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The principle states that the upthrust acting on an object is equal to the weight of the fluid displaced by the object. However, this to me suggests that no matter how deep an object is immersed in the fluid, the upthrust acting on it is the same.

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  • $\begingroup$ The way you wrote the question suggests you think it is wrong that the upthrust is the same at any depth. Why do you think that is wrong? $\endgroup$ – alephzero Apr 11 at 20:59
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    $\begingroup$ Of course the weight of the displaced fluid can be different at different depths, it depends on the compressibility of both the fluid and the object. $\endgroup$ – JTS Apr 11 at 21:13
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    $\begingroup$ No need of FACEPALM :-) I think the question is a very good one, one needs to reflect a bit to come up with the question and another bit to figure out the answer. $\endgroup$ – JTS Apr 11 at 21:26
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    $\begingroup$ @UbaidHassan I find the implications of buoyancy are really cool once you understand them, but it takes a bit to get there. Like I find it really interesting to look into how this still applies for different shapes. That one can be a bit of a mind bender; but once you think of a couple example shapes it becomes pretty clear why, and I find it makes buoyancy really elegant. $\endgroup$ – JMac Apr 11 at 21:34
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Actually the density of the fluid increases with depth, since no fluid is perfectly incompressible. Therefore the pressure difference over the object

$\rho\ g(h2-h1)$

and therefore the buoyancy also increases with depth.

In the above formula, let's take a simplified example of a rectangular object, horizontal area $A$ , and vertical thickness $t$ with the top face a distance h1 from the surface and the bottom face a distance h2 from the surface. We will take the object to be small enough (a few cm) to use an average fluid density over its thickness $t$. Designate the value of that fluid density by $\rho 1$ at depth 1. The buoyant force up on that object is

$\rho 1\ g\ A\ (h2-h1)$

At a deeper point in the fluid, designated as depth 2, the pressure is greater and the fluid density is greater, call it $\rho 2$, the buoyant force assuming the object doesn't compress will be

$\rho 2\ g\ A\ (h2-h1)$

which in general will represent a larger net force upward than at the shallower depth.

The object will of course compress, but in general the compressibility is lower for a solid than for a fluid, since the molecules are generally packed tighter in a solid. Only if the compressibility of the solid and fluid are identical, will the buoyant force remain constant with depth, and in general that will not be true. It is possible for the buoyant force to decrease with depth for a particularly compressible object.

Archimedes' principle still holds because as the density of fluid increases, the weight of the displaced fluid increases accordingly. Archimedes' principle does not imply that buoyant force is constant at all depths and does not require an incompressible fluid.

If the object falls all the way to the bottom and rests there some buoyancy will be lost, and all of it will be lost if the sides of the object are not tilted in a way to create buoyancy. A tapered object could even undergo negative buoyancy. It often happens with stuck drill pipe.

It is of course possible that some objects cannot support pressure and will collapse. I am assuming we are not talking about those.

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  • $\begingroup$ The compressibility of the body is also important. $\endgroup$ – JTS Apr 11 at 21:15
  • $\begingroup$ That is true, although solids are generally less compressible than liquids, but something like a soft rubber could be highly compressible compared to the fluid. $\endgroup$ – Bill Watts Apr 11 at 21:20
  • $\begingroup$ The good answer in this case is the combination of the three answers that have been posted :-) $\endgroup$ – JTS Apr 11 at 21:24
  • $\begingroup$ Just to give some figures: between surface and 5000 m below the sea level, density of sea water increases by about 2%. Small but not negligible. $\endgroup$ – GiorgioP Apr 11 at 21:26
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    $\begingroup$ @UbaidHassan Not necessarily. That's the complication of including the density change. Technically, if density varies as a function of height, then the simple version of Archimedes' principle doesn't apply. Pressure can vary with height, so then even the shape of the container can change how height effects the buoyancy. It's generally approximately the same for most real life situations; but if you want to assume density varies it can make the analysis far more complicated. $\endgroup$ – JMac Apr 15 at 13:18
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For an incompressible liquid and object the depth does not matter. The force is the difference between the force down from the top and the force up from the bottom. After careful consideration you will find the upward force equals the weight of the displaced volume of liquid. Compressibility will influence this result as it will affect the volume of the object and the volume and density of the displaced liquid.

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  • $\begingroup$ so the weight of the volume of fluid displaced always = the force difference between the upthrust and the force acting at an immersed body's top? $\endgroup$ – Ubaid Hassan May 23 at 17:42
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    $\begingroup$ Yes that is correct. $\endgroup$ – my2cts May 23 at 17:45
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You are right, while the object is immersed fully.

If not, the upthrust is smaller, because the weight of displaced fluid is smaller, too.

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no matter how deep an object is immersed in fluid, the upthrust acting on it is the same.

For fully submerged objects in an incompressible fluid (generally a valid approximation for non-extreme depth changes), this is completely true.

It's a consequence of hydrostatic pressure. The pressure in a fluid increases with depth (the hydrostatic pressure formula being $P = \rho g h$, $P$ is pressure, $\rho$ is density, $g$ is the acceleration due to gravity and $h$ is the height of fluid above).

What generates the buoyant force here is the pressure difference. Since hydrostatic pressure varies with depth, a submerged object will always have the same pressure difference between it's top and bottom faces. This means that no matter what depth you go, the pressure difference (and therefore the net force acting on the submerged object), is always the same.

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    $\begingroup$ The validity of the statement is not related to the size of the object (at least for existing man-made objects) but to the hypothesis of small changes of depth in the fluid. Even a steel sphere of the size of a golf ball still experience a significant change of upthrust if moved from the sea level down to 5000 m below. $\endgroup$ – GiorgioP Apr 11 at 21:38
  • $\begingroup$ @GiorgioP Yup, good catch. Brainfarted on that one. $\endgroup$ – JMac Apr 11 at 21:40
  • $\begingroup$ While with the compressibility of the liquid, the quoted statement would need modifications, the Archimedes principle would still remain valid. So, in this sense, the answer to the title of the question would still remain purely affirmative with no qualifications, right? $\endgroup$ – Dvij Mankad Apr 11 at 22:22
  • $\begingroup$ @DvijMankad Presumably Archimedes' principle only applies exactly to incompressible fluids; but I don't know if Archimedes ever stated that, and if it's ever formalized that way (for whatever reason). $\endgroup$ – JMac Apr 12 at 0:24
  • $\begingroup$ @JMac It applies to any fluid in static conditions. The force acting because of pressure on a portion of fluid must be equal and opposite to its weight. $\endgroup$ – JTS Apr 12 at 23:45
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No for completely immersed bodies the buoyant force is independent of depth. To calculate buoyant force on a body with area of cross-section A and height h whose upper surface is at a depth d from the surface of the fluid we first calculate the pressure at the upper and the lower surface of the body .They are P+dpg and P+(h+d)pg where p is the density of the fluid,P is the atmospheric pressure respectively.Now the force on the upper and lower surfaces are (P+ dpg)A directed down and(P+{h+d}pg)A directed up respectively.Hence the net force on the body is hpgA,but V=hA is nothing but the volume of the body,hence the net force on the body from the fluid is Vpg.

Hence we observe that for a completely immersed body the buoyant force is independent of the depth.

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