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Buoyancy force and density

  1. May 26, 2015 #1
    1. The problem statement, all variables and given/known data
    If an object is 500 N normally but has an apparent weight of 300 N when submerged in water, what is the density of the object?

    2. Relevant equations
    ρobjVobjg = ρfluidVobjg

    3. The attempt at a solution
    So the correct answer is 2500 kg/m3 but when I set up the problem, it doesn't turn out correct.
    ρobjVobjg = ρfluidVobjg
    Both volumes and gravity cancel resulting with:
    ρfluidobj
    Is there an error in my set up? Thanks in advance!
     
  2. jcsd
  3. May 26, 2015 #2

    SteamKing

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    You've assumed that the apparent weight is the product of some fictitious density and the volume of the object, neglecting Archimedes' Principle.

    Remember that apparent weight of a submerged object = actual weight of the object in air - buoyant force of the water displaced by the object.
     
  4. May 26, 2015 #3
    Right, 500 N - 200 N = 300 N. Conceptually this makes sense but I am trying to work this out algebraically. Also, just to be clear, you are saying that I cannot use ρobj algebraically this way?
     
  5. May 26, 2015 #4

    SteamKing

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    It depends on what ρobj means. Is it derived from the mass of the object, as in ρobj = mobj / Vobj ?

    It's still not clear why you can't solve this problem algebraically by applying Archimedes' Principle.
     
  6. May 27, 2015 #5
    Can this even be solved algebraically even with an "apparent weight"? I understand apparent weight but in the formula written above: ρobjVobjg=ρfluidVobjg; if I follow math rules, ρobj should be = to ρfluid. Perhaps I am assuming something but I cannot see it.
     
  7. May 27, 2015 #6

    SteamKing

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    Your math is flawless, but it is based on a faulty assumption. Here, let me highlight it for you (again):

    You've assumed that the apparent weight is the product of some fictitious density and the volume of the object, neglecting Archimedes' Principle.

    As I explained in a previous post, that's not how the apparent weight of the object is defined.
     
  8. May 27, 2015 #7
    Oh ok. So, if apparent weight cannot be integrated in the algebra, can this be seen conceptually? Additionally, when ρobjfluid, the object would be seen as submerged but won't sink, yes? I think I am getting the idea now.
     
  9. May 27, 2015 #8

    SteamKing

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    This problem can be solved algebraically, just not in the way you are convinced it should be.

    By ignoring what Archimedes principle is telling you about what the apparent weight of the object is when submerged, I really can't offer you any further guidance, except:

    http://en.wikipedia.org/wiki/Archimedes'_principle

    If the density of the object is the same as the density of the fluid in which it is immersed, the object is said to be neutrally buoyant, which is a fancy way of saying it won't float and it won't sink.

    When submarines submerge, they take in just enough water as ballast to become neutrally buoyant. While submerged, the submarine changes depth by using its control planes to bring the bow up or down when the sub is going ahead. To surface, the submerged sub uses compressed air to remove water from the ballast tanks in order to restore positive buoyancy to the vessel, which then rises to the surface and floats naturally.
     
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