COSMOS Chapter 4 Solution 140 - Edubirdie

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

2023
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112

Pages:

2
Author:

gaarmaakbn9h

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 140. Free-Body Diagram: Express the forces in terms of rectangular components: ( ) W = − ( mg ) j = − ( 3 kg ) 9.81 m/s 2 j = − ( 29.43 N ) j N B = N B ( 0.8j + 0.6k ) ( xB ) 2 + ( LAB = 525 mm = + )2 + ( )2 xB = 325 mm Then, TBC = TBC uuur BC = TBC BC 325i + 400 j − 100k ( 325) 2 2 + ( 400 ) + ( −100 ) 2 = 13 16 4 TBC i + TBC j − TBC k 21 21 21 continued COSMOS: Complete Online Solutions Manual Organization System Equilibrium: ΣM A = 0: rG/ A × W + rB/ A × N B + rC/ A × TBC = 0 i j k i j k i j k T or 162.5 − 200 50 + 325 − 400 100 N B + 650 0 0 BC = 0 21 − 29.43 0 0 0 0.8 0.6 13 16 − 4 Equating the coefficients of the unit vectors to zero: i: 1471.5 − 320 N B = 0 N B = 4.5984 N or N B = ( 3.6787 N ) j + ( 2.7590 N ) k j: − 195 N B + 2600 TBC = 0 21 TBC = 7.2425 N 13 ( 7.2425 N ) = 0 21 Ax = −4.4835 N ΣFx = 0: Ax + ΣFy = 0: Ay − 29.43 N + 3.6787 N + 16 ( 7.2425 N ) = 0 21 Ay = 20.233 N ΣFy = 0: Az + 2.7590 N − 4 ( 7.2425 N ) = 0 21 Az = −1.37948 N Therefore: (a) TBC = 7.24 N (b) A = − ( 4.48 N ) i + ( 20.2 N ) j − (1.379 Ν ) k N B = ( 3.68 N ) j + ( 2.76 N ) k

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