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# Sol Cap 02 - Edicion 8

## COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 60. Note: In problems of this type, P may be directed along one of the cables, with T = Tmax in that cable and T = 0 in the other, or P may be directed in such a way that T is maximum in both cables. The second possibility is investigated first. Free-Body Diagram At C: Force Triangle Force triangle is isoceles with 2β = 180°− 85° β = 47.5° P = 2( 900 N) cos47.5° = 1216 N Since P > 0, solution is correct (a) P = 1216 N ! α = 180°− 55° − 47.5° = 77.5° (b) α = 77.5°! 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 61. Note: Refer to Note in Problem 2.60 Free-Body Diagram At C: Force Triangle (a) Law of Cosines P 2 2 2 ( ) ( ) ( )( ) = 1400 N + 700 N − 2 1400 N 700 N cos85° or P = 1510 N ! (b) Law of Sines sin β sin85° = 1400 N 1510 N sin β = 0.92362 β = 67.461° α = 180° − 55° − 67.461° or α = 57.5°! 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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