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Homework Help: Beams and Wires at an Angle

  1. Mar 5, 2013 #1
    1. The problem statement, all variables and given/known data
    Find cord tension and force on the hinge on this beam given the following:

    Picture of the equation (in attachments)
    beam is 10 kg
    beam is 8.0m long

    2. Relevant equations


    3. The attempt at a solution

    Finding the Torque on the wire...
    I attached my FBD I drew. Before I start, one question? Since the beam is on an angle, would there be a horizontal force on the x-axis resulting from the Mass of the beam? If so, It could be 60 degrees? Mg Sin60?

    That is my attempt using the "Z-rule" in angles


    [T= 0
    Tcw = Tccw
    mg(d) = Tsin( ) * (d)
    (10)(9.8)(4) = T sin30 * (8m) ----> Sin 30* given that directly accross is 90 degrees + 60 degrees given = 30 degrees left at where the beam and wire are attached.

    392N = T sin30 * (8m)

    392N divided by 8m = 49

    49 = T sin30

    49 divided by sin30 --> 98?

    The answer is 49N. Why did I get double the answer? Is there something with my angles?

    Attached Files:

  2. jcsd
  3. Mar 5, 2013 #2
    The weight of the beam acts at angle to the "arm", so its torque must involve the sine of the angle. Intuitively that should be fairly clear: the "more vertical" the beam gets, the easier it is to hold it in place.
  4. Mar 5, 2013 #3
    So instead of "Mg" just being "Mg". It is "Mgsin(30)"? Is that what you are trying to say? If I change that in my equation it works out?

    If that is the case, waste no time to say... Would the horizontal axis on Mg also require a MgCos(30) to workout the force from the hinge in components?
  5. Mar 5, 2013 #4
    It should be fairly obvious that 49 sin 30 = T sin 30 implies T = 49.

    Regarding the forces, the answer is no. If you you use the horizontal and vertical axes, then the weight is purely vertical - it always is; the other forces, however, are not.
  6. Mar 5, 2013 #5
    Figured it out! Thanks
    Last edited: Mar 5, 2013
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