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

2020
Views:

91

Pages:

2
Author:

Tyshawn Shaffer

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 124. Free-Body Diagram: Express tension, weight in terms of rectangular components: uuur EF = ( 300 mm ) i + (1350 mm ) j − ( 700 mm ) k uuur EF 300 i + 1350 j − 700 k =T T=T EF ( 300 )2 + (1350 )2 + ( − 700 )2 = 6 27 14 Ti + T j− Tk 31 31 31 ( ) 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 + ( −100 mm ) i + ( 700 mm ) k  × T = 0 or − ( 750 mm ) Ayk + (125 mm ) Az j + ( 375 mm )( 68.7 N ) k + ( 350 mm )( 68.7 N ) i i j k T + −100 0 700 ( mm ) = 0 31 6 27 −14 continued COSMOS: Complete Online Solutions Manual Organization System Setting the coefficients of the unit vectors equal to zero: (a) i: − 27 T ( 700 mm ) + ( 68.67 N )( 350 mm ) = 0 31 T = 39.422 N (b) k: or T = 39.4N  27  − Ay ( 750 mm ) + ( 68.67 N )( 375 mm ) −  ( 39.422 N )  (100 mm ) = 0 31   Ay = 29.757 N j: 14  6  Az ( 750 mm ) −  ( 39.422 N )  (100 mm ) +  ( 39.422 N )  ( 700 mm ) = 0  31   31  Az = − 4.7476 N ΣFx = 0: Bx + 6 ( 39.422 N ) = 0 31 Bx = − 7.6301 N ΣFy = 0: 29.757 N + B y − 68.67 N + 27 ( 39.422 N ) = 0 31 By = 4.5777 N ΣFz = 0: − 4.7476 N + Bz − 14 ( 39.422 N ) = 0 31 Bz = 22.551 N Therefore: A = ( 29.8 N ) j − ( 4.75 N ) k B = − ( 7.63 N ) i + ( 4.58 N ) j + ( 22.6 N ) k