AS Daily A Level Physics question
A 1.0 m uniform wooden ruler of mass 0.30 kg is supported on a knife edge at the 40 cm mark. A 0.40 kg mass is hung from the 10 cm mark. Where should a 0.20 kg mass be hung on the right-hand side so that the ruler is horizontal in equilibrium? State the mark on the ruler.
Answer
The correct answer is A.
Correct: A — at the 85 cm mark. Taking moments about the 40 cm pivot: anticlockwise from the 0.40 kg at 10 cm is 0.40 g × 0.30 = 0.12 g; clockwise from the ruler’s own weight at 50 cm is 0.30 g × 0.10 = 0.03 g, so the extra clockwise needed is 0.12 g − 0.03 g = 0.09 g; the 0.20 kg must therefore be at distance d = 0.09 g/(0.20 g) = 0.45 m to the right, i.e. 40 cm + 45 cm = 85 cm. B — at the 100 cm mark: ignores the ruler’s own weight, using 0.40×0.30 = 0.20×d so d = 0.60 m, giving 100 cm. C — at the 55 cm mark: inverts the mass–distance relation (making the lighter mass closer), effectively taking d = 0.30/2 = 0.15 m from the pivot. D — at the 62.5 cm mark: mixes up which mass provides the balancing moment or uses the wrong divisor, e.g. using d = (0.12 − 0.03)/0.40 = 0.225 m, placing the 0.20 kg far too close to the pivot.