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diff --git a/Engineering_Mechanics_by_Tayal_A.K./chapter14_4.ipynb b/Engineering_Mechanics_by_Tayal_A.K./chapter14_4.ipynb new file mode 100755 index 00000000..41eaaa76 --- /dev/null +++ b/Engineering_Mechanics_by_Tayal_A.K./chapter14_4.ipynb @@ -0,0 +1,1056 @@ +{
+ "metadata": {
+ "name": "chapter14.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 14: Rectilinear Motion Of A Particle"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14-3,Page No:335"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "a_T=0.18 # m/s^2 # acc of trolley\n",
+ "# Calculations\n",
+ "a_B=-a_T*3**-1 # m/s^2 # from eq'n 4\n",
+ "t=4 # seconds\n",
+ "v_T=a_T*t # m/s # velocity of trolley after 4 seconds\n",
+ "v_B=-v_T*3**-1 # m/s # from eq'n 3\n",
+ "S_T=(0.5)*a_T*t**2 # m # distance moved by trolley in 4 sec\n",
+ "S_B=-S_T*3**-1 # m # from eq'n 2\n",
+ "# Results\n",
+ "\n",
+ "print\"The acceleration of block B is \",round(a_B,2),\"m/s^2\"\n",
+ "print\"The velocity of trolley & the block after 4 sec is \",round(v_T,2),\"m/s\",\"&\",round(v_B,2),\"m/s\"\n",
+ "print\"The distance moved by the trolley & the block is \",round(S_T,2),\"m\",\"&\",round(S_B,2),\"m\"\n",
+ "# The -ve sign indicates that the velocity or the distance travelled is in opposite direction.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The acceleration of block B is -0.06 m/s^2\n",
+ "The velocity of trolley & the block after 4 sec is 0.72 m/s & -0.24 m/s\n",
+ "The distance moved by the trolley & the block is 1.44 m & -0.48 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14-4,Page No:338"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initiliztion of variables\n",
+ "\n",
+ "v_B=12 # cm/s # velocity of block B\n",
+ "u=0\n",
+ "s=24 # cm # distance travelled by bock B\n",
+ "t=5 # seconds\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "a_B=v_B**2/(2*s) # cm/s^2 # using eq'n v^2-u^2=28*a*s for block B. Here u=0\n",
+ "a_A=(1.5)*a_B # cm/s^2 # from eq'n 4 # Here a_A is negative which means acceleration is in opposite direction. However we consider +ve values for further calculations\n",
+ "v_A=u+(a_A*t) # m/s # using eq'n v=u+(a*t)\n",
+ "S_A=(u*t)+((0.5)*a_A*t**2) # m # using eq'n S=(u*t)+((1/2)*a*t^2)\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The acceleration of block A (a_A) is \",round(a_A,1),\"cm/s^2\"\n",
+ "print\"The acceleration of block B (a_B) is \",round(a_B),\"cm/s^2\"\n",
+ "print\"The velocity of block A (v_A) after 5 seconds is \",round(v_A,1),\"m/s\"\n",
+ "print\"The position of block A (S_A) after 5 seconds is \",round(S_A,2),\"m\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The acceleration of block A (a_A) is 4.5 cm/s^2\n",
+ "The acceleration of block B (a_B) is 3.0 cm/s^2\n",
+ "The velocity of block A (v_A) after 5 seconds is 22.5 m/s\n",
+ "The position of block A (S_A) after 5 seconds is 56.25 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14-5,Page No:340"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "u=72*1000*60**-2 # km/hr # speed of the vehicle\n",
+ "s=300 # m # distance where the light is turning is red\n",
+ "t=20 # s # traffic light timed to remain red\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# Now to find the acceleration we use the eq'n s=u*t+(1/2)*a*t^2\n",
+ "a=(((s)-(u*t))*2)*t**-2 # m/s^2 (Deceleration) \n",
+ "v=(u+(a*t))*(60*60*0.001) # km/hr # here we multiply with (60*60)/1000 to convert m/s to km/hr\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"(a) The required uniform acceleration of the car is \",round(a,2),\"m/s^2\"\n",
+ "print\"(b) The speed at which the motorist crosses the traffic light is \",round(v),\"km/hr\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The required uniform acceleration of the car is -0.5 m/s^2\n",
