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Archimede's Principle and Work due to buoyant force

  1. Jun 22, 2015 #1
    1. The problem statement, all variables and given/known data
    A flotation device is in the shape of a right cylinder, with a height of 0.323 m and a face area of 4.81 m2 on top and bottom, and its density is 0.460 times that of fresh water. It is initially held fully submerged in fresh water, with its top face at the water surface. Then it is allowed to ascend gradually until it begins to float. How much work does the buoyant force do on the device during the ascent?
    2. Relevant equations
    Fb=mg
    Rho=M/V
    V=Ah
    W=Fd
    3. The attempt at a solution
    I solved to find that the mass is 714.6698kg through manipulation of the density formula and then I solved to find the Buoyant force to be 7003.76404N. I'm not quite sure how to translate these two values that I found into Work(in joules). I know the equation for work is Force x distance and I know that, fully submerged the distance is at least .323m + whatever portion of the device floats above the surface of the water(my guess is either (.323/2)m or (.460*.323)m. Please help! Thank you!
     
  2. jcsd
  3. Jun 22, 2015 #2

    billy_joule

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    Science Advisor

    No, that's not right, draw pictures of it's start and end position. Where would the cylinder be if the distance is greater than 0.323 m?

    Draw a free body diagram of the floating cylinder, you need to find the submerged volume required to provide a buoyant force equal to force due to gravity.
     
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