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

2023
Views:

144

Pages:

1
Author:

diamundgalx45p

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 94. Let normal force on ∆A be ∆N , and ∆A = r ∆s∆φ ∆N = k ∆A so ∆N = k, ∆A ∆s = ∆r sin θ where φ is the azimuthal angle around the symmetry axis of rotation ∆Fy = ∆N sin θ = kr ∆r ∆φ P = lim Σ∆Fy Total vertical force ∆A → 0 2π P = ∫0 ( (∫ R2 krdr R1 P = π k R22 − R12 Friction force ) ) dφ = 2π k ∫ R2 rdr R1 or k = Total couple M = 2π ( − R12 ) ∆F = µ∆N = µ k ∆A ∆M = r ∆F = r µ kr Moment π P R22 ∆r ∆φ sin θ 2π  R µ k  M = lim Σ∆M = ∫ 0  ∫R 2 r 2dr  dφ ∆A → 0  1 sin θ  µ k R2 2 2 πµ P r dr = ∫ R1 2 sin θ 3 sin θ π R2 − R32 ( ) M = 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. (R 3 2 − R33 ) 2 µ P R23 − R13 3 sin θ R22 − R12