Upthrust and Archimedes' principle
Forces in Action - OCR A-Level Physics
Key Definition
Archimedes' principle
The upthrust exerted by a fluid on a body immersed in it is equal in magnitude to the weight of the fluid displaced by the body. The upthrust always acts vertically upward through the centroid of the displaced fluid.
The upthrust exerted by a fluid on a body immersed in it is equal in magnitude to the weight of the fluid displaced by the body. The upthrust always acts vertically upward through the centroid of the displaced fluid.
$$U = \rho V g$$
- $\rho$ is the density of the fluid in $\text{kg m}^{-3}$, $V$ is the volume of fluid displaced in $\text{m}^{3}$, and $g$ is the gravitational field strength.
- The upthrust on a $0.001 \text{ m}^{3}$ block fully submerged in water is $U = 1000 \times 0.001 \times 9.81 = 9.81 \text{ N}$, regardless of what the block is made of.
- An object floats if its average density is less than the fluid's density; it sinks if denser; it has neutral buoyancy if the two densities are equal.
Floating condition
A floating object has zero resultant force, so $U = W$. The fraction of an object's height (or volume) below the surface equals the ratio of object density to fluid density:
$$\frac{\rho_o}{\rho_f} = \frac{V_{\text{submerged}}}{V_{\text{object}}}$$
Diagram pending
Floating block partially submerged. Weight $W$ acting down through centre of mass; upthrust $U$ acting up through centroid of submerged volume. Waterline cuts the block at fraction $d/h = \rho_o / \rho_f$.
Will be replaced with a GeoGebra SVG in stream 2.
Worked Example [3 marks]
A table-tennis ball has diameter $40 \text{ mm}$ and mass $2.7 \text{ g}$. It is held under water of density $1000 \text{ kg m}^{-3}$. Show that the upthrust acting on the ball is approximately $0.33 \text{ N}$.
Show Solution
1
Radius $r = 0.020 \text{ m}$. Volume $V = \tfrac{4}{3}\pi r^{3} = \tfrac{4}{3}\pi (0.020)^{3} = 3.35 \times 10^{-5} \text{ m}^{3}$.
[1]2
Use $U = \rho V g = 1000 \times 3.35 \times 10^{-5} \times 9.81$.
[1]3
$U = 0.329 \text{ N} \approx 0.33 \text{ N}$, as required.
[1]Answer
$U \approx 0.33 \text{ N}$ upward.
Common Mistake
HIGH
Students often: use the density of the OBJECT in $U = \rho V g$.
Instead: $\rho$ must be the density of the FLUID displaced, and $V$ must be the volume of fluid displaced (which equals the submerged volume of the object, not its total volume if only partly submerged).
Instead: $\rho$ must be the density of the FLUID displaced, and $V$ must be the volume of fluid displaced (which equals the submerged volume of the object, not its total volume if only partly submerged).
Examiner Tips and Tricks
- For a fully submerged object, $V$ in $U = \rho V g$ is the object's full volume; for a floating object, $V$ is only the submerged portion.
- For floating questions, start with $U = W$ and substitute $\rho_f V_{\text{sub}} g = \rho_o V_{\text{obj}} g$ to find the submerged fraction.