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So I understand that buoyancy happens because the fluid exerts larger amounts of pressure from beneath an object compared to above, like in this picture enter image description here

So my question is, what if the object is forced to the bottom of the container, so that no fluid is beneath it. The logic is, if there is no fluid beneath, there is nothing to push it up. So, would that object still experience buoyancy? If yes, why?

edit: interesting to see some answers disagreeing with each other. One thing to note - according to my textbook buoyancy is a force that happens because of hydrostatic pressure - it has nothing to do with density of the objects. So those who say then can will experience buoyancy because it is less dense are wrong I guess, that is not buoyancy.

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  • $\begingroup$ That is a very good question! For my answer I assumed that the can-of-beans would float in the water (experience buoyancy). And you are asking for the edge-case that the container is pressed to the ground first. $\endgroup$
    – rul30
    Jan 30, 2018 at 11:55
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    $\begingroup$ yes that's what I'm talking about $\endgroup$
    – M. Wother
    Jan 30, 2018 at 15:56
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    $\begingroup$ related discussion physics.stackexchange.com/questions/59866/… $\endgroup$
    – agentp
    Jan 30, 2018 at 16:14
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    $\begingroup$ I'm voting to close this question as off-topic because this appears to be about a theoretical edge case of physical laws, and would be more appropriate on physics stack exchange (where it is a duplicate). The question asked does not seem to be focused on real world applications or engineering aspects of the question. $\endgroup$
    – JMac
    Feb 1, 2018 at 14:34
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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Air
    Feb 6, 2018 at 16:52

4 Answers 4

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Yes your can will still have buoyancy when it is submerged to the bottom.

Regardless of the submersion depth, any object will lose weight equal to the weight of water it has displaced, even when held at the bottom. You confuse the hydrostatic pressure with buoyancy.

Hydrostatic pressure will increase with depth, to a point that it may even crush the can in. But the buoyancy exerted by water on the can stays more or less the same, because water is almost incompressible, so its density is more or less the same at shallow and deep water. Hence the displaced water will weight the same on the bottom and buoyancy it causes would be the same.

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    $\begingroup$ Why the downvotes? $\endgroup$
    – user6335
    Jan 30, 2018 at 21:28
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    $\begingroup$ Sad reflection on the quality of the site. Buoyancy happens to even a piece of concrete disegned to be submerged forevr and acting as an abutment for a bridge. You're going to subtract the weight of water to calculate foundation loads. $\endgroup$
    – kamran
    Jan 31, 2018 at 0:09
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    $\begingroup$ 100% correct. The net difference between the gravitational force downwards and the resultant hydrostatic pressure is "buoyancy". When the object is at the surface bobbing happily, there is no buoyancy. If the object is forced above the surface, gravity wins, if it is forced below, buoyancy wins until both forces are equal. Any suggestion that the buoyant object stuck to the bottom of the pond experiences no buoyancy is nonsense! $\endgroup$
    – Donald Gibson
    Jan 31, 2018 at 2:26
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    $\begingroup$ There's a problem with this analysis, that seems to ignore the point of the question. Does the hydrostatic force exist on the bottom when there is no water below it? This really doesn't seem to address that, and just says "buoyancy is the same because difference in hydrostatic pressure is the same"; but you don't address the edge case that this question focuses on; the case when they is no water below. You don't explain how you could manifest a hydrostatic pressure without any fluid in the first place. $\endgroup$
    – JMac
    Feb 1, 2018 at 14:27
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    $\begingroup$ @AsymLabs You're describing the mathematics behind buoyancy; but ignoring why that manifests itself. Buoyancy exists because of the balance of hydrostatic forces on an object; and the increased pressure at increased depth. For any closed surface being affected by hydrostatic force; we get a total buoyancy on the body. For the case of an object with surfaces that are removed from hydrostatic contact with the fluid; it's no longer a closed surface that the pressure acts on; and therefore buoyancy cannot be naively applied in the same way as a submerged object. $\endgroup$
    – JMac
    Feb 5, 2018 at 17:52
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Yes it will - the space the object occupies is lighter than the fluid around it so it wants to rise.

