Buoyancy
- State Archimedes’ principle
- Compute the buoyant force, F_b = ρ_fluid·V_displaced·g
- Predict floating versus sinking from a density comparison
Archimedes’ principle
A fluid pushes up on any object placed in it with a buoyant force equal to the weight of the fluid the object displaces. This is Archimedes’ principle. It arises because pressure increases with depth: the fluid pushes up harder on the bottom of the object than it pushes down on the top, and the difference is a net upward force.
Float or sink: compare densities
Whether an object floats comes down to a density comparison. If the object’s average density is less than the fluid’s, the buoyant force can equal the object’s weight before it is fully submerged, so it floats. If its density is greater, even full submersion cannot lift enough, so it sinks. A steel ship floats because its overall density — steel plus the air inside — is less than water’s.
A rock is fully submerged in water, displacing 0.01 m³ (ρ_water = 1000 kg/m³, g = 10 m/s²). Find the buoyant force on it.
- 1.Use Archimedes’ principle: F_b = ρ_fluid·V·g.
- 2.Substitute: F_b = 1000 × 0.01 × 10.
- 3.Multiply: 1000 × 0.01 = 10, then × 10.
- 4.Result: F_b = 100 N upward.
Buoyant force uses the fluid’s density, not the object’s. A common error is plugging in the object’s density — but Archimedes’ principle is all about the displaced fluid.
A fully submerged object displaces 0.02 m³ of water (ρ = 1000 kg/m³, g = 10 m/s²). What is the buoyant force on it?
An object is placed in water. It will float if:
For a floating object, the buoyant force exactly equals its weight (it is in equilibrium). The fraction submerged equals the ratio of the object’s density to the fluid’s density — a handy shortcut for iceberg-style problems.
Answer the 2 checkpoints as you read.
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