S. dos execicios Beer Johnston

Prévia do material em texto

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 = 
 19α = ° 
 8.4 kN=R 19°W 
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 , 59α β= ° = ° 
(a) 
 
 
 
 
 
 
 
(b) 
 
 
 
 
 
 
 We measure: 575 N, 67α= = °R 
 575 N=R 67°W 
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, 76α= = °R 
 37 lb=R 76°W 
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, 86.5α= = °R 
 61.5 lb=R 86.5°W 
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 sin 38 sin
F R
α β= =° 
90 28 62 , 180 62 38 80α β= ° − ° = ° = ° − ° − ° = ° 
Then: 
2 120 N
sin 62 sin 38 sin80
F R= =° ° ° 
 or (a) 2 107.6 NF = W 
 (b) 75.0 NR = W 
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 
corresponding magnitude of R. 
 
SOLUTION 
 
Using the Law of Sines 
1 80
sin sin 38 sin
F R
α β= =° 
90 10 80 , 180 80 38 62α β= ° − ° = ° = ° − ° − ° = ° 
Then: 
1 80 N
sin80 sin 38 sin 62
F R= =° ° ° 
 or (a) 1 89.2 NF = W 
 (b) 55.8 NR = W 
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 α 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
β °= 
sin 0.44995β = 
 26.74β = ° 
 Then: 40 180α β+ + ° = ° 
 113.3α = °W 
(b) Using the Law of Sines: 
50 lb
sin sin 40
bbF
α
′ = ° 
 71.5 lbbbF ′ = W 
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 α knowing that the 
component along -b b′ is 30 lb. (b) What is the corresponding value of 
the component along - ?a a′ 
 
SOLUTION 
 
 
 
 
 
Using the triangle rule and the Law of Sines 
(a) sin sin 40
30 lb 50 lb
α °= 
sin 0.3857α = 
 22.7α = °W 
(b) 40 180α β+ + ° = ° 
117.31β = ° 
 50 lb
sin sin 40
aaF
β
′ = ° 
 sin50 lb
sin 40
β
′
 =  ° aaF 
 69.1 lbaaF ′ = W 
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 α = 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: ( )180 35 25α = ° − ° + ° 
 120= ° 
Then: 360 N
sin 35 sin120 sin 25
P R= =° ° ° 
 or (a) 489 NP = W 
 (b) 738 NR = W 
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 α if the resultant R of the two forces 
applied at A is to be vertical, (b) the corresponding magnitude of R. 
 
SOLUTION 
 
Using the triangle rule and the Law of Sines 
(a) Have: 360 N 300 N
sin sin 35α = ° 
sin 0.68829α = 
 43.5α = °W 
(b) ( )180 35 43.5β = − ° + ° 
 101.5= ° 
 Then: 300 N
sin101.5 sin 35
R =° ° 
 or 513 NR = W 
 
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 α if the resultant R of the two forces applied to the support is to be 
horizontal, (b) the corresponding magnitude of R. 
 
SOLUTION 
Using the triangle rule and the Law of Sines 
 
(a) Have: 20 lb 14 lb
sin sin 30α = ° 
 sin 0.71428α = 
 45.6α = °W 
(b) ( )180 30 45.6β = ° − ° + ° 
 104.4= ° 
 Then: 14 lb
sin104.4 sin 30
R =° ° 
 27.1 lbR = W 
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 
graphically the magnitude and direction of their resultant using (a) the 
parallelogram law, (b) the triangle rule. 
 
SOLUTION 
Using the triangle rule and the Law of Sines 
 
(a) Have: 25 lb
sin15 sin 30
Q =° ° 
 12.94 lbQ = W 
(b) ( )180 15 30β = ° − ° + ° 
 135= ° 
 Thus: 25 lb
sin135 sin 30
R =° ° 
sin13525 lb 35.36 lb
sin 30
R ° = = °  
 35.4 lbR = W 
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 α 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 sin 30P = ° 
 10 lb= 10 lb=P W 
(b) ( )20 lb cos30R = ° 
 17.32 lb= 17.32 lbR = W 
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 
magnitudeand 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 ,α ° that is, P is horizontal 
 
Then: (a) ( )360 N sin 35P = ° 
 or 206 N=P W 
And: (b) ( )360 N cos35R = ° 
 or 295 NR = W 
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 α = 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 α if the resultant R of the two forces applied to the 
support is to be horizontal, (b) the corresponding magnitude of R. 
 
SOLUTION 
Using the force triangle and the Law of Cosines 
 
( ) ( ) ( )( )2 22 10 lb 20 lb 2 10 lb 20 lb cos110R = + − ° 
 ( ) 2100 400 400 0.342 lb = + − −  
 2636.8 lb= 
 25.23 lbR = 
Using now the Law of Sines 
 10 lb 25.23 lb
sin sin110β = ° 
10 lbsin sin110
25.23 lb
β  = °   
 0.3724= 
So: 21.87β = ° 
Angle of inclination of R, φ is then such that: 
30φ β+ = ° 
 8.13φ = ° 
Hence: 25.2 lb=R 8.13°W 
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 25α = ° − ° + ° 
 105= ° 
Then: ( ) ( ) ( )( )2 22 4.5 kN 6 kN 2 4.5 kN 6 kN cos105R = + − ° 
 270.226 kN= 
or 8.3801 kNR = 
Now: 8.3801 kN 6 kN
sin105 sin β=° 
6 kNsin sin105
8.3801 kN
β  = °   
 0.6916= 
 43.756β = ° 
 8.38 kN=R 18.76°W 
 
16
 
PROBLEM 2.17 
Solve Problem 2.2 using trigonometry 
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 2tan 38.66
2.5
α −= = ° 
1 1.5tan 30.96
2.5
β −= = ° 
Now: ( )180 38.66 30.96 110.38θ = ° − + ° = 
And, using the Law of Cosines: 
( ) ( ) ( )( )2 22 500 N 160 N 2 500 N 160 N cos110.38R = + − ° 
 2331319 N= 
575.6 NR = 
Using the Law of Sines: 
160 N 575.6 N
sin sin110.38γ = ° 
160 Nsin sin110.38
575.6 N
γ  = °   
 0.2606= 
 15.1γ = ° 
( )90 66.44φ α γ= ° − + = ° 
 576 N=R 66.4°W 
 
17
 
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) the triangle rule. 
 
SOLUTION 
 
 
 
 
 
Using the force triangle and the Laws of Cosines and Sines 
We have: 
( )180 15 30γ = ° − ° + ° 
 135= ° 
Then: ( ) ( ) ( )( )2 22 15 lb 25 lb 2 15 lb 25 lb cos135R = + − ° 
 21380.3 lb= 
or 37.15 lbR = 
and 
25 lb 37.15 lb
sin sin135β = ° 
25 lbsin sin135
37.15 lb
β  = °   
 0.4758= 
 28.41β = ° 
Then: 75 180α β+ + ° = ° 
76.59α = ° 
 37.2 lb=R 76.6°W 
 
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 
 
 
 
 
 
 
Using the force triangle and the Laws of Cosines and Sines 
We have: ( )180 45 25 110γ = ° − ° + ° = ° 
Then: ( ) ( ) ( )( )2 22 30 kN 20 kN 2 30 kN 20 kN cos110R = + − ° 
 21710.4 kN= 
 41.357 kNR = 
and 
20 kN 41.357 kN
sin sin110α = ° 
20 kNsin sin110
41.357 kN
α  = °   
 0.4544= 
 27.028α = ° 
Hence: 45 72.028φ α= + ° = ° 
 41.4 kN=R 72.0°W 
 
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 
 
 
 
 
 
Using the force triangle and the Laws of Cosines and Sines 
We have: ( )180 45 25 110γ = ° − ° + ° = ° 
Then: ( ) ( ) ( )( )2 22 30 kN 20 kN 2 30 kN 20 kN cos110R = + − ° 
 21710.4 kN= 
 41.357 kNR = 
and 
30 kN 41.357 kN
sin sin110α = ° 
30 kNsin sin110
41.357 kN
α  = °   
 0.6816= 
 42.97α = ° 
Finally: 45 87.97φ α= + ° = ° 
 41.4 kN=R 88.0°W 
 
 
20
 
 
 
 
PROBLEM 2.21 
Determine the x and y components of each of the forces shown. 
 
SOLUTION 
20 kN Force: 
 ( )20 kN cos 40 ,xF = + ° 15.32 kNxF = W 
 ( )20 kN sin 40 ,yF = + ° 12.86 kNyF = W 
30 kN Force: 
 ( )30 kN cos70 ,xF = − ° 10.26 kNxF = − W 
 ( )30 kN sin 70 ,yF = + ° 28.2 kNyF = W 
42 kN Force: 
 ( )42 kN cos 20 ,xF = − ° 39.5 kNxF = − W 
 ( )42 kN sin 20 ,yF = + ° 14.36 kNyF = W 
21
 
 
 
 
PROBLEM 2.22 
Determine the x and y components of each of the forces shown. 
 
SOLUTION 
40 lb Force: 
 ( )40 lb sin 50 ,xF = − ° 30.6 lbxF = − W 
 ( )40 lb cos50 ,yF = − ° 25.7 lbyF = − W 
60 lb Force: 
 ( )60 lb cos60 ,xF = + ° 30.0 lbxF = W 
 ( )60 lb sin 60 ,yF = − ° 52.0 lbyF = − W 
80 lb Force: 
 ( )80 lb cos 25 ,xF = + ° 72.5 lbxF = W 
 ( )80 lb sin 25 ,yF = + ° 33.8 lbyF = W 
 
22
 
 
 
 
PROBLEM 2.23 
Determine the x and y components of each of the forces shown. 
 
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: 
 ( ) 48102 lb ,
102x
F = − 48.0 lbxF = − W 
 ( ) 90102 lb ,
102y
F = + 90.0 lbyF = W 
212 lb Force: 
 ( ) 56212 lb ,
106x
F = + 112.0 lbxF = W 
 ( ) 90212 lb ,
106y
F = + 180.0 lbyF = W 
400 lb Force: 
 ( ) 80400 lb ,
100x
F = − 320 lbxF = − W 
 ( ) 60400 lb ,
100y
F = − 240 lbyF = − W 
23
 
 
 
PROBLEM 2.24 
Determine the x and y components of each of the forces shown. 
 
SOLUTION 
 
 
 
 
 
We compute the following distances: 
 ( ) ( )2 270 240 250 mmOA = + = 
( ) ( )2 2210 200 290 mmOB = + = 
 ( ) ( )2 2120 225 255 mmOC = + = 
500 N Force: 
 70500 N
250x
F  = −    140.0 NxF = − W 
 240500 N
250y
F  = +    480 NyF = W 
435 N Force: 
 210435 N
290x
F  = +    315 NxF = W 
 200435 N
290y
F  = +    300 NyF = W 
510 N Force: 
 120510 N
255x
F  = +    240 NxF = W 
 225510 N
255y
F  = −    450 NyF = − W 
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) 
cos 40
xPP = ° 
 135 N
cos 40
= ° 
 or 176.2 NP = W 
(b) tan 40 sin 40y xP P P= ° = ° 
 ( )135 N tan 40= °or 113.3 NyP = W 
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) 
sin 35
yPP = ° 
 960 N
sin 35
= ° 
 or 1674 NP = W 
(b) 
tan 35
y
x
P
P = ° 
 960 N
tan 35
= ° 
 or 1371 NxP = W 
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) sin 50xP P= ° 
 
sin 50
xPP = ° 
 260 lb
sin50
= ° 
 339.4 lb= 339 lbP = W 
(b) tan 50x yP P= ° 
 
tan 50
x
y
PP = ° 
 260 lb
tan 50
= ° 
 218.2 lb= 218 lby =P W 
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
yPP = ° 
 25 lb
sin 75
= ° 
 or 25.9 lbP = W 
(b) 
tan 75
y
x
P
P = ° 
 25 lb
tan 75
= ° 
 or 6.70 lbxP = W 
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 = W 
(b) ( )450 N tan 35xP = ° 
 315.1 N= 
 315 NxP = W 
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) 
sin 38
xPP = ° 
 200 N
sin 38
= ° 
 324.8 N= or 325 NP = W 
(b) 
tan 38
x
y
PP = ° 
 200 N
tan 38
= ° 
 255.98 N= 
 or 256 NyP = W 
 
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 N= − +F i j 
( ) ( )425 315 N 300 N= +F i j 
( ) ( )510 240 N 450 N= −F i j 
( ) ( )415 N 330 N= Σ = +R F i j 
Then: 
1 330tan 38.5
415
α −= = ° 
( ) ( )2 2415 N 330 N 530.2 NR = + = 
Thus: 530 N=R 38.5°W 
31
 
 
 
 
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: 
( ) ( )20 15.32 kN 12.86 kN= +F i j 
( ) ( )30 10.26 kN 28.2 kN= − +F i j 
( ) ( )42 39.5 kN 14.36 kN= − +F i j 
( ) ( )34.44 kN 55.42 kN= Σ = − +R F i j 
Then: 
1 55.42tan 58.1
34.44
α −= = °− 
( ) ( )2 255.42 kN 34.44 N 65.2 kNR = + − = 
 65.2 kNR = 58.2°W 
32
 
 
 
 
 
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. 
 
 
 
 
 
 
 
 x yR R= +R i j 
 ( ) ( )71.9 lb 43.86 lb= −i j 
43.86tan
71.9
α = 
 31.38α = ° 
( ) ( )2 271.9 lb 43.86 lbR = + − 
 84.23 lb= 
 84.2 lb=R 31.4°W 
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 
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. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
( ) ( )204 48.0 lb 90.0 lb= − +F i j 
( ) ( )212 112.0 lb 180.0 lb= +F i j 
( ) ( )400 320 lb 240 lb= − −F i j 
Thus 
x y= +R R R 
( ) ( )256 lb 30.0 lb= − +R i j 
Now: 
30.0tan
256
α = 
1 30.0tan 6.68
256
α −= = ° 
and 
( ) ( )2 2256 lb 30.0 lbR = − + 
 257.75 lb= 
 258 lb=R 6.68°W 
34
 
 
 
 
 
PROBLEM 2.35 
Knowing that 35 ,α = ° determine the resultant of the three forces 
shown. 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
300-N Force: 
( )300 N cos 20 281.9 NxF = ° = 
( )300 N sin 20 102.6 NyF = ° = 
400-N Force: 
( )400 N cos55 229.4 NxF = ° = 
( )400 N sin 55 327.7 NyF = ° = 
600-N Force: 
( )600 N cos35 491.5 NxF = ° = 
( )600 N sin 35 344.1 NyF = − ° = − 
and 
1002.8 Nx xR F= Σ = 
86.2 Ny yR F= Σ = 
( ) ( )2 21002.8 N 86.2 N 1006.5 NR = + = 
Further: 
86.2tan
1002.8
α = 
1 86.2tan 4.91
1002.8
α −= = ° 
 1007 N=R 4.91°W 
35
 
 
 
 
PROBLEM 2.36 
Knowing that 65 ,α = ° determine the resultant of the three forces 
shown. 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
300-N Force: 
( )300 N cos 20 281.9 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 sin 5 52.3 NyF = − ° = − 
and 
914.5 Nx xR F= Σ = 
448.8 Ny yR F= Σ = 
( ) ( )2 2914.5 N 448.8 N 1018.7 NR = + = 
Further: 
448.8tan
914.5
α = 
1 448.8tan 26.1
914.5
α −= = ° 
 1019 N=R 26.1°W 
 
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: 
( ) 84145 lb 105 lb
116x
F = − = − 
( ) 80145 lb 100 lb
116y
F = = 
100-lb Force: 
( ) 3100 lb 60 lb
5x
F = − = − 
( ) 4100 lb 80 lb
5y
F = − = − 
156-lb Force: 
( )12156 lb 144 lb
13x
F = = 
( ) 5156 lb 60 lb
13y
F = − = − 
and 
21 lb, 40 lbx x y yR F R F= Σ = − = Σ = − 
( ) ( )2 221 lb 40 lb 45.177 lbR = − + − = 
Further: 
40tan
21
α = 
1 40tan 62.3
21
α −= = ° 
Thus: 45.2 lb=R 62.3°W 
 
37
 
PROBLEM 2.38 
Knowing that 50 ,α = ° determine the resultant of the three forces 
shown. 
 
SOLUTION 
 
 
 
 
 
 
The resultant force R has the x- and y-components: 
( ) ( ) ( )140 lb cos50 60 lb cos85 160 lb cos50x xR F= Σ = ° + ° − ° 
 7.6264 lbxR = − 
and 
( ) ( ) ( )140 lb sin 50 60 lb sin85 160 lb sin 50y yR F= Σ = ° + ° + ° 
 289.59 lbyR = 
Further: 
290tan
7.6
α = 
1 290tan 88.5
7.6
α −= = ° 
Thus: 290 lb=R 88.5°W 
 
 
 
 
 
 
 
 
 
 
 
 
 
38
 
 
PROBLEM 2.39 
Determine (a) the required value of α if the resultant of the three forces 
shown is to be vertical, (b) the corresponding magnitude of the resultant. 
 
SOLUTION 
For an arbitrary angle ,α we have: 
( ) ( ) ( ) ( )140 lb cos 60 lb cos 35 160 lb cosx xR F α α α= Σ = + + ° − 
(a) So, for R to be vertical: 
( ) ( ) ( ) ( )140 lb cos 60 lb cos 35 160 lb cos 0x xR F α α α= Σ = + + ° − = 
 Expanding, 
( )cos 3 cos cos35sin sin 35 0α α α− + ° − ° = 
 Then: 
1
3cos35tan
sin 35
α ° −= ° 
 or 
 
1
1 3cos35tan 40.265
sin 35
α −  ° −= = °  ° 
 40.3α = °W 
(b) Now: 
( ) ( ) ( )140 lb sin 40.265 60 lb sin 75.265 160 lb sin 40.265y yR R F= = Σ = ° + ° + ° 
 252 lbR R= = W 
 
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 3156 lb 100 lb
116 13 5x x BC
R F T= Σ = − + − 
or 0.724 84 lbx BCR T= − + 
and 
( ) ( )80 5 4156 lb 100 lb
116 13 5y y BC
R F T= Σ = − − 
 0.6897 140 lby BCR T= − 
(a) So, for R to be vertical, 
0.724 84 lb 0x BCR T= − + = 
 116.0 lbBCT = W 
(b) Using 
116.0 lbBCT = 
( )0.6897 116.0 lb 140 lb 60 lbyR R= = − = − 
 60.0 lbR R= = W 
 
 
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 cos 25 5.2 kN sin 35 sin 65 0y y AER F T= Σ = ° + ° − ° = 
 or 7.2909 kNAET = 
 7.29 kNAET = W 
(b) xR F= Σ 
 ( ) ( ) ( )4 kN sin 25 5.2 kN cos35 7.2909 kN cos65= − ° − ° − ° 
 9.03 kN= − 
 9.03 kNR = W 
41
 
 
 
 
PROBLEM 2.42 
For the block of Problems 2.35 and 2.36, determine (a) the required value 
of α 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 ,α = ° determine the resultant of the 
three forces shown. 
Problem 2.36: Knowing that 65 ,α = ° 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 α α= Σ = + + (1) 
 ( ) ( )400 N sin 600 N cosy yR F α α= Σ = − (2) 
(a) Setting 0yR = in Equation (2): 
 Thus 600tan 1.5
400
α = = 
 56.3α = °W 
(b) Substituting for α in Equation (1): 
( ) ( )300 N 400 N cos56.3 600 N sin 56.3xR = + ° + ° 
 1021.1 NxR = 
 1021 NxR R= = W 
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: 
( ) ( )2 216 in. 12 in. 20 in.AC = + = 
( ) ( )2 220 in. 21 in. 29 in.BC = + = 
Then, from the Free Body Diagram of point C: 
 16 210: 0
20 29x AC BC
F T TΣ = − + = 
or 29 4
21 5BC AC
T T= × 
and 12 200: 600 lb 0
20 29y AC BC
F T TΣ = + − = 
or 12 20 29 4 600 lb 0
20 29 21 5AC AC
T T + × − =   
Hence: 440.56 lbACT = 
(a) 441 lbACT = W 
(b) 487 lbBCT = W 
43
 
 
 
 
PROBLEM 2.44 
Knowing that 25 ,α = ° determine the tension (a) in cable AC, (b) in 
rope BC. 
 
SOLUTION 
Free-Body Diagram Force Triangle 
 
Law of Sines: 
5 kN
sin115 sin 5 sin 60
AC BCT T= =° ° ° 
(a) 5 kN sin115 5.23 kN
sin 60AC
T = ° =° 5.23 kNACT = W 
(b) 5 kN sin 5 0.503 kN
sin 60BC
T = ° =° 0.503 kNBCT = W 
44
 
 
 
 
 
PROBLEM 2.45 
Knowing that 50α = ° 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: 
 400 lb
sin 25 sin 60 sin 95
AC BCF T= =° ° ° 
(a) 400 lb sin 25 169.69 lb
sin 95AC
F = ° =° 169.7 lbACF = W 
(b) 400 sin 60 347.73 lb
sin 95BC
T = ° =° 348 lbBCT = W 
45
 
 
 
 
PROBLEM 2.46 
Two cables are tied together at C and are loaded as shown. Knowing that 
30 ,α = ° determine the tension (a) in cable AC, (b) in cable BC. 
 
SOLUTION 
Free-Body Diagram Force Triangle 
 
Law of Sines: 
2943 N
sin 60 sin 55 sin 65
AC BCT T= =° ° ° 
(a) 2943 N sin 60 2812.19 N
sin 65AC
T = ° =° 2.81 kNACT = W 
(b) 2943 N sin 55 2659.98 N
sin 65BC
T = ° =° 2.66 kNBCT = W 
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 9.9tan 30.51
16.8AB
θ −= = ° 
1 12tan 22.61
28.8BC
θ −= = ° 
 
 
Thus, in the force triangle, by the Law of Sines: 
1190 N
sin 59.49 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 1.32tan 10.39
7.2CD
θ −= = ° 
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= W 
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 12tan 22.62
28.8EF
θ −= = ° 
1 1.32tan 10.39
7.2DF
θ −= = ° 
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 9.9tan 30.51
16.8AE
θ −= = ° 
Hence, in the force triangle, by the Law of Sines: 
5107.5 N
sin 7.89 sin 59.49
W =° ° 
813.8 NW = 
Finally, the skier weight 813.8 N 300 N 513.8 N= − = 
 skier weight 514 N= W 
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 FΣ = + ° − ° = 
 or 332 lbCF = W 
( ) ( )0: 510 lb cos15 480 lb sin15 0y DF FΣ = − ° + ° = 
 or 368 lbDF = W 
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 FΣ = − ° + + ° = = 
 672 lbBF = W 
( ) ( )0: 420 lb cos15 671.6 lb sin15 0y DF FΣ = − ° + ° = 
 or 232 lbDF = W 
 
50
 
 
 
 
PROBLEM 2.51 
Two forces P and Q are applied as shown to an aircraft connection. 
Knowing thatthe 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 B= + + + =R P Q F F 
Substituting components: 
( ) ( ) ( )400 lb 520 lb cos55 520 lb sin 55   = − + ° − °   R j i j 
 ( ) ( )cos55 sin 55 0B A AF F F+ − ° + ° =i i j 
In the y-direction (one unknown force) 
( )400 lb 520 lb sin 55 sin 55 0AF− − ° + ° = 
Thus, 
( )400 lb 520 lb sin 55 1008.3 lb
sin 55A
F
+ °= =° 
 1008 lbAF = W 
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 = W 
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 B= + + + =R P Q F F 
Substituting components: 
( ) ( ) ( )320 lb 600 lb cos55 600 lb sin 55   = − ° + °   R i i j 
 ( ) ( )cos55 sin 55 0P Q Q+ + ° − ° =i 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
− + °= =° 
 42.1 lbQ = W 
In the y-direction: 
( )600 lb sin 55 sin 55 0P Q° − − ° = 
Thus, 
( )600 lb sin 55 sin 55 457.01 lbP Q= ° − ° = 
 457 lbP = W 
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 15 150: 680 N 0
5 17 17x CA CB
F T TΣ = − + − = 
or 
 1 5 200 N
5 17CA CB
T T− + = (1) 
and 
 ( )4 8 80: 680 N 840 N 05 17 17y CA CBF T TΣ = + − − = 
or 
 1 2 290 N
5 17CA CB
T T+ = (2) 
Solving Equations (1) and (2) simultaneously: 
(a) 750 NCAT = W 
(b) 1190 NCBT = W 
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 15 150: 680 N 0
5 17 17x CA CB
F T TΣ = − + − = 
or 
 1 5 200 N
5 17CA CB
T T− + = (1) 
and 
 ( )4 8 80: 680 N 05 17 17y CA CBF T T WΣ = + − − = 
or 
 1 2 180 N
5 17 4
+ = +CA CBT T W (2) 
Then, from Equations (1) and (2) 
17680 N
28
25
28
CB
CA
T W
T W
= +
=
 
Now, with 1050 NT ≤ 
25: 1050 N
28CA CA
T T W= = 
or 1176 NW = 
and 
17: 1050 N 680 N
28CB CB
T T W= = + 
or 609 NW = 0 609 N∴ ≤ ≤W W 
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 40α = ° and β = 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 cos 40 cos 40 0x ACB DEF T TΣ = ° − ° − ° = 
or 
 0.0531 0.766 0ACB DET T− = (1) 
and 
 ( )0: sin 40 sin 35 sin 40 24.8 kN 0y ACB DEF T TΣ = ° − ° + ° − = 
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) 15.1 kNDET = W 
 (a) 218 kNACBT = W 
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 42α = ° and β = 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 cos 42 20 kN cos 42 0x ACBF TΣ = ° − ° − ° = 
or 
0.1049 14.863 kNACBT = 
 (b) 141.7 kNACBT = W 
Now 
 ( ) ( )0: sin 42 sin 32 20 kN sin 42 0y ACBF T WΣ = ° − ° + ° − = 
or 
 ( )( ) ( )( )141.7 kN 0.1392 20 kN 0.6691 0W+ − = 
 (a) 33.1 kNW = W 
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 20.44 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 20.66 m 0.32 m 0.68 mADL = + = 
 ( )1500 N/m 0.68 m 0.45 mADF = − 
 115 N= 
(a) 
 ( ) ( )4 7 150: 150 N 115 N 0
5 25 17x AC
F TΣ = − + − = 
 or 
 66.18 NACT = 66.2 NACT = W 
57
 
 
 
PROBLEM 2.57 CONTINUED 
(b) and 
 ( ) ( ) ( )3 24 80: 150 N 66.18 N 115 N 05 25 17yF WΣ = + + − = 
 or 208 N=W W 
 
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 35 120: 3 0
5 37 37x s
F W W FΣ = − + + = 
or 
4.4833sF W= 
and 
 ( ) ( )3 12 350: 3 400 N 05 37 37y sF W W FΣ = + + − = 
Then: 
( ) ( ) ( )3 12 353 4.4833 400 N 0
5 37 37
W W W+ + − = 
or 
 62.841 NW = 
and 
281.74 NsF = 
or 
(a) 62.8 NW = W 
 
59
 
 
 
PROBLEM 2.58 CONTINUED 
(b) Have spring force 
( )s AB oF k L L= − 
 Where 
( )AB AB AB oF k L L= − 
 and 
( ) ( )2 20.360 m 1.050 m 1.110 mABL = + = 
 So: 
( )0281.74 N 800 N/m 1.110 mL= − 
 or 0 758 mmL = W 
 
60
 
 
 
 
 
PROBLEM 2.59 
For the cables and loading of Problem 2.46, determine (a) the value of α 
for which the tension in cable BC is as small as possible, (b) the 
corresponding value ofthe tension. 
 
SOLUTION 
The smallest BCT is when BCT is perpendicular to the direction of ACT 
Free-Body Diagram At C Force Triangle 
 
(a) 55.0α = °W 
(b) ( )2943 N sin 55BCT = ° 
 2410.8 N= 
 2.41 kNBCT = W 
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 TΣ = − = 
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 m∴ =BC 
( )2 4.14 mL BC= = 
 4.14 mL = W 
 
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 α. 
 
SOLUTION 
 Free-Body Diagram: C Force Triangle 
 
 
 
 
 
 
Force triangle is isoceles with 
2 180 85β = ° − ° 
47.5β = ° 
(a) ( )2 200 lb cos 47.5 270 lbP = ° = 
 Since 0,P > the solution is correct. 270 lbP = W 
(b) 180 55 47.5 77.5α = ° − ° − ° = ° 77.5α = °W 
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 α. 
 
