The floating cube Consider a cubeshaped block floating in wa

The floating cube. Consider a cube-shaped block floating in water. Assume its top and bottom surfaces are parallel with the surface of the water. The length of an edge of the cube is L and the cube\'s density is rho. Since the cube floats, rho, must be less than the density of water, rho When floating \"at rest\", that is, at equilibrium, the height of the cube above the water surface is h Derive an expression for h involving L, p_f and p_c, using the Archimedes\'s Principle result that for an object floating at equilibrium, the net force (buoyant force minus weight) on the cube is zero. If you dunk the cube deeper into the water by an amount Deltah, the net force on the block changes an amount DeltaF. Show that AF is proportional to, but oppositely directed from, the \"dunk depth\" Deltah. (Clearly, your relation will only hold when |Delta h |

Solution

a.) Here, the weight of the cube is balanced by the buoyant force acting on it which in turn is equal to the weight of water displaced by the cube. That is mg = V*Pf

Pc*L^3*g = L*L*(L-h)*Pf

Hence, h = L[Pf - gLPc]/Pf

b.) Now we dip the cube further by a distance x, hence the net unbalance force would be the difference of the new buoyant force and the weight.

That is L*L*(L-h + x)*Pf - L*L*(L-h)*Pf [Since from part a, we know that the weight is equal to the buoyant force acting in the initial position]

That is, the net force is L*L*x*Pf acting upwards, while displacement is downwards.

c) The situation above is same as that for an simple harmonic motion, where spring constant k = L*L*Pf

and mass of the block is L^3 *Pc

hence frequency = (1/2*pi) * sqrt(k/m)

Frequency = (1/2*pi) * sqrt(Pf / L*Pc)

d.) Yes, the frequency depends on the density of the cube. The limitation it imposes is that the frequency can never be greater than one.

e.) Yes, the frequency also depends on the size of the cube. And the cubes with larger side will have smaller frequency.