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Sol Cap 02 - Edicion 8

COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 137. ( 500 lb)[ cos30 sin15 sin 30 cos30 cos15 ] ( 500 lb)[ 0.2241i 0.50j 0.8365k ] ( 112.05 lb) i ( 250 lb) j ( 418.25 lb) k ( 600 lb)[ cos 40 cos 20 sin 40 cos40 sin 20 ] ( 600 lb)[ 0.71985i 0.64278j 0.26201k ] ( 431.91 lb) i ( 385.67 lb) j ( 157.206 lb) k ( 319.86 lb) ( 635.67 lb) ( 261.04 lb) P = − ° ° i + ° j + ° ° k = − + + =− + + Q = ° ° i + ° j − ° ° k = + − = + − R = P + Q = i + j + k ( ) ( ) ( ) 2 2 2 R = 319.86 lb + 635.67 lb + 261.04 lb = 757.98 lb R = 758 lb ! Rx 319.86 lb cosθ x = = = 0.42199 R 757.98 lb θ = 65.0°! Ry 635.67 lb cosθ y = = = 0.83864 R 757.98 lb θ = 33.0°! Rz 261.04 lb cosθ z = = = 0.34439 R 757.98 lb θ = 69.9°! x y z 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 138. Equating to zero the coefficients of i, j, k: The forces applied at A are: TAB , TAC, TAD and P where P = Pj . To express the other forces in terms of the unit vectors i, j, k, we write uuur AB =− ( 0.72 m) i + ( 1.2 m) j −( 0.54 m ) k, AB = 1.5 m uuur AC = ( 1.2 m) j + ( 0.64 m ) k, AC = 1.36 m uuur AD = ( 0.8 m) i + ( 1.2 m) j − ( 0.54 m ) k, AD = 1.54 m uuur and TAB = TABλ AB = T AB AB = ( − 0.48i + 0.8j − 0.36k) TAB AB uuur TAC = TACλ AC = T AC AC = ( 0.88235j + 0.47059k) TAC AC uuur TAD = TADλ AD = T AD AD = ( 0.51948i + 0.77922j − 0.35065k) TAD AD Equilibrium Condition with W =−Wj Σ F = 0: T + T + T − Wj = 0 AB AC AD Substituting the expressions obtained for TAB, TAC, and TADand factoring i, j, and k: ( − 0.48TAB + 0.51948TAD ) i + ( 0.8TAB + 0.88235TAC + 0.77922TAD −W) + ( − 0.36T + 0.47059T − 0.35065T ) k = 0 AB AC AD − 0.48T + 0.51948T = 0 AB AD 0.8T + 0.88235T + 0.77922T − W = 0 AB AC AD − 0.36T + 0.47059T − 0.35065T = 0 AB AC AD Substituting T AB = 3 kN in Equations (1), (2) and (3) and solving the resulting set of equations, using conventional algorithms for solving linear algebraic equations, gives T AC = T AD = 4.3605 kN 2.7720 kN W = 8.41 kN ⊳ j 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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