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Archimedes principle problem? *1 fluid, 1 solid*

  1. Mar 29, 2008 #1
    I'm having trouble arriving at a finalized answer dealing with the sum of the forces with this problem dealing with Archimedes principle (buoyancy is weight of the displacement).

    Here's the question:

    A cube of wood whose edge is 12 mm is floating in a liquid in a glass with one of its faces parallel to the liquid surface. The density of wood is 762 kg/m^3, that of liquid is 1296 kg/m^3. How far (h) below the liquid surface is the bottom face of the cube?


    Relevant equations

    [tex]\Sigma[/tex] F = ma = 0 (it's floating)

    [tex]\rho[/tex] = F / A

    Here's what I tried doing.

    [tex]\Sigma[/tex] F_y = B - F_l - F_w = 0 , with B as Buoyancy, F_l is the liquids force, and the F_w is the forceo f the wood . I ended up using Volume and density to plug in for Forces obtaining this....

    [tex]\rho[/tex] * V_l * g - [tex]\rho[/tex] * V_w = [tex]\rho[/tex] * V_w * H

    I'm not sure if this is near the correct approach.
     
  2. jcsd
  3. Mar 29, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The buoyant force is the force of the liquid on the wood. Only two forces act on the wood: its weight and the buoyant force.

    Hint: What volume of liquid must be displaced to support the wood? How does that compare to the volume of the wood?
     
  4. Mar 30, 2008 #3
    Oh, so it's similar to the case of the fully sugmerged object, only with different densities given and g is negligible.
     
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