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

2021
Views:

74

Pages:

2
Author:

strofamay53

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 113. Free-Body Diagram: Express the forces in terms of their rectangular components: uuur FB 12i − 18j + 36k 2 3 6 TFB = TFB = TFB = TFB i − TFB j + TFBk 2 2 2 FB 7 7 (12 ) + ( −18) + ( 36 ) 7 TFC = TFC uuur FC = TFC FC −12i − 18j + 36k ( −12 ) 2 2 + ( −18 ) + ( 36 ) 2 2 3 6 = − TFBi − TFB j + TFBk 7 7 7 From the free-body diagram, note that zE 12 = 27 18 z E = 18.00 ft continued COSMOS: Complete Online Solutions Manual Organization System Then, using TED = 2720 lb TED = 2720 − 27 j + 57.75k ( − 27 )2 + (18 + 39.75)2 TED = ( 32 lb )( − 36j + 77k ) (a) ΣM A = 0: rAD × TED + rAF × TFB + rAE × W = 0 i j k i j k i j k i j k TFB TFC 0 18 −12 + 0 18 −12 + 2720 0 27 −18 = 0 or 32 0 0 39.75 + 7 7 0 − 36 77 2 −3 6 − 2 −3 6 0 −1 0 Equating the coefficients of the unit vectors to zero i: 32 ( 39.75 )( 36 ) + TFB T ( 72 ) + FC ( 72 ) + 2720 ( −18) = 0 7 7 72 72 TFB + TFC − 3168 = 0 7 7 j: (1) TFB T ( − 24 ) + FC ( 24 ) = 0 7 7 TFB = TFC (2) Substituting Eq. (2) in (1) gives  72  2   TFB − 3168 = 0  7  or TFB = 154.0 lb TFC = 154.0 lb (b) ΣFx = 0: Ax + 2 2 (154.0) − (154.0 ) = 0 7 7 Ax = 0 ΣFy = 0: Ay − 3 3 (154.0) − (154.0 ) − ( 32 )( 36 ) − 2720 = 0 7 7 Ay = 4004 lb ΣFz = 0: Az + 6 6 (154.0) + (154.0 ) + ( 32 )( 77 ) = 0 7 7 Az = − 2728 lb Therefore: A = ( 4.00 kips ) j − ( 2.73 kips ) k