An ideal hydraulic press can produce a larger force at the wider piston. The trade-off is that the wider piston moves a shorter distance. Find the force using the pressure transmitted through the liquid, then find the movement using the volume displaced.
SEAB’s 2027 SEC G3 Physics syllabus (K323) includes explaining pressure transmission in a hydraulic press. This original worked problem treats the liquid as incompressible and ignores friction, leakage, piston weights and changes in pressure due to height.
- Use force divided by piston area to find pressure.
- Use equal displaced liquid volumes to find the distances.
- Check the result using force multiplied by distance.
Worked example | One piston has five times the area
The small piston has an area of 20 cm² and the large piston has an area of 100 cm². Push the small piston with 80 N. Its area is 0.0020 m², so the applied pressure increase is 80 ÷ 0.0020 = 40,000 Pa.
The same pressure increase acts at the large piston. Its area is 0.0100 m², so the ideal output force is 40,000 × 0.0100 = 400 N. The area is five times as large and so is the output force.
Move the small piston through 25 cm. It displaces 20 × 25 = 500 cm³ of liquid. The large piston moves 500 ÷ 100 = 5 cm. The force increases fivefold and the movement is one fifth as large.
Compare the work at each piston
In the ideal model, the input work is 80 × 0.25 = 20 J. The output work is 400 × 0.05 = 20 J. Use metres when calculating work in joules.
When finding the distance, both piston areas can stay in cm² if the movements are in cm. When finding pressure in pascals from newtons, use m². Label the conversion at the point you need it.
Try a different pair of areas
The piston areas are 12 cm² and 36 cm². The input force is 50 N and the small piston moves 9 cm. The output force is 150 N and the large piston moves 3 cm. Check both sides give 4.5 J.
Ask a study partner why the larger area increases force in one calculation and reduces distance in the other. The force calculation holds pressure fixed. The distance calculation holds displaced volume fixed.
Frequently asked questions
Do both pistons experience the same force?
The same pressure increase can act on different areas. Multiply by each area to find the corresponding force.
What happens if liquid leaks?
The ideal equality of displaced volumes may no longer apply. Follow any loss information supplied by the question.
Official information
SEAB’s 2027 SEC G3 Physics syllabus (K323), Topic 5 Pressure. Checked 9 November 2026.