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Statics - Method of Sections

 
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Nov10-12, 02:09 AM   #1
 

Statics - Method of Sections

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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Nov10-12, 04:40 AM   #2
 
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Nov10-12, 04:53 AM   #3
 
Sorry man, its up now
 
Nov10-12, 11:45 AM   #4
 
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Statics - Method of Sections

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