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Sol Cap 02 - Edicion 8

COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 133. Free-Body Diagram First, consider the sum of forces in the x-direction because there is only one unknown force: or Now or ( ) ( ) Σ F = 0: T cos32° − cos42° − 20 kN cos 42° = 0 x ACB 0.104903T ACB = 14.8629 kN T ACB = 141.682 kN ( ) ( ) Σ F = 0: T sin 42°− sin 32° + 20 kN sin 42°− W = 0 y ACB ( )( ) ( )( ) 141.682 kN 0.139211 + 20 kN 0.66913 − W = 0 (b) T = 141.7 kN ! (a) ACB W = 33.1 kN ! 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 134. Free-Body Diagram: Pulley A Σ F = 0: 2Psin 25° − Pcosα = 0 x and cosα = 0.8452 or α = ± 32.3° For α =+ 32.3° Σ F = 0: 2Pcos 25°+ Psin 32.3°− 350 lb = 0 y or P = 149.1 lb 32.3° ⊳ For α =− 32.3° Σ F = 0: 2Pcos 25° + Psin − 32.3° − 350 lb = 0 y or P = 274 lb 32.3° ⊳ 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.

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