# Tension force question

1. Feb 11, 2013

### cbchapm2

1. The problem statement, all variables and given/known data
See attached figure below for problem:

I'm designating the "horizontal cable" T1, the "incline cable" T2, and the "vertical cable" T3.

2. Relevant equations
I used the following to find x and y components of each force:
x comp=Tncos(theta)
y comp=Tnsin(theta)

3. The attempt at a solution

I made a chart for T1, T2, and T3 for the respective x and y components.
T1:
x=T1cos(180)
y=0 (because it's only the x direction)
I didn't know which angle to use for theta, I used 180 because of where I put the x and y axis in my free body diagram.

T2:
x=T2cos(38)
y=T2sin(38)
Again, I didn't know if I should use the 38* or add 180 to it...

T3:
x=0 (because it's only in the y direction?)
y=532 N (given)

I started to sum my x components and then solve for T2, but the answer that I got was wrong. I did this:
Sun of x components: 0=T1cos(180)+T2cos(38)
T2=T1(cos(180)/cos(38))
T2=1.27 N
I tried both 38* and 218* for the theta of T2 and both answers were wrong.

Sum of y components: 0=T1sin(180)+T2sin(38)+532N
Once I got the correct value for T2, I would plug it into this equation to find T1. And T3 will equal the weight (532 N) because of equilibrium, right?

Can anyone help me get on the right track?

Thanks!

Last edited: Feb 11, 2013
2. Feb 11, 2013

### haruspex

You've switched the sign there.
Where did that come from?
To be consistent with the signs, that should be -532N (sin(270)=-1).

3. Feb 12, 2013

### SteamKing

Staff Emeritus
You have some confusion with angles. The convention is that the positive x- axis has angle 0 degrees, and the angle increases as one goes counterclockwise. Therefore, positive y is 90 degrees, negative x is 180 degrees and negative y is 270 degrees.

4. Feb 12, 2013

### cbchapm2

I got the right answer! When I solved the first equation for T2, I dropped the T1 and it threw off everything else too.