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Horizontal Tension

  1. Jun 22, 2015 #1
    Hi guys, I came across this question and I was wondering if someone could explain how we get:
    TLsinO - mg(1/2LcosO) = 0
    Your help would be much appreciated.

    MFoA4hb.jpg
     
  2. jcsd
  3. Jun 22, 2015 #2
    Do you know how to calculate a torque?
     
  4. Jun 22, 2015 #3
    t=fd and t-wa ?
     
  5. Jun 22, 2015 #4
    What does d represent there?
     
  6. Jun 22, 2015 #5
    distance....
     
  7. Jun 22, 2015 #6
    Not just any distance. It's specifixally the perpendicular distance from the wall. Since the beam is at an angle, you need the trig equations to find the perpendicular distance.
     
  8. Jun 22, 2015 #7
    If you understand this, what's the problem to explain it in a couple sentences instead of playing games? If you don't want to explain, it's totally fine.
    I am so confused what these two parts mean:
    1) TLsinO
    2) mg(1/2LcosO)
     
  9. Jun 23, 2015 #8

    haruspex

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    If you are familiar with torque then you should know this: if a force F acts through a point P, the torque the force has about a point Q is F x distance PQ x sin(angle between PQ and the direction of the force).
    Can you relate that to the diagram and see how the expressions you quote correspond to the torque exerted by T about the hinge, and the torque the weight of the arm exerts about the hinge?
     
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