Question

A cube of wood having an edge dimension of 18.7 cm and a density of 645 kg/m3 floats on water. (a) What is the distance from the horizontal top surface of the cube to the water level? cm (b) What mass of lead should be placed on the cube so that the top of the cube will be just level with the water surface?

Answer 1

a) let y is the distance from the horizontal top surface of the cube to the water level

In the equilibrium

Buoyant fore = weight of the cube

weight of the dispalced water = m*g

rho_water*V_displaced*g = rho_cube*V_cube*g

1000*18.7^2*y*g = 645*18.7^3*g

==> y = (645/1000)*18.7

= 12.06 cm <<<<<<<<<<<------------------Answer

b) let M is the mass of lead is required

In the equilibrium

Buoyant fore = weight of the cube + lead

weight of the dispalced water = m*g + M*g

rho_water*V_cube*g = m*g + M*g

rho_water*V_cube = m + M

M = rho_water*V_cube - rho_cube*V_cube

= 1000*0.187^3 - 645*0.187^3

= 2.32 kg <<<<<<<<<<<------------------Answer