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Sol Cap 02 - Edicion 8

COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 111. Note that because the line of action of each of the cords passes through the vertex A of the cone, the cords all have the same length, and the unit vectors lying along the cords are parallel to the unit vectors lying along the generators of the cone. Thus, for example, the unit vector along BE is identical to the unit vector along the generator AB. Hence: It follows that: λ = λ = AB BE cos 45° i + 8j − sin 45° k 65 ⎛cos 45° i + 8j − sin 45° k ⎞ TBE = TBEλ BE = TBE ⎜ ⎟ ⎝ 65 ⎠ ⎛cos30° i + 8j + sin 30° k ⎞ TCF = TCFλ CF = TCF ⎜ ⎟ ⎝ 65 ⎠ At A: Σ F = 0: T + T + T + W + P = 0 BE CF DG ⎛− cos15° i + 8j − sin15° k ⎞ TDG = TDGλ DG = TDG ⎜ ⎟ ⎝ 65 ⎠ Then, isolating the factors of i, j, and k, we obtain three algebraic equations: TBE TCF TDG i : cos 45°+ cos30°− cos15°+ P = 0 65 65 65 or T cos 45°+ T cos30°− T cos15°+ P 65 = 0 (1) BE CF DG 8 8 8 j : TBE + TCF + TDG − W = 0 65 65 65 or 65 TBE + TCF + TDG − W = 0 (2) 8 TBE TCF TDG k : − sin 45° + sin 30° − sin15° = 0 65 65 65 or − T sin 45° + T sin 30° − T sin15° = 0 (3) With P = 0 and the tension in cord BE = 0.2 lb: BE CF DG Solving the resulting Equations (1), (2), and (3) using conventional methods in Linear Algebra (elimination, matrix methods or iteration – with MATLAB or Maple, for example), we obtain: T CF = T DG = 0.669 lb 0.746 lb W = 1.603 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 112. See Problem 2.111 for the Figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: i : T cos 45°+ T cos30°− T cos15°+ 65P = 0 (1) BE CF DG 65 j : TBE + TCF + TDG − W = 0 (2) 8 k : − T sin 45° + T sin 30° − T sin15° = 0 (3) BE CF DG With W = 1.6 lb , the range of values of P for which the cord CF is taut can found by solving Equations (1), (2), and (3) for the tension T CF as a function of P and requiring it to be positive ( > 0). Solving (1), (2), and (3) with unknown P, using conventional methods in Linear Algebra (elimination, matrix methods or iteration – with MATLAB or Maple, for example), we obtain: TCF ( P ) = − 1.729 + 0.668 lb Hence, for T CF > 0 − 1.729P + 0.668 > 0 or P < 0.386 lb ∴ φ ≤ P < 0.386 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.

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