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Hi, i'm not sure how to prove this about buoyance

  1. Jan 1, 2006 #1
  2. jcsd
  3. Jan 1, 2006 #2
    See i've understand that
    F=1000*9.8*3*15
    F=441000N

    But how do I prove that this is the difference between the upward and downward force?
     
  4. Jan 1, 2006 #3

    Fermat

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    Well, this is the difference between the upward and the downward force.
    You have to show that this is the same as the buoyant force.

    Hmm. What is your definition of buoyant force ?
     
  5. Jan 1, 2006 #4
    the weight of the liquid displaced
     
  6. Jan 1, 2006 #5

    Fermat

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    That's right, but from your attachment you wrote,

    [tex]\Delta P = \rho g \Delta h[/tex]

    which looks like you are caculating the force due to the pressure difference.

    Have you actually calculated this force ?
     
  7. Jan 1, 2006 #6
    I'm not exactly sure which force you're talking about. I found this formula in my textbook. I scanned the formulas we have for buoyancy, I got it from the one on the right of this page
    http://img474.imageshack.us/img474/641/buoyancy1rq.jpg

    We have another formula as well which is not on that page.
    Fb = F2 - F1
    F2 = upward force applied on the lower surface
    F1 = downward force applied to the upper surface

    But that's all i've got :s
     
  8. Jan 1, 2006 #7

    Fermat

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    OK.
    I can't see the bit on the right very well.

    I'll post a few bits to clear things up.
    However I'm half-way through my lunch at the moment and my wife wants my attention, so I may be a few minutes gettting back.

    Ciao.
     
  9. Jan 1, 2006 #8

    Fermat

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    The buoyant force, call it BF, is equal to the weight of the fluid displaced.

    [tex]BF = \rho gV[/tex]

    where V is the volume of fluid displaced.

    When a body is completely immersed in a fluid, as shown in your original problem, then the difference in pressure between the top and bottom faces is given as follows,

    [tex]P_0 = \mbox{ pressure on upper surface}[/tex]
    [tex]P_1 = \mbox{ pressure on lower surface}[/tex]

    [tex]P_0 = \rho g h_0[/tex]
    [tex]P_1 = \rho g h_1[/tex]

    [tex]\Delta P = \rho g \Delta h[/tex]

    [tex]\mbox{where } \Delta h \mbox{ is the difference in height between the top and bottom surfaces.}[/tex]

    The upward force due to this difference in pressure is,

    [tex]F = \Delta P A[/tex]
    [tex]F = \rho g \Delta h A[/tex]
    [tex]F = \rho gV[/tex]

    [tex]\mbox{since } \Delta h A \mbox{ is just the volume of displaced fluid.}[/tex]

    So, the force, F, due to the difference in pressures, is equal to the buoyant force, giving

    [tex]BF = \rho gV = F = \rho g \Delta h A[/tex]
    =====================
     
    Last edited: Jan 1, 2006
  10. Jan 1, 2006 #9

    Fermat

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    I managed to see that bit on the right hand side of the page.

    That formula,

    ΔP = ρgΔh

    is for the difference in pressure and gives the upward force as,

    F = ΔP*A = ρgΔh*A

    Now, this is equal to the buoyant force, but the buoyant force is defined as the weight of the fluid displaced. i.e.

    BF = ρgV - by definition
    =======
     
  11. Jan 1, 2006 #10
    Thanks Fermat, forgive the weird question, i've got a book full of them. Tip to anyone who reads this: "never choose to take distant courses when you can take them at school"
     
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