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

2021
Views:

316

Pages:

2
Author:

crawniadxfo

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 151. Free-Body Diagram: (a) The location of D follows from the geometry of the problem. Since the steel plate is rectangular rD/ A is perpendicular to rB/ A and therefore: rD/ A ⋅ rB/ A = 0 Denoting the coordinates of D by (0, y, z): rD/ A = − ( 0.1 m ) i + yj + ( z − 0.7 m ) k rB/ A = ( 0.3 m ) i − ( 0.4 m ) k and Thus, rD/ A ⋅ rB/ A = − 0.03 − 0.4 z + 0.28 = 0 or z = 0.625 m. rD/ A = ( − 0.1 m )2 + 2 y 2 + ( 0.625 m − 0.7 m ) = 0.75 m continued COSMOS: Complete Online Solutions Manual Organization System Solving for y: y = 0.73951 m x = 0, y = 0.740 m, z = 0.625 m Location of D is therefore: (b) Consider moment equilibrium about axis AB: λ AB uuur AB = = AB 0.3i − 0.4k ( 0.3)2 + ( − 0.4 )2 = 0.6i − 0.8k rD/ A = − ( 0.1 m ) i + ( 0.73951 m ) j − ( 0.075 m ) k rD/B = − ( 0.4 m ) i + ( 0.73951 m ) j + ( 0.625 m − 0.3 m ) k N D = N Di ( ) W = − ( mg ) j = − ( 40 kg ) 9.81 m/s 2 j = − ( 392.4 N ) j Then, ΣM AB = 0: ( ) ( ) λ AB ⋅ rD/ A × N D + λ AB ⋅ rG/B × W = 0 − 0.8 −0.8 0.6 0 0.6 0 − 0.1 0.73951 − 0.075 + − 0.2 0.36976 0.1625 = 0 − 392.4 ND 0 0 0 0 0.59161 N D − 24.525 = 0 N D = 41.455 N or N D = ( 41.455 N ) i

COSMOS Chapter 4 Solution 151

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