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Homework Help: Tension Problem

  1. Dec 2, 2007 #1
    This is quiet simple problem, but I am a bit confused.

    I have a bridge of S metres length attached to a wall at one end, and a cable at the other. The cable will break under a particular tension force T. If I walk out along it from the wall end, how far can I go before the cable breaks.

    So first of all, the max tension that I can exert on it must be the max tension T minus the tension the bridge puts on the cable right?
    The bridge's weight acts from its center, ie. center of gravity. Its weight is mg, but do I need to take into account its distance from the wall to its center of gravity???
  2. jcsd
  3. Dec 5, 2007 #2

    Shooting Star

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    When you of weight W are at a dist x from the wall, then taking moment about the point where the bridge is fixed to the wall, we have,

    Wx + mgS/2 = TS => T = mg/2 + Wx/S.

    Depending on both W and x, T increases. But T doesn't depend on the dist of the CM from the wall. But your weight and dist from the wall are indeed important factors.
  4. Dec 10, 2007 #3
    I don't want to complicate the problem but the answer depends on how the bridge is fixed to the wall. Is it just sitting "PINNED" on the wall or is it "FIXED".

    If the bridge is indeed "fixed" we would need more information, namely, about the materials of both the cable and bridge to determine an answer.
  5. Dec 10, 2007 #4

    Shooting Star

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    The question here is not about the bridge breaking away from the wall but the breaking of the cable supporting the other end. It doesn't matter if the bridge is 'fixed', 'pinned' or 'glued' to the wall, or if somebody is holding up the bridge where it joins the wall, as long as it's kept horizontal. Of course, it's assumed here that the bridge cannot break away from the wall. In a real bridge, many things have to be taken under consideration, as you have pointed out.
  6. Dec 17, 2007 #5
    The slope of the elastic curve of the bridge depends on the fixity of the ends. The cable only supplies one reaction. In a 'pinned' supported bridge the wall will supply either 2 reactions, horizontal and vertical, while a 'fixed' support supplies 3 reactions, horizontal, vertical and a rotational restraint. In short, if the end of bridge is 'pinned' your equation for moments is valid. If the end of the bridge is 'fixed' at the other end your moment equation is not valid, in otherwords, your missing a force in the equation.
  7. Dec 18, 2007 #6

    Shooting Star

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    I was under the impression that it would be just the other way round.

    This is a high school problem, and it’s tacitly implied that the bridge is horizontal at all points. In this problem, the rotational restraint is assumed. I am not an engineer, but is your pinned bridge horizontal at the fixed end. It is hanging more like a catenary, perhaps?

    I would appreciate it if you would give a more realistic solution to the problem, given the data supplied by the OP
  8. Dec 18, 2007 #7
    Shooting Star, I click around in the forums and I am impressed with the range of complexity of the problems and responses. I believe I missed the real intent of the problem. My mistake for confusing the issue. Your original moment equation and response is correct.

    The original post uses the word 'attached'. In the engineering world, the word "attached" implies reactions; begging the question how many reactions?

    A first or second year college student in engineering or physics would be required to "enforce deflection compatibility" at the rod --> the stretch of the rod must equal the "sag" of the bridge at that point. Now I realize that this is not the intent of the question. Again, I reget the confusion.
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