Statics - Method of Sections


by Arty7
Tags: method, sections, statics
Arty7
Arty7 is offline
#1
Nov10-12, 02:09 AM
P: 7
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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SteamKing
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#2
Nov10-12, 04:40 AM
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Thanks
P: 5,529
No attachments provided.
Arty7
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#3
Nov10-12, 04:53 AM
P: 7
Sorry man, its up now

SteamKing
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#4
Nov10-12, 11:45 AM
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P: 5,529

Statics - Method of Sections


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


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