Archimedes' Principle — Question 2

Question 2

An object can float at an interface while fully submerged

A rigid object of volume $V$ and density $900\,\mathrm{kg/m^3}$ lies wholly below the free surface across an oil–water interface. Oil density is $800$ and water density $1000\,\mathrm{kg/m^3}$. Ignore surface tension and assume sufficiently deep layers.

1. Find the fraction of object volume in water at vertical equilibrium.

2. Explain the vertical response to a small upward displacement that transfers a positive volume $\delta V$ from water into oil. Is that equilibrium restoring?

Question 2 - Solution

An object can float at an interface while fully submerged

(a) Derivation and calculation

Let $fV$ be in water and $(1-f)V$ in oil. Then$$B=gV[\rho_wf+\rho_o(1-f)],\quad W=\rho_bgV.$$$$\rho_b=\rho_o+(\rho_w-\rho_o)f\Rightarrow\boxed{f=\frac{900-800}{1000-800}=\frac12}.$$

(b) Result and interpretation

An upward displacement transfers $\delta V>0$ from water to oil:$$\delta B=\rho_og\delta V-\rho_wg\delta V=\boxed{-(\rho_w-\rho_o)g\delta V<0}.$$$$\delta W=0\Rightarrow\text{net force becomes downward}.$$The equilibrium is vertically restoring while the body still crosses the interface; rotational stability is a separate question.

Physical check

A physical fraction $0<f<1$ requires the object density to lie between the two liquid densities.