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Sol Cap 02 - Edicion 8

COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 58. Free-Body Diagram At A: First Note ... 2 2 With L = ( 22 in. ) + ( 16.5 in. ) AB L AB = AD L AD = F = k L − L Then AB AB ( AB O ) (a) F x 0: 27.5 in. ( 30 in. ) ( 16 in. ) L = + 34 in. 2 2 = ( 9 lb/in. )( 27.5 in. − 22.5 in. ) = 45 lb = ( − ) = ( 3 lb/in. )( 34 in. − 22.5 in. ) F k L L AD AD AD O = 34.5 lb 4 7 15 − 45 lb + T AC + 34.5lb = 0 5 25 17 or T = 19.8529 lb Σ = ( ) ( ) (b) F y 0: Σ = ( ) ( ) ( ) AC 3 24 8 45 lb + 19.8529 lb + 34.5 lb − W = 0 5 25 17 T = 19.85 lb ! AC W = 62.3 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 59. (a) For T AB to be a minimum TAB must be perpendicular to ∴ α + 10° = 60° TAC T = 70 lb sin 30° (b) Then ( ) AB or α = 50.0° or T = 35.0 lb AB 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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