A fluid at rest exerts pressure on every surface it touches. Because hydrostatic pressure increases with depth, an immersed body generally experiences a larger pressure force on its lower surfaces than on its upper surfaces. The resulting net fluid force is the buoyant force.

For a simple rectangular body of horizontal area $A$ and submerged height $h$ in a fluid of constant density $\rho_f$, the pressure difference between its bottom and top is

$$p_{\rm bottom}-p_{\rm top}=\rho_fgh.$$

The horizontal pressure forces cancel in pairs, while the vertical pressure forces give

$$F_B=(p_{\rm bottom}-p_{\rm top})A =\rho_fghA.$$

Since $hA$ is the displaced volume $V_{\rm disp}$,

$$F_B=\rho_fgV_{\rm disp}.$$

The quantity $\rho_fV_{\rm disp}$ is the mass of the fluid that would occupy the displaced volume. Therefore the buoyant force equals the weight of the displaced fluid:

$$\boxed{F_B=\rho_fgV_{\rm disp}}.$$

This is Archimedes' principle. The same result applies to an arbitrarily shaped body surrounded by a hydrostatic fluid: adding the pressure forces over its surface produces an upward resultant equal to the weight of the displaced fluid. For a fluid whose density is effectively uniform over the displaced volume, the compact expression above applies directly.

Partially and fully immersed bodies

The relevant volume is always the volume of fluid actually displaced.

• For a fully submerged rigid body, $V_{\rm disp}$ is the body's submerged volume.
• For a body crossing the free surface, only the portion below the fluid surface contributes to $V_{\rm disp}$.

The buoyant force does not depend directly on the object's own mass or density. Those properties matter when the buoyant force is compared with the object's weight.

Apparent weight in a fluid

A force sensor supporting an immersed object does not usually read $mg$. If a fully submerged object is held at rest by an upward tension $T$, vertical force balance gives

$$T+F_B-mg=0,$$

so

$$T=mg-F_B.$$

The reduced support force is sometimes called the object's apparent weight in the fluid.

Example

A $5.0,\mathrm{kg}$ object of volume $2.0\times10^{-3},\mathrm{m^3}$ is completely submerged in water with $\rho_f=1000,\mathrm{kg/m^3}$.

Its buoyant force is

$$F_B=(1000)(9.81)(2.0\times10^{-3})\approx19.6,\mathrm N.$$

Its weight is

$$mg=(5.0)(9.81)\approx49.1,\mathrm N.$$

A spring scale holding it at rest therefore reads

$$T=49.1-19.6\approx29.5,\mathrm N.$$

The fluid has not changed the object's gravitational weight; the pressure gradient supplies an additional upward force, reducing the support force required.