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Sol Cap 02 - Edicion 8

COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 72. First replace 30 lb forces by their resultant Q: Q = 230lbcos25 ( ) ° Q = 54.378 lb Equivalent loading at A: Law of Cosines: 2 2 2 ( ) ( ) ( ) ( )( ) ( α) ( α) 120 lb = 100 lb + 54.378 lb − 2 100 lb 54.378 lb cos 125° − cos 125° − = − 0.132685 This gives two values: 125°− α = 97.625° Thus for R < 120 lb: α = 27.4° 125°− α =− 97.625° α = 223° 27.4°< α < 223°! 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 73. (a) ( ) F = 950 lb sin 50° cos 40° x = 557.48 lb F y =− ( 950 lb) cos50° F = 557 lb ! x =− 610.65 lb F =−611 lb ! y ( ) F = 950 lb sin50° sin 40° z = 467.78 lb F = 468 lb ! z (b) cosθ = x 557.48 lb 950 lb or θ x = 54.1° ! cosθ y = −610.65 lb 950 lb or θ y = 130.0° ! cosθ = z 467.78 lb 950 lb or θ z = 60.5° ! 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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