1. PROBLEM 2.1Two forces are applied to an eye bolt fastened to a beam. Determinegraphically the magnitude and direction of their resultant using (a) theparallelogram law, (b) the triangle rule.SOLUTION(a)(b)We measure: R = 8.4 kN = 19R = 8.4 kN 19W1 2. PROBLEM 2.2The cable stays AB and AD help support pole AC. Knowing that thetension is 500 N in AB and 160 N in AD, determine graphically themagnitude and direction of the resultant of the forces exerted by the staysat A using (a) the parallelogram law, (b) the triangle rule.SOLUTIONWe measure: = 51.3, = 59(a)(b)We measure: R = 575 N, = 67R = 575 N 67W2 3. PROBLEM 2.3Two forces P and Q are applied as shown at point A of a hook support.Knowing that P = 15 lb and Q = 25 lb, determine graphically themagnitude and direction of their resultant using (a) the parallelogram law,(b) the triangle rule.SOLUTION(a)(b)We measure: R = 37 lb, = 76R = 37 lb 76W3 4. PROBLEM 2.4Two forces P and Q are applied as shown at point A of a hook support.Knowing that P = 45 lb and Q = 15 lb, determine graphically themagnitude and direction of their resultant using (a) the parallelogram law,(b) the triangle rule.SOLUTION(a)(b)We measure: R = 61.5 lb, = 86.5R = 61.5 lb 86.5W4 5. PROBLEM 2.5Two control rods are attached at A to lever AB. Using trigonometry andknowing that the force in the left-hand rod is F1 = 120 N, determine(a) the required force F2 in the right-hand rod if the resultant R of theforces exerted by the rods on the lever is to be vertical, (b) thecorresponding magnitude of R.SOLUTIONGraphically, by the triangle lawWe measure: F2 108 NR 77 NBy trigonometry: Law of SinesF R 2 = =120sin sin 38 sin = 90 28 = 62, = 180 62 38 = 80Then:F = R =2 120 Nsin 62 sin 38 sin80 or (a) F2 = 107.6 N W(b) R = 75.0 NW5 6. PROBLEM 2.6Two control rods are attached at A to lever AB. Using trigonometry andknowing that the force in the right-hand rod is F2 = 80 N, determine(a) the required force F1 in the left-hand rod if the resultant R of theforces exerted by the rods on the lever is to be vertical, (b) thecorresponding magnitude of R.SOLUTIONUsing the Law of SinesF R 1 = =80sin sin 38 sin = 90 10 = 80, = 180 80 38 = 62Then:F = R =1 80 Nsin80 sin 38 sin 62 or (a) F1 = 89.2 N W(b) R = 55.8 N W6 7. PROBLEM 2.7The 50-lb force is to be resolved into components along lines a-a andb-b. (a) Using trigonometry, determine the angle knowing that thecomponent along a-a is 35 lb. (b) What is the corresponding value ofthe component along b-b?SOLUTIONUsing the triangle rule and the Law of Sines(a) sin sin 40 =35 lb 50 lbsin = 0.44995 = 26.74Then: + + 40 = 180 = 113.3W(b) Using the Law of Sines:50 lbFbb =sin sin 40Fbb = 71.5 lb W7 8. PROBLEM 2.8The 50-lb force is to be resolved into components along lines a-a andb-b. (a) Using trigonometry, determine the angle knowing that thecomponent along b-b is 30 lb. (b) What is the corresponding value ofthe component along a-a?SOLUTIONUsing the triangle rule and the Law of Sines(a) sin sin 40 =30 lb 50 lbsin = 0.3857 = 22.7W(b) + + 40 = 180 = 117.3150 lbFaa =sin sin 4050 lb sinFaa= sin 40 Faa = 69.1 lbW8 9. PROBLEM 2.9To steady a sign as it is being lowered, two cables are attached to the signat A. Using trigonometry and knowing that = 25, determine (a) therequired magnitude of the force P if the resultant R of the two forcesapplied at A is to be vertical, (b) the corresponding magnitude of R.SOLUTIONUsing the triangle rule and the Law of SinesHave: = 180 (35 + 25)= 120P = R =Then: 360 Nsin 35 sin120 sin 25 or (a) P = 489 N W(b) R = 738 NW9 10. PROBLEM 2.10To steady a sign as it is being lowered, two cables are attached to the signat A. Using trigonometry and knowing that the magnitude of P is 300 N,determine (a) the required angle if the resultant R of the two forcesapplied at A is to be vertical, (b) the corresponding magnitude of R.SOLUTIONUsing the triangle rule and the Law of Sines(a) Have: 360 N =300 Nsin sin 35sin = 0.68829 = 43.5W(b) = 180 (35 + 43.5)= 101.5R =Then: 300 Nsin101.5 sin 35 or R = 513 NW10 11. PROBLEM 2.11Two forces are applied as shown to a hook support. Using trigonometryand knowing that the magnitude of P is 14 lb, determine (a) the requiredangle if the resultant R of the two forces applied to the support is to behorizontal, (b) the corresponding magnitude of R.SOLUTIONUsing the triangle rule and the Law of Sines(a) Have: 20 lb =14 lbsin sin 30sin = 0.71428 = 45.6W(b) = 180 (30 + 45.6)= 104.4R =Then: 14 lbsin104.4 sin 30 R = 27.1 lbW11 12. PROBLEM 2.12For the hook support of Problem 2.3, using trigonometry and knowingthat the magnitude of P is 25 lb, determine (a) the required magnitude ofthe force Q if the resultant R of the two forces applied at A is to bevertical, (b) the corresponding magnitude of R.Problem 2.3: Two forces P and Q are applied as shown at point A of ahook support. Knowing that P = 15 lb and Q = 25 lb, determinegraphically the magnitude and direction of their resultant using (a) theparallelogram law, (b) the triangle rule.SOLUTIONUsing the triangle rule and the Law of SinesQ =(a) Have: 25 lbsin15 sin 30 Q = 12.94 lbW(b) = 180 (15 + 30)= 135R =Thus: 25 lbsin135 sin 30 25 lb sin135 35.36 lbR = = sin 30 R = 35.4 lbW12 13. PROBLEM 2.13For the hook support of Problem 2.11, determine, using trigonometry,(a) the magnitude and direction of the smallest force P for which theresultant R of the two forces applied to the support is horizontal,(b) the corresponding magnitude of R.Problem 2.11: Two forces are applied as shown to a hook support. Usingtrigonometry and knowing that the magnitude of P is 14 lb, determine(a) the required angle if the resultant R of the two forces applied to thesupport is to be horizontal, (b) the corresponding magnitude of R.SOLUTION(a) The smallest force P will be perpendicular to R, that is, verticalP = (20 lb)sin 30= 10 lb P = 10 lb W(b) R = (20 lb)cos30= 17.32 lb R = 17.32 lbW13 14. PROBLEM 2.14As shown in Figure P2.9, two cables are attached to a sign at A to steadythe sign as it is being lowered. Using trigonometry, determine (a) themagnitude and direction of the smallest force P for which the resultant Rof the two forces applied at A is vertical, (b) the corresponding magnitudeof R.SOLUTIONWe observe that force P is minimum when is 90, that is, P is horizontalThen: (a) P = (360 N)sin 35or P = 206 N WAnd: (b) R = (360 N)cos35or R = 295 NW14 15. PROBLEM 2.15For the hook support of Problem 2.11, determine, using trigonometry, themagnitude and direction of the resultant of the two forces applied to thesupport knowing that P = 10 lb and = 40.Problem 2.11: Two forces are applied as shown to a hook support. Usingtrigonometry and knowing that the magnitude of P is 14 lb, determine(a) the required angle if the resultant R of the two forces applied to thesupport is to be horizontal, (b) the corresponding magnitude of R.SOLUTIONUsing the force triangle and the Law of CosinesR2 = (10 lb)2 + (20 lb)2 2(10 lb)(20 lb)cos110= 100 + 400 400(0.342) lb2= 636.8 lb2R = 25.23 lbUsing now the Law of Sines10 lb =25.23 lbsin sin110sin 10 lb sin110 = 25.23 lb = 0.3724So: = 21.87Angle of inclination of R, is then such that: + = 30 = 8.13Hence: R = 25.2 lb 8.13W15 16. PROBLEM 2.16Solve Problem 2.1 using trigonometryProblem 2.1: Two forces are applied to an eye bolt fastened to a beam.Determine graphically the magnitude and direction of their resultantusing (a) the parallelogram law, (b) the triangle rule.SOLUTIONUsing the force triangle, the Law of Cosines and the Law of SinesWe have: = 180 (50 + 25)= 105Then: R2 = (4.5 kN)2 + (6 kN)2 2(4.5 kN)(6 kN)cos105= 70.226 kN2or R = 8.3801 kNNow: 8.3801 kN =6 kNsin105 sin sin 6 kN sin105 = 8.3801 kN = 0.6916 = 43.756R = 8.38 kN 18.76W16 17. PROBLEM 2.17Solve Problem 2.2 using trigonometryProblem 2.2: The cable stays AB and AD help support pole AC. Knowingthat the tension is 500 N in AB and 160 N in AD, determine graphicallythe magnitude and direction of the resultant of the forces exerted by thestays at A using (a) the parallelogram law, (b) the triangle rule.SOLUTIONFrom the geometry of the problem:tan 1 2 38.66 = = 2.5tan 1 1.5 30.96 = = 2.5Now: = 180 (38.66 + 30.96) = 110.38And, using the Law of Cosines:R2 = (500 N)2 + (160 N)2 2(500 N)(160 N)cos110.38= 331319 N2R = 575.6 NUsing the Law of Sines:160 N =575.6 Nsin sin110.38sin 160 N sin110.38 = 575.6 N = 0.2606 = 15.1 = (90 ) + = 66.44R = 576 N 66.4W17 18. PROBLEM 2.18Solve Problem 2.3 using trigonometryProblem 2.3: Two forces P and Q are applied as shown at point A of ahook support. Knowing that P = 15 lb and Q = 25 lb, determinegraphically the magnitude and direction of their resultant using (a) theparallelogram law, (b) the triangle rule.SOLUTIONUsing the force triangle and the Laws of Cosines and SinesWe have: = 180 (15 + 30)= 135Then: R2 = (15 lb)2 + (25 lb)2 2(15 lb)(25 lb)cos135= 1380.3 lb2or R = 37.15 lband25 lb =37.15 lbsin sin135sin 25 lb sin135 = 37.15 lb = 0.4758 = 28.41Then: + + 75 = 180 = 76.59R = 37.2 lb 76.6W18 19. PROBLEM 2.19Two structural members A and B are bolted to a bracket as shown.Knowing that both members are in compression and that the force is30 kN in member A and 20 kN in member B, determine, usingtrigonometry, the magnitude and direction of the resultant of the forcesapplied to the bracket by members A and B.SOLUTIONUsing the force triangle and the Laws of Cosines and SinesWe have: = 180 (45 + 25) = 110Then: R2 = (30 kN)2 + (20 kN)2 2(30 kN)(20 kN)cos110= 1710.4 kN2R = 41.357 kNand20 kN =41.357 kNsin sin110sin 20 kN sin110 = 41.357 kN = 0.4544 = 27.028Hence: = + 45 = 72.028R = 41.4 kN 72.0W19 20. PROBLEM 2.20Two structural members A and B are bolted to a bracket as shown.Knowing that both members are in compression and that the force is20 kN in member A and 30 kN in member B, determine, usingtrigonometry, the magnitude and direction of the resultant of the forcesapplied to the bracket by members A and B.SOLUTIONUsing the force triangle and the Laws of Cosines and SinesWe have: = 180 (45 + 25) = 110Then: R2 = (30 kN)2 + (20 kN)2 2(30 kN)(20 kN)cos110= 1710.4 kN2R = 41.357 kNand30 kN =41.357 kNsin sin110sin 30 kN sin110 = 41.357 kN = 0.6816 = 42.97Finally: = + 45 = 87.97R = 41.4 kN 88.0W20 21. PROBLEM 2.21Determine the x and y components of each of the forces shown.SOLUTION20 kN Force:Fx = +(20 kN)cos 40, Fx = 15.32 kN WFy = +(20 kN)sin 40, Fy = 12.86 kNW30 kN Force:Fx = (30 kN)cos70, Fx = 10.26 kN WFy = +(30 kN)sin 70, Fy = 28.2 kNW42 kN Force:Fx = (42 kN)cos 20, Fx = 39.5 kNWFy = +(42 kN)sin 20, Fy = 14.36 kNW21 22. PROBLEM 2.22Determine the x and y components of each of the forces shown.SOLUTION40 lb Force:Fx = (40 lb)sin 50, Fx = 30.6 lbWFy = (40 lb)cos50, Fy = 25.7 lbW60 lb Force:Fx = +(60 lb)cos60, Fx = 30.0 lbWFy = (60 lb)sin 60, Fy = 52.0 lbW80 lb Force:Fx = +(80 lb)cos 25, Fx = 72.5 lbWFy = +(80 lb)sin 25, Fy = 33.8 lb W22 23. PROBLEM 2.23Determine the x and y components of each of the forces shown.SOLUTIONWe compute the following distances:2 2( ) ( )( ) ( )( ) ( )48 90 102 in.56 2 90 2106 in.80 2 60 2100 in.OAOBOC= + == + == + =Then:204 lb Force:(102 lb) 48 ,x 102 F = Fx = 48.0 lbW(102 lb) 90 ,y 102 F = + Fy = 90.0 lb W212 lb Force:(212 lb) 56 ,x 106 F = + Fx = 112.0 lbW(212 lb) 90 ,y 106 F = + Fy = 180.0 lbW400 lb Force:(400 lb) 80 ,x 100 F = Fx = 320 lb W(400 lb) 60 ,y 100 F = Fy = 240 lbW23 24. PROBLEM 2.24Determine the x and y components of each of the forces shown.SOLUTIONWe compute the following distances:( )2 ( )2 OA = 70 + 240 = 250 mm( )2 ( )2 OB = 210 + 200 = 290 mm( )2 ( )2 OC = 120 + 225 = 255 mm500 N Force:500 N 70x 250 F = Fx = 140.0 N W500 N 240y 250 F = + Fy = 480 N W435 N Force:435 N 210x 290 F = + Fx = 315 N W435 N 200y 290 F = + Fy = 300 NW510 N Force:510 N 120x 255 F = + Fx = 240 NW510 N 225y 255 F = Fy = 450 N W24 25. PROBLEM 2.25While emptying a wheelbarrow, a gardener exerts on each handle AB aforce P directed along line CD. Knowing that P must have a 135-Nhorizontal component, determine (a) the magnitude of the force P, (b) itsvertical component.SOLUTION(a)P = Pxcos 40135 Ncos 40=or P = 176.2 NW(b) Py = Px tan 40 = Psin 40= (135 N)tan 40or Py = 113.3 NW25 26. PROBLEM 2.26Member BD exerts on member ABC a force P directed along line BD.Knowing that P must have a 960-N vertical component, determine (a) themagnitude of the force P, (b) its horizontal component.SOLUTION(a)Pysin 35P =960 Nsin 35=or P = 1674 N W(b)ytan 35xPP =960 Ntan 35=or Px = 1371N W26 27. PROBLEM 2.27Member CB of the vise shown exerts on block B a force P directed alongline CB. Knowing that P must have a 260-lb horizontal component,determine (a) the magnitude of the force P, (b) its vertical component.SOLUTIONWe note:CB exerts force P on B along CB, and the horizontal component of P is Px = 260 lb.Then:(a) Px = Psin 50P = Pxsin 50260 lbsin50== 339.4 lb P = 339 lbW(b) Px = Py tan 50P = Pxtan 50y260 lbtan 50== 218.2 lb Py = 218 lb W27 28. PROBLEM 2.28Activator rod AB exerts on crank BCD a force P directed along line AB.Knowing that P must have a 25-lb component perpendicular to arm BC ofthe crank, determine (a) the magnitude of the force P, (b) its componentalong line BC.SOLUTIONUsing the x and y axes shown.(a) Py = 25 lbThen:Pysin 75P =25 lbsin 75=or P = 25.9 lbW(b)ytan 75xPP =25 lbtan 75=or Px = 6.70 lb W28 29. PROBLEM 2.29The guy wire BD exerts on the telephone pole AC a force P directedalong BD. Knowing that P has a 450-N component along line AC,determine (a) the magnitude of the force P, (b) its component in adirection perpendicular to AC.SOLUTIONNote that the force exerted by BD on the pole is directed along BD, and the component of P along ACis 450 N.Then:(a) P = 450 N =549.3 Ncos35P = 549 NW(b) Px = (450 N)tan 35= 315.1NPx = 315 N W29 30. PROBLEM 2.30The guy wire BD exerts on the telephone pole AC a force P directedalong BD. Knowing that P has a 200-N perpendicular to the pole AC,determine (a) the magnitude of the force P, (b) its component alongline AC.SOLUTION(a)P = Pxsin 38200 Nsin 38== 324.8 N or P = 325 NW(b)P = Pxtan 38y200 Ntan 38== 255.98 Nor Py = 256 N W30 31. PROBLEM 2.31Determine the resultant of the three forces of Problem 2.24.Problem 2.24: Determine the x and y components of each of the forcesshown.SOLUTIONFrom Problem 2.24:F500 = (140 N)i + (480 N) j( ) ( ) F425 = 315 N i + 300 N j( ) ( ) F510 = 240 N i 450 N jR = F = (415 N)i + (330 N) jThen:tan 1 330 38.5 = = 415( )2 ( )2 R = 415 N + 330 N = 530.2 NThus: R = 530 N 38.5W31 32. PROBLEM 2.32Determine the resultant of the three forces of Problem 2.21.Problem 2.21: Determine the x and y components of each of the forcesshown.SOLUTIONFrom Problem 2.21:F20 = (15.32 kN)i + (12.86 kN) j( ) ( ) F30 = 10.26 kN i + 28.2 kN j( ) ( ) F42 = 39.5 kN i + 14.36 kN jR = F = (34.44 kN)i + (55.42 kN) jThen:tan 1 55.42 58.1 = = 34.44( )2 ( )2 R = 55.42 kN + 34.44 N = 65.2 kNR = 65.2 kN 58.2W32 33. PROBLEM 2.33Determine the resultant of the three forces of Problem 2.22.Problem 2.22: Determine the x and y components of each of the forcesshown.SOLUTIONThe components of the forces were determined in 2.23.R = Rxi + Ry j= (71.9 lb)i (43.86 lb) jtan 43.8671.9 = = 31.38( )2 ( )2 R = 71.9 lb + 43.86 lb= 84.23 lbR = 84.2 lb 31.4WForce x comp. (lb) y comp. (lb)40 lb 30.6 25.760 lb 30 51.9680 lb 72.5 33.8Rx = 71.9 Ry = 43.8633 34. PROBLEM 2.34Determine the resultant of the three forces of Problem 2.23.Problem 2.23: Determine the x and y components of each of the forcesshown.SOLUTIONThe components of the forces weredetermined in Problem 2.23.F204 = (48.0 lb)i + (90.0 lb) j( ) ( ) F212 = 112.0 lb i + 180.0 lb j( ) ( ) F400 = 320 lb i 240 lb jThusR = Rx + RyR = (256 lb)i + (30.0 lb) jNow:tan 30.0256 =tan 1 30.0 6.68 = = 256and( )2 ( )2 R = 256 lb + 30.0 lb= 257.75 lbR = 258 lb 6.68W34 35. PROBLEM 2.35Knowing that = 35, determine the resultant of the three forcesshown.SOLUTION300-N Force:Fx = (300 N)cos 20 = 281.9 NFy = (300 N)sin 20 = 102.6 N400-N Force:Fx = (400 N)cos55 = 229.4 NFy = (400 N)sin 55 = 327.7 N600-N Force:Fx = (600 N)cos35 = 491.5 NFy = (600 N)sin 35 = 344.1 NandRx = Fx = 1002.8 NRy = Fy = 86.2 N( )2 ( )2 R = 1002.8 N + 86.2 N = 1006.5 NFurther:tan 86.21002.8 =tan 1 86.2 4.91 = = 1002.8R = 1007 N 4.91W35 36. PROBLEM 2.36Knowing that = 65, determine the resultant of the three forcesshown.SOLUTION300-N Force:Fx = (300 N)cos 20 = 281.9 NFy = (300 N)sin 20 = 102.6 N400-N Force:Fx = (400 N)cos85 = 34.9 NFy = (400 N)sin85 = 398.5 N600-N Force:Fx = (600 N)cos5 = 597.7 NFy = (600 N)sin 5 = 52.3 NandRx = Fx = 914.5 NRy = Fy = 448.8 N( )2 ( )2 R = 914.5 N + 448.8 N = 1018.7 NFurther:tan 448.8914.5 =tan 1 448.8 26.1 = = 914.5R = 1019 N 26.1W36 37. PROBLEM 2.37Knowing that the tension in cable BC is 145 lb, determine the resultant ofthe three forces exerted at point B of beam AB.SOLUTIONCable BC Force:(145 lb) 84 105 lbx 116 F = = (145 lb) 80 100 lby 116 F = =100-lb Force:(100 lb) 3 60 lbx 5 F = = (100 lb) 4 80 lby 5 F = = 156-lb Force:(156 lb)12 144 lbx 13 F = =(156 lb) 5 60 lby 13 F = = andRx = Fx = 21 lb, Ry = Fy = 40 lb( )2 ( )2 R = 21 lb + 40 lb = 45.177 lbFurther:tan 4021 =tan 1 40 62.3 = = 21Thus: R = 45.2 lb 62.3W37 38. PROBLEM 2.38Knowing that = 50, determine the resultant of the three forcesshown.SOLUTIONThe resultant force R has the x- and y-components:Rx = Fx = (140 lb)cos50 + (60 lb)cos85 (160 lb)cos50Rx = 7.6264 lbandRy = Fy = (140 lb)sin 50 + (60 lb)sin85 + (160 lb)sin 50Ry = 289.59 lbFurther:tan 2907.6 =tan 1 290 88.5 = = 7.6Thus: R = 290 lb 88.5W38 39. PROBLEM 2.39Determine (a) the required value of if the resultant of the three forcesshown is to be vertical, (b) the corresponding magnitude of the resultant.SOLUTIONFor an arbitrary angle , we have:Rx = Fx = (140 lb)cos + (60 lb)cos( + 35) (160 lb)cos(a) So, for R to be vertical:Rx = Fx = (140 lb)cos + (60 lb)cos( + 35) (160 lb)cos = 0Expanding,cos + 3(cos cos35 sin sin 35) = 0Then:13 cos35tansin 35 =or cos35 1 = tan 1 3 = 40.265 sin 35 = 40.3W(b) Now:R = Ry = Fy = (140 lb)sin 40.265 + (60 lb)sin 75.265 + (160 lb)sin 40.265R = R = 252 lbW39 40. PROBLEM 2.40For the beam of Problem 2.37, determine (a) the required tension in cableBC if the resultant of the three forces exerted at point B is to be vertical,(b) the corresponding magnitude of the resultant.Problem 2.37: Knowing that the tension in cable BC is 145 lb, determinethe resultant of the three forces exerted at point B of beam AB.SOLUTIONWe have:84 12 (156 lb) 3 (100 lb)x x 116 BC 13 5 R = F = T + or Rx = 0.724TBC + 84 lband80 5 (156 lb) 4 (100 lb)y y 116 BC 13 5 R = F = T Ry = 0.6897TBC 140 lb(a) So, for R to be vertical,Rx = 0.724TBC + 84 lb = 0TBC = 116.0 lbW(b) UsingTBC = 116.0 lbR = Ry = 0.6897(116.0 lb) 140 lb = 60 lbR = R = 60.0 lbW40 41. PROBLEM 2.41Boom AB is held in the position shown by three cables. Knowing that thetensions in cables AC and AD are 4 kN and 5.2 kN, respectively,determine (a) the tension in cable AE if the resultant of the tensionsexerted at point A of the boom must be directed along AB,(b) the corresponding magnitude of the resultant.SOLUTIONChoose x-axis along bar AB.Then(a) RequireRy = Fy = 0: (4 kN)cos 25 + (5.2 kN)sin 35 TAE sin 65 = 0or TAE = 7.2909 kNTAE = 7.29 kN W(b) R = Fx= (4 kN)sin 25 (5.2 kN)cos35 (7.2909 kN)cos65= 9.03 kNR = 9.03 kN W41 42. PROBLEM 2.42For the block of Problems 2.35 and 2.36, determine (a) the required valueof of the resultant of the three forces shown is to be parallel to theincline, (b) the corresponding magnitude of the resultant.Problem 2.35: Knowing that = 35, determine the resultant of thethree forces shown.Problem 2.36: Knowing that = 65, determine the resultant of thethree forces shown.SOLUTIONSelecting the x axis along aa, we writeRx = Fx = 300 N + (400 N)cos + (600 N)sin (1)Ry = Fy = (400 N)sin (600 N)cos (2)(a) Setting Ry = 0 in Equation (2):Thus tan = 600 =1.5400 = 56.3W(b) Substituting for in Equation (1):Rx = 300 N + (400 N)cos56.3 + (600 N)sin 56.3Rx = 1021.1 NR = Rx = 1021 N W42 43. PROBLEM 2.43Two cables are tied together at C and are loaded as shown. Determine thetension (a) in cable AC, (b) in cable BC.SOLUTIONFree-Body DiagramFrom the geometry, we calculate the distances:( )2 ( )2 AC = 16 in. + 12 in. = 20 in.