K323 / 2027
Pressure overview

Lesson 4 of 6

Hydraulic systems

  • I can explain pressure transmission in hydraulic systems, particularly the hydraulic press.Syllabus

    Syllabus K323, 5(c). Explain pressure transmission in hydraulic systems, particularly the hydraulic press.

An applied increase in pressure is transmitted through a confined liquid. Acting over a larger piston area, that increase can produce a larger force.

Use p = F / A and convert square units carefully. An increase in pressure is different from the total pressure already present in the liquid.

A hydraulic press connects a small input piston to a larger output piston through liquid. Pushing the input piston increases the pressure in the liquid. The pressure increase is transmitted throughout the confined liquid, including to the output piston.

For the simple model here, the piston faces are at the same height, the same atmospheric pressure acts above both pistons, and piston weights and friction are neglected. Compare the additional forces caused by the applied pressure increase.

Δp = F1 / A1 = F2 / A2F1 is the extra input force; A1 is the input piston area. F2 is the extra output force; A2 is the output piston area. The transmitted pressure increase is the same.

Equal pressure increase, different forces

The piston faces are at the same height. Ignore piston weights and friction; both external sides have the same atmospheric pressure.

An ideal hydraulic press with unequal piston areasA small input piston and a larger output piston enclose one continuous body of liquid. Their lower faces are at the same horizontal level. The input area is 4.0 square centimetres and the applied input force is 80 N downwards. The output area is 200 square centimetres and the additional upward output force is 4000 N. The transmitted pressure increase is 200000 Pa. The area and force ratios are both 50. The piston widths and arrow lengths are schematic, not quantitative scales; the output force arrow is deliberately longer. Piston weights and friction are ignored and the outside atmospheric pressures are equal.Input4.0 cm2Output200 cm280 N4000 NSame piston-face levelConnected liquid

Schematic: piston widths and force-arrow lengths are not to scale. Use the labelled areas and forces.

4.0 cm2 = 0.00040 m2; 200 cm2 = 0.020 m2.

Pressure increase = 80 / 0.00040 = 200000 Pa.

Additional output force = 200000 x 0.020 = 4000 N.

The smaller piston has area 4.0 cm2; the larger has area 200 cm2. An 80 N input produces a 200000 Pa pressure increase and a 4000 N additional output force in the stated ideal model. Use the labelled areas, not measurements of the drawing.

Worked example

A larger area gives a larger output force

The input force is 80 N, the input area is 4.0 cm2 and the output area is 200 cm2.

  1. Convert the areas: A1 = 0.00040 m2 and A2 = 0.020 m2.
  2. Calculate the pressure increase: Δp = 80 / 0.00040 = 200000 Pa.
  3. Use that increase at the output: F2 = ΔpA2 = 200000 x 0.020 = 4000 N.

The area ratio is 200 / 4.0 = 50, so the output force is 50 times the input force. The pressure increase itself is not multiplied by 50; it is the same 200000 Pa at both pistons.

Try it step by step

Work it out in steps: Input area in m2, then Pressure increase in Pa, then Output force in N.

  1. Problem 1. An input force of 50 N acts on a piston of area 2.0 cm2. The output piston has area 100 cm2. Find the output force.
  2. Problem 2. An input force of 120 N acts on a piston of area 5.0 cm2. The output piston has area 150 cm2. Find the output force.
  3. Problem 3. An input force of 90 N acts on a piston of area 3.0 cm2. The output piston has area 180 cm2. Find the output force.
Show the answer

Problem 1.

Input area in m2: A1 = 2.0 cm2 = 2.0 / 10000 = 0.00020 m2

Pressure increase in Pa: change in p = F1 / A1 = 50 / 0.00020 = 250000 Pa

Output force in N: A2 = 0.010 m2, so F2 = 250000 x 0.010 = 2500 N (2 sf)

Problem 2.

Input area in m2: A1 = 5.0 / 10000 = 0.00050 m2

Pressure increase in Pa: change in p = 120 / 0.00050 = 240000 Pa

Output force in N: A2 = 0.015 m2, so F2 = 240000 x 0.015 = 3600 N (2 sf)

Problem 3.

Input area in m2: A1 = 3.0 / 10000 = 0.00030 m2

Pressure increase in Pa: change in p = 90 / 0.00030 = 300000 Pa

Output force in N: A2 = 0.018 m2, so F2 = 300000 x 0.018 = 5400 N (2 sf)

The atmosphere also exerts pressure above the pistons. In this model its equal contribution is accounted for on both sides, leaving the extra force due to the applied increase. The 4000 N is the force available to balance an additional load at the output, with the piston in equilibrium.

The larger force comes with a smaller movement

Treat the liquid as incompressible, with no leaks. The volume pushed in by the small piston must be accommodated at the other piston. The large-area piston therefore rises a shorter distance for that displaced volume.

The press trades movement for force; it does not create energy. The force comparison above does not mean the large piston can move the same distance as the small one while providing the larger force in this ideal system.

Equal pressure increase does not mean equal force. Multiply that increase by each piston area. Check that the piston levels and the stated assumptions justify the simple force comparison.

Check your understanding

In an ideal hydraulic press, the output piston area is four times the input area. The pistons are at the same level, with the same atmosphere above them; ignore piston weights and friction. An extra 60 N input force is applied. What extra output force does this produce?

  1. 240 N
  2. 60 N
  3. 15 N
Show the answer

240 N

F2 / A2 = F1 / A1. Since A2 = 4A1, the extra output force is 4 x 60 = 240 N.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

In a hydraulic jack, a force of 200 N is applied to a small piston of area 0.0020 m2. The large piston has an area of 0.050 m2. Explain how the jack produces a larger force, and calculate the force on the large piston. [3]

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