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 = W 
(b) Law of Sines: 
sin sin85
300 323.5
β °= ° 
sin 0.9238β = 
 or 67.49β = ° 
180 55 67.49 57.5α = ° − ° − ° = ° 
 57.5α = °W 
64
 
 
 
 
 
PROBLEM 2.63 
For the structure and loading of Problem 2.45, determine (a) the value of 
α 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: 55α = ° 55α = °W 
(b) ( )400 lb sin 60BCT = ° 
 or 346.4 lbBCT = 346 lbBCT = W 
 
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 α 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 BC+ + =T 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 90.0α = °W 
(b) Have ( ) ( )70 lb sin 180 70 30BCT = ° − ° − ° 
 68.93 lb= 
 68.9 lbBCT = W 
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 20.3 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: 
 ( )40: 49.986 N 0
5y
F WΣ = − + = 
 or 40.0 NW = W 
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 20: 00.3y s
hF W F
h
Σ = − + =
+
 
or 240 0.09shF h= + 
Now.. ( )s AB ABF k L L′= − 
where ( )2 20.3 m 0.3 2 mAB ABL h L′ = + = 
Then: ( )2 2560 0.09 0.3 2 40 0.09h h h + − = +   
or ( ) 214 1 0.09 4.2 2 mh h h h− + = ∼ 
Solving numerically, 
 415 mmh = W 
 
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) 
 
 
 
 
 
 
 ( )( )20: 2 280 kg 9.81 m/s 0yF TΣ = − = 
( )1 2746.8 N
2
T = 
 1373 NT = W 
 ( )( )20: 2 280 kg 9.81 m/s 0yF TΣ = − = 
( )1 2746.8 N
2
T = 
 1373 NT = W 
 ( )( )20: 3 280 kg 9.81 m/s 0yF TΣ = − = 
( )1 2746.8 N
3
T = 
 916 NT = W 
 ( )( )20: 3 280 kg 9.81 m/s 0yF TΣ = − = 
( )1 2746.8 N
3
T = 
 916 NT = W 
 
 ( )( )20: 4 280 kg 9.81 m/s 0yF TΣ = − = 
( )1 2746.8 N
4
T = 
 687 NT = W 
 
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) 
 
 
 
 
 ( )( )20: 3 280 kg 9.81 m/s 0yF TΣ = − = 
( )1 2746.8 N
3
T = 
 916 NT = W 
 
 
 ( )( )20: 4 280 kg 9.81 m/s 0yF TΣ = − = 
( )1 2746.8 N
4
T = 
 687 NT = W 
 
 
 
 
 
 
 
 
 
 
 
 
 
70
 
 
PROBLEM 2.69 
A 350-lb load is supported by the rope-and-pulley arrangement shown. 
Knowing that β = 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 αΣ = ° − = 
and 
cos 0.8452 or 32.3α α= = ± ° 
For 32.3α = + ° 
 0: 2 cos 25 sin 32.3 350 lb 0yF P PΣ = ° + ° − = 
 or 149.1 lb=P 32.3°W 
For 32.3α = − ° 
 0: 2 cos 25 sin 32.3 350 lb 0yF P PΣ = ° + − ° − = 
 or 274 lb=P 32.3°W 
 
71
 
PROBLEM 2.70 
A 350-lb load is supported by the rope-and-pulley arrangement shown. 
Knowing that 35 ,α = ° determine (a) the angle β, (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 cos 25 0xF P PβΣ = − ° = 
Hence: 
(a) 1sin cos 25
2
β = ° or 24.2β = °W 
(b) 0: 2 cos sin 35 350 lb 0yF P PβΣ = + ° − = 
 Hence: 
2 cos 24.2 sin 35 350 lb 0P P° + ° − = 
 or 145.97 lbP = 146.0 lbP = W 
 
 
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 shownby 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 TΣ = ° − ° − ° = 
 Hence 2303.5 NACBT = 
 2.30 kN=ACBT W 
(b) ( ) ( )0: sin 30 sin 50 800 N sin 50 0y ACBF T QΣ = ° + ° + ° − = 
 ( )( ) ( )2303.5 N sin 30 sin 50 800 N sin 50 0Q° + ° + ° − = 
 or 3529.2 NQ = 3.53 kN=Q W 
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 PΣ = ° − ° − ° = 
or 0.3473 ACBP T= (1) 
 ( )0: sin 30 sin 50 sin 50 2000 N 0y ACBF T PΣ = ° + ° + ° − = 
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 = W 
(b) Using (1) 
 ( )0.3473 1306 N 453.57 NP = = 
 454 NP = W 
74
 
 
 
 
PROBLEM 2.73 
Determine (a) the x, y, and z components of the 200-lb force, (b) the 
angles θx, θy, and θz that the force forms with the coordinate axes. 
 
SOLUTION 
(a) ( )200 lb cos30 cos 25 156.98 lbxF = ° ° = 
 157.0 lbxF = + W 
( )200 lb sin 30 100.0 lbyF = ° = 
 100.0 lbyF = + W 
( )200 lb cos30 sin 25 73.1996 lbzF = − ° ° = − 
 73.2 lbzF = − W 
(b) 156.98cos
200x
θ = or 38.3xθ = °W 
 100.0cos
200y
θ = or 60.0yθ = °W 
 73.1996cos
200z
θ −= or 111.5zθ = °W 
75
 
 
 
 
PROBLEM 2.74 
Determine (a) the x, y, and z components of the 420-lb force, (b) the 
angles θx, θy, and θz that the force forms with the coordinate axes. 
 
SOLUTION 
(a) ( )420 lb sin 20 sin 70 134.985 lbxF = − ° ° = − 
 135.0 lbxF = − W 
( )420 lb cos 20 394.67 lbyF = ° = 
 395 lbyF = + W 
( )420 lb sin 20 cos70 49.131 lbzF = ° ° = 
 49.1 lbzF = + W 
(b) 134.985cos
420x
θ −= 
 108.7xθ = °W 
394.67cos
420y
θ = 
 20.0yθ = °W 
49.131cos
420z
θ = 
 83.3zθ = °W 
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 θx, θy, and θz that the force forms with axes at A which 
are parallel to the coordinate axes. 
 
SOLUTION 
 
(a) ( )4.2 kN sin 50 cos 40 2.4647 kNxF = ° ° = 
 2.46 kNxF = + W 
 ( )4.2 kN cos50 2.6997 kNyF = − ° = − 
 2.70 kNyF = − W 
 ( )4.2 kN sin 50 sin 40 2.0681 kNzF = ° ° = 
 2.07 kNzF = + W 
(b) 2.4647cos
4.2x
θ = 
 54.1xθ = °W 
77
 
PROBLEM 2.75 CONTINUED 
 2.7cos
4.2y
θ −= 
 130.0yθ = °W 
 2.0681cos
4.0z
θ = 
 60.5zθ = °W 
 
78
 
 
 
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) the components of the force exerted by this cable on the 
tree, (b) the angles θx, θy, and θz that the force forms with axes at A which 
are parallel to the coordinate axes. 
 
SOLUTION 
 
(a) ( )3.6 kN cos 45 sin 25 1.0758 kNxF = − ° ° = − 
 1.076 kNxF = − W 
( )3.6 kN sin 45 2.546 kNyF = − ° = − 
 2.55 kNyF = − W 
( )3.6 kN cos 45 cos 25 2.3071 kNzF = ° ° = 
 2.31 kNzF = + W 
(b) 1.0758cos
3.6x
θ −= 
 107.4xθ = °W 
79
 
PROBLEM 2.76 CONTINUED 
2.546cos
3.6y
θ −= 
 135.0yθ = °W 
2.3071cos
3.6z
θ = 
 50.1zθ = °W 
 
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 θx, 
θy, and θz that the force exerted at A forms with the coordinate axes. 
 
SOLUTION 
(a) sin 30 sin 50 220.6 NxF F= ° ° = (Given) 
220.6 N 575.95 N
sin30 sin50
= =° °F 
 576 N=F W 
(b) 220.6cos 0.3830
575.95
θ = = =xx FF 
 67.5xθ = °W 
cos30 498.79 NyF F= ° = 
498.79cos 0.86605
575.95
y
y
F
F
θ = = = 
 30.0yθ = °W 
 sin 30 cos50zF F= − ° ° 
 ( )575.95 N sin 30 cos50= − ° ° 
 185.107 N= − 
185.107cos 0.32139
575.95
z
z
F
F
θ −= = = − 
 108.7zθ = °W 
81
 
 
 
 
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) the tension in wire BD, (b) the angles θx, 
θy, and θz that the force exerted at B forms with the coordinate axes. 
 
SOLUTION 
(a) sin 30 sin 40 64.28 NzF F= − ° ° = − (Given) 
 64.28 N 200.0 N
sin30 sin40
= =° °F 200 NF = W 
(b) sin 30 cos 40xF F= − ° ° 
 ( )200.0 N sin 30 cos 40= − ° ° 
 76.604 N= − 
 76.604cos 0.38302
200.0
x
x
F
F
θ −= = = − 112.5xθ = °W 
 cos30 173.2 NyF F= ° = 
 173.2cos 0.866
200
y
y
F
F
θ = = = 30.0yθ = °W 
 64.28 NzF = − 
 64.28cos 0.3214
200
z
z
F
F
θ −= = = − 108.7zθ = °W 
82
 
 
 
 
 
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) the 
components of the force exerted by this wire on the plate, (b) the angles 
θx, θy, and θz that the force forms with the coordinate axes. 
 
SOLUTION 
(a) ( )120 lb sin 30 cos60 30 lbxF = − ° ° = − 
 30.0 lbxF = − W 
( )120 lb cos30 103.92 lbyF = ° = 
 103.9 lb= +yF W 
( )120 lb sin 30 sin 60 51.96 lbzF = ° ° = 
 52.0 lbzF = + W 
(b) 30.0cos 0.25
120
x
x
F
F
θ −= = = − 
 104.5xθ = °W 
103.92cos 0.866
120
y
y
F
F
θ = = = 
 30.0yθ = °W 
51.96cos 0.433
120
z
z
F
F
θ = = = 
 64.3zθ = °W 
83
 
 
 
 
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) the tension in wire CD, (b) the angles θx, θy, 
and θz that the force exerted at C forms with the coordinate axes. 
 
SOLUTION 
(a) sin 30 cos60 40 lbxF F= − ° ° = − (Given) 
40 lb 160 lb
sin30 cos60
= =° °F 
 160.0 lbF = W 
(b) 40cos 0.25
160
x
x
F
F
θ −= = = − 
 104.5xθ = °W 
( )160 lb cos30 103.92 lbyF = ° = 
 103.92cos 0.866
160
y
y
F
F
θ = = = 
 30.0yθ = °W 
( )160 lb sin 30 sin 60 69.282 lbzF = ° ° = 
69.282cos 0.433
160
z
z
F
F
θ = = = 
 64.3zθ = °W 
 
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 = W 
 800cos 0.8889
900
x
x
F
F
θ = = = 27.3xθ = °W 
 260cos 0.2889
900
y
y
F
F
θ = = = 73.2yθ = °W 
 320cos0.3555
900
z
z
F
F
θ −= = = − 110.8zθ = °W 
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 = W 
 400cos 0.30769
1300
x
x
F
F
θ = = = 72.1xθ = °W 
 1200cos 0.92307
1300
y
y
F
F
θ −= = = − 157.4yθ = °W 
 300cos 0.23076
1300
z
z
F
F
θ = = = 76.7zθ = °W 
86
 
 
 
 
 
 
PROBLEM 2.83 
A force acts at the origin of a coordinate system in a direction defined by 
the angles θx = 64.5° and θz = 55.9°. Knowing that the y component of 
the force is –200 N, determine (a) the angle θy, (b) the other components 
and the magnitude of the force. 
 
SOLUTION 
(a) We have 
( ) ( ) ( ) ( ) ( ) ( )2 2 22 2 2cos cos cos 1 cos 1 cos cosx y z y y zθ θ θ θ θ θ+ + = ⇒ = − − 
 Since 0yF < we must have cos 0yθ < 
 Thus, taking the negative square root, from above, we have: 
 ( ) ( )2 2cos 1 cos64.5 cos55.9 0.70735yθ = − − ° − ° = − 135.0yθ = °W 
(b) Then: 
 200 N 282.73 N
cos 0.70735
y
y
F
F θ
−= = =− 
 and ( )cos 282.73 N cos64.5x xF F θ= = ° 121.7 NxF = W 
 ( )cos 282.73 N cos55.9z zF F θ= = ° 158.5 NyF = W 
 283 NF = W 
87
 
 
 
 
 
PROBLEM 2.84 
A force acts at the origin of a coordinate system in a direction defined by 
the angles θx = 75.4° and θy = 132.6°. Knowing that the z component of 
the force is –60 N, determine (a) the angle θz, (b) the other components 
and the magnitude of the force. 
 
SOLUTION 
(a) We have 
( ) ( ) ( ) ( ) ( ) ( )2 2 22 2 2cos cos cos 1 cos 1 cos cosx y z y y zθ θ θ θ θ θ+ + = ⇒ = − − 
 Since 0zF < we must have cos 0zθ < 
 Thus, taking the negative square root, from above, we have: 
 ( ) ( )2 2cos 1 cos75.4 cos132.6 0.69159zθ = − − ° − ° = − 133.8zθ = °W 
(b) Then: 
 60 N 86.757 N
cos 0.69159
z
z
FF θ
−= = =− 86.8 NF = W 
 and ( )cos 86.8 N cos75.4x xF F θ= = ° 21.9 NxF = W 
 ( )cos 86.8 N cos132.6y yF F θ= = ° 58.8 NyF = − W 
88
 
 
 
 
 
 
PROBLEM 2.85 
A force F of magnitude 400 N acts at the origin of a coordinate system. 
Knowing that θx = 28.5°, Fy = –80 N, and Fz > 0, determine (a) the 
components Fx and Fz, (b) the angles θy and θz. 
 
SOLUTION 
(a) Have 
 ( )cos 400 N cos 28.5x xF F θ= = ° 351.5 NxF = W 
 Then: 
2 2 2 2
x y zF F F F= + + 
 So: ( ) ( ) ( )2 2 2 2400 N 352.5 N 80 N zF= + − + 
 Hence: 
 ( ) ( ) ( )2 2 2400 N 351.5 N 80 NzF = + − − − 173.3 NzF = W 
(b) 
 80cos 0.20
400
y
y
F
F
θ −= = = − 101.5yθ = °W 
 173.3cos 0.43325
400
z
z
F
F
θ = = = 64.3zθ = °W 
89
 
 
 
 
 
PROBLEM 2.86 
A force F of magnitude 600 lb acts at the origin of a coordinate system. 
Knowing that Fx = 200 lb, θz = 136.8°, Fy < 0, determine (a) the 
components Fy and Fz, (b) the angles θx and θy. 
 
SOLUTION 
(a) ( )cos 600 lb cos136.8z zF F θ= = ° 
 437.4 lb= − 437 lbzF = − W 
 Then: 
2 2 2 2
x y zF F F F= + + 
 So: ( ) ( ) ( ) ( )22 2 2600 lb 200 lb 437.4 lbyF= + + − 
 Hence: ( ) ( ) ( )2 2 2600 lb 200 lb 437.4 lbyF = − − − − 
 358.7 lb= − 359 lbyF = − W 
(b) 
 200cos 0.333
600
x
x
F
F
θ = = = 70.5xθ = °W 
 358.7cos 0.59783
600
y
y
F
F
θ −= = = − 126.7yθ = °W 
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 kJJJG 
( ) ( ) ( )2 2 24 m 20 m 5 m 21 mBA = + + − = 
( ) ( ) ( )2100 N 4 m 20 m 5 m
21 mBA
BAF F
BA
 = = = + − F i j k
JJJG
λ 
( ) ( ) ( )400 N 2000 N 500 N= + −F i j k 
 400 N, 2000 N, 500 Nx y zF F F= + = + = − W 
91
 
 
 
92
 
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 kJJJG 
( ) ( ) ( )2 2 24 m 20 m 14.8 m 25.2 mDA = + + = 
( ) ( ) ( )1260 N 4 m 20 m 14.8 m
25.2 mDA
DAF F
DA
 = = = + + F i j k
JJJG
λ 
( ) ( ) ( )200 N 1000 N 740 N= + +F i j k 
 200 N, 1000 N, 740 Nx y zF F F= + = + = + W 
 
 
 
 
 
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 kJJJG 
 ( ) ( ) ( )2 2 232 in. 48 in. 36 in. 68 in.BA = + + − = 
( ) ( ) ( )204 lb 32 in. 48 in. 36 in.
68 in.BA
BAF F
BA
 = = = + − F i j k
JJJG
λ 
( ) ( ) ( )96 lb 144 lb 108 lb= + −F i j k 
 96.0 lb, 144.0 lb, 108.0 lbx y zF F F= + = + = − W 
93
 
 
 
 
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 
( ) ( ) ( )25 in. 48 in. 36 in.DA = − + +i j kJJJG 
 ( ) ( ) ( )2 2 225 in. 48 in. 36 in. 65 in.DA = − + + = 
( ) ( ) ( )195 lb 25 in. 48 in. 36 in.
65 in.DA
DAF F
DA
 = = = − + + F i j k
JJJG
λ 
( ) ( ) ( )75 lb 144 lb 108 lb= − + +F i j k 
 75.0 lb, 144.0 lb, 108.0 lbx y zF F F= − = + = + W 
 
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 kJJJG 
( ) ( ) ( )2 2 20.96 m 1.12 m 0.96 m 1.76 mDB = + − + − = 
( ) ( ) ( )220 N 0.96 m 1.12 m 0.96 m
1.76 mDB DB
DBT T
DB
 = = = − − T i j k
JJJG
λ 
( ) ( ) ( )120 N 140 N 120 NDB = − −T i j k 
 ( ) ( ) ( )120.0 N, 140.0 N, 120.0 NDB DB DBx y zT T T= + = − = − W 
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 kJJJG 
( ) ( ) ( )2 2 20.96 m 1.20 m 1.28 m 2.00 mEB = + − + = 
( ) ( ) ( )250 N 0.96 m 1.20 m 1.28 m
2.00 mEB EB
EBT T
EB
 = = = − + T i j k
JJJG
λ 
( ) ( ) ( )120 N 150 N 160 NEB = − +T i j k 
 ( ) ( ) ( )120.0 N, 150.0 N, 160.0 NEB EB EBx y zT T T= + = − = + W 
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 sin 30 cos30 cos15= − ° ° + ° + ° °P i j k 
 ( )[ ]500 lb 0.2241 0.50 0.8365= − + +i j k 
 ( ) ( ) ( )112.05 lb 250 lb 418.25 lb= − + +i j k 
( )[ ]600 lb cos 40 cos 20 sin 40 cos 40 sin 20= ° ° + ° − ° °Q i j k 
 ( )[ ]600 lb 0.71985 0.64278 0.26201= + −i j k 
 ( ) ( ) ( )431.91 lb 385.67 lb 157.206 lb= + −i j k 
( ) ( ) ( )319.86 lb 635.67 lb 261.04 lb= + = + +R P Q i j k 
( ) ( ) ( )2 2 2319.86 lb 635.67 lb 261.04 lb 757.98 lbR = + + = 
 758 lbR = W 
319.86 lbcos 0.42199
757.98 lb
x
x
R
R
θ = = = 
 65.0xθ = °W 
635.67 lbcos 0.83864
757.98 lb
y
y
R
R
θ = = = 
 33.0yθ = °W 
261.04 lbcos 0.34439
757.98 lb
z
z
R
R
θ = = =69.9zθ = °W 
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 0.8365= − + +P i j k 
 ( ) ( ) ( )134.46 lb 300 lb 501.9 lb= − + +i j k 
( )[ ]400 lb 0.71985 0.64278 0.26201= + −Q i j k 
 ( ) ( ) ( )287.94 lb 257.11 lb 104.804 lb= + −i j k 
( ) ( ) ( )153.48 lb 557.11 lb 397.10 lb= + = + +R P Q i j k 
( ) ( ) ( )2 2 2153.48 lb 557.11 lb 397.10 lb 701.15 lbR = + + = 
 701 lbR = W 
153.48 lbcos 0.21890
701.15 lb
x
x
R
R
θ = = = 
 77.4xθ = °W 
557.11 lbcos 0.79457
701.15 lb
y
y
R
R
θ = = = 
 37.4yθ = °W 
397.10 lbcos 0.56637
701.15 lb
z
z
R
R
θ = = = 
 55.5zθ = °W 
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 kJJJG 
( ) ( ) ( )2 2 2400 mm 450 mm 600 mm 850 mmAB = + − + = 
( ) ( ) ( )1000 mm 450 mm 600 mmAC = − +i j kJJJG 
( ) ( ) ( )2 2 21000 mm 450 mm 600 mm 1250 mmAC = + − + = 
( ) ( ) ( ) ( )400 mm 450 mm 600 mm850 N
850 mmAB AB ABAB
ABT T
AB
 − += = =   
i j k
T
JJJG
λ 
( ) ( ) ( )400 N 450 N 600 NAB = − +T i j k 
( ) ( ) ( ) ( )1000 mm 450 mm 600 mm1020 N
1250 mmAC AC ACAC
ACT T
AC
 − += = =   
i j k
T
JJJG
λ 
( ) ( ) ( )816 N 367.2 N 489.6 NAC = − +T i j k 
( ) ( ) ( )1216 N 817.2 N 1089.6 NAB AC= + = − +R T T i j k 
Then: 1825.8 NR = 1826 NR = W 
and 1216cos 0.66601
1825.8x
θ = = 48.2xθ = °W 
 817.2cos 0.44758
1825.8y
θ −= = − 116.6yθ = °W 
 1089.6cos 0.59678
1825.8z
θ = = 53.4zθ = °W 
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 kJJJG 
( ) ( ) ( )2 2 2400 mm 450 mm 600 mm 850 mmAB = + − + = 
( ) ( ) ( )1000 mm 450 mm 600 mmAC = − +i j kJJJG 
( ) ( ) ( )2 2 21000 mm 450 mm 600 mm 1250 mmAC = + − + = 
( ) ( ) ( ) ( )400 mm 450 mm 600 mm1020 N
850 mmAB AB AB AB
ABT T
AB
 − += = =   
i j k
T
JJJG
λ 
( ) ( ) ( )480 N 540 N 720 NAB = − +T i j k 
( ) ( ) ( ) ( )1000 mm 450 mm 600 mm850 N
1250 mmAC AC AC AC
ACT T
AC
 − += = =   
i j k
T
JJJG
λ 
( ) ( ) ( )680 N 306 N 408 NAC = − +T i j k 
( ) ( ) ( )1160 N 846 N 1128 NAB AC= + = − +R T T i j k 
Then: 1825.8 NR = 1826 NR = W 
and 1160cos 0.6353
1825.8x
θ = = 50.6xθ = °W 
 846cos 0.4634
1825.8y
θ −= = − 117.6yθ = °W 
 1128cos 0.6178
1825.8z
θ = = 51.8zθ = °W 
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 NBD = − + +T i j k 
( ) ( ) ( )120 N 150 N 160 NBE = − + −T i j k 
Then: 
 B BD BE= +R T T 
 ( ) ( ) ( )240 N 290 N 40 N= − + −i j k 
and 378.55 NR = 379 NBR = W 
240cos 0.6340
378.55x
θ = − = − 
 129.3xθ = °W 
290cos 0.7661
378.55y
θ = = − 
 40.0yθ = °W 
40cos 0.1057
378.55z
θ = − = − 
 96.1zθ = °W 
 
101
 
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) the tension in AC, (b) the magnitude and direction of 
the resultant of the two forces. 
 
SOLUTION 
Have 
( )( )920 lb sin 50 cos 40 cos50 sin 50 sin 40AB = ° ° − ° + ° °T i j j 
( )cos 45 sin 25 sin 45 cos 45 cos 25AC ACT= − ° ° − ° + ° °T i j j 
(a) 
A AB AC= +R T T 
( ) 0A xR = 
∴ ( ) ( )0: 920 lb sin 50 cos 40 cos 45 sin 25 0A x ACxR F T= Σ = ° ° − ° ° = 
 or 
 1806.60 lbACT = 1807 lbACT = W 
(b) 
( ) ( ) ( ): 920 lb cos50 1806.60 lb sin 45A yyR F= Σ − ° − ° 
( ) 1868.82 lbA yR = − 
( ) ( ) ( ): 920 lb sin 50 sin 40 1806.60 lb cos 45 cos 25A zzR F= Σ ° ° + ° ° 
( ) 1610.78 lbA zR = 
∴ ( ) ( )1868.82 lb 1610.78 lbAR = − +j k 
 Then: 
 2467.2 lbAR = 2.47 kipsAR = W 
102
 
 
PROBLEM 2.98 CONTINUED 
 and 
 0cos 0
2467.2x
θ = = 90.0xθ = °W 
 1868.82cos 0.7560
2467.2y
θ −= = − 139.2yθ = °W 
 1610.78cos 0.65288
2467.2z
θ = = 49.2zθ = °W 
 
103
 
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 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 
( )sin 50 cos 40 cos50 sin 50 sin 40AB ABT= ° ° − ° + ° °T i j j 
( )( )850 lb cos 45 sin 25 sin 45 cos 45 cos 25AC = − ° ° − ° + ° °T i j j 
(a) 
( ) 0A xR = 
∴ ( ) ( )0: sin 50 cos 40 850 lb cos 45 sin 25 0A x ABxR F T= Σ = ° ° − ° ° = 
 432.86 lbABT = 433 lbABT = W 
(b) 
( ) ( ) ( ): 432.86 lb cos50 850 lb sin 45A yyR F= Σ − ° − ° 
( ) 879.28 lbA yR = − 
( ) ( ) ( ): 432.86 lb sin 50 sin 40 850 lb cos 45 cos 25A zzR F= Σ ° ° + ° ° 
( ) 757.87 lbA zR = 
∴ ( ) ( )879.28 lb 757.87 lbA = − +R j k 
 1160.82 lbAR = 1.161 kipsAR = W 
 0cos 0
1160.82x
θ = = 90.0xθ = °W 
 879.28cos 0.75746
1160.82y
θ −= = − 139.2yθ = °W 
 757.87cos 0.65287
1160.82z
θ = = 49.2zθ = °W 
 
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 kJJJG 
( ) ( ) ( )2 2 245 in. 48 in. 36 in. 75 in.AC = + − + = 
( ) ( ) ( )27 lb 45 in. 48 in. 36 in.
75 in.AC AC AC AC
ACT T
AC
 = = = − + T i j k
JJJG
λ 
( ) ( ) ( )16.2 lb 17.28 lb 12.96AC = − +T i j k 
and 
( ) ( ) ( )32 in. 48 in. 36 in.AB = − − +i j kJJJG 
( ) ( ) ( )2 2 232 in. 48 in. 36 in. 68 in.AB = − + − + = 
( ) ( ) ( )32 in. 48 in. 36 in.
68 in.
AB
AB AB AB AB
AB TT T
AB
 = = = − − + T i j k
JJJG
λ 
( )0.4706 0.7059 0.5294AB ABT= − − +T i j k 
and 
( ) ( ) ( )25 in. 48 in. 36 in.AD = − −i j kJJJG 
( ) ( ) ( )2 2 225 in. 48 in. 36 in. 65 in.AD = + − + = 
( ) ( ) ( )25 in. 48 in. 36 in.
65 in.
AD
AD AD AD AD
AD TT T
AD
 = = = − − T i j k
JJJG
λ 
( )0.3846 0.7385 0.5538AD ADT= − −T i j k 
 
105
 
PROBLEM 2.100 CONTINUED 
Now 
 AB AD AD= + +R 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 = W 
 257 lbADT = W 
 
