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# Sol Cap 02 - Edicion 8

## COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 54. From Similar Triangles we have: L 2 ( 3 m) ( 8 L) ( 4.95 m) 2 2 2 − = − − − 9 = 64 −16 L − 24.5025 or L = 3.0311 m Then 4.95 m cos β = 8 m − 3.0311 m or β = 4.9989° And 3m cosα = 3.0311 m or α = 8.2147° Free-Body Diagram At B: (a) Σ F x = 0: −T cosα − T cosα + T cos β = 0 or ABC DE ABC T DE cos β − cosα = T cosα Σ F y = 0: ABC ( ) T sinα + T sinα + T sin β − 720 N = 0 ABC DE ABC Substituting for α and β gives T ABC T ABC ⎡ ⎛cos β − cosα ⎞ ⎤ ⎢sinα + sinα⎜ ⎟ + sin β⎥= 720 ⎣ ⎝ cosα ⎠ ⎦ T ABC = ( 720) cos sin ( α + β ) ( 720) cos8.2147° ( °+ ° ) = sin 8.2147 4.9989 α T ABC = 3117.5 N or T = 3.12 kN " ABC cos 4.9989°− cos8.2147° 3117.5 N cos8.2147° (b) = ( ) T DE T DE = 20.338 N 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. or T = 20.3 N " DE

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 55. Free-Body Diagram At C: 3 15 15 Σ F x = 0: − TAC + TBC − ( 150 lb ) = 0 5 17 17 or 17 − TAC + 5TBC = 750 (1) 5 4 8 8 Σ F y = 0: TAC + TBC − ( 150 lb ) − 190 lb = 0 5 17 17 or 17 T 5 + 2T = 1107.5 (2) AC BC Then adding Equations (1) and (2) 7T BC = 1857.5 or T BC = 265.36 lb Therefore (a) T AC = 169.6 lb ! (b) T = 265 lb ! BC 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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