# Chapter 8:Friction¶

## Example 8.1,Page No.235¶

In [1]:
import math

#Declaration Of Variables

W=100 #N #Weight of Body
P=F=60 #N #Horizontal Force

#Calculations

#Normal Reaction Force
R=W=100 #N

#Coefficient of Friction
mu=F*R**-1

#Result
print"Coefficient of Friction is",round(mu,2)

Coefficient of Friction is 0.6


## Example 8.2,Page No.236¶

In [2]:
import math

#Declaration Of Variables

W=200 #N #Weight of Body
mu=0.3 #Coefficient of Friction

#Calculations

#Normal Reaction
R=W=200 #N

#Horizontal Force
F=mu*R #N

#Result
print"Horizontal Force is",round(F,2),"N"

Horizontal Force is 60.0 N


## Example 8.3,Page No.236¶

In [3]:
import math
from math import sin, cos, tan, radians, pi

#Declaration Of Variables

W=50 #N #WEight
F=15 #N #Force required to pull
theta=15 #Degree #Angle made by Force

#Calculations

#Normal Reaction
R=W-F*sin(theta*pi*180**-1) #N

#Coefficient of friction
mu=F*cos(theta*pi*180**-1)*R**-1 #N

#Result
print"Coefficient of Friction is",round(mu,2)

Coefficient of Friction is 0.31


## Example 8.4,Page No.236¶

In [4]:
import math
from math import sin, cos, tan, radians, pi

#Declaration Of Variables

W=70 #N #Weight of Body
F=20 #N #force applied
theta=20 #degrees #Angle made by Force

#Calculations

#resolving Forces Normal to plane
R=W+F*sin(20*pi*180**-1) #N

#Resolving Forces along the plane
mu=F*cos(20*pi*180**-1)*R**-1 #N

#Result
print"coefficient of Friction is",round(mu,2)

coefficient of Friction is 0.24


## Example 8.5,Page No.237¶

In [5]:
import math

#Declaration Of Variables

P1=20 #N #pull
P2=25 #N #Required push
theta2=25 #Inclination of push

#Calculations

#Case-1 (When body is pulled)

#Resolving Forces along the plane
#mu*R1=P1*cos(theta*pi*180**-1) #N    .........1

#Resolving Force normal to plane
#R1=W-P1*sin(theta*pi*180**-1) #N

#Sub value of mu*R1 in above Equation we get
#mu*(W-8.452)=18.126      ......................2

#Case-2 (When body is pushed)

#Resolving Forces along the plane
#mu*R2=P2*cos(theta2*pi*180**-1) #N    .........3

#Resolving Force normal to plane
#R2=W-P2*cos(theta*pi*180**-1) #N

#Sub value of mu*R2 in above Equation we get
#mu*(W-10.565)=22.657      .................4

#dividing equation 2 by 4 and Further simplifying we get

#Weight of body
W=383*4.53**-1

#Sub value of W in Equation 2
mu=18.126*(W-8.452)**-1

#Result
print"Weight of Body is",round(W,2),"N"
print"Coefficient of Friction is",round(mu,2)

Weight of Body is 84.55 N
Coefficient of Friction is 0.24


## Example 8.7,Page No.239¶

In [1]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np

#Declaration Of Variables

W=1000 #N #Weight of stone Block
mu=0.6 #Coefficient of Friction
theta=20 #Degrees #Angle with Horizontal

#Calculations

#PArt-1
#resolving Horizontal Forces
#P*cos(theta)=mu*R    ...........................1

#Resolving Verticla Forces
#R+P*sin(theta)=W    .............................2

#Sub value of P from equation 2,we get
P=mu*W*(cos(theta*pi*180**-1)+mu*sin(theta*pi*180**-1))**-1 #N   ........3

#PArt-2
#Let phi=Angle of Friction

#Form Equation 3 ,angle 20 is replaced by angle phi
#Force required to pull the body
#P2=mu*W*(cos(phi)+mu*sin(phi))**-1

