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Equilibrium of systems in acccelerating frame

  1. May 5, 2013 #1
    69g5g8.png

    I've drawn the following FBD (using D'alembert prinicple)
    [The N in my drawing is the F_A stated in the question]

    579kn.jpg
    ----------------

    Assuming my FBD has nothing wrong, am I right to say that I am free to choose my torque equation about ANY POINT?

    somehow my solutions vary depending on my point of consideration.

    Torque eq about CM:
    N + 2Fy cos(30) -2Fx sin(30) = 0

    -or-
    or eq about pivot:
    N - 2mg cos(30) - 2ma sin(30) = ma

    It just doesnt give the same solution.
    --------------------------------------------
    My other translational equilibrium eqs are:

    ma - Nsin(30) + Fx = 0
    Fy + Ncos(30) - mg = 0
     
  2. jcsd
  3. May 5, 2013 #2
    solved.

    my mistakes when pressing the calculator )=
     
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