+ "(b) The speed at which the motorist crosses the traffic light is 36.0 km/hr\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14-6,Page No:340"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "S=50 # m # height of the tower\n",
+ "v=25 # m/s # velocity at which the stone is thrown up from the foot of the tower\n",
+ "g=9.81 # m/s^2 # acc due to graity\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# The equation of time for the two stones to cross each other is given as,\n",
+ "t=S/v # seconds\n",
+ "S_1=(0.5)*g*t**2 # m # from the top\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The time (t) at which the two stones cross each other is \",round(t),\"seconds\"\n",
+ "print\"The two stones cross each other (from top) at a distance of \",round(S_1,1),\"m\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The time (t) at which the two stones cross each other is 2.0 seconds\n",
+ "The two stones cross each other (from top) at a distance of 19.6 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14-7,Page No:341"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Intilization of variables\n",
+ "\n",
+ "acc=0.5 # m/s^2 # acceleration of the elevator\n",
+ "s=25 # m # distance travelled by the elevator from the top\n",
+ "u=0 # m/s\n",
+ "g=9.81 # m/s^2 # acc due to gravity\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# Using eq'n the eq'n v^2-u^2=2*a*s, solving for v we get,\n",
+ "v=sqrt((2*acc*s)+(u^2)) # m/s \n",
+ "# Now solving eq'n 1 & 2 for t we get, (4.655*t^2)-(5*t)+(25)=0\n",
+ "# Find the roots of the eq'n using the eq'n,t=(-b+sqrt(b^2-(4*a*c)))/(2*a).In this eq'n the values of a,b & c are,\n",
+ "a=4.655\n",
+ "b=-5\n",
+ "c=-25\n",
+ "t=(-b+((b**2)-(4*a*c))**0.5)/(2*a) # seconds\n",
+ "\n",
+ "# Let S_1 be the distance travelled by the elevator after it travels 25 m from top when the stone hits the elevator,This disance S_1 is given as,\n",
+ "S_1=(v*t)+((1/2)*acc*t**2) # m\n",
+ "\n",
+ "# Let S be the total dist from top when the stone hits the elevator,\n",
+ "S=S_1+s # m\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The time taken by the stone to hit the elevator is \",round(t,3),\"second\"\n",
+ "print\"The distance (S)travelled by the elevator at the time of impact is \",round(S,2),\"m\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The time taken by the stone to hit the elevator is 2.916 second\n",
+ "The distance (S)travelled by the elevator at the time of impact is 40.15 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14-9,Page No:343"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "v=60 # km/hr # velocity of the train\n",
+ "d1=15 # km # Distance travelled by the local train from the velocity-time graph (here d1= Area OED)\n",
+ "d2=12 # km # from the velocity-time graph (here d2= Area OABB')\n",
+ "d3=3 # km # from the velocity-time graph (here d3= Area BB'C)\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "t_1=d2*v**-1 # hr # time of travel for first 12 kms\n",
+ "t_2=(2*d3)*v**-1 # hr # time of for next 3 kms\n",
+ "\n",
+ "# Total time of travel for passenger train is given by eq'n\n",
+ "t=t_1+t_2 # hr\n",
+ "\n",
+ "# Now time of travel of the local train (let it be T) is given as,\n",
+ "T=2*t # hr\n",
+ "V_max=2*d1/T # km/hr\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The maximum speed of the local train is \",round(V_max),\"km/hr\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum speed of the local train is 50.0 km/hr\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14-10,Page No:345"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "a=10 # m/s^2 # acceleration of the particle\n",
+ "S_5th=50 # m # distance travelled by the particle during the 5th second\n",
+ "t=5 # seconds\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# The distance travelled by the particle in time t is given by, S=(u*t)+(1/2)*a*t^2.....(consider this as eq'n 1)\n",
