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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?

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  • $\begingroup$ Yes! y is the distance from the surface of the water to the base of the cube $\endgroup$ Apr 18 at 0:56
  • $\begingroup$ Your equation is assuming the buoyant force changes even after it is fully submerged and keeps moving deeper into the fluid? $\endgroup$ Apr 18 at 1:27
  • $\begingroup$ If body average density equals to fluid density, then yes - body will float immersed at any depth, as surrounding fluid volume. This principle is used in U-Boats, by filling containers with additional weight, when bigger depth is needed, or removing some if uplift force required. Also if required UBoat can float at any depth. $\endgroup$ Apr 21 at 16:39
  • $\begingroup$ Neutral buoyancy is an unstable condition. $\endgroup$ May 26 at 11:44
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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.

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  • $\begingroup$ I keep flip flopping. Now I think that the problem is what you think it is, and what I initially thought it was but the way the OP asks it makes it masquerade as if it were something else. $\endgroup$
    – DKNguyen
    Apr 18 at 1:43
  • $\begingroup$ @DKNguyen We can wait for the OP to respond in order to clarify. $\endgroup$
    – Mark H
    Apr 18 at 1:46
  • $\begingroup$ Part of the problem is I keep forgetting that it's a solid block and my brain keeps wanting to think about boats where the hull density is greater than water even though the ship itself isn't/ $\endgroup$
    – DKNguyen
    Apr 18 at 1:46
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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 $\rho\leq\rho_f$. If $\rho\geq\rho_f$, 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 $\rho\leq\rho_f$.

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.

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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.

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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.

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    $\begingroup$ 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.. This is incorrect. $\endgroup$ Apr 21 at 15:40
  • $\begingroup$ @ BioPhysicist Please shine a light of wisdom on my ignorance. How is that incorrect? $\endgroup$
    – Berend
    Apr 21 at 15:44
  • $\begingroup$ Toss a static body in a static liquid and wait for the waves to settle down and you will find it will experience a static buoyancy. Regardless whether it is laying on the bottom or floating on the surface. How can that be incorrect? $\endgroup$
    – Berend
    Apr 21 at 15:59
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    $\begingroup$ The buoyant force will be equal to the weight of the displaced fluid. If the object is only partially submerged the buoyant force will be less because less fluid is displaced. $\endgroup$ Apr 21 at 16:01
  • $\begingroup$ Yes, but a static body in a static liquid will always displace the same amount of liquid, resulting in always the same buoyancy. What would make it sink deeper? It is static. $\endgroup$
    – Berend
    Apr 21 at 16:04

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