taking moment about point D\n",
- "CE=(A*(l+(l/2)))/FG #N\n",
- "theta=arctan((l/2)/FG) #degrees \n",
- "#Summing forces in the vertical direction\n",
- "CD=(A+(BD*cos60))/cos(theta) #N\n",
- "\n",
- "#Result\n",
- "print'The values of the forces are as follows'\n",
- "print'A=',round(A),\"N\",',G_x=',round(G_x),\"N\",',G_y=',round(G_y),\"N\",',BD=',round(BD),\"N (C)\",',CE=',round(CE),\"N (T)\",'and CD=',round(CD),\"N(T)\"\n",
- "#Decimal Accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values of the forces are as follows\n",
- "A= 1488.0 N ,G_x= 2000.0 N ,G_y= 2976.0 N ,BD= -2976.0 N (C) ,CE= 2577.0 N (T) and CD= 0.0 N(T)\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-5, Page no 103"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "A=2000 #lb\n",
- "E=2000 #lb\n",
- "# as theta=60 degrees and theta1=30 degrees, which means:\n",
- "sin60=sqrt(3)*2**-1\n",
- "cos60=2**-1\n",
- "sin30=2**-1\n",
- "cos30=sqrt(3)*2**-1\n",
- "\n",
- "#Sign convention positive means Tension and negative means Compression\n",
- "#Taking sum of forces along x and y direction in fig7-13\n",
- "AB=-A/sin60 #lb\n",
- "AG=-AB*cos60 #lb\n",
- "#Taking sum of forces along x and y direction in fig7-14\n",
- "BG=((-AB*cos30)-1000)/(cos30) #lb\n",
- "BC=((AB*sin30)-(BG*sin30)) #lb\n",
- "#Taking sum of forces along x and y direction in fig7-15\n",
- "GC=-(BG*sin60)/sin60 #lb\n",
- "GF=AG+BG*cos60-GC*(cos60) #lb\n",
- "#By symmetry of structure\n",
- "DE=AB #lb\n",
- "FE=AG #lb\n",
- "DF=BG #lb\n",
- "CD=BC #lb\n",
- "\n",
- "#Result\n",
- "print'The forces in the truess are'\n",
- "print'AB=DE=',round(AB),\"lb (C)\",',AG=FE=',round(AG),\"lb (T)\",',BG=DF=',round(BG),\"lb (T)\",',BC=CD=',round(BC),\"lb (C)\",'and CG=CF=',round(GC),\"lb (C)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces in the truess are\n",
- "AB=DE= -2309.0 lb (C) ,AG=FE= 1155.0 lb (T) ,BG=DF= 1155.0 lb (T) ,BC=CD= -1732.0 lb (C) and CG=CF= -1155.0 lb (C)\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-6, Page no 104"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=500 #N\n",
- "A=1000 #N\n",
- "# as theta=60 degrees,\n",
- "sin60=sqrt(3)*2**-1\n",
- "l=20 #m\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point G\n",
- "R_c=(20*3*A+50*F+30*F+10*F)/40 #N\n",
- "#Returning to fig7-17\n",
- "#Taking moment about point C\n",
- "BD=(l*A+(l/2)*F)/(l*sin60) #N\n",
- "#Taking sum of forces in vertical direction\n",
- "CD=(A+F-R_c)/sin60 #N\n",
- "\n",
- "#Result\n",
- "print'The forces in the members are as follows'\n",
- "print'BD=',round(BD),\"N (T)\",'and CD=',round(CD),\"N (C).\",'Also the reaction at C is',round(R_c),\"N\"\n",
- "#Decimal accuracy causes discrepancey in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces in the members are as follows\n",
- "BD= 1443.0 N (T) and CD= -1299.0 N (C). Also the reaction at C is 2625.0 N\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-7, Page no 104"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=800 #lb/ft\n",
- "a=600 #ft\n",
- "d=40 #ft\n",
- "\n",
- "#Calculations\n",
- "T=0.5*w*a*(sqrt(1+(a**2/(16*d**2)))) #lb\n",
- "H=(w*a**2)/(8*d) #lb\n",
- "#Taking the first two terms of the series\n",
- "l=a*(1+(8/3)*(d*a**-1)**2-(32/5)*0.00002) #ft\n",
- "\n",
- "#Result\n",
- "print'The value of T=',round(T),\"lb\",'and that of H=',round(H),\"lb.\",'Also l=',round(l),\"ft\"\n",
- "#Deciaml accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of T= 929516.0 lb and that of H= 900000.0 lb. Also l= 605.0 ft\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-8,Page no 105"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=800*300 #lb\n",
- "\n",
- "#Calculations\n",
- "#Summing forces in horizontal and vertical direction\n",
- "theta=arctan(40*150**-1) #degrees\n",
- "H=l/tan(theta) #lb\n",
- "T_max=sqrt(l**2+H**2) #lb\n",
- "\n",
- "#Result\n",
- "print'The maximun tension is',round(T_max),\"lb\",'and H=',round(H),\"lb\"\n",
- "#Decimal accuracy causes discrepancy in answers"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximun tension is 931450.0 lb and H= 900000.0 lb\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-9,Page no 105"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#For simplicity a1 and a2 values are being considered as constant free of H\n",
- "a_1=sqrt(10*14.7**-1)\n",
- "a_2=sqrt(30/14.7)\n",
- "y=10 #m\n",
- "\n",
- "#Calculations\n",
- "H=(300/(a_1+a_2))**2 #N\n",
- "#Now reconsidering a1 and a2 actual values\n",
- "a1=a_1*sqrt(H) #m\n",
- "a2=a_2*sqrt(H) #m\n",
- "#Theta calculations\n",
- "x=a1\n",
- "theta=arctan(2*y/x)\n",
- "#T calculations\n",
- "T=sqrt((864*a2**2)+H**2) #N\n",
- "\n",
- "#Result\n",
- "print'The tension in the cable is',round(T,2),\"*10**-3 kN\"\n",
- "# The answer may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in the cable is 18585.57 *10**-3 kN\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-10, Page no 106"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "T=140000 #N\n",
- "w=2000 #N/m\n",
- "a=20 #m\n",
- "\n",
- "#Calculations\n",
- "#Calculation step by step\n",
- "lhs=(140000*2)*(2000*20)**-1 \n",
- "d=sqrt(1/((((lhs**2)-1)*16)/(20**2))) #m\n",
- "l=a*(1+(8/3)*(d/a)**2) #m\n",
- "\n",
- "#Result\n",
- "print'The sag in the cable is',round(d,2),\"m\"\n",
- "print'The required length is',round(l,2),\"m\"\n",
- "\n",
- "# Value of l is off by 0.2 m"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The sag in the cable is 0.72 m\n",
- "The required length is 20.05 m\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-11, Page no 106"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=10*16**-1 #lb/ft\n",
- "a=80 #ft\n",
- "P=500 #lb\n",
- "\n",
- "#Calculations\n",
- "lhs=(P*2)/(w*a)\n",
- "d=sqrt(1*((((lhs**2)-1)*16)/(80**2))**-1) #ft\n",
- "\n",
- "#Result\n",
- "print'The sag in the cable is',round(d),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The sag in the cable is 1.0 ft\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-12, Page no 107"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=0.518 #lb/ft\n",
- "d=50 #ft\n",
- "l=500 #ft\n",
- "#Plot coding\n",
- "A=linspace(0,800,9) #defined x axis\n",
- "B=A+50\n",
- "C=[50000,500*(2*100)**-1,500*(2*200)**-1,500*(2*300)**-1,500*(2*400)**-1,500*(2*500)**-1,500*(2*600)**-1,500*(2*700)**-1,500*(2*800)**-1]\n",
- "D=cosh(C)\n",
- "E=([D[0]*A[0],D[1]*A[1],D[2]*A[2],D[3]*A[3],D[4]*A[4],D[5]*A[5],D[6]*A[6],D[7]*A[7],D[8]*A[8]])\n",
- "plot(A,B,A,E) #plotting two lines on the same plot\n",
- "\n",
- "#Calculations\n",
- "#By close observation of plot taking c around 650\n",
- "#consider c=635\n",
- "c=635\n",
- "T_max=w*(c+d) #lb\n",
- "a=c+d\n",
- "l=(4*(a*a-c*c))**0.5\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The maximum tension is',round(T_max),\"lb\",'and length required is',round(l),\"ft respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum tension is 355.0 lb and length required is 514.0 ft respectively.\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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Zv4RPkz6lk10nJveYzEDXgTd1Ebas4mL49FOYORMef9zUo765LJIS9ZyM8C1Q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\n",
- "text": [
- "<matplotlib.figure.Figure at 0x5cb46f0>"
- ]
- }
- ],
- "prompt_number": 70
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-13, Page no 108"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=0.6 #kg/m\n",
- "l=240 #m\n",
- "d=24 #m\n",
- "\n",
- "#Calculations\n",
- "c=((((1*4**-1)*(l**2))-(24*24)))/(2*d)\n",
- "T_max=9.8*m*(c+d) #N\n",
- "a=arcsinh((l)/(2*c))*576\n",
- "\n",
- "#Result\n",
- "print'The maximun tension is',round(T_max),\"N\"\n",
- "print'The value of a is',round(a)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximun tension is 1835.0 N\n",
- "The value of a is 234.0\n"
- ]
- }
- ],
- "prompt_number": 25
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter7_2.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter7_2.ipynb deleted file mode 100755 index 1343d7ec..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter7_2.ipynb +++ /dev/null @@ -1,668 +0,0 @@ -{
- "metadata": {
- "name": "chapter7.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 7: Trusses And Cables"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-1, Page no 99"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "F1=2000 #lb\n",
- "F2=4000 #lb\n",
- "l1=10 #ft\n",
- "l2=30 #ft\n",
- "l3=20 #ft\n",
- "l4=40 #ft\n",
- "# as t=60 degrees\n",
- "sint=sqrt(3)*2**-1\n",
- "cost=2**-1\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point B and A\n",
- "Ra=(F1*l2+F2*l1)/l4 \n",
- "Rb=(F2*l2+F1*l1)/l4\n",
- "#Consider fig 7-4(c)\n",
- "A=np.array([[1,-cost],[0,-sint]])\n",
- "B=np.array([[0],[-2500]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Consider figure 7-4(d)\n",
- "A1=np.array([[1,cost],[0,-sint]])\n",
- "B1=np.array([-C[1]*cost,-C[1]*sint+F1])\n",
- "C1=np.linalg.solve(A1,B1)\n",
- "#Consider figure 7-4(e)\n",
- "CD=577\n",
- "CE=C[0]+C1[1]*cost+CD*cost\n",
- "#Consider figure 7-4(f)\n",
- "DE=Rb/sint\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The reactions are:Ra=',round(Ra),\"lb\",'and Rb=',round(Rb),\"lb\"\n",
- "print'Force in member AB=',round(C[1]),\"lb (C)\",'and AC=',round(C[0]),\"lb (T)\"\n",
- "print'Force in member BC=',round(C1[1]),\"lb (T)\",'and BD=',round(C1[0]),\"lb (-ve sign indicates compression)\"\n",
- "print'Force in member CD=',round(CD),\"lb (C)\",'and CE=',round(CE),\"lb (T)\"\n",
- "print'Force in member DE=',round(DE),\"lb (C)\"\n",
- "\n",
- "#Decimal Accuracy causes discrepancy in answers.Thus answers wary as compared to the textbook.\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reactions are:Ra= 2500.0 lb and Rb= 3500.0 lb\n",
- "Force in member AB= 2887.0 lb (C) and AC= 1443.0 lb (T)\n",
- "Force in member BC= 577.0 lb (T) and BD= -1732.0 lb (-ve sign indicates compression)\n",
- "Force in member CD= 577.0 lb (C) and CE= 2021.0 lb (T)\n",
- "Force in member DE= 4041.0 lb (C)\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-2, Page no 101"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "s=4 #m length of sides\n",
- "l=2 #kN load acting on each node\n",
- "r=7 #kN by inspection reaction at A\n",
- "sin60=sqrt(3)*2**-1\n",
- "tan30=sqrt(3)**-1\n",
- "tan60=sqrt(3)\n",
- "\n",
- "#Calculation\n",
- "#Taking Moment about point G\n",
- "FH=(-r*12+2*10+2*6+2*2)/(2*tan60) #kN Compressive\n",
- "#Taking moment about point H\n",
- "GI=(r*14-2*12-2*8-2*4)/(2*tan30) #kN Tension\n",
- "#Summing forces in the vertical direction\n",
- "HG=-(r-(l*3))/sin60 #kN Compression\n",
- "#Taking moment about point J yields\n",
- "IK=(-2*4-2*8+r*10)/(2*tan60) #kN\n",
- "\n",
- "#Result\n",
- "print'The value of the forces in the components are as follows'\n",
- "print'FH=',round(FH,1),\"kN (C)\",',GI=',round(GI,1),\"kN (T)\",',HG=',round(HG,2),\"kN (C)\",'and IK=',round(IK,1),\"kN (T)\"\n",
- "print'The answer in the text book for GI is wrong'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of the forces in the components are as follows\n",
- "FH= -13.9 kN (C) ,GI= 43.3 kN (T) ,HG= -1.15 kN (C) and IK= 13.3 kN (T)\n",
- "The answer in the text book for GI is wrong\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-3, Page no 101"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=30 degrees,\n",
- "sin30=2**-1\n",
- "EF=40000 #lb\n",
- "l=36 #feet\n",
- "\n",
- "#Calculation\n",
- "#Taking moment about point D and setting EF=40000lbs\n",
- "P=-(EF*sin30*l)/l #lb\n",
- "\n",
- "#Result\n",
- "print'The maximum value of P is',round(P),\"lb\"\n",
- "print'The negative sign indicates the downward direction'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum value of P is -20000.0 lb\n",
- "The negative sign indicates the downward direction\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-4, Page no 102"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=12 #m\n",
- "# as theta1=30 degrees\n",
- "cos30=sqrt(3)*2**-1\n",
- "cos60=2**-1\n",
- "sin30=2**-1\n",
- "\n",
- "F1=1000 #N\n",
- "F2=2000 #N\n",
- "\n",
- "#Calculation\n",
- "FG=l*cos30 #m\n",
- "DG=(l+(l/2))/cos30 #m\n",
- "#Taking moment about point G\n",
- "A=(F1*l+F2*FG+F1*DG)/(l*3) #N\n",
- "#Summing forces in horizontal direction\n",
- "G_x=(2*F1+F2)*sin30 #N\n",
- "#Summing forces in the vertical direction\n",
- "G_y=(2*F1+F2)*cos30+F1-A #N\n",
- "#Taking moment about point C\n",
- "BD=-(A*l)/(l/2) #N\n",
- "#Taking moment about point D\n",
- "CE=(A*(l+(l/2)))/FG #N\n",
- "theta=arctan((l/2)/FG) #degrees \n",
- "#Summing forces in the vertical direction\n",
- "CD=(A+(BD*cos60))/cos(theta) #N\n",
- "\n",
- "#Result\n",
- "print'The values of the forces are as follows'\n",
- "print'A=',round(A),\"N\",',G_x=',round(G_x),\"N\",',G_y=',round(G_y),\"N\",',BD=',round(BD),\"N (C)\",',CE=',round(CE),\"N (T)\",'and CD=',round(CD),\"N(T)\"\n",
- "#Decimal Accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values of the forces are as follows\n",
- "A= 1488.0 N ,G_x= 2000.0 N ,G_y= 2976.0 N ,BD= -2976.0 N (C) ,CE= 2577.0 N (T) and CD= 0.0 N(T)\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-5, Page no 103"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "A=2000 #lb\n",
- "E=2000 #lb\n",
- "# as theta=60 degrees and theta1=30 degrees, which means:\n",
- "sin60=sqrt(3)*2**-1\n",
- "cos60=2**-1\n",
- "sin30=2**-1\n",
- "cos30=sqrt(3)*2**-1\n",
- "\n",
- "#Sign convention positive means Tension and negative means Compression\n",
- "#Taking sum of forces along x and y direction in fig7-13\n",
- "AB=-A/sin60 #lb\n",
- "AG=-AB*cos60 #lb\n",
- "#Taking sum of forces along x and y direction in fig7-14\n",
- "BG=((-AB*cos30)-1000)/(cos30) #lb\n",
- "BC=((AB*sin30)-(BG*sin30)) #lb\n",
- "#Taking sum of forces along x and y direction in fig7-15\n",
- "GC=-(BG*sin60)/sin60 #lb\n",
- "GF=AG+BG*cos60-GC*(cos60) #lb\n",
- "#By symmetry of structure\n",
- "DE=AB #lb\n",
- "FE=AG #lb\n",
- "DF=BG #lb\n",
- "CD=BC #lb\n",
- "\n",
- "#Result\n",
- "print'The forces in the truess are'\n",
- "print'AB=DE=',round(AB),\"lb (C)\",',AG=FE=',round(AG),\"lb (T)\",',BG=DF=',round(BG),\"lb (T)\",',BC=CD=',round(BC),\"lb (C)\",'and CG=CF=',round(GC),\"lb (C)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces in the truess are\n",
- "AB=DE= -2309.0 lb (C) ,AG=FE= 1155.0 lb (T) ,BG=DF= 1155.0 lb (T) ,BC=CD= -1732.0 lb (C) and CG=CF= -1155.0 lb (C)\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-6, Page no 104"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=500 #N\n",
- "A=1000 #N\n",
- "# as theta=60 degrees,\n",
- "sin60=sqrt(3)*2**-1\n",
- "l=20 #m\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point G\n",
- "R_c=(20*3*A+50*F+30*F+10*F)/40 #N\n",
- "#Returning to fig7-17\n",
- "#Taking moment about point C\n",
- "BD=(l*A+(l/2)*F)/(l*sin60) #N\n",
- "#Taking sum of forces in vertical direction\n",
- "CD=(A+F-R_c)/sin60 #N\n",
- "\n",
- "#Result\n",
- "print'The forces in the members are as follows'\n",
- "print'BD=',round(BD),\"N (T)\",'and CD=',round(CD),\"N (C).\",'Also the reaction at C is',round(R_c),\"N\"\n",
- "#Decimal accuracy causes discrepancey in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces in the members are as follows\n",
- "BD= 1443.0 N (T) and CD= -1299.0 N (C). Also the reaction at C is 2625.0 N\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-7, Page no 104"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=800 #lb/ft\n",
- "a=600 #ft\n",
- "d=40 #ft\n",
- "\n",
- "#Calculations\n",
- "T=0.5*w*a*(sqrt(1+(a**2/(16*d**2)))) #lb\n",
- "H=(w*a**2)/(8*d) #lb\n",
- "#Taking the first two terms of the series\n",
- "l=a*(1+(8/3)*(d*a**-1)**2-(32/5)*0.00002) #ft\n",
- "\n",
- "#Result\n",
- "print'The value of T=',round(T),\"lb\",'and that of H=',round(H),\"lb.\",'Also l=',round(l),\"ft\"\n",
- "#Deciaml accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of T= 929516.0 lb and that of H= 900000.0 lb. Also l= 605.0 ft\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-8,Page no 105"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=800*300 #lb\n",
- "\n",
- "#Calculations\n",
- "#Summing forces in horizontal and vertical direction\n",
- "theta=arctan(40*150**-1) #degrees\n",
- "H=l/tan(theta) #lb\n",
- "T_max=sqrt(l**2+H**2) #lb\n",
- "\n",
- "#Result\n",
- "print'The maximun tension is',round(T_max),\"lb\",'and H=',round(H),\"lb\"\n",
- "#Decimal accuracy causes discrepancy in answers"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximun tension is 931450.0 lb and H= 900000.0 lb\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-9,Page no 105"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#For simplicity a1 and a2 values are being considered as constant free of H\n",
- "a_1=sqrt(10*14.7**-1)\n",
- "a_2=sqrt(30/14.7)\n",
- "y=10 #m\n",
- "\n",
- "#Calculations\n",
- "H=(300/(a_1+a_2))**2 #N\n",
- "#Now reconsidering a1 and a2 actual values\n",
- "a1=a_1*sqrt(H) #m\n",
- "a2=a_2*sqrt(H) #m\n",
- "#Theta calculations\n",
- "x=a1\n",
- "theta=arctan(2*y/x)\n",
- "#T calculations\n",
- "T=sqrt((864*a2**2)+H**2) #N\n",
- "\n",
- "#Result\n",
- "print'The tension in the cable is',round(T,2),\"*10**-3 kN\"\n",
- "# The answer may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in the cable is 18585.57 *10**-3 kN\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-10, Page no 106"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "T=140000 #N\n",
- "w=2000 #N/m\n",
- "a=20 #m\n",
- "\n",
- "#Calculations\n",
- "#Calculation step by step\n",
- "lhs=(140000*2)*(2000*20)**-1 \n",
- "d=sqrt(1/((((lhs**2)-1)*16)/(20**2))) #m\n",
- "l=a*(1+(8/3)*(d/a)**2) #m\n",
- "\n",
- "#Result\n",
- "print'The sag in the cable is',round(d,2),\"m\"\n",
- "print'The required length is',round(l,2),\"m\"\n",
- "\n",
- "# Value of l is off by 0.2 m"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The sag in the cable is 0.72 m\n",
- "The required length is 20.05 m\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-11, Page no 106"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=10*16**-1 #lb/ft\n",
- "a=80 #ft\n",
- "P=500 #lb\n",
- "\n",
- "#Calculations\n",
- "lhs=(P*2)/(w*a)\n",
- "d=sqrt(1*((((lhs**2)-1)*16)/(80**2))**-1) #ft\n",
- "\n",
- "#Result\n",
- "print'The sag in the cable is',round(d),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The sag in the cable is 1.0 ft\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-12, Page no 107"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=0.518 #lb/ft\n",
- "d=50 #ft\n",
- "l=500 #ft\n",
- "#Plot coding\n",
- "A=linspace(0,800,9) #defined x axis\n",
- "B=A+50\n",
- "C=[50000,500*(2*100)**-1,500*(2*200)**-1,500*(2*300)**-1,500*(2*400)**-1,500*(2*500)**-1,500*(2*600)**-1,500*(2*700)**-1,500*(2*800)**-1]\n",
- "D=cosh(C)\n",
- "E=([D[0]*A[0],D[1]*A[1],D[2]*A[2],D[3]*A[3],D[4]*A[4],D[5]*A[5],D[6]*A[6],D[7]*A[7],D[8]*A[8]])\n",
