1. 4.4 Principles of Moments Also known as Varignon’s Theorem
“Moment of a force about a point is equal to the sum of the moments of the forces’ components about the point”
For F = F1 + F2,
MO = r X F1 + r X F2
= r X (F1 + F2)
= r X F

2. 4.4 Principles of Moments
The guy cable exerts a force F on the pole and creates a moment about the base at A
MA = Fd
If the force is replaced by Fx and Fy at point B where the cable acts on the pole, the sum of moment about point A yields the same resultant moment

3. 4.4 Principles of Moments Fy create zero moment about A MA = Fxh
Apply principle of transmissibility and slide the force where line of action intersects the ground at C, Fx create zero moment about A
MA = Fyb

4. 4.4 Principles of Moments Example 4.6
A 200N force acts on the bracket. Determine
the moment of the force about point A.

5. 4.4 Principles of Moments Solution Method 1:
From trigonometry using triangle BCD,
CB = d = 100cos45° = 70.71mm
= m
Thus,
MA = Fd = 200N( m)
= 14.1N.m (CCW)
As a Cartesian vector,
MA = {14.1k}N.m

6. 4.4 Principles of Moments Solution Method 2:
Resolve 200N force into x and y components
Principle of Moments
MA = ∑Fd
MA = (200sin45°N)(0.20m) – (200cos45°)(0.10m)
= 14.1 N.m (CCW)
Thus,
MA = {14.1k}N.m

7. 4.4 Principles of Moments Example 4.7
The force F acts at the end of the angle
bracket. Determine the moment of the force
about point O.

8. 4.4 Principles of Moments Solution Method 1
View Free Body Diagram
Solution
Method 1
MO = 400sin30°N(0.2m)-400cos30°N(0.4m)
= -98.6N.m
= 98.6N.m (CCW)
As a Cartesian vector,
MO = {-98.6k}N.m

9. 4.4 Principles of Moments Solution Method 2:
Express as Cartesian vector
r = {0.4i – 0.2j}N
F = {400sin30°i – 400cos30°j}N
= {200.0i – 346.4j}N
For moment,

10. 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

11. 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

12. 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

13. 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

14. 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

15. 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

16. 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

17. 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

18. 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,

19. 4.5 Moment of a Force about a Specified Axis
Vector Analysis
- Or expressed as,
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

20. 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)

21. 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)

22. 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.

23. 4.5 Moment of a Force about a Specified Axis
Solution
Method 1
Negative sign indicates that sense of Mx is
opposite to i

24. 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

25. 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

26. 4.5 Moment of a Force about a Specified Axis
Solution

27. 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.

28. 4.5 Moment of a Force about a Specified Axis
View Free Body
Diagram
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,
For simplicity, choose rD

29. 4.5 Moment of a Force about a Specified Axis
Solution
For force,
In determinant form,
Negative sign indicates MAB is opposite to uB

30. 4.5 Moment of a Force about a Specified Axis
Solution
In Cartesian form,
*Note: if axis AB is defined
using unit vector directed
from B towards A, the above
formulation –uB should be used.
MAB = MAB(-uB)