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Resultant Force

  1. Oct 19, 2013 #1

    Jud

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    I'm sure this problem is pretty trivial to most but apologies in advance, I'm just starting out sorry.

    1. The problem statement, all variables and given/known data
    F2-3.jpg

    I'm not sure how to go about working this out due to the 800N

    3. The attempt at a solution

    F1 (800N) x-pos = 800 tan (90)?? , y-pos = ???
    F2 (600N) x-pos = 600 cos(30) , y-pos = -600 sin(30)
    1. The problem statement, all variables and given/known data
     
  2. jcsd
  3. Oct 19, 2013 #2

    UltrafastPED

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    You can draw the parallelogram of forces, and the diagonal length (from your starting point) is the net force.

    Or you can complete their conversion to vectors; F1 = 800 j, where j is the unit vector in the y-direction. Your F2 looks correct ... so now add them together!
     
  4. Oct 19, 2013 #3

    Jud

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    so,

    F1 (800N) x-pos = 0 , y-pos = 800
    F2 (600N) x-pos = 600 cos(30) , y-pos = -600 sin (30)

    sqrt (519.615)^2 + (500)^2 = 721.1 N

    is that correct?
     
  5. Oct 19, 2013 #4

    UltrafastPED

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    It's certainly in the correct range ... if you are unsure you should recheck your calculations.
     
  6. Oct 19, 2013 #5

    Jud

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    Ok Thank you for the help, greatly appreciated.
     
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