- "plot(A,B,A,E) #plotting two lines on the same plot\n",
- "\n",
- "#Calculations\n",
- "#By close observation of plot taking c around 650\n",
- "#consider c=635\n",
- "c=635\n",
- "T_max=w*(c+d) #lb\n",
- "a=c+d\n",
- "l=(4*(a*a-c*c))**0.5\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The maximum tension is',round(T_max),\"lb\",'and length required is',round(l),\"ft respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum tension is 355.0 lb and length required is 514.0 ft respectively.\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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Zv4RPkz6lk10nJveYzEDXgTd1Ebas4mL49FOYORMef9zUo765LJIS9ZyM8C1Q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\n",
- "text": [
- "<matplotlib.figure.Figure at 0x5cb46f0>"
- ]
- }
- ],
- "prompt_number": 70
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-13, Page no 108"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=0.6 #kg/m\n",
- "l=240 #m\n",
- "d=24 #m\n",
- "\n",
- "#Calculations\n",
- "c=((((1*4**-1)*(l**2))-(24*24)))/(2*d)\n",
- "T_max=9.8*m*(c+d) #N\n",
- "a=arcsinh((l)/(2*c))*576\n",
- "\n",
- "#Result\n",
- "print'The maximun tension is',round(T_max),\"N\"\n",
- "print'The value of a is',round(a)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximun tension is 1835.0 N\n",
- "The value of a is 234.0\n"
- ]
- }
- ],
- "prompt_number": 25
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter7_3.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter7_3.ipynb deleted file mode 100755 index 297dc26a..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter7_3.ipynb +++ /dev/null @@ -1,669 +0,0 @@ -{
- "metadata": {
- "name": "chapter7.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 7: Trusses And Cables"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-1, Page no 99"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "F1=2000 #lb\n",
- "F2=4000 #lb\n",
- "l1=10 #ft\n",
- "l2=30 #ft\n",
- "l3=20 #ft\n",
- "l4=40 #ft\n",
- "# as t=60 degrees\n",
- "sint=sqrt(3)*2**-1\n",
- "cost=2**-1\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point B and A\n",
- "Ra=(F1*l2+F2*l1)/l4 \n",
- "Rb=(F2*l2+F1*l1)/l4\n",
- "#Consider fig 7-4(c)\n",
- "A=np.array([[1,-cost],[0,-sint]])\n",
- "B=np.array([[0],[-2500]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Consider figure 7-4(d)\n",
- "A1=np.array([[1,cost],[0,-sint]])\n",
- "B1=np.array([-C[1]*cost,-C[1]*sint+F1])\n",
- "C1=np.linalg.solve(A1,B1)\n",
- "#Consider figure 7-4(e)\n",
- "CD=577\n",
- "CE=C[0]+C1[1]*cost+CD*cost\n",
- "#Consider figure 7-4(f)\n",
- "DE=Rb/sint\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The reactions are:Ra=',round(Ra),\"lb\",'and Rb=',round(Rb),\"lb\"\n",
- "print'Force in member AB=',round(C[1]),\"lb (C)\",'and AC=',round(C[0]),\"lb (T)\"\n",
- "print'Force in member BC=',round(C1[1]),\"lb (T)\",'and BD=',round(C1[0]),\"lb (-ve sign indicates compression)\"\n",
- "print'Force in member CD=',round(CD),\"lb (C)\",'and CE=',round(CE),\"lb (T)\"\n",
- "print'Force in member DE=',round(DE),\"lb (C)\"\n",
- "\n",
- "#Decimal Accuracy causes discrepancy in answers.Thus answers wary as compared to the textbook.\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reactions are:Ra= 2500.0 lb and Rb= 3500.0 lb\n",
- "Force in member AB= 2887.0 lb (C) and AC= 1443.0 lb (T)\n",
- "Force in member BC= 577.0 lb (T) and BD= -1732.0 lb (-ve sign indicates compression)\n",
- "Force in member CD= 577.0 lb (C) and CE= 2021.0 lb (T)\n",
- "Force in member DE= 4041.0 lb (C)\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-2, Page no 101"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "s=4 #m length of sides\n",
- "l=2 #kN load acting on each node\n",
- "r=7 #kN by inspection reaction at A\n",
- "sin60=sqrt(3)*2**-1\n",
- "tan30=sqrt(3)**-1\n",
- "tan60=sqrt(3)\n",
- "\n",
- "#Calculation\n",
- "#Taking Moment about point G\n",
- "FH=(-r*12+2*10+2*6+2*2)/(2*tan60) #kN Compressive\n",
- "#Taking moment about point H\n",
- "GI=(r*14-2*12-2*8-2*4)/(2*tan30) #kN Tension\n",
- "#Summing forces in the vertical direction\n",
- "HG=-(r-(l*3))/sin60 #kN Compression\n",
- "#Taking moment about point J yields\n",
- "IK=(-2*4-2*8+r*10)/(2*tan60) #kN\n",
- "\n",
- "#Result\n",
- "print'The value of the forces in the components are as follows'\n",
- "print'FH=',round(FH,1),\"kN (C)\",',GI=',round(GI,1),\"kN (T)\",',HG=',round(HG,2),\"kN (C)\",'and IK=',round(IK,1),\"kN (T)\"\n",
- "print'The answer in the text book for GI is wrong'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of the forces in the components are as follows\n",
- "FH= -13.9 kN (C) ,GI= 43.3 kN (T) ,HG= -1.15 kN (C) and IK= 13.3 kN (T)\n",
- "The answer in the text book for GI is wrong\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-3, Page no 101"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=30 degrees,\n",
- "sin30=2**-1\n",
- "EF=40000 #lb\n",
- "l=36 #feet\n",
- "\n",
- "#Calculation\n",
- "#Taking moment about point D and setting EF=40000lbs\n",
- "P=-(EF*sin30*l)/l #lb\n",
- "\n",
- "#Result\n",
- "print'The maximum value of P is',round(P),\"lb\"\n",
- "print'The negative sign indicates the downward direction'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum value of P is -20000.0 lb\n",
- "The negative sign indicates the downward direction\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-4, Page no 102"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=12 #m\n",
- "# as theta1=30 degrees\n",
- "cos30=sqrt(3)*2**-1\n",
- "cos60=2**-1\n",
- "sin30=2**-1\n",
- "\n",
- "F1=1000 #N\n",
- "F2=2000 #N\n",
- "\n",
- "#Calculation\n",
- "FG=l*cos30 #m\n",
- "DG=(l+(l/2))/cos30 #m\n",
- "#Taking moment about point G\n",
- "A=(F1*l+F2*FG+F1*DG)/(l*3) #N\n",
- "#Summing forces in horizontal direction\n",
- "G_x=(2*F1+F2)*sin30 #N\n",
- "#Summing forces in the vertical direction\n",
- "G_y=(2*F1+F2)*cos30+F1-A #N\n",
- "#Taking moment about point C\n",
- "BD=-(A*l)/(l/2) #N\n",
- "#Taking moment about point D\n",
- "CE=(A*(l+(l/2)))/FG #N\n",
- "theta=arctan((l/2)/FG) #degrees \n",
- "#Summing forces in the vertical direction\n",
- "CD=(A+(BD*cos60))/cos(theta) #N\n",
- "\n",
- "#Result\n",
- "print'The values of the forces are as follows'\n",
- "print'A=',round(A),\"N\",',G_x=',round(G_x),\"N\",',G_y=',round(G_y),\"N\",',BD=',round(BD),\"N (C)\",',CE=',round(CE),\"N (T)\",'and CD=',round(CD),\"N(T)\"\n",
- "#Decimal Accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values of the forces are as follows\n",
- "A= 1488.0 N ,G_x= 2000.0 N ,G_y= 2976.0 N ,BD= -2976.0 N (C) ,CE= 2577.0 N (T) and CD= 0.0 N(T)\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-5, Page no 103"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "A=2000 #lb\n",
- "E=2000 #lb\n",
- "# as theta=60 degrees and theta1=30 degrees, which means:\n",
- "sin60=sqrt(3)*2**-1\n",
- "cos60=2**-1\n",
- "sin30=2**-1\n",
- "cos30=sqrt(3)*2**-1\n",
- "\n",
- "#Sign convention positive means Tension and negative means Compression\n",
- "#Taking sum of forces along x and y direction in fig7-13\n",
- "AB=-A/sin60 #lb\n",
- "AG=-AB*cos60 #lb\n",
- "#Taking sum of forces along x and y direction in fig7-14\n",
- "BG=((-AB*cos30)-1000)/(cos30) #lb\n",
- "BC=((AB*sin30)-(BG*sin30)) #lb\n",
- "#Taking sum of forces along x and y direction in fig7-15\n",
- "GC=-(BG*sin60)/sin60 #lb\n",
- "GF=AG+BG*cos60-GC*(cos60) #lb\n",
- "#By symmetry of structure\n",
- "DE=AB #lb\n",
- "FE=AG #lb\n",
- "DF=BG #lb\n",
- "CD=BC #lb\n",
- "\n",
- "#Result\n",
- "print'The forces in the truess are'\n",
- "print'AB=DE=',round(AB),\"lb (C)\",',AG=FE=',round(AG),\"lb (T)\",',BG=DF=',round(BG),\"lb (T)\",',BC=CD=',round(BC),\"lb (C)\",'and CG=CF=',round(GC),\"lb (C)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces in the truess are\n",
- "AB=DE= -2309.0 lb (C) ,AG=FE= 1155.0 lb (T) ,BG=DF= 1155.0 lb (T) ,BC=CD= -1732.0 lb (C) and CG=CF= -1155.0 lb (C)\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-6, Page no 104"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=500 #N\n",
- "A=1000 #N\n",
- "# as theta=60 degrees,\n",
- "sin60=sqrt(3)*2**-1\n",
- "l=20 #m\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point G\n",
- "R_c=(20*3*A+50*F+30*F+10*F)/40 #N\n",
- "#Returning to fig7-17\n",
- "#Taking moment about point C\n",
- "BD=(l*A+(l/2)*F)/(l*sin60) #N\n",
- "#Taking sum of forces in vertical direction\n",
- "CD=(A+F-R_c)/sin60 #N\n",
- "\n",
- "#Result\n",
- "print'The forces in the members are as follows'\n",
- "print'BD=',round(BD),\"N (T)\",'and CD=',round(CD),\"N (C).\",'Also the reaction at C is',round(R_c),\"N\"\n",
- "#Decimal accuracy causes discrepancey in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces in the members are as follows\n",
- "BD= 1443.0 N (T) and CD= -1299.0 N (C). Also the reaction at C is 2625.0 N\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-7, Page no 104"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=800 #lb/ft\n",
- "a=600 #ft\n",
- "d=40 #ft\n",
- "\n",
- "#Calculations\n",
- "T=0.5*w*a*(sqrt(1+(a**2/(16*d**2)))) #lb\n",
- "H=(w*a**2)/(8*d) #lb\n",
- "#Taking the first two terms of the series\n",
- "l=a*(1+(8/3)*(d*a**-1)**2-(32/5)*0.00002) #ft\n",
- "\n",
- "#Result\n",
- "print'The value of T=',round(T),\"lb\",'and that of H=',round(H),\"lb.\",'Also l=',round(l),\"ft\"\n",
- "#Deciaml accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of T= 929516.0 lb and that of H= 900000.0 lb. Also l= 605.0 ft\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-8,Page no 105"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=800*300 #lb\n",
- "\n",
- "#Calculations\n",
- "#Summing forces in horizontal and vertical direction\n",
- "theta=arctan(40*150**-1) #degrees\n",
- "H=l/tan(theta) #lb\n",
- "T_max=sqrt(l**2+H**2) #lb\n",
- "\n",
- "#Result\n",
- "print'The maximun tension is',round(T_max),\"lb\",'and H=',round(H),\"lb\"\n",
- "#Decimal accuracy causes discrepancy in answers"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximun tension is 931450.0 lb and H= 900000.0 lb\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-9,Page no 105"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#For simplicity a1 and a2 values are being considered as constant free of H\n",
- "a_1=sqrt(10*14.7**-1)\n",
- "a_2=sqrt(30/14.7)\n",
- "y=10 #m\n",
- "\n",
- "#Calculations\n",
- "H=(300/(a_1+a_2))**2 #N\n",
- "#Now reconsidering a1 and a2 actual values\n",
- "a1=a_1*sqrt(H) #m\n",
- "a2=a_2*sqrt(H) #m\n",
- "#Theta calculations\n",
- "x=a1\n",
- "theta=arctan(2*y/x)\n",
- "#T calculations\n",
- "T=sqrt((864*a2**2)+H**2) #N\n",
- "\n",
- "#Result\n",
- "print'The tension in the cable is',round(T,2),\"*10**-3 kN\"\n",
- "# The answer may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in the cable is 18585.57 *10**-3 kN\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-10, Page no 106"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "T=140000 #N\n",
- "w=2000 #N/m\n",
- "a=20 #m\n",
- "\n",
- "#Calculations\n",
- "#Calculation step by step\n",
- "lhs=(140000*2)*(2000*20)**-1 \n",
- "d=sqrt(1/((((lhs**2)-1)*16)/(20**2))) #m\n",
- "l=a*(1+(8/3)*(d/a)**2) #m\n",
- "\n",
- "#Result\n",
- "print'The sag in the cable is',round(d,2),\"m\"\n",
- "print'The required length is',round(l,2),\"m\"\n",
- "\n",
- "# Value of l is off by 0.2 m"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The sag in the cable is 0.72 m\n",
- "The required length is 20.05 m\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-11, Page no 106"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=10*16**-1 #lb/ft\n",
- "a=80 #ft\n",
- "P=500 #lb\n",
- "\n",
- "#Calculations\n",
- "lhs=(P*2)/(w*a)\n",
- "d=sqrt(1*((((lhs**2)-1)*16)/(80**2))**-1) #ft\n",
- "\n",
- "#Result\n",
- "print'The sag in the cable is',round(d),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The sag in the cable is 1.0 ft\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-12, Page no 107"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "w=0.518 #lb/ft\n",
- "d=50 #ft\n",
- "l=500 #ft\n",
- "#Plot coding\n",
- "A=linspace(0,800,9) #defined x axis\n",
- "B=A+50\n",
- "C=[50000,500*(2*100)**-1,500*(2*200)**-1,500*(2*300)**-1,500*(2*400)**-1,500*(2*500)**-1,500*(2*600)**-1,500*(2*700)**-1,500*(2*800)**-1]\n",
- "D=cosh(C)\n",
- "E=([D[0]*A[0],D[1]*A[1],D[2]*A[2],D[3]*A[3],D[4]*A[4],D[5]*A[5],D[6]*A[6],D[7]*A[7],D[8]*A[8]])\n",
- "plot(A,B,A,E) #plotting two lines on the same plot\n",
- "\n",
- "#Calculations\n",
- "#By close observation of plot taking c around 650\n",
- "#consider c=635\n",
- "c=635\n",
- "T_max=w*(c+d) #lb\n",
- "a=c+d\n",
- "l=(4*(a*a-c*c))**0.5\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The maximum tension is',round(T_max),\"lb\",'and length required is',round(l),\"ft respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum tension is 355.0 lb and length required is 514.0 ft respectively.\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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Zv4RPkz6lk10nJveYzEDXgTd1Ebas4mL49FOYORMef9zUo765LJIS9ZyM8C1Q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\n",
- "text": [
- "<matplotlib.figure.Figure at 0x62f93d0>"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-13, Page no 108"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=0.6 #kg/m\n",
- "l=240 #m\n",
- "d=24 #m\n",
- "\n",
- "#Calculations\n",
- "c=((((1*4**-1)*(l**2))-(24*24)))/(2*d)\n",
- "T_max=9.8*m*(c+d) #N\n",
- "a=arcsinh((l)/(2*c))*576\n",
- "\n",
- "#Result\n",
- "print'The maximun tension is',round(T_max),\"N\"\n",
- "print'The value of a is',round(a)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximun tension is 1835.0 N\n",
- "The value of a is 234.0\n"
- ]
- }
- ],
- "prompt_number": 25
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter7_4.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter7_4.ipynb deleted file mode 100755 index 297dc26a..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter7_4.ipynb +++ /dev/null @@ -1,669 +0,0 @@ -{
- "metadata": {
- "name": "chapter7.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 7: Trusses And Cables"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-1, Page no 99"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "F1=2000 #lb\n",
- "F2=4000 #lb\n",
- "l1=10 #ft\n",
- "l2=30 #ft\n",
- "l3=20 #ft\n",
- "l4=40 #ft\n",
- "# as t=60 degrees\n",
- "sint=sqrt(3)*2**-1\n",
- "cost=2**-1\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point B and A\n",
- "Ra=(F1*l2+F2*l1)/l4 \n",
- "Rb=(F2*l2+F1*l1)/l4\n",
- "#Consider fig 7-4(c)\n",
- "A=np.array([[1,-cost],[0,-sint]])\n",
- "B=np.array([[0],[-2500]])\n",
- "C=np.linalg.solve(A,B)\n",
- "#Consider figure 7-4(d)\n",
- "A1=np.array([[1,cost],[0,-sint]])\n",
- "B1=np.array([-C[1]*cost,-C[1]*sint+F1])\n",
- "C1=np.linalg.solve(A1,B1)\n",
- "#Consider figure 7-4(e)\n",
- "CD=577\n",
- "CE=C[0]+C1[1]*cost+CD*cost\n",
- "#Consider figure 7-4(f)\n",
- "DE=Rb/sint\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The reactions are:Ra=',round(Ra),\"lb\",'and Rb=',round(Rb),\"lb\"\n",
- "print'Force in member AB=',round(C[1]),\"lb (C)\",'and AC=',round(C[0]),\"lb (T)\"\n",
- "print'Force in member BC=',round(C1[1]),\"lb (T)\",'and BD=',round(C1[0]),\"lb (-ve sign indicates compression)\"\n",
- "print'Force in member CD=',round(CD),\"lb (C)\",'and CE=',round(CE),\"lb (T)\"\n",
- "print'Force in member DE=',round(DE),\"lb (C)\"\n",
- "\n",
- "#Decimal Accuracy causes discrepancy in answers.Thus answers wary as compared to the textbook.\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reactions are:Ra= 2500.0 lb and Rb= 3500.0 lb\n",
- "Force in member AB= 2887.0 lb (C) and AC= 1443.0 lb (T)\n",
- "Force in member BC= 577.0 lb (T) and BD= -1732.0 lb (-ve sign indicates compression)\n",
- "Force in member CD= 577.0 lb (C) and CE= 2021.0 lb (T)\n",
- "Force in member DE= 4041.0 lb (C)\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-2, Page no 101"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "s=4 #m length of sides\n",
- "l=2 #kN load acting on each node\n",
- "r=7 #kN by inspection reaction at A\n",
- "sin60=sqrt(3)*2**-1\n",
- "tan30=sqrt(3)**-1\n",
- "tan60=sqrt(3)\n",
- "\n",
- "#Calculation\n",
- "#Taking Moment about point G\n",
- "FH=(-r*12+2*10+2*6+2*2)/(2*tan60) #kN Compressive\n",
- "#Taking moment about point H\n",
- "GI=(r*14-2*12-2*8-2*4)/(2*tan30) #kN Tension\n",
- "#Summing forces in the vertical direction\n",
- "HG=-(r-(l*3))/sin60 #kN Compression\n",
- "#Taking moment about point J yields\n",
- "IK=(-2*4-2*8+r*10)/(2*tan60) #kN\n",
- "\n",
- "#Result\n",
- "print'The value of the forces in the components are as follows'\n",
- "print'FH=',round(FH,1),\"kN (C)\",',GI=',round(GI,1),\"kN (T)\",',HG=',round(HG,2),\"kN (C)\",'and IK=',round(IK,1),\"kN (T)\"\n",
- "print'The answer in the text book for GI is wrong'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of the forces in the components are as follows\n",
- "FH= -13.9 kN (C) ,GI= 43.3 kN (T) ,HG= -1.15 kN (C) and IK= 13.3 kN (T)\n",
- "The answer in the text book for GI is wrong\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-3, Page no 101"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=30 degrees,\n",
- "sin30=2**-1\n",
- "EF=40000 #lb\n",
- "l=36 #feet\n",
- "\n",
- "#Calculation\n",
- "#Taking moment about point D and setting EF=40000lbs\n",
- "P=-(EF*sin30*l)/l #lb\n",
- "\n",
- "#Result\n",
- "print'The maximum value of P is',round(P),\"lb\"\n",
- "print'The negative sign indicates the downward direction'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum value of P is -20000.0 lb\n",
- "The negative sign indicates the downward direction\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-4, Page no 102"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=12 #m\n",
- "# as theta1=30 degrees\n",
- "cos30=sqrt(3)*2**-1\n",
- "cos60=2**-1\n",
- "sin30=2**-1\n",
- "\n",
- "F1=1000 #N\n",
- "F2=2000 #N\n",
- "\n",
- "#Calculation\n",
- "FG=l*cos30 #m\n",
- "DG=(l+(l/2))/cos30 #m\n",
- "#Taking moment about point G\n",
- "A=(F1*l+F2*FG+F1*DG)/(l*3) #N\n",
- "#Summing forces in horizontal direction\n",
- "G_x=(2*F1+F2)*sin30 #N\n",
- "#Summing forces in the vertical direction\n",
- "G_y=(2*F1+F2)*cos30+F1-A #N\n",
- "#Taking moment about point C\n",
- "BD=-(A*l)/(l/2) #N\n",
- "#Taking moment about point D\n",
- "CE=(A*(l+(l/2)))/FG #N\n",
- "theta=arctan((l/2)/FG) #degrees \n",
- "#Summing forces in the vertical direction\n",
- "CD=(A+(BD*cos60))/cos(theta) #N\n",
- "\n",
- "#Result\n",
- "print'The values of the forces are as follows'\n",
- "print'A=',round(A),\"N\",',G_x=',round(G_x),\"N\",',G_y=',round(G_y),\"N\",',BD=',round(BD),\"N (C)\",',CE=',round(CE),\"N (T)\",'and CD=',round(CD),\"N(T)\"\n",
