CHAPTER 2

Copyright ÂŠ 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

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

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

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

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°

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°

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°

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)

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

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°

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

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°

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°

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°

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

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

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

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

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

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

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

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

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

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

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

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°

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

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

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

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 ÂŠ McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 2.31 Determine the resultant of the three forces of Problem 2.21. PROBLEM 2.21 Determine the x and y components of each of the forces shown.

SOLUTION Components of the forces were determined in Problem 2.21: Force

x Comp. (lb)

y Comp. (lb)

29 lb

+21.0

+20.0

50 lb

–14.00

+48.0

51 lb

+24.0

–45.0

Rx = +31.0

Ry = +23.0

R = Rx i + R y j = (31.0 lb)i + (23.0 lb) j Ry tan α = Rx 23.0 = 31.0 α = 36.573° 23.0 lb R= sin (36.573°) = 38.601 lb

R = 38.6 lb

36.6° W

PROBLEM 2.32 Determine the resultant of the three forces of Problem 2.23. PROBLEM 2.23 Determine the x and y components of each of the forces shown.

SOLUTION Components of the forces were determined in Problem 2.23: Force

x Comp. (N)

y Comp. (N)

80 N

+61.3

+51.4

120 N

+41.0

+112.8

150 N

–122.9

+86.0

Rx = −20.6

Ry = +250.2

R = Rx i + R y j = (−20.6 N)i + (250.2 N) j R tan α = y Rx 250.2 N tan α = 20.6 N tan α = 12.1456 α = 85.293° 250.2 N R= sin 85.293°

R = 251 N

85.3° W

PROBLEM 2.33 Determine the resultant of the three forces of Problem 2.24. PROBLEM 2.24 Determine the x and y components of each of the forces shown.

SOLUTION Force

x Comp. (lb)

y Comp. (lb)

40 lb

+20.00

–34.64

50 lb

–38.30

–32.14

60 lb

+54.38

+25.36

Rx = +36.08

Ry = −41.42

R = Rx i + Ry j = (+36.08 lb)i + (−41.42 lb) j Ry tan α = Rx 41.42 lb tan = 36.08 lb tan α = 1.14800 = 48.942° R=

41.42 lb sin 48.942°

R = 54.9 lb

48.9° W

PROBLEM 2.34 Determine the resultant of the three forces of Problem 2.22. PROBLEM 2.22 Determine the x and y components of each of the forces shown.

SOLUTION Components of the forces were determined in Problem 2.22: Force

x Comp. (N)

y Comp. (N)

800 lb

+640

+480

424 lb

–224

–360

408 lb

+192

–360

Rx = +608

Ry = −240

R = Rx i + Ry j = (608 lb)i + (−240 lb) j R tan α = y Rx 240 = 608 α = 21.541° 240 N R= sin(21.541°) = 653.65 N

R = 654 N

21.5° W

PROBLEM 2.35 Knowing that α = 35°, determine the resultant of the three forces shown.

SOLUTION Fx = +(100 N) cos 35° = +81.915 N

100-N Force:

Fy = −(100 N) sin 35° = −57.358 N Fx = +(150 N) cos 65° = +63.393 N

150-N Force:

Fy = −(150 N) sin 65° = −135.946 N Fx = −(200 N) cos 35° = −163.830 N

200-N Force:

Fy = −(200 N) sin 35° = −114.715 N Force

x Comp. (N)

y Comp. (N)

100 N

+81.915

−57.358

150 N

+63.393

−135.946

200 N

−163.830

−114.715

Rx = −18.522

Ry = −308.02

R = Rx i + Ry j = (−18.522 N)i + (−308.02 N) j Ry tan α = Rx 308.02 = 18.522 α = 86.559° R=

308.02 N sin 86.559

R = 309 N

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

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

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

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

R = 229 N W

PROBLEM 2.40 For the post of Prob. 2.36, determine (a) the required tension in rope AC if the resultant of the three forces exerted at point C is to be horizontal, (b) the corresponding magnitude of the resultant.

SOLUTION 960

R = ΣF = − x

x

R =− x

48

y

(a)

1100 1460

73

4 (500 N) + (200 N) 25 5 (1)

AC

55 R =− T y

24

+ 640 N

T

73

+

AC

1460

R = ΣF = − y

T

+

T AC

7

3 (500 N) − (200 N)

25

5

+ 20 N

For R to be horizontal, we must have Ry = 0. 55 − TAC + 20 N = 0 Set Ry = 0 in Eq. (2): 73

TAC = 26.545 N (b)

(2)

AC

TAC = 26.5 N W

Substituting for TAC into Eq. (1) gives 48

(26.545 N) + 640 N = − x 73 Rx = 622.55 N

R

R = Rx = 623 N

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

R = 95.0 lb W

PROBLEM 2.42 For the block of Problems 2.37 and 2.38, determine (a) the required value of α if the resultant of the three forces shown is to be parallel to the incline, (b) the corresponding magnitude of the resultant.

SOLUTION

Select the x axis to be along a a′. Then

Rx = ΣFx = (60 lb) + (80 lb) cos α + (120 lb) sin α

(1)

Ry = ΣFy = (80 lb) sin α − (120 lb) cos α

(2)

and

(a)

Set Ry = 0 in Eq. (2). (80 lb)sin α − (120 lb) cos α = 0 Dividing each term by cos α gives: (80 lb) tan α = 120 lb 120 lb tanα = 80 lb α = 56.310°

(b)

α = 56.3° W

Substituting for α in Eq. (1) gives:

Rx = 60 lb + (80 lb) cos 56.31° + (120 lb) sin 56.31° = 204.22 lb

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°

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°

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°

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°

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°

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

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

PROBLEM 2.50 Two cables are tied together at C and are loaded as shown. Determine the range of values of P for which both cables remain taut.

SOLUTION Free-Body Diagram

ΣFx = 0

− T sin 30D + TCB sin 30D − P cos 45° − 200N = 0 CA

For TCA = 0 we have,

0.5TCB + 0.70711P − 200 = 0 (1) ΣFy = 0

TCA cos 30° − TCB cos 30D − P sin 45D = 0 ; again setting TCA = 0 yields,

0.86603TCB − 0.70711P = 0 (2) Adding equations (1) and (2) gives,

1.36603TCB = 200 hence TCB = 146.410N and P = 179.315N

Substituting for TCB = 0 into the equilibrium equations and solving simultaneously gives, −0.5TCA + 0.70711P − 200 = 0 0.86603TCA − 0.70711P = 0 And TCA = 546.40N , P = 669.20N Thus for both cables to remain taut, load P must be within the range of 179.315 N and 669.20 N.

179.3 N <P< 669 N W

PROBLEM 2.51 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that P = 500 lb and Q = 650 lb, determine the magnitudes of the forces exerted on the rods A and B.

SOLUTION Free-Body Diagram Resolving the forces into x- and y-directions:

R = P + Q + FA + FB = 0 Substituting components:

R = −(500 lb) j + [(650 lb) cos 50°]i − [(650 lb) sin 50°]j + FB i − (FA cos 50°)i + (FA sin 50°) j = 0

In the y-direction (one unknown force):

−500 lb − (650 lb) sin 50° + FA sin 50° = 0 Thus,

FA =

500 lb + (650 lb) sin 50° sin 50°

= 1302.70 lb In the x-direction: Thus,

FA = 1303 lb W

(650 lb) cos 50° + FB − FA cos 50° = 0 FB = FA cos 50° − (650 lb) cos 50° = (1302.70 lb) cos 50° − (650 lb) cos 50° = 419.55 lb

FB = 420 lb W

PROBLEM 2.52 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that the magnitudes of the forces exerted on rods A and B are FA = 750 lb and FB = 400 lb, determine the magnitudes of P and Q.

SOLUTION Free-Body Diagram Resolving the forces into x- and y-directions:

R = P + Q + FA + FB = 0 Substituting components:

R = −Pj + Q cos 50°i − Q sin 50°j − [(750 lb) cos 50°]i + [(750 lb) sin 50°]j + (400 lb)i

In the x-direction (one unknown force): Q cos 50° − [(750 lb) cos 50°] + 400 lb = 0 (750 lb) cos 50° − 400 lb cos 50° = 127.710 lb

Q=

In the y-direction:

−P − Q sin 50° + (750 lb) sin 50° = 0 P = −Q sin 50° + (750 lb) sin 50° = −(127.710 lb) sin 50° + (750 lb) sin 50° = 476.70 lb

P = 477 lb; Q = 127.7 lb W

PROBLEM 2.53 A welded connection is in equilibrium under the action of the four forces shown. Knowing that FA = 8 kN and

FB = 16 kN, determine the magnitudes of the other two forces.

SOLUTION Free-Body Diagram of Connection

ΣF = 0: x

3 5

3 F −F − F =0 B

C

5

A

FA = 8 kN

With

FB = 16 kN 4 4 F = (16 kN) − (8 kN) C

5

Σ Fy = 0: − FD + 3 With FA and FB as above:

5

F = 6.40 kN W C

3 3 FB − FA = 0 5 5 3

FD = (16 kN) − (8 kN) 5 5

FD = 4.80 kN W

PROBLEM 2.54 A welded connection is in equilibrium under the action of the four forces shown. Knowing that FA = 5 kN and FD = 6 kN, determine the magnitudes of the other two forces.

SOLUTION Free-Body Diagram of Connection

3 3 ΣFy = 0: − FD − FA + FB = 0 5 5 or

3 FB = FD + FA 5

With

FA = 5 kN, FD = 8 kN 5⎡ 3 ⎤ FB = ⎢ 6 kN + (5 kN) ⎥ 3 5 ⎣ ⎦ ΣFx = 0: − FC +

FB = 15.00 kN W

4 4 FB − FA = 0 5 5

4 FC =

(FB − FA ) 5 4 = (15 kN − 5 kN) 5

FC = 8.00 kN W

PROBLEM 2.55 A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that α = 30° and β = 10° and that the combined weight of the boatswain’s chair and the sailor is 200 lb, determine the tension (a) in the support cable ACB, (b) in the traction cable CD.

SOLUTION Free-Body Diagram

ΣFx = 0: TACB cos 10° − TACB cos 30° − TCD cos 30° = 0 TCD = 0.137158TACB

(1)

ΣFy = 0: TACB sin 10° + TACB sin 30° + TCD sin 30° − 200 = 0

0.67365TACB + 0.5TCD = 200

(a)

Substitute (1) into (2):

0.67365TACB + 0.5(0.137158TACB ) = 200 TACB = 269.46 lb

(b)

From (1):

(2)

TACB = 269 lb W

TCD = 0.137158 (269.46 lb)

TCD = 37.0 lb W

PROBLEM 2.56 A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that α = 25° and β = 15° and that the tension in cable CD is 20 lb, determine (a) the combined weight of the boatswain’s chair and the sailor, (b) the tension in the support cable ACB.

SOLUTION Free-Body Diagram

ΣFx = 0: TACB cos 15° − TACB cos 25° − (20 lb) cos 25° = 0 TACB = 304.04 lb ΣFy = 0: (304.04 lb) sin 15° + (304.04 lb) sin 25° + (20 lb) sin 25° − W = 0 W = 215.64 lb (a)

W = 216 lb W

(b) TACB = 304 lb W

PROBLEM 2.57 For the cables of prob. 2.44, find the value of α for which the tension is as small as possible (a) in cable bc, (b) in both cables simultaneously. In each case determine the tension in each cable.

SOLUTION Free-Body Diagram

Force Triangle

(a) For a minimum tension in cable BC, set angle between cables to 90 degrees.

α = 35.0D W

By inspection, TAC = (6 kN) cos 35D

TAC = 4.91 kN W

TBC = (6 kN) sin 35D

TBC = 3.44 kN W

(b) For equal tension in both cables, the force triangle will be an isosceles.

α = 55.0D W

Therefore, by inspection,

T AC

=T BC

= (1 / 2)

6 kN cos 35°

T

=T

AC

BC

= 3.66 kN W

PROBLEM 2.58 For the cables of Problem 2.46, it is known that the maximum allowable tension is 600 N in cable AC and 750 N in cable BC. Determine (a) the maximum force P that can be applied at C, (b) the corresponding value of α.

