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Vectors: Angle Direction

  1. Jul 26, 2012 #1
    1. The problem statement, all variables and given/known data
    I had a problem on my physic hw, and I keep on getting it wrong, so I was wondering:
    in a vector, if the x direction is negative, and the y direction is positive, would the angle direction also be negative? What happens if the x direction is positive, and the y direction is negative? A degree can't be negative right?

    also,

    2) what does it means when the angle direction is "from the positive x axis" or "counterclockwise from the positive x axis is positive"

    2. Relevant equations



    3. The attempt at a solution
    In my hw, I figured out the magnitude already, which is 81.2, and the x direction is -81.22 and y direction is 1.95, so I found the angle direction by tan-1(1.93/-81.22) and got -1.37, but i just ignore the negative sign, since i assume that a degree cant be negative.
     
  2. jcsd
  3. Jul 26, 2012 #2

    PeterO

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    You clearly need to draw your self a picture.

    With a vertical y axis [positive above the x axis, negative below, as per normal] and a horizontal x axis [positive to the right of the y-axis and negative to the left of the y axis, as per normal]

    You should then locate the point (-81.2, 1.95) which would be just above the point -81.2 on the x-axis.

    One common convention for angles - mimicking polar co-ordinates - is to define angles relative to the positive x-axis, travelling anti clockwise [travelling clockwise means a negative angle.
    There is nothing more mysterious about a negative angle than there is with the negative x-axis. It merely means "going the other way"

    The angle you are looking for is just a little less than 180 degrees,

    Lets imagine the angle was 177o. The same point is at angle -183o - but it seems you were asked for the positive answer.

    You will also need Pythagoras to find the actual magnitude of the vector.

    Note: If you are familiar with swapping from rectangular to polar co-ordinates on your calculator, you could just use that transform. Most people are not familiar with that use of a calculator however.
     
  4. Jul 26, 2012 #3

    PeterO

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    Note: your answer of -1.37 is not "entirely" wrong. It means the angle is related to 1.37 degrees, but in a quadrant where the tangent is negative [2nd and 4th]. SO the answer you really want is (180 - 1.37) or (360 - 1.37); meaning 178.63 or 358.63.

    A diagram of the situation will show that the 178.63 is what you are after.
     
  5. Jul 26, 2012 #4
    Thank you for helping me out, but i am still confused on why 180-1.37?
     
  6. Jul 26, 2012 #5
    PeterO answered the question in his post.

     
  7. Jul 26, 2012 #6

    PeterO

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    If you recall your trigonometry: we have expressions like:

    sin(180-θ) = sin(θ) 2nd quadrant
    sin(180+θ) = -sin(θ) 3rd quadrant
    sin(360-θ) = -sin(θ) 4th quadrant

    also
    tan(180-θ) = -tan(θ)
    tan(180+θ) = tan(θ)
    tan(360-θ) = -tan(θ)

    and a set for cosine.

    since the tangent was negative in your problem, it meant you should use either
    tan(180-θ) = -tan(θ)
    or
    tan(360-θ) = -tan(θ)

    A diagram would show your angle was in the 2nd quadrant [between 90 and 180 degrees] so tan(180-θ) was used.
     
  8. Jul 26, 2012 #7
    What would happen if the x direction is +81.22, and the y direction is -1.95?
     
  9. Jul 26, 2012 #8

    PeterO

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    Your diagram would show you it is in the 4th quadrant, so would use (360 - θ) to get 358.63.

    That answer could also be expressed as -1.37
     
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