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 is _ _ _ _ .

Step-by-step solution

A chain of mass m hangs between two fixed points at the same level with inclination 30° at support points. Consider equilibrium of one half of the chain (mass m/2 ): Horizontal equilibrium: T₀ = T₁ (30°) = T₁ 3 2 Vertical equilibrium: T₁ (30°) = m 2 g From vertical equation: T₁ 1 2 = mg 2 , giving T₁ = mg Substituting into horizontal equation: T₀ = mg 3 2 = 3 2 mg