# [Statics] Finding the Tension

## Homework Statement

[PLAIN]http://img830.imageshack.us/img830/1091/unled2cjt.jpg [Broken]

## Homework Equations

I used sum of forces in x y direction and sum of moments

## 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

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Spinnor
Gold Member
Does this help? %^)

#### Attachments

• bridge062.jpg
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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

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

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!

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

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..

Spinnor
Gold Member
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!

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

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

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..

I think that sin must be replaced by cos.

From what I understand there are 2 forces. B and T both pointing in the opposite direction. Is that correct?
One must be careful here. The SIN30 component of B is in the opposite direction of T.