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Beam supported by a pulley

  1. Oct 19, 2014 #1
    1. The problem statement, all variables and given/known data
    Ho3pV26.jpg
    [Ans: FA = -206i + 188j lb, T = 131 lb]

    2. Relevant equations
    Since it's assumed that the system is in equilibrium, so I assume that ΣF=0 and ΣM=0

    3. The attempt at a solution
    I tried to draw a free-body diagram for the system, and set up three equations below:
    nX4yG9y.jpg
    (I don't know the direction of FA yet, so I assume that both components are in positive direction)
    ΣFx = FAx + Tcos45° + Tcos30° = 0
    ΣFy = FAy - 150 - W + Tsin45° + Tsin30° = 0
    ΣMz [Moment centered at A] = (16)(-150 - W) + (40)(Tsin30° + Tsin45°) + (6)(Tcos30° + Tcos45°) = 0

    When I solved the third equation to find tension T, it didn't give me the right answer, I am not sure which part I did wrong, maybe I missed some forces on the free-body diagram, so I made a wrong equation of ΣMz?
    Any hints or approaches are much appreciated.
     
  2. jcsd
  3. Oct 19, 2014 #2

    NTW

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    In the FBD, and when calculating the moments, you don't put W = 20 kg at mid-point along the beam...
     
  4. Oct 19, 2014 #3

    SteamKing

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    To clarify what NTW was trying to tell you, you somehow neglected to include the weight of the beam in your equations of static equilibrium.
     
  5. Oct 19, 2014 #4
    I think I had included the weight of the beam in FBD and equations, but maybe NTW is trying to tell me that the weight should be in another point along the beam?
     
  6. Oct 19, 2014 #5

    SteamKing

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    If the beam has a mass of 20 kg, then the weight in lbs is easy to calculate. (Hint: using 2 lbs/kg is not really acceptable. 2.2 lbs/kg is more accurate)

    If the beam has a uniform cross-section, then using a c.g. of 16 in. from A is not correct either. The c.g. of a uniform bar is going to be half its length, as measured from one end.
     
  7. Oct 19, 2014 #6

    NTW

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    Exactly. Mid-point, as it's a uniform beam...
     
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