From 4a1f703f1c1808d390ebf80e80659fe161f69fab Mon Sep 17 00:00:00 2001 From: Thomas Stephen Lee Date: Fri, 28 Aug 2015 16:53:23 +0530 Subject: add books --- .../chapter14_5.ipynb | 1056 ++++++++++++++++++++ 1 file changed, 1056 insertions(+) create mode 100755 Engineering_Mechanics_by_Tayal_A.K./chapter14_5.ipynb (limited to 'Engineering_Mechanics_by_Tayal_A.K./chapter14_5.ipynb') diff --git a/Engineering_Mechanics_by_Tayal_A.K./chapter14_5.ipynb b/Engineering_Mechanics_by_Tayal_A.K./chapter14_5.ipynb new file mode 100755 index 00000000..41eaaa76 --- /dev/null +++ b/Engineering_Mechanics_by_Tayal_A.K./chapter14_5.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 -- cgit