+ "# Here, The distance travelled by the particle in the 5th second=The distance travelled in 5 seconds - The distance travelled in 4 seconds..... (consider eq'n 2)\n",
+ "# Using eq'n 1: S_(0-5)=(5*u)+(1/2)*10*5^2 = 5*u+125.....(consider eq'n 3)\n",
+ "# again, S_(0-4)=(4*u)+(1/2)*10*4^2 = 4*u+80....(consider eq'n 4)\n",
+ "\n",
+ "# Now,put eq'n 3&4 in eq'n 2 and solve for u. We get, 50=[(5*u+125)-(4*u+80)] i.e 50=u+45\n",
+ "u=(S_5th)-45 # m/s\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The initial velocity of the particle is \",round(u),\"m/s\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The initial velocity of the particle is 5.0 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14-11,Page No:345"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "# Conditions given are\n",
+ "t=1 # s\n",
+ "x=14.75 # m\n",
+ "v=6.33 # m/s\n",
+ "# Calculations\n",
+ "# We use expression 1,2 & 3 to find distance,velocity & acceleration of the particle after 2 sec\n",
+ "T=2 # sec\n",
+ "X=(T**4*12**-1)-(T**3*3**-1)+(T**2)+(5*T)+9 # m # eq'n 3\n",
+ "V=(T**3*3**-1)-(T**2)+(2*T)+5 # m/s \n",
+ "a=(T**2)-(2*T)+2 # m/s^2\n",
+ "# Results\n",
+ "\n",
+ "print\"The distance travelled by the particle is \",round(X,2),\"m\"\n",
+ "print\"The velocity of the particle is \",round(V,2),\"m/s\"\n",
+ "print\"The acceleration of the particle is \",round(a),\"m/s^2\"\n",
+ "# The answer may vary due to decimal point error\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The distance travelled by the particle is 21.67 m\n",
+ "The velocity of the particle is 7.67 m/s\n",
+ "The acceleration of the particle is 2.0 m/s^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14-12,Page No:346"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Calculations\n",
+ "# From eq'n 2 it is clear that velocity of the particle becomes zero at t=3 sec\n",
+ "t=3 # sec .. from eq'n 2\n",
+ "# Position of particle at t=3 sec\n",
+ "x=(t**3)-(3*t**2)-(9*t)+12 # m # from eq'n 1\n",
+ "# Acc of particle at t=3 sec\n",
+ "a=6*(t-1) # m/s^2 # from eq'n 3\n",
+ "# Results\n",
+ "\n",
+ "print\"The time at which the velocity of the particle becomes zero is \",round(t),\"sec\"\n",
+ "print\"The position of the partice at t=3 sec is \",round(x),\"m\"\n",
+ "print\"The acceleration of the particle is \",round(a),\"m/s^2\"\n",
+ "# Ref textbook for the graphs\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The time at which the velocity of the particle becomes zero is 3.0 sec\n",
+ "The position of the partice at t=3 sec is -15.0 m\n",
+ "The acceleration of the particle is 12.0 m/s^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14-15,Page No:354"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of variables\n",
+ "\n",
+ "F=250 # N # Force acting on a body\n",
+ "m=100 # kg # mass of the body\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# Using the eq'n of motion\n",
+ "a=F*m**-1 # m/s^2 \n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The acceleration of the body is \",round(a,2),\"m/s^2\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The acceleration of the body is 2.5 m/s^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 70
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14-16,Page No:354"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "a=1 # m/s^2 # downward/upward acceleration of the elevator\n",
+ "W=500 # N # Weight of man\n",
+ "g=9.81 # m/s^2 # acceleration due to gravity\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# (a) Downward Motion \n",
+ "R_1=W*(1-(a/g)) # N # (Assume pressure as R_1)\n",
+ "\n",
+ "# (b) Upward Motion\n",
+ "R_2=W*(1+(a/g)) # N # (Assume pressure as R_2)\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"(a) The pressure transmitted to the floor by the man for Downward motion of the elevator is \",round(R_1),\"N\"\n",
