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Fluids at rest problem help !

  1. Jan 13, 2008 #1
    Fluids at rest problem help!!!!!!

    1. The problem statement, all variables and given/known data
    How much force does it take to hold 0.5 m^3 of medium density (specific gravity = 0.77) wood under water?

    3. The attempt at a solution
    density of wood is 770 kg/m^3.
    Fb is the force of buoyancy that pushes the wood up.
    F-Fb = mg
    F-9.81(0.5)(1000) = 770*0.5 (9.81)
    F = 8682 N

    I don't know what the answer is, but i don't think that this is right. It might be i don't know. Could someone help.
     
  2. jcsd
  3. Jan 13, 2008 #2

    Kurdt

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    The force of buoyancy is the same as the weight of the water displaced. That force acts upwards on the weight of the wood. Therefore the force required to hold the wood underwater is the excess force from the weight of the water.

    What you have done is worked out both weights but added them when you need to take the difference.
     
  4. Jan 14, 2008 #3
    So Fb - F = mg
    F = 1128 N
     
  5. Jan 14, 2008 #4

    Kurdt

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    Thats correct.
     
  6. Jan 28, 2009 #5
    Re: Fluids at rest problem help!!!!!!

    hi im just looking at this problem....how did you get 770*0.5 for the mass?
     
  7. Jan 28, 2009 #6

    Kurdt

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    Re: Fluids at rest problem help!!!!!!

    The volume of wood is 0.5 cubic metres, the density of wood is 770 kg per cubic metre (at least given in this question).
     
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