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Finding a vector's angle

  1. Aug 27, 2011 #1
    1. The problem statement, all variables and given/known data

    Rewrite the following vector in terms magnitude and angle: A force vector with a magnitude of 58 lb that is in the third quadrant with an x component whose magnitude is 30 lb.

    3. The attempt at a solution

    This is how I drew the problem:

    [PLAIN]http://img535.imageshack.us/img535/4779/vecy.png [Broken]

    Obviously, the answer to the first question asking the vector magnitude is just 58, but I'm having trouble getting the angle.. I tried:

    [itex]cos(\theta) = \frac{-30}{58}[/itex]
    [itex]\theta = cos^{-1}(\frac{-30}{58}) = 121.15^{\circ}[/itex]

    But obviously that cannot be the answer (and it isn't..). So what am I doing wrong?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 27, 2011 #2

    PeterO

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    You are analysing a triangle to use trig to calculate the angle. triangles do not have sides with negative length!
     
    Last edited by a moderator: May 5, 2017
  4. Aug 27, 2011 #3
    But the input of the arccos function is -1<x<1, right? So why would negative input be allowed in the domain if, well, it's not allowed (like your saying)?
     
  5. Aug 28, 2011 #4

    ehild

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    Theta is the angle the force encloses with the positive x axis. There are two angles between 0 and 360° with the same cosine: if one is x, the other is (360°-x).

    ehild
     
  6. Aug 28, 2011 #5
    r* ( costheta i + sintheta j ), since it is in the 3rd quadrant cos turns out to be negative and sin as well.
     
    Last edited by a moderator: May 5, 2017
  7. Aug 28, 2011 #6
    But shouldn't [itex]\theta[/itex] be the angle from the negative x-axis? If I new the y-component of the vector in the picture, then I would use [itex]tan^{-1}(y/x)[/itex] and the angle would be with respect to the negative x-axis; I know, because I've done it before..

    So confused ... :confused:
     
  8. Aug 28, 2011 #7

    vela

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    When you wrote the equation [itex]\cos \theta = -30/58[/itex], you assumed the convention where [itex]\theta[/itex] is the angle measured from the +x-axis. In addition, when you solved for [itex]\theta[/itex], you used the arccos function, which will only give you an angle between 0 and 180 degrees, which corresponds to being above the x-axis.

    Try drawing in the point at 121.15 degrees and distance 58 from the origin. You'll see it also has an x-component equal to -30. Your picture should also make clear the following idea.

    If the point you're working with is below the x-axis, like in this problem, you can find the corresponding angle by subtracting the result of arccos from 360 degrees. So for this problem, you'd find the angle to the point in question is 360-121.15 = 238.85 degrees. But again, this angle will be measured from the +x-axis.

    Now how can you convert that angle from being measured from the +x-axis to being measured from the -x-axis?


    The other approach, which is the one PeterO is alluding to, is to assume the angle is measured from the -x-axis, since that's what you're looking for, and find the equation analogous to [itex]\cos \theta = -30/58[/itex] but based on lengths of the sides of the right triangle.
     
    Last edited: Aug 28, 2011
  9. Aug 28, 2011 #8
    Please re-read my previous post and always take theta the angle with the x axis and do not forget to keep track of which quadrant you are in, for Cos and Sine sign will change in each quadrant; ie: 3rd quadrant cos<0 , sin<0
     
  10. Aug 28, 2011 #9
    Thanks for your explanation! I understand my mistake now. I was trying to measure the angle from the -x axis, but was including the negative sign on the x-component, which I should not have done. I guess I need to review my trig book :blushing:
     
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