The document introduces the concept of moment of a force. It defines moment as a measure of the tendency of a force to cause rotation about a point or axis. It provides methods to calculate the moment of a force in 2D and 3D, including using the cross product and right hand rule. Examples are given to find the moment of individual forces and the resultant moment of a system of forces about a point.

1. Objectives
1. Introduce the concept of the moment of a
force and show how to calculate it in 2 and 3
dimensions.
2. Provide a method for finding the moment of a
force about a specified axis.

2. Moment of a Force
The moment of a force about a
point or an axis provides a
measure of the tendency of the
force to cause a body to rotate
about the point or axis

3. Fx - horizontal force
dy - distance from point O to force
Mo - moment of force about point O
(Mo)z - moment of force about axis z

4. Fz - horizontal force
dy - distance from point O to force
Mo - moment of force about point O
(Mo)x - moment of force about axis x

5. No Moment

6. M Fd
Magnitude of the moment

Direction of the moment
Right Hand Rule

7. Counterclockwise is
positive by
scalar sign convention
Resultant Moment of a System
of Coplanar Forces
O
R
M Fd
  

8. Do not actually need rotation to have a moment.
Moment is the tendency to cause rotation

9. For each case, find the moment
of the force about the point O
Example

10.   
O
M 100N 2m 200N m
  

11.   
O
M 50N 0.75m 75N m
  

12.   
o
O
M 40lb 4 2cos30 ft 229lb ft
   

13.   
o
O
M 60lb 1sin45 ft 42.4lb ft
  

14.   
O
M 7kN 4 1m 21.0kN m
   

15. Determine the
moment of the 800 N
force about points A,
B, C, and D
Example

16. A
B
C
D
M 800 N (2.5 m) 2000 N m
M 800 N (1.5 m) 1200 N m
M 800 N (0 m) 0 N m
M 800 N (0.5 m) 400 N m
  
  
  
  

17. Determine the
resultant moment of
the four forces.
Example

18.  
   
   
 
O
O
O
R
R
o o
R
ccw M Fd
M 50N 2m 60N 0
20N 3sin30 m 40N 3cos30 m
M 334N m 334N m cw
 
  
 
    


19. Cross Product
Another method of vector
multiplication
Read as C equals A cross B
C A B
 
r r r

20. Cross Product
Magnitude:
C ABsin
 

21. Direction:
Right Hand Rule

22. Cross Product
 
A B B A
A B B A
  
   
r r
r r
r r
r r
Not Commutative.

23. Cross Product
2. Scalar Multiplication
   
 
 
a A B aA B
A aB
A B a
  
 
 
r r
r r
r r
r r

24. Cross Product
3. Distributive Law:
     
A B D A B A D
     
r r r
r r r r

25. Unit Vectors
o
90 sin 1
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
i i 0 i j k i k j
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
j i k j j 0 j k i
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
k i j k j i k k 0
    
      
      
      

26. Right Hand Rule

28.    
     
     
     
ˆ ˆ ˆ ˆ ˆ ˆ
A B i j k i j k
ˆ ˆ ˆ ˆ ˆ ˆ
i i i j i k
ˆ ˆ ˆ ˆ ˆ ˆ
j i j j j k
ˆ ˆ ˆ ˆ ˆ ˆ
k i k j k k
x y z x y z
x x x y x z
y x y y y z
z x z y z z
A A A B B B
A B A B A B
A B A B A B
A B A B A B
       
     
     
    
r r
Cartesian Form

29.    
ˆ ˆ ˆ ˆ ˆ ˆ
A B i j k i j k
ˆ ˆ ˆ ˆ ˆ ˆ
k j k i j i
ˆ ˆ ˆ
( )i ( )j ( )k
x y z x y z
x y x z y x y z z x z y
y z z y x z z x x y y x
A A A B B B
A B A B A B A B A B A B
A B -A B A B A B A B A B
       
      
   
r r
Carry Out Operations:

30. ˆ ˆ ˆ
i j k
A B x y z
x y z
Determin
A A
ant fo
B B
m
A
r
B
:
 
r r
Equivalent Formulation

31. î :
ˆ ˆ ˆ
i j k
î ( )
x y z y z z y
x y z
For Element
A A A A B A B
B B B
 
Determinant

32. ĵ
ˆ ˆ ˆ
i j k
ĵ ( )
x y z x z z x
x y z
For Element :
A A A A B A B
B B B
  
Determinant

33. k̂
ˆ ˆ ˆ
i j k
k̂ ( )
x y z x y y x
x y z
For Element :
A A A A B A B
B B B
 
Determinant

34. Moment of a Force -
Vector Formulation
O
M r F
 
r r
r

35. Moment of a Force -
Vector Formulation
O
M rFsin
F(rsin )
Fd
 
 