SOLUTION Free-Body Diagram

(a)

Law of cosines

Force Triangle

P 2 = (600)2 + (750)2 − 2(600)(750) cos (25° + 45°) P = 784.02 N

(b)

Law of sines

P = 784 N W

sin β sin (25° + 45°) = 600 N 784.02 N

β = 46.0°

∴ α = 46.0° + 25°

α = 71.0° W

PROBLEM 2.59 For the situation described in Figure P2.48, determine (a) the value of α for which the tension in rope BC is as small as possible, (b) the corresponding value of the tension.

SOLUTION Free-Body Diagram

Force Triangle

To be smallest, TBC must be perpendicular to the direction of TAC . (a) (b)

Thus,

α = 5.00° TBC = (1200 lb) sin 5°

α = 5.00°

W

TBC = 104.6 lb W

PROBLEM 2.60 Two cables tied together at C are loaded as shown. Determine the range of values of Q for which the tension will not exceed 60 lb in either cable.

SOLUTION ΣFx = 0: −TBC − Q cos 60° + 75 lb = 0

Free-Body Diagram

TBC = 75 lb − Q cos 60°

(1)

ΣFy = 0: TAC − Q sin 60° = 0

TAC = Q sin 60° Requirement: From Eq. (2):

(2)

TAC = 60 lb: Q sin 60° = 60 lb

Requirement:

Q = 69.3 lb TBC = 60 lb:

From Eq. (1):

75 lb − Q cos 60° = 60 lb Q = 30.0 lb 30.0 lb ≤ Q ≤ 69.3 lb W

PROBLEM 2.61 A movable bin and its contents have a combined weight of 2.8 kN. Determine the shortest chain sling ACB that can be used to lift the loaded bin if the tension in the chain is not to exceed 5 kN.

SOLUTION Free-Body Diagram

tan α =

h 0.6 m

(1)

Isosceles Force Triangle

Law of sines:

sin α =

1 2

(2.8 kN) TAC

TAC = 5 kN 1 2

(2.8 kN) 5 kN α = 16.2602°

sin α =

From Eq. (1): tan16.2602° =

h 0.6 m

∴ h = 0.175000 m

Half-length of chain = AC = (0.6 m) 2 + (0.175 m) 2 = 0.625 m Total length:

= 2 × 0.625 m

1.250 m W

PROBLEM 2.62 For W = 800 N, P = 200 N, and d = 600 mm, determine the value of h consistent with equilibrium.

SOLUTION TAC = TBC = 800 N

Free-Body Diagram

(h

AC = BC = ΣFy = 0: 2(800 N)

2

+ d2

h h + d2 2

)

−P=0

2

800 =

Data:

P ⎛d⎞ 1+ ⎜ ⎟ 2 ⎝h⎠ P = 200 N, d = 600 mm and solving for h

800 N =

200 N 2

1+

⎛ ⎞ ⎜ 600 mm ⎟ ⎝

h

2

⎠ h = 75.6 mm W

PROBLEM 2.63 Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the magnitude of the force P required to maintain the equilibrium of the collar when (a) x = 4.5 in., (b) x = 15 in.

SOLUTION (a)

Free Body: Collar A

Force Triangle P

=

4.5

(b)

Free Body: Collar A

50 lb

P = 10.98 lb W

20.5

Force Triangle P 15

=

50 lb 25

P = 30.0 lb W

PROBLEM 2.64 Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the distance x for which the collar is in equilibrium when P = 48 lb.

SOLUTION Free Body: Collar A

Force Triangle

N 2 = (50)2 â&#x2C6;&#x2019; (48)2 = 196 N = 14.00 lb Similar Triangles x 48 lb = 20 in. 14 lb x = 68.6 in. W

PROBLEM 2.65 Three forces are applied to a bracket as shown. The directions of the two 150-N forces may vary, but the angle between these forces is always 50°. Determine the range of values of α for which the magnitude of the resultant of the forces acting at A is less than 600 N.

SOLUTION Combine the two 150-N forces into a resultant force Q:

Q = 2(150 N) cos 25° = 271.89 N Equivalent loading at A:

Using the law of cosines: (600 N) 2 = (500 N) 2 + (271.89 N) 2 + 2(500 N)(271.89 N) cos(55° + α ) cos(55° + α ) = 0.132685 55° + α = 82.375 Two values for α : α = 27.4°

or

55° + α = −82.375° 55° + α = 360° − 82.375° α = 222.6° For R < 600 lb:

27.4° < α < 222.6D W

PROBLEM 2.66 A 200-kg crate is to be supported by the rope-and-pulley arrangement shown. Determine the magnitude and direction of the force P that must be exerted on the free end of the rope to maintain equilibrium. (Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Ch. 4.)

SOLUTION Free-Body Diagram: Pulley A ΣFx = 0: − 2P

cos α = 0.59655 α = ±53.377°

5

+ P cos α = 0

⎜ ⎟ ⎝ 281 ⎠

For α = +53.377°: ⎛ 16 ⎞ ΣF y = 0: 2P ⎜ ⎟ + P sin 53.377° − 1962 N = 0 ⎝ 281 ⎠

P = 724 N

53.4° W

For α = −53.377°: ⎛ 16 ⎞ ΣF y = 0: 2P ⎜ ⎟ + P sin(−53.377°) − 1962 N = 0 ⎝ 281 ⎠

P = 1773

53.4° W

PROBLEM 2.67 A 600-lb crate is supported by several ropeand-pulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Problem 2.66.)

SOLUTION Free-Body Diagram of Pulley

ΣFy = 0: 2T − (600 lb) = 0 T=

1 (600 lb) 2

(a)

T = 300 lb W ΣFy = 0: 2T − (600 lb) = 0 T=

1 (600 lb) 2 T = 300 lb W

(b) ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3

T = 200 lb W

(c) (d )

ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3

(e)

T = 200 lb W

ΣFy = 0: 4T − (600 lb) = 0 T=

1 (600 lb) 4 T = 150.0 lb W

PROBLEM 2.68 Solve Parts b and d of Problem 2.67, assuming that the free end of the rope is attached to the crate. PROBLEM 2.67 A 600-lb crate is supported by several rope-and-pulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Problem 2.66.)

SOLUTION Free-Body Diagram of Pulley and Crate (b)

ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb W

(d )

ΣFy = 0: 4T − (600 lb) = 0 1 T = (600 lb) 4 T = 150.0 lb W

PROBLEM 2.69 A load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Knowing that P = 750 N, determine (a) the tension in cable ACB, (b) the magnitude of load Q.

SOLUTION Free-Body Diagram: Pulley C

ΣFx = 0: TACB (cos 25° − cos 55°) − (750 N) cos 55° = 0

(a)

Hence:

TACB = 1292.88 N TACB = 1293 N W

ΣFy = 0: TACB (sin 25° + sin 55°) + (750 N) sin 55° − Q = 0

(b)

(1292.88 N)(sin 25° + sin 55°) + (750 N) sin 55° − Q = 0

or

Q = 2219.8 N

Q = 2220 N W

PROBLEM 2.70 An 1800-N load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Determine (a) the tension in cable ACB, (b) the magnitude of load P.

SOLUTION Free-Body Diagram: Pulley C ΣFx = 0: TACB (cos 25° − cos 55°) − P cos 55° = 0 P = 0.58010TACB (1)

or

ΣFy = 0: TACB (sin 25° + sin 55°) + P sin 55° − 1800 N = 0

1.24177TACB + 0.81915P = 1800 N (2)

or (a)

Substitute Equation (1) into Equation (2):

1.24177TACB + 0.81915(0.58010TACB ) = 1800 N Hence:

TACB = 1048.37 N TACB = 1048 N W

(b)

Using (1),

P = 0.58010(1048.37 N) = 608.16 N P = 608 N W

PROBLEM 2.71 Determine (a) the x, y, and z components of the 600-N force, (b) the angles θx, θy, and θz that the force forms with the coordinate axes.

SOLUTION

(a)

Fx = (600 N) sin 25° cos 30 D

Fx = 219.60 N

Fx = 220 N W

Fy = (600 N) cos 25° Fy = 543.78 N

Fy = 544 N W

Fz = (380.36 N) sin 25° sin 30 D Fz = 126.785 N (b)

cos θx =

Fx 219.60 N = F 600 N Fy

cos

θy =

F

θz =

F

θx = 68.5° W

543.78 N =

Fz

cos

Fz = 126.8 N W

600 N

θ y = 25.0° W

126.785 N =

600 N

θz = 77.8° W

PROBLEM 2.72 Determine (a) the x, y, and z components of the 450-N force, (b) the angles θx, θy, and θz that the force forms with the coordinate axes.

SOLUTION

(a)

Fx = −(450 N) cos 35°sin 40 D

Fx = −236.94 N

Fx = −237 N W

Fy = (450 N) sin 35° Fy = 258.11 N

Fy = 258 N W

Fz = (450 N) cos 35° cos 40 D Fz = 282.38 N (b)

Fz = 282 N W

cos θ x =

Fx -236.94 N = F 450 N

θx = 121.8° W

cos θ y =

Fy

258.11 N 450 N

θ y = 55.0° W

cos θ z =

Fz 282.38 N = F 450 N

θz = 51.1° W

F

=

Note: From the given data, we could have computed directly θ y = 90D − 35D = 55D , which checks with the answer obtained.

PROBLEM 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z components of that force, (b) the values of the angles θx, θy, and θz defining the direction of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.)

SOLUTION Recoil force

F = 400 N ∴ FH = (400 N) cos 40° = 306.42 N

(a)

Fx = −FH sin 35° = −(306.42 N) sin 35° = −175.755 N

Fx = −175.8 N W

Fy = −F sin 40° = −(400 N) sin 40° = −257.12 N

Fy = −257 N W

Fz = + FH cos 35° = +(306.42 N) cos 35° = +251.00 N (b)

cos θ x =

Fx −175.755 N = 400 N F

Fz = +251 N W

θx = 116.1° W

cos θ y =

Fy

−257.12 N 400 N

θ y = 130.0° W

cos θ z =

Fz 251.00 N = F 400 N

θz = 51.1° W

F

=

PROBLEM 2.74 Solve Problem 2.73, assuming that point A is located 15° north of west and that the barrel of the gun forms an angle of 25° with the horizontal. PROBLEM 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z components of that force, (b) the values of the angles θx, θy, and θz defining the direction of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.)

SOLUTION Recoil force

F = 400 N ∴ FH = (400 N) cos 25° = 362.52 N

(a)

Fx = +FH cos15° = +(362.52 N) cos15° = +350.17 N

Fx = +350 N W

Fy = −F sin 25° = −(400 N) sin 25° = −169.047 N

Fy = −169.0 N W

Fz = +FH sin15° = +(362.52 N) sin15° = +93.827 N (b)

cos θ x =

cos θ y = cos θ z =

Fx +350.17 N = F 400 N Fy F

=

−169.047 N 400 N

Fz +93.827 N = F 400 N

Fz = +93.8 N W

θ x = 28.9° W

θ y = 115.0° W

θz = 76.4° W

PROBLEM 2.75 The angle between spring AB and the post DA is 30°. Knowing that the tension in the spring is 50 lb, determine (a) the x, y, and z components of the force exerted on the circular plate at B, (b) the angles θx, θy, and θz defining the direction of the force at B.

SOLUTION

Fh = F cos 60° = (50 lb) cos 60° Fh = 25.0 lb

Fx = −Fh cos 35° Fx = (−25.0 lb) cos 35 Fx = −20.479 lb

D

Fy = F sin 60D

Fz = −Fh sin 35 D

Fy = (50.0 lb) sin 60D

Fz = (−25.0 lb) sin 35

Fy = 43.301 lb

Fz = −14.3394 lb

D

(a)

Fx = −20.5 lb W Fy = 43.3 lb W

Fz = −14.33 lb W (b)

cos θ x =

Fx −20.479 lb = F 50 lb

cos θ y =

Fy

cos θ z =

Fz -14.3394 lb = F 50 lb

F

=

43.301 lb 50 lb

θ x = 114.2° W θ y = 30.0° W

θ z = 106.7° W

PROBLEM 2.76 The angle between spring AC and the post DA is 30°. Knowing that the tension in the spring is 40 lb, determine (a) the x, y, and z components of the force exerted on the circular plate at C, (b) the angles θx, θy, and θz defining the direction of the force at C.

