Pascal's principle – hydraulic jack, press and brakes
PhysicsForces & DynamicsAges 15–16Virtual lab
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A force F₁ on the small piston raises the pressure of the liquid by p = F₁/A₁ everywhere, so the large piston pushes with F₂ = F₁·A₂/A₁. Change the piston diameters, the applied force and the car's mass to read the pressure in pascals, the force multiplication and the piston travel d₂ = d₁·A₁/A₂, and see that work in equals work out.
Lesson: Fluid pressure; Pascal's principle and hydraulic machines
What it shows
Pascal's principle states that a change of pressure applied to an enclosed liquid is passed on undiminished to every point of the liquid. In a hydraulic machine a small force F₁ on a small piston of area A₁ creates the pressure p = F₁/A₁, which pushes the large piston of area A₂ with F₂ = pA₂ = F₁·A₂/A₁. The liquid is practically incompressible, so the volume leaving the small cylinder enters the large one: A₁d₁ = A₂d₂. The large piston therefore moves less, and for an ideal machine F₁d₁ = F₂d₂. The model ignores friction, leaks, piston weights and the height difference of the liquid.
How to use
Choose Car jack, Press or Brake at the top. Set Small piston diameter D₁, Large piston diameter D₂, Force F₁ and Stroke d₁, then press Pump (Press pedal for the brake) and watch both gauges show the same pressure. Press Record to add a row to the table and compare F₂/F₁ with A₂/A₁ on the graph. Reset opens the release valve.
Parameters you can change
- Device Car jack, Hydraulic press, Hydraulic brake
- Small piston diameter D₁ 1–10 cm
- Large piston diameter D₂ 4–40 cm
- Force on the small piston F₁ 10–500 N
- Car mass 100–3000 kg
- Stroke of the small piston d₁ 2–20 cm
Questions to explore
- Why do both pressure gauges show the same reading while the small piston is pushed?
- How many strokes does the jack need to raise the car by 1 cm with the starting pistons?
- If the force is multiplied by 64, what happens to the distance the large piston moves?
