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📌 Pressure & Hydraulics

Pressure tells you how concentrated a force is. Hydraulic systems multiply that force by changing piston area — the principle behind every jack, brake and press in engineering.

P = F ÷ AF = P × A Pascal's principleF₁/A₁ = F₂/A₂
2Topics
8Examples
4Calculators
12Practice Qs
📌 Engineering context: A small hydraulic ram lifts a 30-tonne excavator because pressure is equal throughout the fluid — but the ram has a huge area. Click any topic to start.

📌PressureP = F ÷ A
💧HydraulicsPascal's principle
⚙️Applications4 workshop calculators
🎮Game12 randomised questions
📝Practice12 graded questions
📌

Pressure

Force per unit area. Spread the same force over a bigger area and pressure drops. Concentrate it on a tiny area and pressure shoots up.

P = F ÷ APascals (Pa)barMPaN/m²

📜 Key rules

Pressure: P = F ÷ A — force (N) divided by area (m²). Unit: pascals (Pa) = N/m².
Rearranged: F = P × A and A = F ÷ P.
Unit conversions: 1 bar = 100 000 Pa  |  1 MPa = 10 bar  |  engineering stresses often in N/mm² = MPa.
Bigger area → lower pressure. Smaller area → higher pressure (same force).
Live ExplorerDrag the sliders — watch pressure change instantly
F = 5000 N A = 0.050 m² P = 100 000 Pa = 1.0 bar
💡 Surface width above shows relative area — narrow = higher pressure
Key IdeaSame force — different pressures
Small area 10 000 N A = 0.01 m² 1 MPa = 10 bar Large area 10 000 N A = 1.0 m² 10 000 Pa = 0.1 bar Same force — pressure depends entirely on area
P = F ÷ A   →   bigger A = lower P
P = F ÷ A   →   smaller A = higher P
📌Try it — Pressure CalculatorChoose what to find, enter values
Enter values above — answer updates as you type.

📚 Key vocabulary

Pressure (P)
force per unit area — how concentrated a force is. Unit: pascal (Pa) = N/m².
Pascal (Pa)
1 N acting on 1 m². Engineering pressures are usually in kPa, MPa or bar.
Bar
1 bar = 100 000 Pa. Workshop air lines run at 6–10 bar.
Gauge pressure
pressure measured above atmospheric — what most workshop gauges display.
N/mm²
equivalent to MPa (1 N/mm² = 1 000 000 Pa). Used for hydraulic and material stresses.
Contact area
the area over which a force acts. Small area = high pressure from the same force.
👁️ Visual aid — hydraulic piston: 500 000 Pa system pressure on 0.008 m² piston produces 4000 N
P in → Fluid pressure = 500 000 Pa throughout F = 4000 N ↑ Piston A = 0.008 m² F = P × A 500 000 × 0.008 = 4000 N ✓
✍️Worked example 1 — pressure under a block  (step through it)
QuestionA metal block weighing 2000 N rests on a surface of area 0.04 m². Calculate the pressure.
FormulaP = F ÷ A
SubstituteP = 2000 ÷ 0.04
Divide2000 ÷ 0.04 = 50 000
AnswerP = 50 000 Pa = 50 kPa = 0.5 bar ✓
Step 1 of 5
✍️Worked example 2 — force from pressure and area
QuestionA hydraulic piston has area 0.008 m² and system pressure is 500 000 Pa. What force does it produce?
FormulaF = P × A
SubstituteF = 500 000 × 0.008
Multiply500 000 × 0.008 = 4000
AnswerF = 4000 N
✍️Worked example 3 — minimum piston area
QuestionA press must produce 15 000 N. Maximum allowable pressure is 300 000 Pa. Find the minimum piston area.
FormulaA = F ÷ P
SubstituteA = 15 000 ÷ 300 000
Divide15 000 ÷ 300 000 = 0.05
AnswerA = 0.05 m²
✍️Worked example 4 — unit conversion
QuestionA workshop air line reads 8 bar. Express this in Pa and in MPa.
Pa8 × 100 000 = 800 000 Pa
MPa800 000 ÷ 1 000 000 = 0.8 MPa
Answer8 bar = 800 000 Pa = 0.8 MPa
💧

Hydraulics & Pascal's Principle

Pressure in an enclosed fluid is the same everywhere. A bigger piston area means a bigger force — that's force multiplication.

