Archimedes' principle¶

Many of us have heard the story of ancient Greek natural philosopher Archimedes sitting in the bath contemplating if the king's crown was made of gold and shouting "Eureka!" when he realised the solution.

When submerged in the bath Archimedes noticed that the water level would rise. He reasoned that the amount it rose would equal the volume of the submerged object. You may have reasoned something similar when considering the change in height $H$ of water in your siphon experiment. Archimedes could weigh the crown using a scales and the volume by observing the change in the water level and then determine the density of the crown.

But there is more to this story. When submerged some objects sink and some float. The fluid is exerting a force on the object in addition to its weight. It seems obvious that the density of the object plays a role. If we make our object more dense it will sink and if we make it less dense it will float upwards. This is how submarines operate; they fill their ballast tanks with water to sink deeper and with air to float upwards. The can achieve a neutral buoyancy when the average density of the vessel is just right.

Take a look at the 3D model below. There is an irregular shaped rock submerged under the water. Vectors show the hydrostatic pressure distribution around the rock. To make the analysis easier we consider the "bounding box" of the irregular rock. This is an imaginary box which the extremities of the rock fit into relative to our coordinate system. The volume of the bounding box is $(h_2 - h_1)A$ .

We can simplify this scenario by considering the lumped forces $F1$ , $F2$ , $F3$ and $F4$ due to the hydrostatic effects and the weight of the water $W$ left over in the bounding box and finally the buoyancy force $F_B$ .

Note that we are going to ignore the object itself. I've removed it from the 3D scene below.

This means we are only considering the effects of the fluid on the object.

If the rock of volume ${\rlap{V}-}$ has a different density to the liquid then it will either sink or float. The magnitude and direction of $F_B - W_{rock}$ will determine this.

Therefore we need to account for the extra buoyancy force $F_B$ when summing the forces due only to the liquid. If we consider only the water in the bounding box

$\begin{align} F_B = F_2 - F_1 - W \end{align}$

Here $W$ is the weight of the fluid in the bounding box and is obtained by subtracting the volume of the rock from the volume of the bounding box.

$\begin{align} W = \gamma((h_2-h1)A - {\rlap{V}-}) \end{align}$

so that

$\begin{align} F_B = \gamma(h_2 - h_1)A - \gamma((h_2 - h_1)A - {\rlap{V}-}) \end{align}$

which reduces to:

$\begin{align} F_B = \gamma {\rlap{V}-} \end{align}$

$F_B$ is the force exerted by the liquid on the object. An equal but opposite force will be exerted by the object on the fluid via Newton's third law. This is a net upward force because the pressure and therefore the force exerted on the object increases with depth.

What is particularly interesting is that the buoyancy force is equal to the weight of water displaced.