Chapter04: Newtons Law

Ex4.1:pg-147

In [1]:
  import math   #Example 4_1
 
  
  #To calculate the force required
vf=12    #units in meters/sec
v0=0     #units in meters/sec
t=8    #units in sec
a=(vf-v0)/t     #units in meters/sec**2
m=900    #units in Kg
F=m*a    #units in Newtons
print "The force required is F=",round(F)," N"
The force required is F= 900.0  N

Ex4.2:pg-147

In [2]:
  import math   #Example 4_2
 
  
  #To find the friction force that opposes the motion
F1=500    #units in Newtons
F2=800    #units in Newtons
theta=30    #units in degrees
Fn=F1+(F2*math.sin(theta*math.pi/180))    #units in Newtons
u=0.6
f=u*Fn     #units in Newtons
print "The Frictional force that is required is f=",round(f)," N"
The Frictional force that is required is f= 540.0  N

Ex4.3:pg-153

In [4]:
  import math   #Example 4_3
 
  
  #To find out at what rate the wagon accelerate and how large a force the ground pushing up on wagon
F1=90     #units in Newtons
F2=60     #units in Newtons
P=F1-F2    #units in Newtons
F3=100     #units in Newtons
F4=math.sqrt(F3**2-F2**2)    #units in Newtons
a=9.8    #units in meters/sec**2
ax=(F4*a)/F1     #units in Meters/sec**2
print "The wagon accelerates at ax=",round(ax,1)," meters/sec**2\n"
print "Force by which the ground pushing is P=",round(P)," N"
The wagon accelerates at ax= 8.7  meters/sec**2

Force by which the ground pushing is P= 30.0  N

Ex4.4:pg-153

In [5]:
  import math   #Example 4_4
 
  
  # To calculate How far does the car goes
w1=3300      #units in lb
F1=4.45     #units in Newtons
w2=1       #units in lb
weight=w1*(F1/w2)    #units in Newtons
g=9.8     #units in meters/sec**2
Mass=weight/g     #units in Kg
speed=38     #units in mi/h
speed=speed*(1.61)*(1/3600)     #units in Km/sec
stoppingforce=0.7*(weight)     #units in Newtons
a=stoppingforce/-(Mass)     #units in meters/sec**2
vf=0
v0=17     #units in meters/sec
x=(vf**2-v0**2)/(2*a)
print "The car goes by x=",round(x,1)," meters"
  #In text book the answer is printed wrong as x=20.9 meters the correct answer is x=21.1 meters
The car goes by x= 21.1  meters

Ex4.5:pg-155

In [6]:
  import math   #Example 4_5
 
  
  #To find the acceleration of the masses
w1=10     #units in Kg
w2=5     #units in Kg
f1=98     #units in Newtons
f2=49     #units in Newtons
w=w1/w2
T=round((f1+(w*f2))/(w+1))     #units in Newtons
a=(f1-T)/w1     #units in meters/sec**2
print "Acceleration is a=",round(a,1)," meters/sec**2"
Acceleration is a= 3.3  meters/sec**2

Ex4.6:pg-156

In [7]:
  import math   #Example 4_6
 

  #To find the acceleration of the objects
w1=0.4     #units in Kg
w2=0.2      #units in Kg
w=w1/w2
a=9.8     #units in meters/sec**2
f=0.098     #units in Newtons
c=w2*a     #units in Newtons
T=((w*c)+f)/(1+w)       #units in Newtons
a=(T-f)/w1     #units in meters/sec**2
print "Acceleration a=",round(a,1)," meters/sec**2"
Acceleration a= 3.1  meters/sec**2

Ex4.7:pg-157

In [8]:
  import math   #Example 4_7
 
  
  #To estimate the lower limit for the speed
  #In a practical situation u should be atleast 0.5
u=0.5
g=9.8     #units in meter/sec**2
x=7     #units in meters
v0=math.sqrt(2*u*g*x)     #units in meters/sec
print "The lower limit of the speed v0=",round(v0,1)," meter/sec"
The lower limit of the speed v0= 8.3  meter/sec

Ex4.9:pg-158

In [9]:
  import math   #Example 4_9
 
  
  #To calculate how large a force must push on car to accelerate
m=1200    #units in Kg
g=9.8     #units in meters/sec**2
d1=4     #units in meters
d2=40     #units in meters
a=0.5     #units in meters/sec**2
P=((m*g)*(d1/d2))+(m*a)     #units in Newtons
print "The force required is P=",round(P)," N"
  #In text book the answer is printed wrong as P=1780 N but the correct answer is P=1776 N
The force required is P= 600.0  N

Ex4.10:pg-159

In [10]:
  import math   #Example 4_10
 
  
  #To calculate the tension in the rope
u=0.7
sintheta=(6.0/10)
w1=50     #units in Kg
g=9.8     #units in meter/sec**2
costheta=(8.0/10)
Fn=w1*g*costheta      #units in Newtons
f=u*Fn     #units in Newtons
T=f+(w1*g*sintheta)
print "The tension in the rope is T=",round(T)," N"
The tension in the rope is T= 568.0  N

Ex4.11:pg-159

In [11]:
  import math   #Example 4_11
 
  
  #To find the acceleration of the system
w1=7.0     #units in Kg
a=9.8     #units in meters/sec**2
w2=5     #units in Kg
w=w1/w2
F1=29.4     #units in Newtons
F2=20     #units in Newtons
f=(F1+F2)      #units in Newtons
T1=w1*a      #units in Newtons
T=(T1+(w*f))/(1+w)      #units in Newtons
a=((w1*a)-T)/w1     #units in meters/sec**2
print "Acceleration a=",round(a,2)," meters/sec**2"
Acceleration a= 1.6  meters/sec**2