Same as pushing a ball to the bottom of the bath - does it stay there?

Edit: for those who say the shape of the ball makes all the difference: try it with a plastic hollow cube (filled with air) so the cube can lay flat on the surface...

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    $\begingroup$ it's not that ease, the question is about a can, given a "perfect" can there would be no pressure force upwards. Your example with a ball is really what makes the difference. Because a ball only touches the bottom with a small area. $\endgroup$
    – rul30
    Jan 30, 2018 at 12:01
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    $\begingroup$ as rul30 mentiones - I'm talking about a theoretically perfect shape, and a theoretical scenario. All of the everyday object have some tiny holes and channels and things (I'm guessing) $\endgroup$
    – M. Wother
    Jan 30, 2018 at 12:16
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    $\begingroup$ Well what difference does it make? According to my textbook the force of buoyancy has nothing to do with object's density, just it's volume $\endgroup$
    – M. Wother
    Jan 30, 2018 at 12:33
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    $\begingroup$ @M.Wother If you are reading it correctly, then your textbook is wrong. it is the mass per a given volume of the object compared to the mass of the surrounding liquid per the same volume that determines whether an object sinks or rises - mass per volume is the definition of density. $\endgroup$
    – AsymLabs
    Feb 5, 2018 at 10:03
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    $\begingroup$ There is an article in a respectable journal addressing this problem: Using surface integrals for checking Archimedes' law of buoyancy F M S Lima Published 22 November 2011 • 2012 IOP Publishing Ltd European Journal of Physics, Volume 33, Number 1 , iopscience.iop.org/article/10.1088/0143-0807/33/1/009 $\endgroup$
    – Andrej Adamenko
    Sep 24, 2020 at 8:44
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I'm not sure why SolarMike deleted his response. The only thing holding the can to the ground ("proud" in naval terms) is the vacuum force, i.e. the same pressure that keeps you from lifting the can off a table if there's a perfect seal to the table.
So long as the can's density is less than that of the surrounding fluid it'll experience a buoyant force. Don't confuse an existing force with the net force . Once there is a channel to allow water to flow under the can, the delta pressure of the water with depth will cause the can to rise to the surface. (It's a matter of the pressure vs. depth, not the density). As shown on the Wikipedia page, the pressure at the bottom of the can (water pressure) is greater than that at the top of the can, thus forcing the can to rise. This pressure differential exists even when the can is proud; it's just the lack of pressure that would result if a vacuum were to form there that holds the can to the ground. So, in sum, the can always sees a buoyant force.

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    $\begingroup$ "perfect seal to the table." As in say a suction cup $\endgroup$
    – joojaa
    Jan 30, 2018 at 16:04
  • $\begingroup$ @joojaa exactly that $\endgroup$
    – Carl Witthoft
    Jan 30, 2018 at 16:10
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    $\begingroup$ @CarlWitthoft I deleted it as I was failing to get others to understand - have un-deleted it now. You made the point better with the buoyant force. $\endgroup$
    – Solar Mike
    Jan 30, 2018 at 17:19
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    $\begingroup$ "So long as the can's density is less than that of the surrounding fluid it'll experience a buoyant force." This should only be true if all sides are actually experiencing a hydrostatic force. As far as I can tell, the point of the question is; "what happens if you don't allow any fluid there". This doesn't seem to address that specific case, or at least not convincingly beyond just asserting it would work that way. $\endgroup$
    – JMac
    Feb 1, 2018 at 14:31
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    $\begingroup$ There is an article in a respectable journal addressing this problem: Using surface integrals for checking Archimedes' law of buoyancy F M S Lima Published 22 November 2011 • 2012 IOP Publishing Ltd European Journal of Physics, Volume 33, Number 1 , iopscience.iop.org/article/10.1088/0143-0807/33/1/009 $\endgroup$
    – Andrej Adamenko
    Sep 24, 2020 at 8:43
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This question is a theoretical/academic edge-case.