+ "print\"(b) The pressure transmitted to the floor by the man for Upward motion of the elevator is \",round(R_2,2),\"N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The pressure transmitted to the floor by the man for Downward motion of the elevator is 449.0 N\n",
+ "(b) The pressure transmitted to the floor by the man for Upward motion of the elevator is 550.97 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 72
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14-17,Page no:355"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "W=5000 # N # Total weight of the elevator\n",
+ "u=0 # m/s\n",
+ "v=2 # m/s # velocity of the elevator\n",
+ "s=2 # m # distance traveled by the elevator\n",
+ "t=2 # seconds # time to stop the lift\n",
+ "w=600 # N # weight of the man\n",
+ "g=9.81 # m/s^2 # acc due to gravity\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# Acceleration acquired by the elevator after travelling 2 m is given by,\n",
+ "a=(((v**2-u**2)**0.5/2)) # m/s^2\n",
+ "\n",
+ "# (a) Let T be the the tension in the cable which is given by eq'n,\n",
+ "T=W*(1+(a/g)) # N\n",
+ "\n",
+ "# (b) Motion of man\n",
+ "# Let R be the pressure experinced by the man.Then from the Eq'n of motion of man pressure is given as,\n",
+ "R=w*(1-(a/g)) # N \n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"(a) The Tensile force in the cable is \",round(T,1),\"N\"\n",
+ "print\"(b) The pressure transmitted to the floor by the man is \",round(R,1),\"N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The Tensile force in the cable is 5509.7 N\n",
+ "(b) The pressure transmitted to the floor by the man is 538.8 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 83
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14-18,Page No:355"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "M_1=10 # kg # mass of the 1st block\n",
+ "M_2=5 # kg # mass of the 2nd block\n",
+ "mu=0.25 # coefficient of friction between the blocks and the surface\n",
+ "g=9.81 # m/s^2 # acc due to gravity\n",
+ "\n",
+ "# Calculations\n",
+ "a=g*(M_2-(mu*M_1))/(M_1+M_2) # m/s^2 # from eq'n 5\n",
+ "T=M_1*M_2*g*(1+mu)/(M_1+M_2) # N # from eq'n 6\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The acceleration of the masses is \",round(a,3),\"m/s^2\"\n",
+ "print\"The tension in the string is \",round(T,3),\"N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The acceleration of the masses is 1.635 m/s^2\n",
+ "The tension in the string is 40.875 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 85
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14-19,Page No:357"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "M_1=150 # kg # mass of the 1st block\n",
+ "M_2=100 # kg # mass of the 2nd block\n",
+ "mu=0.2 # coefficient of friction between the blocks and the inclined plane\n",
+ "g=9.81 # m/s^2 # acc due to gravity\n",
+ "theta=45 # degree # inclination of the surface\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# substuting the value of eq'n 3 in eq'n 1 & solving for T,we get value of T as,\n",
+ "T=((M_1*M_2*g)*(sin(theta*(pi/180))+2-(mu*cos(theta*(pi/180)))))/((4*M_1)+(M_2)) # N\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The tension in the string during the motion of the system is \",round(T,1),\"N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The tension in the string during the motion of the system is 539.3 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 87
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14-20,Page No:358"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "M_1=5 # kg # mass of the 1st block\n",
+ "theta_1=30 # degree # inclination of the 1st plane\n",
+ "M_2=10 # kg # mass of the 2nd block\n",
+ "theta_2=60 # degree # inclination of the 2nd plane\n",
+ "mu=0.33 # coefficient of friction between the blocks and the inclined plane\n",
+ "g=9.81 # m/s^2 # acc due to gravity\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# solving eq'n 1 & 2 for a we get,\n",
