Pressure grows with depth in a resting fluid
You will be able to: Derive and use the pressure–depth relationship for a uniform fluid at rest.
Why does a deeper point in water have higher pressure?
A point two meters below a water surface supports more overlying water than a point one meter below it. The surrounding pressure must increase with depth so a small fluid parcel can remain in vertical force balance.
A useful starting point: Pressure: the same force over a different area →
Words and symbols before equations
- Depth h
- Vertical distance below the chosen surface, positive downward here.
- Surface pressure P_0
- Pressure applied at the fluid’s top boundary.
- Hydrostatic
- Describes a fluid at rest.
- Pressure gradient
- How pressure changes with position; for constant density its depth slope is ρg.
What this picture assumes
Resting uniform liquid, g=10 m/s², surface absolute pressure 100 kPa. The line covers 0–4 m; the teal point marks selected depth. Container shape does not enter this hydrostatic relation.
Connect the picture to the physics
Imagine a vertical fluid column with area A and height h. Its weight is ρAhg. The upward pressure force at the bottom must balance the downward top pressure force plus weight: PA=P_0A+ρAhg. Dividing by A gives P=P_0+ρgh.
At the same depth in one connected uniform liquid at rest, pressure is the same regardless of container width or shape. Area canceled from the force balance. A wider bottom can experience a larger total force even though its pressure at that depth is unchanged.
The P-versus-depth graph is a straight line with intercept P_0 and slope ρg for constant density and g. If density changes significantly with height, one constant-density line is not appropriate over the whole region. Do not use the hydrostatic equation between points in a rapidly accelerating flow without checking the model.
A worked example, step by step
Use ρ=1000 kg/m³, g=10 m/s² and surface pressure 100,000 Pa. Find pressure at depths 1 m and 3 m.
- At 1 m, the increase is ρgh=1000(10)(1)=10,000 Pa.
- P(1 m)=110,000 Pa.
- At 3 m, P=100,000+1000(10)(3)=130,000 Pa.
- The pressure difference is 20,000 Pa, independent of the container’s cross-sectional area.
Use vertical depth below the fluid surface, not distance along a slanted wall or container bottom width.
If container width doubles while h, ρ and P_0 stay fixed, does bottom pressure double?
Compare with an explanation
No. Pressure is unchanged; force on a larger bottom area can increase.
Predict. Change one thing. Explain.
Change depth at fixed density, then change density at fixed depth. Read both the surface-pressure intercept and the slope on the plot. Explain why a narrow and wide vessel can give the same pressure at equal depth.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
At h=2 m, pressure increase=20 kPa; absolute pressure=120 kPa. Slope=10000 Pa/m. Surface intercept stays 100 kPa.
Resting uniform liquid, g=10 m/s², surface absolute pressure 100 kPa. The line covers 0–4 m; the teal point marks selected depth. Container shape does not enter this hydrostatic relation.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant force, motion or energy relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA liquid has ρ=800 kg/m³, g=10 m/s² and surface pressure 100 kPa. (a) Find pressure increase at 2 m. (b) Find absolute pressure there. (c) Give the pressure–depth slope in Pa/m. (d) Predict the effect of a wider container at the same depth.
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Compare with the answer and four-point rubric
- 1 point: 16,000 Pa.
- 1 point: 116,000 Pa=116 kPa.
- 1 point: ρg=8000 Pa/m.
- 1 point: No change in pressure for the same surface pressure and fluid; bottom force also depends on area.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What distance belongs in ρgh?
Vertical depth below the reference surface.
RECALL 2What sets the hydrostatic pressure slope?
ρg.
RECALL 3Does vessel shape change pressure at equal depth in the same connected resting fluid?
No, under the stated constant-density assumptions.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Pressure grows with depth in a resting fluid
- Resting uniform fluid: P=P_0+ρgh.
- Pressure difference between depths: ΔP=ρgΔh.
- P vs h slope=ρg when ρ and g are constant.
Remember: Use vertical depth below the fluid surface, not distance along a slanted wall or container bottom width.
Conditions: Resting uniform liquid, g=10 m/s², surface absolute pressure 100 kPa. The line covers 0–4 m; the teal point marks selected depth. Container shape does not enter this hydrostatic relation.
Refresh Kid · Unit 8 · Objectives 8.2.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 8.2, objectives 8.2.B. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Fall-2026 corrections also checked. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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