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 forcesin 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 2220 mm 192 mm 0 292 mmBAd = − + + = 
 ( ) ( ) ( )2 2 2192 mm 192 mm 96 mm 288 mmDAd = + + = 
 ( ) ( ) ( )2 2 20 192 mm 144 mm 240 mmCAd = + + − = 
and ( ) ( )146 N 220 mm 192 mm
292 mmBA BA BA
F  = = − + F i jλ 
 ( ) ( )110 N 96 N= − +i j 
( ) ( )192 mm 144 mm
240 mm
CA
CA CA CA
FF  = = − F j kλ 
 ( )0.80 0.60CAF= −j k 
( ) ( ) ( )192 mm 192 mm 96 mm
288 mm
DA
DA DA DA
FF  = = + + F i j kλ 
 [ ]0.66667 0.66667 0.33333DAF= + +i j k 
With P= −P j 
At A: 0: 0BA CA DAΣ = + + + =F 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 = W 
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
FF  = = − + F i jλ 
 [ ]0.75342 0.65753 NBAF= − +i j 
( ) ( )192 mm 144 mm
240 mm
CA
CA CA CA
FF  = = − F j kλ 
 ( )0.80 0.60CAF= −j k 
( ) ( ) ( )192 mm 192 mm 96 mm
288 mm
DA
DA DA DA
FF  = = + + F i j kλ 
 [ ]0.66667 0.66667 0.33333DAF= + +i j k 
With: ( )200 N= −P j 
At A: 0: 0BA CA DAΣ = + + + =F 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 = W 
 65.6 NCAF = W 
 118.1 NDAF = W 
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: 
, , and AB AC ADT T T P 
where P=P j . To express the other forces in terms of the unit vectors 
i, j, k, we write 
 ( ) ( )12.6 ft 16.8 ftAB = − −i jJJJG 21 ftAB = 
( ) ( ) ( )7.2 ft 16.8 ft 12.6 ft 22.2 ftAC AC= − + =i j kJJJG 
 ( ) ( )16.8 ft 9.9 ftAD = − −j kJJJG 19.5 ftAD = 
and ( )0.6 0.8AB AB AB AB ABABT T TAB= = = − −T i j
JJJG
λ 
( )0.3242 0.75676 0.56757AC AC AC AC ACACT T TAC= = = − +T i j k
JJJG
λ 
( )0.8615 0.50769AD AD AD AD ADADT T TAD= = = − −T j k
JJJG
λ 
109
 
 
 
 
 
PROBLEM 2.103 CONTINUED 
Equilibrium Condition 
0: 0AB AC ADF PΣ = + + + =T T T j 
Substituting the expressions obtained for , , and AB 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 P− + + − − − +i 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 lb=P W 
 
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 lb=P W 
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: 
, , and AB AC ADT T T P 
where P=P 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 kJJJG 1.5 mAB = 
 ( ) ( )1.2 m 0.64 m ,AC = +j kJJJG 1.36 mAC = 
 ( ) ( ) ( )0.8 m 1.2 m 0.54 m ,AD = + −i j kJJJG 1.54 mAD = 
and ( )0.48 0.8 0.36AB AB AB AB ABABT T TAB= = = − + −T i j k
JJJG
λ 
( )0.88235 0.47059AC AC AC AC ACACT T TAC= = = +T j k
JJJG
λ 
( )0.51948 0.77922 0.35065AD AD AD AD ADADT T TAD= = = + −T i j k
JJJG
λ 
Equilibrium Condition with W= −W j 
0: 0AB AC ADF WΣ = + + − =T T T j 
Substituting the expressions obtained for , , and AB 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 W− + + + + −i j
 ( )0.36 0.47059 0.35065 0AB AC ADT T T+ − + − =k 
112
 
 
 
 
 
PROBLEM 2.105 CONTINUED 
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 = W 
 
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 
See Problem 2.105 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) 
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 = W 
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 
See Problem 2.105 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) 
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 = W 
115
 
 
 
 
 
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: 
 0.48 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 ADTT T− + − = 
Substituting ( )( )2750 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 = W 
 3.82 kNACT = W 
 2.43 kNADT = W 
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: cos 45 8 sin 45
65AB BE
° + − °= = i j kλ λ 
It follows that: cos 45 8 sin 45
65BE BE BE BE
T T ° + − ° = =   
i j kT λ 
 cos30 8 sin 30
65CF CF CF CF
T T ° + + ° = =   
i j kT λ 
 cos15 8 sin15
65DG DG DG DG
T T − ° + − ° = =   
i j kT λ 
117
 
 
PROBLEM 2.109 CONTINUED 
At A: 0: 0BE CF DGΣ = + + + + =F T T T W P 
Then, isolating the factors of i, j, and k, we obtain three algebraic equations: 
 : cos 45 cos30 cos15 0
65 65 65
BE CF DGT T T P° + ° − ° + =i 
or cos 45 cos30 cos15 65 0BE CF DGT T T P° + ° − ° + = (1) 
 8 8 8: 0
65 65 65BE CF DG
T T T W+ + − =j 
or 65 0
8BE CF DG
T T T W+ + − = (2) 
 : sin 45 sin 30 sin15 0
65 65 65
BE CF DGT T T− ° + ° − ° =k 
or sin 45 sin 30 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 = W 
118
 
 
 
 
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: 
 : cos 45 cos30 cos15 65 0BE CF DGT T T P° + ° − ° + =i (1) 
 65: 0
8BE CF DG
T T T W+ + − =j (2) 
 : sin 45 sin 30 sin15 0BE CF DGT T T− ° + ° − ° =k (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∴ < < W 
 
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 kJJJG 
( ) ( ) ( )2 2 218 m 30 m 5.4 m 35.4 mAC = + − + = 
( ) ( ) ( )18 m 30 m 5.4 m
35.4 m
AC
AC AC AC
AC TT T
AC
 = = = − + T i j k
JJJG
λ 
( )0.5085 0.8475 0.1525AC ACT= − +T i j k 
and ( ) ( ) ( )6 m 30 m 7.5 mAB = − − +i j kJJJG 
( ) ( ) ( )2 2 26 m 30 m 7.5 m 31.5 mAB = − + − + = 
( ) ( ) ( )6 m 30 m 7.5 m
31.5 m
AB
AB AB AB
AB TT T
AB
 = = = − − + T i j k
JJJG
λ 
( )0.1905 0.9524 0.2381AB ABT= − − +T i j k 
Finally ( ) ( ) ( )6 m 30 m 22.2 mAD = − − −i j kJJJG 
( ) ( ) ( )2 2 26 m 30 m 22.2 m 37.8 mAD = − + − + − = 
( ) ( ) ( )6 m 30 m 22.2 m
37.8 m
AD
AD AD AD
AD TT T
AD
 = = = − − − T i j k
JJJG
λ 
( )0.1587 0.7937 0.5873AD ADT= − − −T i j k 
120
 
 
 
 
 
PROBLEM 2.111 CONTINUED 
With , at :P A=P j 
 0: 0AB AC AD PΣ = + + + =F 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 T− + − =i (1) 
 : 0.9524 0.8475 0.7937 0AB AC ADT T T P− − − + =j (2) 
 : 0.2381 0.1525 0.5873 0AB AC ADT T T+ − =k (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 kN=P W 
121
 
 
 
 
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 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 kN=P W 
122
 
 
 
 
 
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 
( ) ( ) ( )32 in. 48 in. 36 in.AB = − +i j kJJJG 
( ) ( ) ( )2 2 232 in. 48 in. 36 in. 68 in.AB = − + − + = 
( ) ( ) ( )32 in. 48 in. 36 in.
68 in.
AB
AB AB AB
AB TT T
AB
 = = = − − + T i j k
JJJG
λ 
( )0.4706 0.7059 0.5294AB ABT= − − +T i j k 
and ( ) ( ) ( )45 in. 48 in. 36 in.AC = − +i j kJJJG 
( ) ( ) ( )2 2 245 in. 48 in. 36 in. 75 in.AC = + − + = 
( ) ( ) ( )45 in. 48 in. 36 in.
75 in.
AC
AC AC AC
AC TT T
AC
 = = = − + T i j k
JJJG
λ 
( )0.60 0.64 0.48AC ACT= − +T i j k 
Finally, ( ) ( ) ( )25 in. 48 in. 36 in.AD = − −i j kJJJG 
( ) ( ) ( )2 2 225 in. 48 in. 36 in. 65 in.AD = + − + − = 
123
 
 
 
 
PROBLEM 2.113 CONTINUED 
( ) ( ) ( )25 in. 48 in. 36 in.
65 in.
AD
AD AD AD
AD TT T
AD
 = = = − − T i j k
JJJG
λ 
( )0.3846 0.7385 0.5538AD ADT= − −T i j k 
With ,W=W j at A we have: 
 0: 0AB AC AD WΣ = + + + =F 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 T− + − =i (1) 
 : 0.7059 0.64 0.7385 0AB AC ADT T T W− − − + =j (2) 
 : 0.5294 0.48 0.5538 0AB AC ADT T T+ − =k (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 = W 
124
 
 
 
 
 
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 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 = W125
 
 
 
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: sin 30 sin 50 sin 30 cos 40x AD BDF T TΣ = − ° ° + ° ° 
 sin 30 cos60 0CDT+ ° ° = 
Dividing through by the factor sin 30° and evaluating the trigonometric 
functions gives 
 0.7660 0.7660 0.50 0AD BD CDT T T− + + = (1) 
Similarly, 
0: sin 30 cos50 sin 30 sin 40z AD BDF T TΣ = ° ° + ° ° 
 sin 30 sin 60 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 TΣ = − ° − ° − ° 
 ( )( )228 kg 9.81 m/s 0+ = 
or 317.2 NAD BD CDT T T+ + = 
126
 
 
 
 
 
 
 
 
PROBLEM 2.115 CONTINUED 
Using (3) and (4), above: 
0.3573 317.2 NCD CD CDT T T+ + = 
Then: 135.1 NADT = W 
 46.9 NBDT = W 
 135.1 NCDT = W 
 
127
 
 
 
 
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: 
cos 45 8 sin 45
65AB BE
λ ° + − °= = i j kλ 
It follows that: 
 cos 45 8 sin 45
65BE BE BE BE
T T ° + − ° = =   
i j kT λ 
 cos30 8 sin 30
65CF CF CF CF
T T ° + + ° = =   
i j kT λ 
cos15 8 sin15
65DG DG DG DG
T T − ° + − ° = =   
i j kT λ 
 At A: 0: 0BE CF DGΣ = + + + + =F T T T W P 
132
 
 
PROBLEM 2.119 CONTINUED 
Then, isolating the factors if , , and i j k we obtain three algebraic equations: 
 : cos 45 cos30 cos15 0
65 65 65
BE CF DGT T T° + ° − ° =i 
or cos 45 cos30 cos15 0BE CF DGT T T° + ° − ° = (1) 
 8 8 8: 0
65 65 65BE CF DG
T T T W+ + − =j 
or 2.4 65 0.3 65
8BE CF DG
T T T+ + = = (2) 
: sin 45 sin 30 sin15 0
65 65 65
BE CF DGT T T P− ° + ° − ° − =k 
or sin 45 sin 30 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 = W 
 1.002 lbCFT = W 
 1.117 lbDGT = W 
 
133
 
 
 
 
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 (1), (2), and (3) below: 
 cos 45 cos30 cos15 0BE CF DGT T T° + ° − ° = (1) 
 0.3 65BE CF DGT T T+ + = (2) 
 sin 45 sin 30 sin15 65BE CF DGT T T P− ° + ° − ° = (3) 
With 0.1 lb,=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 = W 
 0.357 lbCFT = W 
 1.056 lbDGT = W 
 
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 N= + = +N 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 kJJJG 
( ) ( ) ( )2 2 20.6 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 TT T
AB
 = = = − + + T i j k
JJJG
λ 
( )0.3436 0.7444 0.5726AB ABT= − + +T i j k 
and ( ) ( ) ( )0.7 m 1.4 m 1 mAC = + −i j kJJJG 
( ) ( ) ( )2 2 20.7 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 TT T
AC
 = = = + − T i j k
JJJG
λ 
( )0.3769 0.7537 0.5384AC ACT= + −T i j k 
Then: 0: 0AB ACΣ = + + + =F N T T W 
135
 
 
 
 
 
 
 
PROBLEM 2.121 CONTINUED 
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 T− + =i (1) 
 : 0.7444 0.7537 0.8944 1962 0AB ACT T N+ + − =j (2) 
 : 0.5726 0.5384 0.4472 0AB ACT T N− − + =k (3) 
Using conventional methods for solving Linear Algebraic Equations 
(elimination, MATLAB or Maple, for example), we obtain 
1311 NN = 
 551 NABT = W 
 503 NACT = W 
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 N= + = +N 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 kJJJG 
( ) ( ) ( )2 2 20.6 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 TT T
AB
 = = = − + + T i j k
JJJG
λ 
( )0.3436 0.7444 0.5726AB ABT= − + +T i j k 
and ( ) ( ) ( )0.7 m 1.4 m 1 mAC = + −i j kJJJG 
( ) ( ) ( )2 2 20.7 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 TT T
AC
 = = = + − T i j k
JJJG
λ 
( )0.3769 0.7537 0.5384AC ACT= + −T i j k 
Then: 0: 0AB ACΣ = + + + + =F N T T P W 
137
 
 
 
 
 
 
 
PROBLEM 2.122 CONTINUED 
Where ( )180 N= −P i 
and ( )( )2200 kg 9.81 m/s = −  W 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 T− + − =i 
 : 0.8944 0.7444 0.7537 1962 0AB ACN T T+ + − =j 
 : 0.4472 0.5726 0.5384 0AB ACN T T− − =k 
Using conventional methods for solving Linear Algebraic Equations 
(elimination, MATLAB or Maple, for example), we obtain 
1302 NN = 
 306 NABT = W 
 756 NACT = W 
 
138PROBLEM 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 kJJJG 
( ) ( ) ( )2 2 29 ft 8 ft 12 ft 17 ftAB = − + + − = 
( ) ( ) ( )9 ft 8 ft 12 ft
17 ft
AB
AB AB AB
AB TT T
AB
 = = = − + − T i j k
JJJG
λ 
 ( )0.5294 0.4706 0.7059AB ABT= − + −T i j k 
and 
 ( ) ( ) ( )0 8 ft 6 ftAC = + +i j kJJJG 
 ( ) ( ) ( )2 2 20 ft 8 ft 6 ft 10 ftAC = + + = 
( ) ( ) ( )0 ft 8 ft 6 ft
10 ft
AC
AC AC AC
AC TT T
AC
 = = = + + T i j k
JJJG
λ 
 ( )0.8 0.6AC ACT= +T j k 
and 
 ( ) ( ) ( )4 ft 8 ft 1 ftAD = + −i j kJJJG 
 ( ) ( ) ( )2 2 24 ft 8 ft 1 ft 9 ftAD = + + − = 
( ) ( ) ( )4 ft 8 ft 1 ft
9 ft
ADE
AD AD ADE
AD TT T
AD
 = = = + − T i j k
JJJG
λ 
 ( )0.4444 0.8889 0.1111AD ADET= + −T i j k 
139
 
PROBLEM 2.123 CONTINUED 
Finally, 
 ( ) ( ) ( )8 ft 8 ft 4 ftAE = − + +i j kJJJG 
 ( ) ( ) ( )2 2 28 ft 8 ft 4 ft 12 ftAE = − + + = 
( ) ( ) ( )8 ft 8 ft 4 ft
12 ft
ADE
AE AE ADE
AE TT T
AE
 = = = − + + T i j k
JJJG
λ 
 ( )0.6667 0.6667 0.3333AE ADET= − + +T i j k 
With the weight of the machinery, ,W= −W j at A, we have: 
0: 2 0AB AC AD WΣ = + + − =F T T T j 
Equating the factors of , , and i 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 = W 
 34.2 lbACT = W 
 110.8 lbADET = W 
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 = W 
 (b) 162.0 lbAET = W 
 (c) 468 lbW = W 
141
 
 
 
 
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 20.84 my z+ = 
680 N
ABT y
= 
680 NQ z
y
= 
With 550 mm 0.55 m,y = = we obtain: 
( )22 20.84 m 0.55 m
0.733 m
= −
∴ =
z
z
 
and 
(a) 680 N 1236.4 N
0.55AB
T = = 
 or 1.236 kNABT = W 
 and 
(b) ( )1236 0.866 N 906 N= =Q 
 or 0.906 kNQ = W 
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) sin 35 3001bP ° = 
300 lb
sin 35
P = ° 
 523 lbP = W 
(b) Vertical Component 
 cos35vP P= ° 
 ( )523 lb cos35= ° 
 428 lb=vP W 
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 
 ( ) ( ) ( )0.78 m 1.6 m 0 mAB = − + +i j kJJJG 
( ) ( ) ( )2 2 20.78 m 1.6 m 0 1.78 mAB = − + + = 
( ) ( ) ( )0.78 m 1.6 m 0 m
1.78 m
AB
AB AB AB
AB TT T
AB
 = = = − + + T i j k
JJJG
λ 
 ( )0.4382 0.8989 0AB ABT= − + +T i j k 
and 
 ( ) ( ) ( )0 1.6 m 1.2 mAC = + +i j kJJJG 
( ) ( ) ( )2 2 20 m 1.6 m 1.2 m 2 mAC = + + = 
( ) ( ) ( )0 1.6 m 1.2 m
2 m
AC
AC AC AC
AC TT T
AC
 = = = + + T i j k
JJJG
λ 
 ( )0.8 0.6AC ACT= +T j k 
and 
 ( ) ( ) ( )1.3 m 1.6 m 0.4 mAD = + +i j kJJJG 
( ) ( ) ( )2 2 21.3 m 1.6 m 0.4 m 2.1 mAD = + + = 
( ) ( ) ( )1.3 m 1.6 m 0.4 m
2.1 m
AD
AD AD AD
AD TT T
AD
 = = = + + T i j k
JJJG
λ 
 ( )0.6190 0.7619 0.1905AD ADT= + +T i j k 
149
 
PROBLEM 2.130 CONTINUED 
Finally, 
 ( ) ( ) ( )0.4 m 1.6 m 0.86 mAE = − + −i j kJJJG 
( ) ( ) ( )2 2 20.4 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 TT T
AE
 = = = − + − T i j k
JJJG
λ 
 ( )0.2151 0.8602 0.4624AE AET= − + −T i j k 
With the weight of the container ,W= −W j at A we have: 
0: 0AB AC AD WΣ = + + − =F T T T j 
Equating the factors of , , and i 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 = W 
 
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 = W 
 (b) 1202 NW = W 
151
 
 
 
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=Q k are applied to the ring to maintain the container is 
the position shown. Knowingthat 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 kJJJG 
( ) ( ) ( )2 2 20.48 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 TT T
AB
 = = = − + − T i j k
JJJG
λ
 ( )0.5455 0.8182 0.1818AB ABT= − + −T i j k 
and 
 ( ) ( ) ( )0.24 m 0.72 m 0.13 mAC = + −i j kJJJG 
( ) ( ) ( )2 2 20.24 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 TT T
AC
 = = = + − T i j k
JJJG
λ 
 ( )0.3177 0.9351 0.1688AC ACT= + −T i j k 
At A: 0: 0AB ACΣ = + + + + =F T T P Q W 
 
142
 
 
 
 
 
 
 
PROBLEM 2.125 CONTINUED 
Noting that AB ACT T= because of the ring A, we equate the factors of 
, , and i j k to zero to obtain the linear algebraic equations: 
( ): 0.5455 0.3177 0T P− + + =i 
or 0.2338P T= 
( ): 0.8182 0.9351 0T W+ − =j 
or 1.7532W T= 
( ): 0.1818 0.1688 0T Q− − + =k 
or 0.356Q T= 
With 1200 N:W = 
1200 N 684.5 N
1.7532
T = = 
160.0 NP = W 
 240 NQ = W 
 
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 P=P i and Q=Q 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 = W 
 107.0 NP = W 
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 2AB x y z= + + 
Here ( ) ( )2 2 2 21 m 0.40 m y z= + + 
or 2 2 20.84 my z+ = 
Thus, with y given, z is determined. 
Now 
( )1 0.40 m 0.4
1 mAB
AB y z y z
AB
= = − + = − +i j k i k k
JJJG
λ 
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 TΣ = + + + =F i k j λ 
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 TΣ = + + − =F i j k λ 
 
 
 
 
 
145
 
 
PROBLEM 2.127 CONTINUED 
Setting the k coefficient to zero gives: 
 0ABQ T z− = 
And using the above result for ABT we have 
680 N
ABQ T z zy
= = 
Then, from the specifications of the problem, 300 mm 0.3 my = = 
( )22 20.84 m 0.3 m= −z 
 0.866 m∴ =z 
and 
(a) 680 N 2266.7 N
0.30AB
T = = 
 or 2.27 kNABT = W 
 and 
(b) ( )2266.7 0.866 1963.2 NQ = = 
 or 1.963 kNQ = W 
146
 
 
 
 
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: 
0.5409ABT P= 
0.295ACT P= 
0.2959ADT P= 
Using 8 kN:P = 
 4.33 kNABT = W 
 2.36 kNACT = W 
 2.37 kNADT = W 
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 = W 
 12.76 lbACT = W 
 121.9 lbADT = W 
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 sin 30 sin15 0BE CF DGT T T− ° + ° − ° = (3) 
 cos 45 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 sin 30 sin 45 sin15 0.2 65 sin 45CF DGT T° + ° + ° − ° = ° 
or 0.9445 0.3714CF DGT T= − (4) 
Multiplying (2 )′ by sin 30° and subtracting (3) from the result: 
( ) ( )sin 30 sin 45 sin 30 sin15 0.2 65 sin 30BE DGT T° + ° + ° + ° = ° 
or 0.6679 0.6286BE DGT T= − (5) 
130
 
 
PROBLEM 2.118 CONTINUED 
Substituting (4) and (5) into (1)′ : 
1.2903 1.7321 65 0DGT P− − = 
 DGT∴ is taut for 1.2903 lb65P < 
 or 0.16000 lbP≤ < W 
 
131
 
 
 
 
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 QΣ = − ° = 
With 60 lbQ = 
(a) ( )( )60 lb 0.866CAT = 
 52.0 lbCAT = W 
(b) 0: sin 30 0x CBF P T QΣ = − − ° = 
 With 75 lbP = 
( )( )75 lb 60 lb 0.50CBT = − 
 or 45.0 lbCBT = W 
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 QΣ = − ° = 
or 0.8660 QCAT = 
Then for 60 lbCAT ≤ 
 0.8660 60 lbQ < 
or 69.3 lbQ ≤ 
From 0: sin 30y CBF T P QΣ = = − ° 
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≤ ≤ W 
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 30: 0
5 5x B C A
F F F FΣ = − − = 
With 8 kN, 16 kNA BF F= = 
( ) ( )4 416 kN 8 kN
5 5C
F = − 
 6.40 kNCF = W 
3 30: 0
5 5y D B A
F F F FΣ = − + − = 
With AF and BF as above: 
( ) ( )3 316 kN 8 kN
5 5D
F = − 
 4.80 kNDF = W 
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 30: 0
5 5y D A B
F F F FΣ = − − + = 
or 3
5B D A
F F F= + 
With 5 kN, 8 kNA DF F= = 
( )5 36 kN 5 kN
3 5B
F  = +   
 15.00 kNBF = W 
4 40: 0
5 5x C B A
F F F FΣ = − + − = 
 ( )4
5C B A
F F F= − 
 ( )4 15 kN 5 kN
5
= − 
 8.00 kNCF = W 
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 
 ( )4.50: 50 lb 0
20.5x
F PΣ = − + = 
 or 10.98 lbP = W 
(b) Triangle Proportions( )150: 50 lb 0
25x
F PΣ = − + = 
 or 30.0 lbP = W 
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
ˆ500: 48 0
ˆ400
x
xF
x
Σ = − + =
+
 
 or 248ˆ ˆ400
50
x x= + 
 ( )2 2ˆ ˆ0.92 lb 400x x= + 
 2 2ˆ 4737.7 inx = 
 ˆ 68.6 in.x = W 
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 kJJJG 
( ) ( ) ( )2 2 2480 510 320 770 mmDB = + + = 
( ) ( ) ( )385 N 480 mm 510 mm 320 mm
770 mmDB
DBF F
DB
 = = = − + F i j k
JJJG
λ 
( ) ( ) ( )240 N 255 N 160 N= − +F i j k 
 240 N, 255 N, 160.0 Nx y zF F F= + = − = + W 
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 
the cable. That is, with 
( ) ( ) ( )0.48 m 0.51 m 0.32 mBD = − + −i j kJJJG 
( ) ( ) ( )2 2 20.48 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 TT T
BD
 = = = − + − T i j k
JJJG
λ 
( )0.6234 0.6623 0.4156BD BDT= − + −T i j k 
and 
( ) ( ) ( )0.27 m 0.40 m 0.6 mBE = − + −i j kJJJG 
( ) ( ) ( )2 2 20.27 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 TT T
BD
 = = = − + − T i j k
JJJG
λ 
( )0.3506 0.5195 0.7792BE BET= − + −T 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 0.7792= − + − − + −F i j k i j k 
 ( ) ( ) ( )375 N 455 N 460 N= − + −i j k 
159
 
PROBLEM 2.139 CONTINUED 
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 = W 
The direction of this force is: 
 1 375cos
747.83x
θ − −= or 120.1xθ = °W 
 1 455cos
747.83y
θ −= or 52.5yθ = °W 
 1 460cos
747.83z
θ − −= or 128.0zθ = °W 
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 
corresponding magnitude of R. 
 
SOLUTION 
Force Triangle 
 
 
(a) For minimum P it must be perpendicular to the vertical resultant R 
( ) 425 lb cos30P∴ = ° 
 or 368 lb=P W 
(b) ( )425 lb sin 30R = ° 
 or 213 lbR = W 
 
 
161
 
 
 
 
 
PROBLEM 3.1 
A 13.2-N force P is applied to the lever which controls the auger of a 
snowblower. Determine the moment of P about A when α is equal to 30°. 
 
SOLUTION 
 
 
First note 
( )sin 13.2 N sin 30 6.60 NxP P α= = ° = 
( )cos 13.2 N cos30 11.4315 Nα= = ° =yP P 
Noting that the direction of the moment of each force component about A 
is counterclockwise, 
 / /A B A y B A xM x P y P= + 
 ( )( ) ( )( )0.086 m 11.4315 N 0.122 m 6.60 N= + 
 1.78831 N m= ⋅ 
 or 1.788 N mA = ⋅M W 
 
 
 
 
 
PROBLEM 3.2 
The force P is applied to the lever which controls the auger of a 
snowblower. Determine the magnitude and the direction of the smallest 
force P which has a 2.20- N m⋅ counterclockwise moment about A. 
 
SOLUTION 
 
 
 
For P to be a minimum, it must be perpendicular to the line joining points 
A and B. 
( ) ( )2 286 mm 122 mm 149.265 mm= + =ABr 
1 1 122 mmtan tan 54.819
86 mm
α θ − −   = = = = °      
y
x
 
Then min=A ABM r P 
or min = A
AB
MP
r
 
 2.20 N m 1000 mm
149.265 mm 1 m
⋅  =    
 14.7389 N= 
min 14.74 N∴ =P 54.8° 
 or min 14.74 N=P 35.2°W 
 
 
 
 
 
PROBLEM 3.3 
A 13.1-N force P is applied to the lever which controls the auger of a 
snowblower. Determine the value of α knowing that the moment of P 
about A is counterclockwise and has a magnitude of 1.95 N m.⋅ 
 
SOLUTION 
 
 
 
By definition / sinθ=A B AM r P 
where ( )90θ φ α= + ° − 
and 1 122 mmtan 54.819
86 mm
φ −  = = °   
Also ( ) ( )2 2/ 86 mm 122 mm 149.265 mm= + =B Ar 
Then ( )( ) ( )1.95 N m 0.149265 m 13.1 N sin 54.819 90 α⋅ = ° + ° − 
or ( )sin 144.819 0.99725α° − = 
or 144.819 85.752α° − = ° 
and 144.819 94.248α° − = ° 
 50.6 , 59.1α∴ = ° °W 
 
 
 
 
 
PROBLEM 3.4 
A foot valve for a pneumatic system is hinged at B. Knowing that 
α = 28°, determine the moment of the 4-lb force about point B by 
resolving the force into horizontal and vertical components. 
 
SOLUTION 
 
 
 
 
 
 
Note that 20 28 20 8θ α= − ° = ° − ° = ° 
and ( )4 lb cos8 3.9611 lb= ° =xF 
 ( )4 lb sin8 0.55669 lb= ° =yF 
Also ( )6.5 in. cos 20 6.1080 in.= ° =x 
 ( )6.5 in. sin 20 2.2231 in.= ° =y 
Noting that the direction of the moment of each force component about B 
is counterclockwise, 
 B y xM xF yF= + 
 ( )( ) ( )( )6.1080 in. 0.55669 lb 2.2231 in. 3.9611 lb= + 
 12.2062 lb in.= ⋅ 
 or 12.21 lb in.B = ⋅M W 
 
 
 
 
 
PROBLEM 3.5 
A foot valve for a pneumatic system is hinged at B. Knowing that 
α = 28°, determine the moment of the 4-lb force about point B by 
resolving the force into components along ABC and in a direction 
perpendicular to ABC. 
 
