Question

A cubical wooden block of specific gravity s1 floats in a liquid of specific gravity s2 as shown .the atmospheric pressure is p0 .find 1.) Force exerted on the face Ad 2).the force exerted on face BC

Answer 1

Given A cubical wooden block of specific gravity s1 floats in a liquid of specific graave to calculate depth of wood under vity s2 as shown .the atmospheric pressure is p0 .find 1.) Force exerted on the face Ad 2).the force exerted on face BC

First we have to calculate depth of wood under water. The depth of surface AD under liquid is the pressure at depth h under water.

We have weight of wood block = wt of the liquid dispersed.

 W₁ x L³ = W₂ x L² x h

h = W₁ x L / W₂

Pressure under depth is m x g x h

     P AD = W₂ x g x h

             = W₂ x g x W₁ x L / W₂

    P AD = g W₁ L

Now F AD

               = g W₁ L x L²

         Force on AD = g W₁ L³

The upper surface BC exposed to atmosphere

So force = pressure / Area

Force on BC = P₀ / L²