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 :

A heavy iron bar (12 kg) has one end on the ground and the other on a man’s shoulder, making 60° with the horizontal. We wish to find the normal force applied by the man.

Let the bar have length $$L$$. The weight $$W = 12$$ kg-wt acts at the center of the bar, which is a distance $$L/2$$ from the ground end.

The component of the weight perpendicular to the bar is $$W\cos 60° = 12 \times \tfrac{1}{2} = 6$$ kg-wt, acting at distance $$L/2$$ from the pivot at the ground end. The normal force $$N$$ applied by the man is also perpendicular to the bar and acts at distance $$L$$ from the pivot.

Balancing torques about the ground end gives $$N \times L = W\cos 60° \times \frac{L}{2}$$ and therefore $$N = \frac{6}{2} = 3$$ kg-wt.

The correct answer is Option C: 3 kg-wt.