( )2 ( )2 BC = 20 in. + 21 in. = 29 in.Then, from the Free Body Diagram of point C:0: 16 21 0x 20 AC 29 BC F = T + T =or 29 4BC 21 5 AC T = Tand 0: 12 20 600 lb 0y 20 AC 29 BC F = T + T =or 12 20 29 4 600 lb 020 AC 29 21 5 AC T + T = Hence: TAC = 440.56 lb(a) TAC = 441 lbW(b) TBC = 487 lb W43 44. PROBLEM 2.44Knowing that = 25, determine the tension (a) in cable AC, (b) inrope BC.SOLUTIONFree-Body Diagram Force TriangleLaw of Sines:5 kNTAC = TBC =sin115 sin 5 sin 60 (a) 5 kN sin115 5.23 kNT AC = =sin 60 TAC = 5.23 kN W(b) 5 kN sin 5 0.503 kNT BC = =sin 60 TBC = 0.503 kN W44 45. PROBLEM 2.45Knowing that = 50 and that boom AC exerts on pin C a forcedirected long line AC, determine (a) the magnitude of that force, (b) thetension in cable BC.SOLUTIONFree-Body Diagram Force TriangleLaw of Sines:400 lbFAC = TBC =sin 25 sin 60 sin 95 (a) 400 lb sin 25 169.69 lbF AC = =sin 95 FAC = 169.7 lbW(b) 400 sin 60 347.73 lbT BC = =sin 95 TBC = 348 lbW45 46. PROBLEM 2.46Two cables are tied together at C and are loaded as shown. Knowing that = 30, determine the tension (a) in cable AC, (b) in cable BC.SOLUTIONFree-Body Diagram Force TriangleLaw of Sines:2943 NTAC = TBC =sin 60 sin 55 sin 65 (a) 2943 N sin 60 2812.19 NT AC = =sin 65 TAC = 2.81 kN W(b) 2943 N sin 55 2659.98 NT BC = =sin 65 TBC = 2.66 kNW46 47. PROBLEM 2.47A chairlift has been stopped in the position shown. Knowing that eachchair weighs 300 N and that the skier in chair E weighs 890 N, determinethat weight of the skier in chair F.SOLUTIONFree-Body Diagram Point BForce TriangleFree-Body Diagram Point CForce TriangleIn the free-body diagram of point B, the geometry gives:tan 1 9.9 30.51AB 16.8 = = tan 1 12 22.61BC 28.8 = = Thus, in the force triangle, by the Law of Sines:1190 NTBC =sin 59.49 sin 7.87 TBC = 7468.6 NIn the free-body diagram of point C (with W the sum of weights of chairand skier) the geometry gives:tan 1 1.32 10.39CD 7.2 = = Hence, in the force triangle, by the Law of Sines:7468.6 NW =sin12.23 sin100.39 W = 1608.5 NFinally, the skier weight = 1608.5 N 300 N = 1308.5 Nskier weight = 1309 N W47 48. PROBLEM 2.48A chairlift has been stopped in the position shown. Knowing that eachchair weighs 300 N and that the skier in chair F weighs 800 N, determinethe weight of the skier in chair E.SOLUTIONFree-Body Diagram Point FForce TriangleFree-Body Diagram Point EForce TriangleIn the free-body diagram of point F, the geometry gives:tan 1 12 22.62EF 28.8 = = tan 1 1.32 10.39DF 7.2 = = Thus, in the force triangle, by the Law of Sines:1100 NTEF =sin100.39 sin12.23 TBC = 5107.5 NIn the free-body diagram of point E (with W the sum of weights of chairand skier) the geometry gives:tan 1 9.9 30.51AE 16.8 = = Hence, in the force triangle, by the Law of Sines:5107.5 NW =sin 7.89 sin 59.49 W = 813.8 NFinally, the skier weight = 813.8 N 300 N = 513.8 Nskier weight = 514 NW48 49. PROBLEM 2.49Four wooden members are joined with metal plate connectors and are inequilibrium under the action of the four fences shown. Knowing thatFA = 510 lb and FB = 480 lb, determine the magnitudes of the other twoforces.SOLUTIONFree-Body DiagramResolving the forces into x and y components:Fx = 0: FC + (510 lb)sin15 (480 lb)cos15 = 0or FC = 332 lb WFy = 0: FD (510 lb)cos15 + (480 lb)sin15 = 0or FD = 368 lbW49 50. PROBLEM 2.50Four wooden members are joined with metal plate connectors and are inequilibrium under the action of the four fences shown. Knowing thatFA = 420 lb and FC = 540 lb, determine the magnitudes of the other twoforces.SOLUTIONResolving the forces into x and y components:Fx = 0: FB cos15 + (540 lb) + (420 lb)cos15 = 0 or FB = 671.6 lbFB = 672 lb WFy = 0: FD (420 lb)cos15 + (671.6 lb)sin15 = 0or FD = 232 lb W50 51. PROBLEM 2.51Two forces P and Q are applied as shown to an aircraft connection.Knowing that the connection is in equilibrium and the P = 400 lb andQ = 520 lb, determine the magnitudes of the forces exerted on the rodsA and B.SOLUTIONFree-Body DiagramResolving the forces into x and y directions:R = P + Q + FA + FB = 0Substituting components:R = (400 lb) j + (520 lb)cos55 i (520 lb)sin 55 j+ FBi (FA cos55)i + (FA sin 55) j = 0In the y-direction (one unknown force)400 lb (520 lb)sin 55 + FA sin 55 = 0Thus,400 lb (520 lb)sin 551008.3 lb+ A sin 55 F= =FA = 1008 lbWIn the x-direction:(520 lb)cos55 + FB FA cos55 = 0Thus,FB = FA cos55 (520 lb)cos55= (1008.3 lb)cos55 (520 lb)cos55= 280.08 lbFB = 280 lbW51 52. PROBLEM 2.52Two forces P and Q are applied as shown to an aircraft connection.Knowing that the connection is in equilibrium and that the magnitudes ofthe forces exerted on rods A and B are FA = 600 lb and FB = 320 lb,determine the magnitudes of P and Q.SOLUTIONFree-Body DiagramResolving the forces into x and y directions:R = P + Q + FA + FB = 0Substituting components:R = (320 lb)i (600 lb)cos55 i + (600 lb)sin 55 j+ Pi + (Qcos55)i (Qsin 55) j = 0In the x-direction (one unknown force)320 lb (600 lb)cos55 + Qcos55 = 0Thus,320 lb (600 lb)cos5542.09 lbcos55Q + = =Q = 42.1 lbWIn the y-direction:(600 lb)sin 55 P Qsin 55 = 0Thus,P = (600 lb)sin 55 Qsin 55 = 457.01 lbP = 457 lb W52 53. PROBLEM 2.53Two cables tied together at C are loaded as shown. Knowing thatW = 840 N, determine the tension (a) in cable AC, (b) in cable BC.SOLUTIONFree-Body DiagramFrom geometry:The sides of the triangle with hypotenuse CB are in the ratio 8:15:17.The sides of the triangle with hypotenuse CA are in the ratio 3:4:5.Thus:0: 3 15 15 (680 N) 0x 5 CA 17 CB 17 F = T + T =or1 5 200 N5 CA 17 CB T + T = (1)and( ) 0: 4 8 8 680 N 840 N 0y 5 CA 17 CB 17 F = T + T =or1 2 290 N5 CA 17 CB T + T = (2)Solving Equations (1) and (2) simultaneously:(a) TCA = 750 NW(b) TCB = 1190 N W53 54. PROBLEM 2.54Two cables tied together at C are loaded as shown. Determine the rangeof values of W for which the tension will not exceed 1050 N in eithercable.SOLUTIONFree-Body DiagramFrom geometry:The sides of the triangle with hypotenuse CB are in the ratio 8:15:17.The sides of the triangle with hypotenuse CA are in the ratio 3:4:5.Thus:0: 3 15 15 (680 N) 0x 5 CA 17 CB 17 F = T + T =or1 5 200 N5 CA 17 CB T + T = (1)and( ) 0: 4 8 8 680 N 0y 5 CA 17 CB 17 F = T + T W =or1 2 80 N 15 17 4TCA + TCB = + W (2)Then, from Equations (1) and (2)680 N 17T W282528CBT WCA= +=Now, withT 1050 N: 1050 N 25CA CA 28 T T = = Wor W = 1176 Nand: 1050 N 680 N 17CB CB 28 T T = = + Wor W = 609 N 0 W 609 N W54 55. PROBLEM 2.55The cabin of an aerial tramway is suspended from a set of wheels that canroll freely on the support cable ACB and is being pulled at a constantspeed by cable DE. Knowing that = 40 and = 35, that thecombined weight of the cabin, its support system, and its passengers is24.8 kN, and assuming the tension in cable DF to be negligible,determine the tension (a) in the support cable ACB, (b) in the tractioncable DE.SOLUTIONNote: In Problems 2.55 and 2.56 the cabin is considered as a particle. Ifconsidered as a rigid body (Chapter 4) it would be found that its center ofgravity should be located to the left of the centerline for the line CD to bevertical.NowFx = 0: TACB (cos35 cos 40) TDE cos 40 = 0or0.0531TACB 0.766TDE = 0 (1)andFy = 0: TACB (sin 40 sin 35) + TDE sin 40 24.8 kN = 0or0.0692TACB + 0.643TDE = 24.8 kN (2)From (1)TACB = 14.426TDEThen, from (2)0.0692(14.426TDE ) + 0.643TDE = 24.8 kNand(b) TDE = 15.1 kNW(a) TACB = 218 kN W55 56. PROBLEM 2.56The cabin of an aerial tramway is suspended from a set of wheels that canroll freely on the support cable ACB and is being pulled at a constantspeed by cable DE. Knowing that = 42 and = 32, that the tensionin cable DE is 20 kN, and assuming the tension in cable DF to benegligible, determine (a) the combined weight of the cabin, its supportsystem, and its passengers, (b) the tension in the support cable ACB.SOLUTIONFree-Body DiagramFirst, consider the sum of forces in the x-direction because there is only one unknown force:Fx = 0: TACB (cos32 cos 42) (20 kN)cos 42 = 0or0.1049TACB = 14.863 kN(b) TACB = 141.7 kN WNowFy = 0: TACB (sin 42 sin 32) + (20 kN)sin 42 W = 0or(141.7 kN)(0.1392) + (20 kN)(0.6691) W = 0(a) W = 33.1 kN W56 57. PROBLEM 2.57A block of weight W is suspended from a 500-mm long cord and twosprings of which the unstretched lengths are 450 mm. Knowing that theconstants of the springs are kAB = 1500 N/m and kAD = 500 N/m,determine (a) the tension in the cord, (b) the weight of the block.SOLUTIONFree-Body Diagram At AFirst note from geometry:The sides of the triangle with hypotenuse AD are in the ratio 8:15:17.The sides of the triangle with hypotenuse AB are in the ratio 3:4:5.The sides of the triangle with hypotenuse AC are in the ratio 7:24:25.Then:FAB = kAB (LAB Lo )and( )2 ( )2 LAB = 0.44 m + 0.33m = 0.55 mSo:FAB = 1500 N/m(0.55 m 0.45 m)= 150 NSimilarly,FAD = kAD (LAD Lo )Then:( )2 ( )2 LAD = 0.66 m + 0.32 m = 0.68 mFAD = 1500 N/m(0.68 m 0.45 m)= 115 N(a)0: 4 (150 N) 7 15 (115 N) 0x 5 25 AC 17 F = + T =orTAC = 66.18 N TAC = 66.2 NW57 58. PROBLEM 2.57 CONTINUED(b) and0: 3 (150 N) 24 (66.18 N) 8 (115 N) 0y 5 25 17 F = + + W =or W = 208 N W58 59. PROBLEM 2.58A load of weight 400 N is suspended from a spring and two cords whichare attached to blocks of weights 3W and W as shown. Knowing that theconstant of the spring is 800 N/m, determine (a) the value of W, (b) theunstretched length of the spring.SOLUTIONFree-Body Diagram At AFirst 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 ratio12:35:37.Then:0: 4 (3 ) 35 ( ) 12 0x 5 37 37 s F = W + W + F =orFs = 4.4833Wand( ) ( ) 0: 3 3 12 35 400 N 0y 5 37 37 s F = W + W + F =Then:3 (3 W ) + 12 ( W ) + 35 (4.4833 W ) 400 N =05 37 37orW = 62.841NandFs = 281.74 Nor(a) W = 62.8 NW59 60. PROBLEM 2.58 CONTINUED(b) Have spring forceFs = k (LAB Lo )WhereFAB = kAB (LAB Lo )and( )2 ( )2 LAB = 0.360 m + 1.050 m = 1.110 mSo:( ) 281.74 N = 800 N/m 1.110 L0 mor L0 = 758 mm W60 61. PROBLEM 2.59For the cables and loading of Problem 2.46, determine (a) the value of for which the tension in cable BC is as small as possible, (b) thecorresponding value of the tension.SOLUTIONThe smallest TBC is when TBC is perpendicular to the direction of TACFree-Body Diagram At C Force Triangle(a) = 55.0W(b) TBC = (2943 N)sin 55= 2410.8 NTBC = 2.41 kN W61 62. PROBLEM 2.60Knowing that portions AC and BC of cable ACB must be equal, determinethe shortest length of cable which can be used to support the load shownif the tension in the cable is not to exceed 725 N.SOLUTIONFree-Body Diagram: C(For T = 725 N) Fy = 0: 2Ty 1000 N = 0Ty = 500 N2 2 2Tx + Ty = TTx2 + (500 N)2 = (725 N)2Tx = 525 NBy similar triangles:1.5 mBC =725 525 BC = 2.07 mL = 2(BC) = 4.14 mL = 4.14 mW62 63. PROBLEM 2.61Two cables tied together at C are loaded as shown. Knowing that themaximum allowable tension in each cable is 200 lb, determine (a) themagnitude of the largest force P which may be applied at C, (b) thecorresponding value of .SOLUTIONFree-Body Diagram: C Force TriangleForce triangle is isoceles with2 = 180 85 = 47.5(a) P = 2(200 lb)cos 47.5 = 270 lbSince P > 0, the solution is correct. P = 270 lbW(b) = 180 55 47.5 = 77.5 = 77.5W63 64. PROBLEM 2.62Two cables tied together at C are loaded as shown. Knowing that themaximum allowable tension is 300 lb in cable AC and 150 lb in cable BC,determine (a) the magnitude of the largest force P which may be appliedat C, (b) the corresponding value of .SOLUTIONFree-Body Diagram: C Force Triangle(a) Law