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

2021
Views:

221

Pages:

2
Author:

Athena Harrison

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 132. Free-Body Diagram: Express tensions, load in terms of rectangular components: uuur BD = − ( 60 in.) i + ( 25 in.) k uuur BE = − ( 60 in.) i + ( 25 in.) j uuur CF = − ( 60 in.) i + ( 25 in.) j BD = BE = CF = TBD TBE TCF = 65 in. uuur BD 12 5 = TBD = − TBD i + TBD k BD 13 13 uuur BE 12 5 = TBE = − TBE i + TBE j BE 13 13 uuur CF 12 5 = TCF = − TCF i + TCF j CF 13 13 ΣM A = 0: or ( −60 )2 + ( 25)2 rB × TBD + rB × TBE + rC × TCF + rG × W = 0 i j k i j k i j k i j k TBD TBE TCF −15 lb ⋅ in. = 0 60 0 0 in. + 60 0 0 in. + 60 0 − 30 in. + 60 0 13 13 13 −12 0 5 −12 5 0 −12 5 0 0 − 500 0 continued COSMOS: Complete Online Solutions Manual Organization System Equating the coefficients of the unit vectors to zero: i: 150 TCF − 7500 = 0 13 TCF = 650.00 lb j: − 300 360 TBD + ( 650 lb ) = 0 13 13 TBD = 780.00 lb k: or TCF = 650 lb or TBD = 780 lb 300 300 TBE − 30000 + ( 650.00 lb ) = 0 13 13 TBE = 650.00 lb ΣFx = 0: Ax − or TBE = 650 lb 12 12 12 ( 780 lb ) − ( 650 lb ) − ( 650 lb ) = 0 13 13 13 Ax = 1920.00 lb ΣFy = 0: Ay + 5 5 ( 780 lb ) + ( 650 lb ) − 500 lb = 0 13 13 Ay = 0 ΣFz = 0: Az + 5 ( 780 lb ) = 0 13 Az = −300.00 lb Therefore, A = (1920 lb ) i − ( 300 lb ) k