Fluid Statics — Pressure at Rest, Hydrostatics and Manometers

Pressure when nothing moves

When a fluid sits still, its pressure has simple, strict rules. At any given depth the pressure is the same everywhere, and it pushes perpendicularly on every surface it touches — the container walls, a submerged plate, your eardrums. There are no sideways shear forces in a static fluid; that is part of what 'fluid' means.

This perpendicular pushing is why a balloon stays round and why a dam feels the water trying to shove it downstream, square-on, at every point of its face.

The hydrostatic equation

Consider the column of fluid above a point at depth $h$. Its weight presses down on the area below, adding $\rho g h$ to whatever pressure $p_0$ sits on the surface:

What the equation implies

Three consequences fall out immediately. First, pressure depends only on depth and density — not on the container's shape. A thimble-wide tube and a swimming pool give the same pressure at the same depth (the hydrostatic paradox). Second, in seawater ($\rho \approx 1025\ \mathrm{kg/m^3}$) every 10 m of depth adds roughly one atmosphere. Third, doubling the depth doubles the gauge pressure — the part above atmospheric.

Absolute vs gauge pressure

Pressure readings come in two flavors. Absolute pressure counts from a perfect vacuum and includes the atmosphere: about $101.3\ \mathrm{kPa}$ at sea level. Gauge pressure counts from atmospheric pressure, so a tire gauge reading $220\ \mathrm{kPa}$ means $220\ \mathrm{kPa}$ above atmosphere — about $321\ \mathrm{kPa}$ absolute.

Most everyday gauges (tires, scuba tanks, boilers) read gauge pressure, while physics equations like $p = p_0 + \rho g h$ use absolute. Converting is just adding or subtracting one atmosphere.

Pascal's principle and hydraulics

Blaise Pascal's 17th-century insight: pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid. Push on a small piston with force $F_1$ over area $A_1$, and a large piston of area $A_2$ pushes back with:

$$\frac{F_1}{A_1} = \frac{F_2}{A_2} \quad\Rightarrow\quad F_2 = F_1\frac{A_2}{A_1}$$

A 100-to-1 area ratio turns your 200 N push into 20,000 N of lifting force — the hydraulic car lift. The catch: the small piston must travel 100 times farther, so no energy is created. Force is multiplied; work is conserved.

Measuring pressure: manometers and barometers

The oldest pressure instruments are fluid columns. A U-tube manometer connects the unknown pressure to a visible liquid column; the height difference $h$ gives the pressure difference directly as $\Delta p = \rho g h$. A mercury barometer is a manometer with a vacuum above the column — standard atmospheric pressure supports $760\ \mathrm{mm}$ of mercury, defining $101.325\ \mathrm{kPa}$.

Modern gauges often use a flexible diaphragm or electronic sensor instead of liquid, but every one of them is ultimately calibrated against the same $\rho g h$ physics.

Why dams are thicker at the bottom

Hydrostatic pressure grows linearly with depth, so the average pressure on a vertical wall of height $H$ is $\tfrac{1}{2}\rho g H$ — and the total force grows as $H^2$. Double a dam's height and the water's total push quadruples.

That $H^2$ scaling is why dams, aquarium walls and ship hulls are built thicker (or more strongly braced) toward the bottom. The water at the base is not just deeper — it is disproportionately more forceful, and the structure must be designed for the bottom, not the average.

Variables and units

Common mistakes

Real-world applications

• Hydraulic car lifts multiply a small force into tonnes using Pascal's principle.

• Dams and aquarium walls are built thicker at the bottom, where hydrostatic pressure is greatest.

• Barometers forecast weather: steadily falling pressure signals an approaching storm system.