Chapter 2

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: 8.4 kNR
19
8.4 kNR 19
1

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 , 59
(a)
(b)
We measure: 575 N, 67R
575 NR 67
2

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)
We measure: 37 lb, 76R
37 lbR 76
3

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: 61.5 lb, 86.5R
61.5 lbR 86.5
4

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 F1 120 N, determine
(a) the required force F2 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: 2 108 NF
77 NR
By trigonometry: Law of Sines
2 120
sin sin38 sin
F R
90 28 62 , 180 62 38 80
Then:
2 120 N
sin 62 sin38 sin80
F R
or (a) 2 107.6 NF
(b) 75.0 NR
5

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 F2 80 N, determine
(a) the required force F1 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
SOLUTION
Using the Law of Sines
1 80
sin sin38 sin
F R
90 10 80 , 180 80 38 62
Then:
1 80 N
sin80 sin38 sin62
F R
or (a) 1 89.2 NF
(b) 55.8 NR
6

PROBLEM 2.7
The 50-lb force is to be resolved into components 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 component along - ?b b
SOLUTION
Using the triangle rule and the Law of Sines
(a)
sin sin 40
35 lb 50 lb
sin 0.44995
26.74
Then: 40 180
113.3
(b) Using the Law of Sines:
50 lb
sin sin 40
bbF
71.5 lbbbF
7

PROBLEM 2.8
The 50-lb force is to be resolved into components 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 component along - ?a a
SOLUTION
(a)
sin sin 40
30 lb 50 lb
sin 0.3857
22.7
(b) 40 180
117.31
50 lb
sin sin 40
aaF
sin
50 lb
sin 40
aaF
69.1lbaaF
8

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
Have: 180 35 25
120
Then:
360 N
sin35 sin120 sin 25
P R
or (a) 489 NP
(b) 738 NR
9

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 vertical, (b) the corresponding magnitude of R.
SOLUTION
(a) Have:
360 N 300 N
sin sin35
sin 0.68829
43.5
(b) 180 35 43.5
101.5
Then:
300 N
sin101.5 sin35
R
or 513 NR
10

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
(a) Have:
20 lb 14 lb
sin sin30
sin 0.71428
45.6
(b) 180 30 45.6
104.4
Then:
14 lb
sin104.4 sin30
R
27.1 lbR
11

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) the required magnitude of
the force Q if the resultant R of the two forces applied at A is to be
vertical, (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
SOLUTION
(a) Have:
25 lb
sin15 sin30
Q
12.94 lbQ
(b) 180 15 30
135
Thus:
25 lb
sin135 sin30
R
sin135
25 lb 35.36 lb
sin30
R
35.4 lbR
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.
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
(a) The smallest force P will be perpendicular to R, that is, vertical
20 lb sin30P
10 lb 10 lbP
(b) 20 lb cos30R
17.32 lb 17.32 lbR
13

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 P is minimum when is 90 , that is, P is horizontal
Then: (a) 360 N sin35P
or 206 NP
And: (b) 360 N cos35R
or 295 NR
14

PROBLEM 2.15
For the hook support of Problem 2.11, 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 .
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 force triangle and the Law of Cosines
2 22
10 lb 20 lb 2 10 lb 20 lb cos110R
2
100 400 400 0.342 lb
2
636.8 lb
25.23 lbR
Using now the Law of Sines
10 lb 25.23 lb
sin sin110
10 lb
sin sin110
25.23 lb
0.3724
So: 21.87
Angle of inclination of R, is then such that:
30
8.13
Hence: 25.2 lbR 8.13
15

PROBLEM 2.16
Solve Problem 2.1 using trigonometry
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
Using the force triangle, the Law of Cosines and the Law of Sines
We have: 180 50 25
105
Then:
2 22
4.5 kN 6 kN 2 4.5 kN 6 kN cos105R
2
70.226 kN
or 8.3801 kNR
Now:
8.3801 kN 6 kN
sin105 sin
6 kN
sin sin105
8.3801 kN
0.6916
43.756
8.38 kNR 18.76
16

PROBLEM 2.17
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
From the geometry of the problem:
1 2
tan 38.66
2.5
1 1.5
tan 30.96
2.5
Now: 180 38.66 30.96 110.38
And, using the Law of Cosines:
2 22
500 N 160 N 2 500 N 160 N cos110.38R
2
331319 N
575.6 NR
Using the Law of Sines:
160 N 575.6 N
sin sin110.38
160 N
sin sin110.38
575.6 N
0.2606
15.1
90 66.44
576 NR 66.4
17

PROBLEM 2.18
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
SOLUTION
Using the force triangle and the Laws of Cosines and Sines
We have:
180 15 30
135
Then:
2 22
15 lb 25 lb 2 15 lb 25 lb cos135R
2
1380.3 lb
or 37.15 lbR
and
25 lb 37.15 lb
sin sin135
25 lb
sin sin135
37.15 lb
0.4758
28.41
Then: 75 180
76.59
37.2 lbR 76.6
18

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
We have: 180 45 25 110
Then:
2 22
30 kN 20 kN 2 30 kN 20 kN cos110R
2
1710.4 kN
41.357 kNR
and
20 kN 41.357 kN
sin sin110
20 kN
sin sin110
41.357 kN
0.4544
27.028
Hence: 45 72.028
41.4 kNR 72.0
19

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
We have: 180 45 25 110
Then:
2 22
30 kN 20 kN 2 30 kN 20 kN cos110R
2
1710.4 kN
41.357 kNR
and
30 kN 41.357 kN
sin sin110
30 kN
sin sin110
41.357 kN
0.6816
42.97
Finally: 45 87.97
41.4 kNR 88.0
20

PROBLEM 2.21
Determine the x and y components of each of the forces shown.
SOLUTION
20 kN Force:
20 kN cos40 ,xF 15.32 kNxF
20 kN sin 40 ,yF 12.86 kNyF
30 kN Force:
30 kN sin70 ,yF 28.2 kNyF
42 kN Force:
42 kN sin 20 ,yF 14.36 kNyF
21

PROBLEM 2.22
SOLUTION
40 lb Force:
40 lb sin50 ,xF 30.6 lbxF
40 lb cos50 ,yF 25.7 lbyF
60 lb Force:
60 lb cos60 ,xF 30.0 lbxF
60 lb sin 60 ,yF 52.0 lbyF
80 lb Force:
80 lb cos25 ,xF 72.5 lbxF
80 lb sin 25 ,yF 33.8 lbyF
22

PROBLEM 2.23
SOLUTION
We compute the following distances:
2 2
2 2
2 2
48 90 102 in.
56 90 106 in.
80 60 100 in.
OA
OB
OC
Then:
204 lb Force:
48
102 lb ,
102
xF 48.0 lbxF
90
102 lb ,
102
yF 90.0 lbyF
212 lb Force:
56
212 lb ,
106
xF 112.0 lbxF
90
212 lb ,
106
yF 180.0 lbyF
400 lb Force:
80
400 lb ,
100
xF 320 lbxF
60
400 lb ,
100
yF 240 lbyF
23

PROBLEM 2.24
SOLUTION
We compute the following distances:
2 2
70 240 250 mmOA
2 2
210 200 290 mmOB
2 2
120 225 255 mmOC
500 N Force:
70
500 N
250
xF 140.0 NxF
240
500 N
250
yF 480 NyF
435 N Force:
210
435 N
290
xF 315 NxF
200
435 N
290
yF 300 NyF
510 N Force:
120
510 N
255
xF 240 NxF
225
510 N
255
yF 450 NyF
24

PROBLEM 2.25
While emptying a wheelbarrow, a gardener exerts on each handle AB a
force P directed along line CD. Knowing that P must have a 135-N
horizontal component, determine (a) the magnitude of the force P, (b) its
vertical component.
SOLUTION
(a)
cos40
xP
P
135 N
cos40
or 176.2 NP
(b) tan 40 sin 40y xP P P
135 N tan 40
or 113.3 NyP
25

PROBLEM 2.26
Member BD exerts on member ABC a force P directed along line BD.
Knowing that P must have a 960-N vertical component, determine (a) the
magnitude of the force P, (b) its horizontal component.
SOLUTION
(a)
sin35
yP
P
960 N
sin35
or 1674 NP
(b)
tan35
y
x
P
P
960 N
tan35
or 1371 NxP
26

PROBLEM 2.27
Member CB of the vise shown exerts on block B a force P directed along
line CB. Knowing that P must have a 260-lb horizontal component,
determine (a) the magnitude of the force P, (b) its vertical component.
SOLUTION
We note:
CB exerts force P on B along CB, and the horizontal component of P is 260 lb.xP
Then:
(a) sin50xP P
sin50
xP
P
260 lb
sin50
339.4 lb 339 lbP
(b) tan50x yP P
tan50
x
y
P
P
260 lb
tan50
218.2 lb 218 lbyP
27

PROBLEM 2.28
Activator rod AB exerts on crank BCD a force P directed along line AB.
Knowing that P must have a 25-lb component perpendicular to arm BC of
the crank, determine (a) the magnitude of the force P, (b) its component
along line BC.
SOLUTION
Using the x and y axes shown.
(a) 25 lbyP
Then:
sin 75
yP
P
25 lb
sin75
or 25.9 lbP
(b)
tan 75
y
x
P
P
25 lb
tan 75
or 6.70 lbxP
28

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 BD, and the component of P along AC
is 450 N.
Then:
(a)
450 N
549.3 N
cos35
P
549 NP
(b) 450 N tan35xP
315.1 N
315 NxP
29

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) the magnitude of the force P, (b) its component along
line AC.
SOLUTION
(a)
sin38
xP
P
200 N
sin38
324.8 N or 325 NP
(b)
tan38
x
y
P
P
200 N
tan38
255.98 N
or 256 NyP
30