+ "a=((((M_2*(sin(theta_2*(pi/180))-(mu*cos(theta_2*(pi/180)))))-(M_1*(sin(theta_1*(pi/180))+(mu*cos(theta_1*(pi/180)))))))*g)/(M_1+M_2) # m/s^2\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The acceleration of the masses is \",round(a,3),\"m/s^2\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The acceleration of the masses is 2.015 m/s^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 89
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14-21,Page No:359"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "S=5 # m # distance between block A&B\n",
+ "mu_A=0.2 # coefficient of friction between the block A and the inclined plane\n",
+ "mu_B=0.1 # coefficient of friction between the block B and the inclined plane\n",
+ "theta=20 # degree # inclination of the pane\n",
+ "g=9.81 # m/s^2 # acc due to gravity\n",
+ "\n",
+ "# Calculatio#\n",
+ "\n",
+ "# EQUATION OF MOTION OF BLOCK A:\n",
+ "# Let a_A & a_B be the acceleration of block A & B.\n",
+ "a_A=(g*sin(theta*(pi/180)))-(mu_A*g*cos(theta*(pi/180))) # m/s^2 # from eq'n 1 & eq'n 2\n",
+ "\n",
+ "# EQUATION OF MOTION OF BLOCK B:\n",
+ "a_B=g*((sin(theta*(pi/180)))-(mu_B*cos(theta*(pi/180)))) # m/s^2 # from eq'n 3 & Rb\n",
+ "\n",
+ "# Now the eq'n for time of collision of the blocks is given as,\n",
+ "t=((S*2)/(a_B-a_A))**0.5 # seconds \n",
+ "S_A=(0.5)*a_A*t**2 # m # distance travelled by block A\n",
+ "S_B=(0.5)*a_B*t**2 # m # distance travelled by block B\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The time before collision is \",round(t,2),\"seconds\"\n",
+ "print\"The distance travelled by block A before collision is \",round(S_A,2),\"m\"\n",
+ "print\"The distance travelled by block B before collision is \",round(S_B,2),\"m\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The time before collision is 3.29 seconds\n",
+ "The distance travelled by block A before collision is 8.2 m\n",
+ "The distance travelled by block B before collision is 13.2 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 96
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14-22,Page No:361"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "P=50 # N # Weight of the car\n",
+ "Q=100 # N # Weight of the rectangular block\n",
+ "g=9.81 # m/s^2 # acc due to gravity\n",
+ "b=25 # cm # width of the rectangular block\n",
+ "d=50 # cm # depth of the block\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "a=(Q*g)/(4*P+2*Q) # m/s^2 # from eq'n 4\n",
+ "W=(Q*(P+Q))/(4*P+Q) # N # from eq'n 6\n",
+ "\n",
+ "# Resuts\n",
+ "\n",
+ "print\"The maximum value of weight (W) by which the car can be accelerated is \",round(W),\"N\"\n",
+ "print\"The acceleration is \",round(a,2),\"m/s^2\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum value of weight (W) by which the car can be accelerated is 50.0 N\n",
+ "The acceleration is 2.45 m/s^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 99
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14-23,Page No:363"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "P=40 # N # weight on puley r_1\n",
+ "Q=60 # N # weight on pulley r_2\n",
+ "g=9.81 # m/s^2 # acc due to gravity\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# The eq'n for acceleration of pulley Pi.e a_p is,\n",
+ "a_p=(2*P-Q)*(4*P+Q)**-1*2*g # m/s^2\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The downward acceleration of P is \",round(a_p,3),\"m/s^2\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The downward acceleration of P is 1.784 m/s^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 112
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14-24,Page No:364"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "M=15 # kg # mass of the wedge\n",
+ "m=6 # kg # mass of the block\n",
+ "theta=30 # degree # angle of the wedge\n",
+ "g=9.81 # m/s^2 # acc due to gravity\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "a_A=((m*g*cos(theta*(pi/180))*sin(theta*(pi/180)))/((M)+(m*(sin(theta*(pi/180)))**2)))/(g) # g # By eliminating R_1 from eq'n 1&3.\n",
