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Simple force problem

  1. Jan 17, 2008 #1
    1. The problem statement, all variables and given/known data

    A baby elephant is stuck in a mud hole. To help pull it out, game keepers use a rope to apply force F(a). By itself,however, force F(a) is insufficient. Therefore, two additional forces F(b) and F(c) are applied. Each of these additional forces has the same magnitude F. The magnitude of the resultant force acting on the elephant in part b of the drawing is twice that in part a. Find the ratio F/F(a)

    2. Relevant equations
    It should be noted that the two end ropes are 40 degrees apart, and F(a) bisects them, creating 20 between each rope.


    3. The attempt at a solution

    I really don't know, i tried 2:1, but that answer didn't work, any thoughts?
     
  2. jcsd
  3. Jan 17, 2008 #2

    Tom Mattson

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    You need to draw a vector diagram, and break the forces down into components using right triangle trigonometry. Do you know what I mean by that?
     
  4. Jan 17, 2008 #3
    ok, i drew the right trangles and got an equation like this:

    cos20 = (F(a)/F) which led me to -

    F(a)- .94F

    and i got a ratio from F/F(a)= 1.06 (i took the inverse)

    when i entered this into the computer, it said it was wrong. Did I miss something that I was supposed to do, am I supposed to multiple that by two because there are two additional ropes ?

    thanks for the help
     
  5. Jan 18, 2008 #4

    Tom Mattson

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    You're on the right track, but you're forgetting that there are 2 of these forces acting on the elephant at 20 degrees (along with Fa).
     
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