Lesson 5: Pascal's Principle and the Hydraulic Press

5.1 Pascal's principle

Consider a liquid in a container fitted with a piston. At a point A at depth h below the piston the pressure is P = P₁ + ρgh, where P₁ is the pressure just under the piston. If we add a weight on the piston, the pressure at the piston surface rises by ΔP. The piston does not move because the liquid is incompressible, but the pressure at every point increases by the same ΔP.

Pascal's principle (law): When pressure is applied to a liquid enclosed in a container, the pressure is transmitted in full to every part of the liquid and to the walls of the container.

(If the pressure is increased too much, the container may burst.)

5.2 The hydraulic press

• It consists of two cylinders connected by a pipe and filled with a suitable liquid (oil).
• One cylinder has a small piston of area a; the other has a large piston of area A.

A force f on the small piston creates a pressure P = f/a. By Pascal's principle this pressure is transmitted to the large piston, which is at the same level, so P = F/A also.

f / a = F / A → F = f · (A / a)

Hence a small force on the small piston produces a large force on the large piston.

Mechanical advantage of the press:

η = F / f = A / a (a ratio, no unit)

Energy conservation. If the small piston moves down a distance y₁ and the large piston moves up y₂, then (ideal press, efficiency 100 %) the work done on the small piston equals the work done by the large piston:

f y₁ = F y₂ → F/f = y₁/y₂ = A/a

So η = F/f = A/a = y₁/y₂. The large piston moves a smaller distance than the small one — we gain force but not work.

Because the area depends on the square of the radius or diameter, if the ratio of the radii is r_large : r_small = n, then η = n². (E.g. a radius ratio of 9 : 1 between the large and the small piston gives η = 81.)

5.3 Applications

1. Hydraulic brakes in cars: the pedal pushes a piston in the master cylinder; the pressure is transmitted through the brake fluid to the pistons of the wheel cylinders, which push brake shoes against the drum (rear drum brakes) or pads against the disc (front disc brakes). A small foot force produces large friction forces at all four wheels, and the pressure is the same to all wheels.
2. Hydraulic lift (car lift in a garage).
3. Hydraulic excavator and other hydraulic machines (jacks, presses, aircraft control systems).

5.4 Worked examples

Example 1. In a hydraulic press the small piston has diameter 4 cm and the large piston has diameter 40 cm. A force of 50 N is applied to the small piston. (g = 10 m/s²)

• Area ratio A/a = (40/4)² = 100, so η = 100.
• F = 50 × 100 = 5000 N; the largest mass it can lift = 5000/10 = 500 kg.
• To raise the large piston by 2 cm: y₁ = η y₂ = 100 × 2 = 200 cm.

Example 2 (car lift). A car of mass 1200 kg (weight 12 000 N) rests on a piston of radius 20 cm; the small piston has a radius of 2 cm. f = F(a/A) = 12 000 × (2/20)² = 12 000/100 = 120 N.

Example 3 (pressure). What pressure is transmitted if 120 N acts on a small piston of area 4 × 10⁻⁴ m²? P = 120/4 × 10⁻⁴ = 3 × 10⁵ N/m².

Example 4 (efficiency idea). A press has η = 40. A force of 250 N acts on the small piston, which is pushed 10 cm. F = 10 000 N; the large piston rises y₂ = 10/40 = 0.25 cm. Work input = 250 × 0.10 = 25 J; work output = 10 000 × 0.0025 = 25 J (equal).

Example 5 (given η and F). A hydraulic press lifts a 2000 N load on the large piston of area 0.05 m² with a small-piston area of 2.5 × 10⁻³ m². η = 0.05/2.5 × 10⁻³ = 20, so f = 2000/20 = 100 N.