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Sol Cap 02 - Edicion 8

COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 56. Free-Body Diagram At C: 3 15 15 Σ F x = 0: − TAC + TBC − ( 150 lb ) = 0 5 17 17 or 17 − TAC 5 + 5TBC = 750 (1) 4 8 8 Σ F y = 0: TAC + TBC − ( 150 lb ) − W = 0 5 17 17 or 17 17 TAC + 2TBC = 300 + W 5 4 (2) Adding Equations (1) and (2) gives 17 7TBC = 1050 + W 4 or 17 TBC = 150 + W 28 Using Equation (1) 17 ⎛ 17 ⎞ − TAC + 5 150 W 750 5 ⎜ + = 28 ⎟ ⎝ ⎠ or 25 TAC = W 28 Now for T ≤ 240 lb ⇒ 25 TAC :240= W 28 or W = 269 lb 17 TBC : 240 = 150 + W 28 or W = 148.2 lb Therefore 0 ≤ W ≤ 148.2 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 57. Free-Body Diagram At A: First note from geometry: The sides of the triangle with hypotenuse AD are in the ratio 12:35:37. The sides of the triangle with hypotenuse AC are in the ratio 3:4:5. The sides of the triangle with hypotenuse AB are also in the ratio 12:35:37. Then: or and Then: or 4 35 12 Σ Fx = 0: − ( 3W) + ( W) + Fs = 0 5 37 37 Fs = 4.4833W 3 12 35 Σ Fy = 0: ( 3W) + ( W) + Fs − 400 N = 0 5 37 37 3 ( 3W) + 12 ( W) + 35 ( 4.4833W) − 400 N = 0 5 37 37 W = 62.841 N and or (a) (b) Have spring force Where and F s = 281.74 N ( ) F = k L − L s AB O ( ) F = k L − L AB AB AB O W = 62.8 N ⊳ ( ) ( ) 2 2 L = 0.360 m + 1.050 m = 1.110 m AB So: L = 758 mm ⊳ O ( ) 281.74 N = 800 N/m 1.110 − m L O or 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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