36. Principle of
Transmissibility

37. Principle of Transmissibility
r vector can be taken to any
point on line of action of F
O
A
B
C
M r F
r F
r F
r F
 
 
 
 
r r
r
r
r
r
r
r
r

38. O
ˆ ˆ ˆ
i j k
M r F x y z
x y z
r r r
F F F
  
r r
r
Cartesian Form

39. O
ˆ
M ( )i
ˆ
( )j
ˆ
( )k
y z z y
x z z x
x y y x
r F -r F
r F r F
r F r F

 
 
r
Cartesian Vector Formulation

40. Resultant Moment of
a System of Forces

41. Resultant Moment of
a System of Forces
 
     
O
R
1 1 2 2 3 3
M r F
r F r F r F
 
     

r r
r
r r r
r r r

42. Example
Find moment about A

43. Solution Steps
1. Find vectors
2. Force vector is 60 N times a unit
vector in direction of
3. Moment
A B
r r
and
r r
CB
û
A A A B
M r F M r F
or
   
r r r r
r r

45. B BA
C CA
CB B C
CB
CB
ˆ ˆ ˆ
r r (1i 3j 2k)m
ˆ ˆ ˆ
r r (3i 4j 0k)m
r r r
ˆ ˆ ˆ
r (1 3)i (3 4)j (2 0)k
ˆ ˆ ˆ
r 2i 1j 2k
   
   
 
     
   
r r
r r
r r r
Position Vectors

46. CB
CB
CB 2 2 2
CB
CB
CB
ˆ ˆ ˆ
r 2i 1j 2k
ˆ ˆ ˆ
r 2i 1j 2k
ˆ ˆ ˆ
û 2i 1j 2k
r ( 2) ( 1) (2)
2 1 2
ˆ ˆ ˆ
û i j k
3 3 3
ˆ
F (60 N) u
ˆ ˆ ˆ
F ( 40i 20j 40k) N
   
  
     
   
   

   
r
r
r
Force Vector

47. B
C
A B
ˆ ˆ ˆ
r (1i 3j 2k)m
ˆ ˆ ˆ
r (3i 4j 0k)m
ˆ ˆ ˆ
F ( 40i 20j 40k) N
ˆ ˆ ˆ ˆ ˆ ˆ
M r F (1i 3j 2k)m ( 40i 20j 40k) N
  
  
   
        
r
r
r
r r
r
Moment Vector

48.  
A B
A
2 2 2
A
ˆ ˆ ˆ ˆ ˆ ˆ
M r F (1i 3j 2k)m ( 40i 20j 40k) N
ˆ ˆ ˆ
i j k
M 1 3 2
-40 -20 40
ˆ ˆ ˆ
[3(40) 2( 20)]i [(1(40) 2( 40)]j [1( 20) 3( 40)]k
ˆ ˆ ˆ
160i 120j 100k N m
M (160) ( 120) (100) 224 N m
        

         
   
     
r r
r
Moment Vector

50. Example
Determine the resultant moment at O and the
coordinate direction angles for the moment.

51. A OA
B OB
ˆ
r r (5j)ft
ˆ ˆ ˆ
r r (4i 5j 2k)ft
 
   
r r
r r
Position
Vectors

52. 1
2
3
ˆ ˆ ˆ
F ( 60i 40j 20k) lb
ˆ
F (50j) lb
ˆ ˆ ˆ
F (80i 40j 30k) lb
   

  
r
r
r
Force Vector

53.        
O
R A 1 A 2 B 3
M r F r F r F r F
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
i j k i j k i j k
0 5 0 0 5 0 4 5 2
-60 40 20 0 50 0 80 40 30
       
   


r r r r r
r r r r
Moment Vector

54.    
 
