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# Sol Cap 02 - Edicion 8

## COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 118. From the solutions to Problems 2.111 and 2.112, have Applying the method of elimination to obtain a desired result: Multiplying (2′) by sin 45° and adding the result to (3): T CF TBE + TCF + TDG = 0.2 65 (2′) − T sin 45° + T sin 30° − T sin15° = 0 (3) BE CF DG T cos45°+ T cos30°− T cos15°− P 65 = 0 (1′ ) BE CF DG ( ) T ( ) sin 45°+ sin 30° + sin 45°− sin15° = 0.2 65 sin 45° DG or T = 0.94455 − 0.37137T Multiplying (2′) by sin 30° and subtracting (3) from the result: T BE ( ) T ( ) sin 30°+ sin 45° + sin 30°+ sin15° = 0.2 65 sin 30° DG or T = 0.66790 − 0.62863T (5) Substituting (4) and (5) into (1′) : BE 1.29028 −1.73205T − P 65 = 0 ∴ T DG is taut for DG P < 1.29028 lb 65 or 0 ≤ P ≤ 0.1600 lb! CF DG DG 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 119. ( ) ( ) ( ) 2 2 2 d = − 30 ft + 24 ft + 32 ft = 50 ft AB ( ) ( ) ( ) 2 2 2 d = − 30 ft + 20 ft + − 12 ft = 38 ft AC T TAB AB = TABλAB = ⎡− ( 30 ft) + ( 24 ft) + ( 32 ft) ⎤ 50 ft ⎣ i j k ⎦ = − 0.6i + 0.48j + 0.64k T AB ( ) T TAC AC = TACλAC = ⎡− ( 30 ft) + ( 20 ft) − ( 12 ft) ⎤ 38 ft ⎣ i j k ⎦ 16 30 N = Ni + Nj 34 34 W =−( 175 lb) j 30 20 12 = T ⎛ AC ⎜− i + j − k ⎞ ⎟ ⎝ 38 38 38 ⎠ At point A: Σ F =0: T + T + N + W 0 i component: j component: AB AC = 30 16 −0.6TAB − TAC + N = 0 (1) 38 34 20 30 0.48TAB + TAC + N − 175 lb = 0 (2) 38 34 12 k component: 0.64TAB − TAC = 0 (3) 38 Solving Equations (1), (2), and (3) simultaneously: T = 30.9 lb ⊳ AB T = 62.5 lb ⊳ AC 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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