1) A moment is a rotational force that tends to cause rotation about an axis. 2) The moment of a force about an axis can be found using either scalar or vector analysis. Scalar analysis involves determining the perpendicular distance from the force to the axis, while vector analysis uses the cross product of the position vector and force vector. 3) No moment is produced about an axis if the force is parallel or passes through the axis, as there is no leverage. The maximum moment occurs when the force is applied at the maximum distance from the axis.

1. 4.5 Moment of a Force about a
Specified Axis
For moment of a force about a point, the
moment and its axis is always perpendicular to
the plane containing the force and the moment
arm
A scalar or vector analysis is used to find the
component of the
moment along a specified
axis that passes through
the point

2. 4.5 Moment of a Force about a
Specified Axis
Scalar Analysis
Example
Consider the pipe assembly that lies in the horizontal
plane and is subjected to the vertical force of F = 20N
applied at point A.
For magnitude of moment,
MO = (20N)(0.5m) = 10N.m
For direction of moment, apply right hand rule

3. 4.5 Moment of a Force about a
Specified Axis
Scalar Analysis
Moment tends to turn pipe about the Ob axis
Determine the component of MO about the y
axis, My since this component tend to
unscrew the pipe from the flange at O
For magnitude of My,
My = 3/5(10N.m) = 6N.m
For direction of My, apply right hand rule

4. 4.5 Moment of a Force about a
Specified Axis
Scalar Analysis
If the line of action of a force F is perpendicular
to any specified axis aa, for the magnitude of the
moment of F about the axis,
Ma = Fda
where da is the perpendicular or shortest
distance from the force line of action to the axis
A force will not contribute a moment about a
specified axis if the force line of action is parallel
or passes through the axis

5. 4.5 Moment of a Force about a
Specified Axis
A horizontal force F applied to
the handle of the flex-headed
wrench, tends to turn the
socket at A about the z-axis
Effect is caused by the moment
of F about the z-axis
Maximum moment occurs when
the wrench is in the horizontal
plane so that full leverage from
the handle is achieved
(MZ)max = Fd

6. 4.5 Moment of a Force about a
Specified Axis
For handle not in the
horizontal position,
(MZ)max = Fd’
where d’ is the perpendicular
distance from the force line
of action to the axis
Otherwise, for moment,
MA = Fd
MZ = MAcosθ

7. 4.5 Moment of a Force about a
Specified Axis
Vector Analysis
Example
MO = rA X F = (0.3i +0.4j) X (-20k)
= {-8i + 6j}N.m
Since unit vector for this
axis is ua = j,
My = MO.ua
= (-8i + 6j)·j = 6N.m

8. 4.5 Moment of a Force about a
Specified Axis
Vector Analysis
Consider body subjected to force
F acting at point A
To determine moment, Ma,
- For moment of F about any
arbitrary point O that lies on the
aa’ axis
MO = r X F
where r is directed from O to A
- MO acts along the moment axis
bb’, so projected MO on the aa’
axis is MA

9. 4.5 Moment of a Force about a
Specified Axis
Vector Analysis
- For magnitude of MA,
MA = MOcosθ = MO·ua
where ua is a unit vector that defines the
direction of aa’ axis
MA = ua·(r X F)
- In determinant form, r r r
i j k
r r r r
M a = (uax i + uay j + uaz k ) ⋅ rx ry rz
Fx Fy Fz

10. 4.5 Moment of a Force about a
Specified Axis
Vector Analysis
- Or expressed as, uax uay uaz
r r r r
M a = u ax ⋅ (r XF ) = rx ry rz
Fx Fy Fz
where uax, uay, uaz represent the x, y, z
components of the unit vector defining the
direction of aa’ axis and rx, ry, rz represent that
of the position vector drawn from any point O on
the aa’ axis and Fx, Fy, Fz represent that of the
force vector

11. 4.5 Moment of a Force about a
Specified Axis
Vector Analysis
The sign of the scalar indicates the direction of
Ma along the aa’ axis
- If positive, Ma has the same sense as ua
- If negative, Ma act opposite to ua
Express Ma as a Cartesian vector,
Ma = Maua = [ua·(r X F)] ua
For magnitude of Ma,
Ma = ∑[ua·(r X F)] = ua·∑(r X F)

12. 4.5 Moment of a Force about a
Specified Axis
Wind blowing on the sign
causes a resultant force F that
tends to tip the sign over due
to moment MA created about
the aa axis
MA = r X F
For magnitude of projection of
moment along the axis whose
direction is defined by unit
vector uA is
MA = ua·(r X F)

13. 4.5 Moment of a Force about a
Specified Axis
Example 4.8
The force F = {-40i + 20j + 10k} N acts on the point
A. Determine the moments of this force about the x
and a axes.

14. 4.5 Moment of a Force about a
Specified Axis
Solution
Method 1
r r r r
rA = {−3i + 4 j + 6k }m
r r
ux = i
1 0 0
r r r r
M x = i ⋅ (rA XF ) = − 3 4 6
− 40 20 10
= −80 N .m
Negative sign indicates that sense of Mx is
opposite to i

15. 4.5 Moment of a Force about a
Specified Axis
Solution
We can also compute Ma using rA as rA extends
from a point on the a axis to the force
r r r
uA = − 3 i +4 j
5 5
−3 4 0
r r r r 5 5
M a = u A ⋅ (rA XF ) = − 3 4 6
− 40 20 10
= −120 N .m

16. 4.5 Moment of a Force about a
Specified Axis
Solution
Method 2
Only 10N and 20N forces contribute moments about
the x axis
Line of action of the 40N is parallel to this axis and
thus, moment = 0
Using right hand rule
Mx = (10N)(4m) – (20N)(6m) = -80N.m
My = (10N)(3m) – (40N)(6m) = -210N.m
Mz = (40N)(4m) – (20N)(3m) = 100N.m

17. 4.5 Moment of a Force about a
Specified Axis
Solution

18. 4.5 Moment of a Force about a
Specified Axis
Example 4.9
The rod is supported by two brackets at A and B.
Determine the moment MAB produced by
F = {-600i + 200j – 300k}N, which tends to rotate
the rod about the AB axis.

19. 4.5 Moment of a Force about a
Specified Axis
Solution
Vector analysis chosen as moment arm from
line of action of F to the AB axis is hard to
determine
For unit vector defining direction of AB axis
of the rod,
r r r r
M AB = u B ⋅ (r XF )
For simplicity, choose rD
r r r
r rB 0.4i + 0.2 j
uB = r =
rB (0.4)2 + (0.2)2
r r
= 0.894i + 0.447 j

20. 4.5 Moment of a Force about a
Specified Axis
Solution
For force,
r r r r
F = {−600i + 200 j − 300k }N
In determinant form,
0.894 0.447 0
r r r r
M AB = u B ⋅ (rD XF ) = 0 0.2 0
− 600 200 − 300
= −53.67 N .m
Negative sign indicates MAB is opposite to uB

21. 4.5 Moment of a Force about a
Specified Axis
Solution
In Cartesian form,
r r r r r
M AB = M AB u B = (−53.67 N .m)(0.894i + 0.447 j )
r r
= {−48.0i − 24.0 j }N .m
*Note: if axis AB is defined
using unit vector directed
from B towards A, the above
formulation –uB should be used.
MAB = MAB(-uB)