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Archimedes' Principle - Submerged object in a fluid

  1. Feb 10, 2008 #1
    1. The problem statement, all variables and given/known data
    A 1.25kg wooden block has an iron ball glued to one side. (a) if the blocks floats in water with the iron ball "high and dry," what is the volume of wood that is submerged? (b) If the block is now inverted, so that the iron ball is completely immersed, does the volume of wood that is submerged increase, decrease or remain the same? (c) Calculate the volume of wood that is submerged when the block is in the inverted position.

    2. Relevant equations
    Volume submerged = V of solid (density of solid/density of fluid)
    Density of water = 1000
    Density of iron = 7860
    Density of wood = 1220

    3. The attempt at a solution
    So, for part a, I was thinking that since the iron ball isn't submerged, the amount of water is displaces is equal to its weight.
    Therefore, the volume of wood that is submerged can be found with (Vsub/Vsolid) = ((density of wood + density of iron ball)/(density of water)), where (Vsub/Vsolid) is the answer.

    For part b and c, I was thinking that the volume of wood will decrease, as when the iron ball is submerged, it displaces a volume of water equal to its own volume.
    I know that volume of a sphere is V = (4/3)(pi)(r^3), however I don't know exactly how I would plug it into the equation: (Vsub/Vsolid) = ((density of wood + density of iron ball)/(density of water))

    thanks in advance
  2. jcsd
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