O
R
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
i j k i j k i j k
M 0 5 0 0 5 0 4 5 2
-60 40 20 0 50 0 80 40 30
ˆ ˆ ˆ
[5(20) 40(0)]i [0]j [0(40) 60(5)]k
ˆ ˆ ˆ
[0]i [0]j [0]k
ˆ ˆ ˆ
[5( 30) 40( 2)]i [4 30 80 2 ]j [4(40) 80(5)]k
ˆ ˆ ˆ
30i 40j 60k lb ft
   

    
  
        
   
r
Moment Vector

55.  
     
 
O
O
O
O
O
R
2 2 2
R
R
R
R
ˆ ˆ ˆ
M 30i 40j 60k lb ft
M 30 40 60 lb ft
M 78.10 lb ft
ˆ ˆ ˆ
30i 40j 60k lb ft
M
û
M 78.10 lb f
ˆ ˆ ˆ
0.3841i 0.5121j 0.7682k
   
    
 
  
 

  
r
r
Moment Vector

56. o
o
o
ˆ ˆ ˆ
û 0.3841i 0.5121j 0.7682k
cos 0.3841 67.4
cos 0.5121 121
cos 0.7682 39.8
  
   
    
   
Direction Angles

58. Principle of Moments
A
F1
F2
F
The moment of a force about a point is
equal to the sum of the moments of the
force’s components about the point.

59. Principle of Moments
 
O
1 2
1 2
M r F
r F r F
r F F
 
   
  
r r
r
r r
r r
r r
r

61. Example
Determine the moment of the force about A.

63.   
 
o
A
A
CB d 100cos45 70.71mm 0.07071m
M Fd 200N 0.07071m 14.1N m
ˆ
M 14.1k N m
   
   
 
r

65.      
 
A
o o
A
M Fd
200sin45 N 0.20m 200cos45 N 0.10m
14.1N m
ˆ
M 14.1k N m

 
 
 

r

66. Example
Determine the moment of the force about 0.

68.  
  
  
 
o
O
o
O
O
ccw
M 400sin30 N 0.2m
400cos30 N 0.4m
98.6N m
M 98.6N m cw
ˆ
M 98.6 k N m



  
  
 
  
 
r

70.     
O
O
ˆ ˆ ˆ
i j k
M r F 0.4 0.2 0
200 346.4 0
ˆ ˆ
0i 0j 0.4 364.4 0.2 200
ˆ
M 98.6 k N m
   

       
 
 
  
 
r r
r
r

71. Moment of a Couple
A couple is two
parallel forces
having the same
magnitude and
opposite
directions
separated by
a distance d.

72. Moment of a Couple
Resultant Force is
zero. Effect of couple
is a moment

73. Moment of a Couple
A Couple consists of two parallel forces,
equal magnitude, opposite directions,
and separated a distant “d” apart.
A Couple Moment about any point O equals
the sum of the moments of both forces.

75. Moment of a Couple
A couple moment about
any Point O equals the
sum of the moments of
both forces
M = r (-F r F = (r r F
But r r = r , and r = (r r ).
M = r F. A couple moment is free vector.
A B B A
A B B A
    
 
 
) ( ) )

76. Moment of Couple
Scalar formulation:
Magnitude of couple
moment is M = Fd. Direction
is perpendicular to plane of
forces. RHR applies

77. Moment of Couple
M = r F

Vector Formulation

78. Example
= =

79. Example

80. Example 4-12
Given: Couple Moment
acting on Pipe OAB.
Find: Determine magnitude
of Couple Moment
acting on pipe. Represent
moment as Cartesian
Vector.
Approach: Use scalar
calculation to calculate
magnitude of couple
moment. M=Fd.

81. Scalar Approach

82. Scalar Approach
  
o
F 25lb
d 6cos30 5.2in
M Fd 25lb 5.2in
M 129.9lb in

 
 
 

83. Scalar Approach

84. Vector Approach
   
 
   
 
O A B
O
o o
ˆ ˆ
M r 25k r 25k
ˆ ˆ
M 8j 25k
ˆ ˆ ˆ ˆ
6cos30 i 8j 6sin30 k 25k
ˆ ˆ ˆ ˆ
M 200i 129.9j 200i 129.9j lb in
     
  
    
      

86. Vector Approach
 
   
 
O AB
o o
O
ˆ
M r 25k
ˆ ˆ ˆ
M 6cos30 i 6sin30 k 25k
ˆ
130j lb in
 
   
  

87. Resultant of a Force
and Couple System
O
R
R c O
Vector :
F F
M M M

 

 

