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A wire exerting force on two poles

  1. Aug 4, 2010 #1
    1. The problem statement, all variables and given/known data
    A wire, weighing 10N, is put on two poles that are 40m apart. Due to its weight, the wire dips for 1m. What are the forces the wire exerts on each of the poles? The wire can be said to take a shape of an arc of a circle.

    2. Relevant equations

    3. The attempt at a solution
    Well, the solution is said to be 50N and, to be honest, I did reach it, but the method seemed a bit brute. Therefore I'm just looking for a confirmation (or refutation, for that matter) on its validity, and even moreso, an alternative one for reaching the same result.

    Basically what I did is I sketched a circle and calculated the angle under which the tangent would hit each of the poles. Because we had that triangle with the cathetis' lengths 1m and 20m, after some manipulation and use of rules on the similarity of triangles, I got to an angle of 2,86º. Then I drew the final triangle with one catheti being 5N - half of the wire's weight, since there are two poles - and just calculated the hypothenuse (tangent) using the cosine function.

    So how does that sound? What I'm "worried" about is that this problem is from a book that we used in our first year of high school and, at least from what I remember, we used simpler maths to reach solutions then (by simpler I mean we probably didn't use that circle thing, but we did use trigonometric functions). So while I'm glad I got to the solution, I'm not sure I should've reached it this way (I'm not saying it's wrong if you can reach a certain solution your own way, but I would like to know the supposedly "proper" way, as well).

    Any thoughts would be greatly appreciated.
  2. jcsd
  3. Aug 4, 2010 #2
    Actually for this particular question I would just approximate the arc to be an isoceles triangle because the 1m dip is quite negligible compared to the length of the wire, so the angle (made with the horizontal) wouldn't change quite a lot.

    http://img801.imageshack.us/img801/1708/screenshot20100804at220.png [Broken]

    Find the vertical component (5N) and use trigo to get the tension. If you're not comfortable with approximating then your method seems correct even though you have to deal with circles which is slightly more troublesome.
    Last edited by a moderator: May 4, 2017
  4. Aug 4, 2010 #3
    Thanks for the answer, but the thing is I tried your suggestion before and came up with an answer of 100N instead of 50N.
  5. Aug 4, 2010 #4
    then we can't approximate (sorry I didn't calculate and check if my approximation is correct or not). my bad. in this case i guess you have to mess with circles by using the arc length (40m) and the rise (1m) to calculate the radius of the circle and finally use trigo to get the tangent to the circle at the point where the wire touches the pole, and then find the angle at that point. did you do it this way? i don't know if there's a shorter way.
  6. Aug 4, 2010 #5

    Doc Al

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    Staff: Mentor

    You cannot approximate the geometry of the circular segment using a triangle. As arkofnoah says, you'll have to dig into the geometry of segments and chords to find the correct angle the wire makes at the attachment points. Then you'll get the expected answer.
  7. Aug 5, 2010 #6
    Like I described above, I did it in a similar way, but I didn't (and still don't) see the need to calculate the radius of the circle. I think you can skip that step by just using rules on (the similarity of) triangles. Correct me if I'm wrong, though.

    What do you mean by geometry of segments and chords? Like I said, I did get the right solution without approximating by using a triangle, but just wanted to see if I did it the proper way and/or if there's a shorter/better one to the one I used. I'm not familiar with the English terms for what I did, but perhaps you can tell me if my proposed method described above actually was using geometry of segments and chords.
  8. Aug 5, 2010 #7

    Doc Al

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    As far as I can see, the method you described in post #1 using triangles is exactly the method described by arkofnoah in post #2. That doesn't work. If you got it to work, please show the details.

    What I mean by "geometry of segments and chords" is that you need to use the information given, that the wire is a segment of a circle, to find the angle that it makes at the point of attachment. You have the length of the chord and the height of the segment. The rest is a bit of geometry. Doing it that way gives you the correct answer.
  9. Aug 5, 2010 #8
    Sorry, I guess I didn't describe my problem-solving process well enough. I did NOT use the method described by arkofnoah, as, like I mentioned, it didn't lead to the right solution. I did sketch a triangle like he did, but I only used it for purposes of getting to the angle of the tangent where it hits the pole. For that I set up a whole system of triangles within the actual circle and used the rules on their similarity to derive the angles not only within those "inner triangles", but also the angle of the tangent in relation to a vertical line (the pole). So I guess I did what you said, found the angle that the circle (which I also sketched) makes at the point of attachment.

    I hope this is now a better description, though I know it's still a bit awkward. I guess it's hard to describe the whole process without being able to sketch it out. Oh, and sorry, I made an error in my first post, the actual angle I got was two times 2,86° (5,72°) in relation to the horizontal line and 90° minus that in relation to the vertical line. So in the end my final equation was set up as:

    [tex]\cos (84.28^\circ) = \frac{5N}{F}[/tex]
  10. Aug 5, 2010 #9

    Doc Al

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    Ah... that makes all the difference.

    Unfortunately, I don't know of a simpler way to solve this problem. (Other than just look up a table of formulas for the geometry of segments and chords. Which is what I would do. :smile:) I consider this more of a geometry exercise than a physics problem.
  11. Aug 5, 2010 #10
    I guess this is why I was a bit apprehensive of whether my tactics in regards to tackling the problem were correct :smile:
  12. Aug 5, 2010 #11


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    An alternative way is to take a free body diagram cut through the mid point (at x= 20 m, with 1 m of sag) and around the left support. You know the vertical left support reaction is 5 N. Now just some moments about the midpoint to solve for the horizontal component of the left end reaction, being sure to place the weight of the wire in that section, 5 N, at the centroid of the arc (approximately at x = 10m). Close enough.
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