CHAPTER 2
PROBLEM 2.1 Two forces are applied as shown to a hook. Determinee graphicall the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. (a) Parallelogram law: (b) Triangle rule: We measure: R =1391kN, α = 47.8 R =1391 N 47.8
PROBLEM 2.22 Two forces are applied as shown to a bracket support. Determine graphicall the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. (a) Parallelogram law: (b) Triangle rule: We measure: R = 906 lb, α = 26.6 R = 906 lb 26.6
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 graphicall the magnitude and direction of the resultant force eerted on the bracket using (a) the parallelogram law, (b) the triangle rule. (a) Parallelogram law: (b) Triangle rule: We measure: R = 20.1 kn, α = 21.2 R = 20.1 kn 21.2
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 graphicall the magnitude and direction of the resultant force eerted on the bracket using (a) the parallelogram law, (b) the triangle rule. (a) Parallelogram law: (b) Triangle rule: We measure: R = 8.03 kips, α = 3.8 R = 8.03 kips 3.8
PROBLEM 2.5 A stake is being pulled out of the ground b means of two ropes as shown. Knowing that α = 30, determine b trigonometr (a) the magnitude of the force P so that the resultant force eerted on the stake is vertical, (b) the corresponding magnitude of the resultant. Using the triangle rule and the law of sines: 120 N P (a) = sin 30 sin 25 (b) 30 + β + 25 = 180 β = 180 25 30 = 125 120 N R = sin 30 sin125 P = 101.4 N R = 196.6 N
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 T 1 = 800 lb, determine b trigonometr (a) the required tension T 2 in the right-hand portion if the resultant R of the forces eerted b the cable at A is to be vertical, (b) the corresponding magnitude of R. Using the triangle rule and the law of sines: (a) 75 + 40 + α = 180 α = 180 75 40 = 65 800 lb T2 = sin 65 sin 75 800 lb R (b) = sin 65 sin 40 T 2 = 853 lb R = 567 lb
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 T 2 = 1000 lb, determine b trigonometr (a) the required tension T 1 in the left-hand portion if the resultant R of the forces eerted b the cable at A is to be vertical, (b) the corresponding magnitude of R. Using the triangle rule and the law of sines: (a) 75 + 40 + β = 180 β = 180 75 40 = 65 1000 lb T1 = sin 75 sin 65 T 1 = 938 lb (b) 1000 lb R = sin 75 sin 40 R = 665 lb
PROBLEM 2.8 A disabled automobile is pulled b 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 ais of the automobile, determine b trigonometr (a) the tension in rope, (b) the magnitude of the resultant of the two forces applied at A. Using the law of sines: T R 2.2 kn = = sin 30 sin125 sin 25 T = 2.603 kn R = 4.264 kn (a) (b) T = 2.60 kn R = 4.26 kn
PROBLEM 2.9 A disabled automobile is pulled b means of two ropes as shown. Knowing that the tension in rope AB is 3 kn, determine b trigonometr the tension in rope and the value of α so that the resultant force eerted at A is a 4.8-kN force directed along the ais of the automobile. Using the law of cosines: Using the law of sines: T 2 2 2 T = (3 kn) + (4.8 kn) 2(3 kn)(4.8 kn)cos 30 = 2.6643 kn sinα sin 30 = 3 kn 2.6643 kn α = 34.3 T = 2.66 kn 34.3
PROBLEM 2.10 Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine b trigonometr (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. Using the triangle rule and law of sines: sinα sin 25 (a) = 50 N 35 N sin α = 0.60374 (b) α = 37.138 α + β + 25 = 180 β = 180 25 37.138 = 117.862 R 35 N = sin117.862 sin 25 α = 37.1 R = 73.2 N
PROBLEM 2.11 A steel tank is to be positioned in an ecavation. Knowing that α = 20, determine b trigonometr (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. Using the triangle rule and the law of sines: (a) β + 50 + 60 = 180 β = 180 50 60 = 70 425 lb P = sin 70 sin 60 P = 392 lb (b) 425 lb R = sin 70 sin 50 R = 346 lb
PROBLEM 2.12 A steel tank is to be positioned in an ecavation. Knowing that the magnitude of P is 500 lb, determine b trigonometr ( 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. 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 90 α = 47.402 R 500 lb (b) = sin (42.598 + 30 ) sin60 α = 42.6 R = 551 lb
PROBLEM 2.13 A steel tank is to be positioned in an ecavation. Determine b trigonometr (a) the magnitude and direction of the smallest force P for whichh the resultantt R of the two forces applied at A is vertical, (b) the corresponding magnitude of R. The smallest force P will be perpendicular to R. (a) P = (425 lb)cos30 P = 368 lb (b) R = (425 lb)sin 30 R = 213 lb
PROBLEM 2.14 For the hook support of Prob. 2.10, determine b trigonometr (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. The smallest force P will be perpendicular to R. (a) P = (50 N)sin 25 P = 21.1 N (b) R = (50 N)cos 25 R = 45.3 N
PROBLEM 2.15 For the hook support shown, determine b trigonometr the magnitude and direction of the resultant of the two forces applied to the support. Using the law of cosines: R 2 2 2 = (200 lb) + (300 lb) 2(200 lb)(300 lb)cos(45 + 65 ) R = 413.57 lb Using the law of sines: sinα sin (45 + 65 ) = 300 lb 413.57 lb α = 42.972 β = 90 + 25 42.972 R = 414 lb 72.0
PROBLEM 2.16 Solve Prob. 2.1 b trigonometr. PROBLEM 2.1 Two forces are applied as shown to a hook. Determine graphicall the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. Using the law of cosines: R 2 2 2 = (900 N) + (600 N ) 2(900 N )(600 N)cos(135 ) R = 1390.57N Using the law of sines: sin( α 30 ) sin (135 ) = 600N 1390.57N α 30 = 17.7642 α = 47.764 R =1391N 47.8
PROBLEM 2.17 Solve Problem 2.4 b trigonometr. PROBLEM 2.4 Two structural members B and C are bolted to bracket A.. Knowing that both members are in tension and thatt P = 6 kips and Q = 4 kips, determine graphicall the magnitude and direction of the resultant force eerted on the bracket using (a) the parallelogram law, (b) the triangle rule. Using the force triangle and the laws of cosines and sines: We have: γ = 180 (50 + 25 ) = 105 Then And R 2 = (4 kips) 2 = 64.423 kips + (6 kips) R = 8.0264 kips 4kips 8.0264 kips = sin(25 + α) sin105 sin(25 + α) = 0.48137 25 + α = 28.775 α = 3.775 2 2 2(4 kips)(6 kips) cos105 R = 8.03 kips 3.8
PROBLEM 2.18 For the stake of Prob. 2.5, knowing that the tension in one rope is 120 N, determine b trigonometr 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 b means of two ropes as shown. Knowing that α = 30, determine b trigonometr (a) the magnitude of the force P so that the resultant force eerted on the stake is vertical, (b) the corresponding magnitude of the resultant. Using the laws of cosines and sines: P 2 2 2 = (120 N) + (160 N) 2(120 N)(160 N)cos 25 P = 72.096 N And sinα sin 25 = 120 N 72.096 N sinα = 0.70343 α = 44.703 P = 72.1 N 44.7
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 b trigonometr the magnitude and direction of the resultant of the two forces. Using the force triangle and the laws of cosines and sines: We have γ = 180 (20 + 10 ) = 150 Then and Hence: R 2 2 = (48 N) + (60 N) 2(48 N)(60 N) cos150 R = 104.366 N 48 N 104.366 N = sinα sin150 sinn α = 0.22996 α = 13.2947 φ = 180 α 80 = 180 13.2947 80 = 86.705 2 R =104. 4 N 86.7
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 b trigonometr the magnitude and direction of the resultant of the two forces. Using the force triangle and the laws of cosines and sines: We have γ = 180 (20 + 10 ) = 150 Then and Hence: 2 R 2 = (60 N) + (48N) 2(60 N)(48 N)cos150 R = 104.366 N 60 N 104.366 N = sinα sin150 sinα = 0.28745 α = 16.7054 φ = 180 α 180 = 180 16.7054 80 = 83.295 2 R =104.44 N 83.3
PROBLEM 2.21 Determine the and components of each of the forces shown. Compute the following distances: OA = (84) = 116 in. 2 2 + (80) OB = (28) = 100 in. 2 2 + (96) OC = (48) = 102 in. 2 2 + (90) 29-lb Force: 84 F =+ (29 lb) 116 F =+ 21.0 lb 80 F =+ (29 lb) 116 F =+ 20.0 lb 50-lb Force: 28 F = (50 lb) 100 F = 14.00 lb 96 F =+ (50 lb) 100 F =+ 48.0 lb 51-lb Force: 48 F =+ (51 lb) 102 F =+ 24.0 lb 90 F = (51 lb) 102 F = 45.0 lb
PROBLEM 2.22 Determine the and components of each of the forces shown. Compute the following distances: 800-N Force: OA = OB = OC = (600) = 1000 mm (560) = 1060 mm (480) 2 2 + (800) 2 2 + (900) 2 2 + (900) = 1020 mm 800 F =+ (800 N) 1000 F =+ 640 N 600 F =+ (800 N) 1000 F =+ 480 N 424-N Force: 560 F = (424 N) 1060 F = 224 N 900 F = (424 N) 1060 F = 360 N 408-N Force: 480 F =+ (408 N) 1020 F =+ 192.0 N 900 F = (408 N) 1020 F = 360 N
PROBLEM 2.23 Determine the and components of each of the forces shown. 80-N Force: F = + (80 N)cos 40 F = 61.3 N F = + (80 N)sin 40 F = 51.4 N 120-N Force: F = + (120 N)cos70 F = 41.0 N F = + (120 N)sin 70 F = 112.8 N 150-N Force: F = (150 N)cos35 F = 122. 9 N F = + (150 N)sin 35 F = 86.0 N
PROBLEM 2.24 Determine the and components of each of the forces shown. 40-lb Force: F = + (40 lb)cos60 F = 20.0 lb F = (40 lb)sin60 F = 34.6 lb 50-lb Force: F = (50 lb)sin50 F = 38.3 lb F = (50 lb)cos50 F = 32.1 lb 60-lb Force: F = + (60 lb)cos25 F = 54.4 lb F = + (60 lb)sin 25 F = 25.4 lb
PROBLEM 2.25 Member BC eerts on member 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. 2 2 BC = (650 mm) + (720 mm) = 970 mm (a) 650 P = P 970 or 970 P = P 650 970 = 325 N 650 = 485 N P = 485 N (b) 720 P = P 970 720 = 485 N 970 = 360 N P = 970 N
PROBLEM 2.26 Member BD eerts 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. (a) (b) Vertical component P sin35 = 300 lb 300 lb P = sin 35 P v = P cos35 = (523 lb) cos 35 P = 523 lb P = 428 lb v
PROBLEM 2.27 The hdraulic clinder BC eerts 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. 180 = 45 + α + 90 + 30 α = 180 45 90 30 = 15 (a) (b) P cosα = P P P = cos α 600 N = cos15 = 621.17 N P tanα = P P = P tanα = (600 N)tan15 = 160.770 N P = 621 N P = 160.8 N
PROBLEM 2.28 Cable eerts on beam AB a force P directed along line. Knowing that P must have a 350-lb vertical component, determine (a) the magnitude of the force P, (b) its horizontal component. (a) P = P cos 55 (b) P 350 lb = cos 55 = 610.21 lb = Psin 55 = (610.211 lb)sin 55 = 499.85 lb P = 610 lb P = 500 lb
PROBLEM 2.29 The hdraulic clinder BD eerts 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. (a) (b) 750 N = Psin 20 P = 2192.9 N PABC = Pcos20 = (2192.9 N)cos 20 P A P = 2190 N ABC = 2060 N