#The Force P will be min if Dericative of (cos(phi)+mu*sin(phi)) is equal to zero
phi=np.arctan(mu)*(180*pi**-1) #degrees

#Force required to pull the body
P2=mu*W*(cos(phi*pi*180**-1)+mu*sin(phi*pi*180**-1))**-1 #N

#Result
print"Minimum Pull necessary is",round(P,2),"N"
print"Pull Required if inclination of rope is equal to angle of friction",round(P,2),"N"

Minimum Pull necessary is 524.06 N
Pull Required if inclination of rope is equal to angle of friction 524.06 N


## Example 8.11,Page No.241¶

In [7]:
import math
from math import sin, cos, tan, radians, pi

#Declaration Of Variables

W=500 #N #weight of Body
P=350 #N #Force applied
alpha=30 #Degrees #Inclination

#Calculations

#Resolving Weights
W1=W*sin(30*pi*180**-1) #N
W2=W*cos(30*pi*180**-1) #N

#Resolving Forces Vertically
#R=W*cos(30)

#Resolving Forces Horizontally
mu=(P-W*sin(alpha*pi*180**-1))*(W*cos(alpha*pi*180**-1))**-1

#Result
print"Coefficient Of Friction",round(mu,2)

Coefficient Of Friction 0.23


## Example 8.12,Page No.246¶

In [8]:
import math
from math import sin, cos, tan, radians, pi

#Declaration Of Variables

W=450 #N #Weight of Body
alpha=30 #Degrees #Inclination of plane
mu=0.25 #coefficient of friction
d=10 #m #Distance travelled by body

#Calculations

#Resolving Force normal to plane
R=W*cos(alpha*pi*180**-1)

#Resolving Forces along the plane
P=W*sin(alpha*pi*180**-1)+mu*R #N

#Work done on the body
W=P*d #J

#Result
print"Force required is",round(P,2),"N"
print"Work done is",round(W,2),"J"

Force required is 322.43 N
Work done is 3224.28 J


## Example 8.13,Page No.246¶

In [9]:
import math
from math import sin, cos, tan, radians, pi

#Declaration Of Variables

#Case-1
P1=200 #N #Force applied
theta1=15 #Degrees #Inclination

P2=230 #N #Force applied
theta2=20 #Degrees #Inclination

#Calculations

#For Case-1,

#W1=W*sin(theta1*pi*180**-1) #N
#W2=W*cos(theta1*pi*180**-1) #N

#Resolving Forces Vertically
#R=W2

#Resolving Foces Horizontally
#mu*W2+W1=P1   .......................1

#For case-2

#W3=W*sin(theta2*pi*180**-1) #N
#W4=W*cos(theta2*pi*180**-1) #N

#Resolving Forces Vertically
#R=W3

#Resolving Foces Horizontally
#mu*W3+W4=P2   .......................2

#After sub values inequations 1 & 2 and dividing equations  2 by 1 we get
mu=mu=(P1*sin(20*pi*180**-1)-P2*sin(15*pi*180**-1))*(P2*cos(15*pi*180**-1)-P1*cos(20*pi*180**-1))**-1

#weight of Body
W=P2*(sin(theta2*pi*180**-1)+mu*cos(theta2*pi*180**-1))**-1 #N

#Result
print"Weight of Body is",round(W,2),"N"
print"Coefficient of Friction is",round(mu,2)

Weight of Body is 392.68 N
Coefficient of Friction is 0.26


## Example 8.15,Page No.249¶

In [10]:
import math
from math import sin, cos, tan, radians, pi

#Declaration Of Variables

W=15 #N #Weight of Block
T=5 #N #Tension in string
alpha=45 #Degrees #Inclination

#Calculations

#Frictional Foce on Block
F=-(T*cos(alpha*pi*180**-1)-W*sin(alpha*pi*180**-1)) #N

