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Fluid question (includes bears)

  1. Feb 8, 2009 #1
    1. The problem statement, all variables and given/known data

    The density of ice is 917 kg/m^3, and the density of sea water is 1025 kg/m^3. A swimming polar bear climbs onto a piece of floating ice that has a volume of 5.2 m^3. what is the weight of the heaviest bear that the ice can support without sinking completely beneath the water?

    2. Relevant equations

    Fb = pVg
    (force buoyancy) = (rho)(volume)(Grav)

    3. The attempt at a solution

    I'm not entirely sure where to begin but I'm thinking that Fb would end up being the weight of the polar bear. Any suggestions? Thanks in advanced!
     
  2. jcsd
  3. Feb 8, 2009 #2
    How much upthrust can each cubic metre of ice provide?

    Multiply that by the number of cubic metres of ice in your floe and that'll be the weight of the maximum bear.
     
  4. Feb 8, 2009 #3
    upthrust? I honestly have never heard of that term.
     
  5. Feb 8, 2009 #4
    The "ice + bear"-system will sink when the gravity acting on them is greater than their buoyancy. The buoyancy is given in the above equation, and this you just have to set equal to the gravity acting on the "ice + bear"-system.

    Solve for the mass of the bear to find its weight.
     
  6. Feb 8, 2009 #5
    robby

    Suggest you check out Archimedes Principle.
     
  7. Feb 8, 2009 #6
    so.. Fb = mg? Would find the Fb of the block of ice in the sea water by doing:
    Fb = (917)(5.2)(9.8) and plug it into Fb = mg?
     
  8. Feb 8, 2009 #7
    yayitsrobby: "Upthrust" and "buoyancy" is the same thing. So we want:

    F_{buoyancy} = F_{gravity},

    which gives us:

    F_{buoyancy} = (m_ice + m_bear)g. And yes, you have found F_{buoyancy} correctly. Now solve for m_bear.
     
  9. Feb 8, 2009 #8
    alright. so the density of the sea water is irrelevant?
     
  10. Feb 8, 2009 #9
    I am wrong here.

    The upward buoyancy force is equal to the magnitude of the weight of fluid displaced by the body. The fluid displaced is just the sea water.

    So we know that the amount displaced has the same volume as the ice, and we know its density. Use this to find F_b.
     
  11. Feb 8, 2009 #10
    alright. so i got Fb = (917)(5.2)(9.8) to equal 46730.32

    46730.32 = (m_ice + m_bear)g

    (rho) = m/v
    917 = m/5.2
    m = 4768.4

    47630.32 = (4768.4 + m_bear)(9.8)
    this solves out to be zero.
     
  12. Feb 8, 2009 #11
    oh so we use do density of sea water in the Fb equation and do (1025)(5.2)(9.8) for Fb? and just go from there?
     
  13. Feb 8, 2009 #12
    Look here

    [tex]
    \begin{array}{l}
    F_{{\rm{buoyancy}}} = m_{displaced\,\,water} g = \rho _{water} V_{ice} g = 52234\,N \\
    \\
    {\rm{and}} \\
    \\
    F_{gravity} = \left( {m_{bear} + m_{ice} } \right)g \\
    \end{array}
    [/tex]

    Solve for m_bear.

    And don't "use it because we have to". You should understand why we use this approach. Perhaps read the first 2-3 lines here: http://en.wikipedia.org/wiki/Buoyancy
     
  14. Feb 8, 2009 #13
    thanks so much you guys!
     
  15. Feb 8, 2009 #14
    Niles,

    Just by way of comparison...

    UpthrustOfIceCompletelyImmersed = (1025 - 917) kg wt/cu.m.
    UpthrustOfIceCompletelyImmersed * 5.2 cu.m. = WeightOfMaximumBear
     
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