Pascal's Principle

Equivalent forms

F_2 = F_1 \frac{A_2}{A_1}

\Delta P = \rho g \Delta h

A ratio of areas becomes a mechanical advantage — a fluid lever.

Symbol	Name	SI unit	Dimension	Range
$F_1$	Input Force Force on the small piston	N	M·L·T⁻²	0 – 1000
$A_1$	Input Area Area of the small piston	m²	L²	0.0001 – 0.1
$A_2$	Output Area Area of the large piston	m²	L²	0.0001 – 1
$F_2$	Output Forceoutput Force on the large piston	N	M·L·T⁻²	0 – 100000

Unit systems

Symbol	SI	CGS	Imperial
$F_1$	N	dyn	lbf
$A_1$	m²	cm²	ft²
$A_2$	m²	cm²	ft²
$F_2$	N	dyn	lbf

Where it holds

Static, incompressible, confined fluid with no external acceleration. Excludes gases under significant compression, free-surface flows, and dynamic regimes where Bernoulli applies.

Dimensional analysis

[F_1]/[A_ $1] = (M\cdot L\cdot T^{-2})/(L^{2}) = M\cdot L^{-1}\cdot T^{-2} \checkmark$ (pressure, same on both sides)

Discovery

Blaise Pascal · 1653

Pascal articulated the principle in his Treatise on the Equilibrium of Liquids, demonstrating it with his famous barrel-burst experiment where a thin vertical tube of water shattered a sealed barrel.

Try this

How can a small car jack lift a 2-ton vehicle with just hand pressure?

A hydraulic press has a small piston of area 0.001 m² and a large piston of area 0.05 m². If you push down with 50 N on the small piston, how much force is exerted on the large one?

Research status: stable

Real-world applications

Hydraulic car lifts and jacks
Disc brakes and clutch systems
Hydraulic presses for metal forming
Backhoes, excavators, and aircraft control surfaces

Common misconceptions

Pascal's principle is about confined fluids — it doesn't apply to open flow
No 'free energy' — you trade distance for force $(F_{1} \times d_{1} = F_{2} \times d_{2})$
It assumes incompressibility — gases violate this strongly

Experimental verification

Bramah's hydraulic press (1795) demonstrated the principle industrially. Every hydraulic brake, jack, and excavator in use today is a continuous verification. The principle has been tested to ppm-level accuracy in pressure standards.

Derivation

For a confined incompressible fluid at static equilibrium, the pressure at any point is uniform (ignoring gravity over small heights).

Therefore

P_{1} = P_{2}

, so

F_{1}

/

A_{1} = F_{2}

/

A_{2}

.

Energy conservation requires

F_{1}\cdot d_{1} = F_{2}\cdot d_{2}

, so the smaller piston must travel a longer distance — the fluid lever trades displacement for force.

Limiting cases

A_

2 = A_1

⟶

F_2 = F_1

Equal pistons give no mechanical advantage.

A_2 ≫ A_1⟶ F_2 ≫ F_1Large area ratio gives huge force amplification (but smaller displacement).

Compressible fluid⟶ Pressure not fully transmittedGases compress, so part of the input force is absorbed; Pascal's principle is exact only for incompressible fluids.

What if…

What if you make the output piston

100\times

larger?

Force multiplies by 100 — but to lift the output piston 1 cm, the input piston must travel 100 cm.

What if the fluid is a gas?

Some input work goes into compressing the gas, not pushing the output piston. Efficiency drops dramatically.

What if there's a leak?

Pressure cannot build up; the output force drops to whatever balances the leak rate.

1

Hydraulic car jack

Given ·

F 1:: 50
A 1:: 0.001
A 2:: 0.05

Find · F_2

Steps

$F_2 = F_1 \times A_2$ /A_1
Area ratio $= 0.05/0.001 = 50$
$F_2 = 50 \times 50 = 2500\,\mathrm{N}$ (enough to lift 250 kg)

Result ·

F_2 = F_1 \times A_2

/A_

1 = 50 \times 50 = 2500\,\mathrm{N}

2

Brake pedal to caliper

Given ·

F 1:: 200
A 1:: 0.0002
A 2:: 0.002

Find · F_2

Steps

Area ratio $= 0.002/0.0002 = 10$
$F_2 = 200 \times 10 = 2000\,\mathrm{N}$ per wheel
$10\times$ amplification of pedal force

Result ·

F_2 = 200 \times 10 = 2000\,\mathrm{N}

Hydrostatic Pressure→archimedes principleBernoulli's Equation→