lOMoARcPSD|6860698 CHAPTER 2 lOMoARcPSD|6860698 lOMoARcPSD|6860698 PRO OBLEM 2.1 1 Twoo forces are applied a as shoown to a hoook. Determinee graphically the magnnitude and dirrection of theiir resultant using (a) the paarallelogram laaw, (b) thhe triangle rulle. TION SOLUT (a) Parallelogram law: l (b) T Triangle rule: W measure: We R = 1391 kN, α = 477.8° R = 1391 N 47.8° lOMoARcPSD|6860698 PROBLEM 2.2 2 Twoo forces are applied as shown s to a bracket b support. Determiine grapphically the magnitude and directioon of their resultant r usiing (a) the paralleloogram law, (b) ( the trianggle rule. TION SOLUT (a) Parallelogram law: l (b) T Triangle rule: W measure: We R = 9006 N, N α = 26.6° 2 R = 9066 N 26.6° lOMoARcPSD|6860698 PROBLEM 2.3 Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 10 kN and Q = 15 kN, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule. SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: R = 20.1 kN, α = 21.2° R = 20.1 kN 21.2° lOMoARcPSD|6860698 PROBLEM 2.4 Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 6 kN and Q = 4 kN, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule. SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: R= 8.03 kN, α = 3.8 ° R = 8.03 kN 3.8° lOMoARcPSD|6860698 PROBLEM 2.5 A stake is being pulled out of the ground by means of two ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the magnitude of the force P so that the resultant force exerted on the stake is vertical, (b) the corresponding magnitude of the resultant. SOLUTION Using the triangle rule and the law of sines: (a) (b) 120 N P = sin 30° sin 25° P = 101.4 N 30° + β + 25° = 180° β = 180° − 25° − 30° = 125° 120 N R = sin 30° sin125° R = 196.6 N lOMoARcPSD|6860698 PROBLEM 2.6 A telephone cable is clamped at A to the pole AB. Knowing that the tension in the left-hand portion of the cable is T1 = 800 N, determine by trigonometry (a) the required tension T2 in the right-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R. SOLUTION Using the triangle rule and the law of sines: (a) 75° + 40° + α = 180° α = 180° − 75° − 40° = 65° (b) T2 800 N = sin 65° sin 75° T2 = 853 N 800 N R = sin 65° sin 40° R = 567 N lOMoARcPSD|6860698 PROBLEM 2.7 A telephone cable is clamped at A to the pole AB. Knowing that the tension in the right-hand portion of the cable is T2 = 1 kN, determine by trigonometry (a) the required tension T1 in the left-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R. SOLUTION Using the triangle rule and the law of sines: (a) 75° + 40° + β = 180° β = 180° − 75° − 40° = 65° (b) T1 1 kN = sin 75° sin 65° T1 = 938 N 1 kN R = sin 75° sin 40° R = 665 N lOMoARcPSD|6860698 PROBLEM 2.8 A disabled automobile is pulled by means of two ropes as shown. The tension in rope AB is 2.2 kN, and the angle α is 25°. Knowing that the resultant of the two forces applied at A is directed along the axis of the automobile, determine by trigonometry (a) the tension in rope AC, (b) the magnitude of the resultant of the two forces applied at A. SOLUTION Using the law of sines: TAC 2.2 kN R = = sin 30° sin125° sin 25D TAC = 2.603 kN R = 4.264 kN (a) TAC = 2.60 kN (b) R = 4.26 kN lOMoARcPSD|6860698 PROBLEM 2.9 A disabled automobile is pulled by means of two ropes as shown. Knowing that the tension in rope AB is 3 kN, determine by trigonometry the tension in rope AC and the value of α so that the resultant force exerted at A is a 4.8-kN force directed along the axis of the automobile. SOLUTION Using the law of cosines: TAC 2 = (3 kN)2 + (4.8 kN)2 − 2(3 kN)(4.8 kN) cos 30° TAC = 2.6643 kN Using the law of sines: sin α sin 30° = 3 kN 2.6643 kN α = 34.3° TAC = 2.66 kN 34.3° lOMoARcPSD|6860698 PROBLEM M 2.10 Two forces are a applied as shown to a hoook support. Knowing K that the magnitude of o P is 35 N, determine d by trigonometry (a) the requirred angle α if thee resultant R of o the two forcces applied to the support iss to be horizontal, (b) the correesponding maggnitude of R. TION SOLUT Using thhe triangle rulee and law of sines: (a) siin α sin 25° = 5 N 50 35 N s α = 0.603774 sin α = 37.1388° (b) α = 37.1° α + β + 25° = 180° β = 180° − 25° − 37.138° = 117.8622° R 35 N = sin1177.862° sin 25° 2 R = 73.2 N lOMoARcPSD|6860698 P PROBLEM 2.11 A steel tank is to be positionned in an excavvation. Knowiing thhat α = 20°, determine d by trigonometry (a) the requirred m magnitude of the force P if i the resultannt R of the two t fo forces applied at A is to be vertical, v (b) thhe correspondiing m magnitude of R. R SOLUT TION Using thhe triangle rulee and the law of sines: (a) β + 50° + 60° = 180° β = 180° − 50° − 60° = 70° (b) 4425 N P = ssin 70° sin 60° P = 392 N 4 N 425 R = ssin 70° sin 50° R = 346 N lOMoARcPSD|6860698 P PROBLEM 2.12 A steel tank is to be positionned in an excavvation. Knowiing thhat the maggnitude of P is 500 N,, determine by trrigonometry (a) the required angle α if thhe resultant R of thhe two forces applied at A is to be vertical, v (b) the c corresponding magnitude off R. SOLUT TION Using thhe triangle rulee and the law of sines: (a) (b) (α + 330°) + 60° + β = 180° β = 180° − (α + 30°) − 60° β = 90° − α sin (90° − α ) sin 60° = 500 N 425 N 90° − α = 47.402° α = 42.6° R 500 N = sin (42.598° + 300°) sin 60° R = 551 N lOMoARcPSD|6860698 P PROBLEM 2.13 A steel tank is to be positioned p in an excavation. D Determine byy trigonometrry (a) the magnitude and a d direction of thee smallest forcce P for whichh the resultantt R o the two forces of f appliedd at A is vertical, v (b) the c corresponding magnitude off R. TION SOLUT The smaallest force P will w be perpenndicular to R. (a) P =(425 N) co os 30° (b) R =(425 N)sin n 30° P = 368 N R =21 13 N lOMoARcPSD|6860698 P PROBLEM 2.14 For the hook support F s of Proob. 2.10, deterrmine by trigoonometry (a) the m magnitude andd direction of the t smallest force fo P for whhich the resulttant R of the twoo forces appliied to the suupport is horrizontal, (b) the corresponding magnitude off R. SOLUT TION The smaallest force P will w be perpenndicular to R. (a) P = (50 N)sin 25° P = 21.1 N (b) R = (50 N) cos 25° R = 45.3 N lOMoARcPSD|6860698 PROBLEM 2.15 For the hook support shown, determine by trigonometry the magnitude and direction of the resultant of the two forces applied to the support. SOLUTION Using the law of cosines: R 2 = (200 N)2 + (300 N)2 − 2(200 N)(300 N) cos (45 D + 65 °) R = 413.57 N Using the law of sines: sin α sin (45D + 65°) = 300 N 413.57 N α = 42.972° β = 90D + 25D − 42.972° R = 414 N 72.0° lOMoARcPSD|6860698 PROBLEM 2.16 Solve Prob. 2.1 by trigonometry. PROBLEM 2.1 Two forces are applied as shown to a hook. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. SOLUTION Using the law of cosines: R 2 = (900 N) 2 + (600 N )2 − 2(900 N )(600 N) cos (135°) R = 1390.57N Using the law of sines: sin(α − 30D ) sin (135°) = 600N 1390.57N D α − 30 = 17.7642° α = 47.764D R = 1391N 47.8° lOMoARcPSD|6860698 PROBLEM 2.17 2 Soolve Problem 2.4 2 by trigonoometry. PR ROBLEM 2.44 Two structuural members B and C are bolted b to bracket A.. Knowing that both members are in tennsion and thatt P = 6 kN and a Q = 4 kN, deetermine graphhically the maagnitude and direction of the resultant force exerted on thhe bracket usinng (a) the parrallelogram laaw, r (bb) the triangle rule. TION SOLUT Using thhe force triang gle and the law ws of cosines and a sines: We have: γ = 1800° − (50° + 25°) = 1055° Then R 2 = (4 kN )2 + (6 k N)2 − 2(4 kN)(6 kN) cos1105° = 64.423 kN 2 R = 8.00264 kN and 4 kN 8.0264 kN = sin105° sin(25° + α ) sin(25° + α ) = 0.481337 25° + α = 28.7755° α = 3.775° R = 8.03 kN 3.8° lOMoARcPSD|6860698 PROBLEM 2.18 For the stake of Prob. 2.5, knowing that the tension in one rope is 120 N, determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 160 N. PROBLEM 2.5 A stake is being pulled out of the ground by means of two ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the magnitude of the force P so that the resultant force exerted on the stake is vertical, (b) the corresponding magnitude of the resultant. SOLUTION Using the laws of cosines and sines: P 2 = (120 N) 2 + (160 N) 2 − 2(120 N)(160 N) cos 25° P = 72.096 N and sin α sin 25° = 120 N 72.096 N sin α = 0.70343 α = 44.703° P = 72.1 N 44.7° lOMoARcPSD|6860698 PROBLEM 2.19 Two forces P and Q are applied a to the lid of a storagge bin as show wn. Knowing thhat P = 48 N and Q = 60 N, N determine by trigonomeetry the magnituude and directiion of the resuultant of the tw wo forces. SOLUT TION Using thhe force triang gle and the law ws of cosines and a sines: We havee γ = 180° − (220° + 10°) = 150° Then and R 2 = (48 N) 2 + (60 N) 2 − 2(48 N)(60 N N) cos1550° R = 104.366 N 48 N 104.366 N = sin150° sin α sinn α = 0.22996 α = 13.2947° Hence: φ = 180° − α − 80° = 180° − 133.2947° − 80° = 86.705° R = 104.4 N 86.7°° lOMoARcPSD|6860698 P PROBLEM M 2.20 Two forces P and Q are appplied to the lid T l of a storagge bin as show wn. K Knowing that P = 60 N andd Q = 48 N, determine d by trigonometry the m magnitude andd direction of the resultant of o the two forcces. SOLUT TION Using thhe force triang gle and the law ws of cosines and a sines: We havee γ = 180° − (20° + 10°) = 150° Then and R 2 = (60 N)2 + (448 N) 2 − 2(60 N)(488 N) cos 150° R = 104.366 N 60 N 104.366 N = sin α sin150° sin α = 0.28745 α = 16.7054° Hence: φ = 180° − α − 180° = 180° − 16.7054° − 80° = 83.2295° R = 104.44 N 83.3° lOMoARcPSD|6860698 PROB BLEM 2.21 Determiine the x and y components of each of thee forces shownn. TION SOLUT Computte the followin ng distances: OA = (84) 2 + (80) 2 =116 mm OB = (28)2 + (96)2 m =100 mm OC = (48)2 + (90)2 = 102 mm 29-N Foorce: 50-N Foorce: 51-N Foorce: Fx = +(29 N) 84 116 Fx = +21.0 N Fy = +(29 N) 80 116 Fy =+20.0 N Fx = −(50 N ) 28 100 Fx =−14.00 N Fy = +(50 N ) 96 100 Fy =+ 48.0 N Fx = +(51 N) 48 102 Fx = +24.0 N Fy = −(51 N) 90 102 Fy =−45.0 N lOMoARcPSD|6860698 PROBL LEM 2.22 Determinne the x and y components of each of the forces shown.. TION SOLUT Computte the followin ng distances: OA = (600) 2 + (800) 2 = 1000 mm m OB = (560) 2 + (900) 2 = 1060 mm m OC = (480) 2 + (900) 2 = 1020 mm m F 800-N Force: F 424-N Force: F 408-N Force: Fx = +(800 N) N 800 1000 Fx = +640 N Fy = +(800 N) N 600 1000 Fy = +480 N Fx = −(424 N) N 560 1060 Fx = −224 N Fy = −(424 N) N 900 1060 Fy = −360 N Fx = +(408 N) N 480 1020 Fx = +192.0 N Fy = −(408 N) N 900 1020 Fy = −360 N lOMoARcPSD|6860698 PROBLEM 2.23 Determine the x and y components of each of the forces shown. SOLUTION 80-N Force: 120-N Force: 150-N Force: Fx = +(80 N)cos 40° Fx = 61.3 N Fy = +(80 N)sin 40° Fy = 51.4 N Fx = +(120 N)cos70° Fx = 41.0 N Fy = +(120 N)sin 70° Fy = 112.8 N Fx = −(150 N)cos35° Fx = −122. 