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Statics - Method of Sections

  1. Nov 10, 2012 #1
    IHi guys with part b of this question im having trouble. I have cut it it at the specified members and used the top half. Now i take the moment at A to get the force in BC. (CW+)M@A=(6*3)+(6cos(30)*3)=0, But when you solve this it dosnt give you the answer its supposed to be, Im probably blind & missing something. Please check out the attachments i provided some more info on it.

    Thankyou in Advance.
     

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    Last edited: Nov 10, 2012
  2. jcsd
  3. Nov 10, 2012 #2

    SteamKing

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    No attachments provided.
     
  4. Nov 10, 2012 #3
    Sorry man, its up now
     
  5. Nov 10, 2012 #4

    SteamKing

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    Where do you get 30 degrees from? Check your triangles.
     
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