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Archimedes' principle with 2 liquids

  1. Jan 13, 2012 #1
    A sphere of density = 500 kg/m^3 floats on water (density of water = 1000 kg/m^3).

    1. What fraction of the volume of the sphere is below the waterline?
    ---I understand this problem and I got the correct answer of 0.5

    2. Another liquid of density 200 kg/m^3 is now added on top of the water. The second liquid will not mix with the water. What fraction of the volume of the sphere is now below the waterline?
    --The answer is supposed to be 3/8.

    I have been trying to find out how to do this for a long time, and I still do not understand how to figure it out but I have an inkling it will be on my final exam in a week! Please help me sound this out. Thank you!





    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 14, 2012 #2

    SammyS

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    How much of the 200 kg/m3 liquid is added? Is it enough so that no portion of the sphere will be above the 200 kg/m3 liquid ?

    If so, it should become a proportion problem.
     
  4. Jan 14, 2012 #3

    ehild

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    Assume that the depth of the second low-density liquid is high enough so the ball is totally immersed.

    The origin of the buoyant force is hydrostatic pressure P which increases with depth and results in an upward force because of Pascal's Law: At a given depth D, the same force acts on a surface of area A, independent of its orientation.

    The formula for the buoyant force is derived for block-shaped objects, but it is true for any shape.

    The hydrostatic force is equal to pressure x area (F=PA) and it is directed opposite to the outer normal of the surface. So the hydrostatic force acts downward at the top upward at the base and horizontally inward at the sides of the block. The horizontal forces cancel so the resultant buoyant force points upward, as the pressure is higher at at the base, at higher depth.

    When you have two liquids, with densities γ1 and γ2 and the block floats between them, the hydrostatic pressure at a depth D is equal to sum of the separate pressures. Taking also the atmospheric pressure Po into account, P2=Po+γ2D2g at the top surface, and P1=Po+γ2D3g +γ1(D2-D3)g at the base. The buoyant force on the block of base area A is
    FB=gA(γ2(D3-D2) +γ1(D1-D3))

    If the height of the block is h and x is the immersion depth in liquid 2, D3-D2=h-x and D1-D3=x.

    So the buoyant force is FB=gA(γ1(x) +γ2(h-x) =g(γ1V12V2)
    where V1 is the volume immersed in liquid 1 and V2 is the volume immersed in liquid 2.
    The formula FB=g(γ1V12V2)
    for the buoyant force in terms of the volumes in the separate liquids is valid for any shape of the immersed object.
    Determine V1/Vsphere.

    ehild
     

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    Last edited: Jan 15, 2012
  5. Jan 20, 2012 #4
    You are amazing. Thank you so much! my final is in two days! this helps so much :)
     
  6. Jan 20, 2012 #5

    ehild

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    Have successful exam.:smile:

    ehild
     
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