Hydraulic Lift: Pascal's Principle

An educational simulation game. Predict the push that lifts a car on a hydraulic lift, or how far the small piston must travel, then push and watch both pistons.

Pascal's principleHydraulics

WORKSHOP LIFT

Lift the car

Ready
Enable JavaScript to work the lift. The instructions and worked example remain readable.
OilPistonsYour predictionTarget height
Pressure in the oilF₁ ÷ A₁ = F₂ ÷ A₂

Time

0.00 s

t

Push

— N

F₁

Oil pressure

— kPa

p = F₁ ÷ A₁

Lift

— kN

F₂ = p·A₂

Load raised

— cm

h₂

Predict before you push

Type your prediction to set up the lift. Push stays off until there is an answer.

Predict what the oil will pass on.

Mission score—

A push or a piston must lift the load with at most 2% to spare, a stroke must be within 2%, and the strokes must be the fewest that do it: each earns 100 points. Getting it right on the first run adds 50. A run counts as soon as it starts.

✓ Right answer① First try

How to play Hydraulic Lift

  1. Choose a job and read the load, the pistons, and the push or the rise on the card.
  2. The pressure is the same in all the oil: F₁ ÷ A₁ = F₂ ÷ A₂. A diameter 10 times bigger is an area 100 times bigger.
  3. For distances, the oil pushed in fills the other cylinder: A₁·d₁ = A₂·h₂. Type your prediction and press Push.
  4. Watch the bars: the lift grows toward the weight, or the two volumes of oil stay equal.

How does a hydraulic lift multiply force?

Squeeze a closed liquid anywhere and the extra pressure reaches every part of it equally: that is Pascal's principle. In a hydraulic lift, a push F₁ on a small piston of area A₁ makes a pressure p = F₁ ÷ A₁ in the oil. The large piston, of area A₂, feels that same pressure over its whole area, so it pushes up with F₂ = p·A₂. With pistons of 10 cm² and 1000 cm², a push of 118 N lifts a 1.2 t car. Areas grow with the square of the diameter, so a piston 10 times wider gives 100 times the force. There's a price. Oil doesn't squeeze, so the oil pushed out of the small cylinder must fill the large one: A₁·d₁ = A₂·d₂. To raise the car 1 cm, the small piston must travel 100 cm, and the work in, F₁·d₁, equals the work out, F₂·d₂. That is why a car jack needs many short strokes, the same trade as a lever's force for distance. Brakes, diggers, and barber chairs all work this way.

A focused learning model

The oil doesn't compress and has no weight of its own to speak of; the pistons slide without friction and the oil never leaks. The pressure is the same everywhere in it at once. The push grows steadily from zero for 2 s; the load lifts the moment p·A₂ reaches its weight (within 0.05% still lifts), and the small piston then travels 30 cm in 1 s. A set distance, or a number of 10 cm jack strokes, is made steadily over 2 s. Widths grow with the square root of the area and are not to scale; the pistons' moves are. A push or a piston size is right if it lifts the load with at most 2% to spare, a distance within 2%, and a number of strokes if it is the fewest that do the job. g = 9.8 m/s². Plays in real time.

Changing language keeps your job, prediction, attempts, and run.