- "#Decimal Accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values of the forces are as follows\n",
- "A= 1488.0 N ,G_x= 2000.0 N ,G_y= 2976.0 N ,BD= -2976.0 N (C) ,CE= 2577.0 N (T) and CD= 0.0 N(T)\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-5, Page no 103"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "A=2000 #lb\n",
- "E=2000 #lb\n",
- "# as theta=60 degrees and theta1=30 degrees, which means:\n",
- "sin60=sqrt(3)*2**-1\n",
- "cos60=2**-1\n",
- "sin30=2**-1\n",
- "cos30=sqrt(3)*2**-1\n",
- "\n",
- "#Sign convention positive means Tension and negative means Compression\n",
- "#Taking sum of forces along x and y direction in fig7-13\n",
- "AB=-A/sin60 #lb\n",
- "AG=-AB*cos60 #lb\n",
- "#Taking sum of forces along x and y direction in fig7-14\n",
- "BG=((-AB*cos30)-1000)/(cos30) #lb\n",
- "BC=((AB*sin30)-(BG*sin30)) #lb\n",
- "#Taking sum of forces along x and y direction in fig7-15\n",
- "GC=-(BG*sin60)/sin60 #lb\n",
- "GF=AG+BG*cos60-GC*(cos60) #lb\n",
- "#By symmetry of structure\n",
- "DE=AB #lb\n",
- "FE=AG #lb\n",
- "DF=BG #lb\n",
- "CD=BC #lb\n",
- "\n",
- "#Result\n",
- "print'The forces in the truess are'\n",
- "print'AB=DE=',round(AB),\"lb (C)\",',AG=FE=',round(AG),\"lb (T)\",',BG=DF=',round(BG),\"lb (T)\",',BC=CD=',round(BC),\"lb (C)\",'and CG=CF=',round(GC),\"lb (C)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces in the truess are\n",
- "AB=DE= -2309.0 lb (C) ,AG=FE= 1155.0 lb (T) ,BG=DF= 1155.0 lb (T) ,BC=CD= -1732.0 lb (C) and CG=CF= -1155.0 lb (C)\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-6, Page no 104"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=500 #N\n",
- "A=1000 #N\n",
- "# as theta=60 degrees,\n",
- "sin60=sqrt(3)*2**-1\n",
- "l=20 #m\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point G\n",
- "R_c=(20*3*A+50*F+30*F+10*F)/40 #N\n",
- "#Returning to fig7-17\n",
- "#Taking moment about point C\n",
- "BD=(l*A+(l/2)*F)/(l*sin60) #N\n",
- "#Taking sum of forces in vertical direction\n",
- "CD=(A+F-R_c)/sin60 #N\n",
- "\n",
- "#Result\n",
- "print'The forces in the members are as follows'\n",
- "print'BD=',round(BD),\"N (T)\",'and CD=',round(CD),\"N (C).\",'Also the reaction at C is',round(R_c),\"N\"\n",
- "#Decimal accuracy causes discrepancey in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces in the members are as follows\n",
- "BD= 1443.0 N (T) and CD= -1299.0 N (C). Also the reaction at C is 2625.0 N\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-7, Page no 104"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=800 #lb/ft\n",
- "a=600 #ft\n",
- "d=40 #ft\n",
- "\n",
- "#Calculations\n",
- "T=0.5*w*a*(sqrt(1+(a**2/(16*d**2)))) #lb\n",
- "H=(w*a**2)/(8*d) #lb\n",
- "#Taking the first two terms of the series\n",
- "l=a*(1+(8/3)*(d*a**-1)**2-(32/5)*0.00002) #ft\n",
- "\n",
- "#Result\n",
- "print'The value of T=',round(T),\"lb\",'and that of H=',round(H),\"lb.\",'Also l=',round(l),\"ft\"\n",
- "#Deciaml accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of T= 929516.0 lb and that of H= 900000.0 lb. Also l= 605.0 ft\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-8,Page no 105"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "l=800*300 #lb\n",
- "\n",
- "#Calculations\n",
- "#Summing forces in horizontal and vertical direction\n",
- "theta=arctan(40*150**-1) #degrees\n",
- "H=l/tan(theta) #lb\n",
- "T_max=sqrt(l**2+H**2) #lb\n",
- "\n",
- "#Result\n",
- "print'The maximun tension is',round(T_max),\"lb\",'and H=',round(H),\"lb\"\n",
- "#Decimal accuracy causes discrepancy in answers"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximun tension is 931450.0 lb and H= 900000.0 lb\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-9,Page no 105"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#For simplicity a1 and a2 values are being considered as constant free of H\n",
- "a_1=sqrt(10*14.7**-1)\n",
- "a_2=sqrt(30/14.7)\n",
- "y=10 #m\n",
- "\n",
- "#Calculations\n",
- "H=(300/(a_1+a_2))**2 #N\n",
- "#Now reconsidering a1 and a2 actual values\n",
- "a1=a_1*sqrt(H) #m\n",
- "a2=a_2*sqrt(H) #m\n",
- "#Theta calculations\n",
- "x=a1\n",
- "theta=arctan(2*y/x)\n",
- "#T calculations\n",
- "T=sqrt((864*a2**2)+H**2) #N\n",
- "\n",
- "#Result\n",
- "print'The tension in the cable is',round(T,2),\"*10**-3 kN\"\n",
- "# The answer may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension in the cable is 18585.57 *10**-3 kN\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-10, Page no 106"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "T=140000 #N\n",
- "w=2000 #N/m\n",
- "a=20 #m\n",
- "\n",
- "#Calculations\n",
- "#Calculation step by step\n",
- "lhs=(140000*2)*(2000*20)**-1 \n",
- "d=sqrt(1/((((lhs**2)-1)*16)/(20**2))) #m\n",
- "l=a*(1+(8/3)*(d/a)**2) #m\n",
- "\n",
- "#Result\n",
- "print'The sag in the cable is',round(d,2),\"m\"\n",
- "print'The required length is',round(l,2),\"m\"\n",
- "\n",
- "# Value of l is off by 0.2 m"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The sag in the cable is 0.72 m\n",
- "The required length is 20.05 m\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-11, Page no 106"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "w=10*16**-1 #lb/ft\n",
- "a=80 #ft\n",
- "P=500 #lb\n",
- "\n",
- "#Calculations\n",
- "lhs=(P*2)/(w*a)\n",
- "d=sqrt(1*((((lhs**2)-1)*16)/(80**2))**-1) #ft\n",
- "\n",
- "#Result\n",
- "print'The sag in the cable is',round(d),\"ft\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The sag in the cable is 1.0 ft\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-12, Page no 107"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "w=0.518 #lb/ft\n",
- "d=50 #ft\n",
- "l=500 #ft\n",
- "#Plot coding\n",
- "A=linspace(0,800,9) #defined x axis\n",
- "B=A+50\n",
- "C=[50000,500*(2*100)**-1,500*(2*200)**-1,500*(2*300)**-1,500*(2*400)**-1,500*(2*500)**-1,500*(2*600)**-1,500*(2*700)**-1,500*(2*800)**-1]\n",
- "D=cosh(C)\n",
- "E=([D[0]*A[0],D[1]*A[1],D[2]*A[2],D[3]*A[3],D[4]*A[4],D[5]*A[5],D[6]*A[6],D[7]*A[7],D[8]*A[8]])\n",
- "plot(A,B,A,E) #plotting two lines on the same plot\n",
- "\n",
- "#Calculations\n",
- "#By close observation of plot taking c around 650\n",
- "#consider c=635\n",
- "c=635\n",
- "T_max=w*(c+d) #lb\n",
- "a=c+d\n",
- "l=(4*(a*a-c*c))**0.5\n",
- "\n",
- "#Result\n",
- "\n",
- "print'The maximum tension is',round(T_max),\"lb\",'and length required is',round(l),\"ft respectively.\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum tension is 355.0 lb and length required is 514.0 ft respectively.\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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Zv4RPkz6lk10nJveYzEDXgTd1Ebas4mL49FOYORMef9zUo765LJIS9ZyM8C1Q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\n",
- "text": [
- "<matplotlib.figure.Figure at 0x62f93d0>"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7-13, Page no 108"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=0.6 #kg/m\n",
- "l=240 #m\n",
- "d=24 #m\n",
- "\n",
- "#Calculations\n",
- "c=((((1*4**-1)*(l**2))-(24*24)))/(2*d)\n",
- "T_max=9.8*m*(c+d) #N\n",
- "a=arcsinh((l)/(2*c))*576\n",
- "\n",
- "#Result\n",
- "print'The maximun tension is',round(T_max),\"N\"\n",
- "print'The value of a is',round(a)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximun tension is 1835.0 N\n",
- "The value of a is 234.0\n"
- ]
- }
- ],
- "prompt_number": 25
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter8.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter8.ipynb deleted file mode 100755 index 06319089..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter8.ipynb +++ /dev/null @@ -1,311 +0,0 @@ -{
- "metadata": {
- "name": "chapter8.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 8: Forces in Beams"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.8-1, Page no 118"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "R_A=100 #N\n",
- "R_B=200 #N\n",
- "\n",
- "#Calculations\n",
- "#Shear force at 2m\n",
- "V=100 #N\n",
- "#Moment at 2m\n",
- "M=R_A*2 #N.m\n",
- "\n",
- "#Result\n",
- "print'The shear force at 2m is +',round(V),\"N\"\n",
- "print'The moment at 2m is +',round(M),\"N-m\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The shear force at 2m is + 100.0 N\n",
- "The moment at 2m is + 200.0 N-m\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.8-2, Page no 118"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import matplotlib.pyplot as plt\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "#length matrix\n",
- "L1=[0,3.99,4,5.99,6] #m\n",
- "#Bending moment matrix\n",
- "B=[0,400,400,0.00001,0] #N.m\n",
- "#Shear force plotting\n",
- "#Here the left side and right side lengths are considered as close as 4 to keep up with right and left distinctions\n",
- "L=[0,3.99,4,5.99,6]\n",
- "S=[100,100,-200,-200,0]\n",
- "g=[0,0,0,0,0]\n",
- "\n",
- "#Calculations cum Result\n",
- "d=transpose(L1)\n",
- "e=transpose(S)\n",
- "plt.plot(d,B)\n",
- "xlabel('distance (m)')\n",
- "ylabel('B.M (N.m)')\n",
- "plt.show()\n",
- "plt.plot(L,e,L,g)\n",
- "xlabel('distance (m)')\n",
- "ylabel('S.F (N)')\n",
- "plt.show()\n",
- "\n",
- "print'The graphs are the solutions'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x514d530>"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x5a12d10>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The graphs are the solutions\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.8-3, Page no 119"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import matplotlib.pyplot as plt\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "w=196 #N/m\n",
- "M_app=4000 #N.m\n",
- "L=6 #m\n",
- "\n",
- "#Calculations\n",
- "#Taking Moment about Point L and equating it to 0\n",
- "R_r=(M_app+w*L*L*0.5)/(3*L) #N\n",
- "#Taking Moment about Point R and equating it to 0\n",
- "R_l= ((((2*L)+(L/2))*(w*L))-(M_app))/(3*L) #N\n",
- "#finding point of zero shear\n",
- "a=R_l*w**-1\n",
- "#defining x\n",
- "x0=[0,18]\n",
- "x=[0,0.5,1,1.5,2,2.5,3,3.5,a,4,4.5,5,5.5,6] #for 0<x<6\n",
- "x1=[6,12] #for6<x<12\n",
- "x2=[12,18] #for 12<x<18\n",
- "xv=[6,12,18] #specially for shear force\n",
- "xo=[12.001,12.002] #Straight line plot\n",
- "#Shear Force Calculations\n",
- "#Summing forces in vertical direction and equating to 0\n",
- "V1=(R_l-w*x[0],R_l-w*x[1],R_l-w*x[2],R_l-w*x[3],R_l-w*x[4],R_l-w*x[5],R_l-w*x[6],R_l-w*x[7],R_l-w*x[8],R_l-w*x[9],R_l-w*x[10],R_l-w*x[11],R_l-w*x[12],R_l-w*x[13]) #N for 0<x<6\n",
- "V2=(R_l)-(w*L) #N for 6<x<18\n",
- "#Bending Moment Calculations\n",
- "M1=(R_l*x[0]-w*x[0]**2*0.5,R_l*x[1]-w*x[1]**2*0.5,R_l*x[2]-w*x[2]**2*0.5,R_l*x[3]-w*x[3]**2*0.5,R_l*x[4]-w*x[4]**2*0.5,R_l*x[5]-w*x[5]**2*0.5,R_l*x[6]-w*x[6]**2*0.5,R_l*x[7]-w*x[7]**2*0.5,R_l*x[8]-w*x[8]**2*0.5,R_l*x[9]-w*x[9]**2*0.5,R_l*x[10]-w*x[10]**2*0.5,R_l*x[11]-w*x[11]**2*0.5,R_l*x[12]-w*x[12]**2*0.5,R_l*x[13]-w*x[13]**2*0.5) #N.m for 0<x<6\n",
- "M2=(R_l*x1[0]-((w*L)*(x1[0]-3)),R_l*x1[1]-((w*L)*(x1[1]-3))) #N.m for 6<x<12\n",
- "M3=(R_l*x2[0]-((w*L)*(x2[0]-3))+M_app,R_l*x2[1]-((w*L)*(x2[1]-3))+M_app) #N.m for 12<x<18\n",
- "Mo=[-1464.8652,2509.3333]\n",
- "#Maximum bending moment\n",
- "M_max=R_l*a*0.5 #N.m\n",
- "\n",
- "#Plotting SFD & BMD\n",
- "p=[0,a,5.99,6,11.99,12,17.99,18]\n",
- "y=[0,1467,1020,1020,-1486,2514,0,0]\n",
- "z=[0,a,5.99,6,11.99,12,17.99,18]\n",
- "b=[758,0,-418,-418,-418,-418,-418,0]\n",
- "g=[0,0,0,0,0,0,0,0]\n",
- "d=transpose(p)\n",
- "e=transpose(b)\n",
- "plt.plot(d,y)\n",
- "xlabel('distance (m)')\n",
- "ylabel('B.M (N.m)')\n",
- "plt.show()\n",
- "xlabel('distance (m)')\n",
- "ylabel('S.F (N)')\n",
- "plt.plot(z,e,z,g)\n",
- "plt.show()\n",
- "\n",
- "#Result\n",
- "print'The value of reactions are: R_l=',round(R_l),\"N\",'and R_r=',round(R_r),\"N\"\n",
- "print'The point of maximum bending moment is',round(a,2),\"meters from left support\",'and maximum bending moment is',round(M_max),\"N.m\"\n",
- "print'The bending moment and shear force diagrams have been plotted'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x5d1cd10>"
- ]
- },
- {
- "metadata": {},
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- "text": [
- "<matplotlib.figure.Figure at 0x60b29d0>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of reactions are: R_l= 757.0 N and R_r= 418.0 N\n",
- "The point of maximum bending moment is 3.86 meters from left support and maximum bending moment is 1462.0 N.m\n",
- "The bending moment and shear force diagrams have been plotted\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.8-4, Page no 121"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initlization of variables\n",
- "F1=2000 #lb\n",
- "w=100 #lb/ft\n",
- "\n",
- "#Calculations\n",
- "R_r=(-F1*5+w*14*13)/20 #lb\n",
- "R_l=(F1*25+w*14*7)/20 #lb\n",
- "#Shear Force matrix\n",
- "V=[-2000,-2000,990,990,-410,0] #lb\n",
- "#Bending Moment matrix\n",
- "B=[0,-10000,-10000,-4060,840,0]\n",
- "#Length matrix for shear force\n",
- "X_v=[0,5,5.0001,11,20.89999,20.9]\n",
- "#Length matrix for bendimg moment\n",
- "X_b=[0,4.99,5,11,19.9,20.9]\n",
- "g=[0,0,0,0,0,0]\n",
- "\n",
- "#Plotting of SFD & BMD.\n",
- "d=transpose(X_v)\n",
- "e=transpose(V)\n",
- "plt.plot(d,B)\n",
- "xlabel('distance (ft)')\n",
- "ylabel('B.M (lb.ft)')\n",
- "plt.show()\n",
- "plt.plot(X_b,e,X_b,g)\n",
- "xlabel('distance (ft)')\n",
- "ylabel('S.F (lb)')\n",
- "plt.show()\n",
- "\n",
- "#Result\n",
- "print'The bending Moment and Shear Force diagrams have been plotted'\n",
- "#Note\n",
- "#The textbook does not specify the span and hence there seems to be a disagreement between the textbook and python solution.here the values have just been plotted\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x5c00290>"
- ]
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Z2YwaNYrQ0FDCw8OJiIhg27ZtlJaWUlFRQUJCAgBjx45l9erVAKxZs4a0tDQA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\nRMRjFCoiIuIxChUREfEYhYqIiHjM/wMjUibUs6KdhgAAAABJRU5ErkJggg==\n",
- "text": [
- "<matplotlib.figure.Figure at 0x5a4e330>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The bending Moment and Shear Force diagrams have been plotted\n"
- ]
- }
- ],
- "prompt_number": 6
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter8_1.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter8_1.ipynb deleted file mode 100755 index 06319089..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter8_1.ipynb +++ /dev/null @@ -1,311 +0,0 @@ -{
- "metadata": {
- "name": "chapter8.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 8: Forces in Beams"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.8-1, Page no 118"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "R_A=100 #N\n",
- "R_B=200 #N\n",
- "\n",
- "#Calculations\n",
- "#Shear force at 2m\n",
- "V=100 #N\n",
- "#Moment at 2m\n",
- "M=R_A*2 #N.m\n",
- "\n",
- "#Result\n",
- "print'The shear force at 2m is +',round(V),\"N\"\n",
- "print'The moment at 2m is +',round(M),\"N-m\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The shear force at 2m is + 100.0 N\n",
- "The moment at 2m is + 200.0 N-m\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.8-2, Page no 118"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import matplotlib.pyplot as plt\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "#length matrix\n",
- "L1=[0,3.99,4,5.99,6] #m\n",
- "#Bending moment matrix\n",
- "B=[0,400,400,0.00001,0] #N.m\n",
- "#Shear force plotting\n",
- "#Here the left side and right side lengths are considered as close as 4 to keep up with right and left distinctions\n",
- "L=[0,3.99,4,5.99,6]\n",
- "S=[100,100,-200,-200,0]\n",
- "g=[0,0,0,0,0]\n",
- "\n",
- "#Calculations cum Result\n",
- "d=transpose(L1)\n",
- "e=transpose(S)\n",
- "plt.plot(d,B)\n",
- "xlabel('distance (m)')\n",
- "ylabel('B.M (N.m)')\n",
- "plt.show()\n",
- "plt.plot(L,e,L,g)\n",
- "xlabel('distance (m)')\n",
- "ylabel('S.F (N)')\n",
- "plt.show()\n",
- "\n",
- "print'The graphs are the solutions'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x514d530>"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x5a12d10>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The graphs are the solutions\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.8-3, Page no 119"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import matplotlib.pyplot as plt\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "w=196 #N/m\n",
- "M_app=4000 #N.m\n",
- "L=6 #m\n",
- "\n",
- "#Calculations\n",
- "#Taking Moment about Point L and equating it to 0\n",
- "R_r=(M_app+w*L*L*0.5)/(3*L) #N\n",
- "#Taking Moment about Point R and equating it to 0\n",
- "R_l= ((((2*L)+(L/2))*(w*L))-(M_app))/(3*L) #N\n",
- "#finding point of zero shear\n",
- "a=R_l*w**-1\n",
- "#defining x\n",
- "x0=[0,18]\n",
- "x=[0,0.5,1,1.5,2,2.5,3,3.5,a,4,4.5,5,5.5,6] #for 0<x<6\n",
- "x1=[6,12] #for6<x<12\n",
- "x2=[12,18] #for 12<x<18\n",
- "xv=[6,12,18] #specially for shear force\n",
- "xo=[12.001,12.002] #Straight line plot\n",
- "#Shear Force Calculations\n",
- "#Summing forces in vertical direction and equating to 0\n",
- "V1=(R_l-w*x[0],R_l-w*x[1],R_l-w*x[2],R_l-w*x[3],R_l-w*x[4],R_l-w*x[5],R_l-w*x[6],R_l-w*x[7],R_l-w*x[8],R_l-w*x[9],R_l-w*x[10],R_l-w*x[11],R_l-w*x[12],R_l-w*x[13]) #N for 0<x<6\n",
- "V2=(R_l)-(w*L) #N for 6<x<18\n",
- "#Bending Moment Calculations\n",
- "M1=(R_l*x[0]-w*x[0]**2*0.5,R_l*x[1]-w*x[1]**2*0.5,R_l*x[2]-w*x[2]**2*0.5,R_l*x[3]-w*x[3]**2*0.5,R_l*x[4]-w*x[4]**2*0.5,R_l*x[5]-w*x[5]**2*0.5,R_l*x[6]-w*x[6]**2*0.5,R_l*x[7]-w*x[7]**2*0.5,R_l*x[8]-w*x[8]**2*0.5,R_l*x[9]-w*x[9]**2*0.5,R_l*x[10]-w*x[10]**2*0.5,R_l*x[11]-w*x[11]**2*0.5,R_l*x[12]-w*x[12]**2*0.5,R_l*x[13]-w*x[13]**2*0.5) #N.m for 0<x<6\n",
- "M2=(R_l*x1[0]-((w*L)*(x1[0]-3)),R_l*x1[1]-((w*L)*(x1[1]-3))) #N.m for 6<x<12\n",
- "M3=(R_l*x2[0]-((w*L)*(x2[0]-3))+M_app,R_l*x2[1]-((w*L)*(x2[1]-3))+M_app) #N.m for 12<x<18\n",
- "Mo=[-1464.8652,2509.3333]\n",
- "#Maximum bending moment\n",
- "M_max=R_l*a*0.5 #N.m\n",
- "\n",
- "#Plotting SFD & BMD\n",
- "p=[0,a,5.99,6,11.99,12,17.99,18]\n",
- "y=[0,1467,1020,1020,-1486,2514,0,0]\n",
- "z=[0,a,5.99,6,11.99,12,17.99,18]\n",
- "b=[758,0,-418,-418,-418,-418,-418,0]\n",
- "g=[0,0,0,0,0,0,0,0]\n",
- "d=transpose(p)\n",
- "e=transpose(b)\n",
- "plt.plot(d,y)\n",
- "xlabel('distance (m)')\n",
- "ylabel('B.M (N.m)')\n",
- "plt.show()\n",
- "xlabel('distance (m)')\n",
- "ylabel('S.F (N)')\n",
- "plt.plot(z,e,z,g)\n",
- "plt.show()\n",
- "\n",
- "#Result\n",
- "print'The value of reactions are: R_l=',round(R_l),\"N\",'and R_r=',round(R_r),\"N\"\n",
- "print'The point of maximum bending moment is',round(a,2),\"meters from left support\",'and maximum bending moment is',round(M_max),\"N.m\"\n",
- "print'The bending moment and shear force diagrams have been plotted'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x5d1cd10>"
- ]
- },
- {
- "metadata": {},
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- "text": [
