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Archimedes had a different problem. He needed to measure the volume of a crown, a complex shape. He solved the problem by measuring the volume of water that the crown displaced when placed in a vat of water.

Here we have a simple cube shape which will be released into a tall narrow liquid container of square cross section. So we already know the volume of the cube but we need to compute the height change of the liquid as the cube either reaches a depth at which it floats or sinks to the bottom of the container. To do this we will compute the liquid volume that the cube displaces. We also want to estimate the speed of the cube as it moves downward into the liquid to its final position. The cube's downward speed will be proportional to the difference between its mass and the mass of the liquid that it has displaced. We'll also compute the liquid pressure Vs depth in the container.

As a warmup we'll first discuss what happens when we put a very small cube into a liquid container of almost infinite cross sectional area, `A_("container")`. First, the liquid level will rise only infinitesimally. If the density (`e.g. "grams/cm"^3`), `rho_(cube)`, of the cube is less than the density of the liquid , `rho_(liquid)` then the cube will float with reach a wetted depth, `d`, below the liquid surface where

`d=s rho_(cube)/rho_(liquid)\ \ \ \(1)`

and where `s` is the length of a side of the cube. If the density difference, `rho_(cube)-rho_(liquid)gt0`, then the cube will sink to the bottom of the container at a speed dependent on the density difference.The container we will use is shown in Canvas 2 and Canvas 3. If we release a cube into this container and the cube side length is a large fraction of the container width (see Canvas 3), then we can expect the liquid level to first rise fast and then slow down when the cube becomes partially immersed (floats) of fully immersed (sinks). To describe the liquid rise, we need the cross sectional area, `A_(cube)`, of the cube and the cross sectional area, `A_("container")`, of the liquid container. When the cube is partially immersed in the liquid, the liquid level in the gap area (see Canvas 3)

`A_(gap)=A_("container")-A_(cube) \ \ \ \ (2)`

will rise. As the starting condition, the bottom of the container will have coordinate `y=-h/2` and the top will be `h=h/2 ` where `h` is the container height (see Canvas 2). Since we have two media here, a solid and a liquid, we will need to distinguish the y coordinate of both bottom of the cube, `cube_y`, and the top surface of the liquid, `surface_y`. An important property of a liquid is that it is essentially incompressible so the total volume of the liquid in our container is constant regardless of the position of the cube. When the cube moves downward, it displaces a volume `V_(displaced)` of liquid equal to`V_(displaced)=-cube_y A_(cube)`

where `cube_y` and is the coordinate of the bottom of the cube relative to `y=0`. The liquid level in the gap will have this same volume:`V_(gap)=V_(displaced)=-cube_y A_(cube) \ \ \ \ (3)`

Then the y coordinate of the liquid level in the gap will be:`y_(gap)=V_(displaced)/(A_(gap)) \ \ \ \(4)`

The gap liquid surface coordinate relative to the bottom of the cube, `y_("gap bottom")`, will be:`y_("gap bottom")=V_("displaced")/(A_(gap))-cube_y \ \ \ \(6)`

Then the `y` coordinate of the gap liquid surface relative to the top of the cube, `y_("gap top")` is`y_("gap top")=V_("displaced")/(A_(gap))-cube_y-s \ \ \ \ (7)`

A threshold between floating and sinking occurs when `y_("gap top")=0`
`-cube_y A_(cube)/(A_(gap))-cube_y=s`

`cube_y=-s/(1+(A_(cube))/A_(gap))
\ \ \ \ (8)`

` surface_(yt)=s+cube_(y)=s-s/(1+(A_(cube))/A_(gap))`

Some important changes occur when ``y_("gap top")= 0`` where the liquid surface is at the top of the cube or at the top of the gap.1. The cube descent slows to zero since it almost floats.

2. When `y_("gap top")lt 0` the cube will float. After release its downdward speed will decrease linearly to zero until the expected fraction (see equation 1) of the surface is immersed .