9 N Fy = +(150 N)sin 35° Fy = 86.0 N lOMoARcPSD|6860698 PROBLEM 2.24 Determine the x and y components of each of the forces shown. SOLUTION 40-N Force: 50-N Force: 60-N Force: Fx = +(40 N)cos60° Fx = 20.0 N Fy = −(40 N)sin 60° Fy = −34.6 N Fx = −(50 N)sin 50° Fx = −38.3 N Fy = −(50 N)cos50° Fy = −32.1 N Fx = +(60 N)cos 25° Fx = 54.4 N Fy = +(60 N)sin 25° Fy = 25.4 N lOMoARcPSD|6860698 PROBLEM 2.25 Member BC exerts on member AC a force P directed along line BC. Knowing that P must have a 325-N horizontal component, determine (a) the magnitude of the force P, (b) its vertical component. SOLUTION BC = (650 mm) 2 + (720 mm) 2 = 970 mm ⎛ 650 ⎞ Px = P ⎜ ⎟ ⎝ 970 ⎠ (a) or ⎛ 970 ⎞ P = Px ⎜ ⎟ ⎝ 650 ⎠ ⎛ 970 ⎞ = 325 N ⎜ ⎟ ⎝ 650 ⎠ = 485 N P = 485 N (b) ⎛ 720 ⎞ Py = P ⎜ ⎟ ⎝ 970 ⎠ ⎛ 720 ⎞ = 485 N ⎜ ⎟ ⎝ 970 ⎠ = 360 N Py = 970 N lOMoARcPSD|6860698 PRO OBLEM 2.26 6 Membber BD exertss on memberr ABC a forcce P directed along line BD. B Know wing that P muust have a 300--N horizontal component, determine d (a) the t magniitude of the foorce P, (b) its vertical v compoonent. TION SOLUT P sin 35° = 300 N (a) P= (b) V Vertical compo onent 300 N sin 35° P = 523 N Pv = P cos35° =(523 N) cos 355° Pv = 428 N lOMoARcPSD|6860698 PR ROBLEM 2.27 2 Thhe hydraulic cyylinder BC exxerts on membber AB a forcee P dirrected along liine BC. Know wing that P muust have a 6000-N com mponent perppendicular to member m AB, determine d (a) the maagnitude of thee force P, (b) its componentt along line AB B. TION SOLUT 180° = 45° + α + 90° + 300° α = 180° − 45° − 90° − 30° = 15° (a) Px P P P= x coss α 6000 N = coss15° = 6211.17 N cos α = P = 621 N (b) tan α = Py Px Py = Px tan t α = (6000 N) tan15° = 1600.770 N Py = 160.8 N lOMoARcPSD|6860698 PRO OBLEM 2.2 28 Cablee AC exerts on o beam AB a force P directed along linne AC. Knowiing that P must have a 350-N verticcal componentt, determine (aa) the magnituude of thee force P, (b) its horizontal component. TION SOLUT (a) P= = Py cos 55° 350 N cos 55° = 610.21 N (b) P = 610 N Px = P sin 555° =(610.211 N)sin 55° = 499.85 N Px = 500 N lOMoARcPSD|6860698 P PROBLEM 2 2.29 Thhe hydraulic cylinder BD exerts e on mem mber ABC a force P directed allong line BD D. Knowing that P mustt have a 750-N component peerpendicular too member ABC, determine (a) the magnitude of the force P, (b) its compoonent parallel to ABC. TION SOLUT (a) 750 N = P sinn 20° P = 21922.9 N (b) P = 2190 N PABC = P coss 20° = (2192.9 N) cos 20°° PABC = 2060 N A lOMoARcPSD|6860698 PROBLEM 2.30 The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P must have a 720-N component perpendicular to the pole AC, determine (a) the magnitude of the force P, (b) its component along line AC. SOLUTION (a) 37 Px 12 37 = (720 N) 12 = 2220 N P= P = 2.22 kN (b) 35 Px 12 35 = (720 N) 12 = 2100 N Py = Py = 2.10 kN lOMoARcPSD|6860698 PROBLE EM 2.31 Determine the t resultant of o the three forrces of Problem m 2.21. PROBLEM M 2.21 Determ mine the x andd y componennts of each of the forces show wn. TION SOLUT Componnents of the fo orces were deteermined in Prooblem 2.21: Force F x Comp. (N) y Compp. (N) 29 2 N +21.0 +200.0 50 5 N –14.00 +48.0 51 5 N +24.0 –45.0 Rx = +31.0 Ry = + 23.0 R = Rx i + Ry j = (31.0 N)i + (23.0 N) j tann α = Ry Rx 23.0 31.0 α = 36.573° = R= 23.0bN sin (36.5773°) = 38.601 N R = 38.66 N 36.6° lOMoARcPSD|6860698 PROBLE EM 2.32 Determine the resultant of o the three forrces of Probleem 2.23. M 2.23 Determ mine the x andd y componennts of each of the PROBLEM forces show wn. TION SOLUT Componnents of the fo orces were dettermined in Prroblem 2.23: y Com mp. (N) Force x Comp. (N) 80 N +61.3 + +51.4 120 N +41.0 +1112.8 150 N –122.9 + +86.0 Rx = −20.6 Ry = + 250.2 2 R = Rx i + R y j = (−20..6 N)i + (250.22 N)j tan α = Ry Rx 250.22 N 20.66 N tan α = 12.14456 tan α = α = 85.2993° R= 2500.2 N sin 855.293° R = 251 N 85.3°° lOMoARcPSD|6860698 PROBL LEM 2.33 Determine the resultantt of the three forces f of Problem 2.24. PROBLE EM 2.24 Deteermine the x and y componnents of eachh of the forcess shown. TION SOLUT Force x Comp. (N) y Comp. (N) 40 4 N +20.00 –34.664 50 5 N –38.30 –32.114 60 6 N +54.38 +25.336 Rx = +36.08 Ry = −41.442 R = Rx i + Ry j = (+36.08 N) i + (− 41.42 N) j tan α = Ry Rx 41.42 N 36.08 N tan α = 1.14800 tan α = α = 48.942° R= 41.42 N sin 48.942° R = 54.99 N 48.9° lOMoARcPSD|6860698 PROBL LEM 2.34 Determinee the resultantt of the three forces f of Probllem 2.22. PROBLE EM 2.22 Deterrmine the x annd y componennts of each of the forces shoown. SOLUT TION Componnents of the fo orces were determined in Problem 2.22: Force x Comp. (N) ( y Comp. (N) 80 00 N +640 +4800 42 24 N –224 –3600 40 08 N +192 –3600 Rx = +608 Ry = −2400 R = Rx i + Ry j = (608 N)i + (−240 N) j Ry tann α = Rx 240 608 α = 21.541° 240 N R= sin(21.5441°) = 653.65 N = R = 6544 N 21.5° lOMoARcPSD|6860698 PRO OBLEM 2.35 5 Know wing that α = 35°, determiine the resulttant of the thhree forcess shown. SOLUT TION Fx = +(100 N) cos35° = +81.915 N 100-N Force: F Fy = −(100 N)sin 35° = −57.358 N Fx = +(150 N) cos 65° = +63.393 N 150-N Force: F Fy = −(150 N)sin 65° = −135.946 N Fx = −(200 N) cos35° = −163.830 N 200-N Force: F Fy = −(200 N)sin 35° = −114.715 N Force x Com mp. (N) N) y Comp. (N 100 N +81.915 −57.3558 150 N +63.393 −135.9446 200 N −163.830 −114.7115 Rx = −18.522 Ry = −308.022 R = Rx i + Ry j = (−18.5222 N)i + (−3088.02 N) j taan α = Ry Rx 308.02 18.522 α = 86.559° = R= 308.02 N sin 86.559 R = 3099 N 86.6° lOMoARcPSD|6860698 PR ROBLEM 2.36 Knoowing that the tension in roppe AC is 365 N, N determine the resuultant of the thhree forces exeerted at point C of post BC. TION SOLUT Determiine force comp ponents: Cable foorce AC: 500-N Force: F 200-N Force: F and 960 = −2440 N 1460 1100 = −2775 N Fy = −(365 N) N 1460 Fx = −(365 N) N 24 = 480 N 25 7 Fy = (500 N)) = 140 N 25 Fx = (500 N)) 4 = 160 N 5 3 Fy = −(200 N) N = −120 N 5 Fx = (200 N)) Rx = ΣFx = −240 N + 480 N + 160 N = 400 4 N Ry = ΣFy = −275 N + 140 N − 120 N = −255 N R = Rx2 + Ry2 = (400 N) N 2 + (−255 N N) 2 = 474.37 N Further: 255 400 α = 32.5° tan α = R = 4744 N 32.5° lOMoARcPSD|6860698 PROBLEM 2.37 7 Knowiing that α = 40°, determiine the resultant of the thhree forces shown. SOLUT TION 60-N Foorce: Fx = (60 N) cos 20° = 56.382 N Fy = (60 N)sin 20° = 200.521 N 80-N Foorce: Fx = (80 N) cos 60° = 40.000 N Fy = (80 N)sin 60° = 69.282 9 N 120-N Force: F Fx = (1200 N) cos30° = 1103.923 N Fy = −(1220 N)sin 30° = −60.000 N and Rx = ΣFx = 200.305 N Ry = ΣFy = 29.803 N R= (2000.305 N) 2 + (29.803 ( N) 2 = 202..510 N Further: tan α = 29.803 200.305 α = tan −1 = 8.46° 29.803 200.305 R = 2033 N 8.46° lOMoARcPSD|6860698 PRO OBLEM 2.3 38 Know wing that α = 75°, determ mine the resulttant of the three forcees shown. TION SOLUT 60-N Foorce: Fx = (60 N) coos 20° = 56.3882 N Fy = (60 N)siin 20° = 20.521 N 80-N Foorce: Fx = (80 N) coos 95° = −6.97725 N Fy = (80 N)sin 95 ° = 79.6966 N 120-N Force: F Fx = (120 N) cos c 5° = 119.5443 N Fy = (120 N)ssin 5 ° = 10.4599 N Then Rx = ΣFx = 1668.953 N Ry = ΣFy = 1110.676 N and R= (168.9553 N)2 + (110..676 N)2 = 201.976 N 110.676 168.953 tann α = 0.65507 α = 33.228° tann α = R =2022 N 33.2° lOMoARcPSD|6860698 PROBLEM 2.39 For the collar of Problem 2.35, determine (a) the required value of α if the resultant of the three forces shown is to be vertical, (b) the corresponding magnitude of the resultant. SOLUTION Rx = ΣFx = (100 N) cos α + (150 N) cos (α + 30°) − (200 N) cos α Rx = −(100 N) cos α + (150 N) cos (α + 30°) (1) Ry = ΣFy = −(100 N)sin α − (150 N)sin (α + 30°) − (200 N)sin α Ry = −(300 N)sin α − (150 N)sin (α + 30°) (a) (2) For R to be vertical, we must have Rx = 0. We make Rx = 0 in Eq. (1): −100 cos α + 150 cos (α + 30°) = 0 −100 cos α + 150 (cos α cos 30° − sin α sin 30°) = 0 29.904 cos α = 75sin α 29.904 75 = 0.39872 α = 21.738° tan α = (b) α = 21.7° Substituting for α in Eq. (2): Ry = −300sin 21.738° − 150sin 51.738° = −228.89 N R = | Ry | = 228.89 N R = 229 N lOMoARcPSD|6860698 PROBLEM 2.40 For the post of Prob. 2.36, determine (a) the required tension in rope AC if the resultant of the three forces exerted at point C is to be horizontal, (b) the corresponding magnitude of the resultant. SOLUTION Rx = ΣFx = − Rx = − 48 TAC + 640 N 73 Ry = ΣFy = − Ry = − (a) 960 24 4 (500 N) + (200 N) TAC + 1460 25 5 1100 7 3 TAC + (500 N) − (200 N) 1460 25 5 55 TAC + 20 N 73 (2) For R to be horizontal, we must have Ry = 0. Set Ry = 0 in Eq. (2): − 55 TAC + 20 N = 0 73 TAC = 26.545 N (b) (1) TAC = 26.5 N Substituting for TAC into Eq. (1) gives 48 (26.545 N) + 640 N 73 Rx = 622.55 N Rx = − R = Rx = 623 N R = 623 N lOMoARcPSD|6860698 PRO OBLEM 2.4 41 Deteermine (a) thee required tenssion in cable AC A , knowing that t the resulttant of thhe three forcess exerted at Point P C of booom BC must be b directed aloong BC, (b) the correspponding magnnitude of the resultant. TION SOLUT Using thhe x and y axess shown: Rx = ΣFx = TAC sin10° + (50 N) cos355° + (75 N) coos 60° = TAC sin10° + 78.458 N (1) Ry = ΣFy = (50 N)sin 35 ° + (75 N)sinn 60° − TAC cos10° Ry= 93.631 N − TAC cos10° (a ) (2) Seet Ry = 0 in Eq. E (2): 93.631 N −TAC cos10° = 0 TAC = 95.075 N (b ) TAC = 95.1 N Suubstituting forr TAC in Eq. (11): Rx = (95.075 N) sin10 ° + 78.4558 N = 94.968 N R = Rx R = 95.0 N lOMoARcPSD|6860698 PRO OBLEM 2.42 2 For thhe block of Problems P 2.377 and 2.38, deetermine (a) the required value of α if the resultannt of the threee forces shownn is to be parallel p to the incline, (b) thhe correspondiing magnitudee of the ressultant. SOLUT TION Select thhe x axis to bee along a a′. Then Rx = ΣFx = (60 N) + (80 N)cos o α + (120 N) sin α (1) Ry = ΣFy = (80 N)sin α − (12 0 N)cos α (2) and (a ) Seet Ry = 0 in Eq. E (2). (80 N )sin α − (120 N) cos α = 0 D Dividing each term t by cos α gives: (800 N) tan α =1220 N 1220 N 8 N 80 α = 566.310° tanα = (b ) α = 56.3° Suubstituting forr α in Eq. (1)) gives: Rx = 60 N + (800 N)cos56.31° + (120 N)sinn 56.31° = 204. 22 N Rx = 204 N lOMoARcPSD|6860698 PROBLEM 2.43 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram Force Triangle Law of sines: TAC T 400 N = BC = sin 60° sin 40° sin 80° (a) TAC = 400 N (sin 60°) sin 80° TAC = 352 N (b) TBC = 400 N (sin 40°) sin 80° TBC = 261 N lOMoARcPSD|6860698 PROBLEM 2.44 Two cables are tied together at C and are loaded as shown. Knowing that α = 30°, determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram Force Triangle Law of sines: TAC T 6 kN = BC = sin 60° sin 35° sin 85° (a) TAC = 6 kN (sin 60°) sin 85° TAC = 5.22 kN (b) TBC = 6 kN (sin 35°) sin 85° TBC = 3.45 kN lOMoARcPSD|6860698 PROBLEM 2.45 Two cables are tied together at C and loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram 1.4 4.8 α = 16.2602° 1.6 tan β = 3 β = 28.073° tan α = Force Triangle Law of sines: TAC TBC 1.98 kN = = sin 61.927° sin 73.740° sin 44.333° (a) TAC = 1.98 kN sin 61.927° sin 44.333° TAC = 2.50 kN (b) TBC = 1.98 kN sin 73.740° sin 44.333° TBC = 2.72 kN lOMoARcPSD|6860698 PROBLEM 2.46 Two cabbles are tied together at C and are looaded as show wn. Knowingg that P = 5000 N and α = 60°, determinne the tensionn in (a) in caable AC, (b) inn cable BC. SOLUT TION Free-Bod dy Diagram Law of sines: s Forrce Triangle TAC T 500 N = BC = sin 35° sin s 75° sin 700° (a ) TAC = 5500 N sin 35° sin 70° TAC = 305 N (b ) TBC = 5500 N sin 75° siin 70° TBC = 514 N lOMoARcPSD|6860698 PRO OBLEM 2.4 47 Two cables are tieed together at C and are loaaded as shownn. Determine the tensiion (a) in cable AC, (b) in caable BC. TION SOLUT Free-Bod dy Diagram F Force Trianggle W = mg = (20 00 kg)(9.81 m//s 2 ) = 196 62 N s Law of sines: TAC TBC 1962 N A = = sin 15° sin 105° sin 60° (a ) TAC = (1962 N N) sin 15° sinn 60° TAC = 586 N (b ) TBC = (1962 N)sin N 105° sinn 60° TBC = 2190 N lOMoARcPSD|6860698 P PROBLEM 2 2.48 Knowing that α = 20°, determine the tension (a) in cable K AC C, (b) in rope BC. TION SOLUT Freee-Body Diagraam s Law of sines: Force Trianggle TAC T 12 kN = BC = sinn 110° sin 5° sin 65° (a ) TAC = 12 0 kN sin 110° sin 65 6 ° TAC = 1.244 kN (b ) TBC = 12 kN sin 5° sin 65 6 ° TBC = 115.4 N lOMoARcPSD|6860698 PROBLEM 2.49 9 Two cables are tied togethher at C annd are loadded as show wn. Know wing that P = 300 N, deetermine the tension in cables c AC and a BC. TION SOLUT Free-B Body Diagram m ΣFx = 0 − TCA sin 30D + TCB sin 30D − P coss 45° − 200N = 0 For P = 200N we have, −0.5TCAA + 0.5TCB + 2112.13 − 200 = 0 (1) ΣFy = 0 TCA coos30° − TCB cos30D − P sin 45D = 0 0.8 86603TCA + 0.886603TCB − 2112.13 = 0 (2)) multaneously gives, Solvinng equations (1) and (2) sim TCA = 134.6 N TCB = 110.4 N lOMoARcPSD|6860698 PROBLEM 2.50 0 Two cables are tied togethher