Numericals on resultant of con-current 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 Con-current force system
(Numericals on Triangle Law and Parallelogram Law)
Presented By,
Miss. Shinde Bharti M. (Assistant Professor)
Department of Civil Engineering
Email- shindebharticivil@sanjivani.org.in
1

2. 2
1. A disabled automobile is pulled by means of ropes subjected to
the forces as shown. Determine the magnitude and direction of
resultant using triangle law and parallelogram law.
Soln: P=4KN, Q=2KN
Using parallelogram law,
300
2KN
4KN
250
A
B
C
2 2
2 2
1
1 0
0
2
4 2 2x 4x 2 55
5 4
w.r..t.P
2 55
17 65
4 2 55
Angle w.r.t. X-axis=30-17.65=12.34
R P Q PQcos
R cos
R . KN
Q sin
tan ....................
P Qcos
sin
tan .
cos







  
  

 
  

 
 
 
 

 


3. 3
Using triangle law,
Using sine rule,
   
   
   
 
0
4 2
125 25 30
4 2
25 30
4 30x cos 30x 2 25x cos 25x
1 155 5 2766
12 34
4
125 25
4
125 37 34
5 4
R
sin sin sin
sin sin
sin cos sin sin cos sin
. cos . sin
.
R
sin sin
R
sin sin .
R . KN
 
 
   
 


 
 
 
 
   
 
 



 
300


4KN
2KN
R
300
250
1250
25+
30-
12.340
A
5.4 KN

4. 4
2. Determine the magnitude and direction of two forces as shown in
fig. Take P=100N.
Soln:
Using parallelogram law,
1
1
2 2
2 2
w.r..t.P
45
15
100 45
45
15
100 45
26 79 0 1894 0 707
51 76
2
100 51 76 2x100x51 76 45
141 41
Q sin
tan ....................
P Qcos
Q sin
tan
Qcos
Q sin
tan
Qcos
. . Q . Q
Q . N
R P Q PQcos
R . . cos
R . N






 
  

 
 
  

 
 

  
 
  
  

300
P Q
150
R

5. 5
Using triangle law,
Using sine rule,
100
135 30 15
100
135 30
141 42
100
30 15
51 76
R Q
sin sin sin
R
sin sin
R . N
Q
sin sin
Q . N
 
 
 

 
R
Q
P=100N
1350
300
150
150
300

6. 6
4. Determine the components F1 and F2 so that the resultant of two
forces is 20KN vertically upward
Soln:
Given: R=20 vertically upward
Using parallelogram law,
1 2
1
1 2
1 2
1 2
1 2
1 2
2 2
1 2 1 2
2 2
1 2 1 2
2 2
1 2
w.r..t.F
90
60
90
1 732 0
1 732 0 (1)
2
20 2 90
400
400
F sin
tan ....................
F F cos
F sin
tan
F F cos
. F F
. F F ............................
R F F F F cos
F F F F cos
F F






 
  

 
 
  

 
  
  
  
  
 
 2 2
1 1
1
2
(1.732F ) from (1)
10
17 32
F ...................
F KN
F . KN

 
 
300
F2
F1
600

7. 7
Using triangle law,
Using sine rule,
1 2
1
1
1 2
2
90 30 60
20
90 30
10
30 60
17 32
F F
R
sin sin sin
F
sin sin
F KN
F F
sin sin
F . KN
 
 
 

 
R
F1
F2
900
600
300

8. 8
4. Two forces are applied as shown using trigonometry and knowing
that the magnitude of P is 60N determine angle  if R is
horizontal determine corresponding magnitude of resultant
Soln: Given: R is horizontal
Using triangle law,

90N
P
300
R
P=60N
90N
150-

300
 
 
0
60 90
150 30
60 90
30
48 59
60
150 48 59 30
60
101 4 30
117 63
R
sin sin sin
sin sin
.
R
sin . sin
R
sin . sin
R . N
 


 

 
 
 

 
 

9. 9
5. Two forces P=20N, Q=30N making an angle 1200 with each other
what will be the value of resultant.
Soln: Given: P=20N, Q=30N , = 1200
Using parallelogram law,
2 2
2 2
1
1
0
2
20 30 1200 120
26 45
w.r..t.
30 120
20 30 120
79 1
R P Q PQcos
R cos
R . N
Q sin
tan .................... P
P Qcos
sin
tan
cos
.








  
  

 
  

 
 
  

 
 

10. 10