A 10 kg load hangs 2 m from the pivot of a light uniform lever. What force at 0.5 m on the other side of the pivot balances it? (take g = 10 m/s²)

Answer and explanation

Correct answer: C. 400 N

  1. Step 1: Take moments about the pivot: F × 0.5 = 10 × 10 × 2. Step 2. F = 400 N. Answer: 400 N. Why not the others: 100 N is what you get if you ignore the distances. 25 N is what you get if you swap the two distances. 200 N is what you get if you forget to divide by the second distance.

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