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Simple Archimedes Principle

  1. Feb 15, 2009 #1

    djeitnstine

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    Gold Member

    This (textbook) question seems so simple yet I have been having the hardest time solving it :S I know there's something completely obvious I'm missing.

    1. The problem statement, all variables and given/known data

    A styrofoam slab has a thickness h and density [tex]\rho_{s}[/tex]. When a swimmer of mass m is resting on it, the slab floats in fresh water with its tip at the same level as the water surface. Find the area of the slab.

    2. Relevant equations

    [tex]\Sigma F=F_{buoyant}-Mg=0[/tex]
    [tex]F_{buoyant}=Mg[/tex]
    [tex]F_{buoyant} = \rho_{f}ghA[/tex]

    (Archimedes Principal: Any object completely or partially submerged in a fluid experiences an upward buoyant force whose magnitude is equal to the weight of the fluid displaced by the object)

    When totally submerged [tex]\Sigma F= (\rho_{f}-\rho_{o})V_{o}g[/tex]

    Where [tex]\rho_{o}[/tex] is the density of the object

    3. The attempt at a solution

    I don't even know. I wrote out a bunch of stuff. I know the density of water is [tex]1.00 (10^{3} \frac{kg}{m^{3}})[/tex]. Of course I tried substituting and that gets me no where. I tried making a free body diagram and all that says is that the Buoyant force is equal to weight of the swimmer and the board (duh).

    [tex]\Sigma F= F_{buoy} - F_{board} - F_{swimmer} = 0[/tex]

    Also a simple manipulation showed that [tex](\rho_{f}-\rho_{s}) \Delta h A = m_{s}[/tex]. I think is right?

    Honestly I think some more numbers are missing :S
     
  2. jcsd
  3. Feb 15, 2009 #2

    djeitnstine

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    Gold Member

    Ok I read the solution and it was only a formula that they wanted... So I was correct.
     
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