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

2021
Views:

323

Pages:

2
Author:

Caroline Mcfarland

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 116. Note: In all three cases, the right member has only three forces acting, two of which are parallel. Thus the third force, at B, must be parallel to the link forces. (a) FBD whole: ΣM A = 0: − 2aP − a 4 1 B + 5a B=0 4 17 17 4 B=0 17 ΣFx = 0: Ax − ΣFy = 0: Ay − P + 1 B=0 17 B = 2.06 P B = 2.06P 14.04° ! A = 2.06P 14.04° ! A x = 2P Ay = P 2 rigid ! (b) FBD whole: Since B passes through A, ΣM A = 2aP = 0 only if P = 0 ∴ no equilibrium if P ≠ 0 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. not rigid ! COSMOS: Complete Online Solutions Manual Organization System (c) FBD whole: ΣM A = 0: 5a 1 3a 4 B+ B − 2aP = 0 4 17 17 ΣFx = 0: Ax + 4 B=0 17 ΣFy = 0: Ay − P + 1 B=0 17 B= 17 P 4 B = 1.031P 14.04° ! A = 1.250P 36.9° ! Ax = − P Ay = P − P 3P = 4 4 System 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.