SOLUTION 
 
 
First resolve the 4-lb force into components P and Q, where 
( )4.0 lb sin 28 1.87787 lbQ = ° = 
Then /=B A BM r Q 
 ( )( )6.5 in. 1.87787 lb= 
 12.2063 lb in.= ⋅ 
 or 12.21 lb in.B = ⋅M W 
 
 
 
 
PROBLEM 3.6 
It is known that a vertical force of 800 N is required to remove the nail at 
C from the board. As the nail first starts moving, determine (a) the 
moment about B of the force exerted on the nail, (b) the magnitude of the 
force P which creates the same moment about B if α = 10°, (c) the 
smallest force P which creates the same moment about B. 
 
SOLUTION 
 
 
 
 
 (a) Have /=B C B NM r F 
 ( )( )0.1 m 800 N= 
 80.0 N m= ⋅ 
 or 80.0 N mB = ⋅M W 
(b) By definition 
/ sinθ=B A BM r P 
 where ( )90 90 70θ α= ° − ° − ° − 
 90 20 10= ° − ° − ° 
 60= ° 
( ) 80.0 N m 0.45 m sin 60P∴ ⋅ = ° 
 205.28 N=P 
 or 205 NP = W 
(c) For P to be minimum, it must be perpendicular to the line joining 
 points A and B. Thus, P must be directed as shown. 
 Thus min / min= =B A BM dP r P 
 or ( ) min80.0 N m 0.45 m⋅ = P 
min 177.778 N∴ =P 
 or min 177.8 N=P 20°W 
 
 
 
 
 
PROBLEM 3.7 
A sign is suspended from two chains AE and BF. Knowing that the 
tension in BF is 45 lb, determine (a) the moment about A of the force 
exert by the chain at B, (b) the smallest force applied at C which creates 
the same moment about A. 
 
SOLUTION 
 
 
 
 
 
 (a) Have /A B A BF=M r T× 
 Noting that the direction of the moment of each force componentabout A is counterclockwise, 
 = +A BFy BFxM xT yT 
 ( )( ) ( )( )6.5 ft 45 lb sin 60 4.4 ft 3.1 ft 45 lb cos60= ° + − ° 
 282.56 lb ft= ⋅ 
 or 283 lb ftA = ⋅M W 
(b) Have ( )/ minA C A C=M r F× 
 For CF to be minimum, it must be perpendicular to the 
 line joining points A and C. 
( )minA CM d F∴ = 
 where ( ) ( )2 2/ 6.5 ft 4.4 ft 7.8492 ftC Ad r= = + = 
( )( )min 282.56 lb ft 7.8492 ft CF∴ ⋅ = 
( )min 35.999 lbCF = 
1 4.4 fttan 34.095
6.5 ft
φ −  = = °   
90 90 34.095 55.905θ φ= ° − = ° − ° = ° 
 or ( )min 36.0 lbC =F 55.9°W 
 
 
 
 
 
PROBLEM 3.8 
A sign is suspended from two chains AE and BF. Knowing that the 
tension in BF is 45 lb, determine (a) the moment about A of the force 
exerted by the chain at B, (b) the magnitude and sense of the vertical 
force applied at C which creates the same moment about A, (c) the 
smallest force applied at B which creates the same moment about A. 
 
SOLUTION 
 
 
 
 
 
 
 (a) Have /A B A BF=M r T× 
 Noting that the direction of the moment of each force component 
 about A is counterclockwise, 
 = +A BFy BFxM xT yT 
 ( )( ) ( )( )6.5 ft 45 lb sin 60 4.4 ft 3.1 ft 45 lb cos60= ° + − ° 
 282.56 lb ft= ⋅ 
 or 283 lb ftA = ⋅M W 
(b) Have /A C A C=M r F× 
 or =A CM xF 
282.56 lb ft 43.471 lb
6.5 ft
A
C
MF
x
⋅∴ = = = 
 or 43.5 lbC =F W 
(c) Have ( )/ minA B A B=M r F× 
 For BF to be minimum, it must be perpendicular to the line joining 
 points A and B. 
( )min ∴ =A BM d F 
 where ( ) ( )2 26.5 ft 4.4 ft 3.1 ft 6.6287 ftd = + − = 
( )min 282.56 lb ft 42.627 lb6.6287 ft
⋅∴ = = =AB MF d 
 and 1 6.5 fttan 78.690
4.4 ft 3.1 ft
θ −  = = ° −  
 or ( )min 42.6 lb=BF 78.7°W 
 
 
 
 
 
PROBLEM 3.9 
The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts 
a 125-N force directed along its center line on the ball and socket at B, 
determine the moment of the force about A. 
 
SOLUTION 
 
 
 
 
 
First note ( ) ( )2 2240 mm 46.6 mmCBd = + 
 244.48 mm= 
Then 240 mmcos
244.48 mm
θ = 
 46.6 mmsin
244.48 mm
θ = 
and cos sinCB CB CBF Fθ θ= −F i j 
 ( ) ( )125 N 240 mm 46.6 mm
244.48 mm
 = − i j 
Now /A B A CB=M r F× 
where ( ) ( )/ 306 mm 240 mm 46.6 mmB A = − +r i j 
 ( ) ( )306 mm 286.6 mm= −i j 
Then ( ) ( ) ( )125 N306 mm 286.6 mm 240 46.6
244.48A
 = − − M i j i j× 
 ( ) ( )27878 N mm 27.878 N m= ⋅ = ⋅k k 
 or 27.9 N mA = ⋅M W 
 
 
 
 
 
 
PROBLEM 3.10 
The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts 
a 125-N force directed along its center line on the ball and socket at B, 
determine the moment of the force about A. 
 
SOLUTION 
 
 
 
 
 
 
First note ( ) ( )2 2344 mm 152.4 mm 376.25 mmCBd = + = 
Then 344 mmcos
376.25 mm
θ = 152.4 mmsin
376.25 mm
θ = 
and ( ) ( )cos sinCB CB CBF Fθ θ= −F i j 
 ( ) ( )125 N 344 mm 152.4 mm
376.25 mm
 = + i j 
Now /A B A CB=M r F× 
where ( ) ( )/ 410 mm 87.6 mmB A = −r i j 
Then ( ) ( ) ( )125 N410 mm 87.6 mm 344 152.4
376.25A
 = − − M i j i j× 
 ( )30770 N mm= ⋅ k 
 ( )30.770 N m= ⋅ k 
 or 30.8 N mA = ⋅M W 
 
 
 
 
 
PROBLEM 3.11 
A winch puller AB is used to straighten a fence post. Knowing that the 
tension in cable BC is 260 lb, length a is 8 in., length b is 35 in., and 
length d is 76 in., determine the moment about D of the force exerted by 
the cable at C by resolving that force into horizontal and vertical 
components applied (a) at point C, (b) at point E. 
 
SOLUTION 
 
 
 
 
 
(a) Slope of line 35 in. 5
76 in. 8 in. 12
EC = =+ 
 Then ( )12
13ABx AB
T T= 
 ( )12 260 lb 240 lb
13
= = 
 and ( )5 260 lb 100 lb
13ABy
T = = 
 Then ( ) ( )35 in. 8 in.D ABx AByM T T= − 
 ( )( ) ( )( )240 lb 35 in. 100 lb 8 in.= − 
 7600 lb in.= ⋅ 
 or 7600 lb in.D = ⋅M W 
(b) Have ( ) ( )D ABx AByM T y T x= + 
 ( )( ) ( )( )240 lb 0 100 lb 76 in.= + 
 7600 lb in.= ⋅ 
 or 7600 lb in.D = ⋅M W 
 
 
 
 
 
PROBLEM 3.12 
It is known that a force with a moment of 7840 lb in.⋅ about D is required 
to straighten the fence post CD. If 8a = in., 35b = in., and 112d = 
in., determine the tension that must be developed in the cable of winch 
puller AB to create the required moment about point D. 
 
SOLUTION 
 
Slope of line 35 in. 7
112 in. 8 in. 24
EC = =+ 
Then 24
25ABx AB
T T= 
and 7
25ABy AB
T T= 
Have ( ) ( )D ABx AByM T y T x= + 
( ) ( )24 77840 lb in. 0 112 in.
25 25AB AB
T T∴ ⋅ = + 
250 lbABT = 
 or 250 lbABT = W 
 
 
 
 
 
PROBLEM 3.13 
It is known that a force with a moment of 1152 N m⋅ about D is required 
to straighten the fence post CD. If the capacity of the winch puller AB is 
2880 N, determine the minimum value of distance d to create the 
specified moment about point D knowing that 0.24a = m and 
1.05b = m. 
 
SOLUTION 
 
The minimum value of d can be found based on the equation relating the moment of the force ABT about D: 
( ) ( )maxD AB yM T d= 
where 1152 N mDM = ⋅ 
 ( ) ( )max max sin 2880 N sinAB AByT T θ θ= = 
Now ( ) ( )2 2
1.05 msin
0.24 1.05 md
θ =
+ +
 
 ( ) ( ) ( )2 2
1.051152 N m 2880 N
0.24 1.05
d
d
  ∴ ⋅ =  + +  
 
or ( ) ( )2 20.24 1.05 2.625d d+ + = 
or ( ) ( )2 2 20.24 1.05 6.8906d d+ + = 
or 25.8906 0.48 1.1601 0d d− − = 
Using the quadratic equation, the minimum values of d are 0.48639 m and −0.40490 m. 
Since only the positive value applies here, 0.48639 md = 
 or 486 mmd = W 
 
 
 
 
 
PROBLEM 3.14 
A mechanic uses a piece of pipe AB as a lever when tightening an 
alternator belt. When he pushes down at A, a force of 580 N is exerted on 
the alternator B. Determine the moment of that force about bolt C if its 
line of action passes through O. 
 
SOLUTION 
 
 
 
 
 
 
Have /C B C B=M r F× 
Noting the direction of the moment of each force component about C is 
clockwise, 
C By BxM xF yF= + 
where 144 mm 78 mm 66 mmx = − = 
 86 mm 108 mm 194 mmy = + = 
and ( ) ( ) ( )2 2
78 580 N 389.65 N
78 86
BxF = =+
 
 ( ) ( ) ( )2 2
86 580 N 429.62 N
78 86
ByF = =+
 
( )( ) ( )( ) 66 mm 429.62 N 194 mm 389.65 NCM∴ = + 
 103947 N mm= ⋅ 
 103.947 N m= ⋅ 
 or 103.9 N mC = ⋅M W 
 
 
 
 
 
PROBLEM 3.15 
Form the vector products B C× and ,′B C× where ,B B′= and use the 
results obtained to prove the identity 
( ) ( )1 12 2sin cos sin sin .α β α β α β= + + − 
 
SOLUTION 
 
 
 
 
 
First note ( )cos sinB β β= +B i j 
 ( )cos sinB β β′ = −B i j 
 ( )cos sinC α α= +C i j 
By definition ( )sinBC α β= −B C× (1) 
 ( )sinBC α β′ = +B C× (2) 
Now ( ) ( )cos sin cos sinB Cβ β α α= + +B C i j i j× × 
 ( )cos sin sin cosBC β α β α= − k (3) 
 ( ) ( )cos sin cos sinB Cβ β α α= − +B C i j i j× × 
 ( )cos sin sin cosBC β α β α= + k (4) 
Equating magnitudes of B C× from Equations (1) and (3), (5) 
( )sin cos sin sin cosα β β α β α− = − 
Similarly, equating magnitudes of ′B C× from Equations (2) and (4), 
 ( )sin cos sin sin cosα β β α β α+ = + (6) 
Adding Equations (5) and (6) 
( ) ( )sin sin 2cos sinα β α β β α− + + = 
 ( ) ( )1 1 sin cos sin sin
2 2
α β α β α β∴ = + + − W 
 
 
 
 
 
PROBLEM 3.16 
A line passes through the points (420 mm, −150 mm) and (−140 mm, 
180 mm). Determine the perpendicular distance d from the line to the 
origin O of the system of coordinates. 
 
SOLUTION 
 
 
 
 
 
 
Have /AB O Ad = rλ × 
where /
/
B A
AB
B A
= rr
λ 
and ( ) ( )/ 140 mm 420 mm 180 mm 150 mmB A  = − − + − − r i j 
 ( ) ( )560 mm 330 mm= − +i j 
 ( ) ( )2 2/ 560 330 mm 650 mmB A = − + =r 
( ) ( ) ( )560 mm 330 mm 1 56 33
650 mm 65AB
− +∴ = = − +i j i jλ 
( ) ( ) ( ) ( )/ 0 0 420 mm 150 mmO A A Ax y= − + − = − +r i j i j 
( ) ( ) ( )1 56 33 420 mm 150 mm 84.0 mm
65
d  ∴ = − − − + = i j i j× 
 84.0 mmd = W 
 
 PROBLEM 3.17 
A plane contains the vectors A and B. Determine the unit vector normal 
to the plane when A and B are equal to, respectively, (a) 4i − 2j + 3k and 
−2i + 6j − 5k, (b) 7i + j − 4k and −6i − 3k + 2k. 
 
SOLUTION 
(a) Have = A B
A B
×λ × 
 where 4 2 3= − +A i j k 
 2 6 5= − + −B i j k 
 Then ( ) ( ) ( ) ( )4 2 3 10 18 6 20 24 4 2 4 7 10
2 6 5
= − = − + − + + − = − + +
− −
i j k
A B i j k i j k× 
 and ( ) ( ) ( )2 2 22 4 7 10 2 165= − + + =A B× 
 ( )2 4 7 10 
2 165
− + +∴ = i j kλ or ( )1 4 7 10
165
= − + +i j kλ W 
(b) Have = A B
A B
×λ × 
 where 7 4= + −A i j k 
 6 3 2= − − +B i j k 
 Then ( ) ( ) ( ) ( )7 1 4 2 12 24 14 21 6 5 2 2 3
6 3 2
= − = − + − + − + = − + −
− −
i j k
A B i j k i j k× 
 and ( ) ( ) ( )2 2 25 2 2 3 5 17= − + + − =A B× 
 ( )5 2 2 3 
5 17
− + −∴ = i j kλ or ( )1 2 2 3
17
= − + −i j kλ W 
 
 PROBLEM 3.18 
The vectors P and Q are two adjacent sides of a parallelogram. 
Determine the area of the parallelogram when (a) P = (8 in.)i + (2 in.)j − 
(1 in.)k and Q = −(3 in.)i + (4 in.)j + (2 in.)k, (b) P = −(3 in.)i + (6 in.)j + 
(4 in.)k and Q = (2 in.)i + (5 in.)j − (3 in.)k. 
 
SOLUTION 
(a) Have A = P Q× 
 where ( ) ( ) ( )8 in. 2 in. 1 in.= + −P i j k 
 ( ) ( ) ( )3 in. 4 in. 2 in.= − + +Q i j k 
 Then ( ) ( ) ( )2 28 2 1 in 4 4 3 16 32 6 in
3 4 2
 = − = + + − + + 
−
i j k
P Q i j k× 
 ( ) ( ) ( )2 2 28 in 13 in 38 in= − +i j k 
 ( ) ( ) ( )2 2 2 2 2 8 13 38 in 40.951 inΑ∴ = + − + = or 241.0 inA = W 
(b) Have A = P Q× 
 where ( ) ( ) ( )3 in. 6 in. 4 in.= − + +P i j k 
 ( ) ( ) ( )2 in. 5 in. 3 in.= + −Q i j k 
 Then ( ) ( ) ( )2 23 6 4 in 18 20 8 9 15 12 in
2 5 3
 = − = − − + − + − − 
−
i j k
P Q i j k× 
 ( ) ( ) ( )2 2 238 in 1 in 27 in= − − −i j k 
 ( ) ( ) ( )2 2 2 2 2 38 1 27 in 46.626 inΑ∴ = − + − + − = or 246.6 inA = W 
 
 PROBLEM 3.19 
Determine the moment about the origin O of the force F = −(5 N)i − (2 
N)j + (3 N)k which acts at a point A. Assume that the position vector of 
A is (a) r = (4 m)i − (2 m)j − (1 m)k, (b) r = −(8 m)i + (3 m)j + (4 m)k, 
(c) r = (7.5 m)i + (3 m)j − (4.5 m)k. 
 
SOLUTION 
(a) Have O =M r F× 
 where ( ) ( ) ( )5 N 2 N 3 N= − − +F i j k 
 ( ) ( ) ( )4 m 2 m 1 m= − −r i j k 
 ( ) ( ) ( ) 4 2 1 N m 6 2 5 12 8 10 N m
5 2 3
O  ∴ = − − ⋅ = − − + − + − − ⋅ 
− −
i j k
M i j k 
 ( )8 7 18 N m= − − − ⋅i j k 
 or ( ) ( ) ( )8 N m 7 N m 18 N mO = − ⋅ − ⋅ − ⋅M i j k W 
(b) Have O =M r F× 
 where ( ) ( ) ( )5 N 2 N 3 N= − − +F i j k 
 ( ) ( ) ( )8 m 3 m 4 m= − + −r i j k 
 ( ) ( ) ( ) 8 3 4 N m 9 8 20 24 16 15 N m
5 2 3
O  ∴ = − ⋅ = + + − + + + ⋅ 
− −
i j k
M i j k 
 ( )17 4 31 N m= + + ⋅i j k 
 or ( ) ( ) ( )17 N m 4 N m 31 N mO = ⋅ + ⋅ + ⋅M i j k W 
(c) Have O =M r F× 
 where ( ) ( ) ( )5 N 2 N 3 N= − − +F i j k 
 ( ) ( ) ( )7.5 m 3 m 4.5 m= + −r i j k 
 
 PROBLEM 3.19 CONTINUED 
( ) ( ) ( ) 7.5 3 4.5 N m 9 9 22.5 22.5 15 15 N m
5 2 3
O  ∴ = − ⋅ = − + − + − + ⋅ 
− −
i j k
M i j k 
or 0O =M W 
This answer is expected since r and F are proportional 2 .
3
− =  F r Therefore, vector F has a line of action 
passing through the origin at O. 
 
 PROBLEM 3.20 
Determine the moment about the origin O of the force F = −(1.5 lb)i + 
(3 lb)j − (2 lb)k which acts at a point A. Assume that the position vector 
of A is (a) r = (2.5 ft)i − (1 ft)j + (2 ft)k, (b) r = (4.5 ft)i − (9 ft)j + 
(6 ft)k, (c) r = (4 ft)i − (1 ft)j + (7 ft)k. 
 
SOLUTION 
(a) Have O =M r F× 
 where ( ) ( ) ( )1.5 lb 3 lb 2 lb= − + +F i j k 
 ( ) ( ) ( )2.5 ft 1 ft 2 ft= − +r i j k 
 Then ( ) ( ) ( )2.5 1 2 lb ft 2 6 3 5 7.5 1.5 lb ft
1.5 3 2
O  = − ⋅ = − + − + + − ⋅ 
− −
i j k
M i j k 
 or ( ) ( ) ( )4 lb ft 2 lb ft 6 lb ftO = − ⋅ + ⋅ + ⋅M i j kW 
(b) Have O =M r F× 
 where ( ) ( ) ( )1.5 lb 3 lb 2 lb= − + −F i j k 
 ( ) ( ) ( )4.5 ft 9 ft 6 ft= − +r i j k 
 Then ( ) ( ) ( )4.5 9 6 lb ft 18 18 9 9 13.5 13.5 lb ft
1.5 3 2
O  = − ⋅ = − + − + + − ⋅ 
− −
i j k
M i j k 
 or 0O =M W 
 This answer is expected since r and F are proportional 1 .
3
− =  F r 
 Therefore, vector F has a line of action passing through the origin at O. 
 (c) Have O =M r F× 
 where ( ) ( ) ( )1.5 lb 3 lb 2 lb= − − −F i j k 
 ( ) ( ) ( )4 ft 1 ft 7 ft= − +r i j k 
 Then ( ) ( ) ( )4 1 7 lb ft 2 21 10.5 8 12 1.5 lb ft
1.5 3 2
O  = − ⋅ = − + − + + − ⋅ 
− −
i j k
M i j k 
 or ( ) ( ) ( )19 lb ft 2.5 lb ft 10.5 lb ftO = − ⋅ − ⋅ + ⋅M i j kW 
 
 
 
 
 
 
PROBLEM 3.21 
Before the trunk of a large tree is felled, cables AB and BC are attached as 
shown. Knowing that the tension in cables AB and BC are 777 N and 
990 N, respectively, determine the moment about O of the resultant force 
exerted on the tree by the cables at B. 
 
SOLUTION 
 
Have /O B O B=M r F× 
where ( )/ 8.4 mB O =r j 
 B AB BC= +F T T 
( ) ( ) ( )
( ) ( ) ( ) ( )2 2 2
0.9 m 8.4 m 7.2 m
777 N
0.9 8.4 7.2 m
AB BA ABT
− − += =
+ +
i j k
T λ 
( ) ( ) ( )
( ) ( ) ( ) ( )2 2 2
5.1 m 8.4 m 1.2 m
990 N
5.1 8.4 1.2 m
BC BC BCT
− += =
+ +
i j k
T λ 
 
 PROBLEM 3.21 CONTINUED 
( ) ( ) ( ) ( ) ( ) ( ) 63.0 N 588 N 504 N 510 N 840 N 120 NB    ∴ = − − + + − +   F i j k i j k 
 ( ) ( ) ( )447 N 1428 N 624 N= − +i j k 
and ( ) ( )0 8.4 0 N m 5241.6 N m 3754.8 N m
447 1428 624
O = ⋅ = ⋅ − ⋅
−
i j k
M i k 
or ( ) ( )5.24 kN m 3.75 kN mO = ⋅ − ⋅M i k W 
 
 
 
 
 
PROBLEM 3.22 
Before a telephone cable is strung, rope BAC is tied to a stake at B and is 
passed over a pulley at A. Knowing that portion AC of the rope lies in a 
plane parallel to the xy plane and that the tension T in the rope is 124 N, 
determine the moment about O of the resultant force exerted on the 
pulley by the rope. 
 
SOLUTION 
 
 
 
 
 
 
Have /O AO=M r R× 
where ( ) ( ) ( )/ 0 m 9 m 1 mAO = + +r i j k 
1 2= +R T T 
( ) ( )1 124 N cos10 124 N sin10   = − ° − °   T i j 
 ( ) ( )122.116 N 21.532 N= − −i j 
( ) ( ) ( )
( ) ( ) ( ) ( )2 2 2 2 2
1.5 m 9 m 1.8 m
124 N
1.5 m 9 m 1.8 m
T
 − + = =  + +  
i j k
T λ 
 ( ) ( ) ( )20 N 120 N 24 N= − +i j k 
( ) ( ) ( ) 102.116 N 141.532 N 24 N∴ = − − +R i j k 
 0 9 1 N m
102.116 141.532 24
O = ⋅
− −
i j k
M 
 ( ) ( ) ( )357.523 N m 102.116 N m 919.044 N m= ⋅ − ⋅ + ⋅i j k 
 or ( ) ( ) ( )358 N m 102.1 N m 919 N mO = ⋅ − ⋅ + ⋅M i j kW 
 
 
 
 
 
PROBLEM 3.23 
An 8-lb force is applied to a wrench to tighten a showerhead. Knowing 
that the centerline of the wrench is parallel to the x axis, determine the 
moment of the force about A. 
 
SOLUTION 
 
 
 
Have /A C A=M r F× 
where ( ) ( ) ( )/ 8.5 in. 2.0 in. 5.5 in.C A = − +r i j k 
 ( )8cos 45 sin12 lbxF = − ° ° 
 ( )8sin 45 lbyF = − ° 
 ( )8cos 45 cos12 lbzF = − ° ° 
( ) ( ) ( ) 1.17613 lb 5.6569 lb 5.5332 lb∴ = − − −F i j k 
and 8.5 2.0 5.5 lb in.
1.17613 5.6569 5.5332
A = − ⋅
− − −
i j k
M 
 ( ) ( ) ( )42.179 lb in. 40.563 lb in. 50.436 lb in.= ⋅ + ⋅ − ⋅i j k 
 or ( ) ( ) ( )42.2 lb in. 40.6 lb in. 50.4 lb in.A = ⋅ + ⋅ − ⋅M i j k W 
 
 
 
 
 
PROBLEM 3.24 
A wooden board AB, whichis used as a temporary prop to support a 
small roof, exerts at point A of the roof a 228 N force directed along BA. 
Determine the moment about C of that force. 
 
SOLUTION 
 
 
 
 
Have /C AC BA=M r F× 
where ( ) ( ) ( )/ 0.96 m 0.12 m 0.72 mAC = − +r i j k 
and BA BA BAF=F λ 
 ( ) ( ) ( )( ) ( ) ( ) ( )2 2 2
0.1 m 1.8 m 0.6 m
228 N
0.1 1.8 0.6 m
 − + − =  + +  
i j k
 
 ( ) ( ) ( )12.0 N 216 N 72 N= − + −i j k 
 0.96 0.12 0.72 N m
12.0 216 72
C∴ = − ⋅
− −
i j k
M 
 ( ) ( ) ( )146.88 N m 60.480 N m 205.92 N m= − ⋅ + ⋅ + ⋅i j k 
 or ( ) ( ) ( )146.9 N m 60.5 N m 206 N mC = − ⋅ + ⋅ + ⋅M i j kW 
 
 
 
 
 
PROBLEM 3.25 
The ramp ABCD is supported by cables at corners C and D. The tension 
in each of the cables is 360 lb. Determine the moment about A of the 
force exerted by (a) the cable at D, (b) the cable at C. 
 
SOLUTION 
 
 
 
 
 
 (a) Have /A E A DE=M r T× 
 where ( )/ 92 in.E A =r j 
 DE DE DET=T λ 
 ( ) ( ) ( )( ) ( ) ( ) ( )2 2 2
24 in. 132 in. 120 in.
360 lb
24 132 120 in.
+ −=
+ +
i j k
 
 ( ) ( ) ( )48 lb 264 lb 240 lb= + −i j k 
( ) ( ) 0 92 0 lb in. 22,080 lb in. 4416 lb in
48 264 240
A∴ = ⋅ = − ⋅ − ⋅
−
i j k
M i k 
 or ( ) ( )1840 lb ft 368 lb ftA = − ⋅ − ⋅M i k W 
(b) Have /A G A CG=M r T× 
 where ( ) ( )/ 108 in. 92 in.G A = +r i j 
( ) ( ) ( )
( ) ( ) ( ) ( )2 2 2
24 in. 132 in. 120 in.
360 lb
24 132 120 in.
CG CG CGT
− + −= =
+ +
i j k
T λ 
 ( ) ( ) ( )48 lb 264 lb 240 lb= − + −i j k 
 108 92 0 lb in.
48 264 240
A∴ = ⋅
− −
i j k
M 
 ( ) ( ) ( )22,080 lb in. 25,920 lb in. 32,928 lb in.= − ⋅ + ⋅ + ⋅i j k 
 or ( ) ( ) ( )1840 lb ft 2160 lb ft 2740 lb ftA = − ⋅ + ⋅ + ⋅M i j kW 
 
 
 
 
 
PROBLEM 3.26 
The arms AB and BC of a desk lamp lie in a vertical plane that forms an 
angle of o30 with the xy plane. To reposition the light, a force of 
magnitude 8 N is applied at C as shown. Determine the moment of the 
force about O knowing that 450=AB mm, 325=BC mm, and line 
CD is parallel to the z axis. 
 