- "<matplotlib.figure.Figure at 0x60b29d0>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of reactions are: R_l= 757.0 N and R_r= 418.0 N\n",
- "The point of maximum bending moment is 3.86 meters from left support and maximum bending moment is 1462.0 N.m\n",
- "The bending moment and shear force diagrams have been plotted\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.8-4, Page no 121"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initlization of variables\n",
- "F1=2000 #lb\n",
- "w=100 #lb/ft\n",
- "\n",
- "#Calculations\n",
- "R_r=(-F1*5+w*14*13)/20 #lb\n",
- "R_l=(F1*25+w*14*7)/20 #lb\n",
- "#Shear Force matrix\n",
- "V=[-2000,-2000,990,990,-410,0] #lb\n",
- "#Bending Moment matrix\n",
- "B=[0,-10000,-10000,-4060,840,0]\n",
- "#Length matrix for shear force\n",
- "X_v=[0,5,5.0001,11,20.89999,20.9]\n",
- "#Length matrix for bendimg moment\n",
- "X_b=[0,4.99,5,11,19.9,20.9]\n",
- "g=[0,0,0,0,0,0]\n",
- "\n",
- "#Plotting of SFD & BMD.\n",
- "d=transpose(X_v)\n",
- "e=transpose(V)\n",
- "plt.plot(d,B)\n",
- "xlabel('distance (ft)')\n",
- "ylabel('B.M (lb.ft)')\n",
- "plt.show()\n",
- "plt.plot(X_b,e,X_b,g)\n",
- "xlabel('distance (ft)')\n",
- "ylabel('S.F (lb)')\n",
- "plt.show()\n",
- "\n",
- "#Result\n",
- "print'The bending Moment and Shear Force diagrams have been plotted'\n",
- "#Note\n",
- "#The textbook does not specify the span and hence there seems to be a disagreement between the textbook and python solution.here the values have just been plotted\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x5c00290>"
- ]
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Z2YwaNYrQ0FDCw8OJiIhg27ZtlJaWUlFRQUJCAgBjx45l9erVAKxZs4a0tDQA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\nRMRjFCoiIuIxChUREfEYhYqIiHjM/wMjUibUs6KdhgAAAABJRU5ErkJggg==\n",
- "text": [
- "<matplotlib.figure.Figure at 0x5a4e330>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The bending Moment and Shear Force diagrams have been plotted\n"
- ]
- }
- ],
- "prompt_number": 6
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter8_2.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter8_2.ipynb deleted file mode 100755 index 06319089..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter8_2.ipynb +++ /dev/null @@ -1,311 +0,0 @@ -{
- "metadata": {
- "name": "chapter8.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 8: Forces in Beams"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.8-1, Page no 118"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "R_A=100 #N\n",
- "R_B=200 #N\n",
- "\n",
- "#Calculations\n",
- "#Shear force at 2m\n",
- "V=100 #N\n",
- "#Moment at 2m\n",
- "M=R_A*2 #N.m\n",
- "\n",
- "#Result\n",
- "print'The shear force at 2m is +',round(V),\"N\"\n",
- "print'The moment at 2m is +',round(M),\"N-m\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The shear force at 2m is + 100.0 N\n",
- "The moment at 2m is + 200.0 N-m\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.8-2, Page no 118"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import matplotlib.pyplot as plt\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "#length matrix\n",
- "L1=[0,3.99,4,5.99,6] #m\n",
- "#Bending moment matrix\n",
- "B=[0,400,400,0.00001,0] #N.m\n",
- "#Shear force plotting\n",
- "#Here the left side and right side lengths are considered as close as 4 to keep up with right and left distinctions\n",
- "L=[0,3.99,4,5.99,6]\n",
- "S=[100,100,-200,-200,0]\n",
- "g=[0,0,0,0,0]\n",
- "\n",
- "#Calculations cum Result\n",
- "d=transpose(L1)\n",
- "e=transpose(S)\n",
- "plt.plot(d,B)\n",
- "xlabel('distance (m)')\n",
- "ylabel('B.M (N.m)')\n",
- "plt.show()\n",
- "plt.plot(L,e,L,g)\n",
- "xlabel('distance (m)')\n",
- "ylabel('S.F (N)')\n",
- "plt.show()\n",
- "\n",
- "print'The graphs are the solutions'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x514d530>"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x5a12d10>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The graphs are the solutions\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.8-3, Page no 119"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import matplotlib.pyplot as plt\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "w=196 #N/m\n",
- "M_app=4000 #N.m\n",
- "L=6 #m\n",
- "\n",
- "#Calculations\n",
- "#Taking Moment about Point L and equating it to 0\n",
- "R_r=(M_app+w*L*L*0.5)/(3*L) #N\n",
- "#Taking Moment about Point R and equating it to 0\n",
- "R_l= ((((2*L)+(L/2))*(w*L))-(M_app))/(3*L) #N\n",
- "#finding point of zero shear\n",
- "a=R_l*w**-1\n",
- "#defining x\n",
- "x0=[0,18]\n",
- "x=[0,0.5,1,1.5,2,2.5,3,3.5,a,4,4.5,5,5.5,6] #for 0<x<6\n",
- "x1=[6,12] #for6<x<12\n",
- "x2=[12,18] #for 12<x<18\n",
- "xv=[6,12,18] #specially for shear force\n",
- "xo=[12.001,12.002] #Straight line plot\n",
- "#Shear Force Calculations\n",
- "#Summing forces in vertical direction and equating to 0\n",
- "V1=(R_l-w*x[0],R_l-w*x[1],R_l-w*x[2],R_l-w*x[3],R_l-w*x[4],R_l-w*x[5],R_l-w*x[6],R_l-w*x[7],R_l-w*x[8],R_l-w*x[9],R_l-w*x[10],R_l-w*x[11],R_l-w*x[12],R_l-w*x[13]) #N for 0<x<6\n",
- "V2=(R_l)-(w*L) #N for 6<x<18\n",
- "#Bending Moment Calculations\n",
- "M1=(R_l*x[0]-w*x[0]**2*0.5,R_l*x[1]-w*x[1]**2*0.5,R_l*x[2]-w*x[2]**2*0.5,R_l*x[3]-w*x[3]**2*0.5,R_l*x[4]-w*x[4]**2*0.5,R_l*x[5]-w*x[5]**2*0.5,R_l*x[6]-w*x[6]**2*0.5,R_l*x[7]-w*x[7]**2*0.5,R_l*x[8]-w*x[8]**2*0.5,R_l*x[9]-w*x[9]**2*0.5,R_l*x[10]-w*x[10]**2*0.5,R_l*x[11]-w*x[11]**2*0.5,R_l*x[12]-w*x[12]**2*0.5,R_l*x[13]-w*x[13]**2*0.5) #N.m for 0<x<6\n",
- "M2=(R_l*x1[0]-((w*L)*(x1[0]-3)),R_l*x1[1]-((w*L)*(x1[1]-3))) #N.m for 6<x<12\n",
- "M3=(R_l*x2[0]-((w*L)*(x2[0]-3))+M_app,R_l*x2[1]-((w*L)*(x2[1]-3))+M_app) #N.m for 12<x<18\n",
- "Mo=[-1464.8652,2509.3333]\n",
- "#Maximum bending moment\n",
- "M_max=R_l*a*0.5 #N.m\n",
- "\n",
- "#Plotting SFD & BMD\n",
- "p=[0,a,5.99,6,11.99,12,17.99,18]\n",
- "y=[0,1467,1020,1020,-1486,2514,0,0]\n",
- "z=[0,a,5.99,6,11.99,12,17.99,18]\n",
- "b=[758,0,-418,-418,-418,-418,-418,0]\n",
- "g=[0,0,0,0,0,0,0,0]\n",
- "d=transpose(p)\n",
- "e=transpose(b)\n",
- "plt.plot(d,y)\n",
- "xlabel('distance (m)')\n",
- "ylabel('B.M (N.m)')\n",
- "plt.show()\n",
- "xlabel('distance (m)')\n",
- "ylabel('S.F (N)')\n",
- "plt.plot(z,e,z,g)\n",
- "plt.show()\n",
- "\n",
- "#Result\n",
- "print'The value of reactions are: R_l=',round(R_l),\"N\",'and R_r=',round(R_r),\"N\"\n",
- "print'The point of maximum bending moment is',round(a,2),\"meters from left support\",'and maximum bending moment is',round(M_max),\"N.m\"\n",
- "print'The bending moment and shear force diagrams have been plotted'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x5d1cd10>"
- ]
- },
- {
- "metadata": {},
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- "text": [
- "<matplotlib.figure.Figure at 0x60b29d0>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of reactions are: R_l= 757.0 N and R_r= 418.0 N\n",
- "The point of maximum bending moment is 3.86 meters from left support and maximum bending moment is 1462.0 N.m\n",
- "The bending moment and shear force diagrams have been plotted\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.8-4, Page no 121"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initlization of variables\n",
- "F1=2000 #lb\n",
- "w=100 #lb/ft\n",
- "\n",
- "#Calculations\n",
- "R_r=(-F1*5+w*14*13)/20 #lb\n",
- "R_l=(F1*25+w*14*7)/20 #lb\n",
- "#Shear Force matrix\n",
- "V=[-2000,-2000,990,990,-410,0] #lb\n",
- "#Bending Moment matrix\n",
- "B=[0,-10000,-10000,-4060,840,0]\n",
- "#Length matrix for shear force\n",
- "X_v=[0,5,5.0001,11,20.89999,20.9]\n",
- "#Length matrix for bendimg moment\n",
- "X_b=[0,4.99,5,11,19.9,20.9]\n",
- "g=[0,0,0,0,0,0]\n",
- "\n",
- "#Plotting of SFD & BMD.\n",
- "d=transpose(X_v)\n",
- "e=transpose(V)\n",
- "plt.plot(d,B)\n",
- "xlabel('distance (ft)')\n",
- "ylabel('B.M (lb.ft)')\n",
- "plt.show()\n",
- "plt.plot(X_b,e,X_b,g)\n",
- "xlabel('distance (ft)')\n",
- "ylabel('S.F (lb)')\n",
- "plt.show()\n",
- "\n",
- "#Result\n",
- "print'The bending Moment and Shear Force diagrams have been plotted'\n",
- "#Note\n",
- "#The textbook does not specify the span and hence there seems to be a disagreement between the textbook and python solution.here the values have just been plotted\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x5c00290>"
- ]
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Z2YwaNYrQ0FDCw8OJiIhg27ZtlJaWUlFRQUJCAgBjx45l9erVAKxZs4a0tDQA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\nRMRjFCoiIuIxChUREfEYhYqIiHjM/wMjUibUs6KdhgAAAABJRU5ErkJggg==\n",
- "text": [
- "<matplotlib.figure.Figure at 0x5a4e330>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The bending Moment and Shear Force diagrams have been plotted\n"
- ]
- }
- ],
- "prompt_number": 6
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter8_3.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter8_3.ipynb deleted file mode 100755 index 05d402a2..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter8_3.ipynb +++ /dev/null @@ -1,320 +0,0 @@ -{
- "metadata": {
- "name": "chapter8.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 8: Forces in Beams"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.8-1, Page no 118"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "R_A=100 #N\n",
- "R_B=200 #N\n",
- "\n",
- "#Calculations\n",
- "#Shear force at 2m\n",
- "V=100 #N\n",
- "#Moment at 2m\n",
- "M=R_A*2 #N.m\n",
- "\n",
- "#Result\n",
- "print'The shear force at 2m is +',round(V),\"N\"\n",
- "print'The moment at 2m is +',round(M),\"N-m\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The shear force at 2m is + 100.0 N\n",
- "The moment at 2m is + 200.0 N-m\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.8-2, Page no 118"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import matplotlib.pyplot as plt\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "#length matrix\n",
- "L1=[0,3.99,4,5.99,6] #m\n",
- "#Bending moment matrix\n",
- "B=[0,400,400,0.00001,0] #N.m\n",
- "#Shear force plotting\n",
- "#Here the left side and right side lengths are considered as close as 4 to keep up with right and left distinctions\n",
- "L=[0,3.99,4,5.99,6]\n",
- "S=[100,100,-200,-200,0]\n",
- "g=[0,0,0,0,0]\n",
- "\n",
- "#Calculations cum Result\n",
- "d=transpose(L1)\n",
- "e=transpose(S)\n",
- "plt.plot(d,B)\n",
- "xlabel('Span (m)')\n",
- "ylabel('B.M (N.m)')\n",
- "plt.show()\n",
- "plt.plot(L,e,L,g)\n",
- "xlabel('Span (m)')\n",
- "ylabel('S.F (N)')\n",
- "plt.show()\n",
- "\n",
- "print'The graphs are the solutions'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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2nxUWFhIYGFjrMaoHgmi4uXPhnXfUMOjQQe9qhHBfTz8N118Pf/87tGmjdzU1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\n9m7zt1vfm9aE6yovL2fkyJH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- "text": [
- "<matplotlib.figure.Figure at 0x50dd510>"
- ]
- },
- {
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- "output_type": "display_data",
- "png": 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BAAgIgQEACAiBAQAICIEBAAjI/wHIbwwoarZYdwAAAABJRU5ErkJggg==\n",
- "text": [
- "<matplotlib.figure.Figure at 0x58cf070>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The graphs are the solutions\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.8-3, Page no 119"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import matplotlib.pyplot as plt\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "w=196 #N/m\n",
- "M_app=4000 #N.m\n",
- "L=6 #m\n",
- "\n",
- "#Calculations\n",
- "#Taking Moment about Point L and equating it to 0\n",
- "R_r=(M_app+w*L*L*0.5)/(3*L) #N\n",
- "#Taking Moment about Point R and equating it to 0\n",
- "R_l= ((((2*L)+(L/2))*(w*L))-(M_app))/(3*L) #N\n",
- "#finding point of zero shear\n",
- "a=R_l*w**-1\n",
- "#defining x\n",
- "x0=[0,18]\n",
- "x=[0,0.5,1,1.5,2,2.5,3,3.5,a,4,4.5,5,5.5,6] #for 0<x<6\n",
- "x1=[6,12] #for6<x<12\n",
- "x2=[12,18] #for 12<x<18\n",
- "xv=[6,12,18] #specially for shear force\n",
- "xo=[12.001,12.002] #Straight line plot\n",
- "#Shear Force Calculations\n",
- "#Summing forces in vertical direction and equating to 0\n",
- "V1=(R_l-w*x[0],R_l-w*x[1],R_l-w*x[2],R_l-w*x[3],R_l-w*x[4],R_l-w*x[5],R_l-w*x[6],R_l-w*x[7],R_l-w*x[8],R_l-w*x[9],R_l-w*x[10],R_l-w*x[11],R_l-w*x[12],R_l-w*x[13]) #N for 0<x<6\n",
- "V2=(R_l)-(w*L) #N for 6<x<18\n",
- "#Bending Moment Calculations\n",
- "M1=(R_l*x[0]-w*x[0]**2*0.5,R_l*x[1]-w*x[1]**2*0.5,R_l*x[2]-w*x[2]**2*0.5,R_l*x[3]-w*x[3]**2*0.5,R_l*x[4]-w*x[4]**2*0.5,R_l*x[5]-w*x[5]**2*0.5,R_l*x[6]-w*x[6]**2*0.5,R_l*x[7]-w*x[7]**2*0.5,R_l*x[8]-w*x[8]**2*0.5,R_l*x[9]-w*x[9]**2*0.5,R_l*x[10]-w*x[10]**2*0.5,R_l*x[11]-w*x[11]**2*0.5,R_l*x[12]-w*x[12]**2*0.5,R_l*x[13]-w*x[13]**2*0.5) #N.m for 0<x<6\n",
- "M2=(R_l*x1[0]-((w*L)*(x1[0]-3)),R_l*x1[1]-((w*L)*(x1[1]-3))) #N.m for 6<x<12\n",
- "M3=(R_l*x2[0]-((w*L)*(x2[0]-3))+M_app,R_l*x2[1]-((w*L)*(x2[1]-3))+M_app) #N.m for 12<x<18\n",
- "Mo=[-1464.8652,2509.3333]\n",
- "#Maximum bending moment\n",
- "M_max=R_l*a*0.5 #N.m\n",
- "\n",
- "#Plotting SFD & BMD\n",
- "p=[0,a,5.99,6,11.99,12,17.99,18]\n",
- "y=[0,1467,1020,1020,-1486,2514,0,0]\n",
- "z=[0,a,5.99,6,11.99,12,17.99,18]\n",
- "b=[758,0,-418,-418,-418,-418,-418,0]\n",
- "g=[0,0,0,0,0,0,0,0]\n",
- "d=transpose(p)\n",
- "e=transpose(b)\n",
- "plt.plot(d,y,d,g)\n",
- "xlabel('Span (m)')\n",
- "ylabel('B.M (N.m)')\n",
- "plt.show()\n",
- "xlabel('Span (m)')\n",
- "ylabel('S.F (N)')\n",
- "plt.plot(z,e,z,g)\n",
- "plt.show()\n",
- "\n",
- "#Result\n",
- "print'The value of reactions are: R_l=',round(R_l),\"N\",'and R_r=',round(R_r),\"N\"\n",
- "print'The point of maximum bending moment is',round(a,2),\"meters from left support\",'and maximum bending moment is',round(M_max),\"N.m\"\n",
- "print'The bending moment and shear force diagrams have been plotted'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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/CzF7jf4qKYHISLUU+2OP2fzpRB08+SRs2gTbtsH11+udxrXt3w/jxsHXX+ud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\nGEiPHj04fPiw3V6HEM6qY0f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- "text": [
- "<matplotlib.figure.Figure at 0x58e3370>"
- ]
- },
- {
- "metadata": {},
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- "text": [
- "<matplotlib.figure.Figure at 0x5b14af0>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of reactions are: R_l= 757.0 N and R_r= 418.0 N\n",
- "The point of maximum bending moment is 3.86 meters from left support and maximum bending moment is 1462.0 N.m\n",
- "The bending moment and shear force diagrams have been plotted\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.8-4, Page no 121"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initlization of variables\n",
- "F1=2000 #lb\n",
- "w=100 #lb/ft\n",
- "\n",
- "#Calculations\n",
- "R_r=(-F1*5+w*14*13)/20 #lb\n",
- "R_l=(F1*25+w*14*7)/20 #lb\n",
- "#Shear Force matrix\n",
- "V=[-2000,-2000,990,990,-410,0] #lb\n",
- "#Bending Moment matrix\n",
- "B=[0,-10000,-10000,-4060,840,0]\n",
- "#Length matrix for shear force\n",
- "X_v=[0,5,5.0001,11,20.89999,20.9]\n",
- "#Length matrix for bendimg moment\n",
- "X_b=[0,4.99,5,11,19.9,20.9]\n",
- "g=[0,0,0,0,0,0]\n",
- "\n",
- "#Plotting of SFD & BMD.\n",
- "d=transpose(X_v)\n",
- "e=transpose(V)\n",
- "plt.plot(d,B,d,g)\n",
- "xlabel('Span (ft)')\n",
- "ylabel('B.M (lb.ft)')\n",
- "plt.show()\n",
- "plt.plot(X_b,e,X_b,g)\n",
- "xlabel('Span (ft)')\n",
- "ylabel('S.F (lb)')\n",
- "plt.show()\n",
- "\n",
- "#Result\n",
- "print'The bending Moment and Shear Force diagrams have been plotted'\n",
- "#Note\n",
- "#The textbook does not specify the span and hence there seems to be a disagreement between the textbook and python solution.here the values have just been plotted\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x5a0a670>"
- ]
- },
- {
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- "text": [
- "<matplotlib.figure.Figure at 0x5a2a550>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The bending Moment and Shear Force diagrams have been plotted\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [],
- "language": "python",
- "metadata": {},
- "outputs": []
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter8_4.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter8_4.ipynb deleted file mode 100755 index fe432847..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter8_4.ipynb +++ /dev/null @@ -1,312 +0,0 @@ -{
- "metadata": {
- "name": "chapter8.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 8: Forces in Beams"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.8-1, Page no 118"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "R_A=100 #N\n",
- "R_B=200 #N\n",
- "\n",
- "#Calculations\n",
- "#Shear force at 2m\n",
- "V=100 #N\n",
- "#Moment at 2m\n",
- "M=R_A*2 #N.m\n",
- "\n",
- "#Result\n",
- "print'The shear force at 2m is +',round(V),\"N\"\n",
- "print'The moment at 2m is +',round(M),\"N-m\" \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The shear force at 2m is + 100.0 N\n",
- "The moment at 2m is + 200.0 N-m\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.8-2, Page no 118"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import matplotlib.pyplot as plt\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "#length matrix\n",
- "L1=[0,3.99,4,5.99,6] #m\n",
- "#Bending moment matrix\n",
- "B=[0,400,400,0.00001,0] #N.m\n",
- "#Shear force plotting\n",
- "#Here the left side and right side lengths are considered as close as 4 to keep up with right and left distinctions\n",
- "L=[0,3.99,4,5.99,6]\n",
- "S=[100,100,-200,-200,0]\n",
- "g=[0,0,0,0,0]\n",
- "\n",
- "#Calculations cum Result\n",
- "d=transpose(L1)\n",
- "e=transpose(S)\n",
- "plt.plot(d,B)\n",
- "xlabel('Span (m)')\n",
- "ylabel('B.M (N.m)')\n",
- "plt.show()\n",
- "plt.plot(L,e,L,g)\n",
- "xlabel('Span (m)')\n",
- "ylabel('S.F (N)')\n",
- "plt.show()\n",
- "\n",
- "print'The graphs are the solutions'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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2nxUWFhIYGFjrMaoHgmi4uXPhnXfUMOjQQe9qhHBfTz8N118Pf/87tGmjdzU1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\n9m7zt1vfm9aE6yovL2fkyJH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- "text": [
- "<matplotlib.figure.Figure at 0x50dd510>"
- ]
- },
- {
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- "output_type": "display_data",
- "png": 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BAAgIgQEACAiBAQAICIEBAAjI/wHIbwwoarZYdwAAAABJRU5ErkJggg==\n",
- "text": [
- "<matplotlib.figure.Figure at 0x58cf070>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The graphs are the solutions\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.8-3, Page no 119"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import matplotlib.pyplot as plt\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "w=196 #N/m\n",
