# Chapter 3 : Rigid bodies equivalent systems of forces¶

## Example 3.1 Page No : 85¶

In [1]:
import math

# Given Data
F = 100.; 			# lb , Vertical force applied to end of lever
theta = 60.;			# degree, angle made by lever with +ve X axis
l = 24.; 			#  , length of lever

# Calculations and Results
# a ) Momemt about O
d = l*math.cos(math.radians(theta));			# mm ,perpendicular dismath.tance from  o to the line of action

Mo = F*d;			# N.m, Magnitude of moment about O
print "Magnitude of moment about O of the 500 N is %d lb.in and it is in clockwise direction\
\n as force tends to rotate lever clockwise"%(Mo);

# b) Horizontal force

d = l*math.sin(math.radians(theta));			#in, perpendicular dismath.tance from  o to the line of action

F = Mo/d;			# N, Horizontal Force at A required to produce same Moment about O
print "Magnitude of Horizontal Force at A required to produce same Moment about O is %f lb "%(F);

# c)Smallest force

# F is smaller when d is maximum in expression Mo = F*d, so we choose force perpendicular to OA
Mo = 1200.			#in lb
d = 24.			# in ,perpendicular dismath.tance from  o to the line of action
F = Mo/d;			# N, Smallest Force at A required to produce same Moment about O
print "Magnitude of smallest Force at A required to produce same Moment about O is %f lb "%(F);

#d) 1200 N vertical force
Mo = 1200.;			# lb-in,
F = 240.			#in lb
d = Mo/F;			# m, perpendicular dismath.tance from  o to the line of action of force
OB = d/math.cos(math.radians(theta));			#m, dismath.tance of point B from O

print "Verical force of 1200 N must act at %f in far from the shaft to create same moment about O"%(OB);

Magnitude of moment about O of the 500 N is 1200 lb.in and it is in clockwise direction
as force tends to rotate lever clockwise
Magnitude of Horizontal Force at A required to produce same Moment about O is 57.735027 lb
Magnitude of smallest Force at A required to produce same Moment about O is 50.000000 lb
Verical force of 1200 N must act at 10.000000 in far from the shaft to create same moment about O


## Example 3.2 Page No : 86¶

In [2]:
import math
from numpy.linalg import det

# Given data
F = 800.; 			# N , Force applied on bracket
theta = 60.;			# degree, angle made by lever with +ve X axis
theta = theta*math.pi/180;			# Conversion of angle into radian
r_AB = [-0.2, 0.16];			#m vector drawn from B to A resolved in recmath.tangular component

# Calculations
F = [F*math.cos(theta), F*math.sin(theta)]			#N , vector F resolved in recmath.tangular component
k = 1;			# Unit vector along Z axis
# M_B = r_AB * F relation 3.7 from section 3.5
M_B = det([r_AB, F])*k;			# N.m

# Results
print "The moment of force 800 N about B is %.2f N.m . -ve sign shows its acting clockwise"%(M_B);

The moment of force 800 N about B is -202.56 N.m . -ve sign shows its acting clockwise


## Example 3.3 Page No : 86¶

In [4]:
import math

# Given data
P = 30.; 			# lb, Force applied to shift lever
alpha = 20.;			# degree, angle made by force P with -ve X axis

# Calculations
d = 3			#in ft
M_o = Q*d			#N.m , here negative signs are taken as each component creates moment clockwise

# Results
print "The moment of force P about B is %.2f lb-ft . -ve sign  shows its acting clockwise"%(M_o);

The moment of force P about B is 30.78 lb-ft . -ve sign  shows its acting clockwise


## Example 3.4 Page No : 87¶

In [6]:
import math
from numpy.linalg import norm,det

# Given data
# M_A = r_CA * F relation 3.7 from section 3.5
f = 200.; 			# N , Magnitude of Force directed along CD
r_CA = [0.3,0, 0.08];			#m, vector AC reprecsented in recmath.tangular component
#lambda = CD/norm(CD)-m, Unit vector along CD
#F = f*lambda;			#m, Force
CD = [-0.3, 0.24, -0.32];			#Vector CD resolved into recmath.tangular component
# norm(CD); m, magnitude of vector CD

# Calculations
lambda1 = CD/norm(CD);			#m, Unit vector along CD
F = f*lambda1;			#m, Force
# M_A = r_CA * F relation 3.7 from section 3.5
#i = 1; j = 1; k = 1; Unit vectors along X, Y and Z direction respectively

# Componenets of moment M_A along X,Y and Z direction respectively
M_Ax = det([[r_CA[1],r_CA[2]],[F[1], F[2]]]);			#N.m
M_Ay = -det([[r_CA[0],r_CA[2]] , [F[0],F[2]]]);			#N.m
M_Az = det([[r_CA[0],r_CA[1]],[F[0], F[1]]]);			# N.m

# Results
print "Answer can be written as M_B  =  %.2f N.m i + %.2f N.m j + %.2f N.m k "%(M_Ax,M_Ay,M_Az);

Answer can be written as M_B  =  -7.68 N.m i + 28.80 N.m j + 28.80 N.m k


## Example 3.6 Page No : 0¶

In [7]:
#Given data
# Moment arms
Fx = -30.;			#in lb
Fy = 20.;			#in lb
Fz = 20.;			#in lb

#couple Forces
x = 18.;			#iN
y = 12.;			#iN
z = 9.;			#iN

# Calculations
Mx = Fx*x;			#N.m, Component of Moment along X axis
My = Fy*y;			#N.m, Component of Moment along Y axis
Mz = Fz*z;			#N.m, Component of Moment along Z axis

# Results
#This three moments represent component of math.single couple M
print "Couple M equivalent to two couple can be written as  M  =  %.2f lb-in i + %.2f lb-in j + %.2f lb-in k "%(Mx,My,Mz);

Couple M equivalent to two couple can be written as  M  =  -540.00 lb-in i + 240.00 lb-in j + 180.00 lb-in k


## Example 3.7 Page No : 113¶

In [8]:
import math

# Given Data
Mo = 24.;			#N.m *k, Couple of moment
f = -400.;			#N, Magnitude of force
OB = 300.;			#mm,Dismath.tance of force from point O
theta = 60.;			# degree, angle made by lever with +ve X axis

# Calculations and Results

120.0