Consider the tension in the cables AC and BC asshown below:
Vector Mechanics for Engineers Statics and Dynamics 11th Edition Beer SOLUTIONS MANUAL Download at: http://testbanklive.com/download/vector-mechanics-for-engineers-staticsand-dynamics-11th-edition-beer-solutions-manual/
CHAPTER 2 Copyright © McGraw-Hill Education. Permission required for reproduction or display. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 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 (a) Parallelogram law: (b) Triangle rule: We measure: R = 1391 kN, α = 47.8° R = 1391 N Copyright © McGraw-Hill Education. Permission required for reproduction or display. 47.8° W PROBLEM 2.2 Two forces are applied as shown to a bracket support. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: R = 906 lb, α = 26.6° R = 906 lb Copyright © McGraw-Hill Education. Permission required for reproduction or display. 26.6° W 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 Copyright © McGraw-Hill Education. Permission required for reproduction or display. 21.2° W 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 kips and Q = 4 kips, 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 kips, α = 3.8° R = 8.03 kips Copyright © McGraw-Hill Education. Permission required for reproduction or display. 3.8° W 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 W 30° + β + 25° = 180° β = 180° − 25° − 30° = 125° 120 N R = sin 30° sin125° Copyright © McGraw-Hill Education. Permission required for reproduction or display. R = 196.6 N W 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 lb, 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° 800 lb = sin 75° sin 65° (b) 800 lb sin 65° T2 = R sin 40° Copyright © McGraw-Hill Education. Permission required for reproduction or display. T = 853 lb W 2 R = 567 lb W 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 = 1000 lb, 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° 1000 lb (b) = T1 sin 75° sin 65° 1000 lb R = sin 75° sin 40° Copyright © McGraw-Hill Education. Permission required for reproduction or display. T = 938 lb W 1 R = 665 lb W 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 R 2.2 kN = = sin 30° sin125° sin 25D TAC = 2.603 kN R = 4.264 kN (a) TAC = 2.60 kN W (b) Copyright © McGraw-Hill Education. Permission required for reproduction or display. R = 4.26 kN W 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 Copyright © McGraw-Hill Education. Permission required for reproduction or display. 34.3° W PROBLEM 2.10 Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R. SOLUTION Using the triangle rule and law of sines: (a) sin α sin 25° = 50 N 35 N sin α = 0.60374 α = 37.138° (b) α = 37.1° W α + β + 25° = 180° β = 180° − 25° − 37.138° = 117.862° 35 N R = sin117.862° sin 25° Copyright © McGraw-Hill Education. Permission required for reproduction or display. R = 73.2 N W PROBLEM 2.11 A steel tank is to be positioned in an excavation. Knowing that α = 20°, determine by trigonometry (a) the required magnitude of the force P if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R. SOLUTION Using the triangle rule and the law of sines: (a) β + 50° + 60° = 180° β = 180° − 50° − 60° = 70° (b) P 425 lb = sin 70° sin 60° P = 392 lb W 425 lb = R sin 70° sin 50° R = 346 lb W Copyright © McGraw-Hill Education. Permission required for reproduction or display. PROBLEM 2.12 A steel tank is to be positioned in an excavation. Knowing that the magnitude of P is 500 lb, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R. SOLUTION Using the triangle rule and the law of sines: (a) (α + 30°) + 60° + β = 180° β = 180° − (α + 30°) − 60° β = 90° − α sin (90° −α ) sin 60° = 425 lb 500 lb (b) 90° − α = 47.402° α = 42.6° W R R = 551 lb W sin (42.598° + 30°) = 500 lb sin 60° Copyright © McGraw-Hill Education. Permission required for reproduction or display. PROBLEM 2.13 A steel tank is to be positioned in an excavation. Determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied at A is vertical, (b) the corresponding magnitude of R. SOLUTION The smallest force P will be perpendicular to R. (a) P = (425 lb) cos 30° (b) R = (425 lb) sin 30° Copyright © McGraw-Hill Education. Permission required for reproduction or display. P = 368 lb R = 213 lb PROBLEM 2.14 For the hook support of Prob. 2.10, determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied to the support is horizontal, (b) the corresponding magnitude of R. SOLUTION The smallest force P will be perpendicular to R. (a) P = (50 N) sin 25° (b) R = (50 N) cos 25° Copyright © McGraw-Hill Education. Permission required for reproduction or display. P = 21.1 N W R = 45.3 N W 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 lb) 2 + (300 lb) 2 − 2(200 lb)(300 lb) cos (45D + 65°) R = 413.57 lb Using the law of sines: sin α = sin (45D +65°) 300 lb 413.57 lb α = 42.972° β = 90D + 25D − 42.972° R = 414 lb Copyright © McGraw-Hill Education. Permission required for reproduction or display. 