- "M_app=4000 #N.m\n",
- "L=6 #m\n",
- "\n",
- "#Calculations\n",
- "#Taking Moment about Point L and equating it to 0\n",
- "R_r=(M_app+w*L*L*0.5)/(3*L) #N\n",
- "#Taking Moment about Point R and equating it to 0\n",
- "R_l= ((((2*L)+(L/2))*(w*L))-(M_app))/(3*L) #N\n",
- "#finding point of zero shear\n",
- "a=R_l*w**-1\n",
- "#defining x\n",
- "x0=[0,18]\n",
- "x=[0,0.5,1,1.5,2,2.5,3,3.5,a,4,4.5,5,5.5,6] #for 0<x<6\n",
- "x1=[6,12] #for6<x<12\n",
- "x2=[12,18] #for 12<x<18\n",
- "xv=[6,12,18] #specially for shear force\n",
- "xo=[12.001,12.002] #Straight line plot\n",
- "#Shear Force Calculations\n",
- "#Summing forces in vertical direction and equating to 0\n",
- "V1=(R_l-w*x[0],R_l-w*x[1],R_l-w*x[2],R_l-w*x[3],R_l-w*x[4],R_l-w*x[5],R_l-w*x[6],R_l-w*x[7],R_l-w*x[8],R_l-w*x[9],R_l-w*x[10],R_l-w*x[11],R_l-w*x[12],R_l-w*x[13]) #N for 0<x<6\n",
- "V2=(R_l)-(w*L) #N for 6<x<18\n",
- "#Bending Moment Calculations\n",
- "M1=(R_l*x[0]-w*x[0]**2*0.5,R_l*x[1]-w*x[1]**2*0.5,R_l*x[2]-w*x[2]**2*0.5,R_l*x[3]-w*x[3]**2*0.5,R_l*x[4]-w*x[4]**2*0.5,R_l*x[5]-w*x[5]**2*0.5,R_l*x[6]-w*x[6]**2*0.5,R_l*x[7]-w*x[7]**2*0.5,R_l*x[8]-w*x[8]**2*0.5,R_l*x[9]-w*x[9]**2*0.5,R_l*x[10]-w*x[10]**2*0.5,R_l*x[11]-w*x[11]**2*0.5,R_l*x[12]-w*x[12]**2*0.5,R_l*x[13]-w*x[13]**2*0.5) #N.m for 0<x<6\n",
- "M2=(R_l*x1[0]-((w*L)*(x1[0]-3)),R_l*x1[1]-((w*L)*(x1[1]-3))) #N.m for 6<x<12\n",
- "M3=(R_l*x2[0]-((w*L)*(x2[0]-3))+M_app,R_l*x2[1]-((w*L)*(x2[1]-3))+M_app) #N.m for 12<x<18\n",
- "Mo=[-1464.8652,2509.3333]\n",
- "#Maximum bending moment\n",
- "M_max=R_l*a*0.5 #N.m\n",
- "\n",
- "#Plotting SFD & BMD\n",
- "p=[0,a,5.99,6,11.99,12,17.99,18]\n",
- "y=[0,1467,1020,1020,-1486,2514,0,0]\n",
- "z=[0,a,5.99,6,11.99,12,17.99,18]\n",
- "b=[758,0,-418,-418,-418,-418,-418,0]\n",
- "g=[0,0,0,0,0,0,0,0]\n",
- "d=transpose(p)\n",
- "e=transpose(b)\n",
- "plt.plot(d,y,d,g)\n",
- "xlabel('Span (m)')\n",
- "ylabel('B.M (N.m)')\n",
- "plt.show()\n",
- "xlabel('Span (m)')\n",
- "ylabel('S.F (N)')\n",
- "plt.plot(z,e,z,g)\n",
- "plt.show()\n",
- "\n",
- "#Result\n",
- "print'The value of reactions are: R_l=',round(R_l),\"N\",'and R_r=',round(R_r),\"N\"\n",
- "print'The point of maximum bending moment is',round(a,2),\"meters from left support\",'and maximum bending moment is',round(M_max),\"N.m\"\n",
- "print'The bending moment and shear force diagrams have been plotted'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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/CzF7jf4qKYHISLUU+2OP2fzpRB08+SRs2gTbtsH11+udxrXt3w/jxsHXX+ud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\nGEiPHj04fPiw3V6HEM6qY0f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- "text": [
- "<matplotlib.figure.Figure at 0x58e3370>"
- ]
- },
- {
- "metadata": {},
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- "text": [
- "<matplotlib.figure.Figure at 0x5b14af0>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of reactions are: R_l= 757.0 N and R_r= 418.0 N\n",
- "The point of maximum bending moment is 3.86 meters from left support and maximum bending moment is 1462.0 N.m\n",
- "The bending moment and shear force diagrams have been plotted\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.8-4, Page no 121"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initlization of variables\n",
- "F1=2000 #lb\n",
- "w=100 #lb/ft\n",
- "\n",
- "#Calculations\n",
- "R_r=(-F1*5+w*14*13)/20 #lb\n",
- "R_l=(F1*25+w*14*7)/20 #lb\n",
- "#Shear Force matrix\n",
- "V=[-2000,-2000,990,990,-410,0] #lb\n",
- "#Bending Moment matrix\n",
- "B=[0,-10000,-10000,-4060,840,0]\n",
- "#Length matrix for shear force\n",
- "X_v=[0,5,5.0001,11,20.89999,20.9]\n",
- "#Length matrix for bendimg moment\n",
- "X_b=[0,4.99,5,11,19.9,20.9]\n",
- "g=[0,0,0,0,0,0]\n",
- "\n",
- "#Plotting of SFD & BMD.\n",
- "d=transpose(X_v)\n",
- "e=transpose(V)\n",
- "plt.plot(d,B,d,g)\n",
- "xlabel('Span (ft)')\n",
- "ylabel('B.M (lb.ft)')\n",
- "plt.show()\n",
- "plt.plot(X_b,e,X_b,g)\n",
- "xlabel('Span (ft)')\n",
- "ylabel('S.F (lb)')\n",
- "plt.show()\n",
- "\n",
- "#Result\n",
- "print'The bending Moment and Shear Force diagrams have been plotted'\n",
- "#Note\n",
- "#The textbook does not specify the span and hence there seems to be a disagreement between the textbook and python solution.here the values have just been plotted\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x5a0a670>"
- ]
- },
- {
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ZGUlUVBR5eXmUlpZSXl5OYmIiAKNHj2b16tUArFmzhvT0dAAGDx7M5s2bLVij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\ngg==\n",
- "text": [
- "<matplotlib.figure.Figure at 0x5a2a550>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The bending Moment and Shear Force diagrams have been plotted\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter9.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter9.ipynb deleted file mode 100755 index b8787d4f..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter9.ipynb +++ /dev/null @@ -1,889 +0,0 @@ -{
- "metadata": {
- "name": "chapter9.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 9: Friction"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-1, Page no 130"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Calculations\n",
- "#Simplifying equation (3) after substituting value of Nb in it we get\n",
- "#m_u**2+m_u*2*tan(50)-1=0\n",
- "#Solution of the equation\n",
- "a=1\n",
- "b=2*1.19175 # here 1.19175 is value of tan(50)\n",
- "c=-1\n",
- "g=(b**2-(4*a*c))**0.5\n",
- "\n",
- "#solution\n",
- "x1=(-b+g)/(2*a)\n",
- "x2=(-b-g)/(2*a)\n",
- "#As x2 does not make any physical sense x1 is the answer\n",
- "\n",
- "#Result\n",
- "print'The value of mu is',round(x1,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of mu is 0.36\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-3, Page no 131"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=70 #kg\n",
- "g=9.81 #m/s**2\n",
- "# as theta=20 degrees, we have\n",
- "sintheta=0.3420\n",
- "costheta=0.9396\n",
- "\n",
- "#Calculations\n",
- "#Solving by martix method\n",
- "#Taking sum along vertical and horizontal direction and equating them to zero\n",
- "A=np.array([[sintheta,1,0],[-costheta,0,1],[0,-4**-1,1]])\n",
- "#RHS matrix\n",
- "R=np.array([[m*g],[0],[0]])\n",
- "ans1=np.linalg.solve(A,R) #force vector N\n",
- "#Calculation part 2\n",
- "#Similar solution by matrix method\n",
- "#Taking moment about point O and summing forces in horizontal and vertical direction and equating all to zero\n",
- "B=np.array([[4*costheta,0,0],[-costheta,1,0],[sintheta,0,1]])\n",
- "#RHS matrix\n",
- "J=np.array([[m*g*1.5],[0],[m*g]])\n",
- "ans2=np.linalg.solve(B,J) #force Vector N\n",
- "\n",
- "#Result\n",
- "print'The value of P in first case is',round(ans1[0]),\"N\",'and that in second case is',round(ans2[0]),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P in first case is 167.0 N and that in second case is 274.0 N\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-4, Page no 132"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=200 #lb\n",
- "Fapp=300 #lb\n",
- "mu=0.3 #coefficient of friction\n",
- "# as theta=30 degrees, we have\n",
- "sintheta=2**-1\n",
- "costheta=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Summing forces in the plane parallel to the slope\n",
- "F=-(W*sintheta-Fapp*costheta) #lb\n",
- "N1=(W*costheta+Fapp*sintheta) #lb\n",
- "#Max value obtained\n",
- "Fprime= mu*N1\n",
- "\n",
- "#Result\n",
- "print'The value of F is',round(F),\"lb\"\n",
- "print'The value of N1 is',round(N1),\"lb\"\n",
- "print'The value of Fprime is',round(Fprime),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of F is 160.0 lb\n",
- "The value of N1 is 323.0 lb\n",
- "The value of Fprime is 97.0 lb\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-5, Page no 132"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mu1=0.2 #coefficient of friction between wedges and A\n",
- "mu2=4**-1 #coefficient of friction between wedges \n",
- "F=20 #tonnes\n",
- "\n",
- "#Calculations\n",
- "#Using the matrix method to solve\n",
- "#Summing forces in vertical and horizontal direction\n",
- "A=np.array([[1,-(mu1*10+1)/(101)**0.5],[0,(10-mu1*1)/101**0.5]]) #force matrix\n",
- "B=np.array([[mu2*F*1000],[F*1000]]) #lb\n",
- "#Solving both matrices\n",
- "R=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The forces N2 and P are:',round(R[1]),\"lb\",'and',round(R[0]),\"lb\"\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces N2 and P are: 20510.0 lb and 11122.0 lb\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-6, Page no 133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=45 degrees,we have\n",
- "sintheta=sqrt(2)**-1\n",
- "costheta=sqrt(2)**-1\n",
- "mu1=4**-1 #coefficient of friction between A and B\n",
- "mu2=3**-1 #coefficient of friction between A and Floor\n",
- "ma=14 #kg\n",
- "mb=9 #kg\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Summing forces in vertical direction\n",
- "Nb=mb*g #N\n",
- "#Also\n",
- "Fprimeb=mu1*Nb #N\n",
- "#Summing forces in direction\n",
- "T=Fprimeb #N\n",
- "#Considering the fig(c)\n",
- "#Summing forces in the horizontal direction and vertical direction and solving by matrix method \n",
- "A=np.array([[-costheta,mu2],[sintheta,1]]) #N\n",
- "B=np.array([[-Fprimeb],[(mb*g+ma*g)]]) #N\n",
- "R=np.linalg.solve(A,B) #N\n",
- "\n",
- "#Result\n",
- "print'The value of P and Na are:',round(R[0]),\"N\",'and',round(R[1]),\"N respectively.\"\n",
- "\n",
- "# The ans may wary due o decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P and Na are: 103.0 N and 153.0 N respectively.\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-7, Page no 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=40 #kg\n",
- "m2=13.5 #kg\n",
- "mu=3**-1 #coefficient of friction\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Solving by substitution\n",
- "#After simplification we get\n",
- "x=mu*m2*g\n",
- "y=mu*(m1*g+m2*g)\n",
- "theta=arctan((x+y)/(m1*g))*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The value of the angle is',round(theta,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of the angle is 29.2 degrees\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-8, Page no 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=350 #lb\n",
- "# as theta=30 degrees, we have\n",
- "sintheta=2**-1\n",
- "costheta=sqrt(3)*2**-1\n",
- "# and phi=15 degrees,thus\n",
- "sinphi=0.2588\n",
- "cosphi=0.9659\n",
- "\n",
- "#Calculations\n",
- "#Solving by the matrix method\n",
- "A=np.array([[costheta,sinphi],[-sintheta,cosphi]])\n",
- "B=np.array([[W*sintheta],[W*costheta]])\n",
- "an=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The value of P and R are:',round(an[0],1),\"lb\",'and',round(an[1],1),\"lb respectively.\"\n",
- "\n",
- "# The ans may wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P and R are: 93.8 lb and 362.4 lb respectively.\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-9, Page no 135"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=45 degrees, we have\n",
- "sintheta=sqrt(2)**-1\n",
- "costheta=sqrt(2)**-1\n",
- "m1=45 #kg\n",
- "m2=135 #kg\n",
- "g=9.81 #m/s**2\n",
- "mu=0.25 #coefficient of riction\n",
- "\n",
- "#Calculations\n",
- "N2=m2*g #N\n",
- "T=mu*N2 #N\n",
- "N1=m1*g*costheta #N\n",
- "Fprime1=N1*mu #N\n",
- "P=T+Fprime1-(m1*g*sintheta) #N\n",
- "\n",
- "#Result\n",
- "print'The values are'\n",
- "print'N2=',round(N2),\"N\"\n",
- "print'T=',round(T),\"N\"\n",
- "print'N1=',round(N1),\"N\"\n",
- "print'Fprime1=',round(Fprime1),\"N\"\n",
- "print'P=',round(P,1),\"N\"\n",
- "\n",
- "# The ans may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are\n",
- "N2= 1324.0 N\n",
- "T= 331.0 N\n",
- "N1= 312.0 N\n",
- "Fprime1= 78.0 N\n",
- "P= 97.0 N\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-10, Page no 135"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "mu=0.2 #coefficient of friction\n",
- "F1=150 #lb\n",
- "F2=100 #lb\n",
- "# as theta=60 degrees\n",
- "costheta=2**-1\n",
- "# also theta1=30 degrees\n",
- "costheta1=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "N1=F1*costheta #lb\n",
- "T=(mu*N1)+(F1*(costheta1)) #lb considering positive\n",
- "#Equilibrium for 100lb\n",
- "#Eliminating N2 from both equations\n",
- "#Taking derivative we get\n",
- "theta2=arctan(mu)*(180/pi) #degrees\n",
- "#Hence P becomes\n",
- "# in calculation of P we use the values of sin(theta2) & cos(theta2) as,\n",
- "sintheta2=0.196\n",
- "costheta2=0.98\n",
- "P=(F2*mu+T)*(costheta2+(mu*sintheta2))**-1 #lb\n",
- "\n",
- "#Result\n",
- "print'The minimum value of P is',round(P),\"lb\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The minimum value of P is 162.0 lb\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-11, Page no 136"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=180 #N\n",
- "m=100 #kg\n",
- "g=9.81 #m/s**2\n",
- "mu=0.25 #coeffiecient of friction\n",
- "\n",
- "#Calculations\n",
- "#Assuming F2 is maximum\n",
- "N2=F*2/(1+mu) #N\n",
- "F2=mu*N2 #N\n",
- "N1=m*g-F2 #N\n",
- "F1=F-F2 #N\n",
- "\n",
- "#Result\n",
- "print'The vaules are'\n",
- "print'N2=',round(N2,3),\"N\"\n",
- "print'F2=',round(F2,3),\"N\"\n",
- "print'N1=',round(N1,3),\"N\"\n",
- "print'F1=',round(F1,3),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The vaules are\n",
- "N2= 288.0 N\n",
- "F2= 72.0 N\n",
- "N1= 909.0 N\n",
- "F1= 108.0 N\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-13, Page no 138"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "mu_ca=0.3 #ceofficient of friction between copper block A and aluminium block B\n",
- "mu_af=0.2 #coefficient of friction between aluminium block B and Floor\n",
- "ma=3 #kg\n",
- "mb=2 #kg\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#For A\n",
- "#Taking sum of forces along X and Y direction\n",
- "Na=ma*g #N\n",
- "P=mu_ca*Na #N\n",
- "#For B\n",
- "#Taking sum of forces along X and Y direction\n",
- "Nb=Na+mb*g #N\n",
- "Fb=mu_ca*Na #N\n",
- "#Now largest value of friction before slip is \n",
- "Fprimeb=mu_af*Nb #N\n",
- "#Now as Fb<F'b hence initial assumption is incorrect and P=Fb\n",
- "P=Fb #N\n",
- "\n",
- "#Result\n",
- "print'The value of force that will cause motion is',round(P,2),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of force that will cause motion is 8.83 N\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-15, Page no 141"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d_m=2 #in mean diameter of the screw\n",
- "p=4**-1 #in\n",
- "mu=0.15 #coefficient of friction\n",
- "l=2 #ft\n",
- "L=4000 #lb\n",
- "\n",
- "#Calculations\n",
- "phi=arctan(mu)*(180/pi) #degrees\n",
- "beta=arctan(p*(pi*l)**-1)*(180/pi) #degrees\n",
- "x=phi+beta # degrees\n",
- "#Force to raise the load\n",
- "# Here the value of x=10.77 degrees, thus\n",
- "tanx=0.19\n",
- "P=(L*tanx)/(d_m*12) #lb\n",
- "#Force to lower the load\n",
- "# Also,\n",
- "y=phi-beta\n",
- "# Thus y yeilds 6.23 degrees, thus\n",
- "tany=0.109\n",
- "P2=(L*tany)/(d_m*12) #lb\n",
- "\n",
- "#Result\n",
- "print'The force to raise the load is',round(P,1 ),\"lb\"\n",
- "print'The force to lower the load is',round(P2,1),\"lb\"\n",
- "\n",
- "# The answer waries due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force to raise the load is 31.7 lb\n",
- "The force to lower the load is 18.2 lb\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-16, Page no 141"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r_m=2.338 #in\n",
- "d_m=3.25 #in\n",
- "mu=0.06 #coefficient of friction\n",
- "P=1500 #lb\n",
- "p=4**-1 #pitch\n",
- "\n",
- "#Calculation\n",
- "phi=arctan(mu)*(180/pi) #degrees\n",
- "beta=round(arctan(p/(2*pi*r_m))*(180/pi)) #degrees\n",
- "x=phi+beta\n",
- "# Thus tanx yeilds\n",
- "tanx=0.077\n",
- "M=P*r_m*tanx+mu*P*(d_m/2) #lb.in\n",
- "\n",
- "#Result\n",
- "print'The moment required is',round(M),\"lb.in\"\n",
- "\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment required is 416.0 lb.in\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-17, Page no 141"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=750 #mm diameter\n",
- "alpha=pi #wrap angle radians\n",
- "mu=0.25 #coefficient of friction\n",
- "T_t=200 #N tension on the tight side\n",
- "\n",
- "#Calculation\n",
- "T2=T_t/(exp(mu*alpha)) #N\n",
- "\n",
- "#Result\n",
- "print'The tension of the slack side is',round(T2,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension of the slack side is 91.2 N\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-18, Page no 141"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=635 #mm diameter of the drum\n",
- "P=178 #N\n",
- "mu=3**-1 #coefficient of friction\n",
- "l1=100 #mm \n",
- "l2=660 #mm\n",
- "# as theta1=60 degrees\n",
- "sintheta1=sqrt(3)*2**-1\n",
- "costheta1=2**-1\n",
- "# as theta2=30 degrees,\n",
- "sintheta2=2**-1\n",
- "costheta2=sqrt(3)*2**-1\n",
- "GD=d/2 #mm\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point C\n",
- "Tb=(P*(l1+l2))/(l1*sintheta1) #N\n",
- "CD=((d/2)-(l1*costheta2))/sintheta2 #mm\n",
- "#from fig 9-22(b) \n",
- "theta=arcsin(GD/CD)*(180/pi) #degrees\n",
- "#from fig9-22(c)\n",
- "w_d=(180+30+theta) #degrees\n",
- "w=(w_d)*(pi/180) #radians\n",
- "#As Tc is greater than Tb\n",
- "Tc=Tb*(exp(mu*w)) #N\n",
- "M=(Tc-Tb)*GD #N.mm\n",
- "an=M/1000 #N.m\n",
- "\n",
- "#Result\n",
- "print'The braking moment required is',round(an),\"N.m\"\n",
- "\n",
- "#Note the unit of the final enswer carefully\n",
- "# The answer is off by 2 N.m"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The braking moment required is 1668.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-19, Page no 142"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "L=1000 #lb\n",
- "P=10 #lb\n",
- "\n",
- "#Calculations\n",
- "mu=log(L/P)/(4*2*pi) \n",
- "\n",
- "#Result\n",
- "print'The coefficient of friction is',round(mu,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The coefficient of friction is 0.18\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-20, Page no 142"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=900 #kg\n",
- "mu=0.2 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "T2=m*g/(exp(2*2*pi*mu)) #N\n",
- "\n",
- "#Result\n",
- "print'The force needed to hold the mass is',round(T2),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force needed to hold the mass is 714.0 N\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-21, Page no 143"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=760 #mm\n",
- "W=500 #N\n",
- "a=0.305 #mm coefficient of rolling resisatnce\n",
- "r=d/2 #mm\n",
- "\n",
- "#Calculations\n",
- "P=(W*a)/r #N\n",
- "\n",
- "#Result\n",
- "print'The force necessary is',round(P,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force necessary is 0.4 N\n"
- ]
- }
- ],
- "prompt_number": 33
