Atmospheric Pressure and Archimedes' Principle

Lessons

Senior 3 · Physics

Atmospheric Pressure and Archimedes' Principle

5 lessons

Lesson 4 of 5

Archimedes' Principle and Upthrust

25 min

The bucket that gets heavy

Pull a full bucket up from a well on a rope. While the bucket is still under the water it rises easily. The moment it breaks the surface it suddenly becomes heavy, although it holds exactly the same water. Under the water, something was helping you: the water itself was pushing the bucket up.

That upward push of a fluid on anything placed in it is called upthrust. More than 2,200 years ago the Greek scientist Archimedes found a simple rule for its size. This lesson tests the rule, explains where the force comes from, and uses it to measure densities.

Definition

Upthrust (buoyant force, $U$ ): the upward force that a fluid (liquid or gas) exerts on a body wholly or partly immersed in it. SI unit: the newton (N).

Apparent weight: the reading of a spring balance holding a body in a fluid, equal to the true weight minus the upthrust.

Experiment: how big is the upthrust?

You need: a spring balance (0–10 N), a stone on a thread, a displacement (Eureka) can, a small beaker and a second balance to weigh the beaker.

Hang the stone from the spring balance in air and read its weight $W$ (here 5.0 N).
Fill the displacement can until water runs out of the spout; when it stops dripping, put the empty, weighed beaker under the spout.
Lower the stone until it is completely under water, without touching the can. Read the balance again: the apparent weight $W^{'}$ (here 3.0 N).
Weigh the beaker with the water that overflowed, and subtract the beaker's own weight.

Result: the loss in weight is $W - W' = 5.0 - 3.0 = 2.0\ \text{N}$ , and the water pushed out by the stone also weighs 2.0 N. Repeat with other objects and other liquids, and the two always agree.

Important

Archimedes' principle: when a body is wholly or partly immersed in a fluid, it experiences an upthrust equal to the weight of the fluid it displaces.

U = W - W' = \text{weight of fluid displaced} = \rho_{fluid}\,V_{displaced}\,g

Where does upthrust come from?

Pressure in a fluid increases with depth ( $p = h\rho g$ , S2). Imagine a block of cross-section $A$ held under water, its top face at depth $h_{1}$ and its bottom face at depth $h_{2}$ :

The sideways pushes on opposite sides are equal and cancel. The bottom face is deeper, so the water pushes up on it harder than it pushes down on the top face. The resultant is upwards:

U = p_2 A - p_1 A = (h_2 - h_1)\,\rho\,g\,A = \rho\,g\,V

since $(h_{2} - h_{1}) A$ is the volume $V$ of the block. And $\rho V g$ is exactly the weight of water that would fill the space the block occupies: Archimedes' principle follows from the way pressure grows with depth.

What does the upthrust depend on?

Factor	Effect on upthrust
Volume of the body under the fluid	More volume immersed, more fluid displaced, larger upthrust
Density of the fluid	Denser fluid, larger upthrust (salt water more than fresh water, water much more than air)
Gravitational field strength $g$	Upthrust is a weight, so it is proportional to $g$
Depth, once fully immersed	No effect: both faces go deeper together, so the pressure difference stays the same
The body's own weight or material	No effect on the upthrust itself; it only decides whether the body floats or sinks (lesson 5)

Worked example 1: the stone's volume and density

The stone in the experiment weighs 5.0 N in air and 3.0 N in water. Find (a) the upthrust; (b) its volume; (c) its density ( $g = 9.8\ \mathrm{N\,kg^{-1}}$ ).

\text{(a) } U = 5.0 - 3.0 = 2.0\ \text{N}

\text{(b) } V = \frac{U}{\rho_w\,g} = \frac{2.0}{1000 \times 9.8} \approx 2.04 \times 10^{-4}\ \mathrm{m^3} \approx 204\ \mathrm{cm^3}

\text{(c) } m = \frac{5.0}{9.8} \approx 0.510\ \text{kg} \qquad \rho = \frac{m}{V} = \frac{0.510}{2.04 \times 10^{-4}} \approx 2500\ \mathrm{kg\,m^{-3}}

Definition

Relative density (RD) of a substance: its density divided by the density of water; it has no unit. Because the upthrust in water equals the weight of an equal volume of water,

\text{RD} = \frac{\text{weight in air}}{\text{weight of an equal volume of water}} = \frac{W}{W - W'}

Example: for the stone above, $\text{RD} = \frac{5.0}{2.0} = 2.5$ , so its density is $2.5 \times 1000 = 2500\ \mathrm{kg\,m^{-3}}$ .

Worked example 2: the same stone in cooking oil

What would the balance read with the stone fully immersed in cooking oil of density $920\ \mathrm{kg\,m^{-3}}$ ?