72.0° W 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 1390.57 N D α − 30 = 17.7642° α = 47.764D R = 1391N Copyright © McGraw-Hill Education. Permission required for reproduction or display. 47.8° W PROBLEM 2.17 Solve Problem 2.4 by trigonometry. 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 kips and Q = 4 kips, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule. SOLUTION Using the force triangle and the laws of cosines and sines: We have: Then γ = 180° − (50° + 25°) = 105° R 2 = (4 kips) 2 + (6 kips) 2 − 2(4 kips)(6 kips) cos105° = 64.423 kips 2 R = 8.0264 kips And 4 kips = 8.0264 kips sin(25° + α ) sin105° sin(25° + α ) = 0.48137 25° + α = 28.775° α = 3.775° R = 8.03 kips Copyright © McGraw-Hill Education. Permission required for reproduction or display. 3.8° W 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 Copyright © McGraw-Hill Education. Permission required for reproduction or display. 44.7° W PROBLEM 2.19 Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 48 N and Q = 60 N, determine by trigonometry the magnitude and direction of the resultant of the two forces. SOLUTION Using the force triangle and the laws of cosines and sines: We have Then and γ = 180° − (20° + 10°) = 150° R 2 = (48 N) 2 + (60 N) 2 − 2(48 N)(60 N) cos150° R = 104.366 N 48 N sin α = 104.366 N sin150° sin α = 0.22996 α = 13.2947° Hence: φ = 180° − α − 80° = 180° − 13.2947° − 80° = 86.705° R = 104.4 N Copyright © McGraw-Hill Education. Permission required for reproduction or display. 86.7° W PROBLEM 2.20 Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 60 N and Q = 48 N, determine by trigonometry the magnitude and direction of the resultant of the two forces. SOLUTION Using the force triangle and the laws of cosines and sines: We have Then and γ = 180° − (20° + 10°) = 150° R 2 = (60 N) 2 + (48 N) 2 −2(60 N)(48 N) cos 150° R = 104.366 N 60 N sin α = 104.366 N sin150° sin α = 0.28745 α = 16.7054° Hence: φ = 180° − α − 180° = 180° − 16.7054° − 80° = 83.295° R = 104.4 N Copyright © McGraw-Hill Education. Permission required for reproduction or display. 83.3° W PROBLEM 2.21 Determine the x and y components of each of the forces shown. SOLUTION Compute the following distances: OA = (84) 2 + (80) 2 = 116 in. OB = (28) 2 + (96) 2 = 100 in. OC = (48) 2 + (90) 2 = 102 in. 29-lb Force: 50-lb Force: Fx = +(29 lb) 84 116 Fx = +21.0 lb W Fy = +(29 lb) 80 116 Fy = +20.0 lb W Fx = −(50 lb) 28 100 Fx = −14.00 lb W 96 Fy = +(50 lb) 100 Fy = +48.0 lb W 48 51-lb Force: Fx = +(51 lb) Fy = −(51 lb) 102 90 102 Copyright © McGraw-Hill Education. Permission required for reproduction or display. Fx = +24.0 lb W Fy = −45.0 lb W PROBLEM 2.22 Determine the x and y components of each of the forces shown. SOLUTION Compute the following distances: OA = (600) 2 + (800) 2 = 1000 mm OB = (560) 2 + (900) 2 = 1060 mm OC = (480) 2 + (900) 2 = 1020 mm 800-N Force: 424-N Force: 408-N Force: Fx = +(800 N) 800 1000 Fx = +640 N W Fy = +(800 N) 600 1000 Fy = +480 N W Fx = −(424 N) 560 1060 Fx = −224 N W Fy = −(424 N) 900 1060 Fy = −360 N W Fx = +(408 N) 480 1020 Fx = +192.0 N W Fy = −(408 N) 900 1020 Fy = −360 N W Copyright © McGraw-Hill Education. Permission required for reproduction or display. 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 W Fy = +(80 N) sin 40° Fy = 51.4 N W Fx = +(120 N) cos 70° Fx = 41.0 N W Fy = +(120 N) sin 70° Fy = 112.8 N W Fx = −(150 N) cos 35° Fx = −122. 9 N W Fy = +(150 N) sin 35° Fy = 86.0 N W Copyright © McGraw-Hill Education. Permission required for reproduction or display. PROBLEM 2.24 Determine the x and y components of each of the forces shown. SOLUTION 40-lb Force: 50-lb Force: 60-lb Force: Fx = +(40 lb) cos 60° Fx = 20.0 lb W Fy = −(40 lb) sin 60° Fy = −34.6 lb W Fx = −(50 lb) sin 50° Fx = −38.3 lb W Fy = −(50 lb) cos 50° Fy = −32.1 lb W Fx = +(60 lb) cos 25° Fx = 54.4 lb W Fy = +(60 lb) sin 25° Fy = 25.4 lb W Copyright © McGraw-Hill Education. Permission required for reproduction or display. 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 W (b) ⎛ 720 ⎞ Py = P ⎜ ⎟ ⎝ 970 ⎠ ⎛ 720 ⎞ = 485 N ⎜ ⎟ ⎝ 970 ⎠ = 360 N Py = 970 N W Copyright © McGraw-Hill Education. Permission required for reproduction or display. PROBLEM 2.26 Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 300-lb horizontal component, determine (a) the magnitude of the force P, (b) its vertical component. SOLUTION P sin 35° = 300 lb (a) P= (b) Vertical component 300 lb sin 35° P = 523 lb W Pv = P cos 35° = (523 lb) cos 35° Copyright © McGraw-Hill Education. Permission required for reproduction or display. Pv = 428 lb W PROBLEM 2.27 The hydraulic cylinder BC exerts on member AB a force P directed along line BC. Knowing that P must have a 600-N component perpendicular to member AB, determine (a) the magnitude of the force P, (b) its component along line AB. SOLUTION 180° = 45° + α + 90° + 30° α = 180° − 45° − 90° − 30° = 15° (a) Px P P P= x cos α 600 N = cos15° = 621.17 N cos α = P = 621 N W (b) Py Px Py = Px tan α tan α = = (600 N) tan15° = 160.770 N Py = 160.8 N W Copyright © McGraw-Hill Education. Permission required for reproduction or display. PROBLEM 