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:
500 140 N 480 NF i j
425 315 N 300 NF i j
510 240 N 450 NF i j
415 N 330 NR F i j
Then:
1 330
tan 38.5
415
2 2
415 N 330 N 530.2 NR
Thus: 530 NR 38.5
31

PROBLEM 2.32
shown.
SOLUTION
From Problem 2.21:
20 15.32 kN 12.86 kNF i j
30 10.26 kN 28.2 kNF i j
42 39.5 kN 14.36 kNF i j
34.44 kN 55.42 kNR F i j
Then:
1 55.42
tan 58.1
34.44
2 2
55.42 kN 34.44 N 65.2 kNR
65.2 kNR 58.2
32

PROBLEM 2.33
shown.
SOLUTION
The components of the forces were determined in 2.23.
x yR RR i j
71.9 lb 43.86 lbi j
43.86
tan
71.9
31.38
2 2
71.9 lb 43.86 lbR
84.23 lb
84.2 lbR 31.4
Force comp. (lb)x comp. (lb)y
40 lb 30.6 25.7
60 lb 30 51.96
80 lb 72.5 33.8
71.9xR 43.86yR
33

PROBLEM 2.34
shown.
SOLUTION
The components of the forces were
determined in Problem 2.23.
204 48.0 lb 90.0 lbF i j
212 112.0 lb 180.0 lbF i j
400 320 lb 240 lbF i j
Thus
x yR R R
256 lb 30.0 lbR i j
Now:
30.0
tan
256
1 30.0
tan 6.68
256
and
2 2
256 lb 30.0 lbR
257.75 lb
258 lbR 6.68
34

PROBLEM 2.35
Knowing that 35 , determine the resultant of the three forces
shown.
SOLUTION
300-N Force:
300 N cos20 281.9 NxF
300 N sin 20 102.6 NyF
400-N Force:
400 N cos55 229.4 NxF
400 N sin55 327.7 NyF
600-N Force:
600 N cos35 491.5 NxF
600 N sin35 344.1 NyF
and
1002.8 Nx xR F
86.2 Ny yR F
2 2
1002.8 N 86.2 N 1006.5 NR
Further:
86.2
tan
1002.8
1 86.2
tan 4.91
1002.8
1007 NR 4.91
35

PROBLEM 2.36
shown.
SOLUTION
300-N Force:
300 N cos20 281.9 NxF
300 N sin 20 102.6 NyF
400-N Force:
400 N cos85 34.9 NxF
400 N sin85 398.5 NyF
600-N Force:
600 N cos5 597.7 NxF
600 N sin5 52.3 NyF
and
914.5 Nx xR F
448.8 Ny yR F
2 2
914.5 N 448.8 N 1018.7 NR
Further:
448.8
tan
914.5
1 448.8
tan 26.1
914.5
1019 NR 26.1
36

PROBLEM 2.37
Knowing that the tension in cable BC is 145 lb, determine the resultant of
the three forces exerted at point B of beam AB.
SOLUTION
Cable BC Force:
84
145 lb 105 lb
116
xF
80
145 lb 100 lb
116
yF
100-lb Force:
3
100 lb 60 lb
5
xF
4
100 lb 80 lb
5
yF
156-lb Force:
12
156 lb 144 lb
13
xF
5
156 lb 60 lb
13
yF
and
21 lb, 40 lbx x y yR F R F
2 2
21 lb 40 lb 45.177 lbR
Further:
40
tan
21
1 40
tan 62.3
21
Thus: 45.2 lbR 62.3
37

PROBLEM 2.38
shown.
SOLUTION
The resultant force R has the x- and y-components:
140 lb cos50 60 lb cos85 160 lb cos50x xR F
7.6264 lbxR
and
140 lb sin 50 60 lb sin85 160 lb sin50y yR F
289.59 lbyR
Further:
290
tan
7.6
1 290
tan 88.5
7.6
Thus: 290 lbR 88.5
38

PROBLEM 2.39
Determine (a) the required value of if the resultant of the three forces
shown is to be vertical, (b) the corresponding magnitude of the resultant.
SOLUTION
For an arbitrary angle , we have:
140 lb cos 60 lb cos 35 160 lb cosx xR F
(a) So, for R to be vertical:
140 lb cos 60 lb cos 35 160 lb cos 0x xR F
Expanding,
cos 3 cos cos35 sin sin35 0
Then:
1
3
cos35
tan
sin35
or
1
1 3
cos35
tan 40.265
sin35
40.3
(b) Now:
140 lb sin 40.265 60 lb sin 75.265 160 lb sin 40.265y yR R F
252 lbR R
39

PROBLEM 2.40
For the beam of Problem 2.37, determine (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.
Problem 2.37: Knowing that the tension in cable BC is 145 lb, determine
the resultant of the three forces exerted at point B of beam AB.
SOLUTION
We have:
84 12 3
156 lb 100 lb
116 13 5
x x BCR F T
or 0.724 84 lbx BCR T
and
80 5 4
156 lb 100 lb
116 13 5
y y BCR F T
0.6897 140 lby BCR T
(a) So, for R to be vertical,
0.724 84 lb 0x BCR T
116.0 lbBCT
(b) Using
116.0 lbBCT
0.6897 116.0 lb 140 lb 60 lbyR R
60.0 lbR R
40

PROBLEM 2.41
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) the tension in cable AE if the resultant of the tensions
exerted at point A of the boom must be directed along AB,
(b) the corresponding magnitude of the resultant.
SOLUTION
Choose x-axis along bar AB.
Then
(a) Require
0: 4 kN cos25 5.2 kN sin35 sin65 0y y AER F T
or 7.2909 kNAET
7.29 kNAET
(b) xR F
4 kN sin 25 5.2 kN cos35 7.2909 kN cos65
9.03 kN
9.03 kNR
41

PROBLEM 2.42
For the block of Problems 2.35 and 2.36, determine (a) the required value
of of the resultant of the three forces shown is to be parallel to the
incline, (b) the corresponding magnitude of the resultant.
Problem 2.35: Knowing that 35 , determine the resultant of the
three forces shown.
Problem 2.36: Knowing that 65 , determine the resultant of the
three forces shown.
SOLUTION
Selecting the x axis along ,aa we write
300 N 400 N cos 600 N sinx xR F (1)
400 N sin 600 N cosy yR F (2)
(a) Setting 0yR in Equation (2):
Thus
600
tan 1.5
400
56.3
(b) Substituting for in Equation (1):
300 N 400 N cos56.3 600 N sin56.3xR
1021.1 NxR
1021 NxR R
42

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
From the geometry, we calculate the distances:
2 2
16 in. 12 in. 20 in.AC
2 2
20 in. 21 in. 29 in.BC
Then, from the Free Body Diagram of point C:
16 21
0: 0
20 29
x AC BCF T T
or
29 4
21 5
BC ACT T
and
12 20
0: 600 lb 0
20 29
y AC BCF T T
or
12 20 29 4
600 lb 0
20 29 21 5
AC ACT T
Hence: 440.56 lbACT
(a) 441 lbACT
(b) 487 lbBCT
43

PROBLEM 2.44
Knowing that 25 , determine the tension (a) in cable AC, (b) in
rope BC.
SOLUTION
Free-Body Diagram Force Triangle
Law of Sines:
5 kN
sin115 sin5 sin 60
AC BCT T
(a)
5 kN
sin115 5.23 kN
sin60
ACT 5.23 kNACT
(b)
5 kN
sin5 0.503 kN
sin60
BCT 0.503 kNBCT
44

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
Law of Sines:
400 lb
sin 25 sin60 sin95
AC BCF T
(a)
400 lb
sin 25 169.69 lb
sin95
ACF 169.7 lbACF
(b)
400
sin 60 347.73 lb
sin95
BCT 348 lbBCT
45

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
Law of Sines:
2943 N
sin 60 sin55 sin65
AC BCT T
(a)
2943 N
sin60 2812.19 N
sin 65
ACT 2.81 kNACT
(b)
2943 N
sin55 2659.98 N
sin 65
BCT 2.66 kNBCT
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
Force Triangle
Free-Body Diagram Point C
Force Triangle
In the free-body diagram of point B, the geometry gives:
1 9.9
tan 30.51
16.8
AB
1 12
tan 22.61
28.8
BC
Thus, in the force triangle, by the Law of Sines:
1190 N
sin59.49 sin 7.87
BCT
7468.6 NBCT
In the free-body diagram of point C (with W the sum of weights of chair
and skier) the geometry gives:
1 1.32
tan 10.39
7.2
CD
Hence, in the force triangle, by the Law of Sines:
7468.6 N
sin12.23 sin100.39
W
1608.5 NW
Finally, the skier weight 1608.5 N 300 N 1308.5 N
skier weight 1309 N
47

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
Force Triangle
Free-Body Diagram Point E
Force Triangle
In the free-body diagram of point F, the geometry gives:
1 12
tan 22.62
28.8
EF
1 1.32
tan 10.39
7.2
DF
Thus, in the force triangle, by the Law of Sines:
1100 N
sin100.39 sin12.23
EFT
5107.5 NBCT
In the free-body diagram of point E (with W the sum of weights of chair
and skier) the geometry gives:
1 9.9
tan 30.51
16.8
AE
Hence, in the force triangle, by the Law of Sines:
5107.5 N
sin 7.89 sin59.49
W
813.8 NW
Finally, the skier weight 813.8 N 300 N 513.8 N
skier weight 514 N
48

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
FA 510 lb and FB 480 lb, determine the magnitudes of the other two
forces.
SOLUTION
Free-Body Diagram
Resolving the forces into x and y components:
0: 510 lb sin15 480 lb cos15 0x CF F
or 332 lbCF
0: 510 lb cos15 480 lb sin15 0y DF F
or 368 lbDF
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
FA 420 lb and FC 540 lb, determine the magnitudes of the other two
forces.
SOLUTION
Resolving the forces into x and y components:
0: cos15 540 lb 420 lb cos15 0 or 671.6 lbx B BF F F
672 lbBF
0: 420 lb cos15 671.6 lb sin15 0y DF F
or 232 lbDF
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:
0A BR P Q F F
Substituting components:
400 lb 520 lb cos55 520 lb sin55R j i j
cos55 sin55 0B A AF F Fi i j
In the y-direction (one unknown force)
400 lb 520 lb sin55 sin55 0AF
Thus,
400 lb 520 lb sin55
1008.3 lb
sin55
AF
1008 lbAF
In the x-direction:
520 lb cos55 cos55 0B AF F
Thus,
cos55 520 lb cos55B AF F
1008.3 lb cos55 520 lb cos55
280.08 lb
280 lbBF
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 FA 600 lb and FB 320 lb,
determine the magnitudes of P and Q.
SOLUTION
Free-Body Diagram Resolving the forces into x and y directions:
0A BR P Q F F
Substituting components:
320 lb 600 lb cos55 600 lb sin55R i i j
cos55 sin55 0P Q Qi i j
In the x-direction (one unknown force)
320 lb 600 lb cos55 cos55 0Q
Thus,
320 lb 600 lb cos55
42.09 lb
cos55
Q
42.1lbQ
In the y-direction:
600 lb sin55 sin55 0P Q
Thus,
600 lb sin55 sin55 457.01lbP Q
457 lbP
52

PROBLEM 2.53
Two cables tied together at C are loaded as shown. Knowing that
W 840 N, determine the tension (a) in cable AC, (b) in 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:
3 15 15
0: 680 N 0
5 17 17
x CA CBF T T
or
1 5
200 N
5 17
CA CBT T (1)
and
4 8 8
0: 680 N 840 N 0
5 17 17
y CA CBF T T
or
1 2
290 N
5 17
CA CBT T (2)
Solving Equations (1) and (2) simultaneously:
(a) 750 NCAT
(b) 1190 NCBT
53

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:
3 15 15
0: 680 N 0
5 17 17
x CA CBF T T
or
1 5
200 N
5 17
CA CBT T (1)
and
4 8 8
0: 680 N 0
5 17 17
y CA CBF T T W
or
1 2 1
80 N
5 17 4
CA CBT T W (2)
Then, from Equations (1) and (2)
17
680 N
28
25
28
CB
CA
T W
T W
Now, with 1050 NT
25
: 1050 N
28
CA CAT T W
or 1176 NW
and
17
: 1050 N 680 N
28
CB CBT T W
or 609 NW 0 609 NW
54

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) in the support cable ACB, (b) in the traction
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
0: cos35 cos40 cos40 0x ACB DEF T T
or
0.0531 0.766 0ACB DET T (1)
and
0: sin 40 sin35 sin 40 24.8 kN 0y ACB DEF T T
or
0.0692 0.643 24.8 kNACB DET T (2)
From (1)
14.426ACB DET T
Then, from (2)
0.0692 14.426 0.643 24.8 kNDE DET T
and
(b) 15.1kNDET
(a) 218 kNACBT
55

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) the combined weight of the cabin, its support
system, and its passengers, (b) the tension in the support cable ACB.
SOLUTION
Free-Body Diagram
First, consider the sum of forces in the x-direction because there is only one unknown force:
0: cos32 cos42 20 kN cos42 0x ACBF T
or
0.1049 14.863 kNACBT
(b) 141.7 kNACBT
Now
0: sin 42 sin32 20 kN sin 42 0y ACBF T W
or
141.7 kN 0.1392 20 kN 0.6691 0W
(a) 33.1kNW
56

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 kAB 1500 N/m and kAD 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:
AB AB AB oF k L L
and
2 2
0.44 m 0.33 m 0.55 mABL
So:
1500 N/m 0.55 m 0.45 mABF
150 N
Similarly,
AD AD AD oF k L L
Then:
2 2
0.66 m 0.32 m 0.68 mADL
1500 N/m 0.68 m 0.45 mADF
115 N
(a)
4 7 15
0: 150 N 115 N 0
5 25 17
x ACF T
or
66.18 NACT 66.2 NACT
57

PROBLEM 2.57 CONTINUED
(b) and
3 24 8
0: 150 N 66.18 N 115 N 0
5 25 17
yF W
or 208 NW
58

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 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 in the ratio 3:4:5.
The sides of the triangle with hypotenuse AB are also in the ratio
12:35:37.
Then:
4 35 12
0: 3 0
5 37 37
x sF W W F
or
4.4833sF W
and
3 12 35
0: 3 400 N 0
5 37 37
y sF W W F
Then:
3 12 35
3 4.4833 400 N 0
5 37 37
W W W
or
62.841 NW
and
281.74 NsF
or
(a) 62.8 NW
59

(b) Have spring force
s AB oF k L L
Where
AB AB AB oF k L L
and
2 2
0.360 m 1.050 m 1.110 mABL
So:
0281.74 N 800 N/m 1.110 mL
or 0 758 mmL
60

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 BCT is when BCT is perpendicular to the direction of ACT
Free-Body Diagram At C Force Triangle
(a) 55.0
(b) 2943 N sin55BCT
2410.8 N
2.41kNBCT
61

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 725 NT 0: 2 1000 N 0y yF T
500 NyT
2 2 2
x yT T T
2 22
500 N 725 NxT
525 NxT
By similar triangles:
1.5 m
725 525
BC
2.07 mBC
2 4.14 mL BC
4.14 mL
62

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 .
SOLUTION
Free-Body Diagram: C Force Triangle
Force triangle is isoceles with
2 180 85
47.5
(a) 2 200 lb cos47.5 270 lbP
Since 0,P the solution is correct. 270 lbP
(b) 180 55 47.5 77.5 77.5
63

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 .
SOLUTION
Free-Body Diagram: C Force Triangle
(a) Law of Cosines:
2 22
300 lb 150 lb 2 300 lb 150 lb cos85P
323.5 lbP
Since 300 lb,P our solution is correct. 324 lbP
(b) Law of Sines:
sin sin85
300 323.5
sin 0.9238
or 67.49
180 55 67.49 57.5
57.5
64

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
BCT must be perpendicular to ACF to be as small as possible.
Free-Body Diagram: C Force Triangle is
a right triangle
(a) We observe: 55 55
(b) 400 lb sin 60BCT
or 346.4 lbBCT 346 lbBCT
65

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: 0BD AB BCT F T
where magnitude and direction of BDT are known, and the direction
of ABF is known.
Then, in a force triangle:
By observation, BCT is minimum when 90.0
(b) Have 70 lb sin 180 70 30BCT
68.93 lb
68.9 lbBCT
66

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
Have: s AB ABF k L L
where:
2 2
0.3 m 0.4 m 0.3 2 mAB ABL L
0.5 m
Then: 660 N/m 0.5 0.3 2 msF
49.986 N
For the collar:
4
0: 49.986 N 0
5
yF W
or 40.0 NW
67

PROBLEM 2.66
The 40-N 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 for which the system is in equilibrium.
SOLUTION
Free-Body Diagram: Collar A 2 2
0: 0
0.3
y s
h
F W F
h
or 2
40 0.09shF h
Now.. s AB ABF k L L
where
2 2
0.3 m 0.3 2 mAB ABL h L
Then: 2 2
560 0.09 0.3 2 40 0.09h h h
or 2
14 1 0.09 4.2 2 mh h h h
Solving numerically,
415 mmh
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:
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
(a)
(b)
(c)
(d)
(e)
2
0: 2 280 kg 9.81 m/s 0yF T
1
2746.8 N
2
T
1373 NT
2
0: 2 280 kg 9.81 m/s 0yF T
1
2746.8 N
2
T
1373 NT
2
0: 3 280 kg 9.81 m/s 0yF T
1
2746.8 N
3
T
916 NT
2
0: 3 280 kg 9.81 m/s 0yF T
1
2746.8 N
3
T
916 NT
2
0: 4 280 kg 9.81 m/s 0yF T
1
2746.8 N
4
T
687 NT
69

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: 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)
(d)
2
0: 3 280 kg 9.81 m/s 0yF T
1
2746.8 N
3
T
916 NT
2
0: 4 280 kg 9.81 m/s 0yF T
1
2746.8 N
4
T
687 NT
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 0: 2 sin 25 cos 0xF P P
and
cos 0.8452 or 32.3
For 32.3
0: 2 cos25 sin32.3 350 lb 0yF P P
or 149.1 lbP 32.3
For 32.3
0: 2 cos25 sin 32.3 350 lb 0yF P P
or 274 lbP 32.3
71

PROBLEM 2.70
A 350-lb load is supported by the rope-and-pulley arrangement shown.
Knowing that 35 , determine (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: 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 0: 2 sin cos25 0xF P P
Hence:
(a)
1
sin cos25
2
or 24.2
(b) 0: 2 cos sin35 350 lb 0yF P P
Hence:
2 cos24.2 sin35 350 lb 0P P
or 145.97 lbP 146.0 lbP
72

PROBLEM 2.71
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 800 N,
determine (a) the tension in cable ACB, (b) the magnitude of load Q.
SOLUTION
Free-Body Diagram: Pulley C
(a) 0: cos30 cos50 800 N cos50 0x ACBF T
Hence 2303.5 NACBT
2.30 kNACBT
(b) 0: sin30 sin50 800 N sin50 0y ACBF T Q
2303.5 N sin30 sin50 800 N sin50 0Q
or 3529.2 NQ 3.53 kNQ
73

PROBLEM 2.72
A 2000-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 the cable ACB, (b) the magnitude of load P.
SOLUTION
Free-Body Diagram: Pulley C
0: cos30 cos50 cos50 0x ACBF T P
or 0.3473 ACBP T (1)
0: sin30 sin50 sin50 2000 N 0y ACBF T P
or 1.266 0.766 2000 NACBT P (2)
(a) Substitute Equation (1) into Equation (2):
1.266 0.766 0.3473 2000 NACB ACBT T
Hence: 1305.5 NACBT
1306 NACBT
(b) Using (1)
0.3473 1306 N 453.57 NP
454 NP
74

PROBLEM 2.73
Determine (a) the x, y, and z components of the 200-lb force, (b) the
angles x, y, and z that the force forms with the coordinate axes.
SOLUTION
(a) 200 lb cos30 cos25 156.98 lbxF
157.0 lbxF
200 lb sin30 100.0 lbyF
100.0 lbyF
200 lb cos30 sin 25 73.1996 lbzF
73.2 lbzF
(b)
156.98
cos
200
x or 38.3x
100.0
cos
200
y or 60.0y
73.1996
cos
200
z or 111.5z
75

PROBLEM 2.74
Determine (a) the x, y, and z components of the 420-lb force, (b) the
angles x, y, and z that the force forms with the coordinate axes.
SOLUTION
(a) 420 lb sin 20 sin70 134.985 lbxF
135.0 lbxF
420 lb cos20 394.67 lbyF
395 lbyF
420 lb sin 20 cos70 49.131 lbzF
49.1 lbzF
(b)
134.985
cos
420
x
108.7x
394.67
cos
420
y
20.0y
49.131
cos
420
z
83.3z
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) the components of the force exerted by this cable on the
tree, (b) the angles x, y, and z that the force forms with axes at A which
are parallel to the coordinate axes.
SOLUTION
(a) 4.2 kN sin50 cos40 2.4647 kNxF
2.46 kNxF
4.2 kN cos50 2.6997 kNyF
2.70 kNyF
4.2 kN sin50 sin 40 2.0681 kNzF
2.07 kNzF
(b)
2.4647
cos
4.2
x
54.1x
77

2.7
cos
4.2
y
130.0y
2.0681
cos
4.0
z
60.5z
78

PROBLEM 2.76
anchored in the ground. Knowing that the tension in cable AC is 3.6 kN,
determine (a) the components of the force exerted by this cable on the
tree, (b) the angles x, y, and z that the force forms with axes at A which
are parallel to the coordinate axes.
SOLUTION
(a) 3.6 kN cos45 sin 25 1.0758 kNxF
1.076 kNxF
3.6 kN sin 45 2.546 kNyF
2.55 kNyF
3.6 kN cos45 cos25 2.3071 kNzF
2.31 kNzF
(b)
1.0758
cos
3.6
x
107.4x
79

2.546
cos
3.6
y
135.0y
2.3071
cos
3.6
z
50.1z
80

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) the tension in wire AD, (b) the angles x,
y, and z that the force exerted at A forms with the coordinate axes.
SOLUTION
(a) sin30 sin50 220.6 NxF F (Given)
220.6 N
575.95 N
sin30 sin50
F
576 NF
(b)
220.6
cos 0.3830
575.95
x
x
F
F
67.5x
cos30 498.79 NyF F
498.79
cos 0.86605
575.95
y
y
F
F
30.0y
sin30 cos50zF F
575.95 N sin30 cos50
185.107 N
185.107
cos 0.32139
575.95
z
z
F
F
108.7z
81

PROBLEM 2.78
Knowing that the z component of the force exerted by wire BD on the
plate is –64.28 N, determine (a) the tension in wire BD, (b) the angles x,
y, and z that the force exerted at B forms with the coordinate axes.
SOLUTION
(a) sin30 sin 40 64.28 NzF F (Given)
64.28 N
200.0 N
sin30 sin40
F 200 NF
(b) sin30 cos40xF F
200.0 N sin30 cos40
76.604 N
76.604
cos 0.38302
200.0
x
x
F
F
112.5x
cos30 173.2 NyF F
173.2
cos 0.866
200
y
y
F
F
30.0y
64.28 NzF
64.28
cos 0.3214
200
z
z
F
F
108.7z
82

PROBLEM 2.79
Knowing that the tension in wire CD is 120 lb, determine (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) 120 lb sin30 cos60 30 lbxF
30.0 lbxF
120 lb cos30 103.92 lbyF
103.9 lbyF
120 lb sin30 sin60 51.96 lbzF
52.0 lbzF
(b)
30.0
cos 0.25
120
x
x
F
F
104.5x
103.92
cos 0.866
120
y
y
F
F
30.0y
51.96
cos 0.433
120
z
z
F
F
64.3z
83

PROBLEM 2.80
Knowing that the x component of the forces exerted by wire CD on the
plate is –40 lb, determine (a) the tension in wire CD, (b) the angles x, y,
and z that the force exerted at C forms with the coordinate axes.
SOLUTION
(a) sin30 cos60 40 lbxF F (Given)
40 lb
160 lb
sin30 cos60
F
160.0 lbF
(b)
40
cos 0.25
160
x
x
F
F
104.5x
160 lb cos30 103.92 lbyF
103.92
cos 0.866
160
y
y
F
F
30.0y
160 lb sin30 sin 60 69.282 lbzF
69.282
cos 0.433
160
z
z
F
F
64.3z
84

PROBLEM 2.81
Determine the magnitude and direction of the force
800 lb 260 lb 320 lb .F i j k
SOLUTION
2 2 22 2 2
800 lb 260 lb 320 lbx y zF F F F 900 lbF
800
cos 0.8889
900
x
x
F
F
27.3x
260
cos 0.2889
900
y
y
F
F
73.2y
320
cos 0.3555
900
z
z
F
F
110.8z
85

PROBLEM 2.82
Determine the magnitude and direction of the force
400 N 1200 N 300 N .F i j k
SOLUTION
2 2 22 2 2
400 N 1200 N 300 Nx y zF F F F 1300 NF
400
cos 0.30769
1300
x
x
F
F
72.1x
1200
cos 0.92307
1300
y
y
F
F
157.4y
300
cos 0.23076
1300
z
z
F
F
76.7z
86

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) the angle y, (b) the other components
and the magnitude of the force.
SOLUTION
(a) We have
2 2 22 2 2
cos cos cos 1 cos 1 cos cosx y z y y z
Since 0yF we must have cos 0y
Thus, taking the negative square root, from above, we have:
2 2
cos 1 cos64.5 cos55.9 0.70735y 135.0y
(b) Then:
200 N
282.73 N
cos 0.70735
y
y
F
F
and cos 282.73 N cos64.5x xF F 121.7 NxF
cos 282.73 N cos55.9z zF F 158.5 NyF
283 NF
87

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 of
the force is –60 N, determine (a) the angle z, (b) the other components
and the magnitude of the force.
SOLUTION
(a) We have
2 2 22 2 2
cos cos cos 1 cos 1 cos cosx y z y y z
Since 0zF we must have cos 0z
Thus, taking the negative square root, from above, we have:
2 2
cos 1 cos75.4 cos132.6 0.69159z 133.8z
(b) Then:
60 N
86.757 N
cos 0.69159
z
z
F
F 86.8 NF
and cos 86.8 N cos75.4x xF F 21.9 NxF
cos 86.8 N cos132.6y yF F 58.8 NyF
88

PROBLEM 2.85
A force F of magnitude 400 N acts at the origin of a coordinate system.
Knowing that x 28.5 , Fy –80 N, and Fz 0, determine (a) the
components Fx and Fz, (b) the angles y and z.
SOLUTION
(a) Have
cos 400 N cos28.5x xF F 351.5 NxF
Then:
2 2 2 2
x y zF F F F
So:
2 2 2 2
400 N 352.5 N 80 N zF
Hence:
2 2 2
400 N 351.5 N 80 NzF 173.3 NzF
(b)
80
cos 0.20
400
y
y
F
F
101.5y
173.3
cos 0.43325
400
z
z
F
F
64.3z
89

PROBLEM 2.86
A force F of magnitude 600 lb acts at the origin of a coordinate system.
Knowing that Fx 200 lb, z 136.8 , Fy 0, determine (a) the
components Fy and Fz, (b) the angles x and y.
SOLUTION
(a) cos 600 lb cos136.8z zF F
437.4 lb 437 lbzF
Then:
2 2 2 2
x y zF F F F
So:
22 2 2
600 lb 200 lb 437.4 lbyF
Hence:
2 2 2
600 lb 200 lb 437.4 lbyF
358.7 lb 359 lbyF
(b)
200
cos 0.333
600
x
x
F
F
70.5x
358.7
cos 0.59783
600
y
y
F
F
126.7y
90

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
4 m 20 m 5 mBA i j k
2 2 2
4 m 20 m 5 m 21 mBA
2100 N
4 m 20 m 5 m
21 m
BA
BA
F F
BA
F i j k
400 N 2000 N 500 NF i j k
400 N, 2000 N, 500 Nx y zF F F
91

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
4 m 20 m 14.8 mDA i j k
2 2 2
4 m 20 m 14.8 m 25.2 mDA
1260 N
4 m 20 m 14.8 m
25.2 m
DA
DA
F F
DA
F i j k
200 N 1000 N 740 NF i j k
200 N, 1000 N, 740 Nx y zF F F
92

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
32 in. 48 in. 36 in.BA i j k
2 2 2
32 in. 48 in. 36 in. 68 in.BA
204 lb
32 in. 48 in. 36 in.
68 in.
BA
BA
F F
BA
F i j k
96 lb 144 lb 108 lbF i j k
96.0 lb, 144.0 lb, 108.0 lbx y zF F F
93

PROBLEM 2.90
the tension in cable AD is 195 lb, determine the components of the force
exerted on the plate at D.
SOLUTION
25 in. 48 in. 36 in.DA i j k
2 2 2
25 in. 48 in. 36 in. 65 in.DA
195 lb
25 in. 48 in. 36 in.
65 in.
DA
DA
F F
DA
F i j k
75 lb 144 lb 108 lbF i j k
75.0 lb, 144.0 lb, 108.0 lbx y zF F F
94

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
0.96 m 1.12 m 0.96 mDB i j k
2 2 2
0.96 m 1.12 m 0.96 m 1.76 mDB
220 N
0.96 m 1.12 m 0.96 m
1.76 m
DB DB
DB
T T
DB
T i j k
120 N 140 N 120 NDBT i j k
120.0 N, 140.0 N, 120.0 NDB DB DBx y z
T T T
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
0.96 m 1.20 m 1.28 mEB i j k
2 2 2
0.96 m 1.20 m 1.28 m 2.00 mEB
250 N
0.96 m 1.20 m 1.28 m
2.00 m
EB EB
EB
T T
EB
T i j k
120 N 150 N 160 NEBT i j k
120.0 N, 150.0 N, 160.0 NEB EB EBx y z
T T T
96

PROBLEM 2.93
Find the magnitude and direction of the resultant of the two forces shown
knowing that 500 NP and 600 N.Q
SOLUTION
500 lb cos30 sin15 sin30 cos30 cos15P i j k
500 lb 0.2241 0.50 0.8365i j k
112.05 lb 250 lb 418.25 lbi j k
600 lb cos40 cos20 sin 40 cos40 sin 20Q i j k
600 lb 0.71985 0.64278 0.26201i j k
431.91 lb 385.67 lb 157.206 lbi j k
319.86 lb 635.67 lb 261.04 lbR P Q i j k
2 2 2
319.86 lb 635.67 lb 261.04 lb 757.98 lbR
758 lbR
319.86 lb
cos 0.42199
757.98 lb
x
x
R
R
65.0x
635.67 lb
cos 0.83864
757.98 lb
y
y
R
R
33.0y
261.04 lb
cos 0.34439
757.98 lb
z
z
R
R
69.9z
97

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:
600 lb 0.2241 0.50 0.8365P i j k
134.46 lb 300 lb 501.9 lbi j k
400 lb 0.71985 0.64278 0.26201Q i j k
287.94 lb 257.11 lb 104.804 lbi j k
153.48 lb 557.11 lb 397.10 lbR P Q i j k
2 2 2
153.48 lb 557.11 lb 397.10 lb 701.15 lbR
701 lbR
153.48 lb
cos 0.21890
701.15 lb
x
x
R
R
77.4x
557.11 lb
cos 0.79457
701.15 lb
y
y
R
R
37.4y
397.10 lb
cos 0.56637
701.15 lb
z
z
R
R
55.5z
98

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
400 mm 450 mm 600 mmAB i j k
2 2 2
400 mm 450 mm 600 mm 850 mmAB
1000 mm 450 mm 600 mmAC i j k
2 2 2
1000 mm 450 mm 600 mm 1250 mmAC
400 mm 450 mm 600 mm
850 N
850 mm
AB AB ABAB
AB
T T
AB
i j k
T
400 N 450 N 600 NAB
T i j k
1000 mm 450 mm 600 mm
1020 N
1250 mm
AC AC ACAC
AC
T T
AC
i j k
T
816 N 367.2 N 489.6 NAC
T i j k
1216 N 817.2 N 1089.6 NAB ACR T T i j k
Then: 1825.8 NR 1826 NR
and
1216
cos 0.66601
1825.8
x 48.2x
817.2
cos 0.44758
1825.8
y 116.6y
1089.6
cos 0.59678
1825.8
z 53.4z
99

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
400 mm 450 mm 600 mmAB i j k
2 2 2
400 mm 450 mm 600 mm 850 mmAB
1000 mm 450 mm 600 mmAC i j k
2 2 2
1000 mm 450 mm 600 mm 1250 mmAC
400 mm 450 mm 600 mm
1020 N
850 mm
AB AB AB AB
AB
T T
AB
i j k
T
480 N 540 N 720 NABT i j k
1000 mm 450 mm 600 mm
850 N
1250 mm
AC AC AC AC
AC
T T
AC
i j k
T
680 N 306 N 408 NACT i j k
1160 N 846 N 1128 NAB ACR T T i j k
Then: 1825.8 NR 1826 NR
and
1160
cos 0.6353
1825.8
x 50.6x
846
cos 0.4634
1825.8
y 117.6y
1128
cos 0.6178
1825.8
z 51.8z
100

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
120 N 140 N 120 NBDT i j k
120 N 150 N 160 NBET i j k
Then:
B BD BER T T
240 N 290 N 40 Ni j k
and 378.55 NR 379 NBR
240
cos 0.6340
378.55
x
129.3x
290
cos 0.7661
378.55
y
40.0y
40
cos 0.1057
378.55
z
96.1z
101

PROBLEM 2.98
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) the tension in AC, (b) the magnitude and direction of
the resultant of the two forces.
SOLUTION
Have
920 lb sin50 cos40 cos50 sin50 sin 40ABT i j j
cos45 sin 25 sin 45 cos45 cos25AC ACTT i j j
(a)
A AB ACR T T
0A x
R
0: 920 lb sin50 cos40 cos45 sin 25 0A x ACx
R F T
or
1806.60 lbACT 1807 lbACT
(b)
: 920 lb cos50 1806.60 lb sin 45A yy
R F
1868.82 lbA y
R
: 920 lb sin50 sin 40 1806.60 lb cos45 cos25A zz
R F
1610.78 lbA z
R
1868.82 lb 1610.78 lbAR j k
Then:
2467.2 lbAR 2.47 kipsAR
102

and
0
cos 0
2467.2
x 90.0x
1868.82
cos 0.7560
2467.2
y 139.2y
1610.78
cos 0.65288
2467.2
z 49.2z
103

PROBLEM 2.99
anchored in the ground. Knowing that the tension in AC is 850 lb and that
the resultant of the forces exerted at A by cables AB and AC lies in the yz
plane, determine (a) the tension in AB, (b) the magnitude and direction of
the resultant of the two forces.
SOLUTION
Have
sin50 cos40 cos50 sin50 sin 40AB ABTT i j j
850 lb cos45 sin 25 sin 45 cos45 cos25ACT i j j
(a)
0A x
R
0: sin50 cos40 850 lb cos45 sin 25 0A x ABx
R F T
432.86 lbABT 433 lbABT
(b)
: 432.86 lb cos50 850 lb sin 45A yy
R F
879.28 lbA y
R
: 432.86 lb sin50 sin 40 850 lb cos45 cos25A zz
R F
757.87 lbA z
R
879.28 lb 757.87 lbAR j k
1160.82 lbAR 1.161 kipsAR
0
cos 0
1160.82
x 90.0x
879.28
cos 0.75746
1160.82
y 139.2y
757.87
cos 0.65287
1160.82
z 49.2z
104

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. 48 in. 36 in.AC i j k
2 2 2
45 in. 48 in. 36 in. 75 in.AC
27 lb
45 in. 48 in. 36 in.
75 in.
AC AC AC AC
AC
T T
AC
T i j k
16.2 lb 17.28 lb 12.96ACT i j k
and
32 in. 48 in. 36 in.AB i j k
2 2 2
32 in. 48 in. 36 in. 68 in.AB
32 in. 48 in. 36 in.
68 in.
AB
AB AB AB AB
AB T
T T
AB
T i j k
0.4706 0.7059 0.5294AB ABTT i j k
and
25 in. 48 in. 36 in.AD i j k
2 2 2
25 in. 48 in. 36 in. 65 in.AD
25 in. 48 in. 36 in.
65 in.
AD
AD AD AD AD
AD T
T T
AD
T i j k
0.3846 0.7385 0.5538AD ADTT i j k
105

Now
AB AD ADR T T T
0.4706 0.7059 0.5294 16.2 lb 17.28 lb 12.96ABT i j k i j k
0.3846 0.7385 0.5538ADT i j k
Since R must be vertical, the i and k components of this sum must be zero.
Hence:
0.4706 0.3846 16.2 lb 0AB ADT T (1)
0.5294 0.5538 12.96 lb 0AB ADT T (2)
Solving (1) and (2), we obtain:
244.79 lb, 257.41 lbAB ADT T
245 lbABT
257 lbADT
106

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
2 2 2
220 mm 192 mm 0 292 mmBAd
2 2 2
192 mm 192 mm 96 mm 288 mmDAd
2 2 2
0 192 mm 144 mm 240 mmCAd
and
146 N
220 mm 192 mm
292 mm
BA BA BAFF i j
110 N 96 Ni j
192 mm 144 mm
240 mm
CA
CA CA CA
F
FF j k
0.80 0.60CAF j k
192 mm 192 mm 96 mm
288 mm
DA
DA DA DA
F
FF i j k
0.66667 0.66667 0.33333DAF i j k
With PP j
At A: 0: 0BA CA DAF F F F P
i-component: 110 N 0.66667 0DAF or 165 NDAF
j-component: 96 N 0.80 0.66667 165 N 0CAF P (1)
k-component: 0.60 0.33333 165 N 0CAF (2)
Solving (2) for CAF and then using that result in (1), gives 279 NP
107

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:
220 mm 192 mm
292 mm
BA
BA BA BA
F
FF i j
0.75342 0.65753 NBAF i j
192 mm 144 mm
240 mm
CA
CA CA CA
F
FF j k
0.80 0.60CAF j k
192 mm 192 mm 96 mm
288 mm
DA
DA DA DA
F
FF i j k
0.66667 0.66667 0.33333DAF i j k
With: 200 NP j
At A: 0: 0BA CA DAF F F F P
Hence, equating the three (i, j, k) components to 0 gives three equations
i-component: 0.75342 0.66667 0BA DAF F (1)
j-component: 0.65735 0.80 0.66667 200 N 0BA CA DAF F F (2)
k-component: 0.60 0.33333 0CA DAF F (3)
Solving (1), (2), and (3), gives
DA104.5 N, 65.6 N, 118.1 NBA CAF F F
104.5 NBAF
65.6 NCAF
118.1 NDAF
108

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:
, , andAB AC ADT T T P
where PP j . To express the other forces in terms of the unit vectors
i, j, k, we write
12.6 ft 16.8 ftAB i j 21 ftAB
7.2 ft 16.8 ft 12.6 ft 22.2 ftAC ACi j k
16.8 ft 9.9 ftAD j k 19.5 ftAD
and 0.6 0.8AB AB AB AB AB
AB
T T T
AB
T i j
0.3242 0.75676 0.56757AC AC AC AC AC
AC
T T T
AC
T i j k
0.8615 0.50769AD AD AD AD AD
AD
T T T
AD
T j k
109

Equilibrium Condition
0: 0AB AC ADF PT T T j
Substituting the expressions obtained for , , andAB AC ADT T T and
factoring i, j, and k:
0.6 0.3242 0.8 0.75676 0.8615AB AC AB AC ADT T T T T Pi j
0.56757 0.50769 0AC ADT T k
Equating to zero the coefficients of i, j, k:
0.6 0.3242 0AB ACT T (1)
0.8 0.75676 0.8615 0AB AC ADT T T P (2)
0.56757 0.50769 0AC ADT T (3)
Setting 60 lbABT in (1) and (2), and solving the resulting set of
equations gives
111 lbACT
124.2 lbADT
239 lbP
110

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:
0.6 0.3242 0AB ACT T (1)
0.8 0.75676 0.8615 0AB AC ADT T T P (2)
0.56757 0.50769 0AC ADT T (3)
Substituting 100 lbACT in Equations (1), (2), and (3) above, and solving the resulting set of equations
using conventional algorithms gives
54 lbABT
112 lbADT
215 lbP
111

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 is 3 kN.
SOLUTION
The forces applied at A are:
, , andAB AC ADT T T P
where PP j . To express the other forces in terms of the unit vectors
i, j, k, we write
0.72 m 1.2 m 0.54 m ,AB i j k 1.5 mAB
1.2 m 0.64 m ,AC j k 1.36 mAC
0.8 m 1.2 m 0.54 m ,AD i j k 1.54 mAD
and 0.48 0.8 0.36AB AB AB AB AB
AB
T T T
AB
T i j k
0.88235 0.47059AC AC AC AC AC
AC
T T T
AC
T j k
0.51948 0.77922 0.35065AD AD AD AD AD
AD
T T T
AD
T i j k
Equilibrium Condition with WW j
0: 0AB AC ADF WT T T j
Substituting the expressions obtained for , , andAB AC ADT T T and
factoring i, j, and k:
0.48 0.51948 0.8 0.88235 0.77922AB AD AB AC ADT T T T T Wi j
0.36 0.47059 0.35065 0AB AC ADT T T k
112

Equating to zero the coefficients of i, j, k:
0.48 0.51948 0AB ADT T
0.8 0.88235 0.77922 0AB AC ADT T T W
0.36 0.47059 0.35065 0AB AC ADT T T
Substituting 3 kNABT in Equations (1), (2) and (3) and solving the
resulting set of equations, using conventional algorithms for solving
linear algebraic equations, gives
4.3605 kNACT
2.7720 kNADT
8.41 kNW
113

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.
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 is 3 kN.
SOLUTION
below:
0.48 0.51948 0AB ADT T
0.8 0.88235 0.77922 0AB AC ADT T T W
0.36 0.47059 0.35065 0AB AC ADT T T
Substituting 2.8 kNADT in Equations (1), (2), and (3) above, and solving the resulting set of equations
using conventional algorithms, gives
3.03 kNABT
4.40 kNACT
8.49 kNW
114

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.
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 is 3 kN.
SOLUTION
below:
0.48 0.51948 0AB ADT T
0.8 0.88235 0.77922 0AB AC ADT T T W
0.36 0.47059 0.35065 0AB AC ADT T T
Substituting 2.4 kNACT in Equations (1), (2), and (3) above, and solving the resulting set of equations
using conventional algorithms, gives
1.651 kNABT
1.526 kNADT
4.63 kNW
115

PROBLEM 2.108
A 750-kg crate is supported by three cables as shown. Determine the
tension in each cable.
SOLUTION
below:
0.48 0.51948 0AB ADT T
0.8 0.88235 0.77922 0AB AC ADT T T W
0.36 0.47059 0.35065 0AB AC ADT T T
Substituting 2
750 kg 9.81 m/s 7.36 kNW in Equations (1), (2), and (3) above, and solving the
resulting set of equations using conventional algorithms, gives
2.63 kNABT
3.82 kNACT
2.43 kNADT
116

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:
cos45 8 sin 45
65
AB BE
i j k
It follows that:
cos45 8 sin 45
65
BE BE BE BET T
i j k
T
cos30 8 sin30
65
CF CF CF CFT T
i j k
T
cos15 8 sin15
65
DG DG DG DGT T
i j k
T
117

At A: 0: 0BE CF DGF T T T W P
Then, isolating the factors of i, j, and k, we obtain three algebraic equations:
: cos45 cos30 cos15 0
65 65 65
BE CF DGT T T
Pi
or cos45 cos30 cos15 65 0BE CF DGT T T P (1)
8 8 8
: 0
65 65 65
BE CF DGT T T Wj
or
65
0
8
BE CF DGT T T W (2)
: sin 45 sin30 sin15 0
65 65 65
BE CF DGT T T
k
or sin 45 sin30 sin15 0BE CF DGT T T (3)
With 0P and the tension in cord 0.2 lb:BE
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:
0.669 lbCFT
0.746 lbDGT
1.603 lbW
118

PROBLEM 2.110
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:
: cos45 cos30 cos15 65 0BE CF DGT T T Pi (1)
65
: 0
8
BE CF DGT T T Wj (2)
: sin 45 sin30 sin15 0BE CF DGT T Tk (3)
With 1.6 lbW , 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 CFT as a function of P and requiring it to be 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:
1.729 0.668 lbCFT P
Hence, for 0CFT 1.729 0.668 0P
or 0.386 lbP
0 0.386 lbP
119

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 30 m 5.4 mAC i j k
2 2 2
18 m 30 m 5.4 m 35.4 mAC
18 m 30 m 5.4 m
35.4 m
AC
AC AC AC
AC T
T T
AC
T i j k
0.5085 0.8475 0.1525AC ACTT i j k
and 6 m 30 m 7.5 mAB i j k
2 2 2
6 m 30 m 7.5 m 31.5 mAB
6 m 30 m 7.5 m
31.5 m
AB
AB AB AB
AB T
T T
AB
T i j k
0.1905 0.9524 0.2381AB ABTT i j k
Finally 6 m 30 m 22.2 mAD i j k
2 2 2
6 m 30 m 22.2 m 37.8 mAD
6 m 30 m 22.2 m
37.8 m
AD
AD AD AD
AD T
T T
AD
T i j k
0.1587 0.7937 0.5873AD ADTT i j k
120

With , at :P AP j
0: 0AB AC AD PF T T T j
Equating the factors of i, j, and k to zero, we obtain the linear algebraic
equations:
: 0.1905 0.5085 0.1587 0AB AC ADT T Ti (1)
: 0.9524 0.8475 0.7937 0AB AC ADT T T Pj (2)
: 0.2381 0.1525 0.5873 0AB AC ADT T Tk (3)
In Equations (1), (2) and (3), set 3.6 kN,ABT and, using conventional
methods for solving Linear Algebraic Equations (MATLAB or Maple,
for example), we obtain:
1.963 kNACT
1.969 kNADT
6.66 kNP
121

PROBLEM 2.112
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 2.6 kNACT
and solve the three resulting linear equations using conventional tools for solving Linear Algebraic Equations
(MATLAB or Maple, for example), to obtain
4.77 kNABT
2.61 kNADT
8.81 kNP
122

PROBLEM 2.113
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
32 in. 48 in. 36 in.AB i j k
2 2 2
32 in. 48 in. 36 in. 68 in.AB
32 in. 48 in. 36 in.
68 in.
AB
AB AB AB
AB T
T T
AB
T i j k
0.4706 0.7059 0.5294AB ABTT i j k
and 45 in. 48 in. 36 in.AC i j k
2 2 2
45 in. 48 in. 36 in. 75 in.AC
45 in. 48 in. 36 in.
75 in.
AC
AC AC AC
AC T
T T
AC
T i j k
0.60 0.64 0.48AC ACTT i j k
Finally, 25 in. 48 in. 36 in.AD i j k
2 2 2
25 in. 48 in. 36 in. 65 in.AD
123

25 in. 48 in. 36 in.
65 in.
AD
AD AD AD
AD T
T T
AD
T i j k
0.3846 0.7385 0.5538AD ADTT i j k
With ,WW j at A we have:
0: 0AB AC AD WF T T T j
Equating the factors of i, j, and k to zero, we obtain the linear algebraic
equations:
: 0.4706 0.60 0.3846 0AB AC ADT T Ti (1)
: 0.7059 0.64 0.7385 0AB AC ADT T T Wj (2)
: 0.5294 0.48 0.5538 0AB AC ADT T Tk (3)
In Equations (1), (2) and (3), set 15 lb,ACT and, using conventional
methods for solving Linear Algebraic Equations (MATLAB or Maple,
for example), we obtain:
136.0 lbABT
143.0 lbADT
211 lbW
124

PROBLEM 2.114
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 120 lbADT and
solve the three resulting linear equations using conventional tools for solving Linear Algebraic Equations
(MATLAB or Maple, for example), to obtain
12.59 lbACT
114.1 lbABT
177.2 lbW
125

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
0: sin30 sin50 sin30 cos40x AD BDF T T
sin30 cos60 0CDT
Dividing through by the factor sin30 and evaluating the trigonometric
functions gives
0.7660 0.7660 0.50 0AD BD CDT T T (1)
Similarly,
0: sin30 cos50 sin30 sin 40z AD BDF T T
sin30 sin60 0CDT
or 0.6428 0.6428 0.8660 0AD BD CDT T T (2)
From (1) 0.6527AD BD CDT T T
Substituting this into (2):
0.3573BD CDT T (3)
Using ADT from above:
AD CDT T (4)
Now,
0: cos30 cos30 cos30y AD BD CDF T T T
2
28 kg 9.81 m/s 0
or 317.2 NAD BD CDT T T
126

Using (3) and (4), above:
0.3573 317.2 NCD CD CDT T T
Then: 135.1 NADT
46.9 NBDT
135.1 NCDT
127

PROBLEM 2.116
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:
0.5409ABT P
0.295ACT P
0.2959ADT P
Using 8 kN:P
4.33 kNABT
2.36 kNACT
2.37 kNADT
128

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:
0.6440ABT P
0.0709ACT P
0.6771ADT P
Using 180 lb:P
115.9 lbABT
12.76 lbACT
121.9 lbADT
129

PROBLEM 2.118
For the cone of Problem 2.110, determine the range of values of P for
which cord DG is taut if P is directed in the –x direction.
SOLUTION
From the solutions to Problems 2.109 and 2.110, have
0.2 65BE CF DGT T T (2 )
sin 45 sin30 sin15 0BE CF DGT T T (3)
cos45 cos30 cos15 65 0BE CF DGT T T P (1 )
Applying the method of elimination to obtain a desired result:
Multiplying (2 ) by sin 45 and adding the result to (3):
sin 45 sin30 sin 45 sin15 0.2 65 sin 45CF DGT T
or 0.9445 0.3714CF DGT T (4)
Multiplying (2 ) by sin30 and subtracting (3) from the result:
sin30 sin 45 sin30 sin15 0.2 65 sin30BE DGT T
or 0.6679 0.6286BE DGT T (5)
130

Substituting (4) and (5) into (1) :
1.2903 1.7321 65 0DGT P
DGT is taut for
1.2903
lb
65
P
or 0.16000 lbP
131

PROBLEM 2.119
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:
cos45 8 sin 45
65
AB BE
i j k
It follows that:
cos45 8 sin 45
65
BE BE BE BET T
i j k
T
cos30 8 sin30
65
CF CF CF CFT T
i j k
T
cos15 8 sin15
65
DG DG DG DGT T
i j k
T
At A: 0: 0BE CF DGF T T T W P
132

Then, isolating the factors if , , andi j k we obtain three algebraic equations:
: cos45 cos30 cos15 0
65 65 65
BE CF DGT T T
i
or cos45 cos30 cos15 0BE CF DGT T T (1)
8 8 8
: 0
65 65 65
BE CF DGT T T Wj
or
2.4
65 0.3 65
8
BE CF DGT T T (2)
: sin 45 sin30 sin15 0
65 65 65
BE CF DGT T T
Pk
or sin 45 sin30 sin15 65BE CF DGT T T P (3)
With 0,P 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
0.299 lbBET
1.002 lbCFT
1.117 lbDGT
133

PROBLEM 2.120
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 (1), (2), and (3) below:
cos45 cos30 cos15 0BE CF DGT T T (1)
0.3 65BE CF DGT T T (2)
sin 45 sin30 sin15 65BE CF DGT T T P (3)
With 0.1lb,P solving (1), (2), and (3), using conventional methods in Linear Algebra (elimination, matrix
methods or iteration–with MATLAB or Maple, for example), we obtain
1.006 lbBET
0.357 lbCFT
1.056 lbDGT
134

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 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 to the chute.)
SOLUTION
From the geometry of the chute:
2 0.8944 0.4472
5
N
NN j k j k
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 1.3 m 1 mAB i j k
2 2 2
0.6 m 1.3 m 1 m 1.764 mAB
0.6 m 1.3 m 1 m
1.764 m
AB
AB AB AB
AB T
T T
AB
T i j k
0.3436 0.7444 0.5726AB ABTT i j k
and 0.7 m 1.4 m 1 mAC i j k
2 2 2
0.7 m 1.4 m 1 m 1.8574 mAC
0.7 m 1.4 m 1 m
1.764 m
AC
AC AC AC
AC T
T T
AC
T i j k
0.3769 0.7537 0.5384AC ACTT i j k
Then: 0: 0AB ACF N T T W
135

With 200 kg 9.81 m/s 1962 N,W and equating the factors of i, j,
and k to zero, we obtain the linear algebraic equations:
: 0.3436 0.3769 0AB ACT Ti (1)
: 0.7444 0.7537 0.8944 1962 0AB ACT T Nj (2)
: 0.5726 0.5384 0.4472 0AB ACT T Nk (3)
Using conventional methods for solving Linear Algebraic Equations
(elimination, MATLAB or Maple, for example), we obtain
1311 NN
551 NABT
503 NACT
136

PROBLEM 2.122
Solve Problem 2.121 assuming that a third worker is exerting a force
(180 N)P 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 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 to the chute.)
SOLUTION
From the geometry of the chute:
2 0.8944 0.4472
5
N
NN j k j k
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 1.3 m 1 mAB i j k
2 2 2
0.6 m 1.3 m 1 m 1.764 mAB
0.6 m 1.3 m 1 m
1.764 m
AB
AB AB AB
AB T
T T
AB
T i j k
0.3436 0.7444 0.5726AB ABTT i j k
and 0.7 m 1.4 m 1 mAC i j k
2 2 2
0.7 m 1.4 m 1 m 1.8574 mAC
0.7 m 1.4 m 1 m
1.764 m
AC
AC AC AC
AC T
T T
AC
T i j k
0.3769 0.7537 0.5384AC ACTT i j k
Then: 0: 0AB ACF N T T P W
137

Where 180 NP i
and 2
200 kg 9.81 m/sW j
1962 N j
Equating the factors of i, j, and k to zero, we obtain the linear equations:
: 0.3436 0.3769 180 0AB ACT Ti
: 0.8944 0.7444 0.7537 1962 0AB ACN T Tj
: 0.4472 0.5726 0.5384 0AB ACN T Tk
Using conventional methods for solving Linear Algebraic Equations
(elimination, MATLAB or Maple, for example), we obtain
1302 NN
306 NABT
756 NACT
138

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
9 ft 8 ft 12 ftAB i j k
2 2 2
9 ft 8 ft 12 ft 17 ftAB
9 ft 8 ft 12 ft
17 ft
AB
AB AB AB
AB T
T T
AB
T i j k
0.5294 0.4706 0.7059AB ABTT i j k
and
0 8 ft 6 ftAC i j k
2 2 2
0 ft 8 ft 6 ft 10 ftAC
0 ft 8 ft 6 ft
10 ft
AC
AC AC AC
AC T
T T
AC
T i j k
0.8 0.6AC ACTT j k
and
4 ft 8 ft 1ftAD i j k
2 2 2
4 ft 8 ft 1ft 9 ftAD
4 ft 8 ft 1ft
9 ft
ADE
AD AD ADE
AD T
T T
AD
T i j k
0.4444 0.8889 0.1111AD ADETT i j k
139

Finally,
8 ft 8 ft 4 ftAE i j k
2 2 2
8 ft 8 ft 4 ft 12 ftAE
8 ft 8 ft 4 ft
12 ft
ADE
AE AE ADE
AE T
T T
AE
T i j k
0.6667 0.6667 0.3333AE ADETT i j k
With the weight of the machinery, ,WW j at A, we have:
0: 2 0AB AC AD WF T T T j
Equating the factors of , , andi j k to zero, we obtain the following linear algebraic equations:
0.5294 2 0.4444 0.6667 0AB ADE ADET T T (1)
0.4706 0.8 2 0.8889 0.6667 0AB AC ADE ADET T T T W (2)
0.7059 0.6 2 0.1111 0.3333 0AB AC ADE ADET T T T (3)
Knowing that 320 lb,W we can solve Equations (1), (2) and (3) using conventional methods for solving
Linear Algebraic Equations (elimination, matrix methods via MATLAB or Maple, for example) to obtain
46.5 lbABT
34.2 lbACT
110.8 lbADET
140

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 (1), (2), and (3), below:
0.5294 2 0.4444 0.6667 0AB ADE ADET T T (1)
0.4706 0.8 2 0.8889 0.6667 0AB AC ADE ADET T T T W (2)
0.7059 0.6 2 0.1111 0.3333 0AB AC ADE ADET T T T (3)
Knowing that the tension in cable AB is 68 lb, we can solve Equations (1), (2) and (3) using conventional
methods for solving Linear Algebraic Equations (elimination, matrix methods via MATLAB or Maple, for
example) to obtain
(a) 50.0 lbACT
(b) 162.0 lbAET
(c) 468 lbW
141

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
PP i and QQ k are applied to the ring to maintain the container is
the position shown. Knowing that 1200W 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
0.48 m 0.72 m 0.16 mAB i j k
2 2 2
0.48 m 0.72 m 0.16 m 0.88 mAB
0.48 m 0.72 m 0.16 m
0.88 m
AB
AB AB AB
AB T
T T
AB
T i j k
0.5455 0.8182 0.1818AB ABTT i j k
and
0.24 m 0.72 m 0.13 mAC i j k
2 2 2
0.24 m 0.72 m 0.13 m 0.77 mAC
0.24 m 0.72 m 0.13 m
0.77 m
AC
AC AC AC
AC T
T T
AC
T i j k
0.3177 0.9351 0.1688AC ACTT i j k
At A: 0: 0AB ACF T T P Q W
142

Noting that AB ACT T because of the ring A, we equate the factors of
, , andi j k to zero to obtain the linear algebraic equations:
: 0.5455 0.3177 0T Pi
or 0.2338P T
: 0.8182 0.9351 0T Wj
or 1.7532W T
: 0.1818 0.1688 0T Qk
or 0.356Q T
With 1200 N:W
1200 N
684.5 N
1.7532
T
160.0 NP
240 NQ
143

PROBLEM 2.126
For the system of Problem 2.125, determine W and P knowing that
160Q 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 PP i and QQ k are applied to the ring to
maintain the container is the position shown. Knowing that 1200W N,
determine P and Q. (Hint: The tension is the same in both 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
160 NQ to obtain
160 N
456.3 N
0.3506
T
800 NW
107.0 NP
144

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 (680 N)P j is applied at A, determine
(a) the tension in the wire when 300y mm, (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:
2 2 2 2
AB x y z
Here
2 2 2 2
1m 0.40 m y z
or 2 2 2
0.84 my z
Thus, with y given, z is determined.
Now
1
0.40 m 0.4
1m
AB
AB
y z y z
AB
i j k i k k
Where y and z are in units of meters, m.
From the F.B. Diagram of collar A:
0: 0x z AB ABN N P TF i k j
Setting the jcoefficient to zero gives:
0ABP yT
With 680 N,P
680 N
ABT
y
Now, from the free body diagram of collar B:
0: 0x y AB ABN N Q TF i j k
145

Setting thek coefficient to zero gives:
0ABQ T z
And using the above result for ABT we have
680 N
ABQ T z z
y
Then, from the specifications of the problem, 300 mm 0.3 my
22 2
0.84 m 0.3 mz
0.866 mz
and
(a)
680 N
2266.7 N
0.30
ABT
or 2.27 kNABT
and
(b) 2266.7 0.866 1963.2 NQ
or 1.963 kNQ
146

PROBLEM 2.128
Solve Problem 2.127 assuming 550y 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 (680 N)P j is applied at
A, determine (a) the tension in the wire when 300y mm, (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:
2 2 2
0.84 my z
680 N
ABT
y
680 N
Q z
y
With 550 mm 0.55 m,y we obtain:
22 2
0.84 m 0.55 m
0.733 m
z
z
and
(a)
680 N
1236.4 N
0.55
ABT
or 1.236 kNABT
and
(b) 1236 0.866 N 906 NQ
or 0.906 kNQ
147

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) sin35 3001bP
300 lb
sin35
P
523 lbP
(b) Vertical Component
cos35vP P
523 lb cos35
428 lbvP
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
0.78 m 1.6 m 0 mAB i j k
2 2 2
0.78 m 1.6 m 0 1.78 mAB
0.78 m 1.6 m 0 m
1.78 m
AB
AB AB AB
AB T
T T
AB
T i j k
0.4382 0.8989 0AB ABTT i j k
and
0 1.6 m 1.2 mAC i j k
2 2 2
0 m 1.6 m 1.2 m 2 mAC
0 1.6 m 1.2 m
2 m
AC
AC AC AC
AC T
T T
AC
T i j k
0.8 0.6AC ACTT j k
and
1.3 m 1.6 m 0.4 mAD i j k
2 2 2
1.3 m 1.6 m 0.4 m 2.1mAD
1.3 m 1.6 m 0.4 m
2.1m
AD
AD AD AD
AD T
T T
AD
T i j k
0.6190 0.7619 0.1905AD ADTT i j k
149

Finally,
0.4 m 1.6 m 0.86 mAE i j k
2 2 2
0.4 m 1.6 m 0.86 m 1.86 mAE
0.4 m 1.6 m 0.86 m
1.86 m
AE
AE AE AE
AE T
T T
AE
T i j k
0.2151 0.8602 0.4624AE AETT i j k
With the weight of the container ,WW j at A we have:
0: 0AB AC AD WF T T T j
Equating the factors of , , andi j k to zero, we obtain the following linear algebraic equations:
0.4382 0.6190 0.2151 0AB AD AET T T (1)
0.8989 0.8 0.7619 0.8602 0AB AC AD AET T T T W (2)
0.6 0.1905 0.4624 0AC AD AET T T (3)
Knowing that 1000 NW and that because of the pulley system at B ,AB ADT T P where P is the
externally applied (unknown) force, we can solve the system of linear equations (1), (2) and (3) uniquely
for P.
378 NP
150

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
AB ADT T P
and using the linear algebraic equations of Problem 2.131 with 150 N,ACT we obtain
(a) 454 NP
(b) 1202 NW
151

PROBLEM 2.132
Two cables tied together at C are loaded as shown. Knowing that
Q 60 lb, determine the tension (a) in cable AC, (b) in cable BC.
SOLUTION
0: cos30 0y CAF T Q
With 60 lbQ
(a) 60 lb 0.866CAT
52.0 lbCAT
(b) 0: sin30 0x CBF P T Q
With 75 lbP
75 lb 60 lb 0.50CBT
or 45.0 lbCBT
152

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 0: cos30 0x CAF T Q
or 0.8660 QCAT
Then for 60 lbCAT
0.8660 60 lbQ
or 69.3 lbQ
From 0: sin30y CBF T P Q
or 75 lb 0.50CBT Q
For 60 lbCBT
75 lb 0.50 60 lbQ
or 0.50 15 lbQ
Thus, 30 lbQ
Therefore, 30.0 69.3 lbQ
153

PROBLEM 2.134
A welded connection is in equilibrium under the action of the four forces
shown. Knowing that 8 kNAF and 16 kN,BF determine the
magnitudes of the other two forces.
SOLUTION
Free-Body Diagram of
Connection
3 3
0: 0
5 5
x B C AF F F F
With 8 kN, 16 kNA BF F
4 4
16 kN 8 kN
5 5
CF
6.40 kNCF
3 3
0: 0
5 5
y D B AF F F F
With AF and BF as above:
3 3
16 kN 8 kN
5 5
DF
4.80 kNDF
154

PROBLEM 2.135
A welded connection is in equilibrium under the action of the four forces
shown. Knowing that 5 kNAF and 6 kN,DF determine the
magnitudes of the other two forces.
SOLUTION
Free-Body Diagram of
Connection
3 3
0: 0
5 5
y D A BF F F F
or
3
5
B D AF F F
With 5 kN, 8 kNA DF F
5 3
6 kN 5 kN
3 5
BF
15.00 kNBF
4 4
0: 0
5 5
x C B AF F F F
4
5
C B AF F F
4
15 kN 5 kN
5
8.00 kNCF
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
4.5
0: 50 lb 0
20.5
xF P
or 10.98 lbP
(b) Triangle Proportions
15
0: 50 lb 0
25
xF P
or 30.0 lbP
156

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:
2
ˆ50
0: 48 0
ˆ400
x
x
F
x
or 248
ˆ ˆ400
50
x x
2 2
ˆ ˆ0.92 lb 400x x
2 2
ˆ 4737.7 inx
ˆ 68.6 in.x
157

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
480 mm 510 mm 320 mmDB i j k
2 2 2
480 510 320 770 mmDB
385 N
480 mm 510 mm 320 mm
770 mm
DB
DB
F F
DB
F i j k
240 N 255 N 160 NF i j k
240 N, 255 N, 160.0 Nx y zF F F
158

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
0.48 m 0.51 m 0.32 mBD i j k
2 2 2
0.48 m 0.51 m 0.32 m 0.77 mBD
0.48 m 0.51 m 0.32 m
0.77 m
BD
BD BD BD
BD T
T T
BD
T i j k
0.6234 0.6623 0.4156BD BDTT i j k
and
0.27 m 0.40 m 0.6 mBE i j k
2 2 2
0.27 m 0.40 m 0.6 m 0.770 mBE
0.26 m 0.40 m 0.6 m
0.770 m
BE
BE BE BE
BD T
T T
BD
T i j k
0.3506 0.5195 0.7792BE BETT i j k
Now, because of the frictionless ring at B, 385 NBE BDT T and the force on the support due to the two
cables is
385 N 0.6234 0.6623 0.4156 0.3506 0.5195 0.7792F i j k i j k
375 N 455 N 460 Ni j k
159

The magnitude of the resultant is
2 2 22 2 2
375 N 455 N 460 N 747.83 Nx y zF F F F
or 748 NF
The direction of this force is:
1 375
cos
747.83
x or 120.1x
1 455
cos
747.83
y or 52.5y
1 460
cos
747.83
z or 128.0z
160

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
SOLUTION
Force Triangle (a) For minimum P it must be perpendicular to the vertical resultant R
425 lb cos30P
or 368 lbP
(b) 425 lb sin30R
or 213 lbR
161

Chapter 2

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