of Cosines:P2 = (300 lb)2 + (150 lb)2 2(300 lb)(150 lb)cos85P = 323.5 lbSince P > 300 lb, our solution is correct. P = 324 lbW(b) Law of Sines:sin sin85300 323.5 =sin = 0.9238or = 67.49 = 180 55 67.49 = 57.5 = 57.5W64 65. PROBLEM 2.63For the structure and loading of Problem 2.45, determine (a) the value of for which the tension in cable BC is as small as possible, (b) thecorresponding value of the tension.SOLUTIONTBC must be perpendicular to FAC to be as small as possible.Free-Body Diagram: C Force Triangle isa right triangle(a) We observe: = 55 = 55W(b) TBC = (400 lb)sin 60or TBC = 346.4 lb TBC = 346 lbW65 66. PROBLEM 2.64Boom AB is supported by cable BC and a hinge at A. Knowing that theboom exerts on pin B a force directed along the boom and that the tensionin rope BD is 70 lb, determine (a) the value of for which the tension incable BC is as small as possible, (b) the corresponding value of thetension.SOLUTIONFree-Body Diagram: B (a) Have: TBD + FAB + TBC = 0where magnitude and direction of TBD are known, and the directionof FAB is known.Then, in a force triangle:By observation, TBC is minimum when = 90.0W(b) Have TBC = (70 lb)sin (180 70 30)= 68.93 lbTBC = 68.9 lbW66 67. PROBLEM 2.65Collar A shown in Figure P2.65 and P2.66 can slide on a frictionlessvertical rod and is attached as shown to a spring. The constant of thespring is 660 N/m, and the spring is unstretched when h = 300 mm.Knowing that the system is in equilibrium when h = 400 mm, determinethe weight of the collar.SOLUTIONFree-Body Diagram: Collar AHave: Fs = k (LAB LAB )where:( )2 ( )2 LAB = 0.3 m + 0.4 m LAB = 0.3 2 m= 0.5 mThen: Fs = 660 N/m(0.5 0.3 2 )m= 49.986 NFor the collar:0: 4 (49.986 N) 0y 5 F = W + =or W = 40.0 N W67 68. PROBLEM 2.66The 40-N collar A can slide on a frictionless vertical rod and is attachedas shown to a spring. The spring is unstretched when h = 300 mm.Knowing that the constant of the spring is 560 N/m, determine the valueof h for which the system is in equilibrium.SOLUTIONFree-Body Diagram: Collar AF W h F0: 0 = + =y s( 0.3)2 +h2or hFs = 40 0.09 + h2Now.. Fs = k (LAB LAB )where LAB = (0.3)2 + h2 m LAB = 0.3 2 mThen: h 560( 0.09 + h2 0.3 2 ) = 40 0.09 + h2 or (14h 1) 0.09 + h2 = 4.2 2h hmSolving numerically,h = 415 mm W68 69. PROBLEM 2.67A 280-kg crate is supported by several rope-and-pulley arrangements asshown. Determine for each arrangement the tension in the rope. (Hint:The tension in the rope is the same on each side of a simple pulley. Thiscan be proved by the methods of Chapter 4.)SOLUTIONFree-Body Diagram of pulley(a)(b)(c)(d)(e)Fy = 0: 2T (280 kg)(9.81 m/s2 ) = 01 (2746.8 N)2T =T = 1373 NWFy = 0: 2T (280 kg)(9.81 m/s2 ) = 01 (2746.8 N)2T =T = 1373 NWFy = 0: 3T (280 kg)(9.81 m/s2 ) = 01 (2746.8 N)3T =T = 916 NWFy = 0: 3T (280 kg)(9.81 m/s2 ) = 01 (2746.8 N)3T =T = 916 NWFy = 0: 4T (280 kg)(9.81 m/s2 ) = 01 (2746.8 N)4T =T = 687 NW69 70. PROBLEM 2.68Solve parts b and d of Problem 2.67 assuming that the free end of therope is attached to the crate.Problem 2.67: A 280-kg crate is supported by several rope-and-pulleyarrangements as shown. Determine for each arrangement the tension inthe rope. (Hint: The tension in the rope is the same on each side of asimple pulley. This can be proved by the methods of Chapter 4.)SOLUTIONFree-Body Diagram of pulleyand crate(b)(d)Fy = 0: 3T (280 kg)(9.81 m/s2 ) = 01 (2746.8 N)3T =T = 916 NWFy = 0: 4T (280 kg)(9.81 m/s2 ) = 01 (2746.8 N)4T =T = 687 NW70 71. PROBLEM 2.69A 350-lb load is supported by the rope-and-pulley arrangement shown.Knowing that = 25, determine the magnitude and direction of theforce P which should be exerted on the free end of the rope to maintainequilibrium. (Hint: The tension in the rope is the same on each side of asimple pulley. This can be proved by the methods of Chapter 4.)SOLUTIONFree-Body Diagram: Pulley A Fx = 0: 2Psin 25 Pcos = 0andcos = 0.8452 or = 32.3For = +32.3Fy = 0: 2Pcos 25 + Psin 32.3 350 lb = 0or P = 149.1 lb 32.3WFor = 32.3Fy = 0: 2Pcos 25 + Psin 32.3 350 lb = 0or P = 274 lb 32.3W71 72. PROBLEM 2.70A 350-lb load is supported by the rope-and-pulley arrangement shown.Knowing that = 35, determine (a) the angle , (b) the magnitude ofthe force P which should be exerted on the free end of the rope tomaintain equilibrium. (Hint: The tension in the rope is the same on eachside of a simple pulley. This can be proved by the methods of Chapter 4.)SOLUTIONFree-Body Diagram: Pulley A 0: 2 sin cos 25 0 Fx = P P =Hence:(a) sin 1 cos 25 = or = 24.2W2(b) Fy = 0: 2Pcos + Psin 35 350 lb = 0Hence:2Pcos 24.2 + Psin 35 350 lb = 0or P = 145.97 lb P = 146.0 lbW72 73. PROBLEM 2.71A load Q is applied to the pulley C, which can roll on the cable ACB. Thepulley is held in the position shown by a second cable CAD, which passesover the pulley A and supports a load P. Knowing that P = 800 N,determine (a) the tension in cable ACB, (b) the magnitude of load Q.SOLUTIONFree-Body Diagram: Pulley C(a) Fx = 0: TACB (cos30 cos50) (800 N)cos50 = 0Hence TACB = 2303.5 NTACB = 2.30 kN W(b) Fy = 0: TACB (sin 30 + sin 50) + (800 N)sin 50 Q = 0(2303.5 N)(sin 30 + sin 50) + (800 N)sin 50 Q = 0or Q = 3529.2 N Q = 3.53 kN W73 74. PROBLEM 2.72A 2000-N load Q is applied to the pulley C, which can roll on the cableACB. The pulley is held in the position shown by a second cable CAD,which passes over the pulley A and supports a load P. Determine (a) thetension in the cable ACB, (b) the magnitude of load P.SOLUTIONFree-Body Diagram: Pulley CFx = 0: TACB (cos30 cos50) Pcos50 = 0or P = 0.3473TACB (1)Fy = 0: TACB (sin 30 + sin 50) + Psin 50 2000 N = 0or 1.266TACB + 0.766P = 2000 N (2)(a) Substitute Equation (1) into Equation (2):1.266TACB + 0.766(0.3473TACB ) = 2000 NHence: TACB = 1305.5 NTACB = 1306 NW(b) Using (1)P = 0.3473(1306 N) = 453.57 NP = 454 N W74 75. zyxPROBLEM 2.73Determine (a) the x, y, and z components of the 200-lb force, (b) theangles , , and that the force forms with the coordinate axes.SOLUTION(a) Fx = (200 lb)cos30cos 25 = 156.98 lbFx = +157.0 lbWFy = (200 lb)sin 30 = 100.0 lbFy = +100.0 lb WFz = (200 lb)cos30sin 25 = 73.1996 lbFz = 73.2 lb W(b) cos 156.98x 200 = or x = 38.3Wcos 100.0y 200 = or y = 60.0Wcos 73.1996= or z = 111.5Wz 200 75 76. zyxPROBLEM 2.74Determine (a) the x, y, and z components of the 420-lb force, (b) theangles , , and that the force forms with the coordinate axes.SOLUTION(a) Fx = (420 lb)sin 20sin 70 = 134.985 lbFx = 135.0 lbWFy = (420 lb)cos 20 = 394.67 lbFy = +395 lb WFz = (420 lb)sin 20cos70 = 49.131 lbFz = +49.1 lbW(b) cos =134.985x 420 x = 108.7Wcos 394.67y 420 = y = 20.0Wcos 49.131z 420 = z = 83.3W76 77. zyPROBLEM 2.75To stabilize a tree xpartially uprooted in a storm, cables AB and AC areattached to the upper trunk of the tree and then are fastened to steel rodsanchored in the ground. Knowing that the tension in cable AB is 4.2 kN,determine (a) the components of the force exerted by this cable on thetree, (b) the angles , , and that the force forms with axes at A whichare parallel to the coordinate axes.SOLUTION(a) Fx = (4.2 kN)sin 50cos 40 = 2.4647 kNFx = +2.46 kN WFy = (4.2 kN)cos50 = 2.6997 kNFy = 2.70 kNWFz = (4.2 kN)sin 50sin 40 = 2.0681 kNFz = +2.07 kN W(b) cos 2.4647x 4.2 = x = 54.1W77 78. PROBLEM 2.75 CONTINUEDcos =2.7y 4.2 y = 130.0Wcos 2.0681z 4.0 = z = 60.5W78 79. zyPROBLEM 2.76To stabilize a tree xpartially uprooted in a storm, cables AB and AC areattached to the upper trunk of the tree and then are fastened to steel rodsanchored in the ground. Knowing that the tension in cable AC is 3.6 kN,determine (a) the components of the force exerted by this cable on thetree, (b) the angles , , and that the force forms with axes at A whichare parallel to the coordinate axes.SOLUTION(a) Fx = (3.6 kN)cos 45sin 25 = 1.0758 kNFx = 1.076 kNWFy = (3.6 kN)sin 45 = 2.546 kNFy = 2.55 kNWFz = (3.6 kN)cos 45cos 25 = 2.3071 kNFz = +2.31 kN W(b) cos =1.0758x 3.6 x = 107.4W79 80. PROBLEM 2.76 CONTINUEDcos =2.546y 3.6 y = 135.0Wcos 2.3071z 3.6 = z = 50.1W80 81. xPROBLEM 2.77A horizontal circular plate is suspended as shown from three wires whichare attached to a support at D and form 30 angles with the vertical.Knowing that the x component of the force exerted by wire AD on theplate is 220.6 N, determine (a) the tension in wire AD, (b) the angles ,, and zythat the force exerted at A forms with the coordinate axes.SOLUTION(a) Fx = F sin 30sin 50 = 220.6 N (Given)220.6 N 575.95 Nsin30 sin50= = FF = 576 NW(b) cos 220.6 0.3830 = x = =x575.95FF x = 67.5WFy = F cos30 = 498.79 Ncos 498.79 0.86605575.95yyFF = = = y = 30.0WFz = F sin 30cos50= (575.95 N)sin 30cos50= 185.107 Ncos 185.107 0.32139575.95zzFF= = = z = 108.7W81 82. xPROBLEM 2.78A horizontal circular plate is suspended as shown from three wires whichare attached to a support at D and form 30 angles with the vertical.Knowing that the z component of the force exerted by wire BD on theplate is 64.28 N, determine (a) the tension in wire BD, (b) the angles ,, and zythat the force exerted at B forms with the coordinate axes.SOLUTION(a) Fz = F sin 30sin 40 = 64.28 N (Given)64.28 N 200.0 Nsin30 sin40F = =F = 200 N W (b) Fx = F sin 30cos 40= (200.0 N)sin 30cos 40= 76.604 Ncos 76.604 0.38302200.0xxFF= = = x = 112.5WFy = F cos30 = 173.2 Ncos 173.2 0.866200yyFF = = = y = 30.0WFz = 64.28 Ncos 64.28 0.3214200zzFF= = = z = 108.7W82 83. PROBLEM 2.79A horizontal circular plate is suspended as shown from three wires whichare attached to a support at D and form 30 angles with the vertical.Knowing that the tension in wire CD is 120 lb, determine (a) thecomponents of the force exerted by this wire on the plate, (b) the angles, yx, and zthat the force forms with the coordinate axes.SOLUTION(a) Fx = (120 lb)sin 30cos60 = 30 lbFx = 30.0 lbWFy = (120 lb)cos30 = 103.92 lbFy = +103.9 lb WFz = (120 lb)sin 30sin 60 = 51.96 lbFz = +52.0 lb W(b) cos 30.0 0.25120xxFF= = = x = 104.5Wcos 103.92 0.866120yyFF = = = y = 30.0Wcos 51.96 0.433120zzFF = = = z = 64.3W83 84. yxPROBLEM 2.80A horizontal circular plate is suspended as shown from three wires whichare attached to a support at D and form 30 angles with the vertical.Knowing that the x component of the forces exerted by wire CD on theplate is 40 lb, determine (a) the tension in wire CD, (b) the angles , ,and zthat the force exerted at C forms with the coordinate axes.SOLUTION(a) Fx = F sin 30cos60 = 40 lb (Given)40 lb 160 lb= =sin30 cos60 FF = 160.0 lb W(b) cos 40 0.25160xxFF= = = x = 104.5WFy = (160 lb)cos30 = 103.92 lbcos 103.92 0.866160yyFF = = = y = 30.0WFz = (160 lb)sin 30sin 