2.28 Cable AC exerts on beam AB a force P directed along line AC. Knowing that P must have a 350-lb vertical component, determine (a) the magnitude of the force P, (b) its horizontal component. SOLUTION (a) P= Py cos 55° = 350 lb cos 55° = 610.21 lb (b) P = 610 lb W Px = P sin 55° = (610.21 lb) sin 55° = 499.85 lb Copyright © McGraw-Hill Education. Permission required for reproduction or display. Px = 500 lb W PROBLEM 2.29 The hydraulic cylinder BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 750-N component perpendicular to member ABC, determine (a) the magnitude of the force P, (b) its component parallel to ABC. SOLUTION (a) 750 N = P sin 20° P = 2190 N W P = 2192.9 N (b) PABC = P cos 20° = (2192.9 N) cos 20° PABC = 2060 N W Copyright © McGraw-Hill Education. Permission required for reproduction or display. 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) P= 37 P x 12 37 = (720 N) 12 = 2220 N P = 2.22 kN W (b) P = y 35 P x 12 35 = (720 N) 12 = 2100 N Py = 2.10 kN W Copyright © McGraw-Hill Education. Permission required for reproduction or display. PROBLEM 2.31 Determine the resultant of the three forces of Problem 2.21. PROBLEM 2.21 Determine the x and y components of each of the forces shown. SOLUTION Components of the forces were determined in Problem 2.21: Force x Comp. (lb) y Comp. (lb) 29 lb +21.0 +20.0 50 lb –14.00 +48.0 51 lb +24.0 –45.0 Rx = +31.0 Ry = +23.0 R = Rx i + R y j = (31.0 lb)i + (23.0 lb) j Ry tan α = Rx 23.0 = 31.0 α = 36.573° 23.0 lb R= sin (36.573°) = 38.601 lb R = 38.6 lb Copyright © McGraw-Hill Education. Permission required for reproduction or display. 36.6° W PROBLEM 2.32 Determine the resultant of the three forces of Problem 2.23. PROBLEM 2.23 Determine the x and y components of each of the forces shown. SOLUTION Components of the forces were determined in Problem 2.23: Force x Comp. (N) y Comp. (N) 80 N +61.3 +51.4 120 N +41.0 +112.8 150 N –122.9 +86.0 Rx = −20.6 Ry = +250.2 R = Rx i + R y j = (−20.6 N)i + (250.2 N) j R tan α = y Rx 250.2 N tan α = 20.6 N tan α = 12.1456 α = 85.293° 250.2 N R= sin 85.293° R = 251 N Copyright © McGraw-Hill Education. Permission required for reproduction or display. 85.3° W PROBLEM 2.33 Determine the resultant of the three forces of Problem 2.24. PROBLEM 2.24 Determine the x and y components of each of the forces shown. SOLUTION Force x Comp. (lb) y Comp. (lb) 40 lb +20.00 –34.64 50 lb –38.30 –32.14 60 lb +54.38 +25.36 Rx = +36.08 Ry = −41.42 R = Rx i + Ry j = (+36.08 lb)i + (−41.42 lb) j Ry tan α = Rx 41.42 lb tan = 36.08 lb tan α = 1.14800 = 48.942° R= 41.42 lb sin 48.942° R = 54.9 lb Copyright © McGraw-Hill Education. Permission required for reproduction or display. 48.9° W PROBLEM 2.34 Determine the resultant of the three forces of Problem 2.22. PROBLEM 2.22 Determine the x and y components of each of the forces shown. SOLUTION Components of the forces were determined in Problem 2.22: Force x Comp. (N) y Comp. (N) 800 lb +640 +480 424 lb –224 –360 408 lb +192 –360 Rx = +608 Ry = −240 R = Rx i + Ry j = (608 lb)i + (−240 lb) j R tan α = y Rx 240 = 608 α = 21.541° 240 N R= sin(21.541°) = 653.65 N R = 654 N Copyright © McGraw-Hill Education. Permission required for reproduction or display. 21.5° W PROBLEM 2.35 Knowing that α = 35°, determine the resultant of the three forces shown. SOLUTION Fx = +(100 N) cos 35° = +81.915 N 100-N Force: Fy = −(100 N) sin 35° = −57.358 N Fx = +(150 N) cos 65° = +63.393 N 150-N Force: Fy = −(150 N) sin 65° = −135.946 N Fx = −(200 N) cos 35° = −163.830 N 200-N Force: Fy = −(200 N) sin 35° = −114.715 N Force x Comp. (N) y Comp. (N) 100 N +81.915 −57.358 150 N +63.393 −135.946 200 N −163.830 −114.715 Rx = −18.522 Ry = −308.02 R = Rx i + Ry j = (−18.522 N)i + (−308.02 N) j Ry tan α = Rx 308.02 = 18.522 α = 86.559° R= 308.02 N sin 86.559 R = 309 N Copyright © McGraw-Hill Education. Permission required for reproduction or display. 86.6° W PROBLEM 2.36 Knowing that the tension in rope AC is 365 N, determine the resultant of the three forces exerted at point C of post BC. SOLUTION Determine force components: Cable force AC: Fx = −(365 N) 960 = −240 N 1460 1100 Fy = −(365 N) 500-N Force: Fx = (500 N) Fy = (500 N) 200-N Force: and 1460 = −275 N 24 = 480 N 25 7 25 = 140 N 4 = 160 N 5 3 Fy = −(200 N) = −120 N 5 Fx = (200 N) Rx = ΣFx = −240 N + 480 N + 160 N = 400 N R y = ΣFy = −275 N + 140 N − 120 N = −255 N R = Rx2 + Ry2 = (400 N)2 + (−255 N) 2 = 474.37 N Further: 255 400 α = 32.5° tan α = R = 474 N Copyright © McGraw-Hill Education. Permission required for reproduction or display. 32.5° W PROBLEM 2.37 Knowing that α = 40°, determine the resultant of the three forces shown. SOLUTION 60-lb Force: Fx = (60 lb) cos 20° = 56.382 lb Fy = (60 lb) sin 20° = 20.521 lb 80-lb Force: Fx = (80 lb) cos 60° = 40.000 lb Fy = (80 lb) sin 60° = 69.282 lb 120-lb Force: Fx = (120 lb) cos 30° = 103.923 lb Fy = −(120 lb) sin 30° = −60.000 lb and Rx = ΣFx = 200.305 lb Ry = ΣFy = 29.803 lb R = (200.305 lb) 2 + (29.803 lb) 2 Further: = 202.510 lb 29.803 tan α = 200.305 α = tan −1 = 8.46° 29.803 200.305 R = 203 lb Copyright © McGraw-Hill Education. Permission required for reproduction or display. 