#Normal Reaction of inclined plane
R=(W*cos(alpha*pi*180**-1)+T*sin(alpha*pi*180**-1)) #N

#Coefficient of friction
mu=F*R**-1

#Result
print"Frictional Force on Block is",round(F,2),"N"
print"Normal Reaction of inclined plane",round(R,2),"N"
print"Coefficient of friction",round(mu,2)

Frictional Force on Block is 7.07 N
Normal Reaction of inclined plane 14.14 N
Coefficient of friction 0.5


## Example 8.16,Page No.250¶

In [11]:
import math
from math import sin, cos, tan, radians, pi

#Declaration Of Variables

mu_s=0.4 #Coefficient of static Friction
mu_k=0.3 #Coefficient of Kinetic friction
M=40 #Kg #MAss of body
g=9.81 #acceleration due to gravity
W=M*g #N
theta=40
alpha=30 #Inclination
P=800 #N

#Calculations

#normal Reaction
R=P*sin(theta*pi*180**-1)+M*g*cos(alpha*pi*180**-1) #N

#Max Frictional Force
F=mu_s*R #N

#Total Force along plane
F=P*cos(theta*pi*180**-1)-M*g*sin(alpha*pi*180**-1) #N

#Magntude of Frictional Force
F2=mu_k*R #N

#Result
print"Magnitude of Friction Force",round(F2,2),"N"

Magnitude of Friction Force 256.22 N


## Example 8.17,Page No.252¶

In [2]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np

#Declaration Of Variables

W=30 #N #Weight acting vertically downward
P2=6 #N #Force at angle of 30 with inclined plane
theta=30 #Degrees #Inclination of force with the inclined plane
mu=0.3 #Coefficient of friction

#Calculations

#Part-1

#Reaction Force is given by
#R=W*cos(alpha)-P2*sin(theta)

#Now resolving Force we get
alpha=np.arcsin(P2*cos(theta*pi*180**-1)*W**-1)
alpha2=180*pi**-1*alpha #degrees

#PArt-2

#resolving Forces normal to inclined plane
R=W*cos(round(alpha2,2)*pi*180**-1) #N

#Resolving Forces normal to inclined plane
P1=W*sin(round(alpha2,2)*pi*180**-1)+mu*round(R,2)

#Result
print"Force required to move a load 30 N up a rough plane is",round(P1,2),"N"

Force required to move a load 30 N up a rough plane is 14.06 N


## Example 8.18,Page No.253¶

In [13]:
import math

#Declaration Of Variables

W1=400 #N #Weight of first body
W2=800 #N #Weight of second body
mu1=0.15 #Coefficient of friction of first body
mu2=0.40 #Coefficient of friction of seconnd body

#Calculations

#Forces acting on the first body

#Resolving force along plane
#W1*sin(alpha)=T+mu1*R1   ............................1

#Resolving Forces normal to plane
#W1*cos(alpha)=R1

#Sub value of R1 in equation1,we get
#T=400*sin(alpha)-60*cos(alpha)  .......................2

#Forces on second body

#Resolving forces along the plane
#W2*sin(alpha)+T=mu2*R2   .........................3

#Resolving forces normal to plane
#R2=W2*cos(alpha)

#sub value of R2 in equation3
#T=320*cos(alpha)-W2*sin(alpha)  ...............4

#Equating values of T,given by equation 2 and 3
#W1*sin(alpha)-60*cos(alpha)=320*cos(alpha)-W2*sin(alpha)
#Further simplifying we get
alpha=arctan(380*1200**-1)*(180*pi**-1) #Degrees

#Sub value of alpha in equation 2
T=W1*sin(round(alpha,2)*pi*180**-1)-60*cos(round(alpha,2)*pi*180**-1)

#Result
print"Inclination of the plane to the horizontal is",round(alpha,2),"Degrees"
print"Tension in the cord is",round(T,2),"N"

