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Enviado por:  Ninoka  23 junio 2011Tags: Palabras: 21564   |   Páginas: 87Views: 329COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 1.(a)(b)We measure:R = 37 lb,α = 76° R = 37 lb76° !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 SystemChapter 2, Solution 2.(a)(b)We measure:R = 57 lb, α = 86° R = 57 lb86° !Vector Mechanics forEngineers: 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 SystemChapter 2, Solution 3.(a)Parallelogram law:(b)Triangle rule:We measure:R = 10.5 kNα = 22.5°R = 10.5 kN22.5° !Vector Mechanics for Engineers: Statics andDynamics, 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 SystemChapter 2, Solution 4.(a)Parallelogram law: We measure:R = 5.4 kN α = 12° R = 5.4 kN12° !(b)Triangle rule:We measure:R = 5.4 kN α = 12° R = 5.4 kN12° !Vector Mechanics forEngineers: 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 SystemChapter 2, Solution 5.Using the triangle rule and the Law of Sines (a)sin β sin 45° = 150 N 200 N sin β = 0.53033β = 32.028° α + β + 45° = 180°α= 103.0° !(b) Using the Law of SinesFbb′ 200 N = sin α sin 45°Fbb′ = 276 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 SystemChapter 2, Solution 6.Using the trianglerule and the Law of Sines (a)sin α sin 45° = 120 N 200 N sin α = 0.42426α = 25.104°orα = 25.1° !(b)β + 45° + 25.104° = 180° β = 109.896°Using the Law of SinesFaa′ 200 N = sin β sin 45° Faa′ 200 N = sin109.896° sin 45°orFaa′ = 266 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 SystemChapter 2, Solution 7.Using the triangle rule and the Law of Cosines, Have: β = 180° − 45°β = 135°Then:R 2 = ( 900 ) + ( 600 ) − 2 ( 900 )( 600 ) cos 135°2 2or R = 1390.57 NUsing the Law of Sines,600 1390.57 = sin γ sin135° or γ = 17.7642° and α = 90° − 17.7642°α = 72.236°(a) (b)α = 72.2° !R =1.391 kN !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 SystemChapter 2, Solution 8.By trigonometry: Law of SinesF2 R 30 = = sin α sin 38° sin βα = 90° − 28° = 62°, β = 180° −62° − 38° = 80°Then:F2 R 30 lb = = sin 62° sin 38° sin 80°or (a) F2 = 26.9 lb ! (b) R = 18.75 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 SystemChapter 2, Solution...