# Chapter 4 : Equilibrium of rigid bodies¶

## Example 4.1 Page No : 166¶

In :
import math
from numpy.linalg import norm

# Calculations and Results
#Determination of B
#At equillibrium +sum(M_A) = 0
#B*1.5m-(9.81kN)(2 m)-(23.5 kN)(6 m) = 0, B assumed to be in +ve X direction
B = (9.81*2+23.5*6)/1.5			#kN
print "B = %.2f kN  +ve sign shows reaction is directed as assumed "%(B);
#Determination of Ax
#Sum Fx = 0
#Ax+B = 0
Ax = -B;			#kN
print "Ax = %.2f kN"%(Ax);
#Determination of Ay
#Sum Fy = 0
#Ay-9.81 kN-23.5kN = 0
Ay = 9.81+23.5;			#kN
print "Ay = %.2f kN"%(Ay);
A = [Ax,Ay];			#kN Adding component
A = norm(A);			#Magnitude of force A
theta = theta*180/math.pi;			#degrees, conversion into degrees
print "Reaction at A is A = %.2f kN making angle %.2f degrees  with + ve x axis "%(A,theta);
#Slight variation in the answer because of roundoff error

B = 107.08 kN  +ve sign shows reaction is directed as assumed
Ax = -107.08 kN
Ay = 33.31 kN
Reaction at A is A = 112.14 kN making angle -17.28 degrees  with + ve x axis


## Example 4.2 Page No : 148¶

In :
import math

# Given Data
#At equillibrium equations are +-> sum Fx = 0, +sum(M_A) = 0, +sum(M_B) = 0
#Sum Fx = 0 gives
Bx = 0;			#kN

# Calculations and Results
print "Bx = %.0f kN "%(Bx);
#+sum(M_A) = 0 gives -(70kN)(0.9m)+By(2.7m)-(27kN)(3.3m)-(27kN)(3.9m) = 0, B assumed to be in +ve Y direction
By = (70*0.9+27*3.3+27*3.9)/2.7			#kN
print "By = %.2f kN  +ve sign shows reaction is directed as assumed "%(By);

#+sum(M_B) = 0 gives -A(2.7m)+(70kN)(1.8m)-(27kN)(0.6m)-(27kN)(1.2m) = 0, A assumed to be in +ve Y direction
A = (70*1.8-27*0.6-27*1.2)/2.7			#kN
print "A = %.2f kN  +ve sign shows reaction is directed as assumed "%(A);

Bx = 0 kN
By = 95.33 kN  +ve sign shows reaction is directed as assumed
A = 28.67 kN  +ve sign shows reaction is directed as assumed


## Example 4.3 Page No : 168¶

In :
import math

# Given Data
#Take x axis parallel to track and Y axis perpendicular to track
W = 25.;			#kN
# Resolving weight
Wx = W*math.cos(25*math.pi/180);			#kN
Wy = -W*math.sin(25*math.pi/180);			#kN
#At equillibrium equations are +-> sum Fx = 0, +sum(M_A) = 0, +sum(M_B) = 0

# Calculations and Results
#+sum(M_A) = 0 gives -(10.5kN)(625 mm)-(22.65 kN)(150 mm)+ R2(1250 mm) = 0, R2 assumed to be in +ve Y direction
R2 = (10.5*625+22.65*150.)/1250;			#kN
print "R2 = %.0f kN  +ve sign shows reaction is directed as assumed "%(R2);

#+sum(M_B) = 0 gives (10.5kN)(625 mm)-(22.65 kN)(150 mm)+ R1(1250 mm) = 0, R1 assumed to be in +ve Y direction
R1 = (10.5*625-22.65*150)/1250;			#kN
print "R1 = %.1f kN  +ve sign shows reaction is directed as assumed "%(R1);

#Sum Fx = 0 gives, 22.65 N-T = 0
T = 22.65;			#kN
print "T = %.2f kN  +ve sign shows reaction is directed as assumed "%(T);

R2 = 8 kN  +ve sign shows reaction is directed as assumed
R1 = 2.5 kN  +ve sign shows reaction is directed as assumed
T = 22.65 kN  +ve sign shows reaction is directed as assumed


## Example 4.4 Page No : 168¶

In :
import math

# Given Data
Ax = 4.5			#in m
Ay = 6.			#in m

# Calculations and Results
DF = math.sqrt((Ax**2)+(Ay**2))
F = 150.			#in KN
Ex = -(Ax/DF)*F
print "Ex = %.2f kN "%(Ex);
Ey = ((Ay/DF)*F)+(4*20)
print "Ey = %.2f kN "%(Ey);

M_E = -((20*7.2)+(20*5.4)+(20*3.6)+(20*1.8)-((Ay/DF)*F*Ax))
print "M_E = %.0f kN  +ve sign shows reaction is directed as assumed "%(M_E);

Ex = -90.00 kN
Ey = 200.00 kN
M_E = 180 kN  +ve sign shows reaction is directed as assumed


## Example 4.5 Page No : 169¶

In :
import math
from numpy import arange

# Given Data
#At equillibrium  +sum(Mo) = 0,
#s = r*theta;
#F = k*s = k*r*theta;
k = 45.;			#N/mm
r = 75.;			#mm
W = 1800.;			#N
l = 200.;			#mm

# Calculations and Results
# trial and error
print "Probable answers by trial and error method are ";
for i in arange(0,0.1+math.pi/2,.1): 			# from 0 to 90 degrees
difference = (math.sin(i)-k*r**2*(i)/(W*l));
if difference<0.01:   			# Approximation
theta = i;
theta = theta*180/math.pi;			#Degrees , conversion into degrees
print "Theta = %.2f degrees"%(theta);