SOLUTION

Fh = F cos 60° = (40 lb) cos 60° Fh = 20.0 lb (a)

Fx = Fh cos 35° = (20.0 lb) cos 35°

Fx = 16.3830 lb

Fy = F sin 60° = (40 lb) sin 60° Fy = 34.641 lb

Fz = −Fh sin 35° = −(20.0 lb) sin 35°

Fz = −11.4715 lb Fx = 16.38 lb W Fy = 34.6 lb W

Fz = −11.47 lb W (b)

cos θ x =

Fx 16.3830 lb = 40 lb F

cos θ y =

Fy

cos θ z =

Fz = -11.4715 lb 40 lb F

F

=

34.641 lb 40 lb

θ x = 65.8° W θ y

= 30.0° W θ z

= 106.7° W

PROBLEM 2.77 Cable AB is 65 ft long, and the tension in that cable is 3900 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor B, (b) the angles θ x , θ y , and θ z defining the direction of that force.

SOLUTION cos θ y =

From triangle AOB:

56 ft 65 ft

= 0.86154

θ y = 30.51° Fx = −F sin θ y cos 20°

(a)

= −(3900 lb) sin 30.51° cos 20°

Fx = −1861 lb W Fy = +F cos θ y = (3900 lb)(0.86154)

Fz = +(3900 lb) sin 30.51° sin 20°

(b)

cos θ = x

From above:

Fx

=−

F

1861 lb

= −0.4771

Fz F

=+

Fz = +677 lb W

θ = 118.5° W x

3900 lb

θ y = 30.51° cos θz =

Fy = +3360 lb W

θ y = 30.5° W 677 lb 3900 lb

= +0.1736

θz = 80.0° W

PROBLEM 2.78 Cable AC is 70 ft long, and the tension in that cable is 5250 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor C, (b) the angles θx, θy, and θz defining the direction of that force.

SOLUTION AC = 70 ft

In triangle AOB:

OA = 56 ft F = 5250 lb

cos θ y =

56 ft 70 ft

θ y = 36.870° FH = F sin θ y = (5250 lb) sin 36.870° = 3150.0 lb (a)

Fx = −FH sin 50° = −(3150.0 lb) sin 50° = −2413.0 lb Fx = −2410 lb W Fy = +F cos θ y = +(5250 lb) cos 36.870° = +4200.0 lb Fy = +4200 lb W

Fz = −FH cos 50° = −3150 cos 50° = −2024.8 lb

(b)

cos θ x =

From above:

Fx −2413.0 lb = F 5250 lb

θ y = 36.870° θz =

Fz −2024.8 lb = F 5250 lb

Fz = −2025 lb W

θx = 117.4° W θ y = 36.9° W

θz = 112.7° W

PROBLEM 2.79 Determine the magnitude and direction of the force F = (240 N)i – (270 N)j + (680 N)k.

SOLUTION F=

Fx2 + Fy2 + Fz2

F = (240 N) 2 + (−270 N) 2 + (−680 N) 2

cos θ x =

Fx 240 N = F 770 N

θy =

F

=

θy =

F

770 N

θ y = 110.5° W

680 N

Fz

cos

θ x = 71.8° W

−270 N

Fy

cos

F = 770 N W

=

770 N

θ z = 28.0° W

PROBLEM 2.80 Determine the magnitude and direction of the force F = (320 N)i + (400 N)j − (250 N)k.

SOLUTION F=

Fx2 + Fy2 + Fz2

F = (320 N) 2 + (400 N) 2 + (−250 N) 2

F = 570 N W

cos θ x =

Fx 320 N = F 570 N

θ x = 55.8° W

cos θ y =

Fy

θ y = 45.4° W

cos

θy =

F

400 N 570 N

Fz

−250 N

F

=

=

570 N

θ z = 116.0° W

PROBLEM 2.81 A force acts at the origin of a coordinate system in a direction defined by the angles θx = 69.3° and θz = 57.9°. Knowing that the y component of the force is –174.0 lb, determine (a) the angle θy, (b) the other components and the magnitude of the force.

SOLUTION cos 2 θ + cos 2 θ + cos 2 θ = 1 x

y

z

cos (69.3°) + cos θ y + cos (57.9°) = 1 2

2

2

cos θ y = ±0.7699

(a) (b)

Since Fy < 0, we choose cos θ y = −0.7699

∴ θ y = 140.3° W

Fy = F cos θ y −174.0 lb = F (−0.7699) F = 226.0 lb

F = 226 lb W

Fx = F cosθ x = (226.0 lb) cos 69.3°

Fx = 79.9 lb W

Fz = F cosθ z = (226.0 lb) cos 57.9°

Fz = 120.1 lb W

PROBLEM 2.82 A force acts at the origin of a coordinate system in a direction defined by the angles θx = 70.9° and θy = 144.9°. Knowing that the z component of the force is –52.0 lb, determine (a) the angle θz, (b) the other components and the magnitude of the force.

SOLUTION cos 2 θ + cos 2 θ + cos 2 θ = 1 x

y

z

cos 70.9 + cos 144.9° + cos θ z ° = 1 2

D

2

2

cos θ z = ±0.47282 (a) (b)

Since Fz < 0, we choose cosθ z = −0.47282

∴ θz = 118.2° W

Fz = F cos θ z −52.0 lb = F (−0.47282) F = 110.0 lb

F = 110.0 lb W

Fx = F cosθ x = (110.0 lb) cos 70.9°

Fx = 36.0 lb W

Fy = F cos θ y = (110.0 lb) cos144.9°

Fy = −90.0 lb W

PROBLEM 2.83 A force F of magnitude 210 N acts at the origin of a coordinate system. Knowing that Fx = 80 N, θz = 151.2°, and Fy < 0, determine (a) the components Fy and Fz, (b) the angles θx and θy.

SOLUTION Fz = F cosθ z = (210 N) cos151.2°

(a)

= −184.024 N Then: So:

Hence:

F 2 = F 2x + F 2y + F 2 z (210 N) 2 = (80 N) 2 + (Fy ) 2 + (184.024 N) 2 y

F = − (210 N) 2 − (80 N) 2 − (184.024 N) 2

Fy = −62.0 lb W

= −61.929 N (b)

Fz = −184.0 N W

cos θ = x

cos θ y =

Fx F

=

80 N

= 0.38095

210 N

θ = 67.6° W x

Fy 61.929 N = = −0.29490 F 210 N

θ y = 107.2° W

PROBLEM 2.84 A force F of magnitude 1200 N acts at the origin of a coordinate system. Knowing that θx = 65°, θy = 40°, and Fz > 0, determine (a) the components of the force, (b) the angle θz.

SOLUTION cos 2 θ + cos 2 θ + cos 2 θ = 1 x

y

z

cos 65 + cos 40° + cos θz ° = 1 2

D

2

2

cos θ z = ±0.48432 (b) (a)

Since Fz > 0, we choose cos θz = 0.48432, or θz = 61.032D

∴ θz = 61.0° W

F = 1200 N

Fx = F cosθx = (1200 N) cos65D

Fx = 507 N W

Fy = F cos θ y = (1200 N) cos 40°

Fy = 919 N W

Fz = F cos θ z = (1200 N) cos 61.032°

Fz = 582 N W

PROBLEM 2.85 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D.

SOLUTION JJJG DB = (480 mm)i − (510 mm) j + (320 mm)k DB = (480 mm) 2 + (510 mm2 ) + (320 mm) 2 = 770 mm F = Fλ DB JJJG DB =F DB 385 N = [(480 mm)i − (510 mm)j + (320 mm)k] 770 mm = (240 N)i − (255 N) j + (160 N)k Fx = +240 N, Fy = −255 N, Fz = +160.0 N W

PROBLEM 2.86 For the frame and cable of Problem 2.85, determine the components of the force exerted by the cable on the support at E. PROBLEM 2.85 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D.

SOLUTION JJJG EB = (270 mm)i − (400 mm) j + (600 mm)k EB = (270 mm)2 + (400 mm)2 + (600 mm)2 = 770 mm F = Fλ EB JJJG EB =F EB 385 N = [(270 mm)i − (400 mm)j + (600 mm)k] 770 mm F = (135 N)i − (200 N) j + (300 N)k Fx = +135.0 N, Fy = −200 N, Fz = +300 N W

PROBLEM 2.87 In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension in cable AB is 2 kips, determine the components of the force exerted at A by the cable.

SOLUTION

Cable AB:

=

λ AB

JJJG AB

=

(−46.765 ft )i +(45 ft ) j +(36 ft )k

AB

TAB = TAB λAB =

74.216 ft −46.765i + 45j + 36k 74.216

(TAB ) x = −1.260 kips W (TAB ) y = +1.213 kips W

(TAB ) z = +0.970 kips W

PROBLEM 2.88 In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension in cable AC is 1.5 kips, determine the components of the force exerted at A by the cable.

SOLUTION

Cable AB:

=

λ AC

T AC

JJJG AC

=

(−46.765 ft )i +(55.8 ft ) j +(−45 ft )k 85.590 ft

AC =T

= (1.5 kips)

λ AC

AC

−46.765i +55.8j −45k 85.590

(TAC ) x = −0.820 kips W (TAC ) y = +0.978 kips W

(TAC ) z = −0.789 kips W

PROBLEM 2.89 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AB is 408 N, determine the components of the force exerted on the plate at B.

SOLUTION

We have: JJJG BA = +(320 mm)i + (480 mm)j - (360 mm)k Thus:

BA = 680 mm

JJJG

B

BA BA

⎛ 8 ⎜ 17 ⎝

BA

=T

F =T λ

BA

T BA

⎞ ⎟ ⎠

i+

⎛ 8

=T

⎛ 12 ⎜ 17 ⎝

⎜ 17 i + 17 ⎝

BA

BA

T BA

12

⎞ ⎟ ⎠

j−

⎛ 9 ⎜ 17 ⎝

T BA

j-

⎞ ⎟ ⎠

9

k

17 ⎟ ⎠

k=0

Setting TBA = 408 N yields, Fx = +192.0 N, Fy = +288 N, Fz = −216 N W

PROBLEM 2.90 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 429 N, determine the components of the force exerted on the plate at D.

SOLUTION

We have: JJJG DA = −(250 mm)i + (480 mm)j + (360 mm)k Thus:

DA = 650 mm

JJJG F =T λ D

DA DA

⎛ 5 ⎜ 13 ⎝

DA

=T DA

T DA

⎞ ⎟ ⎠

DA

i+

=T

⎛ 48 ⎜ 65 ⎝

DA

DA

5

⎜ 13 ⎝ ⎞

T

⎟ ⎠

j+

i+

⎛ 36 ⎜ 65 ⎝

48

j+

65

T DA

⎞ ⎟ ⎠

36 ⎞ k 65 ⎟ ⎠

k=0

Setting TDA = 429 N yields, Fx = −165.0 N, Fy = +317 N, Fz = +238 N W

PROBLEM 2.91 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 300 N and Q = 400 N.

SOLUTION P = (300 N)[− cos 30° sin15°i + sin 30°j + cos 30° cos15°k ] = − (67.243 N)i + (150 N) j + (250.95 N)k Q = (400 N)[cos 50° cos 20°i + sin 50°j − cos 50° sin 20°k ] = (400 N)[0.60402i + 0.76604j − 0.21985] = (241.61 N)i + (306.42 N) j − (87.939 N)k R = P+Q = (174.367 N)i + (456.42 N) j + (163.011 N)k R = (174.367 N) 2 + (456.42 N) 2 + (163.011 N) 2 = 515.07 N cos θ = x

cos θ = y

cos θ = z

Rx R Ry R Rz R

=

= =

R = 515 N W

174.367 N 515.07 N 456.42 N 515.07 N 163.011 N 515.07 N

= 0.33853

θ = 70.2° W x

= 0.88613

θ = 27.6° W y

= 0.31648

θ = 71.5° W z

PROBLEM 2.92 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 400 N and Q = 300 N.

