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Unformatted text preview: -.86/1.86 .181AB-.215TAE=0 1.661AB+.8AC+.86AE-1000=0 .19AB+.16AC-.462AE=0 P=AB AB=377.32 AC=125.1 AE=317.65 Matt Richards EGR 201-61 Assignment 2.123 4/21/2011 Problem 123, pg 63 Page 2 of 2 Summary: P=377.32N...
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This note was uploaded on 04/20/2011 for the course EGR 201 taught by Professor Dd during the Spring '11 term at University of Dayton.
- Spring '11