1. A disabled vehicle is pulled by ropes with forces of 4 kN and 2 kN. Using the triangle law and parallelogram law, the magnitude of the resultant force is calculated to be 5.4 kN at an angle of 12.34° from the x-axis. 2. Given a force P of 100 N, the magnitude and direction of two forces are determined. Using the parallelogram law, the magnitudes are found to be 100 N and 150 N, with an angle of 45° between them. 3. The components of two forces are determined such that the resultant is 20 kN vertically upward. Using the parallelogram law, the components are calculated to

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