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

2024
Views:

297

Pages:

2
Author:

Apolenajz

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 120. Free-Body Diagram: Geometry: Using triangle ACD and the law of sines sin α sin 50° = or α = 20.946° 15 in. 7 in. β = 50° + 20.946° = 70.946° Expressing FCD in terms of its rectangular coordinates: FCD = FCD sin β j + FCD cos β k = FCD sin 70.946° j + FCD cos 70.946°k FCD = 0.94521FCD j + 0.32646 FCDk ΣM B = 0: ( − 26 in.) i × A + ( −13 in.) i + (16 in.) sin 50°j + (16 in.) cos 50°k  × ( − 75 lb ) j + ( − 26 in.) i + ( 7 in.) k  × FCD = 0 or − ( 26 in.) Ayk + ( 26 in.) Az j + (13 in.)( 75 lb ) k + (16 in.)( 75 lb ) cos 50°i − ( 26 in.) ( 0.94521FCD ) k + ( 26 in.) ( 0.32646 FCD ) j − ( 7 in.) ( 0.94521FCD ) i = 0 continued COSMOS: Complete Online Solutions Manual Organization System (a) Setting the coefficients of the unit vectors to zero: i: ( 75 lb ) (16 in.) cos 50° − ( 0.94521FCD ) ( 7 in.) = 0 FCD = 116.6 lb (b) ΣFx = 0: k: Ax = 0 − 0.94521(116.580 lb )  ( 26 in.) + ( 75 lb )(13 in.) − Ay ( 26 in.) = 0 Ay = − 72.693 lb j:  0.32646 (116.580 lb )  ( 26 in.) + Az ( 26 in.) = 0 Az = − 38.059 lb ΣFy = 0: − 72.693 lb + 0.94521(116.580 lb ) − 75 lb + By = 0 By = 37.500 lb ΣFz = 0: − 38.059 lb + 0.32646 (116.580 lb ) + Bz = 0 Bz = 0 Therefore: A = − ( 72.7 lb ) j − ( 38.1 lb ) k B = ( 37.5 lb ) j

COSMOS Chapter 4 Solution 120

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