The big idea: Push a beach ball into water and it fights back, shoving upward. The more of it you push under, the harder it shoves.
Buoyancy and Archimedes: That upward push is buoyancy (or upthrust). To fit in, the ball has to shove water out of the way. Archimedes' principle: weigh the water it shoved aside, and that is exactly how big the upthrust is.
Floats
- Buoyancy ≥ weight
- Object is less dense than the fluid
- e.g. a cork on water
Sinks
- Buoyancy < weight
- Object is more dense than the fluid
- e.g. a steel ball in water
Why the fluid pushes up: A fluid presses on every surface of the object, and it presses harder lower down.
So the push on the bottom is bigger than the push on the top. The difference is the upthrust.
Spot it: Buoyancy depends on the fluid's density and the volume pushed aside — not on the object's own density or what it is made of.
More of the object underwater → more fluid pushed aside → bigger upthrust.
Same ball, same tank — only how far in it goes changes.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
Free preview
This is the free notes preview
You're reading the free notes. Aimnova Pro unlocks the full study experience — and you can try it with your first topic free to keep:
- FlashcardsLock in vocabulary and key terms with spaced repetition.
- Practice questionsAnswer exam-style questions and get instant AI marking.
- Mock exams & past-paper vaultSit full mocks and see exactly how examiners award marks.
- Personalised study planA daily plan built around your exam date and weak areas.
Archimedes' principle in symbols: the upthrust equals the weight of fluid displaced, which is the fluid's density × the volume pushed aside × g.
- buoyancy (upthrust) force (N)
- density of the fluid (kg m⁻³)
- volume of fluid pushed aside (m³)
- gravitational field strength (9.8 N kg⁻¹)
Two things to get right: 1. Use the fluid's density for ρ — not the object's.
2. V is only the submerged volume (the part actually under the surface), not always the whole object.
You also need density = mass ÷ volume to swap between an object's mass and its size. It's a simple quotient, so use a formula triangle:
- density (kg m⁻³)
- mass (kg)
- volume (m³)
m = mass, ρ = density, V = volume. Cover the one you want: two side by side → multiply; one above the other → divide.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
Two forces on an object in a fluid: weight (mg) pulls down, buoyancy (Fb) pushes up. Floating ⇒ they balance.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
A metal sphere of volume 2.0 × 10⁻³ m³ is held fully underwater. Water has density 1.0 × 10³ kg m⁻³ and g = 9.8 N kg⁻¹. Find the buoyancy force on it.
Model answer plan
See the answer broken into clear points, with guidance on how each point earns marks.
Learn what examiners really want
See exactly what to write to score full marks. Our AI shows you model answers and the key phrases examiners look for.
How this is tested — buoyancy almost always comes paired with a force balance. Two flavours:
Paper 1A
- Compare the upthrust on two objects in the same fluid (Fb ∝ V).
- Does it float or sink?
Paper 2
- 'Show that' on floating objects — set buoyancy = weight.
- Find a submerged fraction, a mass, or a density.
The classic trap: Using the object's density for ρ instead of the fluid's — or using the whole volume when only part is submerged.
The floating rule: A floating object is in equilibrium: the upward buoyancy exactly balances its weight.
Fb = weight, so ρfluid × Vsubmerged × g = ρobject × Vtotal × g.
Cancel g from both sides → the fraction submerged equals the density ratio ρobject ÷ ρfluid.
A wooden block floats in water. The wood has density 6.0 × 10² kg m⁻³; water 1.0 × 10³ kg m⁻³. Show that the fraction of the block below the surface is 0.60.
Model answer plan
See the answer broken into clear points, with guidance on how each point earns marks.
The Paper 1B version: weighing it as it goes in
A cylinder hangs from a force sensor and is lowered into a liquid. The sensor reads F = weight − upthrust, so F falls as more of the cylinder goes under. The cylinder's cross-sectional area is 8.0 × 10⁻⁴ m².
Animated graph
Watch the graph build step by step in study mode.
Reading this graph: The intercept is the weight (no upthrust yet). The gradient is −ρAg: divide its size by A and g for the liquid's density.
The weight ÷ g gives the cylinder's mass; divide by its volume (A × length) for the cylinder's own density.