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Tension of wire

  1. Sep 19, 2004 #1
    I am having a little difficulty with a physics problem, hopefully someone can help!

    Q. The distance between two telephone poles is 50.0m. When a 1.00kg bird lands on the telephone wire midway between the poles, the wire sags .200m. Ignore the weight of the wire.

    Using the data given I used 25.0m and .200m as legs of a right triangle and solved for the hypotenuse which I found to be 25.0007999. Then I found the angles that made up that triangle.

    Then I did the sum of Force x = T1 sin89.54 + T2sign89.54 = 0
    and the sum of Force y = -T1cos89.54 + -T2cos89.54 = 9.8
    Then solving for T2 I get approximately 610 N as well as 610 N for T1. So a net Force Y of -9.8.

    What am i doing wrong? I know the answer is 613 N, but I am lost.

    Thanks!
     
  2. jcsd
  3. Sep 19, 2004 #2

    Pyrrhus

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    Actually

    [tex] \tan\theta = \frac{0.2}{25} = 0.458^o [/tex]

    Also i notice on your Y-axis analysis you should have

    [tex] T_{1}sin(0.458) + T_{2}sin(0.458) = mg [/tex]

    and on your x-axis

    [tex] T_{1}cos(0.458) = T_{2}cos(0.458) [/tex]

    Hint: Tensions are equal!
     
    Last edited: Sep 19, 2004
  4. Sep 19, 2004 #3
    venom, what program do u use to typ in the math symbols? how can i do that so I can type things easier?
     
  5. Sep 19, 2004 #4

    Pyrrhus

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    The program is called LaTeX, and this forum has it integrated.

    Check this link
    https://www.physicsforums.com/misc/howtolatex.pdf
     
  6. Sep 19, 2004 #5
    Thank you for your help Cyclovenom! I was able to get the correct answer using the information you gave me.

    I do have 1 other question though,
    In the work I did, I was using T1 Cos 89.54 + T2cos 89.54 = 9.8
    This should give the same answer I would think, however it gives 610 N rather than the 613 N. Why is there a difference?

    To get the angle of 89.54 I simply did ArcTan 25/.2

    Thanks again,
    Joshua
     
  7. Sep 19, 2004 #6

    Pyrrhus

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    Well if you see the triangle actually the oposite side will be 0.2 and the adjacent side will be 25 m.

    so the x-component of a tension like T will hace been Tcos(angle) and the Y component Tsin(angle), also weight points directly to the center of the Earth, so it will be pointing down the bird while the tensions point in the direction of the wire, so the y-components of the tensions will relate with weight in the equation above. Remember we are considering equilibrium situation.
     
  8. Sep 19, 2004 #7

    Pyrrhus

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    Also Imagine you put a coordinate system with origin in the bird's location. You can brind the angle you found between the space the straight line and x-axis make, that is a geometry postulate very useful, If i remember correctly it was altenate inner angles are equal.
     
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