Q 11-06-087JEE MainJEE Main 2024 (27 Jan, Shift 2)Medium
A heavy iron bar of weight $12\ \text{kg-wt}$ is having its one end on the ground and the other on the shoulder of a man. The rod makes an angle $60^\circ$ with the horizontal. The normal force applied by the man on the bar is:
Answer: (C) $3\ \text{kg-wt}$
The man's normal force $N$ acts perpendicular to the bar at its upper end. Taking torques about the end on the ground (bar length $L$, weight at the midpoint):
$$N\cdot L = W\cdot\frac L2\cos60^\circ \;\Rightarrow\; N = \frac{12}{2}\times\frac12 = 3\ \text{kg-wt}$$
Solution by Sreeraj P, M.Sc Physics