)(1000)(10) ⇒F=20secθ Weight of rod, w = 20 × 10 = 20 N For rotational equilibrium of rod, net torque about O should be zero. ∴F(
secθ
2
)(sinθ)=w=(1sinθ) ⇒
20
2
sec2θ=20 ⇒θ=45° ⇒F=20sec45° F=20√2N For vertical equilibrium of rod, force exerted by the hinge on the rod will be (20√2−20) N downwards i.e. 8.3 N downwards