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 --- .../chapter20_9.ipynb | 546 +++++++++++++++++++++ 1 file changed, 546 insertions(+) create mode 100755 Engineering_Mechanics_by_Tayal_A.K./chapter20_9.ipynb (limited to 'Engineering_Mechanics_by_Tayal_A.K./chapter20_9.ipynb') diff --git a/Engineering_Mechanics_by_Tayal_A.K./chapter20_9.ipynb b/Engineering_Mechanics_by_Tayal_A.K./chapter20_9.ipynb new file mode 100755 index 00000000..7b30b675 --- /dev/null +++ b/Engineering_Mechanics_by_Tayal_A.K./chapter20_9.ipynb @@ -0,0 +1,546 @@ +{ + "metadata": { + "name": "chapter20.ipynb" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 20: Motion Of Projectile" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.20-1,Page No:518" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of variables\n", + "\n", + "v_o=500 # m/s # velocity of the projectile\n", + "alpha=30 # angle at which the projectile is fired\n", + "t=30 # seconds\n", + "g=9.81 # m/s^2 # acc due to gravity\n", + "\n", + "# Calculations\n", + "\n", + "v_x=v_o*cos(alpha*(pi/180)) # m/s # Initial velocity in the horizontal direction\n", + "v_y=v_o*sin(alpha*(pi/180)) # m/s # Initial velocity in the vertical direction\n", + "\n", + "# MOTION IN HORIZONTA DIRECTION:\n", + "V_x=v_x # m/s # V_x=Horizontal velocity after 30 seconds\n", + "\n", + "# MOTION IN VERTICAL DIRECTION: # using the eq'n v=u+a*t\n", + "V_y=v_y-(g*t) # m/s # -ve sign denotes downward motion\n", + "\n", + "# Let the Resultant velocity be v_R. It is given as,\n", + "v_R=((V_x)**2+(-V_y)**2)**0.5# m/s\n", + "theta=arctan((-V_y)/V_x)*(180/pi) # degree # direction of the projectile\n", + "\n", + "# Results\n", + "\n", + "print\"The velocity of the projectile is \",round(v_R,2),\"m/s\" # The answer of velocity is wrong in the text book.\n", + "print\"The direction of the projectile is \",round(theta,2),\"degree\" # -ve value of theta indicates that the direction is in downward direction\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The velocity of the projectile is 435.27 m/s\n", + "The direction of the projectile is 5.84 degree\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.20-2,Page no:519" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initilization of variables\n", + "\n", + "v_A=10 # m/s # velocity of body A\n", + "alpha_A=60 # degree # direction of body A\n", + "alpha_B=45 # degree # direction of body B\n", + "\n", + "# Calculations\n", + "\n", + "# (a) The velocity (v_B) for the same range is given by eq'n;\n", + "v_B=((v_A**2*sin(2*alpha_A*(pi/180)))/(sin(2*alpha_B*(pi/180))))**0.5 # m/s\n", + "\n", + "# (b) Now velocity v_B for the same maximum height is given as,\n", + "v_b=((v_A**2)*((sin(alpha_A*(pi/180)))**2/(sin(alpha_B*(pi/180)))**2))**0.5 # m/s\n", + "\n", + "# (c) Now the velocity (v) for the equal time of flight is;\n", + "v=(v_A*sin(alpha_A*(pi/180)))/(sin(alpha_B*(pi/180))) # m/s\n", + "\n", + "# Results\n", + "\n", + "print\"(a) The velocity of body B for horizontal range is \",round(v_B,1),\"m/s\"\n", + "print\"(b) The velocity of body B for the maximum height is \",round(v_b,2),\"m/s\"\n", + "print\"(c) The velocity of body B for equal time of flight is \",round(v,2),\"m/s\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a) The velocity of body B for horizontal range is 9.3 m/s\n", + "(b) The velocity of body B for the maximum height is 12.25 m/s\n", + "(c) The velocity of body B for equal time of flight is 12.25 m/s\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.20-3,Page No:520 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initilization of variables\n", + "\n", + "y=3.6 # m # height of the wall\n", + "x_1=4.8 # m # position of the boy w.r.t the wall\n", + "x_2=3.6 # m # distance from the wall where the ball hits the ground\n", + "g=9.81 # m/s^2 # acc due to gravity\n", + "\n", + "# Calculations\n", + "\n", + "# The range of the projectile is r, given as,\n", + "r=x_1+x_2 # m\n", + "\n", + "# Let the angle of the projection be alpha, which is derived and given as,\n", + "alpha=arctan((y)/(x_1-(x_1**2/r)))*(180/pi) # degree\n", + "\n", + "# Now substuting the value of alpha in eq'n 3 we get the least velocity (v_o) as;\n", + "v_o=((g*r)/(sin(2*alpha*(pi/180))))**0.5 # m/s\n", + "\n", + "# Results\n", + "\n", + "print\"The least velocity with which the ball can be thrown is \",round(v_o,2),\"m/s\"\n", + "print\"The angle of projection for the same is \",round(alpha,1),\"degree\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The least velocity with which the ball can