Question

A flexible chain of mass \(m\) hangs between two fixed points at the same level. The inclination of the chain with the horizontal at the two points of support is \(30^\circ\). Considering the equilibrium of each half of the chain, the tension of the chain at the lowest point is ________.

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

Hint:

For hanging chains, tension at the lowest point equals the horizontal component of tension at supports.

Answer 1

The Correct Option is B

To find the tension at the lowest point of a flexible chain hanging between two fixed points, we need to consider the equilibrium condition of the chain. Each half of the chain experiences forces due to gravity and tension at the endpoints.

Let's break down the forces acting on one half of the chain:

1. The weight of half the chain is \(\dfrac{mg}{2}\), acting downwards.
2. The tension in the chain at the lowest point is horizontal, denoted as \(T\).
3. The tension at the point of support makes an angle of \(30^\circ\) with the horizontal, denoted as \(T_s\).

Considering the vertical component of the tension at the support point, we have:

\(T_s \sin 30^\circ = \dfrac{mg}{2}\)

The sine of \(30^\circ\) is \(\dfrac{1}{2}\), so:

\(T_s \cdot \dfrac{1}{2} = \dfrac{mg}{2}\)

Solving this, we find that:

\(T_s = mg\)

Now, consider the horizontal component of the tension at the support point:

\(T_s \cos 30^\circ = T\)

The cosine of \(30^\circ\) is \(\dfrac{\sqrt{3}}{2}\), so we substitute \(T_s = mg\):

\(mg \cdot \dfrac{\sqrt{3}}{2} = T\)

Simplifying, the tension at the lowest point of the chain is:

\(T = \dfrac{\sqrt{3}}{2}mg\)

Thus, the correct answer is \(\dfrac{\sqrt{3}}{2}mg\).