Archimedes' Principle — Buoyancy Formula & Examples

Use Archimedes' principle whenever an object sits in a fluid and you need the upward push the fluid gives it, or whether the object floats, sinks, or hangs in balance. The principle says a partly or fully immersed object experiences an upward force, the buoyant force, equal to the weight of the fluid it displaces. The steps below apply it without the usual slips.

Step 1: Identify The Fluid

Use the density of the fluid, not the density of the object. In many introductory problems the buoyant force is

where $\rho_{\mathrm{fluid}}$ 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. The object's density matters for whether it floats, but it does not appear in this formula.

Step 2: Find The Displaced Volume

Use the submerged volume. It equals the full object volume only when the object is fully submerged. For a floating object, the displaced volume is just the submerged part.

Step 3: Apply The Buoyancy Formula

Compute $F_b = \rho_{\mathrm{fluid}} g V_{\mathrm{displaced}}$ with consistent SI units. The reason this points up is physical: fluid pressure usually increases with depth, so the bottom of an immersed object is pushed harder than the top, and that pressure difference is a net upward force. Archimedes' principle is the shortcut that finds it without summing pressure over every surface.

Step 4: Compare Forces

Compare the buoyant force with the object's weight to decide what happens. If buoyancy exceeds weight, the object tends to rise; if it is smaller, the object tends to sink; if they 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, which is why a floating object settles at a depth where it displaces exactly enough fluid.

Full Worked Example: A Submerged Block

A metal block is fully submerged in fresh water and displaces $0.005\ \mathrm{m^3}$ of water. Use $\rho_{\mathrm{water}} = 1000\ \mathrm{kg/m^3}$ and $g = 9.8\ \mathrm{m/s^2}$.

Step 1: the relevant density is the water's. Step 2: fully submerged, so $V_{\mathrm{displaced}} = 0.005\ \mathrm{m^3}$. Step 3:

so the water pushes up with $49\ \mathrm{N}$. Step 4: if the block weighs $60\ \mathrm{N}$, weight exceeds buoyancy, so it tends to sink; if it weighs $49\ \mathrm{N}$, the forces balance and it can be in equilibrium. The buoyant force is set by the fluid and the displaced volume; whether the object rises or sinks still depends on its weight. This is also why a steel ship floats even though steel is denser than water: its shape displaces a large volume before the hull is fully under.

Where Each Step Trips People Up

Step 1 (fluid): Using the object's density in the buoyancy formula. The formula uses the fluid's density. Self-check: did I plug in $\rho_{\mathrm{fluid}}$, not $\rho_{\mathrm{object}}$?

Step 2 (displaced volume): Using total volume when the object is only partly submerged. For a floating object, only the submerged part counts. Self-check: is the object fully under, or just partly?

Step 3 (formula and conditions): Forgetting the model conditions. School problems usually treat fluid density as constant; strongly varying density with depth needs more careful treatment from pressure ideas. Self-check: is constant $\rho_{\mathrm{fluid}}$ reasonable here?

Step 4 (compare): Treating buoyant force as the net force. Buoyancy is one force; motion depends on the net force after comparing it with weight and any other forces. Self-check: have I added up all the forces, not just buoyancy?

Where Archimedes' Principle Is Used

It appears 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, and a practical shortcut: knowing the displaced volume and the fluid density lets you estimate the support force without modeling the full pressure field.

To practice, keep the displaced volume at $0.005\ \mathrm{m^3}$ but switch the fluid from water to oil or seawater. Since only $\rho_{\mathrm{fluid}}$ changes, you can see at once how fluid density changes the upward force, then decide whether the object rises, sinks, or stays in equilibrium.