Tension woes simple mass problem

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veitch
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I thought I had this stuff figured out, but my answer is wrong. Any tips?
Thanks

Homework Statement



A sign hangs precariously from your prof's office door. Calculate the magnitude of the tension in string 1 if theta-1 = 26.10degrees, theta-2 = 54.87degrees, and the mass of the sign is 6.50 kg.

http://img89.imageshack.us/img89/1089/prob05signly6.th.gif http://g.imageshack.us/thpix.php

Homework Equations



[tex]\sum[/tex]Fx = T1x - T2x = 0
[tex]\sum[/tex]Fy = T1y + T2y - Fg = 0
Fg = mg

The Attempt at a Solution



X components:

T1x = T1 Cos 26.1
T2x = T2 Cos 54.87

Y components:

T1y = T1 Sin 26.1
T2y = T2 Sin 54.87

[tex]\sum[/tex]Fx = T1Cos 26.1 - T2Cos 54.87 = 0
So...
T1Cos 26.1 = T2Cos 54.87
T1 = T2(Cos 54.87/Cos 26.1)

and

[tex]\sum[/tex]F = T1Sin 26.1 + T2 Sin 54.87 = (6.5 kg)(9.8 m/s2)

Then I tried plugging that T1 in there...

T2(Cos 54.87/Cos 26.1)Sin 26.1 + T2Sin 54.87 = 24.7 N
T2(0.6408)Sin 26.1 + T2Sin 54.87 = 24.7 N
T2(0.2819) + T2(0.8178) = 24.7 N
T2(0.2819 + 0.8178) = 24.7 N
T2 = 22.5 N

I can't attempt any more solutions as the assignment was due 20 minutes ago (it's online)... but I did poorly on my midterm yesterday so I really need to get a better understand of the material before it's too late... any advice would be greatly appreciated!
 
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veitch said:
[tex]\sum[/tex]F = T1Sin 26.1 + T2 Sin 54.87 = (6.5 kg)(9.8 m/s2)

Then I tried plugging that T1 in there...

T2(Cos 54.87/Cos 26.1)Sin 26.1 + T2Sin 54.87 = 24.7 N
Your method looks fine, just be careful with the arithmetic: 6.5*9.8 ≠ 24.7.