8.46° W PROBLEM 2.38 Knowing that α = 75°, determine the resultant of the three forces shown. SOLUTION 60-lb Force: Fx = (60 lb) cos 20° = 56.382 lb Fy = (60 lb) sin 20° = 20.521 lb 80-lb Force: Fx = (80 lb) cos 95° = −6.9725 lb Fy = (80 lb) sin 95° = 79.696 lb 120-lb Force: Fx = (120 lb) cos 5° = 119.543 lb Fy = (120 lb) sin 5° = 10.459 lb Then Rx = ΣFx = 168.953 lb Ry = ΣFy = 110.676 lb and R = (168.953 lb) 2 + (110.676 lb)2 = 201.976 lb 110.676 168.953 tan α = 0.65507 α = 33.228° tan α = R = 202 lb Copyright © McGraw-Hill Education. Permission required for reproduction or display. 33.2° W 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 = 75 sin α 29.904 75 = 0.39872 α = 21.738° tan α = (b) α = 21.7° W Substituting for α in Eq. (2): Ry = −300 sin 21.738° − 150 sin 51.738° = −228.89 N R = | Ry | = 228.89 N Copyright © McGraw-Hill Education. Permission required for reproduction or display. R = 229 N W 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 960 R = ΣF = − x x R =− x 48 y (a) 1100 1460 73 4 (500 N) + (200 N) 25 5 (1) AC 55 R =− T y 24 + 640 N T 73 + AC 1460 R = ΣF = − y T + T AC 7 3 (500 N) − (200 N) 25 5 + 20 N For R to be horizontal, we must have Ry = 0. 55 − TAC + 20 N = 0 Set Ry = 0 in Eq. (2): 73 TAC = 26.545 N (b) (2) AC TAC = 26.5 N W Substituting for TAC into Eq. (1) gives 48 (26.545 N) + 640 N = − x 73 Rx = 622.55 N R R = Rx = 623 N Copyright © McGraw-Hill Education. Permission required for reproduction or display. R = 623 N W PROBLEM 2.41 Determine (a) the required tension in cable AC, knowing that the resultant of the three forces exerted at Point C of boom BC must be directed along BC, (b) the corresponding magnitude of the resultant. SOLUTION Using the x and y axes shown: Rx = ΣFx = TAC sin10° + (50 lb) cos 35° + (75 lb) cos 60° = TAC sin10° + 78.458 lb (1) R y = ΣFy = (50 lb) sin 35° + (75 lb) sin 60° − TAC cos10° R y = 93.631 lb − TAC cos10° (a) (2) Set Ry = 0 in Eq. (2): 93.631 lb − TAC cos10° = 0 TAC = 95.075 lb (b) TAC = 95.1 lb W Substituting for TAC in Eq. (1): Rx = (95.075 lb) sin10° + 78.458 lb = 94.968 lb R = Rx Copyright © McGraw-Hill Education. Permission required for reproduction or display. R = 95.0 lb W PROBLEM 2.42 For the block of Problems 2.37 and 2.38, determine (a) the required value of α if the resultant of the three forces shown is to be parallel to the incline, (b) the corresponding magnitude of the resultant. SOLUTION Select the x axis to be along a a′. Then Rx = ΣFx = (60 lb) + (80 lb) cos α + (120 lb) sin α (1) Ry = ΣFy = (80 lb) sin α − (120 lb) cos α (2) and (a) Set Ry = 0 in Eq. (2). (80 lb)sin α − (120 lb) cos α = 0 Dividing each term by cos α gives: (80 lb) tan α = 120 lb 120 lb tanα = 80 lb α = 56.310° (b) α = 56.3° W Substituting for α in Eq. (1) gives: Rx = 60 lb + (80 lb) cos 56.31° + (120 lb) sin 56.31° = 204.22 lb Copyright © McGraw-Hill Education. Permission required for reproduction or display. Rx = 204 lb W 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 lb = BC = sin 60° sin 40° sin 80° (a) T = AC (b) T BC 400 lb (sin 60°) AC sin 80° = 400 lb = 352 lb W T (sin 40°) sin 80° Copyright © McGraw-Hill Education. Permission required for reproduction or display. T BC = 261 lb W 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) T = AC (b) T BC 6 kN (sin 60°) AC sin 85° = 6 kN = 5.22 kN W T (sin 35°) T sin 85° Copyright © McGraw-Hill Education. Permission required for reproduction or display. BC = 3.45 kN W 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) T = AC (b) T BC 1.98 kN sin 61.927° AC sin 44.333° = 1.98 kN = 2.50 kN W T sin 73.740° T sin 44.333° Copyright © McGraw-Hill Education. Permission required for reproduction or display. BC = 2.72 kN W PROBLEM 2.46 Two cables are tied together at C and are loaded as shown. Knowing that P = 500 N and α = 60°, determine the tension in (a) in cable AC, (b) in cable BC. SOLUTION Force Triangle Free-Body Diagram Law of sines: (a) 500 N T TAC = BC = sin 35° sin 75° sin 70° = T AC (b) T BC 500 N sin 35° AC sin 70° = 500 N = 305 N W T sin 75° sin 70° Copyright © McGraw-Hill Education. Permission required for reproduction or display. T BC = 514 N W PROBLEM 2.47 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 W = mg = (200 kg)(9.81 m/s 2 ) = 1962 N Law of sines: TBC 1962 N TAC = = sin 105° sin 60° sin 15° (a) TAC = (b) T TBC = (1962 N) sin 15° sin 60° (1962 N) sin 105° TAC = 586 N W T sin 60° Copyright © McGraw-Hill Education. Permission required for reproduction or display. BC = 2190 N W PROBLEM 2.48 Knowing that α = 20°, determine the tension (a) in cable AC, (b) in rope BC. SOLUTION Free-Body Diagram Law of sines: Force Triangle TAC T 1200 lb = BC = sin 110° sin 5° sin 65° (a) TAC = 1200 lb sin 110° sin 65° TAC = 1244 lb W (b) TBC = 1200 lb sin 5° sin 65° TBC = 115.4 lb W Copyright © McGraw-Hill Education. Permission required for reproduction or display. PROBLEM 2.49 Two cables are tied together at C and are loaded as shown. Knowing that P = 300 N, determine the tension in cables AC and BC. SOLUTION Free-Body Diagram ΣFx = 0 − TCA sin 30D + TCB sin 30D − P cos 45° − 200N = 0 For P = 200N we have, −0.5TCA + 0.5TCB + 212.13 − 200 = 0 (1) ΣFy = 0 TCA cos 30° − TCB cos 30D − P sin 45D = 0 0. 