Pascal's principleF₁/A₁ = F₂/A₂Force multiplicationMechanical advantage

📜 Key rules

Pascal's principle: pressure is the same everywhere in an enclosed fluid.
Hydraulic equation: F₁ ÷ A₁ = F₂ ÷ A₂ — rearranged to find any unknown.
Output force: F₂ = (A₂ ÷ A₁) × F₁. If output piston is 10× bigger, force is 10× bigger.
Trade-off: the larger piston moves a shorter distance — volume of fluid is conserved.
Live ExplorerHydraulic force multiplier — drag sliders to explore
F₁=200N A₁=20cm² = pressure = F₂=2000N A₂=200cm² F₂ = (A₂÷A₁) × F₁ = ×10 → 2000 N MA = 10.0
💡 Mechanical advantage = A₂ ÷ A₁ — double the output area, double the force
💧Try it — Hydraulic Force CalculatorEnter values — answer updates as you type
Enter values above — answer updates as you type.

📚 Key vocabulary

Pascal's principle
pressure applied to a confined fluid transmits undiminished in all directions.
Mechanical advantage
output force ÷ input force — equal to the area ratio A₂ ÷ A₁.
Incompressible fluid
a fluid whose volume stays constant under pressure — hydraulic oil behaves this way.
Volume conservation
if A₂ is 10× larger, the output piston moves 10× less — same fluid volume moved.
Piston
the sliding component inside a cylinder that converts fluid pressure into linear force.
Hydraulic system
a system using pressurised liquid (usually oil) to transmit and multiply force.
👁️ Visual aid — hydraulic jack: 200 N on 0.002 m² → 2000 N on 0.02 m² (same pressure throughout)
F₁=200 N ↓ A₁ = 0.002 m² P = 100 000 Pa everywhere F₂ = 2000 N ↑ A₂ = 0.02 m² (10× bigger → 10× force) A₂ ÷ A₁ = 10 F₂ = 2000 N
✍️Worked example 5 — hydraulic output force  (step through it)
QuestionA hydraulic jack: input area 0.002 m², output area 0.02 m², input force 200 N. Find the output force.
FormulaF₂ = (A₂ ÷ A₁) × F₁
Area ratio0.02 ÷ 0.002 = 10
Multiply10 × 200 = 2000
AnswerF₂ = 2000 N   (mechanical advantage = 10) ✓
Step 1 of 5
✍️Worked example 6 — input force needed
QuestionOutput force needed = 8000 N. Output area = 0.04 m², input area = 0.005 m². Find the input force.
FormulaF₁ = (A₁ ÷ A₂) × F₂
Ratio0.005 ÷ 0.04 = 0.125
Multiply0.125 × 8000 = 1000
AnswerF₁ = 1000 N
✍️Worked example 7 — system pressure then output force
QuestionInput: 600 N on 0.003 m² piston. Output piston = 0.015 m². Find system pressure and output force.
Step 1P = 600 ÷ 0.003 = 200 000 Pa
Step 2F₂ = 200 000 × 0.015 = 3000
AnswerP = 200 000 Pa; F₂ = 3000 N
✍️Worked example 8 — finding output piston area
QuestionInput: 300 N on 0.006 m². Required output: 2500 N. Find the output piston area needed.
Step 1P = 300 ÷ 0.006 = 50 000 Pa
Step 2A₂ = 2500 ÷ 50 000 = 0.05
AnswerA₂ = 0.05 m²
⚙️

Engineering Applications

Four live calculators — results update as you type.

📌 Floor loading — contact pressure

Check the floor pressure when a machine is installed — must stay below the floor slab rating.

P = F ÷ A
🧮 Calculator
Enter values above.

