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# Sol Cap 02 - Edicion 8

## COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 42. (a) Require R =Σ F = 0: y y ( ) ( ) 900 lb cos 25°+ 1200 lb sin 35°− sin 65°= 0 or T = 1659.45 lb AE T AE T = 1659 lb ! AE (b) R =Σ F x R =− ( 900 lb) sin 25°− ( 1200 lb) cos35°− ( 1659.45 lb) cos65° R = 2060 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.

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 43. Free-Body Diagram Force Triangle Law of Sines: FAC TBC 400 lb = = sin 25° sin 60° sin 95° (a) 400 lb F AC = sin 25 ° = 169.691 lb sin 95° F = 169.7 lb ! AC (b) 400 T BC = sin 60 ° = 347.73 lb sin 95° T = 348 lb ! BC 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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