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Balanced Forces: Hanging Sign

  1. Nov 18, 2015 #1
    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!
     
  2. jcsd
  3. Nov 18, 2015 #2
    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.
     
  4. Nov 18, 2015 #3
    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.
     
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