Basics, Moment, Varignon theorem, Numericals on resultant of non-concurrent force system

1. Sanjivani Rural Education Society’s
Sanjivani College of Engineering, Kopargaon
(An Autonomous Institute, Affiliated to Savitribai Phule Pune University, Pune)
NAAC ‘A’ Grade Accredited , ISO 9001:2015 certified
Subject: Engineering Mechanics
Unit 1: Resultant of Non-concurrent force System
Presented By,
Miss. Shinde Bharti M. (Assistant Professor)
Department of Structural Engineering
Email- shindebhartist@sanjivani.org.in
1

2. 2 Resultant of Non-concurrent force system-
• Moment
It is the turning effect produce by a force. Moment of force
about any point is the product of magnitude of the force and
perpendicular distance about that point.
 MA= F * d
S.I Unit = N.m or KN.m
Sign Conventions- (+ve ) ( -ve)
F
d

3. 3 • Couple
Two unlike parallel forces having same magnitude forms a
couple.
P
P
d
• Resultant of couple is always zero.
• Moment of couple is the product of magnitude of one force and lever arm.

4. 4
• Varignon’s Theorm
The moment of resultant of a force system about any point
is equal to the algebric sum of moment of all forces about the
same point.
where, x – is the perpendicular distance of resultant R about pt.
where moment is taken
 
 


R x M
M
x
R

5. 5
1. Determine the moment of force about point B as shown in fig.
Soln :
600
160 mm
200 mm
A
B
800 60x 0 16 800 60x 0 2
202 56
B
M cos . sin .
. N.m
 

800 cos60
800 sin60

6. 6
2. Determine the moment of force about point A and B as shown in fig.
Soln :
10N 20N 30N
1m 1.5m 2m 1m
25N
3.5m
RA RB
25x 2 10x1 20x 2 5 30x 4 5
145
25x3 5 30x1 20x3 10x 4 5
47 5
A
B
M . .
Nm
M . .
. Nm
    

   


7. 7
3. A bracket is subjected to forces and couple as shown in fig.
Find the value of M and its direction if resultant is passing
through D.
Soln :
2 2
1 0
60 25 85
20 80 60
104 04
35 21
Using Varignon's theorm,
25x1 3 80x1 5 60x 0 7 400 104 01x 0 0
354 5
x
y
x y
y
x
D
F N
F N
R F F . N
F
tan .
F
M Rx
. . . M . .
M . Nm
 
    
   
  
 
 
 
 
 

     
 


 



60 N
20 N
25 N
80 N
0.4 m 0.5 m 0.6 m
0.7 m
0.6 m
M 400 Nm
D
B
A
C

8. 8
4. Lever OA of length 600mm making an angle of 600 with respect of
+ve x-axis what will be small force applied at A to produce
moment of 150 Nm clockwise.
Soln : smallest force when perpendicular distance is maximum
F is perpendicular to OA
A
O
0.6m
F
600
x 0 6
150 0 6
250
O
M F .
. F
F N
 
 
 
150 Nm

9. 9
5. Four forces acts tangentially to a 8cm diameter circle what is
the magnitude of resultant.
Soln :
6.Four sides of rectangle ABCD represents four Ab=5N, BC=10N,
CD=15N, DA=20N Determine the magnitude of resultant.
Soln :
100 N
200 N
100 N
200 N
100 100 0
200 200 0
0
x
y
F
F
R
  
  
 


20 N
5 N
10 N
15 N
A B
C
D
2 2
1 0
5 15 10
10 20 10
10 10 14 14
10
45
10
x
y
F
F
R . N
tan
 
   
   
   
 
  
 

 



10. 10
13 2 30 0 122 13 2 30 0 086
1 78
o
M . sin x . . cos x .
.
  
 

11. 11