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Three forces on a box in different directions what is the acceleration

  1. Sep 27, 2012 #1
    1. The problem statement, all variables and given/known data

    Force A 250 degrees @ 200 N
    Force B 150 degrees @ 300 N
    Force C 50 degrees @ 400 N
    mass = 40kg
    mew = .14
    what is acceleration

    2. Relevant equations

    Fa^2 = Fy net^2 x Fx net^2
    Fn = Fg
    Ff = Fn x mew
    Fnet = Fa - Ff
    F= ma


    3. The attempt at a solution

    Fax=Cos 20 (200) = -187.93
    Fay = Sin 20 (200) = -68.40

    Fbx = Sin 30 (300) = 150
    Fby = Cos 30 (300) = -259.81

    Fcx = Cos 50 (400) = 257.11
    Fcy = Sin 50 (400) = 306.42

    Fx net = 219.18 N
    Fynet = -21.79 N

    Fa^2 = Fxnet^2 + Fynet^2
    Fa^2 = 219.18^2 + -21.79^2
    Fa^2 = 48039.87 + -474.80
    Fa^2 = 47565.07
    Fa = 218.09 N

    Fn = 9.81 m/s^2 x 40Kg = 392.40N

    Ff = Fn (mew) = 392.40N (.14) = 54.94 N

    Fnet = Fa - Ff = 218.09 - 54.94 = 163.15 N
    a = F/m = 163.15/40 = 4.08 m/s

    I'm just not sure Ive done this correct
     
  2. jcsd
  3. Sep 27, 2012 #2
    The components of Fa and Fb are not computed correctly. BTW, you could make your life easier by choosing x to coincide with one of the forces, say Fc, then Fcx = 400 N and Fcy = 0, and the angles of the other forces are relative to Fc, i.e., 200 and 100.
     
  4. Sep 27, 2012 #3
    Sorry voko I dont follow. I drew my x,y graph and entered the force vectors. I then used the angles relative to the x axis, (as the text says) hence F ax = Cos 20 and F bx = Sin 30. I'm not sure about your meaning to "choose x to coincide with one of the forces." I'm in grade 11 and this is what we were taught to do.
     
  5. Sep 27, 2012 #4
    No problem, you can ignore this part. But the components of Fa and Fb are still not computed properly. For example, Fax = Fa cos 250 = -68.4 N.
     
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