PROBLEM 2.30 The gu wire BD eerts on the telephone pole a force P directed along BD. Knowing that P must have a 720-N component perpendicular to the pole, determine (a) the magnitude of the force P, (b) its component along line. (a) 37 P= P 12 37 = (720 N) 12 = 2220 N (b) P 35 = P 12 35 = (720 N) 12 = 2100 N P = 2.22 kn P = 2.10 kn
PROBLEM 2.31 Determine the resultant of the three forces of Problem 2.21. PROBLEMM 2.21 Determine the and components of each of the forces shown. Components of the forces were determined in Problem 2.21: Force 29 lb 50 lb 51 lb Comp. (lb) +21.0 14.00 +24.0 R =+ 31.0 Comp. (lb) +20.0 +48.0 45.0 R = + 23.0 R = R i+ R j R tan α = R = (31.0 lb) i + (23.0 lb) j 23.0 = 31.0 α = 36.573 23.0 lb R = sin (36.573 ) = 38.601 lb R = 38.6 lb 36.6
PROBLEM 2.32 Determine the resultant of the three forces of Problem 2.23. PROBLEMM 2.23 Determine the and components of each of the forces shown. Components of the forces were determined in Problem 2.23: Force 80 N 120 N 150 N Comp. (N) +61.3 +41.0 122.9 R = 20.6 Comp. (N) +51.4 +112.8 +86.0 R = + 250.2 R = Ri+ R j = ( 20..6 N) i + (250.22 N)j R tanα = R 250.22 N tanα = 20.6 N tanα = 12.1456 α = 85.293 250.2 N R = sin85.293 R = 251N 85.3
PROBLEM 2.33 Determine the resultantt of the three forces of Problem 2.24. PROBLEM 2.24 Determine the the forces shown. and components of each of Force 40 lb 50 lb 60 lb Comp. (lb) +20.00 38.30 +54.38 R =+ 36.08 Comp. (lb) 34.64 32.14 +25.36 R = 41.42 R = Ri+ R j = ( + 36.08 lb) i + ( 41.42 lb)j R tan α = R 41.42 lb tan α = 36.08 lb tan α = 1.14800 α = 48.942 41.42 lb R = sin 48.942 R = 54.9 lb 48.9
PROBLEM 2.34 Determinee the resultantt of the three forces of Problem 2.22. PROBLEM 2.22 Determine the and components of each of the forces shown. Components of the forces were determined in Problem 2.22: Force 800 lb 424 lb 408 lb Comp. (N) +640 224 +192 R =+ 608 Comp. (N) +4800 360 360 R = 240 R = R i+ R j R tan α = R = (608 lb) i + ( 240 lb) j 240 = 608 α = 21.541 240 N R = sin(21.541 ) = 653.65 N R = 654 N 21.5
PROBLEM 2.35 Knowing that α = forces shown. 35, determine the resultant of the three 100-N Force: 150-N Force: 200-N Force: F = + (100 N)cos35 =+ 81.915 N F = (100 N)sin35 = 57.358 N F = + (150 N)cos65 =+ 63.393 N F = (150 N)sin65 = 135.946 N F = (200 N)cos35 = 163.830 N F = (200 N)sin35 = 114.715 N Force 100 N 150 N 200 N Comp. (N) + 81.915 + 63.393 163.830 R = 18.522 Comp. (N) 57.358 135.946 114.715 R = 308.02 R = R i+ R j R tanα = R = ( 18.522 N) i + ( 308.02 N)j 308.02 = 18.522 α = 86.559 308.02 N R = sin 86.559 R = 309 N 86.6
PROBLEM 2. 36 Knowing that the tension in rope is 365 N, determine the resultant of the three forces eerted at point C of post BC. Determine force components: 960 Cable force : F = (365 N) = 240 N 1460 1100 F = (365 N) = 275N 1460 24 500-N Force: F = (500 N) = 480 N 25 7 F = (500 N) = 140 N 25 200-N Force: and 4 F = (200 N) = 160 N 5 3 F = (200 N) = 120 N 5 R R =Σ F = 240 N + 480 N+ 160 N = 400 N =Σ F = 275 N + 140 N 120 N = 255 N Further: 2 R = R + R = 2 R (400 N) + ( 255N) = 474.37 N 255 tanα = 400 α = 32.5 2 2 R = 4744 N 32.5
PROBLEM 2.37 Knowing that α = 40, determine the resultant of the three forces shown. 60-lb Force: 80-lb Force: 120-lb Force: and F = (60 lb)cos20 = 56.382 lb F = (60 lb)sin 20 = 20.521 lb F = (80 lb)cos60 = 40.000 lb F = (80 lb)sin 60 = 69.282 lb F = (120 lb)cos30 = 103.923 lb F R R = (120lb)sin30 = 60.000 lb =ΣF =ΣF = 200.305 lb = 29.803 lb Further: R = (200.305 lb) + (29.803 lb) = 202..510 lb 29. 803 tanα = 200.305 1 29.803 α = tan 200.305 = 8.46 2 2 R = 203 lb 8.46
PROBLEM 2.38 Knowing that α = 75, determine the resultant of the three forces shown. 60-lb Force: 80-lb Force: 120-lb Force: Then and F = (60 lb)cos 20 = 56.382 lb F = (60 lb)sin 20 = 20.521 lb F = (80 lb)cos95 = 6.9725 lb F = (80 lb)sin 95 = 79.6966 lb F = (120 lb) cos 5 = 119.543 lb F R R = (120 lb)sin 5 = 10.459 lb =Σ F = 168.953 lb =Σ F = 110.676 lb R = (168.953 lb) + (110.676 lb) = 201.976 lb 110.676 tan α = 168.953 tan α = 0.65507 α = 33.228 2 2 R = 2022 lb 33.2
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. R R R R =ΣF = (100 N)cos α + (150 N)cos( α + 30 ) (200 N)cosα = (100 N)cos α + (150 N)cos( α + 30 ) (1) =ΣF = (100 N)sin α (150 N)sin ( α + 30 ) (200 N)sinα = (300 N)sin α (150 N)sin ( α + 30 ) (2) (a) For R to be vertical, we must have R = 0. We make R = 0 in Eq. (1): 100cos α + 150cos( α + 30 ) = 0 100cos α + 150(cos α cos 30 sin α sin 30 ) = 0 29.904 cos α = 75sinα (b) Substituting for α in Eq. (2): R = 300sin 21.738 150sin 51.738 = 228.89 N 29.904 tanα = 75 = 0.39872 α = 21.738 α = 21.7 R= R = 228.89 N R = 229 N
PROBLEM 2.40 For the post of Prob. 2.36, determine (a) the required tension in rope if the resultant of the three forces eerted at point C is to be horizontal, (b) the corresponding magnitude of the resultant. 960 24 4 R =Σ F = T + (500 N) + (200 N) 1460 25 5 48 R = T + 640 N 73 (1) 1100 7 3 R =Σ F = T + (500 N) (200 N) 1460 25 5 55 R = T + 20 N 73 (2) (a) For R to be horizontal, we must have R = 0. Set R = 0 in Eq. (2): 55 T + 20 N = 0 73 T = 26.545 N T = 26.5 N (b) Substituting for T into Eq. (1) gives R R 48 = (26.545 N) + 640 N 73 = 622.55 N R= R = 623 N R = 623 N
PROBLEM 2.41 Determine (a) the required tension in cable, knowing that the resultant of the three forces eerted at Point C of boom BC must be directed along BC, (b) the corresponding magnitude of the resultant. Using the and aes shown: R R R = ΣF = ΣF = T sin10 + (50 lb) cos35 + (75 lb)cos60 = T sin10 + 78.458 lb = (50 lb)sin 35 + (75 lb)sin 60 T cos10 = 93.631 lb cos10 T (1) (2) (a) Set R = 0 in Eq. (2): 93.631 lb T cos10 = 0 T = 95.075 lb T = 95.1 lb (b) Substituting for T in Eq. (1): r R = (95.075 lb) sin10 + 78.458 lb = 94.968 lb R = R R = 95.0 lb
PROBLEM 2.422 For the block of Problems 2.37 and 2.38, determine (a) the required value of α if the resultant of the threee forces shown is to be parallel to the incline, (b) the corresponding magnitude of the resultant. Select the ais to be along a a. Then R =Σ = (60 lb) + (80 lb)cos α + (120 lb) sinα F (1) and R =Σ = (80 lb)sin α (120 lb)cosα F (2) (a) Set R = 0 in Eq. (2). (80 lb)sin α (120 lb) cosα = 0 Dividing each term b cosα gives: (80 lb) tanα = 120 lb 120 lb tanα = 80 lb α = 56.310 α = 56.3 (b) Substituting for α in Eq. (1) gives: R = 60 lb + (80 lb)cos56.31 + (120lb) sin56.31 = 204..22 lb R = 204 lb
PROBLEM 2.43 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable, (b) in cable BC. Free-Bod Diagram Force Triangle Law of sines: T TBC 400 lb = = sin 60 sin 40 sin80 (a) (b) 400 lb T = (sin 60 ) sin80 400 lb T BC = (sin 40 ) sin80 T = 352 lb T = 261 lb BC
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, (b) in cable BC. Free-Bod Diagram Force Triangle Law of sines: T TBC 6 kn = = sin 60 sin 35 sin85 (a) (b) 6 kn T = (sin 60 ) sin85 6 kn T BC = (sin 35 ) sin85 T = 5.22 kn T = 3.45 kn BC
PROBLEM 2.45 Two cables are tied together at C and loaded as shown. Determine the tension (a) in cable, (b) in cable BC. Free-Bod Diagram 1.4 tanα = 4.8 α = 16.2602 1.6 tan β = 3 β = 28.073 Force Triangle Law of sines: T TBC 1.98 kn = = sin 61.927 sin 73.740 sin 44.333 (a) (b) 1.98 kn T = sin 61.927 sin 44.333 1.98 kn T BC = sin 73.740 sin 44.333 T = 2.50 kn T = 2.72 kn BC
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 tensionn in (a) in cable, (b) in cable BC. Free-Bod Diagram Force Triangle Law of sines: T sin 35 T BC 500 N = = sin 75 sin 70 (a) 500 N T = sin 35 sin 70 T = 305 N (b) 500 N T BC = sin 75 sin70 T = 514 N BC
PROBLEM 2.47 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable, (b) in cable BC. Free-Bod Diagram Force Triangle W = mg = (200kg)(9.81m/s ) = 1962 N 2 Law of sines: (a) T A TBC 1962 N = = sin 15 sin 105 sin 60 (1962 N) sin 15 T = sin 60 T = 586 N (b) (1962 N)sin 105 T BC = sin 60 T = 2190 N BC
PROBLEM 2.48 Knowing that α = 20, determine the tension (a) in cable, (b) in rope BC. Free-Bod Diagram Force Triangle Law of sines: T TBC 1200 lb = = sin 110 sin 5 sin 65 (a) 12000 lb T = sin 110 sin 65 T = 1244 lb (b) 1200 lb T BC = sin 5 sin 65 T = 115.4 lb BC
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 and BC. Free-Bod Diagram ΣF = 0 TCA sin 30 + TCB sin 30 P cos 45 200N = 0 For P = 200N we have, 0.5T CA + 0.5T CB + 212.13 200 = 0 (1) ΣF = 0 T CA cos30 T CB cos30 Psin45 = 0 0.86603T + 0.86603T 212.13 = 0 (2) Solving equations ( 1) and (2) simultaneousl gives, CA CB T = 134.6 N CA T = 110.4 N CB
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. Free-Bod Diagram ΣF = 0 TCA sin 30 + TCB sin 30 P cos45 200N = 0 For TCA = 0 we have, 0.5T CB + 0.70711P 200 = 0 (1) ΣF = 0 T CA cos30 T CB cos30 Psin 45 = 0 ; again setting TCA A = 0 ields, 0.86603T 0.70711P = 0 (2) CB Adding equations (1) and (2) gives, 1.36603T CB = 200 henc ce T CB = 146.410N and P = 179.315N Substituting for T CB = 0 into the equilibrium equations and solving simultaneousl gives, 0.5 T CA 0.86603T + 0.70711P 200 = 0 CA 0.707111 P = 0 And T CA = 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 < P< 669 N
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 eerted on the rods A and B. Resolving the forces into - and -directions: Thus, R = P+ Q+ F + F = 0 Substituting components: R = (500 lb) j + [(650 lb)cos50 ] i [(650 lb)sin50 ] j In the -direction (one unknown force): 500 lb (650 lb)sin 50 + F A sin50 = 0 F A B A A 500 lb + (650 lb)sin 50 = sin 50 =1302.70 lb B + F i ( F cos50 ) i + F sin 50 ) j = 0 ( A Free-Bod Diagram F = 1303 lb A In the -direction: Thus, (650lb)cos50 + FB FA cos5 50 = 0 F B = F cos50 (650 lb)cos50 A = (1302.700 lb)cos50 (650 lb) cos50 = 419.55 lb F = 420 lb B
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 eerted on rods A and B are F A = 750 lb and F B = 400 lb, determine the magnitudes of P and Q. Resolving the forces into - and -directions: Substituting components: R = P + Q+ FA + F B = 0 R = Pj + Q cos 50 i Q sin 50 j [(750 lb) cos 50 ] i + [(750 lb)sin 50 ] j + (400 lb) i Free-Bod Diagram In the -direction (one unknown force): Q cos 50 [(750 lb) cos 50 + ] 4000 lb = 0 In the -direction: P (750 lb)cos 50 400 lb Q = cos 50 = 127.710 lb 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
