University Physics I · Optional: Fluids · 15.4
Pascal's Principle & Hydraulics
Squeeze an enclosed fluid and every point gains the same rise in pressure. Two pistons of different area then feel forces in the ratio of their areas, and the small one travels further by exactly that ratio.
Build the model
Connect the measurement to the mechanism.
Pascal's principle follows from two facts already in hand: a confined liquid barely changes volume, and hydrostatic pressure differences depend only on density and height. Raise the pressure anywhere in an enclosed fluid at rest and every point rises by the same ΔP, because no height term has changed. A hydraulic system converts that transmitted pressure into force by collecting it over area, F = PA, so a piston a hundred times larger returns a hundred times the force.
Nothing is created. The same incompressibility that carries the pressure also conserves volume, so the large piston moves one hundredth as far and F₁d₁ = F₂d₂ exactly. Hydraulics trades distance for force; the price of a large force is a long, slow stroke, plus whatever real oil, hoses, seals, and trapped air take on top.
- Simple definition
- Pascal's principle: a pressure increase applied to an enclosed fluid at rest reaches every point of that fluid, and every wall confining it, at full strength.
- Example
- Pressing a car's brake pedal raises the pressure in the master cylinder by several megapascals, and every caliper fed by that circuit sees the same rise within milliseconds.
A piston's force spread over its face sets the pressure it adds to the trapped fluid.
Pa = N m⁻²; use gauge pressure on both pistons
Same pressure, more area, more force: the ratio of piston areas is the force multiplier.
Areas in m²; for round pistons A₂/A₁ = (d₂/d₁)²
The oil leaving one cylinder must fill the other, so the wider piston moves the shorter distance.
d is piston displacement in m; incompressible fluid, no leaks
Force up and distance down by the same factor leaves the energy you supplied unchanged.
Ideal case: no friction, no leaks, both pistons at the same height
A lower output cylinder carries the weight of the oil column above it as extra pressure.
y measured upward; matters when ρgΔh is a real fraction of the working pressure
Pressure times volume flow is the power delivered, equal to force times piston speed.
Q in m³ s⁻¹; 1 Pa × 1 m³ s⁻¹ = 1 W
The change is what travels
In a fluid at rest the pressure difference between two points depends only on their height difference: P₁ − P₂ = ρg(y₂ − y₁). Nothing in that relation knows about a piston. So if you push on one boundary and raise the pressure there by ΔP, every other point must rise by the same ΔP, or the height relation would no longer hold. Pascal's principle is therefore a corollary of hydrostatics plus incompressibility, not a separate law. Notice exactly what is transmitted: the increment. The bottom of a hydraulic reservoir still sits at higher absolute pressure than the top, by ρgh, before the pump runs and after it.
Area turns pressure into force
Pressure acts normally on every surface it touches, so a piston of area A held against gauge pressure P feels F = PA. Two pistons in the same enclosed oil at the same height share one pressure, and their forces differ purely by area: F₂/F₁ = A₂/A₁. A garage lift with a 25 cm output piston has A₂ = π(0.125 m)² = 4.91 × 10⁻² m². Holding a 1400 kg car, weight 1.37 × 10⁴ N with g = 9.81 m s⁻², needs a gauge pressure of 13 734/0.0491 = 2.80 × 10⁵ Pa, about 2.8 bar. Feed that same pressure to a 2.5 cm input piston of area 4.91 × 10⁻⁴ m² and only 137 N is needed. A diameter ratio of 10 gives an area ratio of 100, and so a force ratio of 100.
The bill arrives as distance
Oil leaving the small cylinder has to appear in the large one, so A₁d₁ = A₂d₂. The same factor of 100 that multiplied the force divides the displacement. Raising that car by 1.8 m sweeps 0.0491 × 1.8 = 0.0884 m³, about 88 litres, which the small piston must supply over a stroke of 180 m — some 600 pump strokes of 0.30 m each. Multiply out both sides: 137.3 N × 180 m = 24.7 kJ in, and 13 734 N × 1.8 m = 24.7 kJ out. A hydraulic system is a force amplifier and an energy conduit. Any claim of amplified work is an arithmetic error.
Power decides how fast
Divide the work equality by time and it becomes a power equality: F₁v₁ = F₂v₂. Since v = Q/A at each piston, both sides equal ΔP × Q. The area ratio fixes the force; the pump's volume flow fixes the speed. A machine at 20 MPa fed by a 12 L min⁻¹ pump — 2.0 × 10⁻⁴ m³ s⁻¹ — delivers 2.0 × 10⁻⁴ × 20 × 10⁶ = 4.0 kW to the load, whatever the cylinder areas are. That is why a hand jack replaces one impossible 180 m stroke with hundreds of short ones, check valves holding the gain between strokes, and why an industrial press is specified by a pressure and a flow rate together.
