1) Moment of force is calculated using the cross product of the position vector r and force vector F. 2) The magnitude of the moment is equal to the force F multiplied by the perpendicular distance d between the line of action of F and the point of reference. 3) The direction of the moment is determined by the right-hand rule applied to r and F.

1. 4.3 Moment of Force
- Vector Formulation
Moment of force F about point O can
be expressed using cross product
MO = r X F
where r represents position
vector from O to any point
lying on the line of action
of F

2. 4.3 Moment of Force
- Vector Formulation
Magnitude
For magnitude of cross product,
MO = rF sinθ
where θ is the angle measured
between tails of r and F
Treat r as a sliding vector. Since d = r
sinθ,
MO = rF sinθ = F (rsinθ) = Fd

3. 4.3 Moment of Force
- Vector Formulation
Direction
Direction and sense of MO are determined by
right-hand rule
- Extend r to the dashed position
- Curl fingers from r towards F
- Direction of MO is the same
as the direction of the thumb

4. 4.3 Moment of Force
- Vector Formulation
Direction
*Note:
- “curl” of the fingers indicates the sense of
rotation
- Maintain proper order of r
and F since cross product
is not commutative

5. 4.3 Moment of Force
- Vector Formulation
Principle of Transmissibility
For force F applied at any point A,
moment created about O is MO = rA x
F
F has the properties of a sliding vector
and
therefore act at any point
along its line of action and
still create the same
moment about O

6. 4.3 Moment of Force
- Vector Formulation
Cartesian Vector Formulation
For force expressed in Cartesian
form, r r r
i j k
r r r
M O = r XF = rx ry rz
Fx Fy Fz
where rx, ry, rz represent the x, y, z
components of the position vector
and Fx, Fy, Fz represent that of the
force vector

7. 4.3 Moment of Force
- Vector Formulation
Cartesian Vector Formulation
With the determinant expended,
MO = (ryFz – rzFy)i – (rxFz - rzFx)j + (rxFy – yFx)k
MO is always perpendicular to
the plane containing r and F
Computation of moment by cross
product is better than scalar for
3D problems

8. 4.3 Moment of Force
- Vector Formulation
Cartesian Vector Formulation
Resultant moment of forces about point
O can be determined by vector addition
MRo = ∑(r x F)

9. 4.3 Moment of Force
- Vector Formulation
Moment of force F about point
A, pulling on cable BC at any
point along its line of action,
will remain constant
Given the perpendicular
distance from A to cable is rd
MA = rdF
In 3D problems,
MA = rBC x F

10. 4.3 Moment of Force
- Vector Formulation
Example 4.4
The pole is subjected to a 60N force that is
directed from C to B. Determine the magnitude
of the moment created by this force about the
support at A.

11. 4.3 Moment of Force
- Vector Formulation
Solution
Either one of the two position vectors can be
used for the solution, since MA = rB x F or MA
= rC x F
Position vectors are represented as
rB = {1i + 3j + 2k} m and
rC = {3i + 4j} m
Force F has magnitude 60N
and is directed from C to B

12. 4.3 Moment of Force
- Vector Formulation
Solution
r r
F = (60 N )u F
r
 (1 − 3)i + 93 − 4) r + 92 − 0) k 
r
j
= (60 N )  

 (−2) + (−1) + ( 2)
2 2 2


{ }
r r r
= − 40i − 20 j + 40k N
Substitute into determinant formulation
r r r
i j k
r r r
M A = rB XF = 1 3 2
− 40 − 20 40
{ }
r r r
= [3(40) − 2(−20)]i − [1(40) − 2(−40)] j + [1(−20) − 3( 40)]k

13. 4.3 Moment of Force
- Vector Formulation
Solution r r r
i j k
Or r r r
M A = rC XF = 3 4 0
− 40 − 20 40
{ }
r r r
= [4(40) − 0(−20)]i − [3(40) − 0(−40)] j + [3(−20) − 4(40)]k
Substitute into determinant formulation
{ }
r r r
r
M A = 160i − 120 j + 100k N .m
For magnitude,
r
M A = (160) 2 + (120) 2 + (100) 2
= 224 N .m

14. 4.3 Moment of Force
- Vector Formulation
Example 4.5
Three forces act on the rod. Determine the
resultant moment they create about the flange
at O and determine the coordinate direction
angles of the moment axis.

15. 4.3 Moment of Force
- Vector Formulation
Solution
Position vectors are directed from point
O to each force
rA = {5j} m and
rB = {4i + 5j - 2k} m
For resultant moment about O,
r r r r r r r r
M Ro = Σ( r XF ) = rA XF1 + rB XF2 + rC XF3
r r r r r r r r r
i j k i j k i j k
r r r
= 0 5 0 + 0 5 0 + 4 5 − 2 = {30i − 40 j + 60k }N .m
− 60 40 20 0 50 0 80 40 − 30

16. 4.3 Moment of Force
- Vector Formulation
Solution
For magnitude
r
M Ro = (30) 2 + (−40) 2 + (60) 2 = 78.10 N .m
For unit vector defining the direction of moment axis,
r r r r
r M Ro 30i − 40 j + 60k
u= r =
M Ro 78.10
r r r
= 0.3941i − 0.5121 j + 0.76852k

17. 4.3 Moment of Force
- Vector Formulation
Solution
For the coordinate angles of the moment axis,
cos α = 0.3841; α = 67.4o
cos β = −0.5121; β = 121o
cos γ = 0.7682; γ = 39.8o