Chapter 4:Conditions Of Equilibrium

Example 4.1,Page No.68

In [1]:
import math

#Declaration Of Variables
F1=100 #N #Force acting on body

#Calculations

#As the Force F1 & F2 are acting on the same body and at same point but in opposite directions
#These two forces will be equal
F2=F1

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

Magnitude of Force F2 is 100.0 N

Example 4.2,Page No.68

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

#Declaration Of Variables

#Force
F3=400 #N
theta1=30 #Degree #Angle made by forces F2 & F3

#Calculations

#By Lami's Theorem
#F1*sin(120)**-1=F2*sin(120)**-1=F3*sin(120)**-1
F2=F3*sin(120*pi*180**-1)**-1*sin(120*pi*180**-1)
F1=F2*sin(120*pi*180**-1)**-1*sin(120*pi*180**-1)

#Result
print"Magnitude of Forces:F1",round(F1,2),"N"
print"                   :F2",round(F2,2),"N"

Magnitude of Forces:F1 400.0 N
                   :F2 400.0 N

Example 4.3,Page No.69

In [3]:
import math

#Declaration Of Variables

#FOrces
F1=250 #N
F3=1000 #N
L_AB=1 #m #Length of AB

#Calculations

#Sum of forces in y direction
F2=F1+F3 #N

#Moment at pt A
#-F2*L_AB+F3*(L_AB+x)=0
#After further simplifying we get
x=F2*L_AB*F3**-1-L_AB

#Result
print"Magnitude of Force F2 is",round(F2,2),"N"
print"Distance of F2 From F3 is",round(x,2),"N"

Magnitude of Force F2 is 1250.0 N
Distance of F2 From F3 is 0.25 N

Example 4.4,Page No.69

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

#Declaration Of Variables

#Forces
F1=18 #N
F2=22.5 #N
F3=15 #N
F4=30 #N

#Angles
theta2=45 #Degree
theta3=90 #Degree
theta4=30 #Degree

#Calculations

#Sum of Forces in x-direction
#F1+F2*cos(45)-F4*cos(30)-F5*cos(theta5)=0
#After further simplifying we get
#F5*cos(theta5)=F1+F2*cos(45)-F4*cos(30)....................1

#Sum of Forces in y-direction
#F3+F2*sin(45)-F4*sin(30)-F5*sin(theta5)=0
#After further simplifying we get
#F5*sin(theta5)=F3+F2*sin(45)-F4*sin(30)....................2

#Dividing equation 2 and 1 we get
X=F3+F2*sin(45*pi*180**-1)-F4*sin(30*pi*180**-1)
Y=F1+F2*cos(45*pi*180**-1)-F4*cos(30*pi*180**-1)
theta=np.arctan((X)*(Y)**-1)*(pi**-1*180)

F5=(F1+F2*cos(45*pi*180**-1)-F4*cos(30*pi*180**-1))*(cos(theta*pi*180**-1))**-1


#Result
print"Magnitude of force F5 is",round(F5,2),"N"
print"Direction of F5 is",round(theta,2),"Degrees"

Magnitude of force F5 is 17.78 N
Direction of F5 is 63.51 Degrees

Example 4.5,Page No.71

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

#Declaration Of Variables 

#Forces
F_C=1500 #N
theta_C=60 #degrees

F_B=1805 #N
theta_B=33.67 #Degrees

F_A=2240 #N
theta_A=63.43 #Degrees

#Distances of forces from D
L_DC=2 #m
L_DB=3 #m
L_DE=4 #m
L_DO=3 #m


#Calculations

#NEt forces along Y-axis
R_y=-F_C*cos(theta_C*pi*180**-1)-F_B*cos(theta_B*pi*180**-1)+F_A*cos(theta_A*pi*180**-1)

#Net Forces along x-axis
R_x=F_C*sin(theta_C*pi*180**-1)-F_B*sin(theta_B*pi*180**-1)-F_A*sin(theta_A*pi*180**-1)

#Resultant Forces
R=(R_x**2+R_y**2)**0.5 #N

#Angle made by resultant
theta=np.arctan(R_y*R_x**-1)*(pi**-1*180) #Degrees

#Net Moment about point O
M_O=-F_C*cos(theta_C*pi*180**-1)*L_DO-F_B*cos(theta_B*pi*180**-1)*L_DO-F_C*sin(theta_C*pi*180**-1)*L_DC+F_B*sin(theta_B*pi*180**-1)*L_DB+F_A*sin(theta_A*pi*180**-1)*L_DE


