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

2020
Views:

56

Pages:

2
Author:

Athena Harrison

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 162. First note Free-Body Diagram: TBG = λBGTBG = − (18 in.) i + (13.5 in.) k 2 = TBG ( −0.8i + TDH = λDH TDH = 2 (18) + (13.5) in. k) − (18 in.) i + ( 24 in.) j (18)2 + ( 24 )2 in. = TDH ( −0.6i + TBG TDH j) Since λFJ = λDH , TFJ = TFJ ( −0.6i + 0.8j) From free-body diagram of member ABF ΣM A( x-axis ) = 0: ( 0.8TFJ ) ( 48 in.) + ( 0.8TDH )( 24 in.) − (120 lb )( 36 in.) − (120 lb )(12 in.) = 0 ∴ 3.2TFJ + 1.6TDH = 480 ΣM A( z -axis ) = 0: (1) ( 0.8TFJ ) (18 in.) + ( 0.8TDH )(18 in.) − (120 lb )(18 in.) − (120 lb )(18 in.) = 0 ∴ − 3.2TFJ − 3.2TDH = −960 (2) Equation (1) + Equation (2) Substituting in Equation (1) ΣM A( y -axis ) = 0: TDH = 300 lb TFJ = 0 ( 0.6TFJ ) ( 48 in.) + 0.6 ( 300 lb ) ( 24 in.) − ( 0.6TBG ) (18 in.) = 0 ∴ TBG = 400 lb continued COSMOS: Complete Online Solutions Manual Organization System ΣFx = 0: − 0.6TFJ − 0.6TDH − 0.8TBG + Ax = 0 −0.6 ( 300 lb ) − 0.8 ( 400 lb ) + Ax = 0 ∴ Ax = 500 lb ΣFy = 0: 0.8TFJ + 0.8TDH − 240 lb + Ay = 0 0.8 ( 300 lb ) − 240 + Ay = 0 ∴ Ay = 0 ΣFz = 0: 0.6TBG + Az = 0 0.6 ( 400 lb ) + Az = 0 ∴ Az = −240 lb Therefore, A = ( 500 lb ) i − ( 240 lb ) k