6603TCA + 0.86603TCB − 212.13 = 0 (2) Solving equations (1) and (2) simultaneously gives, TCA = 134.6 N W TCB = 110.4 N W Copyright © McGraw-Hill Education. Permission required for reproduction or display. PROBLEM 2.50 Two cables are tied together at C and are loaded as shown. Determine the range of values of P for which both cables remain taut. SOLUTION Free-Body Diagram ΣFx = 0 − T sin 30D + TCB sin 30D − P cos 45° − 200N = 0 CA For TCA = 0 we have, 0.5TCB + 0.70711P − 200 = 0 (1) ΣFy = 0 TCA cos 30° − TCB cos 30D − P sin 45D = 0 ; again setting TCA = 0 yields, 0.86603TCB − 0.70711P = 0 (2) Adding equations (1) and (2) gives, 1.36603TCB = 200 hence TCB = 146.410N and P = 179.315N Substituting for TCB = 0 into the equilibrium equations and solving simultaneously gives, −0.5TCA + 0.70711P − 200 = 0 0.86603TCA − 0.70711P = 0 And TCA = 546.40N , P = 669.20N Thus for both cables to remain taut, load P must be within the range of 179.315 N and 669.20 N. 179.3 N <>< 669=' n='> Copyright © McGraw-Hill Education. Permission required for reproduction or display. PROBLEM 2.51 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that P = 500 lb and Q = 650 lb, determine the magnitudes of the forces exerted on the rods A and B. SOLUTION Free-Body Diagram Resolving the forces into x- and y-directions: R = P + Q + FA + FB = 0 Substituting components: R = −(500 lb) j + [(650 lb) cos 50°]i − [(650 lb) sin 50°]j + FB i − (FA cos 50°)i + (FA sin 50°) j = 0 In the y-direction (one unknown force): −500 lb − (650 lb) sin 50° + FA sin 50° = 0 Thus, FA = 500 lb + (650 lb) sin 50° sin 50° = 1302.70 lb In the x-direction: Thus, FA = 1303 lb W (650 lb) cos 50° + FB − FA cos 50° = 0 FB = FA cos 50° − (650 lb) cos 50° = (1302.70 lb) cos 50° − (650 lb) cos 50° = 419.55 lb Copyright © McGraw-Hill Education. Permission required for reproduction or display. FB = 420 lb W PROBLEM 2.52 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that the magnitudes of the forces exerted on rods A and B are FA = 750 lb and FB = 400 lb, determine the magnitudes of P and Q. SOLUTION Free-Body Diagram Resolving the forces into x- and y-directions: R = P + Q + FA + FB = 0 Substituting components: R = −Pj + Q cos 50°i − Q sin 50°j − [(750 lb) cos 50°]i + [(750 lb) sin 50°]j + (400 lb)i In the x-direction (one unknown force): Q cos 50° − [(750 lb) cos 50°] + 400 lb = 0 (750 lb) cos 50° − 400 lb cos 50° = 127.710 lb Q= In the y-direction: −P − Q sin 50° + (750 lb) sin 50° = 0 P = −Q sin 50° + (750 lb) sin 50° = −(127.710 lb) sin 50° + (750 lb) sin 50° = 476.70 lb P = 477 lb; Q = 127.7 lb W Copyright © McGraw-Hill Education. Permission required for reproduction or display. PROBLEM 2.53 A welded connection is in equilibrium under the action of the four forces shown. Knowing that FA = 8 kN and FB = 16 kN, determine the magnitudes of the other two forces. SOLUTION Free-Body Diagram of Connection ΣF = 0: x 3 5 3 F −F − F =0 B C 5 A FA = 8 kN With FB = 16 kN 4 4 F = (16 kN) − (8 kN) C 5 Σ Fy = 0: − FD + 3 With FA and FB as above: 5 F = 6.40 kN W C 3 3 FB − FA = 0 5 5 3 FD = (16 kN) − (8 kN) 5 5 Copyright © McGraw-Hill Education. Permission required for reproduction or display. FD = 4.80 kN W PROBLEM 2.54 A welded connection is in equilibrium under the action of the four forces shown. Knowing that FA = 5 kN and FD = 6 kN, determine the magnitudes of the other two forces. SOLUTION Free-Body Diagram 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 W 4 4 FB − FA = 0 5 5 4 FC = (FB − FA ) 5 4 = (15 kN − 5 kN) 5 Copyright © McGraw-Hill Education. Permission required for reproduction or display. FC = 8.00 kN W 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 200 lb, 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° − 200 = 0 0.67365TACB + 0.5TCD = 200 (a) Substitute (1) into (2): 0.67365TACB + 0.5(0.137158TACB ) = 200 TACB = 269.46 lb (b) From (1): (2) TACB = 269 lb W TCD = 0.137158 (269.46 lb) Copyright © McGraw-Hill Education. Permission required for reproduction or display. TCD = 37.0 lb W 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 20 lb, determine (a) the combined weight of the boatswain’s chair and the sailor, (b) the tension in the support cable ACB. SOLUTION Free-Body Diagram ΣFx = 0: TACB cos 15° − TACB cos 25° − (20 lb) cos 25° = 0 TACB = 304.04 lb ΣFy = 0: (304.04 lb) sin 15° + (304.04 lb) sin 25° + (20 lb) sin 25° − W = 0 W = 215.64 lb (a) W = 216 lb W (b) TACB = 304 lb W Copyright © McGraw-Hill Education. Permission required for reproduction or display. 