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 --- .../chapter16_6.ipynb | 471 --------------------- 1 file changed, 471 deletions(-) delete mode 100755 Engineering_Mechanics_by_Tayal_A.K./chapter16_6.ipynb (limited to 'Engineering_Mechanics_by_Tayal_A.K./chapter16_6.ipynb') diff --git a/Engineering_Mechanics_by_Tayal_A.K./chapter16_6.ipynb b/Engineering_Mechanics_by_Tayal_A.K./chapter16_6.ipynb deleted file mode 100755 index 7bd00420..00000000 --- a/Engineering_Mechanics_by_Tayal_A.K./chapter16_6.ipynb +++ /dev/null @@ -1,471 +0,0 @@ -{ - "metadata": { - "name": "chapter16.ipynb" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 16: Kinetics Of A Particle : Work And Energy" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.16-1,Page No:432" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "# Initilization of variables\n", - "\n", - "k=1000 # N/m # stiffness of spring\n", - "x_1=0.1 # m # distance upto which the spring is stretched\n", - "x_2=0.2 # m \n", - "x_0=0 # initial position of spring\n", - "\n", - "# Calculations\n", - "\n", - "# Work required to stretch the spring by 10 cm from undeformed position is given as,\n", - "U_10=-(k/2)*(x_1**2-x_0**2) # N-m \n", - "# Work required to stretch from 10 cm to 20 cm is,\n", - "U=-(0.5)*k*(x_2**2-x_1**2) # N-m\n", - "\n", - "# Results\n", - "\n", - "print\"The work of the spring force is \",round(U_10),\"N-m\"\n", - "print\"The work required to stretch the spring by 20 cm is \",round(U),\"N-m\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The work of the spring force is -5.0 N-m\n", - "The work required to stretch the spring by 20 cm is -15.0 N-m\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.16-3,Page No:436" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "# Initilization of variables\n", - "\n", - "M_A=100 # kg # mass of block A\n", - "M_B=150 # kg # mass of block B\n", - "mu=0.2 # coefficient of friction between the blocks and the surface\n", - "x=1 # m # distance by which block A moves\n", - "g=9.81 # m/s^2 # acc due to gravity\n", - "\n", - "# Calculations\n", - "\n", - "# Consider the respective F.B.D\n", - "# Applying the principle of work and energy to the system of blocks A&B and on simplifying we get the value of v as,\n", - "v=(((-mu*M_A*g)+(M_B*g))/(125))**0.5 # m/s \n", - "\n", - "# Results\n", - "\n", - "print\"The velocity of block A is \",round(v,2),\"m/s\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The velocity of block A is 3.19 m/s\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.16-4,Page No:440" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "# Initilization of variables\n", - "\n", - "M=500*10**3 # kg # mass of the train\n", - "u=0 # m/s # initial speed\n", - "v=90*1000*3600**-1 # m/s # final speed\n", - "t=50 # seconds\n", - "F_r=15*10**3 # N # Frictioal resistance to motion\n", - "\n", - "# Calculations\n", - "\n", - "# Acceleration is given as,\n", - "a=v*t**-1 # m/s^2\n", - "# The total force required to accelerate the train is,\n", - "F=M*a # N\n", - "# The maximum power required is at, t=50s & v=25 m/s\n", - "P=(F+F_r)*v*(10**-6) # MW\n", - "# At any time after 50 seconds, the force required only to overcome the frictional resistance of 15*10^3 N is,\n", - "P_req=F_r*v*(10**-3) # kW\n", - "\n", - "# Results\n", - "\n", - "print\"(a) The maximum power required is \",round(P,3),\"MW\"\n", - "print\"(b) The power required to maintain a speed of 90 km/hr is \",round(P_req,2),\"kW\" \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "(a) The maximum power required is 6.625 MW\n", - "(b) The power required to maintain a speed of 90 km/hr is 375.0 kW\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.16-5,Page No:440" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "# Initilization of variables\n", - "\n", - "W=50 # N # Weight suspended on spring\n", - "k=10 # N/cm # stiffness of the spring\n", - "x_2=15 # cm # measured extensions\n", - "h=10 # cm # height for position 2\n", - "\n", - "# Calculations\n", - "\n", - "# Consider the required F.B.D.\n", - "\n", - "# POSITION 1: The force exerted by the spring is,\n", - "F_1=W # N\n", - "\n", - "# Extension of spring from undeformed position is x_1,\n", - "x_1=F_1/k # cm\n", - "\n", - "# POSITION 2: When pulled by 10 cm to the floor. P.E of weight is,\n", - "P_E_g=-W*h # N-cm # (P_E_g= P_E_gravity)\n", - "\n", - "# P.E of the spring with respect to position 1\n", - "P_E_s=(0.5)*k*(x_2**2-x_1**2) # N-cm # (P_E_s= P_E_ spring)\n", - "\n", - "# Total P.E of the system with respect to position 1\n", - "P_E_t=P_E_g+P_E_s # N-cm # (P_E_t= P_E_total)\n", - "\n", - "# Total energy of the system,\n", - "E_2=P_E_t # N-cm\n", - "\n", - "# Total energy of the system in position 3 w.r.t position 1 is:\n", - "x=-(100)**0.5 # cm\n", - "x=+(100)**0.5 # cm\n", - "\n", - "# Results\n", - "\n", - "print\"The potential energy of the system is \",round(E_2),\"N-cm\"\n", - "print\"The maximum height above the floor that the weight W will attain after release is \",round(x),\"cm\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The potential energy of the system is 500.0 N-cm\n", - "The maximum height above the floor that the weight W will attain after release is 10.0 cm\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.16-6,Page No:442" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "# Initilization of variables\n", - "\n", - "m=5 # kg # mass of the ball\n", - "k=500 # N/m # stiffness of the spring\n", - "h=10 # cm # height of drop\n", - "g=9.81 # m/s^2 # acc due to gravity\n", - "\n", - "# Calculations\n", - "\n", - "# Consider the respective F.B.D.