COSMOS Chapter 6 Solution 42 | Michigan State University - 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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2020
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COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 42. FBD Truss: ΣFx = 0: Kx = 0 ( ) 6a K y − 125 lb − 5a ( 250 lb ) ΣM A = 0: − 4a ( 250 lb ) − 3a ( 375 lb ) − 2a ( 500 lb ) − a ( 500 lb ) = 0 K y = 937.5 lb Ay − 3 ( 250 lb ) − 2 ( 500 lb ) ΣFy = 0: − 375 lb − 125 lb + 937.5 lb = 0 A y = 1312.5 lb FBD Section ABEC: ( 2 ft ) FCF + ( 4 ft )( 500 lb ) ΣM E = 0: + ( 8 ft )( 250 lb − 1312.5 lb ) = 0 FCF = 3250 lb, ΣFy = 0: FCF = 3.25 kips T W 1312.5 lb − 250 lb − 2 ( 500 lb ) − FEF = 62.5 5 lb, ΣFx = 0: 3250 lb + ( 1 FEF = 0 5 FEF = 139.8 lb T W ) 2 62.5 5 lb − FEG = 0 5 FEG = 3375 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. FEG = 3.38 kips C W

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