U = \rho_{oil}\,V\,g = 920 \times 2.04 \times 10^{-4} \times 9.8 \approx 1.84\ \text{N}

W' = 5.0 - 1.84 \approx 3.2\ \text{N}

The oil is less dense than water, so the same displaced volume weighs less and the upthrust is smaller.

Worked example 3: finding the density of a liquid

A metal block weighs 12.0 N in air, 10.5 N in water and 10.8 N in paraffin. Find the density of the paraffin.

The block displaces the same volume in both liquids, so the upthrusts are in the ratio of the densities:

\frac{\rho_{paraffin}}{\rho_{water}} = \frac{12.0 - 10.8}{12.0 - 10.5} = \frac{1.2}{1.5} = 0.80 \quad\Rightarrow\quad \rho_{paraffin} = 800\ \mathrm{kg\,m^{-3}}

Warning

The fluid's density, not the body's. In $U = \rho V g$ , $\rho$ is the density of the fluid, and $V$ is the volume under the fluid. A common mistake is to use the density of the object, which gives its weight instead of the upthrust.

Tip

Loss in weight = weight of fluid displaced. Whatever the object, the balance reading drops by exactly the weight of the fluid it pushes aside. Measure that drop in water, and you have the object's volume: $V = \frac{W - W'}{\rho_w g}$ .

Upthrust and Archimedes' principle (Physics Online). Watch how the pressure difference between the top and bottom of a submerged object leads to the rule that upthrust equals the weight of fluid displaced.

✦

Practice

State Archimedes' principle.
A metal block weighs 12.0 N in air and 10.5 N in water. Find (a) the upthrust; (b) the block's volume; (c) its density. Which metal might it be?
A stone weighs 6.0 N in air and 3.6 N when fully immersed in water. Find its relative density.
Find the upthrust on a 0.020 m³ log held completely under water.
Explain, using pressure, why the upthrust on a fully immersed stone does not change as it sinks deeper.
A pupil of volume 0.065 m³ stands in air of density 1.05 kg m⁻³. Find the upthrust from the air on her. Why do we usually ignore it?
A ball of iron and a ball of wood of the same volume are held completely under water. Compare the upthrusts on them. Why does one sink and the other float when released?
A pupil says: "A heavier object always receives a bigger upthrust." Evaluate this statement with an example.

Show answersHide answers

A body wholly or partly immersed in a fluid experiences an upthrust equal to the weight of the fluid it displaces.
(a) $U = 1.5\ \text{N}$ ; (b) $V = \frac{1.5}{1000 \times 9.8} \approx 1.53 \times 10^{-4}\ \mathrm{m^3}$ ; (c) $m = \frac{12.0}{9.8} \approx 1.22\ \text{kg}$ , so $\rho = \frac{1.22}{1.53 \times 10^{-4}} \approx 8000\ \mathrm{kg\,m^{-3}}$ : steel or iron.
$\text{RD} = \frac{6.0}{6.0 - 3.6} = \frac{6.0}{2.4} = 2.5$ .
$U = 1000 \times 0.020 \times 9.8 = 196\ \text{N}$ .
Both the top and bottom faces go deeper by the same amount, so both pressures rise by the same amount; the difference $(h_2 - h_1)\rho g$ stays the same, and so does the upthrust (as long as the water's density is constant).
$U = 1.05 \times 0.065 \times 9.8 \approx 0.67\ \text{N}$ . Her weight is several hundred newtons, so 0.67 N is too small to notice.
The upthrusts are equal, because they displace equal volumes of water. The iron ball's weight is greater than this upthrust, so it sinks; the wooden ball's weight is less, so it rises and floats.
Wrong. Upthrust depends on the volume displaced and the fluid's density, not on the object's weight. For example, a 1 kg block of polystyrene held under water displaces far more water, and so receives a far bigger upthrust, than a 10 kg lead weight.

Key points

Upthrust $U$ = weight of fluid displaced = $\rho_{fluid} V g$ = loss in weight $W - W^{'}$ .
It arises because pressure increases with depth, so the fluid pushes up on the bottom harder than down on the top.
It depends on the volume immersed, the fluid's density and $g$ ; not on depth (once fully immersed) or on the body's weight.
Relative density = $\frac{W}{W - W'}$ ; a body's volume in water is $\frac{W - W'}{\rho_w g}$ .

Sources and credits

11.7: Archimedes' Principle, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Consulted for the physics; no text was copied.
Video: Upthrust and Archimedes' Principle, Physics Online, embedded from YouTube.

Written by Ganzaa: the lesson text, the two diagrams, the table, worked examples and practice exercises.

1The bucket that gets heavy

2Experiment: how big is the upthrust?

3Where does upthrust come from?

4What does the upthrust depend on?

The bucket that gets heavy

Experiment: how big is the upthrust?

Where does upthrust come from?

What does the upthrust depend on?