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Hydraulic systems

Key idea: Learn how hydraulic presses and brakes transmit pressure, multiply force, and trade force for distance, with O Level calculations and exam tips.

  • SEC G3 Physics 2027
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Learning objectives

  • Define pressure as force per unit area
  • Apply pressure = force ÷ area
  • Explain pressure transmission in a hydraulic press
  • Apply density = mass ÷ volume
  • Apply liquid-column pressure = height × density × gravitational field strength
  • Explain how liquid-column height measures atmospheric pressure
  • Explain how a manometer measures pressure difference

1. Definition

A. Hydraulic system (hydraulic press idea)

A hydraulic system uses an enclosed, approximately incompressible liquid to transmit a pressure change from one part of the system to another.

2. Key Ideas

  • Pressure is transmitted through an enclosed liquid (hydraulic system idea).
  • Use p = F/A.
  • In an ideal hydraulic system with the piston faces at the same level (or with height differences neglected):
    • pressure is the same at both pistons: p₁ = p₂
    • so F₁/A₁ = F₂/A₂ and F₂ = F₁(A₂/A₁)
  • For circular pistons, A ∝ d², so: F₂/F₁ = (d₂/d₁)²
  • Convert units carefully (especially cm² to m²):
    • 1 cm² = 1 × 10⁻⁴ m²
  • Force is multiplied, but energy is not. For an ideal system:
    • equal displaced volumes: A₁x₁ = A₂x₂
    • equal input and output work: F₁x₁ = F₂x₂

3. Detailed Explanations

A. Why a hydraulic press can multiply force

Hydraulic force multiplication and distance trade-offA small input piston and larger output piston are joined by enclosed liquid. Equal pressure gives a larger output force on the larger piston. A long input movement corresponds to a shorter output movement because displaced liquid volumes are equal.Ideal hydraulic pressarea A1larger area A2input F1output F2p1 = p2same level; losses neglectedx1x2F2/F1 = A2/A1 and A1x1 = A2x2
Scroll diagram horizontally to read all labels.
At the same level in an ideal system, both pistons receive the same pressure. A larger output area gives a larger force, but the output piston moves a smaller distance.
  1. A force F₁ acts on a small piston of area A₁.

The pressure on the liquid is: p₁ = F₁/A₁

  1. This pressure change is transmitted through the enclosed liquid. For piston faces at the same level in the ideal model: p₂ = p₁

  2. The output force on the large piston (area A₂) is: F₂ = p₂A₂

Putting these together:

F₁/A₁ = F₂/A₂,; F₂ = F₁(A₂/A₁).

So if A₂ is larger than A₁, the output force F₂ is larger than the input force F₁.

Force multiplication is not free energy

The large piston moves a shorter distance. Since the liquid volume displaced is the same on both sides, A₁x₁ = A₂x₂. In an ideal system, the input work equals the output work: F₁x₁ = F₂x₂.

B. Quick recipe for calculations

  1. Write F₁/A₁ = F₂/A₂.
  2. Convert areas into the same units.
  3. If pistons are circular and you are given diameters, use: A₂/A₁ = (d₂/d₁)²

C. Applications

Car lifts / hydraulic jacks: a small input force produces a large output force to lift heavy loads.

Hydraulic brakes: pressing the brake pedal increases pressure in brake fluid, which transmits the pressure to pistons at the wheels. The pistons push the brake pads against the disc/drum.

Simplified hydraulic brake systemA pedal force pushes a master-cylinder piston. A fluid-filled brake line connects it to two wheel pistons that press brake pads against opposite sides of a brake disc.Hydraulic brake: pressure transmitted through enclosed fluidmaster cylinderpedal forcebrake linebrake discwheel pistons push the padsonto both sides of the disc
Scroll diagram horizontally to read all labels.
The master piston creates a pressure change in the enclosed brake fluid. That pressure acts at the wheel pistons, which push the pads onto the rotating disc.
Safety: hydraulic brakes

Brake fluid is designed to be (nearly) incompressible. Air bubbles make the brakes less effective and can be dangerous.

4. Common Mistakes

  • Using diameter ratio instead of area ratio (remember A ∝ d² for circles).
  • Mixing units (e.g. A₁ in cm² and A₂ in m²).
  • Forgetting that forces must be in newtons (use W = mg if the load is given as a mass).
  • Stating “force is transmitted” instead of “pressure is transmitted”.
  • Assuming force multiplication also multiplies energy; the larger piston moves a shorter distance.
  • Ignoring piston height in a setup where the hydrostatic pressure difference is not negligible.

5. Exam Tips

  • Start with p = F/A and then set p₁ = p₂.
  • If the question says “force is multiplied by 10”, it means: F₂/F₁ = 10 ⇒ A₂/A₁ = 10
  • For circular pistons, you can skip π: A₂/A₁ = (π (d₂/2)²)/(π (d₁/2)²) = (d₂/d₁)²
  • State the idea clearly: “A pressure change applied to an enclosed liquid is transmitted throughout the liquid.”

6. Worked Examples

Modelled example 1

Force multiplication using areas

Core

Problem

A hydraulic press has input piston area A₁ = 2.0 cm² and output piston area A₂ = 50 cm². A force F₁ = 200 N is applied to the input piston. Find the output force F₂.

