# Chapter 8 : Friction¶

## Example 8.1 Page No : 396¶

In [2]:
import math

# Given Data
h = 100.;			#lb, horizontal force
W = 300.;			#lb, weight of block
us = 0.2;			# Coeffiecient of static friction
uk = 0.20;			#Co = efficient of kinetic friction

# Calculations and Results
#Applying sumFx  = 0 , we get
F = h-3./5*W;			#lb, Force along plane
F = -F
#Applying sumFy = 0, we get
N = 4./5*W			#lb, Normal force to the plane
print "Force F required to maintain the equillibrium is thus %.0f lb, up and to right"%(F);

# Maximum friction force
Fm = us*N;			#lb,Maximum friction force
print " Maximum friction force is %.2f lb is less than that of required to maintain equillibrium\
\n that is %.2f lb  So, equillibrium will nat maintain and block wil move down"%(Fm,F);
# Actual value of friction force
Fk = (0.6*300)-(h)-(Fm);			#lb, Actual value of friction force
print "Actual value of friction force is %.2f lb directed up and to the right"%(Fk);

Force F required to maintain the equillibrium is thus 80 lb, up and to right
Maximum friction force is 48.00 lb is less than that of required to maintain equillibrium
that is 80.00 lb  So, equillibrium will nat maintain and block wil move down
Actual value of friction force is 32.00 lb directed up and to the right


## Example 8.2 Page No : 397¶

In [3]:
import math

# Given Data
F = 800.;			#N Firce in verical direction
us = 0.35;			# Coeffiecient of static friction
uk = 0.25;			#Co = efficient of kinetic friction
theta = 25.;			#degree, angle of inclination

# Calculations and Results
# Force P start block moving up
# At static equillibrium Tan(Theta_s) = us
P = F*math.tan(theta+theta_s);			#N,Force P to start block moving up
print "Force P to start block moving up is %.0f N"%(P);

# Force P to keep block moving up
# At kinetic equillibrium Tan(Theta_k) = uk
P = F*math.tan(theta+theta_k);			#N,Force P to keep block moving up
print "Force P to keep block moving up is %.0f N"%(P);

# Force P to prevent  block from sliding down

P = F*math.tan(theta-theta_s);			#N,Force P to prevent  block from sliding down
print "Force P to prevent  block from sliding down is %.0f N"%(P);

Force P to start block moving up is 780 N
Force P to keep block moving up is 649 N
Force P to prevent  block from sliding down is 80 N


## Example 8.3 Page No : 398¶

In [4]:
import math

# Given Data
us = 0.25;			# Coeffiecient of static friction

# Calculations and Results
#Applying equillibrium equation we get relation in x
print "Apply equillibrium equations. It is theoritical part. ";
x = 12-(.75*2)+1.5			#in, Dismath.tance at which the applied load can be supported
print "Minimum dismath.tance at which the applied load can be supported is %.0f in"%(x);

Apply equillibrium equations. It is theoritical part.
Minimum dismath.tance at which the applied load can be supported is 12 in


## Example 8.4 Page No : 411¶

In [5]:
import math

# Given Data
F = 400.;			#lb, force exerte
us = 0.35;			# Coeffiecient of static friction
phi = math.degrees(math.atan(us));			#rad, angle of friction
#print (phi)
theta = 8.;			#degree, angle of inclination

# Calculations and Results
#Umath.sing math.sine rule
#force p to raise block
#free body , block B
#free body wedge A
print " force required to raise block is P = %.0f lb"%(P);

#force to lower block
#free body , block B
#free body wedge A
print " force required to lower block is P = %.0f lb"%(P);

 force required to raise block is P = 423 lb
force required to lower block is P = 206 lb


## Example 8.5 Page No : 412¶

In [6]:
import math

# Given Data
pitch = 2.;			#mm, pitch of screw
d = 10.;			#mm, mean diameter of thread
us = 0.30;			# Coeffiecient of static friction
M = 40.;			#kN.m , Maximum couple

# Calculations and Results
#Force exerted by clamp
L = 2*pitch;			#mm, as screw is double threaded
theta = math.atan(L/(2*math.pi*r));			#rad, angle of inclination
phi = math.atan(us);			#rad, angle of friction
Q = M/r*1000;			#N, Force applied to block representing screw
Q = Q/1000			#kN, Conversion into kN
W = Q/math.tan(theta+phi);			#kN, Magnitude of force exerted on the piece of wood
print "Magnitude of force exerted on the piece of wood  is W =  %.2f kN "%(W);
#Couple required to loosen clamp
Q = W*math.tan(phi-theta);			#kN, Force required to loosen clamp
Couple = Q*r;			#N.m, Couple required to loosen clamp
print "Couple  required to loosen clamp is %.2f N.m"%(Couple);

Magnitude of force exerted on the piece of wood  is W =  18.01 kN
Couple  required to loosen clamp is 14.97 N.m


## Example 8.6 Page No : 423¶

In [11]:
import math
from sympy.mpmath import cot
#from numpy import cot

# Given Data
r = 1.			#in in
us = 0.20;			# Coeffiecient of static friction between shaft and pully

# Calculations and Results
#Vertical Force required to raise load
rf = r*us;			#in, Perpendicular dismath.tance from the center Of pully to line of action
P1 = (2.20*500)/1.8			#lb , downward Force required to raise load
print "Force required to raise load is %f lb in downward direction"%(P1);

#Vertcal Force required to hold load

P = (1.80*500)/2.20			#lb , downward Force required to hold load
print "Force required to hold load is %.0f lb in downward direction"%(P);

#Horizontal force P to start raimath.sing the load
OE = rf;			#mm,
OD = math.sqrt(2)*2;			#in, pythagorus theorm

# from force triangle
P = 500*cot(math.radians(40.9));			#lb, Horizontal force P to start raimath.sing the load
print "Horizontal force P required to start raising the load is %.0f lb"%(P);

Force required to raise load is 611.111111 lb in downward direction
Force required to hold load is 409 lb in downward direction
Horizontal force P required to start raising the load is 577 lb


## Example 8.7 Page No : 431¶

In [12]:
import math

# Given Data
T1 = 150.;			#N, Force on free end of hawser
T2 = 7500.;			#N, Force on other end of hawser

# Calculations and Results
#a, coefficient of friction
bta = 2*2*math.pi;			#rad, angle of contact, 2 turns
#By equation 8.13
us = math.log(T2/T1)/bta;			# Co-efficient of static friction
print "Coefficient of static friction between hawser and ballard is us =  %0.3f "%(us);

#Number of wraps when tension in hawser = 75 kN

#One turn  =  2* pi angle, bta corresponds to
ten = T1*math.exp(bta*us)
print "Tension  is %f N  "%(ten);

Coefficient of static friction between hawser and ballard is us =  0.311
Tension  is 53033.008589 N


## Example 8.8 Page No : 432¶

In [13]:
import math

#Given
T2 = 600.;			#lb, Tension from side 2
us = 0.25;			# Coeffiecient of static friction between pulley and belt
bta = (2*math.pi)/3;			#Co = efficient of kinetic friction between pulley and belt
r1 = 8.			#in in
#Pulley B

# Calculations
T1 = T2/(math.exp(us*bta))			#N, Tension from side 1
#print (T1)

#Pulley A

The largest torque which can be exerted by belt on pulley A is MA =  1957 lb-in