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Young's Modulus of aquarium

  1. Apr 2, 2008 #1
    1. The problem statement, all variables and given/known data

    A large 3.00×10^4 L aquarium is supported by four wood posts (Douglas fir) at the corners. Each post has a square 3.80 cm x 3.80 cm cross section and is 80.0 cm tall.
    By how much is each post compressed by the weight of the aquarium?

    2. Relevant equations

    deltaL = F*L_0/(Y*A)

    Y (for Douglas fir) = 1 * 10^10 N/m^2

    3. The attempt at a solution

    This seems very straight forward, I just need to determine F.
    F= ma = m(aquarium)g
    but, what is the mass of a 3.00×10^4 L aquarium?
    once I determine that, I figure it'll just be:

    deltaL = ( m * 9.8 * .8)/(.144 * 10^10) meters
    => 5.44*10^10*m meters

    How do I find m ??

    Thanks for the help!
     
  2. jcsd
  3. Apr 2, 2008 #2

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    Since nothing else is given, consider the significant portion of the weight of the aquarium to be due to just water.
     
  4. Apr 2, 2008 #3
    So I took the mass of water to be 1Kg per liter.
    So the m = 3 * 10^4

    Then the equation comes out as:


    deltaL = ( 3 * 10^4 * 9.8 * .8)/(.144 * 10^10) meters
    = 1.63 * 10^-4 m

    But I am told this is the incorrect answer.

    Any help???

    Thanks!
     
  5. Apr 2, 2008 #4

    Mapes

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    What's the total area on which the weight acts, in m2?
     
  6. Apr 2, 2008 #5
    4*.144 = .576 m^2

    but, isn't it asking the deltaL for an individual post?

    I thought that I might have to divide the force by 4, but that answer was also incorrect.

    Thanks
     
  7. Apr 2, 2008 #6

    Mapes

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    Isn't it 0.0144 m2? Also, it's the pressure (stress) that causes the strain; you need to find the pressure on a single leg before you can calculate the strain. The way you originally had it, the compression distance was independent of the number of legs.

    EDIT: Right, 0.00144 m2
     
    Last edited: Apr 2, 2008
  8. Apr 2, 2008 #7
    Hang on, I see the mistake I made earlier: 3.8 cm x 3.8 cm = .038m x .038m = .00144 m^2 (thanks btw), but wouldn't then total area then be 4 * .00144 = .00576 m^2 ?

    Then the pressure would be 3 * 10^4 * 9.8 / .00576.

    Then plug in that pressure for F in
    deltaL = F*L_0/(Y*A)
    Is that right?
     
    Last edited: Apr 2, 2008
  9. Apr 2, 2008 #8
    Although, it retrospect this can't be right since

    deltaL = ( (3 * 10^4 * 9.8) / .00576 * .8) / (.00144 * 10^10) meters

    = 2.84 m

    and that is way to large to be realistic
     
  10. Apr 2, 2008 #9

    Mapes

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    You divided by area twice.
     
  11. Apr 2, 2008 #10
    oh yeah, duh.

    Thank you so much for your help!!! Seriously.
     
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