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

2023
Views:

191

Pages:

3
Author:

Campbell Wagner

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 134. First note Free-Body Diagram: TDI = λ DI TDI = = TEH = λ EH TEH = = TFG = λ FGTFG = = − ( 0.65 m ) i + ( 0.2 m ) j − ( 0.44 m ) k ( 0.65)2 + ( 0.2 )2 + ( 0.44 )2 m TDI TDI ( − 0.65i + 0.2j − 0.44k ) 0.81 − ( 0.45 m ) i + ( 0.24 m ) j ( 0.45)2 + ( 0.24 )2 m TEH TEH ( − 0.45 i ) + ( 0.24 j) 0.51 − ( 0.45 m ) i + ( 0.2 m ) j + ( 0.36 m ) k ( 0.45 ) 2 2 2 + ( 0.2 ) + ( 0.36 ) m TFG TFG ( −0.45i + 0.2j + 0.36k ) 0.61 From free-body diagram of frame ΣM A = 0: rD/ A × TDI + rC/ A × ( −280 N ) j + rH / A × TEH + rF / A × TFG + rF / A × ( −360 N ) j = 0 or i j k i j k i j k i j k  TDI   TEH   TFG  0.65 0.2 0  0.32 0   + 0.65 0 0 ( 280 N ) + 0  + 0.45 0 0.06    0.81   0.51   0.61  −0.65 0.2 −0.44 −0.45 0.24 0 −0.45 0.2 0.36 0 −1 0 i j k + 0.45 0 0.06 ( 360 N ) = 0 0 −1 0 or ( − 0.088 i + 0.286 j + 0.26 k ) + ( − 0.012 i − 0.189 j + 0.09 k ) TDI T + ( − 0.65 k ) 280 N + ( 0.144 k ) EH 0.81 0.51 TFG + ( 0.06 i − 0.45 k )( 360 N ) = 0 0.61 continued COSMOS: Complete Online Solutions Manual Organization System From i-coefficient T  T  −0.088  DI  − 0.012  FG  + 0.06 ( 360 N ) = 0  0.81   0.61  ∴ 0.108642TDI + 0.0196721TFG = 21.6 From j-coefficient (1) T  T  0.286  DI  − 0.189  FG  = 0 0.81    0.61  ∴ TFG = 1.13959TDI (2) From k-coefficient T  T  T  0.26  DI  − 0.65 ( 280 N ) + 0.144  EH  + 0.09  FG  0.81 0.51       − 0.45 ( 360 N ) = 0 ∴ 0.32099TDI + 0.28235TEH + 0.147541TFG = 344 N (3) Substitution of Equation (2) into Equation (1) 0.108642TDI + 0.0196721(1.13959TDI ) = 21.6 ∴ TDI = 164.810 N TDI = 164.8 N or Then from Equation (2) TFG = 1.13959 (164.810 N ) = 187.816 N TFG = 187.8 N or And from Equation (3) 0.32099 (164.810 N ) + 0.28235TEH + 0.147541(187.816 N ) = 344 N ∴ TEH = 932.84 N TEH = 933 N or The vector forms of the cable forces are: TDI = 164.810 N ( −0.65i + 0.2j − 0.44k ) 0.81 = − (132.25 N ) i + ( 40.694 N ) j − ( 89.526 N ) k TEH = 932.84 N ( − 0.45i + 0.24 j) = − (823.09 N ) i + ( 438.98 N ) j 0.51 187.816 N ( − 0.45i + 0.2 j + 0.36k ) 0.61 = − (138.553 N ) i + ( 61.579 N ) j + (110.842 N ) k TFG = continued COSMOS: Complete Online Solutions Manual Organization System Then, from free-body diagram of frame ΣFx = 0: Ax − 132.25 − 823.09 − 138.553 = 0 ∴ Ax = 1093.89 N ΣFy = 0: Ay + 40.694 + 438.98 + 61.579 − 360 − 280 = 0 ∴ Ay = 98.747 N ΣFz = 0: Az − 89.526 + 110.842 = 0 ∴ Az = −21.316 N or A = (1094 N ) i + ( 98.7 N ) j − ( 21.3 N ) k