Chapter 9: Friction

Example 9.9-1, Page no 130

In [3]:
import math

#Calculations
#Simplifying  equation (3) after substituting value of Nb in it we get
#m_u**2+m_u*2*tan(50)-1=0
#Solution of the equation
a=1
b=2*1.19175 # here 1.19175 is value of tan(50)
c=-1
g=(b**2-(4*a*c))**0.5

#solution
x1=(-b+g)/(2*a)
x2=(-b-g)/(2*a)
#As x2 does not make any physical sense x1 is the answer

#Result
print'The value of mu is',round(x1,2)

The value of mu is 0.36

Example 9.9-3, Page no 131

In [5]:
import math
import numpy as np

#Initilization of variables
m=70 #kg
g=9.81 #m/s**2
# as theta=20 degrees, we have
sintheta=0.3420
costheta=0.9396

#Calculations
#Solving by martix method
#Taking sum along vertical and horizontal direction and equating them to zero
A=np.array([[sintheta,1,0],[-costheta,0,1],[0,-4**-1,1]])
#RHS matrix
R=np.array([[m*g],[0],[0]])
ans1=np.linalg.solve(A,R) #force vector N
#Calculation part 2
#Similar solution by matrix method
#Taking moment about point O and summing forces in horizontal and vertical direction and equating all to zero
B=np.array([[4*costheta,0,0],[-costheta,1,0],[sintheta,0,1]])
#RHS matrix
J=np.array([[m*g*1.5],[0],[m*g]])
ans2=np.linalg.solve(B,J) #force Vector N

#Result
print'The value of P in first case is',round(ans1[0]),"N",'and that in second case is',round(ans2[0]),"N"

The value of P in first case is 167.0 N and that in second case is 274.0 N

Example 9.9-4, Page no 132

In [7]:
import math

#Initilization of variables
W=200 #lb
Fapp=300 #lb
mu=0.3 #coefficient of friction
# as theta=30 degrees, we have
sintheta=2**-1
costheta=sqrt(3)*2**-1

#Calculations
#Summing forces in the plane parallel to the slope
F=-(W*sintheta-Fapp*costheta) #lb
N1=(W*costheta+Fapp*sintheta) #lb
#Max value obtained
Fprime= mu*N1

#Result
print'The value of F is',round(F),"lb"
print'The value of N1 is',round(N1),"lb"
print'The value of Fprime is',round(Fprime),"lb"

The value of F is 160.0 lb
The value of N1 is 323.0 lb
The value of Fprime is 97.0 lb

Example 9.9-5, Page no 132

In [9]:
import math
import numpy as np

#Initilization of variables
mu1=0.2 #coefficient of friction between wedges and A
mu2=4**-1 #coefficient of friction between wedges 
F=20 #tonnes

#Calculations
#Using the matrix method to solve
#Summing forces in vertical and horizontal direction
A=np.array([[1,-(mu1*10+1)/(101)**0.5],[0,(10-mu1*1)/101**0.5]]) #force matrix
B=np.array([[mu2*F*1000],[F*1000]]) #lb
#Solving both matrices
R=np.linalg.solve(A,B) #lb

#Result
print'The forces N2 and P are:',round(R[1]),"lb",'and',round(R[0]),"lb"
#Decimal accuracy causes discrepancy in answers

The forces N2 and P are: 20510.0 lb and 11122.0 lb

Example 9.9-6, Page no 133

In [11]:
import math
import numpy as np

#Initilization of variables
# as theta=45 degrees,we have
sintheta=sqrt(2)**-1
costheta=sqrt(2)**-1
mu1=4**-1 #coefficient of friction between A and B
mu2=3**-1 #coefficient of friction between A and Floor
ma=14 #kg
mb=9 #kg
g=9.81 #m/s**2

#Calculations
#Summing forces in vertical direction
Nb=mb*g #N
#Also
Fprimeb=mu1*Nb #N
#Summing forces in direction
T=Fprimeb #N
#Considering the fig(c)
#Summing forces in the horizontal direction and vertical direction and solving by matrix method 
A=np.array([[-costheta,mu2],[sintheta,1]]) #N
B=np.array([[-Fprimeb],[(mb*g+ma*g)]]) #N
R=np.linalg.solve(A,B) #N

#Result
print'The value of P and Na are:',round(R[0]),"N",'and',round(R[1]),"N respectively."

# The ans may wary due o decimal point descrepancy.

The value of P and Na are: 103.0 N and 153.0 N respectively.

