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

2021
Views:

93

Pages:

2
Author:

Dixie Summers

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 125. Free-Body Diagram: Express tension, weight in terms of rectangular components: uur IF = ( 75 mm ) i + (1350 mm ) j − ( 250 mm ) k uur IF =T T=T IF = 75i + 1350 j − 250 k ( 75)2 + (1350 )2 + ( − 250 )2 3 54 10 Ti + Tj− Tk 55 55 55 ( ) W = − (mg ) j = − ( 7 kg ) 9.81 m/s 2 j = − (68.67 N)j ΣM B = 0: − ( 750 mm ) i × A +  − ( 375 mm ) i + ( 350 mm ) k  × ( − 68.7 N ) j + (125 mm ) i + ( 250 mm ) k  × T = 0 or − ( 750 mm ) Ay k + (125 mm ) Az j + ( 375 mm )( 68.7 N ) k + ( 350 mm )( 68.7 N ) i i j k T + 125 0 250 ( mm ) = 0 55 3 54 −10 continued COSMOS: Complete Online Solutions Manual Organization System Setting the coefficients of the unit vectors equal to zero: (a) i: − 54 T ( 250 mm ) + ( 68.67 N )( 350 mm ) = 0 55 T = 97.918 N (b) k: or T = 97.9 N  54  − Ay ( 750 mm ) + ( 68.67 N )( 375 mm ) −  ( 97.918 N )  (125 mm ) = 0  55  Ay = 50.358 N j:  10  Az ( 750 mm ) −  ( 97.918 N )  (125 mm ) + 55   3   55 ( 97.918 N )  ( 250 mm ) = 0   Az = − 4.7475 N ΣFx = 0: Bx + 3 ( 97.918 N ) = 0 55 Bx = − 5.3410 N ΣFy = 0: 50.358 N + B y − 68.67 N + 54 ( 97.918 N ) = 0 55 By = − 77.826 N ΣFz = 0: − 4.7475 N + Bz − 10 ( 97.918 N ) = 0 55 Bz = 22.551 N Therefore: A = ( 50.4 N ) j − ( 4.75 N ) k B = − ( 5.34 N ) i − ( 77.8 N ) j + ( 22.6 N ) k

COSMOS Chapter 4 Solution 125

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