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Two forces each of magnitude 10 N act vertically upwards and downwards respectively at the two ends of a uniform road of length 4 m which is pivoted at its mid point as shown in fig 1.34. Determine the magnitude of resultant moment of forces about the pivot O. |
Answer» Given, AB = 4 m hence, OA = 2m and OB = 2m Moment of force F (=10N) at A about the point O = F × OA = 10 × 2 = 20 Nm (clockwise) Moment of force F (=10N) at point B about the point O = F × OB= 10 × 2 = 20 Nm (clockwise) Total moment of forces about the mid-point O= = 20 + 20 = 40 Nm(clockwise) |
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