Question

A flexible chain of weight W hangs between two fixed points A & B which are the same horizontal level. The inclination of the chain with the horizontal at both the points of support is $\theta$. What is the tension of the chain at the mid point ?

• A. $\frac{W}{2}cosec\theta$
• B. $\frac{W}{2}tan\theta$
• C. $\frac{W}{2}cot\theta$
• D. none

Answer 1

The correct option is A $\frac{W}{2}cosec\theta$

The weight of the chain is a downward acting force at the centre of mass of the chain which less at the mid point$.$

The tension in the chain is the same throughout the chain and is a force that equals the net force acting on the chain$.$

If the tension is $F,$ this can be divided into components in the horizontal direction and the vertical direction$.$

The force in the horizontal direction is $F\cdot \cos \theta$ this components cancels out$.$

The vertical components is equal to $2\cdot F\cdot \sin \theta$ and this is equal to the weight of the chain $W.$

$2\cdot F\cdot \sin \theta =W$

$\begin{array}{c}\Rightarrow F=\frac{W}{2\cdot \sin \theta }\\ \Rightarrow F=\left(\frac{W}{2}\right)\cdot \cos ec\theta \end{array}$

Hence,

option $\left(A\right)$ is correct answer.