Chapter 5: Equilibrium of Coplanar Force Systems.

Example 5.5-1, Page no 58

In [11]:
import math
import numpy as np

#Initilization of variables
# From eqn's 1&2
D=np.array([[6/sqrt(40),-4/sqrt(20)],[2/sqrt(40),2/sqrt(20)]])
B=np.array([0,25]) #lb

#Calculations

X=np.linalg.solve(D,B)

#Result

print'The tension in  cable AB is',round(X[1],1),"lb"
print'The tension in cable AC is',round(X[0],1),"lb"

# The tensions in the cable AB & AC is off by 0.1 lb

The tension in  cable AB is 33.5 lb
The tension in cable AC is 31.6 lb

Example 5.5-2, Page no 59

In [15]:
import math

#Initilization of variables
F1=100 #lb
R=16 #in

#Calculations
theta=arcsin(14*R**-1)*(180/pi) #degrees
# since theta=61 degrees,
sin61=0.8746
cos61=0.4848
N=F1/sin61 #lb
P=N*cos61 #lb

#Result

print'The value of normal reaction offered is',round(N,1),"lb"
print'The push required is',round(P,1),"lb"

The value of normal reaction offered is 114.3 lb
The push required is 55.4 lb

Example 5.5-3,Page no 59

In [45]:
import math

#Initilization of variables
L=20 #m
M=1200 #kg
g=9.81 #m/s**2
H=10 #m

#Calculations

AB=sqrt(L**2-H**2) #Applying Pythagoras Theorem
costheta=17.3/20
F1=M*g*H/AB #N
F2=M*g/costheta #N

#Result

print'Force F1 is',round(F1),"N"
print'Force F2 is',round(F2),"N"

#Decimal accuracy causes discrepancy in answers compared to the textbook answers

Force F1 is 6797.0 N
Force F2 is 13609.0 N

Example 5.5-4, Page No 60

In [17]:
import math
import numpy as np

#Initilization of variables
Fx=1000 #lb
Fy=1000 #lb
costheta=9*15**-1
cosbeta=12*15**-1
sintheta=4*5**-1
sinbeta=3*5**-1

#Calculations
#Matrix solution
A=np.array([[costheta,-cosbeta],[sintheta,sinbeta]]) 
B=np.array([-1000,1000])
X=np.linalg.solve(A,B)

#Result
print'The force in AB is',round(X[0]),"lb compression"
print'The force in BC is',round(X[1]),"lb compression"

The force in AB is 200.0 lb compression
The force in BC is 1400.0 lb compression

Example 5.5-5, Page no 61

In [24]:
import math
import numpy as np

#Initilization of variables
w=10 #lb/ft
L=12 #ft
# as theta=30 degrees,
sin30=2**-1
cos30=sqrt(3)*2**-1

#Calculation
#Matrix Calculations
A=np.array([[cos30,-cos30],[sin30,sin30]]) 
B=np.array([0,120]) 
X=np.linalg.solve(A,B)

#Result
print'The tension in the cable is,T=',round(X[0]),"lb"
print'The reaction at B is,R',round(X[1]),"lb"

The tension in the cable is,T= 120.0 lb
The reaction at B is,R 120.0 lb

Example 5.5-6,Page no 61

In [33]:
import math

#Initilization of variables
W1=40 #lb
W2=30 #lb
# as theta1=30 degrees,
sin30=2**-1

#Calculations
#Summing the forces parallel to 30 degree plane
T=W1*sin30
theta=arcsin(T/W2)*(180/pi)

#Result
print'The tension in the cable is',round(T),"lb"
print'The angle is',round(theta,1),"degrees"

The tension in the cable is 20.0 lb
The angle is 41.8 degrees

Example 5.5-8,Page no 62

In [5]:
import math

#Initilization of variables
F1=125 #N
F2=200 #N
F3=340 #N
F4=180 #N
x1=4 #m
x2=3 #m
x3=10 #m
x4=15 #m
x5=17 #m

#Calculations
Rb=(-F1*x1+F2*x2+F3*x3+F4*x4)/x5 #moment about point A
Ra=(F1*(x1+x5)+F3*(x5-x3)+F2*(x5-x2)+F4*(x5-x4))/x5 #moment about point B

