A hydraulic system transmits pressure through a liquid, so a small force on a small piston can balance a large force on a large piston. To read the data, find the area ratio first and compare it with what was measured.
This lesson belongs to force and pressure in SPM Science. If pressure itself is still shaky, go back to calculating pressure from force and area.
What does Pascal’s principle say about the two pistons?
Pressure applied to an enclosed liquid is transmitted equally to every part of the liquid. So the pressure under the small piston equals the pressure under the large piston.
Write it as F₁ ÷ A₁ = F₂ ÷ A₂. The force grows by the same factor as the area. The liquid does not change the pressure; the larger area turns the same pressure into a larger force.
Worked example: a table of trials
A student tests a model lift. The small piston has an area of 2.5 cm² and the large piston has an area of 50 cm². Assume the liquid cannot be compressed.
The area ratio is 50 ÷ 2.5 = 20. So an ideal system multiplies the input force by 20.
| Trial | Input force (N) | Ideal output (N) | Measured output (N) |
|---|---|---|---|
| 1 | 10 | 200 | 188 |
| 2 | 20 | 400 | 372 |
| 3 | 30 | 600 | 552 |
Every measured value is lower than the ideal value. Trial 2 shows the pattern: 372 ÷ 400 = 0.93, so the system delivers 93% of the ideal force. Trial 3 gives 552 ÷ 600 = 0.92, which is 92%.
A good SPM answer states the trend and a reason. “The measured output is always less than the ideal output, because some force is lost to friction at the pistons or to air trapped in the liquid.”
Force gained, distance lost
The large piston gives more force, yet it moves less. The liquid volume leaving the small cylinder equals the volume entering the large one.
If the small piston is pushed down 20 cm, the volume moved is 2.5 × 20 = 50 cm³. On the large piston, 50 cm³ over 50 cm² is a rise of 1 cm. The force rose by 20 times and the distance fell by 20 times, so no energy is created.
The mistake that loses the mark
A common slip is to say the pressure is larger at the large piston, because the force is larger. The table seems to support that idea, so the error is easy to miss.
| Statement | Wrong | Right |
|---|---|---|
| Pressure at the large piston | Bigger, since force is bigger | Equal to the small piston |
| Why force is bigger | Liquid multiplies pressure | Same pressure acts on a larger area |
| Energy | Output work is greater | Output work is at most equal to input work |
The fix is to calculate the pressure at both pistons. In Trial 1, the small piston gives 10 ÷ 2.5 = 4 N/cm², and the ideal large piston gives 200 ÷ 50 = 4 N/cm². The same number shows the principle working.
Stating assumptions in a data answer
When a question says the system is ideal, use these assumptions and name them in your answer.
- The liquid cannot be compressed.
- There is no air bubble in the liquid.
- There is no leak.
- Friction at the pistons is negligible.
If the measured value is below the ideal one, pick the assumption that failed. That is the link between the calculation and the real apparatus.
Check yourself
A hydraulic jack has a small piston of area 5 cm² and a large piston of area 80 cm². It must lift a load of 1 600 N. Find the ideal force on the small piston, and the distance the large piston rises if the small piston moves 8 cm.
Answer
The area ratio is 80 ÷ 5 = 16. The pressure under the large piston is 1 600 ÷ 80 = 20 N/cm².
The small piston needs 20 × 5 = 100 N. Check: 1 600 ÷ 16 = 100.
The liquid volume moved is 5 × 8 = 40 cm³. Over 80 cm², the large piston rises 40 ÷ 80 = 0.5 cm.
What to study next
Test the whole cluster with the force and pressure practice set. The graph evidence and fair-comparison lab helps with reading a table against a stated assumption.
If you want a teacher to go through hydraulic and pressure data with you, see online one-to-one Science tuition.