Oil having a density of 930 kg/m^3 floats on

water. A rectangular block of wood 4.93 cm

high and with a density of 968 kg/m^3 floats

partly in the oil and partly in the water. The

oil completely covers the block.

How far below the interface between the

two liquids is the bottom of the block?

Answer in units of m

0.0268 m

Explanation:

Draw a free body diagram of the block. There are three forces: weight force mg pulling down, buoyancy of the oil B₁ pushing up, and buoyancy of the water B₂ pushing up.

Sum of forces in the y direction:

∑F = ma

B₁ + B₂ - mg = 0

ρ₁V₁g + ρ₂V₂g - mg = 0

ρ₁V₁ + ρ₂V₂ = m

ρ₁V₁ + ρ₂V₂ = ρV

ρ₁Ah₁ + ρ₂Ah₂ = ρAh

ρ₁h₁ + ρ₂h₂ = ρh

(930 kg/m³) h₁ + (1000 kg/m³) h₂ = (968 kg/m³) (4.93 cm)

Since the block is fully submerged, h₁ + h₂ = 4.93 cm.

(930 kg/m³) (4.93 cm - h₂) + (1000 kg/m³) h₂ = (968 kg/m³) (4.93 cm)

h₂ = 2.68 cm

h₂ = 0.0268 m