Question

Determine the point of application of third force for which body is in equillibrium when forces of 20 N & 30 N are acting on the rod as shown in figure.

Answer 1

Let the magnitude of third force is F, is applied in upward direction then the body is in the equilibrium when 

(i) Vector F_net =  0 (Translational Equillibrium)

⇒ 20 + F = 30 ⇒ F = 10 N

So the body is in translational equilibrium when 10 N force act on it in upward direction.

(ii) Let us assume that this 10 N force act.

Then keep the body in rotational equilibrium

So Torque about C = 0

so 10 N force is applied at 70 cm from point A to keep the body in equilibrium.