Pressure, force & Pascal's law

The first of the three master relations. Pressure on an area makes a force — and because pressure spreads equally through trapped oil, a gentle push on a small piston can lift a car.

Source: Rabie, Fluid Power Engineering, Ch. 1.

Before you start

What you need first

What fluid power is — pressure does the work (Topic 1).
Area of a circle — $A=\tfrac{\pi}{4}D^2$ from a diameter.

What you'll be able to do

State Pascal's law and what it means.
Use $p=\tfrac{F}{A}$ and $F=pA$ both ways.
Explain the force multiplier — why a small pressure on a big area gives a big force.
Explain why the jack multiplies force but not energy — work in = work out.
Convert bar ↔ pascals safely.

Start here · the principle

Pascal's law

In a confined fluid at rest, a pressure applied at any point is transmitted equally, in all directions, to every part of the fluid.

So the pressure is the same everywhere in the trapped oil. Push on the oil here, and that same pressure pushes on every wall and piston touching it — including a big one far away.

Pressure = force ÷ area

Pressure measures how concentrated a force is — the amount of force pressing on each unit of area it acts over:

$$p = \dfrac{F}{A}$$

where:
Symbol	Meaning	SI unit
p	pressure	Pa (N/m²)
F	force on the piston	N
A	piston area	m²

Pressure $p$ acts over area $A$ to balance the force $F$.

Turn it around — this is the muscle

Rearranged, the same relation tells us the force the oil can produce:

$$F = p\,A$$

For a round piston, $A=\tfrac{\pi}{4}D^2$, so the force grows with the square of the diameter. A little more bore buys a lot more force.

The key idea · intuition

A small push, a big lift

Press on a small piston with a modest force. By Pascal's law the same pressure appears under a big piston — and pressure times the bigger area is a much bigger force.

That is the hydraulic press / jack: the pressure is equal on both sides, so the force is multiplied by the ratio of the areas.

Same pressure, bigger area → bigger force. A hand pump really can lift a car — but only because the oil barely compresses, as we see next.

The jack: equal pressure on both pistons, so $F_2/F_1 = A_2/A_1$.

The catch · conservation of energy

Force is multiplied — energy is not

The jack looks like something for nothing, but it cannot be. Oil is very nearly incompressible, so the volume pushed out of the small cylinder reappears, unchanged, under the big one — it cannot squash away or vanish:

$$A_1 d_1 = A_2 d_2$$

So the small piston must travel a long way (its stroke $d_1$) to raise the big piston only a little ($d_2$) — the distance is divided by exactly the ratio the force was multiplied. Combine $F_2/F_1 = A_2/A_1$ with $d_1/d_2 = A_2/A_1$:

$$F_1\,d_1 = F_2\,d_2$$

The work you put in — force × distance — equals the work you take out. That is conservation of energy: hydraulics multiplies force, never energy. You buy the bigger force by moving a longer stroke (many pump strokes for one slow lift).

Oil is only almost incompressible; that tiny squash is real, and we measure it later as the bulk modulus (Topic 10). Here it is far too small to matter.

A note on pressure units

The SI unit is the pascal (Pa), but it is tiny. In the workshop we talk in bar:

$$1~\text{bar} = 10^{5}~\text{Pa} = 100~\text{kPa}$$

Always convert bar → Pa before using a formula. Typical hydraulic machines run at 100–350 bar — that is $1\!-\!3.5\times10^{7}$ Pa.

✏️ Try it yourself — no numbers needed

Two cylinders are fed the same pressure. Cylinder B's piston has twice the diameter of cylinder A's. How many times more force does B give — and why?

Four times. Force is $F=pA$, and for a round piston $A=\tfrac{\pi}{4}D^2$. At the same pressure, $F\propto D^2$. Why: doubling the diameter makes the area $2^2 = 4$ times bigger, so the force is 4× bigger — not 2×. The diameter counts twice.

Common mistakes to avoid

Mistake	Fix
Leaving pressure in bar inside $F=pA$	Convert first: $1~\text{bar}=10^5~\text{Pa}$.
Putting the radius into $\tfrac{\pi}{4}D^2$	That formula uses the diameter; with radius it is $\pi r^2$.
Thinking the big piston "feels more pressure"	Pascal: pressure is equal everywhere. It is the larger area that makes the larger force.
Thinking the jack gives "free" force/energy	It only trades force for distance: $F_1d_1=F_2d_2$. The small piston moves much farther than the load rises.

Recap — the whole topic on one screen

$$p=\dfrac{F}{A}\qquad F=pA \qquad \dfrac{F_2}{F_1}=\dfrac{A_2}{A_1}$$

Idea	What you own now
Pascal's law	Pressure is equal everywhere in trapped oil
Force from pressure	$F=pA$; for a circle $F\propto D^2$
Force multiplier	Same pressure, bigger area → bigger force
No free lunch	Force ×, distance ÷; $F_1d_1=F_2d_2$ (energy conserved)
Units	Convert bar → Pa (×10⁵) before any formula

Next topic

Flow rate & piston speed

Pressure made the force. Next we ask what makes the piston move, and how fast — that is the job of flow.

→ Flow rate & piston speed

Mistake	Fix
Leaving pressure in bar inside \(F=pA\)	Convert first: \(1~\text{bar}=10^5~\text{Pa}\).
Putting the radius into \(\tfrac{\pi}{4}D^2\)	That formula uses the diameter; with radius it is \(\pi r^2\).
Thinking the big piston "feels more pressure"	Pascal: pressure is equal everywhere. It is the larger area that makes the larger force.
Thinking the jack gives "free" force/energy	It only trades force for distance: \(F_1d_1=F_2d_2\). The small piston moves much farther than the load rises.