A heavy iron bar of weight 12kg is having its one end on the ground...

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

${τ}_{a} = {F}_{2}L . . . \left(1\right)$

And, the net clockwise torque about the same point is

${τ}_{c} = Wcos\theta \times \dfrac{L}{2} . . . \left(2\right)$

Equations (1) and (2) imply that

Wcos\theta \times \dfrac{L}{2} = {F}_{2}L ⇒ 12 \times 10 \times \cos 60 ° \times \left(\dfrac{L}{2}\right) = {F}_{2}L ⇒ {F}_{2} = \dfrac{120 \times 10 \times \cos 60 °}{2} = 30$</p><p>Hence, the required weight can be expressed as</p><p>${F}_{2} = 3kg - wt$