#Moment of R about O

#X-intercept
x=M_O*-R_x**-1

#Y-intercept
y=M_O*-R_y**-1

#Result
print"Resultant is",round(R,2),"N"
print"X intercept is",round(x,2),"m"
print"Y intercept is",round(y,2),"m"

Resultant is 2114.37 N
X intercept is 0.97 m
Y intercept is 1.33 m

Example 4.6,Page No.73

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

#Declaration Of Variables
theta=60 #Degrees #Angle made by chain with ceiling
W=5 #N #Weight of lamp
theta2=120 #Degree #Angle made by chain with cord
theta3=150 #Degree #Angle made by chain with wire holding lamp
theta4=90 #Degree #Angle made by cord with wire holding lamp

#Calculations

#LEt T1=tension in cord
#T2=tension in chain

#By lami's theorem
#T1*sin(theta3)**-1=T2*sin(theta4)**-1=W*sin(theta2)**-1

T1=W*sin(theta2*pi*180**-1)**-1*sin(theta3*pi*180**-1) #N
T2=W*sin(theta2*pi*180**-1)**-1*sin(theta4*pi*180**-1) #N

#Result
print"Tension in chain is",round(T2,2),"N"
print"Tension in cord is",round(T1,2),"N"

Tension in chain is 5.77 N
Tension in cord is 2.89 N

Example 4.7,Page No.74

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

#Declaration Of Variables

#Forces
F_P=1000 #N
F_Q=1500 #N
F_R=1000 #N
F_S=500  #N

theta=90 #Degree  #Angle made by F_P with PS
theta2=60 #Degree #Angle made by F_Q with QS
theta3=45 #Degree #Angle made by F_R with RS
theta4=30 #Degree #Angle made by F_S with PS

#Calculations

#Resultant of forces along x-axis
R_x=-F_Q*cos(theta2*pi*180**-1)-F_R*cos(theta3*pi*180**-1)-F_S*cos(theta4*pi*180**-1)

#Resultant of forces along y-axis
R_y=-F_P*sin(theta*pi*180**-1)-F_Q*sin(theta2*pi*180**-1)-F_R*sin(theta3*pi*180**-1)-F_S*sin(theta4*pi*180**-1)

#Resultant
R=(R_x**2+R_y**2)**0.5 #N

#Direction of resultant
theta=np.arctan(R_y*R_x**-1)*(180*pi**-1) #Degree

#Result
print"Magnitude of Resultant is",round(R,2),"N"
print"Direction of Resultant is",round(theta,2),"degree"

Magnitude of Resultant is 3764.97 N
Direction of Resultant is 59.87 degree

Example 4.9,Page No.78

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

#Declaration Of Variables
W=100 #N #Weight of roller
L_BC=10 #cm #Radius of roller
L_AB=20 #cm #Length of tie rod

#Calculations

theta=np.arcsin(L_BC*L_AB**-1)*(pi**-1*180) #Degrees

F=W*cos(theta*pi*180**-1)**-1 #N #Force in tie rod
R_C=F*sin(theta*pi*180**-1)

#Result 
print"Force in tie rod is",round(F,2),"N"
print"Reaction at C is",round(R_C,2),"N"

Force in tie rod is 115.47 N
Reaction at C is 57.74 N

Example 4.11,Page No.79

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

#Declaration Of Variables
W=120 #N #Weight of ball
theta=30 #Degrees #angle made by groove
theta2=60 #Degrees #Angle made by groove

#Calculations

#LEt R_A and R_B be the reactions at A and B respectively
#by LAmi's theorem
#R_B*sin(120)**-1=R_A*sin(150)**-1=W*sin(90)**-1

R_C=W*sin(90*pi*180**-1)**-1*sin(120*pi*180**-1) #N
R_A=W*sin(90*pi*180**-1)**-1*sin(150*pi*180**-1) #N


#Result
print"Reaction at A is",round(R_A,2),"N"
print"Reaction at c is",round(R_C,2),"N"

Reaction at A is 60.0 N
Reaction at c is 103.92 N

Example 4.11(A),Page No.81

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

#Declaration Of Variables
W=100 #N #weight of roller

#Calculations

#LEt R_A and R_B be the reactions at A and B respectively
theta=45 #Degrees #Angle made by R_B with horizontal
theta2=30 #Degrees  #Angle made by R_A with horizontal

#by LAmi's theorem
#R_B*sin(120)**-1=R_A*sin(135)**-1=W*sin(105)**-1

R_B=W*sin(105*pi*180**-1)**-1*sin(120*pi*180**-1) #N
R_A=W*sin(105*pi*180**-1)**-1*sin(135*pi*180**-1) #N

