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

2019
Views:

81

Pages:

2
Author:

Sean Horton

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 117. (a) Member FBDs: FBD I: ΣM A = 0: aF1 − 2aP = 0 FBD II: ΣM B = 0: − aF2 = 0 F1 = 2P; ΣFy = 0: Ay − P = 0 F2 = 0 ΣFx = 0: Bx + F1 = 0, Bx = − F1 = −2P B x = 2P ΣFy = 0: By = 0 FBD I: Ay = P ΣFx = 0: − Ax − F1 + F2 = 0 Ax = F2 − F1 = 0 − 2P so B = 2P ! A x = 2P so A = 2.24P 26.6° ! Frame is rigid ! (b) FBD left: FBD I: FBD II: FBD whole: ΣM E = 0: a a 5a P + Ax − Ay = 0 2 2 2 ΣM B = 0: 3aP + aAx − 5aAy = 0 Ax − 5 Ay = − P Ax − 5 Ay = −3P This is impossible unless P = 0 ∴ not rigid ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System (c) Member FBDs: FBD I: ΣFy = 0: A − P = 0 FBD II: ΣM D = 0: aF1 − 2aA = 0 F1 = 2 P ΣFx = 0: F2 − F1 = 0 F2 = 2 P ΣM B = 0: 2aC − aF1 = 0 C = ΣFx = 0: F1 − F2 + Bx = 0 ΣFx = 0: By + C = 0 F1 = P 2 A= P ! C= P ! Bx = P − P = 0 By = −C = − P B= P ! Frame is rigid ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.