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

2023
Views:

400

Pages:

1
Author:

Apolenajz

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 26. Free-Body Diagram: (a) a = 2 in. ΣΜ A = 0: ( 2 in.)( 20 lb ) − (1.5 in.)(16 lb ) − 32 lb ⋅ in. − ( 2 in.) E sin 60° − ( 5.5 in.) E cos 60° = 0 E = −3.5698 lb ΣFx = 0: or E = 3.57 lb 60.0° Ax − 16 lb + ( 3.5698 lb ) cos 60° = 0 Ax = 14.2151 lb ΣFy = 0: Ay − 20 lb − ( 3.5698 lb ) sin 60° = 0 Ay = 23.092 lb A= (14.2151)2 + ( 23.092 )2 θ = tan −1 = 27.117 lb 23.092 = 58.384° 14.2151 A = 27.1 lb Therefore: 58.4° (b) a = 7.5 in. ΣΜ A = 0: ( 7.5 in.)( 20 lb ) − (1.5 in.)(16 lb ) − 32 lb ⋅ in. − ( 2 in.) E sin 60° − ( 5.5 in.) E cos 60° = 0 E = 20.973 lb ΣFx = 0: or E = 21.0 lb 60.0° A = 26.6 lb 3.97° Ax − 16 lb − ( 20.973 lb ) cos 60° = 0 Ax = 26.487 lb ΣFy = 0: Ay − 20 lb + ( 20.973 lb ) sin 60° = 0 Ay = 1.83685 lb A= ( 26.487 )2 + (1.83685)2 θ = tan −1 Therefore: = 26.551 lb 1.83685 = 3.9671° 26.487

COSMOS Chapter 4 Solution 26

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