Pressure and hydraulics explorer
Compare pressure at different liquid depths with a probe and manometer, then explore force and distance in a hydraulic press.
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Learning goals
- Define pressure as force per unit area
- Apply pressure = force ÷ area
- Explain pressure transmission in a hydraulic press
- Apply density = mass ÷ volume
- Apply liquid-column pressure = height × density × gravitational field strength
A probe 0.40 metres below the surface of water in a straight-sided tank reads 3924 pascals. The manometer level difference is 0.400 metres.
- Depth, h
- 0.40 m
- Pressure due to the liquid, hρg
- 3924 Pa
- Total pressure
- 1.049 × 10⁵ Pa
- Manometer height difference
- 0.400 m
- Pressure in the liquid, F₁/A₁
- 75000 Pa
- Force on the large piston, F₂
- 1500 N
- Effort moves down, d₁
- 0.0 cm
- Load rises, d₂
- 0.00 cm
Try this
0 of 4 doneMeasure the pressure at the same depth in two differently shaped tanks. (not done yet)
Pressure depends only on the depth and the density of the liquid, not on the shape of the container or the amount of liquid.
Record the pressure at five different depths in one liquid. (not done yet)
p is proportional to h: the graph is a straight line through the origin with gradient ρg.
Lift the 2000 N load with an effort of 200 N or less. (not done yet)
The same pressure acts on a larger area, so F₂ = F₁ × A₂/A₁. But the large piston rises only d₁ × A₁/A₂: the work done, F₁d₁ = F₂d₂, is not multiplied.
Find the smallest effort, to the nearest 10 N, that lifts the load. (not done yet)
The load just lifts when F₁/A₁ = W/A₂: the pressure from the effort equals the pressure the load exerts on the large piston.
You’ve tried every task.
Your readings
| # | h / m | p / kPa | Remove |
|---|---|---|---|
| No readings yet. Set up a measurement, then record it. | |||