COSMOS Chapter 6 Solution 57 | Answer Key - Edubirdie

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

2023
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65

Pages:

1
Author:

Lily Callahan

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 57. FBD Truss: ΣFx = 0: Ax = 0 ( 4.8 ft )( 300 lb ) + ( 7.2 ft )( 750 lb ) + (11.52 ft )( 750 lb ) + (16.8 ft )(1950 lb ) + (19.2 ft )( 900 lb ) + ( 26.4 ft )(1050 lb ) + ( 33.6 ft ) (150 lb − Ay ) = 0 ΣM N = 0: Ay = 2925 lb FBD Section KJMN: 2925 lb − 2 (150 lb ) − 2 ( 750 lb ) − 1050 lb ΣFy = 0: − 900 lb − 1950 lb − 300 lb + N y = 0 N y = 3075 lb ( 7.2 ft )( 3075 lb − 150 lb ) − ( 2.4 ft )( 300 lb ) ΣM J = 0: − (1.8 ft ) FIK = 0 FIK = 11300 lb, ΣM I = 0: FIK = 11.30 kips T (11.52 ft )( 3075 lb − 150 lb ) − ( 6.72 ft )( 300 lb )  2  − ( 4.32 ft )( 750 lb ) + ( 3.96 ft )  FHJ  = 0  5  ΣFx = 0: FBD Joint I: FHJ = −3590.5 5 lb, FHJ = 8.03 kips C ! 2 12 − −3590.9 5 lb − FIJ − 11300 lb = 0 13 5 ( ) FIJ = − 4461.4 lb, ΣFx = 0: 11300 lb − FIJ = 4.46 kip C ! 12 ( 4461.4 lb ) − FGI = 0 13 FGI = 7181.8 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. FGI = 7.18 kips T !

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