Question

A cube with side length L, and density ρ_C sits at rest on a bathroom scale at the bottom of a large tank. The tank is partially filled with a fluid having density ρ_f, leaving the cube partially submerged. A side profile view of the cube shows the cube face a distance d outside of the fluid, and a distance L-d submerged under the fluid. Determine the distance dwhen ρ_f= 2000 kg/m³, ρ_C= 4000 kg/m³and L = .50 m. Write every step with detail specially when solving for the unknown, since I am trying to learn how to do it and also draw a diagram.

Answer 1

Given,

Density of fluid should be more than that of solid, then only liquid will float otherwise it will sink

Density of the fluid, _f = 4000 kg/m³

Density of the cube, _c = 2000 kg/m³

Side of the cube, L = 0.5 m

Thus, Volume of cube, V = L³ = (0.5)²

                                         = 0.125 m³

Now,

From the figure we, length d is outside the water,

Thus,

Volume of cube submerged, V' = L*L*(L-d)

                                                   = 0.5*0.5*(0.5 - d) = 0.25*(0.5 - d)

Now,

Volume of liquid displaced is equal to volume of cube submerged

Now,

Since system is in equilibrium,

Weight of the cube = Buoyant force

=> _c*V*g = _f*V'*g

=> _c*V = _f*V'

=> 2000*0.125 = 4000*(0.25*(0.5 - d))

=> 250 = 1000*(0.5 - d)

=> 0.5 - d = 250 / 1000 = 0.25

=> d = 0.5 - 0.25

         = 0.25 m

or the length d outside the fluid is 0.25 m