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

2020
Views:

75

Pages:

2
Author:

behend6abs

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 100. Free-Body Diagram: (a) ( 250 mm ) i  × D + (150 mm ) i + ( 50 mm ) j × (150 N ) k ΣM C = 0: + (150 mm ) i + ( − 50 mm ) j × T k + ( 325 mm ) i + ( 55 mm ) j × (− FE ) j = 0 or ( 250 mm ) Dyk − ( 250 mm ) Dz j + (150 mm )(150 N ) j − ( 50 mm )(150 N ) i − (150 mm ) T j − ( 50 mm ) T j − ( 325 mm ) FE k = 0 continued COSMOS: Complete Online Solutions Manual Organization System Setting the coefficients of the unit vectors equal to zero: i: (150 N )( 50 mm ) − T ( 50 mm ) = 0 , j: ( −150 N )(150 mm ) − (150 N )(150 mm ) − Dz ( 250 mm ) = 0, k: ( Dy ) ( 250 mm ) − FE ( 325 mm ) = 0 , or or T = 150 N Dy = T = 150 N or Dz = −180 N 325 FE 250 Spring force FE = kx, where k = 366 N/m elongation of spring x = ( yE )θ =180° − ( yE )θ = 0° = ( 300 + 55 ) mm − ( 300 − 55 ) mm = 110 mm So, FE = ( 366 N/m )( 0.110 m ) = 40.26 N Substituting into expression for Dy: Dy = 52.338 N Force equations: ΣFx = 0: Dx = 0 ΣFy = 0: C y + 52.338 N − 40.26 N = 0, ΣFz = 0: C z + 150 N + 150 N − 180 or C y = −12.078 N N = 0, or C z = −120.000 N Therefore: C = − (12.08 N ) j − (120.0 N ) k D = ( 52.3 N ) j − (180.0 N ) k