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Forces in Equilibrium (Vectors)

  1. Sep 26, 2008 #1
    Hey guys, I am stuck on this question. It is first year engineering vector statics. Here it is.

    [​IMG]

    Thanks for any help!
     
  2. jcsd
  3. Sep 27, 2008 #2

    tiny-tim

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    Hi Oblivion77! :smile:

    Show us what you've tried, and where you're stuck, and then we'll know how to help. :smile:
     
  4. Sep 27, 2008 #3
    Well, i resolved the vectors into the components and added the sums and equaled them to zero (since its in equilibrium)

    [tex] \Sigma Fx = -118.65 -F3Cos[\alpha] + F1 =0[/tex]

    [tex] \Sigma Fy = 68.5 - F3Sin[\alpha] = 0 [/tex]

    I don't know how to go from here, with the "minimum"
     
  5. Sep 28, 2008 #4

    tiny-tim

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    Hi Oblivion77! :smile:

    … just woken up … :zzz:
    Yes, that's right. :smile:

    Tip: when you have two equations with cos and sin,

    put the cos and sin on their own on the left, then square and add (using cos2 + sin2 = 1, of course) :wink:

    That will give you F3 in terms of F1, and then … ? :smile:
     
  6. Sep 28, 2008 #5
    ...after you solve a little bit, you get;


    137/2=F3*Sin(a)

    so for a=>90 degrees sin is max and F3 is min.

    Min F3 is 137/2...And a is 90 degrees...


    and magnitude of F1 is;

    F1=Cos30*137
     
  7. Sep 28, 2008 #6

    tiny-tim

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    Hi MrEnergy! :smile:
    oops! I over-complicated it!! :redface:

    Yes, you're absolutely right. :smile:

    (hmm … not sure how I helped there … :redface:)
     
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