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Torques and a Traffic Light

  1. Apr 2, 2012 #1
    1. The problem statement, all variables and given/known data

    A traffic light hangs from a pole as shown in the figure . The uniform aluminum pole AB is 7.50 m long and has a mass of 12.0 kg. The mass of the traffic light is 21.5 kg.

    2. Relevant equations

    clockwise torques = counterclockwise torques

    3. The attempt at a solution

    This may seem silly but my professor said to use the sin of 53° but one of the tutors told me to use the sin of 37 ° (which ended up being incorrect) because of the parallel line theorem. Could someone explain to me why the tutor was mistaken?

    (12.0 kg)(3.75m)(9.8 m/s^2)(sin 53°) + (21.5 kg)(9.8 m/s^2)(7.50 m)(sin 53°) = F * 3.80 m
     

    Attached Files:

  2. jcsd
  3. Apr 2, 2012 #2

    PhanthomJay

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    The tutor may have just erred in the geometric or trigonometric relationships. The moment, M, of a force, F, about a point, is the cross product of the position vector, r, times the force , that is, M = r X F = (r)(F)(sin theta), where theta is the angle in between the force and position vectors.
     
  4. Apr 2, 2012 #3

    gneill

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    Staff: Mentor

    Is it possible that the tutor suggested the use of cos(37°) rather than sin(37°)?
     
  5. Apr 2, 2012 #4
    Thanks you two! No, he definitely said (sin 37°).
     
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