COSMOS Chapter 6 Solution 84 - Edubirdie

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Michigan State University
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ERG 160 | Success in Science, Technology, Engineering, and Mathematics
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2019
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1
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FrextWhestembki

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 84. (a) FBD AC: Note: CE is a two-force member ΣM A = 0:  FCE   FCE   + ( 2 in.)   − ( 6 in.)( 24 lb ) = 0  2   2  (8 in.)  FCE = 14.4 2 lb, so E x = 14.40 lb E y = 14.40 lb ΣFx = 0: − Ax + 14.4 lb = 0 A x = 14.40 lb ΣFy = 0: Ay − 24 lb + 14.40 lb = 0 A y = 9.60 lb (b) FBD CE: ! ! ! ! Note AC is a two-force member ΣM E = 0: 1  4  FAC + FAC  − (1 in.)( 24 lb ) = 0 17  17  ( 3 in.)  FAC = 1.6 17 lb, A x = 6.40 lb A y = 1.600 lb ( ) ( ) ΣFx = 0: E x − 4 1.6 17 lb = 0, 17 ΣFy = 0: 1 1.6 17 lb − 24 lb = 0 17 Ey + E x = 6.40 lb E y = 22.4 lb 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. ! ! ! !

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