GCSE Physics Principles of hydraulic systems mechanical devices for transmission of forces in liquids calculations of hydraulic force ratios mechanical advantage IGCSE/GCSE Physics revision notes

GCSE level physics exam revision notes all about forces Part 6

Forces and pressure: 6.4

Hydraulic systems - mechanical devices for the transmission of forces in liquids and calculations of hydraulic pressure ratios

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Sub-index physics notes on 6. Forces and Pressure in fluids & calculations

6.4 Hydraulic systems - mechanical devices for the transmission of forces in liquids and calculations

Hydraulics is how we can use fluids (e.g. air or oil) to transfer force to achieve some useful work e.g. operating machinery like car jacks, car brakes or JCB digging machines.

As we have said above, pressure in fluids is caused by particle collisions with themselves and the surface of a container.

These collisions cause a net force at right angles to all surfaces the fluid is in contact with. Hence the equation

pressure = force ÷ area, P = F / A

P, pascals (Pa); F, force in newtons (N); A, area on which force acts in square metres (m²)

A force of 1 N acting on 1 m² creates a pressure of 1 Pa

(I've repeated these details because there are more calculations coming up!)

However, there is an important new idea to understand the applications of hydraulics.

The pressure in a fluid is transmitted equally in all directions.

AND, for the purposes of calculations, you consider the force involved to act at right-angles to the surface.

Since liquids are effectively incompressible, if a force is applied to any point in an enclosed fluid system, the net force is transmitted to any other point in the fluid (gas or liquid). This is the basis of all hydraulic systems.

It is no good using a gas in a hydraulic system, because a gas is readily compressed, affecting the action of the pistons in a hydraulic system - see diagram below.

A lot of the force and energy of the compression action is used to compress the gas and so reducing the transmitted force-pressure.

The principles and calculations involving hydraulic systems.

Diagram of the basic principles of a hydraulic system

The diagram above illustrates the idea of using a hydraulic system to apply a small force to produce a large force - an example of a force multiplier system to effect a mechanical advantage (remember moments and lever systems!)

The formula for pressure is P = F/A (already dealt with in detail on this page).

Therefore: P1 = F1_input/A1 and output P2 = F2_output/A2 (where = pressure, A = area, F = force)

BUT, P1 = P2 because the pressure at any given moment in time is the same throughout the hydraulic system.

Therefore: F1_input/A1 = F2_output/A2, rearranging gives the output force F2_output = F1_input x A2/A1

So the ratio of the two piston areas (A2/A1) gives you the force multiplying effect (as long as A2 > A1).

The mechanical advantage can be simply defined as:

output force / input force = F2 / F1

Note: In hydraulic systems, piston 1 is in the master cylinder and piston 2 is in a slave cylinder (can be several of the latter e.g. in a car braking system of a system for raising a car in a garage).

Some examples of how to do hydraulic system calculations

Q3.1 In a simple hydraulic system piston 1 has a cross-section area of 0.000050 m², and piston 2 has a cross-section area of 0.00025.

If a force of 35 N is applied to piston 1,

(a) What is the force multiplying ratio?

(b) What output force is generated by piston 2?

Worked out ANSWERS to the hydraulic system calculation questions

Q3.2 A medium sized car typically weighs 1000 kg. (gravity force 9.8 N/kg)

A hydraulic system to jack the car up above an inspection pit has four slave cylinders whose pistons have a cross section area of 0.01 m². The cross-section area of the master cylinder piston is 0.00125 m².

(a) What is the weight of the car?

(b) What is the minimum force that must be applied to the master cylinder to raise the car upwards?

Worked out ANSWERS to the hydraulic system calculation questions

diagram explaining how to solve hydraulic pressure system problems mechanical advantage ratio of forces involved calculation questions Q3.3 Solve the following problem with a little help from the diagram on the right of a simple hydraulic system of two pistons and cylinders connected together.

The cross-section area of piston A1 is 0.000400 m².

The cross-section area of piston A2 is 0.00800 m².

(a) If a force of 40 N (F1) is applied to piston 1, calculate the pressure created in the fluid.

(b) Calculate the force F2 created by the force of 40 N on piston A1.

(c) What is the multiplying or mechanical advantage factor in this hydraulic system?

(d) In order to create a force of 2000 N from cylinder 2, what should the cross-section area of piston 2 be if the force applied to piston 1 is still 40 N?

Worked out ANSWERS to the hydraulic system calculation questions

Index physics Forces notes 6. Forces & pressure in fluids, calculations

Key points about forces in fluids - hydraulic systems and calculations

Information sources for Doc Brown's key points: IGCSE-GCSE physics are based on textbooks & syllabus-specifications for students taking the UK AQA, Edexcel, OCR 21st Century Science, OCR Gateway science suite, WJEC, CCEA and CIE GCSE physics 9-1 level science examinations

A structured set of summary revision notes on hydraulic system calculations, tailored to the major UK GCSE/IGCSE physics exam boards: WJEC, CCEA, CIE, AQA, Edexcel, and OCR. These notes cover the physics principles, key formulas, board-specific content, and student tips for mastering hydraulic systems.

