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# Sol Cap 02 - Edicion 8

## COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 20. Using the force triangle and the Laws of Cosines and Sines We have: γ = 180° − ( 45° + 25° ) = 110° 2 2 2 Then: ( ) ( ) ( )( ) and R = 30 kN + 20 kN − 2 30 kN 20 kN cos110° 2 = 1710.42 kN R = 41.357 kN 30 kN 41.357 kN sinα = sin110° ⎛ 30 kN ⎞ sinα = ⎜ sin110° 41.357 kN ⎟ ⎝ ⎠ = 0.68164 α = 42.972° Finally: φ = α + 45° = 87.972° R = 41.4 kN 88.0° ! 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 21. F = 2.4 kN cos50° 2.4 kN Force: ( ) x F = 1.85 kN cos 20° 1.85 kN Force: ( ) x ( ) F = 2.4 kN sin 50° y ( ) F = 1.85 kN sin 20° y F x = 1.543 kN ⊳ F = 1.839 kN ⊳ y F x = 1.738 kN ⊳ F = 1.40 kN cos35° 1.40 kN Force: ( ) x ( ) F =− 1.40 kN sin 35° y F = 0.633 kN ⊳ y F x = 1.147 kN ⊳ F =−0.803 kN ⊳ 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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