EGR 201 3.76 - >> FB=[.15,-.08,-.17] FB = 0.1500...

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SUMMARY M=9.2093 Nm Matt Richards EGR 201-61 Assignment 3.76 07:10:31 Problem 3.76, pg 118 Page 1 of 1 Given: P=0 Find: Replace the two couples with a single equavalent Plan: Cross products, sum of M1 and M2 >> GC=[.3,0,0] GC = 0.3000 0 0 >> F1=[0,0,-18] F1 = 0 0 -18 >> M1=cross(GC,F1) M1 = 0 5.4000
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Unformatted text preview: >> FB=[.15,-.08,-.17] FB = 0.1500 -0.0800 -0.1700 >> FBR=norm(FB) FBR = 0.2404 >> F2=(34/FBR)*FB F2 = 21.2132 -11.3137 -24.0416 >> EF=[0,0,.17] EF = 0 0 0.1700 >> M2=cross(EF,F2) M2 = 1.9233 3.6062 >> M=M1+M2 M = 1.9233 9.0062 >> norm(M) ans = 9.2093...
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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.

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