Chapter 6: Friction¶

import math
import numpy as np

# Initilization of variables

m=5 #kg # mass of the bock
g=9.81 # m/s^2 # acceleration due to gravity
theta=15 # degree # angle made by the forces (P1 & P2) with the horizontal of the block
mu=0.4 #coefficient of static friction

#Calculations

# Case 1. Where P1 is the force required to just pull the bock

# Solving eqn's 1 & 2 using matrix
A=np.array([[cos(theta*(pi/180)), -mu],[sin(theta*(pi/180)), 1]])
B=np.array([0,(m*g)])
C=np.linalg.solve(A,B)

# Calculations 

# Case 2. Where P2 is the force required to push the block

# Solving eqn's 1 & 2 using matrix
P=np.array([[-cos(theta*(pi/180)), mu],[-sin(theta*(pi/180)) ,1]])
Q=np.array([0,(m*g)])
R=np.linalg.solve(P,Q)

# Results

print"The required pull force P1 is ",round(C[0],2),"N" #answer in textbook is wrong
print"The required push force P2 is ",round(R[0]),"N"

import math

# Initilization of variables

W1=50 # N # weight of the first block
W2=50 # N # weight of the second block
mu_1=0.3 # coefficient of friction between the inclined plane and W1
mu_2=0.2 # coefficient of friction between the inclined plane and W2

# Calculations

# On adding eq'ns 1&3 and substuting the values of N1 & N2 from eqn's 2&4 in this and on solving for alpha we get,

alpha=(arctan(((mu_1*W1)+(mu_2*W2))/(W1+W2)))*(180/pi) # degrees

# Results

print"The inclination of the plane is ",round(alpha),"degree"

import math

# Initilization of variables

M=2000 # kg # mass of the car
mu=0.3 # coefficient of static friction between the tyre and the road
g=9.81 # m/s^2 # acc. due to gravity

# Calculations

# Divide eqn 1 by eqn 2, We get
theta=arctan(mu)*(180/pi) #degree

# Results

print"The angle of inclination is ",round(theta,1),"degree"

import math
import numpy as np

# Initilization of variabes

Wa=1000 #N # weight of block A
Wb=500 #N # weight of block B
theta=15 # degree # angle of the wedge
mu=0.2 # coefficient of friction between the surfaces in contact
phi=7.5 # degrees # used in case 2

# Caculations 

# CASE (a)

# consider the equilibrium of upper block A
# rearranging eq'ns 1 &2 and solving them using matrix for N1 & N2
A=np.array([[1 ,-0.4522],[-0.2 ,0.914]])
B=np.array([0,1000])
C=np.linalg.solve(A,B)

# Now consider the equilibrium of lower block B
# From eq'n 4
N3=Wb+(C[1]*cos(theta*(pi/180)))-(mu*C[1]*sin(theta*(pi/180))) #N
# Now from eq'n 3
P=(mu*N3)+(mu*C[1]*cos(theta*(pi/180)))+(C[1]*sin(theta*(pi/180))) # N

# CASE (b)

# The eq'n for required coefficient for the wedge to be self locking is,
mu_req=(theta*pi)/360 # multiplying with (pi/180) to convert it into radians

# Results

print"The minimum horizontal force (P) which should be applied to raise the block is ",round(P),"N"
print"The required coefficient for the wedge to be self locking is ",round(mu_req,4) #answer in textbook is wrong

import math
import numpy as np

# Initilization of variables

P=100 #N # force acting at 0.2 m from A
Q=200 #N # force acting at any distance x from B
l=1 #m # length of the bar
theta=45 #degree #angle made by the normal reaction at A&B with horizontal

# Calculations

#solving eqn's 1 & 2 using matrix for Ra & Rb,
A=np.array([[1, -1],[sin(theta*(pi/180)) ,sin(theta*(pi/180))]])
B=np.array([0,(P+Q)])
C=np.linalg.solve(A,B)

# Now take moment about B
x=((C[0]*l*sin(theta*(pi/180)))-(P*(l-0.2)))/200 #m # here 0.2 is the distance where 100 N load lies from A

# Results

print"The minimum value of x at which the load Q=200 N may be applied before slipping impends is ",round(x,2),"m"