60 = 69.282 lbcos 69.282 0.433160zzFF = = = z = 64.3W84 85. PROBLEM 2.81Determine the magnitude and direction of the forceF = (800 lb)i + (260 lb) j (320 lb)k.SOLUTIONF = Fx2 + Fy2 + Fz2 = (800 lb)2 + (260 lb)2 + (320 lb)2 F = 900 lb Wcos 800 0.8889900xxFF = = = x = 27.3Wcos 260 0.2889900yyFF = = = y = 73.2Wcos 320 0.3555900zzFF= = = z = 110.8W85 86. PROBLEM 2.82Determine the magnitude and direction of the forceF = (400 N)i (1200 N) j + (300 N)k.SOLUTIONF = Fx2 + Fy2 + Fz2 = (400 N)2 + (1200 N)2 + (300 N)2 F = 1300 N Wcos 400 0.307691300xxFF = = = x = 72.1Wcos 1200 0.923071300yyFF= = = y = 157.4Wcos 300 0.230761300zzFF = = = z = 76.7W86 87. zPROBLEM 2.83A force xacts at the origin of a coordinate system in a direction defined bythe angles = 64.5 and = 55.9. Knowing that the y component ofthe force is 200 N, determine (a) the angle y, (b) the other componentsand the magnitude of the force.SOLUTION(a) We have( )2 ( )2 ( )2 ( )2 ( )2 ( )2 cos x + cos y + cos z = 1 cos y = 1 cos y cos zSince Fy < 0 we must have cos y < 0Thus, taking the negative square root, from above, we have:( )2 ( )2 cos y = 1 cos64.5 cos55.9 = 0.70735 y = 135.0W(b) Then:200 N 282.73 NFyycos 0.70735F= = =and Fx = F cos x = (282.73 N)cos64.5 Fx = 121.7 NWFz = F cos z = (282.73 N)cos55.9 Fy = 158.5 NWF = 283 NW87 88. yPROBLEM 2.84A force xacts at the origin of a coordinate system in a direction defined bythe angles = 75.4 and = 132.6. Knowing that the z component ofthe force is 60 N, determine (a) the angle z, (b) the other componentsand the magnitude of the force.SOLUTION(a) We have( )2 ( )2 ( )2 ( )2 ( )2 ( )2 cos x + cos y + cos z = 1 cos y = 1 cos y cos zSince Fz < 0 we must have cos z < 0Thus, taking the negative square root, from above, we have:( )2 ( )2 cos z = 1 cos75.4 cos132.6 = 0.69159 z = 133.8W(b) Then:60 N 86.757 NF Fzzcos 0.69159= = =F = 86.8 N Wand Fx = F cos x = (86.8 N)cos75.4 Fx = 21.9 N WFy = F cos y = (86.8 N)cos132.6 Fy = 58.8 N W88 89. PROBLEM 2.85A force F of magnitude 400 N acts at the origin of a coordinate system.Knowing that x= 28.5, Fy = 80 N, and Fz > 0, determine (a) thecomponents Fx and Fz, (b) the angles yand z.SOLUTION(a) HaveFx = F cos x = (400 N)cos 28.5 Fx = 351.5 N WThen:2 2 2 2F = Fx + Fy + FzSo: (400 N)2 = (352.5 N)2 + (80 N)2 + Fz2Hence:( )2 ( )2 ( )2 Fz = + 400 N 351.5 N 80 N Fz = 173.3 NW(b)cos 80 0.20400yyFF= = = y = 101.5Wcos 173.3 0.43325400zzFF = = = z = 64.3W89 90. zPROBLEM 2.86A force F of magnitude 600 lb acts at the origin of a coordinate system.Knowing that Fx = 200 lb, = 136.8, Fy < 0, determine (a) thecomponents Fy and Fz, (b) the angles xand y.SOLUTION(a) Fz = F cos z = (600 lb)cos136.8= 437.4 lb Fz = 437 lbWThen:2 2 2 2F = Fx + Fy + FzSo: ( ) ( ) ( ) ( ) 2 2 2 2 600 lb = 200 lb + Fy + 437.4 lbHence: ( )2 ( )2 ( )2 Fy = 600 lb 200 lb 437.4 lb= 358.7 lb Fy = 359 lb W(b)cos 200 0.333600xxFF = = = x = 70.5Wcos 358.7 0.59783600yyFF= = = y = 126.7W90 91. PROBLEM 2.87A transmission tower is held by three guy wires anchored by bolts at B,C, and D. If the tension in wire AB is 2100 N, determine the componentsof the force exerted by the wire on the bolt at B.SOLUTIONJJJGBA = (4 m)i + (20 m) j (5 m)k( )2 ( )2 ( )2 BA = 4 m + 20 m + 5 m = 21 m2100 N (4 m) (20 m) (5 m)JJJGF F BAF = = = i + j kBA BA21 mF = (400 N)i + (2000 N) j (500 N)kFx = +400 N, Fy = +2000 N, Fz = 500 NW91 92. PROBLEM 2.88A transmission tower is held by three guy wires anchored by bolts at B,C, and D. If the tension in wire AD is 1260 N, determine the componentsof the force exerted by the wire on the bolt at D.SOLUTIONJJJGDA = (4 m)i + (20 m) j + (14.8 m)k( )2 ( )2 ( )2 DA = 4 m + 20 m + 14.8 m = 25.2 m1260 N (4 m) (20 m) (14.8 m)JJJGF F DAF = = = i + j + kDA DA25.2 mF = (200 N)i + (1000 N) j + (740 N)kFx = +200 N, Fy = +1000 N, Fz = +740 NW 93. PROBLEM 2.89A rectangular plate is supported by three cables as shown. Knowing thatthe tension in cable AB is 204 lb, determine the components of the forceexerted on the plate at B.SOLUTIONJJJGBA = (32 in.)i + (48 in.) j (36 in.)k( )2 ( )2 ( )2 BA = 32 in. + 48 in. + 36 in. = 68 in.204 lb (32 in.) (48 in.) (36 in.)JJJGF F BAF = = = i + j kBA BA68 in.F = (96 lb)i + (144 lb) j (108 lb)kFx = +96.0 lb, Fy = +144.0 lb, Fz = 108.0 lbW93 94. PROBLEM 2.90A rectangular plate is supported by three cables as shown. Knowing thatthe tension in cable AD is 195 lb, determine the components of the forceexerted on the plate at D.SOLUTIONJJJGDA = (25 in.)i + (48 in.) j + (36 in.)k( )2 ( )2 ( )2 DA = 25 in. + 48 in. + 36 in. = 65 in.195 lb ( 25 in.) (48 in.) (36 in.)JJJGF F DAF = = = i + j + kDA DA65 in.F = (75 lb)i + (144 lb) j + (108 lb)kFx = 75.0 lb, Fy = +144.0 lb, Fz = +108.0 lbW94 95. PROBLEM 2.91A steel rod is bent into a semicircular ring of radius 0.96 m and issupported in part by cables BD and BE which are attached to the ring atB. Knowing that the tension in cable BD is 220 N, determine thecomponents of this force exerted by the cable on the support at D.SOLUTIONJJJGDB = (0.96 m)i (1.12 m) j (0.96 m)k( )2 ( )2 ( )2 DB = 0.96 m + 1.12 m + 0.96 m = 1.76 m220 N (0.96 m) (1.12 m) (0.96 m)JJJGT T DBT = = = i j kDB DB DB1.76 mTDB = (120 N)i (140 N) j (120 N)k( DB )x 120.0 N, ( DB )y 140.0 N, ( DB )z 120.0 N T = + T = T = W95 96. PROBLEM 2.92A steel rod is bent into a semicircular ring of radius 0.96 m and issupported in part by cables BD and BE which are attached to the ring atB. Knowing that the tension in cable BE is 250 N, determine thecomponents of this force exerted by the cable on the support at E.SOLUTIONJJJGEB = (0.96 m)i (1.20 m) j + (1.28 m)k( )2 ( )2 ( )2 EB = 0.96 m + 1.20 m + 1.28 m = 2.00 m250 N (0.96 m) (1.20 m) (1.28 m)JJJGT T EBT = = = i j + kEB EB EB2.00 mTEB = (120 N)i (150 N) j + (160 N)k( EB )x 120.0 N, ( EB )y 150.0 N, ( EB )z 160.0 N T = + T = T = + W96 97. PROBLEM 2.93Find the magnitude and direction of the resultant of the two forces shownknowing that P = 500 N and Q = 600 N.SOLUTIONP = (500 lb)[cos30sin15i + sin 30j + cos30cos15k]= (500 lb)[0.2241i + 0.50j + 0.8365k]= (112.05 lb)i + (250 lb) j + (418.25 lb)kQ = (600 lb)[cos 40cos 20i + sin 40j cos 40sin 20k]= (600 lb)[0.71985i + 0.64278j 0.26201k]= (431.91 lb)i + (385.67 lb) j (157.206 lb)kR = P + Q = (319.86 lb)i + (635.67 lb) j + (261.04 lb)k( )2 ( )2 ( )2 R = 319.86 lb + 635.67 lb + 261.04 lb = 757.98 lbR = 758 lbWcos 319.86 lb 0.42199757.98 lbxxRR = = = x = 65.0Wcos 635.67 lb 0.83864757.98 lbyyRR = = = y = 33.0Wcos 261.04 lb 0.34439757.98 lbzzRR = = = z = 69.9W97 98. PROBLEM 2.94Find the magnitude and direction of the resultant of the two forces shownknowing that P = 600 N and Q = 400 N.SOLUTIONUsing the results from 2.93:P = (600 lb)[0.2241i + 0.50j + 0.8365k]= (134.46 lb)i + (300 lb) j + (501.9 lb)kQ = (400 lb)[0.71985i + 0.64278j 0.26201k]= (287.94 lb)i + (257.11 lb) j (104.804 lb)kR = P + Q = (153.48 lb)i + (557.11 lb) j + (397.10 lb)k( )2 ( )2 ( )2 R = 153.48 lb + 557.11 lb + 397.10 lb = 701.15 lbR = 701 lbWcos 153.48 lb 0.21890701.15 lbxxRR = = = x = 77.4Wcos 557.11 lb 0.79457701.15 lbyyRR = = = y = 37.4Wcos 397.10 lb 0.56637701.15 lbzzRR = = = z = 55.5W98 99. PROBLEM 2.95Knowing that the tension is 850 N in cable AB and 1020 N in cable AC,determine the magnitude and direction of the resultant of the forcesexerted at A by the two cables.SOLUTIONJJJGAB = (400 mm)i (450 mm) j + (600 mm)k( )2 ( )2 ( )2 AB = 400 mm + 450 mm + 600 mm = 850 mmJJJGAC = (1000 mm)i (450 mm) j + (600 mm)k( )2 ( )2 ( )2 AC = 1000 mm + 450 mm + 600 mm = 1250 mm( ) (400 mm) (450 mm) (600 mm)850 NT T ABi j kAB AB AB AB AB850 mm + = = = TJJJG(400 N) (450 N) (600 N) ABT = i j + k( ) (1000 mm) (450 mm) (600 mm)1020 NT T ACi j kAC AC AC AC AC1250 mm + = = = TJJJG(816 N) (367.2 N) (489.6 N) ACT = i j + kR = TAB + TAC = (1216 N)i (817.2 N) j + (1089.6 N)kThen: R = 1825.8 N R = 1826 N Wand cos 1216 0.66601x 1825.8 = = x = 48.2Wcos 817.2 0.44758= = y = 116.6Wy 1825.8 cos 1089.6 0.59678z 1825.8 = = z = 53.4W99 100. PROBLEM 2.96Assuming that in Problem 2.95 the tension is 1020 N in cable AB and850 N in cable AC, determine the magnitude and direction of the resultantof the forces exerted at A by the two cables.SOLUTIONJJJGAB = (400 mm)i (450 mm) j + (600 mm)k( )2 ( )2 ( )2 AB = 400 mm + 450 mm + 600 mm = 850 mmJJJGAC = (1000 mm)i (450 mm) j + (600 mm)k( )2 ( )2 ( )2 AC = 1000 mm + 450 mm + 600 mm = 1250 mm( ) (400 mm) (450 mm) (600 mm)1020 NT T ABi j kAB AB AB AB 850 mmAB + = = = TJJJGTAB = (480 N)i (540 N) j + (720 N)k( ) (1000 mm) (450 mm) (600 mm)850 NT T ACi j kAC AC AC AC 1250 mmAC + = = = TJJJGTAC = (680 N)i (306 N) j + (408 N)kR = TAB + TAC = (1160 N)i (846 N) j + (1128 N)kThen: R = 1825.8 N R = 1826 N Wand cos 1160 0.6353x 1825.8 = = x = 50.6Wcos 846 0.4634= = y = 117.6Wy 1825.8 cos 1128 0.6178z 1825.8 = = z = 51.8W100 101. PROBLEM 2.97For the semicircular ring of Problem 2.91, determine the magnitude anddirection of the resultant of the forces exerted by the cables at B knowingthat the tensions in cables BD and BE are 220 N and 250 N, respectively.SOLUTIONFor the solutions to Problems 2.91 and 2.92, we haveTBD = (120 N)i + (140 N) j + (120 N)kTBE = (120 N)i + (150 N) j (160 N)kThen:RB = TBD + TBE= (240 N)i + (290 N) j (40 N)kand R = 378.55 N RB = 379 N Wcos 240 0.6340x 378.55 = = x = 129.3Wcos 290 0.7661y 378.55 = = y = 40.0Wcos 40 0.1057z 378.55 = = z = 96.1W101 102. PROBLEM 2.98To stabilize a tree partially uprooted in a storm, cables AB and AC areattached to the upper trunk of the tree and then are fastened to steel rodsanchored in the ground. Knowing that the tension in AB is 920 lb and thatthe resultant of the forces exerted at A by cables AB and AC lies in the yzplane, determine (a) the tension in AC, (b) the magnitude and direction ofthe resultant of the two forces.SOLUTIONHaveTAB = (920 lb)(sin 50cos 40i cos50j + sin 50sin 40j)TAC = TAC (cos 45sin 25i sin 45j + cos 45cos 25j)(a)RA = TAB + TAC( A )x 0 R = ( A )x x 0: (920 lb)sin 50 cos 40 AC cos 45 sin 25 0 R = F = T =orTAC = 1806.60 lb TAC = 1807 lbW(b)( A )y y : (920 lb)cos50 (1806.60 lb)sin 45 R = F ( A )y 1868.82 lb R = ( A )z z : (920 lb)sin 50 sin 40 (1806.60 lb)cos 45 cos 25 R = F + ( A )z 1610.78 lb R = RA = (1868.82 lb) j + (1610.78 lb)kThen:RA = 2467.2 lb RA = 2.47 kips W102 103. PROBLEM 2.98 CONTINUEDandcos 0 0x 2467.2 = = x = 90.0Wcos 1868.82 0.7560= = y = 139.2Wy 2467.2 cos 1610.78 0.65288z 2467.2 = = z = 49.2W103 104. PROBLEM 2.99To stabilize a tree partially uprooted in a storm, cables AB and AC areattached to the upper trunk of the tree and then are fastened to steel rodsanchored in the ground. Knowing that the tension in AC is 850 lb and thatthe resultant of the forces exerted at A by cables AB and AC lies in the yzplane, determine (a) the tension in AB, (b) the magnitude and direction ofthe resultant of the two forces.SOLUTIONHaveTAB = TAB (sin 50cos 40i cos50j + sin 50sin 40j)TAC = (850 lb)(cos 45sin 25i sin 45j + cos 45cos 25j)(a)( A )x 0 R = ( A )x x 0: AB sin 50 cos 40 (850 lb)cos 45 sin 25 0 R = F = T =TAB = 432.86 lb TAB = 433 lbW(b)( A )y y : (432.86 lb)cos50 (850 lb)sin 45 R = F ( A )y 879.28 lb R = ( A )z z : (432.86 lb)sin 50 sin 40 (850 lb)cos 45 cos 25 R = F + ( A )z 757.87 lb R = RA = (879.28 lb) j + (757.87 lb)kRA = 1160.82 lb RA = 1.161 kips Wcos 0 0x 1160.82 = = x = 90.0Wcos 879.28 0.75746= = y = 139.2Wy 1160.82 cos 757.87 0.65287z 1160.82 = = z = 49.2W104 105. PROBLEM 2.100For the plate of Problem 2.89, determine the tension in cables AB and ADknowing that the tension if cable AC is 27 lb and that the resultant of theforces exerted by the three cables at A must be vertical.SOLUTIONWith:JJJGAC = (45 in.)i (48 in.) j + (36 in.)k( )2 ( )2 ( )2 AC = 45 in. + 48 in. + 36 in. = 75 in.27 lb (45 in.) (48 in.) (36 in.)JJJGT T ACT = = = i j + kAC AC AC AC AC75 in.TAC = (16.2 lb)i (17.28 lb) j + (12.96)kandJJJGAB = (32 in.)i (48 in.) j + (36 in.)k( )2 ( )2 ( )2 AB = 32 in. + 48 in. + 36 in. = 68 in.( 32 in.) (48 in.) (36 in.)JJJGT T AB TABT = = = i j + k68 in.TAB = TAB (0.4706i 0.7059j + 0.5294k)AB AB AB ABABandJJJGAD = (25 in.)i (48 in.) j (36 in.)k( )2 ( )2 ( )2 AD = 25 in. + 48 in. + 36 in. = 65 in.(25 in.) (48 in.) (36 in.)JJJGT T AD TADT = = = i j k65 in.TAD = TAD (0.3846i 0.7385j 0.5538k)AD AD AD ADAD105 106. PROBLEM 2.100 CONTINUEDNowR = TAB + TAD + TAD= TAB (0.4706i 0.7059j + 0.5294k) + (16.2 lb)i (17.28 lb) j + (12.96)k+ TAD (0.3846i 0.7385j 0.5538k)Since R must be vertical, the i and k components of this sum must be zero.Hence:0.4706TAB + 0.3846TAD + 16.2 lb = 0 (1)0.5294TAB 0.5538TAD + 12.96 lb = 0 (2)Solving (1) and (2), we obtain:TAB = 244.79 lb, TAD = 257.41 lbTAB = 245 lbWTAD = 257 lbW106 107. PROBLEM 2.101The support assembly shown is bolted in place at B, C, and D andsupports a downward force P at A. Knowing that the forces in membersAB, AC, and AD are directed along the respective members and that theforce in member AB is 146 N, determine the magnitude of P.SOLUTIONNote that AB, AC, and AD are in compression.Have( )2 ( )2 ( )2 dBA = 220 mm + 192 mm + 0 = 292 mm( )2 ( )2 ( )2 dDA = 192 mm + 192 mm + 96 mm = 288 mm( )2 ( )2 ( )2 dCA = 0 + 192 mm + 144 mm = 240 mmand 146 N ( 220 mm) (192 mm)BA BA BA 292 mm F = F = i + j= (110 N)i + (96 N) jF = F = F CA (192 mm) j (144 mm)k 240 mmCA CA CA= FCA (0.80j 0.60k)F = F = F DA (192 mm) i + (192 mm) j + (96 mm)k 288 mmDA DA DA= FDA [0.66667i + 0.66667j + 0.33333k]With P = PjAt A: F = 0: FBA + FCA + FDA + P = 0i-component: (110 N) + 0.66667FDA = 0 or FDA = 165 Nj-component: 96 N + 0.80FCA + 0.66667(165 N) P = 0 (1)k-component: 0.60FCA + 0.33333(165 N) = 0 (2)Solving (2) for FCA and then using that result in (1), gives P = 279 N W107 108. PROBLEM 2.102The support assembly shown is bolted in place at B, C, and D andsupports a downward force P at A. Knowing that the forces in membersAB, AC, and AD are directed along the respective members and thatP = 200 N, determine the forces in the members.SOLUTIONWith the results of 2.101:F = F = F BA ( 220 mm) i + (192 mm)j 292 mmBA BA BA= FBA [0.75342i + 0.65753j]NF = F = F CA (192 mm) j (144 mm)k 240 mmCA CA CA= FCA (0.80j 0.60k)F = F = F DA (192 mm) i + (192 mm) j + (96 mm)k 288 mmDA DA DA= FDA [0.66667i + 0.66667j + 0.33333k]With: P = (200 N) jAt A: F = 0: FBA + FCA + FDA + P = 0Hence, equating the three (i, j, k) components to 0 gives three equationsi-component: 0.75342FBA + 0.66667FDA = 0 (1)j-component: 0.65735FBA + 0.80FCA + 0.66667FDA 200 N = 0 (2)k-component: 0.60FCA + 0.33333FDA = 0 (3)Solving (1), (2), and (3), givesFBA = 104.5 N, FCA = 65.6 N, FDA = 118.1 NFBA = 104.5 NWFCA = 65.6 NWFDA = 118.1 NW108 109. PROBLEM 2.103Three cables are used to tether a balloon as shown. Determine the verticalforce P exerted by the balloon at A knowing that the tension in cable ABis 60 lb.SOLUTIONThe forces applied at A are:TAB, TAC, TAD and Pwhere P = Pj . To express the other forces in terms of the unit vectorsi, j, k, we writeJJJGAB = (12.6 ft)i (16.8 ft) jAB = 21 ftJJJGAC = (7.2 ft)i (16.8 ft) j + (12.6 ft)k AC = 22.2 ftJJJGAD = (16.8 ft) j (9.9 ft)kAD = 19.5 ftJJJGT T AB Tand T AB = AB AB = AB = ( 0.6 i 0.8 j) ABABJJJGT T AC TAC AC AC AC (0.3242 0.75676 0.56757 ) ACT = = = i j + kACJJJGT T AD TAD AD AD AD ( 0.8615 0.50769 ) ADT = = = j kAD109 110. PROBLEM 2.103 CONTINUEDEquilibrium ConditionF = 0: TAB + TAC + TAD + Pj = 0Substituting the expressions obtained for TAB, TAC, and TAD andfactoring i, j, and k:(0.6TAB + 0.3242TAC )i + (0.8TAB 0.75676TAC 0.8615TAD + P) j+ (0.56757TAC 0.50769TAD )k = 0Equating to zero the coefficients of i, j, k:0.6TAB + 0.3242TAC = 0 (1)0.8TAB 0.75676TAC 0.8615TAD + P = 0 (2)0.56757TAC 0.50769TAD = 0 (3)Setting TAB = 60 lb in (1) and (2), and solving the resulting set ofequations givesTAC = 111 lbTAD = 124.2 lbP = 239 lb W110 111. PROBLEM 2.104Three cables are used to tether a balloon as shown. Determine the verticalforce P exerted by the balloon at A knowing that the tension in cable ACis 100 lb.SOLUTIONSee Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)below:0.6TAB + 0.3242TAC = 0 (1)0.8TAB 0.75676TAC 0.8615TAD + P = 0 (2)0.56757TAC 0.50769TAD = 0 (3)Substituting TAC = 100 lb in Equations (1), (2), and (3) above, and solving the resulting set of equationsusing conventional algorithms givesTAB = 54 lbTAD = 112 lbP = 215 lb W111 112. PROBLEM 2.105The crate shown in Figure P2.105 and P2.108 is supported by threecables. Determine the weight of the crate knowing that the tension incable AB is 3 kN.SOLUTIONThe forces applied at A are:TAB, TAC, TAD and Pwhere P = Pj . To express the other forces in terms of the unit vectorsi, j, k, we writeJJJGAB = (0.72 m)i + (1.2 m) j (0.54 m)k,AB = 1.5 mJJJGAC = (1.2 m) j + (0.64 m)k,AC = 1.36 mJJJGAD = (0.8 m)i + (1.2 m) j (0.54 m)k,AD = 1.54 mJJJGT T AB Tand T AB = AB AB = AB = ( 0.48 i + 0.8 j 0.36 k) ABABJJJGT T AC TAC AC AC AC (0.88235 0.47059 ) ACT = = = j + kACJJJGT T AD TAD AD AD AD (0.51948 0.77922 0.35065 ) ADT = = = i + j kADEquilibrium Condition with W = WjF = 0: TAB + TAC + TAD Wj = 0Substituting the expressions obtained for TAB, TAC, and TAD andfactoring i, j, and k:(0.48TAB + 0.51948TAD )i + (0.8TAB + 0.88235TAC + 0.77922TAD W) j+ (0.36TAB + 0.47059TAC 0.35065TAD )k = 0112 113. PROBLEM 2.105 CONTINUEDEquating to zero the coefficients of i, j, k:0.48TAB + 0.51948TAD = 00.8TAB + 0.88235TAC + 0.77922TAD W = 00.36TAB + 0.47059TAC 0.35065TAD = 0Substituting TAB = 3 kN in Equations (1), (2) and (3) and solving theresulting set of equations, using conventional algorithms for solvinglinear algebraic equations, givesTAC = 4.3605 kNTAD = 2.7720 kNW = 8.41 kN W113 114. PROBLEM 2.106For the crate of Problem 2.105, determine the weight of the crateknowing that the tension in cable AD is 2.8 kN.Problem 2.105: The crate shown in Figure P2.105 and P2.108 issupported by three cables. Determine the weight of the crate knowing thatthe tension in cable AB is 3 kN.SOLUTIONSee Problem 2.105 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)below:0.48TAB + 0.51948TAD = 00.8TAB + 0.88235TAC + 0.77922TAD W = 00.36TAB + 0.47059TAC 0.35065TAD = 0Substituting TAD = 2.8 kN in Equations (1), (2), and (3) above, and solving the resulting set of equationsusing conventional algorithms, givesTAB = 3.03 kNTAC = 4.40 kNW = 8.49 kN W114 115. PROBLEM 2.107For the crate of Problem 2.105, determine the weight of the crateknowing that the tension in cable AC is 2.4 kN.Problem 2.105: The crate shown in Figure P2.105 and P2.108 issupported by three cables. Determine the weight of the crate knowing thatthe tension in cable AB is 3 kN.SOLUTIONSee Problem 2.105 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)below:0.48TAB + 0.51948TAD = 00.8TAB + 0.88235TAC + 0.77922TAD W = 00.36TAB + 0.47059TAC 0.35065TAD = 0Substituting TAC = 2.4 kN in Equations (1), (2), and (3) above, and solving the resulting set of equationsusing conventional algorithms, givesTAB = 1.651 kNTAD = 1.526 kNW = 4.63 kN W115 116. PROBLEM 2.108A 750-kg crate is supported by three cables as shown. Determine thetension in each cable.SOLUTIONSee Problem 2.105 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)below:0.48TAB + 0.51948TAD = 00.8TAB + 0.88235TAC + 0.77922TAD W = 00.36TAB + 0.47059TAC 0.35065TAD = 0Substituting W = (750 kg)(9.81 m/s2 ) = 7.36 kN in Equations (1), (2), and (3) above, and solving theresulting set of equations using conventional algorithms, givesTAB = 2.63 kN WTAC = 3.82 kNWTAD = 2.43 kNW116 117. PROBLEM 2.109A force P is applied as shown to a uniform cone which is supported bythree cords, where the lines of action of the cords pass through the vertexA of the cone. Knowing that P = 0 and that the tension in cord BE is0.2 lb, determine the weight W of the cone.SOLUTIONNote that because the line of action of each of the cords passes through the vertex A of the cone, the cords allhave the same length, and the unit vectors lying along the cords are parallel to the unit vectors lying along thegenerators of the cone.Thus, for example, the unit vector along BE is identical to the unit vector along the generator AB. = = i j kHence: cos 45 8 sin 45 + 65 AB BET i j kIt follows that: cos 45 8 sin 45TBE + BE = BE = TBE 65 T cos30 i 8 j sin 30kTCF + + CF = CF = TCF 65 T cos15 i 8 j sin15kTDG + DG = DG = TDG 65 117 118. PROBLEM 2.109 CONTINUEDAt A: F = 0: TBE + TCF + TDG + W + P = 0Then, isolating the factors of i, j, and k, we obtain three algebraic equations:i TBE + TCF TDG + P =: cos 45 cos30 cos15 065 65 65or TBE cos 45 + TCF cos30 TDG cos15 + P 65 = 0 (1): 8 8 8 065 65 65 j TBE + TCF + TDG W =or 65 0BE CF DG 8 T + T + T W = (2)k TBE + TCF TDG =: sin 45 sin 30 sin15 065 65 65or TBE sin 45 + TCF sin 30 TDG sin15 = 0 (3)With P = 0 and the tension in cord BE = 0.2 lb:Solving the resulting Equations (1), (2), and (3) using conventional methods in Linear Algebra (elimination,matrix methods or iteration with MATLAB or Maple, for example), we obtain:TCF = 0.669 lbTDG = 0.746 lbW = 1.603 lbW118 119. PROBLEM 2.110A force P is applied as shown to a uniform cone which is supported bythree cords, where the lines of action of the cords pass