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Sol Cap 02 - Edicion 8

COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 129. Using the triangle rule and the Law of Sines (a) Have: 20 lb 14 lb sinα = sin 30° sinα = 0.71428 (b) β = 180° − ( 30° + 45.6° ) α = 45.6°⊳ = 104.4° Then: R 14 lb = sin104.4° sin 30° R = 27.1 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.

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 130. We compute the following distances: 500 N Force: ( ) ( ) 2 2 OA = 70 + 240 = 250 mm ( ) ( ) 2 2 OB = 210 + 200 = 290 mm ( ) ( ) 2 2 OC = 120 + 225 = 255 mm F x F y ⎛ 70 ⎞ =−500 N⎜ 250 ⎟ ⎝ ⎠ ⎛240 ⎞ =+ 500 N⎜ 250 ⎟ ⎝ ⎠ F =− 140.0 N ! x F = 480 N ! y 435 N Force: 510 N Force: F x F y F x F y ⎛210 ⎞ =+ 435 N⎜ 290 ⎟ ⎝ ⎠ ⎛200 ⎞ =+ 435 N⎜ 290 ⎟ ⎝ ⎠ ⎛120 ⎞ =+ 510 N⎜ 255 ⎟ ⎝ ⎠ ⎛225 ⎞ =−510 N⎜ 255 ⎟ ⎝ ⎠ F = 315 N ! x F = 300 N ! y F = 240 N ! x F =− 450 N ! 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.

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