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

2024
Views:

396

Pages:

2
Author:

Dixie Summers

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 126. Free-Body Diagram: Express forces, weight in terms of rectangular components: uuur CE = ( 3 ft ) i + ( 4 ft ) j − ( 2 ft ) k FCE = FCE uuur CE = FCE CE 3i + 4 j + 2k ( 3) 2 + ( 4 )2 + ( 2 ) 2 = 0.55709 FCE i + 0.74278 FCE j + 0.37139 FCE k W = − (mg ) j = − (300 lb)j ΣM B = 0: or ( 4 ft ) k × A + (1.5 ft ) i + ( 2 ft ) k  × ( − 300 lb ) j + ( 3 ft ) i + ( 4 ft ) k  × FCE = 0 − ( 4 ft ) Ay i + ( 4 ft ) Az j − (1.5 ft )( 300 lb ) k + ( 2 ft )( 300 lb ) i i j k + 3 0 4 FCE ( ft ) = 0 0.55709 0.74278 0.37139 continued COSMOS: Complete Online Solutions Manual Organization System Setting the coefficients of the unit vectors equal to zero: k: ( 0.74278 FCE ) ( 3 ft ) − ( 300 lb )(1.5 ft ) = 0 FCE = 201.94 lb or FCE = 202 lb j: Ax ( 4 ft ) + 0.55709 ( 201.94 lb )  ( 4 ft ) − 0.37139 ( 201.94 lb )  ( 3 ft ) = 0 Ax = − 56.250 lb i: − Ay ( 4 ft ) −  0.74278 ( 201.94 lb )  ( 4 ft ) + ( 300 lb )( 2 ft ) = 0 Ay = 0 ΣFx = 0: − 56.250 lb + Bx + 0.55709 ( 201.94 lb ) = 0 Bx = − 56.249 lb ΣFy = 0: 0 + By − 300 lb + 0.74278 ( 201.94 lb ) = 0 By = 150.003 lb ΣFz = 0: Bz + 0.371391( 201.94 lb ) = 0 Bz = − 74.999 lb Therefore: A = − ( 56.3 lb ) i B = − ( 56.2 lb ) i + (150.0 lb ) j − ( 75.0 lb ) k