Question

A flexible chain of mass $m$ is hanging as shown. Find tension at the lowest point. 

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

Hint:

For symmetric hanging systems, analyze equilibrium of one half to simplify calculations.

Answer 1

The Correct Option is A

Step 1: Understand the equilibrium of the chain

The flexible chain is hanging symmetrically, and each side makes an angle

θ = 30° with the vertical.

At the lowest point, the chain is in equilibrium under the action of tension and weight.

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Step 2: Resolve the tension forces

Let T be the tension in each side of the chain at the lowest point.

The vertical component of tension on each side is:

T cosθ

Since the chain is symmetric, there are two such vertical components acting upward.

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Step 3: Apply equilibrium condition

The total upward force balances the weight of the chain:

2T cosθ = mg

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Step 4: Solve for the tension T

T = mg / (2 cosθ)

Substituting θ = 30°:

cos 30° = √3 / 2

T = mg / (2 × √3/2)

T = mg / √3

T = (√3 / 2) mg

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Final Answer:

The tension at the lowest point of the chain is
(√3 / 2) mg