Chapter 7 Application of friction

Example 7.1 Application of Friction

In [1]:
import math
# Initilization of variables
d1=24 # cm # diameter of larger pulley
d2=12 # cm # diameter of smaller pulley
d=30 #cm # seperation betweem 1st & the 2nd pulley
# Calcuations
r1=d1/2 #cm # radius of 1st pulley
r2=d2/2 #cm # radius of 2nd pulley
theta=math.degrees(math.asin((r1-r2)/d)) #degrees 
# Angle of lap
beta_1=180+(2*theta) #degree # for larger pulley
beta_2=180-(2*theta) #degree #for smaller pulley
L=math.pi*(r1+r2)+(2*d)+((r1-r2)**2/d) #cm # Length of the belt
# Results
print('The angle of lap for the larger pulley is %f degree'%beta_1)
print('The angle of lap for the smaller pulley is %f degree'%beta_2)
print('The length of pulley required is %f cm'%L)

The angle of lap for the larger pulley is 203.073918 degree
The angle of lap for the smaller pulley is 156.926082 degree
The length of pulley required is 117.748668 cm

Example 7.2 Application of Friction

In [2]:
from __future__ import division
import math
# Initilization of variables
d1=0.6 #m # diameter of larger pulley
d2=0.3 #m # diameter of smaller pulley
d=3.5 #m # separation between the pulleys
# Calculations
r1=d1/2 #m # radius of larger pulley
r2=d2/2 #m # radius of smaller pulley
theta=math.degrees(math.asin((r1+r2)/d)) #degree
# Angle of lap for both the pulleys is same, i.e
beta=180+(2*theta) # degree
L=((math.pi*(r1+r2))+(2*d)+((r1-r2)**2/d)) #cm # Length of the belt
# Results
print('The angle of lap for the pulley is %f degree'%beta)
print('The length of pulley required is %f m'%L)

The angle of lap for the pulley is 194.774097 degree
The length of pulley required is 8.420145 m

Example 7.4 Belt friction

In [5]:
from __future__ import division
import math
# Initilization of variables
W1=1000 #N
mu=0.25 #coefficient of friction
T1=W1 # Tension in the 1st belt carrying W1
e=2.718 #constant
# Calculations
T2=T1/(e**(mu*math.pi)) #N # Tension in the 2nd belt
W2=T2 #N
# Results
print('The minimum weight W2 to keep W1 in equilibrium is %f N'%W2)

The minimum weight W2 to keep W1 in equilibrium is 455.975258 N

Example 7.5 Belt friction

In [6]:
from __future__ import division
import math
# Initilization of variables
mu=0.5 # coefficient of friction between the belt and the wheel
W=100 #N
theta=45 #degree
e=2.718
Lac=0.75 #m # ength of the lever
Lab=0.25 #m
Lbc=0.50 #m
r=0.25 #m
# Calculations
beta=((180+theta)*math.pi)/180 # radian # angle of lap
# from eq'n 2
T1=(W*Lbc)/Lab #N 
T2=T1/(e**(mu*beta)) #N # from eq'n 1
# consider the F.B.D of the pulley and take moment about its center, we get Braking Moment (M)
M=r*(T1-T2) #N-m
# Results
print('The braking moment (M) exerted by the vertical weight W is %f N-m'%M)

The braking moment (M) exerted by the vertical weight W is 42.980225 N-m

Example 7.6 Belt friction

In [8]:
from __future__ import division
import math
# Initiization of variables
W= 1000 #N # or 1kN
mu=0.3 # coefficient of friction between the rope and the cylinder
e=2.718 # constant
alpha=90 # degree # since 2*alpha=180 egree
# Calculations
beta=2*math.pi*3 # radian # for 3 turn of the rope
# Here T1 is the larger tension in that section of the rope which is about to slip
T1=W #N
F=W/e**(mu*(1/(math.sin(alpha*math.pi/180)))*(beta)) #N  Here T2=F
# Results
print('The  force required to suport the weight of 1000 N i.e 1kN is %f N'%F)

The  force required to suport the weight of 1000 N i.e 1kN is 3.502492 N

Example 7.7 Belt friction

In [13]:
from __future__ import division
import math
# Initilization of variables
Pw=50 #kW
T_max=1500 #N
v=10 # m/s # velocity of rope
w=4 # N/m # weight of rope
mu=0.2 # coefficient of friction 
g=9.81 # m/s**2 # acceleration due to gravity
e=2.718 # constant
alpha=30 # degree # since 2*alpha=60 
# Calcuations
T_e=(w*v**2)/g # N # where T_e is the centrifugal tension
T1=(T_max)-(T_e) #N
T2=T1/(e**(mu*(1/math.sin(alpha*math.pi/180))*(math.pi))) #N # From eq'n T1/T2=e^(mu*cosec(alpha)*beta)
P=(T1-T2)*v*(10**-3) #kW # power transmitted by a single rope
N=Pw/P # Number of ropes required
# Results
print('The number of ropes required to transmit 50 kW is %f or ~ %.0f'%(N,N))

