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 ____ .

Solution

Physics sir · 4 steps

Cut the chain at its lowest point and balance the forces on one half.

1. Forces on one half. Take the left half of the chain, of mass m/2. Three forces act on it: its weight mg/2 downward, the tension T at the support along the chain at 30° above the horizontal, and the tension T₀ at the lowest point.
2. Direction at the bottom. At the lowest point the chain is horizontal, so T₀ acts horizontally. That means only the support tension T has a vertical component.
3. Vertical balance. The vertical part of T holds up the half's weight: T sin 30° = mg/2. With sin 30° = 1/2, this gives T = mg.

   T₀ = T cos 30° = mg × √3/2

4. So. Horizontal balance gives the tension at the lowest point as T₀ = √3/2mg, which is option (b).

The idea to keep. For a hanging chain or rope, cut it at the lowest point. There the tension is horizontal, so each half's weight is carried only by the vertical component of the support tension.

Answer from the NTA final answer key. Solution by PrepWiser.