EGR 201 2.111 - Sum of forces...

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Matt Richards EGR 201-61 Assignment 2.111 4/21/2011 Problem 111, pg 61 Page 1 of 1 Given: Tension in AC is 60 N Find: Determine the weight of the plate Plan: Three dimensional equilibrium AB AC AD X -320 450 250 Y -480 -480 -480 Z 360 360 -360 Sqrt(X^2+Y^2+Z^2) 680 750 650 i j k FAB -320/680 -480/620 360/680 FAC 60*450/750 60*-480/750 60*360/750 FAD 250/650 -480/650 -360/650 Sum of forces X= -320/680(TAB)+450/75(60)+250/650(TAB)=0
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Unformatted text preview: Sum of forces Y=-480/680(TAB)+480/750(60)+-480/750(TAD) +w=0 Sum of forces Z= 360/680(TAB)+360/750(60)-360/650(FAD)=0-320/680TAB+250/650TAD=-36 360/680TAB-360/650TAD=-28.8 TAB=541.9N TAD=569.4N TAB = 541.9000 >> TAD=569.4 TAD = 569.4000 >> W=.706*TAB+.738*TAD+38.4 W = 841.1986 Summary W=841.1986 N...
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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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