Inclination of the plane to the horizontal is 17.57 Degrees
Tension in the cord is 63.55 N


## Example 8.19,Page No.255¶

In [14]:
import math

#Declaration Of Variables

W_B=1500 #N #Weight of block B
mu_A=0.25 #Coefficint of friction of block A
mu_B=0.35 #Coefficient of Friction of block B
alpha=60 #Degrees

#Calculations

#BLock A
#F_A=mu_A*W_A #Force of friction

#Block B
#HOrizontal Force of friction of block A is transmitted through rod to block B

#Force of friction of block B
#F_B=mu2*R_B

#Resolving Horizontal Forces
#mu1*W_A+F_B*cos(alpha)=R_B*cos(30)
#After further simplifying we get
#mu_A*W_A=0.691*R_B   ...........................1

#Resolving Forces vertically
#R_B*sin(30)+F_B*sin(alpha)=W_B
#After further simplifying we get
R_B=W_B*0.803**-1 #N

#Sub value of R_b in equation 1 we get
W_A=0.691*R_B*mu_A**-1

#Result
print"Smallest Weight of Block A is",round(W_A,2),"N"

Smallest Weight of Block A is 5163.14 N


## Example 8.20,Page No.257¶

In [15]:
import math
from math import sin, cos, tan, radians, pi

#Declaration Of Variables

W_A=100 #N Weight of block A
W_B=300 #N #Weight of block B
alpha=45 #Degrees #Inclination of plane
phi=30 #Degrees #Inclination of rigid bar with horizontal
mu=tan(15*pi*180**-1) #Degrees

#Calculations

#LEt R_A and R_B be the reactions at A And B respectively and t be the thrust in rod

#Equilibrium of bLoack A
#Resolving forces along plane
#W_A*sin(alpha)+F_A=T*cos(15)
#After further simplifying we get
#70.7+0.2679*R_A=0.969*T    .................1

#Resolving force snormal to plane
#R_A=W_A*cos(alpha)+T*sin(15)=R_A

#Now sub value of R_A in equation 1 we get
#70.7+0.269*(100*0.707+T*0.2588)=0.9659*T
#After further simplifying we get
T=89.64*0.8966**-1 #N

#Equilibrium of block B
#Resolving forces normal to plane
R_B=W_B+T*sin(phi*180**-1*pi)
#Resolving forces along plane
P=T*cos(phi*pi*180**-1)+mu*R_B

#Result
print"HOrizontal Force required to be apllied to block B to just move block A is",round(P,2),"N"

HOrizontal Force required to be apllied to block B to just move block A is 180.36 N


## Example 8.21,Page No.258¶

In [16]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np

#Declaration Of Variables

mu=0.2 #Coefficient of friction
W1=100 #N #weight of Block1
W2=150 #N #Weight of Block2
theta=60 #Degrees

#Calculations

#Case-1

#Reaction Force
R=W2*cos(theta*pi*180**-1) #N

#Tension in the string
T=W2*sin(theta*pi*180**-1)+mu*R #N

#Case-2

theta2=np.arctan(mu)*(pi**-1*180) #Angle made by Force with horizontal acting on block 1

#Force on block with weight 100 N
P=164.9*((cos(theta2*pi*180**-1)+mu*sin(theta2*pi*180**-1))**-1)

#Result
print"Least Value of Force P to cause motion to impend rightwards is",round(P,2),"N"

Least Value of Force P to cause motion to impend rightwards is 161.7 N


## Example 8.22,Page No.260¶

In [17]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np

#Declaration Of Variables

W1=90 #N #Weight of Block 1
W2=30 #N #Weight of Block2
mu=1*3**-1 #Coefficient of friction

#Calculations

#Considering Equilibrium of weight W2

#Tension in the string
#T=W2*sin(theta)+mu*R1 .....................................1

#Normal reaction to the plane
#R1=W2*cos(theta)    .......................................2

