Capitulo 2 SOlucionario libro Estatica

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Transcript of Capitulo 2 SOlucionario libro Estatica

  • 1.COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 1.(a)(b) We measure:R = 37 lb, = 76R = 37 lb 76 !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.

2. COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 2.(a)(b)We measure:R = 57 lb, = 86R = 57 lb 86 !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. 3. COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 3.(a) Parallelogram law:(b) Triangle rule: We measure:R = 10.5 kN = 22.5R = 10.5 kN 22.5 !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. 4. COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 4.(a) Parallelogram law: We measure: R = 5.4 kN = 12R = 5.4 kN 12 !(b) Triangle rule: We measure: R = 5.4 kN = 12R = 5.4 kN 12 !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. 5. COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 5.Using the triangle rule and the Law of Sinessin sin 45(a) = 150 N 200 N sin = 0.53033 = 32.028 + + 45 = 180 = 103.0 !(b) Using the Law of SinesFbb 200 N = sin sin 45Fbb = 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. 6. COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 6.Using the triangle rule and the Law of Sinessin sin 45(a) = 120 N 200 N sin = 0.42426 = 25.104 or = 25.1 !(b) + 45 + 25.104 = 180 = 109.896Using the Law of SinesFaa 200 N = sin sin 45Faa 200 N =sin109.896 sin 45orFaa = 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. 7. COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 7.Using the triangle rule and the Law of Cosines,Have: = 180 45 = 135 Then: R 2 = ( 900 ) + ( 600 ) 2 ( 900 )( 600 ) cos 1352 2 or R = 1390.57 NUsing the Law of Sines,600 1390.57 = sin sin135or = 17.7642and = 90 17.7642 = 72.236(a) = 72.2 !(b)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. 8. COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 8.By trigonometry: Law of Sines F2 R30 == sin sin 38 sin = 90 28 = 62, = 180 62 38 = 80Then:F2 R 30 lb = =sin 62 sin 38 sin 80or (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. 9. COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 9.Using the Law of Sines F1 R 20 lb == sin sin 38 sin = 90 10 = 80, = 180 80 38 = 62Then:F1R 20 lb= = sin 80 sin 38 sin 62or (a) F1 = 22.3 lb !(b) R = 13.95 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. 10. COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 10.60 N 80 NUsing the Law of Sines:=sin sin10 or = 7.4832 = 180 (10 + 7.4832 ) = 162.517Then: R 80 N =sin162.517 sin10or R = 138.405 N(a) = 7.48 !(b) R = 138.4 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. 11. COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 11.Using the triangle rule and the Law of SinesHave: = 180 ( 35 + 25 )= 120P R 80 lbThen: = = sin 35 sin120 sin 25or (a) P = 108.6 lb !(b) R = 163.9 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. 12. COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 12.Using the triangle rule and the Law of Sines 80 lb70 lb(a) Have:= sin sin 35 sin = 0.65552 = 40.959 or = 41.0 !(b) = 180 ( 35 + 40.959 ) = 104.041R 70 lbThen: = sin104.041 sin 35or R = 118.4 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. 13. COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 13.We observe that force P is minimum when = 90.Then:(a) P = ( 80 lb ) sin 35or P = 45.9 lb !And:(b) R = ( 80 lb ) cos 35or R = 65.5 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. 14. COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 14.For TBC to be a minimum,R and TBC must be perpendicular.Thus TBC = ( 70 N ) sin 4 = 4.8829 NAnd R = ( 70 N ) cos 4 = 69.829 N(a) TBC = 4.88 N6.00 !(b)R = 69.8 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. 15. COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 15.Using the force triangle and the Laws of Cosines and SinesWe have: = 180 (15 + 30 )= 135 R 2 = (15 lb ) + ( 25 lb ) 2 (15 lb )( 25 lb ) cos1352 2Then:= 1380.33 lb2or R = 37.153 lband 25 lb 37.153 lb = sin sin135 25 lb sin = sin135 37.153 lb = 0.47581 = 28.412Then: + + 75 = 180 = 76.588R = 37.2 lb 76.6 !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. 16. COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 16.Using the Law of Cosines and the Law of Sines,R 2 = ( 45 lb ) + (15 lb ) 2 ( 45 lb )(15 lb ) cos1352 2or R = 56.609 lb56.609 lb 15 lb = sin135 sinor = 10.7991R = 56.6 lb85.8 !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. 17. COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 17. = 180 25 50 = 105 Using the Law of Cosines:R 2 = ( 5 kN ) + ( 8 kN ) 2 ( 5 kN )( 8 kN ) cos1052 2or R = 10.4740 kN Using the Law of Sines:10.4740 kN 8 kN=sin105 sin or = 47.542and = 47.542 25 = 22.542R = 10.47 kN 22.5 "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. 18. COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 19.Using the force triangle and the Laws of Cosines and SinesWe have: = 180 ( 45 + 25 ) = 110R 2 = ( 30 kN ) + ( 20 kN ) 2 ( 30 kN )( 20 kN ) cos1102 2Then: = 1710.42 kN 2 R = 41.357 kNand 20 kN41.357 kN=sin sin110 20 kN sin = sin110 41.357 kN = 0.45443 = 27.028Hence: = + 45 = 72.028R = 41.4 kN 72.0 !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. 19. COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 19.Using the force triangle and the Laws of Cosines and SinesWe have: = 180 ( 45 + 25 ) = 110R 2 = ( 30 kN ) + ( 20 kN ) 2 ( 30 kN )( 20 kN ) cos1102 2Then: = 1710.42 kN 2 R = 41.357 kNand 20 kN41.357 kN=sin sin110 20 kN sin = sin110 41.357 kN = 0.45443 = 27.028Hence: = + 45 = 72.028R = 41.4 kN 72.0 !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. 20. COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 20.Using the force triangle and the Laws of Cosines and SinesWe have: = 180 ( 45 + 25 ) = 110R 2 = ( 30 kN ) + ( 20 kN ) 2 ( 30 kN )( 20 kN ) cos1102 2Then: = 1710.42 kN 2 R = 41.357 kNand 30 kN41.357 kN=sin sin110 30 kN sin = sin110 41.357 kN = 0.68164 = 42.972Finally: = + 45 = 87.972R = 41.4 kN 88.0 !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. 21. COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 21.2.4 kN Force: Fx = ( 2.4 kN ) cos 50Fx = 1.543 kN Fy = ( 2.4 kN ) sin 50Fy = 1.839 kN1.85 kN Force: Fx = (1.85 kN ) cos 20Fx = 1.738 kNFy = (1.85 kN ) sin 20Fy =