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

2020
Views:

426

Pages:

2
Author:

Nicole Wang

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 119. Free-Body Diagram: Express all forces in terms of rectangular components: rB = (120 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 uuuur ur FH = TFH T FH = TFH FH uur ur CI T CI = TCI = TCI CI ( −120 ) 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 + 120 0 −10 + 72 0 0 CI − 19600 lb ⋅ in. = 0 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 − 16800 lb ⋅ in. = 0 or TCI = TFH = 442 lb TEG = 377 lb Force equations: ΣFx = 0: Dx − 20 45 4 ( 376.72 lb ) − ( 441.67 lb ) − ( 441.67 lb ) = 0, 29 53 5 ΣFy = 0: Dy + 28 ( 441.67 lb ) − 140 lb = 0, 53 ΣFz = 0: Dz + 21 3 ( 376.72 lb ) − ( 441.67 lb ) = 0, 29 5 or or Dx = 988.15 lb Dy = − 93.335 lb or Dz = − 7.7952 lb D = ( 998 lb ) i − ( 93.3 lb ) j − ( 7.80 lb ) k