Density and buoyancy: Archimedes' principle explained

Key ideas

\rho = \frac{m}{V}

$\rho$ (rho) is density, in kg/m³ or g/cm³. $1\ \text{g/cm}^3 = 1000\ \text{kg/m}^3$ .
$m$ is mass (kg or g) and $V$ is volume (m³ or cm³). 1 mL = 1 cm³ and 1 L = 1000 cm³.

F_b = \rho_{\text{fluid}}\, V_{\text{displaced}}\, g

$F_b$ is the buoyant force, in newtons (N), pointing up.
$\rho_{\text{fluid}}$ is the density of the fluid, in kg/m³ (water: 1000 kg/m³).
$V_{\text{displaced}}$ is the volume of fluid pushed aside, in m³. For a fully submerged object, it is the object's whole volume.
$g = 9.8\ \text{m/s}^2$ .

A floating object sinks until it displaces its own weight of fluid. That gives a quick rule: the fraction of a floating object under the surface equals its density divided by the fluid's density.

Densities of some materials (OpenStax College Physics 2e, Table 11.1)
Material	Density (g/cm³)	In water
Air (0 °C)	0.00129	Rises
Ice	0.917	Floats
Water (4 °C)	1.000	Neutral
Seawater	1.025	Neutral (in seawater)
Aluminum	2.7	Sinks
Iron or steel	7.8	Sinks
Gold	19.32	Sinks

Worked examples

Example 1: identify a metal by its density

Problem A metal block has a mass of 54.0 g and a volume of 20.0 cm³. What is its density, and what metal could it be?

Divide mass by volume.
$\rho = \frac{54.0\ \text{g}}{20.0\ \text{cm}^3} = 2.70\ \text{g/cm}^3$
Compare with the table: aluminum is 2.7 g/cm³.

Answer 2.70 g/cm³, which matches aluminum

Example 2: buoyant force on a sunken rock

Problem A rock with a volume of 0.0030 m³ is fully under water. What buoyant force acts on it?

Fully submerged, so the displaced volume is the rock's volume.
$F_b = (1000\ \text{kg/m}^3)(0.0030\ \text{m}^3)(9.8\ \text{m/s}^2)$
Multiply.
$F_b = 29.4\ \text{N}$

Answer 29 N upward

Example 3: apparent weight

Problem That rock has a mass of 8.0 kg. How much does it seem to weigh while hanging from a spring scale under water?

Its real weight.
$W = mg = (8.0\ \text{kg})(9.8\ \text{m/s}^2) = 78.4\ \text{N}$
The buoyant force from Example 2 pushes up, so the scale reads less.
$W_{\text{apparent}} = 78.4\ \text{N} - 29.4\ \text{N} = 49.0\ \text{N}$

Answer 49 N

Example 4: how much of a block floats under water

Problem A wooden block has a density of 600 kg/m³. What fraction of it is under water when it floats?

Floating means buoyant force equals weight.
$\rho_{\text{water}} V_{\text{sub}}\, g = \rho_{\text{wood}} V\, g$
Solve for the fraction under water.
$\frac{V_{\text{sub}}}{V} = \frac{\rho_{\text{wood}}}{\rho_{\text{water}}} = \frac{600}{1000} = 0.60$

Answer 0.60, so 60% of the block is under water

Common mistakes and how to fix them

Using the object's density in the buoyancy formula. Fix: $F_b$ uses the fluid's density and the volume of fluid pushed aside.
Thinking heavy things always sink. A steel ship floats because its hollow shape gives it a low average density. Fix: compare densities, not weights.
Mixing g/cm³ and kg/m³. Fix: multiply g/cm³ by 1000 to get kg/m³.
Using the whole volume for a floating object. Fix: only the part under the surface displaces water.

Practice problems

A sample has a mass of 250 g and a volume of 100. cm³. What is its density?
1. 0.400 g/cm³
2. 2.50 g/cm³
3. 25.0 g/cm³
4. 350 g/cm³
Show answer
Answer: 2.50 g/cm³
$\rho = m/V = 250 \div 100. = 2.50$ g/cm³.
Ice has a density of 0.917 g/cm³. Does it float in fresh water?
1. Yes, it floats
2. No, it sinks
Show answer
Answer: Yes, it floats
0.917 g/cm³ is less than water's 1.00 g/cm³, so ice floats with most of its volume under the surface.
What fraction of an iceberg (0.917 g/cm³) floating in seawater (1.025 g/cm³) is under the surface?
Show answer
Answer: 0.895 (about 89%)
Fraction under = $\rho_{\text{ice}} / \rho_{\text{sea}} = 0.917 \div 1.025 = 0.895$ . Only about a tenth shows above the water.
A 5.00 × 10⁻⁴ m³ object is fully under water. What is the buoyant force on it?
Show answer
Answer: 4.90 N
$F_b = 1000 \times (5.00 \times 10^{-4}) \times 9.8 = 4.90$ N.
What is the mass of 2.0 L of water?
1. 0.50 kg
2. 2.0 kg
3. 20. kg
4. 2000 kg
Show answer
Answer: 2.0 kg
2.0 L = 2000 cm³, and $m = \rho V = 1.00 \times 2000 = 2000$ g = 2.0 kg.

Frequently asked questions

Why does a steel ship float?

A ship's hull is mostly air inside, so the ship's average density, steel plus air, is less than water's. It sinks until the water it pushes aside weighs as much as the whole ship. If the hull fills with water, its average density rises above water's and it sinks.

Why is it easier to float in the ocean than in a pool?

Seawater is denser than fresh water because of the dissolved salt, about 1.025 g/cm³ instead of 1.00. A denser fluid gives more buoyant force for the same displaced volume, so you float a little higher.

Does buoyancy happen in air too?

Yes. Air is a fluid, so it pushes up on everything with a force equal to the weight of air displaced. It is tiny for most objects, but a helium balloon rises because its total density is less than the air's.

Sources

OpenStax College Physics 2e, 11.2 Density, accessed October 1, 2026
OpenStax College Physics 2e, 11.7 Archimedes' Principle, accessed October 1, 2026

Key ideas

Worked examples

Common mistakes and how to fix them

Frequently asked questions

Sources

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