Question

An Aluminum cube with size L is immersed in a box containing water and Mercury. Assuming that the cube is in equilibrium in the horizontal position, compute the height of the centre of mass of the cube with respect to the surface separating water and Mercury.

density of water ρw = 1gcm−3
density of Mercury ρHg = 13.6gcm−3
density of Aluminum ρAl = 2.7gcm−3
L = 10cm.

Answer 1

B = (dm v1 + dw v2) g where B = buoyant force and d the density

B = da ( v1 + v2) g where v1 is volume of Hg displaced and v2 the water

(dm v1 + dw v2) = da ( v1 + v2)

v1 ( dm - da) = v2 ( da - dw)

v1 / v2 = (2.7 - 1) / (13.6 - 2.7) = .156

L1 = .156 L2 where L1 is length of side in Hg and L2 length of side in water

Since the center of mass is at L / 2 and L1 + L2 = L

L1 = .156 (L - L1)

L1 = .156 L / 1.156 = .135 L

L / 2 - .135 L = .365 L above the interface of the liquids