SOLUTION 
 
 
 
 
 
 
 
 Have /O C O C=M r F× 
 where ( ) ( )/ cos30C O xz xzxr AB BC= + ° 
( )0.450 m sin 45 0.31820 mxzAB = ° = 
( )0.325 m sin 50 0.24896 mxzBC = ° = 
( ) ( ) ( )/ 0.150 m 0.450 m cos 45C O y y yyr OA AB BC= + − = + ° 
 ( )0.325 m cos50 0.25929 m− ° = 
 ( ) ( )/ sin 30C O xz xzzr AB BC= + ° 
 ( )0.31820 m 0.24896 m sin 30 0.28358 m= + ° = 
or ( ) ( ) ( )/ 0.49118 m 0.25929 m 0.28358 mC O = + +r i j k 
( ) ( )8 N cos 45 sin 20 1.93476 NC xF = − ° ° = − 
 ( ) ( )8 N sin 45 5.6569 NC yF = − ° = − 
 ( ) ( )8 N cos 45 cos 20 5.3157 NC zF = ° ° = 
or ( ) ( ) ( )1.93476 N 5.6569 N 5.3157 NC = − − +F i j k 
 0.49118 0.25929 0.28358 N m
1.93476 5.6569 5.3157
O∴ = ⋅
− −
i j k
M 
 ( ) ( ) ( )2.9825 N m 3.1596 N m 2.2769 N m= ⋅ − ⋅ − ⋅i j k 
 or ( ) ( ) ( )2.98 N m 3.16 N m 2.28 N mO = ⋅ − ⋅ − ⋅M i j kW 
 
 
 
 
PROBLEM 3.27 
In Problem 3.21, determine the perpendicular distance from point O to 
cable AB. 
Problem 3.21: Before the trunk of a large tree is felled, cables AB and 
BC are attached as shown. Knowing that the tension in cables AB and BC 
are 777 N and 990 N, respectively, determine the moment about O of the 
resultant force exerted on the tree by the cables at B. 
 
SOLUTION 
 
Have | O BAT d=M | 
where perpendicular distance from O to line .d AB= 
Now /O B O BA=Μ r T× 
and ( )/ 8.4 mB O =r j 
( ) ( ) ( )
( ) ( ) ( ) ( )2 2 2
0.9 m 8.4 m 7.2 m
777 N
0.9 8.4 7.2 m
BA BA ABT
− − += =
+ +
i j k
T λ 
 ( ) ( ) ( )63.0 N 588 N 504 N= − − +i j k 
( ) ( )0 8.4 0 N m 4233.6 N m 529.2 N m
63.0 588 504
O∴ = ⋅ = ⋅ + ⋅
− −
i j k
M i k 
and ( ) ( )2 2| | 4233.6 529.2 4266.5 N mO = + = ⋅M 
( )4266.5 N m 777 N d∴ ⋅ = 
or 5.4911 md = 
 or 5.49 md = W 
 
 
 
 
 
PROBLEM 3.28 
In Problem 3.21, determine the perpendicular distance from point O to 
cable BC. 
Problem 3.21: Before the trunk of a large tree is felled, cables AB and 
BC are attached as shown. Knowing that the tension in cables AB and BC 
are 777 N and 990 N, respectively, determine the moment about O of the 
resultant force exerted on the tree by the cables at B. 
 
SOLUTION 
 
 
 
 
 
 
Have | |O BCT d=M 
where perpendicular distance from to line .d O BC= 
/O B O BC=M r T× 
/ 8.4 mB O =r j 
( ) ( ) ( )
( ) ( ) ( ) ( )2 2 2
5.1 m 8.4 m 1.2 m
990 N
5.1 8.4 1.2 m
BC BC BCT
− += =
+ +
i j k
T λ 
 ( ) ( ) ( )510 N 840 N 120 N= − +i j k 
( ) ( ) 0 8.4 0 1008 N m 4284 N m
510 840 120
O∴ = = ⋅ − ⋅
−
i j k
M i k 
and ( ) ( )2 2| | 1008 4284 4401.0 N mO = + = ⋅M 
( )4401.0 N m 990 N d∴ ⋅ = 
4.4454 md = 
 or 4.45 md = W 
 
 
 
 
 
 
PROBLEM 3.29 
In Problem 3.24, determine the perpendicular distance from point D to a 
line drawn through points A and B. 
Problem 3.24: A wooden board AB, which is used as a temporary prop to 
support a small roof, exerts at point A of the roof a 228 N force directed 
along BA. Determine the moment about C of that force. 
 
SOLUTION 
 
 
 
 
 
Have | |D BAF d=M 
where perpendicular distance from to line .d D AB= 
/D A D BA=M r F× 
( ) ( )/ 0.12 m 0.72 mA D = − +r j k 
( ) ( ) ( )( )
( ) ( ) ( ) ( )2 2 2
0.1 m 1.8 m 0.6 m
228 N
0.1 1.8 0.6 m
BA BA BAF
− + −= =
+ +
i j k
F λ 
 ( ) ( ) ( )12.0 N 216 N 72 N= − + −i j k 
 0 0.12 0.72 N m
12.0 216 72
D∴ = − ⋅
− −
i j k
M 
 ( ) ( ) ( )146.88 N m 8.64 N m 1.44 N m= − ⋅ − ⋅ − ⋅i j k 
and ( ) ( ) ( )2 2 2| | 146.88 8.64 1.44 147.141 N mD = + + = ⋅M 
( )147.141 N m 228 N d∴ ⋅ = 
0.64536 md = 
 or 0.645 md = W 
 
 
 
 
 
PROBLEM 3.30 
In Problem 3.24, determine the perpendicular distance from point C to a 
line drawn through points A and B. 
Problem 3.24: A wooden board AB, which is used as a temporary prop to 
support a small roof, exerts at point A of the roof a 228 N force directed 
along BA. Determine the moment about C of that force. 
 
SOLUTION 
 
 
 
 
 
 
Have | |C BAF d=M 
where perpendicular distance from to line .d C AB= 
/C AC BA=M r F× 
( ) ( ) ( )/ 0.96 m 0.12 m 0.72 mAC = − +r i j k 
( ) ( ) ( )( )
( ) ( ) ( ) ( )2 2 2
0.1 m 1.8 m 0.6
228 N
0.1 1.8 0.6 m
BA BA BAF
− + −= =
+ +
i j k
F λ 
 ( ) ( ) ( )12.0 N 216 N 72 N= − + −i j k 
 0.96 0.12 0.72 N m
12.0 216 72
C∴ = − ⋅
− −
i j k
M 
 ( ) ( ) ( )146.88 N m 60.48 N m 205.92 N m= − ⋅ − ⋅ + ⋅i j k 
and ( ) ( ) ( )2 2 2| | 146.88 60.48 205.92 260.07 N mC = + + = ⋅M 
( )260.07 N m 228 N d∴ ⋅ = 
1.14064 md = 
 or 1.141 md = W 
 
 
 
 
 
 
PROBLEM 3.31 
In Problem 3.25, determine the perpendicular distance from point A to 
portion DE of cable DEF. 
Problem 3.25: The ramp ABCD is supported by cables at corners C 
and D. The tension in each of the cables is 360 lb. Determine the moment 
about A of the force exerted by (a) the cable at D, (b) the cable at C. 
 
SOLUTION 
 
Have A DET d=M 
where perpendicular distance from to line .d A DE= 
/A E A DE=M r T× 
 ( )/ 92 in.E A =r j 
( ) ( ) ( )
( ) ( ) ( ) ( )2 2 2
24 in. 132 in. 120 in.
360 lb
24 132 120 in.
DE DE DET
+ −= =
+ +
i j k
T λ 
 ( ) ( ) ( )48 lb 264 lb 240 lb= + −i j k 
 0 92 0 N m
48 264 240
A∴ = ⋅
−
i j k
M 
 ( ) ( )22,080 lb in. 4416 lb in.= − ⋅ − ⋅i k 
 
 PROBLEM 3.31 CONTINUED 
and ( ) ( )2 222,080 4416 22,517 lb in.A = + = ⋅M 
( )22,517 lb in. 360 lb d∴ ⋅ = 
 62.548 in.d = 
 or 5.21 ftd = W 
 
 
 
 
 
PROBLEM 3.32 
In Problem 3.25, determinethe perpendicular distance from point A to a 
line drawn through points C and G. 
Problem 3.25: The ramp ABCD is supported by cables at corners C 
and D. The tension in each of the cables is 360 lb. Determine the moment 
about A of the force exerted by (a) the cable at D, (b) the cable at C. 
 
SOLUTION 
 
Have A CGT d=M 
where perpendicular distance from to line .d A CG= 
/A G A CG=M r T× 
( ) ( )/ 108 in. 92 in.G A = +r i j 
 CG CG CGT=T λ 
( ) ( ) ( )
( ) ( ) ( ) ( )2 2 2
24 in. 132 in. 120 in.
360 lb
24 132 120 in.
− + −=
+ +
i j k
 
 ( ) ( ) ( )48 lb 264 lb 240 lb= − + −i j k 
 108 92 0 lb in.
48 264 240
A∴ = ⋅
− −
i j k
M 
 ( ) ( ) ( )22,080 lb in. 25,920 lb in. 32,928 lb in.= − ⋅ + ⋅ + ⋅i j k 
and ( ) ( ) ( )2 2 222,080 25,920 32,928 47,367 lb in.A = + + = ⋅M 
( )47,367 lb in. 360 lb d∴ ⋅ = 
 131.575 in.d = or 10.96 ftd = W 
 
 
 
PROBLEM 3.33 
In Problem 3.25, determine the perpendicular distance from point B to a 
line drawn through points D and E. 
Problem 3.25: The ramp ABCD is supported by cables at corners C 
and D. The tension in each of the cables is 360 lb. Determine the moment 
about A of the force exerted by (a) the cable at D, (b) the cable at C. 
 
SOLUTION 
 
Have B DET d=M 
where perpendicular distance from to line .d B DE= 
/B E B DE=M r T× 
( ) ( )/ 108 in. 92 in.E B = − +r i j 
( ) ( ) ( )
( ) ( ) ( ) ( )2 2 2
24 in. 132 in. 120 in.
360 lb
24 132 120 in.
DE DE DET
+ −= =
+ +
i j k
T λ 
 ( ) ( ) ( )48 lb 264 lb 240 lb= + −i j k 
 108 92 0 lb in.
48 264 240
B∴ = − ⋅
−
i j k
M 
 ( ) ( ) ( )22,080 lb in. 25,920 lb in. 32,928 lb in.= − ⋅ − ⋅ − ⋅i j k 
and ( ) ( ) ( )2 2 222,080 25,920 32,928 47,367 lb in.B = + + = ⋅M 
( )47,367 lb in. 360 lb d∴ ⋅ = 
131.575 in.d = 
 or 10.96 ftd = W 
 
 
 
 
 
PROBLEM 3.34 
Determine the value of a which minimizes the perpendicular distance 
from point C to a section of pipeline that passes through points A and B. 
 
SOLUTION 
 
 
 
Assuming a force F acts along AB, 
( )/C AC F d= =M r F× 
where 
perpendicular distance from to line d C AB= 
( ) ( ) ( )
( ) ( ) ( )2 2 2
8 m 7 m 9 m
8 7 9 m
ABF F
+ −= =
+ +
i j k
F λ 
 ( ) ( ) ( )0.57437 0.50257 0.64616F= + −i j k 
( ) ( ) ( )/ 1 m 2.8 m 3 mAC a= − − −r i j k 
 1 2.8 3
0.57437 0.50257 0.64616
C a F∴ = − −
−
i j k
M 
 ( ) ( )0.30154 0.50257 2.3693 0.57437a a= + + − i j 
 ]2.1108 F+ k 
Since ( )22 2/ /or C AC AC dF= × =M r F r F× 
( ) ( ) ( )2 2 2 20.30154 0.50257 2.3693 0.57437 2.1108a a d∴ + + − + = 
Setting ( )2 0d dda = to find a to minimize d 
 ( )( )2 0.50257 0.30154 0.50257a+ 
 ( )( )2 0.57437 2.3693 0.57437 0a+ − − = 
Solving 2.0761 ma = 
 or 2.08 ma = W 
 
 PROBLEM 3.35 
Given the vectors P = 7i − 2j + 5k, Q = −3i − 4j + 6k, and S = 8i + 
j − 9k, compute the scalar products ,P Q⋅ ,P S⋅ and .Q S⋅ 
 
SOLUTION 
( ) ( )7 2 5 3 4 6= − + − − +P Q i j k i j k⋅ ⋅ 
 ( )( ) ( )( ) ( )( )7 3 2 4 5 6= − + − − + 
 17= 
 or 17=P Q⋅ W 
( ) ( )7 2 5 8 9= − + + −P S i j k i j k⋅ ⋅ 
 ( )( ) ( )( ) ( )( )7 8 2 1 5 9= + − + − 
 9= 
 or 9=P S⋅ W 
( ) ( )3 4 6 8 9= − − + + −Q S i j k i j k⋅ ⋅ 
 ( )( ) ( )( ) ( )( )3 8 4 1 6 9= − + − + − 
 82= − 
 or 82= −Q S⋅ W 
 
 
 
 
 
PROBLEM 3.36 
Form the scalar products B C⋅ and ,′B C⋅ where ,B B′= and use the 
results obtained to prove the identity 
( ) ( )1 12 2cos cos cos cos .α β α β α β= + + − 
 
SOLUTION 
 
 
 
By definition 
( )cosBC α β= −B C⋅ 
where ( ) ( )cos sinB β β = + B i j 
 ( ) ( )cos sinC α α = + C i j 
( )( ) ( )( ) ( )cos cos sin sin cosB C B C BCβ α β α α β∴ + = − 
or ( )cos cos sin sin cosβ α β α α β+ = − (1) 
By definition 
( )cosBC α β′ = +B C⋅ 
where ( ) ( )cos sinβ β′  = − B i j 
( )( ) ( )( ) ( )cos cos sin sin cosB C B C BCβ α β α α β∴ + − = + 
or ( )cos cos sin sin cosβ α β α α β− = + (2) 
Adding Equations (1) and (2), 
( ) ( )2 cos cos cos cosβ α α β α β= − + + 
 or ( ) ( )1 1cos cos cos cos
2 2
α β α β α β= + + − W 
 
 
 
 
 
 
PROBLEM 3.37 
Consider the volleyball net shown. Determine the angle formed by guy 
wires AB and AC. 
 
SOLUTION 
 
 
 
First note 
( ) ( ) ( )2 2 2/ 1.95 m 2.4 m 0.6 m B AAB = = − + − +r 
 3.15 m= 
( ) ( ) ( )2 2 2/ 0 m 2.4 m 1.8 m C AAC = = + − +r 
 3.0 m= 
and 
( ) ( ) ( )/ 1.95 m 2.40 m 0.6 m B A = − − +r i j k 
 ( ) ( )/ 2.40 m 1.80 m C A = − +r j k 
By definition 
/ / / / cosB A C A B A C A θ=r r r r⋅ 
or ( ) ( ) ( )( )1.95 2.40 0.6 2.40 1.80 3.15 3.0 cosθ− − + − + =i j k j k⋅ 
( )( ) ( )( ) ( )( )1.95 0 2.40 2.40 0.6 1.8 9.45cosθ− + − − + = 
cos 0.72381θ∴ = 
and 43.630θ = ° 
 or 43.6θ = °W 
 
 
 
 
 
 
PROBLEM 3.38 
Consider the volleyball net shown. Determine the angle formed by guy 
wires AC and AD. 
 
SOLUTION 
 
 
 
First note 
( ) ( )2 2/ 2.4 1.8 m 3 mC AAC = = − + =r 
( ) ( ) ( )2 2 2/ 1.2 2.4 0.3 m 2.7 mD AAD = = + − + =r 
and ( ) ( )/ 2.4 m 1.8 m C A = − +r j k 
( ) ( ) ( )/ 1.2 m 2.4 m 0.3 m D A = − +r i j k 
By definition 
/ / / / cosC A D A C A D A θ=r r r r⋅ 
or ( ) ( ) ( )( )2.4 1.8 1.2 2.4 0.3 3 2.7 cosθ− + − + =j k i j k⋅ 
 ( )( ) ( )( ) ( )( )0 1.2 2.4 2.4 1.8 0.3 8.1cosθ+ − − + = 
and 6.3cos 0.77778
8.1
θ = = 
 38.942θ = ° 
 or 38.9θ = °W 
 
 
 
 
 
PROBLEM 3.39 
Steel framing members AB, BC, and CD are joined at B and C and are 
braced using cables EF and EG. Knowing that E is at the midpoint of BC 
and that the tension in cable EF is 330 N, determine (a) the angle 
between EF and member BC, (b) the projection on BC of the force 
exerted by cable EF at point E. 
 
SOLUTION 
 
(a) By definition ( )( )1 1 cosBC EF θ=λ ⋅ λ 
 where ( ) ( ) ( )( ) ( ) ( ) ( )2 2 2
16 m 4.5 m 12 m 1 16 4.5 12
20.516 4.5 12 m
BC
− −= = − −
+ +
i j k
i j kλ 
 ( ) ( ) ( )( ) ( ) ( ) ( )2 2 2
7 m 6 m 6 m 1 7 6 6
11.07 6 6 m
EF
− − += = − − +
+ +
i j k
i j kλ 
( ) ( )16 4.5 12 7 6 6 cos
20.5 11.0
θ− − − − +∴ =i j k i j k⋅ 
( )( ) ( )( ) ( )( ) ( )( )16 7 4.5 6 12 6 20.5 11.0 cosθ− + − − + − = 
and 1 157cos 134.125
225.5
θ − − = = °   
 or 134.1θ = °W 
(b) By definition ( ) cosEF EFBCT T θ= 
 ( )330 N cos134.125= ° 
 229.26 N= − 
 or ( ) 230 NEF BCT = − W 
 
 
 
 
PROBLEM 3.40 
Steel framing members AB, BC, and CD are joined at B and C and are 
braced using cables EF and EG. Knowing that E is at the midpoint of BC 
and that the tension in cable EG is 445 N, determine (a) the angle 
between EG and member BC, (b) the projection on BC of the force 
exerted by cable EG at point E. 
 
SOLUTION 
 
(a) By definition ( )( )1 1 cosBC EG θ=λ ⋅ λ 
 where ( ) ( ) ( )( ) ( ) ( )2 2 2
16 m 4.5 m 12 m 16 4.5 12
20.516 m 4.5 12 m
BC
− − − −= =
+ +
i j k i j kλ 
 0.78049 0.21951 0.58537= − −i j k 
 ( ) ( ) ( )( ) ( ) ( )2 2 2
8 m 6 m 4.875 m 8 6 4.875
11.1258 6 4.875 m
EG
− + − += =
+ +
i j k i j kλ 
 0.71910 0.53933 0.43820= − +i j k 
( ) ( )( ) ( )( )
( )( )
16 8 4.5 6 12 4.875
cos
20.5 11.25BC EG
θ+ − − + −∴ = =λ ⋅ λ 
and 1 96.5cos 64.967
228.06
θ −  = = °   
 or 65.0θ = °W 
(b) By definition ( ) cosEG EGBCT T θ= 
 ( )445 N cos64.967= ° 
 188.295 N= 
 or ( ) 188.3 NEG BCT = W 
 
 
 
 
 
PROBLEM 3.41 
Slider P can move along rod OA. An elastic cord PC is attached to the 
slider and to the vertical member BC. Knowing that the distance from O 
to P is 0.12 m and the tension in the cord is 30 N, determine (a) the angle 
between the elastic cord and the rod OA, (b) the projection on OA of theforce exerted by cord PC at point P. 
 
SOLUTION 
(a) By definition ( )( )1 1 cosOA PC θ=λ ⋅ λ 
 where ( ) ( ) ( )( ) ( ) ( )2 2 2
0.24 m 0.24 m 0.12 m
0.24 0.24 0.12 m
OA
+ −=
+ +
i j kλ 
 2 2 1
3 3 3
= + −i j k 
 Knowing that /| | 0.36 mA O OAL= =r and that P is located 0.12 m from O, it follows that the coordinates 
 of P are 1
3
 the coordinates of A. 
( )0.08 m, 0.08 m, 0.040 mP∴ − 
 Then ( ) ( ) ( )( ) ( ) ( )2 2 2
0.10 m 0.22 m 0.28 m
0.10 0.22 0.28 m
PC
+ +=
+ +
i j kλ 
 0.27037 0.59481 0.75703= + +i j k 
( )2 2 1 0.27037 0.59481 0.75703 cos
3 3 3
θ ∴ + − + + =  i j k i j k⋅ 
 and ( )1cos 0.32445 71.068θ −= = ° 
 or 71.1θ = °W 
(b) ( ) ( )cos 30 N cos71.068PC PCOAT T θ= = ° 
 ( ) 9.7334 NPC OAT = 
 or ( ) 9.73 NPC OAT = W 
 
 
 
 
 
PROBLEM 3.42 
Slider P can move along rod OA. An elastic cord PC is attached to the 
slider and to the vertical member BC. Determine the distance from O to P 
for which cord PC and rod OA are perpendicular. 
 
SOLUTION 
The requirement that member OA and the elastic cord PC be perpendicular implies that 
0OA PC =λ ⋅ λ or / 0OA C P =rλ ⋅ 
where ( ) ( ) ( )( ) ( ) ( )2 2 2
0.24 m 0.24 m 0.12 m
0.24 0.24 0.12 m
OA
+ −=
+ +
i j kλ 
 2 2 1
3 3 3
= + −i j k 
Letting the coordinates of P be ( ), , ,P x y z we have 
( ) ( ) ( )/ 0.18 0.30 0.24 mC P x y z = − + − + − r i j k 
 ( ) ( ) ( )2 2 1 0.18 0.30 0.24 0
3 3 3
x y z   ∴ + − − + − + − =    i j k i j k⋅ (1) 
Since ( )/ 2 2 ,3OPP O OA OP
dd= = + −r i j kλ 
Then ,
2 2 1,
3 3 3OP OP OP
x d y d z d−= = = (2) 
Substituting the expressions for x, y, and z from Equation (2) into Equation (1), 
( )1 2 2 12 2 0.18 0.30 0.24 0
3 3 3 3OP OP OP
d d d      + − − + − + + =            i j k i j k⋅ 
or 3 0.36 0.60 0.24 0.72OPd = + − = 
 0.24 mOPd∴ = 
 or 240 mmOPd = W 
 
 
 
 
 
PROBLEM 3.43 
Determine the volume of the parallelepiped of Figure 3.25 when 
(a) P = −(7 in.)i − (1 in.)j + (2 in.)k, Q = (3 in.)i − (2 in.)j + (4 in.)k, and 
S = −(5 in.)i + (6 in.)j − (1 in.)k, (b) P = (1 in.)i + (2 in.)j − (1 in.)k, 
Q = −(8 in.)i − (1 in.)j + (9 in.)k, and S = (2 in.)i + (3 in.)j + (1 in.)k. 
 
SOLUTION 
Volume of a parallelepiped is found using the mixed triple product. 
(a) ( )Vol = P Q S⋅ × 
 ( )3 3
7 1 2
3 2 4 in 14 168 20 3 36 20 in
5 6 1
− −
= − = − + + − + −
− −
 
 3187 in= 
 3or Volume 187 in= W 
(b) ( )Vol = P Q S⋅ × 
 ( )3 3
1 2 1
8 1 9 in 1 27 36 16 24 2 in
2 3 1
−
= − − = − − + + + − 
 346 in= 
 or 3Volume 46 in= W 
 
 PROBLEM 3.44 
Given the vectors P = 4i − 2j + Pzk, Q = i + 3j − 5k, and S = −6i + 
2j − k, determine the value of Pz for which the three vectors are coplanar. 
 
SOLUTION 
For the vectors to all be in the same plane, the mixed triple product is zero. 
 ( ) 0=P Q S⋅ × 
( )
4 2
1 3 5 12 40 60 2 2 18
6 2 1
z
z
P
O P
−
∴ = − = − + − − + +
− −
 
so that 34 1.70
20z
P = = 
 or 1.700zP = W 
 
 
 
 
 
PROBLEM 3.45 
The 0.732 1.2-m× lid ABCD of a storage bin is hinged along side AB 
and is held open by looping cord DEC over a frictionless hook at E. If the 
tension in the cord is 54 N, determine the moment about each of the 
coordinate axes of the force exerted by the cord at D. 
 
SOLUTION 
 
 
First note ( ) ( )2 20.732 0.132 mz = − 
 0.720 m= 
Then ( ) ( ) ( )2 2 20.360 0.720 0.720 mDEd = + + 
 1.08 m= 
and ( ) ( ) ( )/ 0.360 m 0.720 m 0.720 mE D = + −r i j k 
Have ( )/OEDE E D
DE
T
d
=T r 
 ( )54 N 0.360 0.720 0.720
1.08
= + −i j k 
 ( ) ( ) ( )18.0 N 36.0 N 36.0 N= + −i j k 
Now /A D A DE=M r T× 
where ( ) ( )/ 0.132 m 0.720 mD A = +r j k 
Then 0 0.132 0.720 N m
18.0 36.0 36.0
A = ⋅
−
i j k
M 
 
 PROBLEM 3.45 CONTINUED 
( )( ) ( )( ) ( )( ){ 0.132 36.0 0.720 36.0 0.720 18.0 0A    ∴ = − − + −   M i j 
 ( )( ) }0 0.132 18.0 N m + − ⋅ k 
or ( ) ( ) ( )30.7 N m 12.96 N m 2.38 N mA = − ⋅ + ⋅ − ⋅M i j k 
 30.7 N m, 12.96 N m,x yM M∴ = − ⋅ = ⋅ 2.38 N mzM = − ⋅ W 
 
 
 
 
 
PROBLEM 3.46 
The 0.732 1.2-m× -m lid ABCD of a storage bin is hinged along side AB 
and is held open by looping cord DEC over a frictionless hook at E. If the 
tension in the cord is 54 N, determine the moment about each of the 
coordinate axes of the force exerted by the cord at C. 
 
SOLUTION 
 
 
 
 
 
 
 
First note ( ) ( )2 20.732 0.132 mz = − 
 0.720 m= 
Then ( ) ( ) ( )2 2 20.840 0.720 0.720 mCEd = + + 
 1.32 m= 
and ( )/E CCE CE
CE
T
d
= rT 
 ( ) ( ) ( ) ( )0.840 m 0.720 m 0.720 m 54 N
1.32 m
− + −= i j k 
 ( ) ( ) ( )36.363 N 29.454 N 29.454 N= − + −i j k 
Now /A E A CE=M r T× 
where ( ) ( )/ 0.360 m 0.852 mE A = +r i j 
Then 0.360 0.852 0 N m
34.363 29.454 29.454
A = ⋅
− −
i j k
M 
 ( ) ( ) ( )25.095 N m 10.6034 N m 39.881 N m= − ⋅ + ⋅ + ⋅i j k 
 25.1 N m, 10.60 N m, 39.9 N mx y zM M M∴ = − ⋅ = ⋅ = ⋅ W 
 
 
 
 
PROBLEM 3.47 
A fence consists of wooden posts and a steel cable fastened to each post 
and anchored in the ground at A and D. Knowing that the sum of the 
moments about the z axis of the forces exerted by the cable on the posts 
at B and C is −66 N · m, determine the magnitude CDT when 56 N.BAT = 
 
SOLUTION 
 
 
 
 
 
Based on ( ) ( )| |z B BA C CDy y  = +   M k r T k r T⋅ × ⋅ × 
where 
( )66 N mz = − ⋅M k 
( ) ( ) ( )1 mB Cy y= =r r j 
 BA BA BAT=T λ 
 ( ) ( ) ( ) ( )1.5 m 1 m 3 m 56 N
3.5 m
− += i j k 
 ( ) ( ) ( )24 N 16 N 48 N= − +i j k 
 CD CD CDT=T λ 
 ( ) ( ) ( )2 m 1 m 2 m
3.0 m CD
T
− −= i j k 
 ( )1 2 2
3 CD
T= − −i j k 
 ( ) ( ) ( ) ( ) ( ){ }66 N m 1 m 24 N 16 N 48 N ∴ − ⋅ = − + k j i j k⋅ × 
 ( ) ( )11 m 2 2
3 CD
T  + − −    k j i j k⋅ × 
or 266 24
3 CD
T− = − − 
( )3 66 24 N
2CD
T∴ = − 
 or 63.0 NCDT = W 
 
 
 
 
 
PROBLEM 3.48 
A fence consists of wooden posts and a steel cable fastened to each post 
and anchored in the ground at A and D. Knowing that the sum of the 
moments about the y axis of the forces exerted by the cable on the posts 
at B and C is 212 N · m, determine the magnitude of BAT when 
33 N.CDT = 
 
SOLUTION 
 
 
 
 
 
 
Based on ( ) ( )| |y B BA C CDz z = + M j r T r T⋅ × × 
where 
( )212 N my = ⋅M j 
( ) ( )8 mB z =r k 
( ) ( )2 mC z =r k 
 BA BA BAT=T λ 
 ( ) ( ) ( )1.5 m 1 m 3 m
3.5 m BA
T
− −= i j k 
 ( )1.5 3
3.5
BAT= − +i j k 
 CD CD CDT=T λ 
 ( ) ( ) ( ) ( )2 m 1 m 2 m 33 N
3.0 m
− −= i j k 
 ( )22 11 22 N= − −i j k 
 ( ) ( ) ( )212 N m 8 m 1.5 3
3.5
BAT  ∴ ⋅ = − +    j k i j k⋅ × 
 ( ) ( )2 m 22 11 22 N + − − j k i j k⋅ × 
or ( ) ( )8 1.5212 2 22
3.5 BA
T= + 
 168
18.6667BA
T∴ = 
 or 49.0 NBAT = W 
 
 
 
 
 
PROBLEM 3.49 
To lift a heavy crate, a man uses a block and tackle attached to the bottom 
of an I-beam at hook B. Knowing that the moments about the y and z axes 
of the force exerted at B by portion AB of the rope are, respectively, 
100 lb ft⋅ and 400 lb ft− ⋅ , determine the distance a. 
 
