A uniform meter rule is balanced horizontally at its midpoint pivot (the 50 cm mark). A 4.0 N weight is suspended from the 20 cm mark. What force must be applied downward at the 90 cm mark to maintain horizontal equilibrium?

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A 2.0 N
B 3.0 N
C 4.0 N
D 6.0 N

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Answer: B. 3.0 N

By the principle of moments, anticlockwise moment = clockwise moment. The 4.0 N weight is 30 cm from the pivot: 4.0 × 30 = 120 N cm. The 90 cm mark is 40 cm from the pivot, so F × 40 = 120, giving F = 3.0 N.

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A uniform meter rule is balanced horizontally at its midpoint pivot (the 50 cm mark). A 4.0 N weight is suspended from the 20 cm mark. What force must be applied downward at the 90 cm mark to maintain horizontal equilibrium?

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