through the vertexA of the cone. Knowing that the cone weighs 1.6 lb, determine the rangeof values of P for which cord CF is taut.SOLUTIONSee Problem 2.109 for the Figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)below:i: TBE cos 45 + TCF cos30 TDG cos15 + 65P = 0 (1): 65 0BE CF DG 8 j T + T + T W = (2)k: TBE sin 45 + TCF sin 30 TDG sin15 = 0 (3)With W = 1.6 lb , the range of values of P for which the cord CF is taut can found by solving Equations (1),(2), and (3) for the tension TCF as a function of P and requiring it to be positive (>0).Solving (1), (2), and (3) with unknown P, using conventional methods in Linear Algebra (elimination, matrixmethods or iteration with MATLAB or Maple, for example), we obtain:TCF = (1.729P + 0.668)lbHence, for TCF > 0 1.729P + 0.668 > 0or P < 0.386 lb 0 < P < 0.386 lbW119 120. PROBLEM 2.111A transmission tower is held by three guy wires attached to a pin at A andanchored by bolts at B, C, and D. If the tension in wire AB is 3.6 kN,determine the vertical force P exerted by the tower on the pin at A.SOLUTIONThe force in each cable can be written as the product of the magnitude ofthe force and the unit vector along the cable. That is, withJJJGAC = (18 m)i (30 m) j + (5.4 m)k( )2 ( )2 ( )2 AC = 18 m + 30 m + 5.4 m = 35.4 m(18 m) (30 m) (5.4 m)JJJGT T AC TACT = = = i j + k35.4 mAC AC ACACTAC = TAC (0.5085i 0.8475j + 0.1525k)JJJGand AB = (6 m)i (30 m) j + (7.5 m)k( )2 ( )2 ( )2 AB = 6 m + 30 m + 7.5 m = 31.5 m(6 m) (30 m) (7.5 m)JJJGT T AB TABT = = = i j + k31.5 mAB AB ABABTAB = TAB (0.1905i 0.9524j + 0.2381k)JJJGFinally AD = (6 m)i (30 m) j (22.2 m)k( )2 ( )2 ( )2 AD = 6 m + 30 m + 22.2 m = 37.8 m(6 m) (30 m) (22.2 m)JJJGT T AD TADT = = = i j k37.8 mAD AD ADADTAD = TAD (0.1587i 0.7937j 0.5873k)120 121. PROBLEM 2.111 CONTINUEDWith P = Pj, at A:F = 0: TAB + TAC + TAD + Pj = 0Equating the factors of i, j, and k to zero, we obtain the linear algebraicequations:i: 0.1905TAB + 0.5085TAC 0.1587TAD = 0 (1)j: 0.9524TAB 0.8475TAC 0.7937TAD + P = 0 (2)k: 0.2381TAB + 0.1525TAC 0.5873TAD = 0 (3)In Equations (1), (2) and (3), set TAB = 3.6 kN, and, using conventionalmethods for solving Linear Algebraic Equations (MATLAB or Maple,for example), we obtain:TAC = 1.963 kNTAD = 1.969 kNP = 6.66 kN W121 122. PROBLEM 2.112A transmission tower is held by three guy wires attached to a pin at A andanchored by bolts at B, C, and D. If the tension in wire AC is 2.6 kN,determine the vertical force P exerted by the tower on the pin at A.SOLUTIONBased on the results of Problem 2.111, particularly Equations (1), (2) and (3), we substitute TAC = 2.6 kNand solve the three resulting linear equations using conventional tools for solving Linear Algebraic Equations(MATLAB or Maple, for example), to obtainTAB = 4.77 kNTAD = 2.61 kNP = 8.81 kN W122 123. PROBLEM 2.113A rectangular plate is supported by three cables as shown. Knowing thatthe tension in cable AC is 15 lb, determine the weight of the plate.SOLUTIONThe (vector) force in each cable can be written as the product of the(scalar) force and the unit vector along the cable. That is, withJJJGAB = (32 in.)i (48 in.) j + (36 in.)k( )2 ( )2 ( )2 AB = 32 in. + 48 in. + 36 in. = 68 in.(32 in.) (48 in.) (36 in.)JJJGT T AB TABT = = = i j + k68 in.AB AB ABABTAB = TAB (0.4706i 0.7059j + 0.5294k)JJJGand AC = (45 in.)i (48 in.) j + (36 in.)k( )2 ( )2 ( )2 AC = 45 in. + 48 in. + 36 in. = 75 in.(45 in.) (48 in.) (36 in.)JJJGT T AC TACT = = = i j + k75 in.AC AC ACACTAC = TAC (0.60i 0.64j + 0.48k)JJJGFinally, AD = (25 in.)i (48 in.) j (36 in.)k( )2 ( )2 ( )2 AD = 25 in. + 48 in. + 36 in. = 65 in.123 124. PROBLEM 2.113 CONTINUED(25 in.) (48 in.) (36 in.)JJJGT T AD TADT = = = i j k65 in.AD AD ADADTAD = TAD (0.3846i 0.7385j 0.5538k)With W = Wj, at A we have:F = 0: TAB + TAC + TAD + Wj = 0Equating the factors of i, j, and k to zero, we obtain the linear algebraicequations:i: 0.4706TAB + 0.60TAC 0.3846TAD = 0 (1)j: 0.7059TAB 0.64TAC 0.7385TAD + W = 0 (2)k: 0.5294TAB + 0.48TAC 0.5538TAD = 0 (3)In Equations (1), (2) and (3), set TAC = 15 lb, and, using conventionalmethods for solving Linear Algebraic Equations (MATLAB or Maple,for example), we obtain:TAB = 136.0 lbTAD = 143.0 lbW = 211 lb W124 125. PROBLEM 2.114A rectangular plate is supported by three cables as shown. Knowing thatthe tension in cable AD is 120 lb, determine the weight of the plate.SOLUTIONBased on the results of Problem 2.111, particularly Equations (1), (2) and (3), we substitute TAD = 120 lb andsolve the three resulting linear equations using conventional tools for solving Linear Algebraic Equations(MATLAB or Maple, for example), to obtainTAC = 12.59 lbTAB = 114.1 lbW = 177.2 lbW125 126. PROBLEM 2.115A horizontal circular plate having a mass of 28 kg is suspended as shownfrom three wires which are attached to a support D and form 30 angleswith the vertical. Determine the tension in each wire.SOLUTIONFx = 0: TAD sin 30sin 50 + TBD sin 30cos 40+ TCD sin 30cos60 = 0Dividing through by the factor sin 30 and evaluating the trigonometricfunctions gives0.7660TAD + 0.7660TBD + 0.50TCD = 0 (1)Similarly,Fz = 0: TAD sin 30cos50 + TBD sin 30sin 40 TCD sin 30sin 60 = 0or 0.6428TAD + 0.6428TBD 0.8660TCD = 0 (2)From (1) TAD = TBD + 0.6527TCDSubstituting this into (2):TBD = 0.3573TCD (3)Using TAD from above:TAD = TCD (4)Now,Fy = 0: TAD cos30 TBD cos30 TCD cos30+ (28 kg)(9.81 m/s2 ) = 0or TAD + TBD + TCD = 317.2 N126 127. PROBLEM 2.115 CONTINUEDUsing (3) and (4), above:TCD + 0.3573TCD + TCD = 317.2 NThen: TAD = 135.1 N WTBD = 46.9 NWTCD = 135.1 NW127 128. PROBLEM 2.119A force P is applied as shown to a uniform cone which is supported bythree cords, where the lines of action of the cords pass through the vertexA of the cone. Knowing that the cone weighs 2.4 lb and that P = 0,determine the tension in each cord.SOLUTIONNote that because the line of action of each of the cords passes through the vertex A of the cone, the cords allhave the same length, and the unit vectors lying along the cords are parallel to the unit vectors lying along thegenerators of the cone.Thus, for example, the unit vector along BE is identical to the unit vector along the generator AB.Hence: = = cos 45 i + 8 j sin 45k65 AB BEIt follows that:T cos 45 i 8 j sin 45kTBE + BE = BE = TBE 65 T cos30 i 8 j sin 30kTCF + + CF = CF = TCF 65 T cos15 i 8 j sin15kTDG + DG = DG = TDG 65 At A: F = 0: TBE + TCF + TDG + W + P = 0132 129. PROBLEM 2.119 CONTINUEDThen, isolating the factors if i, j, and k we obtain three algebraic equations:i TBE + TCF TDG =: cos 45 cos30 cos15 065 65 65or TBE cos 45 + TCF cos30 TDG cos15 = 0 (1): 8 8 8 065 65 65 j TBE + TCF + TDG W =or 2.4 65 0.3 65BE CF DG 8 T + T + T = = (2)k TBE + TCF TDG P =: sin 45 sin 30 sin15 065 65 65or TBE sin 45 + TCF sin 30 TDG sin15 = P 65 (3)With P = 0, the tension in the cords can be found by solving the resulting Equations (1), (2), and (3) usingconventional methods in Linear Algebra (elimination, matrix methods or iterationwith MATLAB or Maple,for example). We obtainTBE = 0.299 lbWTCF = 1.002 lb WTDG = 1.117 lbW133 130. PROBLEM 2.120A force P is applied as shown to a uniform cone which is supported bythree cords, where the lines of action of the cords pass through the vertexA of the cone. Knowing that the cone weighs 2.4 lb and that P = 0.1 lb,determine the tension in each cord.SOLUTIONSee Problem 2.121 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:TBE cos 45 + TCF cos30 TDG cos15 = 0 (1)TBE + TCF + TDG = 0.3 65 (2)TBE sin 45 + TCF sin 30 TDG sin15 = P 65 (3)With P = 0.1 lb, solving (1), (2), and (3), using conventional methods in Linear Algebra (elimination, matrixmethods or iterationwith MATLAB or Maple, for example), we obtainTBE = 1.006 lbWTCF = 0.357 lb WTDG = 1.056 lb W134 131. PROBLEM 2.121Using two ropes and a roller chute, two workers are unloading a 200-kgcast-iron counterweight from a truck. Knowing that at the instant shownthe counterweight is kept from moving and that the positions of points A,B, and C are, respectively, A(0, 0.5 m, 1 m), B(0.6 m, 0.8 m, 0), andC(0.7 m, 0.9 m, 0), and assuming that no friction exists between thecounterweight and the chute, determine the tension in each rope. (Hint:Since there is no friction, the force exerted by the chute on thecounterweight must be perpendicular to the chute.)SOLUTIONFrom the geometry of the chute:N = N j + k = N j + k(2 ) (0.8944 0.4472 )5As in Problem 2.11, for example, the force in each rope can be written asthe product of the magnitude of the force and the unit vector along thecable. Thus, withJJJGAB = (0.6 m)i + (1.3 m) j + (1 m)k( )2 ( )2 ( )2 AB = 0.6 m + 1.3 m + 1 m = 1.764 m(0.6 m) (1.3 m) (1 m)JJJGT T AB TABT = = = i + j + k1.764 mAB AB ABABTAB = TAB (0.3436i + 0.7444j + 0.5726k)JJJGand AC = (0.7 m)i + (1.4 m) j (1 m)k( )2 ( )2 ( )2 AC = 0.7 m + 1.4 m + 1 m = 1.8574 m(0.7 m) (1.4 m) (1 m)JJJGT T AC TACT = = = i + j k1.764 mAC AC ACACTAC = TAC (0.3769i + 0.7537j 0.5384k)Then: F = 0: N + TAB + TAC + W = 0135 132. PROBLEM 2.121 CONTINUEDWith W = (200 kg)(9.81 m/s) = 1962 N, and equating the factors of i, j,and k to zero, we obtain the linear algebraic equations:i: 0.3436TAB + 0.3769TAC = 0 (1)j: 0.7444TAB + 0.7537TAC + 0.8944N 1962 = 0 (2)k: 0.5726TAB 0.5384TAC + 0.4472N = 0 (3)Using conventional methods for solving Linear Algebraic Equations(elimination, MATLAB or Maple, for example), we obtainN = 1311 NTAB = 551 N WTAC = 503 NW136 133. PROBLEM 2.122Solve Problem 2.121 assuming that a third worker is exerting a forceP = (180 N)i on the counterweight.Problem 2.121: Using two ropes and a roller chute, two workers areunloading a 200-kg cast-iron counterweight from a truck. Knowing that atthe instant shown the counterweight is kept from moving and that thepositions of points A, B, and C are, respectively, A(0, 0.5 m, 1 m),B(0.6 m, 0.8 m, 0), and C(0.7 m, 0.9 m, 0), and assuming that no frictionexists between the counterweight and the chute, determine the tension ineach rope. (Hint: Since there is no friction, the force exerted by the chuteon the counterweight must be perpendicular to the chute.)SOLUTIONFrom the geometry of the chute:N = N j + k = N j + k(2 ) (0.8944 0.4472 )5As in Problem 2.11, for example, the force in each rope can be written asthe product of the magnitude of the force and the unit vector along thecable. Thus, withJJJGAB = (0.6 m)i + (1.3 m) j + (1 m)k( )2 ( )2 ( )2 AB = 0.6 m + 1.3 m + 1 m = 1.764 m(0.6 m) (1.3 m) (1 m)JJJGT T AB TABT = = = i + j + k1.764 mAB AB ABABTAB = TAB (0.3436i + 0.7444j + 