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Finding the Tensions using the sum of moments

  1. Apr 14, 2010 #1
    1. The problem statement, all variables and given/known data
    The tower is 70 m tall. If the tension in cable BC is
    2 kN, and you want to adjust the tensions in cable
    BD and BE so that the couple exerted on the tower
    by the fixed support at A is zero.

    A) Determine the tensions in cable BD and BE
    by applying the sum of the moments about
    point A .

    The Coordinates in meters

    2. Relevant equations


    3. The attempt at a solution

    Well I know that I have to solve by taking the sum of the moments about A. That means that the Moment about A of Tension BC + Moment about A of Tension BD + Moment about A of Tension BE = 0
    and The moment about A of Tension BC is nothing more than the Fx of Tension BC about the distance from point A to B. So That means I find the angle that Tension BC makes to the tower get my Fx and times it by the tower distance. So now what do I need to do next. Looking for the next step.
  2. jcsd
  3. Apr 15, 2010 #2


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    You are correct so far. Note that the moment about A from the cable tension BC is about the z axis (the axis into the plane), that is, M_z (from the force BC) is (F_x)(y). Also note that for equilibrium, the sum of all moments M_z from all the cable forces must be 0, and the sum of all moments M_x from all the cable forces about the x axis must be zero. In general, M_z = F_x(y) + (F_y)(x), and M_x = (F_y)z + F_z(y), (and M_y = F_x(z) + F_z(x), which you don't need in this problem, since x and z are 0). So just determine those F_x and F_z components of each cable force, and solve the resulting equilibrium equations for the unknown values.
  4. Apr 15, 2010 #3
    So what I need to do to find the Tensions of BE and BD I need to set up the Tension times the Perpendicular distance to point A?
    Do I need to take the tensions and place them as if they are only in the x-z plane?
    I guess what I am trying to figure out is how do I need to go about to get the distance relative to the tensions?
    I can see that I have two equations from M about the z axis and M about the x axis. Requiring the use of the substitution method of one of the moments above.
  5. Apr 15, 2010 #4


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    Yes, that's a good way of doing it. Look at the guyed tower in a plan view from the top. Each cable tension BE and BD will have an x and z component, which are geometrically/trigonometrically related. The equations will simplify a bit since the x and z lever arms are 0. When you solve the simultaneous equilibrium of moments equations, you'll get the x and z components of the unknown cable tensions. Then you'll have to use more geometry/trig to get the cable tensions in the guys. It's kind of tough to keep it straight. And watch your plus and minus signs.
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