SOLUTION P = (400 N)[− cos 30° sin15°i + sin 30°j + cos 30° cos15°k ] = − (89.678 N)i + (200 N) j + (334.61 N)k Q = (300 N)[cos 50° cos 20°i + sin 50°j − cos 50° sin 20°k ] = (181.21 N)i + (229.81 N)j − (65.954 N)k R = P+Q = (91.532 N)i + (429.81 N) j + (268.66 N)k R = (91.532 N) 2 + (429.81 N) 2 + (268.66 N) 2 = 515.07 N cos θ = x

cos θ = y

cos θ = z

Rx R Ry R Rz R

= = =

91.532 N 515.07 N 429.81 N 515.07 N 268.66 N 515.07 N

R = 515 N W = 0.177708

θ = 79.8° W x

= 0.83447

θ = 33.4° W y

= 0.52160

θ = 58.6° W z

PROBLEM 2.93 Knowing that the tension is 425 lb in cable AB and 510 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

SOLUTION JJJG AB = (40 in.)i − (45 in.) j + (60 in.)k AB = (40 in.) 2 + (45 in.) 2 + (60 in.) 2 = 85 in. JJJG AC = (100 in.)i − (45 in.) j + (60 in.)k 2 2 2 AC = (100 in.) + (45 in.) + (60 in.) = 125 in. JJJG ⎡ (40in.)i −(45 in.) j +(60in.)k ⎤ AB T =T λ =T = (425 lb) AB AB AB AB ⎢ ⎥ AB 85 in. ⎣ ⎦ TAB = (200 lb)i − (225 lb) j + (300 lb)k JJJG ⎡ (100 in.)i −(45 in.) j +(60 in.)k ⎤ AC T =T λ =T = (510 lb) AC AC AC AC ⎢ ⎥ AC 125 in. ⎣ ⎦ TAC = (408 lb)i − (183.6 lb) j + (244.8 lb)k

R = TAB + TAC = (608)i − (408.6 lb) j + (544.8 lb)k

Then: and

R = 912.92 lb cos θ = x

cos θ = y

cos θ = z

608 lb 912.92 lb 408.6 lb 912.92 lb 544.8 lb 912.92 lb

R = 913 lb W = 0.66599 = −0.44757

θ = 48.2° W x

θ = 116.6° W y

= 0.59677

θ = 53.4° W z

PROBLEM 2.94 Knowing that the tension is 510 lb in cable AB and 425 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

SOLUTION JJJG AB = (40 in.)i − (45 in.) j + (60 in.)k AB = (40 in.) 2 + (45 in.) 2 + (60 in.) 2 = 85 in. JJJG AC = (100 in.)i − (45 in.) j + (60 in.)k 2 2 2 AC = (100 in.) + (45 in.) + (60 in.) = 125 in. JJJG ⎡ (40in.)i −(45 in.) j +(60in.)k ⎤ AB T =T λ =T = (510 lb) AB AB AB AB ⎢ ⎥ AB 85 in. ⎣ ⎦ TAB = (240 lb)i − (270 lb) j + (360 lb)k JJJG ⎡ (100 in.)i −(45 in.) j +(60 in.)k ⎤ AC T =T λ =T = (425 lb) AC AC AC AC ⎢ ⎥ AC 125 in. ⎣ ⎦ TAC = (340 lb)i − (153 lb) j + (204 lb)k

R = TAB + TAC = (580 lb)i − (423 lb) j + (564 lb)k

Then: and

R = 912.92 lb cos θ = x

cos θ y = cos θ = z

580 lb

R = 913 lb W = 0.63532

−423 lb = −0.46335 912.92 lb 564 lb 912.92 lb

θ = 50.6° W x

912.92 lb

= 0.61780

θ y = 117.6° W

θ = 51.8° W z

PROBLEM 2.95 For the frame of Problem 2.85, determine the magnitude and direction of the resultant of the forces exerted by the cable at B knowing that the tension in the cable is 385 N. PROBLEM 2.85 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D.

SOLUTION JJJG BD = −(480 mm)i + (510 mm) j − (320 mm)k BD = (480 mm) 2 + (510 mm) 2 + (320 mm) 2 = 770 mm

=T λ

F BD

BD

=T BD

BD

JJJG BD BD

(385 N)

[−(480 mm)i + (510 mm) j − (320 mm)k ] (770 mm) = −(240 N)i + (255 N) j − (160 N)k

=

JJJG BE = −(270 mm)i + (400 mm) j − (600 mm)k 2 2 2 BE = (270 mm) + (400 mm) + (600 mm) = 770 mm

=T λ

F BE

BE

=T BE

BE

JJJG BE BE

(385 N)

[−(270 mm)i + (400 mm) j − (600 mm)k ] (770 mm) = −(135 N)i + (200 N) j − (300 N)k

=

R = FBD + FBE = −(375 N)i + (455 N)j − (460 N)k R = (375 N)2 + (455 N)2 + (460 N)2 = 747.83 N −375 N cos θ x = 747.83 N cos θ = y

455 N 747.83 N

−460 N cos θ z = 747.83 N

R = 748 N W

θ x = 120.1° W

θ = 52.5° W y

θ z = 128.0° W

PROBLEM 2.96 For the plate of Prob. 2.89, determine the tensions in cables AB and AD knowing that the tension in cable AC is 54 N and that the resultant of the forces exerted by the three cables at A must be vertical.

SOLUTION

We have: JJJG AB = −(320 mm)i − (480 mm)j + (360 mm)k JJJG AC = (450 mm)i − (480 mm) j + (360 mm)k JJJG AD = (250 mm)i − (480 mm) j − ( 360 mm ) k Thus:

AB = 680 mm AC = 750 mm AD = 650 mm

JJJG TAB = TAB λ AB = TAB

T AC

=T

λ AC

AC

AB TAB = ( −320i − 480j + 360k ) AB 680 JJJG AC

=T AC

=

54

( 450i

− 480j + 360k )

Substituting into the Eq. R = ΣF and factoring i, j, k : 250 ⎛ 320 ⎞ R= − T + 32.40 + T i AD ⎟ ⎜ 680 AB 650 ⎝ ⎠ 480 480 ⎛ ⎞ + − T − 34.560 − T j AD ⎟ ⎜ 680 AB 650 ⎝ ⎠ 360 ⎛ 360 ⎞ + T + 25.920 − T k AD ⎟ ⎜ 680 AB 650 ⎝ ⎠

PROBLEM 2.96 (Continued) Since R is vertical, the coefficients of i and k are zero: i:

320

680 360

k:

680

+ 32.40 +

T AB

250 650

+ 25.920 −

T AB

=0

T

360 650

(1)

=0

T

(2)

Multiply (1) by 3.6 and (2) by 2.5 then add: −

252 TAB + 181.440 = 0 680

TAB = 489.60 N TAB = 490 N W Substitute into (2) and solve for TAD : 360 360 (489.60 N) + 25.920 − T AD = 0 680 650

PROBLEM 2.97 The boom OA carries a load P and is supported by two cables as shown. Knowing that the tension in cable AB is 183 lb and that the resultant of the load P and of the forces exerted at A by the two cables must be directed along OA, determine the tension in cable AC.

SOLUTION

Cable AB:

TAB = 183 lb T AB

Cable AC:

=T λ AB

=T AB

= (183 lb)

(−48 in.)i +(29 in.) j +(24 in.)k

AB

TAB

AB = −(144 lb)i + (87 lb)j + (72 lb)k JJJG

T

=T λ

AC

AC

JJJG AB

AC

=T AC

AC

(−48 in.)i +(25 in.) j +(−36 in.)k

=T AC

AC

61 in.

65 in.

48 25 36 TAC i + TAC j − TAC k 65 65 65

P = Pj

For resultant to be directed along OA, i.e., x-axis R = 0: ΣF = (72 lb) − z

z

36 65

T′ = 0 AC

T AC

= 130.0 lb W

PROBLEM 2.98 For the boom and loading of Problem. 2.97, determine the magnitude of the load P. PROBLEM 2.97 The boom OA carries a load P and is supported by two cables as shown. Knowing that the tension in cable AB is 183 lb and that the resultant of the load P and of the forces exerted at A by the two cables must be directed along OA, determine the tension in cable AC.

SOLUTION See Problem 2.97. Since resultant must be directed along OA, i.e., the x-axis, we write R = 0: ΣF = (87 lb) + y

y

25

−P=0

T

65

AC

TAC = 130.0 lb from Problem 2.97.

Then

(87 lb) +

25 (130.0 lb) − P = 0 65

P = 137.0 lb W

PROBLEM 2.99 A container is supported by three cables that are attached to a ceiling as shown. Determine the weight W of the container, knowing that the tension in cable AB is 6 kN.

SOLUTION Free-Body Diagram at A:

The forces applied at A are:

TAB , TAC , TAD , and W

where W = W j. To express the other forces in terms of the unit vectors i, j, k, we write JJJG AB = − (450 mm)i + (600 mm) j AB = 750 mm JJJG AC = + (600 mm) j − (320 mm)k AC = 680 mm JJJG AD = + (500 mm)i + (600 mm) j + (360 mm)k AD = 860 mm JJJG and

=λ T

T AB

AB

=T

AB AB

AB

=T AB

AB

(−450 mm)i + (600 mm)j 750 mm

60 ⎞ ⎛ 45 = ⎜⎝ − i + j ⎠TAB 75 75 ⎟ JJJG T AC

=λ T AC AC

AC

=T AC

AC

(600 mm)i −(320 mm) j

=T AC

680 mm

32 ⎞ ⎛ 60 =⎜ j − k ⎟ TAC 68 68 ⎝ ⎠ JJJG AD (500 mm)i +(600 mm) j +(360 mm)k =T AD

860 mm

60 36 ⎞ ⎛ 50 =⎜ i+ j + k ⎟ TAD 86 86 86 ⎝ ⎠

PROBLEM 2.99 (Continued) ΣF = 0: ∴ TAB + TAC + TAD + W = 0

Equilibrium condition:

Substituting the expressions obtained for TAB , TAC , and TAD ; factoring i, j, and k; and equating each of the coefficients to zero gives the following equations: −

From i:

45

AB

75

From j:

60 75

From k:

T AB

+

60

60

+

T AC

68 −

68

T AC

=0

(1)

−W = 0

(2)

=0

(3)

86

86

32

50

+

T

+

36 86

Setting TAB = 6 kN in (1) and (2), and solving the resulting set of equations gives TAC = 6.1920 kN TAC = 5.5080 kN

W = 13.98 kN W

PROBLEM 2.100 A container is supported by three cables that are attached to a ceiling as shown. Determine the weight W of the container, knowing that the tension in cable AD is 4.3 kN.

SOLUTION See Problem 2.99 for the figure and analysis leading to the following set of linear algebraic equations: −

45

AB

75 60 75

+

T AB

60

60

+

T AC

68 −

32 68

50

+

T

AC

(1)

−W = 0

(2)

=0

(3)

86

86 T

=0

T

+

36 86

Setting TAD = 4.3 kN into the above equations gives TAB = 4.1667 kN TAC = 3.8250 kN

W = 9.71 kN W

PROBLEM 2.101 Three cables are used to tether a balloon as shown. Determine the vertical force P exerted by the balloon at A knowing that the tension in cable AD is 481 N.

SOLUTION

The forces applied at A are:

FREE-BODY DIAGRAM AT A

TAB , TAC , TAD , and P

where P = Pj. To express the other forces in terms of the unit vectors i, j, k, we write JJJG AB = − (4.20 m)i − (5.60 m) j JJJG AC = (2.40 m)i − (5.60 m) j + (4.20 m)k JJJG AD = − (5.60 m)j − (3.30 m)k

AB = 7.00 m AC = 7.40 m AD = 6.50 m

JJJG AB and

TAB = TAB λ AB = TAB TAC = TAC λ AC TAD = TAD λ AD

= (− 0.6i − 0.8j)TAB AB JJJG AC = TAC = (0.32432i − 0.75676j + 0.56757k )TAC AC JJJG AD = TAD = (− 0.86154j − 0.50769k )TAD AD

PROBLEM 2.101 (Continued)

Equilibrium condition:

ΣF = 0: TAB + TAC + TAD + Pj = 0

Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k: (− 0.6TAB + 0.32432TAC )i + (−0.8TAB − 0.75676TAC − 0.86154TAD + P)j + (0.56757TAC − 0.50769TAD )k = 0 Equating to zero the coefficients of i, j, k: − 0.6TAB + 0.32432TAC = 0

(1)

− 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0

(2)

(3)

Setting TAD = 481 N in (2) and (3), and solving the resulting set of equations gives TAC = 430.26 N TAD = 232.57 N

P = 926 N W

PROBLEM 2.102 Three cables are used to tether a balloon as shown. Knowing that the balloon exerts an 800-N vertical force at A, determine the tension in each cable.

