Archimedes' Principle — Question 8

Question 8

Stable stratification can select a submerged resting depth

A small rigid object of volume $V$ and density $\rho_b$ is immersed in liquid whose density varies with downward depth $h$ as $\rho(h)=\rho_0+\alpha h$, with $\alpha>0$. The object is small enough that density variation across it is negligible. No tether or contact acts.

1. Find its neutral-buoyancy depth and the condition for that depth to lie in a layer $0<h<H$.

2. If displaced downward by a small $\delta h$, find the resulting vertical force with downward positive and decide the stability.

Question 8 - Solution

Stable stratification can select a submerged resting depth

(a) Derivation and calculation

Using the local-density approximation,$$B\simeq(\rho_0+\alpha h)gV,\qquad W=\rho_bgV.$$$$B=W\Rightarrow\boxed{h_*=(\rho_b-\rho_0)/\alpha}.$$$$0<h_*<H\iff\boxed{\rho_0<\rho_b<\rho_0+\alpha H}.$$

(b) Result and interpretation

With downward force positive and $h=h_*+\delta h$,$$F_h=gV[\rho_b-\rho_0-\alpha(h_*+\delta h)].$$$$\boxed{F_h=-\alpha gV\delta h},\qquad\frac{dF_h}{d(\delta h)}=-\alpha gV<0.$$The force opposes vertical displacement; the neutral depth is restoring.

Physical check

The conclusion assumes a rigid fixed-volume object and negligible density variation across the object itself.