Compare force and area
Change one property at a time and use perpendicular force per area to explain contact pressure.
AP Physics 1 · Unit 8 · Fluids
Pressure
Change force and area, measure pressure with depth, and connect atmospheric, absolute and gauge pressure using a movable sensor.
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Learn by doing
Three ideas you’ll put into motion
Measure pressure with depth
Move a sensor through liquids, read pressure–depth graphs and develop the hydrostatic model from force balance.
Choose a pressure reference
Distinguish absolute and gauge pressure, investigate a vacuum seal and add contributions through oil and water.
The big idea
The key ideas, at a glance.
Pressure is force per area
Pressure is a scalar: P = F⊥/A. Forces act perpendicular to a surface; equal loads produce greater pressure on smaller areas.
Start from a reference pressure
In a stationary uniform liquid, P = Psurface + ρgh. Gauge pressure subtracts atmospheric pressure. Through layered liquids, add ρgh for each layer above the sensor.
A quick comparison
Keep the perpendicular force fixed while doubling its contact area.
The average pressure becomes half as large.
Teaching this topic?Classroom fit, AP connections, and support
- Classroom fit
- Use for a concept introduction, guided independent practice, or a projected class discussion.
- Before students begin
- Density, force, weight, area, basic algebra and graph slope. The lesson introduces pressure and its reference scales.
- Built-in support
- Prepared scenes, guided prompts, and feedback on lesson questions.
- AP Physics 1 connections
- Unit 8, Topic 8.2: pressure. Relate pressure to perpendicular force per area, microscopic interactions, depth and density, and distinguish absolute and gauge pressure.
- Check for understanding
- Ask why equal depths in the same stationary liquid have equal pressure, why a vacuum can have negative gauge pressure, and where a layered-liquid pressure graph changes slope.
Ready to test your thinking?
The first scene is set up. Start the lesson and explore.