#Sub value of R1 in equation 1 we get
#T=W2*sin(theta)+10*cos(theta)    ........................3

#Considering Equilibrium of weight W1

#Resolving Forces along the plane
#W1*sin(theta)=10*cos(theta)+mu*R2    ..................4

#Resolving Forces normal to plane
#R2=120*cos(theta)   ...........5

#Sub value of R2 in equation 4 we get
theta=np.arctan(0.5555)*(pi**-1*180) #Degrees

#Result
print"Value of angle theta should be",round(theta,2),"degrees"

Value of angle theta should be 29.05 degrees


## Example 8.23,Page No.262¶

In [18]:
import math

#Declaration Of Variables

L_AC=10 #m #Length of AC
L_BC=8  #m #Length of BC
W=20 #N #weight

#Calculations

L_AB=(L_AC**2-L_BC**2)**0.5 #m #Length of AB
L_CD=L_BC*2**-1 #m

#Now Resolving Forces

#Vertically
#Reaction Force at C
R_C=W #N

#Horizontally
#R_A=F_C=mu*R_C

#Taking Moment at pt C
#Coefficient of friction
mu=W*L_CD*(R_C*L_AB)**-1

#Frictional Force acting at C
F_C=round(mu,2)*R_C #N

#Result
print"Coefficient of Friction",round(mu,2)
print"Frictional Force acting at pt C",round(F_C,2),"N"

Coefficient of Friction 0.67
Frictional Force acting at pt C 13.4 N


## Example 8.24,Page No.263¶

In [19]:
import math

#Declaration Of Variables

L_AB=13 #m #Length of AB
L_AC=5 #m #Distance of lower ladder from wall
mu=0.3 #Coefficient of friction

#Calculations

#Vertical Forces
R_A=W #N

#Horizontal Forces
R_B=F_A=mu*R_A #N

#MAx amount of frictional Force availale at A
F_A #N

L_BC=(L_AB**2-L_AC**2)**0.5 #m

#Moment at pt A

#Horizontal Forces
F_A2=R_B2 #N

#Result
print"Frictional Force acting on ladder is",round(F_A,2),"N"

Frictional Force acting on ladder is 7.5 N


## Example 8.25,Page No.264¶

In [20]:
import math
from math import sin, cos, tan, radians, pi

#Declaration Of Variables

L_AB=14 #m #Length of AB
L_BD=6 #m #LEngth of BD
mu=1*3**-1 #Coefficient of friction
CBA=60 #Degrees

#Calculations

#Resolving forces

#Vertically
R_B=W #N
#Actual Force of friction at pt B
F_B=mu*R_B #N

#Horizontally
R_A=F_B #N

L_BE=L_BD*cos(CBA*pi*180**-1) #m
L_AC=L_AB*sin(CBA*pi*180**-1) #m

R_A2=R_B*L_BE*L_AC**-1
F_B2=R_A2 #N

#Result
print"Force available at B the force required force for equilibrium,the ladder will be stable:F_B is",round(F_B,2),"N"

Force available at B the force required force for equilibrium,the ladder will be stable:F_B is 200.0 N


## Example 8.26,Page No.265¶

In [21]:
import math
from math import sin, cos, tan, radians, pi

#Declaration Of Variables

L=L_AB=6 #m #Length of AB
W1=750 #N #Weight of man
L1=4 #m #Distance of man from top of ladder
L2=L-L1 #m #Distance of man form foot of ladder
L_BE=4 #m #Length of BE

#Calculations

#Resolving Forces Vertically
R_A=W+W1 #N

#Horizontally
#R_B=F_A=mu*R_A #N

L_BC=L_AB*sin(alpha*pi*180**-1) #m #Length of BC
L_AC=L_AB*cos(alpha*pi*180**-1) #m #Length of AC