PROBLEM 2. 53 A welded connection is in equilibrium under the action of the four forces shown. Knowing that F A = 8 kn and F B = 16 kn, determine the magnitudes of the other two forces. Free-Bod Diagram of Connection ΣFF = 0: 3 F 5 B 3 FC FA = 0 5 With F F A B = 8 kn = 16 kn 4 4 F C = (16 kn) (8 kn) 5 5 F = 6.40 kn C Σ F = 0: F D 3 3 + FB F 5 5 A = 0 With F A and F B as above: 3 3 F D = (16 kn) (8 kn) 5 5 F = 4.80 kn D
PROBLEM 2.54 A welded connection is in equilibrium under the action of the four forces shown. Knowing that F A = 5 kn and F D = 6 kn, determine the magnitudes of the other two forces. Free-Bod Diagram of Connection or With ΣF F B F A = 0: FD 3 = FD + F 5 = 5kN, A FD 3 3 FA + FB = 0 5 5 = 8kN F B ΣF 5 3 = 6kN+ 3 (5 kn) 5 = 0: FC + 4 4 FB FA = 0 5 5 F = 15.00 kn B F C 4 = ( FB FA) 5 4 = (15 kn 5kN) 5 F = 8.00 kn C
PROBLEM 2.55 A sailor is being rescued using a boatswain s chair that is suspended from a pulle that can roll freel on the support cable B and is pulled at a constant speed b 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 B, (b) in the traction cable CD. Free-Bod Diagram Σ F = 0: T B cos 10 T B cos 30 T CD cos 30 = 0 T CD = 0.137158T B (1) Σ F = 0: T B sin 10 + T B sin 30 +T CD sin 30 200 = 0 0.67365T A CB + 0.5T = 200 CD (2) (a) Substitute (1) into (2): 0.67365T B + 0.5(0.137158TT B ) = 200 T A 269.46 lb T = 269 lb B (b) From (1): CB = 0.137158 T = CD 8(269.46 lb) T = 37.0 lb CD
PROBLEM 2.56 A sailor is being rescued using a boatswain s chair thatt is suspended from a pulle thatt can roll freel on the support cable B and is pulled at a constant speed b cable CD. Knowing that α = 25 and β = 15 and that the tensionn in cable CD is 20 lb, determine (a) the combined weight of the boatswain ss chair and the sailor, (b) the tension in the support cable B. Free-Bod Diagram Σ F = 0: T B cos 15 T cos 25 (20 lb)cos 25 = 0 B T A Σ = 0: (304.04 lb) sin 15 + (304.04 lb)sin 25 F B = 304.04 lb +(20 lb)sin 25 W = 0 W = 215.64 lb (a) W = 216 lb (b) T = 304 lb B
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 simultaneousl. In each case determine the tension in each cable. Free-Bod Diagram Force Triangle (a) For a minimum tension in cable BC, set angle between cables to 90 degrees. B inspection, α = 35.0 T = (6 kn)cos35 T = (6 kn)sin35 BC T = 4.91 kn T = 3.44 kn BC (b) For equal tension in both cables, the force triangle will be an isosceles. Therefore, b inspection, α = 55.0 T 6 kn = TBC = (1 / 2) cos35 T = T = 3.66 kn BC
PROBLEM 2.58 For the cables of Problem 2.46, it is known that the maimum allowable tension is 600 N in cable and 750 N in cable BC. Determine (a) the maimum force P that can be applied at C, (b) the corresponding value of α. Free-Bod Diagram Force Triangle (a) (b) Law of cosines Law of sines P 2 2 2 = (600) + (750) 2(600)(750) cos(25 + 45 ) P = 784.02 N sin β sin (25 + 45 ) = 600 N 784.02 N β = 46.0 α = 46.0 + 25 P = 784 N α = 71.0
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. Free-Bod Diagram Force Triangle To be smallest, T BC must be perpendicular to the direction of T. (a) (b) Thus, α = 5.00 T = (1200 lb)sin 5 BC α = 5.00 T = 104.6 lb BC
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 eceed 60 lb in either cable. Free-Bod Diagram Σ = 0: F T BC Qcos60 + 75 lb = 0 T = 75 lb Q cos60 BC (1) Σ = 0: F T Q sin 60 = 0 T = Q sin60 (2) Requirement: Requirement: T = 60 lb: From Eq. (2): Q sin 60 = 60 lb Q = 69.3 lb T = 60 lb: BC From Eq. (1): 75 lb Qcos60 = 60lb Q = 30.0 lb 30.0 lb Q 69.3 lb
PROBLEM 2.61 A movable bin and its contents have a combined weight of 2.8 kn. Determine the shortest chain sling B that can be used to lift the loaded bin if the tension in the chain is not to eceed 5 kn. Free-Bod Diagram h tanα = 0.6 m (1) Isosceles Force Triangle h From Eq. (1): tan16.2602 = h = 0.175000 m 0.6 m Law of sines: sinα = T sinα = 1 2 = 5kN 1 2 (2.8 kn) T (2.8 kn) 5kN α = 16.2602 Half-length of chain 2 2 = = (0.6 m) + (0.175 m) = 0.625 m Total length: = 2 0.625 m 1.250 m
PROBLEM 2.62 For W = 800 N, P = 200 N, and d = 600 mm, determine the value of h consistent with equilibrium. Free-Bod Diagram T = T = 800 N BC 2 2 ( ) = BC = h + d P d 800 = 1+ 2 h h Σ F = 0: 2(800 N) P= 0 2 2 h + d 2 Data: P= 200 N, d = 600 mm and solving for h 200 N 600 mm 800 N = 1+ 2 h 2 h = 75.6 mm
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) = 15 in. = 4.5 in., (b) (a) Free Bod: Collar A Force Triangle P 50 lb = 4.5 20.5 P =10.98 lb (b) Free Bod: Collar A Force Triangle P 15 = 50 lb 25 P = 30.0 lb
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 for which the collar is in equilibrium when P = 48 lb. Free Bod: Collar A Force Triangle N 2 = (50) (48) = 196 N = 14.00 lb 2 2 Similar Triangles 48 lb = 20 in. 14 lb = 68.6 in.
PROBLEM 2.65 Three forces are applied to a bracket as shown. The directions of the two 150-N forces ma var, but the angle between these forces is alwas 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. Combine the two 150-N forces into a resultant force Q: Q = 2(150 N)cos25 = 271.89 N Equivalent loading at A: Using the law of cosines: 2 2 2 (600 N) = (500 N) + (271.89 N) + 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 < 222.6
PROBLEM 2.66 A 200-kg crate is to bee supported b the rope-and-pulle arrangement shown. Determinee the magnitude and direction of the force P that must be eerted 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 pulle. This can be proved b the methods of Ch. 4.) Free-Bod Diagram: Pulle A 5 Σ F = 0: 2P + P cosα = 0 281 cos α = 0.59655 α = ± 53.377 For α =+ 53.377 : Σ F = 0: 2P 16 + P sin 53.377 1962 281 N = 0 P = 724 N 53.4 For α = 53.377 : Σ F = 0: 2P 16 + P sin( 53.377 ) 1962 N= 0 281 P =1773 53.4
PROBLEM 2.67 A 600-lb crate is supported b several rope- for each arrangement the tension in the rope. (See the hint for Problem 2.66.) and-pulle arrangements as shown. Determine Free-Bod Diagram of Pulle (a) ΣFF = 0: 2T (600 lb) = 0 1 T = (600 lb) 2 T = 300 lb (b) ΣFF = 0: 2T (600 lb) = 0 1 T = (600 lb) 2 T = 300 lb (c) (d) (e) ΣFF ΣFF ΣFF = 0: 3T (600 lb) = 0 = 0: 3T (600 lb) = 0 = 0: 4T (600 lb) = 0 1 T = (600 lb) 3 1 T = (600 lb) 3 1 T = (600 lb) 4 T = 200 lb T = 200 lb T =150.0 lb
PROBLEM 2..68 Solve Parts b and d of Problem 2.67, assuming thatt the free end of the rope is attached to the crate. PROBLEM 2.67 A 600-lb crate is supported b several rope-and-pulle arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Problem 2.66.) Free-Bod Diagram of Pulle and Crate (b) Σ = 0: F 3 T (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb (d) Σ = 0: F 4 T (600 lb) = 0 1 T = (600 lb) 4 T =150.0 lb
PROBLEM 2.69 A load Q is applied to the pulle C, whichh can roll on the cable B. The pulle is held in the position shown b a second cable CAD, which passes over the pulle A and supports a load P. Knowing that P = 750 N, determine (a) the tension in cable B, (b) the magnitude of load Q. Free-Bod Diagram: Pulle C (a) Σ F = 0: Hence: T B B (cos 25 cos55 ) (750 N)cos55 = 0 T = 1292.88 N B T A B =1293 N (b) Σ F = 0: T (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
PROBLEM 2.70 An 1800-N load Q is applied to the pulle C, which can roll on the cable B. The pulle is held in the position shown b a second cable CAD, which passes over the pulle A and supports a load P. Determine (a) the tensionn in cable B, (b) the magnitude of load P. Free-Bod Diagram: Pulle C Σ = 0: (cos25 cos55 ) Pcos55 = 0 F T B or or (a) Σ = 0: (sin25 + sin55 ) + Psin 55 1800 N = 0 F T B Substitute Equation (1) into Equation (2): 1.24177T + 0. 81915(0.58010 T ) = 18000 N B P = 0.58010T B 1.24177TT + 0.81915 P = 1800 N B B (1) (2) Hence: T = 1048.37 N B (b) Using (1), P = 0. 58010(1048.37 N) = 608.166 N P = 608 N T A B =1048 N
PROBLEM 2.71 Determine (a) the,, and z components of the 600-N force, (b) the angles θ, θ, and θ z that the force forms with the coordinate aes. (a) F = (600 N)sin 25 cos30 F = 219.60 N F = 220 N F F = (600 N)cos 25 = 543.78 N F = 544 N (b) F = (380.36 N)sin 25 sin 30 z F = 126.785 N z F = 126.8 N F 219.60 N cos θ = = θ = 68.5 F 600 N F 543.78 N cos θ = = θ = 25.0 F 600 N Fz 126.785 N cos θ z = = θ z = 77.8 F 600 N z
PROBLEM 2.72 Determine (a) the,, and z components of the 450-N force, (b) the angles θ, θ, and θ z that the force forms with the coordinate aes. (a) F = (450 N)cos 35 sin 40 F = 236.94 N F = 237 N F F z z = (450 N)sin 35 = 258.11 N F = 258 N F = (450 N)cos35 cos40 F = 282.38 N F = 282 N F -236.94 N (b) cos θ = = θ = 121.8 F 450 N F 258.11 N cos θ = = θ = 55.0 F 450 N Fz 282.38 N cos θ z = = θ z = 51.1 F 450 N Note: From the given data, we could have computed directl θ = 90 35 = 55, which checks with the answer obtained. z
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 maimum recoil force is 400 N, determine (a) the,, and z components of that force, (b) the values of the angles θ, θ, and θ z defining the direction of the recoil force. (Assume that the,, and z aes are directed, respectivel, east, up, and south.) Recoil force F = 400 N = (400 N)cos40 F H = 306.42 N (a) F = FH sin 35 = (306.42 N) sin 35 = 175.755 N F = 175.8 N F = Fsin 40 = (400 N)sin 40 = 257.12 N F = 257 N (b) F =+ F cos35 z H =+ (306.42 N) cos 35 =+ 251.00 N F =+ 251 N F 175.755 N cosθ = = θ = 116.1 F 400 N z F 257.12 N cosθ = = θ = 130.0 F 400 N Fz 251.00 N cosθ z = = θ z = 51.1 F 400 N
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 maimum recoil force is 400 N, determine (a) the,, and z components of that force, (b) the values of the angles θ, θ, and θ z defining the direction of the recoil force. (Assume that the,, and z aes are directed, respectivel, east, up, and south.) Recoil force F = 400 N F H = (400 N)cos25 = 362.52 N (a) F =+ F cos15 H =+ (362.52 N)cos15 =+ 350.17 N F =+ 350 N F = Fsin 25 = (400 N)sin 25 = 169.047 N F = 169.0 N (b) F =+ F sin15 z H =+ (362.52 N)sin15 =+ 93.827 N F =+ 93.8 N F + 350.17 N cosθ = = θ = 28.9 F 400 N z F 169.047 N cosθ = = θ = 115.0 F 400 N Fz + 93.827 N cosθz = = θ z = 76.4 F 400 N
