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°. Considering the equilibrium of each half of the chain, the tension of the chain at the lowest point is

Q. 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°. Considering the equilibrium of each half of the chain, the tension of the chain at the lowest point is ____ .

(A) m g

(B) ^√3/₂ m g

(D) √3 m g

Correct Answer: ^√3/₂ m g

Explanation

The chain is symmetric about its lowest point. Hence, the mass of the chain is equally divided on both sides. Each half of the chain has mass m/2.

Consider the equilibrium of one half of the chain. Three forces act on this half:

The weight of the half chain acting vertically downward = (m/2)g

The tension T at the lowest point acting horizontally

The tension at the support acting along the chain, making an angle of 30° with the horizontal

Resolve the tension at the support into horizontal and vertical components.

Vertical component of tension = T_s sin 30°

For vertical equilibrium:

T_s sin 30° = (m/2) g

T_s × 1/2 = (m/2) g

T_s = m g

Now consider horizontal equilibrium. The horizontal component of the support tension balances the tension at the lowest point.

T = T_s cos 30°

T = m g × (√3 / 2)

T = (√3 / 2) m g

Hence, the tension of the chain at the lowest point is (√3 / 2) m g.

Explanation

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