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# A chain of mass $m$ is attached to point A and B of two fixed walls as shown in the figure. Find the tension in the chain at A.A) $\dfrac{1}{2}mg\cos \theta$B) $mg\sin \theta$C) $\dfrac{1}{2}mg$ $\mathrm{cosec}\theta$D) $mg\tan \theta$

Last updated date: 10th Aug 2024
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Hint: In the given setup, the forces acting on the chain are the tension and the weight of the chain. Here the weight of the chain will be directed downwards. By symmetry, the tension at A and B will be equal and will be directed along the length of the chain and so will have a horizontal component and a vertical component. The force balance equation in the vertical direction will provide us with the tension at A.

Step 1: Sketch a free body diagram of the chain and depict the forces acting on the chain.
The above diagram represents the free body diagram of the chain.

As seen from the figure the weight of the body $W = mg$ is directed downwards at the point C. At points A and B, the tension $T$ is directed along the length of the chain making an angle $\theta$ with the horizontal.
Step 2: Resolve the tension at A and B into its horizontal component and its vertical component.

The above figure shows the resolution of the tension $T$ into its horizontal component $T\cos \theta$ and its vertical component $T\sin \theta$ .
So the forces acting in the vertical direction are the weight of the chain, the vertical components of tension at A and B.
The forces acting in the horizontal direction are the horizontal components of tension at A and B.
Step 3: Express the force balance equation in the vertical direction.
The force balance equation in the vertical direction can be expressed as
$T\sin \theta + T\sin \theta = mg$ ------------- (1)
$\Rightarrow 2T\sin \theta = mg$
And finally, we have $T = \dfrac{{mg}}{{2\sin \theta }} = \dfrac{1}{2}mg$ $\mathrm{cosec}\theta$
So the tension at A is obtained as $T = \dfrac{1}{2}mg$ $\mathrm{cosec}\theta$

So the correct option is C.

Note: The force balance equation in the vertical direction is obtained by expressing the total forces acting on the chain i.e., $T\sin \theta + T\sin \theta - mg = 0$ . The weight has a direction opposite to that of the vertical components of the tension at A and B. We take the upward direction to be positive and so weight is negative.