A heavy iron bar, of weight W is having its one end on the ground and the other on the shoulder of a person. The bar makes an angle θ with the horizontal. The weight experienced by the person is: (a) Wsin θ (b) W (c) W/2 (d) Wcos θ

Answer: (c). Take the shoulder's support force N as vertical. Moments about the end on the ground. Weight W acts at the middle: horizontal arm L/2cos θ; N acts at the far end: arm Lcos θ. N Lcos θ = WL/2cos θ ⇒ N = W/2 So the answer is (c) W/2.

A heavy iron bar, of weight W is having its one end on the ground and… | JEE Main 2024 (9 Apr, Shift 1)

Q13JEE Main 2024 · 9 APR · SHIFT 1Medium

Torque, Couple & Equilibrium of Rigid Bodies

A heavy iron bar, of weight $W$ is having its one end on the ground and the other on the shoulder of a person. The bar makes an angle $\theta$ with the horizontal. The weight experienced by the person is:

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Answer: (c)

Take the shoulder's support force $N$ as vertical. Moments about the end on the ground.
Weight $W$ acts at the middle: horizontal arm $\frac{L}{2}\cos\theta$ ; $N$ acts at the far end: arm $L\cos\theta$ .
$N L\cos\theta = W\frac{L}{2}\cos\theta \Rightarrow N = \frac{W}{2}$
So the answer is (c) $\frac{W}{2}$ .

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