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,, and z components of the force eerted on the circular plate at B, (b) the angles θ, θ, and θ z defining the direction of the force at B. F F h h = Fcos 60 = (50 lb)cos 60 = 25.0 lb h z h F = F cos 35 F = Fsin 60 F = F sin35 z F = ( 25.0 lb)cos35 F = (50.0 lb)sin 60 F = ( 25.0 lb)sin35 F = 20.479 lb F = 43.301 lb F = 14.3394 lb (a) z F = 20.5 lb F = 43.3 lb F = 14.33 lb z (b) F 20.479 lb cos θ = = θ = 114.2 F 50 lb F 43.301 lb cos θ = = θ = 30.0 F 50 lb Fz -14.3394 lb cos θ z = = θ z = 106.7 F 50 lb
PROBLEM 2.76 The angle between spring and the post DA is 30. Knowing that the tension in the spring is 40 lb, determine (a) the,, and z components of the force eerted on the circular plate at C, (b) the angles θ, θ, and θ z defining the direction of the force at C. (a) h F = Fcos 60 h h = (40 lb)cos 60 F = 20.0 lb F = F cos 35 F = Fsin 60 F = F sin 35 z h = (20.0 lb)cos 35 = (40 lb)sin 60 = (20.0 lb)sin 35 F = 16.3830 lb F = 34.641 lb F = 11.4715 lb z F = 16.38 lb F = 34.6 lb F = 11.47 lb z (b) F 16.3830 lb cos θ = = θ = 65.8 F 40 lb F 34.641 lb cos θ = = θ = 30.0 F 40 lb Fz -11.4715 lb cos θ z = = θ z = 106.7 F 40 lb
PROBLEM 2.77 Cable AB is 65 ft long, and the tension in that cable is 3900 lb. Determine (a) the,, and z components of the force eerted b the cable on the anchor B, (b) the angles θ, θ, and θ z defining the direction of that force. From triangle AOB: (a) 56 ft cosθ = 65 ft = 0.86154 θ = 30.51 F = F sinθ cos 20 = (3900 lb)sin30.51 cos20 F = + F cosθ = (3900 lb)(0.86154) F =+ 3360 lb F = 1861 lb F =+ (3900 lb)sin 30.51 sin20 z F =+ 677 lb z (b) F 1861 lb cosθθ = = F 3900 lb = 0.47 771 θ = 118.5 From above: θ = 30.51 θ = 30.5 Fz cosθ z = F 677 lb =+ = + 3900 lb 0.1736 θ = 80.0 z
PROBLEM 2.78 Cable is 70 ft long, and the tension in that cable is 5250 lb. Determine (a) the,, and z components of the force eerted b the cable on the anchor C, (b) the angles θ, θ, and θ z defining the direction of that force. In triangle AOB: = 70 ft OA = 56 ft F = 5250 lb 56 ft cosθ = 70 ft θ = 36.870 F H = Fsinθ = (5250 lb)sin 36.870 = 3150.0 lb (a) F = F sin 50 = (3150.0 lb)sin 50 = 2413.0 lb H (b) F =+ F cos θ =+ (5250 lb) cos36.870 =+ 4200.0 lb z H F = 2410 lb F =+ 4200 lb F = F cos50 = 3150cos50 = 2024.8 lb F = 2025 lb F 2413.0 lb cosθ = = θ = 117.4 F 5250 lb From above: θ = 36.870 θ = 36.9 Fz 2024.8 lb θz = = θ z = 112.7 F 5250 lb z
PROBLEM 2.79 Determine the magnitude and direction of the force F = (240 N)i (270 N)j + (680 N)k. 2 2 2 z F = F + F + F F F = 2 2 2 = (240 N) + ( 270 N) + ( 680 N) 770 N F 240 N cos θ = = θ = 71.8 F 770 N F 270 N cos θ = = θ = 110.5 F 770 N Fz 680 N cos θ = = θ z = 28.0 F 770 N
PROBLEM 2.80 Determine the magnitude and direction of the force F = (320 N)i + (400 N)j (250 N)k. 2 2 2 z F = F + F + F F F = 2 2 2 = (320 N) + (400 N) + ( 250 N) 570 N F 320 N cos θ = = θ = 55.8 F 570 N F 400 N cos θ = = θ = 45.4 F 570 N Fz 250 N cos θ = = θ z = 116.0 F 570 N
PROBLEM 2.81 A force acts at the origin of a coordinate sstem in a direction defined b the angles θ = 69.3 and θ z = 57.9. Knowing that the component of the force is 174.0 lb, determine (a) the angle θ, (b) the other components and the magnitude of the force. 2 2 2 θ θ θz cos + cos + cos = 1 2 2 2 θ cos (69.3 ) + cos + cos (57.9 ) = 1 cosθ =± 0.7699 (a) Since F < 0, we choose cosθ = 0.7699 θ = 140.3 (b) F = Fcosθ 174.0 lb = F( 0.7699) F = 226.0 lb F = 226 lb F = Fcos θ = (226.0 lb) cos69.3 F = 79.9 lb F = Fcos θ = (226.0 lb)cos57.9 F = 120.1lb z z z
PROBLEM 2.82 A force acts at the origin of a coordinate sstem in a direction defined b the angles θ = 70.9 and θ = 144.9. Knowing that the z component of the force is 52.0 lb, determine (a) the angle θ z, (b) the other components and the magnitude of the force. 2 2 2 θ θ θz cos + cos + cos = 1 2 2 2 cos 70.9 + cos 144.9 + cos θ = 1 z cosθ =± 0.47282 (a) Since F z < 0, we choose cosθ z = 0.47282 θ z = 118.2 z (b) F = Fcosθ z 52.0 lb = F( 0.47282) z F = 110.0 lb F = 110.0 lb F = Fcos θ = (110.0 lb)cos70.9 F = 36.0 lb F = Fcos θ = (110.0 lb)cos144.9 F = 90.0 lb
PROBLEM 2.83 A force F of magnitude 210 N acts at the origin of a coordinate sstem. Knowing that F = 80 N, θ z = 151.2, and F < 0, determine (a) the components F and F z, (b) the angles θ and θ. (a) F = Fcos θ = (210 N) cos151.2 z z = 184.024 N F = 184.0 N z Then: So: Hence: 2 2 2 2 = + + z F F F F 2 2 2 2 = + F + (210 N) (80 N) ( ) (184.024 N) 2 2 2 F = (210 N) (80 N) (184.024 N) = 61.929 N F = 62.0 lb (b) F 80 N cosθ = = = 0.38095 θ = 67.6 F 210 N F 61.929 N cosθ = = = 0.29490 F 210 N θ = 107.2
PROBLEM 2.84 A force F of magnitude 1200 N acts at the origin of a coordinate sstem. Knowing that θ = 65, θ = 40, and F z > 0, determine (a) the components of the force, (b) the angle θ z. 2 2 2 θ θ θz cos + cos + cos = 1 2 2 2 cos 65 + cos 40 + cos θ = 1 z cosθ =± 0.48432 z (b) Since F z > 0, we choose cosθ = 0.48432, or θ = 61.032 (a) F = 1200 N z F = Fcos θ = (1200 N) cos65 z = 61.0 θ z F = 507 N F = Fcos θ = (1200 N)cos 40 F = 919 N F = Fcos θ = (1200 N)cos61.032 F = 582 N z z z
PROBLEM 2.85 A frame ABC is supported in part b 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 eerted b the cable on the support at D. DB = (480 mm) i (510 mm) j+ (320 mm) k 2 2 2 DB = (480 mm) + (510 mm ) + (320 mm) = 770 mm F = Fλ DB 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 F =+ 240 N, F = 255 N, F =+ 160.0 N z
PROBLEM 2.86 For the frame and cable of Problem 2.85, determine the components of the force eerted b the cable on the support at E. PROBLEM 2.85 A frame ABC is supported in part b 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 eerted b the cable on the support at D. EB = (270 mm) i (400 mm) j+ (600 mm) k 2 2 2 EB = (270 mm) + (400 mm) + (600 mm) = 770 mm F = Fλ EB 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 F =+ 135.0 N, F = 200 N, F =+ 300 N z
PROBLEM 2.87 In order to move a wrecked truck, two cables are attached at A and pulled b winches B and C as shown. Knowing that the tension in cable AB is 2 kips, determine the components of the force eerted at A b the cable. Cable AB: λ T AB AB ( 46.765 ft) i+ (45 ft) j+ (36 ft) k = = AB 74.216 ft 46.765i+ 45j+ 36k = T λ = 74.216 AB AB AB ( T AB) = 1.260 kips ( T ) =+ 1.213 kips AB ( T AB) z =+ 0.970 kips
PROBLEM 2.88 In order to move a wrecked truck, two cables are attached at A and pulled b winches B and C as shown. Knowing that the tension in cable is 1.5 kips, determine the components of the force eerted at A b the cable. Cable AB: λ T ( 46.765 ft) i+ (55.8 ft) j+ ( 45 ft) k = = 85.590 ft 46.765i+ 55.8j 45k = T λ = (1.5 kips) 85.590 ( T ) = 0.820kips ( T ) =+ 0.978 kips ( T ) = 0.789 kips z
PROBLEM 2.89 A rectangular plate is supported b three cables as shown. Knowing that the tension in cable AB is 408 N, determine the components of the force eerted on the plate at B. We have: Thus: BA =+ (320 mm) i+ (480 mm) j-(360 mm) k BA = 680 mm F BA 8 12 9 = T λ = T = T BA i+ j- k 17 17 17 B BA BA BA BA 8 12 9 TBA TBA TBA 0 17 i+ = 17 j 17 k Setting T = 408 N ields, BA F =+ 192.0 N, F =+ 288 N, F = 216 N z
PROBLEM 2.90 A rectangular plate is supported b three cables as shown. Knowing that the tension in cable AD is 429 N, determine the components of the force eerted on the plate at D. We have: Thus: DA = (250 mm) i+ (480 mm) j+ (360 mm) k DA = 650 mm F DA 5 48 36 = T λ = T = T DA i+ j+ k 13 65 65 D DA DA DA DA 5 48 36 TDA TDA TDA 0 13 + + = 65 65 i j k Setting T = 429 N ields, DA F = 165.0 N, F =+ 317 N, F =+ 238 N z
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. P= (300 N)[ cos30 sin15 i+ sin 30 j+ cos30 cos15 k] = (67.243 N) i+ (150 N) j+ (250.95 N) k Q= (400 N)[cos50 cos 20 i+ sin 50 j cos50 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 2 2 2 R = (174.367 N) + (456.42 N) + (163.011 N) = 515.07 N R = 515 N R 174.367 N cosθ = = = 0.33853 θ = 70.2 R 515.07 N R 456.42 N cosθ = = = 0.88613 θ = 27.6 R 515.07 N Rz 163.011 N cosθ z = = = 0.31648 θ z = 71.5 R 515.07 N
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. P = (400 N)[ cos30 sin15 i+ sin 30 j+ cos30 cos15 k] = (89.678 N) i+ (200 N) j+ (334.61 N) k Q= (300 N)[cos50 cos 20 i+ sin 50 j cos50 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 = 2 2 2 (91.532 N) + (429.81 N) + (268.66 N) = 515.07 N R = 515 N R 91.532 N cosθ = = = 0.177708 R 515.07 N θ = 79.8 R 429.81 N cosθ = = = 0.83447 θ = 33.4 R 515.07 N Rz 268.66 N cosθ z = = = 0.52160 R 515.07 N θ z = 58.6
PROBLEM 2.93 Knowing that the tension is 425 lb in cable AB and 510 lb in cable, determine the magnitude and direction of the resultant of the forces eerted at A b the two cables. Then: and AB = (40in.) i (45in.) j+ (60in.) k 2 2 2 AB = (40 in.) + (45 in.) + (60 in.) = 85 in. = (100 in.) i (45 in.) j+ (60 in.) k 2 2 2 = (100 in.) + (45 in.) + (60 in.) = 125 in. AB (40in.) i (45in.) j+ (60in.) k TAB = TAB λab = TAB = (425 lb) AB 85 in. TAB = (200 lb) i (225 lb) j+ (300 lb) k (100 in.) i (45 in.) j+ (60 in.) k T = T λ = T = (510 lb) 125 in. T = (408 lb) i (183.6 lb) j+ (244.8 lb) k R = T + T = (608) i (408.6 lb) j+ (544.8 lb) k AB R = 912.92 lb R = 913 lb 608 lb cosθ = = 0.66599 θ = 48.2 912.92 lb 408.6 lb cosθ = = 0.44757 912.92 lb θ = 116.6 544.8 lb cosθ z = = 0.59677 912.92 lb θ z = 53.4
PROBLEM 2.94 Knowing that the tension is 510 lb in cable AB and 425 lb in cable, determine the magnitude and direction of the resultant of the forces eerted at A b the two cables. Then: and AB = (40in.) i (45in.) j+ (60in.) k 2 2 2 AB = (40 in.) + (45 in.) + (60 in.) = 85 in. = (100 in.) i (45 in.) j+ (60 in.) k 2 2 2 = (100 in.) + (45 in.) + (60 in.) = 125 in. AB (40in.) i (45in.) j+ (60in.) k TAB = TAB λab = TAB = (510 lb) AB 85 in. TAB = (240 lb) i (270 lb) j+ (360 lb) k (100 in.) i (45 in.) j+ (60 in.) k T = T λ = T = (425 lb) 125 in. T = (340 lb) i (153 lb) j+ (204 lb) k R = T + T = (580 lb) i (423 lb) j+ (564 lb) k AB R = 912.92 lb R = 913 lb 580 lb cosθ = = 0.63532 θ = 50.6 912.92 lb 423 lb cosθ = = 0.46335 θ = 117.6 912.92 lb 564 lb cosθ z = = 0.61780 θ z = 51.8 912.92 lb