#Result
print'The reaction at A is',round(Ra),"N"
print'The reaction at B is',round(Rb),"N"

# The ans for B is off by 1 N

The reaction at A is 480.0 N
The reaction at B is 364.0 N

Example 5.5-9, Page no 63

In [6]:
import math

#Initilization of variables
F1=1000 #lb
F2=1200 #lb
F3=2000 #lb
x1=1 #ft
x2=7 #ft
x4=2 #ft
x3=6 #ft

#Calculation
#Equilibrium equations
Rn=(F3*(x1+x2+x3)+F2*(x1+x2)+F1*x1)/(x1+x3+x2+x4) #Moment about point M
Rm=(F1*(x2+x3+x4)+F2*(x3+x4)+F3*x4)/(x1+x2+x3+x4) #Moment about point N

#Result
print'The reaction at M is',round(Rm),"lb"
print'The reaction at N is',round(Rn),"lb"

#Decimal Accuracy causes discrepancy in answers between computation and textbook

The reaction at M is 1787.0 lb
The reaction at N is 2412.0 lb

Example 5.5-10, Page no 64

In [34]:
import math

#Initilization of variables
P=10 #kg
g=9.81 #m/s**2

#Calculations
# equilibrium at fig b
T1=P*g/2 #N
# equilibrium at fig c
T2=T1/2 #N
#equilibrium at fig d
P=T2

#Result
print'The force P is',round(P,1),"N"

The force P is 24.5 N

Example 5.5-11, Page no 64

In [7]:
import math

#Initilization of variables
k=20 # lb/in
w=20 # lb/ft
x1=4 #ft
x2=10 # ft
x3=8 #ft
x4=6 #ft
x5=9 #ft
F1=1920 #lb.rad
F2=3360 #lb.rad

#calculations
theta=(w*x2*x5)*(F1*x3+F2*(x3+x4))**-1 #radians
FB=F1*theta
FC=F2*theta
A=(w*x2)-FB-FC

#Result
print'The force in spring B is',round(FB,1),"lb"
print'The force in spring C is',round(FC,1),"lb"
print'The reaction at A is',round(A,1),"lb up"

 # The answer waries slightly due to decimal point discrepancy

The force in spring B is 55.4 lb
The force in spring C is 96.9 lb
The reaction at A is 47.7 lb up

Example 5.5-12, Page no 65

In [35]:
import math

#Initilization of variables
L=3.8 #m
w=10 # kg/m
P=1000 #N
t=0.8 #m
g=9.81 #m/s**2

#Calculations
Gf=L*w*g #N
A=(P*L+Gf*L*0.5)/t #N Taking moment about point B
B=(P*(L-t)+Gf*(0.5*L-t))/t #N Taking moment about point A

#Result
print'The reaction at point A is',round(A),"N"
print'The reaction at point B is',round(B),"N"

# The answers in the textbook are incorrect

The reaction at point A is 5635.0 N
The reaction at point B is 4263.0 N

Example 5.5-13, Page no 65

In [37]:
import math

#Initilization of variables
Wa=400 #lb
Wb=200 #lb
# as theta=30 degrees,
sin30=2**-1

#Calculations
Ta=Wa*sin30 #lb
Tb=Wb*sin30 #lb
#Taking moment about point O
P=(Tb*12+Ta*6)/24 #lb

#Result
print'The value of Ta is',round(Ta,3),"lb"
print'The value of Tb is',round(Tb,3),"lb"
print'The value of P is',round(P,3),"lb"

The value of Ta is 200.0 lb
The value of Tb is 100.0 lb
The value of P is 100.0 lb

Example 5.5-15, Page no 66

In [4]:
import math
import numpy as np

#Initilization of variables
F=np.array([5,2,3,1.5]) #kN Forces are defined as a cloumn matrix
theta=(pi*np.array([90,60,45,80]))/180 #degrees angles are also defined as a column matrix
d=np.array([2,6,13,17]) #distances from point C of each force
c=np.array([17,15,11,4]) #distance form point D of each force
#Calculations

