Floating, Sinking and Upthrust - Physics for Pilot Assessments

Anything submerged in a fluid is pushed upwards, and this lesson is about where that push comes from, how big it is, and how it decides whether an object floats. By the end you should be able to calculate the upthrust on a submerged object, work out an object's volume and density from how much weight it appears to lose in water, and say why a steel hull floats when a steel bolt does not.

Where the push comes from

Take a block held under water with its top face 1 metre down and its bottom face 1.5 metres down. From the previous lessons, the water presses perpendicular to every face it touches, and the pressure it presses with depends on the depth of that face. Consider the four side faces first. Each is vertical, so the water presses horizontally on it, and for every point on the left face there is a point on the right face at exactly the same depth pressing back just as hard the other way. Those all cancel in pairs, and the block is not pushed sideways. Now the top and bottom. The top face is pressed downwards and the bottom face upwards — and the bottom face is deeper, so it is pressed harder. Those two do not cancel, and what is left over is a net upward force. That is upthrust, and nothing more mysterious than the depth-pressure gradient produces it.

The side pushes cancel in pairs. The base is deeper than the top, so it is pushed harder, and the difference is the upthrust — which turns out to equal the weight of the fluid the block shoved out of the way.

How big it is

Put symbols to that. Let the block have a flat top and bottom of area A and a height H, with its top at depth h₁, so its bottom is at depth h₂ = h₁ + H. The pressure on the top face is ρgh₁ and the pressure on the bottom face is ρgh₂, where ρ is the density of the fluid. Force is pressure multiplied by area, so the downward force on the top is ρgh₁A and the upward force on the bottom is ρgh₂A. The net upward force is the difference between them, ρgh₂A − ρgh₁A, which can be written ρgA(h₂ − h₁). But h₂ − h₁ is just the block's own height H, so the upthrust is ρgAH — and A multiplied by H is the block's volume.

So the upthrust equals the density of the fluid, times g, times the volume of the object that is under the fluid. Notice what dropped out on the way: the depth. The two depths appeared only as a difference, so an object 1 metre down and the same object 30 metres down feel the same upthrust, provided both are fully submerged and the fluid's density has not changed. Notice also what never appeared at all: what the object is made of. Upthrust depends on the fluid and on the volume displaced, and not in any way on the object's own density.

Upthrust

F_\text{up} = \rho g V

$F_\text{up}$: the upthrust — the net upward force from the fluid, in newtons (N)
$\rho$: the density of the fluid, not of the object, in kilograms per cubic metre (kg/m³)
$g$: gravitational field strength, about 10 newtons per kilogram (N/kg) on Earth
$V$: the volume of fluid displaced — for a fully submerged object, the object's own volume, in cubic metres (m³)

Read that expression again and it says something neater than it looks. A density multiplied by a volume is a mass, so ρV is the mass of a volume V of the fluid; and a mass multiplied by g is a weight. So ρgV is the weight of the fluid that used to occupy the space the object now occupies — the fluid the object displaced, meaning pushed out of the way. That is Archimedes' principle: the upthrust on a body in a fluid equals the weight of the fluid it displaces. It is the same statement as the formula, and it is worth holding both forms, because items are set both ways round.

Example 2 — working backwards from an apparent weight loss

An object hangs from a spring balance. In air the balance reads 50 newtons; with the object fully submerged in water it reads 30 newtons. Find the upthrust, then the object's volume, then its density.

A balance reads the force it has to supply. Submerged, the water is taking part of the load, so the upthrust is the difference between the two readings: $F_\text{up} = 50 - 30 = 20\ \text{N}$
Now use upthrust = density of water × g × volume, and solve for the volume by dividing both sides by density × g, which leaves $\text{volume} = \frac{\text{upthrust}}{\text{density} \times g}$ .
The bottom of that fraction is 1000 × 10 = 10 000, so the volume is $\frac{20}{10\,000} = 0.002$ cubic metres.
For the density you also need the mass, which comes from the true weight: weight = mass × g, so $\text{mass} = \frac{\text{weight}}{g} = \frac{50}{10} = 5$ kilograms.
$\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{5}{0.002}$ . Multiply both numbers by 1000: $\frac{5000}{2} = 2500\ \text{kg/m}^3$
2500 is well above water's 1000, so this object sinks — which is consistent with its 50-newton weight beating the 20-newton upthrust.

Float, sink, or hover

Two forces act on a submerged object: its weight pulling down and the upthrust pushing up. From the mechanics module, if two opposing forces are equal the object stays as it is, and if they are not it moves in the direction of the larger. So compare them. The object's weight is its own density times g times its volume, $\rho_{\text{object}} \, g V$ . The upthrust while it is fully submerged is the fluid's density times g times that same volume, $\rho_{\text{fluid}} \, g V$ . Both expressions contain g and both contain V, so the comparison reduces to nothing but the two densities. If the object's density is the greater, its weight wins and it sinks; if the fluid's is the greater, the upthrust wins and it rises; if the two are equal it hovers wherever you leave it. That is where the floating rule from the density lesson comes from.

