Pascal's Law and Hydraulics - Physics for Pilot Assessments

A hydraulic jack lets you lift a car with one hand. It is not a trick and it is not free: it follows from two facts about liquids, that they barely compress and that they pass a pressure on undiminished. By the end of this lesson you should be able to work out the output force of a hydraulic system from its two piston areas, run that calculation backwards to size a piston for a force you need, and say exactly what you give up in exchange for the multiplication.

Liquids barely compress

Squeeze a syringe full of air with your finger over the nozzle and the plunger moves a long way; do the same with a syringe full of water and it hardly moves at all. The reason is the particle arrangement from the first lesson of this module. In a gas the particles are far apart with a great deal of empty space between them, so there is something to remove, and a modest push halves the volume without difficulty. In a liquid the particles are already touching, so compressing it means pushing them into one another against a very steep resistance — water at the pressure a kilometre of ocean produces shrinks by roughly half a per cent. For every purpose in this course a liquid is incompressible, which has a consequence worth stating on its own: whatever volume you push in at one end must appear at the other, and it must appear immediately.

Pascal's law

Now enclose that liquid — fill a sealed vessel completely, with no air pockets anywhere in it — and push on it somewhere with a piston. Because the liquid cannot be squashed into a smaller space and cannot escape, it has no option but to pass the push on; and because a liquid presses perpendicular to whatever surface it meets rather than along the line you pushed, it passes that push on in every direction. The result is Pascal's law: a pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid and to the walls of its container. Undiminished is the word that matters. The pressure does not weaken with distance, or around bends, or on its way through a narrow connecting pipe — the far end of a ten-metre brake line sees the same pressure as the cylinder your foot is pressing on.

Two honest qualifications. The liquid's own weight still contributes its ρgh, so a point low in the system sits at a slightly higher pressure than a point at the top; in a jack the applied pressures are so much larger than that contribution that it is ignored, but it is genuinely there. And real systems lose a little pressure to friction while the fluid is actually moving along pipes. Pascal's law describes what is transmitted when the fluid is standing still and being pressed, which is exactly the condition a jack or a brake operates in.

Multiplying a force

Here is where it becomes useful. Pressure is force divided by area, so a force applied to a small piston produces a certain pressure in the fluid. That pressure is now present everywhere in the fluid, including underneath a second and much larger piston. And force = pressure × area, so at the larger piston the same pressure produces a proportionally larger force. Nothing has been created: the pressure is identical at both ends, exactly as Pascal's law says it must be. It is the area that differs, and the force is that one pressure multiplied by whatever area you give it. Ten times the area, ten times the force.

Force multiplication in a hydraulic system

F_2 = F_1 \times \frac{A_2}{A_1}

$F_1$: the input force pushing on the small piston, in newtons (N)
$A_1$: the area of the small piston, in square metres (m²)
$F_2$: the output force produced at the large piston, in newtons (N)
$A_2$: the area of the large piston, in square metres (m²)

That expression is only the two steps run together, and in a calculation it is usually clearer to keep them apart: find the pressure from the input piston first, then apply that pressure to the output piston. What the formula makes obvious is that the ratio of the areas is the whole story, so a system's multiplication is fixed by geometry alone. Since pistons are normally circular, and the area of a circle grows with the square of its diameter, doubling a piston's diameter quadruples its area and therefore quadruples its force — which is a favourite trap, because the tempting answer is that it doubles.

One pressure throughout the fluid. Both pistons are circular, and the wide one is drawn just over three times as wide, so its face has ten times the area — and the same pressure on it produces ten times the force.

Example 1 — finding the output force

You push on a small piston of area 0.01 square metres with a force of 100 newtons. The output piston has an area of 0.05 square metres. What force does it produce?

Find the pressure in the fluid from the input side: $\text{pressure} = \frac{\text{force}}{\text{area}} = \frac{100}{0.01}$
Dividing by 0.01 is the same as multiplying by 100, so the pressure is 10 000 pascals.
That pressure is now everywhere in the fluid, including under the output piston, so use force = pressure × area: $F_2 = 10\,000 \times 0.05 = 500\ \text{N}$
Check it with the area ratio instead: $\frac{A_2}{A_1} = \frac{0.05}{0.01} = 5$ So the output piston has five times the area, and five times 100 newtons is 500 newtons. The same answer, as it has to be.

Example 3 — pistons quoted by diameter

A hydraulic system's input piston is 2 centimetres across and its output piston 6 centimetres across. You push on the input piston with 50 newtons. What force appears at the output?