PROBLEM 2.95 For the frame of Problem 2.85, determine the magnitude and direction of the resultant of the forces eerted b the cable at B knowing that the tension in the cable is 385 N. PROBLEM 2.85 A frame ABC is supported in part b 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 eerted b the cable on the support at D. BD = (480 mm) i+ (510 mm) j (320 mm) k 2 2 2 BD = (480 mm) + (510 mm) + (320 mm) = 770 mm BD FBD = TBDλ 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 BE = (270 mm) i+ (400 mm) j (600 mm) k 2 2 2 BE = (270 mm) + (400 mm) + (600 mm) = 770 mm BE FBE = TBEλ 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= F + F = (375 N) i+ (455 N) j (460 N) k BD BE 2 2 2 R = (375 N) + (455 N) + (460 N) = 747.83 N R = 748 N 375 N cos θ = 747.83 N θ = 120.1 455 N cos θ = 747.83 N θ = 52.5 460 N cos θ z = 747.83 N θ z = 128.0
PROBLEM 2.96 For the plate of Prob. 2.89, determine the tensions in cables AB and AD knowing that the tension in cable is 54 N and that the resultant of the forces eerted b the three cables at A must be vertical. We have: Thus: AB = (320 mm) i (480 mm) j+ (360 mm) k AB = 680 mm = (450 mm) i (480 mm) j+ (360 mm) k = 750 mm AD = (250 mm) i (480 mm) j 360 mm k AD = 650 mm ( ) Substituting into the Eq. R =ΣF and factoring i, j, k : AB TAB TAB = TAB λab = TAB = i j+ k AB 680 54 T = T λ = T = ( 450 i 480 j+ 360 k) 750 AD TAD TAD = TAD λad = TAD = i j k AD 650 ( 320 480 360 ) ( 250 480 360 ) 320 250 R = TAB + 32.40 + TAD 680 650 i 480 480 + TAB 34.560 TAD 680 650 j 360 360 + TAB + 25.920 TAD 680 650 k
PROBLEM 2.96 (Continued) Since R is vertical, the coefficients of i and k are zero: i : k : Multipl (1) b 3.6 and (2) b 2.5 then add: 252 T 181.440 0 680 AB + = 320 250 TAB + 32.40 + TAD = 0 (1) 680 650 360 360 TAB + 25.920 TAD = 0 (2) 680 650 T = 489.60 N AB Substitute into (2) and solve for T AD: 360 360 (489.60 N) + 25.920 T 0 680 650 AD = T = 514.80 N AD T = 490 N AB T = 515 N AD
PROBLEM 2.97 The boom OA carries a load P and is supported b two cables as shown. Knowing that the tension in cable AB is 183 lb and that the resultant of the load P and of the forces eerted at A b the two cables must be directed along OA, determine the tension in cable. Cable AB: Cable : Load P: T AB = 183 lb AB ( 48 in.) i+ (29 in.) j+ (24 in.) k TAB = TABλ AB = TAB = (183 lb) AB 61in. TAB = (144 lb) i+ (87 lb) j+ (72 lb) k ( 48 in.) i+ (25 in.) j+ ( 36 in.) k T = Tλ = T = T 65 in. 48 25 36 T = Ti+ T j Tk 65 65 65 P= P j For resultant to be directed along OA, i.e., -ais 36 Rz = 0: Σ Fz = (72 lb) T = 0 T = 130.0 lb 65
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 b two cables as shown. Knowing that the tension in cable AB is 183 lb and that the resultant of the load P and of the forces eerted at A b the two cables must be directed along OA, determine the tension in cable. See Problem 2.97. Since resultant must be directed along OA, i.e., the -ais, we write 25 R = 0: Σ F = (87 lb) + T P= 0 65 T = 130.0 lb from Problem 2.97. 25 Then (87 lb) + (130.0 lb) P = 0 P = 137.0 lb 65
PROBLEM 2.99 A container is supported b 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. Free-Bod Diagram at A: The forces applied at A are: TAB, T, TAD, and W where W= W j. To epress the other forces in terms of the unit vectors i, j, k, we write AB = (450 mm) i+ (600 mm) j AB = 750 mm =+ (600 mm) j (320 mm) k = 680 mm AD =+ (500 mm) i+ (600 mm) j+ (360 mm) k AD = 860 mm ( 450mm) (600mm) and AB i+ j TAB = λabtab = TAB = TAB AB 750 mm 45 60 = i+ j T AB 75 75 (600 mm) i (320 mm) j T = λt = T = T 680 mm T 60 32 = j k 68 68 T AD (500 mm) i+ (600 mm) j+ (360 mm) k = λ T = T = T AD 860 mm AD AD AD AD AD 50 60 36 = i+ j+ k 86 86 86 T AD
PROBLEM 2.99 (Continued) Equilibrium condition: Σ F = 0: T + T + T + W = 0 AB AD Substituting the epressions obtained for TAB, T,and T AD; factoring i, j, and k; and equating each of the coefficients to zero gives the following equations: From i: 45 50 TAB + TAD = 0 75 86 (1) From j: 60 60 60 TAB + T + TAD W = 0 75 68 86 (2) From k: 32 36 T + TAD = 0 68 86 (3) Setting T = 6kN in (1) and (2), and solving the resulting set of equations gives AB T T = 6.1920 kn = 5.5080 kn W = 13.98 kn
PROBLEM 2.100 A container is supported b 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. See Problem 2.99 for the figure and analsis leading to the following set of linear algebraic equations: 45 50 TAB + TAD = 0 (1) 75 86 60 60 60 TAB + T + TAD W = 0 (2) 75 68 86 32 36 T + TAD = 0 (3) 68 86 Setting T = 4.3 kn into the above equations gives AD T T AB = 4.1667 kn = 3.8250 kn W = 9.71 kn
PROBLEM 2.1011 Three cables are used to tether a balloon as shown. Determine the vertical force P eerted b the balloon at A knowing that the tension in cable AD is 481 N. FREE-BODY DIAGRAM AT A The forces applied at A are: T AB, T, T, and P where P = Pj. To epress the other forces in terms of the unit vectors i, j, k, we write AB = (4.20 m) i (5.60 m) j AB = 7.00 m = (2.40 m) i (5.60 m) j + (4.20 m) k = 7.40 m AD = (5.60 m) j (3.30 m) k AD = 6.50 m AB and TAB = TAB λab = TAB = ( 0.6i 0.8 j) T AB AB T = T λ = T = (0.32432i 0.75676j + 0.56757 k) T AD TAD = TAD λad = T AD = ( 0.86154j 0.50769 k) T AD AD AD
PROBLEM 2.101 (Continued) Equilibrium condition: Σ F = T 0: AB + T + TAD + Pj = 0 Substituting the epressions obtained for T, T AB,andT and factoring i,, j, and k: AD ( 0.6 + 0.32432 T ) i + ( 0.8T 0.75676T 0.86154 T + P )j T AB Equating to zero the coefficients of i, j, k: 0.8T AB + (0.567577 T 0.50769 T ) k = 0 0.6TT 0.75676T 0.56757TT AB AB + 0.32432 T = 0 0.86154T AD + P = 0 AD 0.50769 T = 0 AD AD (1) (2) (3) Setting T = 481 N AD in (2) and (3), and solving the resulting set of equations gives T T AD = 430.26 N = 232.57 N P = 926 N
PROBLEM 2.102 Three cables are used to tether a balloon as shown. Knowing that the balloon eerts an 800-N vertical force at A, determine the tension in each cable. See Problem 2.101 for the figure and analsis leading to the linear algebraic Equations (1), (2), and (3). 0.6T + 0.32432T = 0 (1) AB 0.8T 0.75676T 0.86154T + P = 0 (2) AB AD From Eq. (1): T = 0.54053T From Eq. (3): T = 1.11795T Substituting for T AB and T AD in terms of T into Eq. (2) gives AB AD 0.56757T 0.50769T = 0 (3) 0.8(0.54053 T ) 0.75676T 0.86154(1.11795 T ) + P = 0 AD 2.1523 T = P; P = 800 N 800 N T = 2.1523 = 371.69 N Substituting into epressions for T AB and T AD gives TAB = 0.54053(371.69 N) T = 1.11795(371.69 N) AD T = 201 N, T = 372 N, T = 416 N AB AD
PROBLEM 2.103 A 36-lb triangular plate is supported b three wires as shown. Determine the tension in each wire, knowing that a = 6 in. B Smmetr TDB = TDC Free-Bod Diagram of Point D: The forces applied at D are: TDB, TDC, TDA, and P where P= Pj= (36 lb) j. To epress the other forces in terms of the unit vectors i, j, k, we write DA = (16 in.) i (24 in.) j DA = 28.844 in. DB = (8in.) i (24in.) j+ (6in.) k DB = 26.0in. DC = (8 in.) i (24 in.) j (6 in.) k DC = 26.0 in. DA and TDA = TDA λda = TDA = (0.55471i 0.83206 j) TDA DA DB TDB = TDBλDB = TDB = ( 0.30769i 0.92308j+ 0.23077 k) TDB DB DC T = T λ = T = ( 0.30769i 0.92308j 0.23077 k) T DC DC DC DC DC DC
PROBLEM 2.103 (Continued) Equilibrium condition: Σ F = 0: T + T + T + (36 lb) j = 0 DA DB DC Substituting the epressions obtained for TDA, TDB,andT DC and factoring i, j, and k: (0.55471T 0.30769T 0.30769 T ) i+ ( 0.83206T 0.92308T 0.92308T + 36 lb) j DA DB DC DA DB DC Equating to zero the coefficients of i, j, k: + (0.23077T 0.23077 T ) k = 0 DB DC 0.55471T 0.30769T 0.30769T = 0 (1) DA DB DC 0.83206T 0.92308T 0.92308T + 36 lb = 0 (2) DA DB DC Equation (3) confirms thattdb = TDC. Solving simultaneousl gives, 0.23077T 0.23077T = 0 (3) DB DC T = 14.42 lb; T = T = 13.00 lb DA DB DC
PROBLEM 2.104 Solve Prob. 2.103, assuming that a = 8 in. PROBLEM 2.103 A 36-lb triangular plate is supported b three wires as shown. Determine the tension in each wire, knowing that a = 6 in. B Smmetr TDB = TDC Free-Bod Diagram of Point D: The forces applied at D are: TDB, TDC, TDA, and P where P= Pj= (36 lb) j. To epress the other forces in terms of the unit vectors i, j, k, we write DA = (16 in.) i (24 in.) j DA = 28.844 in. DB = (8in.) i (24 in.) j+ (8 in.) k DB = 26.533 in. DC = (8 in.) i (24 in.) j (8 in.) k DC = 26.533 in. DA and TDA = TDAλDA = TDA = (0.55471i 0.83206 j) TDA DA DB TDB = TDBλDB = TDB = ( 0.30151i 0.90453j+ 0.30151 k) TDB DB DC TDC = TDCλDC = TDC = ( 0.30151i 0.90453j 0.30151 k) TDC DC
PROBLEM 2.104 (Continued) Equilibrium condition: Σ F = 0: T + T + T + (36 lb) j = 0 DA DB DC Substituting the epressions obtained for TDA, TDB,andT DC and factoring i, j, and k: (0.55471T 0.30151T 0.30151 T ) i+ ( 0.83206T 0.90453T 0.90453T + 36 lb) j DA DB DC DA DB DC Equating to zero the coefficients of i, j, k: + (0.30151T 0.30151 T ) k = 0 DB DC 0.55471T 0.30151T 0.30151T = 0 (1) DA DB DC 0.83206T 0.90453T 0.90453T + 36 lb = 0 (2) DA DB DC Equation (3) confirms thattdb = TDC. Solving simultaneousl gives, 0.30151T 0.30151T = 0 (3) DB DC T = 14.42 lb; T = T = 13.27 lb DA DB DC
PROBLEM 2.105 A crate is supported b threee cables as shown. Determine the weight of the crate knowing that the tension in cable is 544 lb. Solution The forces applied at A are: where P = Pj. To epress the other forces in terms of the unit vectors i, j, k, we write AB = (36 in.) i + (60 in.) j (27 in.) k AB = 75 in. = (60 in.) j + (32in.) k = 68 in. AD = (40 in.) i + (60 in.) j (27 in.) k AD = 77 in. AB and TAB = TAB λab = TAB AB = ( 0.48i + 0.8 j 0.36 k) TAB T = T λ = T = (0.88235j + 0.47059 k) T AD TAD = TAD λad = T AD AD = (0.51948i + 0.77922j 0.35065 k) Equilibrium Conditionn with W = Wj T, T, T AB AD Σ F = T + T and W 0: AB + TAD W j= 0 T AD
PROBLEM 2.105 (Continued) Substituting the epressions obtained for TAB, T,and T AD and factoring i, j, and k: ( 0.48TAB + 0.51948 TAD ) i+ (0.8TAB + 0.88235T + 0.77922 TAD W) j + ( 0.36T + 0.47059T 0.35065 T ) k = 0 Equating to zero the coefficients of i, j, k: AB AD 0.48T + 0.51948T = 0 (1) AB AD 0.8T + 0.88235T + 0.77922T W = 0 (2) AB AD 0.36T + 0.47059T 0.35065T = 0 (3) AB AD Substituting T = 544 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives: TAB = 374.27 lb T = 345.82 lb W = 1049 lb AD