An object that rises does not leave the water — it rises until part of it is above the surface, and then something worth following happens. Only the submerged part displaces any fluid, so as the object emerges the displaced volume falls and the upthrust falls with it. It settles at the depth where the upthrust has dropped to exactly the object's weight, and no further. So a floating object displaces its own weight of fluid rather than its own volume, and the fraction of it sitting below the surface is simply the ratio of the two densities: a block of density 600 in water of density 1000 floats with six-tenths of itself submerged. Ice at about 920 floats with roughly 92 per cent of itself below the surface, which is why an iceberg shows so little of what is there.

Example 3 — how much of it is under water

A block of wood of density 600 kilograms per cubic metre and volume 0.01 cubic metres floats in water. What does it weigh, and what fraction of it is submerged?

First its weight: $\text{mass} = \text{density} \times \text{volume} = 600 \times 0.01 = 6\ \text{kg},\ \text{and}\ \text{weight} = \text{mass} \times g = 6 \times 10 = 60\ \text{N}$
Floating means the upthrust exactly equals that weight, so 60 = 1000 × 10 × (submerged volume).
The right-hand side is 10 000 × (submerged volume), so divide both sides by 10 000: $\text{submerged volume} = \frac{60}{10\,000} = 0.006\ \text{m}^3$
As a fraction of the whole block: $\frac{0.006}{0.01} = 0.6$ So 60 per cent of it is under water.
Note the shortcut that falls out: $\frac{600}{1000}$ is also 0.6. The submerged fraction of a floating object is always its own density divided by the fluid's.

Ships, and what shape actually does

Steel has a density of about 7800 kilograms per cubic metre, so a solid lump of it sinks in water without hesitation. A ship built of the same steel floats, and the reason is not that its shape defies the rule but that the shape changes what you are taking the density of. A hull encloses a very large volume of air, and the object the water has to support is the whole hull — steel, air, cargo and all. Divide that total mass by that total volume and the figure comes out well below 1000, so it floats. Crush the same hull into a solid block and the air is gone, the volume collapses, the density shoots past water's, and it sinks. Shape matters, then, but only through how much fluid it lets the object push aside — which is why the same sheet of steel can be made to float or sink depending on what you fold it into, why a ship sits lower in the water as it is loaded, and why it sits lower still in fresh water than in denser sea water.

Example 4 — the same steel, two shapes

A small barge is welded from 4000 kilograms of steel, and its hull encloses a total volume of 5 cubic metres. Does it float in water, and how much of it sits below the surface? And what happens if the same steel is crushed into a solid block?

The object the water has to support is the whole hull, so take the density of the whole thing: $\frac{\text{mass}}{\text{volume}} = \frac{4000}{5} = 800\ \text{kg/m}^3$
800 is less than water's 1000, so the barge floats — even though every solid part of it is steel at 7800.
The submerged fraction of a floating object is its own density divided by the fluid's: $\frac{800}{1000} = 0.8$ So it rides with 80 per cent of its volume below the surface.
Check it through the displaced weight instead: 80 per cent of 5 cubic metres is 4 cubic metres of water, which has a mass of 4000 kilograms — exactly the barge's mass, so the upthrust exactly carries its weight.
Crush the hull into a solid block and the enclosed air is gone: the object is now nothing but steel, its density is steel's own 7800, and it sinks. Nothing about the material changed — only how much space the same 4000 kilograms was spread across.

Why it matters in aviation

Upthrust is how the first aircraft flew at all. A balloon or an airship has no wing and needs no aerodynamic lift to stay up: the envelope displaces a very large volume of air, and the weight of that displaced air exceeds the combined weight of the gas inside, the fabric, the structure and the load. The excess is the upward force that supports the whole thing, which is why a lifting gas is chosen for its low density and why such a craft is trimmed by changing either the weight it carries or the volume it displaces.

The water cases are the same principle in a different fluid. An aircraft that comes down on water floats for as long as its hull encloses enough intact volume to displace its own weight, and rafts and life jackets take over the job of displacing water once it does not. A seaplane's floats are sized so that the water they displace supports the whole aircraft with some freeboard left over, and it settles deeper as fuel and payload are added. In a much quieter application, a float-type fuel gauge reads what it does because the float rests where upthrust balances its weight, so it follows the surface of the fuel.

What the test asks

The distractor to watch is the one that credits shape with the whole job — an option saying an object floats "because it is hollow" or "because it is flat", with no mention of density or of displaced weight anywhere in it. Hollowness matters only because it lowers the overall density; flatness matters only because it increases the volume displaced for a given weight. Expect also an item on apparent weight, where the trap is thinking the object's actual weight changed when it was lowered into the water. It did not: the mass is the same and gravity is the same, so the weight is the same, and all that has fallen is the force the balance has to supply, because the water is now carrying part of the load. And expect a depth item, where the correct answer is that the upthrust on a fully submerged object does not change as it goes deeper.