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

2024
Views:

321

Pages:

2
Author:

Quintin Ware

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 122. Free-Body Diagram: ( ) W = mg = (15 kg ) 9.81 m/s 2 = 147.15 N First note (a)  ( 0.08 m ) i + ( 0.25 m ) j − ( 0.2 m ) k  T TEF = λ EF TEF =  TEF = EF ( 0.08i + 0.25j − 0.2k )  0.33 ( 0.08)2 + ( 0.25)2 + ( 0.2 )2 m   From free-body diagram of rectangular plate ΣM x = 0: (147.15 N )( 0.1 m ) − (TEF ) y ( 0.2 m ) = 0  0.25   14.715 N ⋅ m −   TEF  ( 0.2 m ) = 0  0.33   or TEF = 97.119 N or or TEF = 97.1 N (b) ΣFx = 0: Ax + (TEF ) x = 0  0.08  Ax +   ( 97.119 N ) = 0  0.33  ∴ Ax = −23.544 N continued COSMOS: Complete Online Solutions Manual Organization System ΣM B( z -axis ) = 0: − Ay ( 0.3 m ) − (TEF ) y ( 0.04 m ) + W ( 0.15 m ) = 0  0.25   − Ay ( 0.3 m ) −   97.119 N  ( 0.04 m ) + 147.15 N ( 0.15 m ) = 0  0.33   or ∴ Ay = 63.765 N ΣM B( y -axis ) = 0: Az ( 0.3 m ) + (TEF ) x ( 0.2 m ) + (TEF ) z ( 0.04 m ) = 0  0.08    0.2   Az ( 0.3 m ) +   TEF  ( 0.2 m ) −   TEF  ( 0.04 m ) = 0 0.33 0.33       ∴ Az = −7.848 N and A = − ( 23.5 N ) i + ( 63.8 N ) j − ( 7.85 N ) k ΣFy = 0: Ay − W + (TEF ) y + By = 0  0.25  63.765 N − 147.15 N +   ( 97.119 N ) + By = 0  0.33  ∴ By = 9.81 N ΣFz = 0: Az − (TEF ) z + Bz = 0  0.2  −7.848 N −   ( 97.119 N ) + Bz = 0  0.33  ∴ Bz = 66.708 N and B = ( 9.81 N ) j + ( 66.7 N ) k