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:

Question

Aicharya · Accepted Answer

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

$τ_{a} = F_{2} L . . . (1)$

And, the net clockwise torque about the same point is

$τ_{c} = W \cos θ \times \frac{L}{2} . . . (2)$

Equations (1) and (2) imply that

$W \cos θ \times \frac{L}{2} = F_{2} L \Rightarrow 12 \times 10 \times \cos 60 ° \times (\frac{L}{2}) = F_{2} L \Rightarrow F_{2} = \frac{120 \times 10 \times \cos 60 °}{2} = 30$

Hence, the required weight can be expressed as

$F_{2} = 3 kg - wt$

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.…

Solution