Question

Oil having a density of 923 kg/m3 floats on
water. A rectangular block of wood 3.45 cm
high and with a density of 963 kg/m3 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.

Answer 1

Answer:

0.0179 m

Explanation:

Draw a free body diagram of the block.  There are 3 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

(923 kg/m³) h₁ + (1000 kg/m³) h₂ = (963 kg/m³) (3.45 cm)

Since the block is fully submerged, h₁ + h₂ = 3.45 cm.

(923 kg/m³) (3.45 cm − h₂) + (1000 kg/m³) h₂ = (963 kg/m³) (3.45 cm)

3184.35 − 923 h₂ + 1000 h₂ = 3322.35

77 h₂ = 138

h₂ = 1.79 cm

h₂ = 0.0179 m