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Raft buoyancy problem

  1. Apr 24, 2013 #1
    1. The problem statement, all variables and given/known data

    A raft is constructed of wood having a density of 625 kg/m^3. It's surface area is 5.5 m^2, and it's volume is 1.5 m^3. When the raft is placed in fresh water (denistiy = 1000 kg/m^3), to what depth, h, is the raft submerged under the water?

    2. Relevant equations

    P = ρgh
    Where P = pressure, ρ = density

    3. The attempt at a solution

    P = ρgh
    Sub, F /A = P
    F/A = ρgh

    F = mg (g will cancel when we rearrange to solve for h)
    m here is the mass of wood which is the density of wood times it's volume.
    So, V x ρ_wood = M_wood.
    Putting this all together.
    And solving for h

    M_wood / ( A(ρ_h20) = h
    With values I got.

    .1704 meters.

    What do you think? Right wrong please confirm my process.
    Thanks
     
  2. jcsd
  3. Apr 25, 2013 #2

    rl.bhat

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    Check your calculation.
    625x5.5x1.5x9.8 = 5.5xhx1000x9.8
    Find h.
     
  4. Apr 25, 2013 #3
    I didn't get the same answer when I substituted h back in to 625x5.5x1.5x9.8 = 5.5xhx1000x9.8. The equations were not equivalent. Where did I mess up?Thanks
     
  5. Apr 25, 2013 #4
    No I didn't . Is it the right height though?
     
    Last edited: Apr 25, 2013
  6. Apr 25, 2013 #5

    rl.bhat

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    Calculate h from my post#2. What is the value of h?
     
  7. Apr 25, 2013 #6
    625x5.5x1.5x9.8 = 5.5xhx1000x9.8

    625 x 5.5 x 1.5 x 9.8/5.5 x 1000 x 9.8 = h
    h = I got .9375???
     
  8. Apr 25, 2013 #7

    rl.bhat

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    Put the unit. That is the correct answer.
     
  9. Apr 25, 2013 #8
    What? I know what you mean put the unit but what did I do wrong?
     
  10. Apr 25, 2013 #9

    rl.bhat

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    h = 0.9375 m.
     
  11. Apr 25, 2013 #10
    I know the unit. I was just asking from my original set up what did I do wrong? I don't understand why my setup is wrong?
     
  12. Apr 25, 2013 #11

    rl.bhat

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    Show your calculation. How did you get your answer?
     
  13. Apr 25, 2013 #12
    (1.5 m^3 x 625 kg/ m^3 ) / (5.5m^2 x 1000 kg/m^3) = .1704m

    ???? This is what I'm asking why is this the wrong approach to the problem?
    Mass of wood / area times density of water
    Thanks
     
  14. Apr 25, 2013 #13

    rl.bhat

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    Mass of the wood = density of the wood x area of the raft x height of the raft. In your calculation, you have left the area.
     
  15. Apr 25, 2013 #14
    I think I just used the wrong equation well sort of.
    I have F/A = ρgh
    Should be m_wg = ρgh
    m_w = mass of wood.
    Because if I do F/A then I have an extra multiplication by 1/A. In your equation that cancelled.

    Yeah thanks
     
  16. Apr 25, 2013 #15

    rl.bhat

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    You are welcome.
     
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