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Bouyant force with water and in equilibrium

  1. Nov 9, 2012 #1
    1. The problem statement, all variables and given/known data
    A cube of wood having an edge dimension of 20.9 cm and a density of 651 kg/m3 floats on water.

    what is the distance between the top of the cube to the water level?


    2. Relevant equations

    F = density*g*Volume

    3. The attempt at a solution

    I believe i have the right concept down in which i did the sum of the forces so it would be:

    -(density of cube*g*Vtotal) + (density of water*g*Vdisplaced) = 0

    I would like someone to confirm this for me and i also wanted to ask...couldn't i just use mg for the force of the cube downward and then add the buoyant force of water to it? (if not then why not?)

    for example: -mcube*g + (density of water*g*Vdisplaced)
     
  2. jcsd
  3. Nov 9, 2012 #2
    nvm...i complete forgot that d = m/v, therefor the force due to gravity is the same for mg or density*g*volume....sorry about that
     
  4. Nov 9, 2012 #3

    SteamKing

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    Hint: check out Archimedes principle.
     
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