KING FAHD UNIVERSITY OF PETROLEUM
& MINERALS, DHAHRAN, SAUDI ARABIA
ENGINEERING MECHANICS STATICS

MOMENT OF A FORCE USING THE CROSS PRODUCT,
AND PRINCIPLE OF MOMENTS
To express the moment, M , of a force, F, about a point O, using the vector Cross Product
To express the moment, M , of a force, F, about a point O, in the Cartesian vector form
To explain the Principle of Transmissibility of a
Force
To explain resultant moment, M of forces (F , , --F

Let us consider a force F acting through a point
A
and a position vector, r, for a line joining points
O and
A
, as shown in the following figure:
The moment of the force F about point
O
, M
O
, may be determined using the cross-product of the position vector and the force, as follows:
M
O
= r F
Note: It is very important to note that the position vector should start from the point about which the moment has to be calculated. For example, in the above case, the position vector r = r
OA moment is being calculated about the point
O
.
, because the

Referring to the following figure, the magnitude, direction and sense of the moment M
O
may be explained as follows:
Magnitude:
O
=
sin
=
(
=
Direction (i.e., Moment Axis):
Direction of the moment M
O will be the about axis perpendicular to the plane of the force and position vector, as shown in the above figure.
Sense:
Using the right-hand rule, the sense of the moment M
O
“counterclockwise”, as shown in the above figure. will be

Let us consider a position vector r
OA application
A
of a forces F.
from a point
O
to the point of
The position vector r
OA and the force vector F may be expressed in
Cartesian vector form, as follows: r
OA
= r x i + r y j + r z k and F =
F x i +
F y j +
F z k
The moment of the force about point O , M
O
Cartesian vector form, as follows:
, may be expressed in
M
O
= r
OA
× = i j k r x r y r z
F x
F y
F z
= ( r y
F z
– r z
F y
) i – ( r x
F z
– r z
F x
) j + ( r x
F y
– r y
F x
) k

According t o the principle of transmissibility of a force, “the force may act at any point A or
B
or
C along its line of action and still create same moment about a point
O
”.
Let us consider a force F acting a line passing through points A , B , and
C , as shown in the following figure:
The moment of the force about a point
O
, M
O
, may be determined as:
M
O
= r
A
× F r
B
× F r
C
F

Let us consider forces F
1 position vectors r
1 figure:
, r
2
, r
3
, F
2
, F
3
, ------ positioned with the help of
-------- from a point
O
, as shown in the following
The resultant moment, M
RO
, of the forces may be obtained by taking algebraic sum of the cross-products of the position vectors and forces, as given below:
M
R
O
( r F )

PRINCIPLE OF MOMENTS OR VARIGNON ’ ’ S THEOREM
The principle of moments or Varignon’s theorem states that “the moment of a force about a point is equal to the sum of moments of the force’s components about the point”.
Let us consider a force F positioned with the help of position vector r from a point O , as shown in the following figure:
The moment of the force about point
O
may be taken as the sum of the moments of the components of the force F
1
M
O
= r × F
O
, M
and F
1
+ r × F
2
= r × (F
1
+ F
2
2
, as given below:
) = r × F

PROBLEM SOLVING:
Determine the moment of the force at
A about point
O , as shown in the following figure. Express the result as a Cartesian vector.

PROBLEM SOLVING:
From above figure, the co-ordinates of points
O
and
A
may be obtained as:
O
(0, 0, 0) m and
A
( − 3, − 7, 4) m
The position vector for
OA may be written as: r
OA
{
3 i 7 j 4 k
} m
The moment of the force F about point O may be obtained as:
M
O
= r
OA
F
{
3 i 7 j 4 k
} {
60 i −
30 j −
20 k
} { k −
60 j +
420 k +
140 i +
240 j + i
}
⇒ M
O
= {
260 i +
180 j +
510 k
}
N-m
Magnitude of the moment M
O
is
M
O
=
260
2 +
180
2 + 2
510
=
A s

PROBLEM SOLVING:
As shown in the following figure, the applied force F creates a moment about point P , M
P
= { − 1i – 3j – 9k} lb-in. If a = 3 in, determine the dimensions b and c of the block.

PROBLEM SOLVING:
From the above figure, the co-ordinates of points
A and
P
may be obtained as:
A ( a , b , 0) in and P (0, 0, c ) in
The position vector for PA is given as: r
PA
= { a i b j c k
} in
The moment of the force F about point
P may be expressed as:
M
P
=
= r
PA
× F
{ a i b j c k
} {
3 i
2 j
1 k
} lb-in.
= {
2 k + −
3 k − −
3 c j +
2 c i
} lb-in.
= { − i +
( a − c j +
(2 a − b k
} lb-in.
Equating the above expression of × ⇒
6 c / −
3 b = −
3 moment M
{
(2 c b i + a
P
−
to its given value: c j +
(2 a − k
} {
9 k
}
× ⇒
2 a −
6 c / = −
6
2 a −
3 b = −
Comparing the coefficients of i, j, and k on both sides of the above equation:
2 c – b
= − 1 --------- (1)
a –
3 c
= − 3 --------- (2)
2 a –
3 b
= − 9 --------- (3)
Substituting
=
3 in. (given) in Eq.(3)
− = − − = −
∴ b =
5 in. Ans.
From Eq. (1),
2
= − +
5
=
4
∴ c =
2 in. Ans.

1.
A force F is acting at a point
A
as shown in the following figure. The moment, M
O the force F about point
O
is
, of
(a) {20i - 30k} N-m (b) {60i - 30k} N-m
(c) {60i - 30j} N-m (d) {60i + 40j - 30k} N-m
Ans: (b)
Feedback:
The co-ordinates of points
O
and
A
may be obtained as:
O
(0, 0, 0)m &
A
(1, 2, 2) m.
The position vector for OA may be written as: r
OA
= {
1 i + j +
2 k
} m
The moment of the force F about point
O
may be obtained as:
M
O
= r
OA
F
{
1 i
2
= {
60 i −
30 k
}
N-m j
2 k
} {
10 i −
10 j +
20 k
} { k −
20 j −
20 k +
40 i +
20 j + i
}