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What is the tension in the wire?

  1. Nov 3, 2012 #1
    There is a street lamp that weighs 150N. The two wires that are attached to it form at 120°. What is the tension on the wires?

    So i drew a diagram. Since the force of gravity acting on the lamp is 150N down, there must be an equal opposite force acting at 150N up.
    Then I drew a right triangle and it a 30-60-90. (The 120° was bisected by the upward force).
    Then I said that sin(30)=150/T. Then T=150/sin(30) which equals 300N. Is this right?
     
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  3. Nov 3, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi Psyguy22! Welcome to PF! :smile:
    Almost. :rolleyes:

    there are two wires … :wink:
     
  4. Nov 3, 2012 #3
    so would I divide 300 by 2? That would give me 150N again though?
     
  5. Nov 3, 2012 #4

    tiny-tim

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    Hi Psyguy22! :smile:
    Let's see …

    2Tsin30° = 150 N

    so T = 150N …​

    what's worrying you about that? :confused:
     
  6. Nov 3, 2012 #5
    From the triangles I drew, I thought that each wire had 300N
     
  7. Nov 3, 2012 #6

    tiny-tim

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    not sure how you get that from triangles :confused:

    remember, each triangle only gets half of the 150 N weight :wink:
     
  8. Nov 3, 2012 #7
    HOw do I post attachments to show what I drew. I dont know if I'm completly understanding what your saying
     
  9. Nov 4, 2012 #8

    haruspex

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    Each wire has tension T and makes an angle of 60 degrees to the vertical (at least, that has been assumed; it's not clear in the problem statement). What contribution does each wire make to the vertical force needed to support the lamp?
     
  10. Nov 4, 2012 #9
    Wouldn't they both have to contribute 150N? But they don't act directly upwards. Wouldn't there combined tension be more that 150N
    I tried watching khanacademy and both his examples, the tension in the wires was more then the actual weight of the object.
     
  11. Nov 4, 2012 #10

    haruspex

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    They only have to match 150N in total, so why would each need to contribute 150N vertically?
     
  12. Nov 4, 2012 #11
    I guess I do not know. So would I put that each one only contributes 75N vertically?
    Then 75/sin(30)= 150N is this right?
     
  13. Nov 4, 2012 #12

    haruspex

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    Sounds right to me. Look at what that gives you in your force diagram. Three forces each 120 degrees from its neighbours, all pulling with the same force.
     
  14. Nov 4, 2012 #13
    Ok. Thank you! I really like to make sure I understand what I am doing so thanks for all your help!:)
     
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