💧 Hydraulic ram — output force

Find the output force of a hydraulic ram from its operating pressure and piston area.

F = P × A
🧮 Calculator
Enter values above.

🔨 Pneumatic clamp — bore area

Find the cylinder bore area needed to produce a clamping force at the workshop air line pressure.

A = F ÷ P
🧮 Calculator
Enter values above.

🔌 Hydraulic ratio — force multiplication

Find the mechanical advantage and output force of a two-piston hydraulic system.

F₂ = (A₂ ÷ A₁) × F₁
🧮 Calculator
Enter values above.
🎮

Pressure Challenge

12 randomised questions per round. New numbers every time.

📌 Quick-fire round

Question 1 / 12 Score: 0 Streak: 0 🔥
Press Start to begin!
📝

Practice Questions

12 graded questions. Type your answer to check it, then reveal the full working.

📌 Show your formula, substitution and every step. 1 bar = 100 000 Pa. 1 MPa = 1 000 000 Pa.
Question 1 · Pressure1 mark
A force of 6000 N acts on an area of 0.3 m². Calculate the pressure in Pa.
P = F ÷ A = 6000 ÷ 0.3 = 20 000 Pa (= 20 kPa = 0.2 bar)
Question 2 · Pressure2 marks
A machine weighing 36 000 N stands on a base of 1.2 m². Calculate the contact pressure in kPa.
P = 36 000 ÷ 1.2 = 30 000 Pa = 30 kPa
Question 3 · Force from pressure2 marks
A hydraulic cylinder has area 0.015 m² at 400 000 Pa. Calculate the output force.
F = P × A = 400 000 × 0.015 = 6000 N
Question 4 · Area from pressure2 marks
A pneumatic clamp must produce 2400 N at 600 000 Pa. Find the minimum piston area.
A = F ÷ P = 2400 ÷ 600 000 = 0.004 m²
Question 5 · Unit conversion1 mark
Express 12 bar in Pa.
12 × 100 000 = 1 200 000 Pa (= 1.2 MPa)
Question 6 · Hydraulics2 marks
A hydraulic jack: input area 0.001 m², output area 0.01 m², input force 500 N. Find the output force.
F₂ = (0.01 ÷ 0.001) × 500 = 10 × 500 = 5000 N
Question 7 · Hydraulics2 marks
A press needs 12 000 N output. Output area = 0.06 m², input area = 0.004 m². Find the input force.
F₁ = (0.004 ÷ 0.06) × 12 000 = 0.0667 × 12 000 = 800 N
Question 8 · System pressure2 marks
A 400 N input acts on a 0.004 m² piston. What pressure does this create in the fluid?
P = 400 ÷ 0.004 = 100 000 Pa (= 1 bar)
Question 9 · Output area2 marks
System pressure = 250 000 Pa. Required output force = 5000 N. Find the output piston area.
A = F ÷ P = 5000 ÷ 250 000 = 0.02 m²
Question 10 · Combined2 marks
A 250 N force on 0.005 m² drives a 0.05 m² piston. Find the system pressure and the output force.
P = 250 ÷ 0.005 = 50 000 Pa
F₂ = 50 000 × 0.05 = 2500 N
Question 11 · Combined3 marks
A hydraulic press: input 180 N on 0.003 m², output area 0.045 m². Find (a) system pressure, (b) output force, (c) mechanical advantage.
P = 180 ÷ 0.003 = 60 000 Pa
F₂ = 60 000 × 0.045 = 2700 N
MA = 2700 ÷ 180 = 15
Question 12 · Combined3 marks
A 4000 kg machine stands on feet with total contact area 0.16 m². Find (a) the weight, (b) floor pressure in kPa, (c) does it exceed a 300 kPa floor rating?
W = 4000 × 9.81 = 39 240 N
P = 39 240 ÷ 0.16 = 245 250 Pa = 245.25 kPa
245.25 < 300 → within rating ✓
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SkillLondon · Level 2 Engineering Science · Pressure & Hydraulics · skilllondon.co.uk