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

2024
Views:

65

Pages:

2
Author:

hekeldigzjbz

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 101. Free-Body Diagram: Start by determining the spring force, FE : FE = − FE cosθ j + FE sin θ k The magnitude FE = kx, where k = 366 N/m, and elongation of spring x = ( yE )θ = 90° − ( yE )θ = 0° = ( 300 mm )2 + ( 55 mm )2 − ( 300 mm − 55mm ) = 305 mm − 245 mm = 60 mm So, FE = ( 366 N/m )( 0.06 m ) = 21.96 N. Note that the length of the spring at θ = 90° is therefore 305 mm. Then  300   55  FE = − ( 21.96 N )   j + ( 21.96 N )  k 305    305  or FE = − ( 21.6 N ) j + ( 3.96 N ) k continued COSMOS: Complete Online Solutions Manual Organization System ( 250 mm ) i  × D + (150 mm ) i + ( 50 mm ) j × (150 N ) k + (150 mm ) i + ( −50 mm ) j × T k ΣM C = 0: + ( 325 mm ) i + ( −55 mm ) k  ×  − ( 21.6 N ) j + ( 3.96 N ) k  = 0 ( 250 mm ) Dyk − ( 250 mm ) Dz j + (150 mm )(150 N ) j or − ( 50 mm )(150 N ) i − (150 mm ) T j − ( 50 mm ) T j − ( 325 mm )( 21.6 N ) k − ( 325 mm )( 3.96 N ) j − ( 55 mm )( 21.6 N ) i = 0 Setting the coefficients of the unit vectors equal to zero: (a) i: (150 N )( 50 mm ) − T ( 50 mm ) − ( 21.6 N )( 55 mm ) = 0, or T = 126.240 N T = 126.2 N j: ( −150 N )(150 mm ) − (126.24 N )(150 mm ) − ( 3.96 N )( 325 mm ) − Dz ( 250 mm ) = 0, or Dz = −170.892 N k: ( Dy ) ( 250 mm ) − ( 21.6 N )(325 mm ) = 0, Dy = 28.080 N or Force equations: ΣFx = 0: Dx = 0 ΣFy = 0: C y + 28.080 N − 21.6 N = 0, ΣFz = 0: C z + 126.240 N + 150 N − 170.892 N + 3.96 N = 0, or C y = −6.4800 N or C z = −109.308 N Therefore: C = − ( 6.48 N ) j − (109.3 N ) k D = ( 28.1 N ) j − (170.9 N ) k