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter9_1.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter9_1.ipynb deleted file mode 100755 index b8787d4f..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter9_1.ipynb +++ /dev/null @@ -1,889 +0,0 @@ -{
- "metadata": {
- "name": "chapter9.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 9: Friction"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-1, Page no 130"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Calculations\n",
- "#Simplifying equation (3) after substituting value of Nb in it we get\n",
- "#m_u**2+m_u*2*tan(50)-1=0\n",
- "#Solution of the equation\n",
- "a=1\n",
- "b=2*1.19175 # here 1.19175 is value of tan(50)\n",
- "c=-1\n",
- "g=(b**2-(4*a*c))**0.5\n",
- "\n",
- "#solution\n",
- "x1=(-b+g)/(2*a)\n",
- "x2=(-b-g)/(2*a)\n",
- "#As x2 does not make any physical sense x1 is the answer\n",
- "\n",
- "#Result\n",
- "print'The value of mu is',round(x1,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of mu is 0.36\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-3, Page no 131"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=70 #kg\n",
- "g=9.81 #m/s**2\n",
- "# as theta=20 degrees, we have\n",
- "sintheta=0.3420\n",
- "costheta=0.9396\n",
- "\n",
- "#Calculations\n",
- "#Solving by martix method\n",
- "#Taking sum along vertical and horizontal direction and equating them to zero\n",
- "A=np.array([[sintheta,1,0],[-costheta,0,1],[0,-4**-1,1]])\n",
- "#RHS matrix\n",
- "R=np.array([[m*g],[0],[0]])\n",
- "ans1=np.linalg.solve(A,R) #force vector N\n",
- "#Calculation part 2\n",
- "#Similar solution by matrix method\n",
- "#Taking moment about point O and summing forces in horizontal and vertical direction and equating all to zero\n",
- "B=np.array([[4*costheta,0,0],[-costheta,1,0],[sintheta,0,1]])\n",
- "#RHS matrix\n",
- "J=np.array([[m*g*1.5],[0],[m*g]])\n",
- "ans2=np.linalg.solve(B,J) #force Vector N\n",
- "\n",
- "#Result\n",
- "print'The value of P in first case is',round(ans1[0]),\"N\",'and that in second case is',round(ans2[0]),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P in first case is 167.0 N and that in second case is 274.0 N\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-4, Page no 132"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=200 #lb\n",
- "Fapp=300 #lb\n",
- "mu=0.3 #coefficient of friction\n",
- "# as theta=30 degrees, we have\n",
- "sintheta=2**-1\n",
- "costheta=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Summing forces in the plane parallel to the slope\n",
- "F=-(W*sintheta-Fapp*costheta) #lb\n",
- "N1=(W*costheta+Fapp*sintheta) #lb\n",
- "#Max value obtained\n",
- "Fprime= mu*N1\n",
- "\n",
- "#Result\n",
- "print'The value of F is',round(F),\"lb\"\n",
- "print'The value of N1 is',round(N1),\"lb\"\n",
- "print'The value of Fprime is',round(Fprime),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of F is 160.0 lb\n",
- "The value of N1 is 323.0 lb\n",
- "The value of Fprime is 97.0 lb\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-5, Page no 132"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mu1=0.2 #coefficient of friction between wedges and A\n",
- "mu2=4**-1 #coefficient of friction between wedges \n",
- "F=20 #tonnes\n",
- "\n",
- "#Calculations\n",
- "#Using the matrix method to solve\n",
- "#Summing forces in vertical and horizontal direction\n",
- "A=np.array([[1,-(mu1*10+1)/(101)**0.5],[0,(10-mu1*1)/101**0.5]]) #force matrix\n",
- "B=np.array([[mu2*F*1000],[F*1000]]) #lb\n",
- "#Solving both matrices\n",
- "R=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The forces N2 and P are:',round(R[1]),\"lb\",'and',round(R[0]),\"lb\"\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces N2 and P are: 20510.0 lb and 11122.0 lb\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-6, Page no 133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=45 degrees,we have\n",
- "sintheta=sqrt(2)**-1\n",
- "costheta=sqrt(2)**-1\n",
- "mu1=4**-1 #coefficient of friction between A and B\n",
- "mu2=3**-1 #coefficient of friction between A and Floor\n",
- "ma=14 #kg\n",
- "mb=9 #kg\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Summing forces in vertical direction\n",
- "Nb=mb*g #N\n",
- "#Also\n",
- "Fprimeb=mu1*Nb #N\n",
- "#Summing forces in direction\n",
- "T=Fprimeb #N\n",
- "#Considering the fig(c)\n",
- "#Summing forces in the horizontal direction and vertical direction and solving by matrix method \n",
- "A=np.array([[-costheta,mu2],[sintheta,1]]) #N\n",
- "B=np.array([[-Fprimeb],[(mb*g+ma*g)]]) #N\n",
- "R=np.linalg.solve(A,B) #N\n",
- "\n",
- "#Result\n",
- "print'The value of P and Na are:',round(R[0]),\"N\",'and',round(R[1]),\"N respectively.\"\n",
- "\n",
- "# The ans may wary due o decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P and Na are: 103.0 N and 153.0 N respectively.\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-7, Page no 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=40 #kg\n",
- "m2=13.5 #kg\n",
- "mu=3**-1 #coefficient of friction\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Solving by substitution\n",
- "#After simplification we get\n",
- "x=mu*m2*g\n",
- "y=mu*(m1*g+m2*g)\n",
- "theta=arctan((x+y)/(m1*g))*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The value of the angle is',round(theta,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of the angle is 29.2 degrees\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-8, Page no 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=350 #lb\n",
- "# as theta=30 degrees, we have\n",
- "sintheta=2**-1\n",
- "costheta=sqrt(3)*2**-1\n",
- "# and phi=15 degrees,thus\n",
- "sinphi=0.2588\n",
- "cosphi=0.9659\n",
- "\n",
- "#Calculations\n",
- "#Solving by the matrix method\n",
- "A=np.array([[costheta,sinphi],[-sintheta,cosphi]])\n",
- "B=np.array([[W*sintheta],[W*costheta]])\n",
- "an=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The value of P and R are:',round(an[0],1),\"lb\",'and',round(an[1],1),\"lb respectively.\"\n",
- "\n",
- "# The ans may wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P and R are: 93.8 lb and 362.4 lb respectively.\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-9, Page no 135"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=45 degrees, we have\n",
- "sintheta=sqrt(2)**-1\n",
- "costheta=sqrt(2)**-1\n",
- "m1=45 #kg\n",
- "m2=135 #kg\n",
- "g=9.81 #m/s**2\n",
- "mu=0.25 #coefficient of riction\n",
- "\n",
- "#Calculations\n",
- "N2=m2*g #N\n",
- "T=mu*N2 #N\n",
- "N1=m1*g*costheta #N\n",
- "Fprime1=N1*mu #N\n",
- "P=T+Fprime1-(m1*g*sintheta) #N\n",
- "\n",
- "#Result\n",
- "print'The values are'\n",
- "print'N2=',round(N2),\"N\"\n",
- "print'T=',round(T),\"N\"\n",
- "print'N1=',round(N1),\"N\"\n",
- "print'Fprime1=',round(Fprime1),\"N\"\n",
- "print'P=',round(P,1),\"N\"\n",
- "\n",
- "# The ans may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are\n",
- "N2= 1324.0 N\n",
- "T= 331.0 N\n",
- "N1= 312.0 N\n",
- "Fprime1= 78.0 N\n",
- "P= 97.0 N\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-10, Page no 135"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "mu=0.2 #coefficient of friction\n",
- "F1=150 #lb\n",
- "F2=100 #lb\n",
- "# as theta=60 degrees\n",
- "costheta=2**-1\n",
- "# also theta1=30 degrees\n",
- "costheta1=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "N1=F1*costheta #lb\n",
- "T=(mu*N1)+(F1*(costheta1)) #lb considering positive\n",
- "#Equilibrium for 100lb\n",
- "#Eliminating N2 from both equations\n",
- "#Taking derivative we get\n",
- "theta2=arctan(mu)*(180/pi) #degrees\n",
- "#Hence P becomes\n",
- "# in calculation of P we use the values of sin(theta2) & cos(theta2) as,\n",
- "sintheta2=0.196\n",
- "costheta2=0.98\n",
- "P=(F2*mu+T)*(costheta2+(mu*sintheta2))**-1 #lb\n",
- "\n",
- "#Result\n",
- "print'The minimum value of P is',round(P),\"lb\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The minimum value of P is 162.0 lb\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-11, Page no 136"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=180 #N\n",
- "m=100 #kg\n",
- "g=9.81 #m/s**2\n",
- "mu=0.25 #coeffiecient of friction\n",
- "\n",
- "#Calculations\n",
- "#Assuming F2 is maximum\n",
- "N2=F*2/(1+mu) #N\n",
- "F2=mu*N2 #N\n",
- "N1=m*g-F2 #N\n",
- "F1=F-F2 #N\n",
- "\n",
- "#Result\n",
- "print'The vaules are'\n",
- "print'N2=',round(N2,3),\"N\"\n",
- "print'F2=',round(F2,3),\"N\"\n",
- "print'N1=',round(N1,3),\"N\"\n",
- "print'F1=',round(F1,3),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The vaules are\n",
- "N2= 288.0 N\n",
- "F2= 72.0 N\n",
- "N1= 909.0 N\n",
- "F1= 108.0 N\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-13, Page no 138"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "mu_ca=0.3 #ceofficient of friction between copper block A and aluminium block B\n",
- "mu_af=0.2 #coefficient of friction between aluminium block B and Floor\n",
- "ma=3 #kg\n",
- "mb=2 #kg\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#For A\n",
- "#Taking sum of forces along X and Y direction\n",
- "Na=ma*g #N\n",
- "P=mu_ca*Na #N\n",
- "#For B\n",
- "#Taking sum of forces along X and Y direction\n",
- "Nb=Na+mb*g #N\n",
- "Fb=mu_ca*Na #N\n",
- "#Now largest value of friction before slip is \n",
- "Fprimeb=mu_af*Nb #N\n",
- "#Now as Fb<F'b hence initial assumption is incorrect and P=Fb\n",
- "P=Fb #N\n",
- "\n",
- "#Result\n",
- "print'The value of force that will cause motion is',round(P,2),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of force that will cause motion is 8.83 N\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-15, Page no 141"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d_m=2 #in mean diameter of the screw\n",
- "p=4**-1 #in\n",
- "mu=0.15 #coefficient of friction\n",
- "l=2 #ft\n",
- "L=4000 #lb\n",
- "\n",
- "#Calculations\n",
- "phi=arctan(mu)*(180/pi) #degrees\n",
- "beta=arctan(p*(pi*l)**-1)*(180/pi) #degrees\n",
- "x=phi+beta # degrees\n",
- "#Force to raise the load\n",
- "# Here the value of x=10.77 degrees, thus\n",
- "tanx=0.19\n",
- "P=(L*tanx)/(d_m*12) #lb\n",
- "#Force to lower the load\n",
- "# Also,\n",
- "y=phi-beta\n",
- "# Thus y yeilds 6.23 degrees, thus\n",
- "tany=0.109\n",
- "P2=(L*tany)/(d_m*12) #lb\n",
- "\n",
- "#Result\n",
- "print'The force to raise the load is',round(P,1 ),\"lb\"\n",
- "print'The force to lower the load is',round(P2,1),\"lb\"\n",
- "\n",
- "# The answer waries due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force to raise the load is 31.7 lb\n",
- "The force to lower the load is 18.2 lb\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-16, Page no 141"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r_m=2.338 #in\n",
- "d_m=3.25 #in\n",
- "mu=0.06 #coefficient of friction\n",
- "P=1500 #lb\n",
- "p=4**-1 #pitch\n",
- "\n",
- "#Calculation\n",
- "phi=arctan(mu)*(180/pi) #degrees\n",
- "beta=round(arctan(p/(2*pi*r_m))*(180/pi)) #degrees\n",
- "x=phi+beta\n",
- "# Thus tanx yeilds\n",
- "tanx=0.077\n",
- "M=P*r_m*tanx+mu*P*(d_m/2) #lb.in\n",
- "\n",
- "#Result\n",
- "print'The moment required is',round(M),\"lb.in\"\n",
- "\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment required is 416.0 lb.in\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-17, Page no 141"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=750 #mm diameter\n",
- "alpha=pi #wrap angle radians\n",
- "mu=0.25 #coefficient of friction\n",
- "T_t=200 #N tension on the tight side\n",
- "\n",
- "#Calculation\n",
- "T2=T_t/(exp(mu*alpha)) #N\n",
- "\n",
- "#Result\n",
- "print'The tension of the slack side is',round(T2,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension of the slack side is 91.2 N\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-18, Page no 141"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=635 #mm diameter of the drum\n",
- "P=178 #N\n",
- "mu=3**-1 #coefficient of friction\n",
- "l1=100 #mm \n",
- "l2=660 #mm\n",
- "# as theta1=60 degrees\n",
- "sintheta1=sqrt(3)*2**-1\n",
- "costheta1=2**-1\n",
- "# as theta2=30 degrees,\n",
- "sintheta2=2**-1\n",
- "costheta2=sqrt(3)*2**-1\n",
- "GD=d/2 #mm\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point C\n",
- "Tb=(P*(l1+l2))/(l1*sintheta1) #N\n",
- "CD=((d/2)-(l1*costheta2))/sintheta2 #mm\n",
- "#from fig 9-22(b) \n",
- "theta=arcsin(GD/CD)*(180/pi) #degrees\n",
- "#from fig9-22(c)\n",
- "w_d=(180+30+theta) #degrees\n",
- "w=(w_d)*(pi/180) #radians\n",
- "#As Tc is greater than Tb\n",
- "Tc=Tb*(exp(mu*w)) #N\n",
- "M=(Tc-Tb)*GD #N.mm\n",
- "an=M/1000 #N.m\n",
- "\n",
- "#Result\n",
- "print'The braking moment required is',round(an),\"N.m\"\n",
- "\n",
- "#Note the unit of the final enswer carefully\n",
- "# The answer is off by 2 N.m"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The braking moment required is 1668.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-19, Page no 142"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "L=1000 #lb\n",
- "P=10 #lb\n",
- "\n",
- "#Calculations\n",
- "mu=log(L/P)/(4*2*pi) \n",
- "\n",
- "#Result\n",
- "print'The coefficient of friction is',round(mu,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The coefficient of friction is 0.18\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-20, Page no 142"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=900 #kg\n",
- "mu=0.2 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "T2=m*g/(exp(2*2*pi*mu)) #N\n",
- "\n",
- "#Result\n",
- "print'The force needed to hold the mass is',round(T2),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force needed to hold the mass is 714.0 N\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-21, Page no 143"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=760 #mm\n",
- "W=500 #N\n",
- "a=0.305 #mm coefficient of rolling resisatnce\n",
- "r=d/2 #mm\n",
- "\n",
- "#Calculations\n",
- "P=(W*a)/r #N\n",
- "\n",
- "#Result\n",
- "print'The force necessary is',round(P,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force necessary is 0.4 N\n"
- ]
- }
- ],
- "prompt_number": 33
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter9_2.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter9_2.ipynb deleted file mode 100755 index b8787d4f..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter9_2.ipynb +++ /dev/null @@ -1,889 +0,0 @@ -{
- "metadata": {
- "name": "chapter9.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 9: Friction"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-1, Page no 130"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Calculations\n",
- "#Simplifying equation (3) after substituting value of Nb in it we get\n",
- "#m_u**2+m_u*2*tan(50)-1=0\n",
- "#Solution of the equation\n",
- "a=1\n",
- "b=2*1.19175 # here 1.19175 is value of tan(50)\n",
- "c=-1\n",
- "g=(b**2-(4*a*c))**0.5\n",
- "\n",
- "#solution\n",
- "x1=(-b+g)/(2*a)\n",
- "x2=(-b-g)/(2*a)\n",
- "#As x2 does not make any physical sense x1 is the answer\n",
- "\n",
- "#Result\n",
- "print'The value of mu is',round(x1,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of mu is 0.36\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-3, Page no 131"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=70 #kg\n",
- "g=9.81 #m/s**2\n",
- "# as theta=20 degrees, we have\n",
- "sintheta=0.3420\n",
- "costheta=0.9396\n",
- "\n",
- "#Calculations\n",
- "#Solving by martix method\n",
- "#Taking sum along vertical and horizontal direction and equating them to zero\n",
- "A=np.array([[sintheta,1,0],[-costheta,0,1],[0,-4**-1,1]])\n",
- "#RHS matrix\n",
- "R=np.array([[m*g],[0],[0]])\n",
- "ans1=np.linalg.solve(A,R) #force vector N\n",
- "#Calculation part 2\n",
- "#Similar solution by matrix method\n",
- "#Taking moment about point O and summing forces in horizontal and vertical direction and equating all to zero\n",
- "B=np.array([[4*costheta,0,0],[-costheta,1,0],[sintheta,0,1]])\n",
- "#RHS matrix\n",
- "J=np.array([[m*g*1.5],[0],[m*g]])\n",
- "ans2=np.linalg.solve(B,J) #force Vector N\n",
- "\n",
- "#Result\n",
- "print'The value of P in first case is',round(ans1[0]),\"N\",'and that in second case is',round(ans2[0]),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P in first case is 167.0 N and that in second case is 274.0 N\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-4, Page no 132"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=200 #lb\n",
- "Fapp=300 #lb\n",
- "mu=0.3 #coefficient of friction\n",
- "# as theta=30 degrees, we have\n",
- "sintheta=2**-1\n",
- "costheta=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Summing forces in the plane parallel to the slope\n",
- "F=-(W*sintheta-Fapp*costheta) #lb\n",
- "N1=(W*costheta+Fapp*sintheta) #lb\n",
- "#Max value obtained\n",
- "Fprime= mu*N1\n",
- "\n",
- "#Result\n",
- "print'The value of F is',round(F),\"lb\"\n",
- "print'The value of N1 is',round(N1),\"lb\"\n",
- "print'The value of Fprime is',round(Fprime),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of F is 160.0 lb\n",
- "The value of N1 is 323.0 lb\n",
- "The value of Fprime is 97.0 lb\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-5, Page no 132"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mu1=0.2 #coefficient of friction between wedges and A\n",
- "mu2=4**-1 #coefficient of friction between wedges \n",
- "F=20 #tonnes\n",
- "\n",
- "#Calculations\n",
- "#Using the matrix method to solve\n",
- "#Summing forces in vertical and horizontal direction\n",
- "A=np.array([[1,-(mu1*10+1)/(101)**0.5],[0,(10-mu1*1)/101**0.5]]) #force matrix\n",
- "B=np.array([[mu2*F*1000],[F*1000]]) #lb\n",
- "#Solving both matrices\n",
- "R=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The forces N2 and P are:',round(R[1]),\"lb\",'and',round(R[0]),\"lb\"\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces N2 and P are: 20510.0 lb and 11122.0 lb\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-6, Page no 133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=45 degrees,we have\n",
- "sintheta=sqrt(2)**-1\n",
- "costheta=sqrt(2)**-1\n",
- "mu1=4**-1 #coefficient of friction between A and B\n",
- "mu2=3**-1 #coefficient of friction between A and Floor\n",
- "ma=14 #kg\n",
- "mb=9 #kg\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Summing forces in vertical direction\n",
- "Nb=mb*g #N\n",
- "#Also\n",
- "Fprimeb=mu1*Nb #N\n",
- "#Summing forces in direction\n",
- "T=Fprimeb #N\n",
- "#Considering the fig(c)\n",
- "#Summing forces in the horizontal direction and vertical direction and solving by matrix method \n",
- "A=np.array([[-costheta,mu2],[sintheta,1]]) #N\n",
- "B=np.array([[-Fprimeb],[(mb*g+ma*g)]]) #N\n",
- "R=np.linalg.solve(A,B) #N\n",
- "\n",
- "#Result\n",
- "print'The value of P and Na are:',round(R[0]),\"N\",'and',round(R[1]),\"N respectively.\"\n",
- "\n",
