Question

A flexible chain of mass $m$ is hanging as shown. Find tension at the lowest point. 

• A. $\dfrac{\sqrt{3}}{2}mg$
• B. $\dfrac{1}{2}mg$
• C. $\dfrac{\sqrt{2}}{3}mg$
• D. $\sqrt{2}mg$

Hint:

For symmetric hanging systems, analyze equilibrium of one half to simplify calculations.

Answer 1

The Correct Option is A

Step 1: Consider half of the chain.
Due to symmetry, tension at lowest point acts equally on both halves.
Step 2: Free body diagram of half chain.
Vertical equilibrium:
\[ T \sin 30^\circ = \dfrac{mg}{2} \]
Step 3: Solving for tension.
\[ T \times \dfrac{1}{2} = \dfrac{mg}{2} \Rightarrow T = \dfrac{\sqrt{3}}{2}mg \]