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 mid-point of the chain is

• A. $\frac{Mg\cot \theta }{2}$
• B. $Mg\cot \theta$
• C. $zero$
• D. $\frac{Mg}{2}(sin\theta +cos\theta )$

Answer 1

The correct option is A $\frac{Mg\cot \theta }{2}$

Let Tension, T, at one end A.
Vertical component i.e. $Tsin\theta$ will be balance by half the weight of chain at one end
Therefore, $Tsin\theta =Mg/2$ ...........(i)
Also, Let T' be the horizontal component such that
$Tcos\theta =T^{\prime }$ .......... (ii)
From (i) & (ii)
$T^{\prime }=\frac{Mgcot\theta }{2}$
Vertical component is balanced by weight of the chain and therefore tension in the chain is due to the horizontal component