SOLUTION 
 
 
 
 
 
Based on /O A O BA=M r T× 
where 
 O x y zM M M= + +M i j k 
 ( ) ( )100 lb ft 400 lb ftxM= + ⋅ − ⋅i j k 
( ) ( )/ 6 ft 4 ftA O = +r i j 
 BA BA BAT=T λ 
 ( ) ( ) ( )6 ft 12 ft BA
BA
a
T
d
− −= i j k 
 100 400 6 4 0
6 12
BA
x
BA
TM
d
a
∴ + − =
− −
i j k
i j k 
 ( ) ( ) ( )4 6 96BA
BA
T a a
d
 = − + − i j k 
From -coefficient:j100 6AB BAd aT= 100or 6BA BAT da= (1) 
From -coefficient:k 400 96AB BAd T− = − 400or 96BA BAT d= (2) 
Equating Equations (1) and (2) yields ( )( )
100 96
6 400
a = 
 or 4.00 fta = W 
 
 
 
 
PROBLEM 3.50 
To lift a heavy crate, a man uses a block and tackle attached to the bottom 
of an I-beam at hook B. Knowing that the man applies a 200-lb force to 
end A of the rope and that the moment of that force about the y axis is 
175 lb ft⋅ , determine the distance a. 
 
SOLUTION 
 
 
 
 
 
 
Based on ( )/| |y A O BA=M j r T⋅ × 
where ( ) ( )/ 6 ft 4 ftA O = +r i j 
 /A BBA BA BA BA
BA
T T
d
= = rT λ 
 ( ) ( ) ( ) ( )6 ft 12 ft 200 lb
BA
a
d
− −= i j k 
 ( )200 6 12
BA
a
d
= − −i j k 
 
0 1 0
200175 lb ft 6 4 0
6 12 BA
d
a
∴ ⋅ =
− −
 
 ( ) 200175 0 6
BA
a
d
 = − −  
where ( ) ( ) ( )2 2 26 12 ftBAd a= + + 
 2180 fta= + 
2175 180 1200a a∴ + = 
or 2180 6.8571a a+ = 
Squaring each side 
 2 2180 47.020a a+ = 
Solving 1.97771 fta = 
 or 1.978 fta = W 
 
 
 
 
 
PROBLEM 3.51 
A force P is applied to the lever of an arbor press. Knowing that P lies in 
a plane parallel to the yz plane and that 230 lb in.,xM = ⋅ 
200 lb in.,yM = − ⋅ and 35 lb in.,zM = − ⋅ determine the magnitude of P 
and the values of φ and θ. 
 
SOLUTION 
 
 
 
 
 
Based on ( ) ( ) ( ) ( )cos 9 in. sin sin 9 in. cosxM P Pφ θ φ θ   = −    (1) 
 ( )( )cos 5 in.yM P φ= − (2) 
 ( )( )sin 5 in.zM P φ= − (3) 
Then ( )( )
sin (5)Equation (3) :
Equation (2) cos (5)
z
y
PM
M P
φ
φ
−= − 
or 35tan 0.175 9.9262
200
φ φ−= = = °− 
 or 9.93φ = °W 
Substituting into Equation (2)φ 
( )200 lb in. cos9.9262 (5 in.)P− ⋅ = − ° 
 40.608 lbP = 
 or 40.6 lbP = W 
Then, from Equation (1) 
( ) ( )230 lb in. 40.608 lb cos9.9262 9 in. sinθ   ⋅ = °    
 ( ) ( )40.608 lb sin 9.9262 9 in. cosθ   − °    
or 0.98503sin 0.172380cos 0.62932θ θ− = 
Solving numerically, 48.9θ = °W 
 
 
 
 
 
 
PROBLEM 3.52 
A force P is applied to the lever of an arbor press. Knowing that P lies in 
a plane parallel to the yz plane and that 180 lb in.yM = − ⋅ and 
30 lb in.,zM = − ⋅ determine the moment xM of P about the x axis when 
60 .θ = ° 
 
SOLUTION 
 
 
 
 
 
 
Based on ( ) ( ) ( ) ( )cos 9 in. sin sin 9 in. cosxM P Pφ θ φ θ   = −    (1) 
 ( )( )cos 5 in.yM P φ= − (2) 
 ( )( )sin 5 in.zM P φ= − (3) 
Then ( )( )( )( )
sin 5Equation (3) :
Equation (2) cos 5
z
y
PM
M P
φ
φ
−= − 
or 30 tan
180
φ− =− 
 9.4623φ∴ = ° 
From Equation (3), 
 ( )( )30 lb in. sin 9.4623 5 in.P− ⋅ = − ° 
 36.497 lbP∴ = 
From Equation (1), 
 ( )( )( )36.497 lb 9 in. cos9.4623 sin 60 sin 9.4623 cos60xM = ° ° − ° ° 
 253.60 lb in.= ⋅ 
 or 254 lb in.xM = ⋅ W 
 
 
 
 
 
PROBLEM 3.53 
The triangular plate ABC is supported by ball-and-socket joints at B and 
D and is held in the position shown by cables AE and CF. If the force 
exerted by cable AE at A is 220 lb, determine the moment of that force 
about the line joining points D and B. 
 
SOLUTION 
 
 
 
 
 
Have ( )/DB DB A D AEM = r Tλ ⋅ × 
where ( ) ( )48 in. 14 in. 0.96 0.28
50 in.DB
−= = −i j i jλ 
( ) ( )/ 4 in. 8 in.A D = − +r j k 
 
( ) ( ) ( ) ( )36 in. 24 in. 8 in. 220 lb
44 in.AE AE AE
T
 − + = = i j kT λ 
 ( ) ( ) ( )180 lb 120 lb 40 lb= − +i j k 
 
0.960 0.280 0
0 4 8 lb in.
180 120 40
DBM
−
∴ = − ⋅
−
 
 ( ) ( )( ) ( )( ) ( ) ( )0.960 4 40 8 120 0.280 8 180 0   = − − − + − −    
 364.8 lb in.= ⋅ 
 or 365 lb in.DBM = ⋅ W 
 
 
 
 
 
PROBLEM 3.54 
The triangular plate ABC is supported by ball-and-socket joints at B and 
D and is held in the position shown by cables AE and CF. If the force 
exerted by cable CF at C is 132 lb, determine the moment of that force 
about the line joining points D and B. 
 
SOLUTION 
 
 
 
 
 
 
Have ( )/DB DB C D CFM = r Tλ ⋅ × 
where ( ) ( )48 in. 14 in. 0.96 0.28
50 in.DB
−= = −i j i jλ 
 ( ) ( )/ 8 in. 16 in.C D = −r j k 
 ( ) ( ) ( ) ( )24 in. 36 in. 8 in. 132 lb
44 in.CF CF CF
T
− −= = i j kT λ 
 ( ) ( ) ( )72 lb 108 lb 24 lb= − −i j k 
 
0.96 0.28 0
0 8 16 lb in.
72 108 24
DBM
−
∴ = − ⋅
− −
 
 ( )( ) ( )( ) ( ) ( )( )0.96 8 24 16 108 0.28 16 72 0   = − − − − + − − −    
 1520.64 lb in.= − ⋅ 
 or 1521 lb in.DBM = − ⋅ W 
 
 
 
 
PROBLEM 3.55 
A mast is mounted on the roof of a house using bracket ABCD and is 
guyed by cables EF, EG, and EH. Knowing that the force exerted by 
cable EF at E is 66 N, determine the moment of that force about the line 
joining points D and I. 
 
SOLUTION 
 
 
Have /DI DI F I EFM  =  r Tλ ⋅ × 
where ( ) ( )( ) ( ) ( )2 2
1.6 m 0.4 m 1 4
171.6 0.4 m
DI
−= = −
+
i j
i jλ 
 ( ) ( )/ 4.6 m 0.8 m 5.4 mF I = + =r k k 
 EF EF EFT=T λ 
 ( ) ( ) ( ) ( )1.2 m 3.6 m 5.4 m 66 N
6.6 m
− += i j k 
 ( ) ( ) ( )12 N 36 N 54 N= − +i j k 
 ( ) ( ) ( )6 2 N 6 N 9 N = − + i j k 
 ( )( )
4 1 0
6 N 5.4 m
0 0 1
17 2 6 9
DIM
−
∴ =
−
 
 ( ) ( )7.8582 0 24 2 0 = + + − −  
 172.879 N m= ⋅ 
 or 172.9 N mDIM = ⋅ W 
 
 
 
 
 
 
PROBLEM 3.56 
A mast is mounted on the roof of a house using bracket ABCD and is 
guyed by cables EF, EG, and EH. Knowing that the force exerted by 
cable EG at E is 61.5 N, determine the moment of that force about the 
line joining points D and I. 
 
SOLUTION 
 
 
 
 
Have /DI DI G I EGM  =  r Tλ ⋅ × 
where ( ) ( )1.6 m 0.4 m
0.4 17 mDI
−= i jλ 
 ( )1 4
17
= −i j 
( ) ( )/ 10.9 m 0.8 m 11.7 mG I = − + = −r k k 
 EG EG EGT=T λ 
 ( ) ( ) ( ) ( )1.2 m 3.6 m 11.7 m 61.5 N
12.3 m
− −= i j k 
 ( ) ( ) ( )5 1.2 N 3.6 N 11.7 N = − − i j k 
( ) 4 1 05 N 11.7 m 0 0 1
17
1.2 3.6 11.7
DIM
−
∴ = −
− −
 
 ( ) ( )( )( ) ( )( )( ){ }14.1883 N m 0 4 1 3.6 1 1 1.2 0   = ⋅ − − − + − − −    
 187.286 N m= − ⋅ 
 or 187.3 N mDIM = − ⋅ W 
 
 
 
 
 
 
 
PROBLEM 3.57 
A rectangular tetrahedron has six edges of length a. A force P is directed 
as shown along edge BC. Determine the moment of P about edge OA. 
 
SOLUTION 
 
 
 
 
Have ( )/OA OA C OM = r Pλ ⋅ × 
where 
From triangle OBC 
 ( )
2x
aOA = 
( ) ( ) 1tan30
2 3 2 3z x
a aOA OA  = ° = =   
Since ( ) ( ) ( ) ( )2 2 2 2zx yOA OA OA OA= + + 
or ( ) 22 22
2 2 3y
a aa OA   = + +       
( ) 2 22 2 
4 12 3y
a aOA a a∴ = − − = 
Then /
2
2 3 2 3A O
a aa= + +r i j k 
and 1 2 1
2 3 2 3OA
= + +i j kλ 
 BCP=P λ 
 ( ) ( ) ( )sin30 cos30a a P
a
° − °= i k 
 ( )32P= −i k 
/C O a=r i 
 
 
 
 
 
PROBLEM 3.57 CONTINUED 
( )
1 2 1
2 3 2 3
 1 0 0 2
1 0 3
OA
PM a  ∴ =   
−
 
 ( )( )2 1 32 3aP  = − −    
 
2
aP= 
2OA
aPM = W 
 
 
 
 
 
 
 
PROBLEM 3.58 
A rectangular tetrahedron has six edges of length a. (a) Show that two 
opposite edges, such as OA and BC, are perpendicular to each other. (b) 
Use this property and the result obtained in Problem 3.57 to determine the 
perpendicular distance between edges OA and BC. 
 
SOLUTION 
 
 
 
 
 (a) For edge OA to be perpendicular to edge BC, 
0OA BC =JJJG JJJG⋅ 
 where 
 From triangle OBC 
 ( )
2x
aOA = 
( ) ( ) 1tan30
2 3 2 3z x
a aOA OA  = ° = =   
( ) 
2 2 3y
a aOA OA   ∴ = + +      i j k
JJJG
 
 and ( ) ( )sin 30 cos30BC a a= ° − °i kJJJG 
 3
2 2
a a= −i k 
 ( )32a= −i k 
 Then () ( )3 02 22 3ya a aOA  + + − =    i j k i k⋅ 
 or ( ) ( )2 20 0
4 4y
a aOA+ − = 
 0OA BC∴ =JJJG JJJG⋅ 
 so that OA
JJJG
 is perpendicular to .BC
JJJG W 
 
 
 
 
 
 
PROBLEM 3.58 CONTINUED 
(b) Have ,OAM Pd= with P acting along BC and d the 
 perpendicular distance from to .OA BC
JJJG JJJG
 
 From the results of Problem 3.57, 
2OA
PaM = 
 
2
Pa Pd∴ = 
 or 
2
ad = W 
 
 
 
 
 
 
 
PROBLEM 3.59 
The 8-ft-wide portion ABCD of an inclined, cantilevered walkway is 
partially supported by members EF and GH. Knowing that the 
compressive force exerted by member EF on the walkway at F is 5400 lb, 
determine the moment of that force about edge AD. 
 
SOLUTION 
 
Having ( )/AD AD E A EFM = r Tλ ⋅ × 
where ( ) ( )( ) ( ) ( )2 2
24 ft 3 ft 1 8
6524 3 ft
AD
+= = +
+
i j
i jλ 
( ) ( )/ 7 ft 3 ftE A = −r i j 
( ) ( )( ) ( )
( ) ( ) ( ) ( )
8
24
2 2 2
8 ft 7 ft 3 ft 3 ft 8 ft
5400 lb
1 4 8 ft
EF EF EFT
 − + + + = =
+ +
i j k
T λ 
 ( ) ( ) ( )600 1 lb 4 lb 8 lb = + + i j k 
( )
8 1 0
600 600 7 3 0 lb ft 192 56 lb ft
65 651 4 8
ADM∴ = − ⋅ = − − ⋅ 
 18,456.4 lb ft= − ⋅ or 18.46 kip ftADM = − ⋅ W 
 
 
 
 
 
 
 
PROBLEM 3.60 
The 8-ft-wide portion ABCD of an inclined, cantilevered walkway is 
partially supported by members EF and GH. Knowing that the 
compressive force exerted by member GH on the walkway at H is 
4800 lb, determine the moment of that force about edge AD. 
 
SOLUTION 
 
Having ( )/AD AD G A GHM = r Tλ ⋅ × 
where ( ) ( )( ) ( ) ( )2 2
24 ft 3 ft 1 8
6524 3 ft
AD
+= = +
+
i j
i jλ 
( ) ( ) ( ) ( )/ 20 ft 6 ft 2 10 ft 3 ftG A  = − = − r i j i j 
( ) ( )( ) ( )
( ) ( ) ( ) ( )
16
24
2 2 2
16 ft 20 ft 6 ft 3 ft 8 ft
4800 lb
4 8 8 ft
GH GH GHT
 − + + + = =
+ +
i j k
T λ 
 ( ) ( ) ( )1600 1 lb 2 lb 2 lb = − + + i j k 
( )( ) ( )
8 1 0
1600 lb 2 ft 3200 lb ft 10 3 0 48 20
65 651 2 2
ADM
⋅∴ = − = − −
−
 
 26,989 lb ft= − ⋅ or 27.0 kip ftADM = − ⋅ W 
 
 
 PROBLEM 3.61 
Two forces F1 and F2 in space have the same magnitude F. Prove that the 
moment of F1 about the line of action of F2 is equal to the moment of F2 
about the line of action of F1. 
 
SOLUTION 
 
First note that 1 1 1F=F λ and 2 2 2F=F λ 
Let 1 2 moment of M = F about the line of action of 1M 
and 2 1 moment of M = F about the line of action of 2M 
Now, by definition 
( ) ( )1 1 / 2 1 / 2 2B A B AM F= =r F rλ ⋅ × λ ⋅ × λ 
( ) ( )2 2 / 1 2 / 1 1A B A BM F= =r F rλ ⋅ × λ ⋅ × λ 
Since 1 2F F F= = and / /A B B A= −r r 
 ( )1 1 / 2B AM F= rλ ⋅ × λ 
( )2 2 / 1B AM F= −rλ ⋅ × λ 
Using Equation (3.39) 
( ) ( )1 / 2 2 / 1B A B A= −r rλ ⋅ × λ λ ⋅ × λ 
so that ( )2 1 / 2B AM F= rλ ⋅ × λ 
 12 21 M M∴ = W 
 
 
 
 
 
PROBLEM 3.62 
In Problem 3.53, determine the perpendicular distance between cable AE 
and the line joining points D and B. 
Problem 3.53: The triangular plate ABC is supported by ball-and-socket 
joints at B and D and is held in the position shown by cables AE and CF. 
If the force exerted by cable AE at A is 220 lb, determine the moment of 
that force about the line joining points D and B. 
 
SOLUTION 
 
 
 
 
Have ( )/DB DB A D AEM = r Tλ ⋅ × 
where ( ) ( )48 in. 14 in. 0.96 0.28
50 in.DB
−= = −i j i jλ 
 ( ) ( )/ 4 in. 8 in.A D = − +r j k 
 AE AE AET=T λ 
 ( ) ( ) ( ) ( )36 in. 24 in. 8 in. 220 lb
44 in.
− += i j k 
 ( ) ( ) ( )180 lb 120 lb 40 lb= − +i j k 
0.96 0.28 0
 0 4 8 lb in.
180 120 40
DBM
−
∴ = − ⋅
−
 
 364.8 lb in.= ⋅ 
Only the perpendicular component of AET contributes to the moment of 
AET about line DB. The parallel component of AET will be used to find 
the perpendicular component. 
 
 
 
 
 
 
PROBLEM 3.62 CONTINUED 
Have 
 ( )parallelAE DB AET = Tλ ⋅ 
 ( ) ( ) ( ) ( )0.96 0.28 180 lb 120 lb 40 lb = − − + i j i j k⋅ 
 ( )( ) ( )( ) ( )( )0.96 180 0.28 120 0 40 lb = + − − +  
 ( )172.8 33.6 lb= + 
 206.4 lb= 
Since ( ) ( )perpendicular parallelAE AE AE= +T T T 
( ) ( ) ( )2 2perpendicular parallel AE AE AET T T∴ = − 
 ( ) ( )2 2220 206.41= − 
 76.151 lb= 
Then ( ) ( )perpendicularDB AEM T d= 
( )364.8 lb in. 76.151 lb d⋅ = 
4.7905 in.d = 
 or 4.79 in.d = W 
 
 
 
 
 
 
 
PROBLEM 3.63 
In Problem 3.54, determine the perpendicular distance between cable CF 
and the line joining points D and B. 
Problem 3.54: The triangular plate ABC is supported by ball-and-socket 
joints at B and D and is held in the position shown by cables AE and CF. 
If the force exerted by cable CF at C is 132 lb, determine the moment of 
that force about the line joining points D and B. 
 
SOLUTION 
 
 
 
 
Have ( ) ( )/DB DB C D CFM = r Tλ ⋅ × 
where ( ) ( )48 in. 14 in.
50 in.DB
−= i jλ 
 0.96 0.28= −i j 
( ) ( )/ 8 in. 16 in.C D = −r j k 
 CF CF CFT=T λ 
 ( ) ( ) ( ) ( )24 in. 36 in. 8 in. 132 lb
44 in.
− −= i j k 
 ( ) ( ) ( )72 lb 108 lb 24 lb= − −i j k 
0.96 0.28 0
 0 8 16 lb in
72 108 24
DBM
−
∴ = − ⋅
− −
 
 1520.64 lb in.= − ⋅ 
Only the perpendicular component of CFT contributes to the moment of 
CFT about line DB. The parallel component of CFT will be used to 
obtain the perpendicular component. 
 
 
 
 
 
 
PROBLEM 3.63 CONTINUED 
Have 
 ( )parallelCF DB CFT = Tλ ⋅ 
 ( ) ( ) ( ) ( )0.96 0.28 72 lb 108 lb 24 lb = − − − i j i j k⋅ 
 ( )( ) ( )( ) ( )( )0.96 72 0.28 108 0 24 lb = + − − + −  
 99.36 lb= 
Since ( ) ( )perp. parallelCF CF CF= +T T T 
( ) ( ) ( )2 2perp. parallel CF CF CFT T T∴ = − 
 ( ) ( )2 2132 99.36= − 
 86.900 lb= 
Then ( ) ( )perp.DB CFM T d= 
( )1520.64 lb in. 86.900 lb d− ⋅ = 
17.4988 in.d = 
 or 17.50 in.d = W 
 
 
 
 
 
 
 
PROBLEM 3.64 
In Problem 3.55, determine the perpendicular distance between cable EF 
and the line joining points D and I. 
Problem 3.55: A mast is mounted on the roof of a house using bracket 
ABCD and is guyed by cables EF, EG, and EH. Knowing that the force 
exerted by cable EF at E is 66 N, determine the moment of that force 
about the line joining points D and I. 
 
SOLUTION 
 
 
 
 
Have ( )/DI DI F I EFM = r Tλ ⋅ × 
where ( ) ( ) ( )1.6 m 0.4 m 1 4
0.4 17 m 17DI
−= = −i j i jλ 
( )/ 5.4 mF I =r k 
( ) ( ) ( ) ( )1.2 m 3.6 m 5.4 m 66 N
6.6 mEF EF EF
T
− += = i j kT λ 
 ( ) ( ) ( )6 2 N 6 N 9 N = − + i j k 
( )( ) 4 1 06 N 5.4 m 0 0 1 172.879 N m
17 2 6 9
DIM
−
∴ = = ⋅
−
 
Only the perpendicular component of EFT contributes to the moment of 
EFT about line DI. The parallel component of EFT will be used to find 
the perpendicular component. 
Have 
 ( )parallelEF DI EFT = Tλ ⋅ 
 ( ) ( ) ( ) ( )1 4 12 N 36 N 54 N
17
 = − − + i j i j k⋅ 
 ( )1 48 36 N
17
= + 
 84 N
17
= 
 
 
 
 
 
 
PROBLEM 3.64 CONTINUED 
Since ( ) ( )perp. parallelEF EF EF= +T T T 
 ( ) ( ) ( )2 2perp. parallel EF EF EFT T T∴ = − 
 ( ) 22 8466
17
 = −    
 62.777 N= 
Then ( ) ( )perp.DI EFM T d= 
( )( )172.879 N m 62.777 N d⋅ = 
2.7539 md = 
 or 2.75 md = W 
 
 
 
 
 
 
 
PROBLEM 3.65 
In Problem 3.56, determine the perpendicular distance between cable EG 
and the line joining points D and I. 
Problem 3.56: A mast is mounted on the roof of a house using bracket 
ABCD and is guyed by cables EF, EG, and EH. Knowing that the force 
exerted by cable EG at E is 61.5 N, determine the moment of that force 
about the line joining points D and I. 
 
SOLUTION 
 
 
 
 
Have/DI DI G I EGM  =  r Tλ ⋅ × 
where ( ) ( ) ( )1.6 m 0.4 m 1 4
0.4 17 m 17DI
−= = −i j i jλ 
( ) ( )/ 10.9 m 0.8 m 11.7 mG I = − + = −r k k 
( ) ( ) ( ) ( )1.2 m 3.6 m 11.7 m 61.5 N
12.3 mEG EG EG
T
− −= = i j kT λ 
 ( ) ( ) ( )5 1.2 N 3.6 N 11.7 N = − − i j k 
( )( ) 4 1 05 N 11.7 m 0 0 1
17
1.2 3.6 11.7
DIM
−
∴ = −
− −
 
 187.286 N m= − ⋅ 
Only the perpendicular component of EGT contributes to the moment of 
EGT about line DI. The parallel component of EGT will be used to find 
the perpendicular component. 
Have 
 ( )parallelEG DI EGT = Tλ ⋅ 
 ( ) ( ) ( ) ( )1 4 5 1.2 N 3.6 N 11.7 N
17
 = − − − i j i j k⋅ 
 ( )5 4.8 3.6 N
17
= + 
 42 N
17
= 
 
 
 
 
 
PROBLEM 3.65 CONTINUED 
Since ( ) ( )perp. parallelEF EG EG= +T T T 
 ( ) ( ) ( )2 2perp. parallel EG EG EGT T T∴ = − 
 ( ) 22 4261.5
17
 = −    
 60.651 N= 
Then ( ) ( )perp.DI EGM T d= 
( )( )187.286 N m 60.651 N d⋅ = 
3.0880 md = 
 or 3.09 md = W 
 
 
 
 
 
 
PROBLEM 3.66 
In Problem 3.41, determine the perpendicular distance between post BC 
and the line connecting points O and A. 
Problem 3.41: Slider P can move along rod OA. An elastic cord PC is 
attached to the slider and to the vertical member BC. Knowing that the 
distance from O to P is 0.12 m and the tension in the cord is 30 N, 
determine (a) the angle between the elastic cord and the rod OA, (b) the 
projection on OA of the force exerted by cord PC at point P. 
 
SOLUTION 
Assume post BC is represented by a force of magnitude BCF 
where BC BCF=F j 
Have ( )/OA OA B O BCM = r Fλ ⋅ × 
where ( ) ( ) ( )0.24 m 0.24 m 0.12 m 2 2 1
0.36 m 3 3 3OA
+ −= = + −i j k i j kλ 
( ) ( )/ 0.18 m 0.24 mB O = +r i k 
( )
2 2 1
1 0.18 0 0.24 0.48 0.18 0.22
3 3
0 1 0
BC
OA BC BC
FM F F
−
∴ = = − − = − 
Only the perpendicular component of BCF contributes to the moment of BCF about line OA. The parallel 
component will be found first so that the perpendicular component of BCF can be determined. 
( )parallel
2 2 1
3 3 3OA BC BCBC
F F = = + −  F i j k jλ ⋅ ⋅ 
 2
3 BC
F= 
Since ( ) ( )parallel perp.BC BC BC= +F F F 
( ) ( ) ( ) ( ) 22 2 2perp. parallel 2 3BCBC BC BC BC
FF F F F  = − = −    
 0.74536 BCF= 
Then ( ) ( )perp.OA BCM F d= 
( )0.22 0.74536BC BCF F d= 
 0.29516 md = 
 or 295 mmd = W 
 
 
 
 
 
 
PROBLEM 3.67 
In Problem 3.45, determine the perpendicular distance between cord DE 
and the y axis. 
Problem 3.45: The 0.732 1.2× -m lid ABCD of a storage bin is hinged 
along side AB and is held open by looping cord DEC over a frictionless 
hook at E. If the tension in the cord is 54 N, determine the moment about 
each of the coordinate axes of the force exerted by the cord at D. 
 
SOLUTION 
 
 
 
 
First note 
( ) ( )2 20.732 0.132 mz = − 
 0.720 m= 
Have ( )/y D A DEM = j r T⋅ × 
where ( )/ 0.132 0.720 mD A = +r j k 
 DE DE DET=T λ 
 ( ) ( ) ( ) ( )0.360 m 0.732 m 0.720 m 54 N
1.08 m
+ −= i j k 
 ( ) ( ) ( )18 N 36 N 36 N= + −i j k 
0 1 0
 0 0.132 0.720 12.96 N m
18 36 36
yM∴ = = ⋅
−
 
Only the perpendicular component of DET contributes to the moment of 
DET about the y-axis. The parallel component will be found first so that 
the perpendicular component of DET can be determined. 
( )parallel 36 NDEDET = =j T⋅ 
 
 
 
 
PROBLEM 3.67 CONTINUED 
Since ( ) ( ) ( )parallel perp.DE DE DE= +T T T 
 ( ) ( ) ( )2 2perp. parallelDE DE DET T T= − 
 ( ) ( )2 254 36 40.249 N= − = 
Then ( ) ( )perp.y DEM T d= 
( )( )12.96 N m 40.249 N d⋅ = 
0.32199 md = 
 or 322 mmd = W 
 
 
 
 
 
 
PROBLEM 3.68 
A plate in the shape of a parallelogram is acted upon by two couples. 
Determine (a) the moment of the couple formed by the two 21-N forces, 
(b) the perpendicular distance between the 12-N forces if the resultant of 
the two couples is zero, (c) the value of α if the resultant couple is 
1.8 N m⋅ clockwise and d is 1.05 m. 
 
