In a fluid at rest, pressure changes with depth because lower layers must support the weight of fluid above them.

For a fluid of approximately constant density $\rho$ in a uniform gravitational field,

$$\boxed{p=p_0+\rho gh},$$

where $h$ is depth below a reference level at pressure $p_0$.

Thus moving downward through a static liquid increases pressure linearly with depth when density is nearly constant.

Example

At a depth of $5.0,\mathrm m$ in water, taking

$$\rho\approx1000,\mathrm{kg/m^3},\qquad g\approx9.81,\mathrm{m/s^2},$$

the pressure increase relative to the surface is

$$\Delta p=\rho gh =(1000)(9.81)(5.0) \approx4.91\times10^4,\mathrm{Pa} =49.1,\mathrm{kPa}.$$

If the surface is open to an atmosphere at about $101,\mathrm{kPa}$ absolute pressure, the absolute pressure at that depth is about

$$101+49=150,\mathrm{kPa}.$$

Forces from hydrostatic pressure

Static fluid pressure acts normal to a boundary. If pressure is uniform over a flat area $A$, the resulting normal-force magnitude is

$$F=pA.$$

When pressure varies appreciably over a surface, the local pressure contributions must instead be accumulated over the area.

Hydrostatics describes how pressure varies through a fluid at rest and how that spatially varying pressure produces forces. Other static-fluid ideas, such as the transmission of an externally applied pressure change through a confined fluid, are distinct consequences of fluid pressure and can be applied independently of the pressure-depth relation.