#Result
print"Reaction at A is",round(R_A,2),"N"
print"Reaction at B is",round(R_B,2),"N"

Reaction at A is 73.21 N
Reaction at B is 89.66 N

Example 4.12,Page No.81

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

#Declaration Of Variables
W=100 #N #Weight of roller
F=200 #N #Horizontal Force
L_AB=10 #cm #LEngth of bar AB
L_BC=5  #cm #radius of roller

#Calculations

theta=np.arcsin(L_BC*L_AB**-1)*(pi**-1*180)

#LEt R_C be the reaction at c

#sum of Forces along x-axis
F_AB=F*cos(theta*pi*180**-1)**-1 #N

#sum of forces along y-axis
R_C=W+F_AB*sin(theta*pi*180**-1)

#Result
print"Force in bar AB is",round(F_AB,2),"N"
print"Reaction at C is",round(R_C,2),"N"

Force in bar AB is 230.94 N
Reaction at C is 215.47 N

Example 4.14,Page No.84

In [12]:
import math
from math import sin, cos, tan, radians, pi

#Declaration Of Variables
W=1000 #N #Weight of rollers
theta=30 #Degree #Angle made by groove

#Calculations

#LEt R_A,R_B,R_C,R_D be the reactions at A,B,C,D respectively

#Roller-2
R_D=W*sin(90*pi*180**-1)*sin(150*pi*180**-1) #N
R_A=W*sin(90*pi*180**-1)*sin(120*pi*180**-1) #N

#ROller-1
R_B=(W+R_D*sin(theta*pi*180**-1))*sin(60*pi*180**-1)**-1 #N
R_C=R_B*cos(60*pi*180**-1)+R_D*cos(theta*pi*180**-1)

#Result
print"Reactions at A:R_A",round(R_A,2),"N"
print"             B:R_B",round(R_B,2),"N"
print"             C:R_C",round(R_C,2),"N"

Reactions at A:R_A 866.03 N
             B:R_B 1443.38 N
             C:R_C 1154.7 N

Example 4.15,Page No.85

In [13]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np

#Declaration Of Variables
W=1000 #N #Weight of each sphere
L_AF=L_BF=L_FD=L_DE=L_CE=25 #cm
L=90 #Width of channel
L_FG=40 #cm
L_EF=L_FD+L_DE #cm

#Calculations

theta=np.arcsin(L_FG*L_EF**-1)*(180*pi**-1) #Degrees

#LEt R_A,R_B,R_C,R_D be the reactions at A,B,C,D respectively

#Roller-2
R_D=W*(cos(theta*pi*180**-1))**-1 #N
R_C=R_D*sin(theta*pi*180**-1) #N

#Roller-1
R_A=R_D*sin(theta*pi*180**-1) #N
R_B=R_D*cos(theta*pi*180**-1)+W #N

#Result
print"Reactions at A:R_A",round(R_A,2),"N"
print"             B:R_B",round(R_B,2),"N"
print"             C:R_C",round(R_C,2),"N"

Reactions at A:R_A 1333.33 N
             B:R_B 2000.0 N
             C:R_C 1333.33 N

Example 4.16,Page No.87

In [14]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np

#Declaration Of Variables
W=1000 #N #Weight of 2 circular cyclinders
L_AF=L_FC=L_CG=L_GB=15 #cm
W2=2000 #N #Weight of 3rd cyclinder
r2=15 #cm
L_AB=40 #cm
L_AH=L_HB=20 #cm

#Calculations

theta=np.arcsin(L_AH*(L_AF+L_FC)**-1)*(pi**-1*180) #Degrees

#Let R_G,R_F,R_D,R_E be the reactions at G,F,D,E 

R_F=W2*(2*cos(theta*pi*180**-1))**-1 #N
R_G=R_F #N

#Roller-1

R_D=W+R_F*cos(theta*pi*180**-1) #N
S=R_F*sin(theta*pi*180**-1) #N

#Roller-2

R_E=W+R_G*cos(41.81*pi*180**-1) #N

#Result
print"Reaction at E is ",round(R_E,2),"N"
print"Reactions at D is",round(R_D,2),"N"
print"Force S in the string is ",round(S,2),"N"

Reaction at E is  2000.0 N
Reactions at D is 2000.0 N
Force S in the string is  894.43 N

Example 4.17,Page No.88

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

#Declaration Of Variables
L_OB=L_OC=L_OA=40 #cm #Radius of roller
h=20 #cm #Height of block
L_OD=L_OA-h #cm #Length 
W=3000 #N #Weight of roller