Study the worked solution

  1. Equate the transmitted pressures

    Method

    Set the pressure at the two same-level piston faces equal.

    Reason

    In the ideal enclosed liquid, the applied pressure change is transmitted throughout the liquid.

    Working

    F₁/A₁ = F₂/A₂

  2. Use the area ratio

    Reason

    Both areas are already in the same unit, so their ratio is dimensionless.

    Working

    F₂ = F₁A₂/A₁ = 200(50/2.0) = 5.0 × 10³ N

Guided practice 2

Force multiplication using diameters

About 5 min

Problem

The input piston diameter is 2.0 cm and the output piston diameter is 10 cm. A force of 120 N is applied to the input piston. Find the ideal output force.

Convert the diameter ratio to an area ratio

Unit: N

Hints

Hint 1: convert diameter ratio to area ratio
For circles, A ∝ d², so use (d₂/d₁)².
Hint 2: apply the force ratio
Once the area ratio is known, use F₂/F₁ = A₂/A₁.
View solution step by step

  1. Find the area ratio

    Method

    Square the output-to-input diameter ratio.

    Reason

    Pressure acts over piston area, and circular area varies with the square of diameter.

    Working

    A₂/A₁ = (10/2.0)² = 25

  2. Calculate the output force

    Reason

    The ideal force ratio equals the area ratio.

    Working

    F₂ = 25(120) = 3.0 × 10³ N

Common misconception 3

Output distance

Find and correct the mistake

Learner response

The input piston has area 4.0 cm² and moves down by 15 cm. The output piston has area 60 cm². A student writes: The hydraulic system multiplies force, so the output piston must also move farther than 15 cm.

Locate the first error and calculate the output distance, assuming no leakage.

Diagnose the force-distance claim

Which constraint has been ignored?
Unit: cm

View solution step by step

  1. Locate the conservation error

    Method

    Reject simultaneous force and distance multiplication.

    Reason

    The nearly incompressible liquid transfers volume; force gain is matched by a shorter output movement in the ideal system.

    Working

    A₁x₁ = A₂x₂

  2. Calculate the output distance

    Reason

    The areas share the same unit, so they can be used directly in the volume ratio.

    Working

    x₂ = A₁x₁/A₂ = (4.0)(15)/60 = 1.0 cm

Examiner practice 4

Find fluid pressure, then output force

4 marks

Examination question

A force of 120 N acts on an input piston of area 4.0 cm². The pressure is transmitted to an output piston of area 0.015 m².

  1. Calculate the pressure in the hydraulic fluid.
  2. Calculate the force on the output piston. [4 marks]

Show the area conversion and both calculations

View solution step by step

  1. Convert the input area

    1 mark

    Method

    Convert square centimetres to square metres.

    Reason

    Pressure in pascals uses area in m².

    Working

    4.0 cm² = 4.0 × 10⁻⁴ m²

  2. Calculate the fluid pressure

    1 mark

    Reason

    The input force acts over the converted input-piston area.

    Working

    p = 120/(4.0 × 10⁻⁴) = 3.0 × 10⁵ Pa

  3. Use the transmitted pressure

    1 mark

    Method

    Apply the same ideal pressure at the output piston.

    Reason

    The pressure change is transmitted through the enclosed liquid.

    Working

    F₂ = pA₂

  4. Calculate the output force

    1 mark

    Reason

    The output area is already in m².

    Working

    F₂ = (3.0 × 10⁵)(0.015) = 4.5 × 10³ N

Challenge 5

Required input force to lift a car

Minimal support

Reverse design calculation

A car of weight 1.2 × 10⁴ N is supported by a large piston of area A₂ = 0.020 m². The small piston has area A₁ = 2.0 × 10⁻⁴ m². Find the input force needed to just lift the car.

Work backwards from the required output force

Unit: N

Hints

Hint 1: identify the output requirement
The car’s weight is the output force that the large piston must provide.
Hint 2: make input force the subject
Use F₁ = F₂(A₁/A₂).
View solution step by step

  1. Set equal ideal pressures

    Method

    Use the car’s weight as the required output force.

    Reason

    To just lift the car, the large piston must provide F₂ = 1.2 × 10⁴ N.

    Working

    F₁/A₁ = F₂/A₂

  2. Solve for the input force

    Reason

    The small-to-large area ratio reduces the input force required.

    Working

    F₁ = (1.2 × 10⁴)((2.0 × 10⁻⁴)/0.020) = 120 N

7. Mind Stretchers

Mind stretcher 1: Design for a target forceExtension

A hydraulic press must produce an output force of 8000 N when the input force is 200 N. Find the required area ratio A₂/A₁.

Show Answer

F₂/F₁ = A₂/A₁ ⇒ A₂/A₁ = 8000/200 = 40

Mind stretcher 2: Why air bubbles make brakes “spongy”Extension

Explain why trapped air in hydraulic brake fluid makes braking less effective.

Show Answer

Air is much more compressible than brake fluid. Some master-piston movement compresses trapped bubbles instead of displacing fluid at the wheel pistons. The pedal therefore travels farther and the brakes respond less directly; the required pressure may not be reached before the available pedal travel is used.

8. Practice and next step

Test piston areas and input force in the Pressure and hydraulics explorer, then continue to Barometers and manometers.

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027