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Can a body float in the middle of a fluid?

Physics Asked by Rudy Davids on June 28, 2021

Let’s say we have a cubic body of side $a$ and made of a material with density $rho$ and we measure its immersed height in a fluid of density $rho_f$ by the variable $y$. Then, its potential energy (and considering a gain of potential due to buoyancy) can be written as:

$V = -Mgy + frac{rho_f}{2}a^2y^2g$

To find the system equilibrium points, one can derivate the previous expression in order to y, obtaining:

$begin{equation}
frac{partial V}{partial z} = 0 Longleftrightarrow rho_f a^2gy_{eq} = Mg Longleftrightarrow y_{eq}=frac{Mg}{rho_f a^2 g} = frac{rho a^3 g}{rho_f a^2 g} = frac{rho}{rho_f}a
end{equation}$

Which leads to something that I don’t know how to explain. Having $rho > rho_f$, one will obtain that the body floats mid-water. How is this even possible if, theoretically with the equations obtained, there isn’t any change on the fluid’s density with depth?

4 Answers

Fundamentally, I think it's because you are treating your equation as broader than it really is. It's pretty obvious that it does not apply properly if the RHS is such that V<0 which is the case where the object is floating in the air above the liquid. It very specifically only applies to scenarios where the object actually floats in the liquid partially submerged, that is $rholeqrho_f$. If $rhogeqrho_f$, the dynamics of the system (which are not accounted for in the energy equation) lead it away from this scenario and it no longer applies.

So you either need to restrict that equation so it only applies when $V>0$ which, you know from experience is when $a>y>0$ and $rholeqrho_f$.

The way the energy equations work, they are only applicable under conditions where the object is in contact with the boundary. If you fiddle with parameters so the result in eternally sinking to the bottom or eternal floating to the surface (such as if the object were placed at the boundary between a fluid of higher and lower density) then it no longer applies.

Answered by DKNguyen on June 28, 2021

The buoyant force becomes constant when the body is either fully submerged ($y geq a$) or if the body is completely removed from the fluid ($y leq 0$). So, your potential energy $V$ has to change form to account for this.

Answered by Mark H on June 28, 2021

The buoyancy on an object in a liquid is constant as long as the object and the liquid are constant. Whether the object is completely or only partially submerged is irrelevant. So is whether it is floating on top or in the middle somewhere or laying on the bottom. What makes it tricky in practice, is the fact that it is very hard to get the object and the liquid to stay constant.

The nice side to that is, that it is that instability that generates increase in buoyancy below the surface and thereby enables the body to float in the middle of a fluid.

On paper you need the pressure on the body to change some characteristic of the body. Pressure relates to depth, so there you have your way to control the depth at which the body floats in the liquid. Buoyancy and pressure at depth being the result of the same force.

Answered by Berend on June 28, 2021

An object can remain at a fixed depth, fully submerged as long as the buoyant force equals the weight. This will generally require monitoring and adjustment. A submarine can adjust the amount of compressed air in a tank which admits outside water. A scuba diver can adjust the amount of air in an inflatable vest.

Answered by R.W. Bird on June 28, 2021

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