0.5726k)JJJGand AC = (0.7 m)i + (1.4 m) j (1 m)k( )2 ( )2 ( )2 AC = 0.7 m + 1.4 m + 1 m = 1.8574 m(0.7 m) (1.4 m) (1 m)JJJGT T AC TACT = = = i + j k1.764 mAC AC ACACTAC = TAC (0.3769i + 0.7537j 0.5384k)Then: F = 0: N + TAB + TAC + P + W = 0137 134. PROBLEM 2.122 CONTINUEDWhere P = (180 N)iand W = (200 kg)(9.81 m/s2 ) j= (1962 N) jEquating the factors of i, j, and k to zero, we obtain the linear equations:i: 0.3436TAB + 0.3769TAC 180 = 0j: 0.8944N + 0.7444TAB + 0.7537TAC 1962 = 0k: 0.4472N 0.5726TAB 0.5384TAC = 0Using conventional methods for solving Linear Algebraic Equations(elimination, MATLAB or Maple, for example), we obtainN = 1302 NTAB = 306 N WTAC = 756 N W138 135. PROBLEM 2.123A piece of machinery of weight W is temporarily supported by cables AB,AC, and ADE. Cable ADE is attached to the ring at A, passes over thepulley at D and back through the ring, and is attached to the support at E.Knowing that W = 320 lb, determine the tension in each cable. (Hint:The tension is the same in all portions of cable ADE.)SOLUTIONThe (vector) force in each cable can be written as the product of the (scalar) force and the unit vector alongthe cable. That is, withJJJGAB = (9 ft)i + (8 ft) j (12 ft)k( )2 ( )2 ( )2 AB = 9 ft + 8 ft + 12 ft = 17 ft(9 ft) (8 ft) (12 ft)JJJGT T AB TABT = = = i + j k17 ftAB AB ABABTAB = TAB (0.5294i + 0.4706j 0.7059k)andJJJGAC = (0)i + (8 ft) j + (6 ft)k( )2 ( )2 ( )2 AC = 0 ft + 8 ft + 6 ft = 10 ft(0 ft) (8 ft) (6 ft)JJJGT T AC TACT = = = i + j + k10 ftAC AC ACACTAC = TAC (0.8j + 0.6k)andJJJGAD = (4 ft)i + (8 ft) j (1 ft)k( )2 ( )2 ( )2 AD = 4 ft + 8 ft + 1 ft = 9 ft(4 ft) (8 ft) (1 ft)JJJGT T AD TADET = = = i + j k9 ftAD AD ADEADTAD = TADE (0.4444i + 0.8889j 0.1111k)139 136. PROBLEM 2.123 CONTINUEDFinally,JJJGAE = (8 ft)i + (8 ft) j + (4 ft)k( )2 ( )2 ( )2 AE = 8 ft + 8 ft + 4 ft = 12 ft( 8 ft) (8 ft) (4 ft)JJJGT T AE TADET = = = i + j + k12 ftAE AE ADEAETAE = TADE (0.6667i + 0.6667j + 0.3333k)With the weight of the machinery, W = Wj, at A, we have:F = 0: TAB + TAC + 2TAD Wj = 0Equating the factors of i, j, and k to zero, we obtain the following linear algebraic equations:0.5294TAB + 2(0.4444TADE ) 0.6667TADE = 0 (1)0.4706TAB + 0.8TAC + 2(0.8889TADE ) + 0.6667TADE W = 0 (2)0.7059TAB + 0.6TAC 2(0.1111TADE ) + 0.3333TADE = 0 (3)Knowing that W = 320 lb, we can solve Equations (1), (2) and (3) using conventional methods for solvingLinear Algebraic Equations (elimination, matrix methods via MATLAB or Maple, for example) to obtainTAB = 46.5 lbWTAC = 34.2 lb WTADE = 110.8 lb W140 137. PROBLEM 2.124A piece of machinery of weight W is temporarily supported by cables AB,AC, and ADE. Cable ADE is attached to the ring at A, passes over thepulley at D and back through the ring, and is attached to the support at E.Knowing that the tension in cable AB is 68 lb, determine (a) the tensionin AC, (b) the tension in ADE, (c) the weight W. (Hint: The tension is thesame in all portions of cable ADE.)SOLUTIONSee Problem 2.123 for the analysis leading to the linear algebraic Equations (1), (2), and (3), below:0.5294TAB + 2(0.4444TADE ) 0.6667TADE = 0 (1)0.4706TAB + 0.8TAC + 2(0.8889TADE ) + 0.6667TADE W = 0 (2)0.7059TAB + 0.6TAC 2(0.1111TADE ) + 0.3333TADE = 0 (3)Knowing that the tension in cable AB is 68 lb, we can solve Equations (1), (2) and (3) using conventionalmethods for solving Linear Algebraic Equations (elimination, matrix methods via MATLAB or Maple, forexample) to obtain(a) TAC = 50.0 lb W(b) TAE = 162.0 lb W(c) W = 468 lb W141 138. PROBLEM 2.128Solve Problem 2.127 assuming y = 550 mm.Problem 2.127: Collars A and B are connected by a 1-m-long wire andcan slide freely on frictionless rods. If a force P = (680 N)j is applied atA, determine (a) the tension in the wire when y = 300 mm, (b) themagnitude of the force Q required to maintain the equilibrium of thesystem.SOLUTIONFrom the analysis of Problem 2.127, particularly the results:y2 + z2 = 0.84 m2680 NTABy=Q 680 N zy=With y = 550 mm = 0.55 m, we obtain:2 0.84 m2 (0.55 m)2zz= 0.733m =and(a) 680 N 1236.4 NAB 0.55 T = =or TAB = 1.236 kN Wand(b) Q = 1236(0.866)N = 906 Nor Q = 0.906 kN W147 139. PROBLEM 2.129Member BD exerts on member ABC a force P directed along line BD.Knowing that P must have a 300-lb horizontal component, determine(a) the magnitude of the force P, (b) its vertical component.SOLUTION(a) Psin 35 = 3001b300 lbsin 35P =P = 523 lbW(b) Vertical ComponentPv = Pcos35= (523 lb)cos35Pv = 428 lb W148 140. PROBLEM 2.130A container of weight W is suspended from ring A, to which cables ACand AE are attached. A force P is applied to the end F of a third cablewhich passes over a pulley at B and through ring A and which is attachedto a support at D. Knowing that W = 1000 N, determine the magnitudeof P. (Hint: The tension is the same in all portions of cable FBAD.)SOLUTIONThe (vector) force in each cable can be written as the product of the (scalar) force and the unit vector alongthe cable. That is, withJJJGAB = (0.78 m)i + (1.6 m) j + (0 m)k( )2 ( )2 ( )2 AB = 0.78 m + 1.6 m + 0 = 1.78 m(0.78 m) (1.6 m) (0 m)JJJGT T AB TABT = = = i + j + k1.78 mAB AB ABABTAB = TAB (0.4382i + 0.8989j + 0k)andJJJGAC = (0)i + (1.6 m) j + (1.2 m)k( )2 ( )2 ( )2 AC = 0 m + 1.6 m + 1.2 m = 2 m(0) (1.6 m) (1.2 m)JJJGT T AC TACT = = = i + j + k2 mAC AC ACACTAC = TAC (0.8j + 0.6k)andJJJGAD = (1.3m)i + (1.6 m) j + (0.4 m)k( )2 ( )2 ( )2 AD = 1.3m + 1.6 m + 0.4 m = 2.1m(1.3m) (1.6 m) (0.4 m)JJJGT T AD TADT = = = i + j + k2.1mAD AD ADADTAD = TAD (0.6190i + 0.7619j + 0.1905k)149 141. PROBLEM 2.130 CONTINUEDFinally,JJJGAE = (0.4 m)i + (1.6 m) j (0.86 m)k( )2 ( )2 ( )2 AE = 0.4 m + 1.6 m + 0.86 m = 1.86 m(0.4 m) (1.6 m) (0.86 m)JJJGT T AE TAET = = = i + j k1.86 mAE AE AEAETAE = TAE (0.2151i + 0.8602j 0.4624k)With the weight of the container W = Wj, at A we have:F = 0: TAB + TAC + TAD Wj = 0Equating the factors of i, j, and k to zero, we obtain the following linear algebraic equations:0.4382TAB + 0.6190TAD 0.2151TAE = 0 (1)0.8989TAB + 0.8TAC + 0.7619TAD + 0.8602TAE W = 0 (2)0.6TAC + 0.1905TAD 0.4624TAE = 0 (3)Knowing that W = 1000 N and that because of the pulley system at B TAB = TAD = P, where P is theexternally applied (unknown) force, we can solve the system of linear equations (1), (2) and (3) uniquelyfor P.P = 378 NW150 142. PROBLEM 2.131A container of weight W is suspended from ring A, to which cables ACand AE are attached. A force P is applied to the end F of a third cablewhich passes over a pulley at B and through ring A and which is attachedto a support at D. Knowing that the tension in cable AC is 150 N,determine (a) the magnitude of the force P, (b) the weight W of thecontainer. (Hint: The tension is the same in all portions of cable FBAD.)SOLUTIONHere, as in Problem 2.130, the support of the container consists of the four cables AE, AC, AD, and AB, withthe condition that the force in cables AB and AD is equal to the externally applied force P. Thus, with theconditionTAB = TAD = Pand using the linear algebraic equations of Problem 2.131 with TAC = 150 N, we obtain(a) P = 454 N W(b) W = 1202 NW151 143. PROBLEM 2.125A container of weight W is suspended from ring A. Cable BAC passesthrough the ring and is attached to fixed supports at B and C. Two forcesP = Pi and Q = Qk are applied to the ring to maintain the container isthe position shown. Knowing that W = 1200 N, determine P and Q.(Hint: The tension is the same in both portions of cable BAC.)SOLUTIONThe (vector) force in each cable can be written as the product of the(scalar) force and the unit vector along the cable. That is, withJJJGAB = (0.48 m)i + (0.72 m) j (0.16 m)k( )2 ( )2 ( )2 AB = 0.48 m + 0.72 m + 0.16 m = 0.88 m(0.48 m) (0.72 m) (0.16 m)JJJGT T AB TABT = = = i + j k0.88 mAB AB ABABTAB = TAB (0.5455i + 0.8182j 0.1818k)andJJJGAC = (0.24 m)i + (0.72 m) j (0.13m)k( )2 ( )2 ( )2 AC = 0.24 m + 0.72 m 0.13m = 0.77 m(0.24 m) (0.72 m) (0.13m)JJJGT T AC TACT = = = i + j k0.77 mAC AC ACACTAC = TAC (0.3177i + 0.9351j 0.1688k)At A: F = 0: TAB + TAC + P + Q + W = 0142 144. PROBLEM 2.125 CONTINUEDNoting that TAB = TAC because of the ring A, we equate the factors ofi, j, and k to zero to obtain the linear algebraic equations:i: (0.5455 + 0.3177)T + P = 0or P = 0.2338Tj: (0.8182 + 0.9351)T W = 0or W = 1.7532Tk: (0.1818 0.1688)T + Q = 0or Q = 0.356TWith W = 1200 N:1200 N 684.5 N1.7532T = =P = 160.0 NWQ = 240 NW143 145. PROBLEM 2.126For the system of Problem 2.125, determine W and P knowing thatQ = 160 N.Problem 2.125: A container of weight W is suspended from ring A.Cable BAC passes through the ring and is attached to fixed supports at Band C. Two forces P = Pi and Q = Qk are applied to the ring tomaintain the container is the position shown. Knowing that W = 1200 N,determine P and Q. (Hint: The tension is the same in both portions ofcable BAC.)SOLUTIONBased on the results of Problem 2.125, particularly the three equations relating P, Q, W, and T we substituteQ = 160 N to obtain160 N 456.3 N0.3506T= =W = 800 N WP = 107.0 NW144 146. PROBLEM 2.127Collars A and B are connected by a 1-m-long wire and can slide freely onfrictionless rods. If a force P = (680 N)j is applied at A, determine(a) the tension in the wire when y = 300 mm, (b) the magnitude of theforce Q required to maintain the equilibrium of the system.SOLUTIONFree-Body Diagrams of collars For both Problems 2.127 and 2.128:( AB)2 = x2 + y2 + z2Here (1m)2 = (0.40 m)2 + y2 + z2or y2 + z2 = 0.84 m2Thus, with y given, z is determined.NowJJJGAB 1 (0.40 y z )m 0.4y zAB= = i j + k = i k + kAB 1mWhere y and z are in units of meters, m.From the F.B. Diagram of collar A:F = 0: Nxi + Nzk + Pj + TABAB = 0Setting the j coefficient to zero gives:P yTAB = 0With P = 680 N,680 NTABy=Now, from the free body diagram of collar B:F = 0: Nxi + Ny j + Qk TABAB = 0145 147. PROBLEM 2.127 CONTINUEDSetting the k coefficient to zero gives:Q TABz = 0And using the above result for TAB we have680 NQ TABz zy= =Then, from the specifications of the problem, y = 300 mm = 0.3mz2 = 0.84 m2 (0.3m)2 z = 0.866 mand(a) 680 N 2266.7 NAB 0.30 T = =or TAB = 2.27 kN Wand(b) Q = 2266.7(0.866) = 1963.2 Nor Q = 1.963 kN W146 148. PROBLEM 2.116A transmission tower is held by three guy wires attached to a pin at A andanchored by bolts at B, C, and D. Knowing that the tower exerts on thepin at A an upward vertical force of 8 kN, determine the tension in eachwire.SOLUTIONFrom the solutions of 2.111 and 2.112:TAB = 0.5409PTAC = 0.295PTAD = 0.2959PUsing P = 8 kN:TAB = 4.33 kN WTAC = 2.36 kN WTAD = 2.37 kN W128