Archimedes' Principle — Buoyancy Formula & Examples

Archimedes' principle says that a partly or fully immersed object experiences an upward force equal to the weight of the fluid it displaces. That upward force is called the buoyant force. In many introductory physics problems, you calculate it with

$$F_b = \rho_{\mathrm{fluid}} g V_{\mathrm{displaced}}$$

where $F_b$ is buoyant force, $\rho_{\mathrm{fluid}}$ is the fluid density, $g$ is gravitational acceleration, and $V_{\mathrm{displaced}}$ is the volume of fluid pushed aside. This form works when the fluid density in the displaced region can be treated as known and $g$ is approximately uniform.

What Archimedes' Principle Actually Tells You

The principle does not say that every object floats. It says every immersed object gets an upward buoyant force.

What happens next depends on how that upward force compares with the object's weight:

• buoyant force upward
• weight downward

If the buoyant force is larger than the object's weight, the object tends to rise. If it is smaller, the object tends to sink. If the two are equal and other forces balance, the object can stay in equilibrium.

For a floating object at rest, the buoyant force equals the object's weight. That is why a floating object settles at a depth where it displaces exactly enough fluid.

Why Buoyant Force Points Up

Fluid pressure usually increases with depth. The bottom of an immersed object is therefore pushed harder than the top.

That pressure difference creates a net upward force. Archimedes' principle gives you a shortcut for finding that force without adding up the pressure on every part of the surface.

Worked Example: Buoyant Force On A Submerged Block

A metal block is fully submerged in fresh water and displaces $0.005\ \mathrm{m^3}$ of water. Use

• $\rho_{\mathrm{water}} = 1000\ \mathrm{kg/m^3}$
• $g = 9.8\ \mathrm{m/s^2}$

Then

$$F_b = \rho_{\mathrm{fluid}} g V_{\mathrm{displaced}}$$

becomes

$$F_b = (1000)(9.8)(0.005) = 49\ \mathrm{N}$$

So the water pushes upward on the block with a buoyant force of $49\ \mathrm{N}$.

If the block's weight is $60\ \mathrm{N}$, its weight is larger than the buoyant force, so it tends to sink. If its weight is $49\ \mathrm{N}$, the forces balance, so it can be in equilibrium in that fluid.

This example shows the key idea: the buoyant force is set by the fluid and the displaced volume. Whether the object rises or sinks still depends on its weight.

Why Floating Objects Displace Just Enough Fluid

If an object floats at rest, its weight must equal the buoyant force. That means the object displaces exactly enough fluid for the displaced fluid's weight to match the object's weight.

That is why a steel ship can float even though steel itself is denser than water. The ship's shape lets it displace a large volume of water before the hull is fully submerged.

Common Mistakes With The Buoyancy Formula

Using the object's density in the buoyancy formula

The formula uses the fluid's density. The object's density matters for whether it floats or sinks, but not directly in $F_b = \rho_{\mathrm{fluid}} g V_{\mathrm{displaced}}$.

Using total volume when the object is only partly submerged

For a floating object, the displaced volume is only the submerged part. Total object volume is correct only when the object is fully submerged.

Treating buoyant force as the same thing as net force

Buoyant force is one force. The object's motion depends on the net force after comparing buoyancy with weight and any other relevant forces.

Forgetting the model conditions

In many school problems, the fluid density is treated as constant. In more complex cases, such as strongly varying density with depth, the simple formula still comes from pressure ideas, but the setup may need more careful treatment.

Where Archimedes' Principle Is Used

Archimedes' principle is used in ship design, submarines, hydrometers, hot-air balloons, and fluid statics more broadly. It is one of the quickest ways to connect pressure, density, and equilibrium in a physical system.

It is also a practical shortcut. If you know the displaced volume and the fluid density, you can estimate the support force without modeling the full pressure field.

Try A Similar Buoyancy Problem

Keep the displaced volume at $0.005\ \mathrm{m^3}$, but change the fluid from water to oil or seawater. Since only $\rho_{\mathrm{fluid}}$ changes in the formula, you can see immediately how fluid density changes the upward force. If you want a useful next step, try your own version with new numbers and decide whether the object rises, sinks, or stays in equilibrium.