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# Sol Cap 02 - Edicion 8

## COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 115. From the solutions of 2.107 and 2.108: TAB TAC TAD = 0.5409P = 0.295P = 0.2959P Using P = 8 kN: T = 4.33 kN ! AB T = 2.36 kN ! AC T = 2.37 kN ! AD 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 116. ( ) ( ) ( ) 2 2 2 d = 6m + 6m + 3m = 9m BA ( ) ( ) ( ) 2 2 2 d = − 10.5 m + − 6 m + − 8 m = 14.5 mm AC ( ) ( ) ( ) 2 2 2 d = − 6m + − 6m + 7m = 11mm AD ( ) ( ) 2 2 d = 6m + − 4.5m = 7.5m AE F F BA = FBAλBA = BA ⎡ ( 6m) + ( 6m) + ( 3m) ⎤ 9m ⎣ i j k ⎦ AC = TACλAC 2 2 1 = F ⎛ BA ⎜ i + j + k ⎞ ⎟ ⎝3 3 3 ⎠ T AC T = ⎡−( 10.5 m) − ( 6 m) − ( 8 m) AD = TADλAD 14.5 m ⎣ 21 12 16 = T ⎛ AC ⎜− i − j − k ⎞ ⎟ ⎝ 29 29 29 ⎠ T AD i j k ⎦ ⎤ T = ⎡−( 6m) − ( 6m) + ( 7m) AE = WAEλAE 11 m ⎣ i j k ⎤ ⎦ ⎛ 6 6 7 ⎞ = T AD ⎜− i − j + k ⎟ ⎝ 11 11 11 ⎠ W ⎡ ⎤ 7.5 m ⎣ j ⎦ W = ( 6 m) i − ( 4.5 m) W O ( 0.8i 0.6j ) = W − =−W At point A: Σ F =0: F + T + T + W + W 0 BA AC AD AE O = j i component: 2 21 6 FBA − TAC − TAD + 0.8W = 0 (1) 3 29 11 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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