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Sol Cap 02 - Edicion 8

COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 52. Free-Body Diagram: Σ = 0: − F B cos15° + 2.4 kN + ( 1.9kN) sin15° = 0 F x or F B = 2.9938 kN Σ = 0: F − ( 1.9 kN) cos15° + ( 2.9938 kN) sin15° = 0 F y D F = 2.99 kN ⊳ B F = 1.060 kN ⊳ D 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 53. From Similar Triangles we have: L 2 ( 2.5 m) ( 8 L) ( 5.45 m) 2 2 2 − = − − − 6.25 = 64 −16 L − 29.7025 or L = 2.5342 m And 5.45 m cos β = 8 m − 2.5342 m or β = 4.3576° Then cosα = 2.5 m 2.5342 m or α = 9.4237° Free-Body Diagram At B: Σ F x = 0: or ( ) −T cosα − 35 N cosα + T cos β = 0 T ABC ABC ( ) ABC 35 cos9.4237° = cos4.3576°− cos9.4237° T ABC = 3255.9 N Σ F y = 0: ( ) T sinα + 35 N sinα + T sin β − W = 0 ABC ABC ( ) ( ) sin 9.4237° 3255.9 N + 35 N + 3255.9 N sin 4.3576° − W = 0 or W = 786.22 N (a) (b) W = 786 N " T = 3.26 kN " ABC 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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