SOLUTION 
 
 
 
 (a) Have 1 1 1M d F= 
 where 1 0.4 md = 
 1 21 NF = 
( )( )1 0.4 m 21 N 8.4 N mM∴ = = ⋅ 
 1or 8.40 N m= ⋅M W 
(b) Have 1 2 0+ =M M 
 or ( )28.40 N m 12 N 0d⋅ − = 
 2 0.700 md∴ = W 
(c) Have total 1 2= +M M M 
 or ( )( )( )1.8 N m 8.40 N m 1.05 m sin 12 Nα⋅ = ⋅ − 
 sin 0.52381α∴ = 
 and 31.588α = ° 
 or 31.6α = °W 
 
 
 
 
 
 
 
PROBLEM 3.69 
A couple M of magnitude 10 lb ft⋅ is applied to the handle of a 
screwdriver to tighten a screw into a block of wood. Determine the 
magnitudes of the two smallest horizontal forces that are equivalent to M 
if they are applied (a) at corners A and D, (b) at corners B and C, 
(c) anywhere on the block. 
 
SOLUTION 
 
 
 
 
 
 
 
 
 (a) Have M Pd= 
 or ( ) 1 ft10 lb ft 10 in.
12 in.
P  ⋅ =    
 12 lbP∴ = minor 12.00 lbP = W 
(b) ( ) ( )2 2BCd BE EC= + 
 ( ) ( )2 210 in. 6 in. 11.6619 in.= + = 
 Have M Pd= 
( ) 1 ft10 lb ft 11.6619 in.
12 in.
P  ⋅ =    
 10.2899 lbP = or 10.29 lbP = W 
(c) ( ) ( )2 2ACd AD DC= + 
 ( ) ( )2 210 in. 16 in. 2 89 in.= + = 
 Have ACM Pd= 
( ) 1 ft10 lb ft 2 89 in. 12 in.P  ⋅ =    
 6.3600 lbP = or 6.36 lbP = W 
 
 
 
 
 
 
PROBLEM 3.70 
Two 60-mm-diameter pegs are mounted on a steel plate at A and C, and 
two rods are attached to the plate at B and D. A cord is passed around the 
pegs and pulled as shown, while the rods exert on the plate 10-N forces as 
indicated. (a) Determine the resulting couple acting on the plate when T = 
36 N. (b) If only the cord is used, in what direction should it be pulled to 
create the same couple with the minimum tension in the cord? 
(c) Determine the value of that minimum tension. 
 
SOLUTION 
(a) Have ( )M Fd= Σ 
 ( )( ) ( )( )36 N 0.345 m 10 N 0.380 m= − 
 8.62 N m= ⋅ 
 8.62 N m= ⋅M W 
(b) 
 
 Have 8.62 N mM Td= = ⋅ 
 For T to be minimum, d must be maximum. 
min T∴ must be perpendicular to line AC 
0.380 mtan 1.33333
0.285 m
θ = = 
 and 53.130θ = ° 
 or 53.1θ = °W 
(c) Have min maxM T d= 
 where 8.62 N mM = ⋅ 
( ) ( ) ( )2 2max 0.380 0.285 2 0.030 m 0.535 md  = + + =   
( )min 8.62 N m 0.535 mT∴ ⋅ = 
min 16.1121 NT = 
 minor 16.11 NT = W 
 
 
 
 
 
 
PROBLEM 3.71 
The steel plate shown will support six 50-mm-diameter idler rollers 
mounted on the plate as shown. Two flat belts pass around the rollers, and 
rollers A and D will be adjusted so that the tension in each belt is 45 N. 
Determine (a) the resultant couple acting on the plate if a = 0.2 m, (b) the 
value of a so that the resultant couple acting on the plate is 54 N m⋅ 
clockwise. 
 
SOLUTION 
 
 
 
 
 (a) Note when /0.2 m, C Fa = r is perpendicular to the inclined 45 N 
 forces. 
 Have 
 ( )M Fd= Σ 
 ( ) ( )45 N 0.2 m 2 0.025 ma = − + +  
 ( ) ( )45 N 2 2 2 0.025 ma − +  
 For 0.2 m,a = 
( )( )45 N 0.450 m 0.61569 mM = − + 
 47.956 N m= − ⋅ 
 or 48.0 N m= ⋅M W 
(b) 54.0 N m= ⋅M 
 Moment of couple due to horizontal forces at and M A D= 
 Moment of force-couple systems at and about .C F C+ 
 ( )54.0 N m 45 N 0.2 m 2 0.025 ma − ⋅ = − + +  
 ( ) ( )0.2 m 2C F x yM M F a F a + + + + +  
 where ( )( )45 N 0.025 m 1.125 N mCM = − = − ⋅ 
1.125 N mF CM M= = − ⋅ 
 
 
 
 
 
PROBLEM 3.71 CONTINUED 
45 N
2x
F −= 
45 N
2y
F −= 
( ) 54.0 N m 45 N 0.25 m 1.125 N m 1.125 N ma∴ − ⋅ = − + − ⋅ − ⋅ 
 ( ) ( )45 N 45 N0.2 m 2
2 2
a a− + − 
0.20 21.20 0.25 0.025 0.025
2 2 2
a aa= + + + + + + 
 3.1213 0.75858a = 
0.24303 ma = 
 or 243 mma = WPROBLEM 3.72 
The shafts of an angle drive are acted upon by the two couples shown. 
Replace the two couples with a single equivalent couple, specifying its 
magnitude and the direction of its axis. 
 
SOLUTION 
 
 
 
 
Based on 1 2= +M M M 
where 
( )1 8 N m= − ⋅M j 
( )2 6 N m= − ⋅M k 
( ) ( ) 8 N m 6 N m∴ = − ⋅ − ⋅M j k 
and ( ) ( )2 28 6 10 N m= + = ⋅M 
 or 10.00 N mM = ⋅ W 
( ) ( )8 N m 6 N m 0.8 0.6
10 N m
− ⋅ − ⋅= = = − −⋅
j kM j k
M
λ 
or ( )( )10 N m 0.8 0.6= = ⋅ − −M M j kλ 
 cos 0xθ = 90xθ∴ = ° 
cos 0.8 143.130y yθ θ= − ∴ = ° 
 cos 0.6 126.870z zθ θ= − ∴ = ° 
 or 90.0 , 143.1 , 126.9x y zθ θ θ= ° = ° = °W 
 
 
 
 
 
PROBLEM 3.73 
Knowing that P = 0, replace the two remaining couples with a single 
equivalent couple, specifying its magnitude and the direction of its axis. 
 
SOLUTION 
 
 
 
 
Have 1 2= +M M M 
where 1 / 1C B C=M r P× 
( ) ( )/ 0.96 m 0.40 mC B = −r i j 
( )1 100 NC = −P k 
( ) ( )1 0.96 0.40 0 40 N m 96 N m
0 0 100
∴ = − = ⋅ + ⋅
−
i j k
M i j 
Also, 2 / 2D A E=M r P× 
( ) ( )/ 0.20 m 0.55 mD A = −r j k 
 2 2E ED EP=P λ 
 ( ) ( )( ) ( ) ( )2 2
0.48 m 0.55 m
146 N
0.48 0.55 m
− +=
+
i k
 
 ( ) ( )96 N 110 N= − +i k 
 2 0 0.20 0.55 N m
96 0 110
∴ = − ⋅
−
i j k
M 
 ( ) ( ) ( )22.0 N m 52.8 N m 19.2 N m= ⋅ + ⋅ + ⋅i j k 
 
 
 
 
 
 
PROBLEM 3.73 CONTINUED 
and ( ) ( ) ( )40 N m 96 N m 22.0 N m  = ⋅ + ⋅ + ⋅  M i j i 
 ( ) ( )52.8 N m 19.2 N m + ⋅ + ⋅ j k 
 ( ) ( ) ( )62.0 N m 148.8 N m 19.2 N m= ⋅ + ⋅ + ⋅i j k 
( ) ( ) ( )2 2 22 2 2 62.0 148.8 19.2x y zM M M= + + = + +M 
 162.339 N m= ⋅ 
 or 162.3 N mM = ⋅ W 
 62.0 148.8 19.2
162.339
+ += =M i j k
M
λ 
 0.38192 0.91660 0.118271= + +i j k 
cos 0.38192 67.547x xθ θ= ∴ = ° 
 or 67.5xθ = °W 
cos 0.91660 23.566y yθ θ= ∴ = ° 
 or 23.6yθ = °W 
cos 0.118271 83.208z zθ θ= ∴ = ° 
 or 83.2zθ = °W 
 
 
 
 
 
 
PROBLEM 3.74 
Knowing that P = 0, replace the two remaining couples with a single 
equivalent couple, specifying its magnitude and the direction of its axis. 
 
SOLUTION 
 
 
 
 
Have 4 7= +M M M 
where 4 / 4G C G=M r F× 
( )/ 10 in.G C = −r i 
( )4 4 lbG =F k 
( ) ( ) ( )4 10 in. 4 lb 40 lb in.∴ = − = ⋅M i k j× 
Also, 7 / 7D F D=M r F× 
( ) ( )/ 5 in. 3 in.D F = − +r i j 
 7 7D ED DF=F λ 
 ( ) ( ) ( )( ) ( ) ( ) ( )2 2 2
5 in. 3 in. 7 in.
7 lb
5 3 7 in.
− + +=
+ +
i j k
 
 ( )7 lb 5 3 7
83
= − + +i j k 
( )7 7 lb in. 7 lb in. 5 3 0 21 35 083 835 3 7
⋅ ⋅∴ = − = + +
−
i j k
M i j k 
 ( )0.76835 21 35 lb in.= + ⋅i j 
 
 
 
 
 
 
PROBLEM 3.74 CONTINUED 
and ( ) ( )40 lb in. 0.76835 21 35 lb in.   = ⋅ + + ⋅   M j i j 
 ( ) ( )16.1353 lb in. 66.892 lb in.= ⋅ + ⋅i j 
( ) ( ) ( ) ( )22 2 216.1353 66.892x yM M= + = +M 
 68.811 lb in.= ⋅ 
 or 68.8 lb in.M = ⋅ W 
 ( ) ( )16.1353 lb in. 66.892 lb in.
68.811 lb in.
⋅ + ⋅= = ⋅
i jM
M
λ 
 0.23449 0.97212= +i j 
cos 0.23449 76.438x xθ θ= ∴ = ° 
 or 76.4xθ = °W 
 cos 0.97212 13.5615y yθ θ= ∴ = ° 
 or 13.56yθ = °W 
 cos 0.0zθ = 90zθ∴ = ° 
 or 90.0zθ = °W 
 
 
 
 
 
 
PROBLEM 3.75 
Knowing that P = 5 lb, replace the three couples with a single equivalent 
couple, specifying its magnitude and the direction of its axis. 
 
SOLUTION 
Have 4 7 5= + +M M M M 
where 
( )4 / 4 10 0 0 lb in. 40 lb in.
0 0 4
G C G= = − ⋅ = ⋅
i j k
M r F j× 
( )7 / 7 75 3 0 lb in. 0.76835 21 35 lb in.835 3 7D F D
 = = − ⋅ = + ⋅  −
i j k
M r F i j× 
(See Solution to Problem 3.74.) 
( ) ( )5 / 5 10 6 7 lb in. 35 lb in. 50 lb in.
0 5 0
C A C= = − ⋅ = − ⋅ + ⋅
i j k
M r F i k× 
( ) ( ) ( ) 16.1353 35 40 26.892 50 lb in. ∴ = − + + + ⋅ M i j k 
 ( ) ( ) ( )18.8647 lb in. 66.892 lb in. 50 lb in.= − ⋅ + ⋅ + ⋅i j k 
( ) ( ) ( )2 2 22 2 2 18.8647 66.892 50 85.618 lb in.x y zM M M= + + = + + = ⋅M 
 or 85.6 lb in.M = ⋅ W 
18.8647 66.892 50 0.22034 0.78129 0.58399
85.618
− + += = = − + +M i j k i j k
M
λ 
 cos 0.22034xθ = − 102.729xθ∴ = ° or 102.7xθ = °W 
 cos 0.78129 38.621y yθ θ= ∴ = ° or 38.6yθ = °W 
 cos 0.58399 54.268z zθ θ= ∴ = ° or 54.3zθ = °W 
 
 
 
 
 
PROBLEM 3.76 
Knowing that P = 210 N, replace the three couples with a single 
equivalent couple, specifying its magnitude and the direction of its axis. 
 
SOLUTION 
Have 1 2 P= + +M M M M 
where ( ) ( )1 / 1 0.96 0.40 0 40 N m 96 N m
0 0 100
C B C= = − = ⋅ + ⋅
−
i j k
M r P i j× 
( ) ( ) ( )2 / 2 0 0.20 0.55 22.0 N m 52.8 N m 19.2 N m
96 0 110
D A E= = − = ⋅ + ⋅ + ⋅
−
i j k
M r P i j k× 
(See Solution to Problem 3.73.) 
( ) ( )/ 0.48 0.20 1.10 231 N m 100.8 N m
0 210 0
P E A E= = − = ⋅ + ⋅
i j k
M r P i k× 
( ) ( ) ( ) 40 22 231 96 52.8 19.2 100.8 N m ∴ = + + + + + + ⋅ M i j k 
 ( ) ( ) ( )293 N m 148.8 N m 120 N m= ⋅ + ⋅ + ⋅i j k 
( ) ( ) ( )2 2 22 2 2 293 148.8 120 349.84 N mx y zM M M= + + = + + = ⋅M 
 or 350 N mM = ⋅ W 
293 148.8 120 0.83752 0.42533 0.34301
349.84
+ += = = + +M i j k i j k
M
λ 
 cos 0.83752 33.121x xθ θ= ∴ = ° or 33.1xθ = °W 
 cos 0.42533 64.828y yθ θ= ∴ = ° or 64.8yθ = °W 
 cos 0.34301 69.940z zθ θ= ∴ = ° or 69.9zθ = °W 
 
 
 
 
 
 
PROBLEM 3.77 
In a manufacturing operation, three holes are drilled simultaneously in a 
workpiece. Knowing that the holes are perpendicular to the surfaces of 
the workpiece, replace the couples applied to the drills with a single 
equivalent couple, specifying its magnitude and the direction of its axis. 
 
SOLUTION 
 
 
 
 
Have 1 2 3= + +M M M M 
where ( )( )1 1.1 lb ft cos 25 sin 25= − ⋅ ° + °M j k 
( )2 1.1 lb ft= − ⋅M j 
( )( )3 1.3 lb ft cos 20 sin 20= − ⋅ ° − °M j k 
( ) ( ) 0.99694 1.1 1.22160 0.46488 0.44463∴ = − − − + − +M j k 
 ( ) ( )3.3185 lb ft 0.020254 lb ft= − ⋅ − ⋅j k 
and ( ) ( ) ( )2 2 22 2 2 0 3.3185 0.020254x y zM M M= + + = + +M 
 3.3186 lb ft= ⋅ 
 or 3.32 lb ftM = ⋅ W 
( )0 3.3185 0.020254
3.3186
− −= = i j kM
M
λ 
 0.99997 0.0061032= − −j k 
 cos 0xθ = 90xθ∴ = ° or 90.0xθ = °W 
 cos 0.99997yθ = − 179.555yθ∴ = ° or 179.6yθ = °W 
 cos 0.0061032 90.349z zθ θ= − ∴ = ° or 90.3zθ = °W 
 
 
 
 
 
 
PROBLEM 3.78 
The tension in the cable attached to the end C of an adjustable boom ABC 
is 1000 N. Replace the force exerted by the cable at C with an equivalent 
force-couple system (a) at A, (b) at B. 
 
SOLUTION 
 
 
 
 
 (a) Based on : 1000 NAF F TΣ = = 
 or 1000 NA =F 20°W 
 ( )( ): sin 50A A AM M T dΣ = ° 
 ( ) ( )1000 N sin 50 2.25 m= ° 
 1723.60 N m= ⋅ 
 or 1724 N mA = ⋅M W 
(b) Based on : 1000 NBF F TΣ = = 
 or 1000 NB =F 20°W 
 ( )( ): sin 50B B BM M T dΣ = ° 
 ( ) ( )1000 N sin 50 1.25 m= ° 
 957.56 N m= ⋅ 
 or 958 N mB = ⋅M W 
 
 
 
 
 
 
PROBLEM 3.79 
The 20-lb horizontal force P acts on a bell crank as shown. (a) Replace P 
with an equivalent force-couple system at B. (b) Find the two vertical 
forces at C and D which are equivalent to the couple found in part a. 
 
SOLUTION 
 
 
 
 
 
 
 (a) Based on : 20 lbBF P PΣ = = 
 or 20 lbB =P W 
 : B BM M PdΣ = 
 ( )20 lb 5 in.= 
 100 lb in.= ⋅ 
 or 100 lb in.B = ⋅M W 
(b) If the two vertical forces are to be equivalent to ,BM they must be a 
 couple. Further, the sense of the moment of this couple must be 
 counterclockwise. 
 Then, with CP and DP acting as shown, 
: D CM M P dΣ = 
( )100 lb in. 4 in.CP⋅ = 
 25 lbCP∴ =or 25 lbC =P W 
: 0y D CF P PΣ = − 
 25 lbDP∴ = 
 or 25 lbD =P W 
 
 
 
 
 
 
PROBLEM 3.80 
A 700-N force P is applied at point A of a structural member. Replace P 
with (a) an equivalent force-couple system at C, (b) an equivalent system 
consisting of a vertical force at B and a second force at D. 
 
SOLUTION 
 
 
 
 
 
 (a) Based on : 700 NCF P PΣ = = 
 or 700 NC =P 60°W 
:C C x Cy y CxM M P d P dΣ = − + 
 where ( )700 N cos60 350 NxP = ° = 
( )700 N sin 60 606.22 NyP = ° = 
1.6 mCxd = 
1.1 mCyd = 
( )( ) ( )( ) 350 N 1.1 m 606.22 N 1.6 mCM∴ = − + 
 385 N m 969.95 N m= − ⋅ + ⋅ 
 584.95 N m= ⋅ 
 or 585 N mC = ⋅M W 
(b) Based on : cos60x DxF P PΣ = ° 
 ( )700 N cos60= ° 
 350 N= 
( )( ) ( ): cos60D DA B DBM P d P dΣ ° = 
( ) ( ) ( )700 N cos60 0.6 m 2.4 mBP ° =  
87.5 NBP = 
 or 87.5 NB =P W 
 
 
 
 
 
PROBLEM 3.80 CONTINUED 
: sin 60y B DyF P P PΣ ° = + 
( )700 N sin 60 87.5 N DyP° = + 
518.72 NDyP = 
 ( ) ( )22D Dx DyP P P= + 
 ( ) ( )2 2350 518.72 625.76 N= + = 
1 1 518.72tan tan 55.991
350
Dy
Dx
P
P
θ − −   = = = °      
 or 626 NDP = 56.0°W 
 
 
 
 
 
 
PROBLEM 3.81 
A landscaper tries to plumb a tree by applying a 240-N force as shown. 
Two helpers then attempt to plumb the same tree, with one pulling at B 
and the other pushing with a parallel force at C. Determine these two 
forces so that they are equivalent to the single 240-N force shown in the 
figure. 
 
SOLUTION 
 
 
 
 
Based on 
( ): 240 N cos30 cos cosx B CF F Fα αΣ − ° = − − 
or ( ) ( )cos 240 N cos30B CF F α− + = − ° (1) 
( ): 240 N sin30 sin siny B CF F Fα αΣ ° = + 
or ( ) ( )sin 240 N sin 30B CF F α+ = ° (2) 
From 
( )
Equation (2) : tan tan 30
Equation 1
α = ° 
 30α∴ = ° 
Based on 
( ) ( ) ( ) ( )( ): 240 N cos 30 20 0.25 m cos10 0.60 mC BM F Σ ° − ° = °  
 100 NBF∴ = 
 or 100.0 NB =F 30°W 
From Equation (1), ( )100 N cos30 240cos30CF− + ° = − ° 
140 NCF = 
 or 140.0 NC =F 30°W 
 
 
 
 
 
 
PROBLEM 3.82 
A landscaper tries to plumb a tree by applying a 240-N force as shown. 
(a) Replace that force with an equivalent force-couple system at C. (b) 
Two helpers attempt to plumb the same tree, with one applying a 
horizontal force at C and the other pulling at B. Determine these two 
forces if they are to be equivalent to the single force of part a. 
 
SOLUTION 
 
 
 
 
 
 (a) Based on ( ): 240 N cos30 cos30x CF FΣ − ° = − ° 
 240 NCF∴ = 
 or 240 NC =F 30°W 
( ) ( ): 240 N cos10 0.25 mC A C AM d M d Σ ° = =  
 59.088 N mCM∴ = ⋅ 
 or 59.1 N mC = ⋅M W 
(b) Based on ( ): 240 N sin30 siny BF F αΣ ° = 
 or sin 120BF α = (1) 
( ) ( ) ( ): 59.088 N m 240 N cos10 cos 20B C C CM d F d Σ ⋅ − ° = − °  
( ) ( ) ( )59.088 N m 240 N cos10 0.60 m 0.60 m cos 20CF   ⋅ − ° = − °    
0.56382 82.724CF = 
146.722 NCF = 
 or 146.7 NC =F W 
 and ( ): 240 N cos30 146.722 N cosx BF F αΣ − ° = − − 
 cos 61.124BF α = (2) 
 From 
Equation (1) :
Equation (2)
120tan 1.96323
61.124
α = = 
 63.007α = ° or 63.0α = °W 
 From Equation (1), 120 134.670 N
sin 63.007B
F = =° 
 or 134.7 NB =F 63.0°W 
 
 
 
 
 
PROBLEM 3.83 
A dirigible is tethered by a cable attached to its cabin at B. If the tension 
in the cable is 250 lb, replace the force exerted by the cable at B with an 
equivalent system formed by two parallel forces applied at A and C. 
 
SOLUTION 
 
 
 
 
Require the equivalent forces acting at A and C be parallel and at an angle 
ofα with the vertical. 
Then for equivalence, 
 ( ): 250 lb sin 30 sin sinx A BF F Fα αΣ ° = + (1) 
 ( ): 250 lb cos30 cos cosy A BF F Fα αΣ − ° = − − (2) 
Dividing Equation (1) by Equation (2), 
( )
( )
( )
( )
250 lb sin 30 sin
250 lb cos30 cos
A B
A B
F F
F F
α
α
° +=− ° − + 
Simplifying yields 30α = ° 
Based on 
( ) ( ) ( )( ): 250 lb cos30 12 ft cos30 32 ftC AM F Σ ° = °  
 93.75 lbAF∴ = 
 or 93.8 lbA =F 60°W 
Based on 
( ) ( ) ( )( ): 250 lb cos30 20 ft cos30 32 ftA CM F Σ − ° = °  
 156.25 lbCF∴ = 
 or 156.3 lbC =F 60°W 
 
 
 
 
 
 
PROBLEM 3.84 
Three workers trying to move a 3 3 4-ft× × crate apply to the crate the 
three horizontal forces shown. (a) If 60 lb,P = replace the three forces 
with an equivalent force-couple system at A. (b) Replace the force-couple 
system of part a with a single force, and determine where it should be 
applied to side AB. (c) Determine the magnitude of P so that the three 
forces can be replaced with a single equivalent force applied at B. 
 
SOLUTION 
 
(a) 
 
(b) 
 
(c) 
 
 (a) Based on 
: 50 lb 50 lb 60 lbz AF FΣ − + + = 
60 lbAF = 
 ( )or 60.0 lbA =F kW 
 Based on 
( )( ) ( )( ): 50 lb 2 ft 50 lb 0.6 ftA AM MΣ − = 
70 lb ftAM = ⋅ 
 ( )or 70.0 lb ftA = ⋅M jW 
(b) Based on 
: 50 lb 50 lb 60 lbzF FΣ − + + = 
60 lbF = 
 ( )or 60.0 lb=F kW 
 Based on 
( ): 70 lb ft 60 lbAM xΣ ⋅ = 
1.16667 ftx = 
 or 1.167 ft from along x A AB= W 
(c) Based on 
( ) ( ) ( ) ( ) ( ) ( ): 50 lb 1 ft 50 lb 2.4 ft 3 ft 0BM P RΣ − + − = 
70 23.333 lb
3
P = = 
 or 23.3 lbP = W 
 
 
 
 
 
 
PROBLEM 3.85 
A force and a couple are applied to a beam. (a) Replace this system with 
a single force F applied at point G, and determine the distance d. 
(b) Solve part a assuming that the directions of the two 600-N forces are 
reversed. 
 
SOLUTION 
(a) 
 Have :y C D EF F F F FΣ + + = 
800 N 600 N 600 NF = − + − 
 800 NF = − or 800 N=F W 
 Have ( ) ( ): 1.5 m 2 m 0G C DM F d FΣ − − = 
( )( ) ( )( )800 N 1.5 m 600 N 2 m 0d − − = 
1200 1200
800
d += 
 3 md = or 3.00 md = W 
(b) 
 Changing directions of the two 600 N forces only changes sign of the couple. 
 800 NF∴ = − or 800 N=F W 
 and ( ) ( ): 1.5 m 2 m 0G C DM F d FΣ − + = 
( )( ) ( )( )800 N 1.5 m 600 N 2 md − + 
1200 1200 0
800
d −= = 
 or 0d = W 
 
 
 
 
 
 
PROBLEM 3.86 
Three cables attached to a disk exert on it the forces shown. (a) Replace 
the three forces with an equivalent force-couple system at A. 
(b) Determine the single force which is equivalent to the force-couple 
system obtained in part a, and specify its point of application on a line 
drawn through points A and D. 
 
SOLUTION 
 
(a) Have : B C D AΣ + + =F F F F F 
 Since B D= −F F 
 110 NA C∴ = =F F 20° 
 or 110.0 NA =F 20.0°W 
 Have ( ) ( ) ( ):A BT CT DT AM F r F r F r MΣ − − + = 
( ) ( ) ( ) ( ) ( ) ( )140 N sin15 0.2 m 110 N sin 25 0.2 m 140 N sin 45 0.2 m AM     − ° − ° + ° =      
3.2545 N mAM = ⋅ 
 or 3.25 N mA = ⋅M W 
(b) Have : A EΣ =F F F or 110.0 NE =F 20.0°W 
[ ]( ): cos 20A EM M F aΣ = ° 
( ) ( ) 3.2545 N m 110 N cos 20 a ∴ ⋅ = °  
0.031485 ma = 
 or 31.5 mm below a A= W 
 
 
 
 
 
 
PROBLEM 3.87 
While tapping a hole, a machinist applies the horizontal forces shown to 
the handle of the tap wrench. Show that these forces are equivalent to a 
single force, and specify, if possible, the point of application of the single 
force on the handle. 
 
SOLUTION 
Since the forces at A and B are parallel, the force at B can be replaced with the sum of two forces with one of 
the forces equal in magnitude to the force at A except with an opposite sense, resulting in a force-couple. 
Have 26.5 N 2.5 N,BF = + where the 26.5 N force be part of the couple. Combining the two parallel forces, 
( ) ( )couple 26.5 N 0.080 m 0.070 m cos 25M  = + °  
 3.60 N m= ⋅ 
coupleand, 3.60 N m= ⋅M 
 
 
A single equivalentforce will be located in the negative z-direction. 
Based on ( ) ( ): 3.60 N m 2.5 N cos 25BM a Σ − ⋅ = °  
1.590 ma = − 
( )( )2.5 N cos 25 sin 25′ = ° + °F i j 
 and is applied on an extension of handle BD 
 at a distance of 1.590 m to the right of B W 
 
 
 
 
 
PROBLEM 3.88 
A rectangular plate is acted upon by the force and couple shown. This 
system is to be replaced with a single equivalent force. (a) For 40 ,α = ° 
specify the magnitude and the line of action of the equivalent force. 
(b) Specify the value of α if the line of action of the equivalent force is 
to intersect line CD 12 in. to the right of D. 
 