Example 9.9-7, Page no 134

In [2]:
import math

#Initilization of variables
m1=40 #kg
m2=13.5 #kg
mu=3**-1 #coefficient of friction
g=9.81 #m/s**2

#Calculations
#Solving by substitution
#After simplification we get
x=mu*m2*g
y=mu*(m1*g+m2*g)
theta=arctan((x+y)/(m1*g))*(180/pi) #degrees

#Result
print'The value of the angle is',round(theta,1),"degrees"

The value of the angle is 29.2 degrees

Example 9.9-8, Page no 134

In [6]:
import math
import numpy as np

#Initilization of variables
W=350 #lb
# as theta=30 degrees, we have
sintheta=2**-1
costheta=sqrt(3)*2**-1
# and phi=15 degrees,thus
sinphi=0.2588
cosphi=0.9659

#Calculations
#Solving by the matrix method
A=np.array([[costheta,sinphi],[-sintheta,cosphi]])
B=np.array([[W*sintheta],[W*costheta]])
an=np.linalg.solve(A,B) #lb

#Result
print'The value of P and R are:',round(an[0],1),"lb",'and',round(an[1],1),"lb respectively."

# The ans may wary due to decimal point descrepancy.

The value of P and R are: 93.8 lb and 362.4 lb respectively.

Example 9.9-9, Page no 135

In [11]:
import math

#Initilization of variables
# as theta=45 degrees, we have
sintheta=sqrt(2)**-1
costheta=sqrt(2)**-1
m1=45 #kg
m2=135 #kg
g=9.81 #m/s**2
mu=0.25 #coefficient of riction

#Calculations
N2=m2*g #N
T=mu*N2 #N
N1=m1*g*costheta #N
Fprime1=N1*mu #N
P=T+Fprime1-(m1*g*sintheta) #N

#Result
print'The values are'
print'N2=',round(N2),"N"
print'T=',round(T),"N"
print'N1=',round(N1),"N"
print'Fprime1=',round(Fprime1),"N"
print'P=',round(P,1),"N"

# The ans may wary due to decimal point descrepancy

The values are
N2= 1324.0 N
T= 331.0 N
N1= 312.0 N
Fprime1= 78.0 N
P= 97.0 N

Example 9.9-10, Page no 135

In [9]:
import math

#Initilization of variables
mu=0.2 #coefficient of friction
F1=150 #lb
F2=100 #lb
# as theta=60 degrees
costheta=2**-1
# also theta1=30 degrees
costheta1=sqrt(3)*2**-1

#Calculations
N1=F1*costheta #lb
T=(mu*N1)+(F1*(costheta1)) #lb considering positive
#Equilibrium for 100lb
#Eliminating N2 from both equations
#Taking derivative we get
theta2=arctan(mu)*(180/pi) #degrees
#Hence P becomes
# in calculation of P we use the values of sin(theta2) & cos(theta2) as,
sintheta2=0.196
costheta2=0.98
P=(F2*mu+T)*(costheta2+(mu*sintheta2))**-1 #lb

#Result
print'The minimum value of P is',round(P),"lb"

The minimum value of P is 162.0 lb

Example 9.9-11, Page no 136

In [10]:
import math

#Initilization of variables
F=180 #N
m=100 #kg
g=9.81 #m/s**2
mu=0.25 #coeffiecient of friction

#Calculations
#Assuming F2 is maximum
N2=F*2/(1+mu) #N
F2=mu*N2 #N
N1=m*g-F2 #N
F1=F-F2 #N

#Result
print'The vaules are'
print'N2=',round(N2,3),"N"
print'F2=',round(F2,3),"N"
print'N1=',round(N1,3),"N"
print'F1=',round(F1,3),"N"

The vaules are
N2= 288.0 N
F2= 72.0 N
N1= 909.0 N
F1= 108.0 N

Example 9.9-13, Page no 138

In [12]:
import math

#Initilization of variables
mu_ca=0.3 #ceofficient of friction between copper block A and aluminium block B
mu_af=0.2 #coefficient of friction between aluminium block B and Floor
ma=3 #kg
mb=2 #kg
g=9.81 #m/s**2

#Calculations
#For A
#Taking sum of forces along X and Y direction
Na=ma*g #N
P=mu_ca*Na #N
#For B
#Taking sum of forces along X and Y direction
Nb=Na+mb*g #N
Fb=mu_ca*Na #N
#Now largest value of friction before slip is 
Fprimeb=mu_af*Nb #N
#Now as Fb<F'b hence initial assumption is incorrect and P=Fb
P=Fb #N

