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# Sol Cap 02 - Edicion 8

## COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 84. 2 2 2 x y z (a) We have ( θ ) ( θ ) ( θ ) cos + cos + cos = 1 2 2 ( θ ) = − ( θ ) − ( θ ) 2 or cos 1 cos cos z x y Since F z < 0 we must have cosθ z < 0 2 2 Thus cosθ =− 1 − ( cos113.2° ) − cos( 78.4° ) z cosθ z =− 0.89687 θ = 153.7°! z (b) Then: F Fz −35 lb = = cosθ − 0.89687 z F = 39.025 lb And F = Fcosθ x x ( ) F = 39.025 lb cos113.2° x F =− 15.37 lb ! x F = Fcosθ y y ( ) F = 39.025 lb cos78.4° y F = 7.85 lb ! y F = 39.0 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 85. (a) We have F = Fcosθ y y ( ) F = 250 N cos 72.4° y F y = 75.592 N F = 75.6 N ! y Then 2 2 2 2 = x + y + z F F F F ( ) ( ) ( ) 2 2 2 2 250 N = 80 N + 75.592 N + Fz F z = 224.47 N F = 224 N ! z Fx (b) cosθ x = F cosθ = x 80 N 250 N θ = 71.3°! x cosθ = z cosθ = z Fz F 224.47 N 250 N θ = 26.1°! z 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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