A rod is pivoted at a point. A force of $20 N$ is applied at a distance of $20 c m$ from the pivot perpendicular to the rod as shown in the figure below. Calculate the moment of force about the pivot.

2 Nm

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4 Nm

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6 Nm

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8 Nm

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Solution

The correct option is B
4 Nm

Given,
Force $F$ = $20 N$
Perpendicular distance between line of action of force and axis of rotation $r$ = $20 c m = 0.2 m$
Moment of force= $r \times F = 20 \times 0.2 = 4 N m$ (anti-clockwise)

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