Physics · Materials · AS Level

Density & Upthrust

A steel ship floats; a steel ball bearing sinks. Both are steel, both denser than water, and both experience an upward force from the water they displace — so floating vs sinking isn't about the material alone. This lesson builds density and upthrust from scratch, and derives where that upward force comes from.

8 min read

Density

Density describes how much mass is packed into a given volume of a substance. For a uniform material it's defined as:

$\rho = \dfrac{m}{V}$

where $\rho$ is density, $m$ is mass, and $V$ is volume. The SI unit is kilograms per cubic metre (kg m⁻³), though data tables often use g cm⁻³: 1 g cm⁻³ = 1000 kg m⁻³.

Density is a property of the material, not of any particular sample. A pebble and a boulder of the same rock have wildly different masses and volumes but identical density, since mass and volume scale together. That's what makes density useful for identifying materials — and why it matters for whether something floats.

Substance	Typical density
Air	≈ 1.2 kg m⁻³
Water	≈ 1000 kg m⁻³
Ice	≈ 920 kg m⁻³
Steel	≈ 7800 kg m⁻³
Gold	≈ 19300 kg m⁻³

Pressure

Pressure is force acting per unit area, applied perpendicular to a surface:

$P = \dfrac{F}{A}$

where $F$ is the perpendicular force and $A$ is the area it acts over. The SI unit is the pascal (Pa), where $1\ \text{Pa} = 1\ \text{N m}^{-2}$. Spreading the same force over a larger area reduces the pressure — the idea behind snowshoes and drawing pins.

Pressure in a fluid increases with depth

To understand upthrust, you first need one fact: pressure increases with depth. A column of fluid above a point has weight, which the fluid below must support — so fluid at greater depth is under more pressure than fluid near the surface.

Applied to a submerged object, this single fact explains upthrust.

Hydrostatic pressure

The pressure at depth $h$ below the surface of a fluid of density $\rho$ is $P = \rho g h$, measured in pascals (Pa) with $\rho$ in kg m⁻³, $g$ in m s⁻² and $h$ in m. This is on top of whatever pressure already acts at the surface (atmospheric pressure, if the fluid is open to the air) — it's the extra pressure due to the weight of fluid above that depth.

Upthrust and Archimedes' principle

Any object submerged, partly or fully, in a fluid experiences an upward force from that fluid, called upthrust. It's why objects feel lighter underwater, why a helium balloon rises, and why a steel ship floats despite steel being denser than water. Upthrust arises directly from the pressure difference between the bottom and top of the submerged object.

DerivationUpthrust = ρ_fluid × g × V_displaced (Archimedes' principle)

Consider an object submerged in a fluid of density $\rho_{\text{fluid}}$. Model it, for simplicity, as a rectangular block of cross-sectional area $A$, with its top face at depth $h_1$ and its bottom face at depth $h_2$, where $h_2 > h_1$.

The pressure at depth $h$ in a fluid is $P = \rho_{\text{fluid}} g h$ (plus atmospheric pressure, which acts on both faces equally and will cancel out).

This is the pressure exerted by the weight of fluid above that depth, per unit area.

The fluid pushes down on the top face with pressure $P_1 = \rho_{\text{fluid}} g h_1$, and pushes up on the bottom face with pressure $P_2 = \rho_{\text{fluid}} g h_2$.

Fluid pressure acts perpendicular to a surface from all directions, so it pushes down on the top and up on the bottom of the block.

Since $h_2 > h_1$, the pressure on the bottom face is greater than the pressure on the top face, so the upward force on the bottom face exceeds the downward force on the top face.

Greater depth means greater pressure — this is the whole source of the net upward force.

The forces on the two faces are $F_2 = P_2 A$ (upward) and $F_1 = P_1 A$ (downward). The resultant upward force — the upthrust — is $U = F_2 - F_1 = (P_2-P_1)A = \rho_{\text{fluid}} g (h_2-h_1) A$.

Forces on the side faces act horizontally and cancel by symmetry, so only the top and bottom faces contribute to a net vertical force.

The quantity $(h_2-h_1)A$ is just the height of the block multiplied by its cross-sectional area — the volume of fluid the block displaces, $V_{\text{displaced}}$. So $U = \rho_{\text{fluid}} g V_{\text{displaced}}$.

Whatever volume of space the submerged object occupies is exactly the volume of fluid it has pushed out of the way.

Since $\rho_{\text{fluid}} V_{\text{displaced}}$ is the mass of fluid displaced, $U = m_{\text{displaced}}\,g$ — the upthrust equals the weight of fluid displaced. This holds for any shape of object, not just a rectangular block.

This is Archimedes' principle: the pressure argument that produced it doesn't depend on the block shape used to derive it — any submerged shape can be built up from thin horizontal slabs, each obeying the same reasoning.

▪ derived

So Archimedes' principle: the upthrust on a fully or partially submerged object equals the weight of the fluid it displaces. It applies equally to liquids and gases — the same effect keeps a hot air balloon aloft and a swimmer buoyant.

A rectangular block fully submerged in a fluid, shown in cross-section. The top face of the block is at a shallower depth h1 below the fluid surface, and the bottom face is at a greater depth h2. Downward-pointing arrows labelled P1 push down on the top face, and upward-pointing arrows labelled P2, drawn larger than the P1 arrows, push up on the bottom face, indicating that the pressure and resulting force on the bottom face is greater than on the top face. A net upward arrow labelled upthrust U is shown to the side of the block, representing the resultant force. — The bottom face of a submerged object is at greater depth than the top face, so it experiences greater fluid pressure — the resulting force imbalance is upthrust.

Floating and sinking

Whether an object floats or sinks comes down to comparing the upthrust it would experience if fully submerged with its actual weight.

If full-submersion upthrust is less than weight, the object sinks — the fluid can never push hard enough to support it.
If full-submersion upthrust would exceed weight, the object rises and floats with only part of its volume submerged, sinking in just enough for the upthrust on that fraction to balance its weight.
If the two are exactly equal at full submersion, the object is in equilibrium fully submerged — as with a submarine holding constant depth.

This is why a solid steel ball sinks but a steel ship floats: the ball's full-submersion upthrust (set by steel's volume and water's density) can never match its weight. The ship's hull encloses a huge volume of air, so it displaces far more water for the same mass of steel — enough that upthrust exceeds weight well before full submersion.

Floating condition

For an object floating in equilibrium, the upthrust exactly equals its weight: $\rho_{\text{fluid}}\,g\,V_{\text{submerged}} = mg$, where $V_{\text{submerged}}$ is only the volume of the object actually below the surface, not its total volume.

Key takeaways

Density is $\rho = m/V$, measured in kg m⁻³ — a property of the material, independent of the size of the sample.
Pressure is $P = F/A$, measured in pascals (Pa), where $1\ \text{Pa} = 1\ \text{N m}^{-2}$.
Pressure in a fluid increases with depth according to $P = \rho g h$, because a deeper point has more fluid weight stacked above it.
This depth-pressure difference means the upward force on the bottom of a submerged object exceeds the downward force on its top — the source of upthrust.
Archimedes' principle: upthrust equals the weight of fluid displaced, $U = \rho_{\text{fluid}}\,g\,V_{\text{displaced}}$, derived directly from the pressure difference between top and bottom faces.
An object sinks if the upthrust at full submersion is less than its weight, and floats (partially submerged) if that upthrust would exceed its weight.
A floating object settles at the depth where upthrust on its submerged volume exactly equals its total weight.

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