# [Statics] Finding the Tension

1. Oct 9, 2011

### Ortix

1. The problem statement, all variables and given/known data
[PLAIN]http://img830.imageshack.us/img830/1091/unled2cjt.jpg [Broken]

2. Relevant equations
I used sum of forces in x y direction and sum of moments

3. The attempt at a solution

So my problem here is that I solved for Ay (1275.3N) and now I have to solve for T

The obvious approach would be to use sum of forces in x direction after doing sum of forces in y to find B. Well I did that and found that B = 2550.6 which is double Ay and equal to mg. B is also perpendicular to the bridge thing. Now when I do sum of forces in x direction i do T = Bcos30 but that's wrong apparently.

Which part am I doing wrong? Is my B wrong?

I calculated B like so:
Fy:0 = Ay - mg + Bsin30

Last edited by a moderator: May 5, 2017
2. Oct 10, 2011

### Spinnor

Does this help? %^)

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3. Oct 11, 2011

### Ortix

unfortunately it does not. I figured that much. When i break down the N2 into it's components I still get a wrong answer for T

4. Oct 11, 2011

### grzz

i think that it is important to know whether there is friction with the rollers at A and B.

5. Oct 11, 2011

### Ortix

There is not, so you would think that the x component of the normal force at B would be equal to T, but it's not!

6. Oct 11, 2011

### grzz

If you explain more explicitly how you got B then I can try to give some help.

7. Oct 11, 2011

### Ortix

I calculated B like so:
Fy:0 = Ay - mg + Bsin30

Direct copy from my first post... i don't think i can be more explicit than this..

8. Oct 11, 2011

### Spinnor

It has to be. What other forces in the x direction are there?

9. Oct 11, 2011

### Ortix

From what I understand there are 2 forces. B and T both pointing in the opposite direction. Is that correct?

10. Oct 11, 2011

### grzz

I think that sin must be replaced by cos.

11. Oct 11, 2011

### grzz

One must be careful here. The SIN30 component of B is in the opposite direction of T.