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Statics - Method of Sections |
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| Nov10-12, 02:09 AM | #1 |
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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. |
| Nov10-12, 04:40 AM | #2 |
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No attachments provided.
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| Nov10-12, 04:53 AM | #3 |
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Sorry man, its up now
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| 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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