3. When `y_("gap top")gt=0` the cube is fully immersed and will sink to the bottom of the container.

When `y_("gap top")gt0`, the increased downward pressure on the top of cube is balanced by the increased upward pressure on the bottom of the cube. Therefore the cube floats when `y_("gap top")lt0` and sinks at a rate depending on density difference when `y_("gap top")"gt0`. The force, `F_("sinking")`, that causes sinking, and the speed of sinking depends on the difference between the cube mass, `m_(cube)`, and the mass,`m_(liquid)`, of the liquid that the cube has displaced

`F_("sinking")=(m_(cube)-m_(displaced))g \ \ \ \ (8)`

where `g` is the acceleration of gravity and`m_(displaced)=rho_("liquid")V_(displaced) \ \ \ \ (9)`

If `F_("sinking")lt0` then the cube will float and the liquid never reaches the top of the cube. Since the cube is moving in a liquid, it will assume a speed, `v_(liquid)` proportional to the difference between the cube mass and the liquid gap mass:`v_(liquid)=C(m_(cube)-rho_("liquid")V_("displaced")g \ \ \ \(12)`

where the proportionality constant, `C`, has units speed/force. Equation 12 is analogous to Stokes Law which relates force to speed of a slow object in a viscous fluid.
**Note that the speed in equation 12 is always either positive or zero.
The liquid surface coordinate, `surface_y`, stops increasing when the mass
difference becomes zero which means that cube has reached its floating height and the cube speed has gone to zero.**

To really understand how the cube floats in the gap, we really need to look at the pressure in the container. The pressure, `P`, on a surface is the force on that surface divided by the area of the surface.

`P=("Force")/(Area)\ \ \ \ (13)`

Pressure exerts force on all immersed sides of the cube. Since the cube here is floating, pressure does not exert force on its top or the parts of the sides that are not immersed. Most importantly, pressure exerts upward force on the bottom of the cube and causes it to float. We will now compute the pressure at the level of the bottom of the cube. This can be computed by separately considering the forces due to the gap area and the floated cube. First, the force on the liquid due to the bottom of the cube is its mass times the acceleration of gravity. Therefore its contribution to pressure at its bottom is:`P_(cube)=(m_(cube)g)/A_(cube)=(rho_(cube) s A_(cube)g)/A_(cube)=s rho_(cube) g\ \ \ \ \(14)`

The contribution to the pressure of the gap liquid is its mass times the acceleration of gravity divided by `A_(gap)`:`P_(gap)=(m_(gap) g)/A_(gap) =[(rho_(liquid) gap_h A_(gap))] g/A_(gap) \ \ \ \ (15)`

where `gap_h` is the height of the liquid from the bottom of the cube to the liquid surface in the gap (the wetted depth). From equation 1, the wetted depth, `d`, of a floating object is:`gap_h=d=s rho_(cube)/rho_(liquid) \ \ \ \ (16)`

Substituting `gap_h` from equation 17 into equation 15 we obtain:`P_(gap)=[(rho_(liquid) s rho_(cube)/rho_(liquid))] g=s rho_(cube) g \ \ \ \ (17)`

so `P_(gap)=P_(cube)` Therefore the pressure induced by the cube mass is the same as the pressure induced by the the liquid gap mass and the pressure at level of the bottom of the cube is uniform as expected. For a liquid, all pressures at the same depth must be equal. If there is a temporary pressure inequality at the same depth the properties of a liquid are such that the liquid will flow until the pressures at that depth are equal.
For our cube problem, we've seen that the maximum liquid height in the container is reached when only a small fraction of the
cube is immersed so not much liquid has been displaced at that point. To simulate Archimedes' problem, we use a different setup:
We start with the container full to the brim and measure the volume of the liquid that flows over the side as the cube is lowered.
In that case,
by the point where the cube has lowered by its height, the liquid that has flowed over the side is equal to the volume of the cube which
is the breakthrough that Archimedes made with the crown. On the other hand, if the liquid surface level is at the cube height below its
brim, then after only a small cube depth increase, the liquid would start to over the brim which is similar to this problem. But Archimedes
could take this slight before-overflow volume change into account when he measures the actual overflow and his theory would still be correct.
So the answer to our question is that **Archimedes was correct**.

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