{
"metadata": {
"name": "",
"signature": "sha256:5482cd080863a0bf132cc69662813c918cd153651feccebbb960c53549f6ef87"
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"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": {}
}
]
}