A uniform rod of length 2 . 0 m , specific gravity 0 . 5 and mass 2 kg is hinged at one end to the bottom of a tank of water (specific gravity = 1 . 0 ) filled up to a height of 1 . 0 m as shown in the figure. Taking the case θ ≠ 0 ° the force exerted by the hinge on the rod is: g = 10 m s - 2

Step-by-step solution

Length of rod inside the water = 1.0 s e c θ = s e c θ Upthrust F = 2 2 s e c θ 1 500 1000 ( 10 ) Or F = 20 s e c θ Weight of rod W = 2 × 10 = 20 N For rotational equilibrium of rod, net torque about O should be zero. ∴ F s e c θ 2 sin ⁡ θ = W = 1.0 s i n θ Or 20 2 sec 2 ⁡ θ = 20 Or θ = 45 o ∴ F = 20 sec ⁡ 45 o = 20 2 N The hinge force is F hinge = F - W = 8 . 3 N