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Sol Cap 02 - Edicion 8

COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 131. Note that the force exerted by BD on the pole is directed along BD, and the component of P along AC is 450 N. Then: (a) 450 N P = = 549.3 N cos35° P = 549 N ! (b) ( ) P = 315 N ! x P = 450 N tan 35° x = 315.1 N 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 132. Free-Body Diagram Force Triangle Law of Sines: TAC TBC 5 kN = = sin115° sin 5° sin 60° (a) 5 kN T AC = sin115 ° = 5.23 kN sin 60° T = 5.23 kN ! AC (b) 5 kN T BC = sin 5 ° = 0.503 kN sin 60° T = 0.503 kN ! 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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