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Sol Cap 02 - Edicion 8

COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 38. Σ F x : R x =Σ F x ( ) ( ) ( ) R = 600 N cos50° + 300 N cos85° − 700 N cos50° x R =− 38.132 N x Σ F y : R y =Σ F y ( ) ( ) ( ) R = 600 N sin 50°+ 300 N sin85° + 700 N sin50° y R y = 1294.72 N ( 38.132 N) ( 1294.72 N) R = − + R = 1295 N 1294.72 N tanα = 38.132 N α = 88.3° 2 2 R = 1.295 kN 88.3° ! 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 39. We have: 84 12 3 Rx =Σ Fx =− TBC + − 116 13 5 or R =− 0.72414T + 84 lb and x BC ( 156 lb) ( 100 lb) 80 5 4 Ry =Σ Fy = TBC − − 116 13 5 ( 156 lb) ( 100 lb) R y = 0.68966T − 140 lb BC (a) So, for R to be vertical, R x =− 0.72414T + 84 lb = 0 BC T = 116.0 lb ! BC (b) Using T BC = 116.0 lb ( ) R = R y = 0.68966 116.0 lb − 140 lb = − 60 lb R = R = 60.0 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.

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