SOLUTION See Problem 2.101 for the figure and analysis leading to the linear algebraic Equations (1), (2), and (3). −0.6TAB + 0.32432TAC = 0

(1)

−0.8TAB − 0.75676TAC − 0.86154TAD + P = 0

(2)

(3)

From Eq. (1):

TAB = 0.54053TAC

From Eq. (3):

Substituting for T AB and TAD in terms of TAC into Eq. (2) gives − 0.8(0.54053TAC ) − 0.75676TAC − 0.86154(1.11795TAC ) + P = 0 2.1523TAC = P ;

P = 800 N

800 N 2.1523 = 371.69 N

TAC =

Substituting into expressions for T AB and TAD gives TAB = 0.54053(371.69 N) TAD = 1.11795(371.69 N) TAB = 201 N, TAC = 372 N, TAD = 416 N W

PROBLEM 2.103 A 36-lb triangular plate is supported by three wires as shown. Determine the tension in each wire, knowing that a = 6 in.

SOLUTION By Symmetry TDB = TDC

Free-Body Diagram of Point D:

TDB , TDC , TDA , and P

The forces applied at D are:

where P = Pj = (36 lb)j. To express the other forces in terms of the unit vectors i, j, k, we write JJJG DA = (16 in.)i − (24 in.) j JJJG DB = −(8 in.)i − (24 in.) j + (6 in.)k JJJG DC = −(8 in.)i − (24 in.)j − (6 in.)k

DA = 28.844 in. DB = 26.0 in. DC = 26.0 in.

JJG DA and

TDA = TDA λDA = TDA TDB TDC

= (0.55471i − 0.83206j)TDA DA JJJG DB = TDB λ DB = TDB = (−0.30769i − 0.92308j + 0.23077k )T DB DB JJJG DC = TDC λ DC = TDC = (−0.30769i − 0.92308j − 0.23077k )T DC DC

PROBLEM 2.103 (Continued)

Equilibrium condition:

ΣF = 0: TDA + TDB + TDC + (36 lb)j = 0

Substituting the expressions obtained for TDA , TDB , and TDC and factoring i, j, and k:

(0.55471TDA − 0.30769TDB − 0.30769TDC )i + (−0.83206TDA − 0.92308TDB − 0.92308TDC + 36 lb)j + (0.23077TDB − 0.23077TDC )k = 0 Equating to zero the coefficients of i, j, k: 0.55471TDA − 0.30769TDB − 0.30769TDC = 0

(1)

− 0.83206TDA − 0.92308TDB − 0.92308TDC + 36 lb = 0

(2)

0.23077TDB − 0.23077TDC = 0

(3)

Equation (3) confirms that TDB = TDC . Solving simultaneously gives, TDA = 14.42 lb;

TDB = TDC = 13.00 lb W

PROBLEM 2.104 Solve Prob. 2.103, assuming that a = 8 in. PROBLEM 2.103 A 36-lb triangular plate is supported by three wires as shown. Determine the tension in each wire, knowing that a = 6 in.

SOLUTION By Symmetry TDB = TDC

Free-Body Diagram of Point D:

The forces applied at D are:

TDB , TDC , TDA , and P

where P = Pj = (36 lb)j. To express the other forces in terms of the unit vectors i, j, k, we write JJJG DA = (16 in.)i − (24 in.) j JJJG DB = −(8 in.)i − (24 in.) j + (8 in.)k JJJG DC = −(8 in.)i − (24 in.)j − (8 in.)k

DA = 28.844 in. DB = 26.533 in. DC = 26.533 in.

JJG DA and

TDA = TDA λDA = TDA TDB TDC

= (0.55471i − 0.83206 j)TDA DA JJJG DB = TDB λ DB = TDB = (−0.30151i − 0.90453j + 0.30151k )T DB DB JJJG DC = TDC λ DC = TDC = (−0.30151i − 0.90453j − 0.30151k )T DC DC

PROBLEM 2.104 (Continued)

Equilibrium condition:

ΣF = 0: TDA + TDB + TDC + (36 lb)j = 0

Substituting the expressions obtained for TDA , TDB , and TDC and factoring i, j, and k:

(0.55471TDA − 0.30151TDB − 0.30151TDC )i + (−0.83206TDA − 0.90453TDB − 0.90453TDC + 36 lb)j + (0.30151TDB − 0.30151TDC )k = 0 Equating to zero the coefficients of i, j, k: 0.55471TDA − 0.30151TDB − 0.30151TDC = 0

(1)

− 0.83206TDA − 0.90453TDB − 0.90453TDC + 36 lb = 0

(2)

0.30151TDB − 0.30151TDC = 0

(3)

Equation (3) confirms that TDB = TDC . Solving simultaneously gives, TDA = 14.42 lb;

TDB = TDC = 13.27 lb W

PROBLEM 2.105 A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AC is 544 lb.

Solution The forces applied at A are: TAB , TAC , TAD and W where P = Pj. To express the other forces in terms of the unit vectors i, j, k, we write JJJG AB = − (36 in.)i + (60 in.) j − (27 in.)k AB = 75 in. JJJG AC = (60 in.) j + (32 in.)k AC = 68 in. JJJG AD = (40 in.)i + (60 in.) j − (27 in.)k AD = 77 in. and

TAB = TAB λAB = TAB

TAC

Equilibrium Condition with

JJJG AB AB

= (−0.48i + 0.8j − 0.36k )TAB JJJG AC = T AC λ AC = TAC AC = (0.88235j + 0.47059k )TAC JJJG AD = T AD λ AD = TAD AD = (0.51948i + 0.77922 j − 0.35065k )TAD

W = −Wj ΣF = 0: TAB + TAC + TAD − Wj = 0

PROBLEM 2.105 (Continued)

Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k: (−0.48TAB + 0.51948TAD )i + (0.8TAB + 0.88235TAC + 0.77922TAD − W ) j + (−0.36TAB + 0.47059TAC − 0.35065TAD )k = 0 Equating to zero the coefficients of i, j, k: − 0.48TAB + 0.51948TAD = 0

(1)

0.8TAB + 0.88235TAC + 0.77922TAD − W = 0

(2)

− 0.36TAB + 0.47059TAC − 0.35065TAD = 0

(3)

Substituting TAC = 544 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives: TAB = 374.27 lb TAD = 345.82 lb

W = 1049 lb W

PROBLEM 2.106 A 1600-lb crate is supported by three cables as shown. Determine the tension in each cable.

SOLUTION The forces applied at A are: TAB , TAC , TAD and W where P = Pj. To express the other forces in terms of the unit vectors i, j, k, we write JJJG AB = − (36 in.)i + (60 in.) j − (27 in.)k AB = 75 in. JJJG AC = (60 in.) j + (32 in.)k AC = 68 in. JJJG AD = (40 in.)i + (60 in.) j − (27 in.)k AD = 77 in. and

TAB = TAB λAB = TAB

TAC

Equilibrium Condition with

JJJG AB AB

= (−0.48i + 0.8j − 0.36k )TAB JJJG AC = T AC λ AC = TAC AC = (0.88235j + 0.47059k )TAC JJJG AD = T AD λ AD = TAD AD = (0.51948i + 0.77922 j − 0.35065k )TAD

W = −Wj ΣF = 0: TAB + TAC + TAD − Wj = 0

PROBLEM 2.106 (Continued)

Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k: (−0.48TAB + 0.51948TAD )i + (0.8TAB + 0.88235TAC + 0.77922TAD − W ) j + (−0.36TAB + 0.47059TAC − 0.35065TAD )k = 0 Equating to zero the coefficients of i, j, k: −0.48TAB + 0.51948TAD = 0

(1)

0.8TAB + 0.88235TAC + 0.77922TAD − W = 0

(2)

−0.36TAB + 0.47059TAC − 0.35065TAD = 0

(3)

Substituting W = 1600 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms gives, TAB = 571 lb W TAC = 830 lb W TAD = 528 lb W

PROBLEM 2.107 Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that Q = 0, find the value of P for which the tension in cable AD is 305 N.

SOLUTION ΣFA = 0: TAB + TAC + TAD + P = 0 JJJG

where

P = Pi

AB = −(960 mm)i − (240 mm)j + (380 mm)k JJJG AC = −(960 mm)i − (240 mm) j − (320 mm)k JJJG AD = −(960 mm)i + (720 mm) j − (220 mm)k JJJG T

=T

AB

T

AB

=T

AC

T

AB

AC

AB

=T

λ

=T

AB

=T

λ

AC

AC

AC = 1040 mm AD = 1220 mm

19 ⎞ ⎛ 48 12 − i− j+ k AB ⎜ 53 53 53 ⎟ ⎝ ⎠

=T

AB JJJG AC

3 4 ⎞ ⎛ 12 − i − j− k AC ⎜ 13 13 13 ⎟ ⎝ ⎠

=T

AC

305 N

=

λ

AB = 1060 mm

[(−960 mm)i + (720 mm) j − (220 mm)k ]

1220 mm = −(240 N)i + (180 N) j − (55 N)k

Substituting into ΣFA = 0, factoring i, j, k, and setting each coefficient equal to φ gives: i: P =

48 53

j:

12

+

T AB

AB

53

k:

19

T

T

3

+ 240 N

(1)

= 180 N

(2)

AC

13 +

T

12

T

13

AC

53

AB

4

T

13 AC

= 55 N

(3)

Solving the system of linear equations using conventional algorithms gives: TAB = 446.71 N TAC = 341.71 N

P = 960 N W

PROBLEM 2.108 Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that P = 1200 N, determine the values of Q for which cable AD is taut.

SOLUTION We assume that TAD = 0 and write JJJG

ΣFA = 0: TAB + TAC + Qj + (1200 N)i = 0

AB = −(960 mm)i − (240 mm)j + (380 mm)k JJJG AC = −(960 mm)i − (240 mm) j − (320 mm)k JJJG T

=T λ

AB

T

AB

=T

AC

AC

λ AC

AC = 1040 mm

19 ⎞ ⎛ 48 12 = − i− j+ k T AB AB AB ⎜ 53 53 53 ⎟ ⎝ ⎠ JJJG AC ⎛ 12 3 4 ⎞ =T = − i − j− k T AC AC ⎜ 13 13 13 ⎟ AC ⎝ ⎠ AB

=T

AB

AB = 1060 mm

Substituting into ΣFA = 0, factoring i, j, k, and setting each coefficient equal to φ gives: i: −

48

AB

53 j: −

12 53

k:

19 53

T

AB

AB

3 13

T

T

4 13

+ 1200 N = 0

(1)

+Q =0

(2)

=0

(3)

AC

13 −

T

12

T AC

T AC

Solving the resulting system of linear equations using conventional algorithms gives:

605.71 N TAC = 705.71 N Q = 300.00 N

0 ≤ Q < 300 N W Note: This solution assumes that Q is directed upward as shown (Q ≥ 0), if negative values of Q are considered, cable AD remains taut, but AC becomes slack for Q = −460 N.

PROBLEM 2.109 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AC is 60 N, determine the weight of the plate.

