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Sol Cap 02 - Edicion 8

COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 28. We note: CB exerts force P on B along CB, and the horizontal component of P is Then: (a) P = Psin 50° x P x = 260 lb. P = = P x sin50 ° 260 lb sin50° (b) P = P tan 50° x = 339.40 lb P = 339 lb ! y P = y = Px tan 50° 260 lb tan 50° = 218.16 lb P 218 lb ! y = 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 29. (a) P = 45 N cos 20° or P = 47.9 N ! (b) ( ) P = 47.9 N sin 20° x or P = 16.38 N ! x 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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