at C annd are loadded as show wn. Deterrmine the rannge of valuees of P for which both cables remaain taut. SOLUT TION Free-B Body Diagram m ΣFx = 0 − TCA sin 30D + TCB sin 30D − P coss 45° − 200N = 0 For TCA = 0 we have,, 0.5TCB + 0.70711P − 200 2 = 0 (1) ΣFy = 0 D TCA cos30 c ° − TCB cos30 c − P sin 45D = 0 ; agaain setting TCAA = 0 yields, 0.866603TCB − 0.707711P = 0 (2)) Adding equations (1) and (2) givess, 1.36603TCB 410N and P = 179.315N ce TCB = 146.4 C = 200 henc Substituuting for TCB = 0 into the eqquilibrium equuations and sollving simultanneously gives, −0.5TCA + 0.70711P − 200 = 0 0.86603TCA − 0.707111P = 0 0N , P = 6699.20N Thus for fo both cabless to remain taaut, load P muust be within the Andd TCA = 546.40 range off 179.315 N an nd 669.20 N. 179.3 N <P< 669 N lOMoARcPSD|6860698 P PROBLEM 2 2.51 Tw wo forces P and a Q are appplied as show wn to an aircrraft coonnection. Knnowing that the connection is i in equilibriuum annd that P = 500 N and Q = 650 N, determine the m magnitudes of the t forces exerrted on the rodds A and B. TION SOLUT Freee-Body Diagrram Resolvinng the forces into i x- and y-ddirections: R = P + Q + FA + FB = 0 Substituuting componeents: R = −(500 N) j + [(650 N) cos 50 5 °]i − [(650 N)siin 50°]j c 50°)i + ( FA sin 50°) j = 0 + FB i − ( FA cos In the y-direction (onee unknown forrce): −500 N − (650 N)sin 50° + FA sin 50° = 0 Thus, FA = 500 N + (650 N)sin 500° sin 50° = 1302.70 N In the x-direction: Thus, FA = 1.303 kN (6550 N)cos50° + FB − FA cos550° = 0 FB = FA cos500° − (650 N) cos50 c ° =(1302.700 N) cos50° − (650 N) cos500° = 419.55 N FB = 420 N lOMoARcPSD|6860698 P PROBLEM M 2.52 Two forces P and Q are appplied as show T wn to an aircrraft c connection. Knowing thhat the connnection is in e equilibrium annd that the maagnitudes of thhe forces exerrted o rods A and on a B are FA = 750 N and n FB = 400 N, d determine the magnitudes of o P and Q. TION SOLUT Free-Boody Diagram Resolvinng the forces into i x- and y-ddirections: R = P + Q + FA + FB = 0 Substituuting componeents: R = − Pj + Q cos 500°i − Q sin 50°j ° − [(750 N) cos 50°]i + [(750 N)sin 50 °] j + (400 bN)i In the x-direction (onee unknown forrce): Q cos 50 5 ° − [(750 N) cos 50°] + 4000 N = 0 Q= (750 N) cos 50 ° − 4 00 0 N cos 50° =127.7100 N In the y-direction: − P − Q sin 50° + (750 N)sin 50° = 0 P = −Q sin 50° + (750 N) sin 50 5 ° = −(127.710 N) sin 50° + (7550 N) sin 50° =476.70 N P = 477 N; Q = 127.7 N lOMoARcPSD|6860698 PR ROBLEM 2.53 A welded w connecction is in equuilibrium undder the action of the four forces shown. Knnowing that FA = 8 kN and a FB = 16 kN, deteermine the magnitudes m off the other two t forcces. SOLUT TION F Free-Body Diiagram of Connection ΣFx = 0: With 3 3 FB − FC − FA = 0 5 5 FA = 8 kN FB = 16 kN 4 4 FC = (16 kN)) − (8 kN) 5 5 FC = 6.40 kN 3 3 Σ Fy = 0: − FD + FB − FA = 0 5 5 With FA and FB as abo ove: 3 3 FD = (16 kN)) − (8 kN) 5 5 FD = 4.80 kN lOMoARcPSD|6860698 PROBLEM 2.54 4 A wellded connectioon is in equiliibrium under the t action of the four forces fo shown. Knowing thaat FA = 5 kN and FD = 6 kN, k determ mine the magnnitudes of the other o two forces. SOLUT TION F Free-Body Diiagram of Connection 3 3 ΣFy = 0: − FD − FA + FB = 0 5 5 or 3 FB = FD + FA 5 With FA = 5 kN, FD = 8 kN 5⎡ 3 ⎤ FB = ⎢6 kN + (5 kN) ⎥ 3⎣ 5 ⎦ ΣFx = 0: − FC + FB = 15.00 kN 4 4 FB − FA = 0 5 5 4 ( FB − FA ) 5 4 = (15 kN − 5 kN) 5 FC = FC = 8.00 kN lOMoARcPSD|6860698 PROBLEM 2.55 A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that α = 30° and β = 10° and that the combined weight of the boatswain’s chair and the sailor is 900 N, determine the tension (a) in the support cable ACB, (b) in the traction cable CD. SOLUTION Free-Body Diagram ΣFx = 0: TACB cos 10° − TACB cos 30° − TCD cos 30° = 0 TCD = 0.137158TACB (1) ΣFy = 0: TACB sin 10° + TACB sin 30° + TCD sin 30° − 900 = 0 0.67365TACB + 0.5TCD = 900 (a) Substitute (1) into (2): 0.67365 TACB + 0.5(0.137158 TACB ) = 900 TACB = 1212.56 N (b) From (1): (2) TCD = 0.137158(1212.56 N) TACB = 1.213 kN TCD = 166.3 N lOMoARcPSD|6860698 PROBLEM 2.56 A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that α = 25° and β = 15° and that the tension in cable CD is 80 N, determine (a) the combined weight of the boatswain’s chair and the sailor, (b) in tension in the support cable ACB. SOLUTION Free-Body Diagram ΣFx = 0: TACB cos 15° − TACB cos 25° − (80 N) cos 25° = 0 TACB = 1216.15 N ΣFy = 0: (1216.15 N) sin 15° + (1216.15 N) sin 25° + (80 N) sin 25° − W = 0 W = 862.54 N (a) W = 863 N (b) TACB = 1216 N lOMoARcPSD|6860698 PROBLEM 2.57 For the cables of prob. 2.44, find the value of α for which the tension is as small as possible (a) in cable bc, (b) in both cables simultaneously. In each case determine the tension in each cable. SOLUTION Free-Body Diagram Force Triangle (a) For a minimum tension in cable BC, set angle between cables to 90 degrees. α = 35.0D By inspection, TAC = (6 kN) cos35D TAC = 4.91 kN TBC = (6 kN)sin 35D TBC = 3.44 kN (b) For equal tension in both cables, the force triangle will be an isosceles. α = 55.0D Therefore, by inspection, TAC = TBC = (1 / 2) 6 kN cos35° TAC = TBC = 3.66 kN lOMoARcPSD|6860698 PROBLEM 2.58 For the cables of Prooblem 2.46, itt is known thaat the maximuum allowablle tension is 600 6 N in cable AC and 7500 N in cable BC B . Determine (a) the maaximum forcee P that can be b applied at C, (b) the corresponding c value of α. TION SOLUT Free-Body F Diagram (a ) L of cosines Law Forrce Triangle P 2 = (600)2 + (7750)2 − 2(600)(750) cos (25°° + 45°) P = 784.02 N (b ) L of sines Law P = 784 N sin β sin (25° + 45 4 °) = 600 N 784.02 N β = 46.0° ∴ α = 46.0° + 25° α = 71.0° lOMoARcPSD|6860698 P PROBLEM 2.59 For the situatioon described in Figure P2.48, determine (a) t he value of α for f which the tension in rope p BC is as sm mall ass possible, (b) the corresponnding value off the tension. TION SOLUT Free-Body F Diagram Forrce Triangle To be sm mallest, TBC must m be perpenndicular to thee direction of TAC. (a ) (b ) T Thus, α = 5.00° TBC =(12 kN)sin 5° α = 5.00° TBC = 1.046 kN lOMoARcPSD|6860698 PROBLE EM 2.60 Two cabless tied togetherr at C are loaaded as shownn. Determine the range of vaalues of Q for which the tennsion will nott exceed 60 N in either cablee. TION SOLUT ΣFx = 0: −TBC − Q coos60° + 75 N = 0 Free-Boody Diagram TBC = 75 N − Q cos60° (1) ΣFy = 0: TAC − Q sin 60° = 0 TAC = Q sinn 60° Requiremeent: TAC = 60 N: From Eq. (2): ( Q sin s 60° = 60 N (2) Q = 69.3 N Requiremeent: From Eq. (1): ( TBC = 60 N: 75 N − Q cos 60° = 60 N Q = 300.0 N 30.0 N ≤ Q ≤ 69.3 N lOMoARcPSD|6860698 PROBLEM 2.61 A movable bin and its contents have a combined weight of 2.8 kN. Determine the shortest chain sling ACB that can be used to lift the loaded bin if the tension in the chain is not to exceed 5 kN. SOLUTION Free-Body Diagram tan α = h 0.6 m (1) Isosceles Force Triangle 1 Law of sines: sin α = 2 (2.8 kN) TAC TAC = 5 kN 1 sin α = 2 (2.8 kN) 5 kN α = 16.2602° From Eq. (1): tan16.2602° = h 0.6 m ∴ h = 0.175000 m Half-length of chain = AC = (0.6 m) 2 + (0.175 m) 2 = 0.625 m Total length: = 2 × 0.625 m 1.250 m lOMoARcPSD|6860698 PROBLEM 2.62 For W = 800 N, P = 200 N, and d = 600 mm, determine the value of h consistent with equilibrium. SOLUTION TAC = TBC = 800 N Free-Body Diagram (h + d ) AC = BC = ΣFy = 0: 2(800 N) 800 = Data: P ⎛d ⎞ 1+ ⎜ ⎟ 2 ⎝h⎠ 2 h h + d2 2 2 −P=0 2 P = 200 N, d = 600 mm and solving for h 800 N = 200 N ⎛ 600 mm ⎞ 1+ ⎜ ⎟ 2 h ⎝ ⎠ 2 h = 75.6 mm lOMoARcPSD|6860698 PROBLEM 2.63 Collar A is connected as shown to a 200-N load and can slide on a frictionless horizontal rod. Determine the magnitude of the force P required to maintain the equilibrium of the collar when (a) x = 90 mm, (b) x = 300 mm. SOLUTION (a) Free Body: Collar A Force Triangle P 200 N = 90 410 (b) Free Body: Collar A P = 43.9 N Force Triangle P 200 N = 3 5 P = 120 N lOMoARcPSD|6860698 PROBLEM 2.64 Collar A is connected as shown to a 200-N load and can slide on a frictionless horizontal rod. Determine the distance x for which the collar is in equilibrium when P = 192 N. SOLUTION Free Body: Collar A Force Triangle N 2 = (200)2 − (192)2 = 3136 N = 56 N ngles Similar Triangles x 192 N = 400 mm 56 N x = 1371 mm x = 1371 mm lOMoARcPSD|6860698 PROBLEM 2.65 Three forces are applied to a bracket as shown. The directions of the two 150-N forces may vary, but the angle between these forces is always 50°. Determine the range of values of α for which the magnitude of the resultant of the forces acting at A is less than 600 N. SOLUTION Combine the two 150-N forces into a resultant force Q: Q = 2(150 N) cos 25° = 271.89 N Equivalent loading at A: Using the law of cosines: (600 N) 2 = (500 N) 2 + (271.89 N)2 + 2(500 N)(271.89 N) cos(55° + α ) cos(55° + α ) = 0.132685 Two values for α : 55° + α = 82.375 α = 27.4° or 55° + α = −82.375° 55° + α = 360° − 82.375° α = 222.6° For R < 600 lb: 27.4° < α < 222.6D lOMoARcPSD|6860698 PROBL LEM 2.66 A 200-kgg crate is to bee supported by b the rope-annd-pulley arrangement show wn. Determinee the magnituude and directtion of the forrce P that mu ust be exerted on the free ennd of the rope to maintain equilibrium. (H Hint: The tensiion in the ropee is the same on o each side of o a simple pu ulley. This cann be proved byy the methodss of Ch. 4.) SOLUT TION Free-Boody Diagram:: Pulley A ⎛ 5 ⎞ ΣFx = 0: − 2 P ⎜ c α =0 ⎟ + P cos ⎝ 281 ⎠ cos α = 0.59655 α = ±53.377° For α = +53.377°: ⎛ 166 ⎞ ΣFy = 0: 2 P ⎜ ⎟ + P sin 533.377° − 1962 N = 0 ⎝ 281 ⎠ P = 7244 N 53.4° For α = −53.377°: ⎛ 166 ⎞ ΣFy = 0: 2 P ⎜ 5 °) − 19662 N = 0 ⎟ + P sin( −53.377 ⎝ 281 ⎠ P = 17773 53.4° lOMoARcPSD|6860698 PROBLEM 2.67 A 6-kN crate is supported by several rope-andpulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Problem 2.66.) SOLUTION Free-Body Diagram of Pulley (a) ΣFy = 0: 2T − (6 kN) = 0 T= 1 (6 kN) 2 T = 3 kN (b) ΣFy = 0: 2T − (6 kN) = 0 T= 1 (6 kN) 2 T = 3 kN (c) ΣFy = 0: 3T − (6 kN) = 0 1 T = (6 kN) 3 T = 2 kN (d) ΣFy = 0: 3T − (6 kN) = 0 1 T = (6 kN) 3 T = 2 kN (e) ΣFy = 0: 4T − (6 kN) = 0 T= 1 (6 kN) 4 T = 1.5 kN lOMoARcPSD|6860698 PROBLEM 2.68 Solve Parts b and d of Problem 2.67, assuming that the free end of the rope is attached to the crate. PROBLEM 2.67 A 6-kN crate is supported by several rope-and-pulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Problem 2.66.) SOLUTION Free-Body Diagram of Pulley and Crate (b) ΣFy = 0: 3T − (6 kN) = 0 1 T = (6 kN) 3 T = 2 kN (d) ΣFy = 0: 4T − (6 kN) = 0 T= 1 (6 kN) 4 T = 1.5 kN lOMoARcPSD|6860698 PRO OBLEM 2.6 69 A load Q is appliied to the pullley C, whichh can roll on the cablee ACB. The pulley p is heldd in the position shown byy a seconnd cable CAD D, which paasses over thee pulley A and a 50 N, determ suppoorts a load P. Knowing that P = 75 mine (a) thhe tension in cable c ACB, (b)) the magnitudde of load Q. SOLUT TION Free-Boody Diagram:: Pulley C ΣFx = 0: TACBB (cos 25° − coss55°) − (750 N)cos55° N =0 (a) Hencce: TACB = 12992.88 N TACB = 1293 N A ΣFy = 0: TACB (sin ( 25° + sin 55°) + (750 N) sin 55° − Q = 0 (b) (1292.88 N)(sin 25° + sin 55°) + (750 N) sin 55° − Q = 0 or Q = 2219.8 N Q = 2220 N lOMoARcPSD|6860698 PRO OBLEM 2.7 70 An 1800-N load Q is applied too the pulley C, C which can roll r on thhe cable ACB. The pulley is held in the poosition shown by a seccond cable CAD, CA which passes p over thhe pulley A and a suppoorts a load P. Determine (a) ( the tensionn in cable AC CB, (b) thhe magnitude of o load P. SOLUT TION Free-Boody Diagram:: Pulley C ΣFx = 0: 0 TACB (cos 25 2 ° − cos55°) − P cos55° = 0 P = 0.58010 0 TACB (1) or ΣFy = 0: TACB (sin 25 2 ° + sin 55°) + P sin 55° − 1800 N = 0 1.24177T TACB + 0.819155P = 1800 N (2) or (a) Substittute Equation (1) ( into Equattion (2): 1.244177TACB + 0.81915(0.580110TACB ) = 18000 N Hence: TACB = 10488.37 N TACB = 1048 N A (b) ( Using (1), P = 0.58010(1048.337 N) = 608.166 N P = 608 N lOMoARcPSD|6860698 PROBLEM 2.71 Determine (a) the x, y, and z components of the 600-N force, (b) the angles θx, θy, and θz that the force forms with the coordinate axes. SOLUTION (a) Fx = (600 N)sin 25° cos30D Fx = 219.60 N Fx = 220 N Fy = (600 N) cos 25° Fy = 544 N Fy = 543.78 N Fz = (380.36 N)sin 25° sin 30D Fz = 126.785 N (b) Fz = 126.8 N cos θ x = Fx 219.60 N = F 600 N θ x = 68.5° cos θ y = Fy 543.78 N 600 N θ y = 25.0° cos θ z = Fz 126.785 N = 600 N F θ z = 77.8° F = lOMoARcPSD|6860698 PROBLEM 2.72 Determine (a) the x, y, and z components of the 450-N force, (b) the angles θx, θy, and θz that the force forms with the coordinate axes. SOLUTION (a) Fx = −(450 N)cos 35° sin 40D Fx = −236.94 N Fx = −237 N Fy = (450 N)sin 35° Fy = 258 N Fy = 258.11 N Fz = (450 N) cos35° cos 40D Fz = 282.38 N (b) Fz = 282 N cos θ x = Fx -236.94 N = F 450 N θx = 121.8° cos θ y = Fy 258.11 N 450 N θ y = 55.0° cos θ z = Fz 282.38 N = 450 N F θ z = 51.1° F = Note: From the given data, we could have computed directly θ y = 90D − 35D = 55D , which checks with the answer obtained. lOMoARcPSD|6860698 PROBLEM 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z components of that force, (b) the values of the angles θx, θy, and θz defining the direction of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.) SOLUTION Recoil force F = 400 N ∴ FH = (400 N)cos 40° = 306.42 N (a) Fx = − FH sin 35° = −(306.42 N)sin 35° = −175.755 N Fx = −175.8 N Fy = − F sin 40° = −(400 N)sin 40° = −257.12 N Fy = −257 N Fz = + FH cos 35° = +(306.42 N) cos 35° = +251.00 N (b) cos θ x = Fz = +251 N Fx −175.755 N = F 400 N θ x = 116.1° cos θ y = Fy −257.12 N 400 N θ y = 130.0° cos θ z = Fz 251.00 N = 400 N F θ z = 51.1° F = lOMoARcPSD|6860698 PROBLEM 2.74 Solve Problem 2.73, assuming that point A is located 15° north of west and that the barrel of the gun forms an angle of 25° with the horizontal. PROBLEM 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z components of that force, (b) the values of the angles θx, θy, and θz defining the direction of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.) SOLUTION Recoil force F = 400 N ∴ FH = (400 N)cos 25° = 362.52 N (a) Fx = + FH cos15° = +(362.52 N)cos15° = +350.17 N Fx = +350 N Fy = − F sin 25° = −(400 N)sin 25° = −169.047 N Fy = −169.0 N Fz = + FH sin15° = +(362.52 N)sin15° = +93.827 N (b) cos θ x = cosθ y = cos θ z = Fz = +93.8 N Fx +350.17 N = 400 N F θ x = 28.9° −169.047 N 400 N θ y = 115.0° Fz +93.827 N = 400 N F θ z = 76.4° Fy F = lOMoARcPSD|6860698 PROBLEM 2.75 The angle between spring AB and the post DA is 30°. Knowing that the tension in the spring is 50 N, determine (a) the x, y, and z components of the force exerted on the circular plate at B, (b) the angles θx, θy, and θz defining the direction of the force at B. SOLUTION Fh = F cos 60° = (50 N) cos 60 ° Fh = 25.0 N Fx = − Fh cos 35° Fy = F sin 60D Fz = − Fh sin 35D Fx = (−25.0 N)cos35D Fy = (50.0 N)sin 60D Fz = (−25.0 N)sin 35D Fx = −20.479 N Fy = 43.301 N Fz = −14.3394 N (a) Fx = −20.5 N Fy = 43.3 N Fz = −14.33 N (b) cos θ x = Fx −20.479 N = F 50 N θ x = 114.2° cos θ y = Fy 43.301 N 50 N θ y = 30.0° cos θ z = Fz -14.3394 N = 50 N F θ z = 106.7° F = lOMoARcPSD|6860698 PROBLEM 2.76 The angle between spring AC and the post DA is 30°. Knowing that the tension in the spring is 40 N, determine (a) the x, y, and z components of the force exerted on the circular plate at C, (b) the angles θx, θy, and θz defining the direction of the force at C. SOLUTION Fh = F cos 60° = (40 N) cos 60 ° Fh = 20.0 N (a) Fx = Fh cos 35° = (20.0 N) cos 35° Fx = 16.3830 N Fy = F sin 60° = (40 N)sin 60° Fy = 34.641 N Fz = −Fh sin 35° = −(20.0 N)sin 35° Fz = −11.4715 N Fx = 16.38 N Fy = 34.6 N Fz = −11.47 N (b) cos θ x = Fx 16.3830 N = F 40 N θ x = 65.8° cos θ y = Fy 34.641 N 40 N θ y = 30.0° cos θ z = Fz -11.4715 N = 40 N F θ z = 106.7° F = lOMoARcPSD|6860698 P PROBLEM 2.77 Cable AB is 32.5 m long, and C a the tensioon in that caable is 15 kN. D Determine (a) the x, y, andd z componentts of the forcee exerted by the c cable on the anchor a B, (b) the angles θ x , θ y , and θ z defining the d direction of thaat force. TION SOLUT 28 m 32.5 m = 0.86154 θ y = 30.51° cos θ y = From trianngle AOB: Fx = − F sin θ y cos 20° (a) =−(15 kN)sin 30.51° coos 20° Fx =−7.156 kN (b) Fy =+ F cosθ y = (15 kN)(00.86154) Fy =+12.923 kN Fz =+(15 kN)sin 30.51° sin i 20° Fz = +2.605 kN Fx 7.156 kN = − 0.47771 =− F 15 kN θ x = 118.5° θ y = 30.551° θ y = 30.5° cos θ x = Froom above: cos θ z = Fz 2.605 kN =+ = + 0.1737 15 kN F θ z = 80.0° lOMoARcPSD|6860698 PROBLEM 2.78 Cable AC is 35 m long, and the tension in that cable is 21 kN. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor C, (b) the angles θx, θy, and θz defining the direction of that force. SOLUTION AC = 35 m OA = 28 m F = 21 kN In triangle AOB: 28 m 35 m θ y = 36.870° cos θ y = FH = F sin θ y = (21 kN)sin 36.870° = 12.6 kN = Fx −FH sin 50° =−(12.6 kN)sin 50° =−9.652 kN (a) Fx = −9.652 kN Fy = + F cosθ y = +(21 kN) cos36.870° = + 16.8 kN Fy = +16.8 kN Fz = −FH cos50° = −(12.6 kN ) cos50° = 8.099 kN (b) cos θ x = From above: Fz = −8.099 kN Fx −9.652 kN = F 21 kN θ x = 117.4° θ y = 36.870° θ y = 36.9° cos θ z = Fz −8.099 kN = − 0.3857 = 21 kN F θ z = 112.7° lOMoARcPSD|6860698 PROBLEM 2.79 Determine the magnitude and direction of the force F = (240 N)i – (270 N)j + (680 N)k. SOLUTION F = Fx2 + Fy2 + Fz2 F = (240 N) 2 + (−270 N) 2 + (−680 N) 2 F = 770 N cos θ x = Fx 240 N = F 770 N θ x = 71.8° cos θ y = Fy −270 N 770 N θ y = 110.5° cos θ y = Fz 680 N = F 770 N θ z = 28.0° F = lOMoARcPSD|6860698 PROBLEM 2.80 Determine the magnitude and direction of the force F = (320 N)i + (400 N)j − (250 N)k. SOLUTION F = Fx2 + Fy2 + Fz2 F = (320 N) 2 + (400 N) 2 + (−250 N) 2 F = 570 N cos θ x = Fx 320 N = F 570 N θ x = 55.8° cos θ y = Fy 400 N 570 N θ y = 45.4° Fz −250 N = 570 N F θ z = 116.0° cos θ y = F = lOMoARcPSD|6860698 PROBLEM 2.81 A force acts at the origin of a coordinate system in a direction defined by the angles θx = 69.3° and θz = 57.9°. Knowing that the y component of the force is –174.0 N, determine (a) the angle θ y, (b) the other components and the magnitude of the force. SOLUTION cos 2 θ x + cos 2 θ y + cos 2 θ z = 1 cos 2 (69.3°) + cos 2 θ y + cos 2 (57.9°) = 1 cos θ y = ±0.7699 (a) (b) Since Fy < 0, we choose cosθ y = −0.7699 ∴ θ y = 140.3° Fy = F cos θ y −174.0 N = F ( −0.7699) F = 226.0 N F = 226 N Fx = F cosθ x = (226.0 N)cos69.3° Fx = 79.9 N Fz = F cosθ z = (226.0 N)cos57.9° Fz = 120.1N lOMoARcPSD|6860698 PROBLEM 2.82 A force acts at the origin of a coordinate system in a direction defined by the angles θx = 70.9° and θy = 144.9°. Knowing that the z component of the force is –52.0 N, determine (a) the angle θ z, (b) the other components and the magnitude of the force. SOLUTION cos 2 θ x + cos 2 θ y + cos 2 θ z = 1 cos 2 70.9D + cos 2 144.9° + cos 2 θ z ° = 1 cos θ z = ±0.47282 (a) (b) Since Fz < 0, we choose cosθ z = −0.47282 ∴ θ z = 118.2° Fz = F cos θ z −52.0 N = F (−0.47282) F = 110.0 N F = 110.0 N Fx = F cosθ x = (110.0 N)cos70.9° Fx = 36.0 N Fy = F cosθ y = (110.0 N)cos144.9° Fy = −90.0 N lOMoARcPSD|6860698 PROBLEM 2.83 A force F of magnitude 210 N acts at the origin of a coordinate system. Knowing that Fx = 80 N, θz = 151.2°, and Fy < 0, determine (a) the components Fy and Fz, (b) the angles θx and θy. SOLUTION Fz = F cosθ z = (210 N)cos151.2° (a) = −184.024 N Then: So: Hence: F 2 = Fx2 + Fy2 + Fz2 (210 N) 2 = (80 N) 2 + ( Fy )2 + (184.024 N)2 Fy = − (210 N) 2 − (80 N) 2 − (184.024 N) 2 = −61.929 N (b) Fz = −184.0 N cos θ x = Fx 80 N = = 0.38095 F 210 N cos θ y = Fy F = Fy = −62.0 N θ x = 67.6° 61.929 N = −0.29490 210 N θ y = 107.2° lOMoARcPSD|6860698 PROBLEM 2.84 A force F of magnitude 1200 N acts at the origin of a coordinate system. Knowing that θx = 65°, θy = 40°, and Fz > 0, determine (a) the components of the force, (b) the angle θz. SOLUTION cos 2 θ x + cos 2 θ y + cos 2 θ z = 1 cos 2 65D + cos 2 40° + cos 2 θ z ° = 1 cos θ z = ±0.48432 (b) (a) Since Fz > 0, we choose cosθ z = 0.48432, or θ z = 61.032D ∴ θ z = 61.0° F = 1200 N Fx = F cosθ x = (1200 N) cos65D Fx = 507 N Fy = F cosθ y = (1200 N) cos 40° Fy = 919 N Fz = F cos θ z = (1200 N) cos 61.032° Fz = 582 N lOMoARcPSD|6860698 PROBLEM 2.85 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D. SOLUTION JJJG DB = (480 mm)i − (510 mm) j + (320 mm)k DB = (480 mm)2 + (510 mm2 ) + (320 mm)2 = 770 mm F = F λ DB JJJG DB =F DB 385 N [(480 mm)i − (510 mm)j + (320 mm)k ] = 770 mm = (240 N)i − (255 N) j + (160 N)k Fx = +240 N, Fy = −255 N, Fz = +160.0 N lOMoARcPSD|6860698 PROBLEM 2.86 For the frame and cable of Problem 2.85, determine the components of the force exerted by the cable on the support at E. PROBLEM 2.85 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D. SOLUTION JJJG EB = (270 mm)i − (400 mm) j + (600 mm)k EB = (270 mm)2 + (400 mm)2 + (600 mm)2 = 770 mm F = F λ EB JJJG EB =F EB 385 N [(270 mm)i − (400 mm)j + (600 mm)k ] = 770 mm F = (135 N)i − (200 N) j + (300 N)k Fx = +135.0 N, Fy = −200 N, Fz = +300 N lOMoARcPSD|6860698 PROBLEM 2.87 In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension in cable AB is 10 kN, determine the components of the force exerted at A by the cable. SOLUTION Cable AB: JJJG AB (−15.588 m) i + (15m) j + (12 m)k = λAB = 24.739 m AB −15.588i + 15 j + 12k 10 kN TAB = TAB λAB = 24.739 ( ) (TAB ) x = −6.3 kN (TAB ) y = +6.063 kN (TAB ) z = +4.851 kN lOMoARcPSD|6860698 PROBLEM 2.88 In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension in cable AC is 7.5 kN, determine the components of the force exerted at A by the cable. SOLUTION Cable AB: JJJG AC ( −15.588 m) i + (18.6 m) j + ( −15 m) k = λAC = 28.53 m AC TAC = TAC λAC = (7.5 kN) −15.588i + 18.6 j − 15k 28.53 (TAC ) x = −4.098 kN (TAC ) y = +4.89 kN (TAC ) z = −3.943 kN lOMoARcPSD|6860698 PROBLEM 2.89 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AB is 408 N, determine the components of the force exerted on the plate at B. SOLUTION We have: JJJG BA = +(320 mm)i + (480 mm)j − (360 mm)k Thus: BA = 680 mm JJJG BA 12 9 ⎞ ⎛ 8 = TBA ⎜ i + j - k ⎟ FB = TBA λBA = TBA BA ⎝ 17 17 17 ⎠ ⎛8 ⎞ ⎛ 12 ⎞ ⎛ 9 ⎞ ⎜ 17 TBA ⎟ i + ⎜ 17 TBA ⎟ j − ⎜ 17 TBA ⎟ k = 0 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Setting TBA = 408 N yields, Fx = +192.0 N, Fy = +288 N, Fz = −216 N lOMoARcPSD|6860698 PROBLEM 2.90 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 429 N, determine the components of the force exerted on the plate at D. SOLUTION We have: JJJG DA = −(250 mm)i + (480 mm)j + (360 mm)k Thus: DA = 650 mm JJJG DA 48 36 ⎞ ⎛ 5 j+ k⎟ = TDA ⎜ − i + FD = TDA λDA = TDA DA 65 ⎠ ⎝ 13 65 ⎛5 ⎞ ⎛ 48 ⎞ ⎛ 36 ⎞ − ⎜ TDA ⎟ i + ⎜ TDA ⎟ j + ⎜ TDA ⎟ k = 0 ⎝ 13 ⎠ ⎝ 65 ⎠ ⎝ 65 ⎠ Setting TDA = 429 N yields, Fx = −165.0 N, Fy = +317 N, Fz = +238 N lOMoARcPSD|6860698 PROBLEM 2.91 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 300 N and Q = 400 N. SOLUTION P = (300 N)[− cos 30° sin15°i + sin 30° j + cos 30° cos15°k ] = − (67.243 N)i + (150 N) j + (250.95 N)k Q = (400 N)[cos 50° cos 20°i + sin 50° j − cos 50° sin 20°k ] = (400 N)[0.60402i + 0.76604 j − 0.21985] = (241.61 N)i + (306.42 N) j − (87.939 N)k R =P+Q = (174.367 N)i + (456.42 N) j + (163.011 N)k R = (174.367 N)2 + (456.42 N) 2 + (163.011 N) 2 = 515.07 N R = 515 N cos θ x = Rx 174.367 N = = 0.33853 515.07 N R θ x = 70.2° cos θ y = Ry 456.42 N = 0.88613 515.07 N θ y = 27.6° cos θ z = Rz 163.011 N = = 0.31648 R 515.07 N θ z = 71.5° R = lOMoARcPSD|6860698 PROBLEM 