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

2024
Views:

241

Pages:

3
Author:

teespagmaxrtf

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 135. Free-Body Diagram: First note 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 ( −65i + 20 j − 44k ) 81 − ( 0.45 m ) i + ( 0.24 m ) j ( 0.45)2 + ( 0.24 )2 m TDI TEH TEH ( −15i + 8j) 17 − ( 0.45 m ) i + ( 0.2 m ) j + ( 0.36 m ) k ( 0.45)2 + ( 0.2 )2 + ( 0.36 )2 m TFG TFG ( −45i + 20j + 36k ) 61 From free-body diagram of frame = ΣM A = 0: rD/ A × TDI + rC/ A ×  − ( 280 N ) j + ( 50 N ) k  + rH / A × TEH + rF / A × TFG + rF / A × ( −360 N ) j or i j k i j k i j k  TDI  0.65 0.2 0  0 + 0 0.32 0  + 0.65 0  81  −65 20 −44 −15 8 0 0 −280 50 i j k T + 0.45 0 0.06  FG  61 −45 20 36 and  TEH     17  i j k   + 0.45 0 0.06 ( 360 N ) = 0  0 −1 0 TDI   TEH   + ( −32.5 j − 182k ) + ( 4.8k )    81   17  ( −8.8i + 28.6 j + 26k )  T  + ( −1.2i − 18.9 j + 9.0k )  FG  + ( 0.06i − 0.45k ) ( 360 ) = 0  61  continued COSMOS: Complete Online Solutions Manual Organization System T  T −8.8  DI  − 1.2  FG  81   61 From i-coefficient   + 0.06 ( 360 ) = 0  ∴ 0.108642TDI + 0.0196721TFG = 21.6 (1) T  T  From j-coefficient 28.6  DI  − 32.5 − 18.9  FG  = 0 81    61  ∴ 0.35309TDI − 0.30984TFG = 32.5 (2) From k-coefficient T  T  T  26  DI  − 182 + 4.8  EH  + 9.0  FG  − 0.45 ( 360 ) = 0  81   17    ∴ 0.32099TDI + 0.28235TEH + 0.147541TFG = 344 −3.25 × Equation (1) Add Equation (2) (3) −0.35309TDI − 0.063935TFG = −70.201 0.35309TDI − 0.30984TFG = −0.37378TFG 32.5 = −37.701 ∴ TFG = 100.864 N TFG = 100.9 N or Then from Equation (1) 0.108642TDI + 0.0196721(100.864 ) = 21.6 ∴ TDI = 180.554 N TDI = 180.6 N or and from Equation (3) 0.32099 (180.554 ) + 0.28235TEH + 0.147541(100.864 ) = 344 ∴ TEH = 960.38 N TEH = 960 N or The vector forms of the cable forces are: TDI = 180.554 N ( −65i + 20j − 44k ) 81 = − (144.889 N ) i + ( 44.581 N ) j − ( 98.079 N ) k TEH = 960.38 N ( −15i + 8j) = − (847.39 N ) i + ( 451.94 N ) j 17 100.864 N ( −45i + 20j + 36k ) 61 = − ( 74.409 N ) i + ( 33.070 N ) j + ( 59.527 N ) k TFG = COSMOS: Complete Online Solutions Manual Organization System Then from free-body diagram of frame ΣFx = 0: Ax − 144.889 − 847.39 − 74.409 = 0 ∴ Ax = 1066.69 N ΣFy = 0: Ay + 44.581 + 451.94 + 33.070 − 360 − 280 = 0 ∴ Ay = 110.409 N ΣFz = 0: Az − 98.079 + 59.527 + 50 = 0 ∴ Az = −11.448 N Therefore, A = (1067 N ) i + (110.4 N ) j − (11.45 N ) k