The number of ropes required to transmit 50 kW is 4.789906 or ~ 5

Example 7.8 Belt friction

In [14]:
from __future__ import division
import math
# Initilization of variables
d1=0.45 #m # diameter of larger pulley
d2=0.20 #m # diameter of smaller pulley
d=1.95 #m # separation between the pulley's
T_max=1000 #N # or 1kN which is the maximum permissible tension
mu=0.20 # coefficient of friction
N=100 # r.p.m # speed of larger pulley
e=2.718 # constant
T_e=0 #N # as the data for calculating T_e is not given we assume T_e=0
# Calculations
r1=d1/2 #m # radius of larger pulley
r2=d2/2 #m # radius of smaller pulley
theta=math.degrees(math.asin((r1+r2)/d)) # degree
# for cross drive the angle of lap for both the pulleys is same
beta=((180+(2*(theta)))*math.pi)/180 #radian
T1=T_max-T_e #N
T2=T1/(e**(mu*(beta))) #N # from formulae, T1/T2=e**(mu*beta)
v=((2*math.pi)*N*r1)/60 # m/s # where v=velocity of belt which is given as, v=wr=2*pie*N*r/60
P=(T1-T2)*v*(10**-3) #kW # Power
# Results
print('The power transmitted by the cross belt drive is %f kW'%P)
#answer given in the textbook is incorrect

The power transmitted by the cross belt drive is 1.180543 kW

Example 7.9 Belt friction

In [16]:
from __future__ import division
import math
# Initilization of variabes
b=0.1 #m #width of the belt
t=0.008 #m #thickness of the belt
v=26.67 # m/s # belt speed
beta=165 # radian # angle of lap for the smaller belt
mu=0.3 # coefficient of friction
sigma_max=2 # MN/m**2 # maximum permissible stress in the belt
m=0.9 # kg/m # mass of the belt
g=9.81 # m/s**2
e=2.718 # constant
# Calculations
A=b*t # m**2 # cross-sectional area of the belt
T_e=m*v**2 # N # where T_e is the Centrifugal tension
T_max=(sigma_max)*(A)*(10**6) # N # maximum tension in the belt
T1=(T_max)-(T_e) # N 
T2=T1/(e**((mu*math.pi*beta)/180)) #N # from formulae T1/T2=e**(mu*beta)
P=(T1-T2)*v*(10**-3) #kW # Power transmitted
T_o=(T1+T2)/2 # N # Initial tension
# Now calculations to transmit maximum power
Te=T_max/3 # N # max tension
u=math.sqrt(T_max/(3*m)) # m/s # belt speed for max power
T_1=T_max-Te # N # T1 for case 2
T_2=T_1/(e**((mu*math.pi*beta)/180)) # N 
P_max=(T_1-T_2)*u*(10**-3) # kW # Max power transmitted
# Results
print('The initial power transmitted is %f kW'%P)
print('The initial tension in the belt is %f N'%T_o)
print('The maximum power that can be transmitted is %f kW'%P_max)
print('The maximum power is transmitted at a belt speed of %f m/s'%u)

The initial power transmitted is 14.808043 kW
The initial tension in the belt is 682.223893 N
The maximum power that can be transmitted is 15.020439 kW
The maximum power is transmitted at a belt speed of 24.343225 m/s

Example 7.10 Friction in a square threaded screw

In [30]:
from __future__ import division
import math
# Initilization of variables
p=0.0125 # m # pitch of screw
d=0.1 #m # diameter of the screw
r=0.05 #m # radius of the screw
l=0.5 #m # length of the lever
W=50 #kN # load on the lever
mu=0.20 # coefficient of friction 
# Calculations
theta=math.degrees(math.atan(p/(2*math.pi*r))) #degree # theta is the Helix angle
phi=math.degrees(math.atan(mu)) # degree # phi is the angle of friction
# Taking the leverage due to handle into account,force F1 required is,
a=theta+phi
b=theta-phi
F1=(W*(math.tan(a*math.pi/180)))*(r/l) #kN
# To lower the load
F2=(W*(math.tan(b*math.pi/180)))*(r/l) #kN # -ve sign of F2 indicates force required is in opposite sense
E=(math.tan(theta*math.pi/180)/math.tan((theta+phi)*math.pi/180))*100 # % # here E=eata=efficiency in %
# Results
print('The force required (i.e F1) to raise the weight is %f kN'%F1)
print('The force required (i.e F2) to lower the weight is %f kN'%F2)
print('The efficiency of the jack is %f percent'%E)

The force required (i.e F1) to raise the weight is 1.208561 kN
The force required (i.e F2) to lower the weight is -0.794732 kN
The efficiency of the jack is 16.461202 percent

Example 7.11 Application of Friction

In [33]:
from __future__ import division
import math
# Initilization of variabes
P=20000 #N #Weight of the shaft
D=0.30 #m #diameter of the shaft
R=0.15 #m #radius of the shaft
mu=0.12 # coefficient of friction
# Calculations
# Friction torque T is given by formulae,
T=(2/3)*P*R*mu #N-m
M=T #N-m
# Results
print('The frictional torque is %f N-m'%M)

The frictional torque is 240.000000 N-m