Two vertical steel wires support a 30 kg mass above a 10 kg mass and a pan, as shown in the figure. The upper wire has cross-sectional area 0.008 cm² and the lower wire has cross-sectional area 0.004 cm². The breaking stress of the steel is 12×10⁸ N·m⁻². Taking g = 10 m·s⁻², what is the maximum additional mass that can be placed in the pan without breaking either wire?

Question

G1 Physics · Accepted Answer

Let the additional mass placed in the pan be m kg. The upper wire supports the 30 kg mass, the 10 kg mass and the added mass m. Therefore its tension is T(upper) = (40+m)g. Its area is 0.008 cm² = 8×10⁻⁷ m². At the breaking limit, (40+m)10 ÷ (8×10⁻⁷) = 12×10⁸. This gives 40+m = 96, so m = 56 kg. The lower wire supports the 10 kg mass and the added mass m: T(lower) = (10+m)g. Its area is 0.004 cm² = 4×10⁻⁷ m². At the breaking limit, (10+m)10 ÷ (4×10⁻⁷) = 12×10⁸. Thus, 10+m = 48, so m = 38 kg. Th…