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Tension in wire (equilibrium)

  1. Sep 20, 2014 #1
    I dont manage to get this task right. See the pictures. First i take point D and draw a free body diagram. Then i use the equation for equilibrium. I can then calculate the tension in the wires that goes from D to B and from D to C. I then take point B and draw a free body diagram, but all the answers i get is wrong. Can somebody please see what i have done wrong?

    Attached Files:

  2. jcsd
  3. Sep 20, 2014 #2


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    Method looks ok. Check your working at the bottom of image Task1.jpg. The third line up looks ok. Last two look wrong.

    I got...

    TDB (1.11) = 490
    TDB = 441N
  4. Sep 20, 2014 #3
    Thanks for your reply, and sorry for just putting up pictures. My first time, so i will try to put things together a little nicer. I dont see how to solve that equation. I move 490.5 to the right, divide that with (cos45/cos30)(*sin30) so i stand with TDB+TDB sin45. divide with sin 45 and at last i divide with 2, beacuse TDB+TDB=2TDB. But apparantly somewhere this got terribly wrong, and i dont se where.
  5. Sep 20, 2014 #4
    Ah, now i see what you did, but want you get 2TDB when you add them together?
  6. Sep 20, 2014 #5


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    Here is my working starting with your equation...

    TDB*cos45*sin30/cos30 + TDB*SIN45 - 490.5 = 0

    Take TDB outside brackets..

    TDB*(cos45*sin30/cos30 + SIN45) - 490.5 = 0

    TDB = 490.5/(cos45*sin30/cos30 + SIN45)

    Substitute values..
    = 490.5/(0.707*0.5/0.866 + 0.707)
    = 490.5/(1.12)
    = 438N
    Not quite the 441N I got earlier.
  7. Sep 20, 2014 #6
    Thank you so much!
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