Buoyancy comes from unequal pressure forces
You will be able to: Derive buoyancy from pressure differences and calculate the weight of displaced fluid.
Why does water push a submerged object upward?
A submerged block has water pushing down on its top and up on its bottom. The bottom is deeper, so its upward pressure force is greater. Their difference is an upward buoyant force.
A useful starting point: A fluid parcel still obeys Newton’s laws →
Words and symbols before equations
- Buoyant force B
- Net upward force from a surrounding resting fluid.
- Displaced volume V_disp
- Volume of fluid excluded by the submerged part of the object.
- Fluid density ρ_f
- Density of the surrounding fluid, not the object.
- Archimedes’ principle
- Buoyant-force magnitude equals the weight of displaced fluid.
What this picture assumes
Resting uniform fluid, g=10 m/s². Fluid contacts the submerged surfaces; no sealed contact against the bottom. Displaced volume is the submerged volume, not automatically total object volume.
Connect the picture to the physics
For a fully submerged rectangular block of area A and height H, bottom pressure exceeds top pressure by ρ_f gH. The vertical pressure-force difference is B=(P_bottom−P_top)A=ρ_f gAH. Since AH is displaced volume, B=ρ_f gV_disp.
The same result applies to other shapes surrounded by the fluid. For partial immersion, only the submerged volume counts. Opposing side-force components cancel in a resting fluid; pressure at the surface is not a single upward arrow, but its net effect is buoyancy.
Once a rigid object is fully submerged in a uniform incompressible liquid, lowering it farther does not increase B: top and bottom pressures both rise, leaving their difference unchanged. This assumes fluid contacts the relevant surfaces; an object sealed against a container floor needs separate force analysis.
A worked example, step by step
An object displaces 2 L of water. Use ρ_f=1000 kg/m³ and g=10 m/s². Find the buoyant force and the mass of displaced water.
- V_disp=2 L=0.002 m³.
- Displaced mass=ρ_fV_disp=1000(0.002)=2 kg.
- Buoyant force equals that water’s weight: B=2(10)=20 N upward.
- The result depends on displaced fluid, not directly on the object’s mass.
Use fluid density and submerged volume in B=ρ_f gV_disp. Buoyancy equals object weight only when the relevant vertical forces balance with no other support.
Two fully submerged objects have equal outside volumes but different masses. Same fluid: same B?
Compare with an explanation
Yes, if they exclude equal volumes of the same fluid. Their weights and net forces can differ.
Predict. Change one thing. Explain.
Change displaced volume at fixed fluid density, then change fluid density at fixed volume. Predict the force factor. Imagine lowering a fully submerged rigid object farther without changing either control.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
V_disp=1 L=0.001 m³; displaced mass=1 kg. B=10 N upward. Keeping displaced volume and uniform fluid density fixed keeps B fixed.
Resting uniform fluid, g=10 m/s². Fluid contacts the submerged surfaces; no sealed contact against the bottom. Displaced volume is the submerged volume, not automatically total object volume.
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 block displaces 0.003 m³ of liquid with density 800 kg/m³. Use g=10 m/s². (a) Find displaced-fluid mass. (b) Find B. (c) Predict B if only half that volume is submerged. (d) Explain the source of the upward net pressure force.
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Compare with the answer and four-point rubric
- 1 point: 2.4 kg.
- 1 point: 24 N upward.
- 1 point: 12 N upward.
- 1 point: Pressure is greater on deeper surfaces, producing a larger upward bottom force than downward top force.
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 1Which density enters the buoyancy formula?
The surrounding fluid’s density.
RECALL 2Which volume enters for a floating object?
Its submerged/displaced volume.
RECALL 3Why can equal B coexist with different object weights?
B depends on displaced fluid; weight depends on object mass.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Buoyancy comes from unequal pressure forces
- B=ρ_f gV_disp.
- For a fully submerged rigid body in uniform fluid, B is independent of depth.
Remember: Use fluid density and submerged volume in B=ρ_f gV_disp. Buoyancy equals object weight only when the relevant vertical forces balance with no other support.
Conditions: Resting uniform fluid, g=10 m/s². Fluid contacts the submerged surfaces; no sealed contact against the bottom. Displaced volume is the submerged volume, not automatically total object volume.
Refresh Kid · Unit 8 · Objectives 8.3.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 8.3, objectives 8.3.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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