Worked exercise — A piece of wood in water with petrol

Problem

A container of area $2\;\text{m}^2$ contains $1\;\text{m}$ of water ($\rho_a = 1000\;\text{kg/m}^3$) with $1\;\text{m}$ of petrol ($\rho_b = 600\;\text{kg/m}^3$) on top. A cubic piece of wood of side $0{,}1\;\text{m}$ (of area $A = 0{,}01\;\text{m}^2$) with density $\rho_l = 700\;\text{kg/m}^3$ is dropped in. At what level does it float?

Solution

Step 1 — Where the wood settles. The wood has a higher density than the petrol ($700 > 600$), so it sinks below the petrol layer; but it has a lower density than water ($700 < 1000$), so it does not sink completely. It settles astride the interface. Let $x$ be the fraction of the wood submerged in water and $1-x$ the fraction in petrol.

Step 2 — Force balance. The weight of the wood is balanced by the two Archimedes’ thrusts (one for each displaced fluid):

$\rho_l V g = \rho_a\,x V g + \rho_b\,(1-x) V g$

Step 3 — Simplification. Dividing by $V g$ gives a relation between densities alone:

$\rho_l = \rho_a\,x + \rho_b\,(1-x) \;\Rightarrow\; x = \frac{\rho_l - \rho_b}{\rho_a - \rho_b}$

Step 4 — Calculation.

$x = \frac{700 - 600}{1000 - 600} = \frac{100}{400} = 0{,}25$

$\ev{x = 0{,}25 \;\Rightarrow\; 25\%\ \text{underwater},\ 75\%\ \text{in the petrol}}$

Linked atoms

Argomenti: Fluid statics and dynamics Concetti: Archimedes’ principle · Weight force · Static equilibrium