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What is the draft of the barrel?

  1. Jan 20, 2014 #1
    1. The problem statement, all variables and given/known data

    The oil drum floats in such a way that one circular end is always directly up. The total mass of drum is 200 kilograms. The drum has a cross-sectional area of 0.4 m2, and the North Sea is level.

    What is the draft (the distance from the surface to the deepest point of the barrel at equilibrium) of the barrel?


    2. Relevant equations

    N/A

    3. The attempt at a solution
    I can't begin to solve it because I have no idea how to begin.


    I could use a good initial starting position because I am completely clueless about this problem
     
  2. jcsd
  3. Jan 20, 2014 #2

    SteamKing

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    A good starting point is to remember Archimedes principle: The barrel will sink until the weight of water displaced equals the weight of the barrel.
     
  4. Jan 20, 2014 #3
    So, the weight of the barrel is Fw = ma = 200kg * 9.8 m/s^2

    So the amount of water displaced is equal to that.

    Can I replace Mass of water displaced with m = volume * density (density of water * 200kg of water displaced). Solve for volume, then for volume I can substitute w * l * h? solve for h?
     
  5. Jan 20, 2014 #4

    BvU

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    " m = volume * density "
    Yep, except w*l is no good for a cylinder.

    "(density of water * 200kg of water displaced)" is no good: gets you kg^2/m^3
    to get meters, you want something like kg / (kg/m^3 * m^2)
     
  6. Jan 22, 2014 #5
    The problem statement gives the cross sectional area of the drum. So you need to determine what height of a cylindrical shaped amount of water has this weight. I would think that you need to use the density of sea water though for this calculation.
     
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