88. Resultant of a Force and Couple
System – 2D
Scalar :
x
y
O
R x
R y
R c O
F F
F F
M M M


 


 

89. Replace the forces acting on the
brace shown below with an
equivalent resultant force and
couple moment at point A.
PROBLEM

91. o
1
R
R
1
2
2
R
o
R
y
R
R
o
R
x
R
6
.
66
8
.
382
8
.
882
tan
F
F
tan
N
962
)
8
.
882
(
)
8
.
382
(
N
8
.
882
F
N
8
.
882
45
sin
400
N
600
F
F
F
N
8
.
382
F
N
8
.
382
45
cos
400
N
100
F
F
F
y
y
y
y
y
x
x
x











































R
F


92.    
A
A
A
y
x
R A
o
R
o
R
2 2
R
1 1 o
R
ccw M M ccw
M (100 N)(0) (600 N)(0.4 m) ( 400 sin 45 N)(0.8 m)
( 400 sin45 N)( 0.8 m)
M 551 N m 551 N m (cw)
(382.8) (882.8) 962 N
F 882.8
tan tan 66.6
F 382.8
R
F
 
  
  

    
  
  
 
   
   
  
 
 


94. Concurrent Force Systems

95. Coplanar Systems
Resultant moment MRO =  (r x F) is
 to the resultant force FRO
Therefore FRO can be repositioned a
distance d from point O so as to
create the same moment MRO.

96. =
Coplanar Force Systems

97. Coplanar Force Systems

98. Determine the magnitude, direction, and location on the
beam of the resultant force that is equivalent to the system
of forces shown.

99. o
1
2
2
o
y
y
R
o
x
x
R
7
.
33
350
233
tan
N
5
.
420
)
233
(
)
350
(
N
233
N
200
N
60
sin
500
F
F
N
350
N
100
N
60
cos
500
F
F


























R
F


100.      
     
R E
E
o o
( ccw) M M
500 sin60 4 500 cos60 0
100 0.5 200 2.5
1182.1 N m
 
  

 


101.      
     
m
07
.
5
d
m
N
1
.
1182
)
0
(
350
d
233
m
N
1
.
1182
5
.
2
200
5
.
0
100
0
60
cos
500
4
60
sin
500
M
M
)
ccw
( o
o
E
E
R











 

102. Parallel Force System
1. Assume all forces act in z-direction.
2. Can include couple systems in x-y
plane.
3. Sum Forces and Moments about a
point.
4. Move resultant force a distance d from
point to get same moment.

103. Parallel Force Systems

105. Determine the magnitude, direction, and location on
the slab of the resultant force that is equivalent to
the system of forces shown.

106. m
N
4200
(0)
500
(0)
400
(6)
100
(8)
600
M
m
N
3500
-
(0)
500
(10)
400
-
(5)
100
(0)
600
M
N
-1400
N
500
-
N
400
-
N
100
N
600
-
x
x
O
O














 F
FR

107.  
 
m
3.00
x
m
N
4200
M
N
1400
m
2.50
y
m
N
3500
-
M
N
1400
-
x
x
O
O








x
y

108. a) Replace the force
system with an equivalent
force system
b) specify a location (0,y)
for a single equivalent
force to be applied.
QUESTION

109.  
 
 
 
o o 4
x 5
o o 3
y 5
o
3
O 5
o
F 5(sin40 ) 3cos(60 ) 7.5 1.286 kN
F 5(cos40 ) 3sin(60 ) 7.5 5.732 kN
M 7.5 (3) 5(cos40 )(2)
3cos(60 )(5) 13.34 kN m
1.286 kN y 13.34 kN m
y 10.4 m
y -10.4m
down
    
   
  
   
 






110. 3
4
40o
60o
3 kN
7.5 kN
5 kN
2 m
3 m
5 m
O
5.87 kN
1.286 kN
5.73 kN
77.4o
13.34 kN m

111.  
 
 
o o 3
y 5
o
3
O 5
o
F 5(cos40 ) 3sin(60 ) 7.5 5.732 kN
M 7.5 (3) 5(cos40 )(2)
3cos(60 )(5) 13.34 kN m
1.286 kN y 13.34 kN m
y 10.4 m
y -10.4m
down
   
  
   
 





112. 3
4
40o
60o
3 kN
7.5 kN
5 kN
2 m
3 m
5 m
O
10.4 m
5.87 kN
1.286 kN
5.73 kN
77.4o