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Finding Torque/Vert/Horiz. Forces for a Sign

  1. Mar 27, 2013 #1
    1. The sign Ye Olde Tavern is supported by a horizontal beam (A-C) and a cable (B-C). The sign has a weight of W= 294 N.
    a) Take (Summate) the torques of the system about point A then find the tension in the cable.
    b) Find the vertical force on the wall attachment at point A by taking torque about point C.
    c) Find horizontal force on the wall attachment at point A.


    The Angle is θ
    Total Length of the beam is 4d with the sign being equal to 3d (so there is a gap of 1d)
    A is the base right angle
    B is where the cable and pole meet
    C is where the cable and beam meet.


    T=Ia
    Fx - Tcosx = 0
    Fy + Tsinx -Fg = 0




    I really have no idea how to go about this problem but for
    a) at point A ƩT=∅ along with ƩF=∅
    so I believe that tension would have to equal T=0 for the cord (?)
     
  2. jcsd
  3. Mar 27, 2013 #2
    Hello mmravunac,

    Could you please show me the diagram as I'm not able to see it in my mind.

    Sunil
     
  4. Mar 27, 2013 #3
  5. Mar 27, 2013 #4
    Thanks,

    You are right about a) that the net torque about A is zero but are you sure tension is zero?

    What are the forces acting on the rod? Where are they acting relative to A?
     
  6. Mar 27, 2013 #5
    You need to work out carefully and with patience. Your question itself gives many hints. Normal reaction on the rod by the wall has two components. By taking torque about A, you need not include those forces. Carefully sum anticlockwise and clockwise torques to 0 about A. Do not forget any forces. (Of course torque of normal reactions at beam about A is 0).

    Also, horizontal beam is assumed to be massless, right ? Do not forget to resolve components of tension. Also, make an equation for translational equilibrium along horizontal and vertical directions respectively. It might help you.
     
  7. Mar 27, 2013 #6
    So i think I would have to set the equation to
    Mb is Beam mass which is 0 (?)


    T*4d - Mbg*2.5d - 294N*4d*cos(θ) = 0
    which is set to

    T*4d = 294N*4d*cos
    Dividing each side by 4d gives us

    T= 294Ncos(θ) is this right or can I do more?

    EDIT: on my graphing calculation when i type in 294cos(theta) it gives me 294.
    So does this mean my tension is 294N?
     
    Last edited: Mar 27, 2013
  8. Mar 27, 2013 #7
    Is it T or a component of T? Think about it. (torque= r x F)
     
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