Stevin's law

Consider a vertical cylinder of cross-section $A$ filled with an incompressible liquid of density $\rho$. Let us ideally isolate a column of liquid of height $h$, bounded by two levels. Since this column is in equilibrium, its weight must be balanced by the pressure difference between the lower and upper level, multiplied by the area $A$. The weight of the column is $mg = \rho V g = \rho A h g$; the net push due to the pressure difference is $\Delta P \cdot A$. Equating the two and dividing by $A$, we find that pressure increases by $\rho g h$ on descending by a depth $h$.

Principle — Stevin's law

In a fluid of density $\rho$ at equilibrium in a uniform gravitational field $g$, the pressure increases linearly with depth $h$: $\ev{P(h) = P_0 + \rho\,g\,h}$ where $P_0$ is the pressure at the upper surface. In particular, all points at the same depth have the same pressure, regardless of the shape of the container.

In an open container the free surface is at atmospheric pressure $P_0$; at depth $h$ the pressure equals $P_0 + \rho g h$.

Stevin does not depend on the shape

The pressure at the bottom depends only on the depth, not on the amount of liquid. A $25$ m wide swimming pool and a thin drinking straw, filled with water to the same level, exert the same pressure at the bottom. It is counter-intuitive but true: the bottom bears the weight of the vertical column above it, not of all the liquid present.

This behaviour — pressure depending only on depth and never on the shape of the container — is one of the most characteristic properties of fluids at equilibrium (Lauga 2022; Walker 2013).

Pressure at the bottom of a swimming pool

In a swimming pool $2{,}5$ m deep, what is the total pressure at the bottom? And the gauge pressure, i.e. the component due to the liquid alone?

The gauge pressure is $P_\text{man} = \rho g h = 1000\cdot 9{,}81\cdot 2{,}5 \approx 2{,}45\cdot 10^4\;\text{Pa}$. The total pressure is $P_\text{tot} = P_\text{atm} + P_\text{man} \approx 1{,}013\cdot 10^5 + 2{,}45\cdot 10^4 \approx 1{,}26\cdot 10^5\;\text{Pa}$.

For every additional $10$ m of depth the total pressure increases by about one atmosphere: a rule every diver knows.

Collegamenti

Argomenti: Fluidostatica e fluidodinamica Concetti: Legge di Stevino · Pressione

Esercizi collegati: Esercizio svolto — La colonna d’aria che schiaccia i nostri piedi · Problema — Derivazione di Stevino · Esercizio svolto — Pressa per la frutta