2.92 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 400 N and Q = 300 N. SOLUTION P = (400 N)[− cos 30° sin15°i + sin 30° j + cos 30° cos15°k ] = − (89.678 N)i + (200 N) j + (334.61 N)k Q = (300 N)[cos50° cos 20°i + sin 50° j − cos 50° sin 20°k ] = (181.21 N)i + (229.81 N)j − (65.954 N)k R =P+Q = (91.532 N)i + (429.81 N) j + (268.66 N)k R = (91.532 N)2 + (429.81 N)2 + (268.66 N) 2 = 515.07 N R = 515 N cos θ x = Rx 91.532 N = = 0.177708 R 515.07 N θ x = 79.8° cos θ y = Ry 429.81 N = 0.83447 515.07 N θ y = 33.4° cos θ z = Rz 268.66 N = = 0.52160 R 515.07 N θ z = 58.6° R = lOMoARcPSD|6860698 PROBLEM 2.93 Knowing that the tension is 425 N in cable AB and 510 N in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables. SOLUTION AB = (40 cm)i − (45 cm)j + (60 cm)k AB = (40 cm)2 + (45 cm)2 + (60 cm)2 = 85 cm AC = (100 cm)i − (45 cm)j + (60 cm)k AC = (100 cm)2 + (45 cm)2 + (60 cm)2 = 125 cm TAB = TAB AB = TAB (40 cm)i − (45 cm)j + (60 cm)k AB = (425 N) 85 cm AB TAB = (200 N)i − (225 N)j + (300 N)k TAC = TAC AC = TAC AC (100 cm)i − (45 cm)j + (60 cm)k = (510 N) 125 cm AC TAC = (408 N)i − (183.6 N)j + (244.8 N)k R = TAB + TAC = (608 N)i − (408.6 N)j + (544.8 N)k Then: and R = 912.92 N R = 913 N cos θ x = 608 N = 0.66599 912.92 N θ x = 48.2° cos θ y = −408.6 N = −0.44757 912.92 N θ y = 116.6° cos θ z = 544.8 N = 0.59677 912.92 N θ z = 53.4° lOMoARcPSD|6860698 PROBLEM 2.94 Knowing that the tension is 510 N in cable AB and 425 N in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables. SOLUTION AB = (40 cm)i − (45 cm)j + (60 cm)k AB = (40 cm)2 + (45 cm)2 + (60 cm)2 = 85 cm AC = (100 cm)i − (45 cm)j + (60 cm)k AC = (100 cm)2 + (45 cm)2 + (60 cm)2 = 125 cm TAB = TAB AB = TAB (40 cm)i − (45 cm)j + (60 cm)k AB = (510 N) 85 cm AB TAB = (240 N)i − (270 N)j + (360 N)k TAC = TAC AC = TAC AC (100 cm)i − (45 cm)j + (60 cm)k = (425 N) AC 125 cm TAC = (340 N)i − (153 N)j + (204 N)k R = TAB + TAC = (580 N)i − (423 N)j + (564 N)k Then: and R = 912.92 N R = 913 N cos θ x = 580 N = 0.63532 912.92 N θ x = 50.6° cos θ y = −423 N = −0.46335 912.92 N θ y = 117.6° cos θ z = 564 N = 0.61780 912.92 N θ z = 51.8° lOMoARcPSD|6860698 PROBLEM 2.95 For the frame of Problem 2.85, determine the magnitude and direction of the resultant of the forces exerted by the cable at B knowing that the tension in the cable is 385 N. PROBLEM 2.85 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D. SOLUTION JJJG BD = − (480 mm)i + (510 mm) j − (320 mm)k BD = (480 mm) 2 + (510 mm) 2 + (320 mm) 2 = 770 mm JJJG BD FBD = TBD λ BD = TBD BD (385 N) [−(480 mm)i + (510 mm) j − (320 mm)k ] = (770 mm) = −(240 N)i + (255 N) j − (160 N)k JJJG BE = −(270 mm)i + (400 mm) j − (600 mm)k BE = (270 mm) 2 + (400 mm) 2 + (600 mm) 2 = 770 mm JJJG BE FBE = TBE λ BE = TBE BE (385 N) [−(270 mm)i + (400 mm) j − (600 mm)k ] = (770 mm) = −(135 N)i + (200 N) j − (300 N)k R = FBD + FBE = −(375 N)i + (455 N) j − (460 N)k R = (375 N)2 + (455 N)2 + (460 N)2 = 747.83 N R = 748 N cos θ x = −375 N 747.83 N θ x = 120.1° cos θ y = 455 N 747.83 N θ y = 52.5° cos θ z = −460 N 747.83 N θ z = 128.0° lOMoARcPSD|6860698 PROBLEM 2.96 For the plate of Prob. 2.89, determine the tensions in cables AB and AD knowing that the tension in cable AC is 54 N and that the resultant of the forces exerted by the three cables at A must be vertical. SOLUTION We have: Thus: JJJG AB = −(320 mm)i − (480 mm)j + (360 mm)k AB = 680 mm JJJG AC = (450 mm)i − (480 mm) j + (360 mm)k AC = 750 mm JJJG AD = (250 mm)i − (480 mm) j − ( 360 mm ) k AD = 650 mm JJJG AB TAB TAB = TAB λ AB = TAB = ( −320i − 480 j + 360k ) AB 680 JJJG AC 54 TAC = TAC λAC = TAC = ( 450i − 480 j + 360k ) AC 750 JJJG AD TAD TAD = TAD λAD = TAD = ( 250i − 480 j − 360k ) AD 650 Substituting into the Eq. R = ΣF and factoring i, j, k : 250 ⎛ 320 ⎞ TAB + 32.40 + TAD ⎟ i R = ⎜− 650 ⎝ 680 ⎠ 480 ⎛ 480 ⎞ + ⎜− TAB − 34.560 − TAD ⎟ j 650 ⎝ 680 ⎠ 360 ⎛ 360 ⎞ +⎜ TAB + 25.920 − TAD ⎟ k 650 ⎝ 680 ⎠ lOMoARcPSD|6860698 PROBLEM 2.96 (Continued) Since R is vertical, the coefficients of i and k are zero: i: − 320 250 TAB + 32.40 + TAD = 0 680 650 (1) 360 360 TAB + 25.920 − TAD = 0 680 650 (2) k: Multiply (1) by 3.6 and (2) by 2.5 then add: − 252 TAB + 181.440 = 0 680 TAB = 489.60 N TAB = 490 N Substitute into (2) and solve for TAD : 360 360 (489.60 N) + 25.920 − TAD = 0 680 650 TAD = 514.80 N TAD = 515 N lOMoARcPSD|6860698 PROBLEM 2.97 The boom OA carries a load P and is supported by two cables as shown. Knowing that the tension in cable AB is 183 N and that the resultant of the load P and of the forces exerted at A by the two cables must be directed along OA, determine the tension in cable AC. SOLUTION Cable AB: Cable AC: Load P: TAB = 183 N JJJG AB (−48 cm) i + (29 cm)j + (24 cm) k TAB = TAB λ AB = TAB = (183 N) AB 61 cm TAB = −(144 N)i + (87 N) j + (72 N) k JJJG AC (−48 cm)i + (25 cm)j + (− 36 cm) k TAC = TAC λ AC = TAC = TAC AC 65 cm 48 25 36 TAC = − TAC i + TAC j − TAC k 65 65 65 P = Pj For resultant to be directed along OA, i.e., x-axis Rz = 0: ΣFz = (72 N) − 36 ′ =0 TAC 65 TAC = 130.0 N lOMoARcPSD|6860698 PROBLEM 2.98 For the boom and loading of Problem. 2.97, determine the magnitude of the load P. PROBLEM 2.97 The boom OA carries a load P and is supported by two cables as shown. Knowing that the tension in cable AB is 183 N and that the resultant of the load P and of the forces exerted at A by the two cables must be directed along OA, determine the tension in cable AC. SOLUTION See Problem 2.97. Since resultant must be directed along OA, i.e., the x-axis, we write Ry = 0: ΣFy = (87 N) + 25 TAC − P = 0 65 TAC = 130.0 N from Problem 2.97. Then (87 N) + 25 (130.0 N) − P = 0 65 P = 137.0 N lOMoARcPSD|6860698 PROBLEM 2.99 A container is supported by three cables that are attached to a ceiling as shown. Determine the weight W of the container, knowing that the tension in cable AB is 6 kN. SOLUTION Free-Body Diagram at A: The forces applied at A are: TAB , TAC , TAD , and W where W = W j. To express the other forces in terms of the unit vectors i, j, k, we write JJJG AB = − (450 mm)i + (600 mm) j AB = 750 mm JJJG AC = + (600 mm) j − (320 mm)k AC = 680 mm JJJG AD = + (500 mm)i + (600 mm) j + (360 mm)k AD = 860 mm JJJG (−450 mm)i + (600 mm) j AB TAB = λ ABTAB = TAB = TAB and AB 750 mm ⎛ 45 60 ⎞ = ⎜ − i + j ⎟ TAB ⎝ 75 75 ⎠ JJJG (600 mm)i − (320 mm) j AC TAC = λ AC TAC = TAC = TAC 680 mm AC 32 ⎞ ⎛ 60 = ⎜ j − k ⎟ TAC 68 ⎠ ⎝ 68 JJJG AD (500 mm)i + (600 mm) j + (360 mm)k TAD = λ ADTAD = TAD = TAD AD 860 mm 60 36 ⎞ ⎛ 50 j + k ⎟ TAD =⎜ i+ 86 86 86 ⎝ ⎠ lOMoARcPSD|6860698 PROBLEM 2.99 (Continued) Equilibrium condition: ΣF = 0: ∴ TAB + TAC + TAD + W = 0 Substituting the expressions obtained for TAB , TAC , and TAD ; factoring i, j, and k; and equating each of the coefficients to zero gives the following equations: From i: From j: From k: 45 50 TAB + TAD = 0 75 86 (1) 60 60 60 TAB + TAC + TAD − W = 0 75 68 86 (2) 32 36 TAC + TAD = 0 68 86 (3) − − Setting TAB = 6 kN in (1) and (2), and solving the resulting set of equations gives TAC = 6.1920 kN TAC = 5.5080 kN W = 13.98 kN lOMoARcPSD|6860698 PROBLEM 2.100 A container is supported by three cables that are attached to a ceiling as shown. Determine the weight W of the container, knowing that the tension in cable AD is 4.3 kN. SOLUTION See Problem 2.99 for the figure and analysis leading to the following set of linear algebraic equations: 45 50 TAB + TAD = 0 75 86 (1) 60 60 60 TAB + TAC + TAD − W = 0 75 68 86 (2) 32 36 TAC + TAD = 0 68 86 (3) − − Setting TAD = 4.3 kN into the above equations gives TAB = 4.1667 kN TAC = 3.8250 kN W = 9.71 kN lOMoARcPSD|6860698 PROB BLEM 2.101 1 Three cables c are useed to tether a balloon as shhown. Determine the verttical force P exerted e by the balloon at A knowing k that the tension in cable AD is 481 N. TION SOLUT The forcces applied at A are: FREE-BOD DY DIAGRA AM AT A TAB , TAC , TAD , andd P where P = Pj. To exp press the otherr forces in term ms of the unit vectors i, j, k, we write JJJG AB = 7.00 m AB = − (4.20 m)i − (5.60 m) j JJJG AC = (2.40 m)i − (5.60 ( m) j + (4.20 m)k AC = 7.40 m JJJG AD = − (5.60 m) j − (3.30 m)k D = 6.50 m AD JJJG AB and = (− 0.6i − 0.8 j)TAB TAB = TAB λ AB = TABB AB JJJG AC = (0.322432i − 0.756776 j + 0.56757k )TAC TAC = TAC λ AC = TAC A AC JJJG AD = (− 0.886154 j − 0.507769k )TAD TAD = TAD λ AD = TAD A AD lOMoARcPSD|6860698 P PROBLEM 2 2.101 (Con ntinued) Equilibrrium condition n: ΣF = 0: TABB + TAC + TAD + Pj = 0 Substituuting the expreessions obtaineed for TAB , TAC nd factoring i,, j, and k: A , and TAD an (− − 0.6TAB + 0.32432TAC )i + (−0.8TAB − 0.75676TAC − 0.886154TAD + P)j ) + (0.567577TAC − 0.507699TAD )k = 0 Equatingg to zero the coefficients c off i, j, k: − 0.6TAB A + 0.32432TAC = 0 (1) − 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0 (2) 0.56757T TAC − 0.50769TAD = 0 (3) Setting TAD = 481 N in (2) and (3),, and solving the t resulting seet of equations gives TAC = 4330.26 N TAD = 2332.57 N P = 926 N lOMoARcPSD|6860698 PROBLEM 2.102 Three cables are used to tether a balloon as shown. Knowing that the balloon exerts an 800-N vertical force at A, determine the tension in each cable. SOLUTION See Problem 2.101 for the figure and analysis leading to the linear algebraic Equations (1), (2), and (3). − 0.6TAB + 0.32432TAC = 0 (1) − 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0 (2) 0.56757TAC − 0.50769TAD = 0 (3) From Eq. (1): TAB = 0.54053TAC From Eq. (3): TAD = 1.11795TAC Substituting for TAB and TAD in terms of TAC into Eq. (2) gives − 0.8(0.54053TAC ) − 0.75676TAC − 0.86154(1.11795TAC ) + P = 0 2.1523TAC = P ; P = 800 N 800 N 2.1523 = 371.69 N TAC = Substituting into expressions for TAB and TAD gives TAB = 0.54053(371.69 N) TAD = 1.11795(371.69 N) TAB = 201 N, TAC = 372 N, TAD = 416 N lOMoARcPSD|6860698 PROBLEM 2.103 A 36-N triangular plate is supported by three wires as shown. Determine the tension in each wire, knowing that a = 6 cm. SOLUTION By Symmetry TDB = TDC Free-Body Diagram of Point D: The forces applied at D are: TDB , TDC , TDA , and P where P = Pj = (36 N)j. To express the other forces in terms of the unit vectors i, j, k, we write JJJG DA = (16 cm) i − (24 cm) j DA = 28.844 cm JJJG DB = −(8 cm)i − (24 cm) j + (6 cm) k DB = 26.0 cm JJJG DC = −(8 cm)i − (24 cm)j − (6 cm) k DC = 26.0 cm JJJG DA and TDA = TDA λDA = TDA = (0.55471i − 0.83206 j)TDA DA JJJG DB TDB = TDB λ DB = TDB = (−0.30769i − 0.92308 j + 0.23077k )TDB DB JJJG DC TDC = TDC λ DC = TDC = (−0.30769i − 0.92308 j − 0.23077k )TDC DC lOMoARcPSD|6860698 PROBLEM 2.103 (Continued) Equilibrium condition: ΣF = 0: TDA + TDB + TDC + (36 N) j = 0 Substituting the expressions obtained for TDA , TDB , and TDC and factoring i, j, and k: (0.55471TDA − 0.30769TDB − 0.30769TDC )i + (−0.83206TDA − 0.92308TDB − 0.92308TDC + 36 N) j + (0.23077TDB − 0.23077TDC )k = 0 Equating to zero the coefficients of i, j, k: 0.55471TDA − 0.30769TDB − 0.30769TDC = 0 (1) − 0.83206TDA − 0.92308TDB − 0.92308TDC + 36 N = 0 (2) 0.23077TDB − 0.23077TDC = 0 (3) Equation (3) confirms that TDB = TDC . Solving simultaneously gives, TDA = 14.42 N; TDB = TDC = 13.00 N lOMoARcPSD|6860698 PROBLEM 2.104 Solve Prob. 2.103, assuming that a = 8 cm PROBLEM 2.103 A 36-N triangular plate is supported by three wires as shown. Determine the tension in each wire, knowing that a = 6 cm SOLUTION By Symmetry TDB = TDC Free-Body Diagram of Point D: The forces applied at D are: TDB , TDC , TDA , and P where P = Pj = (36 N)j. To express the other forces in terms of the unit vectors i, j, k, we write JJJG DA = (16 cm) i − (24 cm) j DA = 28.844 cm JJJG DB = −(8 cm)i − (24 cm) j + (8 cm) k DB = 26.533 cm JJJG DC = −(8 cm)i − (24 cm)j − (8 cm)k DC = 