#Result
print'The value of force that will cause motion is',round(P,2),"N"

The value of force that will cause motion is 8.83 N

Example 9.9-15, Page no 141

In [24]:
import math

#Initilization of variables
d_m=2 #in   mean diameter of the screw
p=4**-1 #in
mu=0.15 #coefficient of friction
l=2 #ft
L=4000 #lb

#Calculations
phi=arctan(mu)*(180/pi) #degrees
beta=arctan(p*(pi*l)**-1)*(180/pi) #degrees
x=phi+beta # degrees
#Force to raise the load
# Here the value of x=10.77 degrees, thus
tanx=0.19
P=(L*tanx)/(d_m*12) #lb
#Force to lower the load
# Also,
y=phi-beta
# Thus y yeilds 6.23 degrees, thus
tany=0.109
P2=(L*tany)/(d_m*12) #lb

#Result
print'The force to raise the load is',round(P,1 ),"lb"
print'The force to lower the load is',round(P2,1),"lb"

# The answer waries due to decimal point descrepancy

The force to raise the load is 31.7 lb
The force to lower the load is 18.2 lb

Example 9.9-16, Page no 141

In [32]:
import math

#Initilization of variables
r_m=2.338 #in
d_m=3.25 #in
mu=0.06 #coefficient of friction
P=1500 #lb
p=4**-1 #pitch

#Calculation
phi=arctan(mu)*(180/pi) #degrees
beta=round(arctan(p/(2*pi*r_m))*(180/pi)) #degrees
x=phi+beta
# Thus tanx yeilds
tanx=0.077
M=P*r_m*tanx+mu*P*(d_m/2) #lb.in

#Result
print'The moment required is',round(M),"lb.in"

#Decimal accuracy causes discrepancy in answers

The moment required is 416.0 lb.in

Example 9.9-17, Page no 141

In [33]:
import math

#Initilization of variables
d=750 #mm diameter
alpha=pi #wrap angle    radians
mu=0.25 #coefficient of friction
T_t=200 #N tension on the tight side

#Calculation
T2=T_t/(exp(mu*alpha)) #N

#Result
print'The tension of the slack side is',round(T2,1),"N"

The tension of the slack side is 91.2 N

Example 9.9-18, Page no 141

In [29]:
import math

#Initilization of variables
d=635 #mm diameter of the drum
P=178 #N
mu=3**-1 #coefficient of friction
l1=100 #mm 
l2=660 #mm
# as theta1=60 degrees
sintheta1=sqrt(3)*2**-1
costheta1=2**-1
# as theta2=30 degrees,
sintheta2=2**-1
costheta2=sqrt(3)*2**-1
GD=d/2 #mm

#Calculations
#Taking moment about point C
Tb=(P*(l1+l2))/(l1*sintheta1) #N
CD=((d/2)-(l1*costheta2))/sintheta2 #mm
#from fig 9-22(b) 
theta=arcsin(GD/CD)*(180/pi) #degrees
#from fig9-22(c)
w_d=(180+30+theta) #degrees
w=(w_d)*(pi/180) #radians
#As Tc is greater than Tb
Tc=Tb*(exp(mu*w)) #N
M=(Tc-Tb)*GD #N.mm
an=M/1000 #N.m

#Result
print'The braking moment required is',round(an),"N.m"

#Note the unit of the final enswer carefully
# The answer is off by 2 N.m

The braking moment required is 1668.0 N.m

Example 9.9-19, Page no 142

In [30]:
import math

#Initilization of variables
L=1000 #lb
P=10 #lb

#Calculations
mu=log(L/P)/(4*2*pi) 

#Result
print'The coefficient of friction is',round(mu,2)

The coefficient of friction is 0.18

Example 9.9-20, Page no 142

In [32]:
import math

#Initilization of variables
m=900 #kg
mu=0.2 #coefficient of friction
g=9.8 #m/s**2

#Calculations
T2=m*g/(exp(2*2*pi*mu)) #N

#Result
print'The force needed to hold the mass is',round(T2),"N"

The force needed to hold the mass is 714.0 N

Example 9.9-21, Page no 143

In [33]:
import math

#Initilization of variables
d=760 #mm
W=500 #N
a=0.305 #mm coefficient of rolling resisatnce
r=d/2 #mm

#Calculations
P=(W*a)/r #N

#Result
print'The force necessary is',round(P,1),"N"

The force necessary is 0.4 N