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Very different pressure problem!

  1. Nov 23, 2012 #1
    1. The problem statement, all variables and given/known data

    A rectangular block of wood (M = 316 kg) floats on a calm fresh water lake with [itex]d_{o}[/itex] = 14.1 cm below the water. When a dog steps on the block, the block is pushed downward so that now it floads with d = 17.5 cm beneath the water. Find the mass of the dog.

    2. Relevant equations

    I would assume that I would need to use these equations:

    [itex]P_{2} = P_{1} + \rho hg[/itex]
    [itex]F_{Buoyant} = F_{masses}[/itex]

    3. The attempt at a solution

    Very hard problem here. I would assume that the approach is to write out this form and find the mass of the dog...

    [itex]w_{dog} + w_{box} = \rho_{fluid} v_{box}g[/itex]

    Another expressions I thought of are:

    [itex]P_{2} = 1.013 * 10^{5} + \rho_{mass} * 14.1 * 10^{-2} * 9.81[/itex]
    [itex]P_{3} = 1.013 * 10^{5} + \rho_{mass} * 17.5 * 10^{-2} * 9.81 + \rho_{dog} * 17.5 * 10^{-2} * 9.81[/itex]

    Stuck with the problem here. Don't know where to start.
     
  2. jcsd
  3. Nov 23, 2012 #2
    This is an Archemides principle problem. If the block has a mass of 316 kg, what volume of water is displaced by the block? Knowing the depth of the portion of the block below the water surface and the volume of water displaced, what is the cross sectional area of the block? What is the change in displaced water volume when the dog gets on the block? How much mass of water does this correspond to? What is the mass of the dog?
     
  4. Nov 25, 2012 #3
    I know that the density of the fresh water is 1,000 kg/m³. Then, do I set up the expression like this?

    [itex]\rho_{fluid} v_{displacement}g = m_{block}g[/itex]

    That is the expression without the dog pushing the block down. I have very hard time solving the problem.
     
  5. Nov 25, 2012 #4
    vdisplacement = d0 times what?
     
  6. Nov 25, 2012 #5
    I guess you didn't know.

    It's actually the density * volume of displacement * gravity
     
  7. Nov 25, 2012 #6
    If you thought that it's the pressure equation, then it's...

    [itex] P_{0} + \rho hg = P_{1}[/itex]

    h is the height, and I multiplied that by the density and the gravity. I also included the pressure on the left side.

    I am not quite sure what direction you are leading me to.
     
  8. Nov 25, 2012 #7
    vdisplacement= d0A

    where A is the cross sectional area of the block. This may help you solve the problem with the dog, although it isn't absolutely necessary to get a solution. So what is the cross sectional area of the block that you calculate?
     
  9. Nov 25, 2012 #8
    Ah! Thanks for letting me know that. I am sorry that I didn't see this. I thought something else of what you are trying to show.

    Then, it's:

    [itex]\rho_{fluid} d_{0}A_{block}g = m_{block}g[/itex]

    After solving for the cross-sectional area of the block, I substitute that to this equation:

    [itex]\rho_{fluid} dA_{block}g = m_{block}g + m_{dog}g[/itex]

    Not sure if it's right here. The value of cross-sectional area I have is approximately 2.24 m².
     
  10. Nov 25, 2012 #9
    Actually, I'm correct. Thanks for help!
     
  11. Nov 25, 2012 #10
    My pleasure. Nice job!!!
     
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