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

Explanation

Let the magnitude of third force is F, is applied in upward direction then the body is in the equilibrium when
(i) `vecF_("net")=0` (Translational Equillibrium)
`rArr 20 + F = 30 rArr " "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. Let us assume that this 10 N force act.
Then keep the body in rotational equilibrium So Torque about C = 0
i.e. `tau_(c) =0`

`rArr 30 xx 20 = 10 x`
` x = 60 cm `
so 10 N force is applied at 70 cm from point A to keep the body in equilibrium.