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 W By inspection, TAC = (6 kN) cos 35D TAC = 4.91 kN W TBC = (6 kN) sin 35D TBC = 3.44 kN W (b) For equal tension in both cables, the force triangle will be an isosceles. α = 55.0D W Therefore, by inspection, T AC =T BC = (1 / 2) 6 kN cos 35° T =T AC Copyright © McGraw-Hill Education. Permission required for reproduction or display. BC = 3.66 kN W PROBLEM 2.58 For the cables of Problem 2.46, it is known that the maximum allowable tension is 600 N in cable AC and 750 N in cable BC. Determine (a) the maximum force P that can be applied at C, (b) the corresponding value of α. SOLUTION Free-Body Diagram (a) Law of cosines Force Triangle P 2 = (600)2 + (750)2 − 2(600)(750) cos (25° + 45°) P = 784.02 N (b) Law of sines P = 784 N W sin β sin (25° + 45°) = 600 N 784.02 N β = 46.0° ∴ α = 46.0° + 25° Copyright © McGraw-Hill Education. Permission required for reproduction or display. α = 71.0° W PROBLEM 2.59 For the situation described in Figure P2.48, determine (a) the value of α for which the tension in rope BC is as small as possible, (b) the corresponding value of the tension. SOLUTION Free-Body Diagram Force Triangle To be smallest, TBC must be perpendicular to the direction of TAC . (a) (b) Thus, α = 5.00° TBC = (1200 lb) sin 5° α = 5.00° W TBC = 104.6 lb W Copyright © McGraw-Hill Education. Permission required for reproduction or display. PROBLEM 2.60 Two cables tied together at C are loaded as shown. Determine the range of values of Q for which the tension will not exceed 60 lb in either cable. SOLUTION ΣFx = 0: −TBC − Q cos 60° + 75 lb = 0 Free-Body Diagram TBC = 75 lb − Q cos 60° (1) ΣFy = 0: TAC − Q sin 60° = 0 TAC = Q sin 60° Requirement: From Eq. (2): (2) TAC = 60 lb: Q sin 60° = 60 lb Requirement: Q = 69.3 lb TBC = 60 lb: From Eq. (1): 75 lb − Q cos 60° = 60 lb Q = 30.0 lb 30.0 lb ≤ Q ≤ 69.3 lb W Copyright © McGraw-Hill Education. Permission required for reproduction or display. 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 Law of sines: sin α = 1 2 (2.8 kN) TAC TAC = 5 kN 1 2 (2.8 kN) 5 kN α = 16.2602° sin α = 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 Copyright © McGraw-Hill Education. Permission required for reproduction or display. 1.250 m W 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 AC = BC = ΣFy = 0: 2(800 N) 2 + d2 h h + d2 2 ) −P=0 2 800 = Data: P ⎛d⎞ 1+ ⎜ ⎟ 2 ⎝h⎠ P = 200 N, d = 600 mm and solving for h 800 N = 200 N 2 1+ ⎛ ⎞ ⎜ 600 mm ⎟ ⎝ h 2 ⎠ h = 75.6 mm W PROBLEM 2.63 Collar A is connected as shown to a 50-lb 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 = 4.5 in., (b) x = 15 in. SOLUTION (a) Free Body: Collar A Force Triangle P = 4.5 (b) Free Body: Collar A 50 lb P = 10.98 lb W 20.5 Force Triangle P 15 = 50 lb 25 P = 30.0 lb W PROBLEM 2.64 Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the distance x for which the collar is in equilibrium when P = 48 lb. SOLUTION Free Body: Collar A Force Triangle N 2 = (50)2 − (48)2 = 196 N = 14.00 lb Similar Triangles x 48 lb = 20 in. 14 lb x = 68.6 in. W 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 55° + α = 82.375 Two values for α : 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 W PROBLEM 2.66 A 200-kg crate is to be supported by the rope-and-pulley arrangement shown. Determine the magnitude and direction of the force P that must be exerted on the free end of the rope to maintain equilibrium. (Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Ch. 4.) SOLUTION Free-Body Diagram: Pulley A ΣFx = 0: − 2P cos α = 0.59655 α = ±53.377° ⎛ 5 ⎞ + P cos α = 0 ⎜ ⎟ ⎝ 281 ⎠ For α = +53.377°: ⎛ 16 ⎞ ΣF y = 0: 2P ⎜ ⎟ + P sin 53.377° − 1962 N = 0 ⎝ 281 ⎠ P = 724 N 53.4° W For α = −53.377°: ⎛ 16 ⎞ ΣF y = 0: 2P ⎜ ⎟ + P sin(−53.377°) − 1962 N = 0 ⎝ 281 ⎠ P = 1773 53.4° W PROBLEM 2.67 A 600-lb crate is supported by several ropeand-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 ΣFy = 0: 2T − (600 lb) = 0 T= 1 (600 lb) 2 (a) T = 300 lb W ΣFy = 0: 2T − (600 lb) = 0 T= 1 (600 lb) 2 T = 300 lb W (b) ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb W (c) (d ) ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 (e) T = 200 lb W ΣFy = 0: 4T − (600 lb) = 0 T= 1 (600 lb) 4 T = 150.0 lb W 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 600-lb 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 − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb W (d ) ΣFy = 0: 4T − (600 lb) = 0 1 T = (600 lb) 4 T = 150.0 lb W PROBLEM 2.69 A load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Knowing that P = 750 N, determine (a) the tension in cable ACB, (b) the magnitude of load Q. SOLUTION Free-Body Diagram: Pulley C ΣFx = 0: TACB (cos 25° − cos 55°) − (750 N) cos 55° = 0 (a) Hence: TACB = 1292.88 N TACB = 1293 N W 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 W PROBLEM 2.70 An 1800-N load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Determine (a) the tension in cable ACB, (b) the magnitude of load P. SOLUTION Free-Body Diagram: Pulley C ΣFx = 0: TACB (cos 25° − cos 55°) − P cos 55° = 0 P = 0.58010TACB (1) or ΣFy = 0: TACB (sin 25° + sin 55°) + P sin 55° − 1800 N = 0 1.24177TACB + 0.81915P = 1800 N (2) or (a) Substitute Equation (1) into Equation (2): 1.24177TACB + 0.81915(0.58010TACB ) = 1800 N Hence: TACB = 1048.37 N TACB = 1048 N W (b) Using (1), P = 0.58010(1048.37 N) = 608.16 N P = 608 N W 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° cos 30 D Fx = 219.60 N Fx = 220 N W Fy = (600 N) cos 25° Fy = 543.78 N Fy = 544 N W Fz = (380.36 N) sin 25° sin 30 D Fz = 126.785 N (b) cos θx = Fx 219.60 N = F 600 N Fy cos θy = F θz = F θx = 68.5° W 543.78 N = Fz cos Fz = 126.8 N W 600 N θ y = 25.0° W 126.785 N = 600 N θz = 77.8° W 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 40 D Fx = −236.94 N Fx = −237 N W Fy = (450 N) sin 35° Fy = 258.11 N Fy = 258 N W Fz = (450 N) cos 35° cos 40 D Fz = 282.38 N (b) Fz = 282 N W cos θ x = Fx -236.94 N = F 450 N θx = 121.8° W cos θ y = Fy 258.11 N 450 N θ y = 55.0° W cos θ z = Fz 282.38 N = F 450 N θz = 51.1° W F = Note: From the given data, we could have computed directly θ y = 90D − 35D = 55D , which checks with the answer obtained. 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 W Fy = −F sin 40° = −(400 N) sin 40° = −257.12 N Fy = −257 N W Fz = + FH cos 35° = +(306.42 N) cos 35° = +251.00 N (b) cos θ x = Fx −175.755 N = 400 N F Fz = +251 N W θx = 116.1° W cos θ y = Fy −257.12 N 400 N θ y = 130.0° W cos θ z = Fz 251.00 N = F 400 N θz = 51.1° W F = 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 W Fy = −F sin 25° = −(400 N) sin 25° = −169.047 N Fy = −169.0 N W Fz = +FH sin15° = +(362.52 N) sin15° = +93.827 N (b) cos θ x = cos θ y = cos θ z = Fx +350.17 N = F 400 N Fy F = −169.047 N 400 N Fz +93.827 N = F 400 N Fz = +93.8 N W θ x = 28.9° W θ y = 115.0° W θz = 76.4° W PROBLEM 2.75 The angle between spring AB and the post DA is 30°. Knowing that the tension in the spring is 50 lb, 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 lb) cos 60° Fh = 25.0 lb Fx = −Fh cos 35° Fx = (−25.0 lb) cos 35 Fx = −20.479 lb D Fy = F sin 60D Fz = −Fh sin 35 D Fy = (50.0 lb) sin 60D Fz = (−25.0 lb) sin 35 Fy = 43.301 lb Fz = −14.3394 lb D (a) Fx = −20.5 lb W Fy = 43.3 lb W Fz = −14.33 lb W (b) cos θ x = Fx −20.479 lb = F 50 lb cos θ y = Fy cos θ z = Fz -14.3394 lb = F 50 lb F = 43.301 lb 50 lb θ x = 114.2° W θ y = 30.0° W θ z = 106.7° W PROBLEM 2.76 The angle between spring AC and the post DA is 30°. Knowing that the tension in the spring is 40 lb, 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 lb) cos 60° Fh = 20.0 lb (a) Fx = Fh cos 35° = (20.0 lb) cos 35° Fx = 16.3830 lb Fy = F sin 60° = (40 lb) sin 60° Fy = 34.641 lb Fz = −Fh sin 35° = −(20.0 lb) sin 35° Fz = −11.4715 lb Fx = 16.38 lb W Fy = 34.6 lb W Fz = −11.47 lb W (b) cos θ x = Fx 16.3830 lb = 40 lb F cos θ y = Fy cos θ z = Fz = -11.4715 lb 40 lb F F = 34.641 lb 40 lb θ x = 65.8° W θ y = 30.0° W θ z = 106.7° W PROBLEM 2.77 Cable AB is 65 ft long, and the tension in that cable is 3900 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor B, (b) the angles θ x , θ y , and θ z defining the direction of that force. SOLUTION cos θ y = From triangle AOB: 56 ft 65 ft = 0.86154 θ y = 30.51° Fx = −F sin θ y cos 20° (a) = −(3900 lb) sin 30.51° cos 20° Fx = −1861 lb W Fy = +F cos θ y = (3900 lb)(0.86154) Fz = +(3900 lb) sin 30.51° sin 20° (b) cos θ = x From above: Fx =− F 1861 lb = −0.4771 Fz F =+ Fz = +677 lb W θ = 118.5° W x 3900 lb θ y = 30.51° cos θz = Fy = +3360 lb W θ y = 30.5° W 677 lb 3900 lb = +0.1736 θz = 80.0° W PROBLEM 2.78 Cable AC is 70 ft long, and the tension in that cable is 5250 lb. 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 = 70 ft In triangle AOB: OA = 56 ft F = 5250 lb cos θ y = 56 ft 70 ft θ y = 36.870° FH = F sin θ y = (5250 lb) sin 36.870° = 3150.0 lb (a) Fx = −FH sin 50° = −(3150.0 lb) sin 50° = −2413.0 lb Fx = −2410 lb W Fy = +F cos θ y = +(5250 lb) cos 36.870° = +4200.0 lb Fy = +4200 lb W Fz = −FH cos 50° = −3150 cos 50° = −2024.8 lb (b) cos θ x = From above: Fx −2413.0 lb = F 5250 lb θ y = 36.870° θz = Fz −2024.8 lb = F 5250 lb Fz = −2025 lb W θx = 117.4° W θ y = 36.9° W θz = 112.7° W 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 cos θ x = Fx 240 N = F 770 N θy = F = θy = F 770 N θ y = 110.5° W 680 N Fz cos θ x = 71.8° W −270 N Fy cos F = 770 N W = 770 N θ z = 28.0° W 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 W cos θ x = Fx 320 N = F 570 N θ x = 55.8° W cos θ y = Fy θ y = 45.4° W cos θy = F 400 N 570 N Fz −250 N F = = 570 N θ z = 116.0° W 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 lb, determine (a) the angle θy, (b) the other components and the magnitude of the force. SOLUTION cos 2 θ + cos 2 θ + cos 2 θ = 1 x y z cos (69.3°) + cos θ y + cos (57.9°) = 1 2 2 2 cos θ y = ±0.7699 (a) (b) Since Fy < 0, we choose cos θ y = −0.7699 ∴ θ y = 140.3° W Fy = F cos θ y −174.0 lb = F (−0.7699) F = 226.0 lb F = 226 lb W Fx = F cosθ x = (226.0 lb) cos 69.3° Fx = 79.9 lb W Fz = F cosθ z = (226.0 lb) cos 57.9° Fz = 120.1 lb W 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 lb, determine (a) the angle θz, (b) the other components and the magnitude of the force. SOLUTION cos 2 θ + cos 2 θ + cos 2 θ = 1 x y z cos 70.9 + cos 144.9° + cos θ z ° = 1 2 D 2 2 cos θ z = ±0.47282 (a) (b) Since Fz < 0, we