L_AH=(L_AB-L_BE)*cos(alpha*pi*180**-1) #m

#Coefficient of friction

#Result
print"Coefficient of friction is",round(mu,2)

Coefficient of friction is 0.19


## Example 8.27,Page No.266¶

In [4]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np

#Declaration Of Variables

L=L_AB=4.5 #m
mu=0.4 #Coefficient of friction between ladder and floor
mu2=0.2 #coefficient of frictionbetween LAdder and wall
L_BE=1.2 #m #distance

#Calculations

#Resolving FOrces Vertically
#R_A+F_B=W1+W #N   ................1

#Horizontally
#R_B=mu*R_A  .....................2

#Resolving Force R_B in equation 1 we get
R_A=(W1+W)*1.08**-1 #N

#Reaction at B
R_B=mu*R_A  #N

#Moment at pt A
#After further simplifying we get
alpha=np.arctan(1.665)*(180*pi**-1)

#Result
print"Reaction at the Foot of ladder",round(R_A,2),"N"
print"Reaction at the Foot top of ladder",round(R_B,2),"N"

#Value of alpha is incorrect in book i.e 59degree 65seconds

Angle made by ladder with Horizontal 59.01 Degrees
Reaction at the Foot of ladder 1018.52 N
Reaction at the Foot top of ladder 407.41 N


## Example 8.28,Page No.267¶

In [3]:
import math
from math import sin, cos, tan, radians, pi

#Declaration Of Variables

W2=800 #N #weight of man
mu=0.2 #Coefficinet of friction
L_AG=5*2**-1 #m

#Calculations

#Let R_A and R_B be the reactions at A and b respectively
#F_B=mu*R_B
#F_A=mu*R_A

#Resolving forces vertically
#R_A+F_B=W1+W2 .............1

#Resolving forces horizontally
#R_B=0.2*R_A.....................2

#After sub values and further simplifying we get
R_A=1050*1.04**-1 #N

#Sub value of R_A in equation 2 we get
R_B=0.2*R_A

#Triangle AGD

#TRiangle AEH
#L_AH=x*cos(60)

L_BC=L*cos(alpha*pi*180**-1)
L_AC=L*cos(60*pi*180**-1)

F_B=mu*R_B

#Taking moment at A
#After sub values and further simplifying we get
x=66.276*40**-1

#Result
print"The slipping will be induced at",round(x,2),"m"

The slipping will be induced at 1.66 m


## Example 8.29,Page No.271¶

In [24]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np

#Declaration Of Variables

alpha=10 #Degrees #Angle of Wedge
W=1500 #N #weight of Block
mu=0.3 #Coefficient of friction
phi=np.arctan(mu)*(180*pi**-1)

#Calculations

#Applying Lamis theorem to the point O
#W*(sin(2*phi+90+alpha))**-1=R3*(sin(180-(alpha+phi)))**-1=R2*(sin(90-phi))**-1
#After further simplifying we get
Y=180-(alpha+phi)
Z=sin(Y*pi*180**-1)
Y1=2*phi+90+alpha
Z1=sin(Y1*pi*180**-1)
R3=W*(Z)*(Z1)**-1 #N

Y2=90-phi
Z2=sin(Y2*pi*180**-1)
R2=W*Z2*Z1**-1 #N

#Applying Lamis theorem to the point L
#R1*sin(90+alpha+phi)**-1=R2*sin(90+phi)**-1=P*sin(180-2*phi-alpha)**-1
#After further simplifying we get
Y3=180-(2*phi+alpha)
Z3=sin(Y3*pi*180**-1)
P=Z3*R2*Z2**-1 #N

#Result
print"Minimum Horizontal force be applied on wedge to raise the block is",round(P,2),"N"

