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

2022
Views:

326

Pages:

2
Author:

agatoary6zt

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 133. 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 = TCF = 65 in. uuur BD 12 5 = − 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 TBD = TBD TBE ( −60 )2 + ( 25)2 WG = − ( 500 lb ) j WC = − ( 800 lb ) j COSMOS: Complete Online Solutions Manual Organization System ΣM A = 0: rB × TBD + rB × TBE + rC × TCF + rG × WG + rC × WC = 0 i j k i j k i j k i j k TBD TBE TCF −15 lb ⋅ in. 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 or i j k + 60 0 −30 lb ⋅ in. = 0 0 −800 0 Equating the coefficients of the unit vectors to zero: i: 150 TCF − 7500 − 24000 = 0 13 TCF = 2730 lb j: − or TCF = 2.73 kips 300 360 TBD + ( 2730 lb ) = 0 13 13 TBD = 3276 lb k: or TBD = 3.28 kips 300 300 TBE − 30000 + ( 2730 lb ) − ( 60 )(800 lb ) = 0 13 13 TBE = 650.00 lb ΣFx = 0: Ax − or TBE = 650 lb 12 12 12 ( 3276 lb ) − ( 650 lb ) − ( 2730 lb ) = 0 13 13 13 Ax = 6144 lb ΣFy = 0: Ay + 5 5 ( 2730 lb ) + ( 650 lb ) − 500 lb − 800 lb = 0 13 13 Ay = 0 ΣFz = 0: Az + 5 ( 3276 lb ) = 0 13 Az = −1260.00 lb Therefore, A = ( 6.14 kips ) i − (1.260 kips ) k

COSMOS Chapter 4 Solution 133

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