COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 1. (a) (b) We measure: 37 lb, R = 76 = 37 lb = R 76 ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 2. (a) (b) We measure: 57 lb, R = 86 = 57 lb = R 86 ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 3. (a)Parallelogram law: (b)Triangle rule: We measure: 10.5 kN R =22.5 = 10.5 kN = R 22.5! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 4. (a)Parallelogram law: We measure: 5.4 kN = 12 R = 5.4 kN = R12 ! (b)Triangle rule: We measure: 5.4 kN = 12 R = 5.4 kN = R12! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 5. Using the triangle rule and the Law of Sines (a) sin sin 45150 N 200 N =sin 0.53033 =32.028 = 45 180 + + = 103.0 = ! (b)Using the Law of Sines =45 sinN 200sinb bF 276 NbbF = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 6. Using the triangle rule and the Law of Sines (a) sin sin 45120 N 200 N =sin 0.42426 =25.104 = or 25.1 = ! (b)45 25.104 180 + + = 109.896 = Using the Law of Sines 200 Nsin sin 45aaF= 200 Nsin109.896 sin 45aaF= or 266 NaaF = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 7. Using the triangle rule and the Law of Cosines, Have: 180 45 = 135 = Then: ( ) ( ) ( )( )2 22900 600 2 900 600 cos 135 = + Ror 1390.57 N R = Using the Law of Sines, 600 1390.57sin sin135 = or 17.7642 = and 90 17.7642 = 72.236 = (a) 72.2 = ! (b) 1.391 kN R = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 8. By trigonometry: Law of Sines 230sin sin38 sinF R = = 90 28 62 , 180 62 38 80 = = = = Then: 230 lbsin62 sin38 sin80F R= = or(a)226.9 lb F = ! (b) 18.75 lb R = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 9. Using the Law of Sines 120 lbsin sin38 sinF R = = 90 10 80 , 180 80 38 62 = = = = Then: 120 lbsin80 sin38 sin 62F R= = or(a)122.3 lb F= ! (b)13.95 lb R = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 10. Using the Law of Sines: 60 N 80 Nsin sin10 = or = 7.4832 ( ) 180 10 7.4832 = + 162.517 = Then: 80 Nsin162.517 sin10R= or 138.405 N R =(a)7.48 = ! (b)138.4 N R = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 11. Using the triangle rule and the Law of Sines Have:( ) 180 35 25 = + 120 = Then: 80 lbsin35 sin120 sin 25P R= = or(a) 108.6 lb P = ! (b) 163.9 lb R = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 12. Using the triangle rule and the Law of Sines (a)Have: 80 lb 70 lbsin sin35 = sin 0.65552 =40.959 = or 41.0 = ! (b)( ) 180 35 40.959 = + 104.041 = Then: 70 lbsin104.041 sin35R= or 118.4 lb R = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 13. We observe that force P is minimum when90 . = Then: (a)( ) 80 lb sin35 P = or 45.9 lb = P ! And: (b)( ) 80 lb cos35 R = or 65.5 lb = R ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 14. For BCTto be a minimum, R and BCTmust be perpendicular. Thus( ) 70 N sin 4BCT = 4.8829 N =And( ) 70 N cos 4 R = 69.829 N =(a)4.88 NBCT = 6.00 ! (b)69.8 N R = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 15. Using the force triangle and the Laws of Cosines and Sines We have: ( ) 180 15 30 = + 135 = Then:( ) ( ) ( )( )2 2215 lb 25 lb 2 15 lb 25 lb cos135 R = + 21380.33 lb =or37.153 lb R =and 25 lb 37.153 lbsin sin135 = 25 lbsin sin13537.153 lb = 0.47581 =28.412 = Then:75 180 + + = 76.588 = 37.2 lb = R 76.6 ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 16. Using the Law of Cosines and the Law of Sines, ( ) ( ) ( )( )22 245 lb 15 lb 2 45 lb 15 lb cos135 R = + or 56.609 lb R = 56.609 lb 15 lbsin135 sin= or 10.7991 = 56.6 lb = R 85.8 ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 17. 