choose cosθ z = −0.47282 ∴ θz = 118.2° W Fz = F cos θ z −52.0 lb = F (−0.47282) F = 110.0 lb F = 110.0 lb W Fx = F cosθ x = (110.0 lb) cos 70.9° Fx = 36.0 lb W Fy = F cos θ y = (110.0 lb) cos144.9° Fy = −90.0 lb W 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 = F 2x + F 2y + F 2 z (210 N) 2 = (80 N) 2 + (Fy ) 2 + (184.024 N) 2 y F = − (210 N) 2 − (80 N) 2 − (184.024 N) 2 Fy = −62.0 lb W = −61.929 N (b) Fz = −184.0 N W cos θ = x cos θ y = Fx F = 80 N = 0.38095 210 N θ = 67.6° W x Fy 61.929 N = = −0.29490 F 210 N θ y = 107.2° W 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 + cos 2 θ + cos 2 θ = 1 x y z cos 65 + cos 40° + cos θz ° = 1 2 D 2 2 cos θ z = ±0.48432 (b) (a) Since Fz > 0, we choose cos θz = 0.48432, or θz = 61.032D ∴ θz = 61.0° W F = 1200 N Fx = F cosθx = (1200 N) cos65D Fx = 507 N W Fy = F cos θ y = (1200 N) cos 40° Fy = 919 N W Fz = F cos θ z = (1200 N) cos 61.032° Fz = 582 N W 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 = Fλ 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 W 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 = Fλ 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 W 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 2 kips, determine the components of the force exerted at A by the cable. SOLUTION Cable AB: = λ AB JJJG AB = (−46.765 ft )i +(45 ft ) j +(36 ft )k AB TAB = TAB λAB = 74.216 ft −46.765i + 45j + 36k 74.216 (TAB ) x = −1.260 kips W (TAB ) y = +1.213 kips W (TAB ) z = +0.970 kips W 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 1.5 kips, determine the components of the force exerted at A by the cable. SOLUTION Cable AB: = λ AC T AC JJJG AC = (−46.765 ft )i +(55.8 ft ) j +(−45 ft )k 85.590 ft AC =T = (1.5 kips) λ AC AC −46.765i +55.8j −45k 85.590 (TAC ) x = −0.820 kips W (TAC ) y = +0.978 kips W (TAC ) z = −0.789 kips W 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 B BA BA ⎛ 8 ⎜ 17 ⎝ BA =T F =T λ BA T BA ⎞ ⎟ ⎠ i+ ⎛ 8 =T ⎛ 12 ⎜ 17 ⎝ ⎜ 17 i + 17 ⎝ BA BA T BA 12 ⎞ ⎟ ⎠ j− ⎛ 9 ⎜ 17 ⎝ T BA j- ⎞ ⎟ ⎠ 9 k ⎞ 17 ⎟ ⎠ k=0 Setting TBA = 408 N yields, Fx = +192.0 N, Fy = +288 N, Fz = −216 N W 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 F =T λ D DA DA − ⎛ 5 ⎜ 13 ⎝ DA =T DA T DA ⎞ ⎟ ⎠ DA i+ ⎛ =T ⎛ 48 ⎜ 65 ⎝ DA DA 5 ⎜ 13 ⎝ ⎞ T − ⎟ ⎠ j+ i+ ⎛ 36 ⎜ 65 ⎝ 48 j+ 65 T DA ⎞ ⎟ ⎠ 36 ⎞ k 65 ⎟ ⎠ k=0 Setting TDA = 429 N yields, Fx = −165.0 N, Fy = +317 N, Fz = +238 N W 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 + 0.76604j − 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 cos θ = x cos θ = y cos θ = z Rx R Ry R Rz R = = = R = 515 N W 174.367 N 515.07 N 456.42 N 515.07 N 163.011 N 515.07 N = 0.33853 θ = 70.2° W x = 0.88613 θ = 27.6° W y = 0.31648 θ = 71.5° W z 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)[cos 50° 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 cos θ = x cos θ = y cos θ = z Rx R Ry R Rz R = = = 91.532 N 515.07 N 429.81 N 515.07 N 268.66 N 515.07 N R = 515 N W = 0.177708 θ = 79.8° W x = 0.83447 θ = 33.4° W y = 0.52160 θ = 58.6° W z PROBLEM 2.93 Knowing that the tension is 425 lb in cable AB and 510 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables. SOLUTION JJJG AB = (40 in.)i − (45 in.) j + (60 in.)k AB = (40 in.) 2 + (45 in.) 2 + (60 in.) 2 = 85 in. JJJG AC = (100 in.)i − (45 in.) j + (60 in.)k 2 2 2 AC = (100 in.) + (45 in.) + (60 in.) = 125 in. JJJG ⎡ (40in.)i −(45 in.) j +(60in.)k ⎤ AB T =T λ =T = (425 lb) AB AB AB AB ⎢ ⎥ AB 85 in. ⎣ ⎦ TAB = (200 lb)i − (225 lb) j + (300 lb)k JJJG ⎡ (100 in.)i −(45 in.) j +(60 in.)k ⎤ AC T =T λ =T = (510 lb) AC AC AC AC ⎢ ⎥ AC 125 in. ⎣ ⎦ TAC = (408 lb)i − (183.6 lb) j + (244.8 lb)k R = TAB + TAC = (608)i − (408.6 lb) j + (544.8 lb)k Then: and R = 912.92 lb cos θ = x cos θ = y cos θ = z 608 lb 912.92 lb 408.6 lb 912.92 lb 544.8 lb 912.92 lb R = 913 lb W = 0.66599 = −0.44757 θ = 48.2° W x θ = 116.6° W y = 0.59677 θ = 53.4° W z PROBLEM 2.94 Knowing that the tension is 510 lb in cable AB and 425 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables. SOLUTION JJJG AB = (40 in.)i − (45 in.) j + (60 in.)k AB = (40 in.) 2 + (45 in.) 2 + (60 in.) 2 = 85 in. JJJG AC = (100 in.)i − (45 in.) j + (60 in.)k 2 2 2 AC = (100 in.) + (45 in.) + (60 in.) = 125 in. JJJG ⎡ (40in.)i −(45 in.) j +(60in.)k ⎤ AB T =T λ =T = (510 lb) AB AB AB AB ⎢ ⎥ AB 85 in. ⎣ ⎦ TAB = (240 lb)i − (270 lb) j + (360 lb)k JJJG ⎡ (100 in.)i −(45 in.) j +(60 in.)k ⎤ AC T =T λ =T = (425 lb) AC AC AC AC ⎢ ⎥ AC 125 in. ⎣ ⎦ TAC = (340 lb)i − (153 lb) j + (204 lb)k R = TAB + TAC = (580 lb)i − (423 lb) j + (564 lb)k Then: and R = 912.92 lb cos θ = x cos θ y = cos θ = z 580 lb R = 913 lb W = 0.63532 −423 lb = −0.46335 912.92 lb 564 lb 912.92 lb θ = 50.6° W x 912.92 lb = 0.61780 θ y = 117.6° W θ = 51.8° W z 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 =T λ F BD BD =T BD BD JJJG BD 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 2 2 2 BE = (270 mm) + (400 mm) + (600 mm) = 770 mm =T λ F BE BE =T BE BE JJJG BE BE (385 N) [−(270 mm)i + (400 mm) j ?%8 Comments are closed.
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