+ "\n",
+ "# Here, assume a_r is the acceleration of block B relative to wedge A which is given by substuting a_A in eq'n 2\n",
+ "\n",
+ "a_r=(((g*sin(theta*(pi/180)))*(m+M))/((M)+(m*(sin(theta*(pi/180)))**2)))/(g) # g\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"(a) The acceleration of the wedge is \",round(a_A,3),\"g\"\n",
+ "print\"(b) The acceleration of the bock relative to the wedge is \",round(a_r,3),\"g\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The acceleration of the wedge is 0.157 g\n",
+ "(b) The acceleration of the bock relative to the wedge is 0.636 g\n"
+ ]
+ }
+ ],
+ "prompt_number": 114
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14-25,Page No:366"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "P=30 # N # weight on pulley A\n",
+ "Q=20 # N # weight on pulley B\n",
+ "R=10 # N # weight on puey B\n",
+ "g=9.81 # m/s^2 # acc due to gravity\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# Solving eqn's 6 & 7 using matrix for a & a_1, we get\n",
+ "A=np.array([[70 ,-40],[-10, 30]])\n",
+ "B=np.array([10,-10])\n",
+ "C=np.linalg.solve(A,B)\n",
+ "\n",
+ "# Acceleration of P is given as,\n",
+ "P=C[0] # m/s^2\n",
+ "# Acceleration of Q is given as,\n",
+ "Q=C[1]-C[0] # m/s^2\n",
+ "# Acceleration of R is given as,\n",
+ "R=-(C[1]+C[0]) # m/s^2 # as R is taken to be +ve\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The acceleration of P is \",round(P,2),\"g\"\n",
+ "print\"The acceleration of Q is \",round(Q,2),\"g\"\n",
+ "print\"The acceleration of R is \",round(R,2),\"g\"\n",
+ "# Here the -ve sign indicates deceleration or backward/downward acceleation.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The acceleration of P is -0.06 g\n",
+ "The acceleration of Q is -0.29 g\n",
+ "The acceleration of R is 0.41 g\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14-30,Page No:372"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "W=1 # kg/m # weight of the bar\n",
+ "L_AB=0.6 # m # length of segment AB\n",
+ "L_BC=0.30 # m # length of segment BC\n",
+ "g=9.81 # m/s^2 # acc due to gravity\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# Consider the respective F.B.D.\n",
+ "theta_1=arctan(5/12)*(180/pi) # slope of bar AB # here theta_1= atan(theta)\n",
+ "theta_2=arcsin(5/13)*(180/pi) # theta_2=asin(theta)\n",
+ "theta_3=arccos(12/13)*(180/pi) # theta_3=acos(theta)\n",
+ "M_AB=L_AB*W # kg acting at D # Mass of segment AB\n",
+ "M_BC=L_BC*W # kg acting at E # Mass of segment BC\n",
+ "\n",
+ "# The various forces acting on the bar are:\n",
+ "\n",
+ "# Writing the eqn's of dynamic equilibrium\n",
+ "Y_A=(L_AB*g)+(L_BC*g) # N # sum F_y=0\n",
+ "\n",
+ "# Using moment eq'n Sum M_A=0:Here,in this eq'n the values are as follows,\n",
+ "AF=L_BC*cos(theta_3) \n",
+ "DF=L_BC*sin(theta_2)\n",
+ "AH=(L_AB*cos(theta_3))+((L_BC/2)*sin(theta_2))\n",
+ "IG=(L_AB*sin(theta_2)-((L_BC/2)*cos(theta_3)))\n",
+ "\n",
+ "# On simplifying and solving moment eq'n we get a as,\n",
+ "a=(g*(L_AB*(DF)+L_BC*(IG)))*(L_AB*(AF)+L_BC*(AH))**-1\n",
+ "#a=(2*L_AB*L_BC*g*sin(theta_2))-(L_BC*g*(L_BC/2)*cos(theta_3))/((2*L_AB*L_BC*cos(theta_3*(pi/180)))+(L_BC*(L_BC/2)*sin(theta_2*(pi/180)))) # m/s^2\n",
+ "X_A=0.9*a #N # from eq'n of dynamic equilibrium\n",
+ "R_A=(X_A**2+Y_A**2)**0.5 # N # Resultant of R_A\n",
+ "alpha=arctan(Y_A/X_A)*(180/pi) # degree\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The acceleration is \",round(a,4),\"m/s^2\"\n",
+ "print\"The reaction at A (R_A) is \",round(R_A,3),\"N\"\n",
+ "print\"The angle made by the resultant is \",round(alpha,2),\"degree\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The acceleration is -1.2263 m/s^2\n",
+ "The reaction at A (R_A) is 8.898 N\n",
+ "The angle made by the resultant is -82.87 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |