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# Sol Cap 02 - Edicion 8

## COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 62. Free-Body Diagram At C: Σ F x = 0: 2T − 1200 N = 0 x T x = 600 N ( ) ( ) 2 2 2 x + y = T T T 2 2 2 y ( 600 N) + ( T ) = ( 870 N) T y = 630 N By similar triangles: AC 1.8 m = 870 N 630 N AC = 2.4857 m L = 2( AC ) L = 2( 2.4857 m) L = 4.97 m L = 4.97 m " 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 63. T BC must be perpendicular to F AC to be as small as possible. Free-Body Diagram: C Force Triangle is a Right Triangle (a) We observe: α = 55° α = 55°! (b) ( ) T = 400 lb sin 60° BC or T = 346.41 lb T = 346 lb ! BC 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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