Minimum Horizontal force be applied on wedge to raise the block is 1418.4 N


## Example 8.30,Page No.277¶

In [5]:
import math
from math import sin, cos, tan, radians, pi

#Declaration Of Variables

D=0.1 #m #Diameter
N=150 #r.p.m
mu=0.05 #Coefficient of friction

#Calculations

#Power Loast in friction assuming uniform pressure
T=2*3**-1*mu*W*R #N*m

#Power Lost in Friction
P=2*pi*N*T*60**-1 #W

#Power Lost in friction assuming wear
T2=0.5*mu*W*R #W

#Power Lost in friction
P2=2*pi*N*T2*60**-1 #W

#Result
print"Power Loast in friction assuming uniform pressure is",round(P,2),"W"
print"Power Lost in friction assuming wear is",round(P2,2),"W"

Power Loast in friction assuming uniform pressure is 392.7 W
Power Lost in friction assuming wear is 294.52 W


## Example 8.31,Page No.282¶

In [6]:
import math
from math import sin, cos, tan, radians, pi

#Declaration Of Variables

alpha=60 #Degrees
mu=0.05 #m #coefficient of friction
W=20*10**3 #N
N=210 #r.p.m

#Calculations

#Frictional Torque
T=2*3**-1*mu*W*R*(sin(alpha*pi*180**-1))**-1 #N*m

#Power Lost in Friction for uniform pressure
P=2*pi*N*T*60**-1*10**-3 #KW

#frictional torque
T2=1*2**-1*mu*W*R*(sin(alpha*pi*180**-1))**-1 #N*m

#Power Loast in friction for uniform wear
P2=2*pi*N*T2*60**-1*10**-3 #KW

#Result
print"Power Lost in Friction assuming:Uniforming pressure",round(P,2),"KW"
print"                               :Uniform wear",round(P2,2),"KW"

Power Lost in Friction assuming:Uniforming pressure 2.54 KW
:Uniform wear 1.9 KW


## Example 8.32,Page No.283¶

In [27]:
import math
from math import sin, cos, tan, radians, pi

#Declaration Of Variables

alpha=60 #degrees
p=350*10**3 #N/m**2 #pressure
N=180 #r.p.m
mu=0.05
#r1*2*r2

#Calculations

#From Equation of uniform pressure
r2=(W*(pi*p*3)**-1)**0.5 #m
r1=2*r2 #m

#Frictional Torque
T=2*3**-1*mu*W*(sin(alpha*pi*180**-1))**-1*(r1**3-r2**3)*(r1**2-r2**2)**-1 #m

#Power absorbed in friction
P=2*pi*N*T*60**-1*10**-3 #KW

#Result
print"Power absorbed in friction",round(P,2),"KW"

Power absorbed in friction 3.68 KW


## Example 8.33,Page No.286¶

In [28]:
import math

#Declaration Of Variables

mu=0.05 #Coefficient of friction
N=150 #r.p.m

#Calculations

#Torque
T=2*3**-1*mu*W*((r1**3-r2**3)*(r1**2-r2**2)**-1) #N*m

#Power lost in Frction
P=2*pi*N*T*60**-1*10**-3 #KW

#Result
print"Power lost in Frction is",round(P,2),"KN"

Power lost in Frction is 8.02 KN


## Example 8.34,Page No.287¶

In [10]:
import math
from math import sin, cos, tan, radians, pi

#Declaration Of Variables

mu=0.05 #Coefficient of friction
N=380 #r.p.m
p=350*10**3 #N/m**2 #Intensity of pressure

#Calculations

#Power Loast in Overcoming friction

#Torque
T=2*3**-1*mu*W*((r1**3-r2**3)*(r1**2-r2**2)**-1) #N*m

P=2*pi*N*T*60**-1*10**-3 #KW

#Number of collars required

W2=p*pi*(r1**2-r2**2)

n=W*W2**-1

#Result
print"Power Loast in Overcoming friction is",round(P,2),"KW"
print"Number of collars required for the thrust",round(n,1)

Power Loast in Overcoming friction is 22.22 KW
Number of collars required for the thrust 2.9