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Vector angle

  1. Sep 8, 2010 #1
    1. The problem statement, all variables and given/known data

    a x component of a vector A is -60.0m and the y component is +50.0m. what is the angle between the direction of A and the positive direction of x?

    2. Relevant equations



    3. The attempt at a solution

    i found the magnitude to be 78.1m.

    all i did to solve was do arcsin(50/-60) to get 56.4

    the vector is in the first quadrant with a - x and positive y value so sin would be negative, right?
     
  2. jcsd
  3. Sep 8, 2010 #2

    cepheid

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    This should be arctan (draw a diagram).

    I don't think that's the first quadrant. I also don't think that the sin of that angle would be negative, (not that it is relevant here anyway).
     
  4. Sep 8, 2010 #3

    collinsmark

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    So far so good. :approve:
    Note quite.

    In this interim step, you want to find the angle of a right triangle. The y-component is opposite the angle, and the x-component adjacent.

    [tex] \frac{\mathrm{opposite}}{\mathrm{adjacent}} = \tan \theta [/tex]

    So you need to use arctan to find the angle, not the arcsin.

    (But if you wanted to you could use the arcsin by using [itex] \theta = \arcsin(\mathrm{opposite}/\mathrm{hypotenuse}) [/itex] where the hypotenuse was the magnitude that you found above. But it's usually easier and less prone to error to use the opposite [y-component] and adjacent [x-component] directly and use the arctan.)
    This is the part you need to be careful about. Sometimes drawing the vector on paper makes it easier. If vector A's tail is at the origin, its head is in the second quadrant. You might want to do a quick review of which quadrants are which (quadrant I is in the upper right, 2 in the upper left, 3 in the lower left, and 4 in the lower right).
     
  5. Sep 8, 2010 #4
    This is the part you need to be careful about. Sometimes drawing the vector on paper makes it easier. If vector A's tail is at the origin, its head is in the second quadrant. You might want to do a quick review of which quadrants are which (quadrant I is in the upper right, 2 in the upper left, 3 in the lower left, and 4 in the lower right).[/QUOTE]


    so if the vector is in the first quadrant and i am using the arctan. which is cos/sin and cos and sin are both positive in the first quadrant then the answer would be positive.

    so the arctan(50/60)= 39.8(degrees)
     
  6. Sep 8, 2010 #5

    collinsmark

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    Usually (but not necessarily always), the angle θ is defined with respect to the positive x-axis.

    If it is, the y-component is associated with the sine and the x-component the cosine. But The angle θ is not always defined with respect to the positive x-axis in all problems all the time. So sometimes you'll need to figure this stuff out on a problem-by-problem basis.

    But here is one thing you can and should memorize. The following is always true for all right triangles: In a right triangle where θ is one of the angles (one of the acute angles -- not the 90o one),

    [tex] \sin \theta = \frac{\mathrm{opposite}}{\mathrm{hypotensuse}} [/tex]

    [tex] \cos \theta = \frac{\mathrm{adjacent}}{\mathrm{hypotensuse}} [/tex]

    [tex] \tan \theta = \frac{\mathrm{opposite}}{\mathrm{adjacent}} [/tex]

    and finally the Pythagorean theorem,

    [tex] \left( \mathrm{hypotenuse} \right)^2 = \left( \mathrm{opposite} \right)^2 + \left( \mathrm{adjacent} \right)^2 [/tex]

    Those are the really important ones to memorize.

    Now draw a vector with its tail at the origin and its head in the first quadrant. Draw θ as being the angle between the vector and the x-axis. Actually draw this on paper. Draw a dotted line from the tip of the vector, straight down to the x-axis (the dotted line should be parallel to the y-axis, and the dotted line should intersect the x-axis). You've made a right triangle. Now examine your drawing carefully and determine which component is opposite and which component is adjacent.

    Yes, that's right! :approve: And if you know the vector is in the first quadrant, you can trust that your calculator will give you the correct answer. But you can't trust your calculator to give you the correct, final answer if the vector is in any other quadrant.

    I caution you here that things can get a little tricky with arc tangents, arc sines, etc., when dealing with multiple quadrants. If you're just given the task to calculate the arctan of a negative number on your calculator, the result will be a negative number, but you can't tell if it is in the second or fourth quadrant by that information alone. If you are asked to take the arctan of a positive number, you will get a positive number, but you don't know if the vector is in the first or third quadrant (without more information).

    So as part of the process you need to keep in mind which quadrant the vector is in to begin with. Then, after you use your calculator to get the arctan, adjust the answer accordingly to put it in the correct quadrant (which might involve negating the angle, and may also involve adding or subtracting 180o.)

    But you don't need to take my word for what to do in each particular quadrant. I encourage you to start drawing vectors on paper (with the tails at the origin), and figuring this out for yourself. About all you really need to memorize are the formulas listed above. And don't forget that your calculator will assume you are working with simple right triangles with acute angles. You can figure out the rest (i.e. the rest of the geometry that your calculator does not consider). :wink: [Edit: although you can figure it out for yourself, you will need to practice it. So please, take out some graph paper for this. :smile:]
     
    Last edited: Sep 8, 2010
  7. Sep 8, 2010 #6
    alright, thanks very much, i understand what your saying

    but i submitted that answer of 39.8 and it was wrong. is it in the wrong sig figs(dont think so).
     
    Last edited: Sep 8, 2010
  8. Sep 8, 2010 #7

    cepheid

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    That's because 39.8o is not the correct angle. You have to be careful, because more than one angle can have the same tangent. In other words, arctan doesn't return a unique solution. The correct answer is an angle that has the same tangent as 39.8 degrees and is related to it in a very specific way. If you don't understand what I mean, then consider this hint from your problem statement:

    This is the part where drawing a diagram is extremely helpful.
     
    Last edited: Sep 8, 2010
  9. Sep 8, 2010 #8

    cepheid

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    By the way, ALL of what you've said above is wrong:

    arctan is NOT equal to cos/sin. Arctan is the inverse function of tangent. It takes a number and returns an angle whose tangent is equal to that number.

    Also, as you've been told twice before, the vector is NOT in the first quadrant.
     
  10. Sep 8, 2010 #9
    sorry guys, i thought you said it was in the first quadrant but its in the second quadrant so i would take 39.8 degrees and add 180 degrees because its in the second quadrant to get my answer of 219.8 degrees
     
  11. Sep 9, 2010 #10
    is this correct??
     
  12. Sep 9, 2010 #11

    cepheid

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    No. Angles in the second quadrant have to be between 90 and 180 degrees. Did you draw a diagram? If so, look at it. What is the angle between the vector and the positive x axis?
     
  13. Sep 9, 2010 #12
    would it be 180-39.8 to get 140.2 or 140
     
  14. Sep 9, 2010 #13

    cepheid

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    Yeah exactly. Subtracting the angle from 180 degrees causes the reflection into the second quadrant.
     
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