A flexible chain of mass m hangs between two fixed points at the same level. The inclination of the chain with the horizontal at the two points of support is 30°. Considering the equilibrium of each half of the chain, the tension of the chain at the lowest point i $s _____.$
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Question

A flexible chain of mass m hangs between two fixed points at the same level. The inclination of the chain with the horizontal at the two points of support is 30°. Considering the equilibrium of each half of the chain, the tension of the chain at the lowest point i $s _____.$

Options

G1 Physics · Accepted Answer

Consider one half of the chain. Its mass is m/2. Let T be the tension at the support, where the chain makes 30° with the horizontal. Let T₀ be the horizontal tension at the lowest point. Vertical equilibrium gives T sin30° = (m/2)g. Since sin30° = 1/2, T/2 = mg/2. Hence, T = mg. Horizontal equilibrium gives T₀ = T cos30°. Therefore, T₀ = mg(√3/2) = √3mg/2. Correct option: A.

A flexible chain of mass m hangs between two fixed points at the same level. The inclination of the chain with the horizontal at the two points of support is 30°. Considering the equilibrium of each half of the chain, the tension of the chain at the lowest point i s _____.