Where the ideal model gives out
Four assumptions carry everything above: incompressible fluid, rigid walls, no leaks, pistons at equal height. Hydraulic oil has a bulk modulus near 1.5 GPa, so at 20 MPa a fraction ΔV/V = ΔP/B ≈ 1.3% of the fluid volume disappears into compression. Trapped air is far worse: a 1.0 cm³ bubble at 100 kPa, squeezed slowly to 5 MPa, shrinks to about 0.02 cm³ and swallows 0.98 cm³ of piston sweep — through a 25 mm master cylinder of area 4.91 cm² that is 2.0 mm of dead piston travel, and several times that at the pedal pad once the lever ratio is counted: the spongy feel bleeding cures. Hoses swell, seals rub, and a height offset adds ρgΔh: 1.2 m of oil at 870 kg m⁻³ is 10.2 kPa, negligible beside 20 MPa but 3.7% of the 280 kPa lift above.
Change one variable at a time
Make the relationship visible.
Widen the output piston: its up-arrow grows as the square of the diameter ratio while its rise shrinks by that same square — and the work readout does not move.
AREA RATIO A₂/A₁4.00 ×
OUTPUT FORCE F₂800 N
OUTPUT RISE d₂0.087 m
WORK, EACH SIDE70.0 J
Live interpretationAREA RATIO A₂/A₁: 4.00 ×. OUTPUT FORCE F₂: 800 N. OUTPUT RISE d₂: 0.087 m. WORK, EACH SIDE: 70.0 J
Catch the common trap
Explain before calculating.
A hydraulic lift has A₂/A₁ = 25. The small piston is pushed down 0.50 m by a steady 300 N. With no friction, no leaks, and both pistons at the same height, what does the large piston do?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA bottle jack has a pump piston 12.0 mm in diameter and a ram 45.0 mm in diameter, joined by oil in a sealed circuit at the same height. The handle drives the pump piston with a force of 140 N. Find the pressure increment in the oil and the force at the ram.
- Areas from the diameters: A₁ = π(6.00 × 10⁻³ m)² = 1.131 × 10⁻⁴ m², A₂ = π(22.5 × 10⁻³ m)² = 1.590 × 10⁻³ m².
- The pump piston sets the pressure it adds: ΔP = F₁/A₁ = 140 N ÷ 1.131 × 10⁻⁴ m² = 1.238 × 10⁶ Pa = 1.24 MPa.
- Pascal's principle carries that whole increment to the ram face, so F₂ = ΔP A₂ = 1.238 × 10⁶ Pa × 1.590 × 10⁻³ m² = 1.97 × 10³ N.
- Ratio check, no pressure needed: A₂/A₁ = (45.0/12.0)² = 14.06, and 14.06 × 140 N = 1.97 × 10³ N.
AnswerΔP = 1.24 MPa; F₂ = 1.97 kN — a 14.1× force gain from a 3.75× diameter ratio.
MediumThe same bottle jack raises that 1.97 kN load by 90.0 mm. How far must the pump piston travel in total, how many 20.0 mm strokes of the handle is that, and what is the work done at each end?
- The oil is incompressible, so the swept volumes match: A₁d₁ = A₂d₂, giving d₁ = d₂(A₂/A₁) = 0.0900 m × 14.06 = 1.266 m.
- Handle strokes: 1.266 m ÷ 0.0200 m = 63.3, so 64 strokes with the last one partial.
- Work in: W = F₁d₁ = 140 N × 1.266 m = 177 J.
- Work out: W = F₂d₂ = 1.969 × 10³ N × 0.0900 m = 177 J. The factor 14.06 that multiplied the force divided the distance, and the product came through untouched.
Answerd₁ = 1.27 m, about 64 strokes of 20.0 mm; 177 J in and 177 J out.
HardA press cylinder of diameter 220 mm sits 1.50 m below the pump outlet, where a gauge reads 2.400 × 10⁵ Pa. The oil has density 860 kg m⁻³. Find the force the cylinder can hold, and the percentage error made by ignoring the oil column. Take g = 9.81 m s⁻².
- Piston area: A = π(0.110 m)² = 3.801 × 10⁻² m².
- The cylinder is lower, so it carries the column above it as well: P₂ = P₁ + ρgΔh = 2.400 × 10⁵ Pa + (860 kg m⁻³)(9.81 m s⁻²)(1.50 m) = 2.400 × 10⁵ + 1.265 × 10⁴ = 2.527 × 10⁵ Pa.
- F = P₂A = 2.527 × 10⁵ Pa × 3.801 × 10⁻² m² = 9.604 × 10³ N.
- Dropping the height term: F = 2.400 × 10⁵ Pa × 3.801 × 10⁻² m² = 9.123 × 10³ N.
- Error = (9604 − 9123)/9604 = 5.01%. Pascal's principle still transmits the pump's increment undiminished — the depth term ρgΔh simply sits underneath it, exactly as it did before the pump ran.
AnswerF = 9.60 kN; ignoring the 12.7 kPa oil column understates it by 5.0%, or 481 N.