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A buoyancy problem

  1. Apr 1, 2007 #1
    I'm struggling a bit on a homework problem where a cylindrical rod of radius r is floating in equilibrium in water. The bottom 4 cm of the rod has a density of 2000 kg/m^3 while the other 6 cm of the rod has a density of 900 kg/m^3. The question asks what the depth of the bottom of the rod is.

    Would I be correct in using the following equation to solve for the amount V of water displaced by the rod?

    (900 kg/m^3)(9.8 m/s^2)(0.06 m) - (2000 kg/m^3)(9.8 m/s^2)(0.04 m) + (1000 kg/m^3)(V) = 0

    Am I on the right path here? If not, can someone point me in the right direction?
     
  2. jcsd
  3. Apr 2, 2007 #2
    Your formula is wrong.
    To find the volume of water displaced, you must determine the mass of the rod. The mass of a rod is the product of the density times the volume. The volume is the cross section times the length. In your case you have "two" rods attached at an extremity.
    Once you have the mass of the "double" rod, you can compute the volume of water. You do not need to know the value of g, because the weight is proportional to the mass of the water as well for the rod one.
     
  4. Apr 2, 2007 #3
    Do I have this right?

    Thanks for your response. It got me thinking a bit. Here's what I did.

    Let's say ma = the mass of part "a" of the rod where the length is 6 cm and the density is 900 kg/m^3 and mb = the mass of part "b" of the rod where the length is 4 cm and the density is 2000 kg/m^3.

    ma = (900 kg/m^3)(0.06(pi)(r^2) m^3) = 54(pi)(r^2) kg
    mb = (2000 kg/m^3)(0.04(pi)(r^2) m^3) = 80(pi)(r^2 kg

    Let wr = the weight of the entire rod.

    wr = [54(pi)(r^2)(g) kg] + [80(pi)(r^2)(g)] kg = 134(pi)(r^2)(g) kg

    Let ww = the weight of the water displaced by the rod, pw = the density of water, and Vs = the volume of the rod that is submerged.

    ww = (pw)(Vs)(g)

    Using Archimedes' principle, setting wr = ww:

    134(pi)(r^2)(g) kg = (pw)(Vs)(g)

    Solving for Vs to find the volume of the rod submerged:

    Vs = 134(pi)(r^2)/pw

    The volume of water displaced is (pi)(r^2), r being the radius of the rod, times the height h of the rod that is submerged. So:

    (pi)(r^2)(h) = 134(pi)(r^2)/pw

    Solving for h results in 13.4 cm. This would mean that the rod would sink and not float.

    I'm pretty sure I messed up somewhere along the way, probably in translating the problem into mathematics, mostly because it came to me too easily. So, what am I missing here (if anything)?
     
    Last edited: Apr 2, 2007
  5. Apr 2, 2007 #4
    Yes, you are right. I find the same result. In fact, the mean density of the rod is

    [tex]{{6*900+4*2000}\over 6 + 4}=1340[/tex]
    That is, more than the water: it sinks.
    Bravo!
     
  6. Apr 2, 2007 #5
    Thank you very much for the help!
     
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