180 25 50 = 105 = Using the Law of Cosines: ( ) ( ) ( )( )2 225 kN 8 kN 2 5 kN 8 kN cos105 R = + or 10.4740 kN R = Using the Law of Sines: 10.4740 kN 8 kNsin105 sin = or 47.542 = and = 47.542 25 22.542 = 10.47 kN = R 22.5 " COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 19. Using the force triangle and the Laws of Cosines and Sines We have:( ) 180 45 25 110 = + = Then:( ) ( ) ( )( )2 2230 kN 20 kN 2 30 kN 20 kN cos110 R = + 21710.42 kN =41.357 kN R =and 20 kN 41.357 kNsin sin110 = 20 kNsin sin11041.357 kN = 0.45443 =27.028 = Hence:45 72.028 = + = 41.4 kN = R 72.0 ! !! ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 19. Using the force triangle and the Laws of Cosines and Sines We have:( ) 180 45 25 110 = + = Then:( ) ( ) ( )( )2 2230 kN 20 kN 2 30 kN 20 kN cos110 R = + 21710.42 kN =41.357 kN R =and 20 kN 41.357 kNsin sin110 = 20 kNsin sin11041.357 kN = 0.45443 =27.028 = Hence:45 72.028 = + = 41.4 kN = R 72.0 ! !! ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 20. Using the force triangle and the Laws of Cosines and Sines We have:( ) 180 45 25 110 = + = Then:( ) ( ) ( )( )2 2230 kN 20 kN 2 30 kN 20 kN cos110 R = + 21710.42 kN =41.357 kN R =and 30 kN 41.357 kNsin sin110 = 30 kNsin sin11041.357 kN = 0.68164 =42.972 = Finally:45 87.972 = + = 41.4 kN = R 88.0 ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 21. 2.4 kN Force:( ) 2.4 kNcos50xF = 1.543 kNxF =( ) 2.4 kNsin50yF = 1.839 kNyF =1.85 kN Force:( ) 1.85 kNcos 20xF = 1.738 kNxF =( ) 1.85 kNsin 20yF = 0.633 kNyF =1.40 kN Force:( ) 1.40 kNcos35xF = 1.147 kNxF =( ) 1.40 kNsin35yF = 0.803 kNyF = COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 22. 5 kips:( ) 5 kips cos 40xF = or 3.83 kipsxF =( ) 5 kips sin 40yF = or 3.21 kipsyF =7 kips:( ) 7 kips cos 70xF = or 2.39 kipsxF = ( ) 7 kips sin70yF = or 6.58 kipsyF =9 kips:( ) 9 kips cos 20xF = or 8.46 kipsxF = ( ) 9 kips sin 20yF = or 3.08 kipsyF = COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 23. Determine the following distances: ( ) ( )( ) ( )( ) ( )2 22 22 2160 mm 300 mm 340 mm600 mm 250 mm 650 mm600 mm 110 mm 610 mmOAOBOCddd= + == + == + = 680 NForce: ( ) 160 mm680 N340 mmxF=320 NxF = ! ( ) 300 mm680 N340 mmyF =600 NyF = ! 390 NForce: ( ) 600 mm390 N650 mmxF =360 NxF = ! ( ) 250 mm390 N650 mmyF =150 NyF = ! 610 NForce: ( ) 600 mm610 N610 mmxF =600 NxF = ! ( ) 110 mm610 N610 mmyF=110 NyF = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 24. We compute the following distances: ( ) ( )( ) ( )( ) ( )2 22 22 248 90 102 in.56 90 106 in.80 60 100 in.OAOBOC= + == + == + = Then: 204 lb Force: ( ) 48204 lb ,102xF = 96.0 lbxF = ( ) 90204 lb ,102yF = + 180.0 lbyF =212 lb Force: ( ) 56212 lb ,106xF = + 112.0 lbxF =( ) 90212 lb ,106yF = + 180.0 lbyF =400 lb Force: ( ) 80400 lb ,100xF = 320 lbxF = ( ) 60400 lb ,100yF = 240 lbyF = COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 25. (a) sin35yPP = 960 Nsin35= or 1674 N P =(b) tan35yxPP = 960 Ntan35= or 1371 NxP = COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 26. (a) cos 40xPP = 30 lbcos 40P = or 39.2 lb P = ! (b)tan 40y xP P = ( ) 30 lb tan 40yP = or 25.2 lbyP = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 27. (a)100 NyP = sin 75yPP = 100 Nsin 75P = or 103.5 N P = " (b) tan75yxPP= 100 Ntan 75xP= or 26.8 NxP= " COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 28. We note: CB exerts force P on B along CB, and the horizontal component of P is260 lb.xP =Then: (a)sin50xP P = sin50xPP = 260 lbsin50= 339.40 lb = 339 lb P = ! (b)tan50x yP P = tan50xyPP = 260 lbtan50= 218.16 lb = 218 lby= P! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 29. (a) 45 Ncos 20P = or 47.9 N P = ! (b)( ) 47.9 N sin 20xP= or 16.38 NxP= ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 30. (a) 18 Nsin 20P = or 52.6 N P = ! (b) 18 Ntan 20yP = or 49.5 NyP = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 31. From the solution to Problem 2.21: ( ) ( )2.41.543 kN 1.839 kN = + F i j( ) ( )1.851.738 kN 0.633 kN = + F i j( ) ( )1.401.147 kN 0.803 kN = F i j( ) ( ) 4.428 kN 1.669 kN = = + R F i j( ) ( )2 24.428 kN 1.669 kN R = +4.7321 kN = 1.669 kNtan4.428 kN =20.652 = 4.73 kN = R 20.6 ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 32. From the solution to Problem 2.22: ( ) ( )53.83 kips 3.21 kips = + F i j( ) ( )72.39 kips 6.58 kips = + F i j( ) ( )98.46 kips 3.08 kips = + F i j( ) ( ) 7.02 kips 12.87 = = + R F i j( ) ( )2 27.02 kips 12.87 kips 14.66 kips R = + = 112.87tan 61.47.02 = = 14.66 kips = R 61.4 ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 33. From the solution to Problem 2.24: ( ) ( ) 48.0 lb 90.0 lbOA = + F i j( ) ( ) 112.0 lb 180.0 lbOB = + F i j( ) ( ) 320 lb 240 lbOC= F i j( ) ( ) 256 lb 30 lb = = + R F i j( ) ( )2 2256 lb 30 lb R = +257.75 lb = 30 lbtan256 lb= 6.6839 = 258 lb = R 6.68 ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 34. From Problem 2.23: ( ) ( ) 320 N 600 NOA = + F i j( ) ( ) 360 N 150 NOB = + F i j( ) ( ) 600 N 110 NOC= F i j( ) ( ) 640 N 640 N = = + R F i j( ) ( )2 2640 N 640 N R = +905.097 N = 640 Ntan640 N=45.0 = 905 N = R 45.0 ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 35. Cable BC Force: ( )84145 lb 105 lb116xF = = ( )80145 lb 100 lb116yF = =100-lb Force: ( )3100 lb 60 lb5xF = = ( )4100 lb 80 lb5yF = = 156-lb Force: ( )12156 lb 144 lb13xF = =( )5156 lb 60 lb13yF = = and 21 lb, 40 lbx x y yR F R F = = = = ( ) ( )2 221 lb 40 lb 45.177 lb R = + =Further: 40tan21=140tan 62.321= = Thus:45.2 lb = R 62.3 ! !! ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 36. (a)Since R is to be horizontal, Ry = 0 Then,0y yR F = =( ) ( ) 90 lb 70 lb sin 130 lb cos 0 + =( ) ( ) 13 cos 7 sin 9 = +( )213 1 sin 7 sin 9 = +Squaring both sides: ( )( ) ( )2 2169 1 sin 49 sin 126 sin 81 = + +( ) ( )2218 sin 126 sin 88 0 + =Solving by quadratic formula:sin 0.40899 =or24.1 = ! (b)Since R is horizontal, R = Rx Then, x xR F = = R( ) ( ) 70 cos 24.142 130 sin 24.142xF = + or117.0 lb R = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 37. 