From d36fc3b8f88cc3108ffff6151e376b619b9abb01 Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:40:35 +0530 Subject: Revised list of TBCs --- .../chapter20.ipynb | 546 --------------------- 1 file changed, 546 deletions(-) delete mode 100755 Engineering_Mechanics_by_Tayal_A.K./chapter20.ipynb (limited to 'Engineering_Mechanics_by_Tayal_A.K./chapter20.ipynb') diff --git a/Engineering_Mechanics_by_Tayal_A.K./chapter20.ipynb b/Engineering_Mechanics_by_Tayal_A.K./chapter20.ipynb deleted file mode 100755 index 7b30b675..00000000 --- a/Engineering_Mechanics_by_Tayal_A.K./chapter20.ipynb +++ /dev/null @@ -1,546 +0,0 @@ -{ - "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