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Some Fluid Questions

  1. May 28, 2010 #1
    I have some questions regarding fluids.

    1. This may seem like a dumb question, but what's the explanation (preferrably with calculations) on why a denser object sinks when placed in a less dense (presumably liquid) object in terms of buoyancy?

    2. When I place a solid in a liquid such that it floats, how do I predict beforehand where the water level will be relative to the side of the solid.

    3. I'm really shaky with water pressure and its relation to buoyancy. What's the role of water pressure in keeping something afloat? I know buoyant force keeps things afloat so what does water pressure do? Also, when something is sinking, how do we involve water pressure and buoyant force to calculate how fast it sinks (ignoring drag force).

    Any help is appreciated. Thanks.
     
    Last edited: May 28, 2010
  2. jcsd
  3. May 28, 2010 #2

    Doc Al

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    Consider the forces acting on the body. The buoyant force equal to the weight of the displaced fluid acting upward; the weight of the body acting downward. If the body is more dense than the fluid, there will be a net downward force.

    Compare the density of the solid to the density of the liquid. That will give you the fraction of the body that will be submerged.

    The net force due to water pressure on an object is the buoyant force.
     
  4. May 28, 2010 #3
    Thanks for the response.

    I don't really understand what you mean. Can you do a numeric example?

    So does this mean that if I have a cube, the pressure pushing the bottom of the cube up subtracting the pressure pushing the top of the cube down and multiplied by the area of its face should equal the buoyant force acting on it?
     
  5. May 29, 2010 #4

    Doc Al

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    I'll give you a hint so you can figure it out on your own. You want to compare the volume of liquid displaced to the volume of the body itself. That will tell you what fraction of the body is submerged. (Now set up the equation I described in my last post. You'll need the densities of liquid and body.)

    Yes, assuming the cube is oriented so that its sides are vertical.
     
  6. May 30, 2010 #5
    I did some searching and found that in this statement

    here: http://en.wikipedia.org/wiki/Buoyancy

    Does this mean that the volume of water (or any fluid) displaced is equal to the volume of the object underwater provided that it is floating?

    Thanks in advance
     
  7. May 30, 2010 #6

    Doc Al

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    Yes. That's true even if it's not floating.
     
  8. May 31, 2010 #7
    Is it possible to prove this mathematically or can it only be done experimentally?
     
  9. May 31, 2010 #8

    Doc Al

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    I'm unclear what you're looking to have proven. That the volume of fluid displaced is equal to the volume of the object underwater? That's true by definition.

    If you looking to prove what fraction of a floating object is submerged, then set up the force equation. For a floating object, the upward buoyant force must equal the weight of the object.
     
  10. May 31, 2010 #9

    jack action

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    Whenever an object is put in a container filled with a liquid, it raises the liquid level, like this:

    Submerged-and-Displacing.png

    When the container is an ocean and the object is a pebble (or even a boat), the level difference is not noticeable but it is still there.

    How much liquid is raised? All we know is that the sum of the volumes for both the liquid and the object must be constant, whether there are in separate container or not.

    If there is a volume of liquid that is raised, then you need a force to lift it. That force is the weight of the object. It is like a scale such as this one:

    http://francais.istockphoto.com/file_thumbview_approve/4361116/2/istockphoto_4361116-balance-scale.jpg

    where you put the object on one side and the water displaced on the other. One is the counterweight of the other.

    If the volume of liquid displaced is not large enough to compensate for the weight of the object (object denser than liquid), then the "scale will tip over", i.e. the object will sink to the bottom. If you drop slowly the object in the liquid, and the liquid is denser than the object, then the object will stop sinking (hence it will float) when the weight of the liquid displaced will be equal to the weight of the object.
     
  11. May 31, 2010 #10
    I'm confused as to how the two are different. If I know volume of the object underwater, couldn't I just divide that by the total volume that that should give how much the object is submerged?
     
  12. May 31, 2010 #11
    Thanks for such a detailed illustration. So is this basically the explanation behind Archimedes' principle?
     
  13. May 31, 2010 #12

    Doc Al

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    Sure. But how can you predict how much of an object will be under water?
     
  14. May 31, 2010 #13
    Oh I see now; so basically if an object's less dense than the fluid I just take the gravitational force of the object and divide it by the density of the fluid and g to get volume of fluid displaced (or volume of object under water) and then divide that by the object's total volume to get what fraction of the object is submerged?
     
  15. May 31, 2010 #14

    Doc Al

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    And when you do all that, you end up with a simple result:

    Archimedes' Principle tells us:
    ρobj*g*Vobj = ρfl*g*Vfl

    So, the fraction of the floating object that is submerged is:
    Vfl/Vobj = ρobjfl

    So, if the object has 1/3 the density of water, 1/3 of it will be submerged when floating in water.
     
  16. May 31, 2010 #15

    jack action

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    Yes. Here's the story of how Archimedes found that principle:

    archimedes.jpg
     
    Last edited by a moderator: May 4, 2017
  17. Jun 1, 2010 #16
    lol that's pretty funny. How did he discover buoyant force and that it equals the weight of the water displaced?
     
  18. Jun 2, 2010 #17

    jack action

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    That's Archimedes Principle, that's what he found:

    If something pushes something else, it implies there is a force involved. Today we call this force buoyancy.
     
    Last edited by a moderator: May 4, 2017
  19. Jun 2, 2010 #18
    I was wondering how he figured out that the buoyant force was actually equal to the weight of the water displaced?


    Have a question about this, does this only apply if the object sinks because if an object is floating it shouldn't be possible to push away water of equal volume to itself?
     
  20. Jun 2, 2010 #19

    jack action

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    Like I said earlier, we know that the sum of the volumes for both the liquid and the object must be constant, whether there are in separate container or not. But you can easily measure the volume of water displaced and compare it with the known volume of the object if you want to prove it.

    Only the volume of the object that is submerged will displace an equal volume of water.
     
  21. Jun 2, 2010 #20
    I think we have a misunderstanding here. I was talking about the weight of the displaced water and how to prove that this weight is pushing the object upwards
     
  22. Jun 2, 2010 #21

    Doc Al

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    If you want to understand Archimedes' Principle, study the simple argument on this page: http://hyperphysics.phy-astr.gsu.edu/hbase/pbuoy.html" [Broken]
     
    Last edited by a moderator: May 4, 2017
  23. Jun 2, 2010 #22
    Thanks I think this answers my questions. So it's the fact that pressure changes (due to the displacement of the fluid by the object) that creates this buoyant force, which happens to equal the weight of the displaced fluid.
     
    Last edited by a moderator: May 4, 2017
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