300-N Force: ( ) 300 N cos 20 281.91 NxF = =( ) 300 N sin 20 102.61 NyF = =400-N Force: ( ) 400 N cos85 34.862 NxF = =( ) 400 N sin85 398.48 NyF = =600-N Force: ( ) 600 N cos5 597.72 NxF = =( ) 600 N sin5 52.293 NyF = = and 914.49 Nx xR F = =448.80 Ny yR F = =( ) ( )2 2914.49 N 448.80 N 1018.68 N R = + =Further: 448.80tan914.49=1448.80tan 26.1914.49= = 1019 N = R 26.1 ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 38. xF : x xR F = ( ) ( ) ( ) 600 N cos50 300 N cos85 700 N cos50xR = + 38.132 NxR = yF : y yR F = ( ) ( ) ( ) 600 N sin 50 300 N sin85 700 N sin50yR = + + 1294.72 NyR =( ) ( )2 238.132 N 1294.72 N = + R1295 N = R1294.72 Ntan38.132 N=88.3 = 1.295 kN = R 88.3 ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 39. We have: ( ) ( )84 12 3156 lb 100 lb116 13 5x x BCR F T = = + or0.72414 84 lbx BCR T = +and ( ) ( )80 5 4156 lb 100 lb116 13 5y y BCR F T = = 0.68966 140 lby BCR T = (a)So, for R to be vertical, 0.72414 84 lb 0x BCR T = + =116.0 lbBCT = ! !! ! (b)Using 116.0 lbBCT =( ) 0.68966 116.0 lb 140 lb 60 lbyR R = = = 60.0 lb R R = = ! !! ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 40. (a)Since R is to be vertical, Rx = 0 Then,0x xR F = =( ) ( ) ( ) ( ) 600 N cos 300 N cos 35 700 N cos 0 + + =Expanding:( ) 3 cos cos35 sin sin35 cos 0 =Then: 1cos353tansin35 = 40.265 = 40.3 = ! (b)Since R is vertical, yR R =Then: y yR R F = = ( ) ( ) ( ) 600 N sin 40.265 300 N sin 75.265 700 N sin 40.265 R = + + 1130 N R =1.130 kN R = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 41. Selecting the x axis along, aa we write ( ) ( ) 300 N 400 N cos 600 N sinx xR F = = + + (1) ( ) ( ) 400 N sin 600 N cosy yR F = = (2) (a) Setting0yR =in Equation (2): Thus 600tan 1.5400= = 56.3 = ! (b) Substituting forin Equation (1): ( ) ( ) 300 N 400 N cos56.3 600 N sin56.3xR = + + 1021.11 NxR = 1021 NxR R = = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 42. (a)Require 0:y yR F = =( ) ( ) 900 lb cos 25 1200 lb sin35 sin65 0AET + =or1659.45 lbAET =1659 lbAET = ! (b) xR F = ( ) ( ) ( ) 900 lb sin 25 1200 lb cos35 1659.45 lb cos 65 R = 2060 lb R = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 43. Free-Body DiagramForce Triangle Law of Sines: 400 lbsin 25 sin 60 sin95AC BCF T= = (a) 400 lbsin 25 169.691 lbsin95ACF = =169.7 lbACF = ! (b) 400sin 60 347.73 lbsin95BCT = =348 lbBCT = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 44. Free-Body Diagram: 0:xF =4 2105 29CA CBT T + =or 29 421 5CB CAT T = 0:yF = ( )3 203 kN 05 29CA CBT T + =Then( )3 20 29 43 kN 05 29 21 5CA CAT T + = or2.2028 kNCAT =(a) 2.20 kNCAT = ! (b) 2.43 kNCBT = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 45. Free-Body Diagram: 0:yF = sin50 sin 70 0B CF F + =( )sin50sin70C BF F= 0:xF = cos50 cos 70 940 N 0B CF F + = sin50cos50 cos 70 940sin 70BF | | + = |\ . 1019.96 NBF =( )sin501019.96 Nsin 70CF= or831 NCF = ! 