26.533 cm JJJG DA and TDA = TDA λDA = TDA = (0.55471i − 0.83206 j)TDA DA JJJG DB TDB = TDB λDB = TDB = (−0.30151i − 0.90453j + 0.30151k )TDB DB JJJG DC = (−0.30151i − 0.90453j − 0.30151k )TDC TDC = TDC λDC = TDC DC lOMoARcPSD|6860698 PROBLEM 2.104 (Continued) Equilibrium condition: ΣF = 0: TDA + TDB + TDC + (36 N) j = 0 Substituting the expressions obtained for TDA , TDB , and TDC and factoring i, j, and k: (0.55471TDA − 0.30151TDB − 0.30151TDC )i + (−0.83206TDA − 0.90453TDB − 0.90453TDC + 36 N) j + (0.30151TDB − 0.30151TDC )k = 0 Equating to zero the coefficients of i, j, k: 0.55471TDA − 0.30151TDB − 0.30151TDC = 0 (1) − 0.83206TDA − 0.90453TDB − 0.90453TDC + 36 N = 0 (2) 0.30151TDB − 0.30151TDC = 0 (3) Equation (3) confirms that TDB = TDC . Solving simultaneously gives, TDA = 14.42 N; TDB = TDC = 13.27 N lOMoARcPSD|6860698 PR ROBLEM 2.105 2 A crate is suppoorted by threee cables as shhown. Determine thee weight of thee crate knowinng that the tennsion in cable AC A is 544 5 N. Solutionn The forces ap pplied at A aree: TAB , TAC , TADD and W where P = Pj. To exp press the otherr forces in term ms of the unit vectors i, j, k, we write JJJG AB = − (36 cm) i + (60 cm) j − (27 cm) k AB = 75 cm JJJG AC = (60 cm) j + ( 32 cm)k AC = 68 cm JJJG AD = (40 cm) i + (60 cm) j − (27 cm) k AD = 77 cm JJJG AB and TAB = TAB λAB = TAB AB = (− 0.48i + 0.88 j − 0.36k )TABB JJJG AC TAC = TAC λAC = TAC A AC = (0.88235 j + 0.47059k )TACC JJJG AD TAD = TAD λ AD = TAD A AD = (0.51948i + 0.77922 j − 0.335065k )TAD Equilibrrium Condition n with W = − Wj ΣF = 0: TAB + TAC A + TAD − Wj = 0 lOMoARcPSD|6860698 PROBLEM 2.105 (Continued) Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k: (−0.48TAB + 0.51948TAD )i + (0.8TAB + 0.88235TAC + 0.77922TAD − W ) j + (−0.36TAB + 0.47059TAC − 0.35065TAD )k = 0 Equating to zero the coefficients of i, j, k: − 0.48TAB + 0.51948TAD = 0 (1) 0.8TAB + 0.88235TAC + 0.77922TAD − W = 0 (2) − 0.36TAB + 0.47059TAC − 0.35065TAD = 0 (3) Substituting TAC = 544 N in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives: TAB = 374.27 N TAD = 345.82 N W = 1.049 kN lOMoARcPSD|6860698 P PROBLEM 2 2.106 A 1.6 kN cratte is supporteed by three cables as show wn. Determine the tension t in eachh cable. TION SOLUT The forcces applied at A are: TAB , TAC , TADD and W where P = Pj. To exp press the otherr forces in term ms of the unit vectors i, j, k, we write JJJG AB = − (36 cm) i + (60 cm) j − (27 cm) k AB = 75 cm JJJG AC = (60 cm) j + ( 32 cm)k AC = 68 cm JJJG AD = (40 cm) i + (60 cm) j − (27 cm) k AD = 77 cm JJJG AB and TAB = TAB λAB = TAB AB = (− 0.48i + 0.88 j − 0.36k )TABB JJJG AC TAC = TAC λAC = TAC A AC = (0.88235 j + 0.47059k )TACC JJJG AD TAD = TAD λ AD = TAD A AD = (0.51948i + 0.77922 j − 0.335065k )TAD Equilibrrium Condition n with W = − Wj ΣF = 0: TAB + TAC A + TAD − Wj = 0 lOMoARcPSD|6860698 PROBLEM 2.106 (Continued) Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k: (−0.48TAB + 0.51948TAD )i + (0.8TAB + 0.88235TAC + 0.77922TAD − W ) j + (−0.36TAB + 0.47059TAC − 0.35065TAD )k = 0 Equating to zero the coefficients of i, j, k: −0.48TAB + 0.51948TAD = 0 (1) 0.8TAB + 0.88235TAC + 0.77922TAD − W = 0 (2) −0.36TAB + 0.47059TAC − 0.35065TAD = 0 (3) Substituting W = 1.6 kN in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms gives, TAB = 571 N TAC = 830 N TAD = 528 N lOMoARcPSD|6860698 PROBLEM 2.107 Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that Q = 0, find the value of P for which the tension in cable AD is 305 N. SOLUTION ΣFA = 0: TAB + TAC + TAD + P = 0 where P = Pi JJJG AB = −(960 mm)i − (240 mm)j + (380 mm)k AB = 1060 mm JJJG AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm JJJG AD = −(960 mm)i + (720 mm) j − (220 mm)k AD = 1220 mm JJJG AB 19 ⎞ ⎛ 48 12 = TAB ⎜ − i − j + k ⎟ TAB = TAB λAB = TAB AB 53 53 ⎠ ⎝ 53 JJJG AC 3 4 ⎞ ⎛ 12 = TAC ⎜ − i − j − k ⎟ TAC = TAC λAC = TAC AC ⎝ 13 13 13 ⎠ 305 N [(−960 mm)i + (720 mm) j − (220 mm)k ] TAD = TAD λAD = 1220 mm = −(240 N)i + (180 N) j − (55 N)k Substituting into ΣFA = 0, factoring i, j, k , and setting each coefficient equal to φ gives: i: P = 48 12 TAB + TAC + 240 N 53 13 (1) j: 12 3 TAB + TAC = 180 N 53 13 (2) k: 19 4 TAB − TAC = 55 N 53 13 (3) Solving the system of linear equations using conventional algorithms gives: TAB = 446.71 N TAC = 341.71 N P = 960 N lOMoARcPSD|6860698 PROBLEM 2.108 Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that P = 1200 N, determine the values of Q for which cable AD is taut. SOLUTION We assume that TAD = 0 and write ΣFA = 0: TAB + TAC + Qj + (1200 N)i = 0 JJJG AB = −(960 mm)i − (240 mm)j + (380 mm)k AB = 1060 mm JJJG AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm JJJG AB ⎛ 48 12 19 ⎞ TAB = TAB λAB = TAB = ⎜ − i − j + k ⎟ TAB AB ⎝ 53 53 53 ⎠ JJJG AC ⎛ 12 3 4 ⎞ TAC = TAC λAC = TAC = ⎜ − i − j − k ⎟ TAC AC ⎝ 13 13 13 ⎠ Substituting into ΣFA = 0, factoring i, j, k , and setting each coefficient equal to φ gives: i: − 48 12 TAB − TAC + 1200 N = 0 53 13 (1) j: − 12 3 TAB − TAC + Q = 0 53 13 (2) k: 19 4 TAB − TAC = 0 53 13 (3) Solving the resulting system of linear equations using conventional algorithms gives: TAB = 605.71 N TAC = 705.71 N Q = 300.00 N 0 ≤ Q < 300 N Note: This solution assumes that Q is directed upward as shown (Q ≥ 0), if negative values of Q are considered, cable AD remains taut, but AC becomes slack for Q = −460 N. lOMoARcPSD|6860698 PROB BLEM 2.109 A rectaangular plate is supportedd by three caables as show wn. Knowinng that the teension in cablle AC is 60 N, N determine the weight of the plate. SOLUT TION We notee that the weight of the pllate is equal in magnitude to the force P exerted byy the support on Point A. Freee Body A : ΣF = 0: TAB + TAC + TAD + Pj = 0 We havee: Thus: JJJG AB = −(320 mm)i − (480 mm)j + (360 ( mm)k AB A = 680 mm m JJJG AC = (4 450 mm)i − (4480 mm) j + (3360 mm)k A AC = 750 mm m JJJG AD = (2 250 mm)i − (4480 mm) j − ( 3360 mm ) k AD A = 650 mm m JJJG AB ⎛ 8 12 9 ⎞ = ⎜ − i − j + k ⎟ TAB TAB = TAB λ AB = TAB A AB ⎝ 17 17 17 ⎠ JJJG AC = ( 0.6i − 0.64 j + 0.488k ) TAC TAC = TAC λAC = TAC A AC JJJG 9.6 7.2 ⎞ AD ⎛ 5 =⎜ i− TAD = TAD λAD = TAD j− k TAD A 13 13 ⎟⎠ AD ⎝ 13 Substituuting into the Eq. E ΣF = 0 annd factoring i, j, k : 5 ⎞ ⎛ 8 A ⎟i ⎜ − 17 TAB + 0.6TAC + 13 TAD ⎠ ⎝ 12 9.6 ⎞ ⎛ TAD + P ⎟ j + ⎜ − TABB − 0.64TAC − 17 13 ⎠ ⎝ 7..2 ⎞ ⎛ 9 TAD ⎟ k = 0 + ⎜ TAB + 0.48TAC − 133 ⎝ 17 ⎠ Dim mensions in mm m lOMoARcPSD|6860698 PROBLEM 2.109 (Continued) Setting the coefficient of i, j, k equal to zero: i: − 8 5 TAB + 0.6TAC + TAD = 0 17 13 (1) j: − 12 9.6 TAB − 0.64TAC − TAD + P = 0 7 13 (2) 9 7.2 TAB + 0.48TAC − TAD = 0 17 13 (3) 8 5 TAB + 36 N + TAD = 0 17 13 (1′) 9 7.2 TAB + 28.8 N − TAD = 0 17 13 (3′) k: Making TAC = 60 N in (1) and (3): − Multiply (1′) by 9, (3′) by 8, and add: 554.4 N − 12.6 TAD = 0 TAD = 572.0 N 13 Substitute into (1′) and solve for TAB : TAB = 17 ⎛ 5 ⎞ ⎜ 36 + × 572 ⎟ 8 ⎝ 13 ⎠ TAB = 544.0 N Substitute for the tensions in Eq. (2) and solve for P: 12 9.6 (544 N) + 0.64(60 N) + (572 N) 17 13 = 844.8 N P= Weight of plate = P = 845 N lOMoARcPSD|6860698 PROBLEM 2.110 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 520 N, determine the weight of the plate. SOLUTION See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 8 5 TAB + 0.6TAC + TAD = 0 17 13 (1) 12 9.6 TAB + 0.64 TAC − TAD + P = 0 17 13 (2) 9 7.2 TAB + 0.48TAC − TAD = 0 17 13 (3) − − Making TAD = 520 N in Eqs. (1) and (3): 8 TAB + 0.6TAC + 200 N = 0 17 (1′) 9 TAB + 0.48TAC − 288 N = 0 17 (3′) − Multiply (1′) by 9, (3′) by 8, and add: 9.24TAC − 504 N = 0 TAC = 54.5455 N Substitute into (1′) and solve for TAB : TAB = 17 (0.6 × 54.5455 + 200) TAB = 494.545 N 8 Substitute for the tensions in Eq. (2) and solve for P: 12 9.6 (494.545 N) + 0.64(54.5455 N) + (520 N) 17 13 = 768.00 N P= Weight of plate = P = 768 N lOMoARcPSD|6860698 PROBLEM 2.111 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AB is 840 N, determine the vertical force P exerted by the tower on the pin at A. SOLUTION ΣF = 0: TAB + TAC + TAD + Pj = 0 We write JJJG AB = − 4 i − 20 j + 5 k = AB 21 m JJJG AC = 12i − 20 j + 3.6 k AC = 23.6 m JJJG AD = − 4 i − 20 j −14.8 k AD = 25.2 m JJJG AB TAB = TAB λ AB = TAB AB 20 5 ⎞ ⎛ 4 j + k ⎟ TAB = ⎜− i − 21 ⎠ ⎝ 21 21 JJJG AC TAC = TAC λAC = TAC AC 50 9 ⎞ ⎛ 30 j + k ⎟ TAC =⎜ i− 59 59 ⎠ ⎝ 59 JJJG AD TAD = TAD λAD = TAD AD 50 37 ⎞ ⎛ 10 j − k ⎟ TAD = ⎜− i − 63 63 63 ⎠ ⎝ Substituting into the Eq. ΣF = 0 and factoring i, j, k : Free-Body Diagram at A: lOMoARcPSD|6860698 PROBLEM 2.111 (Continued) 30 10 ⎛ 4 ⎞ ⎜ − 21 TAB + 59 TAC − 63 TAD ⎟ i ⎝ ⎠ 50 50 ⎛ 20 ⎞ + ⎜ − TAB − TAC − TAD + P ⎟ j 59 63 ⎝ 21 ⎠ 9 37 ⎛ 5 ⎞ + ⎜ TAB + TAC − TAD ⎟ k = 0 21 59 63 ⎝ ⎠ Setting the coefficients of i, j, k , equal to zero: i: − 4 30 10 TAB + TAC − TAD = 0 21 59 63 (1) j: − 20 50 50 TAB − TAC − TAD + P = 0 21 59 63 (2) 5 9 37 TAB + TAC − TAD = 0 21 59 63 (3) k: Set TAB = 840 N in Eqs. (1) – (3): 30 10 TAC − TAD = 0 59 63 (1′) 50 50 TAC − TAD + P = 0 59 63 (2′) 9 37 TAC − TAD = 0 59 63 (3′) −160 N + −800 N − 200 N + Solving, TAC = 458.12 N TAD = 459.53 N P = 1.553 kN P = 1.553 kN lOMoARcPSD|6860698 PROBLEM 2.112 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AC is 590 N, determine the vertical force P exerted by the tower on the pin at A. SOLUTION ΣF = 0: TAB + TAC + TAD + Pj = 0 We write JJJG AB = −4i − 20 j + 5 k AB = 21m JJJG AC = i − j + 3.6 k AC = 23.6 m JJJG AD = −4i − 20 j − 14.8 k AD = 25.2 m JJJG AB TAB = TAB λ AB = TAB AB 20 5 ⎞ ⎛ 4 = ⎜− i − j + k ⎟ TAB 21 ⎠ ⎝ 21 21 JJJG AC TAC = TAC λAC = TAC AC 30 50 9 ⎞ ⎛ =⎜ i− j + k ⎟ TAC 59 59 ⎠ ⎝ 59 JJJG AD TAD = TAD λAD = TAD AD 50 37 ⎞ ⎛ 10 = ⎜− i − j − k ⎟ TAD 63 ⎠ ⎝ 63 63 Substituting into the Eq. ΣF = 0 and factoring i, j, k : Free-Body Diagram at A: lOMoARcPSD|6860698 PROBLEM 2.112 (Continued) 30 10 ⎛ 4 ⎞ ⎜ − 21 TAB + 59 TAC − 63 TAD ⎟ i ⎝ ⎠ 50 50 ⎛ 20 ⎞ + ⎜ − TAB − TAC − TAD + P ⎟ j 21 59 63 ⎝ ⎠ 9 37 ⎛ 5 ⎞ + ⎜ TAB + TAC − TAD ⎟ k = 0 59 63 ⎝ 21 ⎠ Setting the coefficients of i, j, k , equal to zero: i: − 4 30 10 TAB + TAC − TAD = 0 21 59 63 (1) j: − 20 50 50 TAB − TAC − TAD + P = 0 21 59 63 (2) 5 9 37 TAB + TAC − TAD = 0 21 59 63 (3) k: Set TAC = 590 N in Eqs. (1) – (3): 4 10 TAB + 300 N − TAD = 0 21 63 (1′) 20 50 TAB − 500 N − TAD + P = 0 21 63 (2′) 5 37 TAB + 90 N − TAD = 0 21 63 (3′) − − Solving, TAB = 1.082 kN TAD = 591.82 N P = 2 kN lOMoARcPSD|6860698 PROBLEM 2.113 y C 1.2 m O B x 2.4 m z 3 4.8 m 9 .6 m .6 m In trying to move across a slippery icy surface, a 720-N man uses two ropes AB and AC. Knowing that the force exerted on the man by the icy surface is perpendicular to that surface, determine the tension in each rope. A 9m SOLUTION Free-Body Diagram at A 10.2 m 4.8 m A 9m N ⎛ 4.8 9 ⎞ N = N ⎜____ i + ____ j⎟ 10.2 10.2 ⎝ ⎠ and W= W j = − (720 N)j → (−9 m)i + (6 m)j − (3.6)k AC ___ TAC = TACλAC = TAC AC = TAC ____________________ 11.4 m ⎛ 9 6 j − ____ 3.6 k⎞ i + ____ = TAC⎜− ____ ⎟ 11.4 11.4 11.4 ⎝ ⎠ → (−9 m)i + (7.2 m)j + (9.6 m)k AB ___ TAB = TABλAB = TAB AB = TAB _______________________ 15 m ⎛ 9 7.2 j + ___ 9.6 k⎞ i + ___ = TAB⎜− ___ 15 15 15 ⎟⎠ ⎝ Equilibrium condition: ΣF = 0 TAB + TAC + N + W = 0 lOMoARcPSD|6860698 PROBLEM 2.113 (Continued) Substituting the expressions obtained for TAB , TAC , N, and W; factoring i, j, and k; and equating each of the coefficients to zero gives the following equations: 9 9 4.8 TAB − TAC + N =0 15 11.4 10.2 (1) From j: 7.2 6 9 TAB + TAC + N − (720 N) = 0 15 11.4 10.2 (2) From k: 9.6 3.6 TAB − TAC = 0 15 11.4 (3) From i: − Solving the resulting set of equations gives: TAB = 126.9 N; TAC = 257.3 N lOMoARcPSD|6860698 PROBLEM 2.114 y C O 1.2 m B x 2.4 m z 3 .6 m PROBLEM 2.113 In trying to move across a slippery icy surface, a 720-N man