#Summing horizontal forces
Ch=F[1]*cos(theta[1])-F[2]*cos(theta[2])+F[3]*cos(theta[3]) #kN "which indidcates that Ch acts to the left instead of the assumed"
#Taking moment about point C
D=(F[0]*d[0]+F[1]*sin(theta[1])*d[1]+F[2]*sin(theta[2])*d[2]+F[3]*sin(theta[3])*d[3])/d[3] #kN
#Taking moment about point D
Cv=(F[0]*c[1]+F[1]*sin(theta[1])*c[2]+F[2]*sin(theta[2])*c[3])/c[1]
#Result

print'The values of Ch,D and Cv are:',round(Ch,2),"kN ,",round(D,1),"kN",'and',round(Cv,2),"kN"

# The ans of Cv is incorrect in textbook

The values of Ch,D and Cv are: -0.86 kN , 4.3 kN and 6.84 kN

Example 5.5-16, Page no 67

In [7]:
import math

#Initilization of variables
w=100 #N/m
F1=200 #N
M=500 #N.m
Lw=2 #m
#Distance from point A
d=np.array([1,2,3,4,5]) #m
#Distance from point B
b=np.array([5,4,3,2,1]) #m

#Calculations
#Taking moment aboout point A
Rb=(w*Lw*d[0]+F1*d[2]-M)/d[3] #N
#Taking moment about point B
Ra=(w*Lw*b[2]+F1*b[4]+M)/b[1] #N

#Result
print'The value of reaction at A is',round(Ra),"N"
print'The value of reaction at B is',round(Rb),"N"

The value of reaction at A is 325.0 N
The value of reaction at B is 75.0 N

Example 5.5-18, Page no 68

In [15]:
import math
import numpy as np

#Initilization of variables
# The values of theta are=[60,60,45] degrees, therefore its values are as,
costheta2=sqrt(2)**-1
sintheta2=sqrt(2)**-1
d=np.array([4.46,3.54,2]) #feet defined as a matrix
F=400 #lb

#Calculations
#Taking moment about point A
Re=(F*(8-d[1]))/8 #lb
Ra=400-Re #lb here i have used the summation of forces in the vertical direction
#Taking moment about point B
Dv=(-F*3.644)*5.77**-1 #lb
#Taking moment about point D
Bv=(F*2.126)/5.77 #lb
#Taking summation of forces in the vertical direction
Cv=-223-Dv #lb
#Taking moment about point D
Ch=((223*d[2]*costheta2)-(Cv*5.173*costheta2))*(5.173*sintheta2)**-1 #lb
#Taking summation of forces in the horizontal direction
Dh=-Ch #lb
#Taking sum of forces in horizontal direction
Bh=-Dh #lb

#Result
print'The Floor reactions are'
print'Ra=',round(Ra),"lb up"
print'Re=',round(Re),"lb up"

print'Pin reaction at C on CE are'
print'Ch=',round(Ch,1),"lb to right"
print'Cv=',round(Cv,1),"lb up"

print'The pin reactions at B on AC are:'
print'Bh=',round(Bh,1),"lb to right"
print'Bv=',round(Bv,1),"lb down" 

The Floor reactions are
Ra= 177.0 lb up
Re= 223.0 lb up
Pin reaction at C on CE are
Ch= 56.6 lb to right
Cv= 29.6 lb up
The pin reactions at B on AC are:
Bh= 56.6 lb to right
Bv= 147.4 lb down

Example 5.5-19, Page no 70

In [52]:
import math

#Initilization of variables
r=0.5 #m
m=10 #kg
g=9.81 #m/s**2
# since theta=60 degrees,
sin30=2**-1
cos30=sqrt(3)*2**-1

#Calculations
#Due to symmetry the reaction will be shared by the structure
A=m*g*r #N
B=A #N
#Vertical forces summed
N1=m*g/(2*sin30) #N
#Taking moment about point C
T=(N1*0.866+B*sin30)*(1.5*cos30)**-1
   
#Result
print'The value of N1 is',round(N1),"N"
print'The value of T is',round(T,1),"N"

# The ans for T is off by 0.1 N

The value of N1 is 98.0 N
The value of T is 84.3 N