A body in the water will experience two forces:

  1. Pressure acting on all surfaces in contact with water
  2. Gravity acting on the mass of the body

The articel on buoyancy over at Wikipedia explains very good how the following equations are set-up. This article also gives the defintion of buoyancy as:

In physics, buoyancy or upthrust, is an upward force exerted by a fluid that opposes the weight of an immersed object.

(The reader has to decide whether a body on the ground is still immersed.)

The buoyancy force, $F_\mathrm{B}$, can be calculated by integrating the stress (here: pressure), $\sigma$ across the whole surface, $A$, of the body:

$F_\mathrm{B} = \oint \sigma\, \mathrm{d}A $

For an immersed body on can use Gauss theorem. This means one can replace the area-integral with a volume-integral. However, in this edge-case the aera-integral of the body is not "closed". As the can sits on the ground there is no water(pressure) at the bottom side of the can (see also the explanation over at Physics.SE 1, 2).

This means for the edge-case, that the body has contact with the ground it is not possible to use the equation based on the volume-integral:

$F_\mathrm{B} = \rho \cdot V_\mathrm{displaced} \cdot g$

The only way to compute the buoyancy force is to integrate the pressure-vectors on the surface of the body.
This means for a perfect flat ground and a perfect can the aera-integral becomes:

$F_\mathrm{B} = -p_\mathrm{at-top-of-can} \cdot A_\mathrm{top}$

The net-force (buoyancy and gravitational force) is:

$F_\mathrm{net} = -p_\mathrm{at-top-of-can} \cdot A_\mathrm{top} - m_\mathrm{can} \cdot g$

Whether $F_\mathrm{B}$ in this case should be called buoyancy needs to be discussed.

A very similar effect are thermals. When sun light wars the air on the ground its density reduces as with your object under water you have no upward (pressure-)force because there is nothing beneath the war air bubble with a higher density. You need a disturbance if this stable system, which brings some higher density fluid underneath the low density area in order to get buoyancy. The following figure from here illustrates these steps. picture of thermals

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    $\begingroup$ @rul30 your arguments make no sense at all ! The net difference between the gravitational force downwards and the resultant hydrostatic pressure is "buoyancy". When the object is at the surface bobbing happily, there is no buoyancy. If the object is forced above the surface, gravity wins, if it is forced below, buoyancy wins until both forces are equal. Any suggestion that the buoyant object stuck to the bottom of the pond experiences no buoyancy is nonsense! – Donald Gibson Jan 31 at 2:26 $\endgroup$
    – Donald Gibson
    Feb 3, 2018 at 16:47
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    $\begingroup$ „When the object is at the surface bobbing happily, there is no buoyancy“ of course there is buoyancy, why should it float? $\endgroup$
    – rul30
    Feb 3, 2018 at 20:31
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    $\begingroup$ I'm a bit concerned that this answer still seems to be getting downvotes even though it's the only one that seems to directly address the point being made. $\endgroup$
    – JMac
    Feb 5, 2018 at 22:08
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    $\begingroup$ @CarlWitthoft What is "vacuum force"? The force that you get when a vacuum is applied to something is because there's no pressure at that location; and pressure everywhere else. The vacuum itself isn't where the force is supplied from. The reason it doesn't float is because you get rid of the hydrostatic pressure below; causing a change in the pressure balance; which now faces downwards instead of the upwards you would usually get with a closed surface under hydrostatic pressure. $\endgroup$
    – JMac
    Feb 5, 2018 at 22:12
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    $\begingroup$ There is an article in a respectable journal addressing this problem: Using surface integrals for checking Archimedes' law of buoyancy F M S Lima Published 22 November 2011 • 2012 IOP Publishing Ltd European Journal of Physics, Volume 33, Number 1 , iopscience.iop.org/article/10.1088/0143-0807/33/1/009 $\endgroup$
    – Andrej Adamenko
    Sep 24, 2020 at 8:44

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