be thrown is 9.78 m/s\n", + "The angle of projection for the same is 60.3 degree\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.20-5,Page No:523 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initilization of variables\n", + "\n", + "v_o=400 # m/s # initial velocity of each gun\n", + "r=5000 # m # range of each of the guns\n", + "g=9.81 # m/s^2 # acc due to gravity\n", + "p=180 # degree \n", + "pi=3.14\n", + "\n", + "# Calculations\n", + "\n", + "# now from eq'n 1\n", + "theta_1=(arcsin(r*g/v_o**2)*(180/pi))/2 # degree # angle at which the 1st gun is fired\n", + "\n", + "# from eq'n 3\n", + "theta_2=(p-2*theta_1)/2 # degree \n", + "\n", + "# For 1st & 2nd gun, s is\n", + "s=r # m\n", + "\n", + "# For 1st gun \n", + "v_x=v_o*cos(theta_1*(pi/180)) # m/s\n", + "\n", + "# Now the time of flight for 1st gun is t_1, which is given by relation,\n", + "t_1=s*(v_x)**-1 # seconds\n", + "\n", + "# For 2nd gun\n", + "V_x=v_o*cos(theta_2*(pi/180))\n", + "\n", + "# Now the time of flight for 2nd gun is t_2\n", + "t_2=s/V_x # seconds\n", + "\n", + "# Let the time difference between the two hits be delta.T. Then,\n", + "T=t_2-t_1 # seconds\n", + "\n", + "# Results\n", + "\n", + "print\"The time difference between the two hits is \",round(T,2),\"seconds\" #answer varies by 0.6 sec due to decimal variance\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The time difference between the two hits is 67.5 seconds\n" + ] + } + ], + "prompt_number": 99 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.20-6,Page No:524" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initilization of variables\n", + "\n", + "h=2000 # m/ height of the plane\n", + "v=540*1000*3600**-1 # m/s # velocity of the plane\n", + "g=9.81 # m/s^2 # acc due to gravity\n", + "pi=3.14\n", + "\n", + "# Calculations\n", + "\n", + "# Time t required to travel down a height 2000 m is given by eq'n,\n", + "u=0 # m/s # initial velocity\n", + "t=(2*h/g)**0.5 # seconds\n", + "\n", + "# Now let s be the horizonta distance travelled by the bomb in time t seconds, then\n", + "s= v*t # m\n", + "\n", + "# angle is given as theta,\n", + "theta=arctan(h/s)*(180/pi) # degree\n", + "\n", + "# Results\n", + "\n", + "print\"The pilot should release the bomb from a distance of \",round(s),\"m\" #the answer varies by 1m due to decimal variance\n", + "print\"The angle at which the target would appear is \",round(theta,1),\"degree\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The pilot should release the bomb from a distance of 3029.0 m\n", + "The angle at which the target would appear is 33.5 degree\n" + ] + } + ], + "prompt_number": 70 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.20-7,Page No:525" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initilization of variables\n", + "\n", + "theta=30 # degree # angle at which the bullet is fired\n", + "s=-50 # position of target below hill\n", + "v=100 # m/s # velocity at which the bullet if fired\n", + "g=9.81 # m/s^2 \n", + "\n", + "# Calculations\n", + "\n", + "v_x=v*cos(theta*(pi/180)) # m/s # Initial velocity in horizontal direction\n", + "v_y=v*sin(theta*(pi/180)) # m/s # Initial velocity in vertical direction\n", + "\n", + "# (a) Max height attained by the bullet\n", + "h=v_y**2/(2*g) # m\n", + "\n", + "# (b)Let the vertical Velocity with which the bullet will hit the target be V_y. Then,\n", + "V_y=((2*-9.81*s)+(v_y)**2)**0.5 # m/s # the value of V_y is +ve & -ve\n", + "\n", + "# Let V be the velocity with wich it hits the target\n", + "V=((v_x)**2+(V_y)**2)**0.5 # m/s\n", + "\n", + "# (c) The time required to hit the target\n", + "a=g # m/s^2\n", + "t=(v_y-(-V_y))/a # seconds\n", + "\n", + "# Results\n", + "\n", + "print\"(a) The maximum height to which the bullet will rise above the soldier is \",round(h,1),\"m\"\n", + "print\"(b) The velocity with which the bullet will hit the target is \",round(V,1),\"m/s\"\n", + "print\"(c) The time required to hit the target is \",round(t,1),\"seconds\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a) The maximum height to which the bullet will rise above the soldier is 127.3 m\n", + "(b) The velocity with which the bullet will hit the target is 104.8 m/s\n", + "(c) The time required to hit the target is 11.1 seconds\n" + ] + } + ], + "prompt_number": 72 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.20-8,Page No:527" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initilization of variables\n", + "\n", + "W=30 # N # Weight of the hammer\n", + "theta=30 # degree # ref fig.20.12\n", + "mu=0.18 # coefficient of friction\n", + "s=10 # m # distance travelled by the hammer # fig 20.12\n", + "g=9.81 # m/s^2 # acc due to gravity\n", + "\n", + "# Calculations\n", + "\n", + "# The acceleration of the hammer is given as,\n", + "a=g*((sin(theta*(pi/180)))-(mu*cos(theta*(pi/180)))) # m/s^2\n", + "\n", + "# The velocity of the hammer at point B is,\n", + "v=(2*a*s)**0.5 # m/s\n", + "\n", + "# Let the initial velocity of the hammer in horizontal direction be v_x & v_y in vertical direction, Then,\n", + "v_x=v*cos(theta*(pi/180)) # m/s\n", + "v_y=v*sin(theta*(pi/180)) # m/s\n", + "\n", + "# MOTION IN VERTICAL DIRECTION\n", + "# Now, let time required to travel vertical distance (i.e BB'=S=5 m) is given by finding the roots of the second degree eq'n as,\n", + "# From the eq'n 4.9*t^2+4.1*t-5=0,\n", + "a=4.9\n", + "b=4.1\n", + "c=-5\n", + "\n", + "# The roots of the eq'n are,\n", + "t=((-b)+((b**2-(4*a*c))**0.5))/(2*a)\n", + "\n", + "# MOTION IN HORIZONTAL DIRECTION\n", + "# Let the horizotal distance travelled by the hammer in time t be s_x.Then,\n", + "s_x=v_x*cos(theta*(pi/180))*t # m\n", + "x=1+s_x # m\n", + "\n", + "# Results\n", + "\n", + "print\"The distance x where the hammer hits the round is \",round(x,2),\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The distance x where the hammer hits the round is 5.16 m\n" + ] + } + ], + "prompt_number": 101 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.20-9,Page no:528" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initilization of variables\n", + "s=1000 # m # distance OB (ref fig.20.13)\n", + "h=19.6 # m # height of shell from ground\n", + "g=9.81 # m/s^2 # acc due to gravity\n", + "\n", + "# Calculations\n", + "\n", + "# MOTION OF ENTIRE SHELL FROM O to A.\n", + "v_y=(2*(g)*h)**0.5 # m/s # initial velocity of shell in vertical direction\n", + "t=v_y/g # seconds # time taken by the entire shell to reach point A\n", + "v_x=s/t # m/s # velocity of shell in vertical direction\n", + "\n", + "# VELOCITIES OF THE TWO PARTS OF THE SHELL AFTER BURSTING AT A:\n", + "\n", + "# Let v_x2 be the horizontal velocity of 1st & the 2nd part after bursting which is given as,\n", + "v_x2=v_x*2 # m/s\n", + "\n", + "# Now distance BC travelled by part 2 is\n", + "BC=v_x2*t # m\n", + "\n", + "# Distance from firing point OC\n", + "OC=s+BC # m\n", + "\n", + "# Results\n", + "\n", + "print\"(a) The velocity of shell just before bursting is \",round(v_x),\"m/s\"\n", + "print\"(b) The velocity of first part immediately before the shell burst is \",round(v_x),\"m/s\"\n", + "print\"(c) The velocity of second part immediately after the shell burst is \",round(v_x2,1),\"m/s\"\n", + "print\"(b) The distance between the firing point & the point where the second part of the shell hit the ground is \",round(OC),\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a) The velocity of shell just before bursting is 500.0 m/s\n", + "(b) The velocity of first part immediately before the shell burst is 500.0 m/s\n", + "(c) The velocity of second part immediately after the shell burst is 1000.5 m/s\n", + "(b) The distance between the firing point & the point where the second part of the shell hit the ground is 3000.0 m\n" + ] + } + ], + "prompt_number": 106 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.20-10,Page No:530" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initilization of variables\n", + "\n", + "v_o=200 # m/s # initial velocity\n", + "theta=60 # degree # angle of the incline\n", + "y=5 # rise of incline\n", + "x=12 # length of incline\n", + "g=9.81 # m/s^2 # acc due to gravity\n", + "\n", + "\n", + "# Calculations\n", + "\n", + "# The angle of the inclined plane with respect to horizontal\n", + "beta=arctan(y*x**-1)*(180/pi) # degree\n", + "\n", + "# The angle of projection with respect to horizontal\n", + "alpha=90-theta # degree\n", + "\n", + "# Range is given by eq'n (ref. fig.20.14)\n", + "AB=(2*v_o**2*(sin((alpha-beta)*(pi/180)))*cos(alpha*(pi/180)))/(g*(cos(beta*(pi/180)))**2) # m\n", + "\n", + "# Range AC when the short is fired down the plane\n", + "AC=(2*v_o**2*(sin((alpha+beta)*(pi/180)))*cos(alpha*(pi/180)))/(g*(cos(beta*(pi/180)))**2) # m\n", + "BC=AB+AC # m\n", + "\n", + "# Results\n", + "\n", + "print\"The range covered (i.e BC) is \",round(BC),\"m\" #due to decimal variancce answer varies by 1m\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The range covered (i.e BC) is 7649.0 m\n" + ] + } + ], + "prompt_number": 138 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit