PROBLEM 2.1
Two forces are applied to an eye bolt fastened to a beam. Determine graphically the magnitude and direction of their resultant using ( a) the parallelogram law, ( b) the triangle rule.
SOLUTION
(a)
(b)
We measure: measure:
R
8.4 kN 19 R
1
8.4 8.4 kN
19
PROBLEM 2.2
The cable stays AB and AD help support pole AC . Knowing that the tension is 500 N in AB and 160 N in AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, ( b) the triangle rule.
SOLUTION
We measure:
51.3,
575 N,
59
(a)
(b)
We measure:
R
67 R
2
575 N
67
PROBLEM 2.3
Two forces P and Q are applied as shown at point A of a hook support. Knowing that P 15 lb and Q 25 lb, determine graphically the magnitude and direction of their resultant using ( a) the parallelogram law, (b) the triangle rule.
SOLUTION
(a)
(b)
R
We measure:
37 lb,
76 R 37 lb
3
76
PROBLEM 2.4
Two forces P and Q are applied as shown at point A of a hook support. Knowing that P 45 lb and Q 15 lb, determine graphically the magnitude and direction of their resultant using ( a) the parallelogram law, (b) the triangle rule.
SOLUTION
(a)
(b)
We measure:
R
61.5 lb,
86.5 R 61.5 61.5 lb
4
86.5
PROBLEM 2.5
Two control rods are attached at A to lever AB. Using trigonometry and knowing that the force in the left-hand rod is F 1 120 N, determine (a) the required force F 2 in the right-hand rod if the resultant R of the forces exerted by the rods on the lever is to be vertical, ( b) the corresponding magnitude of R .
SOLUTION
Graphically, by the triangle law We measure:
F 2
108 N
R
77 N
By trigonometry: Law of Sines F2
sin
90 28
R
sin 38
62,
180
120 sin
62 38
80
Then: F2
sin 62
R
sin 38
120 N sin 80 or (a) (b)
5
F 2
107. 107.6 6N
R
75. 75.0 0 N
PROBLEM 2.6
Two control rods are attached at A to lever AB. Using trigonometry and knowing that the force in the right-hand rod is F 2 80 N, determine (a) the required force F 1 in the left-hand rod if the resultant R of the forces exerted by the rods on the lever is to be vertical, ( b) the corresponding magnitude of R .
SOLUTION
Using the Law of Sines F1
sin
90
10
R
sin 38
80,
180
80 sin
80
38
62
Then: F1
sin 80
R
sin 38
80 N sin 62 or (a) (b)
6
F 1 R
89.2 89.2 N 55. 55.8 8N
PROBLEM 2.7
The 50-lb force is to be resolved into components components along along lines a- a and b-b. (a) Using trigonometry, determine the angle knowing that the component along a-a is 35 lb. (b) What is the corresponding value of the the comp compon onen entt alon along g b-b ?
SOLUTION
Using the triangle rule and the Law of Sines sin
(a)
35 lb
sin 40
50 lb
sin
0.44995
26.74
Then:
40
180 113.3
(b) Using the Law of Sines: F bb
sin
50 lb sin 40 F bb
7
71.5 71.5 lb
PROBLEM 2.8
The 50-lb force is to be resolved into components components along along lines a- a and b-b. (a) Using trigonometry, determine the angle knowing that the component along b-b is 30 lb. (b) What is the corresponding value of the the comp compon onen entt alon along g a-a ?
SOLUTION
Using the triangle rule and the Law of Sines (a)
sin 30 lb
sin 40 50 lb
sin 0.3857 22.7 (b)
40 180 117.31 F aa
sin F aa
50 lb sin 40
sin 50 lb sin 40 sin F aa
8
69.1lb
PROBLEM 2.9
To steady a sign as it is being lowered, two cables are attached to the sign at A. Using trigonometry and knowing that 25, determine (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 Have:
Then:
P
sin 35
9
180
120
R
sin 120
35 25
360 N sin 25 or (a)
P
489 489 N
(b)
R
738 738 N
PROBLEM 2.10
To steady a sign as it is being lowered, two cables are attached to the sign at A. Using trigonometry and knowing that the magnitude of P is 300 N, determine ( a) the required angle if the resultant R of the two forces applied at A is to be verti vertical, cal, (b) the corresponding magnitude of R .
SOLUTION
Using the triangle rule and the Law of Sines (a) Have:
360 N sin sin
300 N sin 35 0.68829
(b)
Then:
43.5
35 43.5
180
101.5
R
sin 101.5
300 N sin 35 35 or
10
R
513 513 N
PROBLEM 2.11
Two forces are applied as shown to a hook support. Using trigonometry and knowing that the magnitude of P is 14 lb, determine ( a) the required angle 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 the Law of Sines
(a) Have:
20 lb sin
sin
14 lb sin 30 0.71428
(b)
Then:
180
104.4
R
sin 104.4
30 45.6
14 lb sin 30 30 R
11
45.6
27.1 lb
PROBLEM 2.12
For the hook support of Problem 2.3, using trigonometry and knowing that the magnitude of P is 25 lb, determine (a ( a) the required magnitude of the force Q if the resultant R of the two forces applied at A is to be vertical, (b (b) the corresponding magnitude of R . Problem 2.3: Two forces P and Q are applied as shown at point A of a
hook support. Knowing that P 15 lb and Q 25 lb, determine graphically the magnitude and direction of their resultant using ( a) the parallelogram law, (b ( b) the triangle rule.
SOLUTION
Using the triangle rule and the Law of Sines
(a) Have:
Q sin 15
25 lb sin 30 Q 12.94 lb
180 15 30
(b)
135 Thus:
R sin 135
25 lb sin 30
sin135 35.36 lb sin30
R 25 lb
R 35.4 lb
12
PROBLEM 2.13
For the hook support of Problem 2.11, determine, using 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 . Two forces are applied as shown to a hook support. Using trigonometry and knowing that the magnitude of P is 14 lb, determine (a) the required angle if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R .
Problem 2.11:
SOLUTION
(a) The smallest force P will be perpendicular to
P
(b)
R , that
is, vertical
lb sin30 20 lb
10 lb
R
os30 20 lb cos30
17.32 lb
13
P
R
10 lb
17.32 lb
PROBLEM 2.14
As shown in Figure P2.9, two cables are attached to a sign at A to steady the sign as it is being lowered. Using trigonometry, determine ( 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
We observe that force
Then:
P
is minimum minimum when
is 90, that is,
(a )
P
P
is horizontal
360 N sin35 or
And:
(b )
R
P
206 N
360 N cos35 or
14
R
295 N
PROBLEM 2.15
For the hook support of Problem 2.11, determine, determine, using trigonometry, the magnitude and direction of the resultant of the two forces applied to the support knowing that P 10 lb and 40. Two forces are applied as shown to a hook support. Using trigonometry and knowing that the magnitude of P is 14 lb, determine (a) the required angle if if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R .
Problem 2.11:
SOLUTION
Using the force triangle and the Law of Cosines
R
2
2
2
10 lb 20 lb 2 10 lb 20 lb cos110 100 400 400 0.342 lb 2
636.8 lb 2 R
25.23 lb
Using now the Law of Sines 10 lb sin
25.23 lb sin 110
10 lb sin 110 25.23 lb
sin
0.3724 So:
21.87
Angle of inclination of R, is then such that: 30 8.13 Hence:
R
15
25.2 lb
8.13
PROBLEM 2.16
Solve Problem 2.1 using trigonometry Two forces are applied to an eye bolt fastened to a beam. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, ( b) the triangle rule.
Problem 2.1:
SOLUTION
Using the force triangle, the Law of Cosines and the Law of Sines
180 50 25
We have:
105 Then:
R
2
2
2
4.5 kN 6 kN 2 4.5 kN 6 kN cos105
70.226 kN 2 or Now:
R
8.3801 kN
8.3801 kN sin 105
6 kN sin
6 kN sin 105 8.3801 kN
sin
0.6916 43.756 R
16
8.38 kN
18.76
PROBLEM 2.17
Solve Problem 2.2 using trigonometry The cable stays AB and AD help support pole AC . Knowing that the tension is 500 N in AB and 160 N in AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, ( b) the triangle rule.
Problem 2.2:
SOLUTION
From the geometry of the problem: 2
tan 1 tan 1
2.5 1.5 2.5
38.66 30.96
180 38.66 30.96 110.38
Now:
And, using the Law of Cosines: R
2
2
2
500 N 160 N 2 500 N 160 N cos110.38
331319 N 2 R
575.6 N
Using the Law of Sines: 160 N sin
575.6 N sin 110.38
160 N sin 110.38 575.6 N
sin
0.2606 15.1 90 66.44 R
17
576 N
66.4
PROBLEM 2.18
Solve Problem 2.3 using trigonometry Problem 2.3: Two forces P and Q are applied as shown at point A of a
hook support. Knowing that P 15 lb and Q 25 lb, determine graphically the magnitude and direction of their resultant using ( a) the parallelogram law, (b ( b) the triangle rule.
SOLUTION
Using the force triangle and the Laws of Cosines and Sines We have: 180 15 30
135 Then:
2
2
R 2 15 lb 25 lb 2 15 lb 25 lb cos135
1380.3 lb 2 R 37.15 lb
or and
25 lb sin
37.15 lb sin 135
25 lb sin 135 37.15 lb
sin
0.4758 28.41 Then:
75 180 76.59 R 37.2 lb
18
76.6
PROBLEM 2.19
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 30 kN in member A and 20 kN in member B, determine, using 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 180 45 25 110
We have: Then:
R
2
2
30 kN 20 kN 2 30 kN 20 kN cos110
2
1710.4 kN 2 R
41.357 kN
and 20 kN sin
41.357 kN sin 110
20 kN sin 110 41.357 kN
sin
0.4544 27.028 45 72.028
Hence:
R
19
41.4 kN
72.0
PROBLEM 2.20
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 20 kN in member A and 30 kN in member B, determine, using 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 180 45 25 110
We have: Then:
R
2
2
2
30 kN 20 kN 2 30 kN 20 kN cos110
1710.4 kN 2 R
41.357 kN
and 30 kN sin
41.357 kN sin 110
30 kN sin 110 41.357 kN
sin
0.6816 42.97 Finally:
45 87.97 R
20
41.4 kN
88.0
PROBLEM 2.21
Determine the x the x and y and y components of each of the forces shown.
SOLUTION
20 kN Force: F x
40, 20 kN cos 40
F x
15.3 15.32 2 kN
F y
40, 20 kN sin 40
F y
12.8 12.86 6 kN
F x
70, 30 kN cos 70
F y
70, 30 kN sin 70
F x
20, 42 kN cos 20
F x
39.5 kN 39.5
F y
20, 42 kN sin 20
F y
30 kN Force: F x
10.26 10.26 kN
F y
28.2 28.2 kN
42 kN Force:
21
14.3 14.36 6 kN
PROBLEM 2.22
Determine the x the x and y and y components of each of the forces shown.
SOLUTION
40 lb Force: F x
50, 40 lb sin 50
F x
30.6 30.6 lb
F y
50, 40 lb cos 50
F y
25.7 25.7 lb
F x
60, 60 lb cos 60
F y
60, 60 lb sin 60
F x
25, 80 lb cos 25
F x
72.5 72.5 lb
F y
25, 80 lb sin 25
F y
33.8 33.8 lb
60 lb Force: F x F y
30.0 30.0 lb
52.0 lb 52.0
80 lb Force:
22
PROBLEM 2.23
Determine the x the x and y and y components of each of the forces shown.
SOLUTION
We compute the following distances: OA
48
2
OB
56
2
OC
80
F x
102 lb
F y
102 lb
F x
212 lb
F y
212 lb
F x
400 lb
F y
400 lb
2
90
2
90
2
60
2
102 in.
106 in.
100 in.
Then: 204 lb Force: Force: 48 102 90 102
,
,
F x
48.0 48.0 lb
F y
90.0 90.0 lb
212 lb Force: Force: 56 106 90 106
,
F x
112.0 112.0 lb
,
F y
180.0 180.0 lb
,
F x
,
F y
400 lb Force: Force:
23
80 100 60 100
320 320 lb
240 240 lb
PROBLEM 2.24
Determine the x and y components of each of the forces shown.
SOLUTION
We compute the following distances: OA
2 2 70 240
OB
2 2 210 200
OC
2 2 120 225
250 mm
290 mm
255 mm
500 N Force:
70 250
F x 140.0 N
240 250
F y 480 N
210 290
F x 315 N
200 290
F y 300 N
120 255
F x 240 N
225 255
F y 450 N
F x 500 N F y 500 N 435 N Force: F x 435 N
F y 435 N 510 N Force: F x 510 N F y 510 N
24
PROBLEM 2.25
While emptying a wheelbarrow, a gardener exerts on each handle AB a force P directed along line CD. CD. Knowing that P must have a 135-N horizontal component, determine ( a) the magnitude of the force P, (b ( b) its vertical component.
SOLUTION
(a)
P
P x cos40 135 N cos40 or P 176.2 176.2 N
(b)
P y
xPtan 40
Psin 4 0
4 0 135 N tan 40 or P y
25
113.3 113.3 N
PROBLEM 2.26
Member BD Member BD exerts on member ABC member ABC a force P directed along line BD. Knowing that P must have a 960-N vertical component, determine (a ( a) the magnitude of the force P, (b (b) its horizontal component.
SOLUTION
(a)
P
P y sin35 960 N sin35 or P 1674 1674 N
(b)
P x
P y tan35 960 N tan35 or P x
26
1371 1371 N
PROBLEM 2.27
Member CB of the vise shown exerts on block B block B a force P directed along line CB. CB. Knowing that P must have a 260-lb horizontal component, determine (a ( a) the magnitude of the force P, (b (b) its vertical component.
SOLUTION
We note: CB exerts force P on B on B along CB, CB, and the horizontal component of P is P x
260 260 lb. lb.
Then: (a)
P x P
(b)
P sin50
P x sin50 260 260 lb sin50 339.4 339.4 lb
P x
P y
P 339 339 lb
P y tan50 P x tan tan 50 260 260 lb tan50 218.2 218.2 lb
27
P y
218 218 lb
PROBLEM 2.28
Activator rod AB exerts on crank BCD crank BCD a force P directed along line AB. Knowing that P must have a 25-lb component perpendicular to arm BC of BC of the crank, determine (a ( a) the magnitude of the force P, (b) its component along line BC .
SOLUTION
Using the x the x and y and y axes shown. (a)
P y
Then:
P
25 lb 25 P y
sin75 25 lb sin75 or P 25.9 25.9 lb
(b )
P x
P y tan75 25 lb tan tan 75 or P x
28
6.70 6.70 lb
PROBLEM 2.29
The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P has a 450-N component along line AC , determine (a) the magnitude of the force P, (b) its component in a direction perpendicular to AC .
SOLUTION
Note that the force exerted by BD on the pole is directed along is 450 N.
BD,
and the component of P along AC
Then: (a)
P
450 N cos35
549. 549.3 3N P
(b)
P x
35 450 N tan 35
315. 315.1 1N P x
29
549 N
315 N
PROBLEM 2.30
The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P has a 200-N perpendicular to the pole AC , determine (a ( a) the magnitude of the force P, (b) its component along line AC line AC .
SOLUTION
(a)
P
(b )
P y
P x sin38 200 N sin38 324. 324.8 8N
or P 325 325 N
P x tan38 200 N tan38 255. 255.98 98 N or P y
30
256 256 N
PROBLEM 2.31
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
From Problem 2.24: F500
R
140 N i 480 N j
F425
315 N i 300 N j
F510
240 N i 450 N j
F
415 N i 330 N j
Then:
R
tan 1
415 N
2
Thus:
330 415
38.5
330 N
2
530.2 N R 530 N
31
38.5
PROBLEM 2.32
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
From Problem 2.21:
R
15.32 kN i 12.86 kN j
F20
F30
10.26 kN i 28.2 kN j
F42
39.5 kN i 14.36 kN j
F
34.44 kN i 55.42 kN j
Then:
R
tan 1
55.42 kN
2
55.42 34.44
58.1
34.44 N
2
65.2 kN R
32
65.2 kN
58.2
PROBLEM 2.33
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
The components of the forces were determined in 2.23. Force
x comp. (lb)
40 lb
30.6
60 lb
30
80 lb
72.5 72.5 R x
R
R
y comp. (lb) 25.7 51.9 51.96 6
33.8 33.8
71.9
R y
43.86
R x i
71.9 lb i 43.86 lb j
Ry j
tan
71.9 lb
2
43.86 71.9 31.38
43.86 lb
2
84.23 lb R 84.2 lb
33
31.4
PROBLEM 2.34
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
The components of the forces were determined in Problem 2.23. F204
F212
F400
48.0 lb i 90.0 lb j
112.0 lb i 180.0 lb j
320 lb i 240 lb j
Thus R R
R x
R y
256 lb i 30.0 lb j
Now: tan
tan 1
30.0 256
30.0 256
6.68
30.0 lb
and R
256 lb
2
2
257.75 lb R 258 lb
34
6.68
PROBLEM 2.35
Knowing that shown.
35, determine the resultant of the three forces
SOLUTION
300-N Force: F x
20 300 N cos 20
F y
20 102.6 N 300 N sin 20
F x
55 400 N cos 55
F y
55 327.7 N 400 N sin 55
F x
35 600 N cos 35
491.5 N
35 600 N sin 35
344.1 N
281.9 N
400-N Force:
229.4 N
600-N Force:
F y
and R x
F x
R y R
1002.8 N
F y
1002.8 N
2
86.2 N
86.2 N
2
1006.5 N
Further: tan
tan 1
86.2 1002.8
86.2 1002.8
4.91 R
35
1007 N
4.91
PROBLEM 2.36
Knowing that shown.
65, determine the resultant of the three forces
SOLUTION
300-N Force: F x
20 300 N cos 20
F y
20 102.6 N 300 N sin 20
F x
400 N cos 85 34.9 N
F y
85 398.5 N 400 N sin 85
F x
600 N cos 5 597.7 N
281.9 N
400-N Force:
600-N Force:
F y
600 N sin 5
52.3
N
and
R
R x
F x
914.5 N
R y
F y
448.8 N
914.5 N
2
448.8 N
2
1018.7 N
Further: tan
tan 1
448.8 914.5
448.8 914.5
26.1 R
36
1019 N
26.1
PROBLEM 2.37
Knowing that the tension in cable BC is BC is 145 lb, determine the resultant of the three forces exerted at point B point B of beam AB. beam AB.
SOLUTION
Cable BC Force: Force: F x
145 lb 145 lb
84
105
116 80
F y
F x
100 lb
F y
100 lb
100 lb
116
lb
100-lb Force: Force: 3 5 4 5
60
lb
80
lb
156-lb Force: Force: F x F y
156 lb
12 13
156 lb
5 13
144 lb
60
lb
and Rx R
Fx 21 lb,
21 lb
2
Ry
Fy 40
40 lb
2
lb
45.177 lb
Further: tan
Thus:
tan 1
40 21
40 21
62.3 R
37
45.2 lb
62.3
PROBLEM 2.38
Knowin Knowing g that that shown.
50, determine the resultant of the three forces
SOLUTION
The resultant force R has the x- and y-components:
140 lb cos 50 60 lb cos 85 160 lb cos 50
R x
F x
R x
7.6264
R y
F y
R y
lb
and 50 60 lb sin 85 160 lb sin 50 50 140 lb sin 50
289.59 lb
Further: tan
Thus:
tan 1
290 7.6
290 7.6
88.5 R
38
290 lb
88.5
PROBLEM 2.39
Determine (a ( a) the required value of if the resultant of the three forces shown is to be vertical, (b (b) the corresponding magnitude of the resultant.
SOLUTION
For an arbitrary angle
, we have:
R x F x 140 lb cos 60 lb cos 35 160 lb cos (a) So, for R R to be vertical: R x F x 140 lb cos 60 lb cos 35 160 lb cos 0 Expanding, c os cos 35 sin sin 35 0 cos 3 co Then: tan
cos35
1 3
sin35
or
cos35 tan 1 sin35
1 3
40.265
40.3
(b) Now: R R y Fy 140 lb sin 40 40.265 60 lb sin 75 75.265 160 lb sin 40 40.265 R R 252 lb
39
PROBLEM 2.40
For the beam of Problem 2.37, determine (a (a) the required tension in cable BC if the resultant of the three forces exerted at point B is to be vertical, (b) the corresponding magnitude of the resultant. Knowing that the tension in cable BC is BC is 145 lb, determine the resultant of the three forces exerted at point B of beam AB. beam AB.
Problem 2.37:
SOLUTION
We have: R x
F x
84 116 R
or
T BC
12
13
3
156 lb 100 lb 5
x0.724T
BC84 lb
and R R
y
F y
80 116
y0.6897T
T BC BC140
5
4
13
156 lb 100 lb 5
lb
(a) So, for R R to be vertical, R
x0.724T
BC84 lb
0 T BC
116.0 lb
(b) Using T BC R
R y
116.0 lb
0.6897 116.0 lb
140
lb
60
lb R
40
R
60.0 lb
PROBLEM 2.41
Boom AB Boom AB is held in the position shown by three cables. Knowing that the tensions in cables AC and AD are 4 kN and 5.2 kN, respectively, determine (a ( a) the tension in cable AE if the resultant of the tensions exerted at point A of the boom must be directed along AB, AB, (b) the corresponding magnitude magnitude of the resultant.
SOLUTION
Choose x-axis along bar AB bar AB.. Then (a) Require R y
F 0 :y
T AE
or
(b)
4 kN cos 25 5.2 kN sin 35
R
sTin 6A5E
0
7.2909 kN T AE
7.29 kN
R
9.03 kN
F x
25 5.2 kN cos 35 7.2909 kN cos 65 4 kN sin 25
9.03
kN
41
PROBLEM 2.42
For the block of Problems 2.35 and 2.36, determine (a (a) the required value of of the resultant of the three forces shown is to be parallel to the incline, (b (b) the corresponding magnitude of the resultant. Problem 2.35:
Knowin Knowing g that that three forces shown.
35, determine the resultant of the
Knowin Knowing g that that three forces shown.
65, determine the resultant of the
Problem 2.36:
SOLUTION
Selecting the x axis along aa, we write
(a) Setting R y
400 N cos
R x
F x
300 N
R y
F y
400 N sin
600 N sin
600 N cos
(1)
(2)
0 in Equation (2):
Thus
tan
600 400
1. 5
56.3
(b) Substituting for in Equation (1): 56.3 400 N cos 56.3 600 N sin 56
R x
300 N
R x
1021.1 N
R
42
R x
1021 N
PROBLEM 2.43
Two cables are tied together at C and are loaded as shown. Determine the tension (a (a) in cable AC , (b (b) in cable BC.
SOLUTION
Free-Body Diagram
From the geometry, we calculate the distances: AC
2 2 16 in. 12 in.
20 in.
BC
2 2 20 in. 21 in.
29 in.
Then, from the Free Body Diagram of point C :
Fx 0:
29
T
and
F y 0 : 12 20
T
21
T AC
20
BC
or
or
16
12 20
4
5
AC
29 21 21
4 5
C 0 T BC B
29
T
AC
T AC
20 29 29
21
20 29
T BC 60 0 lb 0
T A6C00 lb 0
T AC 440.56 lb
Hence: (a)
T AC 441 lb
(b)
T BC 487 lb
43
PROBLEM 2.44
Knowing that rope BC .
25, determine the tension ( a) in cable
AC ,
(b) in
SOLUTION
Free-Body Diagram
Force Triangle
Law of Sines: T
AC
sin 115 (a)
T AC
(b)
T BC
5 kN sin60 5 kN sin60
T
BC
sin 5
5 kN sin 60
sin 115
sin 5
0.503 kN
5.23 kN
44
T AC
T BC
5.23 kN
0.503 kN
PROBLEM 2.45
Knowing that 50 and that boom AC exerts on pin C a force directed long line AC , determine ( a) the magnitude of that force, ( b) the tension in cable BC .
SOLUTION
Free-Body Diagram
Force Triangle
Law of Sines: F
AC
sin 25 (a)
F AC
(b)
T BC
400 lb sin95 400 sin95
T
BC
sin 60
400 lb sin 95
sin 25 25
169.69 lb
sin 60 60
347.73 lb
45
F AC
T BC
169.7 lb
348 lb
PROBLEM 2.46
Two cables are tied together at C and are loaded as shown. Knowing that 30, determine the tension ( a) in cable AC , (b) in cable BC .
SOLUTION
Free-Body Diagram
Force Triangle
Law of Sines: T
AC
sin 60 (a)
T AC
(b)
T BC
2943 N sin sin 65 2943 N sin65
T
BC
sin 55
2943 N sin 65
sin 60 60
2812.19 N
T AC
2.81 kN
sin 55 55
2659.98 N
T BC
2.66 kN
46
PROBLEM 2.47
A chairlift has been stopped in the position shown. Knowing that each chair weighs 300 N and that the skier in chair E weighs 890 N, determine that weight of the skier in chair F .
SOLUTION
Free-Body Diagram Point B
In the free-body diagram of point B, the geometry gives: AB
tan 1
BC
tan 1
9.9 16.8 12 28.8
30.51
22.61
Thus, in the force triangle, by the Law of Sines: Force Triangle
T BC
sin 59 59.49 T BC
Free-Body Diagram Point C
In the free-body diagram of point and skier) the geometry gives: CD
1190 N sin 7. 7.87
7468.6 N
C (with W the
tan 1
1.32 7.2
sum of weights of chair
10.39
Hence, in the force triangle, by the Law of Sines: Force Triangle
W
sin 12.23 W
Fina Finall lly, y, the the skie skierr weig weight ht
7468.6 N sin 10 100.39
1608.5 N
1608 1608.5 .5 N
300 300 N
1308 1308.5 .5 N skier weight
47
1309 N
PROBLEM 2.48
A chairlift has been stopped in the position shown. Knowing that each chair weighs 300 N and that the skier in chair F weighs 800 N, determine the weight of the skier in chair E.
SOLUTION
Free-Body Diagram Point F
In the free-body diagram of point F , the geometry gives: EF
DF
12
tan 1
28.8
tan 1
1.32 7.2
22.62
10.39
Thus, in the force triangle, by the Law of Sines: Force Triangle T EF
sin 100.39 T BC
Free-Body Diagram Point E
1100 N sin 12 12.23
5107.5 N
In the free-body diagram of point E (with and skier) the geometry gives: AE
tan 1
9.9 16.8
W the
sum of weights of chair
30.51
Hence, in the force triangle, by the Law of Sines: W
sin 7. 7.89
Force Triangle
W
Fina Finall lly, y, the skie skierr weig weight ht
813 813.8 N
5107.5 N sin 59 59.49
813.8 N
300 N
513.8 13.8 N skier weight
48
514 N
PROBLEM 2.49
Four wooden members are joined with metal plate connectors and are in equilibrium under the action of the four fences shown. Knowing that F A 510 lb and F B 480 lb, determine the magnitudes of the other two forces.
SOLUTION
Free-Body Diagram
Resolving the forces into x and y components: F x
F y
0: F C
0 : F D
510 lb sin 15 480 lb cos15 0 or F C
332 lb
or F D
368 lb
510 lb cos15 480 lb sin 15 0
49
PROBLEM 2.50
Four wooden members are joined with metal plate connectors and are in equilibrium under the action of the four fences shown. Knowing that F A 420 lb and F C C 540 lb, determine the magnitudes of the other two forces.
SOLUTION
Resolving the forces into x and y components:
F 0x:
cosF1B5
F y
540 lb 420 lb cos15
0 : F D
0
or
F 6B71.6
lb
F B
672 lb
or F D
232 lb
420 lb cos15 671.6 lb sin 15 0
50
PROBLEM 2.51
Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and the P 400 lb and Q 520 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 F A FB 0
Substituting components: R 400 lb j 520 lb cos 55 i 520 lb sin 55 j
FiB FcAos 55 i
FsAin 5 5 j 0
In the y-direction (one unknown force)
400 lb 520 lb sin 55 F A sin 55 0 Thus, F A
400 lb lb 520 lb lb sin 55 55 sin55
1008.3 1008.3 lb 1008 lb F A 1008
In the x-direction: cos 55 F B F A cos 55 520 lb co
0
Thus, lb cos 55 F B F A cos 55 520 lb
1008.3 lb lb cos 55 55 520 lb l b cos 55 55 280.08 280.08 lb F B 280 280 lb
51
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 F A 600 lb and F B 320 lb, determine the magnitudes of P and Q.
SOLUTION
Free-Body Diagram
Resolving the forces into x and y directions: R P Q F A FB 0
Substituting components: R 320 lb i 600 lb cos 55 i 600 lb sin 55 j
P i Qcos 55 i
sin 55 j 0 Q
In the x-direction (one unknown force) 320 lb 600 lb cos 55 Q cos 55 0 Thus, Q
320 lb lb 600 lb lb cos 55 42.09 42.09 lb cos55 Q 42.1lb
In the y-direction: sin 55 P Q sin 55 600 lb si
0
Thus, lb sin 55 Q sin 55 457.01 lb P 600 lb P 457 457 lb
52
PROBLEM 2.53
Two cables tied together at C are loaded as shown. Knowing that W 840 N, determine the tension (a ( a) in cable AC , (b (b) in cable BC cable BC .
SOLUTION
Free-Body Diagram
From geometry: The sides of the triangle with hypotenuse CB are in the ratio 8:15:17. The sides of the triangle with hypotenuse CA are in the ratio 3:4:5. Thus:
F x 0:
3
T
5
CA
15
T
17
15
CB
17
680 N
0
or
1 5
TCA
5 17
T CB
200 N
(1)
and
F y 0:
4 5
T
C A
8
T
17
C B
8 17
680 N 840 N
0
or 1 5
TCA
2 17
T CB
290 N
(2)
Solving Equations (1) and (2) simultaneously: (a)
T CA
(b)
T CB
53
750 N 75
1190 119 0N
PROBLEM 2.54
Two cables tied together at C are loaded as shown. Determine the range of values of W for which the tension will not exceed 1050 N in either cable.
SOLUTION
Free-Body Diagram
From geometry: The sides of the triangle with hypotenuse CB are in the ratio 8:15:17. The sides of the triangle with hypotenuse CA are in the ratio 3:4:5. Thus: F x 0:
3
T
5
A C
15
T
17
15
680 N
CB
17
0
or
1 5
TCA
5
17
T CB
T
C B
200 N
(1)
and
4
F y 0 :
T
5
C A
8 17
8 17
680 N
W 0
or 1 5
TCA
2
TCB
17
80 N
1 4
(2)
W
Then, from Equations (1) and (2)
Now, with T
1050
TCB
TCA
680 N 25 28
17 28
W
W
N TCA : TCA
or
W
1050 N
25 28
W
1176 1176 N
and TCB : TCB
or
W
54
1050 N
609 609 N
680 N
17 28
W
0
W
609 N
PROBLEM 2.55
The cabin of an aerial tramway is suspended from a set of wheels that can roll freely on the support cable ACB and is being pulled at a constant speed by cable DE . Knowing that 40 and 35, that the combined weight of the cabin, its support system, and its passengers is 24.8 kN, and assuming the tension in cable DF to be negligible, determine the tension (a ( a) in the support cable ACB, ACB, (b) in the traction cable DE cable DE .
SOLUTION
Note: In Problems 2.55 and 2.56 the cabin is considered as a particle. If considered as a rigid body (Chapter 4) it would be found that its center of gravity should be located to the left of the centerline for the line CD to be vertical. Now F 0x:
T cAoCsB35
cos 40
TcosD4 E 0 0
or 0.0531T
A CB0.766T
DE0
(1)
and
F 0y:
T sAiCnB40
sin 35
TsinD4E0 24.8 kN
0
or 0.0692T
A CB0.643T
DE2 4.8
kN
(2)
From (1) T
T C14.426 A B
DE
Then, from (2) 0.0692 14.426T
DE 0.643T
DE24 .8
kN
and
55
(b) T DE
15.1 15.1 kN
(a) T ACB
218 218 kN
PROBLEM 2.56
The cabin of an aerial tramway is suspended from a set of wheels that can roll freely on the support cable ACB and is being pulled at a constant speed by cable DE . Knowing that 42 and 32, that the tension in cable DE is 20 kN, and assuming the tension in cable DF to be negligible, determine (a ( a) the combined weight of the cabin, its support system, and its passengers, (b (b) the tension in the support cable ACB. ACB.
SOLUTION
Free-Body Diagram
First, consider the sum of forces in the x the x-direction -direction because there is only one unknown force: F
x 0 :
T
ACcBos 32
cos 42 20 kN co cos 42
0
or 0.1049T ACB
14.863 kN kN (b) T ACB
141. 141.7 7 kN
Now
F y 0:
TACsBin 42
sin 32 20 kN sin 42
W0
or
141.7 kN 0.1392 20 kN 0.6691 W
0 (a) W
56
33.1 33.1 kN
PROBLEM 2.57
A block of weight W is suspended from a 500-mm long cord and two springs of which the unstretched lengths are 450 mm. Knowing that the constants of the springs are k AB 1500 N/m and k AD 500 N/m, determine ( a) the tension in the cord, ( b) the weight of the block.
SOLUTION
Free-Body Diagram At A
First note from geometry: The sides of the triangle with hypotenuse AD are in the ratio 8:15:17. The sides of the triangle with hypotenuse AB are in the ratio 3:4:5. The sides of the triangle with hypotenuse AC are in the ratio 7:24:25. Then: F AB k
L AB AB
AB
Lo
and L AB
0.44 m
2
0.33 m
2
0.55 m
So: F AB
1500 N/m 0.55 m
0.45 m
150 N
Similarly, F AD k
L AD AD
AD
Lo
Then:
0.66 m
L AD F AD
2
0.32 m
1500 N/m 0.68 m
2
0.68 m
0.45 m
115 N
(a) F
0:
x
4 5
150 N
7 25
T
15 17
A1C15 N
0
or T AC
57
66.1 66.18 8N
T AC
66.2 6.2 N
PROBLEM 2.57 CONTINUED
(b) and F y
0:
3 5
150 N
24 25
66.18 N
8 17
115 N W 0 or W
58
208 208 N
PROBLEM 2.58
A load of weight 400 N is suspended from a spring and two cords which are attached to blocks of weights 3W 3 W and W as shown. Knowing that the constant of the spring is 800 N/m, determine ( a) the value of W , (b) the unstretched length of the spring.
SOLUTION
Free-Body Diagram At A
First note from geometry: The sides of the triangle with hypotenuse AD are in the ratio 12:35:37. The sides of the triangle with hypotenuse AC are AC are in the ratio 3:4:5. The sides of the triangle with hypotenuse AB are also in the ratio 12:35:37. Then: F x
0:
4 5
3 W
35 37
W
12 37
F s 0
or F s
4.4833W
and F y
0:
3 5
3 W
12 37
W
35 37
F s 4 00 N
0
Then: 3 5
3W
12 37
W
35 37
4.4833W 400 N
0
or W
62.841 62.841 N
and F s
281. 281.74 74 N
or (a)
W
59
62.8 62. 8 N
PROBLEM 2.58 CONTINUED
(b) Have spring force F
s
k L
AB
L
o
Where F AB k
L AB AB
AB
Lo
and L AB
0.360 m
2
1.050 m
2
1.110 m
So: 281.74 N
800 N/ N/m 1.110
L0
m or
60
L0
758 758 mm
PROBLEM 2.59
For the cables and loading of Problem 2.46, determine ( a) the value of for which the tension in cable BC is as small as possible, ( b) the corresponding value of the tension.
SOLUTION
The smallest
T BC
is when
T BC
is perpendicular to the direction of T AC
Free-Body Diagram At C
Force Triangle
(a) (b)
T BC
2943 N sin55
2410 2410.8 .8 N T BC
61
55.0
2.41kN
PROBLEM 2.60
Knowing that portions AC and BC of cable ACB must be equal, determine the shortest length of cable which can be used to support the load shown if the tension in the cable is not to exceed 725 N.
SOLUTION
Free-Body Diagram: C For T 725 N
F y
0: 2T y T y
T x2 T x2
Ty2
N
0
500 N
500 N T x
1000
2
T 2
725 N
2
525 N
By similar triangles: BC
725
1.5 1.5 m 525
BC 2.07 2.07 m
L 2 BC 4.14 m L
62
4.14 4.14 m
PROBLEM 2.61
Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension in each cable is 200 lb, determine ( a) the magnitude of the largest force P which may be applied at C , (b) the corresponding value of of .
SOLUTION
Free-Body Diagram: C
Force Triangle
Force triangle is isoceles with 2
(a)
P
Since (b)
P
180
85
47.5
2 200 lb cos 47 47.5
0, the solution is correct.
180
270 lb P
55
63
47.5
77.5
270 lb
77.5
PROBLEM 2.62
Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension is 300 lb in cable AC and 150 lb in cable BC , determine ( a) the magnitude of the largest force P which may be applied at C , (b) the corresponding value of of .
SOLUTION
Free-Body Diagram: C
Force Triangle
(a) Law of Cosines: 2
300 lb
P
323.5 lb
P
Since
P
2
150 lb
2
2 300 lb 150 lb cos 85
300 lb, our solution is correct.
P
324 lb
(b) Law of Sines: Sines: sin 300
323.5
sin
0.9238
67.49
or
sin 85
180
55
67.49
57.5
64
57.5
PROBLEM 2.63
For the structure and loading of Problem 2.45, determine ( a) the value of for which the tension in cable BC is as small as possible, ( b) the corresponding value of the tension.
SOLUTION T BC
must be perpendicular to
F AC
to be as small as possible.
Free-Body Diagram: C
(a) We observe: (b) or
Force Triangle is a right triangle
55
T BC
sin 60 400 lb sin
T BC
346.4 lb
65
T BC
55
346 lb
PROBLEM 2.64
Boom AB is supported by cable BC and a hinge at A. Knowing that the boom exerts on pin B a force directed along the boom and that the tension in rope BD is 70 lb, determine ( a) the value of for which the tension in cable BC is as small as possible, (b) the corresponding value of the tension.
SOLUTION
Free-Body Diagram: B
(a) Have:
T
BD
F
AB
T
BC
0
where magnitude and direction of T BD are known, and the direction of F AB is known.
Then, in a force triangle: By observation, (b) Have
T BC
is minimum when
T BC
90.0
70 lb sin 180 70 30 68.93 lb T BC
66
68.9 lb
PROBLEM 2.65
Collar A shown in Figure P2.65 and P2.66 can slide on a frictionless vertical rod and is attached as shown to a spring. The constant of the spring is 660 N/m, and the spring is unstretched when h 300 mm. Knowing that the system is in equilibrium when h 400 mm, determine the weight of the collar.
SOLUTION
Free-Body Diagram: Collar A
F
Have:
s
k L
AB
L
AB
where: L
0. 3 m AB
0.4 m
2
L
0.3 2 Am B
0.5 m F s
Then:
2
660 N/m 0.5
0.3 2 m
49.986 N
For the collar: F y
0:
W
4 5
49.986 N
0 or W
67
40.0 N
PROBLEM 2.66
The 40-N collar A collar A can slide on a frictionless vertical rod and is attached as shown to a spring. The spring is unstretched when h 300 mm. Knowing that the constant of the spring is 560 N/m, determine the value of h of h for which the system is in equilibrium.
SOLUTION
h
yF 0: W
Free-Body Diagram: Collar A
2
0.3 h
2
or
hF s 40 0.09 h 2
Now..
F s k L
L
where
Then:
or
h 560
2 0.A3B
AB
L
s F
0
AB
L 0.3 A2B m
2
hm
0.09 h2 0.3 2 40 0.09 h 2
14h 1
0.09 h 2 4.2 2 h
h m
Solving numerically, numerically, h 415 mm
68
PROBLEM 2.67
A 280-kg crate is supported by several rope-and-pulley arrangements as shown. Determine for each arrangement the tension in the rope. ( Hint: ( Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Chapter 4.)
SOLUTION
Free-Body Diagram of pulley
F y
(a )
0 : 2T 280 kg 9.81 m/s 2 T
1 2
0
2746.8 N T
(b )
F y
0 : 2T 280 kg 9.81 m/s 2 T
1 2
F y
(d )
F y
F y
3
1 3
1 4
0
T
916 N
T
916 N
T
687 N
0
2746.8 N
0 : 4T 280 kg 9.81 m/s 2
1373 N
2746.8 N
T
69
1
0 : 3T 280 kg 9.81 m/s 2 T
(e)
2746.8 N
0 : 3T 280 kg 9.81 m/s 2 T
1373 N
0
T (c)
0
2746.8 N
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 280-kg crate is supported by several rope-and-pulley arrangements as shown. Determine for each arrangement the tension in the rope. ( Hint: ( Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Chapter 4.)
SOLUTION
Free-Body Diagram of pulley and crate
(b ) F y
0 : 3T 280 kg 9.81 m/s 2 T
1 3
0
2746.8 N T
916 N
687 N
(d )
F y
0 : 4T 280 kg 9.81 m/s 2 T
1 4
0
2746.8 N T
70
PROBLEM 2.69
A 350-lb load is supported by the rope-and-pulley arrangement shown. Knowing that 25, determine the magnitude and direction of the force P which should 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 Chapter 4.)
SOLUTION
Free-Body Diagram: Pulley A
xF
0: 2 P sin 25
cos 0 P
and cos
0.8452
For yF
yF
0: 2 P cos 25
32.3
32.3
0: 2 P cos 25
For
or
sin 32 P .3
350 lb
or P
149.1 lb
0 32.3
32.3
sin 32.3 P or P
71
350 lb 274 lb
0 32.3
PROBLEM 2.70
A 350-lb load is supported by the rope-and-pulley arrangement shown. Knowin Knowing g that that 35, determine (a ( a) the angle , (b) the magnitude of the force P which should be exerted on the free end of the rope to maintain equilibrium. ( Hint: ( Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Chapter 4.)
SOLUTION
Free-Body Diagram: Pulley A
xF
0: 2 P sin Pcos 2 5
0
Hence: (a )
sin
(b )
yF
1 2
or
cos 25
0: 2 P cos
P sin 3 5
350 lb
350 lb
0
24.2
0
Hence: 2 P cos 24.2 P sin 35 or
P 145.97 lb
72
P 146.0 lb
PROBLEM 2.71
A load Q is applied to the pulley C , which can roll on the cable ACB cable ACB.. The pulley is held in the position shown by a second cable CAD, CAD, which passes over the pulley A and supports a load P. Knowing that P 800 N, determine (a ( a) the tension in cable ACB, ACB, (b (b) the magnitude of load Q.
SOLUTION
Free-Body Diagram: Pulley C
(a)
F
x 0 :
T
ACcBos 30
T ACB
Hence
cos 50 800 N co c os 50
0
2303.5 N T ACB
(b)
F y 0:
TACsBin 30
sin 50 800 N si sin 50
Q
or
73
3529.2 N
2.30 kN
3.53 kN
Q 0
2303.5 N sin 30 sin 50 800 N sin 50 Q
0 Q
PROBLEM 2.72
A 2000-N load Q is applied to the pulley C , which can roll on the cable ACB. ACB. The pulley is held in the position shown by a second cable CAD, CAD, which passes over the pulley A and supports a load P. Determine (a (a) the tension in the cable ACB, ACB, (b (b) the magnitude of load P.
SOLUTION
Free-Body Diagram: Pulley C
Fx 0:
T ACcBos 30 P
or
Fy 0 :
T ACsBin 30
cos 50
0.3473T ACB
sin 50
1.266T ACB
or
cP os 50 0 (1)
sP in 50 2000 N
0
0.766P 2000 N
(2)
(a) Substitute Equation (1) into into Equation Equation (2): (2): 1.266T Hence:
0A.7 66 CB
T ACB
0.3473T
A2C0B00 N
1305.5 N T ACB
1306 N
(b) Using (1) P 0.3473 1306 N
453.57 N P 454 N
74
PROBLEM 2.73
Determine (a (a) the x, x, y, and z components of the 200-lb force, (b ( b) the angles x, y, and z that the force forms with the coordinate axes.
SOLUTION
(a)
F x
F y
F z
(b)
lb cos 30 30 cos 25 25 156.98 lb lb 200 lb
F x
157.0
lb
F y
100.0
lb
30 100.0 lb 200 lb sin 30
30 sin 25 25 73.1996 lb 200 lb cos 30
cos x
cos y
cos z
75
156.98 200 100.0 200
73.1996
200
lb
F z
73.2
or
x
38.3
or
y
60.0
or
z
111.5
PROBLEM 2.74
Determine (a (a) the x, x, y, and z components of the 420-lb force, (b ( b) the angles x, y, and z that the force forms with the coordinate axes.
SOLUTION
(a)
F x
20 sin 70 70 134.985 lb 420 lb sin 20 F x
F y
F z
(b)
135.0
lb
20 394.67 lb 420 lb cos 20
20 cos 70 70 420 lb sin 20
cos x
F y
395
lb
F z
49.1
lb
49.131 lb
134.985
420 x
cos y
cos z
108.7
394.67 420 y
20.0
z
83.3
49.131 420
76
PROBLEM 2.75
To stabilize a tree partially uprooted in a storm, cables AB and AC are attached to the upper trunk of the tree and then are fastened to steel rods anchored in the ground. Knowing that the tension in cable AB is 4.2 kN, determine (a ( a) the components of the force exerted by this cable on the tree, (b (b) the angles x, y, and z that the force forms with axes at A which are parallel to the coordinate axes.
SOLUTION
(a)
F x
F y
F z
(b)
50 cos 40 40 4.2 kN sin 50
50 4.2 kN cos 50
2.4647 kN
2.6997
F x
2.46
kN
F y
2.70
kN
F z
2.07
kN
kN
50 sin 40 4 0 2.0681 kN 4.2 kN sin 50
cos x
2.4647 4.2 x
77
54.1
PROBLEM 2.75 CONTINUED
cos y
2.7
4.2
y
cos z
130.0
2.0681 4.0
z
78
60.5
PROBLEM 2.76
To stabilize a tree partially uprooted in a storm, cables AB and AC are attached to the upper trunk of the tree and then are fastened to steel rods anchored in the ground. Knowing that the tension in cable AC is 3.6 kN, determine (a ( a) the components of the force exerted by this cable on the tree, (b (b) the angles x, y, and z that the force forms with axes at A which are parallel to the coordinate axes.
SOLUTION
(a)
F x
45 sin 25 25 3.6 kN cos 45
1.0758
kN F x
F y
F z
(b)
45 3.6 kN sin 45
2.546
45 cos 25 25 3.6 kN cos 45
cos x
1.076
kN F y
2.55
kN
F z
2.31
kN
2.3071 kN
1.0758
3.6 x
79
kN
107.4
PROBLEM 2.76 CONTINUED
cos y
2.546
3.6
y
cos z
135.0
2.3071 3.6
z
80
50.1
PROBLEM 2.77
A horizontal circular plate is suspended as shown from three wires which are attached to a support at D and form 30 angles with the vertical. Knowing that the x component of the force exerted by wire AD on the plate is 220.6 N, determine (a ( a) the tension in wire AD, AD, (b) the angles x, y, and z that the force exerted at A forms with the coordinate axes.
SOLUTION
(a)
F x
F sin 30 30 sin 50 50 F
220.6 N (Given)
220.6 N
sin3 sin30 0 sin5 sin50 0
575.95 N F
(b)
cos x
F x F
F y
cos y
F z
220.6
F cos 30 30 F y F
575.95
0.3830 x
67.5
y
30.0
498.79 N
498.79
576 N
575.95
0.86605
F sin30 cos50
.95 N sin30 sin30 cos50 575.95
185.107 N
cos z
F z F
185.107
575.95
0.32139
z
81
108.7
PROBLEM 2.78
A horizontal circular plate is suspended as shown from three wires which are attached to a support at D and form 30 angles with the vertical. Knowing that the z component of the force exerted by wire BD on the plate is –64.28 N, determine (a ( a) the tension in wire BD, BD, (b (b) the angles x, y, and z that the force exerted at B forms with the coordinate axes.
SOLUTION
(a)
F z
F sin 30 30 sin 40 40
F
64.28 N (Given)
64.28 N sin30 sin30 sin4 sin40 0
(b)
F x
200.0 N
F si sin30 cos40
200.0 N sin30 cos40
F
200 N
x
112.5
76.604 N
cos x
F x
F y
F
cos y
200.0
F cos 30 30
F y F
F z cos z
76.604
F z F
173.2 N
173.2
0.38302
200
0.866
y
30.0
64.28 N
64.28
200
0.3214
82
z
108.7
PROBLEM 2.79
A horizontal circular plate is suspended as shown from three wires which are attached to a support at D and form 30 angles with the vertical. Knowing that the tension in wire CD is 120 lb, determine (a ( a) the components of the force exerted by this wire on the plate, ( b) the angles x, y, and z that the force forms with the coordinate axes.
SOLUTION
(a)
F x
30 cos 60 60 120 lb sin 30
30
lb 30.0
lb
103.9
lb
F x F y
30 103.92 lb 120 lb cos 30 F y
F z
30 sin 60 60 51.96 lb 120 lb sin 30 F z
(b)
cos x
F x F
30.0
120
52.0
0.25
x
cos y
cos z
F y
F
F z F
83
lb
103.92 120
51.96 120
104.5
0.866 y
30.0
z
64.3
0.433
PROBLEM 2.80
A horizontal circular plate is suspended as shown from three wires which are attached to a support at D and form 30 angles with the vertical. Knowing that the x component of the forces exerted by wire CD on the plate is –40 lb, determine (a (a) the tension in wire CD, CD, (b (b) the angles x, y, and z that the force exerted at C forms C forms with the coordinate axes.
SOLUTION
(a)
F x
F sin 30 cos 60
F
40 lb (Given)
40 lb
sin3 sin30 0 cos6 cos60 0
160 lb
F (b)
cos x
F x
F
40
160
0.25
x
F y
cos y
F z
160.0 lb
104.5
lb cos 30 30 103.92 lb lb 160 lb
F y F
103.92 160
0.866 y
30.0
z
64.3
30 sin 60 60 69.282 lb 160 lb sin 30
cos z
F z F
69.282 160
84
0.433
PROBLEM 2.81
Determine the magnitude and F 800 lb i 260 lb j 320 lb k.
direction
of
the
force
SOLUTION
F
F2x
F2y F2z
800 lb
cos x
cos y
cos z
F x F F y F
F z F
85
2
260 lb
800 900 260 900
320
900
2
320 lb
2
F
900 lb
0.8889
x
27.3
0.2889
y
73.2
0.3555
z
110.8
PROBLEM 2.82
Determine the magnitude and direction F 400 N i 1200 N j 300 N k.
of
the
force
SOLUTION
F
F2x
F2y F2z
400 N
cos x
cos y
cos z
F x F
F y F
F z F
2
1200 N
400 1300
1200
1300
300 1300
2
0.30769
0.92307
0.23076
86
300 N
2
F
1300 N
x
y
72.1
157.4
z
76.7
PROBLEM 2.83
A force acts at the origin of a coordinate system in a direction defined by the angles x 64.5 and z 55.9. Knowing that the y component of the force is –200 N, determine (a ( a) the angle y, (b ( b) the other components and the magnitude of the force.
SOLUTION
(a) We have
cos x Since F y
2
cos y
2
cos z
0 we must have cos y
2
1 cos
y
2
1 cos
y
2
cos z
2
0
Thus, taking the negative square root, from above, we have: cos y
1 cos 64.5
2
cos 55.9
2
0.70735
y
135.0
(b) Then: F
and
F y
200
N
cos y
0.70735
282.73 N
F x
F cos x
282.73 N cos 64.5
F x
121.7 N
F z
F cos z
282.73 N cos 55.9
F y
158.5 N
F
87
283 N
PROBLEM 2.84
A force acts at the origin of a coordinate system in a direction defined by the angles x 75.4 and y 132.6. Knowing that the z component z component of the force is –60 N, determine (a ( a) the angle z , (b) the other components and the magnitude of the force.
SOLUTION
(a) We have
cos x Since F z
2
cos y
2
cos z
0 we must have cos z
2
1 cos
y
2
1 cos
y
2
cos z
2
0
Thus, taking the negative square root, from above, we have: cos z
1 cos 75.4
2
cos132.6
2
0.69159
z
133.8
F
86.8 N
21.9 N
(b) Then: F
and
F z
F x F y
cos z
F cos x F cos y
60
N
0.69159
86.757 N
86.8 N cos 75.4
86.8 N cos132.6
88
F x F y
58.8
N
PROBLEM 2.85
A force F of magnitude 400 N acts at the origin of a coordinate system. Knowing that x 28.5, F y –80 N, and F z 0, determine (a (a) the components F components F x and F and F z , (b (b) the angles y and z .
SOLUTION
(a) Have F x
F cos x
400 N cos 28.5
F x
351.5 N
F z
173.3 N
Then: F2
400 N
So:
2
F2x F2y F2z
352.5 N
2
80 N
2
2
F z
Hence: F z
400 N
2
351.5 N
2
80
N
2
(b) cos y
cos z
F y F
F z F
89
80
400
173.3 400
0.20
0.43325
y
101.5
z
64.3
PROBLEM 2.86
A force F of magnitude 600 lb acts at the origin of a coordinate system. Knowing that F x 200 lb, z 136.8, F y 0, determine (a (a) the components F components F y and F and F z , (b (b) the angles x and y.
SOLUTION
(a)
F z
F cos z
600 lb cos136.8
437.4
lb
F z
437
lb
F y
359
lb
Then: F2 So:
600 lb
Hence:
F y
2
F2x F2y F2z
200 lb
600 lb
358.7
2
2
F y
2
200 lb
2
437.4 lb
2
2
437.4
lb
lb
(b) cos x
cos y
F y F
F x F
200 600
358.7
600
0.333
0.59783
90
x
y
70.5
126.7
PROBLEM 2.87
A transmission tower is held by three guy wires anchored by bolts at B, C , and D. If the tension in wire AB is 2100 N, determine the components of the force exerted by the wire on the bolt at B.
SOLUTION
BA 4 m i 20 m j 5 m k BA
2 2 2 4 m 20 m 5 m 21 m
F F BA
2100 N BA 4 m i 20 m j 5 m k F BA 21 m F 400 N i 2000 N j 500 N k Fx 400 N,
91
Fy 2000 N,
Fz 500 N
PROBLEM 2.88
A transmission tower is held by three guy wires anchored by bolts at B, C , and D. If the tension in wire AD is 1260 N, determine the components of the force exerted by the wire on the bolt at D.
SOLUTION
DA 4 m i 20 m j 14.8 m k DA
2 2 2 4 m 20 m 14.8 m
25.2 m
F F DA
DA 1260 N 4 m i 20 m j 14.8 m k F DA 25.2 m F 200 N i 1000 N j 740 N k Fx 200 N,
92
Fy 1000 N,
Fz 740 N
PROBLEM 2.89
A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AB is 204 lb, determine the components of the force exerted on the plate at B.
SOLUTION
BA 32 in. i 48 in. j 36 in. k BA
2 2 2 32 in. 48 in. 36 in.
68 in.
F F BA
BA 204 lb 32 in. i 48 in. j 36 in. k F BA 68 in. F 96 lb i 144 lb j 108 lb k Fx 96.0 lb,
93
Fy 144.0 lb,
Fz 108.0 lb
PROBLEM 2.90
A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 195 lb, determine the components of the force exerted on the plate at D.
SOLUTION
DA 25 in. i 48 in. j 36 in. k DA
2 2 2 25 in. 48 in. 36 in.
65 in.
F F DA
DA 195 lb 25 in. i 48 in. j 36 in. k F DA 65 in. F 75 lb i 144 lb j 108 lb k Fx 75.0 lb,
94
Fy 144.0 lb,
Fz 108.0 lb
PROBLEM 2.91
A steel rod is bent into a semicircular ring of radius 0.96 m and is supported in part by cables BD and BE which are attached to the ring at B. Knowing that the tension in cable BD is 220 N, determine the components of this force exerted by the cable on the support at D.
SOLUTION
DB 0.96 m i 1.12 m j 0.96 m k 2 2 2 0.96 m 1.12 m 0.96 m 1.76 m
DB
T
T
DB
T
DB
DB DB
220 N
0.96 m i 1.12 m j 0.96 m k
1.76 m
T DB 120 N i 140 N j 120 N k
T DxB120.0 N, T DyB 140.0 N, T DzB 120.0 N
95
PROBLEM 2.92
A steel rod is bent into a semicircular ring of radius 0.96 m and is supported in part by cables BD and BE which are attached to the ring at B. Knowing that the tension in cable BE is 250 N, determine the components of this force exerted by the cable on the support at E .
SOLUTION
EB 0.96 m i 1.20 m j 1.28 m k 2 2 2 0.96 m 1.20 m 1.28 m
EB
2.00 m
T
T
EB
T
EB
EB EB
250 N
0.96 m i 1.20 m j 1.28 m k
2.00 m
T EB 120 N i 150 N j 160 N k
T ExB120.0 N, T EyB 150.0 N, T EzB 160.0 N
96
PROBLEM 2.93
Find the magnitude and direction of the resultant of the two forces shown knowing that P 500 N and Q 600 N.
SOLUTION
P
500 lb cos 30 sin 15 i sin 30 j cos 30 cos15k
500 lb 0.2241i 0.50 j 0.8365k
Q
R R
112.05 lb i 250 lb j 418.25 lb k
40 j cos 40 40 sin 20 20k 600 lb cos 40 cos 20i sin 40
600 lb 0.71985i 0.64278j 0.26201k
431.91 lb i 385.67 lb j 157.206 lb k
P
Q
319.86 lb i 635.67 lb j 261.04 lb k
319.86 lb
2
635.67 lb
2
261.04 lb
2
757.98 lb R
cos x
cos y
cos z
R x R
R y R
R z R
97
319.86 lb 757.98 lb
635.67 lb 757.98 lb
261.04 lb 757.98 lb
758 lb
0.42199 x
65.0
y
33.0
z
69.9
0.83864
0.34439
PROBLEM 2.94
Find the magnitude and direction of the resultant of the two forces shown knowing that P 600 N and Q 400 N.
SOLUTION
Using the results from 2.93: P
600 lb 0.2241i 0.50j 0.8365k
Q
R R
P
134.46 lb i 300 lb j 501.9 lb k
400 lb 0.71985i 0.64278j 0.26201k
287.94 lb i 257.11 lb j 104.804 lb k
Q
153.48 lb i 557.11 lb j 397.10 lb k
153.48 lb
2
557.11 lb
2
397.10 lb
2
701.15 lb R
cos x
cos y
cos z
R x R
R y R
R z R
153.48 lb 701.15 lb
557.11 lb 701.15 lb
397.10 lb 701.15 lb
98
701 lb
0.21890 x
77.4
y
37.4
z
55.5
0.79457
0.56637
PROBLEM 2.95
Knowing that the tension is 850 N in cable AB and 1020 N in cable AC , determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.
SOLUTION
AB 400 mm i 450 mm j 600 mm k
2 2 2 400 mm 450 mm 600 mm 850 mm
AB
AC 1000 mm i 450 mm j 600 mm k
AC
2 2 2 1000 mm 450 mm 600 mm 1250 mm
400 mm i 450 mm j 600 mm k 850 mm
T
AB
T AB
AB
AB
850 N T AB AB T
AB
400 N i 450 N j 600 N k
1000 mm i 450 mm j 600 mm k 1250 mm
T
AC
T AC
AC
AC
1020 N T AC AC T
AC
R T
816 N i 367.2 N j 489.6 N k
T
AB
1216 N i 817.2 N j 1089.6 N k
AC
R 1825.8 N
Then: and
cos x cos y cos z
1216 1825.8
817.2 1825.8 1089.6 1825.8
99
0.66601 0.44758 0.59678
R 1826 N
48.2
x
y
116.6
z
53.4
PROBLEM 2.96
Assuming that in Problem 2.95 the tension is 1020 N in cable AB and 850 N in cable AC , determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.
SOLUTION
AB 400 mm i 450 mm j 600 mm k
2 2 2 400 mm 450 mm 600 mm 850 mm
AB
AC 1000 mm i 450 mm j 600 mm k
AC
2 2 2 1000 mm 450 mm 600 mm 1250 mm
400 mm i 450 mm j 600 mm k 850 mm
T
AB
T AB
AB
AB
1020 N T AB AB
T AB 480 N i 540 N j 720 N k
1000 mm i 450 mm j 600 mm k 1250 mm
T
AC
T
AC AC
AC
850 N T AC AC
T AC 680 N i 306 N j 408 N k R T
T
AB
R 1825.8 N
cos x cos y
1160 N i 846 N j 1128 N k
AC
Then: and
1160 1825.8
0.6353
846 1825.8
cos z
0.4634
1128 1825.8
0.6178
100
R 1826 N
50.6
x
y
117.6
z
51.8
PROBLEM 2.97
For the semicircular ring of Problem 2.91, determine the magnitude and direction of the resultant of the forces exerted by the cables at B knowing that the tensions in cables BD and BE are 220 N and 250 N, respectively.
SOLUTION
For the solutions to Problems 2.91 and 2.92, we have T BD
120 N i 140 N j 120 N k
T BE
120 N i 150 N j 160 N k
Then: R
B
T
and
T
BD
BE
240 N i 290 N j 40 N k R
cos x
378.55 N
240 378.55
R B
0.6340
x
cos y
cos z
290 378.55
40 378.55
101
379 N
129.3
0.7661
y
40.0
z
96.1
0.1057
PROBLEM 2.98
To stabilize a tree partially uprooted in a storm, cables AB and AC are attached to the upper trunk of the tree and then are fastened to steel rods anchored in the ground. Knowing that the tension in AB is 920 lb and that the resultant of the forces exerted at A by cables AB and AC lies in the yz plane, determine (a ( a) the tension in AC , (b (b) the magnitude and direction of the resultant of the two forces.
SOLUTION
Have 40 j 920 lb sin 50 cos 40i cos 50 j sin 50 sin 40
T AB
T
AC
T
AC cos 45 sin
25 25i
sin 45 45 j cos 45 cos 25 j
(a) R
A
T
R A x
R x A
F
AB
T
AC
0
920 lb sin 50 cos 40
0x:
TcosAC45 s in 25
0
or T AC
T AC
1806.60 lb
1807 lb
(b)
R A y
F y :
45 920 lb cos 50 1806.60 lb sin 45
R A y
1868.82
lb
50 sin 40 40 1806.60 lb cos 45 45 cos 25 25 R A z F z : 920 lb sin 50
R A z 1610.78 lb
R A
1868.82 lb j 1610.78 lb k
Then: R A
R A
2467.2 lb
102
2.47 kips
PROBLEM 2.98 CONTINUED and cos x
cos y
cos z
0 2467.2
1868.82
2467.2 1610.78 2467.2
103
0
0.7560
0.65288
x
y
z
90.0
139.2
49.2
PROBLEM 2.99
To stabilize a tree partially uprooted in a storm, cables AB and AC are attached to the upper trunk of the tree and then are fastened to steel rods anchored in the ground. Knowing that the tension in AC is AC is 850 lb and that the resultant of the forces exerted at A by cables AB and AC lies AC lies in the yz plane, determine (a ( a) the tension in AB, AB, (b (b) the magnitude and direction of the resultant of the two forces.
SOLUTION
Have T T AC
AB
T
ABsin 50 cos 40i
cos 50 j sin 50 sin 40 40 j
25i sin 45 j cos 45 cos 25 j 850 lb cos 45 sin 25
(a)
R A x
R xA
F x 0:
0
T AsBi n 50 cos 40 T AB
850 lb cos 45 sin 25 0 T AB
432.86 lb
433 lb
(b)
R A y
F y :
45 432.86 lb cos 50 850 lb sin 45
R A y
879.28
lb
50 sin 40 40 850 lb cos 45 45 cos 25 25 R A z F z : 432.86 lb sin 50
R A z 757.87 lb
R A
R A
cos z
0 1160.82
879.28
1160.82
R A
1160.82 lb
cos x cos y
879.28 lb j 757.87 lb k
0
0.75746
757.87 1160.82
0.65287
104
1.161 kips
x
y
90.0
139.2
z
49.2
PROBLEM 2.100
For the plate of Problem 2.89, determine the tension in cables AB and AD knowing that the tension if cable AC is 27 lb and that the resultant of the forces exerted by the three cables at A must be vertical.
SOLUTION
With:
45 in. i 48 in. j 36 in. k
AC
AC
2 2 2 45 in. 48 in. 36 in.
75 in.
T
AC
T
AC
AC
AC
T
AC
AC
27 lb
45 in. i 48 in. j 36 in. k
75 in.
T AC 16.2 lb i 17.28 lb j 12.96 k
and
32 in. i 48 in. j 36 in. k
AB
2 2 2 32 in. 48 in. 36 in.
AB
68 in.
T
AB
T
AB
AB
T
AB
T
AB AB
AB
T
AB
32 in. i 48 in. j 36 in. k
68 in.
0.4706i 0.7059 j 0.5294k
T
AB
and
AD
AD
25 in. i 48 in. j 36 in. k
2 2 2 25 in. 48 in. 36 in.
65 in.
T
AD
T
AD
AD
T
T AD
AD
AD
AD
T
T
AD
25 in. i 48 in. j 36 in. k
65 in.
0.3846i 0.7385j 0.5538k
AD
105
PROBLEM 2.100 CONTINUED
Now R T
T
AB
T AB
T
AD
AD
0.4706i 0.7059 j 0.5294k 16.2 lb i 17.28 lb j 12.96 k
T AD 0.3846i 0.7385 j 0.5538k Since R must be vertical, the i and k components of this sum must be zero. Hence:
0.4706T 0.5294T
AB0.3846T AB0.5538T
AD16.2 lb 0
(1)
A1D2.96 lb 0
(2)
Solving (1) and (2), we obtain: T
lb, 244.79 AB
T
lb 257.41 AD
106
T AB
245 lb
T AD
257 lb
PROBLEM 2.101
The support assembly shown is bolted in place at B, C , and D and supports a downward force P at A. Knowing that the forces in members AB, AC , and AD are directed along the respective members and that the force in member AB is 146 N, determine the magnitude of P.
SOLUTION
Note that AB, AC , and AD are in compression. Have
and
d BA
2 2 2 220 mm 192 mm 0
d DA
2 2 2 192 mm 192 mm 96 mm
d CA
2 2 2 0 192 mm 144 mm
F
BA
F BA
BA
146 N
292 mm 288 mm
240 mm
220 mm i 192 mm j
292 mm
110 N i 96 N j FCA F CA CA
F CA
192 mm j 144 mm k
240 mm
F CA 0.80 j 0.60k F
DA
F DA
DA
F DA
192 mm i 192 mm j 96 mm k
288 mm
F DA 0.66667i 0.66667 j 0.33333k P P j
With At A: i-component:
F 0 : F
F
BA
F
CA
110 N 0.66667 F DA 0
P0
DA
or
j-component:
96 N 0.80 FCA 0.66667 16 165 N
-component: k -component:
1 65 N 0 0.60 F CA 0.33333 16
Solving (2) for F CA and then using that result in (1), gives
107
F DA P
165 N
0
(1) (2) P
279 N
PROBLEM 2.102
The support assembly shown is bolted in place at B, C , and D and supports a downward force P at A. Knowing that the forces in members AB, AC , and AD are directed along the respective members and that P 200 N, determine the forces in the members.
SOLUTION
With the results of 2.101: F
BA
F BA
BA
F BA
220 mm i 192 mm j
292 mm
F BA 0.75342i 0.65753j N FCA F CACA
F CA
192 mm j 144 mm k
240 mm
F CA 0.80 j 0.60k F
DA
F DA
DA
F DA
192 mm i 192 mm j 96 mm k
288 mm
F DA 0.66667i 0.66667 j 0.33333k P 200 N j
With:
F 0 : F
At A:
F
BA
F
CA
P0
DA
Hence, equating the three ( i, j, k ) components to 0 gives three equations
0.75342 F
i-component: j-component:
0.65735
F
BA0.80
B0A.66667 F F
CA0.66667
F
D0A
(1)
DA2 00 N 0
(2)
0.60 FCA 0.33333F DA 0
k -component:
(3)
Solving (1), (2), and (3), gives F
104.5 N, BA
F
65.6 N, CA
F DA
118.1 N F BA F CA F DA
108
104.5 N 65.6 N 118.1 N
PROBLEM 2.103
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 AB is 60 lb.
SOLUTION
The forces applied at A are: T AB, T
, T
AC
and P
AD
where P P j . To express the other forces in terms of the unit vectors i, j, k , we write
12.6 ft i 16.8 ft j
AB
AB
7.2 ft i AC16.8 ft j 12.6 ft k
AD
16.8 ft j 9.9 ft k
AD
21 ft 22.2 ft
AC
19.5 ft
and
T
T
AB
AB
AB
T
AB
AB
AB
0.6i 0.8j T
AB
T
T AC
T AC AC
AC AC
AC
0.3242i 0.75676 j 0.56757k T
T
AD T
109
AD
AD T
AD AD
AD
0.8615j 0.50769k T
AD
AC
PROBLEM 2.103 CONTINUED
Equilibrium Condition F
0: T
AB
T
AC
T
AD
P j
0
Substituting the expressions obtained for T A,B T A,C and T
and AD
factoring i, j, and k :
0.6T
AB0.3242T
0.56757T
AiC 0.8T
AB0.75676T
AC0.8615T
DP A
j
k AD 0
A0 C.50769T
Equating to zero the coefficients of i, j, k : 0.6T 0.8T
AB0.75676T
0.56757T Setting T AB
0.3242T AB
AC0
AC0.8615T
A0 C.50769T
(1) ADP
0
(2)
0 AD
(3)
60 lb in (1) and (2), and solving the resulting set of
equations gives T AC T AD
111 lb
124.2 lb P
110
239 lb
PROBLEM 2.104
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 AC is 100 lb.
SOLUTION
See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 0.6T
0.8T
0.75676T
T AC
AC
0.50769T
AC
0
(1)
AC
0.8615T
AB
0.56757T Substituting
0.3242T
AB
P AD
0
(2)
0
(3)
AD
100 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations
using conventional algorithms gives T AB T AD
54 lb 112 lb P
111
215 lb
PROBLEM 2.105
The crate shown in Figure P2.105 and P2.108 is supported by three cables. Determine the weight of the crate knowing that the tension in cable AB cable AB is 3 kN.
SOLUTION
The forces applied at A at A are: T AB, T
AC,
ADand P
T
where P P j . To express the other forces in terms of the unit vectors i, j, k , we write
AB
0.72 m i 1.2 m j 0.54 m k ,
AB
1.5 m
AC 1.2 m j 0.64 m k ,
AD
AC 1.36 m
0.8 m i 1.2 m j 0.54 m k ,
AD
1.54 1.54 m
and
T
B T A
AB
AB A B T
AB
AB
0.48i 0.8j 0.36k T
AB
T
AC T
AC T
AC
AC AC
AC
0.88235j 0.47059k T
AC
T
ADT
AD
AD ADT
AD
AD
Equilibrium Condition with W F
0: T
0.51948i 0.77922 j 0.35065k T
AD
W j AB
T
AC
T
AD
W j
0
Substituting the expressions obtained for T A,B T A,C and T
and AD
factoring i, j, and k :
0.48T
AB0.51948T
0.36T
AiD 0.8T
AB0.47059T
112
AB0.88235T
AC0.35065T
AC0.77922T
k AD 0
ADW
j
PROBLEM 2.105 CONTINUED
Equating to zero the coefficients of i, j, k : 0.48T
0.8T
0.88235T
Substituting
T AB
0.51948T
AB
AB
0.36T
0.77922T
AC
0.47059T
AB
0
AD
W AD
0.35065T
AC
0 0
AD
3 kN in Equations (1), (2) and (3) and solving the
resulting set of equations, using conventional algorithms for solving linear algebraic equations, gives T AC
4.3605 kN
T AD
2.7720 kN W
113
8.41 kN
PROBLEM 2.106
For the crate of Problem 2.105, determine the weight of the crate knowing that the tension in cable AD is 2.8 kN. The crate shown in Figure P2.105 and P2.108 is supported by three cables. Determine the weight of the crate knowing that the tension in cable AB is 3 kN. Problem 2.105:
SOLUTION
See Problem 2.105 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 0.48T
0.8T
Substituting
T AD
0.51948T
AB
0.88235T
0.77922T
AB
0.36T
AC
0.47059T
AB
0
AD
W AD
0.35065T
AC
0
0
AD
2.8 kN in Equations (1), (2), and (3) above, and solving the resulting set of equations
using conventional algorithms, gives T AB
3.03 kN
T AC
4.40 kN W
114
8.49 kN
PROBLEM 2.107
For the crate of Problem 2.105, determine the weight of the crate knowing that the tension in cable AC is 2.4 kN. The crate shown in Figure P2.105 and P2.108 is supported by three cables. Determine the weight of the crate knowing that the tension in cable AB is 3 kN. Problem 2.105:
SOLUTION
See Problem 2.105 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 0.48T
0.8T
Substituting
T AC
0.51948T
AB
0.88235T
0.77922T
AB
0.36T
AC
0.47059T
AB
0
AD
W AD
0.35065T
AC
0
0
AD
2.4 kN in Equations (1), (2), and (3) above, and solving the resulting set of equations
using conventional algorithms, gives T AB
1.651 kN
T AD
1.526 kN W
115
4.63 kN
PROBLEM 2.108
A 750-kg crate is supported by three cables as shown. Determine the tension in each cable.
SOLUTION
See Problem 2.105 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 0.48T
0.8T
Substituting
W
0.88235T
AB
0.36T
0.51948T
AB
0.77922T
AC
0.47059T
AB
kN 750 kg 9.81 m/s 2 7.36 kN
0
AD
W AD
0.35065T
AC
0
0
AD
in Equations (1), (2), and (3) above, and solving the
resulting set of equations using conventional algorithms, algorithms, gives
116
T AB
2.63 kN
T AC
3.82 kN
T AD
2.43 kN
PROBLEM 2.109
A force P is applied as shown to a uniform cone which is supported by three cords, where the lines of action of the cords pass through the vertex A of the cone. Knowing that P 0 and that the tension in cord BE is 0.2 lb, determine the weight W of the cone.
SOLUTION
Note that because the line of action of each of the cords passes through the vertex A of the cone, the cords all have the same length, and the unit vectors lying along the cords are parallel to the unit vectors lying along the generators of the cone. Thus, for example, the unit vector along BE is identical to the unit vector along the generator AB.
Hence: It follows that:
T
BE
TCF
T
DG
T
BE
BE
TCF CF
T
AB
DG
T
T CF
DG
cos 45i 8 j sin 45k
BE
65
cos 45i 8j sin 45k 65
BE
cos 30i 8j sin 30k 65 cos15i 8j sin 15k 65
T DG
117
PROBLEM 2.109 CONTINUED
At A:
0: T
F
T
BE
CF
T
DG
W
P
0
Then, isolating the factors of i, j, and k , we obtain three algebraic equations: T
i:
or
65
cBoEs 45
T
j:
BE
65
or
T
k :
or
8
T
T
With P 0 and the tension in cord
BE
T
65
65
8
T
T
CF
65
CF
sBinE 45
cCoFs 30
cCoFs 30
T
siBnE 45 T
BE
T
cBoEs 45
T
65
W
8 DG
65
65
8
F sCin 30
sCinF 30
65
cDoGs15
cDoGs15
T
T
DG
T
T
T
P
W
P
65
0
0
(1)
0
0
T
65
sDinG15
(2)
sDinG15
0
0
(3)
0.2 lb:
Solving the resulting Equations (1), (2), and (3) using conventional methods in Linear Algebra (elimination, (elimination, matrix methods or iteration – with MATLAB or Maple, for example), we obtain: T CF
0.669 lb
T DG
0.746 lb W
118
1.603 lb
PROBLEM 2.110
A force P is applied as shown to a uniform cone which is supported by three cords, where the lines of action of the cords pass through the vertex A of the cone. Knowing that the cone weighs 1.6 lb, determine the range of values of P for which cord CF is taut.
SOLUTION
See Problem 2.109 for the Figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: i:
T
cBoEs 45
j:
T
k :
T
T
BE
cCoFs 30
T
T
CF
siBnE 45
W
DG
T
cDoGs15
T
65
8
sCinF 30
65 P
0
(1)
0
T
(2) sDinG15
0
(3)
With W 1.6 lb , the range of values of P for which the cord CF is taut can found by solving Equations (1), (2), and (3) for the tension T CF as a function of P and requirin requiring g it it to be positive positive ( 0). Solving (1), (2), and (3) with unknown P , using conventional methods in Linear Algebra (elimination, matrix methods or iteration – with MATLAB or Maple, for example), we obtain: TCF
Hence, for or
T CF
0
1.729P 0.668 lb 1.729 P
P
0.668
0
0.386 lb
119
0
P
0.386 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 3.6 kN, determine the vertical force P exerted by the tower on the pin at A.
SOLUTION
The force in each cable can be written as the product of the magnitude of the force and the unit vector along the cable. That is, with
18 m i 30 m j 5.4 m k
AC
2 2 2 18 m 30 m 5.4 m
AC
35.4 m
T
T
AC
AC
T
AC
T
AC
AC
and
AB
AB
AC
T
18 m i 30 m j 5.4 m k
AC
35.4 m
0.5085i 0.8475j 0.1525k
T
AC
6 m i 30 m j 7.5 m k
2 2 2 6 m 30 m 7.5 m
31.5 m
T
T
AB
AB
T
T
AB
AB
AB
AB
AB
AD
AD
6 m i 30 m j 7.5 m k
31.5 m
0.1905i 0.9524 j 0.2381k
T
Finally
T
AB
6 m i 30 m j 22.2 m k
2 2 2 6 m 30 m 22.2 m
37.8 m
T
T
AD
AD
T
AD
T
AD
AD
AD
T
T
AD
6 m i 30 m j 22.2 m k
37.8 m
0.1587i 0.7937 j 0.5873k
AD
120
PROBLEM 2.111 CONTINUED
With P
Pj,
at A : F
0: T
AB
T
AC
T
P j
AD
0
Equating the factors of i, j, and k to zero, we obtain the linear algebraic equations: i:
0.1905T
0.5085T AB
0.1587T AC
j:
0.9524T
0.8475T AB
0.7937T AC
k : 0.2381T
0.1525T AB
In Equations (1), (2) and (3), set
0.5873T AC
T AB
0 AD
(1)
P AD
0
0 AD
(2) (3)
3.6 kN, and, using conventional
methods for solving Linear Algebraic Equations (MATLAB or Maple, for example), we obtain: T AC
1.963 kN
T AD
1.969 kN P
121
6.66 kN
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 2.6 kN, determine the vertical force P exerted by the tower on the pin at A.
SOLUTION
Based on the results of Problem 2.111, particularly Equations (1), (2) and (3), we substitute
T AC
2.6 kN
and solve the three resulting linear equations using conventional tools for solving Linear Algebraic Equations (MATLAB or Maple, for example), to obtain T AB T AD
4.77 kN 2.61 kN P
122
8.81 kN
PROBLEM 2.113
A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AC is 15 lb, determine the weight of the plate.
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
AB
2 2 2 32 in. 48 in. 36 in.
AB
32 in. i 48 in. j 36 in. k 68 in.
T
T
AB
AB
AB
T
AB
AB
T
AB
AC
AC
T
AB
32 in. i 48 in. j 36 in. k
68 in.
0.4706i 0.7059 j 0.5294k
T
and
AB
45 in. i 48 in. j 36 in. k
2 2 2 45 in. 48 in. 36 in.
75 in.
T
T
AC
AC
T
AC AC
AC
T
AD
AD
123
T
AC
Finally,
T
AC
45 in. i 48 in. j 36 in. k
75 in.
0.60i 0.64j 0.48k
AC
25 in. i 48 in. j 36 in. k
2 2 2 25 in. 48 in. 36 in.
65 in.
PROBLEM 2.113 CONTINUED
T
T
AD
AD
T
With W
W j,
T
AD AD
AD
AD
T
T
AD
25 in. i 48 in. j 36 in. k
65 in.
0.3846i 0.7385j 0.5538k
AD
at A we have:
F 0: T
T
T
AB
AC
W j 0
AD
Equating the factors of i, j, and k to zero, we obtain the linear algebraic equations: i : 0.4706T
AB0.60T
AC
0.3846T
AD
0
(1)
j: 0.7059T
0.64T AB
AC
0.7385T
AD
W 0
(2)
0
(3)
k : 0.5294T
0.48T AB
In Equations (1), (2) and (3), set
0.5538T
AC
T AC
AD
15 lb, and, using conventional
methods for solving Linear Algebraic Equations (MATLAB or Maple, for example), we obtain: T AB
136.0 lb
T AD
143.0 lb W
124
211 lb
PROBLEM 2.114
A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 120 lb, determine the weight of the plate.
SOLUTION
Based on the results of Problem 2.111, particularly Equations (1), (2) and (3), we substitute
T AD
120 lb and
solve the three resulting linear equations using conventional tools for solving Linear Algebraic Equations (MATLAB or Maple, for example), to obtain T AC
12.59 lb
T AB
114.1 lb W
125
177.2 lb
PROBLEM 2.115
A horizontal circular plate having a mass of 28 kg is suspended as shown from three wires which are attached to a support D and form 30 angles with the vertical. Determine the tension in each wire.
SOLUTION
F x0 :
T siAnD30 sin 50
T CD sin 30 cos 60
T siBnD3 0 cos 40
0
Dividing through through by the factor sin 30 and evaluating the trigonometric functions gives 0.7660T
D0.7660T A
D0.50T B
CD 0
(1)
Similarly,
F z0 :
T siAnD30 cos 50
or
0.6428T
From (1)
T CD sin 30 sin 60
AD0.6428T
T
T siBnD3 0 sin 40
A D
T
0
BD0.8660T
B D
0.6527T
CD0
(2)
CD
Substituting this into (2): BD0.3573T
T
(3)
CD
Using T AD from above: T
AD
T
(4)
CD
Now,
F y 0 :
T cAoDs 30
or
T
126
T cBoDs 30
T cCoDs 30
28 kg 9.81 m/s 2 0 AD
T
BD
T CD 317.2 N
PROBLEM 2.115 CONTINUED
Using (3) and (4), above: TCD
Then:
0.3573TCD
T CD
317.2 N T AD T BD T CD
127
135.1 N
46.9 N
135.1 N
PROBLEM 2.116
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 tower exerts on the pin at A an upward vertical force of 8 kN, determine the tension in each wire.
SOLUTION
From the solutions of 2.111 and 2.112: T AB T AC T AD
Using P
0.5409P 0.295P 0.2959P
8 kN:
128
T AB
T AC
T AD
4.33 kN 2.36 kN 2.37 kN
PROBLEM 2.117
For the rectangular plate of Problems 2.113 and 2.114, determine the tension in each of the three cables knowing that the weight of the plate is 180 lb.
SOLUTION
From the solutions of 2.113 and 2.114:
Using
P
T AB
0.6440P
T AC
0.0709P
T AD
0.6771P
180 lb:
129
T AB
T AC
T AD
115.9 lb 12.76 lb 121.9 lb
PROBLEM 2.118
For the cone of Problem 2.110, determine the range of values of P for x direction. which cord DG is taut if P is directed in the – x
SOLUTION
From the solutions to Problems 2.109 and 2.110, have T
T
T
BE
T
sBinE 45
cBoEs 45
T
CF
T
T
DG
sCinF 30 T
cCoFs 30
(2)
0.2 65
T
sDinG15
cDoGs15
P
0 65
(3)
0
(1 )
Applying the method of elimination to obtain a desired result: Multiplyi Multiplying ng (2) by sin sin 45 and adding the result to (3): TCF
sin 45 sin 30 T sin 45 sin 15 0.2 DG
or
TCF
65 sin 45
0.9445 0.37 .3714T DG
(4)
Multiplyi Multiplying ng (2) by sin30 and subtracting (3) from the result: T
or
sin 30 sin 45 T sin 30 sin 15 0.2 BE
DG
T
0.6B6E79 0.6286T 0.
130
65 sin 30
DG
(5)
PROBLEM 2.118 CONTINUED
Substituting (4) (4) and (5) into (1) : 1.2903 1.7321T DG
T DG
P
is taut for P
65
1.2903 65
0 lb or 0
131
P
0.1600 lb
PROBLEM 2.119
A force P is applied as shown to a uniform cone which is supported by three cords, where the lines of action of the cords pass through the vertex A of the cone. Knowing that the cone weighs 2.4 lb and that P 0, determine the tension in each cord.
SOLUTION
Note that because the line of action of each of the cords passes through the vertex A of the cone, the cords all have the same length, and the unit vectors lying along the cords are parallel to the unit vectors lying along the generators of the cone. Thus, for example, the unit vector along BE is identical to the unit vector along the generator AB. Hence:
AB
cos 45i 8 j sin 45k
BE
65
It follows that: T
BE
TCF
T
At A:
DG
T
BE
TCF CF
T
DG
T
T CF
BE
DG
cos 45i 8 j sin 45k 65
BE
cos 30i 8j sin 30k 65 cos15i 8j sin 15k 65
T DG
F 0: T
T
BE
T
CF
132
WP 0
DG
PROBLEM 2.119 CONTINUED
Then, isolating the factors if i, j, and k we obtain three algebraic equations: i:
T
65
or
8
T
BE
65
or
T
k :
or
T
65 T
65
cBoEs 45
T
j:
T
cBoEs 45
CF
65
T
CF
sBinE 45
T
65
T
65
DG
65
2.4
DG
cDoGs15
T
8
T
T
BE
T
cCoFs 30
T
8
T
sBinE 45
cCoFs 30
T
F sCin 30
sCinF 30
T
0
cDoGs15
W
65
8
65
0
(1)
0
0.3 65
(2)
sDinG15 P 0
sDinG15
P
65
(3)
With P 0, the tension in the cords can be found by solving the resulting Equations (1), (2), and (3) using conventional methods in Linear Algebra (elimination, matrix methods or iteration–with MATLAB or Maple, for example). We obtain
133
T BE
0.299 0.299 lb
T CF
1.002 1.002 lb
T DG
1.11 1.117 7 lb
PROBLEM 2.120
A force P is applied as shown to a uniform cone which is supported by three cords, where the lines of action of the cords pass through the vertex A of the cone. Knowing that the cone weighs 2.4 lb and that P 0.1 lb, determine the tension in each cord.
SOLUTION
See Problem 2.121 for the analysis leading to the linear algebraic Equations Equations (1), (2), and (3) below: T
cBoEs 45 T
T
sBinE 45
T
BE
T
T
cCoFs 30 CF
T
sCinF 30
T
DG
T
cDoGs15
0
(1)
0.3 65 sDinG15
(2) P
65
(3)
With P 0.1lb, solving (1), (2), and (3), using conventional methods in Linear Algebra (elimination, matrix methods or iteration–with MATLAB or Maple, for example), we obtain
134
T BE
1.006 1.006 lb
T CF
0.35 0.357 7 lb
T DG
1.056 1.056 lb
PROBLEM 2.121
Using two ropes and a roller chute, two workers are unloading a 200-kg cast-iron counterweight from a truck. Knowing that at the instant shown the counterweight is kept from moving and that the positions of points A, B, and C are, respectively, A(0, –0.5 m, 1 m), B(–0.6 m, 0.8 m, 0), and (0.7 m, 0.9 m, 0), and assuming that no friction exists between the C (0.7 counterweight and the chute, determine the tension in each rope. ( Hint: Since there is no friction, the force exerted by the chute on the counterweight must be perpendicular to the chute.)
SOLUTION
From the geometry of the chute: N
N
5
2 j k N 0.8944 j 0.4472k
As in Problem 2.11, for example, the force in each rope can be written as the product of the magnitude of the force and the unit vector along the cable. Thus, with
0.6 m i 1.3 m j 1 m k
AB
2 2 2 0.6 m 1.3 m 1 m 1.764 m
AB
T
T
AB
AB
T
AB
T
AB
AB
and
AC
AC
T
AB
0.6 m i 1.3 m j 1 m k
1.764 m
0.3436i 0.7444 j 0.5726k
T
AB
AB
0.7 m i 1.4 m j 1 m k
2 2 2 0.7 m 1.4 m 1 m 1.8574 m
T
T
AC
AC
T
AC
T
AC
AC
AC
T
T
AC
0.3769i 0.7537 j 0.5384k
AC
F 0 : N T
Then:
135
0.7 m i 1.4 m j 1 m k
1.764 m
T
AB
W0
AC
PROBLEM 2.121 CONTINUED
With
W
200 kg 9.81 m/s 1962 N,
and equating the factors of i, j,
and k to zero, we obtain the linear algebraic equations: i:
0.3436T
j: 0.7444T k :
0.3769T
0.7537T
0.5726T
(1)
0.8 944 N 1962
AC
0.5384T
AB
AC
AB
0
AB
0.4 472 N
AC
0
(2)
0
(3)
Using conventional methods for solving Linear Algebraic Equations (elimination, (elimination, MATLAB or Maple, for example), we obtain N
136
1311 N T AB
551 N
T AC
503 N
PROBLEM 2.122
Solve Problem 2.121 assuming that a third worker is exerting a force P (180 N)i on the counterweight. Problem 2.121: Using two ropes and a roller chute, two workers are unloading a 200-kg cast-iron counterweight from a truck. Knowing that at the instant shown the counterweight is kept from moving and that the positions of points A, B, and C are, respectively, A(0, –0.5 m, 1 m), B(–0.6 m, 0.8 m, 0), and C (0.7 (0.7 m, 0.9 m, 0), and assuming that no friction exists between the counterweight and the chute, determine the tension in each rope. ( Hint: Since there is no friction, the force exerted by the chute on the counterweight must be perpendicular perpendicular to the chute.)
SOLUTION
From the geometry of the chute: N
N
5
2 j k N 0.8944 j 0.4472k
As in Problem 2.11, for example, the force in each rope can be written as the product of the magnitude of the force and the unit vector along the cable. Thus, with
0.6 m i 1.3 m j 1 m k
AB
2 2 2 0.6 m 1.3 m 1 m 1.764 m
AB
T
T
AB
AB
T
AB
AB
T
AB
AC
AC
T
0.6 m i 1.3 m j 1 m k
AB
1.764 m
0.3436i 0.7444j 0.5726k
T
and
AB
AB
0.7 m i 1.4 m j 1 m k
2 2 2 0.7 m 1.4 m 1 m 1.8574 m
T
T
AC
AC
T
AC
T
AC
AC
T
AC
T
1.764 m
0.3769i 0.7537 j 0.5384k
AC
F 0 : N T
Then:
137
0.7 m i 1.4 m j 1 m k
AC
T
AB
PW0
AC
PROBLEM 2.122 CONTINUED
P 180 N i
Where
.81 m/s 2 j W 200 kg 9.81
and
1962 N j Equating the factors of i, j, and k to zero, we obtain the linear equations: i : 0.3436T
AB0.3769T
AC180 0
j: 0.8944
N
0.7444
T
AB0.7537
T
AC1 96 2 0
k : 0.4472
N
0.5726
T
AB0.5384
T
AC 0
Using conventional methods for solving Linear Algebraic Equations (elimination, (elimination, MATLAB or Maple, for example), we obtain N
138
1302 N T AB
306 N
T AC
756 N
PROBLEM 2.123
A piece of machinery of weight W is temporarily supported by cables AB, AC , and ADE . Cable ADE is attached to the ring at A, passes over the pulley at D and back through the ring, and is attached to the support at E . Knowing that W 320 lb, determine the tension in each cable. ( Hint: The tension is the same in all portions of cable ADE .) .)
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 AB
9 ft i 8 ft j 12 ft k
AB
2 2 2 9 ft 8 ft 12 ft 17 ft
T
T
AB
T
AB
T
T
AB
AB AB
AB
T AB
9 ft i 8 ft j 12 ft k 17 ft
0.5294i 0.4706 j 0.7059k
AB
and
AC
0 i 8 ft j 6 ft k
AC
2 2 2 0 ft 8 ft 6 ft 10 ft
T
T
AC
T
AC
T
AC
T
AC AC
AC
T
AC
0 ft i 8 ft j 6 ft k
10 ft
0.8j 0.6k
AC
and AD
4 ft i 8 ft j 1 ft k
AD
2 2 2 4 ft 8 ft 1 ft 9 ft
T
T
AD
T
AD
T
AD
T
AD ADE
AD
T
4 ft i 8 ft j 1 ft k 9 ft ADE
0.4444i 0.8889 j 0.1111k
ADE
139
PROBLEM 2.123 CONTINUED
Finally, AE
8 ft i 8 ft j 4 ft k
AE
2 2 2 8 ft 8 ft 4 ft 12 ft
T T
T
AE
AE
T
AE
T
AE ADE
AE
T ADE
8 ft i 8 ft j 4 ft k 12 ft
0.6667i 0.6667 j 0.3333k
ADE
With the weight of the machinery, W W j, at A, we have:
F 0 : T
T
AB
2T
AC
W j 0
AD
Equating the factors of i, j, and k to zero, we obtain the following linear algebraic equations:
0.5294T 0.4706T
0.7059T Knowing Knowing that
W
0.4444T A2B 0.
AB0.8T AB0.6T
ADE0.6667T
AC2 0.8889T AC2 0.1111T
0.6667T
ADE
0.3333T
ADE
AD0E
(1)
W 0
(2)
0
(3)
ADE
ADE
320 lb, we can solve Equations (1), (2) and (3) using conventional methods for solving
Linear Algebraic Equations (elimination, (elimination, matrix methods via MATLAB or Maple, for example) to obtain T AB
46.5 lb 46.5
T AC
34.2 lb 34.2
T ADE
140
110.8 lb 110.8
PROBLEM 2.124
A piece of machinery of weight W is temporarily supported by cables AB, AC , and ADE . Cable ADE is attached to the ring at A, passes over the pulley at D and back through the ring, and is attached to the support at E . Knowing that the tension in cable AB is 68 lb, determine ( a) the tension in AC , ( b) the tension in ADE , (c) the weight W . ( Hint: The tension is the same in all portions of cable ADE .) .)
SOLUTION
See Problem 2.123 for the analysis leading to the linear algebraic Equations Equations (1), (2), and (3), below: 0.5294T
0.4706T
0.8T
AB
0.7059T
2 0. 0.4444T
AB
2 0.8889T
AC AC
0.6T
AB
ADE0.6667T
2 0.1111T
AC AC
AD E
0.6667T
ADE
0
ADE
0.3333T
(1)
W A DE
0
(2)
0
(3)
AD E
Knowing that the tension in cable AB is 68 lb, we can solve Equations Equations (1), (2) (2) and (3) (3) using conventional methods for solving Linear Algebraic Equations (elimination, matrix methods via MATLAB or Maple, for example) to obtain (a) (b) (c)
141
T AC T AE
W
50.0 50.0 lb
162.0 162.0 lb
468 468 lb
PROBLEM 2.125
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 P i and Q Qk are applied to the ring to maintain the container is the position shown. Knowing that W 1200 N, determine P and Q. ( Hint: The tension is the same in both portions of cable BAC .) .)
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
AB 0.48 m i 0.72 m j 0.16 m k AB
2 2 2 0.48 m 0.72 m 0.16 m 0.88 m
T
T
AB
T
T
AB
AB
T
AB
AB
AB
T AB
0.48 m i 0.72 m j 0.16 m k
0.88 0.88 m
0.5455i 0.8182 j 0.1818k
AB
and AC 0.24 m i 0.72 m j 0.13 m k
2 2 2 0.24 m 0.72 m 0.13 m 0.77 m
AC
T
T
AC
T
AC
At A:
T
AC
T
AC AC
AC0 .3177i
AC
T AC
0.24 m i 0.72 m j 0.13 m k
0.77 0.77 m
0.9351j 0.1688k
F 0: T
142
T
AB
PQW 0
AC
PROBLEM 2.125 CONTINUED
Noting that T
T A Cbecause of the ring A, we equate the factors of
AB
and k to zero to obtain the linear algebraic equations: i, j, and i:
or
0.5455 0.3177 T P
j:
or
Q
W
0
1.7532T
0.1818 0.1688 T
or With With W
P 0
0.2338T
0.8182 0.9351 T W
k :
Q
0
0.356T
120 1200 0 N: T
1200 1200 N 1.7532
684. 684.5 5N P 160. 160.0 0 N Q
143
240 N
PROBLEM 2.126
For the system of Problem 2.125, determine W and P knowing that Q 160 N.
Problem 2.125: 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 P i and Q Qk are applied to the ring to
maintain the container is the position shown. Knowing that W 1200 N, determine P and Q. ( Hint: The tension is the same in b oth portions of cable BAC .) .)
SOLUTION
Based on the results of Problem 2.125, particularly the three equations relating P , Q, W, and T we substitute Q 160 N to obtain
T
160 N
0.3506
456. 456.3 3N W P
144
800 N
107. 107.0 0 N
PROBLEM 2.127
Collars A and B and B are connected by a 1-m-long wire and can slide freely on frictionless rods. If a force P (680 N) j is applied at A, A, determine (a) the tension in the wire when y
300 mm, (b (b) the magnitude of the
force Q required to maintain the equilibrium of the system.
SOLUTION
Free-Body Diagrams of collars
For both Problems 2.127 and 2.128:
1 m
Here
AB 2
y 2
or
2
x2
0.40 m
z 2
2
y2
z2
y2
z 2
0.84 0.84 m 2
Thus, with y with y given, z given, z is is determined. Now
AB
AB
AB
1 1m
0.40i
yj zk m
0.4i
yk
Where y and z and z are are in units of meters, m. From the F.B. Diagram of collar A collar A:: F
0:
N ix N kz
Pj T A B
AB
0
Setting the j coefficient to zero gives: P yT AB
0
With With P 680 680 N, T AB
680 N y
Now, from the free body diagram of collar B collar B:: F
145
0:
N ix N jy Qk
T A B
AB
0
zk
PROBLEM 2.127 CONTINUED
Setting the k coefficient to zero gives: Q
T AB z 0
And using the above result for T AB we have Q
Then, Then, from from the specif specific icati ations ons of the proble problem, m, y
z 2
T AB z
680 N
z
y
0.84 m 2
0 .3 m
300 mm
0.3 m
2
z 0.86 0.866 6m
and (a)
T AB
680 N
0.30
2266 2266.7 .7 N T AB
or
2.27 2.27 kN
and (b)
Q
2266.7 0.866
1963.2 N Q
or
146
1.963 kN 1.963
PROBLEM 2.128
Solve Problem 2.127 assuming y
550 mm.
Problem 2.127: Collars A and B are connected by a 1-m-long wire and can slide freely on frictionless rods. If a force P (680 N) j is applied at
A, A, determine (a (a) the tension in the wire when y
300 mm, (b (b) the
magnitude of the force Q required to maintain the equilibrium of the system.
SOLUTION
From the analysis of Problem 2.127, particularly the results: y 2
z 2
T AB
Q With y
550 mm
0.84 0.84 m 2
680 N y
680 N y
z
0.55 m, we obtain: z 2
0.84 m2
0.55 m
2
z 0.73 0.733 3m
and (a)
T AB
680 N 0.55
1236 1236.4 .4 N
or
T AB
1.236 1.23 6 kN
Q
0.906 0.90 6 kN
and (b)
Q or
147
1236 0.866 N
906 N
PROBLEM 2.129
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
(a)
35 P sin 35 P
300 1b 1b
300 300 lb sin35 P
523 523 lb
428 428 lb
(b) Vertical Component Pv
P cos35
523 lb cos35 P v
148
PROBLEM 2.130
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 which passes over a pulley at B and through ring A and which 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
AB
0.78 m i 1.6 m j 0 m k
AB
2 2 2 0.78 m 1.6 m 0 1.78 m
T T
T
AB
AB
T
AB
T
AB
AB
AB
T
0.78 m i 1.6 m j 0 m k
AB
1.78 1.78 m
0.4382i 0.8989 j 0k
AB
and
AC
0 i 1.6 m j 1.2 m k
AC
2 2 2 0 m 1.6 m 1.2 m
2m
T
T
AC
T
AC
AC
T
AC AC
AC
T
AC
0 i 1.6 m j 1.2 m k
2m
0.8j 0.6k
T
AC
and AD
1.3 m i 1.6 m j 0.4 m k
AD
2 2 2 1.3 m 1.6 m 0.4 m 2.1 m
T
T
AD
T
AD
T
AD
T
AD AD
AD
T
AD
1.3 m i 1.6 m j 0.4 m k
2.1m
0.6190i 0.7619 j 0.1905k
AD
149
PROBLEM 2.130 CONTINUED
Finally,
AE
0.4 m i 1.6 m j 0.86 m k
AE
2 2 2 0.4 m 1.6 m 0.86 m 1.86 m
T
T
AE
T
AE
T
AE
T
AE AE
AE
T
AE
0.4 m i 1.6 m j 0.86 m k
1.86 1.86 m
0.2151i 0.8602 j 0.4624k
AE
With the weight of the container W W j, at A we have:
F 0: T
T
AB
T
AC
W j 0
AD
Equating Equating the factors factors of i, j, and k to zero, we obtain the following linear algebraic equations:
0.4382T 0.8989T
0.8T
AB
0.6T Knowing Knowing that
W
AB0.6190T
AD0.2151T
0.7619T
AE0
0.8602T
AC
AD
AC0.1905T
AD0.4624T
(1)
W 0
(2)
AE
AE0
1000 N and that because of the pulley system at
(3) B T
AB
T
AD
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
150
378 N
PROBLEM 2.131
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 which passes over a pulley at B and through ring A and which is attached to a support at D. Knowing that the tension in cable AC is 150 N, determine ( a) the magnitude of the force P, (b) the weight W of the container. ( Hint: The tension is the same in all portions of cable FBAD.)
SOLUTION
Here, as in Problem 2.130, 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 T
AB
T
AD
P
and using the linear algebraic equations of Problem 2.131 with
T AC
150 N, we obtain (a) (b)
151
P W
454 N
1202 N
PROBLEM 2.132
Two cables tied together at C are loaded as shown. Knowing that Q 60 lb, determine the tension (a ( a) in cable AC , (b (b) in cable BC .
SOLUTION
F
y
Q
With (a)
T CA
(b)
T
0:
F
0 x:
A C
Qcos 30
0
60 lb
60 lb 0.866
P T
T CA
52.0 lb
or T CB
45.0 lb
CBQsin 30
0
P 75 lb
With T CB
152
75 lb
60 lb 0.50
PROBLEM 2.133
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
Have
F
x
0:
T
T CA
or Then for
From
or
F
y
0:
TCB
75 lb or
Q
Therefore,
0.50Q
60 lb
0.50Q
0.50Q
Thus,
60 lb
P Qsin 30
C B
75 lb
T CB
69.3 lb
T
For
0
60 lb
0.8660Q Q
0.8660 Q
T CA
or
Qcos 30
A C
60 lb
15 lb
30 lb 30.0
153
Q
69.3 lb
PROBLEM 2.134
A welded connection is in equilibrium under the action of the four forces shown. Knowing that F A 8 kN and F B 16 kN, determine the magnitudes of the other two forces.
SOLUTION
Free-Body Diagram of Connection
3
Fx 0:
With
5
F A F C
Fy 0:
4 5
FB FC
8 kN, F B
3
FA 0
5
16 kN
4
16 kN 8 kN 5
FD
3 5
FB
3 5
F C
6.40 kN
F D
4.80 kN
FA 0
With F A and F B as above: F D
154
3 5
3
16 kN 8 kN 5
PROBLEM 2.135
A welded connection is in equilibrium under the action of the four forces shown. Knowing that F A 5 kN and F D 6 kN, determine the magnitudes of the other two forces.
SOLUTION
Free-Body Diagram of Connection
Fy 0: FD
3 5
FA
FB FD
or
3 5
3 5
FB 0
FA
F A 5 kN, F D 8 kN
With
F B
5
3
3
5
6 kN
5 kN F B 15.00 kN
Fx 0: FC F C
4
4
5
5
4 5
FB
4 5
FA 0
FB FA
15 kN 5 kN F C 8.00 kN
155
PROBLEM 2.136
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
Free-Body Diagram of Collar
(a)
Triangle Proportions F x
0:
P
4.5 20.5
50 lb or
(b)
0
P
10.98 lb
Triangle Proportions F x
0:
P
15 25
50 lb or
156
0
P
30.0 lb
PROBLEM 2.137
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 Diagram of Collar
Triangle Proportions
Hence:
F x
0:
50 xˆ
48
400 ˆ x
or
48 50
400
ˆ x
ˆ2
0.92 .92 lb lb 400 400
2
4737.7 in 2
ˆ x
ˆ x
2
0
2
ˆ x
2
ˆ x
157
68.6 in.
PROBLEM 2.138
A frame ABC is supported in part by cable DBE which 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
The force in cable DB can be written as the product of the magnitude of the force and the unit vector along the cable. That is, with DB 480 mm i 510 mm j 320 mm k
DB
2 2 2 480 510 320
770 mm
F F DB
DB 385 N 480 mm i 510 mm j 320 mm k F 770 mm DB F 240 N i 255 N j 160 N k Fx 240 N,
158
Fy 255 N,
N Fz 160.0
PROBLEM 2.139
A frame ABC is supported in part by cable DBE which passes through a frictionless ring at B. 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.
SOLUTION
The force in each cable can be written as the product of the magnitude of the force and the unit vector along the cable. That is, with
0.48 m i 0.51 m j 0.32 m k
BD
BD
2 2 2 0.48 m 0.51 m 0.32 m
0.77 m
T
T
BD
BD
T
BD
BD
BD
T
0.48 m i 0.51 m j 0.32 m k
0.77 m
0.6234i 0.6623j 0.4156k
T
BD
T
BD
BD
and
BE
BE
0.27 m i 0.40 m j 0.6 m k
2 2 2 0.27 m 0.40 m 0.6 m
0.770 m
T
T
BE
BE
T
BD BE
BD
T
BE
T
Now, because of the frictionless ring at B,
T
BE
0.26 m i 0.40 m j 0.6 m k
0.3506i 0.5195j 0.7792k
BE
T
0.770 m
BE
T
385 N and the force on the support due to the two
BD
cables is F 385 N 0.6234i 0.6623j 0.4156k 0.3506i 0.5195j 0.7792k
375 N i 455 N j 460 N k
159
PROBLEM 2.139 CONTINUED
The magnitude of the resultant is F
2
2
2
F x F y F z
375 N
2
455 N
2
460 N
2
747.83 N or F
748 N
The direction of this force is: x
cos1
y
cos 1
z
cos 1
375
or
747.83 455
or
747.83 460
or
747.83
160
x
120.1
y
z
52.5
128.0
PROBLEM 2.140
A steel tank is to be positioned in an excavation. Using trigonometry, determine ( 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
Force Triangle
(a) For minimum P it must be perpendicular to the vertical resultant R P
425 lb cos30 or P
(b)
R
425 lb lb sin30 sin30 425 or
161
368 lb
R
213 lb
