Floating and sinking of objects

Every object placed in a fluid experiences a tug-of-war between two forces.

• Its own weight, $W$, pulling it straight down.
• The buoyant force, $F_B$, pushing it straight up — by Archimedes' Principle, this equals the weight of the fluid the object displaces.

For a fully or partly submerged object, the buoyant force is

where: $\rho_{\text{fluid}}$ is the density of the fluid, $V_{\text{displaced}}$ is the volume of fluid pushed out of the way by the submerged part of the object, and $g$ is the acceleration due to gravity. The object's own weight is $W = \rho_{\text{object}}\, V_{\text{object}}\, g$, using its average density and its total volume. Whether the object floats or sinks now comes down to comparing $\rho_{\text{object}}$ with $\rho_{\text{fluid}}$, because $g$ cancels out of the comparison entirely:

• Denser object sinks. If $\rho_{\text{object}} > \rho_{\text{fluid}}$, even displacing the object's entire volume of fluid cannot produce enough buoyant force to match its weight, so it keeps sinking until it hits the bottom.
• Less dense object floats. If $\rho_{\text{object}} < \rho_{\text{fluid}}$, the object need not be fully submerged — a smaller displaced volume already weighs as much as the object, so it settles into equilibrium ($F_B = W$) with part of it above the surface.
• Equal densities give neutral suspension. If $\rho_{\text{object}} = \rho_{\text{fluid}}$, the object can stay suspended at any depth, neither rising nor sinking.

The one word that resolves every "but steel is heavy" objection is average. A solid lump of steel has an average density equal to steel's own density — much greater than water's — so it sinks. A steel ship is a large hollow shell enclosing decks, cargo holds, and air; its average density is its total mass divided by the entire volume enclosed by the hull, not just the volume of steel in the plates. That average works out well below water's density, so the ship floats even though the steel it is made of would sink on its own.

Explain, from first principles, why a solid lump of a dense material always sinks while the very same material shaped into a large hollow hull can float, using only the comparison of average density to fluid density and no numerical buoyancy calculation.

Step 1. Fix the two competing forces. Every submerged or partly submerged object is pulled down by its own weight $W$ and pushed up by the buoyant force $F_B$, which equals the weight of fluid it displaces. The object's fate — sink, float, or hang neutrally — depends purely on how these two compare once the object has displaced as much fluid as it possibly can.

Step 2. See why volume alone cannot decide it. A large object displaces more fluid than a small one, but a large object also weighs more. Size cancels out of the comparison, which is exactly why the deciding factor is a ratio — mass per unit volume, i.e., density — rather than raw size or raw mass.

Step 3. Apply this to a solid dense material. A solid iron nail has no hollow space: its average density equals iron's own density, roughly 7.8 times water's. Even when the nail is completely submerged, the heaviest possible displaced volume (the nail's own volume) weighs far less than the nail itself, so $F_B$ can never reach $W$, and the nail sinks all the way down.

Step 4. Apply the same reasoning to a hollow shape of the same material. A steel ship encloses a huge volume of air inside its hull — cargo holds, decks, engine rooms. Its average density is total mass divided by the entire enclosed volume, which is now dominated by that empty space rather than by the steel plating alone. This average density drops below water's, so the ship needs to submerge only part of its hull before the displaced water weighs as much as the whole ship, and it settles into floating equilibrium.

Step 5. Generalise. Nothing about "steel" or "wood" as a label decides floating or sinking — the shape and the empty space it encloses determine the average density, and average density compared with the fluid's density is the entire story. This single comparison also explains why an iceberg, made of ice slightly less dense than seawater, floats with most of its bulk submerged, and why a submarine can move between floating, hovering, and sinking simply by changing how much water it lets into its ballast tanks — changing its average density on demand.

A cargo ship has a total mass of 50,000 tonnes. Its hull encloses a total volume (steel, cargo holds, and all the air-filled space inside) of 60,000 cubic metres. Seawater has a density of about 1,025 kilogrammes per cubic metre. Show that the ship floats, and estimate what fraction of the enclosed hull volume must be submerged for it to be in equilibrium.

Step 1. Find the ship's average density. Convert mass to kilogrammes: 50,000 tonnes is $5.0\times10^{7}\ \text{kg}$. Average density is total mass over total enclosed volume: $\rho_{\text{ship}} = \dfrac{5.0\times10^{7}\ \text{kg}}{60{,}000\ \text{m}^3} \approx 833\ \text{kg/m}^3$.

Step 2. Compare with seawater. Since $833\ \text{kg/m}^3$ is less than seawater's $1{,}025\ \text{kg/m}^3$, the ship's average density is lower than the fluid's, so by the density-comparison rule it floats — it need not submerge its entire hull for the displaced water to equal its own weight.

Step 3. Estimate the submerged fraction. At floating equilibrium, $F_B = W$, which means the weight of the displaced seawater equals the ship's weight: $\rho_{\text{seawater}} \times V_{\text{submerged}} \times g = \rho_{\text{ship}} \times V_{\text{total}} \times g$. The $g$ cancels, leaving $V_{\text{submerged}}/V_{\text{total}} = \rho_{\text{ship}}/\rho_{\text{seawater}} = 833/1{,}025 \approx 0.81$.

Step 4. State the result. About 81 percent of the ship's hull volume must sit below the waterline, with roughly 19 percent — the decks and superstructure — riding above the surface, and the ship floats safely because its average density of 833 kg per cubic metre is well below seawater's 1,025 kg per cubic metre.

Competitive stretch

Icebergs follow exactly the same ratio. Ice has a density of about $920\ \text{kg/m}^3$ and seawater about $1{,}025\ \text{kg/m}^3$, so the submerged fraction is $920/1{,}025 \approx 0.90$ — about 90 percent of an iceberg lies hidden underwater, with only around one-tenth showing above the surface, matching the everyday observation that icebergs are far larger than they appear from a ship's deck.

• Judging by material alone. "Steel sinks" is true only for a solid lump; a hollow steel hull can float because its average density (mass over total enclosed volume, including air spaces) is what matters, not the density of the raw material.
• Thinking a floating object feels no buoyant force. A floating object is in equilibrium precisely because $F_B$ is acting and exactly balances $W$ — it is not "resting" on the surface with no upward force at all.
• Assuming most of an iceberg is visible. Since ice is only slightly less dense than seawater, the submerged fraction is close to 90 percent; only a small tip shows above the water.
• Trying to force exact numerical density calculations at Class 9 level beyond the qualitative comparison. The syllabus expects reasoning about denser versus less dense; precise relative-density problem sets belong to Class 10.

Ask the tutor: Want to see how the same average-density idea explains hot-air balloons floating in air, or work through more ship and iceberg style examples with the AI tutor?