uses two ropes AB and AC. Knowing that the force exerted on the man by the icy surface is perpendicular to that surface, determine the tension in each rope. 4.8 m 9.6 m Solve Problem 2.113, assuming that a friend is helping the man at A by pulling on him with a force P = −(240 N)k. A 9m SOLUTION Refer to Problem 2.113 for the figure and analysis leading to the following set of equations, Equation (3) being modified to include the additional force P = (300 N)k. 7.2 T ___ 15 AB 9 T + ____ 4.8 N = 0 9 T − ____ − ___ 15 AB 11.4 AC 10.2 (1) 6 T + ____ 9 N − (720 N) = 0 + ____ 11.4 AC 10.2 (2) 3.6 T − (240 N) = 0 − ____ 11.4 AC (3) 9.6 T ___ 15 AB Solving the resulting set of equations simultaneously gives: TAB = 395.8 N TAC = 42.2 N lOMoARcPSD|6860698 PROBLEM 2.115 For the rectangular plate of Problems 2.109 and 2.110, determine the tension in each of the three cables knowing that the weight of the plate is 792 N. SOLUTION See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below. Setting P = 792 N gives: 8 5 TAB + 0.6TAC + TAD = 0 17 13 (1) 12 9.6 TAB − 0.64TAC − TAD + 792 N = 0 17 13 (2) 9 7.2 TAB + 0.48TAC − TAD = 0 17 13 (3) − − Solving Equations (1), (2), and (3) by conventional algorithms gives TAB = 510.00 N TAB = 510 N TAC = 56.250 N TAC = 56.2 N TAD = 536.25 N TAD = 536 N lOMoARcPSD|6860698 PROBLEM 2.116 For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = 0. SOLUTION ΣFA = 0: TAB + TAC + TAD + P + Q = 0 Where P = Pi and Q = Qj JJJG AB = −(960 mm)i − (240 mm) j + (380 mm)k AB = 1060 mm JJJG AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm JJJG AD = −(960 mm)i + (720 mm) j − (220 mm)k AD = 1220 mm JJJG 19 ⎞ AB ⎛ 48 12 TAB = TAB λAB = TAB = TAB ⎜ − i − j + k ⎟ 53 ⎠ AB ⎝ 53 53 JJJG 3 4 ⎞ AC ⎛ 12 TAC = TAC λAC = TAC = TAC ⎜ − i − j − k ⎟ AC ⎝ 13 13 13 ⎠ JJJG 36 11 ⎞ AD ⎛ 48 TAD = TAD λAD = TAD j− k⎟ = TAD ⎜ − i + AD 61 61 ⎠ ⎝ 61 Substituting into ΣFA = 0, setting P = (2880 N)i and Q = 0, and setting the coefficients of i, j, k equal to 0, we obtain the following three equilibrium equations: i: − 48 12 48 TAB − TAC − TAD + 2880 N = 0 53 13 61 (1) j: − 12 3 36 TAB − TAC + TAD = 0 53 13 61 (2) k: 19 4 11 TAB − TAC − TAD = 0 53 13 61 (3) lOMoARcPSD|6860698 PROBLEM 2.116 (Continued) Solving the system of linear equations using conventional algorithms gives: TAB = 1340.14 N TAC = 1025.12 N TAD = 915.03 N TAB = 1340 N TAC = 1025 N TAD = 915 N lOMoARcPSD|6860698 PROBLEM 2.117 For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = 576 N. SOLUTION See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below: − 48 12 48 TAB − TAC − TAD + P = 0 53 13 61 (1) − 12 3 36 TAB − TAC + TAD + Q = 0 53 13 61 (2) 19 4 11 TAB − TAC − TAD = 0 53 13 61 (3) Setting P = 2880 N and Q = 576 N gives: − 48 12 48 TAB − TAC − TAD + 2880 N = 0 53 13 61 (1′) 12 3 36 TAB − TAC + TAD + 576 N = 0 53 13 61 (2′) 19 4 11 TAB − TAC − TAD = 0 53 13 61 (3′) − Solving the resulting set of equations using conventional algorithms gives: TAB = 1431.00 N TAC = 1560.00 N TAD = 183.010 N TAB = 1431 N TAC = 1560 N TAD = 183.0 N lOMoARcPSD|6860698 PROBLEM 2.118 For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = −576 N. (Q is directed downward). SOLUTION See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below: − 48 12 48 TAB − TAC − TAD + P = 0 53 13 61 (1) − 12 3 36 TAB − TAC + TAD + Q = 0 53 13 61 (2) 19 4 11 TAB − TAC − TAD = 0 53 13 61 (3) Setting P = 2880 N and Q = −576 N gives: − 48 12 48 TAB − TAC − TAD + 2880 N = 0 53 13 61 (1′) 12 3 36 TAB − TAC + TAD − 576 N = 0 53 13 61 (2′) 19 4 11 TAB − TAC − TAD = 0 53 13 61 (3′) − Solving the resulting set of equations using conventional algorithms gives: TAB = 1249.29 N TAC = 490.31 N TAD = 1646.97 N TAB = 1249 N TAC = 490 N TAD = 1647 N lOMoARcPSD|6860698 PROBLEM 2.119 For the transmission tower of Probs. 2.111 and 2.112, determine the tension in each guy wire knowing that the tower exerts on the pin at A an upward vertical force of 1.8 kN. PROBLEM 2.111 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AB is 840 N, determine the vertical force P exerted by the tower on the pin at A. SOLUTION See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: i: − 4 30 10 TAB + TAC − TAD = 0 21 59 63 (1) j: − 20 50 50 TAB − TAC − TAD + P = 0 21 59 63 (2) 5 9 37 TAB + TAC − TAD = 0 21 59 63 (3) k: Substituting for P = 1.8 kN in Equations (1), (2), and (3) above and solving the resulting set of equations using conventional algorithms gives: 4 30 10 TAB + TAC − TAD = 0 21 59 63 (1′) 20 50 50 TAB − TAC − TAD + 1.8 kN = 0 21 59 63 (2′) 5 9 37 TAB + TAC − TAD = 0 21 59 63 (3′) − − TAB = 973.64 N TAC = 531.00 N TAD = 532.64 N TAB = 974 N TAC = 531 N TAD = 533 N lOMoARcPSD|6860698 PROBLEM 2.120 Three wires are connected at point D, which is located 18 cm below the T-shaped pipe support ABC. Determine the tension in each wire when a 180-N cylinder is suspended from point D as shown. SOLUTION Free-Body Diagram of Point D: The forces applied at D are: TDA , TDB , TDC and W where W = −180.0 N j. To express the other forces in terms of the unit vectors i, j, k, we write JJJG DA = (18 cm) j + (22 cm)k DA = 28.425 cm JJJG DB = −(24 cm)i + (18 cm) j − (16 cm)k DB = 34.0 cm JJJG DC = (24 cm)i + (18 cm) j − (16 cm)k DC = 34.0 cm lOMoARcPSD|6860698 PROBLEM 2.120 (Continued) JJJG DA DA = (0.63324 j + 0.77397k )TDA JJJG DB TDB = TDB λDB = TDB DB = (−0.70588i + 0.52941j − 0.47059k )TDB JJJG DC TDC = TDC λDC = TDC DC = (0.70588i + 0.52941j − 0.47059k )TDC TDA = TDa λDA = TDa and Equilibrium Condition with W = − Wj ΣF = 0: TDA + TDB + TDC − Wj = 0 Substituting the expressions obtained for TDA , TDB , and TDC and factoring i, j, and k: (−0.70588TDB + 0.70588TDC )i (0.63324TDA + 0.52941TDB + 0.52941TDC − W ) j (0.77397TDA − 0.47059TDB − 0.47059TDC )k Equating to zero the coefficients of i, j, k: −0.70588TDB + 0.70588TDC = 0 (1) 0.63324TDA + 0.52941TDB + 0.52941TDC − W = 0 (2) 0.77397TDA − 0.47059TDB − 0.47059TDC = 0 (3) Substituting W = 180 N in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms gives, TDA = 119.7 N TDB = 98.4 N TDC = 98.4 N lOMoARcPSD|6860698 PROBLEM 2.121 A container of weight W is suspended from ring A, to which cables AC and AE are attached. A force P is applied to the end F of a third cable that passes over a pulley at B and through ring A and that is attached to a support at D. Knowing that W = 1000 N, determine the magnitude of P. (Hint: The tension is the same in all portions of cable FBAD.) SOLUTION The (vector) force in each cable can be written as the product of the (scalar) force and the unit vector along the cable. That is, with JJJG AB = −(0.78 m)i + (1.6 m) j + (0 m)k AB = (−0.78 m) 2 + (1.6 m) 2 + (0) 2 = 1.78 m and JJJG AB TAB = T λ AB = TAB AB TAB [−(0.78 m)i + (1.6 m) j + (0 m)k ] = 1.78 m TAB = TAB (−0.4382i + 0.8989 j + 0k ) JJJG AC = (0)i + (1.6 m) j + (1.2 m)k and AC = (0 m)2 + (1.6 m) 2 + (1.2 m) 2 = 2 m JJJG AC TAC TAC = T λAC = TAC = [(0)i + (1.6 m) j + (1.2 m)k ] AC 2 m TAC = TAC (0.8 j + 0.6k ) JJJG AD = (1.3 m)i + (1.6 m) j + (0.4 m)k AD = (1.3 m) 2 + (1.6 m)2 + (0.4 m)2 = 2.1 m JJJG T AD = AD [(1.3 m)i + (1.6 m) j + (0.4 m)k ] TAD = T λAD = TAD AD 2.1 m TAD = TAD (0.6190i + 0.7619 j + 0.1905k ) lOMoARcPSD|6860698 PROBLEM 2.121 (Continued) JJJG AE = −(0.4 m)i + (1.6 m) j − (0.86 m)k Finally, AE = (−0.4 m) 2 + (1.6 m) 2 + (−0.86 m) 2 = 1.86 m JJJG AE TAE = T λAE = TAE AE TAE [−(0.4 m)i + (1.6 m) j − (0.86 m)k ] = 1.86 m TAE = TAE (−0.2151i + 0.8602 j − 0.4624k ) With the weight of the container W = −W j, at A we have: ΣF = 0: TAB + TAC + TAD − Wj = 0 Equating 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 the externally applied (unknown) force, we can solve the system of linear Equations (1), (2) and (3) uniquely for P. P = 378 N lOMoARcPSD|6860698 PROBLEM 2.122 Knowing that the tension in cable AC of the system described in Problem 2.121 is 150 N, determine (a) the magnitude of the force P, (b) the weight W of the container. PROBLEM 2.121 A container of weight W is suspended from ring A, to which cables AC and AE are attached. A force P is applied to the end F of a third cable that passes over a pulley at B and through ring A and that is attached to a support at D. Knowing that W = 1000 N, determine the magnitude of P. (Hint: The tension is the same in all portions of cable FBAD.) SOLUTION Here, as in Problem 2.121, the support of the container consists of the four cables AE, AC, AD, and AB, with the condition that the force in cables AB and AD is equal to the externally applied force P. Thus, with the condition TAB = TAD = P and using the linear algebraic equations of Problem 2.131 with TAC = 150 N, we obtain (a) P = 454 N (b) W = 1202 N lOMoARcPSD|6860698 PROBLEM 2.123 Cable BAC passes through a frictionless ring A and is attached to fixed supports at B and C, while cables AD and AE are both tied to the ring and are attached, respectively, to supports at D and E. Knowing that a 200-N vertical load P is applied to ring A, determine the tension in each of the three cables. SOLUTION Free Body Diagram at A: Since TBAC = tension in cable BAC, it follows that TAB = TAC = TBAC TAB = TBAC λ AB = TBAC (−17.5 cm) i + (60 cm) j 60 ⎞ ⎛ −17.5 i+ j = TBAC ⎜ 62.5 cm 62 5. ⎟⎠ ⎝ 62 5. TAC = TBAC λ AC = TBAC (60 cm) i + (25 cm) k 25 ⎞ ⎛ 60 = TBAC ⎜ j + k ⎟ 65 cm 65 65 ⎝ ⎠ TAD = TAD λ AD = TAD (80 cm) i + (60 cm) j 3 ⎞ ⎛4 = TAD ⎜ i + j ⎟ 100 cm 5 5 ⎠ ⎝ TAE = TAE λ AE = TAE (60 cm) j − (45 cm) k 3 ⎞ ⎛4 = TAE ⎜ j − k ⎟ 75 cm 5 ⎠ ⎝5 lOMoARcPSD|6860698 PROBLEM 2.123 (Continued) Substituting into ΣFA = 0, setting P = (−200 N) j, and setting the coefficients of i, j, k equal to φ , we obtain the following three equilibrium equations: 17.5 4 TBAC + TAD = 0 62.5 5 From i: − From 3 4 ⎛ 60 60 ⎞ j: ⎜ + ⎟ TBAC + TAD + TAE − 200 N = 0 5 5 ⎝ 62.5 65 ⎠ From k: (1) 25 3 TBAC − TAE = 0 65 5 (2) (3) Solving the system of linear equations using conventional algorithms gives: TBAC = 76.7 N; TAD = 26.9 N; TAE = 49.2 N lOMoARcPSD|6860698 PROBLEM 2.124 Knowing that the tension in cable AE of Prob. 2.123 is 75 N, determine (a) the magnitude of the load P, (b) the tension in cables BAC and AD. PROBLEM 2.123 Cable BAC passes through a frictionless ring A and is attached to fixed supports at B and C, while cables AD and AE are both tied to the ring and are attached, respectively, to supports at D and E. Knowing that a 200-lb vertical load P is applied to ring A, determine the tension in each of the three cables. SOLUTION Refer to the solution to Problem 2.123 for the figure and analysis leading to the following set of equilibrium equations, Equation (2) being modified to include Pj as an unknown quantity: 17.5 4 TBAC + TAD = 0 62.5 5 (1) 3 4 ⎛ 60 60 ⎞ + ⎟ TBAC + TAD + TAE − P = 0 ⎜ 5 5 ⎝ 62.5 65 ⎠ (2) 25 3 TBAC − TAE = 0 65 5 (3) − Substituting for TAE = 75 N and solving simultaneously gives: (a) P = 305 N (b) TBAC = 117.0 N; TAD = 40.9 N lOMoARcPSD|6860698 P PROBLEM 2 2.125 Collars A and B are connecteed by a 525-m mm-long wire and a caan slide freeely on fricctionless rodds. If a force P = (341 N)j is i applied to collar A, deetermine (a) the teension in the wire w when y = 155 mm, (bb) the magnituude off the force Q required to maintain m the eqquilibrium of the syystem. TION SOLUT For bothh Problems 2.125 and 2.1266: Freee-Body Diagrrams of Collars: ( AB)2 = x2 + y 2 + z 2 Here (00.525 m)2 = (00.20 m)2 + y 2 + z 2 y 2 + z 2 = 0..23563 m2 or Thus, when w y given, z is determinedd, Now JJJG AB λAB = AB 1 (0.20i − yj + zk )m 0.525 m = 0.38095i − 1.90476 yj + 1.90476 zk = Where y and z are in units u of meterrs, m. From thhe F.B. Diagraam of collar A:: ΣF = 0:: N x i + N z k + Pj + TAB λ AB = 0 Setting the t j coefficieent