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Underwater Tension

  1. Mar 26, 2009 #1
    1. The problem statement, all variables and given/known data
    http://img208.imageshack.us/img208/5158/29172992.jpg [Broken]


    So we have some kind of pole floating in water. Area = A. Density = rho. At the bottom we have a point weight M.

    We want to calculate the normal tension inside the pole $\sigma$ at a position x.

    I have the solution and the correct answer which is sigma=-rho*x*g . But will this really be correct below the water?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 29, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi S. Moger! Welcome to PF! :smile:

    (have a rho: ρ and a sigma: σ and type [noparse][tex] … [/tex][/noparse] instead of $ … $ for LaTeX :wink:)
    I don't understand … what's "inside the pole" … and what's "normal tension"? :confused:
     
  4. Mar 29, 2009 #3
    Where is x = 0?

    Maybe you could show some work on this problem?
     
  5. Mar 30, 2009 #4
    With the normal tension I mean the tension in the direction of the pole. There's some isotropic material inside it. (Edit: My bad, it appears it's called normal stress)

    x=0 at the top.

    [itex]\sigma=\frac{P}{A}[/itex]

    So I make a "cut" at position x (above the water). The pole is static which means that the resultant forces acting on this "cut" = 0.

    [itex]x \cdot A \cdot \rho \cdot g-(-P)=0[/itex] (mg down and P up, however P is defined as positive only when it points out from the "cut" area)


    [itex]\sigma=-x \cdot \rho \cdot g[/itex] (so that should be compression)

    But will that formula really work below the water too?
     
    Last edited: Mar 30, 2009
  6. Mar 30, 2009 #5
    Well, you set this up based on a cut above the water. Why don't you try setting up another case based on a cut below the water?

    In your formulation, you have too many uses for P it looks like to me. I think you need to try again, drawing a FBD for the body on one side of the cut and using unique definitions for each of your symbols. Even go so far as to write them down (I know, that sounds awfully academic, doesn't it?)
     
  7. Mar 30, 2009 #6
    If I set it up below the water I seem to get a different result, and the key only mentions one formula (same as above water).

    Those P's are all the same, should they be different?
     
  8. Mar 31, 2009 #7

    nvn

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    S. Moger: Your use of P is correct, and the answer key is correct. There is only one formula, for above and below the water, and it is the one you already derived. Nice work. The water exerts an upward force on the pole only at the bottom of the pole, nowhere else.
     
  9. Mar 31, 2009 #8
    S. Moger & nvn, no, there is a problem with the use of P in the OP, where we see
    sigma = P/A.
    There P is used as a force, elsewhere P is used as a pressure. This is why I said that you need to be careful about using the symbol to mean only one thing.

    Also, S. Moger, did you resolve why, when you put the cut below the water you got a different result?
     
  10. Apr 1, 2009 #9
    I'm aware of the variable name conflict here, but this is the convention of my book. I don't know how they solve the pressure part but probably it doesn't appear.

    Yes, the problem has been resolved.

    Thanks for your inputs.
     
  11. Apr 1, 2009 #10
    It is common practice to write
    sigma = P/A
    in defining axial stress. That does not mean that you must use P for the axial force in all cases. When you work a specific problem, it is important to set things us so that, within the context of that problem, you do not have any variable name conflicts. It is also useful, in that problem, to write down what each variable symbol means for that problem so that there is no ambiguity.

    I'm sure that this seems like a lot of busy work, but from one who has been doing this for half a century, I always do it, every time. It pays off.
     
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