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Question: Answered & Verified by Expert
A heavy iron bar of weight 12 kg is having its one end on the ground and the other on the shoulder of a man. The rod makes an angle 60° with the horizontal, the normal force applied by the man on bar is:
PhysicsRotational MotionJEE MainJEE Main 2024 (27 Jan Shift 2)
Options:
  • A 6 kg-wt
  • B 12 kg-wt
  • C 3 kg-wt
  • D 63 kg-wt
Solution:
1175 Upvotes Verified Answer
The correct answer is: 3 kg-wt

Let's consider the following diagram:

For the equilibrium of the system, the net torque, according to the above figure, on the system about point A should be zero.

With reference to the above figure, the net anti-clockwise torque about A is

τa=F2L   ...1

And, the net clockwise torque about the same point is

τc=Wcosθ×L2    ...2

Equations (1) and (2) imply that

Wcosθ×L2=F2L12×10×cos60°×L2=F2 LF2=120×10×cos60°2=30

Hence, the required weight can be expressed as

F2=3 kg-wt

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