1020 NBF = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 46. Free-Body Diagram: 0:xF = ( ) cos 25 cos 40 70 lb cos10 0AB ACT T + = (1) 0:yF = ( ) sin 25 sin 40 70 lb sin10 0AB ACT T + = (2) Solving Equations (1) and (2) simultaneously: (a)38.6 lbABT = ! (b)44.3 lbACT = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 47. Free-Body Diagram: (a)0:xF = cos30 cos65 0ABT R + = cos30cos65ABR T= 0:yF = ( ) sin30 sin 65 550 N 0ABT R + = cos30sin30 sin 65 550 0cos65ABT + = or405 NABT = ! (b)( )cos30450 Ncos65R= or830 N R = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 48. Free-Body Diagram At B: 12 170: 013 293x BA BCF T T = + =or1.07591BA BCT T =0:yF =5 2300 N = 013 293BA BCT T + 5 29330013 2BC BAT T = 2567.6 3.2918BC BAT T = ( ) 2567.6 3.2918 1.07591BC BCT T = or565.34 NBCT =Free-Body Diagram At C: 17 240: 025 293x BC CDF T T = + =( )17 25565.34 N24 293CDT = 584.86 NCDT = 2 70: 025 293y BC CD CF T T W = + =( ) ( )2 7565.34 N 584.86 N25 293CW = +or97.7 NCW = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 49. Free-Body Diagram: 0:xF =8 kips + 15 kips cos 40 0DT =9.1378 kipsDT =9.14 kipsDT = ! 0:yF = ( ) 9.1378 kips sin 40 0CT =5.87 kipsCT = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 50. Free-Body Diagram: 0:yF =9 kips + T sin 40 0D =14.0015 kipsDT =14.00 kipsDT = 0:xF =( ) 6 kips + 14.0015 kips cos 40 0BT =16.73 kipsBT =16.73 kipsBT = COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 51. Free-Body Diagram: 0:xF = ( ) ( ) 2.3 kNsin15 2.1 kNcos15 0CF + =or1.433 kNCF =0:yF = ( ) ( ) 2.3 kNcos15 2.1 kNsin15 0DF + =or1.678 kNDF = COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 52. Free-Body Diagram: 0:xF = ( ) cos15 2.4 kN +1.9kNsin15 0BF + =or2.9938 kNBF =2.99 kNBF =0:yF = ( ) ( ) 1.9 kNcos15 2.9938 kNsin15 0DF + =1.060 kNDF = COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 53. From Similar Triangles we have: ( ) ( ) ( )2 2 222.5 m 8 5.45 m L L = 6.25 64 16 29.7025 L = or2.5342 m L =And 5.45 mcos8 m 2.5342 m= or4.3576 = Then2.5 mcos2.5342 m =or9.4237 = Free-Body Diagram At B: 0:xF =( ) cos 35 N cos cos 0ABC ABCT T + =or ( ) 35 cos9.4237cos 4.3576 cos9.4237ABCT= 3255.9 NABCT =0:yF =( ) sin 35 N sin sin 0ABC ABCT T W + + =( ) ( ) sin9.4237 3255.9 N + 35 N 3255.9 N sin 4.3576 0 W + =or786.22 N W=(a)786 N W= " (b)3.26 kNABCT = " COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 54. From Similar Triangles we have: ( ) ( ) ( )2 2 223 m 8 4.95 m L L = 9 64 16 24.5025 L = or3.0311 m L =Then 4.95 mcos8 m 3.0311 m= or4.9989 = And 3 mcos3.0311 m =or8.2147 = Free-Body Diagram At B: (a)0:xF =cos cos cos 0ABC DE ABCT T T + =or cos coscosDE ABCT T =0:yF =( ) sin sin sin 720 N 0ABC DE ABCT T T + + = cos cossin sin sin 720cosABCT | |+ + = |\ . ( )( )720 cossinABCT =+ Substituting forandgives ( )( )720 cos8.2147sin 8.2147 4.9989ABCT= + 3117.5 NABCT =or3.12 kNABCT = " (b)( )cos 4.9989 cos8.21473117.5 Ncos8.2147DET = 20.338 NDET =or20.3 NDET = " COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 55. Free-Body Diagram At C: 0:xF = ( )3 15 15150 lb 05 17 17AC BCT T + =or 175 7505AC BCT T + = (1) 0:yF = ( )4 8 8150 lb 190 lb = 05 17 17AC BCT T + or 172 1107.55AC BCT T + = (2) Then adding Equations (1) and (2) 7 1857.5BCT =or265.36 lbBCT =Therefore(a)169.6 lbACT = ! (b)265 lbBCT = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 56. Free-Body Diagram At C: 0:xF = ( )3 15 