Why does upthrust not depend on the object’s material?

Because it arises from the fluid’s pressure difference across the object, which depends on the fluid’s density and the object’s volume. The object’s own density only affects its weight.

Does upthrust change with depth once an object is fully submerged?

No, for an incompressible fluid. Both top and bottom pressures increase equally with depth, so their difference — and hence the upthrust — stays the same.

Do objects in air experience upthrust?

Yes, but it is usually negligible because air is about a thousand times less dense than water. It matters for balloons, where the displaced air’s weight exceeds the balloon’s.

How can a steel ship float?

Its hull encloses a large volume of air, so its average density is lower than water’s. It sinks until it displaces its own weight of water.

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49–50MechanicsFoundation

Density and Upthrust

A steel ship floats and a steel coin sinks. The difference is not the material but the $\textbf{upthrust}$ — the upward force a fluid exerts on anything immersed in it — and upthrust depends on how much fluid the object pushes aside, not on what the object is made of.

30 min Topic 4 · Materials

What you'll be able to do

Use $\rho = m/V$ and convert between density units
Explain upthrust in terms of the pressure difference across an immersed object
State and apply Archimedes’ principle
Determine whether an object floats and what fraction is submerged

Density

Density is mass per unit volume, $\rho = \dfrac{m}{V}$ , in $\mathrm{kg\,m^{-3}}$ . It is a property of the material, independent of the size of the sample.

Unit conversion is the usual trap. $1\ \mathrm{g\,cm^{-3}} = 1000\ \mathrm{kg\,m^{-3}}$ , because a cubic metre is a million cubic centimetres while a kilogram is only a thousand grams. Water is about $1000\ \mathrm{kg\,m^{-3}}$ , steel about $7800$ , air about $1.2$ .

For a regularly shaped solid, measure dimensions and mass. For an irregular one, measure the volume of water it displaces — which is exactly the idea behind upthrust.

\rho = \dfrac{m}{V}

Density.

1\ \mathrm{g\,cm^{-3}} = 1000\ \mathrm{kg\,m^{-3}}

The conversion that catches people out.

Example 1A metal cube of side 2.0 cm has mass 71 g. Find its density in

\mathrm{kg\,m^{-3}}

1Volume:

(2.0\times10^{-2})^{3} = 8.0\times10^{-6}\ \mathrm{m^{3}}

2Mass:

0.071

kg.

\rho = \dfrac{0.071}{8.0\times10^{-6}}

Answer

8.9\times10^{3}\ \mathrm{kg\,m^{-3}}

— consistent with copper.

Tip — Convert lengths to metres before cubing. Cubing centimetres and converting afterwards invites a factor-of-a-million error.

Where upthrust comes from

Pressure in a fluid increases with depth, because the fluid above has weight: $\Delta p = \rho g \Delta h$ .

Consider a block of height $h$ and base area $A$ fully immersed. The fluid pushes down on its top and up on its bottom. The bottom is deeper by $h$ , so the pressure there is greater by $\rho_{f} g h$ , and the net upward force is that pressure difference times the area: $\rho_{f} g h A$ .

But $hA$ is the block’s volume, and $\rho_{f} V$ is the mass of fluid that volume would contain. So the upthrust equals $\rho_{f} V g$ — the weight of the fluid displaced. The sideways pressures cancel in pairs and contribute nothing.

\Delta p = \rho g \Delta h

Pressure grows with depth.

U = \rho_{f} V g

Upthrust — the weight of displaced fluid.

Archimedes’ principle

$\textbf{Archimedes’ principle}$ states that the upthrust on an object in a fluid equals the weight of fluid it displaces. The derivation above shows why for a simple block, and the result holds for any shape.

For a fully submerged object the displaced volume is its whole volume. For a floating object it is only the submerged part.

A spring balance reading drops when an object is lowered into water, and the drop is exactly the upthrust — which is how the principle is checked experimentally.

U = W_{\text{displaced fluid}}

Archimedes’ principle.

\text{apparent weight} = W - U

What a spring balance reads when immersed.

Example 1A

2.0\times10^{-4}\ \mathrm{m^{3}}

stone of density

2500\ \mathrm{kg\,m^{-3}}

hangs from a spring balance, fully submerged in water. Find the balance reading. Take

g = 9.81

and

\rho_{\text{water}} = 1000

1Weight:

mg = 2500 \times 2.0\times10^{-4} \times 9.81 = 4.905

2Upthrust:

1000 \times 2.0\times10^{-4} \times 9.81 = 1.962

3Reading:

4.905 - 1.962

Answer

2.94\ \mathrm{N}

Tip — Use the $\textbf{fluid’s}$ density in the upthrust and the $\textbf{object’s}$ density in its weight. Swapping them is the characteristic error here.

Floating

An object floats when it can displace a weight of fluid equal to its own weight before becoming fully submerged. In equilibrium, upthrust equals weight.

Setting $\rho_{f} V_{\text{sub}} g = \rho_{o} V g$ gives the fraction submerged directly: it equals the ratio of the object’s density to the fluid’s. An object less dense than the fluid floats with that fraction below the surface; denser than the fluid, the fraction would exceed 1, so it sinks.

That is how a steel ship floats: its hull encloses a large volume of air, so its $\textit{average}$ density — total mass over total enclosed volume — is below that of water.

\dfrac{V_{\text{sub}}}{V} = \dfrac{\rho_{o}}{\rho_{f}}

Fraction submerged for a floating object.

Example 1Ice of density

920\ \mathrm{kg\,m^{-3}}

floats in seawater of density

1030\ \mathrm{kg\,m^{-3}}

. What fraction lies above the surface?

1Fraction submerged:

\dfrac{920}{1030} = 0.893

2Fraction above:

1 - 0.893

AnswerAbout 11% — the origin of "the tip of the iceberg".

Equation recap

\rho = \tfrac{m}{V}

Density.

\Delta p = \rho g\Delta h

Pressure with depth.

U = \rho_{f}Vg

Upthrust.

\tfrac{V_{\text{sub}}}{V} = \tfrac{\rho_{o}}{\rho_{f}}

Fraction submerged.

Common mistakes to avoid

Using the object’s density to calculate upthrust.

Upthrust uses the

\textit{fluid’s}

density:

U = \rho_{f}Vg

Taking

1\ \mathrm{g\,cm^{-3}} = 1\ \mathrm{kg\,m^{-3}}

It is

1000\ \mathrm{kg\,m^{-3}}

Saying heavier objects experience more upthrust.

Upthrust depends on displaced volume, not on the object’s weight.

Using the whole volume for a floating object.

Only the submerged volume displaces fluid.

Key takeaways

Density is $m/V$ ; $1\ \mathrm{g\,cm^{-3}} = 1000\ \mathrm{kg\,m^{-3}}$ .
Upthrust comes from greater fluid pressure on the bottom of an object than on its top.
Archimedes: upthrust equals the weight of fluid displaced, $\rho_{f}Vg$ .
A floating object displaces its own weight; fraction submerged is $\rho_{o}/\rho_{f}$ .
Upthrust depends on fluid density and displaced volume, not the object’s material.

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