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Aeroplane follows circular trajectory-Tension? (geometry)

  1. Dec 13, 2016 #1
    1. The problem statement, all variables and given/known data

    So, I have this problem here that's pretty basic, but the solution manual sets different axes, and I'm having a bit of trouble understanding the geometry part, meaning how he applies the given forces to the new axes.

    A model airplane of mass 0.750 kg fl ies with a speed of 35.0 ms in a horizontal circle at the end of a 60.0 m control wire. The forces exerted on the airplane are shown in Figure P6.63: the tension in the control wire, the gravitational force, and aerodynamic lift that acts at θ = 20.0° inward from the vertical. Compute the tension in the wire, assuming it makes a constant angle of 20.0° with the horizontal.


    First of all, here's the main pic:

    OV3OmFy.jpg

    2. Relevant equations

    It's the geometry part that gives me trouble.

    3. The attempt at a solution

    He finds the radius of the circular trajectory that the plane follows:

    6VwAOuT.jpg

    No trouble here, we just take the direction of T as the hypotenouse and going with cosθ = r/l finds the radius.

    Then there's this:

    iwjSW2n.jpg
    English is not my native language, so I'm not sure whether he means that x and y are the crossed lines, or the lines that "fall on top" the directions of T & F respectively. From the test I'm getting the latter, but if I try to apply the usual cosx = adjacent/hypotenouse and sinx = opposite/hypotenouse, I can't figure out how he gets his results.

    I'd appreciate some help!

    PS: I know this isn't exactly a "problem" with statements and such, but it's only this part that troubles me, and I didn't know where else to post it.
     
  2. jcsd
  3. Dec 13, 2016 #2

    haruspex

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    The book defines x as at 20 degrees to the horizontal, along the line of the tether. Correspondingly, y is at 20 degrees to the vertical, along the line of the lift force.
    Please clarify where you have difficulty with the resulting equations.
     
  4. Dec 13, 2016 #3
    How he gets that Fgx = Fg*sin(20) & acx = ac*cos(20). I know that he uses the formulas of cosine and sine, but I can't figure out how he gets these particular results.
     
  5. Dec 13, 2016 #4

    haruspex

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    What would you calculate for the component of Fg in the x direction?
     
  6. Dec 13, 2016 #5
    I worked at it again and realised I was looking at it wrong. The new axes confused me and I lost my train of thought. If I'm not mistaken, it should look something like this:

    ETVFDhk.jpg

    I go by the previous x & y axes, I take the directions the acceleration and Fg had on them, and then I create a net force based on the new axes. Correct? (there might be a misconception or two in my wording, but I'm still trying to get the hang of the english scientific terms).
     
  7. Dec 13, 2016 #6

    haruspex

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    Yes, that looks right. At least, I see you now get Fgx=Fgsin(θ). Is the rest ok now?
     
  8. Dec 13, 2016 #7
    Yeah, that was the only part I had trouble with, not with the formulas themselves. I've got the acceleration vectors on the left as well, so that's all.

    Thanks for taking the time!
     
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