COSMOS Chapter 4 Solution 118

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

2019
Views:

142

Pages:

2
Author:

Lily Callahan

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 118. Free-Body Diagram: Express all forces in terms of rectangular components: rA = (140 in.) i rC = ( 72 in.) i uuur EG = − (120 in.) i + (126 in.) k uuuur FH = − (120 in.) i − ( 90 in.) k uur CI = − ( 72 in.) i + ( 44.8 in.) j W = − (140 lb ) j Then uuur ur EG = TEG T EG = TEG EG −120i + 126k ( −120 ) uuuur ur FH T FH = TFH = TFH FH uur ur CI = TCI T CI = TCI CI 2 + (126 ) 2 =− −120i − 90k ( −120 )2 + ( − 90 )2 − 72i + 44.8j ( − 72 ) 2 + ( 44.8 ) 2 =− 20 21 TEG i + TEGk 29 29 = −0.8TFH i − 0.6TFH k 45 28 TCI i + TCI j 53 53 continued COSMOS: Complete Online Solutions Manual Organization System ΣM D = 0: or rE × TEG + rF × TFH + rC × TCI + rA × W = 0 i j k i j k i j k TEG TFH T + 120 0 −10 + 72 0 0 CI − 19600 lb ⋅ in. = 0 120 0 10 29 5 53 − 20 0 21 −4 0 −3 − 45 28 0 Noting that TCI = TFH and equating the coefficients of the unit vectors to zero: j: k: − 93.793TEG + 80TCI = 0 38.038TCI − 19600 lb ⋅ in. = 0 or TCI = TFH = 515 lb or TEG = 440 lb Force equations: ΣFx = 0: Dx − 20 45 4 ( 439.50 lb ) − ( 515.28 lb ) − ( 515.28 lb ) = 0, 29 53 5 ΣFy = 0: Dy + 28 ( 515.28 lb ) − 140 lb = 0, 53 ΣFz = 0: Dz + 21 3 ( 439.50 lb ) − ( 515.28) lb = 0, 29 5 or or Dx = 1152.83 lb Dy = −132.223 lb or Dz = − 9.0906 lb D = (1153 lb ) i − (132.2 lb ) j − ( 9.09 lb ) k