\n", - "# In eq'n 1 substitute the respective values and simplify it further. In this eq'n of 2nd degree a=1 b=-0.1962 & c=-0.01962. Thus the roots of the eq'n is given as,\n", - "a=1 \n", - "b=-0.1962\n", - "c=-0.01962\n", - "delta=((-b+(((b**2)-(4*a*c))**0.5))/(2*a))*(10**2) # cm # We consider the +ve value of delta\n", - "\n", - "# Results\n", - "\n", - "print\"The maximum deflection of the spring is \",round(delta,2),\"cm\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The maximum deflection of the spring is 26.91 cm\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.16-7,Page No:444" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "# Initilization of variables\n", - "\n", - "m=5 # kg # mass of the ball\n", - "k=500 # N/m # stiffness of the spring\n", - "h=0.1 # m # height of vertical fall\n", - "g=9.81 # m/s^2 # acc due to gravity\n", - "\n", - "# Calculations\n", - "\n", - "# Consider the respective F.B.D\n", - "# On equating the total energies at position 1 & 2 we get eq'n of delta as,\n", - "delta=((2*m*g*h)/(k))**0.5 # m \n", - "\n", - "# Results\n", - "\n", - "print\"The maximum compression of the spring is \",round(delta,3),\"m\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The maximum compression of the spring is 0.14 m\n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.16-9,Page No:445" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "# Initilization of variables\n", - "\n", - "m=5 # kg # mass of the collar\n", - "k=500 # N/m # stiffness of the spring\n", - "AB=0.15 # m # Refer the F.B.D for AB\n", - "AC=0.2 # m # Refer the F.B.D for AC\n", - "g=9.81 # m/s^2 # acc due to gravity\n", - "\n", - "# Calculations\n", - "\n", - "# Consider the respective F.B.D\n", - "\n", - "# POSITION 1: \n", - "P_E_1=m*g*(AB)+0 \n", - "K_E_1=0\n", - "E_1=P_E_1+K_E_1 #\n", - "\n", - "# POSITION 2 : Length of the spring in position 2\n", - "CB=(AB**2+AC**2)**0.5 # m \n", - "# x is the extension in the spring\n", - "x=CB-AC # m\n", - "# On substuting and Equating equations of total energies for position1 & position 2 we get the value of v as,\n", - "v=(((E_1-((0.5)*k*x**2))*2)/m)**0.5 # m/s\n", - "\n", - "# Results\n", - "\n", - "print\"The velocity of the collar will be \",round(v,2),\"m/s\"\n", - "# The answer given in the text book (v=16.4 m/s) is wrong.\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The velocity of the collar will be 1.64 m/s\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.16-10,Page No:446" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "# Initilization of variables\n", - "\n", - "m=5 # kg # mass of the block\n", - "theta=30 # degree # inclination of the plane\n", - "x=0.5 # m # distance travelled by the block\n", - "k=1500 # N/m # stiffness of the spring\n", - "mu=0.2 # coefficient of friction between the block and the surface\n", - "g=9.81 # m/s^2 # acc due to gravity\n", - "\n", - "# Calculations\n", - "\n", - "# Consider the F.B.D of the block\n", - "# Applying the principle of work and energy between the positions 1 & 2 and on further simplification we get the generic eq'n for delta as, 750*delta^2-16.03*delta-8.015=0. From this eq'n e have values of a.b & c as,\n", - "a=750\n", - "b=-16.03\n", - "c=-8.015\n", - "# Thus the roots of the eq'n are given as,\n", - "delta=(-b+((b**2-(4*a*c))**0.5))/(2*a) # m\n", - "\n", - "# Results\n", - "\n", - "print\"The maximum compression of the spring is \",round(delta,3),\"m\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The maximum compression of the spring is 0.115 m\n" - ] - } - ], - "prompt_number": 28 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.16-11,Page No:448" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "# Initilization of variables\n", - "\n", - "M=10 # kg # Here M=M_1=M_2\n", - "g=9.81 # m/s^2 # acc due to gravity\n", - "\n", - "# Calculations\n", - "\n", - "# Consider the respective F.B.D\n", - "# Applying the principle of conservation of energy and by equating the total energies at position 1 & position 2 we get v as,\n", - "v=((M*g*4)/(25))**0.5 # m/s\n", - "\n", - "# Results\n", - "\n", - "print\"The velocity of mass M_2 is \",round(v,2),\"m/s\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The velocity of mass M_2 is 3.96 m/s\n" - ] - } - ], - "prompt_number": 29 - } - ], - "metadata": {} - } - ] -} \ No newline at end of file -- cgit