- "# The ans may wary due o decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P and Na are: 103.0 N and 153.0 N respectively.\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-7, Page no 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=40 #kg\n",
- "m2=13.5 #kg\n",
- "mu=3**-1 #coefficient of friction\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Solving by substitution\n",
- "#After simplification we get\n",
- "x=mu*m2*g\n",
- "y=mu*(m1*g+m2*g)\n",
- "theta=arctan((x+y)/(m1*g))*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The value of the angle is',round(theta,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of the angle is 29.2 degrees\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-8, Page no 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=350 #lb\n",
- "# as theta=30 degrees, we have\n",
- "sintheta=2**-1\n",
- "costheta=sqrt(3)*2**-1\n",
- "# and phi=15 degrees,thus\n",
- "sinphi=0.2588\n",
- "cosphi=0.9659\n",
- "\n",
- "#Calculations\n",
- "#Solving by the matrix method\n",
- "A=np.array([[costheta,sinphi],[-sintheta,cosphi]])\n",
- "B=np.array([[W*sintheta],[W*costheta]])\n",
- "an=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The value of P and R are:',round(an[0],1),\"lb\",'and',round(an[1],1),\"lb respectively.\"\n",
- "\n",
- "# The ans may wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P and R are: 93.8 lb and 362.4 lb respectively.\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-9, Page no 135"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=45 degrees, we have\n",
- "sintheta=sqrt(2)**-1\n",
- "costheta=sqrt(2)**-1\n",
- "m1=45 #kg\n",
- "m2=135 #kg\n",
- "g=9.81 #m/s**2\n",
- "mu=0.25 #coefficient of riction\n",
- "\n",
- "#Calculations\n",
- "N2=m2*g #N\n",
- "T=mu*N2 #N\n",
- "N1=m1*g*costheta #N\n",
- "Fprime1=N1*mu #N\n",
- "P=T+Fprime1-(m1*g*sintheta) #N\n",
- "\n",
- "#Result\n",
- "print'The values are'\n",
- "print'N2=',round(N2),\"N\"\n",
- "print'T=',round(T),\"N\"\n",
- "print'N1=',round(N1),\"N\"\n",
- "print'Fprime1=',round(Fprime1),\"N\"\n",
- "print'P=',round(P,1),\"N\"\n",
- "\n",
- "# The ans may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are\n",
- "N2= 1324.0 N\n",
- "T= 331.0 N\n",
- "N1= 312.0 N\n",
- "Fprime1= 78.0 N\n",
- "P= 97.0 N\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-10, Page no 135"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "mu=0.2 #coefficient of friction\n",
- "F1=150 #lb\n",
- "F2=100 #lb\n",
- "# as theta=60 degrees\n",
- "costheta=2**-1\n",
- "# also theta1=30 degrees\n",
- "costheta1=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "N1=F1*costheta #lb\n",
- "T=(mu*N1)+(F1*(costheta1)) #lb considering positive\n",
- "#Equilibrium for 100lb\n",
- "#Eliminating N2 from both equations\n",
- "#Taking derivative we get\n",
- "theta2=arctan(mu)*(180/pi) #degrees\n",
- "#Hence P becomes\n",
- "# in calculation of P we use the values of sin(theta2) & cos(theta2) as,\n",
- "sintheta2=0.196\n",
- "costheta2=0.98\n",
- "P=(F2*mu+T)*(costheta2+(mu*sintheta2))**-1 #lb\n",
- "\n",
- "#Result\n",
- "print'The minimum value of P is',round(P),\"lb\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The minimum value of P is 162.0 lb\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-11, Page no 136"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=180 #N\n",
- "m=100 #kg\n",
- "g=9.81 #m/s**2\n",
- "mu=0.25 #coeffiecient of friction\n",
- "\n",
- "#Calculations\n",
- "#Assuming F2 is maximum\n",
- "N2=F*2/(1+mu) #N\n",
- "F2=mu*N2 #N\n",
- "N1=m*g-F2 #N\n",
- "F1=F-F2 #N\n",
- "\n",
- "#Result\n",
- "print'The vaules are'\n",
- "print'N2=',round(N2,3),\"N\"\n",
- "print'F2=',round(F2,3),\"N\"\n",
- "print'N1=',round(N1,3),\"N\"\n",
- "print'F1=',round(F1,3),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The vaules are\n",
- "N2= 288.0 N\n",
- "F2= 72.0 N\n",
- "N1= 909.0 N\n",
- "F1= 108.0 N\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-13, Page no 138"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "mu_ca=0.3 #ceofficient of friction between copper block A and aluminium block B\n",
- "mu_af=0.2 #coefficient of friction between aluminium block B and Floor\n",
- "ma=3 #kg\n",
- "mb=2 #kg\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#For A\n",
- "#Taking sum of forces along X and Y direction\n",
- "Na=ma*g #N\n",
- "P=mu_ca*Na #N\n",
- "#For B\n",
- "#Taking sum of forces along X and Y direction\n",
- "Nb=Na+mb*g #N\n",
- "Fb=mu_ca*Na #N\n",
- "#Now largest value of friction before slip is \n",
- "Fprimeb=mu_af*Nb #N\n",
- "#Now as Fb<F'b hence initial assumption is incorrect and P=Fb\n",
- "P=Fb #N\n",
- "\n",
- "#Result\n",
- "print'The value of force that will cause motion is',round(P,2),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of force that will cause motion is 8.83 N\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-15, Page no 141"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d_m=2 #in mean diameter of the screw\n",
- "p=4**-1 #in\n",
- "mu=0.15 #coefficient of friction\n",
- "l=2 #ft\n",
- "L=4000 #lb\n",
- "\n",
- "#Calculations\n",
- "phi=arctan(mu)*(180/pi) #degrees\n",
- "beta=arctan(p*(pi*l)**-1)*(180/pi) #degrees\n",
- "x=phi+beta # degrees\n",
- "#Force to raise the load\n",
- "# Here the value of x=10.77 degrees, thus\n",
- "tanx=0.19\n",
- "P=(L*tanx)/(d_m*12) #lb\n",
- "#Force to lower the load\n",
- "# Also,\n",
- "y=phi-beta\n",
- "# Thus y yeilds 6.23 degrees, thus\n",
- "tany=0.109\n",
- "P2=(L*tany)/(d_m*12) #lb\n",
- "\n",
- "#Result\n",
- "print'The force to raise the load is',round(P,1 ),\"lb\"\n",
- "print'The force to lower the load is',round(P2,1),\"lb\"\n",
- "\n",
- "# The answer waries due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force to raise the load is 31.7 lb\n",
- "The force to lower the load is 18.2 lb\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-16, Page no 141"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r_m=2.338 #in\n",
- "d_m=3.25 #in\n",
- "mu=0.06 #coefficient of friction\n",
- "P=1500 #lb\n",
- "p=4**-1 #pitch\n",
- "\n",
- "#Calculation\n",
- "phi=arctan(mu)*(180/pi) #degrees\n",
- "beta=round(arctan(p/(2*pi*r_m))*(180/pi)) #degrees\n",
- "x=phi+beta\n",
- "# Thus tanx yeilds\n",
- "tanx=0.077\n",
- "M=P*r_m*tanx+mu*P*(d_m/2) #lb.in\n",
- "\n",
- "#Result\n",
- "print'The moment required is',round(M),\"lb.in\"\n",
- "\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment required is 416.0 lb.in\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-17, Page no 141"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=750 #mm diameter\n",
- "alpha=pi #wrap angle radians\n",
- "mu=0.25 #coefficient of friction\n",
- "T_t=200 #N tension on the tight side\n",
- "\n",
- "#Calculation\n",
- "T2=T_t/(exp(mu*alpha)) #N\n",
- "\n",
- "#Result\n",
- "print'The tension of the slack side is',round(T2,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension of the slack side is 91.2 N\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-18, Page no 141"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=635 #mm diameter of the drum\n",
- "P=178 #N\n",
- "mu=3**-1 #coefficient of friction\n",
- "l1=100 #mm \n",
- "l2=660 #mm\n",
- "# as theta1=60 degrees\n",
- "sintheta1=sqrt(3)*2**-1\n",
- "costheta1=2**-1\n",
- "# as theta2=30 degrees,\n",
- "sintheta2=2**-1\n",
- "costheta2=sqrt(3)*2**-1\n",
- "GD=d/2 #mm\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point C\n",
- "Tb=(P*(l1+l2))/(l1*sintheta1) #N\n",
- "CD=((d/2)-(l1*costheta2))/sintheta2 #mm\n",
- "#from fig 9-22(b) \n",
- "theta=arcsin(GD/CD)*(180/pi) #degrees\n",
- "#from fig9-22(c)\n",
- "w_d=(180+30+theta) #degrees\n",
- "w=(w_d)*(pi/180) #radians\n",
- "#As Tc is greater than Tb\n",
- "Tc=Tb*(exp(mu*w)) #N\n",
- "M=(Tc-Tb)*GD #N.mm\n",
- "an=M/1000 #N.m\n",
- "\n",
- "#Result\n",
- "print'The braking moment required is',round(an),\"N.m\"\n",
- "\n",
- "#Note the unit of the final enswer carefully\n",
- "# The answer is off by 2 N.m"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The braking moment required is 1668.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-19, Page no 142"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "L=1000 #lb\n",
- "P=10 #lb\n",
- "\n",
- "#Calculations\n",
- "mu=log(L/P)/(4*2*pi) \n",
- "\n",
- "#Result\n",
- "print'The coefficient of friction is',round(mu,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The coefficient of friction is 0.18\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-20, Page no 142"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=900 #kg\n",
- "mu=0.2 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "T2=m*g/(exp(2*2*pi*mu)) #N\n",
- "\n",
- "#Result\n",
- "print'The force needed to hold the mass is',round(T2),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force needed to hold the mass is 714.0 N\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-21, Page no 143"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=760 #mm\n",
- "W=500 #N\n",
- "a=0.305 #mm coefficient of rolling resisatnce\n",
- "r=d/2 #mm\n",
- "\n",
- "#Calculations\n",
- "P=(W*a)/r #N\n",
- "\n",
- "#Result\n",
- "print'The force necessary is',round(P,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force necessary is 0.4 N\n"
- ]
- }
- ],
- "prompt_number": 33
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter9_3.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter9_3.ipynb deleted file mode 100755 index b8787d4f..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter9_3.ipynb +++ /dev/null @@ -1,889 +0,0 @@ -{
- "metadata": {
- "name": "chapter9.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 9: Friction"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-1, Page no 130"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Calculations\n",
- "#Simplifying equation (3) after substituting value of Nb in it we get\n",
- "#m_u**2+m_u*2*tan(50)-1=0\n",
- "#Solution of the equation\n",
- "a=1\n",
- "b=2*1.19175 # here 1.19175 is value of tan(50)\n",
- "c=-1\n",
- "g=(b**2-(4*a*c))**0.5\n",
- "\n",
- "#solution\n",
- "x1=(-b+g)/(2*a)\n",
- "x2=(-b-g)/(2*a)\n",
- "#As x2 does not make any physical sense x1 is the answer\n",
- "\n",
- "#Result\n",
- "print'The value of mu is',round(x1,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of mu is 0.36\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-3, Page no 131"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=70 #kg\n",
- "g=9.81 #m/s**2\n",
- "# as theta=20 degrees, we have\n",
- "sintheta=0.3420\n",
- "costheta=0.9396\n",
- "\n",
- "#Calculations\n",
- "#Solving by martix method\n",
- "#Taking sum along vertical and horizontal direction and equating them to zero\n",
- "A=np.array([[sintheta,1,0],[-costheta,0,1],[0,-4**-1,1]])\n",
- "#RHS matrix\n",
- "R=np.array([[m*g],[0],[0]])\n",
- "ans1=np.linalg.solve(A,R) #force vector N\n",
- "#Calculation part 2\n",
- "#Similar solution by matrix method\n",
- "#Taking moment about point O and summing forces in horizontal and vertical direction and equating all to zero\n",
- "B=np.array([[4*costheta,0,0],[-costheta,1,0],[sintheta,0,1]])\n",
- "#RHS matrix\n",
- "J=np.array([[m*g*1.5],[0],[m*g]])\n",
- "ans2=np.linalg.solve(B,J) #force Vector N\n",
- "\n",
- "#Result\n",
- "print'The value of P in first case is',round(ans1[0]),\"N\",'and that in second case is',round(ans2[0]),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P in first case is 167.0 N and that in second case is 274.0 N\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-4, Page no 132"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=200 #lb\n",
- "Fapp=300 #lb\n",
- "mu=0.3 #coefficient of friction\n",
- "# as theta=30 degrees, we have\n",
- "sintheta=2**-1\n",
- "costheta=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Summing forces in the plane parallel to the slope\n",
- "F=-(W*sintheta-Fapp*costheta) #lb\n",
- "N1=(W*costheta+Fapp*sintheta) #lb\n",
- "#Max value obtained\n",
- "Fprime= mu*N1\n",
- "\n",
- "#Result\n",
- "print'The value of F is',round(F),\"lb\"\n",
- "print'The value of N1 is',round(N1),\"lb\"\n",
- "print'The value of Fprime is',round(Fprime),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of F is 160.0 lb\n",
- "The value of N1 is 323.0 lb\n",
- "The value of Fprime is 97.0 lb\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-5, Page no 132"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mu1=0.2 #coefficient of friction between wedges and A\n",
- "mu2=4**-1 #coefficient of friction between wedges \n",
- "F=20 #tonnes\n",
- "\n",
- "#Calculations\n",
- "#Using the matrix method to solve\n",
- "#Summing forces in vertical and horizontal direction\n",
- "A=np.array([[1,-(mu1*10+1)/(101)**0.5],[0,(10-mu1*1)/101**0.5]]) #force matrix\n",
- "B=np.array([[mu2*F*1000],[F*1000]]) #lb\n",
- "#Solving both matrices\n",
- "R=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The forces N2 and P are:',round(R[1]),\"lb\",'and',round(R[0]),\"lb\"\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces N2 and P are: 20510.0 lb and 11122.0 lb\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-6, Page no 133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=45 degrees,we have\n",
- "sintheta=sqrt(2)**-1\n",
- "costheta=sqrt(2)**-1\n",
- "mu1=4**-1 #coefficient of friction between A and B\n",
- "mu2=3**-1 #coefficient of friction between A and Floor\n",
- "ma=14 #kg\n",
- "mb=9 #kg\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Summing forces in vertical direction\n",
- "Nb=mb*g #N\n",
- "#Also\n",
- "Fprimeb=mu1*Nb #N\n",
- "#Summing forces in direction\n",
- "T=Fprimeb #N\n",
- "#Considering the fig(c)\n",
- "#Summing forces in the horizontal direction and vertical direction and solving by matrix method \n",
- "A=np.array([[-costheta,mu2],[sintheta,1]]) #N\n",
- "B=np.array([[-Fprimeb],[(mb*g+ma*g)]]) #N\n",
- "R=np.linalg.solve(A,B) #N\n",
- "\n",
- "#Result\n",
- "print'The value of P and Na are:',round(R[0]),\"N\",'and',round(R[1]),\"N respectively.\"\n",
- "\n",
- "# The ans may wary due o decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P and Na are: 103.0 N and 153.0 N respectively.\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-7, Page no 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=40 #kg\n",
- "m2=13.5 #kg\n",
- "mu=3**-1 #coefficient of friction\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Solving by substitution\n",
- "#After simplification we get\n",
- "x=mu*m2*g\n",
- "y=mu*(m1*g+m2*g)\n",
- "theta=arctan((x+y)/(m1*g))*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The value of the angle is',round(theta,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of the angle is 29.2 degrees\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-8, Page no 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=350 #lb\n",
- "# as theta=30 degrees, we have\n",
- "sintheta=2**-1\n",
- "costheta=sqrt(3)*2**-1\n",
- "# and phi=15 degrees,thus\n",
- "sinphi=0.2588\n",
- "cosphi=0.9659\n",
- "\n",
- "#Calculations\n",
- "#Solving by the matrix method\n",
- "A=np.array([[costheta,sinphi],[-sintheta,cosphi]])\n",
- "B=np.array([[W*sintheta],[W*costheta]])\n",
- "an=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The value of P and R are:',round(an[0],1),\"lb\",'and',round(an[1],1),\"lb respectively.\"\n",
- "\n",
- "# The ans may wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P and R are: 93.8 lb and 362.4 lb respectively.\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-9, Page no 135"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=45 degrees, we have\n",
- "sintheta=sqrt(2)**-1\n",
- "costheta=sqrt(2)**-1\n",
- "m1=45 #kg\n",
- "m2=135 #kg\n",
- "g=9.81 #m/s**2\n",
- "mu=0.25 #coefficient of riction\n",
- "\n",
- "#Calculations\n",
- "N2=m2*g #N\n",
- "T=mu*N2 #N\n",
- "N1=m1*g*costheta #N\n",
- "Fprime1=N1*mu #N\n",
- "P=T+Fprime1-(m1*g*sintheta) #N\n",
- "\n",
- "#Result\n",
- "print'The values are'\n",
- "print'N2=',round(N2),\"N\"\n",
- "print'T=',round(T),\"N\"\n",
- "print'N1=',round(N1),\"N\"\n",
- "print'Fprime1=',round(Fprime1),\"N\"\n",
- "print'P=',round(P,1),\"N\"\n",
- "\n",
- "# The ans may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are\n",
- "N2= 1324.0 N\n",
- "T= 331.0 N\n",
- "N1= 312.0 N\n",
- "Fprime1= 78.0 N\n",
- "P= 97.0 N\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-10, Page no 135"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "mu=0.2 #coefficient of friction\n",
- "F1=150 #lb\n",
- "F2=100 #lb\n",
- "# as theta=60 degrees\n",
- "costheta=2**-1\n",
- "# also theta1=30 degrees\n",
- "costheta1=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "N1=F1*costheta #lb\n",
- "T=(mu*N1)+(F1*(costheta1)) #lb considering positive\n",
- "#Equilibrium for 100lb\n",
- "#Eliminating N2 from both equations\n",
- "#Taking derivative we get\n",
- "theta2=arctan(mu)*(180/pi) #degrees\n",
- "#Hence P becomes\n",
- "# in calculation of P we use the values of sin(theta2) & cos(theta2) as,\n",
- "sintheta2=0.196\n",
- "costheta2=0.98\n",
- "P=(F2*mu+T)*(costheta2+(mu*sintheta2))**-1 #lb\n",
- "\n",
- "#Result\n",
- "print'The minimum value of P is',round(P),\"lb\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The minimum value of P is 162.0 lb\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-11, Page no 136"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=180 #N\n",
- "m=100 #kg\n",
- "g=9.81 #m/s**2\n",
- "mu=0.25 #coeffiecient of friction\n",
- "\n",
- "#Calculations\n",
- "#Assuming F2 is maximum\n",
- "N2=F*2/(1+mu) #N\n",
- "F2=mu*N2 #N\n",
- "N1=m*g-F2 #N\n",
- "F1=F-F2 #N\n",
- "\n",
- "#Result\n",
- "print'The vaules are'\n",
- "print'N2=',round(N2,3),\"N\"\n",
- "print'F2=',round(F2,3),\"N\"\n",
- "print'N1=',round(N1,3),\"N\"\n",
- "print'F1=',round(F1,3),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The vaules are\n",
- "N2= 288.0 N\n",
- "F2= 72.0 N\n",
- "N1= 909.0 N\n",
- "F1= 108.0 N\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-13, Page no 138"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "mu_ca=0.3 #ceofficient of friction between copper block A and aluminium block B\n",
- "mu_af=0.2 #coefficient of friction between aluminium block B and Floor\n",
- "ma=3 #kg\n",
- "mb=2 #kg\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#For A\n",
- "#Taking sum of forces along X and Y direction\n",
- "Na=ma*g #N\n",
- "P=mu_ca*Na #N\n",
- "#For B\n",
- "#Taking sum of forces along X and Y direction\n",
- "Nb=Na+mb*g #N\n",
- "Fb=mu_ca*Na #N\n",
- "#Now largest value of friction before slip is \n",
- "Fprimeb=mu_af*Nb #N\n",
- "#Now as Fb<F'b hence initial assumption is incorrect and P=Fb\n",
- "P=Fb #N\n",
- "\n",
- "#Result\n",
- "print'The value of force that will cause motion is',round(P,2),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of force that will cause motion is 8.83 N\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-15, Page no 141"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d_m=2 #in mean diameter of the screw\n",
- "p=4**-1 #in\n",
- "mu=0.15 #coefficient of friction\n",
- "l=2 #ft\n",
- "L=4000 #lb\n",
- "\n",
- "#Calculations\n",
- "phi=arctan(mu)*(180/pi) #degrees\n",
- "beta=arctan(p*(pi*l)**-1)*(180/pi) #degrees\n",
- "x=phi+beta # degrees\n",
- "#Force to raise the load\n",
- "# Here the value of x=10.77 degrees, thus\n",
- "tanx=0.19\n",
- "P=(L*tanx)/(d_m*12) #lb\n",
- "#Force to lower the load\n",
- "# Also,\n",
- "y=phi-beta\n",
- "# Thus y yeilds 6.23 degrees, thus\n",
- "tany=0.109\n",
- "P2=(L*tany)/(d_m*12) #lb\n",
- "\n",
- "#Result\n",
- "print'The force to raise the load is',round(P,1 ),\"lb\"\n",
- "print'The force to lower the load is',round(P2,1),\"lb\"\n",
- "\n",
- "# The answer waries due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force to raise the load is 31.7 lb\n",
- "The force to lower the load is 18.2 lb\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-16, Page no 141"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r_m=2.338 #in\n",
- "d_m=3.25 #in\n",
- "mu=0.06 #coefficient of friction\n",
- "P=1500 #lb\n",
- "p=4**-1 #pitch\n",
- "\n",
- "#Calculation\n",
- "phi=arctan(mu)*(180/pi) #degrees\n",
- "beta=round(arctan(p/(2*pi*r_m))*(180/pi)) #degrees\n",
- "x=phi+beta\n",
- "# Thus tanx yeilds\n",
- "tanx=0.077\n",
- "M=P*r_m*tanx+mu*P*(d_m/2) #lb.in\n",
- "\n",
- "#Result\n",
- "print'The moment required is',round(M),\"lb.in\"\n",
- "\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment required is 416.0 lb.in\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-17, Page no 141"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=750 #mm diameter\n",
- "alpha=pi #wrap angle radians\n",
- "mu=0.25 #coefficient of friction\n",
- "T_t=200 #N tension on the tight side\n",
- "\n",
- "#Calculation\n",
- "T2=T_t/(exp(mu*alpha)) #N\n",
- "\n",
- "#Result\n",
- "print'The tension of the slack side is',round(T2,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension of the slack side is 91.2 N\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-18, Page no 141"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=635 #mm diameter of the drum\n",
- "P=178 #N\n",
- "mu=3**-1 #coefficient of friction\n",
- "l1=100 #mm \n",
- "l2=660 #mm\n",
- "# as theta1=60 degrees\n",
- "sintheta1=sqrt(3)*2**-1\n",
- "costheta1=2**-1\n",
- "# as theta2=30 degrees,\n",
- "sintheta2=2**-1\n",
- "costheta2=sqrt(3)*2**-1\n",
- "GD=d/2 #mm\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point C\n",
- "Tb=(P*(l1+l2))/(l1*sintheta1) #N\n",
- "CD=((d/2)-(l1*costheta2))/sintheta2 #mm\n",
- "#from fig 9-22(b) \n",
- "theta=arcsin(GD/CD)*(180/pi) #degrees\n",
- "#from fig9-22(c)\n",
- "w_d=(180+30+theta) #degrees\n",
- "w=(w_d)*(pi/180) #radians\n",
- "#As Tc is greater than Tb\n",
- "Tc=Tb*(exp(mu*w)) #N\n",
- "M=(Tc-Tb)*GD #N.mm\n",
- "an=M/1000 #N.m\n",
- "\n",
- "#Result\n",
- "print'The braking moment required is',round(an),\"N.m\"\n",
- "\n",
- "#Note the unit of the final enswer carefully\n",
- "# The answer is off by 2 N.m"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The braking moment required is 1668.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-19, Page no 142"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "L=1000 #lb\n",
- "P=10 #lb\n",
- "\n",
- "#Calculations\n",
- "mu=log(L/P)/(4*2*pi) \n",
- "\n",
- "#Result\n",
- "print'The coefficient of friction is',round(mu,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The coefficient of friction is 0.18\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-20, Page no 142"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=900 #kg\n",
- "mu=0.2 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "T2=m*g/(exp(2*2*pi*mu)) #N\n",
- "\n",
- "#Result\n",
- "print'The force needed to hold the mass is',round(T2),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force needed to hold the mass is 714.0 N\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-21, Page no 143"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=760 #mm\n",
- "W=500 #N\n",
- "a=0.305 #mm coefficient of rolling resisatnce\n",
- "r=d/2 #mm\n",
- "\n",
- "#Calculations\n",
- "P=(W*a)/r #N\n",
- "\n",
- "#Result\n",
- "print'The force necessary is',round(P,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force necessary is 0.4 N\n"
- ]
- }
- ],
- "prompt_number": 33
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter9_4.ipynb b/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter9_4.ipynb deleted file mode 100755 index b8787d4f..00000000 --- a/Engineering_Mechanics,_Schaum_Series_by_McLean/chapter9_4.ipynb +++ /dev/null @@ -1,889 +0,0 @@ -{
- "metadata": {
- "name": "chapter9.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 9: Friction"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-1, Page no 130"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Calculations\n",
- "#Simplifying equation (3) after substituting value of Nb in it we get\n",
- "#m_u**2+m_u*2*tan(50)-1=0\n",
- "#Solution of the equation\n",
- "a=1\n",
- "b=2*1.19175 # here 1.19175 is value of tan(50)\n",
- "c=-1\n",
- "g=(b**2-(4*a*c))**0.5\n",
- "\n",
- "#solution\n",
- "x1=(-b+g)/(2*a)\n",
- "x2=(-b-g)/(2*a)\n",
- "#As x2 does not make any physical sense x1 is the answer\n",
- "\n",
- "#Result\n",
- "print'The value of mu is',round(x1,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of mu is 0.36\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-3, Page no 131"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "m=70 #kg\n",
- "g=9.81 #m/s**2\n",
- "# as theta=20 degrees, we have\n",
- "sintheta=0.3420\n",
- "costheta=0.9396\n",
- "\n",
- "#Calculations\n",
- "#Solving by martix method\n",
- "#Taking sum along vertical and horizontal direction and equating them to zero\n",
- "A=np.array([[sintheta,1,0],[-costheta,0,1],[0,-4**-1,1]])\n",
- "#RHS matrix\n",
- "R=np.array([[m*g],[0],[0]])\n",
- "ans1=np.linalg.solve(A,R) #force vector N\n",
- "#Calculation part 2\n",
- "#Similar solution by matrix method\n",
- "#Taking moment about point O and summing forces in horizontal and vertical direction and equating all to zero\n",
- "B=np.array([[4*costheta,0,0],[-costheta,1,0],[sintheta,0,1]])\n",
- "#RHS matrix\n",
- "J=np.array([[m*g*1.5],[0],[m*g]])\n",
- "ans2=np.linalg.solve(B,J) #force Vector N\n",
- "\n",
- "#Result\n",
- "print'The value of P in first case is',round(ans1[0]),\"N\",'and that in second case is',round(ans2[0]),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P in first case is 167.0 N and that in second case is 274.0 N\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-4, Page no 132"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "W=200 #lb\n",
- "Fapp=300 #lb\n",
- "mu=0.3 #coefficient of friction\n",
- "# as theta=30 degrees, we have\n",
- "sintheta=2**-1\n",
- "costheta=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "#Summing forces in the plane parallel to the slope\n",
- "F=-(W*sintheta-Fapp*costheta) #lb\n",
- "N1=(W*costheta+Fapp*sintheta) #lb\n",
- "#Max value obtained\n",
- "Fprime= mu*N1\n",
- "\n",
- "#Result\n",
- "print'The value of F is',round(F),\"lb\"\n",
- "print'The value of N1 is',round(N1),\"lb\"\n",
- "print'The value of Fprime is',round(Fprime),\"lb\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of F is 160.0 lb\n",
- "The value of N1 is 323.0 lb\n",
- "The value of Fprime is 97.0 lb\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-5, Page no 132"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "mu1=0.2 #coefficient of friction between wedges and A\n",
- "mu2=4**-1 #coefficient of friction between wedges \n",
- "F=20 #tonnes\n",
- "\n",
- "#Calculations\n",
- "#Using the matrix method to solve\n",
- "#Summing forces in vertical and horizontal direction\n",
- "A=np.array([[1,-(mu1*10+1)/(101)**0.5],[0,(10-mu1*1)/101**0.5]]) #force matrix\n",
- "B=np.array([[mu2*F*1000],[F*1000]]) #lb\n",
- "#Solving both matrices\n",
- "R=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The forces N2 and P are:',round(R[1]),\"lb\",'and',round(R[0]),\"lb\"\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The forces N2 and P are: 20510.0 lb and 11122.0 lb\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-6, Page no 133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=45 degrees,we have\n",
- "sintheta=sqrt(2)**-1\n",
- "costheta=sqrt(2)**-1\n",
- "mu1=4**-1 #coefficient of friction between A and B\n",
- "mu2=3**-1 #coefficient of friction between A and Floor\n",
- "ma=14 #kg\n",
- "mb=9 #kg\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Summing forces in vertical direction\n",
- "Nb=mb*g #N\n",
- "#Also\n",
- "Fprimeb=mu1*Nb #N\n",
- "#Summing forces in direction\n",
- "T=Fprimeb #N\n",
- "#Considering the fig(c)\n",
- "#Summing forces in the horizontal direction and vertical direction and solving by matrix method \n",
- "A=np.array([[-costheta,mu2],[sintheta,1]]) #N\n",
- "B=np.array([[-Fprimeb],[(mb*g+ma*g)]]) #N\n",
- "R=np.linalg.solve(A,B) #N\n",
- "\n",
- "#Result\n",
- "print'The value of P and Na are:',round(R[0]),\"N\",'and',round(R[1]),\"N respectively.\"\n",
- "\n",
- "# The ans may wary due o decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P and Na are: 103.0 N and 153.0 N respectively.\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-7, Page no 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m1=40 #kg\n",
- "m2=13.5 #kg\n",
- "mu=3**-1 #coefficient of friction\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#Solving by substitution\n",
- "#After simplification we get\n",
- "x=mu*m2*g\n",
- "y=mu*(m1*g+m2*g)\n",
- "theta=arctan((x+y)/(m1*g))*(180/pi) #degrees\n",
- "\n",
- "#Result\n",
- "print'The value of the angle is',round(theta,1),\"degrees\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of the angle is 29.2 degrees\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-8, Page no 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "W=350 #lb\n",
- "# as theta=30 degrees, we have\n",
- "sintheta=2**-1\n",
- "costheta=sqrt(3)*2**-1\n",
- "# and phi=15 degrees,thus\n",
- "sinphi=0.2588\n",
- "cosphi=0.9659\n",
- "\n",
- "#Calculations\n",
- "#Solving by the matrix method\n",
- "A=np.array([[costheta,sinphi],[-sintheta,cosphi]])\n",
- "B=np.array([[W*sintheta],[W*costheta]])\n",
- "an=np.linalg.solve(A,B) #lb\n",
- "\n",
- "#Result\n",
- "print'The value of P and R are:',round(an[0],1),\"lb\",'and',round(an[1],1),\"lb respectively.\"\n",
- "\n",
- "# The ans may wary due to decimal point descrepancy."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of P and R are: 93.8 lb and 362.4 lb respectively.\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-9, Page no 135"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "# as theta=45 degrees, we have\n",
- "sintheta=sqrt(2)**-1\n",
- "costheta=sqrt(2)**-1\n",
- "m1=45 #kg\n",
- "m2=135 #kg\n",
- "g=9.81 #m/s**2\n",
- "mu=0.25 #coefficient of riction\n",
- "\n",
- "#Calculations\n",
- "N2=m2*g #N\n",
- "T=mu*N2 #N\n",
- "N1=m1*g*costheta #N\n",
- "Fprime1=N1*mu #N\n",
- "P=T+Fprime1-(m1*g*sintheta) #N\n",
- "\n",
- "#Result\n",
- "print'The values are'\n",
- "print'N2=',round(N2),\"N\"\n",
- "print'T=',round(T),\"N\"\n",
- "print'N1=',round(N1),\"N\"\n",
- "print'Fprime1=',round(Fprime1),\"N\"\n",
- "print'P=',round(P,1),\"N\"\n",
- "\n",
- "# The ans may wary due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The values are\n",
- "N2= 1324.0 N\n",
- "T= 331.0 N\n",
- "N1= 312.0 N\n",
- "Fprime1= 78.0 N\n",
- "P= 97.0 N\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-10, Page no 135"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "mu=0.2 #coefficient of friction\n",
- "F1=150 #lb\n",
- "F2=100 #lb\n",
- "# as theta=60 degrees\n",
- "costheta=2**-1\n",
- "# also theta1=30 degrees\n",
- "costheta1=sqrt(3)*2**-1\n",
- "\n",
- "#Calculations\n",
- "N1=F1*costheta #lb\n",
- "T=(mu*N1)+(F1*(costheta1)) #lb considering positive\n",
- "#Equilibrium for 100lb\n",
- "#Eliminating N2 from both equations\n",
- "#Taking derivative we get\n",
- "theta2=arctan(mu)*(180/pi) #degrees\n",
- "#Hence P becomes\n",
- "# in calculation of P we use the values of sin(theta2) & cos(theta2) as,\n",
- "sintheta2=0.196\n",
- "costheta2=0.98\n",
- "P=(F2*mu+T)*(costheta2+(mu*sintheta2))**-1 #lb\n",
- "\n",
- "#Result\n",
- "print'The minimum value of P is',round(P),\"lb\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The minimum value of P is 162.0 lb\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-11, Page no 136"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "F=180 #N\n",
- "m=100 #kg\n",
- "g=9.81 #m/s**2\n",
- "mu=0.25 #coeffiecient of friction\n",
- "\n",
- "#Calculations\n",
- "#Assuming F2 is maximum\n",
- "N2=F*2/(1+mu) #N\n",
- "F2=mu*N2 #N\n",
- "N1=m*g-F2 #N\n",
- "F1=F-F2 #N\n",
- "\n",
- "#Result\n",
- "print'The vaules are'\n",
- "print'N2=',round(N2,3),\"N\"\n",
- "print'F2=',round(F2,3),\"N\"\n",
- "print'N1=',round(N1,3),\"N\"\n",
- "print'F1=',round(F1,3),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The vaules are\n",
- "N2= 288.0 N\n",
- "F2= 72.0 N\n",
- "N1= 909.0 N\n",
- "F1= 108.0 N\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-13, Page no 138"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "mu_ca=0.3 #ceofficient of friction between copper block A and aluminium block B\n",
- "mu_af=0.2 #coefficient of friction between aluminium block B and Floor\n",
- "ma=3 #kg\n",
- "mb=2 #kg\n",
- "g=9.81 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "#For A\n",
- "#Taking sum of forces along X and Y direction\n",
- "Na=ma*g #N\n",
- "P=mu_ca*Na #N\n",
- "#For B\n",
- "#Taking sum of forces along X and Y direction\n",
- "Nb=Na+mb*g #N\n",
- "Fb=mu_ca*Na #N\n",
- "#Now largest value of friction before slip is \n",
- "Fprimeb=mu_af*Nb #N\n",
- "#Now as Fb<F'b hence initial assumption is incorrect and P=Fb\n",
- "P=Fb #N\n",
- "\n",
- "#Result\n",
- "print'The value of force that will cause motion is',round(P,2),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of force that will cause motion is 8.83 N\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-15, Page no 141"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d_m=2 #in mean diameter of the screw\n",
- "p=4**-1 #in\n",
- "mu=0.15 #coefficient of friction\n",
- "l=2 #ft\n",
- "L=4000 #lb\n",
- "\n",
- "#Calculations\n",
- "phi=arctan(mu)*(180/pi) #degrees\n",
- "beta=arctan(p*(pi*l)**-1)*(180/pi) #degrees\n",
- "x=phi+beta # degrees\n",
- "#Force to raise the load\n",
- "# Here the value of x=10.77 degrees, thus\n",
- "tanx=0.19\n",
- "P=(L*tanx)/(d_m*12) #lb\n",
- "#Force to lower the load\n",
- "# Also,\n",
- "y=phi-beta\n",
- "# Thus y yeilds 6.23 degrees, thus\n",
- "tany=0.109\n",
- "P2=(L*tany)/(d_m*12) #lb\n",
- "\n",
- "#Result\n",
- "print'The force to raise the load is',round(P,1 ),\"lb\"\n",
- "print'The force to lower the load is',round(P2,1),\"lb\"\n",
- "\n",
- "# The answer waries due to decimal point descrepancy"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force to raise the load is 31.7 lb\n",
- "The force to lower the load is 18.2 lb\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-16, Page no 141"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "r_m=2.338 #in\n",
- "d_m=3.25 #in\n",
- "mu=0.06 #coefficient of friction\n",
- "P=1500 #lb\n",
- "p=4**-1 #pitch\n",
- "\n",
- "#Calculation\n",
- "phi=arctan(mu)*(180/pi) #degrees\n",
- "beta=round(arctan(p/(2*pi*r_m))*(180/pi)) #degrees\n",
- "x=phi+beta\n",
- "# Thus tanx yeilds\n",
- "tanx=0.077\n",
- "M=P*r_m*tanx+mu*P*(d_m/2) #lb.in\n",
- "\n",
- "#Result\n",
- "print'The moment required is',round(M),\"lb.in\"\n",
- "\n",
- "#Decimal accuracy causes discrepancy in answers\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The moment required is 416.0 lb.in\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-17, Page no 141"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=750 #mm diameter\n",
- "alpha=pi #wrap angle radians\n",
- "mu=0.25 #coefficient of friction\n",
- "T_t=200 #N tension on the tight side\n",
- "\n",
- "#Calculation\n",
- "T2=T_t/(exp(mu*alpha)) #N\n",
- "\n",
- "#Result\n",
- "print'The tension of the slack side is',round(T2,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The tension of the slack side is 91.2 N\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-18, Page no 141"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=635 #mm diameter of the drum\n",
- "P=178 #N\n",
- "mu=3**-1 #coefficient of friction\n",
- "l1=100 #mm \n",
- "l2=660 #mm\n",
- "# as theta1=60 degrees\n",
- "sintheta1=sqrt(3)*2**-1\n",
- "costheta1=2**-1\n",
- "# as theta2=30 degrees,\n",
- "sintheta2=2**-1\n",
- "costheta2=sqrt(3)*2**-1\n",
- "GD=d/2 #mm\n",
- "\n",
- "#Calculations\n",
- "#Taking moment about point C\n",
- "Tb=(P*(l1+l2))/(l1*sintheta1) #N\n",
- "CD=((d/2)-(l1*costheta2))/sintheta2 #mm\n",
- "#from fig 9-22(b) \n",
- "theta=arcsin(GD/CD)*(180/pi) #degrees\n",
- "#from fig9-22(c)\n",
- "w_d=(180+30+theta) #degrees\n",
- "w=(w_d)*(pi/180) #radians\n",
- "#As Tc is greater than Tb\n",
- "Tc=Tb*(exp(mu*w)) #N\n",
- "M=(Tc-Tb)*GD #N.mm\n",
- "an=M/1000 #N.m\n",
- "\n",
- "#Result\n",
- "print'The braking moment required is',round(an),\"N.m\"\n",
- "\n",
- "#Note the unit of the final enswer carefully\n",
- "# The answer is off by 2 N.m"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The braking moment required is 1668.0 N.m\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-19, Page no 142"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "L=1000 #lb\n",
- "P=10 #lb\n",
- "\n",
- "#Calculations\n",
- "mu=log(L/P)/(4*2*pi) \n",
- "\n",
- "#Result\n",
- "print'The coefficient of friction is',round(mu,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The coefficient of friction is 0.18\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-20, Page no 142"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "m=900 #kg\n",
- "mu=0.2 #coefficient of friction\n",
- "g=9.8 #m/s**2\n",
- "\n",
- "#Calculations\n",
- "T2=m*g/(exp(2*2*pi*mu)) #N\n",
- "\n",
- "#Result\n",
- "print'The force needed to hold the mass is',round(T2),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force needed to hold the mass is 714.0 N\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9-21, Page no 143"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "d=760 #mm\n",
- "W=500 #N\n",
- "a=0.305 #mm coefficient of rolling resisatnce\n",
- "r=d/2 #mm\n",
- "\n",
- "#Calculations\n",
- "P=(W*a)/r #N\n",
- "\n",
- "#Result\n",
- "print'The force necessary is',round(P,1),\"N\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The force necessary is 0.4 N\n"
- ]
- }
- ],
- "prompt_number": 33
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file |