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Buoyant force problem

  1. Feb 17, 2012 #1
    1. The problem statement, all variables and given/known data

    A less-dense liquid of density ρ1 floats on top of a more-dense liquid of density ρ2 . A uniform cylinder of length l and densityρ, with ρ1< ρ<ρ2, floats at the interface with its long axis vertical.

    What fraction of the length is in the more-dense liquid?

    2. Relevant equations

    F=ρVg


    3. The attempt at a solution

    I thought I could use Vf=ρ/ρ2 but the final answer should include ρ1 and possibly l. Any help would be appreciated.
     
  2. jcsd
  3. Feb 17, 2012 #2

    ehild

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    What is the magnitude of the buoyant force in general?

    ehild
     
  4. Feb 17, 2012 #3
    ρ1Vg + ρ2Vg ?
     
  5. Feb 17, 2012 #4

    ehild

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    What is V?


    ehild
     
  6. Feb 17, 2012 #5
    ∏r^2h...but I don't have r
     
  7. Feb 17, 2012 #6

    ehild

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    Is V the volume of the whole rod?

    ehild
     
  8. Feb 17, 2012 #7
    From what I gathered it is a cylinder.
     
  9. Feb 17, 2012 #8

    ehild

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    The buoyant force is equal to the weight of the fluid displaced. The rod is immersed partly in one fluid and the other part is in the other fluid. It can not displace its whole volume from both fluids. Assume that x length is immersed in fluid 2, and the other piece is in fluid 1. What is the total buoyant force in terms of the volume of the rod?


    ehild
     
  10. Feb 17, 2012 #9
    I'm still not sure how to calculate that with the given variables.
     
  11. Feb 17, 2012 #10

    ehild

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    Write up Archimedes principle.

    ehild
     
  12. Feb 17, 2012 #11
    F=weight of the fluid displaced.
     
  13. Feb 17, 2012 #12

    ehild

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    The rod floats, it is in equilibrium. How is F related to the weight of the rod?



    ehild
     
    Last edited: Feb 18, 2012
  14. Oct 9, 2012 #13
    Hello, I am kind of stuck on this question too (well I have an answer which I am not too sure about)

    http://img27.imageshack.us/img27/3685/lratio.jpg [Broken] this is a simple diagram that I drew.

    Let F1 be the Buoyant force of less dense liquid and it is pointing downwards
    while F2 is the buoyant force of more dense liquid is pointing upwards.

    Since it is floating in middle I have assumed it to be at equilibrium, and hence F2= F1+ FG

    I have used A as the area of cylinder since it is same for all regions. l1 for length of cylinder in less dense liquid and l2 for the length of cylinder in more dense liquid.

    Since g appears on both sides of equation I have ignored it ( by divide the whole thing by g)

    my equation ended up looking like the following;

    ρ2Al21Al1+ρAl

    I have divided both sides by A leading to


    ρ2l21l1+ρl

    I have divided both sides by ρ2 which leads to

    l2=(ρ1l1+ρl)/ρ2

    the question asks what fraction is in more dense liquid which is l2/l

    1l1+ρl)/ρ2l

    simplifying this leads to
    1l12l)+(ρ/ρ2)

    Does this seem correct?

    Regards
     
    Last edited by a moderator: May 6, 2017
  15. Oct 10, 2012 #14
    any help would be appreciated

    Regards
     
  16. Oct 10, 2012 #15

    ehild

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    The hydrostatic pressure must be the same along the red line in the figure. The red line is at depth l=l1+l2 with respect to the top of block. The change of the hydrostatic pressure from zero to depth l is either ρgl or ρ1gl12gl2. The two pressures must be equal.

    You have to express l2/l in terms of the densities. For that, use that l1=l-l2.

    ehild
     

    Attached Files:

  17. Oct 10, 2012 #16

    Hi and thank you for your reply.

    I understand what you mean, but I can't seem to solve l2/l easily, the answer I got was l2/l=(ρ-ρ1l+ρ1l2)/(ρ2l2)

    I think I have made a mistake somewhere. I started by setting ρgl = ρ1gl12gl2

    divided both sides by g, took ρ1l1 to other side of equation giving me ρl1l1 = ρ2l2, then used l1=l-l2
    and divided both sides by l to get l2/l.

    That lead me to l2/l=(ρ-ρ1l+ρ1l2)/(ρ2l2) but I still have l2 on both sides of the equation.

    Any obvious error sources that you can see?

    Regards
     
    Last edited: Oct 10, 2012
  18. Oct 10, 2012 #17

    ehild

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    It is wrong from here. When you take a term to the other side of the equation, the sign will change. Substitute l-l2 for l1 first, collect the terms with l1 at one side of the equation, and the terms with l on the other side.

    ehild
     
  19. Oct 10, 2012 #18

    Hi, sorry I had - minus sign in my hand written workout, for some reason I must have put a plus sign,


    another one where I only put l1=l-l2 at the end lead me to l2/l=(ρ-ρ1(l-l2))/(ρ2)

    Still end up with l & l2 on both sides for some reason

    I will upload hand written workout in a few hours
     
  20. Oct 10, 2012 #19

    ehild

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    You must collect the l2 terms on one side of the equation.

    You have the equation

    ρl=ρ1l12l2.

    Substitute l1=l-l2:

    ρl=ρ1(l-l2)+ρ2l2

    Expand:

    ρl=ρ1l-ρ1l22l2

    Collect the terms with l2:

    ρl=ρ1l+(ρ21)l2

    Bring ρ1l to the left-hand side.

    Can you proceed?


    ehild
     
  21. Oct 11, 2012 #20
    Hi and thank you for your reply.

    I ended up with the following

    ρ-ρ121 = l2/l

    Does that seem ok?
     
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