1.

The diagram in fig.1.33 shows two forces F1 = 5 N and F2 = 3N acting at points A and B of a rod pivoted at a point O, such that OA = 2m and OB = 4m.Calculate:(i) Moment of force F1 about O.(ii) Moment of force F2 about O.(iii) Total moment of the two forces about O.

Answer»

Given AO = 2m and OB = 4m

(i) Moment of force F1(= 5N) at A about the point O

= F1 × OA

= 5 × 2 = 10Nm (anticlockwise)

(ii) Moment of force F2 (= 3N) at B about the point O

= F2 × OB

= 3 × 4 =12 Nm(clockwise)

(iii) Total moment of forces about the mid-point O =

= 12 – 10 = 2 Nm (clockwise)



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