PROBLEM 2.106 A 1600-lb crate is supported b three cables as shown. Determine the tension in each cable. The forces applied at A are: where P = Pj. To epress the other forces in terms of the unit vectors i, j, k, we write AB = (36 in.) i + (60 in.) j (27 in.) k AB = 75 in. = (60 in.) j + (32in.) k = 68 in. AD = (40 in.) i + (60 in.) j (27 in.) k AD = 77 in. AB and TAB = TAB λab = TAB AB = ( 0.48i + 0.8 j 0.36 k) TAB T = T λ = T = (0.88235j + 0.47059 k) T AD TAD = TAD λad = T AD AD = (0.51948i + 0.77922j 0.35065 k) Equilibrium Conditionn with W = Wj T, T, T AB AD Σ F = T + T and W 0: AB + TAD W j= 0 T AD
PROBLEM 2.106 (Continued) Substituting the epressions obtained for TAB, T,and T AD and factoring i, j, and k: ( 0.48TAB + 0.51948 TAD ) i+ (0.8TAB + 0.88235T + 0.77922 TAD W) j + ( 0.36T + 0.47059T 0.35065 T ) k = 0 Equating to zero the coefficients of i, j, k: AB AD 0.48T + 0.51948T = 0 (1) AB AD 0.8T + 0.88235T + 0.77922T W = 0 (2) AB AD 0.36T + 0.47059T 0.35065T = 0 (3) AB AD Substituting W = 1600 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms gives, T = 571 lb AB T = 830 lb T = 528 lb AD
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. Σ FA = 0: TAB + T + TAD + P = 0 where P = Pi AB = (960 mm) i (240 mm) j+ (380 mm) k AB = 1060 mm = (960 mm) i (240 mm) j (320 mm) k = 1040 mm AD = (960 mm) i+ (720 mm) j (220 mm) k AD = 1220 mm AB 48 12 19 TAB = TAB λab = TAB = TAB AB i j+ k 53 53 53 12 3 4 T = T λ = T = T i j k 13 13 13 305 N TAD = TAD λad = [( 960 mm) i+ (720 mm) j (220 mm) k ] 1220 mm = (240 N) i+ (180 N) j (55 N) k Substituting into Σ F = 0, factoring i, j, k, and setting each coefficient equal to φ gives: A 48 12 i : P = TAB + T + 240 N (1) 53 13 j : k : 12 3 TAB + T = 180 N (2) 53 13 19 4 TAB T = 55 N (3) 53 13 Solving the sstem of linear equations using conventional algorithms gives: TAB = 446.71 N T = 341.71 N P = 960 N
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. We assume that T AD = 0 and write Σ FA = 0: TAB + T + Qj+ (1200 N) i = 0 AB = (960 mm) i (240 mm) j+ (380 mm) k AB = 1060 mm = (960 mm) i (240 mm) j (320 mm) k = 1040 mm AB 48 12 19 TAB = TABλAB = TAB = i j+ k TAB AB 53 53 53 12 3 4 T = Tλ = T = T i j k 13 13 13 Substituting into Σ F = 0, factoring i, j, k, and setting each coefficient equal to φ gives: A 48 12 i : TAB T + 1200 N = 0 (1) 53 13 12 3 j : TAB T + Q = 0 (2) 53 13 19 4 k : TAB T = 0 (3) 53 13 Solving the resulting sstem of linear equations using conventional algorithms gives: TAB = 605.71 N T = 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 becomes slack for Q = 460 N.
PROBLEM 2.109 A rectangular plate is supported b three cables as shown. Knowing that the tension in cable is 60 N, determine the weight of the plate. We note that the weight of the plate is equal in magnitude to the force P eerted b the support on Point A. We have: Thus: ΣFF = T + T + T + AB = (320 mm) i (480 mm) j + (360 mm) k = (450 mm) i (480 mm) j + (360 mm) k AD = (250 mm) i (480 mm) j 360 mm k Substituting into the Eq. Σ F = 0 and factoring i, jk, : T T T AB AD 0: AB ( = T λ = T AB AB AB = T λ = T = T λ = T AD AD AD AD 8 5 TAB + 0.6T + T AD 17 13 i 12 9.6 + TAB 0.64T T 17 13 9 7..2 + TAB + 0.48T T 17 13 ) P j = 0 AB = 680 mmm = 750 mmm AD = 650 mmm AB 8 12 9 = i j+ k T AB 17 17 17 = ( 0.6i 0.64j + 0.48 k) T AD 5 9.6 7.2 = AD i j k T 13 13 13 AD AD + P j k = 0 AB AD Freee Bod A: Dimensions in mm
PROBLEM 2.109 (Continued) Setting the coefficient of i, j, k equal to zero: i : j : k : Making T = 60 N in (1) and (3): Multipl (1 ) b 9, (3 ) b 8, and add: Substitute into (1 ) and solve for T AB: Substitute for the tensions in Eq. (2) and solve for P: T 8 5 TAB + 0.6T + TAD = 0 (1) 17 13 12 9.6 TAB 0.64T TAD + P = 0 (2) 7 13 9 7.2 TAB + 0.48T TAD = 0 (3) 17 13 8 5 TAB + 36 N + TAD = 0 (1 ) 17 13 9 7.2 TAB + 28.8 N TAD = 0 (3 ) 17 13 AB 12.6 554.4 N TAD = 0 TAD = 572.0 N 13 17 5 = 36 + 572 TAB = 544.0 N 8 13 12 9.6 P = (544N) + 0.64(60 N) + (572 N) 17 13 = 844.8 N Weight of plate = P = 845 N
PROBLEM 2.110 A rectangular plate is supported b three cables as shown. Knowing that the tension in cable AD is 520 N, determine the weight of the plate. See Problem 2.109 for the figure and the analsis leading to the linear algebraic Equations (1), (2), and (3) below: 8 5 TAB + 0.6T + TAD = 0 17 13 (1) 12 9.6 TAB + 0.64T TAD + P = 0 17 13 (2) 9 7.2 TAB + 0.48T TAD = 0 17 13 (3) Making T = 520 N in Eqs. (1) and (3): AD 8 TAB + 0.6T + 200 N = 0 17 (1 ) 9 TAB + 0.48T 288 N = 0 17 (3 ) Multipl (1 ) b 9, (3 ) b 8, and add: 9.24T 504 N = 0 T = 54.5455 N Substitute into (1 ) and solve for T AB: 17 TAB = (0.6 54.5455 + 200) TAB = 494.545 N 8 Substitute for the tensions in Eq. (2) and solve for P: 12 9.6 P = (494.545 N) + 0.64(54.5455 N) + (520 N) 17 13 = 768.00 N Weight of plate = P = 768 N
PROBLEM 2.111 A transmission tower is held b three gu wires attached to a pin at A and anchored b bolts at B, C, and D. If the tension in wire AB is 840 lb, determine the vertical force P eerted b the tower on the pin at A. Σ F = 0: T + T + T + j = 0 Free-Bod Diagram at A: AB AD P We write AB = 20i 100j+ 25k AB = 105 ft = 60i 100j+ 18k = 118 ft AD = 20i 100j 74k AD = 126 ft AB TAB = TABλ AB = TAB AB 4 20 5 = i j+ k TAB 21 21 21 T = T λ = T 30 50 9 = i j+ k T 59 59 59 AD TAD = TAD λad = TAD AD 10 50 37 = i j k TAD 63 63 63 Substituting into the Eq. Σ F = 0 and factoring i, j, k:
PROBLEM 2.111 (Continued) 4 30 10 TAB + T TAD 21 59 63 i 20 50 50 + TAB T TAD + P 21 59 63 j 5 9 37 + TAB + T TAD k = 0 21 59 63 Setting the coefficients of i, j, k, equal to zero: i : j : k : Set T = 840 lb in Eqs. (1) (3): AB 4 30 10 TAB + T TAD = 0 (1) 21 59 63 20 50 50 TAB T TAD + P = 0 (2) 21 59 63 5 9 37 TAB + T TAD = 0 (3) 21 59 63 30 10 160 lb + T TAD = 0 (1 ) 59 63 50 50 800 lb T TAD + P = 0 (2 ) 59 63 9 37 200 lb + T TAD = 0 (3 ) 59 63 Solving, T = 458.12 lb T = 459.53 lb P = 1552.94 lb P = 1553 lb AD
PROBLEM 2.112 A transmission tower is held b three gu wires attached to a pin at A and anchored b bolts at B, C, and D. If the tension in wire is 590 lb, determine the vertical force P eerted b the tower on the pin at A. Σ F = 0: T + T + T + j = 0 Free-Bod Diagram at A: AB AD P We write AB = 20i 100j+ 25k AB = 105 ft = 60i 100j+ 18k = 118 ft AD = 20i 100j 74k AD = 126 ft AB TAB = TABλ AB = TAB AB 4 20 5 = i j+ k TAB 21 21 21 T = T λ = T 30 50 9 = i j+ k T 59 59 59 AD TAD = TAD λad = TAD AD 10 50 37 = i j k TAD 63 63 63 Substituting into the Eq. Σ F = 0 and factoring i, j, k :
PROBLEM 2.112 (Continued) 4 30 10 TAB + T TAD 21 59 63 i 20 50 50 + TAB T TAD + P 21 59 63 j 5 9 37 + TAB + T TAD k = 0 21 59 63 Setting the coefficients of i, j, k, equal to zero: i : j : k : Set T = 590 lb in Eqs. (1) (3): 4 30 10 TAB + T TAD = 0 (1) 21 59 63 20 50 50 TAB T TAD + P = 0 (2) 21 59 63 5 9 37 TAB + T TAD = 0 (3) 21 59 63 4 10 TAB + 300 lb TAD = 0 (1 ) 21 63 20 50 TAB 500 lb TAD + P = 0 (2 ) 21 63 5 37 TAB + 90 lb TAD = 0 (3 ) 21 63 Solving, T = 1081.82 lb T = 591.82 lb P = 2000 lb AB AD
PROBLEM 2.113 In tring to move across a slipper ic surface, a 175-lb man uses two ropes AB and. Knowing that the force eerted on the man b the ic surface is perpendicular to that surface, determine the tension in each rope. Free-Bod Diagram at A T T Equilibrium condition: Σ F = 0 = T λ = T = T AB AB AB AB AB ( 30 ft) i+ (20 ft) j (12 ft) k 38 ft 15 10 6 = T i+ j k 19 19 19 AB ( 30 ft) i+ (24 ft) j+ (32 ft) k = T λ = T = T AB 50 ft 15 12 16 = TAB i+ j+ k 25 25 25 T + T + N+ W = 0 AB 16 30 N = N i+ j 34 34 and W = W j= (175 lb) j
PROBLEM 2.113 (Continued) Substituting the epressions obtained for TAB, T, N, and W; factoring i, j, and k; and equating each of the coefficients to zero gives the following equations: From i: 15 15 16 TAB T + N = 0 25 19 34 (1) From j: 12 10 30 TAB + T + N (175 lb) = 0 25 19 34 (2) From k: 16 6 TAB T 25 19 = 0 (3) Solving the resulting set of equations gives: T AB = 30.8 lb; T = 62.5 lb
PROBLEM 2.114 Solve Problem 2.113, assuming that a friend is helping the man at A b pulling on him with a force P = (45 lb)k. PROBLEM 2.113 In tring to move across a slipper ic surface, a 175-lb man uses two ropes AB and. Knowing that the force eerted on the man b the ic surface is perpendicular to that surface, determine the tension in each rope. Refer to Problem 2.113 for the figure and analsis leading to the following set of equations, Equation (3) being modified to include the additional force P= ( 45lb) k. 15 15 16 TAB T + N = 0 (1) 25 19 34 12 10 30 TAB + T + N (175 lb) = 0 (2) 25 19 34 16 6 TAB T (45 lb) = 0 (3) 25 19 Solving the resulting set of equations simultaneousl gives: T = 81.3 lb AB T = 22.2 lb
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. See Problem 2.109 for the figure and the analsis leading to the linear algebraic Equations (1), (2), and (3) below. Setting P = 792 N gives: 8 5 TAB + 0.6T + TAD = 0 (1) 17 13 12 9.6 TAB 0.64T TAD + 792 N = 0 (2) 17 13 Solving Equations (1), (2), and (3) b conventional algorithms gives 9 7.2 TAB + 0.48T TAD = 0 (3) 17 13 T = 510.00 N T = 510 N AB T = 56.250 N T = 56.2 N T = 536.25 N T = 536 N AD AB AD
PROBLEM 2.116 For the cable sstem of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = 0. Where Σ F = 0: T + T + T + P+ Q = 0 A AB AD P = Pi and Q= Qj AB = (960 mm) i (240 mm) j+ (380 mm) k AB = 1060 mm = (960 mm) i (240 mm) j (320 mm) k = 1040 mm AD = (960 mm) i+ (720 mm) j (220 mm) k AD = 1220 mm AB 48 12 19 TAB = TAB λab = TAB = TAB AB i j+ k 53 53 53 12 3 4 T = T λ = T = T i j k 13 13 13 AD 48 36 11 TAD = TAD λad = TAD = TAD + AD i j k 61 61 61 Substituting into Σ F A = 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 T TAD + 2880 N = 0 53 13 61 (1) j : 12 3 36 TAB T + TAD = 0 53 13 61 (2) k : 19 4 11 TAB T TAD = 0 53 13 61 (3)
PROBLEM 2.116 (Continued) Solving the sstem of linear equations using conventional algorithms gives: T T T AB AD = 1340.14 N = 1025.12 N = 915.03 N T = 1340 N AB T = 1025 N T = 915 N AD
PROBLEM 2.117 For the cable sstem of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = 576 N. See Problem 2.116 for the analsis leading to the linear algebraic Equations (1), (2), and (3) below: 48 12 48 TAB T TAD + P = 0 53 13 61 (1) 12 3 36 TAB T + TAD + Q = 0 53 13 61 (2) 19 4 11 TAB T TAD = 0 53 13 61 (3) Setting P = 2880 N and Q = 576 N gives: 48 12 48 TAB T TAD + 2880 N = 0 53 13 61 (1 ) 12 3 36 TAB T + TAD + 576 N = 0 53 13 61 (2 ) 19 4 11 TAB T TAD = 0 53 13 61 (3 ) Solving the resulting set of equations using conventional algorithms gives: T T T AB AD = 1431.00 N = 1560.00 N = 183.010 N T = 1431 N AB T = 1560 N T = 183.0 N AD
