Buoyancy and Archimedes' principle

Drop a stone into a pond and it sinks; drop a block of wood and it bobs back up. Lift the same stone underwater and it suddenly feels lighter in your hand than it did in air. Both effects trace to one cause: a fluid pushes back on anything immersed in it, and that push is always directed upward. This upward push is the buoyant force, and its size is fixed by a beautifully simple rule discovered over two thousand years ago by Archimedes while stepping into a bath. This article builds the idea from the pressure difference between the top and bottom of a submerged body, states Archimedes' principle precisely, and uses it to decide when something floats, sinks, or hangs in equilibrium. It continues directly from Pascal's law and the fact that fluid pressure grows with depth, $P = P_0 + \rho g h$ — the same fact that explains ships, icebergs, hydrometers, and why your ears feel different pressure at the bottom of a swimming pool.

The core insight

A body immersed in a fluid feels an upward buoyant force equal to the weight of the fluid it displaces, $F_B = \rho_{\text{fluid}} V_{\text{displaced}} g$ , because fluid pressure — which grows with depth — pushes harder on the body's lower surface than on its upper surface.

The core idea

Where the push comes from. Pressure in a static fluid increases with depth: $P = P_0 + \rho g h$ . Every surface of a submerged body feels this pressure pushing perpendicular to it. The side pushes cancel by symmetry, but the bottom face sits deeper than the top face, so it feels a larger pressure pushing up than the top face feels pushing down. The two do not cancel — their net effect is a single upward force, the buoyant force or upthrust.

Archimedes' principle. A body partly or fully immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid it displaces:

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

where: $\rho_{\text{fluid}}$ is the density of the fluid, $V_{\text{displaced}}$ is the volume of fluid pushed out of the way by the immersed part of the body, and $g$ is the acceleration due to gravity. Only the fluid's density and the displaced volume matter — the body's own material, shape, and weight never enter this formula directly.

Figure

A submerged cube showing greater upward pressure force on its deeper bottom face than the downward force on its shallower top face, with the difference equal to the buoyant force

Floating, sinking, or hovering. Compare the body's average density $\rho_{\text{body}}$ with the fluid's density $\rho_{\text{fluid}}$ : if $\rho_{\text{body}} < \rho_{\text{fluid}}$ the body floats, submerging only enough of its volume that the displaced weight balances its own weight; if $\rho_{\text{body}} = \rho_{\text{fluid}}$ it can rest in equilibrium fully submerged at any depth; if $\rho_{\text{body}} > \rho_{\text{fluid}}$ it sinks, since even full immersion cannot displace enough fluid weight to hold it up. For a floating body, equating weight to buoyant force gives:

\frac{V_{\text{submerged}}}{V_{\text{body}}} = \frac{\rho_{\text{body}}}{\rho_{\text{fluid}}}

where: $V_{\text{submerged}}$ is the volume below the fluid surface and $V_{\text{body}}$ is the body's total volume. An object weighed while immersed also shows an apparent weight loss equal to the buoyant force, $W_{\text{apparent}} = W_{\text{true}} - F_B$ .

Derivation ABuoyant force from the pressure on a submerged cube

The goal is to show, using nothing beyond the depth-pressure relation $P = P_0 + \rho g h$ and the definition of pressure as force per unit area, that the net upward force on any fully submerged body equals exactly the weight of the fluid it displaces — the result Archimedes stated without this algebra, and the one every buoyancy calculation in this article rests on.

Step 1. Model the body as a cube of face area $A$ with its top face at depth $h_1$ below the fluid surface and its bottom face at depth $h_2 = h_1 + L$ , where $L$ is the cube's side length. The pressures at these two depths are $P_1 = P_0 + \rho_{\text{fluid}} g h_1$ on top and $P_2 = P_0 + \rho_{\text{fluid}} g h_2$ on the bottom, both following directly from the depth-pressure law.

Step 2. Convert each pressure to a force. Pressure acts perpendicular to a surface, so the downward force pressing on the top face is $F_1 = P_1 A$ and the upward force pressing on the bottom face is $F_2 = P_2 A$ . The four side faces feel horizontal pressure forces that cancel in pairs by symmetry, so they contribute nothing to the vertical balance.

Step 3. Subtract to get the net vertical force. The net upward force is $F_B = F_2 - F_1 = (P_2 - P_1)A = (\rho_{\text{fluid}} g h_2 - \rho_{\text{fluid}} g h_1)A = \rho_{\text{fluid}} g (h_2 - h_1) A$ . Since $h_2 - h_1 = L$ , this becomes $F_B = \rho_{\text{fluid}} g L A$ .

Step 4. Recognise $LA$ as the cube's volume $V$ , which for a fully submerged cube equals the displaced fluid volume $V_{\text{displaced}}$ . So $F_B = \rho_{\text{fluid}} V_{\text{displaced}} g$ — exactly Archimedes' principle, and the quantity $\rho_{\text{fluid}} V_{\text{displaced}} g$ is precisely the weight of the fluid the cube pushed out of its way. Although this derivation used a cube for convenient geometry, the same pressure-integration argument holds for any shape, because any solid can be built up from thin vertical slabs, each obeying the same $P_2 > P_1$ logic. Verified: for the general result, integrating pressure over any closed surface of a body immersed in a fluid of uniform density gives net vertical force $\rho_{\text{fluid}} g V_{\text{displaced}}$ , independent of shape — the standard result underlying Archimedes' principle.

Worked example

A solid block of volume $2.0\times10^{-3}\ \text{m}^3$ and density $2500\ \text{kg/m}^3$ is fully submerged in water ( $\rho_{\text{water}} = 1000\ \text{kg/m}^3$ ). Find its true weight, the buoyant force on it, and its apparent weight in water. Take $g = 9.8\ \text{m/s}^2$ .

Step 1. Find the mass and true weight. Mass $= \rho_{\text{body}} V = 2500 \times 2.0\times10^{-3} = 5.0\ \text{kg}$ , so $W_{\text{true}} = mg = 5.0 \times 9.8 = 49\ \text{N}$ .

Step 2. Find the buoyant force. The block is fully submerged, so $V_{\text{displaced}} = V_{\text{body}} = 2.0\times10^{-3}\ \text{m}^3$ , giving $F_B = \rho_{\text{water}} V g = 1000 \times 2.0\times10^{-3} \times 9.8 = 19.6\ \text{N}$ .

Step 3. Find the apparent weight. $W_{\text{apparent}} = W_{\text{true}} - F_B = 49 - 19.6 = \mathbf{29.4\ N}$ . The block feels about $40\%$ lighter underwater, consistent with its density being $2.5$ times that of water — heavier bodies still lose only the weight of the fluid they displace, never their full weight.

Competitive stretch

An iceberg (density $\rho_{\text{ice}} = 920\ \text{kg/m}^3$ ) floats in seawater (density $\rho_{\text{sea}} = 1025\ \text{kg/m}^3$ ). What fraction of its volume stays above the surface? Using the floating-body relation, $V_{\text{submerged}}/V_{\text{body}} = \rho_{\text{ice}}/\rho_{\text{sea}} = 920/1025 \approx 0.898$ , so $89.8\%$ is submerged and $\mathbf{10.2\%}$ rises above the water — the familiar "tip of the iceberg."

Common traps

Using the body's own density instead of the fluid's density in $F_B = \rho_{\text{fluid}} V_{\text{displaced}} g$ — buoyant force depends only on what the fluid weighs, not what the body is made of.
Assuming $V_{\text{displaced}} = V_{\text{body}}$ for a floating object — that equality holds only when the body is fully submerged; a floating body displaces just its submerged fraction.
Forgetting that apparent weight loss equals the buoyant force, not the whole weight — a dense object still feels most of its true weight underwater.
Thinking buoyancy needs a special shape — Archimedes' principle applies to any shape, since only the displaced volume matters, as the cube derivation generalises.

Ask the tutor: Want to practice more apparent-weight or floating-fraction problems, or see why a hydrometer works using this same principle?

Why does a heavy steel ship float when a small steel nail sinks?

Density decides this, not the material alone. A solid steel nail has the density of steel throughout, about $7850\ \text{kg/m}^3$ , far above water, so it sinks. A ship's hull encloses a huge volume of air, so its average density — total mass divided by the full hull volume — is much less than water's, even though the steel itself is dense. The ship settles low enough that the weight of water it displaces equals its total weight, exactly as Archimedes' principle predicts, and it floats.

Does buoyant force depend on how deep the object is submerged?

Once a body is fully submerged, no — the buoyant force stays the same at any depth, because it depends only on the displaced volume and the fluid's density, both of which stay fixed as long as the body's volume and the fluid's density do not change. What does change with depth is the absolute pressure on the body's surfaces, but the difference between top and bottom pressure, which is what produces the net upward force, stays constant. This is why a fully submerged body of density equal to the fluid can rest in equilibrium at any depth.

If I weigh an object in air and then in water, is the difference always the buoyant force?

Yes, exactly — that is the standard way buoyant force is measured in a lab. The spring balance reading in water, the apparent weight, is less than the reading in air, the true weight, and the difference $W_{\text{true}} - W_{\text{apparent}}$ equals $F_B$ by definition. This trick also gives a quick way to find a body's density: since $F_B = \rho_{\text{water}} V g$ and $W_{\text{true}} = \rho_{\text{body}} V g$ , dividing the two weights gives $\rho_{\text{body}}/\rho_{\text{water}}$ without ever measuring volume directly.

Buoyancy and Archimedes' principle

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads