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Addition of vectors and direction

  1. Sep 10, 2009 #1
    1. The problem statement, all variables and given/known data

    A vector A has a magnitude of 56.0 m and points in a direction 20.0° below the positive x axis. A second vector, B, has a magnitude of 70.0 m and points in a direction 45.0° above the positive x axis.

    Using the component method of vector addition, find the magnitude and direction of the vector C (counterclockwise from the positive x-axis).

    2. Relevant equations

    a) Cx = Ax + Bx
    b) Cy = Ay + By
    c) Ax = ACos[tex]\Theta[/tex]
    d) Ay = ASin[tex]\Theta[/tex]
    e) Bx = BCos[tex]\Theta[/tex]
    f) By = BSin[tex]\Theta[/tex]

    3. The attempt at a solution

    I have calculated out all the relevant equations:
    a) Cx = 52.6 + 49.5 = 102.1m
    b) Cy = 49.5 - 19.2 = 30.3m
    c) Ax = 56Cos20 = 52.6m
    d) Ay = 56Sin20 = -19.2m (negative due to y in the negative direction per my written graph)
    e) Bx = 70Cos45 = 49.5m
    f) By = 70Sin45 = 49.5m

    I have come up with vector C to be 106.5m using the Pythagorean theorem, but I cannot seem to get the correct direction of vector C. I have tried using the equation tan[tex]^{-1}[/tex](Cy/Cx) which gives me tan[tex]^{-1}[/tex](30.3/102.1) = 16.5, but when I plug it into the answer it tells me to check my syntax. By my graph, Vector C is in Q1 so I would assume from x-axis counter clockwise the answer should be 16.5deg. What I am unsure of is the fact that Vector A is in QIV and therefore the angle is intersected by the x-axis, so maybe that is where I am having my problem. My professor didn't really cover anything to this degree, so I have been relying on the book and it hasn't been much help either. Your help is greatly appreciated.

    As this is my first post to this forum, I was trying to follow the rules to the letter, but if I don't need to type everything I did, would someone please let me know. Thank you very much.

    Ahhh....The joy of returning to school after a 10 year hiatus!!
  2. jcsd
  3. Sep 10, 2009 #2


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    Homework Helper
    Gold Member

    Your answer looks fine to me, so it must be that the automated system you are plugging it into requires it to be in a certain form (syntax).
  4. Sep 10, 2009 #3
    What other form could there be. It is asking for the direction counterclockwise from the x-axis and is asking for the answer in degrees.
  5. Sep 10, 2009 #4
    Nevermind, I was racking my brain to try and figure out what the problem was, so I just decided to forget it and try re-inputting my answer again after about 4 different times of working out the problem and it finally took my answer @ 16.5 deg which is what I had put into the automated system originally. Thanks for your help.
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