# Suspended Sign

1. Sep 28, 2009

### chrisfnet

1. The problem statement, all variables and given/known data

Attached is a diagram of the problem.

A 28kg sign is suspended by two massless cables. Find the tension in each.

2. Relevant equations

3. The attempt at a solution

I'll call the cable at 57 degrees T1 and the other cable T2.

T1∙cos 57 + T2∙cos 36 = 9.8m/s2∙28kg
T1∙cos 57 + T2∙cos 36 = 274.4N

T1∙sin 57 + T2∙sin 36 = 0N

This is where I'm getting confused - rusty on my algebra with simultaneous equations. Am I on track so far though, then just solve for T1 and T2?

#### Attached Files:

• ###### Physics-Problem-Two.jpg
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2. Sep 29, 2009

### chrisfnet

T1 = (-T2∙sin 36)/sin 57

Substitute that into the first equation...

(-T2∙sin 36∙cos 57)/sin 57 + T2∙cos 36 = 274.4N
T2 = 143.05N

Then solve for T1.

T1∙cos 57 + (143.05N)∙cos 36 = 274.4N
T1 = 291.33N

3. Sep 29, 2009

### rl.bhat

From third equation you can write
T1 = - T2*sin36/sin57.
Substitute this value in eq. 2 and solve for T2.

4. Sep 29, 2009

### chrisfnet

Thanks - that's exactly what I ended up doing. Not sure why I was over-complicating it. Could anyone confirm my solutions are correct?

Thanks!

5. Sep 29, 2009

### chrisfnet

Can anyone verify that my work is correct? Thanks!