Floatation

Why some bodies float instead of sinking. A body floats in a liquid when its average density is less than the density of the liquid. Such a body cannot remain fully submerged: if it were pushed entirely under, the upthrust on its whole volume would exceed its weight, and the net force would shove it back up. It keeps rising until only part of it stays below the surface — just enough that the upthrust from that immersed part alone matches the weight. This balance is floating equilibrium:

where: $F_B$ is the upthrust (buoyant force) on the immersed part of the body, and $W$ is the total weight of the body.

Turning the balance into a density ratio. Let $\rho_1$ be the body's density, $V_2$ its total volume, $\rho_2$ the liquid's density, and $V_1$ the volume of the body that is submerged. The weight is $\rho_1 V_2 g$, and the upthrust equals the weight of liquid displaced by the immersed part, $\rho_2 V_1 g$. This ratio is worked out in full below as the node's main derivation.

Reading the ratio. The relation says the fraction of the body submerged equals the ratio of its density to the liquid's density — nothing more, nothing less. A body with density close to the liquid's floats almost fully submerged, riding low with only a sliver above the surface. A body much lighter than the liquid floats high, with most of its bulk poking out. Because the body is in equilibrium, $F_B = W$ means its apparent weight is zero — a floating object placed on a hypothetical spring balance dipped in the liquid would read zero, since the liquid's upthrust already carries its entire weight.

Everyday applications. A ship is hollow, enclosing a large volume of air; its average density (steel plus cargo plus enclosed air, divided by the total outer volume) is less than water's, even though solid steel itself sinks. A submarine adjusts its average density with ballast tanks — filling them with water increases average density to submerge, emptying them lowers it to rise. An iceberg floats mostly submerged because ice is only slightly less dense than seawater, so a large fraction of its volume must stay underwater to displace enough water to balance its weight. A balloon filled with a gas lighter than air floats in air by the same density-ratio logic, treating air itself as the surrounding fluid.

A block of wood has density $600\ \text{kg/m}^3$ and floats in water of density $1000\ \text{kg/m}^3$. What fraction of the block is above the surface, and if the block has total volume $5 \times 10^{-3}\ \text{m}^3$, what volume of it is submerged?

Step 1. Apply the floatation relation to find the submerged fraction. $\dfrac{\rho_1}{\rho_2} = \dfrac{V_1}{V_2} = \dfrac{600}{1000} = 0.6$. So $60\%$ of the block's volume stays submerged.

Step 2. Find the fraction above the surface. Since the whole volume is either above or below the surface, the fraction above is $1 - 0.6 = 0.4$, i.e. $40\%$ of the block floats above the water.

Step 3. Convert the submerged fraction into an actual volume using the given total volume. $V_1 = 0.6 \times V_2 = 0.6 \times 5\times10^{-3}\ \text{m}^3 = 3\times10^{-3}\ \text{m}^3$.

The block floats with 40% of its volume above water and 60%, or $3\times10^{-3}\ \text{m}^3$, submerged.

Competitive stretch

A solid of volume $2\times10^{-4}\ \text{m}^3$ floats in a liquid of density $1200\ \text{kg/m}^3$ with $1.5\times10^{-4}\ \text{m}^3$ of its volume immersed. What is the density of the solid, and does that value make physical sense for a floating body?

Step 1. Substitute directly into the floatation relation rearranged for $\rho_1$: $\rho_1 = \rho_2 \times \dfrac{V_1}{V_2} = 1200 \times \dfrac{1.5\times10^{-4}}{2\times10^{-4}} = 1200 \times 0.75 = 900\ \text{kg/m}^3$.

Step 2. Check the answer against the physics. $900\ \text{kg/m}^3$ is less than the liquid's $1200\ \text{kg/m}^3$, which is exactly what floating requires — a density greater than $1200\ \text{kg/m}^3$ would sink the solid entirely, and the relation would have no valid answer with $V_1 < V_2$.

The density of the solid is $900\ \text{kg/m}^3$, consistent with it floating rather than sinking.

• Thinking heavy objects always sink. It is average density, not total weight, that decides floating — a hollow steel ship floats while a small solid steel ball sinks.
• Using the total volume as the displaced volume. Only the immersed volume $V_1$ displaces liquid and produces upthrust for a floating body — not the whole volume $V_2$, unlike a fully submerged body where $V_1 = V_2$.
• Believing the upthrust on a floating body exceeds its weight. At equilibrium $F_B = W$ exactly; if upthrust were greater the body would keep rising indefinitely, which never happens.
• Forgetting the apparent weight of a freely floating object is zero, since the liquid's upthrust already balances its entire weight.

Ask the tutor: Want to see how changing the liquid's density (say, floating the same block in oil instead of water) shifts the immersed fraction? Ask the AI tutor to walk through it with you.