The big idea: Any object in a fluid (a liquid or a gas) feels an upward push called the buoyancy force (or upthrust).
The push happens because the fluid presses harder underneath the object than on top.
Archimedes' principle: that upward push equals the weight of the fluid the object pushes out of the way.
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
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.
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)
[Diagram: phys-formula-triangle] - Available in full study mode
[Diagram: phys-free-body] - Available in full study mode
Worked example — buoyancy on a submerged sphere
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.
Solution
- Start with the given formula:
- Put in the numbers (ρ = 1.0 × 10³, V = 2.0 × 10⁻³, g = 9.8):
- Work it out — keep the unit:
Final answer
Fb = 20 N (about 19.6 N), directed upward.
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How this is tested: Buoyancy almost always comes paired with a force balance.
- Paper 1A: quick ratio MCQs — e.g. compare the upthrust on two objects of different size in the same fluid (Fb ∝ V). - Paper 2: 'show that' questions on floating objects — set buoyancy = weight to find a submerged fraction, a mass, or a density.
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.
F_b = weight, so ρ_fluid × V_{submerged} × g = ρ_object × V_{total} × g.
Cancel g from both sides → the fraction submerged equals the density ratio ρobject ÷ ρfluid.
IB-style question — fraction of a block that floats underwater
A wooden block floats in water. The wood has density 6.0 × 10² kg m⁻³; water has density 1.0 × 10³ kg m⁻³. Show that the fraction of the block below the surface is 0.60.
Solution
- Floating ⇒ equilibrium: buoyancy = weight.
- Cancel g, then divide by Vtotal to get the submerged fraction:
- Put in the densities:
Final answer
0.60 — 60% of the block sits below the surface. The submerged fraction is just the density ratio (object ÷ fluid).