The formula wants areas, and the question gives diameters. The pistons are circular, and a circle's area grows with the square of its diameter.
The output diameter is $\frac{6}{2} = 3$ times the input diameter, so the output area is 3 × 3 = 9 times the input area. Neither area itself is ever needed — only their ratio enters the formula, and everything else cancels.
Nine times the area means nine times the force: $F_2 = 9 \times 50 = 450\ \text{N}$
The tempting wrong answer is 150 newtons, from tripling the force along with the diameter. Force follows area, and area follows the square of the diameter — three times the width is nine times the force.

What you pay for it

The force multiplication is real, and it is not free — you pay for it in distance. Because the liquid is incompressible, the volume that leaves the small cylinder must arrive in the large one, and a volume is an area multiplied by a length. Push the small piston down by some distance and the volume you displace is its area multiplied by that distance; that same volume, spreading out under a piston thirty times wider, can only raise it by a thirtieth as much. Thirty times the force, a thirtieth of the movement. That is why a bottle jack needs dozens of strokes of its handle to raise a car a few centimetres, and why the handle's travel is long while the ram's is short.

The volume moved is the same at both pistons

A_1 d_1 = A_2 d_2

$A_1$: the area of the small piston, in square metres (m²)
$d_1$: the distance the small piston moves, in metres (m)
$A_2$: the area of the large piston, in square metres (m²)
$d_2$: the distance the large piston moves, in metres (m)

Put the two relationships side by side and the trade is exact. Multiply the force by thirty and divide the distance by thirty, and the force multiplied by the distance comes out the same at both ends: 200 newtons through 0.6 metres on one side, 6000 newtons through 0.02 metres on the other. Nothing in a hydraulic system makes that product bigger; the system only trades one of its two factors for the other. Hold on to this, because it is the difference between a machine and a magic trick.

Example 4 — how far the load actually rises

In the jack from Example 2, the small piston is 0.002 square metres and the large one 0.06 square metres. You push the small piston down by 60 centimetres. How far does the load rise?

Convert to metres first: 60 centimetres is $\frac{60}{100} = 0.6$ metres.
The volume displaced by the small piston is its area × its travel: $A_1 d_1 = 0.002 \times 0.6 = 0.0012\ \text{m}^3$
All of that volume arrives under the large piston, so its $\text{travel} = \frac{\text{volume}}{\text{area}} = \frac{0.0012}{0.06}$ .
Multiply both numbers by 10 000 to clear the decimals: $d_2 = \frac{12}{600} = 0.02\ \text{m}$ That is 2 centimetres.
So 60 centimetres of handle travel gives 2 centimetres of lift — a thirtieth, matching the thirty-fold force gain exactly.

Where it turns up

Brakes are the most familiar case: your foot's force on the pedal pressurises the fluid in a master cylinder, and that pressure acts on the larger pistons at each wheel. The same two facts also explain the one hydraulic fault everybody has heard of. Air in a brake line ruins it, because air is compressible in the way the fluid is not: press the pedal and much of your travel goes into squashing the bubble smaller rather than into moving fluid along, so the pedal feels soft and long and the pressure that actually reaches the wheels is a fraction of what it should be. Bleeding the system is nothing more than getting the compressible part out so that Pascal's law has an incompressible fluid to work on. Jacks, presses, digger arms and landing-gear retraction all run on the same principle.

Why it matters in aviation

Aircraft hydraulic systems are Pascal's law doing heavy work. A pump holds the fluid at high pressure — commonly around 3000 pounds per square inch on transport aircraft — and because that pressure is transmitted undiminished, it is available at every actuator on the circuit. A modest flow of fluid moving a piston a short distance therefore becomes a large force at the far end: enough to raise the landing gear against the airflow, drive the flaps out along their tracks, clamp the wheel brakes, and on larger aircraft move the flight control surfaces, where the aerodynamic loads are far beyond what a pilot could balance by muscle alone.

Two features of the law shape how those systems are built. Since the pressure travels through fluid in pipes rather than along a mechanical linkage, one pump can operate services spread the length of the airframe and the routing can follow whatever path the structure allows. For the same reason, a leak anywhere lowers the pressure everywhere on that circuit, which is why important services are fed by more than one independent system. And the incompressibility the law depends on is why any air trapped in a line makes the affected service feel soft and slow to respond.

What the test asks

The trap is treating the force multiplication as a gain rather than a trade. An option will suggest that a hydraulic system gives you more push over the same distance, or that a small enough input piston could lift anything with no penalty at all — and the second half of that is subtly true and therefore dangerous: the force really is unlimited in principle, but so is the number of strokes you will be pumping. Any option describing an output that has been multiplied with nothing given up is wrong. The other reliable distractor concerns the pressure itself: the pressures at the two pistons are equal and it is the forces that differ, so an option claiming that the pressure is multiplied by the area ratio has swapped the two quantities over.