SOLUTION We note that the weight of the plate is equal in magnitude to the force P exerted by the support on Point A. Free Body A : ΣF = 0: TAB + TAC + TAD + Pj = 0 We have: JJJG AB = −(320 mm)i − (480 mm)j + (360 mm)k JJJG AC = (450 mm)i − (480 mm) j + (360 mm)k JJJG AD = (250 mm)i − (480 mm) j − ( 360 mm ) k Thus:

AB = 680 mm AC = 750 mm AD = 650 mm

JJJG AB TAB = TAB λ AB = TAB

TAC = TAC λAC = TAC

⎛ 8 12 9 ⎞ = − i − j + k T AB ⎜⎝ 17 17 17 ⎟⎠ AB JJJG AC AC JJJG

= ( 0.6i − 0.64 j + 0.48k ) TAC

AD ⎛ 5 9.6 7.2 ⎞ =⎜ i− j− k ⎟ TAD 13 13 ⎠ AD ⎝ 13

Substituting into the Eq. ΣF = 0 and factoring i, j, k : ⎛ − 8 T + 0.6T + 5 T ⎞ i AB AC AD ⎟ ⎜ 13 ⎝ 17 ⎠ 9.6 ⎛ 12

+ ⎜ − TAB − 0.64TAC − TAD + P ⎟ j 13 ⎝ 17 ⎠ 7.2 ⎛ 9 ⎞ + ⎜ TAB + 0.48TAC − TAD ⎟ k = 0 ⎠ 13 ⎝ 17

Dimensions in mm

PROBLEM 2.109 (Continued) Setting the coefficient of i, j, k equal to zero: i:

j:

8 17 12

− 0.64T

T AB

7 9

k:

5 T AD = 0 13

TAB + 0.6TAC +

9.6

AC

AB

T

13 7.2

+ 0.48T

T

17

AC

13

(1)

+P=0

(2)

=0

(3)

Making TAC = 60 N in (1) and (3): 8

AB

17 9

T

7.2

AB

=0

(1′)

=0

(3′)

13

+ 28.8 N −

T

17

5

+ 36 N +

T

13

Multiply (1′) by 9, (3′) by 8, and add: 554.4 N −

12.6

=0 T

13

Substitute into (1′) and solve for TAB : =

T AB

17 ⎛ 8⎜ ⎝

36 +

5

× 572

13

⎞ ⎟ ⎠

T

= 544.0 N

AB

Substitute for the tensions in Eq. (2) and solve for P : 12 9.6 (544 N) + 0.64(60 N) + (572 N) 17 13 = 844.8 N

P=

Weight of plate = P = 845 N W

PROBLEM 2.110 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 520 N, determine the weight of the plate.

SOLUTION See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 8

12

17

+ 0.64T

T

17

AB

9

− AC

(1)

+P=0

(2)

=0

(3)

+ 200 N = 0

(1′)

− 288 N = 0

(3′)

9.6

AB

T

13

7.2

+ 0.48T

T

17

5 T AD = 0 13

TAB + 0.6TAC +

AC

13

Making TAD = 520 N in Eqs. (1) and (3): 8

17 9

+ 0.6T

T AB

+ 0.48T

T

17

AC

AB

AC

Multiply (1′) by 9, (3′) by 8, and add: 9.24TAC − 504 N = 0

TAC = 54.5455 N

Substitute into (1′) and solve for TAB : TAB =

17 8

(0.6 × 54.5455 + 200) TAB = 494.545 N

Substitute for the tensions in Eq. (2) and solve for P: P=

12 9.6 (494.545 N) + 0.64(54.5455 N) + (520 N) 17 13

= 768.00 N

Weight of plate = P = 768 N W

PROBLEM 2.111 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AB is 840 lb, determine the vertical force P exerted by the tower on the pin at A.

SOLUTION ΣF = 0: TAB + TAC + TAD + Pj = 0

JJJG AB = −20i − 100j + 25k AB = 105 ft JJJG AC = 60i − 100j + 18k AC = 118 ft JJJG AD = −20i − 100j − 74k AD = 126 ft JJG We write

T AB

T AC

AB

AB

=T

=T λ AB

AB

AB

4 20 5 = ⎜⎛ − i − j + k ⎟ T⎞ AB 21 ⎠ ⎝ 21 21 JJJG AC =T λ =T AC

AC

AC

AC 50 9 ⎞ ⎛ 30 =⎜ i− j + k ⎟ TAC 59 ⎠ ⎝ 59 59 JJJG AD =T λ =T AD 50 37 ⎞ ⎛ 10 = ⎜− i − j − k ⎟ TAD 63 ⎠ ⎝ 63 63

Substituting into the Eq. ΣF = 0 and factoring i, j, k :

Free-Body Diagram at A:

PROBLEM 2.111 (Continued) ⎛ − 4 T + 30T − 10 ⎞ ⎜ 21 AB 59 AC 63 TAD ⎟ i ⎝ ⎠ 20 50 50 + ⎛⎜ − TAB − TAC − TAD + P ⎟ j⎞ 59 63 ⎝ 21 ⎠ 9 37 ⎛ 5 ⎞ + T + T − T k=0 AC AB AD ⎜ 21 ⎟ 59 63 ⎝ ⎠

Setting the coefficients of i, j, k, equal to zero: i:

4

AB

21

j:

20

+

T

k:

5

− AB

AB

21

AC

50

9

− AC

T

AC

=0

(1)

+P=0

(2)

=0

(3)

50

63 −

T

59

10 63

T

59 +

T

T

59

T

21

30

37

63

Set TAB = 840 lb in Eqs. (1) – (3): −160 lb +

30

−800 lb −

50

(1′)

+P=0

(2′)

200 lb +

9 37 TAC − T AD = 0 59 63

(3′)

Solving,

TAC = 458.12 lb

AC

T

59

10

=0

59

T

AC

50 63

63

T

P = 1552.94 lb

P = 1553 lb W

PROBLEM 2.112 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AC is 590 lb, determine the vertical force P exerted by the tower on the pin at A.

SOLUTION ΣF = 0: TAB + TAC + TAD + Pj = 0

JJJG AB = −20i − 100j + 25k AB = 105 ft JJJG AC = 60i − 100j + 18k AC = 118 ft JJJG AD = −20i − 100j − 74k AD = 126 ft JJG We write

T AB

T AC

AB

AB

=T

=T λ AB

AB

AB

4 20 5 = ⎛⎜ − i − j + k ⎟ T⎞ AB 21 ⎠ ⎝ 21 21 JJJG AC =T λ =T AC

AC

AC

AC 50 9 ⎞ ⎛ 30 =⎜ i− j + k ⎟ TAC 59 59 ⎠ ⎝ 59 JJJG AD =T λ =T AD 10 50 37 ⎞ ⎛ = ⎜− i − j − k ⎟ TAD 63 ⎠ ⎝ 63 63

Substituting into the Eq. ΣF = 0 and factoring i, j, k :

Free-Body Diagram at A:

PROBLEM 2.112 (Continued) ⎛ − 4 T + 30T − 10 ⎞ ⎜ 21 AB 59 AC 63 TAD ⎟ i ⎝ ⎠ 20 50 50 ⎛ + ⎜ − TAB − TAC − TAD + P ⎟ j⎞ 59 63 ⎝ 21 ⎠ 9 37 ⎛ 5 ⎞ + T + T − T k=0 ⎜ 21 AB 59 AC 63 AD ⎟ ⎝ ⎠ Setting the coefficients of i, j, k, equal to zero: −

i:

4

AB

21 −

j:

+

T

20

AB

21 5

k:

AB

21

AC

50

9

− AC

T

− AC

=0

(1)

+P=0

(2)

=0

(3)

50

63

T

59

10 63

T

59 +

T

T

59

T

30

37

63

Set TAC = 590 lb in Eqs. (1) – (3): −

4

20 21

AB

5 21

Solving,

50 63

+ 90 lb −

T AB

=0

(1′)

+P=0

(2′)

T

63

− 500 lb −

T

10

AB

21

+ 300 lb −

T

T

37 63

=0

T

(3′)

TAB = 1081.82 lb TAD = 591.82 lb

P = 2000 lb W

PROBLEM 2.113 In trying to move across a slippery icy surface, a 175-lb man uses two ropes AB and AC. Knowing that the force exerted on the man by the icy surface is perpendicular to that surface, determine the tension in each rope.

SOLUTION Free-Body Diagram at A

N=N

30 ⎞ ⎛ 16 i+ j ⎜ 34 ⎠⎟ ⎝ 34 and W = W j = −(175 lb) j

T

=T

=T λ

AC

AC

JJJG AC

AC

AC

AC

JJJG T

=T λ

AB

AB

AB

=T AB

AB

AB

(−30 ft)i +(20 ft)j −(12 ft)k

=T AC

38 ft 6 ⎞ ⎛ 15 10 = TAC ⎜ − i + j − k ⎟ ⎝ 19 19 19 ⎠ (−30 ft)i +(24 ft)j +(32 ft)k

=T AB

50 ft 12 16 ⎞ ⎛ 15 = TAB ⎜ − i + j+ k⎟ 25 25 ⎠ ⎝ 25

Equilibrium condition: ΣF = 0 TAB + TAC + N + W = 0

PROBLEM 2.113 (Continued) Substituting the expressions obtained for TAB , TAC , N, and W; factoring i, j, and k; and equating each of the coefficients to zero gives the following equations: −

From i:

15

AB

25

From j:

12 25

From k:

+

T AB

10 19

T AC

+

15

(1)

N − (175 lb) = 0

(2)

6 16 T AB − T AC = 0 25 19

(3)

19 30

T AC

+

16

N =0

T

34

34

Solving the resulting set of equations gives: TAB = 30.8 lb; TAC = 62.5 lb W

PROBLEM 2.114 Solve Problem 2.113, assuming that a friend is helping the man at A by pulling on him with a force P = −(45 lb)k. PROBLEM 2.113 In trying to move across a slippery icy surface, a 175-lb man uses two ropes AB and AC. Knowing that the force exerted on the man by the icy surface is perpendicular to that surface, determine the tension in each rope.

SOLUTION Refer to Problem 2.113 for the figure and analysis leading to the following set of equations, Equation (3) being modified to include the additional force P = (−45 lb)k. 15

15

(1)

12 10 30 T AB + TAC + N − (175 lb) = 0 25 19 34

(2)

25

16 25

T AB

T AB

T

19

6 19

+

16

N =0

AC

34

− (45 lb) = 0

T

(3)

AC

Solving the resulting set of equations simultaneously gives: TAB = 81.3 lb W TAC = 22.2 lb W

PROBLEM 2.115 For the rectangular plate of Problems 2.109 and 2.110, determine the tension in each of the three cables knowing that the weight of the plate is 792 N.

SOLUTION See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below. Setting P = 792 N gives: 8

AB

17

12 17

− 0.64T

T AB

17

9.6

AB

+ 792 N = 0

(2)

+ 0.48T

T

(1)

13

T

13

5

=0

AC

− AC

9

+

+ 0.6T

T

AC

7.2 13

=0

T

(3)

Solving Equations (1), (2), and (3) by conventional algorithms gives TAB = 510.00 N

TAB = 510 N W

TAC = 56.250 N

TAC = 56.2 N W

PROBLEM 2.116 For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = 0.

SOLUTION ΣFA = 0: TAB + TAC + TAD + P + Q = 0

Where

P = Pi and Q = Qj JJJG AB = −(960 mm)i − (240 mm) j + (380 mm)k AB = 1060 mm JJJG AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm JJJG AD = −(960 mm)i + (720 mm) j − (220 mm)k AD = 1220 mm JJJG AB 12 19 ⎞ ⎛ 48 T =T λ =T =T − i− j+ k AB AB AB AB AB ⎜ AB 53 53 53 ⎟ ⎝ ⎠ JJJG AC 3 4 ⎞ ⎛ 12 =T T =T λ =T − i − j− k AC AC AC AC AC ⎜ AC 13 13 13 ⎟ ⎝ ⎠ JJJG AD 11 ⎞ ⎛ 48 36 =T − i+ j− k T =T λ =T AD AD AD AD AD ⎜ AD 61 61 61 ⎟ ⎝ ⎠

Substituting into ΣFA = 0, setting P = (2880 N)i and Q = 0, and setting the coefficients of i, j, k equal to 0, we obtain the following three equilibrium equations: i: −

48

AB

53 j: −

12

T

T

12

T AC

13 −

3

T

48

T

36

(1)

=0

(2)

61 +

+ 2880 N = 0

T

53

k:

19 53

AB

T AB

13 4 13

AC

T AC

61 11 61

=0

(3)

PROBLEM 2.116 (Continued) Solving the system of linear equations using conventional algorithms gives: TAB = 1340.14 N TAC = 1025.12 N TAD = 915.03 N

TAB = 1340 N W TAC = 1025 N W TAD = 915 N W

PROBLEM 2.117 For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = 576 N.

