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

2022
Views:

67

Pages:

1
Author:

Lamghari6kx

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 106. Horizontal pipe Vertical pipe π Contact angles β H = Contact angle βV = π 2 µ sH = 0.30 µ sV = ? For Pmin , the 100 lb force impends downward, and µ 100 lb =  e  sH π 2 R =  e µ     sH π 2 (  eµ   sV π )Q =  e µ s π π2 (  eµ   sV π ) e µ sH π2 P   π 0.30 + µ sV )  100 lb = e ( ( 20 lb ) , so eπ ( 0.30 + µsV ) = 5   (a) For Pmax the force P impends downward, and the ratios are reversed, so Pmax = 5 (100 lb ) = 500 lb (b) π ( 0.30 + µ sV ) = ln 5 µ sV = 1 π ln 5 − 0.30 = 0.21230 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. µ sV = 0.212

COSMOS Chapter 8 Solution 106

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