COSMOS Chapter 4 Solution 51 - Edubirdie

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

2019
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456

Pages:

1
Author:

omiliar74be

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 51. Free-Body Diagram: ΣM E = 0: ME + (16.2 kN)x + (5.4 kN)(2.6 m) – T(1.5 m) = 0 ME = (1.5 T − 16.2 x – 14.04) kN ⋅ m (1) For x = 0.6 m, (1) gives: (ME )1 = (1.5 T − 23.76) kN ⋅ m For x = 7 m, (1) gives: (ME )2 = (1.5 T − 127.44) kN ⋅ m (a) The maximum absolute value of ME is obtained when (ME )1 = − (ME ) and 1.5 T − 23.76 kN = − (1.5 T – 127.44 kN) T = 50.400 kN (b) or T = 50.4 kN For this value of T: ME = 1.5(50.400) kN ⋅ m − 23.76 kN ⋅ m = 51.84 kN ⋅ m or ME = 51.8 kN ⋅ m

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