15150 lb 05 17 17AC BCT T + =or 175 7505AC BCT T + = (1) 0:yF = ( )4 8 8150 lb = 05 17 17AC BCT T W + or 17 172 3005 4AC BCT T W + = + (2) Adding Equations (1) and (2) gives 177 10504BCT W = +or 1715028BCT W = +Using Equation (1) 17 175 150 7505 28ACT W + + = or 2528ACT W =Now for 240 lb T 25: 24028ACT W =or269 lb W= 17: 240 15028BCT W = +or148.2 lb W=Therefore0 148.2 lb W ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 57. Free-Body Diagram At A: First note from geometry: The sides of the triangle with hypotenuse AD are in the ratio 12:35:37. The sides of the triangle with hypotenuse AC are in the ratio 3:4:5. The sides of the triangle with hypotenuse AB are also in the ratio 12:35:37. Then: ( ) ( )4 35 120: 3 05 37 37x sF W W F = + + =or 4.4833sF W =and ( ) ( )3 12 350: 3 400 N 05 37 37y sF W W F = + + =Then: ( ) ( ) ( )3 12 353 4.4833 400 N 05 37 37W W W + + =or 62.841 N W=and 281.74 NsF =or (a)62.8 N W=(b)Have spring force ( )s AB OF k L L = Where ( )AB AB AB OF k L L = and ( ) ( )2 20.360 m 1.050 m 1.110 mABL = + =So: ( ) 281.74 N 800 N/m 1.110 mOL = or758 mmOL = COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 58. Free-Body Diagram At A: First Note ... With( ) ( )2 222 in. 16.5 in.ABL = + 27.5 in.ABL = ( ) ( )2 230 in. 16 in.ADL = + 34 in.ADL = Then ( )AB AB AB OF k L L = ( )( ) 9 lb/in. 27.5 in. 22.5 in. = 45 lb = ( )AD AD AD OF k L L = ( )( ) 3 lb/in. 34 in. 22.5 in. = 34.5 lb = (a) 0:xF = ( ) ( )4 7 1545 lb 34.5 lb 05 25 17ACT + + =or 19.8529 lbACT =19.85 lbACT = ! (b) 0:yF = ( ) ( ) ( )3 24 845 lb 19.8529 lb 34.5 lb 05 25 17W + + =62.3 lb W= ! COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 59. (a)For to be a minimum ABT must be perpendicular to ABTACT10 60 + = or50.0 = W (b)Then( ) 70 lb sin30ABT = or35.0 lbABT = W

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 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. Chapter 2, Solution 60. Note:In problems of this type,P may be directed along one of the cables, with maxT T =in that cable and 0 T=in the other, orPmay be directed in such a way that T is maximum in both cables. The second possibility is investigated first. Free-Body Diagram At C: Force Triangle Force triangle is isoceles with2 180 85 = 47.5 = ( ) 2 900 N cos 47.5 1216 N P = = Since0, P >solution is correct (a)1216 N P = ! 180 55 47.5 77.5 = = (b)77.5 = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 61. Note: Refer to Note in Problem 2.60 Free-Body Diagram At C: Force Triangle (a)Law of Cosines ( ) ( ) ( )( )2 221400 N 700 N 2 1400 N 700 N cos85 = + Por1510 N = P ! (b)Law of Sines sin sin851400 N 1510 N =sin 0.92362 =67.461 = 180 55 67.461 = or57.5 = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 62. Free-Body Diagram At C: 0:xF =2 1200 N 0 =xT600 N =xT( ) ( )222+ =x yT T T( ) ( ) ( )22 2600 N 870 N + =yT630 N =yTBy similar triangles: 1.8 m870 N 630 N=AC 2.4857 m AC =2( ) = L AC( ) 2 2.4857 m = L4.97 m L =4.97 m L = " COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 63. BCTmust be perpendicular to ACFto be as small as possible. Free-Body Diagram: CForce Triangle is a Right Triangle (a)We observe:55 = 55 = ! (b)( ) 400 lb sin 60BCT = or346.41 lb =BCT 346 lbBCT = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 64. At Collar A ... Have( ) = s AB ABF k L LFor stretched length ( ) ( )2 212 in. 16 in. = +ABL20 in.ABL =For unstretched length 12 2 in.ABL =Then ( )4 lb/in. 20 12 2 in.sF= 12.1177 lb =sFFor the collar ... 