PROBLEM 2.118 For the cable sstem 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). See Problem 2.116 for the analsis leading to the linear algebraic Equations (1), (2), and (3) below: Setting P = 2880 N and Q = 576 N gives: 48 12 48 TAB T TAD + P = 0 53 13 61 (1) 12 3 36 TAB T + TAD + Q = 0 53 13 61 (2) 19 4 11 TAB T TAD = 0 53 13 61 (3) 48 12 48 TAB T TAD + 2880 N = 0 53 13 61 (1 ) 12 3 36 TAB T + TAD 576 N = 0 53 13 61 (2 ) 19 4 11 TAB T TAD = 0 53 13 61 (3 ) Solving the resulting set of equations using conventional algorithms gives: T T T AB AD = 1249.29 N = 490.31 N = 1646.97 N T = 1249 N AB T = 490 N T = 1647 N AD
PROBLEM 2.119 For the transmission tower of Probs. 2.111 and 2.112, determine the tension in each gu wire knowing that the tower eerts on the pin at A an upward vertical force of 1800 lb. PROBLEM 2.111 A transmission tower is held b three gu wires attached to a pin at A and anchored b bolts at B, C, and D. If the tension in wire AB is 840 lb, determine the vertical force P eerted b the tower on the pin at A. See Problem 2.111 for the figure and the analsis leading to the linear algebraic Equations (1), (2), and (3) below: i : j : k : 4 30 10 TAB + T TAD = 0 21 59 63 (1) 20 50 50 TAB T TAD + P = 0 21 59 63 (2) 5 9 37 TAB + T TAD = 0 21 59 63 (3) Substituting for P = 1800 lb in Equations (1), (2), and (3) above and solving the resulting set of equations using conventional algorithms gives: 4 30 10 TAB + T TAD = 0 21 59 63 (1 ) 20 50 50 TAB T TAD + 1800 lb = 0 21 59 63 (2 ) 5 9 37 TAB + T TAD = 0 21 59 63 (3 ) TAB = 973.64 lb T T = 531.00 lb = 532.64 lb T = 974 lb AD AB T = 531 lb T = 533 lb AD
PROBLEM 2.120 Three wires are connected at point D, which is located 18 in. below the T-shaped pipe support ABC. Determine the tension in each wire when a 180-lb clinder is suspended from point D as shown. Free-Bod Diagram of Point D: The forces applied at D are: TDA, TDB, TDC and W where W= 180.0 lb j. To epress the other forces in terms of the unit vectors i, j, k, we write DA = (18 in.) j+ (22 in.) k DA = 28.425 in. DB = (24 in.) i+ (18 in.) j (16 in.) k DB = 34.0 in. DC = (24in.) i+ (18in.) j (16in.) k DC = 34.0 in.
and Equilibrium Condition with T T T PROBLEM 2.120 (Continued) DA = T λ = T DA = (0.63324j+ 0.77397 k) TDA DB = T λ = T DB = ( 0.70588i+ 0.52941j 0.47059 k) T DC = T λ = T DC = (0.70588i+ 0.52941j 0.47059 k) T DA Da DA Da DB DB DB DB DC DC DC DC W= Wj DC DB Σ F = 0: T + T + T Wj = 0 DA DB DC Substituting the epressions obtained for TDA, TDB,andT DC and factoring i, j, and k: ( 0.70588TDB + 0.70588 TDC ) i (0.63324TDA + 0.52941TDB + 0.52941 TDC W) j (0.77397T 0.47059T 0.47059 T ) k Equating to zero the coefficients of i, j, k: DA DB DC 0.70588T + 0.70588T = 0 (1) DB 0.63324T + 0.52941T + 0.52941T W = 0 (2) DA DB DC 0.77397T 0.47059T 0.47059T = 0 (3) DA DB DC Substituting W = 180 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms gives, DC T = 119.7 lb DA T = 98.4 lb DB T = 98.4 lb DC
PROBLEM 2.121 A container of weight W is suspended from ring A, to which cables and AE are attached. A force P is applied to the end F of a third cable that passes over a pulle 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.) 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 AB = (0.78 m) i+ (1.6 m) j+ (0 m) k 2 2 2 AB = ( 0.78m) + (1.6 m) + (0) = 1.78 m AB TAB = TλAB = TAB AB TAB = [ (0.78 m) i+ (1.6 m) j+ (0 m) k] 1.78 m TAB = TAB ( 0.4382i+ 0.8989j+ 0 k) and = (0) i+ (1.6 m) j+ (1.2 m) k 2 2 2 = (0 m) + (1.6 m) + (1.2 m) = 2 m T T = Tλ = T = [(0) i+ (1.6 m) j+ (1.2 m) k] 2 m T = T (0.8j+ 0.6 k) and AD = (1.3 m) i+ (1.6 m) j+ (0.4 m) k 2 2 2 AD = (1.3 m) + (1.6 m) + (0.4 m) = 2.1 m AD TAD TAD = TλAD = TAD = [(1.3 m) i+ (1.6 m) j+ (0.4 m) k] AD 2.1 m T = T (0.6190i+ 0.7619j+ 0.1905 k) AD AD
PROBLEM 2.121 (Continued) Finall, AE = (0.4 m) i+ (1.6 m) j (0.86 m) k 2 2 2 AE = ( 0.4 m) + (1.6 m) + ( 0.86 m) = 1.86 m AE TAE = TλAE = TAE AE TAE = [ (0.4 m) i+ (1.6m) j (0.86m) k] 1.86 m T = T ( 0.2151i+ 0.8602j 0.4624 k) AE AE With the weight of the container W= W j, at A we have: Σ F = 0: T + T + T j = 0 AB AD W Equating the factors of i, j, and k to zero, we obtain the following linear algebraic equations: 0.4382T + 0.6190T 0.2151T = 0 (1) AB AD AE 0.8989T + 0.8T + 0.7619T + 0.8602T W = 0 (2) AB AD AE 0.6T + 0.1905T 0.4624T = 0 (3) AD AE Knowing that W = 1000 N and that because of the pulle sstem at B TAB = TAD = P, where P is the eternall applied (unknown) force, we can solve the sstem of linear Equations (1), (2) and (3) uniquel for P. P = 378 N
PROBLEM 2.122 Knowing that the tension in cable of the sstem 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 and AE are attached. A force P is applied to the end F of a third cable that passes over a pulle 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.) Here, as in Problem 2.121, the support of the container consists of the four cables AE,, AD, and AB, with the condition that the force in cables AB and AD is equal to the eternall applied force P. Thus, with the condition T = T = P AB AD and using the linear algebraic equations of Problem 2.131 with T = 150 N, we obtain (a) (b) P = 454 N W = 1202 N
PROBLEM 2.123 Cable B passes through a frictionless ring A and is attached to fied supports at B and C, while cables AD and AE are both tied to the ring and are attached, respectivel, 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. Free Bod Diagram at A: Since T = tension in cable B, it follows that B TAB = T = TB ( 17.5 in.) i+ (60 in.) j 17.5 60 TAB = TB λab = TB = TB 62.5 in. i+ j 62.5 62.5 (60 in.) i+ (25 in.) k 60 25 T = TB λ = TB = TB j+ k 65 in. 65 65 (80 in.) i+ (60 in.) j 4 3 TAD = TAD λad = TAD = TAD i+ j 100 in. 5 5 T = T λ = T AE AE AE AE (60 in.) j (45 in.) k 4 3 = T AE j k 75 in. 5 5
PROBLEM 2.123 (Continued) Substituting into Σ F A = 0, setting P= ( 200 lb) j, and setting the coefficients of i, j, k equal to φ, we obtain the following three equilibrium equations: From i : 17.5 4 TB + TAD = 0 62.5 5 (1) From j : 60 60 3 4 + TB + TAD + TAE 200lb= 0 62.5 65 5 5 (2) From k : 25 3 TB TAE 65 5 = 0 (3) Solving the sstem of linear equations using conventional algorithms gives: T = 76.7 lb; T = 26.9 lb; T = 49.2 lb B AD AE
PROBLEM 2.124 Knowing that the tension in cable AE of Prob. 2.123 is 75 lb, determine (a) the magnitude of the load P, (b) the tension in cables B and AD. PROBLEM 2.123 Cable B passes through a frictionless ring A and is attached to fied supports at B and C, while cables AD and AE are both tied to the ring and are attached, respectivel, 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. Refer to the solution to Problem 2.123 for the figure and analsis leading to the following set of equilibrium equations, Equation (2) being modified to include Pj as an unknown quantit: 17.5 4 TB + TAD = 0 (1) 62.5 5 60 60 3 4 + TB + TAD + TAE P = 0 62.5 65 5 5 25 3 TB TAE = 0 (3) 65 5 Substituting for T = 75 lb and solving simultaneousl gives: AE (2) (a) P = 305 lb (b) T = 117.0 lb; T = 40.9 lb B AD
PROBLEM 2.125 Collars A and B are connected b a 525-mm-long wire and can slide freel on frictionless rods. If a force P = (341 N) j is applied to collar A, determine (a) the tension in the wire when = 155 mm, (b) the magnitude of the force Q required to maintain the equilibrium of the sstem. For both Problems 2. 125 and 2.126: Free-Bod Diagrams of Collars: ( AB) 2 2 2 2 = + + z Here or 2 (0.525 m) = (0.20 m) + 2 2 + z = 0..23563 m 2 2 2 2 + z Thus, when given, z is determined, Now Where and z are in units of meters, m. From the F.B. Diagram of collar A: : Σ F = 0: Setting the j coefficient to zero gives P (1.90476 T ) AB = 0 λ AB AB = AB 1 = (0.20i j+ zk)m 0.525 m = 0.38095 i 1.90476j + 1.90476zk : Ni Nz + k + Pj + T λ AB AB = 0 With T P = 341 N AB 341 N = 1.90476 Now, from the free bod diagram of collar B: Setting the k coefficient to zero gives And using the above result for TAB, we have Σ F = 0: N i+ N j+ Q k λ = 0 Q T AB (1.90476 z ) = 0 T AB AB 341 N (341N)( z) Q= TABz = (1..90476 z ) = (1.90476)
PROBLEM 2.125 (Continued) Then from the specifications of the problem, = 155 mm = 0.155 m and (a) (b) or and or 2 2 2 z = 0.23563 m (0.155 m) z = 0.46 m 341 N T AB = 0.155(1.90476) = 1155.00 N Q = = 341 N(0.46 m)(0.866) (0.155 m) (1012.00 N) T = 1155 N AB Q = 1012 N
PROBLEM 2.126 Solve Problem 2.125 assuming that = 275 mm. PROBLEM 2.125 Collars A and B are connected b a 525-mm-long wire and can slide freel on frictionless rods. If a force P= (341 N) j is applied to collar A, determine (a) the tension in the wire when = 155 mm, (b) the magnitude of the force Q required to maintain the equilibrium of the sstem. From the analsis of Problem 2.125, particularl the results: With = 275 mm = 0.275 m, we obtain: 2 2 2 + z = 0.23563 m T AB 341 N = 1.90476 341 N Q = z and (a) (b) or and or 2 2 2 z = 0.23563 m (0.275 m) z = 0.40 m 341 N T AB = = 651.00 (1.90476)(0.275 m) Q = 341 N(0.40 m) (0.275 m) T = 651 N AB Q = 496 N
PROBLEM 2.127 Two structural members A and B are bolted to a bracket as shown. Knowing that both members are in compression and that the force is 15 kn in member A and 10 kn in member B, determine b trigonometr the magnitude and direction of the resultant of the forces applied to the bracket b members A and B. Using the force triangle and the laws of cosines and sines, we have γ = 180 (40 + 20 ) = 120 Then and Hence: R 2 2 = (15 kn) + (10 kn) 2(15 kn)(10 kn)cos120 = 475 kn R = 21.794 kn 10 kn 21.794 kn = sin α sin120 10 kn sin α = sin120 21.794 kn = 0.39737 α = 23.414 φ = α + 50 = 73.414 2 2 R = 21.8 kn 73.4
PROBLEM 2.128 Determine the and components of each of the forces shown. Compute the following distances: 102-lb Force: 2 2 OA = (24 in.) + (45 in.) = 51.0 in. 2 2 OB = (28 in.) + (45 in.) = 53.0 in. 2 2 OC = (40 in.) + (30 in.) = 50.0 in. 24 in. F = 102 lb F = 48.0 lb 51.0 in. 45 in. F =+ 102 lb F =+ 90.0 lb 51.0 in. 106-lb Force: 28 in. F =+106 lb F =+ 56.0 lb 53.0 in. 45 in. F =+ 106 lb F =+ 90.0 lb 53.0 in. 200-lb Force: 40 in. F = 200 lb F = 160.0 lb 50.0 in. 30 in. F = 200 lb F = 120.0 lb 50.0 in.
