David Bauer Physics & Astronomy · UCLA

Physics 5B Chapter 10

Buoyancy

Buoyancy

The buoyant force

Because pressure increases with depth, a submerged object feels a larger upward push on its bottom than the downward push on its top. The net upward force from the fluid pressure is the buoyant force $\vec{F}_B$. The buoyant force is what lifts a helium balloon and what makes you feel lighter in a swimming pool.

An open tank of water with a solid cylinder fully submerged. Red pressure-force arrows act on the cylinder's faces: a short arrow pushes down on the top face, labeled p sub top times A, and a longer arrow pushes up on the bottom face, labeled p sub bot times A, longer because the pressure is greater at the greater depth. A pair of equal horizontal arrows push inward on the two sides at the same height and cancel. Dimension arrows mark the depth d from the surface to the top of the cylinder and the cylinder's height h. Because the bottom face sits deeper than the top face, the upward force wins, and the net upward buoyant force equals rho sub f g h A, the weight of the displaced fluid.

Derivation: The buoyant force

Consider a cylinder of height $h$ and face area $A$, with its top submerged at depth $d$. The horizontal pressure forces on its sides cancel. The net upward force is the difference between the pressure forces on its bottom and top:

\[F_B = p_{\mathrm{bot}}A - p_{\mathrm{top}}A = \bigl[p_0 + \rho_f g (d + h)\bigr] A - \bigl(p_0 + \rho_f g \, d\bigr) A = \rho_f g \underbrace{hA}_{V}.\]

Notice that the surface pressure $p_0$ and the depth $d$ both cancel. The buoyant force depends on the volume of fluid displaced, not on how far below the surface the cylinder is.

Archimedes’ principle

This result holds for any shape. The quantity $\rho_f V g$ is the weight of the displaced fluid, so we can state the result as

The magnitude of the buoyant force on an object equals the weight of the fluid it displaces.

This is called Archimedes’ principle. As an equation,

\[\boxed{F_B = \rho_f \, V_{\mathrm{disp}} \, g}\]

In this equation, $\rho_f$ is the density of the fluid, and $V_{\mathrm{disp}}$ is the volume of the object below the fluid surface.

Float or sink?

Whether an object floats depends on its density compared to the fluid’s. For an object of volume $V$ and density $\rho_o$ fully submerged in a fluid of density $\rho_f$, the net force is

\[F_{\mathrm{net},y} = F_B - mg = \rho_f V g - \rho_o V g = (\rho_f - \rho_o) V g.\]

The direction of the net force depends on the relative densities:

  • If $\rho_o > \rho_f$, the net force is downward, so the object sinks.
  • If $\rho_o = \rho_f$, the object is in equilibrium while fully submerged. This is called neutral buoyancy.
  • If $\rho_o < \rho_f$, the net force is upward, so the object rises and floats at the surface.

Floating, sinking, and neutral buoyancy

Three tanks of water side by side, each holding one solid block, with two red force arrows drawn from each block's center: an upward arrow for the buoyant force F sub B and a downward arrow for the weight m g. In the left tank, labeled sinks, rho sub o greater than rho sub f, the block sits deep in the water and the weight arrow is longer than the buoyant-force arrow, so the net force is downward. In the middle tank, labeled neutral, rho sub o equals rho sub f, the block hovers fully submerged at mid depth and the two arrows have equal length. In the right tank, labeled floats, rho sub o less than rho sub f, the block sits at the surface only partly submerged, and the two arrows again have equal length: the block sinks just deep enough that the weight of the displaced water balances its own weight.

A floating object is in equilibrium when the buoyant force equals its weight. By Archimedes’ principle, this means that the displaced fluid weighs as much as the object: $F_B = \rho_f V_{\mathrm{disp}}\, g = mg$.

Apparent weight

Even when a submerged object is dense enough to sink, the buoyant force supports part of its weight. The additional force that must be applied to keep an object in equilibrium is called the apparent weight:

\[w_{\mathrm{app}} = mg - F_B.\]

This apparent weight is what a scale supporting an object measures.

This is the principle behind hydrostatic weighing for body composition. Weighing a person in air and again underwater gives their volume, and hence their average density. Since fat is less dense than muscle and bone, the average density can be used to estimate body fat.

Example: A floating cube

A solid cube with side length $10\ \mathrm{cm}$ floats in water with the bottom face of the cube $4.0\ \mathrm{cm}$ below the surface. What is the density of the cube?

An open tank of water with a solid cube floating at the surface. The cube's side length is marked as 10 centimeters by a dimension arrow across its top. The waterline crosses the cube as a dashed line, and a dimension line on the right marks the submerged depth: the bottom face of the cube sits 4.0 centimeters below the water surface, so less than half of the cube is under water. The question is to find the density of the cube from how deep it floats.

Solution

The cube floats in equilibrium, so the buoyant force equals its weight, $F_B = mg$. Let $A$ be the area of the cube’s bottom face. The submerged part of the cube has volume $V_{\mathrm{disp}} = (0.040\ \mathrm{m})A$, and the whole cube has volume $(0.10\ \mathrm{m})A$, so

\[\rho_w g\,(0.040\ \mathrm{m})A = \rho_c g\,(0.10\ \mathrm{m})A.\]

Canceling $g$ and $A$ and solving for the density of the cube,

\[\rho_c = (0.40)(1000\ \mathrm{kg/m^3}) = \boxed{400\ \mathrm{kg/m^3}}.\]