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Question
The diagram shows two forces F1 = 5 N and F2 = 3N acting at point A and B of a rod pivoted at a point O, such that OA = 2m and OB = 4m
Calculate:
- Moment of force F1 about O
- Moment of force F2 about O
- Total moment of the two forces about O.
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Solution
Given AO = 2m and OB = 4m
1) Moment of Force F1 (= 5N) at A about the point O
= F1 × OA
= 5 × 2
= 10 Nm (anticlockwise)
2) Moment of force F2 = (= 3N) at B about the point O
= F2 × OB
= 3 × 4
= 12 Nm (clockwise)
3) Total moment of forces at O
= clockwise moments - anticlockwise moments
= 12 - 10
= 2 Nm (clockwise)
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