Question

A cube of wood having an edge dimension of 18.9 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

Here,

a = 0.189 m

a) let the distance from the top surface is x

balancing the forces on the cube

buoyant force - weight of the wood = 0

0.189^2 * (0.189 - x) * 1000 * 9.8 - 0.189^3 * 645 * 9.8 = 0

solving for

x = 0.067 m = 6.7 cm

the distance from the horizontal top to water surface is 6.7 cm

b)

let the mass that can be placed is m

extra weight = weight of extra water displaced

m * 9.8 = 1000 * 0.189^2 * 0.067 * 9.8

solving for m

m = 2.39 Kg

the extra mass that can be placed is 2.39 Kg