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Resultant Vectors with angles and magnitude

  1. Nov 25, 2008 #1
    1. What is the resultant vector, and it's angle when the forces move 0.45newtons North, then 0.85 newtons East. Where does the resultant angle go, and what would the direction be?



    2. I used the Pythagorean Theorom to find the length of the resultant vector.

    The length of my resultant vector using the Pythagorean Theorm was ~0.96newtons. Now, to find the resultant vector angle, I know I use Tan-1(opposite/adjacent). I just don't know which one is opposite and which one is adjacent.

    Once I find that, what would the direction be in the final result?
     
  2. jcsd
  3. Nov 25, 2008 #2

    G01

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    In this case, drawing a vector diagram will really help you.

    Draw the triangle made by the forces and the resultant force and mark the angle in question. The force that is "opposite" is the vector forming the side of this triangle that is opposite that angle.
     
  4. Nov 25, 2008 #3
    I drew the vector diagram. So, would the angle that I am solving for be the angle that the resultant vector is coming from? (the tail)

    So then would I use the cotangent (0.85/0.45)?
     
  5. Nov 25, 2008 #4

    G01

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    The angle could be any angle, just make sure you point out which angle it is.

    For instance, you could pick the angle between the resultant and the x axis. Or you could pick the angle between the resultant and the y-axis. Either one is correct, but the opposite and adjacent sides will switch depending on which angle you pick.

    Just remember which angle you found and mark it appropriately.

    If by "angle the resultant is coming from" you mean the angle between the resultant and the x-axis, then yes you are correct.
     
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