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Homework Help: Buoyant Force Problem

  1. Feb 8, 2010 #1
    1. The problem statement, all variables and given/known data

    Suppose a buoy is made of a sealed steel tube of mass 5 kg with a diameter D = 7 cm and a length of 6 meters. At the end of the buoy is a spherical weight of galvanized steel (specific gravity=7.85). If the buoy floats in fresh water, what must be the mass of the steel M at the bottom to make the distance h=195 cm?

    2. Relevant equations

    FB = W
    F = [tex]\rho[/tex]gV

    3. The attempt at a solution

    I know that in order for this object to float the buoyant force must equal the mass of the submerged object. So

    FB = Wcyl + Wsph = Wwater which is also

    Vcyl[tex]\rho[/tex]sg +Vsph[tex]\rho[/tex]sg = Vwater[tex]\rho[/tex]g

    This is where I get confused. In order to find the mass of the sphere I need to find its volume since I have the density, but how do I determine the volume of water displaced if I dont know the volume of the sphere. Hopefully my reasoning is correct. Any help would be great! I've also attached a copy of the picture provided.

    http://i429.photobucket.com/albums/qq12/ACE_99_photo/ps-222-1-q6-1.jpg" [Broken]
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 8, 2010 #2


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    Homework Helper

    Mass of tube is given.
    Mass of the sphere = ρs*V.
    weight of the displaced liquid = (Volume of the immersed tube + Volume of the sphere)*ρw
    Volume of the immersed tube = π*D^2/4*(L-h)
    From these information find the volume of the sphere and then mass of the sphere.
  4. Feb 8, 2010 #3
    Based on what rl.bhat stated I managed to figure out the following.

    mtube + [tex]\rho[/tex]sVsph = [Vcyl sub + Vsph][tex]\rho[/tex]w

    isolate for Vsph to get

    Vsph = Vcyl[tex]\rho[/tex]w - 5 kg / [tex]\rho[/tex]w + [tex]\rho[/tex]w

    Solving for Vsphere I get V = 0.011962 therefore making the mass 9.39 kg.
  5. Feb 8, 2010 #4


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    Homework Helper

    The equation should be
    Vs = (Vc*ρw - 5 kg)/(ρw + ρs).
    Μay be typo.
    Last edited: Feb 8, 2010
  6. Feb 8, 2010 #5
    Ya that was just a typo. Thanks for your help
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