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Standard VIII

Physics

Tension

A heavy rod o...

Question

A heavy rod of length $L$ and weight $W$ is suspended horizontally by two vertical ropes as shown. The first rope is attached to the left end of the rod while the second rope is attached a distance $\frac{L}{4}$ from right end. The tension in the second rope is :

\frac{W}{2}

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\frac{W}{4}

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\frac{W}{3}

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\frac{2 W}{3}

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W

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Solution

The correct option is D $\frac{2 W}{3}$
The torque about the rope 1 due to weight mg is balances toque about rope 1 due to $T_{2}$
W will act at the center of rod.
$∴ W \times L / 2 = T_{2} \times 3 L / 4$
So, $T_{2} = W \times 2 / 3$

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A heavy rod of length L and weight W is suspended horizontally by two vertical ropes as shown. The first rope is attached to the left end of the rod while the second rope is attached a distance L4 from right end. The tension in the second rope is :

The correct option is D 2W3The torque about the rope 1 due to weight mg is balances toque about rope 1 due to T2W will act at the center of rod.∴W×L/2=T2×3L/4So, T2=W×2/3

A heavy rod of length $L$ and weight $W$ is suspended horizontally by two vertical ropes as shown. The first rope is attached to the left end of the rod while the second rope is attached a distance $\frac{L}{4}$ from right end. The tension in the second rope is :

The correct option is D $\frac{2 W}{3}$
The torque about the rope 1 due to weight mg is balances toque about rope 1 due to $T_{2}$
W will act at the center of rod.
$∴ W \times L / 2 = T_{2} \times 3 L / 4$
So, $T_{2} = W \times 2 / 3$