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Understanding Buoyancy: The Limits of Archimedes

  1. Dec 15, 2015 #1

    jfizzix

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  2. jcsd
  3. Dec 15, 2015 #2

    Ken G

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    Although I agree there is always value in seeing a mathematical derivation, because it informs us about how to do other similar problems, I'd say there is a much simpler and more elegant proof that the buoyancy force is the weight of the fluid displaced. One simply asserts that the total pressure force on the object does not depend on the object itself, only on the size and shape of its boundary (something used in the above derivation as well). This means that the object can be made of anything-- including whatever is the fluid. In other words, the object could be the fluid, with an imaginary boundary of arbitrary shape. Of course any such object must be in force balance, and has a weight, so the buoyancy force must equal that weight.
     
  4. Dec 16, 2015 #3
    I have also realised that if a light object is pulled under the surface and then released, the maximum upward acceleration it can have is -g, because water has to fall under it to create the uplift.
     
  5. Dec 16, 2015 #4
    This is not a correct mechanistic explanation of what is happening.
     
  6. Dec 16, 2015 #5
    The derivation in this article is not strictly correct mathematically, unless one imagines that the space that is occupied by the solid is replaced by fluid of the same density as the surrounding fluid. Otherwise, the divergence theorem could not be applied inside the solid, which has a different density. The state of stress inside an elastic solid would be different from that of a liquid filling the space, and the elastic solid would have to deform (slightly) and develop stress in order to match the pressure distribution at its surface. In the end, consideration of the stress distribution within the solid would deliver the same result that we obtain if we assume that the space is filled with the original fluid, but it would not be as straightforward as this derivation suggests.
     
  7. Dec 16, 2015 #6

    cjl

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    This is in fact not correct - the fluid acceleration does not need to match the upward acceleration of the object, and the object can accelerate faster than g.
     
  8. Dec 16, 2015 #7

    jfizzix

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    Yes, in this derivation, one must consider the fluid without an object in it in order for the divergence theorem to work.
    Doing that, one finds that the force obtained by integrating the pressure over the surface [itex]/partial\Omega[/itex] is equal to integrating the weight of the fluid elements over the volume [itex]\Omega[/itex].
     
  9. Dec 16, 2015 #8
    Yes. I totally agree. I just wish this had been specifically discussed in the article. Is it too late to edit in a little extra discussion?
     
  10. Dec 16, 2015 #9

    jfizzix

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    I can do that.. One sec..


    ..there we go!
     
    Last edited: Dec 16, 2015
  11. Dec 17, 2015 #10
    Its the displacement of mass of the fluid not its weight which causes buoyancy.
     
  12. Dec 17, 2015 #11
    Awesome derivation.
     
  13. Dec 17, 2015 #12

    cjl

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    Nope, it's definitely the weight. In half the gravitational field, the buoyant force of an object will be cut in half (all else equal). In zero G, there is no buoyant force at all.
     
  14. Dec 17, 2015 #13
    Thank you for the comment but can you elaborate a little, as I felt that to accelerate faster than -g it would have no fluid touching the underside to create uplift.
     
  15. Dec 17, 2015 #14
    What makes you feel that way? For a fluid to cavitate, the pressure at the location in question must drop below the equilibrium vapor pressure of the liquid. Bodies can easily accelerate upward within a liquid with an acceleration than g without causing cavitation in the liquid. In my expert mentor judgment, the arguments you have been putting forth make no sense. Please read the article and the other posts within this thread to improve your background. All the posts in this thread are focusing on the static situation, where the object is not accelerating. So consideration of an acceleration of the object is not on topic. Plus, speculating and guessing in the way that you have done without the solid background is the same as espousing a personal theory, which constitutes misinformation. I strongly urge you to cease such speculation.

    Chet
     
  16. Dec 20, 2015 #15
    A submarine in port at the surface displaces its own weight of water. The the ballast tanks are opened, and water floods in. The sub begins to sink until it is completely submerged. It then displaces its own volume of water. While sinking it displaces a mix of its weight and volume (which could be described mathematically.) These are known facts without the math.
    A Spanish marine engineer at Navantia left off a zero somewhere, and the new subs Spain was building would not be able to surface once submerged. A US sub maker had to be consulted to remedy the situation. Presumably there is an opening for a marine engineer at Navantia. I thought this little story about practical effects of fluid statics would be interesting
     
  17. Dec 20, 2015 #16

    jfizzix

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    I would wonder how it could not resurface if it was able to float in the first place. I guess the ballast tanks couldn't be completely purged again?
     
  18. Jan 6, 2016 #17
    Water density increases with increasing depth. The water that initially filled the ballast tanks is not as dense as the water deep down around the ocean floor level. Once the sub is deep underneath, you need more than just purging water to overcome those tons of water pressing the sub down. I assume if you analyze the water at a deeper level and figure out how it behaves, use action/reaction mechanism to put it to your use to push the submarine upward.
     
  19. Jan 6, 2016 #18
    This is not quite correct. The buoyant force depends only on the local density at the depth of the sub. Even through there are tons of water pressing down from above, the pressure underneath the sub presses up with nearly equal force. If one takes into account the directionality of the water pressure acting on the sub from all sides, the net force depends only on the weight of the water that is displaced at depth.
     
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