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:

Step-by-step solution

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 × L 2 . . . 2 Equations (1) and (2) imply that W cos θ × L 2 = F 2 L ⇒ 12 × 10 × cos 60 ° × L 2 = F 2 L ⇒ F 2 = 120 × 10 × cos 60 ° 2 = 30 Hence, the required weight can be expressed as F 2 =