Hydraulic Systems: Core Concept

Definition: Hydraulic systems use incompressible liquids to transmit force through pressure.
Pascal’s Principle: Pressure applied to a confined fluid is transmitted equally in all directions.
Formula:
p = F/A and F = pA
where:
- ( p ) = pressure (Pa or N/m²)
- ( F ) = force (N)
- ( A ) = area (m²)

Force Multiplication in Hydraulic Systems

Input piston: Small area → small force
Output piston: Large area → large force

Pressure remains constant:

P1 = F1_input/A1 and P2 = F2_output/A2

Therefore F1_input/A1 = F2_output/A2

Typical Board-Specific Specification Content

Detailed treatment of hydraulic systems, force multiplication, and pressure
Requires understanding of equal pressure transmission
Includes worked examples and real-world applications

Triple science includes hydraulic systems and pressure calculations
Focus on input/output force relationships
Higher tier includes force multiplication and mechanical advantage
Covers hydraulic systems, pressure-area-force relationships
Includes calculations and real-world examples
Emphasises Pascal’s Law and force transmission
Includes hydraulic systems, pressure calculations, and force multiplication
Focus on formula use, units, and practical applications
Higher tier includes comparative force analysis

Student Tips for Exam Success

Memorise key formulas: p = F / A
Understand Pascal’s Principle: pressure is constant in a confined liquid
Convert units carefully: cm² to m², N/cm² to Pa
Use diagrams to visualise piston sizes and pressure flow
Practice rearranging formulas for pressure, force, or area
Apply to real-world systems: brakes, lifts, presses
Use past papers to identify common question formats

Keywords, phrases and learning objectives for forces involving pressure situations

Be able to describe and explain the principles of hydraulic systems and how they produce a mechanical advantage by increasing an applied force by transmission of the force through an enclosed liquid system.

Be able to define the mechanical advantage of a given hydraulic system e.g. in terms of the input and output force ratio..

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Index physics Forces notes 6. Forces & pressure in fluids, calculations

Worked out ANSWERS to the hydraulic system calculation Qs

Q3.1 In a simple hydraulic system piston 1 has a cross-section area of 0.000050 m², and piston 2 has a cross-section area of 0.00025.

If a force of 35 N is applied to piston 1,

(a) What is the force multiplying ratio?

force ratio = A2/A1 = 0.00025/0.000050 = 5.0

(b) What output force is generated by piston 2?

F2 = F1 x A2/A1 = 35 x 0.00025/0.000050 = 35 x 5 = 175 N

Q3.2 A medium sized car typically weighs 1000 kg. (gravity force 9.8 N/kg)

A hydraulic system to jack the car up above an inspection pit has four slave cylinders whose pistons have a cross section area of 0.01 m². The cross-section area of the master cylinder piston is 0.00125 m².

(a) What is the weight of the car?

w = mg = 1000 x 9.8 = 9800 N

(b) What is the minimum force that must be applied to the master cylinder to raise the car upwards?

F1/A1 = F2/A2, rearranging gives F1_input = F2_output x A1/A2

The output force must at least match the weight of the car, so F2 = 9800 N

A1 = 0.00125 m², A2 = 4 x 0.01 = 0.04 m² (remember there are 4 connected cylinders).

Therefore F1 = 9800 x 0.00125/0.04 = 306 N (3 sf)

So relatively small force can be used to raise a much larger weight by means of a hydraulic system.

pressure in liquid fluids Q3.3 Solve the following problem with a little help from the diagram on the right of a simple hydraulic system of two pistons and cylinders connected together.

The cross-section area of piston A1 is 0.000400 m².

The cross-section area of piston A2 is 0.00800 m².

(a) If a force of 40 N (F1) is applied to piston 1, calculate the pressure created in the fluid.

P = F / a = 40 / 0.0004 = 100 000 Pa

(b) Calculate the force F2 created by the force of 40 N on piston A1.

Since the pressure is the same throughout the hydraulic system, we can say the pressure acting on piston 2 is also 100 000 Pa.

P = F / A, F = P x A = 100 000 x 0.008 = 800 N

(c) What is the multiplying or mechanical advantage factor in this hydraulic system?

force multiplying factor = final force / initial force = 800 / 40 = 20

(d) In order to create a force of 2000 N from cylinder 2, what should the cross-section area of piston 2 be if the force applied to piston 1 is still 40 N?

The pressure P2 remains the same at 100 000 N

F2 is now 2000 N, need to solve for A2

P = F / A, A = F / P = 2000 / 100 000 = 0.02 m²

Index physics Forces notes 6. Forces & pressure in fluids, calculations