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Sol Cap 02 - Edicion 8

COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 6. (a) Using the triangle rule and the Law of Sines sinα sin 45° = 120 N 200 N sinα = 0.42426 α = 25.104° or α = 25.1°! (b) β + 45° + 25.104° = 180° β = 109.896° Using the Law of Sines F aa ′ 200 N = sin β sin 45° F aa ′ 200 N = sin109.896° sin 45° or F ′ = 266 N ! aa 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 7. Using the triangle rule and the Law of Cosines, Have: β = 180° − 45° Then: or β = 135° R 2 2 2 ( ) ( ) ( )( ) = 900 + 600 − 2 900 600 cos 135° R = 1390.57 N Using the Law of Sines, 600 1390.57 sinγ = sin135° or γ = 17.7642° and α = 90° − 17.7642° α = 72.236° (a) α = 72.2°! (b) R = 1.391 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.

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