to zero givees P − (1.990476 y )TAB = 0 P = 341 N With TAB = 341 N 1.90476 y Now, frrom the free bo ody diagram of o collar B: ΣF = 0: N x i + N y j + Qk − TAB λAB = 0 Setting the t k coefficieent to zero givves Q − TAB (1.90476 z ) = 0 And usiing the above result r for TAB , we have Q = TAB z = 341 N (3441 N)( z ) (1..90476 z ) = y ( (1.90476) y lOMoARcPSD|6860698 PROBLEM 2.125 (Continued) Then from the specifications of the problem, y = 155 mm = 0.155 m z 2 = 0.23563 m 2 − (0.155 m)2 z = 0.46 m and 341 N 0.155(1.90476) = 1155.00 N TAB = (a) TAB = 1155 N or and Q= (b) 341 N(0.46 m)(0.866) (0.155 m) = (1012.00 N) or Q = 1012 N lOMoARcPSD|6860698 PROBLEM 2.126 Solve Problem 2.125 assuming that y = 275 mm. PROBLEM 2.125 Collars A and B are connected by a 525-mm-long wire and can slide freely on frictionless rods. If a force P = (341 N)j is applied to collar A, determine (a) the tension in the wire when y = 155 mm, (b) the magnitude of the force Q required to maintain the equilibrium of the system. SOLUTION From the analysis of Problem 2.125, particularly the results: y 2 + z 2 = 0.23563 m 2 341 N 1.90476 y 341 N Q= z y TAB = With y = 275 mm = 0.275 m, we obtain: z 2 = 0.23563 m 2 − (0.275 m)2 z = 0.40 m and TAB = (a) 341 N = 651.00 (1.90476)(0.275 m) TAB = 651 N or and Q= (b) or 341 N(0.40 m) (0.275 m) Q = 496 N lOMoARcPSD|6860698 PROB BLEM 2.12 27 Two sttructural mem mbers A and B are bolted to a bracket as shown.. Knowing thhat both mem mbers are in compression c a and that thee force is 15 kN in membeer A and 10 kN k in memberr B, determ mine by trigonometry the magnitude m and direction of the resultannt of the forcees applied to the t bracket byy members A and a B. TION SOLUT Using thhe force triang gle and the law ws of cosines and a sines, we havee γ = 180° − (400° + 20°) = 120° Then R 2 = (15 kN)2 + (10 kN) 2 − 2(15 kN N)(10 kN) cos120° = 475 kN 2 R = 21.794 kN N and Hence: 10 kN N 21.794 kN N = sin α sin120° N ⎞ ⎛ 10 kN sin120° sin α = ⎜ 21.794 k ⎟⎠ kN ⎝ = 0.39737 α = 23.414 φ = α + 50° = 73.414 R = 21.88 kN 73.4° lOMoARcPSD|6860698 y PROBLEM 2.12888 8 480 mm 560 mm Determine the x and y components of each of the forces shown. 900 mm 102 N 106 N x O 200 N 600 mm 800 mm SOLUTION We compute the following distances: 48 cm ___________ 56 cm OA = √ (48)2 + (90)2 = 102 cm. B A ___________ OB = √(56)2 + (90)2 = 106 cm. 90 cm ___________ OC = √(80)2 + (60)2 = 100 cm. 106 N 102 N Then: O 60 cm 102 N Force: 200 N C 48 , Fx = − (102 N) ___ 102 Fx = −48.0 N 90 , Fy = + (102 N) ___ 102 Fy = 90.0 N 56 , Fx = + (106 N) ___ 106 Fx = 56.0 N 90 , Fy = + (106 N) ___ 106 Fy = 90.0 N 80 , Fx = − (200 N) ___ 100 Fx = −160 N 60 , Fy = − (200 N) ___ 100 Fy = −120 N 80 cm 106 N Force: 200 N Force: lOMoARcPSD|6860698 PROBLEM M 2.129 A hoist trollley is subjecteed to the threee forces show wn. Knowing that t α = 40°, determine (a) thhe required maagnitude of thhe force P if the resultant of the three forcces is to be vertical, v (b) thhe correspondiing magnitude of the resultantt. SOLUT TION Rx = ΣFx = P + (200 N)sin 40° − (400 N) cos 40° 4 Rx = P − 177.860 1 N Ry = ΣFy = (200 N) cos c 40° + (400 N)sin 40° Ry = 410.32 N (a) (2) Foor R to be verrtical, we mustt have Rx = 0.. Set Rx = 0 in Eq. E (1) 0 = P − 1777.860 N P =177.8660 N (b) (1) P = 177.9 N Siince R is to bee vertical: R = Ry = 410 N R = 410 N lOMoARcPSD|6860698 PRO OBLEM 2.130 Know wing that α = 55° and that boom AC exerts on pin C a force directed alongg line AC, deteermine (a) thee magnitude of o that force, (b) ( the tensionn in cable BC. TION SOLUT Freee-Body Diagraam s Law of sines: Force Trianggle FAC T 300 N = BC = sinn 35° sin 500° sin 95° (a) FAC = 300 N sin 35° sin 955° FAC = 172.7 N (b) TBC = 300 N sin 50° sin 955° TBC = 231 N lOMoARcPSD|6860698 PR ROBLEM 2.131 Twoo cables are tied together at C and looaded as show wn. Knoowing that P = 360 N, determ mine the tenssion (a) in caable AC, (b) in cable BC. B SOLUT TION Free Body: C (a) ΣFx = 0: − (b) ΣFy = 0: 12 4 TAC + (3360 N) = 0 13 5 TAC = 312 N 5 3 N − 480 N = 0 (312 N) + TBC + (360 N) 113 5 TBC = 480 N − 120 N − 216 N TBC = 144.0 N lOMoARcPSD|6860698 PROBLEM 2.132 2 Two cabbles tied togethher at C are looaded as show wn. Knowing that t the maxximum allow wable tensionn in each caable is 800 N, determinne (a) the maagnitude of thhe largest forcce P that can be applied at a C, (b) the coorresponding value of α. TION SOLUT Free-Body Diagram: D C Force Trian ngle Force triiangle is isoscceles with 2β = 1880° − 85° β = 47.5° P = 2(800 2 N)cos 477.5° = 1081 N (a) he solution is correct. c Since P > 0, th (b) P = 1081 N α = 180° − 50° − 477.5° = 82.5° α = 82.5° lOMoARcPSD|6860698 PROBLEM 2.133 The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in wire AC is 120 N, determine (a) the components of the force exerted by this wire on the pole, (b) the angles θx, θy, and θz that the force forms with the coordinate axes. SOLUTION (a) Fx = (120 N) cos 60 ° cos 20 ° Fx = 56.382 N Fx = +56.4 N Fy = −(120 N) sin 60 ° Fy = −103.923 N Fy = −103.9 N Fz = −(120 N) cos 60° sin 20° Fz = −20.521 N (b) cos θ x = Fx 56.382 N = F 120 N cos θ y = Fy cos θ z = F = −103.923 N 120 N Fz −20.52 N = F 120 N Fz = −20.5 N θ x = 62.0° θ y = 150.0° θ z = 99.8° lOMoARcPSD|6860698 PROBLEM 2.134 Knowing that the tension in cable AC is 2130 N, determine the components of the force exerted on the plate at C. SOLUTION JJJG CA = −(900 mm)i + (600 mm) j − (920 mm)k CA = (900 mm)2 + (600 mm)2 + (920 mm)2 = 1420 mm TCA = TCA λ CA JJJG CA = TCA CA 2130 N [−(900 mm)i + (600 mm) j − (920 mm)k ] TCA = 1420 mm = −(1350 N)i + (900 N) j − (1380 N)k (TCA ) x = −1350 N, (TCA ) y = 900 N, (TCA ) z = −1380 N lOMoARcPSD|6860698 PROBLEM 2.135 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 600 N and Q = 450 N. SOLUTION P = (600 N)[sin 40° sin 25°i + cos 40° j + sin 40° cos 25°k ] = (162.992 N)i + (459.63 N) j + (349.54 N)k Q = (450 N)[cos 55° cos 30°i + sin 55° j − cos 55° sin 30°k ] = (223.53 N)i + (368.62 N) j − (129.055 N)k R =P+Q = (386.52 N)i + (828.25 N) j + (220.49 N)k R = (386.52 N) 2 + (828.25 N) 2 + (220.49 N) 2 = 940.22 N R = 940 N cos θ x = Rx 386.52 N = R 940.22 N θ x = 65.7° cosθ y = Ry 828.25 N 940.22 N θ y = 28.2° cos θ z = Rz 220.49 N = R 940.22 N θ z = 76.4° R = lOMoARcPSD|6860698 PROBLE EM 2.136 A containeer of weight W is suspendded from ring A. Cable BAC C passes throuugh the ring annd is attachedd to fixed suppports at B andd C. Two forrces P = Pi and a Q = Qk are a applied to t the ring to t maintain the container in i the positioon shown. Knowing K that W = 376 N, determine P and Q. (Hintt: The tensionn is the same inn both portionss of cable BAC C.) TION SOLUT TAB = T λ AB JJJK AB =T AB (−130 mm)i + (4000 mm) j + (1600 mm)k =T 4500 mm 40 16 ⎞ ⎛ 133 j+ k⎟ =T ⎜− i + 45 45 ⎠ ⎝ 455 Free-Bod dy A: TAC = T λ AC JJJK AC =T AC (−150 mm)i + (4000 mm) j + (−2440 mm)k =T 4990 mm 40 24 ⎞ ⎛ 155 j− k⎟ =T ⎜− i + 49 49 ⎠ ⎝ 499 ΣF = 0: TABB + TAC + Q + P + W = 0 t zero: Setting coefficients off i, j, k equal to i: − 13 15 T − T +P=0 45 49 0.59501T = P (1) j: + 40 40 T + T −W = 0 45 49 1.70521T = W (2) k: + 16 24 T − T +Q =0 45 49 0.1342440T = Q (3) lOMoARcPSD|6860698 PROBLEM 2.136 (Continued) Data: W = 376 N 1.70521T = 376 N T = 220.50 N 0.59501(220.50 N) = P P = 131.2 N 0.134240(220.50 N) = Q Q = 29.6 N lOMoARcPSD|6860698 PROBLEM 2.137 7 Collars A and B are coonnected by a 25-cm-long wire w and can slide freely onn frictionless rods. If a 60-lN force Q is applied to colllar B as shhown, determ mine (a) the tension in the wire whhen x =9 cm , (b) the coorresponding magnitude of o the force P requiredd to maintain thhe equilibrium m of the system m. SOLUT TION Free-Body Diagrams D of Collars: C A: B: JJJG AB − xi − (20 cm) j + zk λAB = = 25 cm AB ΣF = 0: Pi + N y j + N z k + TAB λ AB = 0 A Collar A: Substituute for λAB an nd set coefficieent of i equal to t zero: P− B Collar B: TAB x =0 25 cm (1) ΣF = 0: (660 N)k + N x′ i + N y′ j − TAB λ ABB = 0 Substituute for λ AB and d set coefficiennt of k equal too zero: 60 N − x = 9 cm (a) TAB z =0 25 cm (2) (9 cm) 2 + (20 cm) 2 + z 2 = (25 cm)2 z =12 cm (b) Frrom Eq. (2): 60 N − T AAB (12 cm) 25 cm F From Eq. (1): P= (1255.0 N)(9 cm) 25 cm TAB = 125.0 N P = 45.0 N lOMoARcPSD|6860698 PROBLEM 2.138 Collars A and B are connected by a 25-cm-long wire and can slide freely on frictionless rods. Determine the distances x and z for which the equilibrium of the system is maintained when P = 120 N and Q = 60 N. SOLUTION See Problem 2.137 for the diagrams and analysis leading to Equations (1) and (2) below: P= TAB x =0 25 cm (1) 60 N − TAB z =0 25 cm (2) For P = 120 N, Eq. (1) yields T= (25 cm)(20 N) AB x (1′) From Eq. (2): T= (25 cm)(60 N) AB z (2′) x =2 z Dividing Eq. (1′) by (2′), Now write x 2 + z 2 + (20 cm)2 = (25 cm) 2 (3) (4) Solving (3) and (4) simultaneously, 4 z 2 + z 2 + 400 = 625 z 2 = 45 z = 6.7082 cm From Eq. (3): x = 2 z = 2(6.7082 cm) = 13.4164 cm x = 13.42 cm, z =6.71 cm lOMoARcPSD|6860698 PROBLEM 2.F1 Two cables are tied together at C and loaded as shown. Draw the free-body diagram needed to determine the tension in AC and BC. SOLUTION Free-Body Diagram of Point C: W = (1600 kg)(9.81 m/s 2 ) W = 15.6960(103 ) N W =15.696 kN lOMoARcPSD|6860698 PROBLEM 2.F2 Two forces of magnitude TA = 8 kN and T B= 15 kN are applied as shown to a welded connection. Knowing that the connection is in equilibrium, draw the free-body diagram needed to determine the magnitudes of the forces TC and TD. SOLUTION Free-Body Diagram of Point E: lOMoARcPSD|6860698 PROBLEM 2.F3 The 60-N collar A can slide on a frictionless vertical rod and is connected as shown to a 65-N counterweight C. Draw the free-body diagram needed to determine the value of h for which the system is in equilibrium. SOLUTION Free-Body Diagram of Point A: lOMoARcPSD|6860698 P PROBLEM 2 2.F4 A chairlift has been stoppedd in the positiion shhown. Knowinng that each chair weighs 250 2 N and that the skier in chair E weighs 765 N, drraw the freee-body diagraams needed to deetermine the weight w of the skier s in chair F. F SOLUT TION Free-Boody Diagram of Point B: WE = 250 N + 765 N = 10115 N 88.25 = 30.510° 14 1 10 θ BC = tan −1 = 22.620° 2 24 θ AB = tan −1 Use this free body to determ mine TAB and TBC. Free-Boody Diagram of Point C: θCD = tan −1 11.1 = 10.3889° 6 Use this free body to determ mine TCD and WF. Then weight of skier WS is found by WS = WF − 250 N lOMoARcPSD|6860698 PROBLEM 2.F5 Three cables are used to tether a balloon as shown. Knowing that the tension in cable AC is 444 N, draw the free-body diagram needed to determine the vertical force P exerted by the balloon at A. SOLUTION Free-Body Diagram of Point A: lOMoARcPSD|6860698 PROBLEM 2.F6 A container of mass m = 120 kg is supported by three cables as shown. Draw the free-body diagram needed to determine the tension in each cable. SOLUTION Free-Body Diagram of Point A: W = (120 kg)(9.81 m/s 2 ) = 1177.2 N lOMoARcPSD|6860698 PROBLEM 2.F7 A 150-N cylinder is supported by two cables AC and BC that are attached to the top of vertical posts. A horizontal force P, which is perpendicular to the plane containing the posts, holds the cylinder in the position shown. Draw the free-body diagram needed to determine the magnitude of P and the force in each cable. SOLUTION Free-Body Diagram of Point C: lOMoARcPSD|6860698 PROBLEM 2.F8 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. Knowing that the tension in wire AB is 2.5 kN, draw the free-body diagram needed to determine the vertical force P exerted by the tower on the pin at A. SOLUTION Free-Body Diagram of point A:
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