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Sol Cap 02 - Edicion 8

COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 36. (a) Since R is to be horizontal, R y = 0 Then, R =Σ F = 0 y y ( ) α ( ) ( 13) cosα = ( 7) sinα + 9 90 lb + 70 lb sin − 130 lb cosα = 0 2 ( ) 13 1 − sin α = 7 sinα + 9 2 2 Squaring both sides: ( ) ( ) ( ) 169 1 − sin α = 49 sin α + 126 sinα + 81 2 ( ) α ( ) 218 sin + 126 sinα − 88 = 0 Solving by quadratic formula: sinα = 0.40899 or α = 24.1°! (b) Since R is horizontal, R = R x Then, R = Rx = ΣFx F x ( ) ( ) Σ = 70 cos 24.142° + 130 sin 24.142° or R = 117.0 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 37. 300-N Force: 400-N Force: 600-N Force: and ( ) F = 300 N cos 20° = 281.91 N x ( ) F = 300 N sin 20° = 102.61 N y ( ) F = 400 N cos85° = 34.862 N x ( ) F = 400 N sin85° = 398.48 N y ( ) F = 600 N cos5° = 597.72 N x ( ) F =− 600 N sin 5°=− 52.293 N y R R x y =Σ F = 914.49 N x =Σ F = 448.80 N y ( ) ( ) 2 2 R = 914.49 N + 448.80 N = 1018.68 N Further: tanα = 448.80 914.49 −1 448.80 α = tan = 26.1° 914.49 R = 1019 N 26.1° ! 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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