Question

A flexible chain of mass $m$ hangs between two fixed points $A$ and $B$ at the same level. The inclination of the chain with the horizontal at the two points of support is $\theta$. The tension at the midpoint $C$ of the chain is

• A. $\frac{mg}{\tan \theta }$
• B. $\frac{mg}{2\tan \theta }$
• C. Zero
• D. $\left(mg\right)\left(\frac{\sin \theta +\cos \theta }{2}\right)$

Answer 1

The correct option is B $\frac{mg}{2\tan \theta }$

Let $T$ be the tension in the chain at the ends and $T^{\prime }$ be the tension in the chain at the midpoint $C$.
Resolve components of $T$ in vertical and horizontal direction (as shown in the figure).


From the figure,
$2T\sin \theta =mg$
$\therefore$ Tension at the ends of the chain $T=\frac{mg}{2\sin \theta }$

Now, consider the half portion of the hanging chain
At the midpoint $C$, tension in the chain is horizontal

$T^{\prime }=T\cos \theta ⟹T^{\prime }=\frac{mg}{2}\cot \theta$
or Tension at the mid-point $C$
$=\frac{mg}{2\tan \theta }$