# Tension in wire (equilibrium)

1. Sep 20, 2014

### blubbulf

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?

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2. Sep 20, 2014

### CWatters

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
so
TDB = 441N

3. Sep 20, 2014

### blubbulf

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.

4. Sep 20, 2014

### blubbulf

Ah, now i see what you did, but want you get 2TDB when you add them together?

5. Sep 20, 2014

### CWatters

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

Rearrange..
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.

6. Sep 20, 2014

### blubbulf

Thank you so much!