University:
Michigan State University
Course:
ERG 160 | Success in Science, Technology, Engineering, and Mathematics
Academic year:

2019
Views:

426

Pages:

2
Author:

Madelyn Bradshaw

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 130. Free-Body Diagram: Express forces and moments in terms of rectangular components: FDE = FDE − 40 i − 70 j + 40k ( − 40 ) 2 2 + ( − 70 ) + ( 40 ) 2 = FDE (− 4i − 7 j + 4k ) 9 FA = ( 24 N )( sin 20° i − cos 20° j) B = By j + Bzk , (a) ΣFx = 0: M B = M By j + M Bzk 4 − FDE + 24sin 20° = 0 9 FDE = 18.4691 N ΣM B = 0: or FCF or FDE = 18.47 N rBC × FCF + rBD × FDE + rBA × FA + M B = 0 i j k i j k i j k 18.4691 −1 +  M B y j + M B zk  0 − 48 36 + 0 − 80 60 + ( 80 )( 24 ) 0 0 9 0 −1 0 −4 −7 4 sin 20° − cos 20° 0 continued COSMOS: Complete Online Solutions Manual Organization System Equating the coefficients of the unit vectors to zero: i: 36FCF + 18.4691 (100 ) + (80 )( 24 )( − cos 20° ) = 0 9 FCF = 44.417 N or FCF = 44.4 N (b) j: 18.4691 ( − 240 ) + (80 )( 24 )( − sin 20° ) + M By = 0 9 M By = 1149.19 N ⋅ mm k: 18.4691 ( − 320 ) + M Bz = 0 9 M Bz = 656.68 N ⋅ mm ΣFy = 0: By − 44.417 − 7 (18.4691) − 24 cos 20° = 0 9 By = 81.3 N ΣFz = 0: Bz + 4 (18.4691) = 0 9 Bz = −8.21 N Therefore: B = ( 81.3 N ) j − ( 8.21 N ) k M B = (1.149 N ⋅ m ) j + ( 0.657 N ⋅ m ) k

COSMOS Chapter 4 Solution 130

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