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

2024
Views:

84

Pages:

1
Author:

Nossardisd7p

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 43. FBD Truss: ΣFx = 0: ΣM A = 0: Kx = 0 ( ) 6a K y − 125 lb − 5a ( 250 lb ) − 4a ( 250 lb ) − 3a ( 375 lb ) − 2a ( 500 lb ) − a ( 500 lb ) = 0 K y = 937.5 lb ΣFy = 0: Ay − 3 ( 250 lb ) − 2 ( 500 lb ) − 375 lb − 125 lb + 937.5 lb = 0 A y = 1312.5 lb FBD Section IJLK: ΣM z = 0: ( 4 ft )( 937.5 lb −125 lb ) − ( 2 ft ) FHJ = 0 FHJ = 1625 lb, ΣFy = 0: 937.5 lb − 125 lb − 250 lb − FHI = 562.5 5 lbs, ΣFx = 0: 1.625 kips + ( FHJ = 1.625 kips C W 1 FHI = 0 5 FHI = 1.258 kips C W ) 2 0.5625 5 kips − FFI = 0 5 FFI = 2.75 kips T W 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.

COSMOS Chapter 6 Solution 43

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