HW_10 - Problem 3.120(a Have R = F = FF FE(1.1 lb where FF = 10 sin 60 j cos 60 k = 8.6603 j 5.0k FE = 5k lb(1.2(1.3 Thus R = 8.66 j lb(1.4 Have R

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Problem 3.120: (a) Have F E RF FF = ∑= + (1.1) where () ( ) F F 10 sin60 j cos60 k 8.6603 j 5.0k ( lb ) ⎡⎤ =+ = + ⎣⎦ DD (1.2) ( ) E F 5k lb =− (1.3) Thus ( ) R 8.66 j lb = (1.4) Have ( ) R C F/C F E/C E R C Mr F r F r r9 i 4 j r1 8 i 1 3 j ijk i j k M 9 4 0 18 13 0 08 . 6 6 0 35 0 0 5 45i 45 j 77.9k lb in =− +− =++ F (1.5) (b) To determine which direction duct section CD has a tendency to turn, have RR CD
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This note was uploaded on 02/12/2010 for the course EGM 2511 taught by Professor Jenkins during the Spring '08 term at University of Florida.

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HW_10 - Problem 3.120(a Have R = F = FF FE(1.1 lb where FF = 10 sin 60 j cos 60 k = 8.6603 j 5.0k FE = 5k lb(1.2(1.3 Thus R = 8.66 j lb(1.4 Have R

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