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Net force of a vector

  1. Dec 5, 2008 #1
    Consider 3 force vectors F1, F2, and F3. The vector F1 has magnitude F1 = 36N and direction θ = 110°; the vector F2 has magnitude F2 = 22N and direction θ = - 140°; and the vector F3 has magnitude F3 = 28N and direction θ = 20°. All the direction angles θ are measured from the positive x axis: counter-clockwise for θ > 0 and clockwise for θ < 0.

    What is the magnitude F or the net force vector F = F1+F2+F3? Answer in units of N.

    I found the x resultants to be -12.31 + -14.14 + 26.31 = -0.14
    I found the y resultants to be 33.83 + 16.85 + 9.58 = 60.26

    (sqrt)-0.14^2 + 60.26^2 = 60.26

    Can someone tell me were Iam going wrong??????
     
  2. jcsd
  3. Dec 5, 2008 #2
    Can you show us detail about how you derived the x- and y- components for each vector?

    I think you have switched x and y components on one of your vectors, and have a sign wrong. I think it's best to draw each vector separately and write out all the steps -- that way you can check if these intermediate parts are correct.
     
  4. Dec 5, 2008 #3
    ok here is what i did

    F1x = 36N cos110 = -12.31
    F1y = 36N sin110 = 33.83
    F2x = 22N cos130 = -14.14
    F2y = 22N sin130 = 16.85
    F3x = 28N cos20 = 26.31
    F3y = 28N sin20 = 9.58

    then i found the resultant for the x and y component
     
  5. Dec 5, 2008 #4
    Check your vector 2... why do you have 130 degrees?
     
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