Vector Mechanics Statics 9th Solution Manual

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Vector Mechanics for Engineers Statics and Dynamics 11th Edition Beer SOLUTIONS MANUAL Download at: 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) 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) 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) 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) 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) 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.

Engineering Mechanics Statics 9th Edition Rc Hibbeler Solution Manual Pdf

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.

Vector Mechanics Statics 10th Edition

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. (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.

Engineering Mechanics Statics Solutions Pdf

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.