Hanging Sign Equations: Finding Tension and Force Exerted by Beam

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Olivia Carey
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upload_2015-11-18_3-27-35.png


Equations I used:
Fty = Ft(sinθ)
Ftx = Ft(cosθ)

My attempt:

I drew a free body diagram that looked like this (the red are just components of the tension, I know they wouldn't usually be included on a free body).
upload_2015-11-18_3-33-20.png


Finding Magnitude of Tension:
Fg = 516 (I gathered from the problem)
Fg = Fty
Fty = 516
Fty = Ft(sinθ)
516 = Ft (sin35)
Ft = 899.6

Finding Magnitude of Force Exerted By Beam:
Ftx = F
Ftx = Ft(cosθ)
F = Ft(cosθ)
F = 516 (cos35)
F = 422.7

I don't know where I went wrong, but I could really use some help figuring it out! Thanks!
 
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Olivia Carey said:
View attachment 92005
Finding Magnitude of Force Exerted By Beam:
Ftx = F
Ftx = Ft(cosθ)
F = Ft(cosθ)
F = 516 (cos35)
F = 422.7

Are you aware that you can see, that your answer can't be correct as F < Fty? By just comparing the lengths of the vectors it is obvious, that there must be a mistake.

Regarding the mistake: Your formulas are correct, you just plugged in a wrong value for Ft.
 
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stockzahn said:
Are you aware that you can see, that your answer can't be correct as F < Fty? By just comparing the lengths of the vectors it is obvious, that there must be a mistake.

Regarding the mistake: Your formulas are correct, you just plugged in a wrong value for Ft.

Thank you so much! I have no idea how I didn't catch that, but I guess that's why having a fresh pair of eyes always helps! I plugged in the correct value for Ft and got 736.9, which was correct.