0yF =( )412.1177 lb 05W + =9.69 lb = W ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 65. At Collar A ... 0:yF = 2 29 lb 012shFh + =+ or 29144 = +shF hNow( ) = s AB ABF k L LWhere the stretched length ( )2212 in. = +ABL h12 2 in. =ABL Then 29144 = +shF hBecomes ( )2 23 lb/in. 144 12 2 9144 + = + h h hor( )23 144 12 2 + = h h hSolving Numerically ...16.81 in. = h COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 66. Free-Body Diagram: B (a)Have:0BD AB BC+ + = T F Twhere magnitude and direction of BDTare known, and the direction of ABFis known. Then, in a force triangle: By observation, BCTis minimum when 90.0 = (b)Have( ) ( ) 310 N sin 180 70 30 = BCT305.29 N =305 N =BCT COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 67. Free-Body Diagram At C: Since140 lb,AB BCT T = =Force triangle is isosceles: With 2 75 180 + = 52.5 = Then90 52.5 30 = 7.50 = ( ) 140 lbcos52.52= P 170.453 lb = P170.5 lb = P 7.50 COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 68. Free-Body Diagram of Pulley (a) (b) (c) (d) (e) ( )( )20: 2 280 kg 9.81 m/s 0yF T = =( )12746.8 N2T=1373 N T= ( )( )20: 2 280 kg 9.81 m/s 0yF T = =( )12746.8 N2T=1373 N T= ( )( )20: 3 280 kg 9.81 m/s 0yF T = =( )12746.8 N3T=916 N T= ( )( )20: 3 280 kg 9.81 m/s 0yF T = =( )12746.8 N3T=916 N T= ( )( )20: 4 280 kg 9.81 m/s 0yF T = =( )12746.8 N4T=687 N T= COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 69. Free-Body Diagram of Pulley and Crate (b) (d) ( )( )20: 3 280 kg 9.81 m/s 0yF T = =( )12746.8 N3T=916 N T= ( )( )20: 4 280 kg 9.81 m/s 0yF T = =( )12746.8 N4T=687 N T= COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 70. Free-Body Diagram: Pulley C (a)( ) ( ) 0: cos30 cos50 800 Ncos50 0x ACBF T = =Hence2303.5 NACBT =2.30 kN =ACBT(b)( ) ( ) 0: sin30 sin50 800 Nsin50 0y ACBF T Q = + + = ( )( ) ( ) 2303.5 N sin30 sin50 800 N sin50 0 Q + + =or3529.2 N Q = 3.53 kN = Q COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 71. Free-Body Diagram: Pulley C ( ) 0: cos30 cos50 cos50 0x ACBF T P = =or0.34730 =ACBP T (1) ( ) 0: sin30 sin50 sin50 2000 N 0y ACBF T P = + + =or1.26604 0.76604 2000 N + =ACBT P (2) (a) Substitute Equation (1) into Equation (2): ( ) 1.26604 0.766040.34730 2000 N + =ACB ACBT THence:1305.41 N =ACBT 1305 N =ACBT(b) Using (1) ( ) 0.34730 1305.41 N 453.37 N = = P 453 N = P COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 72. First replace 30 lb forces by their resultant Q: ( ) 2 30 lb cos 25 Q = 54.378 lb Q = Equivalent loading at A: Law of Cosines: ( ) ( ) ( ) ( )( ) ( ) ( )2 2 2120 lb 100 lb 54.378 lb 2 100 lb 54.378 lb cos 125 cos 125 0.132685 = + = This gives two values:125 97.625 = 27.4 = 125 97.625 = 223 = Thus for 120 lb: < R27.4 223 < < ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 73. (a)( ) 950 lb sin50 cos 40xF = 557.48 lb =557 lb =xF ! ( ) 950 lb cos50 = yF610.65 lb = 611 lb = yF ! ( ) 950 lb sin50 sin 40zF = 467.78 lb =468 lb =zF ! (b) 557.48 lbcos950 lb =x or 54.1 = x! 610.65 lbcos950 lb=y or 130.0 = y! 467.78 lbcos950 lb =z or 60.5 = z! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 74. (a)( ) 810 lb cos 45 sin 25 = xF242.06 lb = 242 lb = xF ! ( ) 810 lb sin 45 = yF572.76 lb = 573 lb = yF ! ( ) 810 lb cos 45 cos 25 = zF519.09 lb =519 lb =zF ! (b) 242.06 lbcos810 lb=x or 107.4 = x! 572.76 lbcos810 lb=y or 135.0 = y! 519.09 lbcos810 lb =z or 50.1 = z! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 75. (a)( ) 900 N cos30 cos 25 = xF706.40 N =706 N =xF ! ( ) 900 N sin30 = yF450.00 N =450 N =yF ! ( ) 900 N cos30 sin 25 = zF329.04 N = 329 N = zF ! (b) 706.40 Ncos900 N =x or 38.3 = x! 450.00 Ncos900 N =y or 60.0 = y! 329.40 Ncos900 N=z or 111.5 = z! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 76. (a)( ) 1900 N sin 20 sin 70xF = 610.65 N = 611 NxF = ! ( ) 1900 N cos 20yF = 1785.42 N =1785 NyF = ! ( ) 1900 N sin 20 cos 70zF = 222.26 N =222 NzF = ! (b) 610.65 Ncos1900 Nx=or 108.7x = ! 1785.42 Ncos1900 Ny =or 20.0y = ! 222.26 Ncos1900 Nz =or 83.3z = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 77. (a)( ) 180 lb cos35 sin 20xF = 50.430 lb =50.4 lbxF = ! ( ) 180 lb sin35yF = 103.244 lb = 103.2 lbyF = ! ( ) 180 lb cos35 cos 20zF = 138.555 lb =138.6 lbzF = ! (b) 50.430 lbcos180 lbx =or 73.7x = ! 103.244 lbcos180 lby=or 125.0y = ! 138.555 lbcos180 lbz =or 39.7z = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 78. (a)( ) 180 lb cos30 cos 25xF = 141.279 lb =141.3 lbxF = ! ( ) 180 lb sin30yF = 90.000 lb = 90.0 lbyF = ! ( ) 180 lb cos30 sin 25zF = 65.880 lb =65.9 lbzF = ! (b) 141.279 lbcos180 lbx =or 38.3x = ! 90.000 lbcos180 lby=or 120.0y = ! 65.880 lbcos180 lbz =or 68.5z = ! COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 79. (a) ( ) 220 N cos60 cos35xF = 90.107 N = 90.1 NxF = W( ) 220 N sin 60yF = 190.526 N =190.5 NyF = W( ) 220 N cos 60 sin35zF = 63.093 N = 63.1 NzF = W(b) 90 107 cos220 Nx .=114.2x = W 190.526 Ncos220 Ny =30.0y = W 63.093 Ncos220 Nz=106.7z = W

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 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. Chapter 2, Solution 80. (a)180 NxF =Withcos60 cos35xF F = 180 N = cos60 cos35 F or 439.38 N F =439 N F = ! (b) 180 Ncos439.48 Nx =65.8x = ! ( ) 439.48 N sin 60yF = 380.60 NyF = 380.60 Ncos439.48 Ny =30.0y = ! ( ) 439.48 N cos60 sin35zF = 126.038 NzF = 126.038 Ncos439.48 Nz=106.7z = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 81. 2 2 2x y zF F F F = + +( ) ( ) ( )2 2 265 N 80 N 200 N F = + + 225 N F= ! 65 Ncos225 NxxFF = =73.2x = ! 80 Ncos225 NyyFF= =110.8y = ! 200 Ncos225 NzzFF= =152.7z = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 82. 2 2 2x y zF F F F = + +( ) ( ) ( )2 2 2450 N 600 N 1800 N F= + + 1950 N F= ! 450 Ncos1950 NxxFF = =76.7x = ! 600 Ncos1950 NyyFF = =72.1y = ! 1800 Ncos1950 NzzFF= =157.4z = ! COSMOS: Complete Online Solutions Manual Organization System 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. Chapter 2, Solution 83. (a)We have( ) ( ) ( )22 2cos cos cos 1x y z + + = ( ) ( ) ( )22 2cos 1 cos cosy x z = Since0yF