Fluid pressure and Pascal's law

Step into a swimming pool and your ears feel it before your eyes do — the water pressing in harder the deeper you go. That squeeze is fluid pressure, and it obeys a beautifully simple rule: it grows steadily with depth, and it does not care what shape the container is. A second, equally simple rule governs what happens when you push on an enclosed fluid from outside — a car mechanic's hydraulic jack, a dentist's chair, and the brakes in every car all lean on it. This article builds both ideas from the definition of pressure as force per area, derives the depth formula $P = P_0 + \rho g h$ from first principles, and uses Pascal's law to explain how a small hand-force on a narrow piston lifts a two-tonne car on a wide one.

The core insight

Pressure in a static fluid increases linearly with depth as $P = P_0 + \rho g h$ , and by Pascal's law any extra pressure applied anywhere in an enclosed fluid is transmitted undiminished everywhere else — which is why a small force on a small piston produces a large force on a large piston.

The core idea

Pressure is defined as the normal force acting per unit area:

P = \frac{F}{A}

where: $F$ is the force acting perpendicular to a surface and $A$ is the area over which it acts, giving units of pascal ( $1\ \text{Pa} = 1\ \text{N/m}^2$ ). Pressure itself has no direction — at any point inside a fluid at rest, it pushes equally in every direction. It is a scalar; only the force it produces on a specific surface has a direction, namely normal to that surface.

Pressure builds with depth. Imagine slicing a still fluid into thin horizontal layers. Each layer must hold up the weight of every layer sitting above it, so a layer deeper down carries a bigger load and is under greater pressure. This single idea — weight stacking up with depth — is the entire physical content of the fluid-column formula derived below. It also explains why pressure at a given depth is the same everywhere in a connected fluid regardless of the container's shape: two points at equal depth have equal weight of fluid above them, so they feel equal pressure, even if one sits under a narrow neck and the other under a wide basin (the "hydrostatic paradox").

Pascal's law extends this to enclosed fluids: a pressure change applied at any point of an enclosed, incompressible fluid is transmitted undiminished to every other point of the fluid and to the walls of the vessel. Push harder anywhere, and the whole fluid — and everything it touches — feels that same extra push.

Figure

A U shaped tube with two pistons of different area connected by fluid, a small force pushing down the narrow piston and a large force pushing up the wide piston, labelled with the equal pressure relation

Why fluid pressure grows with depth

The formula for pressure at depth is not a rule to memorise — it comes directly from balancing the weight of the fluid against the pressure holding it up, and this is worth deriving once carefully so the result feels earned rather than handed down.

Derivation APressure at a depth in a fluid, and the piston relation that follows from it

The goal here is to build the depth formula $P = P_0 + \rho g h$ from a simple force balance on a slab of fluid, and then to show why Pascal's law forces the same pressure onto two connected pistons of different area — together these two results are the working machinery behind everything from a swimming pool's water pressure to a hydraulic car lift.

Step 1. Picture a thin horizontal slab of fluid at depth $h$ below the free surface, with cross-sectional area $A$ and thickness $dh$ , sitting inside a larger body of fluid of density $\rho$ . This slab is in equilibrium: it is not accelerating, so the net force on it is zero.

Step 2. List the vertical forces on the slab. Pushing down from above is the pressure $P_0$ at the surface acting over area $A$ , giving force $P_0 A$ , plus the slab's own weight $\rho A\,dh\,g$ (density times volume times $g$ ). Pushing up from below is the pressure $P$ at depth $h$ acting over the same area $A$ , giving force $PA$ . Since the slab does not move, these balance: $PA = P_0 A + \rho A\,dh\,g$ .

Step 3. Divide through by the common area $A$ : $P = P_0 + \rho g\,dh$ . Extending this slab-by-slab summation from the surface down to depth $h$ (each layer adding its own weight-per-area) gives the total pressure at depth $h$ :

P = P_0 + \rho g h

where: $P_0$ is the pressure at the free surface (often atmospheric pressure, $1.013\times10^5\ \text{Pa}$ ), $\rho$ is the fluid's density, $g$ is the acceleration due to gravity, and $h$ is the depth below the surface. Notice that $A$ cancelled out entirely — the result depends only on depth, density and $g$ , never on the container's shape or cross-section, exactly as the "why it works" argument above predicted.