SOLUTION See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 48

AB

53 −

12

12

T

13 3

T AB

53

AC

AC

AB

53

(1)

+Q =0

(2)

=0

(3)

+ 2880 N = 0

(1′)

+ 576 N = 0

(2′)

36

+

T

+P=0

T

61

T

13

19

48

T

4

61 −

T

13

AC

11 61

Setting P = 2880 N and Q = 576 N gives: −

48 53 −

T

12 53

AB

12 13 −

T AB

T AC

3 13 19 53

48 61 +

T AC

AB

36

61 −

T

T

4 13

T AC

11 61

=0

Solving the resulting set of equations using conventional algorithms gives:

(3′)

TAB = 1431.00 N TAC = 1560.00 N TAD = 183.010 N TAB = 1431 N W TAC = 1560 N W TAD = 183.0 N W

PROBLEM 2.118 For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = −576 N. (Q is directed downward).

SOLUTION See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below: −

48 53

12

12

T AB

3

− AB

53

AC

+Q =0

(2)

=0

(3)

+ 2880 N = 0

(1′)

− 576 N = 0

(2′)

61

4

AB

53

(1)

36

+

T

+P=0

T

61

T

13

19

48

AC

13

T

T

T AC

13

11

61

Setting P = 2880 N and Q = −576 N gives: −

48 53 −

T

12 53

AB

12

AB

3

T

13 19 53

48

AC

13 −

T

T

+

36

T

61 −

AB

61

AC

T

T

4 13

T AC

11 61

=0

Solving the resulting set of equations using conventional algorithms gives:

(3′)

1249.29 N TAC = 490.31 N TAD = 1646.97 N TAB = 1249 N W TAC = 490 N W TAD = 1647 N W

PROBLEM 2.119 For the transmission tower of Probs. 2.111 and 2.112, determine the tension in each guy wire knowing that the tower exerts on the pin at A an upward vertical force of 1800 lb. PROBLEM 2.111 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AB is 840 lb, determine the vertical force P exerted by the tower on the pin at A.

SOLUTION See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: −

i:

4

AB

21 −

j:

+

T

20

5

k:

− AB

AB

21

AC

50

9

− AC

AC

59

T

50

37

=0

(1)

+P=0

(2)

=0

(3)

63 −

T

10 63

T

59 +

T

T

59

T

21

30

63

Substituting for P = 1800 lb in Equations (1), (2), and (3) above and solving the resulting set of equations using conventional algorithms gives: −

4

AB

21

20 21

T AB

50

+

T

59 5 21

AC

AB

T

9 59

10

=0

(1′)

+ 1800 lb = 0

(2′)

AC

63

+

T

50

T

59

T

30

63

T AC

37 63

=0

(3′)

= 973.64 lb TAC = 531.00 lb TAD = 532.64 lb

TAB = 974 lb W TAC = 531 lb W TAD = 533 lb W

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 cylinder is suspended from point D as shown.

SOLUTION Free-Body Diagram of Point D:

The forces applied at D are: TDA , TDB , TDC and W

where W = −180.0 lbj. To express the other forces in terms of the unit vectors i, j, k, we write JJJG DA = (18 in.) j + (22 in.)k DA = 28.425 in. JJJG DB = −(24 in.)i + (18 in.) j − (16 in.)k DB = 34.0 in. JJJG DC = (24 in.)i + (18 in.) j − (16 in.)k DC = 34.0 in.

PROBLEM 2.120 (Continued) =T

T

and

DA

T DB

T DC

Equilibrium Condition with

=T

λ

Da

JJJG DA

DA

Da

DB DB

DB

DC

DC

DA = (0.63324 j + 0.77397k )TDA JJJG DB =T λ =T DB = (−0.70588i + 0.52941j − 0.47059k )TDB JJJG DC =T λ =T DC

DC = (0.70588i + 0.52941j − 0.47059k )TDC

W = −Wj ΣF = 0: TDA + TDB + TDC − Wj = 0

Substituting the expressions obtained for TDA , TDB , and TDC and factoring i, j, and k: (−0.70588TDB + 0.70588TDC )i (0.63324TDA + 0.52941TDB + 0.52941TDC − W ) j (0.77397TDA − 0.47059TDB − 0.47059TDC )k Equating to zero the coefficients of i, j, k: −0.70588TDB + 0.70588TDC = 0

(1)

0.63324TDA + 0.52941TDB + 0.52941TDC − W = 0

(2)

0.77397TDA − 0.47059TDB − 0.47059TDC = 0

(3)

Substituting W = 180 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms gives, TDA = 119.7 lb W TDB = 98.4 lb W TDC = 98.4 lb W

PROBLEM 2.121 A container of weight W is suspended from ring A, to which cables AC and AE are attached. A force P is applied to the end F of a third cable that passes over a pulley at B and through ring A and that is attached to a support at D. Knowing that W = 1000 N, determine the magnitude of P. (Hint: The tension is the same in all portions of cable FBAD.)

SOLUTION The (vector) force in each cable can be written as the product of the (scalar) force and the unit vector along the cable. That is, with JJJG AB = −(0.78 m)i + (1.6 m) j + (0 m)k AB = (−0.78 m) 2 + (1.6 m) 2 + (0) 2 = 1.78 m T AB

and

= Tλ

JJJG AB

=T AB

AB

AB TAB = [−(0.78 m)i + (1.6 m) j + (0 m)k ] 1.78 m = TAB (−0.4382i + 0.8989j + 0k)

TAB JJJG AC = (0)i + (1.6 m) j + (1.2 m)k

AC = (0 m) 2 + (1.6 m) 2 + (1.2 m) 2 = 2 m JJJG T AC = AC [(0)i + (1.6 m) j + (1.2 m)k ] TAC = T λAC = TAC AC 2 m

and

TAC = TAC (0.8j + 0.6k) JJJG AD = (1.3 m)i + (1.6 m) j + (0.4 m)k AD = (1.3 m) 2 + (1.6 m) 2 + (0.4 m) 2 = 2.1 m JJJG AD T = AD [(1.3 m)i + (1.6 m) j + (0.4 m)k ] TAD = T λAD = TAD AD 2.1 m TAD = TAD (0.6190i + 0.7619j + 0.1905k )

PROBLEM 2.121 (Continued) JJJG AE = −(0.4 m)i + (1.6 m) j − (0.86 m)k

Finally,

AE = (−0.4 m) 2 + (1.6 m) 2 + (−0.86 m) 2 = 1.86 m JJJG AE T = Tλ = T AE

AE

AE

AE

TAE [−(0.4 m)i + (1.6 m) j − (0.86 m)k ] 1.86 m = TAE (−0.2151i + 0.8602 j − 0.4624k ) =

TAE With the weight of the container

W = −W j, at A we have: ΣF = 0: TAB + TAC + TAD − Wj = 0

Equating the factors of i, j, and k to zero, we obtain the following linear algebraic equations: −0.4382TAB + 0.6190TAD − 0.2151TAE = 0

(1)

0.8989TAB + 0.8TAC + 0.7619TAD + 0.8602TAE − W = 0

(2)

0.6TAC + 0.1905TAD − 0.4624TAE = 0

(3)

Knowing that W = 1000 N and that because of the pulley system at B TAB = TAD = P, where P is the externally applied (unknown) force, we can solve the system of linear Equations (1), (2) and (3) uniquely for P. P = 378 N W

PROBLEM 2.122 Knowing that the tension in cable AC of the system described in Problem 2.121 is 150 N, determine (a) the magnitude of the force P, (b) the weight W of the container. PROBLEM 2.121 A container of weight W is suspended from ring A, to which cables AC and AE are attached. A force P is applied to the end F of a third cable that passes over a pulley at B and through ring A and that is attached to a support at D. Knowing that W = 1000 N, determine the magnitude of P. (Hint: The tension is the same in all portions of cable FBAD.)

SOLUTION Here, as in Problem 2.121, the support of the container consists of the four cables AE, AC, AD, and AB, with the condition that the force in cables AB and AD is equal to the externally applied force P. Thus, with the condition TAB = TAD = P

and using the linear algebraic equations of Problem 2.131 with TAC = 150 N, we obtain (a)

P = 454 N W

(b)

W = 1202 N W

PROBLEM 2.123 Cable BAC passes through a frictionless ring A and is attached to fixed supports at B and C, while cables AD and AE are both tied to the ring and are attached, respectively, to supports at D and E. Knowing that a 200-lb vertical load P is applied to ring A, determine the tension in each of the three cables.

SOLUTION Free Body Diagram at A: Since TBAC = tension in cable BAC, it follows that TAB = TAC = TBAC

T AB

T AC

= TBAC

= TBAC

(−17.5 in.)i +(60 in.) j

λ AB

= TBAC

62.5 in. (60 in.)i +(25 in.)k

λ AC

= TBAC

65 in.

(80 in.)i +(60 in.) j

TAE = TAE λ AE = TAE

⎛ −17.5 = TBAC ⎜ ⎝ ⎛ 60

= TBAC ⎜ ⎝

65

62.5

i

60 + 62.5

j

⎞ ⎟ ⎠

j

+

25 ⎞ k 65 ⎟ ⎠

⎛4 3 ⎞ i j = TAD ⎜ + ⎟ 5 5 100 in. ⎝ ⎠ 4 3 (60 in.) j −(45 in.)k ⎛ ⎞ j k = TAE ⎜ − 75 in. 5 5 ⎟ ⎝ ⎠

PROBLEM 2.123 (Continued) Substituting into ΣFA = 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 62.5

From

j:

⎛ 60

BAC

+

⎜ ⎝ 62.5

From

k:

25 65

+

T

4

60 ⎞

BAC

5

3 + T

T

⎟ 65 ⎠

(1)

5

BAC

3 − T

T

=0

T

5

5

− 200 lb = 0

(2)

AE

=0

(3)

AE

Solving the system of linear equations using conventional algorithms gives: TBAC = 76.7 lb; TAD = 26.9 lb; TAE = 49.2 lb W

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 BAC and AD. PROBLEM 2.123 Cable BAC passes through a frictionless ring A and is attached to fixed supports at B and C, while cables AD and AE are both tied to the ring and are attached, respectively, to supports at D and E. Knowing that a 200-lb vertical load P is applied to ring A, determine the tension in each of the three cables.

SOLUTION Refer to the solution to Problem 2.123 for the figure and analysis leading to the following set of equilibrium equations, Equation (2) being modified to include Pj as an unknown quantity: −

17.5

BAC

62.5

⎛ 60

+

T

+

60 ⎞

5

⎟ 65 ⎠

25

3 − T

65

BAC

5

=0

T

(1)

3 + T

T

⎜ ⎝ 62.5

T

4

BAC

=0

5

5

−P=0

(2)

AE

(3)

AE

Substituting for TAE = 75 lb and solving simultaneously gives: (a)

P = 305 lb W

(b)

TBAC = 117.0 lb; TAD = 40.9 lb W

PROBLEM 2.125 Collars A and B are connected by a 525-mm-long wire and can slide freely on frictionless rods. If a force P = (341 N)j is applied to collar A, determine (a) the tension in the wire when y = 155 mm, (b) the magnitude of the force Q required to maintain the equilibrium of the system.

SOLUTION For both Problems 2.125 and 2.126:

Free-Body Diagrams of Collars: ( AB)2 = x2 + y 2 + z 2

Here

(0.525 m)2 = (0.20 m)2 + y 2 + z 2 y 2 + z 2 = 0.23563 m2

or Thus, when y given, z is determined, Now

λAB

JJJG AB = AB =

1

(0.20i − yj + zk )m

0.525 m = 0.38095i − 1.90476 yj + 1.90476zk Where y and z are in units of meters, m. From the F.B. Diagram of collar A:

ΣF = 0: N x i + N z k + Pj + TAB λ AB = 0

Setting the j coefficient to zero gives

P − (1.90476 y)TAB = 0 P = 341 N

With

TAB =

341 N 1.90476 y

Now, from the free body diagram of collar B:

ΣF = 0: N x i + N y j + Qk − TAB λAB = 0

Setting the k coefficient to zero gives

Q − TAB (1.90476 z) = 0

And using the above result for TAB , we have

Q =T z =

341 N

AB

(1.90476) y (1.90476z) =

(341 N)( z) y

PROBLEM 2.125 (Continued) Then from the specifications of the problem, y = 155 mm = 0.155 m z 2 = 0.23563 m2 â&#x2C6;&#x2019; (0.155 m) 2 z = 0.46 m and TAB =

(a)

341 N 0.155(1.90476)

= 1155.00 N

TAB = 1155 N W

or and Q=

(b)

341 N(0.46 m)(0.866) (0.155 m)

= (1012.00 N) or

Q = 1012 N W

PROBLEM 2.126 Solve Problem 2.125 assuming that y = 275 mm. PROBLEM 2.125 Collars A and B are connected by a 525-mm-long wire and can slide freely on frictionless rods. If a force P = (341 N)j is applied to collar A, determine (a) the tension in the wire when y = 155 mm, (b) the magnitude of the force Q required to maintain the equilibrium of the system.

