COSMOS Chapter 4 Solution 160 - Edubirdie

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

2024
Views:

11

Pages:

1
Author:

strofamay54

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 160. Free-Body Diagram: First note ( ) W = mg = ( 30 kg ) 9.81 m/s 2 = 294.3 N FEC = λ EC FEC = ( sin15° ) i + ( cos15° ) j FEC From free-body diagram of cover (a) ΣM z = 0: ( FEC cos15° ) (1.0 m ) − W ( 0.5 m ) = 0 or FEC cos15° (1.0 m ) − ( 294.3 N )( 0.5 m ) = 0 ∴ FEC = 152.341 N (b) or FEC = 152.3 N ΣM x = 0: W ( 0.4 m ) − Ay ( 0.8 m ) − ( FEC cos15° ) ( 0.8 m ) = 0 or ( 294.3 N )( 0.4 m ) − Ay ( 0.8 m ) − (152.341 N ) cos15° ( 0.8 m ) = 0 ∴ Ay = 0 ΣM y = 0: Ax ( 0.8 m ) + ( FEC sin15° ) ( 0.8 m ) = 0 or Ax ( 0.8 m ) + (152.341 N ) sin15° ( 0.8 m ) = 0 ∴ Ax = −39.429 N ΣFx = 0: Ax + Bx + FEC sin15° = 0 −39.429 N + Bx + (152.341 N ) sin15° = 0 ∴ Bx = 0 ΣFy = 0: FEC cos15° − W + By = 0 or (152.341 N ) cos15° − 294.3 N + B y = 0 ∴ By = 147.180 N or A = − ( 39.4 N ) i B = (147.2 N ) j

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