{ "metadata": { "name": "", "signature": "sha256:5482cd080863a0bf132cc69662813c918cd153651feccebbb960c53549f6ef87" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter04: Newtons Law" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.1:pg-147" ] }, { "cell_type": "code", "collapsed": false, "input": [ " #Example 4_1\n", " \n", " \n", " #To calculate the force required\n", "vf=12 #units in meters/sec\n", "v0=0 #units in meters/sec\n", "t=8 #units in sec\n", "a=(vf-v0)/t #units in meters/sec**2\n", "m=900 #units in Kg\n", "F=m*a #units in Newtons\n", "print \"The force required is F=\",round(F),\" N\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The force required is F= 900.0 N\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.2:pg-147" ] }, { "cell_type": "code", "collapsed": false, "input": [ " #Example 4_2\n", " \n", " \n", " #To find the friction force that opposes the motion\n", "F1=500 #units in Newtons\n", "F2=800 #units in Newtons\n", "theta=30 #units in degrees\n", "Fn=F1+(F2*math.sin(theta*math.pi/180)) #units in Newtons\n", "u=0.6\n", "f=u*Fn #units in Newtons\n", "print \"The Frictional force that is required is f=\",round(f),\" N\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The Frictional force that is required is f= 540.0 N\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.3:pg-153" ] }, { "cell_type": "code", "collapsed": false, "input": [ " #Example 4_3\n", " \n", " \n", " #To find out at what rate the wagon accelerate and how large a force the ground pushing up on wagon\n", "F1=90 #units in Newtons\n", "F2=60 #units in Newtons\n", "P=F1-F2 #units in Newtons\n", "F3=100 #units in Newtons\n", "F4=sqrt(F3**2-F2**2) #units in Newtons\n", "a=9.8 #units in meters/sec**2\n", "ax=(F4*a)/F1 #units in Meters/sec**2\n", "print \"The wagon accelerates at ax=\",round(ax,1),\" meters/sec**2\\n\"\n", "print \"Force by which the ground pushing is P=\",round(P),\" N\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The wagon accelerates at ax= 8.7 meters/sec**2\n", "\n", "Force by which the ground pushing is P= 30.0 N\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.4:pg-153" ] }, { "cell_type": "code", "collapsed": false, "input": [ " #Example 4_4\n", " \n", " \n", " # To calculate How far does the car goes\n", "w1=3300 #units in lb\n", "F1=4.45 #units in Newtons\n", "w2=1 #units in lb\n", "weight=w1*(F1/w2) #units in Newtons\n", "g=9.8 #units in meters/sec**2\n", "Mass=weight/g #units in Kg\n", "speed=38 #units in mi/h\n", "speed=speed*(1.61)*(1/3600) #units in Km/sec\n", "stoppingforce=0.7*(weight) #units in Newtons\n", "a=stoppingforce/-(Mass) #units in meters/sec**2\n", "vf=0\n", "v0=17 #units in meters/sec\n", "x=(vf**2-v0**2)/(2*a)\n", "print \"The car goes by x=\",round(x,1),\" meters\"\n", " #In text book the answer is printed wrong as x=20.9 meters the correct answer is x=21.1 meters\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The car goes by x= 21.1 meters\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.5:pg-155" ] }, { "cell_type": "code", "collapsed": false, "input": [ " #Example 4_5\n", " \n", " \n", " #To find the acceleration of the masses\n", "w1=10 #units in Kg\n", "w2=5 #units in Kg\n", "f1=98 #units in Newtons\n", "f2=49 #units in Newtons\n", "w=w1/w2\n", "T=round((f1+(w*f2))/(w+1)) #units in Newtons\n", "a=(f1-T)/w1 #units in meters/sec**2\n", "print \"Acceleration is a=\",round(a,1),\" meters/sec**2\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Acceleration is a= 3.3 meters/sec**2\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.6:pg-156" ] }, { "cell_type": "code", "collapsed": false, "input": [ " #Example 4_6\n", " \n", "\n", " #To find the acceleration of the objects\n", "w1=0.4 #units in Kg\n", "w2=0.2 #units in Kg\n", "w=w1/w2\n", "a=9.8 #units in meters/sec**2\n", "f=0.098 #units in Newtons\n", "c=w2*a #units in Newtons\n", "T=((w*c)+f)/(1+w) #units in Newtons\n", "a=(T-f)/w1 #units in meters/sec**2\n", "print \"Acceleration a=\",round(a,1),\" meters/sec**2\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Acceleration a= 3.1 meters/sec**2\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.7:pg-157" ] }, { "cell_type": "code", "collapsed": false, "input": [ " #Example 4_7\n", " \n", " \n", " #To estimate the lower limit for the speed\n", " #In a practical situation u should be atleast 0.5\n", "u=0.5\n", "g=9.8 #units in meter/sec**2\n", "x=7 #units in meters\n", "v0=math.sqrt(2*u*g*x) #units in meters/sec\n", "print \"The lower limit of the speed v0=\",round(v0,1),\" meter/sec\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The lower limit of the speed v0= 8.3 meter/sec\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.9:pg-158" ] }, { "cell_type": "code", "collapsed": false, "input": [ " #Example 4_9\n", " \n", " \n", " #To calculate how large a force must push on car to accelerate\n", "m=1200 #units in Kg\n", "g=9.8 #units in meters/sec**2\n", "d1=4 #units in meters\n", "d2=40 #units in meters\n", "a=0.5 #units in meters/sec**2\n", "P=((m*g)*(d1/d2))+(m*a) #units in Newtons\n", "print \"The force required is P=\",round(P),\" N\"\n", " #In text book the answer is printed wrong as P=1780 N but the correct answer is P=1776 N\n" ], "language": "python", "metadata": {}, "outputs": [] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.10:pg-159" ] }, { "cell_type": "code", "collapsed": false, "input": [ " #Example 4_10\n", " \n", " \n", " #To calculate the tension in the rope\n", "u=0.7\n", "sintheta=(6.0/10)\n", "w1=50 #units in Kg\n", "g=9.8 #units in meter/sec**2\n", "costheta=(8.0/10)\n", "Fn=w1*g*costheta #units in Newtons\n", "f=u*Fn #units in Newtons\n", "T=f+(w1*g*sintheta)\n", "print \"The tension in the rope is T=\",round(T),\" N\"\n" ], "language": "python", "metadata": {}, "outputs": [] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.11:pg-159" ] }, { "cell_type": "code", "collapsed": false, "input": [ " #Example 4_11\n", " \n", " \n", " #To find the acceleration of the system\n", "w1=7.0 #units in Kg\n", "a=9.8 #units in meters/sec**2\n", "w2=5 #units in Kg\n", "w=w1/w2\n", "F1=29.4 #units in Newtons\n", "F2=20 #units in Newtons\n", "f=(F1+F2) #units in Newtons\n", "T1=w1*a #units in Newtons\n", "T=(T1+(w*f))/(1+w) #units in Newtons\n", "a=((w1*a)-T)/w1 #units in meters/sec**2\n", "print \"Acceleration a=\",round(a,2),\" meters/sec**2\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Acceleration a= 1.6 meters/sec**2\n" ] } ], "prompt_number": 8 } ], "metadata": {} } ] }