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Smallest possible tension and its angle

  1. Sep 18, 2011 #1
    1. The problem statement, all variables and given/known data
    determine the value of Beta for which the tension in cable B is as small as possible, then find the corresponding tension. Given Alpha = 5 degrees and the weight of the object is 12kN
    This is an image of it http://imageshack.us/photo/my-images/801/unledgsn.png/


    2. Relevant equations
    Fy=0 Fx=0 (at equilibrium)


    3. The attempt at a solution
    Fx= -TAsin5 + TBcos(Beta)
    Fy=TAcos5 - TBsin(Beta) - 12

    from Fx i get TA = (TBcos(Beta))/Sin5
    and i plug it into Fy to get 12=TB(11.4cos(Beta)-sin(Beta))

    but this doesnt answer my first question of what is the angle, only helps with part 2 of the question..
     
  2. jcsd
  3. Sep 18, 2011 #2
    Have you learned about torque? If so measure the torque about some point along the arrow labeled tension in A, say the arrow head of tension A. In equilibrium the sum of the torques will be zero. Then it is clear what the angle beta should be so as to maximize the torque about that point due to the tension in B and thereby minimizing the tension in B.
     
  4. Sep 18, 2011 #3
    thanks for the reply, but well we havn't studied torque yet so we are supposed to use another way that depends on components of force vectors, we are currently studying statics in particles. so is there a way that is similar the solution Im using?
     
  5. Sep 18, 2011 #4
    Let us assume what you wrote is correct,

    12=TB(11.4cos(Beta)-sin(Beta))

    To make TB minimum we want (11.4cos(Beta)-sin(Beta)) to be maximum. So set

    d (11.4cos(Beta)-sin(Beta)) / d beta equal 0 ?
     
  6. Sep 19, 2011 #5
    i thought so too but then if I use -5 for beta ud get lower tension.. and for -10 you'd get about the same amount as of -5.
     
  7. Sep 19, 2011 #6
    Near the angle of minimum tension (-5 degrees), tension varies slowly with change in angle.
     
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