Pascal's law: hydraulic lift, press and brakes
Pressure applied to an enclosed liquid is passed on undiminished to every point. So a small force on a small piston gives a large force on a large piston.
What Pascal's law says
Poke a few holes in a plastic bottle full of water and close its cap. Squeeze it anywhere, and water spurts harder from every hole. Liquids are almost incompressible: their volume hardly changes when pressed. So pressure added at one point travels through the whole liquid. Pascal's law: pressure applied to any part of an enclosed liquid is passed on undiminished (without loss). It reaches every point of the liquid and the walls of its container. Machines that use liquid pressure in this way are called hydraulic machines.
How a hydraulic lift multiplies force
Two cylinders filled with oil are joined at the bottom. The narrow one has a piston of area A₁; the wide one has a piston of area A₂.
- 1Push the small piston with force F₁. The pressure on the oil is P = F₁/A₁.
- 2By Pascal's law, the same pressure P acts on the large piston.
- 3Upward force on the large piston: F₂ = P × A₂ = F₁ × A₂/A₁.
- 4A₂ is larger than A₁, so F₂ is larger than F₁ by the ratio A₂/A₁.
- 5The oil pushed out of the small cylinder fills the large one: A₁d₁ = A₂d₂. So the small piston moves much farther than the large one.
- 6★So, a hydraulic lift gives a large force for a small force, but not more work: F₁d₁ = F₂d₂ in an ideal machine.
★ key step
How to draw a hydraulic lift
- 1Draw a U-shaped vessel with a narrow limb on the left and a wide limb on the right. Shade it and label it oil (liquid).
- 2Draw a small piston in the narrow limb. Label its area A₁ and draw a downward arrow F₁ on it.
- 3Draw a large piston in the wide limb with a platform on top. Draw a car or a heavy load on the platform.
- 4Label the large piston's area A₂ and draw an upward arrow F₂ under the platform.
- 5Write beside the figure: F₂ = F₁ × A₂/A₁.
Uses of Pascal's law
| Device | Where you see it | How Pascal's law is used |
|---|---|---|
| Hydraulic lift or jack | Car and bus service stations; a jack to change a truck tyre | A small push on a small piston raises a car resting on a large piston |
| Hydraulic press | Packing cotton bales, squeezing oil from seeds, shaping metal sheets | The large force from the big piston squeezes the material between two plates |
| Hydraulic brakes | Cars, buses and trucks | The pedal pushes a piston in the master cylinder. Brake oil carries the pressure to pistons at all four wheels, which press the brake shoes against the drums together |
Worked example: holding up a car
A hydraulic jack has a small piston of area 25 cm² and a large piston of area 0.50 m². What force on the small piston will hold up a car of mass 1000 kg? (g = 9.8 m s⁻²)
- Weight of the car: F₂ = mg = 1000 kg × 9.8 m s⁻² = 9800 N.
- A₁ = 25 cm² = 25 × 10⁻⁴ m² = 2.5 × 10⁻³ m²; A₂ = 0.50 m².
- Area ratio A₂/A₁ = 0.50 ÷ 2.5 × 10⁻³ = 200.
- F₁ = F₂ × A₁/A₂ = 9800 N ÷ 200 = 49 N.
49 N, about the weight of a 5 kg bag. To raise the car by 10 cm, the small piston must move 200 × 10 cm = 20 m in all, over many strokes
Worked example: pistons given by radius
In a car's brakes, the master piston has radius 1.0 cm and each wheel piston has radius 3.0 cm. The pedal pushes the master piston with a force of 100 N. Find the force on each wheel piston.
- Area = πr², so A₂/A₁ = (r₂/r₁)² = (3.0 ÷ 1.0)² = 9.
- F₂ = F₁ × A₂/A₁ = 100 N × 9 = 900 N.
900 N on each wheel piston. When radii or diameters are given, square their ratio
At a car service station, a mechanic raises a whole car on a hydraulic lift using oil pressure from a small pump. In a bus, brake oil carries one light press of the pedal to all four wheels at once.
3 marks: “State Pascal's law. Explain, with a labelled diagram, how a hydraulic lift raises a heavy load with a small force.” Statement: pressure on an enclosed liquid is transmitted undiminished to all points. Diagram: U-tube with oil, pistons A₁ and A₂. P = F₁/A₁ reaches the large piston, so F₂ = F₁A₂/A₁. A large A₂ gives a large F₂.
2 marks: “The pistons of a hydraulic press have areas in the ratio 1 : 50. What force on the small piston can lift a load of 2000 N?” F₁ = F₂ × A₁/A₂ = 2000 N ÷ 50 = 40 N.
2 marks: “Does a hydraulic lift give more work out than you put in? Explain.” No. The force is multiplied by A₂/A₁, but the large piston moves less, since A₁d₁ = A₂d₂. So F₁d₁ = F₂d₂: work out equals work in, at best.
Pascal passes pressure: same P, bigger A, bigger F.