Question

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 $60^\circ$ with the horizontal, the weight experienced by the man is :

• A. 6 kg
• B. 12 kg
• C. 3 kg
• D. $6\sqrt{3}$ kg

Answer 1

Answer: (C)

3 kg

Answer 2

Given:

- Weight of the bar $W = 12 \, \text{kg}$,
- The bar makes an angle $\theta = 60^\circ$ with the horizontal,
- Let $N_1$ be the normal force at the ground and $N_2$ be the force experienced by the man’s shoulder.

Condition for equilibrium about the pivot:

Torque about $O = 0$.

Taking moments about $O$:

$120 \left(\frac{L}{2} \cos 60^\circ \right) - N_2 L = 0$

Simplifying:

$N_2 = \frac{120}{2} \cdot \frac{1}{2} = 30 \, \text{N}.$

Converting to weight experienced:

$\text{Weight} = \frac{30 \, \text{N}}{10 \, \text{m/s}^2} = 3 \, \text{kg}.$

The Correct answer is: 3 kg