#Calculations

#LEt R_B be the reaction at B
L_BD=((L_OB**2-L_OD**2)**0.5) #cm
theta=np.arctan(L_BD*(L_OC+L_OD)**-1)*(180*pi**-1) #degree

#Sum of all vertical Forces
R_B=W*(cos(theta*pi*180**-1))**-1 #N

#Sum of all horizontal Forces
P=R_B*sin(theta*pi*180**-1) #N

#Result
print"Reaction at B is",round(R_B,2),"N"
print"Horizontal Reaction at C is",round(P,2),"N"

Reaction at B is 3464.1 N
Horizontal Reaction at C is 1732.05 N

Example 4.18,Page No.89

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

#Declaration Of Variables
#Declaration Of Variables
L_OB=L_OC=L_OA=40 #cm #Radius of roller
h=20 #cm #Height of block
L_OD=L_OA-h #cm #Length 
W=3000 #N #Weight of roller

#Calculations

theta=np.arccos(L_OD*L_OB**-1)*(pi**-1*180) #N

R_B=W*(cos(theta*pi*180**-1))**-1 #N
P=R_B*sin(theta*pi*180**-1) #N

#LEt P_min be the least Force applied
#let alpha be the angle made by least force
#P_min=W*L_BD*L_BC**-1

L_BD=((L_OB**2-L_OD**2)**0.5) #cm

#But L_BC=L_BO*sin(alpha) 

#Force P will be min when sin(Alpha) is max.
#thus sin(alpha)=90 or sin(alpha)=0. therefore sub value in above equation,we get min Force
#LEt P_min be the Least Force to be applied
P_min=W*L_BD*(L_OB*1)**-1 #N

#Direction of least force is right angle to L_BO

#Result
print"Minimum least force is",round(P_min,2),"N"
print"Magnitude of force applied horizontally at centre of roller",round(P,2),"N"

Minimum least force is 2598.08 N
Magnitude of force applied horizontally at centre of roller 5196.15 N

Example 4.19,Page No.91

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

#Declaration Of Variables
L_BC=25 #cm 
L_AB=40 #cm

#Calculations

#Let alpha be the angle made by force F so that body will be in equilibrium
#Theta is angle made by R_A with horizontal,so Force F has to make same angle with horizontal
alpha=np.arctan(L_AB*L_BC**-1)*(pi**-1*180) #degrees
theta=alpha

#Result
print"Angle theta is",round(theta,2),"Degrees"

Angle theta is 57.99 Degrees

Example 4.20,Page No.91

In [18]:
import math
from math import sin, cos, tan, radians, pi

#Declaration Of Variables
L_AB=1.6 #m
L_BD=1.2 #m
L_BC=0.8 #m
L_CD=0.4 #m
F_C=200 #N #Force t C
theta=60 #Degrees

#Calculations



#Sum of all forces in x-direction
R_Bx=F_C #N

L_BD2=L_BD*sin(theta*pi*180**-1) #m
L_DD=L_BD*cos(theta*pi*180**-1) #m
L_BC2=sin(theta*pi*180**-1)*L_BC #m

R_D=F_C*L_BC2*L_DD**-1 #N
R_By=R_D

#Resultant reaction at B
R_B=(R_Bx**2+R_By**2)**0.5 #N

#Sum of moments at A
M_A=R_By*L_AB


#Result
print"Couple to be apllied to hold the system is",round(M_A,2),"N"
print"Magnitude of pin reaction at B",round(R_B,2),"N"

Couple to be apllied to hold the system is 369.5 N
Magnitude of pin reaction at B 305.51 N

Example 4.21,Page No.93

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

#Declaration Of Variables
W=2000 #N #Weight of chain
L_AB=2 #m
F_B=320 #N

#Calculations

#By LAmi's theorem
#F_A*sin(90)=W*sin(180-theta)**-1=F_B*sin(90+theta)**-1
#But sin(180-theta)=sin(theta),sin(90+theta)=cos(theta)
#Tan(theta)=W*F_B**-1
theta=np.arctan(W*F_B**-1)*(180*pi**-1) #Degrees

F_A=W*(sin(theta*pi*180**-1))**-1 #N
#Let x be the lateral distance
x=cos(theta*pi*180**-1)*2

#Result
print"Force in the chain is",round(F_A,2),"N"
print"Horizontal displacement is",round(x,2),"m"

Force in the chain is 2025.44 N
Horizontal displacement is 0.32 m