Virtual laboratory/Physics/Hydrostatics/Pascal's law — the hydraulic press

Pascal's law — the hydraulic press

Press the small piston — the pressure is transmitted through the liquid and lifts a large load on the large piston. Because F₁/S₁=F₂/S₂: the larger the ratio of areas, the greater the force multiplication. Change the piston areas and observe.

Pascal's law — the hydraulic press
Physics · Hydrostatics · Pascal's law (FIZ.HYDRO.PP.04)
Small piston area S1S_15 cm²
Large piston area S2S_225 cm²
A hydraulic press lets a small force on the small piston:
force F₁
100 N
pressure p
20.0 N/cm²
force F₂
500 N
ratio S₂/S₁
×5.0
🎯 Challenge: hit force F₂
streak: 0

I'll give you a target force on the large piston — set the piston areas to reach it. F₂ = F₁·S₂/S₁.

What just happened?

You enlarged the large piston, and the force F₂ lifting the load grew. The liquid transmits pressure equally in every direction, so a larger area means a larger force.

Pascal's law:

p=F1S1=F2S2p = \dfrac{F_1}{S_1} = \dfrac{F_2}{S_2}

— pressure applied to an enclosed liquid is transmitted unchanged to every point. F — force (N), S — piston area (cm² or m²), p — pressure (N/cm² or Pa).

What does that mean in practice?

Since the pressure is the same on both sides, you get the force on the large piston from the ratio of areas:

F2=F1⋅S2S1F_2 = F_1\cdot\dfrac{S_2}{S_1}

For free? No — the cost is distance.

However many times you win on force, you lose the same number of times on distance: the small piston must move S2/S1S_2/S_1 times farther than the large one rises. The work is the same F1h1=F2h2F_1 h_1 = F_2 h_2 — this is how hydraulic jacks and brakes work.

Why you need this for the exam:

p = F₁/S₁ = F₂/S₂ plus F₂ = F₁·S₂/S₁ solve almost every problem about a hydraulic press and jack. The trick: write down that the pressure is equal on both sides, and translate that into the ratio of areas.

Remember: the liquid transmits pressure equally, so a larger piston gives a larger force — F₂ = F₁·S₂/S₁. This is the foundation of hydraulics.

Illustrative model · NexTutor Virtual laboratory

Enjoyed the experiment? Share it.

It's free — send it to friends from school.

Theory & formulas

Pascal's law

Pressure applied to an enclosed liquid is transmitted equally in every direction. In a hydraulic press the same pressure acts on both pistons: F1S1=F2S2\dfrac{F_1}{S_1} = \dfrac{F_2}{S_2}

Force multiplication

Hence the force on the large piston: F2=F1 S2S1F_2 = F_1\,\dfrac{S_2}{S_1} The larger the ratio of areas, the greater the force multiplication.

No free energy

The large piston moves as many times less as the force is greater (F1s1=F2s2F_1 s_1 = F_2 s_2) — the work on both sides is equal. Applications: brakes, presses, jacks.

Symbols: F1, F2F_1,\ F_2 — forces on the pistons, S1, S2S_1,\ S_2 — piston areas, s1, s2s_1,\ s_2 — piston displacements.

Frequently asked questions

›

How does a hydraulic press multiply force?

The pressure p=F/S is the same everywhere in the liquid, so a larger piston (larger S) produces a proportionally larger force F₂=F₁·S₂/S₁.

›

Does a hydraulic press give free energy?

No — the large piston moves as many times less as the force is greater; the work (F·s) on both sides is equal.

›

Where is Pascal's law applied?

Hydraulic brakes, presses, car jacks, machine control systems.

Teaching? Embed this experiment on your site.

Free widget — paste it on your school site, blog or notes. It links to the full version.

Teaching a class or tutoring? Guide students in NexTutor →