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# Sol Cap 02 - Edicion 8

## COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 135. (a) F = Fsin 30° sin 50° = 220.6 N (Given) x F 220.6 N = = 575.95 N sin30 ° sin50° F = 576 N ! (b) Fx 220.6 cosθ x = = = 0.38302 F 575.95 θ = 67.5°! x Fy = Fcos30° = 498.79 N Fy 498.79 cosθ y = = = 0.86605 F 575.95 θ = 30.0°! y Fz =− Fsin 30° cos50° =− ( 575.95 N) sin 30° cos50° =− 185.107 N Fz −185.107 cosθ z = = = − 0.32139 F 575.95 θ = 108.7°! 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.

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 136. (a) F Fcosθ ( ) Then: z = = 600 lb cos136.8° z =− 437.38 lb F =− 437 lb ! 2 2 2 2 = x + y + z F F F F z 2 2 2 2 F y So: ( 600 lb) = ( 200 lb) + ( ) + ( − 437.38 lb) 2 2 2 Hence: F =− ( 600 lb) −( 200 lb) −( − 437.38 lb) y =− 358.75 lb F =− 359 lb ! y (b) Fx 200 cosθ x = = = 0.33333 θ x = 70.5°! F 600 Fy −358.75 cosθ y = = = − 0.59792 θ y = 126.7°! F 600 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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