Chapter 3:Coplanar Parallel forces¶

Example 3.1,Page No.48¶

In [1]:
import math

#Declaration Of Variables

#Lengths
L_AB=L_BC=L_CD=L_DA=2 #m

F_B=10 #N
F_C=20 #N
F_D=30 #N
F_A=40 #N

#Calculations

#Taking Moment at point A
#As the Forces F_A & F_B pass through point A,these Forces will be zero
#Resultant Moment of all Forces
M_A=-(-F_D*L_DA-F_C*L_CD) #N.m

#Result
print"Resultant Moment about point A is",round(M_A,2),"N.m (Anticlockwise)"

Resultant Moment about point A is 100.0 N.m (Anticlockwise)


Example 3.2,Page No.50¶

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

#Declaration Of Variables
F_A=100 #N
OCB=60 #Degrees
L_OC=3 #m

#Calculations

#Triangle OBC is a right Angled triangle
L_OB=L_OC*sin(60*pi*180**-1)

#Moment of force 100 #N about o
M_O=F_A*L_OB #Nm

#Result
print"Moment of Force about O is",round(M_O,2),"Nm"

Moment of Force about O is 259.81 Nm


Example 3.3,Page No.54¶

In [3]:
import math

#Declaration Of Variables

#Lengths
L_AB=30 #cm
L_BC=40 #cm

F_A=100 #N
F_B=200 #N
F_C=300 #N

#Calculations

#resultant of all forces
R=F_A+F_B+F_C #N

#Let resultant be acting at distance x from point A

#Now taking moments at point A
M_A=-(-F_C*(L_AB+L_BC)-F_B*L_AB) #N.m

#Moment of resultant R about A
#M_R=R*x

#But algebraic sum of moments of all forces about A = Moment of resultant about A
x=M_A*R**-1

#Result
print"Resultant is",round(R,2),"N"
print"Distance of resultant from point A",round(x,2),"cm"

Resultant is 600.0 N
Distance of resultant from point A 45.0 cm


Example 3.4,Page No.54¶

In [1]:
import math

#Declaration Of Variables

#Lengths
L_AC=4 #m
L_CD=3 #m

#Forces
F_A=50  #N
F_D=100 #N
F_B=200 #N
R=250 #N #Resultant

#Calculations

#Part-1
#Magnitude of Force F_B
F=R-F_A-F_D #N

#Part-2
#Distance from pt A
#Taking Moment of all forces at pt A
#M_A=0
#Now moments of all forces = Moment of Resultant
#After simplifying  further we get

#Result
print"Magnitude of Force F is",round(F,2),"N"
print"Distance of force f from A is",round(x,2),"m"

Magnitude of Force F is 100.0 N
Distance of force f from A is 1.5 m


Example 3.5,Page No.55¶

In [5]:
import math

#Declaration Of Variables

#Lengths
L_AB=0.9 #m
L_BC=1.2 #m
L_CD=0.75 #m
L=L_AB+L_BC+L_CD #m

#Forces
F_A=100 #N
F_B=150 #N
F_C=25 #N
F_D=200 #N

#Calculations

#Part-1
#Magnitude of Resultant
R=F_A-F_B-F_C+F_D #N

#Part-2
#Let x be the distance of Resultant from A
x=-(F_B*L_AB+F_C*(L_AB+L_BC)-F_D*L)*R**-1 #m

#Result
print"Magnitude of Resultant",round(R,2),"N"
print"Distance of Resultant from x is",round(x,2),"m"

Magnitude of Resultant 125.0 N
Distance of Resultant from x is 3.06 m


Example 3.6,Page No.57¶

In [6]:
import math

#Declaration Of Variables

#Lengths
L_AC=L_CD=1 #m
L_DB=1.5 #m
L=3.5 #m

#Forces
F_A=32.5 #N
F_C=150 #N
F_D=67.5 #N
F_B=10 #N

#Calculations

#Part-1

#Single Force ststem
R=-(F_A-F_C+F_D-F_B) #N

#Let x be the distance of Resultant from A
x=-(F_C*L_AC-F_D*(L_AC+L_CD)+F_B*L)*R**-1 #m

