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Homework Help: Fluid dynamics and an iceberg

  1. Jul 11, 2010 #1
    1. The problem statement, all variables and given/known data
    An iceberg is floating on sea water such that only 10% of its volume is above the water level. What is the density of sea water expressed in terms of the density of ice.

    2. Relevant equations

    [tex]\frac{Vf}{Vi}[/tex] = [tex]\frac{\rho}{\rho}[/tex]

    [tex]\rho[/tex]f = density of sea water
    [tex]\rho[/tex]i = density of ice

    3. The attempt at a solution
    I know this is a rather simple problem but where I am getting hung up is the ratio of the visible portion of the iceberg. I hate to admit but the simple maths trip me up. I began by isolating the eq so that:

    [tex]\rho[/tex]f = [tex]\rho[/tex]i [tex]\frac{Vi}{Vf}[/tex]

    from which I get [tex]\frac{9}{10}[/tex] [tex]\rho[/tex]i

    By the way this is an even number problem so I cannot look up the answer, so is this correct?

    Thanks in advance.

  2. jcsd
  3. Jul 11, 2010 #2


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    You left out the subscipts on the ρ terms in your equation.

    Think about your answer. It says that the sea water is (more, less?) dense than the ice, and therefore the iceberg should (float, sink?).
  4. Jul 11, 2010 #3
    As you said, It's a very simple problem.

    [tex]\rho_{0}V_{0}g=\rho g (V_{0}-V)[/tex]

    [tex]\rho_{0}[/tex] and [tex]V_{0}[/tex] are density and volume of the iceberg and [tex]\rho[/tex] is density of water and [tex]V[/tex] is visible volume of the iceberg.

    You can easily get the sought ratio.

  5. Jul 11, 2010 #4
    From my answer it says that the sea water is less dense than that of ice, which is actually incorrect when I think about a glass of ice water. So then my answer ought to be:

    [tex]\rho[/tex]f = [tex]\frac{10}{9}[/tex] x [tex]\rho[/tex]i

    which correctly shows that the sea water is more dense than ice, is this the correct answer?

    Thanks redbelly. Also I didn't put the subscripts in the original EQ because I have issues when I apply subs scripts in a fraction for some silly reason.

  6. Jul 11, 2010 #5


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    Looks good. One way to remember this is that the total mass of the iceberg (or other floating object) equals the mass of the displaced water (or other fluid). Since mass is equal to ρV, you can set up the equation by equating the two masses.
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