Step 4. Now apply Pascal's law to two connected pistons. Because a pressure change is transmitted undiminished through an enclosed fluid, the pressure at the small piston (area $A_1$ , force $F_1$ ) must equal the pressure at the large piston (area $A_2$ , force $F_2$ ):

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

Since $A_2 > A_1$ in a hydraulic lift, $F_2$ is greater than $F_1$ — a small input force becomes a large output force. Energy is still conserved: the large piston moves a correspondingly smaller distance, so the work done, $F_1 d_1 = F_2 d_2$ , balances on both sides.

Worked example

A hydraulic car lift has a small piston of area $A_1 = 5.0\times10^{-3}\ \text{m}^2$ and a large piston of area $A_2 = 0.25\ \text{m}^2$ , and must support a car of mass $1200\ \text{kg}$ ( $g = 9.8\ \text{m/s}^2$ ). Find the force needed on the small piston, and compare it with the pressure at the bottom of a $3.0\ \text{m}$ deep swimming pool of water ( $\rho = 1000\ \text{kg/m}^3$ , $P_0 = 1.013\times10^5\ \text{Pa}$ ).

Step 1. Find the weight to be supported by the large piston: $F_2 = mg = 1200 \times 9.8 = 11760\ \text{N}$ .

Step 2. Apply the piston relation from the derivation, solved for $F_1$ : $F_1 = F_2\,\dfrac{A_1}{A_2} = 11760 \times \dfrac{5.0\times10^{-3}}{0.25} = 11760 \times 0.02 = 235.2\ \text{N}$ . A modest push of about $235\ \text{N}$ — roughly the weight of a $24\ \text{kg}$ mass — is enough to lift the $1200\ \text{kg}$ car, because the area ratio ( $50\times$ ) is exactly the mechanical advantage.

Step 3. For comparison, find the pressure at the bottom of the pool using $P = P_0 + \rho g h$ : $P = 1.013\times10^5 + (1000)(9.8)(3.0) = 1.013\times10^5 + 2.94\times10^4 = 1.307\times10^5\ \text{Pa}$ .

The small piston needs only $235.2\ \text{N}$ to lift a $1200\ \text{kg}$ car, while the total pressure at the bottom of the pool is $1.31\times10^5\ \text{Pa}$ , of which $2.94\times10^4\ \text{Pa}$ is the gauge pressure due to the water column alone.

Common traps

Thinking pressure at a point depends on the container's shape or the total volume of fluid — for a static, connected fluid it depends only on depth, density and $g$ .
Forgetting atmospheric pressure $P_0$ when a question asks for total (absolute) pressure rather than gauge pressure ( $\rho g h$ alone).
Believing a hydraulic lift creates energy for free — it multiplies force, not work; the large piston moves a proportionally smaller distance.
Treating pressure as a vector with one fixed direction — pressure itself is a scalar; only the force it produces on a chosen surface has a direction, normal to that surface.

Ask the tutor: Want to see how the same $P = P_0 + \rho g h$ formula explains why your ears pop when you dive, or work through another hydraulic-lift numerical step by step?

Why does pressure not depend on the shape of the container?

Because the derivation only ever used the weight of the fluid directly above a point and the area it acts over, and that area cancelled out of the final formula. Two points at the same depth always have the same weight of fluid pressing down per unit area, whether that fluid sits in a narrow tube or a wide tank above them. This is exactly why connected vessels of very different shapes, filled to the same height, show identical pressure at the base — the classic "hydrostatic paradox."

Does the hydraulic lift break the rule that you cannot get something for nothing?

No. Pascal's law multiplies force, but work (force times distance) is conserved: the small piston must be pushed down a much larger distance than the large piston rises, in exactly the same ratio as the area ratio, so $F_1 d_1 = F_2 d_2$ . You trade distance for force, the same bargain every simple machine makes — nothing is created.

Is gauge pressure the same as the pressure I should use in P equals rho g h?

$\rho g h$ by itself is the gauge pressure — the extra pressure due only to the fluid column, ignoring the atmosphere pushing down on the surface. The total, or absolute, pressure is $P = P_0 + \rho g h$ , adding atmospheric pressure back in. Whether a question wants gauge or absolute pressure depends on its wording, so always check which one is being asked for before reporting a final answer.

Fluid pressure and Pascal's law

The core idea

Why fluid pressure grows with depth

Worked example

Common traps

Before this

Where this leads