#Part-2

#Single Force is given By R

#Now moment of couple at pt A
M_A=-R*round(x,2) #N.m

#Part-3

#Now couple at B
L_BE=L+x
M_B=R*round(L_BE,3) #N.m

#Result
print"Single Force is",round(R,2),"N"
print"Couple at A",round(M_A,2),"N.m"
print"Couple at B",round(M_B,2),"N.m"

Single Force is 60.0 N
Couple at A 49.8 N.m
Couple at B 160.02 N.m


Example 3.7,Page No.59¶

In [7]:
import math

#Declaration Of Variables

#Lengths
L_AB=L_DE=0.6 #m
L_BC=0.9 #m
L_CD=1.2 #m

#Forces
F_A=4 #N
F_B=8 #N
F_C=8 #N
F_D=16 #N
F_E=12 #N

#Calculations

#Resultant Of All Forces
R=-F_A+F_B-F_C+F_D-F_E #N

#As the Resultatn Force is zero,tere will be two possibilities.The system will have a resultant coup
#Algebraic sum of moments of all forces about A
M_A=-F_B*L_AB+F_C*(L_AB+L_BC)-F_D*(L_AB+L_BC+L_CD)+F_E*(L_AB+L_BC+L_CD+L_DE) #N.m

#As the algebraic sum of moments of all forces is not zero,the ststem will have couple of magnitude 3.6 #N.m

#Result
print"Resultant of Parallel Forces is",round(R,2),"N"

Resultant of Parallel Forces is 0.0 N


Example 3.8,Page No.59¶

In [8]:
import math

#Declaration Of Variables

L_AB=L_DE=2 #m
L_BC=0.5 #m
L_CD=0.5 #m

#Forces
F_A=20 #N
F_B=20 #N
F_C=40 #N
F_D=30 #N
F_E=10 #N

#Calculations

#Resultant Of All Forces
R=-F_A+F_B+F_C-F_D-F_E #N

#As the Resultant is zero and also the resultant force on  the body is zero,the body will be in equilibrium

#Result
print"Resultant of Parallel Forces is",round(R,2),"N"

Resultant of Parallel Forces is 0.0 N


Example 3.9,Page No.60¶

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

#Declaration Of Variables

#Distances
L_OA=200 #mm
L_OB=100 #mm
L_BC=200 #mm

COA=90 #Degrees

#Forces
F_A=2000 #N
F_B=1500 #N
F_C=1000 #N

#Calculations

#Resolving FOrce A in two components
F_A1=F_A*cos(30*pi*180**-1) #Component along x-axis
F_A2=F_A*sin(30*pi*180**-1) #Component along y-axis

#Resolving all forces along X-axis
F_x=F_A1-F_B-F_C
F_y=F_A2

#Resultant
R=(F_x**2+F_y**2)**0.5 #N

#Taking Moments of all forces about pt O
M_o=-(F_y*L_OA-F_B*L_OB-F_C*(L_BC+L_OB)) #N.mm

#Result
print"Equivalent system through point O is:Resultant",round(R,2),"N"
print"                                    :Moment",round(M_o,2),"N.mm"

Equivalent system through point O is:Resultant 1260.85 N
:Moment 250000.0 N.mm


Example 3.10,Page No.61¶

In [10]:
import math

#Declaration Of Variables

#Lengths
L_AC=1 #m
L_CB=1.5 #m
L_CD=0.8 #m
L_DB=0.7 #m
L=2.5 #m

#Forces
F_C=4000 #N
F_B=2500 #N
M_D=2000 #N*m

#Calculations

#Resultant of all forces
R=-F_C+F_B #N

#As the Force is acting in downward direction,so negative sign
R2=-R

#Sum of Moments of all Forces
M=(F_C*L_AC+M_D-F_B*L)
M2=-M #Anticlockwise

#Now distance of resultant from x is
x=(F_C*L_AC+M_D-F_B*L)*R2**-1

x2=-x

#Result
print"Resultant of the system",round(R2,2),"N"
print"Equivalent system through A",round(M2,2),"N.m (Anticlockwise)"

Resultant of the system 1500.0 N
Equivalent system through A 250.0 N.m (Anticlockwise)