Probable answers by trial and error method are
Theta = 0.00 degrees
Theta = 80.21 degrees
Theta = 85.94 degrees
Theta = 91.67 degrees


## Example 4.6 Page No : 185¶

In :
import math

# Given Data
m = 10.;			#kg mass of joist
g = 9.81;			#m/s**2 gravitational acceleration
W = m*g;			#N
AB = 4.;			#m

# Calculations
# Three force body
BF = AB*math.cos(45*math.pi/180);			#m
AF = BF;			#m

AE = 1./2*AF;			#m
EF = AE;			#m
CD = AE;			#m
BD = CD/math.tan((45.+25)*math.pi/180);			#m
DF = BF-BD;			#m
CE = DF;			#m
alpha = alpha*180/math.pi;			#degrees

#From geometry

G = 90-alpha;			#degrees
B = alpha-(90-(45+25));			#degrees
C = 180-(G+B);			#Degrees

#Force triangle
#T/math.sin(G) = R/math.sin(C) = W/math.sin(B)..... math.sine law

T = W/math.sin(B*math.pi/180)*math.sin(G*math.pi/180);			#N
R = W/math.sin(B*math.pi/180)*math.sin(C*math.pi/180);			#N

# Results
print "Tension in cable T =  %.1f N \
\nReaction At A is  R =  %.1f N with angle alpha =  %.1f degrees with +ve X axis"%(T,R,alpha);

Tension in cable T =  82.1 N
Reaction At A is  R =  147.9 N with angle alpha =  58.6 degrees with +ve X axis


## Example 4.7 Page No : 194¶

In :
import math

# Given Data
m1 = 80.;			#kg mass of man
m2 = 20.;			#kg, mass of ladder
m = m1+m2;			#kg
g = 9.81;			#m/s**2 gravitational acceleration
W = -m*g;			#N, j

# Calculations and Results
C = -0.6*W/3;			#N
Bz = -0.6*C/1.2;			#N
By = -0.9*W/1.2;			#N

print " Reaction At B is B =  %.0f) N j +%.1f N)k"%(By,Bz);
print " Reaction At C is C =  %.2f) N k"%(C);
Ay = -W-By;			#N
Az = -C-Bz;			#N

print " Reaction At A is A =  %.0f) N j +%.1f N)k "%(Ay,Az);

 Reaction At B is B =  736) N j +-98.1 N)k
Reaction At C is C =  196.20) N k
Reaction At A is A =  245) N j +-98.1 N)k


## Example 4.8 Page No : 0¶

In :
import math
from numpy.linalg import solve

# Given Data
W = -1200.;			#N,j  Weight
BD = [-2.4,1.2,-2.4];			#m, Vector BD
EC = [-1.8,0.9,0.6];			#m, Vector EC
#T_BD = norm(T_BD)*BD/norm(BD);			# m, vector of tension in BD
#T_EC = norm(T_EC)*EC/norm(EC);			# m, vector of tension in EC
# Applying equillibrium conditions we get
# Sum_F = 0, and Sum(M_A) = 0 and setting co-efficient equal to zero
A = [[0.8,0.771],[1.6,-0.514]];			#MAtrix of co-efficient
b = [[-1440],];			#matrix b

# Calculations and Results
x = solve(A,b);			# solution matrix
T_BD = x;			# N,Tension in BD
T_EC = x;			#N, Tension in EC
print "T_BD =  %.0f N and T_EC =  %.0f N "%(x,x);

Ax = 2./3*T_BD+6./7*T_EC;			#N, x component of reaction at A
Ay = -(1./3*T_BD+3./7*T_EC+W);			#N, Y component of rection at A
Az = 2./3*T_BD-2./7*T_EC;			#N, z component of reaction at A
print "Reaction at A is A = %.0f N)i +%.0f N)j +%.1f N)k "%(Ax,Ay,Az);
#Answe in Newton instead of lbs
#1lbs = 4.44N

T_BD =  -450 N and T_EC =  -1401 N
Reaction at A is A = -1501 N)i +1950 N)j +100.2 N)k


## Example 4.9 Page No : 198¶

In :
import math
from numpy.linalg import norm
from numpy import array

# Given Data
#Free body diagram
m = 30.			#in kg
g = 9.81			#in m/s2
w = -m*g			#in J

# Calculations and Results
DC = array([-480, 240, -160])			#in mm
X = norm(DC)
T = DC/X
print ("Tension in the vector form = ")
print (T)
#Equilibrium equations
#From equation 2, setting unit vector = 0
Ax = 49			#in N
Ay = 73.5			#in N
A = [Ax, Ay]
y = norm(A)
print ("Tension in the vector form in N = ")
print (y)

Tension in the vector form =
[-0.85714286  0.42857143 -0.28571429]
Tension in the vector form in N =
88.3360062489


## Example 4.10 Page No : 197¶

In :
import math
from numpy import array
#page 197

# Given Data
Tmin = 300.			#lb
AC = array([12, 12, 0])
w = array([,[-450],])
x1 = AC*w
print (x1)
x = x1#array([0, 0, x1])

# Calculations
lambda1 = array([2./3, 2./3, -1./3])*-x1#array([,,[-x1]])
y = x*lambda1
print (y)

#Location of G
#EG and Tmin are having same direction, so their component should be in proportion
x = -1.8/Tmin*Tmin+1.8;			#m, X co-ordinate of G
y = -1.8/Tmin*Tmin+3.6;			#m, Y co-ordinate of G

# Results
print "Co-ordinates of G are x = %.0f m and y =  %.1f m"%(x,y);

[[    0     0     0]
[-5400 -5400     0]
[    0     0     0]]
[[        0.         0.        -0.]
[-19440000. -19440000.        -0.]
[        0.         0.        -0.]]
Co-ordinates of G are x = 0 m and y =  1.8 m