SOLUTION From the analysis of Problem 2.125, particularly the results:

With y = 275 mm = 0.275 m, we obtain:

y 2 + z 2 = 0.23563 m 2 341 N TAB = 1.90476 y 341 N Q= z y z 2 = 0.23563 m2 â&#x2C6;&#x2019; (0.275 m)2 z = 0.40 m

and (a)

=

T AB

341 N

= 651.00

(1.90476)(0.275 m) TAB = 651 N W

or and Q=

(b) or

341 N(0.40 m) (0.275 m) Q = 496 N W

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 by trigonometry the magnitude and direction of the resultant of the forces applied to the bracket by members A and B.

SOLUTION Using the force triangle and the laws of cosines and sines, we have

γ = 180° − (40° + 20°) = 120°

Then

R 2 = (15 kN)2 + (10 kN) 2 − 2(15 kN)(10 kN) cos120° = 475 kN 2 R = 21.794 kN

and

10 kN

=

sin α sin α =

21.794 kN sin120° ⎛ 10 kN

sin120°

⎜ ⎟ ⎝ 21.794 kN ⎠ = 0.39737 α = 23.414 Hence:

φ = α + 50° = 73.414

R = 21.8 kN

73.4° W

PROBLEM 2.128 Determine the x and y components of each of the forces shown.

SOLUTION Compute the following distances: OA = (24 in.) 2 + (45 in.) 2 = 51.0 in. OB = (28 in.) 2 + (45 in.) 2 = 53.0 in. OC = (40 in.) 2 + (30 in.) 2 = 50.0 in. 102-lb Force:

F = −102 lb x

F = +102 lb y

106-lb Force:

51.0 in. 45 in.

F = −48.0 lb W x

F = +90.0 lb W y

51.0 in.

Fx = +106 lb

28 in. 53.0 in.

Fx = +56.0 lb W

F = +106 lb

45 in.

F = +90.0 lb W

y

200-lb Force:

24 in.

F = −200 lb x

F = −200 lb y

53.0 in. 40 in. 50.0 in. 30 in. 50.0 in.

y

F = −160.0 lb W x

F = −120.0 lb W y

PROBLEM 2.129 A hoist trolley is subjected to the three 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.

SOLUTION Rx =

ΣFx = P + (200 lb) sin 40° − (400 lb) cos 40°

Rx = P − 177.860 lb Ry =

ΣFy = (200 lb) cos 40° + (400 lb) sin 40°

Ry = 410.32 lb (a)

(1)

(2)

For R to be vertical, we must have Rx = 0. Set

Rx = 0 in Eq. (1) 0 = P − 177.860 lb P = 177.860 lb

(b)

P = 177.9 lb W

Since R is to be vertical: R = Ry = 410 lb

R = 410 lb W

PROBLEM 2.130 Knowing that α = 55° and that boom AC exerts on pin C a force directed along line AC, determine (a) the magnitude of that force, (b) the tension in cable BC.

SOLUTION Free-Body Diagram

Law of sines:

FAC T = BC = 300 lb sin 35° sin 50° sin 95° 300 lb

(a) FAC = (b)

Force Triangle

TBC =

sin 35°

FAC = 172.7 lb W

300 lb sin 50° sin 95°

TBC = 231 lb W

sin 95°

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 AC, (b) in cable BC.

SOLUTION Free Body: C

(a)

ΣFx = 0: −

(b)

ΣF = 0: y

12 4 T AC + (360 N) = 0 13 5

5 13

(312 N) + T

+ BC

3

TAC = 312 N W

(360 N) − 480 N = 0

5

TBC = 480 N − 120 N − 216 N

TBC = 144.0 N W

PROBLEM 2.132 Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension 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.

SOLUTION Free-Body Diagram: C

Force Triangle

Force triangle is isosceles with 2β = 180° − 85°

β = 47.5° P = 2(800 N)cos 47.5° = 1081 N

(a) Since P > 0, the solution is correct. (b)

P = 1081 N W

α = 180° − 50° − 47.5° = 82.5°

α = 82.5° W

PROBLEM 2.133 The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in wire AC is 120 lb, determine (a) the components of the force exerted by this wire on the pole, (b) the angles θx, θy, and θz that the force forms with the coordinate axes.

SOLUTION (a)

Fx = (120 lb) cos 60° cos 20° Fx = 56.382 lb

Fx = +56.4 lb W

Fy = −(120 lb) sin 60° Fy = −103.923 lb

Fy = −103.9 lb W

Fz = −(120 lb) cos 60° sin 20° Fz = −20.521 lb (b)

cos θ x =

Fx 56.382 lb = F 120 lb

cos θ y =

Fy

cos θ z =

Fz −20.52 lb = F 120 lb

F

=

−103.923 lb 120 lb

Fz = −20.5 lb W

θ x = 62.0° W θ y = 150.0° W θ z = 99.8° W

PROBLEM 2.134 Knowing that the tension in cable AC is 2130 N, determine the components of the force exerted on the plate at C.

SOLUTION JJJG CA = −(900 mm)i + (600 mm) j − (920 mm)k CA = (900 mm)2 + (600 mm) 2 + (920 mm) 2

TCA

TCA

= 1420 mm = TCA λ CA JJJG CA = TCA CA 2130 N = [−(900 mm)i + (600 mm) j − (920 mm)k ] 1420 mm = −(1350 N)i + (900 N) j − (1380 N)k (TCA ) x = −1350 N, (TCA ) y = 900 N, (TCA ) z = −1380 N W

PROBLEM 2.135 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 600 N and Q = 450 N.

SOLUTION P = (600 N)[sin 40° sin 25°i + cos 40°j + sin 40° cos 25°k ] = (162.992 N)i + (459.63 N) j + (349.54 N)k Q = (450 N)[cos 55° cos 30°i + sin 55°j − cos 55° sin 30°k ] = (223.53 N)i + (368.62 N) j − (129.055 N)k R =P+Q = (386.52 N)i + (828.25 N) j + (220.49 N)k R = (386.52 N) 2 + (828.25 N) 2 + (220.49 N) 2 = 940.22 N cos θ x =

Rx 386.52 N = R 940.22 N Ry

cos

θy = cos θ z =

R

R = 940 N W

θ x = 65.7° W

828.25 N =

940.22 N

Rz 220.49 N = R 940.22 N

θ y = 28.2° W θ z = 76.4° W

PROBLEM 2.136 A container of weight W is suspended from ring A. Cable BAC passes through the ring and is attached to fixed 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 tension is the same in both portions of cable BAC.)

SOLUTION TAB = T λ AB JJJK AB =T AB (−130 mm)i + (400 mm) j + (160 mm)k =T 450 mm 40 16 ⎞ ⎛ 13 =T − i+ j+ k ⎜ ⎟ ⎝ 45 45 45 ⎠

Free-Body A:

TAC = Tλ AC JJJK AC =T AC (−150 mm)i + (400 mm) j + (−240 mm)k =T 490 mm 40 24 ⎞ ⎛ 15 = T ⎜− i + j− k ⎟ 49 49 ⎠ ⎝ 49 ΣF = 0: TAB + TAC + Q + P + W = 0 Setting coefficients of i, j, k equal to zero: i: −

j: +

13

T−

T +P=0

0.59501T = P

(1)

T −W = 0

1.70521T = W

(2)

0.134240 T = Q

(3)

45

49

40

40

T+

45 k: +

15

16

49 T−

24

T +Q =0

45

49

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 W

0.134240(220.50 N) = Q

Q = 29.6 N W

PROBLEM 2.137 Collars A and B are connected by a 25-in.-long wire and can slide freely on frictionless rods. If a 60-lb force Q is applied to collar B as shown, determine (a) the tension in the wire when x = 9 in., (b) the corresponding magnitude of the force P required to maintain the equilibrium of the system.

SOLUTION Free-Body Diagrams of Collars:

A:

B:

λAB

JJJG AB − xi − (20 in.) j + zk = = AB 25 in.

ΣF = 0: Pi + N y j + N z k + TAB λ AB = 0

Collar A:

Substitute for λAB and set coefficient of i equal to zero: P−

Collar B:

TAB x 25 in.

=0

(1)

ΣF = 0: (60 lb)k + N x′ i + N ′y j − TAB λ AB = 0

Substitute for λ AB and set coefficient of k equal to zero: 60 lb − x = 9 in.

(a)

TAB z 25 in.

=0

(2)

(9 in.) 2 + (20 in.) 2 + z 2 = (25 in.) 2 z = 12 in.

From Eq. (2):

(b)

From Eq. (1):

60 lb − TAB (12 in.) 25 in. P=

(125.0 lb)(9 in.)

TAB = 125.0 lb W 25 in.

P = 45.0 lb W

PROBLEM 2.138 Collars A and B are connected by a 25-in.-long wire and can slide freely on frictionless rods. Determine the distances x and z for which the equilibrium of the system is maintained when P = 120 lb and Q = 60 lb.

SOLUTION See Problem 2.137 for the diagrams and analysis leading to Equations (1) and (2) below: P= 60 lb −

TAB x =0 25 in. TAB z 25 in.

=0

(1) (2)

For P = 120 lb, Eq. (1) yields

TAB x = (25 in.)(20 lb)

(1′)

From Eq. (2):

TAB z = (25 in.)(60 lb)

(2′)

x =2 z

Dividing Eq. (1′) by (2′), Now write

x 2 + z 2 + (20 in.)2 = (25 in.) 2

(3) (4)

Solving (3) and (4) simultaneously, 4 z 2 + z 2 + 400 = 625 z 2 = 45 z = 6.7082 in. From Eq. (3):

x = 2 z = 2(6.7082 in.) = 13.4164 in. x = 13.42 in., z = 6.71 in. W

PROBLEM 2F1 Two cables are tied together at C and loaded as shown. Draw the free-body diagram needed to determine the tension in AC and BC.

SOLUTION Free-Body Diagram of Point C:

W = (1600 kg)(9.81 m/s 2 ) W = 15.6960(103 ) N W =15.696 kN

PROBLEM 2.F2 Two forces of magnitude TA = 8 kips and TB = 15 kips are applied as shown to a welded connection. Knowing that the connection is in equilibrium, draw the free-body diagram

needed to determine the magnitudes of the forces TC and TD.

SOLUTION Free-Body Diagram of Point E:

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-body diagram needed to determine the value of h for which the system is in equilibrium.

SOLUTION Free-Body 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-body diagrams needed to determine the weight of the skier in chair F.

SOLUTION Free-Body Diagram of Point B: WE = 250 N + 765 N = 1015 N 8.25 θ AB = tan −1 = 30.510° 14 10 = 22.620° θ BC = tan −1 24 Use this free body to determine TAB and TBC.

Free-Body Diagram of Point C:

θ CD = tan −1

1.1 6

= 10.3889°

Use this free body to determine TCD and WF. Then weight of skier WS is found by

WS = WF − 250 N W

PROBLEM 2.F5 Three cables are used to tether a balloon as shown. Knowing that the tension in cable AC is 444 N, draw the free-body diagram needed to determine the vertical force P exerted by the balloon at A.

SOLUTION Free-Body Diagram of Point A:

PROBLEM 2.F6 A container of mass m = 120 kg is supported by three cables as shown. Draw the free-body diagram needed to determine the tension in each cable

SOLUTION Free-Body Diagram of Point A:

W = (120 kg)(9.81 m/s 2 ) = 1177.2 N

PROBLEM 2.F7 A 150-lb cylinder is supported by two cables AC and BC that are attached to the top of vertical posts. A horizontal force P, which is perpendicular to the plane containing the posts, holds the cylinder in the position shown. Draw the free-body diagram needed to determine the magnitude of P and the force in each cable.

SOLUTION Free-Body Diagram of Point C:

PROBLEM 2.F8 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. Knowing that the tension in wire AB is 630 lb, draw the free-body diagram needed to determine the vertical force P exerted by the tower on the pin at A.

SOLUTION Free-Body Diagram of point A: