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Sol Cap 02 - Edicion 8

COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 126. See Problem 2.125 for the analysis leading to the linear algebraic Equations ( 1 ′), ( 2 ′), and ( 3′ ) below: i component: 81T 34T 0 − + = ( 1′ ) AB j component: 72 T 122.4 T 37.4 T 153 W ADE + + = ( 2′ ) AB AC ADE k component: 108 T 91.8 T 17 T 0 − + + = ( 3′ ) AB AC ADE Setting T AB = 300 N and solving (1), (2), (3) simultaneously: (a) (b) (c) T = 221 N ! AC T = 715 N ! ADE W = 2060 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.

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 127. Free-Body Diagrams of collars For both Problems 2.127 and 2.128: Here ( ) ( ) ( ) 2 2 2 2 AB = x + y + z 2 2 2 2 1m = 0.40m + y + z or y 2 2 2 + z = 0.84 m Thus, with y given, z is determined. Now uuur AB 1 λ AB = = ( 0.40i − yj + zk) m = 0.4i − yk + zk AB 1m Where y and z are in units of meters, m. From the F.B. Diagram of collar A: Σ F = 0: N i + N k + Pj + T λ = 0 Setting the jcoefficient to zero gives: With P = 680 N, x z AB AB P − yT AB = 0 T AB = 680 N y Now, from the free body diagram of collar B: Σ F = 0: N i + N j + Qk − T λ = 0 x y AB AB continued 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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