A flexible chain of mass $m$ hangs between two fixed points at the same level. The inclination of th

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 $30^{\circ}$. Considering the equilibrium of each half of the chain, the tension of the chain at the lowest point is $\_\_\_\_$

Professor PC Thomas Classes · Answer

The chain is in equilibrium under its total weight $mg$ and the tensions $T$ at the two supports. The upward vertical components of the tension at the supports balance the total weight: $2T \sin(30^\circ) = mg \implies 2T(\frac{1}{2}) = mg \implies T = mg$. The horizontal component of the tension must be uniform throughout the entire chain to maintain horizontal equilibrium. Therefore, the tension at the lowest point, which is entirely horizontal, equals the horizontal component of the tension at the supports: $T_0 = T \cos(30^\circ) = mg \frac{\sqrt{3}}{2}$.