PROBLEM 2.129 A hoist trolle is subjected to the threee forces shown. Knowing that α = 40, determine (a) the required magnitude of the force P if the resultant of the three forces is to be vertical, (b) the corresponding magnitude of the resultant. R = Σ F = P+ (200 lb)sin 40 (400 lb)cos 40 R = P 177.860 lb R = Σ F = (200 lb)cos 40 + (400 R = 410. 32 lb lb)sin 40 (1) (2) (a) For R to be vertical, we must have R = 0. (b) Set Since R is to be vertical: R = 0 in Eq. (1) 0= P 177.860 lb P = 177.860lb R = = 410 lb R P = 177.9 lb R = 410 lb
PROBLEM 2.130 Knowing that α = 55 and that boom eerts on pin C a force directed along line, determine (a) the magnitude of that force, (b) the tensionn in cable BC. Free-Bod Diagram Force Triangle Law of sines: F TBC 300 lb = = sin 35 sin 50 sin 95 (a) 300 lb F = sin 35 sin 95 F = 172.7 lb (b) 300 lb T BC = sin 50 sin 95 T BC = 231 lb
PROBLEM 2. 131 Two cables are tied together at C and loaded as shown. Knowing that P = 360 N, determine the tension (a) in cable, (b) in cable BC. Free Bod: C (a) 12 4 Σ F = 0: T + (3 360 N) = 0 13 5 T = 312 N (b) Σ F = 0: 5 3 (312 N) + T BC + (360 N) 480 N = 0 13 5 T = 480 N 120 N 216 N BC T = 144.0 N BC
PROBLEM 2.1322 Two cables tied together at C are loaded as shown. Knowing that the maimum allowable tensionn in each cable is 800 N, determine (a) the magnitude of the largest force P that can be applied at C, (b) the corresponding value of α. Free-Bod Diagram: C Force Triangle Force triangle is isosceles with (a) 2β = 180 85 β = 47.5 P = 2(800 N)cos 47.5 = 1081 N (b) Since P > 0, the solution is correct. α = 180 50 47.5 = 82.5 P = 1081 N α = 82.5
PROBLEM 2.133 The end of the coaial cable AE is attached to the pole AB, which is strengthened b the gu wires and AD. Knowing that the tension in wire is 120 lb, determine (a) the components of the force eerted b this wire on the pole, (b) the angles θ, θ, and θ z that the force forms with the coordinate aes. (a) F = (120 lb)cos 60 cos 20 F = 56.382 lb F =+ 56.4 lb F F = (120 lb)sin 60 = 103.923 lb F = 103.9 lb (b) F = (120 lb)cos 60 sin 20 z F = 20.521 lb F = 20.5 lb z F 56.382 lb cos θ = = θ = 62.0 F 120 lb F 103.923 lb cos θ = = θ = 150.0 F 120 lb Fz 20.52 lb cos θz = = θ z = 99.8 F 120 lb z
PROBLEM 2.134 Knowing that the tension in cable is 2130 N, determine the components of the force eerted on the plate at C. CA = (900 mm) i+ (600 mm) j (920 mm) k 2 2 2 CA = (900 mm) + (600 mm) + (920 mm) = 1420 mm TCA = TCAλCA CA = TCA CA 2130 N TCA = [ (900 mm) i+ (600 mm) j (920 mm) k ] 1420 mm = (1350 N) i+ (900 N) j (1380 N) k ( T ) = 1350 N, ( T ) = 900 N, ( T ) = 1380 N CA CA CA z
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. 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)[cos55 cos30 i+ sin 55 j cos55 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 = 2 2 2 (386.52 N) + (828.25 N) + (220.49 N) = 940.22 N R = 940 N R 386.52 N cosθ = = θ = 65.7 R 940.22 N R 828.25 N cosθ = = θ = 28.2 R 940.22 N Rz 220.49 N cosθ z = = θ z = 76.4 R 940.22 N
PROBLEM 2.136 A container of weight W is suspended from ring A. Cable B passes through the ring and is attached to fied supports at B and C. Two forces P = Pi and Q = Qk are applied to the ring to maintain the container in the position shown. Knowing that W = 376 N, determine P and Q. (Hint: The tensionn is the same in both portions of cable B.) T AB = TλAB AB = T AB Free-Bod A: ( 130 mm) i + (4000 mm) j + (160 mm)k = T 450 mm 13 40 16 = T i+ j+ k 45 45 45 T = Tλ = T Setting coefficients of i, j, k equal to zero: ( 150 mm) i + (4000 mm) j + ( 240 mm)k = T 490 mm 15 40 24 = T i+ j k 49 49 49 Σ F = 0: T + T + Q + P+ W = 0 i : j : k : AB 13 45 T 15 0 49 T + P = + 40 45 T + 40 0 49 T W = 0.59501T = P 1.70521T = W + 16 45 T 24 49 T + Q = 0 0.134240T = Q (1) (2) (3)
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
PROBLEM 2.137 Collars A and B are connected b a 25-in.-long wire and can slide freel on frictionless rods. If a 60-lb force Q is applied to collar B as shown, determine (a) the tension in the wire when = 9 in.., (b) the corresponding magnitude of the force P required to maintain the equilibrium of the sstem. A: Free-Bod Diagrams of Collars: B: Collar A: AB i (20 in.) j + zk = = AB 25 in. Σ F= 0: Pi+ N j+ N k+ T λ = 0 λ A B z AB AB Substitute for λ AB and set coefficient of i equal to zero: TAB P 0 25 in. Collar B: Σ F =0: (60 lb) k+ N i + N j TABλ AB 0 Substitute for λab and set coefficient of k equal to zero: TAB z 60 lb = 0 25 in. (a) = 9 in. 2 2 2 (9 in.) + (20 in.) + z = (25 in.) z = 12 in. From Eq. (2): 60 lb T AB (12 in.) 25 in. (b) From Eq. (1): (125.0 lb)(9 in.) P = 25 in. 2 AB (1) (2) T = 125.0 lb P = 45.0 lb
PROBLEM 2.138 Collars A and B are connected b a 25-in.-long wire and can slide freel on frictionless rods. Determine the distances and z for which the equilibrium of the sstem is maintained when P = 120 lb and Q = 60 lb. See Problem 2.137 for the diagrams and analsis leading to Equations (1) and (2) below: TAB P = = 0 25 in. (1) TAB z 60 lb = 0 25 in. (2) For P = 120 lb, Eq. (1) ields T = (25 in.)(20 lb) (1 ) From Eq. (2): AB T z = (25 in.)(60 lb) (2 ) AB Dividing Eq. (1 ) b (2 ), 2 z = (3) Now write 2 2 2 2 + z + (20 in.) = (25 in.) (4) Solving (3) and (4) simultaneousl, 2 2 4z + z + 400 = 625 z 2 = 45 z = 6.7082 in. From Eq. (3): = 2z = 2(6.7082 in.) = 13.4164 in. = 13.42 in., z = 6.71 in.
PROBLEM 2F1 Two cables are tied together at C and loaded as shown. Draw the free-bod diagram needed to determine the tension in and BC. Free-Bod Diagram of Point C: W = (1600 kg)(9.81 m/s ) W = 15.6960(10 ) N W =15.696 kn 3 2
PROBLEM 2.F2 Two forces of magnitude T A = 8 kips and T B = 15 kips are applied as shown to a welded connection. Knowing that the connection is in equilibrium, draw the free-bod diagram needed to determine the magnitudes of the forces T C and T D. Free-Bod Diagram of Point E:
PROBLEM 2.F3 The 60-lb collar A can slide on a frictionless vertical rod and is connected as shown to a 65-lb counterweight C. Draw the free-bod diagram needed to determine the value of h for which the sstem is in equilibrium. Free-Bod Diagram of Point A:
PROBLEM 2.F4 A chairlift has been stopped in the position shown. Knowing that each chair weighs 250 N and that the skier in chair E weighs 765 N, draw the free-bod diagrams needed to determine the weight of the skier in chair F. Free-Bod Diagram of Point B: W θ θ E AB BC = 250 N + 765 N = 1015N = tan = tan 1 1 8.25 = 30.510 14 10 = 22.620 24 Use this free bod to determine T AB and T BC. Free-Bod Diagram of Point C: θ CD = tan 1.1 10.3889 6 = 1 1 Use this free bod to determine T CD and W F. Then weight of skier W S is found b W = 250 N S W F
PROBLEM 2.F5 Three cables are used to tether a balloon as shown. Knowing that the tension in cable is 444 N, draw the free-bod diagram needed to determine the vertical force P eerted b the balloon at A. Free-Bod Diagram of Point A:
PROBLEM 2.F6 A container of mass m = 120 kg is supported b three cables as shown. Draw the free-bod diagram needed to determine the tension in each cable Free-Bod Diagram of Point A: W = (120 kg)(9.81 m/s ) = 1177.2 N 2
PROBLEM 2.F7 A 150-lb clinder is supported b two cables 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 clinder in the position shown. Draw the free-bod diagram needed to determine the magnitude of P and the force in each cable. Free-Bod Diagram of Point C:
PROBLEM 2.F8 A transmission tower is held b three gu wires attached to a pin at A and anchored b bolts at B, C, and D. Knowing that the tension in wire AB is 630 lb, draw the free-bod diagram needed to determine the vertical force P eerted b the tower on the pin at A. Free-Bod Diagram of point A: