Pascal's Principle: Hydraulic Force Multiplication
1 · Predict
A hydraulic lift has a small input piston and a much larger output piston, connected by a sealed fluid. If you push on the small piston, why does the large piston lift a much heavier load?
- The output force is multiplied in proportion to how much larger the output piston's area is.
- The output force equals the input force no matter the piston sizes.
- The output force is smaller because the fluid absorbs some of the push.
2 · Set Up
- Open the fluid-pascal preset and press Reset. Two connected pistons of different areas share a sealed, incompressible fluid.
- Enable the input/output pressure and force readouts.
- Set the input piston area, output piston area, and input force for each trial, and record the output-force reading.
3 · Collect Data
| Input piston area A₁ (m²) | Output piston area A₂ (m²) | Input force F₁ (N) | Output force F₂ (N) |
|---|---|---|---|
| 0.001 | 0.01 | 50 | |
| 0.002 | 0.02 | 80 | |
| 0.001 | 0.05 | 20 |
Plot output force F₂ (y-axis) against the area ratio A₂/A₁ (x-axis) for your three trials.
4 · Analyze
- For one trial, compute F₂ = F₁ · (A₂/A₁) and compare it to the reading.
- Pascal's principle says pressure applied to an enclosed fluid is transmitted equally everywhere. Use P = F/A on both pistons to explain why F₂/F₁ must equal A₂/A₁.
5 · Extend
- A car hydraulic jack uses a small hand-pump piston to lift an entire car. Using your data, explain roughly how much larger the jack's output piston area must be than its input piston area, relative to the force you can apply by hand and the car's weight.
- If the output piston lifts a load with 10× the input force, how far does the output piston move compared to how far you push the input piston? (Hint: think about conservation of fluid volume.)