SOLUTION 
 
 
 
 
 
 
 
 
 (a) Have ( ) ( ): 3 lb sin 40 3 lb sin 40x xF FΣ − ° + ° = 
 0xF∴ = 
 Have ( ) ( ): 3 lb cos40 10 lb 3 lb cos40y yF FΣ − ° − + ° = 
 10 lbyF∴ = − 
 or 10.00 lbF = W 
 Note: The two 3-lb forces form a couple 
 and / / /:A C A C B A B X AΣ + =M r P r P r F× × × 
3 16 10 0 160 1 0 0 10 0 0
sin 40 cos 40 0 0 1 0 0 1 0
d− + =
° ° − −
i j k i j k i j k
 
( ) ( ): 3 16 cos 40 10 3sin 40 160 10d° − − ° − = −k 
36.770 19.2836 160 10d+ − = − 
 10.3946 in.d∴ = 
 or 10.00 lb=F at 10.39 in. right of A or at 5.61 in. left of B W 
(b) From part (a), 10.00 lb=F 
 Have ( )/ /: 12 in.A C A C B A BΣ + =M r P r P i F× × × 
3 16 10 0 160 1 0 0 120 1 0 0
sin cos 0 0 1 0 0 1 0α α
− + =
− −
i j k i j k i j k
 
: 48cos 30sin 160 120α α+ − = −k 
24cos 20 15sinα α= − 
 
 
 
PROBLEM 3.88 CONTINUED 
 Squaring both sides of the equation, and 
 using the identity 2 2cos 1 sin ,α α= − results in 
2sin 0.74906sin 0.21973 0α α− − = 
 Using quadratic formula 
sin 0.97453 sin = 0.22547α α= − 
 
 so that 
 77.0 and 13.03α α= ° = − °W 
 
 
 
 
 
PROBLEM 3.89 
A hexagonal plate is acted upon by the force P and the couple shown. 
Determine the magnitude and the direction of the smallest force P for 
which this system can be replaced with a single force at E. 
 
SOLUTION 
 
 
 
 
Since the minimum value of P acting at B is realized when minP is 
perpendicular to a line connecting B and E, 30α = ° 
Then, 
/ min /: 0E B E D A DΣ + =M r P r P× × 
where 
( ) ( )/ 0.30 m 2 0.30 m cos30B E  = − + ° r i j 
 ( ) ( )0.30 m 0.51962 m= − +i j 
( )/ 0.30 m 2 0.3 m sin 30D A  = + ° r i 
 ( )0.60 m= i 
( )450 ND =P j 
( ) ( )min min cos30 sin 30P  = ° + ° P i j 
min 0.30 0.51962 0 0.60 0 0 N m 0
0.86603 0.50 0 0 450 0
P∴ − + ⋅ =
i j k i j k
 
( ) ( )min 0.15 m 0.45 m 270 N m 0P − − + ⋅ =k k 
min 450 NP∴ = 
 minor 450 N=P 30°W 
 
 
 
 
 
PROBLEM 3.90 
An eccentric, compressive 270-lb force P is applied to the end of a 
cantilever beam. Replace P with an equivalent force-couple system at G. 
 
SOLUTION 
 
 
 
Have 
( ): 270 lbΣ − =F i F 
 ( ) 270 lb∴ = −F iW 
Also, have 
/:G A GΣ =M r P M× 
270 0 4 2.4 lb in.
1 0 0
− − ⋅ =
−
i j k
M 
( ) ( )( ) ( )( ) 270 lb in. 2.4 1 4 1 ∴ = ⋅ − − − − − M j k 
 ( ) ( )or 648 lb in. 1080 lb in.= ⋅ − ⋅M j kW 
 
 
 
 
 
PROBLEM 3.91 
Two workers use blocks and tackles attached to the bottom of an I-beam 
to lift a large cylindrical tank. Knowing that the tension in rope AB is 
324 N, replace the force exerted at A by rope AB with an equivalent 
force-couple system at E. 
 
SOLUTION 
 
 
 
Have : ABΣ =F T F 
where 
 AB AB ABT=T λ 
 ( ) ( ) ( ) ( )0.75 m 6.0 m 3.0 m 324 N
6.75 m
− += i j k 
( ) 36 N 8 4AB∴ = − +T i j k 
so that ( ) ( ) ( )36.0 N 288 N 144.0 N= − +F i j kW 
Have 
/:E A E ABΣ =M r T M× 
or ( )( )7.5 m 36 N 0 1 0
1 8 4
=
−
i j k
M 
( )( ) 270 N m 4∴ = ⋅ −M i k 
 ( ) ( )or 1080 N m 270 N m= ⋅ − ⋅M i kW 
 
 
 
 
 
PROBLEM 3.92 
Two workers use blocks and tackles attached to the bottom of an I-beam 
to lift a large cylindrical tank. Knowing that the tension in rope CD is 
366 N, replace the force exerted at C by rope CD with an equivalent 
force-couple system at O. 
 
SOLUTION 
 
 
 
Have : CDΣ =F T F 
where 
 CD CD CDT=T λ 
 ( ) ( ) ( ) ( )0.3 m 5.6 m 2.4 m 366 N
6.1 m
− − += i j k 
( )( ) 6.0 N 3 56 24CD∴ = − − +T i j k 
so that ( ) ( ) ( )18.00 N 336 N 144.0 N= − − +F i j kW 
Have 
/:O C O CDΣ =M r T M× 
or ( )( )7.5 m 6 N 0 1 0
3 56 24
=
− −
i j k
M 
( )( ) 45 N m 24 3∴ = ⋅ +M i k 
 ( ) ( )or 1080 N m 135.0 N m= ⋅ + ⋅M i kW 
 
 
 
 
 
PROBLEM 3.93 
To keep a door closed, a wooden stick is wedged between the floor and 
the doorknob. The stick exerts at B a 45-lb force directed along line AB. 
Replace that force with an equivalent force-couple system at C. 
 
SOLUTION 
 
 
 
 
Have 
: AB CΣ =F P F 
where 
 AB AB ABP=P λ 
 ( ) ( ) ( ) ( )2.0 in. 38 in. 24 in. 45 lb
44.989 in.
+ −= i j k 
 ( ) ( ) ( )or 2.00 lb 38.0 lb 24.0 lbC = + −F i j k W 
Have 
/:C B C AB CΣ =M r P M× 
 2 29.5 33 0 lb in.
1 19 12
C = − ⋅
−
i j k
M 
 ( ) ( )( ) ( )( ){2 lb in. 33 12 29.5 12= ⋅ − − − −i j 
 ( )( ) ( )( ) }29.5 19 33 1 + − − k 
 ( ) ( ) ( )or 792 lb in. 708 lb in. 1187 lb in.C = ⋅ + ⋅ + ⋅M i j k W 
 
 
 
 
 
PROBLEM 3.94 
A 25-lb force acting in a vertical plane parallel to the yz plane is applied 
to the 8-in.-long horizontal handle AB of a socket wrench. Replace the 
force with an equivalent force-couple system at the origin O of the 
coordinate system. 
 
SOLUTION 
 
 
 
Have 
: BΣ =F P F 
where 
( ) ( )25 lb sin 20 cos20B  = − ° + ° P j k 
 ( ) ( )8.5505 lb 23.492 lb= − +j k 
 ( ) ( )or 8.55 lb 23.5 lb= − +F j k W 
Have 
/:O B O B OΣ =M r P M× 
where 
( ) ( ) ( )/ 8cos30 15 8sin 30 in.B O  = ° + − ° r i j k 
 ( ) ( ) ( )6.9282 in. 15 in. 4 in.= + −i j k 
 6.9282 15 4 lb in.
0 8.5505 23.492
O∴ − ⋅ =
−
i j k
M 
( ) ( ) ( )318.18 162.757 59.240 lb in.O  = − − ⋅ M i j k 
 ( ) ( ) ( )or 318 lb in. 162.8 lb in. 59.2 lb in.O = ⋅ − ⋅ − ⋅M i j kW 
 
 
 
 
 
PROBLEM 3.95 
A 315-N force F and 70-N · m couple M are applied to corner A of the 
block shown. Replace the given force-couple system with an equivalent 
force-couple system at corner D. 
 
SOLUTION 
 
 
 
 
Have 
 : DΣ =F F F 
 AI F= λ 
 ( ) ( ) ( ) ( )0.360 m 0.120 m 0.180 m 315 N
0.420 m
− += i j k 
 ( )( )750 N 0.360 0.120 0.180= − +i j k 
 ( ) ( ) ( )or 270 N 90.0 N 135.0 ND = − +F i j kW 
Have 
/:D I D DΣ + =M M r F M× 
where 
 ACM=M λ 
 ( ) ( ) ( )0.240 m 0.180 m 70.0 N m
0.300 m
−= ⋅i k 
 ( )( )70.0 N m 0.800 0.600= ⋅ −i k 
( )/ 0.360 mI D =r k 
( )( ) ( ) 70.0 N m 0.8 0.6 0 0 0.36 750 N m
0.36 0.12 0.18
D∴ = ⋅ − + ⋅
−
i j k
M i k 
 ( ) ( )56.0 N m 42.0 N m= ⋅ − ⋅i k ( ) ( )32.4 N m 97.2 N m + ⋅ + ⋅ i j 
 ( ) ( ) ( )or 88.4 N m 97.2 N m 42.0 N mD = ⋅ + ⋅ − ⋅M i j k W 
 
 
 
 
 
PROBLEM 3.96 
The handpiece of a miniature industrial grinder weighs 2.4 N, and its 
center of gravity is located on the y axis. The head of the handpiece is 
offset in the xz plane in such a way that line BC forms an angle of 25° 
with the x direction. Show that the weight of the handpiece and the two 
couples 1M and 2M can be replaced with a single equivalent force. 
Further assuming that 1 0.068 N mM = ⋅ and 2 0.065 N m,M = ⋅ 
determine (a) the magnitude and the direction of the equivalent force, 
(b) the point where its line of action intersects the xz plane. 
 
SOLUTION 
 
 
 
 
First assume that the given force W and couples 1M and 2M act at the 
origin. 
Now W= − jW 
and ( ) ( )1 2 2 1 2cos 25 sin 25M M M= + = − ° + − °M M M i k 
Note that since W and M are perpendicular,it follows that they can be 
replaced with a single equivalent force. 
(a) Have ( ) or 2.4 NF W= = − = −jW F j 
( )or 2.40 N= −F jW 
(b) Assume that the line of action of F passes through point P (x, 0, z). 
 Then for equivalence 
/P O= ×M r F 
 where /P O x z= +r i k 
 ( ) ( )2 1 2 cos25 sin 25M M M∴ − ° + − °i k 
 ( ) ( )0
0 0
x z Wz Wx
W
= = −
−
i j k
i k 
 
 
 
 
PROBLEM 3.96 CONTINUED 
 Equating the i and k coefficients, 
1 2cos25 sin 25 and zM M Mz x
W W
− ° − ° = = −   
(b) For 1 22.4 N, 0.068 N m, 0.065 N mW M M= = ⋅ = ⋅ 
0.068 0.065sin 25 0.0168874 m
2.4
x − °= = −− 
 or 16.89 mmx = − W 
0.065cos25 0.024546 m
2.4
z − °= = − 
 or 24.5 mmz = − W 
 
 
 
 
 
PROBLEM 3.97 
A 20-lb force 1F and a 40- lb ft⋅ couple 1M are applied to corner E of the 
bent plate shown. If 1F and 1M are to be replaced with an equivalent 
force-couple system ( )2 2, F M at corner B and if ( )2 0,zM = determine 
(a) the distance d, (b) 2F and 2.M 
 
SOLUTION 
 
(a) Have 2: 0Bz zM MΣ = 
 ( )/ 1 1 0H B zM+ =k r F⋅ × (1) 
 where ( ) ( )/ 31 in. 2 in.H B = −r i j 
1 1EH F=F λ 
 ( ) ( ) ( ) ( )6 in. 6 in. 7 in. 20 lb
11.0 in.
+ −= i j k 
 ( )20 lb 6 6 7
11.0
= + −i j k 
1 1zM = k M⋅ 
 1 1EJ M=M λ 
 ( ) ( ) ( )
2
3 in. 7 in.
480 lb in.
58 in.
d
d
− + −= ⋅
+
i j k
 
 Then from Equation (1), 
( )( )
2
0 0 1
7 480 lb in.20 lb in.31 2 0 0
11.0 586 6 7 d
− ⋅⋅− + =
+−
 
 
 PROBLEM 3.97 CONTINUED 
 Solving for d, Equation (1) reduces to 
( )
2
20 lb in. 3360 lb in.186 12 0
11.0 58d
⋅ ⋅+ − =
+
 
 From which 5.3955 in.d = 
or 5.40 in.d = W 
(b) ( )2 1 20 lb 6 6 711.0= = + −F F i j k 
 ( )10.9091 10.9091 12.7273 lb= + −i j k 
( ) ( ) ( )2or 10.91 lb 10.91 lb 12.73 lb= + −F i j k W 
 2 / 1 1H B= +M r F M× 
 ( ) ( )5.3955 3 720 lb in.31 2 0 480 lb in.
11.0 9.3333
6 6 7
− + −⋅= − + ⋅
−
i j k
i j k
 
 ( )25.455 394.55 360 lb in.= + + ⋅i j k 
 ( )277.48 154.285 360 lb in.+ − + − ⋅i j k 
( ) ( )2 252.03 lb in. 548.84 lb in.= − ⋅ + ⋅M i j 
( ) ( )2or 21.0 lb ft 45.7 lb ft= − ⋅ + ⋅M i jW 
 
 
 
 
 
PROBLEM 3.98 
A 4-ft-long beam is subjected to a variety of loadings. (a) Replace each 
loading with an equivalent force-couple system at end A of the beam. 
(b) Which of the loadings are equivalent? 
 
SOLUTION 
 (a) 
 
 
 
 (a) Have : 200 lb 100 lby aF RΣ − − = 
 or 300 lba =R W 
 and ( )( ): 900 lb ft 100 lb 4 ftA aM MΣ ⋅ − = 
 or 500 lb fta = ⋅M W 
(b) Have : 300 lby bF RΣ − = 
 or 300 lbb =R W 
 and : 450 lb ftA bM MΣ − ⋅ = 
 or 450 lb ftb = ⋅M W 
(c) Have : 150 lb 450 lby cF RΣ − = 
 or 300 lbc =R W 
 and ( )( ): 2250 lb ft 450 lb 4 ftA cM MΣ ⋅ − = 
 or 450 lb ftc = ⋅M W 
(d) Have : 200 lb 400 lby dF RΣ − + = 
 or 200 lbd =R W 
 and ( )( ): 400 lb 4 ft 1150 lb ftA dM MΣ − ⋅ = 
 or 450 lb ftd = ⋅M W 
(e) Have : 200 lb 100 lby eF RΣ − − = 
 or 300 lbe =R W 
 and ( )( ): 100 lb ft 200 lb ft 100 lb 4 ftA eM MΣ ⋅ + ⋅ − = 
 or 100 lb fte = ⋅M W 
 
 
 
 
 
 
(b) 
PROBLEM 3.98 CONTINUED 
(f) Have : 400 lb 100 lby fF RΣ − + = 
 or 300 lbf =R W 
 and ( )( ): 150 lb ft 150 lb ft 100 lb 4 ftA fM MΣ − ⋅ + ⋅ + = 
 or 400 lb ftf = ⋅M W 
(g) Have : 100 lb 400 lby gF RΣ − − = 
 or 500 lbg =R W 
 and ( )( ): 100 lb ft 2000 lb ft 400 lb 4 ftA gM MΣ ⋅ + ⋅ − = 
 or 500 lb ftg = ⋅M W 
(h) Have : 150 lb 150 lby hF RΣ − − = 
 or 300 lbh =R W 
 and ( )( ): 1200 lb ft 150 lb ft 150 lb 4 ftA hM MΣ ⋅ − ⋅ − = 
 or 450 lb fth = ⋅M W 
 Therefore, loadings (c) and (h) are equivalent W 
 
 
 
 
 
 
PROBLEM 3.99 
A 4-ft-long beam is loaded as shown. Determine the loading of Problem 
3.98 which is equivalent to this loading. 
 
SOLUTION 
 
 
 
Have : 100 lb 200 lby RΣ − − =F 
 or 300 lb=R 
and ( )( ): 200 lb ft 1400 lb ft 200 lb 4 ftAM MΣ − ⋅ + ⋅ − = 
 or 400 lb ft= ⋅M 
 Equivalent to case (f) of Problem 3.98 W 
Problem 3.98 Equivalent force-couples at A 
 case R M 
 ( )a 300 lb 500 lb ft⋅ 
 ( )b 300 lb 450 lb ft⋅ 
 ( )c 300 lb 450 lb ft⋅ 
 ( )d 200 lb 450 lb ft⋅ 
 ( )e 300 lb 100 lb ft⋅ 
 ( )f 300 lb 400 lb ft⋅ W 
 ( )g 500 lb 500 lb ft⋅ 
 ( )h 300 lb 450 lb ft⋅ 
 
 
 
 
 
 
PROBLEM 3.100 
Determine the single equivalent force and the distance from point A to its 
line of action for the beam and loading of (a) Problem 3.98b, 
(b) Problem 3.98d, (c) Problem 3.98e. 
Problem 3.98: A 4-ft-long beam is subjected to a variety of loadings. 
(a) Replace each loading with an equivalent force-couple system at end A 
of the beam. (b) Which of the loadings are equivalent? 
 
SOLUTION 
 (a) 
 
(b) 
 
(c) 
 
 
For equivalent single force at distance d from A 
Have : 300 lbyF RΣ − = 
 or 300 lb=R W 
and ( )( ): 300 lb 450 lb ft 0CM dΣ − ⋅ = 
 or 1.500 ftd = W 
Have : 200 lb 400 lbyF RΣ − + = 
 or 200 lb=R W 
and ( )( ) ( )( ): 200 lb 400 lb 4 1150 lb ft 0CM d dΣ + − − ⋅ = 
 or 2.25 ftd = W 
Have : 200 lb 100 lbyF RΣ − − = 
 or 300 lb=R W 
and ( )( ) ( )( ): 100 lb ft 200 lb 100 lb 4 200 lb ft 0CM d dΣ ⋅ + − − + ⋅ = 
 or 0.333 ftd = W 
 
 
 
 
 
PROBLEM 3.101 
Five separate force-couple systems act at the corners of a metal block, 
which has been machined into the shape shown. Determine which of 
these systems is equivalent to a force ( )10 N=F j and a couple of 
moment ( ) ( )6 N m 4 N m= ⋅ + ⋅M i k located at point A. 
 
SOLUTION 
 
 
 
 
The equivalent force-couple system at A for each of the five force-couple 
systems will be determined. Each will then be compared to the given 
force-couple system to determine if they are equivalent. 
Force-couple system at B 
Have ( ): 10 NΣ =F j F 
or ( )10 N=F j 
and ( )/:A B B AΣ Σ + =M M r F M× 
( ) ( ) ( ) ( )4 N m 2 N m 0.2 m 10 N⋅ + ⋅ + =i k i j M× 
( ) ( )4 N m 4 N m= ⋅ + ⋅M i k 
 Comparing to given force-couple system at A, 
 Is Not Equivalent W 
Force-couple system at C 
Have ( ): 10 NΣ =F j F 
or ( )10 N=F j 
and ( )/:A C C AΣ + =M M r F M× 
( ) ( ) ( ) ( )8.5 N m 0.2 m 0.25 m 10 N ⋅ + + = i i k j M× 
( ) ( )6 N m 2.0 N m= ⋅ + ⋅M i k 
 Comparing to given force-couple system at A, 
 Is Not Equivalent W 
 
 
 
 
 
 
 
 
 
PROBLEM 3.101 CONTINUED 
Force-couple system at E 
Have ( ): 10 NΣ =F j F 
or ( )10 N=F j 
and ( )/:A E E AMΣ + =M r F M× 
( ) ( ) ( ) ( )6 N m 0.4 m 0.08 m 10 N ⋅ + − = i i j j M× 
( ) ( )6 N m 4 N m= ⋅ + ⋅M i k 
 Comparing to given force-couple system at A, 
 Is Equivalent W 
Force-couple system at G 
Have ( ) ( ): 10 N 10 NΣ + =F i j F 
or ( ) ( )10 N 10 N= +F i j 
F has two force components 
 force-couple system at G∴ 
 Is Not Equivalent W 
Force-couple system at I 
Have ( ): 10 NΣ =F j F 
or ( )10 N=F j 
and ( )/:A I I AΣ Σ + =M M r F M× 
 ( ) ( )10 N m 2 N m⋅ − ⋅i k 
 ( ) ( ) ( ) ( )0.4 m 0.2 m 0.4 m 10 N + − + = i j k j M× 
or ( ) ( )6 N m 2 N m= ⋅ + ⋅M i k 
 Comparing to given force-couple system at A, 
 Is Not EquivalentW 
 
 
 
 
 
PROBLEM 3.102 
The masses of two children sitting at ends A and B of a seesaw are 38 kg 
and 29 kg, respectively. Where should a third child sit so that the 
resultant of the weights of the three children will pass through C if she 
has a mass of (a) 27 kg, (b) 24 kg. 
 
SOLUTION 
 
 
 
 
First ( )38 kgA AW m g g= = 
 ( )29 kgB BW m g g= = 
(a) ( )27 kgC CW m g g= = 
 For resultant weight to act at C, 0CMΣ = 
 Then ( ) ( ) ( ) ( ) ( ) ( )38 kg 2 m 27 kg 29 kg2 m 0g g d g     − − =      
76 58 0.66667 m
27
d −∴ = = 
 or 0.667 md = W 
(b) ( )24 kgC CW m g g= = 
 For resultant weight to act at C, 0CMΣ = 
 Then ( ) ( ) ( ) ( ) ( ) ( )38 kg 2 m 24 kg 29 kg 2 m 0g g d g     − − =      
76 58 0.75 m
24
d −∴ = = 
 or 0.750 md = W 
 
 
 
 
 
PROBLEM 3.103 
Three stage lights are mounted on a pipe as shown. The mass of each 
light is 1.8 kgA Bm m= = and 1.6 kgCm = . (a) If 0.75d = m, 
determine the distance from D to the line of action of the resultant of the 
weights of the three lights. (b) Determine the value of d so that the 
resultant of the weights passes through the midpoint of the pipe. 
 
SOLUTION 
 
 
 
 
First ( )1.8 kgA B AW W m g g= = = 
 ( )1.6 kgC CW m g g= = 
(a) 0.75 md = 
 Have A B CR W W W= + + 
( )1.8 1.8 1.6 kgR g = + +  
 or ( )5.2 Ng=R 
 Have 
( ) ( ) ( ) ( ): 1.8 0.3 m 1.8 1.3 m 1.6 2.05 m 5.2DM g g g g DΣ − − − = − 
 1.18462 mD∴ = 
 or 1.185 mD = W 
(b) 1.25 m
2
LD = = 
 Have 
( )( ) ( )( ) ( )( ): 1.8 0.3 m 1.8 1.3 m 1.6 1.3 mDM g g g dΣ − − − + 
 ( )( )5.2 1.25 mg= − 
 0.9625 md∴ = 
 or 0.963 md = W 
 
 
 
 
 
PROBLEM 3.104 
Three hikers are shown crossing a footbridge. Knowing that the weights 
of the hikers at points C, D, and E are 800 N, 700 N, and 540 N, 
respectively, determine (a) the horizontal distance from A to the line of 
action of the resultant of the three weights when 1.1 m,a = (b) the value 
of a so that the loads on the bridge supports at A and B are equal. 
 
SOLUTION 
 
(a) 
 
(b) 
 
 
 (a) 1.1 ma = 
 Have : C D EF W W W RΣ − − − = 
 800 N 700 N 540 NR∴ = − − − 
2040 NR = 
 or 2040 N=R 
 Have 
( )( ) ( )( ) ( )( ): 800 N 1.5 m 700 N 2.6 m 540 N 4.25 mAMΣ − − − 
 ( )R d= − 
( ) 5315 N m 2040 N d∴ − ⋅ = − 
 and 2.6054 md = 
 or 2.61 m to the right of d A= W 
(b) For equal reaction forces at A and B, the resultant, ,R must act at the 
 center of the span. 
 From 
2A
LM R  Σ = −    
( )( ) ( )( ) ( )( ) 800 N 1.5 m 700 N 1.5 m 540 N 1.5 m 2.5a a∴ − − + − + 
 ( )( )2040 N 3 m= − 
3060 2050 6120a+ = 
 and 1.49268 ma = 
 or 1.493 ma = W 
 
 
 
 
 
PROBLEM 3.105 
Gear C is rigidly attached to arm AB. If the forces and couple shown can 
be reduced to a single equivalent force at A, determine the equivalent 
force and the magnitude of the couple M. 
 
SOLUTION 
 
For equivalence 
( ) ( ): 90 N sin30 125 N cos40x xF RΣ − ° + ° = 
 or 50.756 NxR = 
( ) ( ): 90 N cos30 200 N 125 N sin 40y yF RΣ − ° − − ° = 
 or 358.29 NyR = − 
Then ( ) ( )2 250.756 358.29 361.87 NR = + − = 
and 358.29tan 7.0591 81.937
50.756
y
x
R
R
θ θ−= = = − ∴ = − ° 
 or 362 N=R 81.9°W 
Also 
( ) ( ) ( ) ( ) ( ) ( ): 90 N sin 35 0.6 m 200 N cos 25 0.85 m 125 N sin 65 1.25 m 0AM M      Σ − ° − ° − ° =      
 326.66 N mM∴ = ⋅ 
 or 327 N mM = ⋅ W 
 
 
 
 
 
PROBLEM 3.106 
To test the strength of a 25 20-in.× suitcase, forces are applied as shown. 
If 18 lb,P = (a) determine the resultant of the applied forces, (b) locate 
the two points where the line of action of the resultant intersects the edge 
of the suitcase. 
 
SOLUTION 
 
 
(a) 18 lbP = 
 Have ( ) ( ) ( ) ( )42 lb: 20 lb + 3 2 18 lb 36 lb
13 x y
R RΣ − − + + + = +F i i j j i i j 
( ) ( )18.9461 lb 41.297 lb x yR R∴ − + = +i j i j 
 or ( ) ( )18.95 lb 41.3 lb= − +R i j 
( ) ( )2 22 2 18.9461 41.297 45.436 lbx yR R R= + = + = 
1 1 41.297tan tan 65.355
18.9461
y
x
x
R
R
θ − −   = = = − °   −  
 
 or 45.4 lb=R 65.4°W 
(b) Have B BΣ =M M 
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )42 lb4 in. 20 lb 21 in. 3 2 12 in. 36 lb 3 in. 18 lb
13B
 = − + − + + +  M j i i i j j i i j× × × × 
( ) 191.246 lb in.B∴ = ⋅M k 
 
 
 PROBLEM 3.106 CONTINUED 
Since B B=M r R× 
( ) ( )191.246 lb in. 0 41.297 18.9461
18.9461 41.297 0
x y x y∴ ⋅ = = +
−
i j k
k k 
For 191.2460, 4.6310 in.
41.297
y x= = = or 4.63 in.x = W 
For 191.2460, 10.0942 in.
18.9461
x y= = = or 10.09 in.y = W 
 
 
 
 
 
 
PROBLEM 3.107 
Solve Problem 3.106 assuming that 28 lb.P = 
Problem 3.106: To test the strength of a 25 20-in.× suitcase, forces are 
applied as shown. If 18 lb,P = (a) determine the resultant of the applied 
forces, (b) locate the two points where the line of action of the resultant 
intersects the edge of the suitcase. 
 
SOLUTION 
 
 
(a) P = 28 lb 
 Have ( ) ( ) ( ) ( )42: 20 lb 3 2 28 lb 36 lb
13 x y
R RΣ − + − + + + = +F i i j j i i j 
( ) ( )18.9461 lb 51.297 lb x yR R∴ − + = +i j i j 
 or ( ) ( )18.95 lb 51.3 lb= − +R i j 
( ) ( )2 22 2 18.9461 51.297 54.684 lbx yR R R= + = + = 
1 1 51.297tan tan 69.729
18.9461
y
x
x
R
R
θ − −   = = = − °   −  
 
 or 54.7 lb=R 69.7°W 
(b) Have B BΣ =M M 
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )42 lb4 in. 20 lb 21 in. 3 2 12 in. 36 lb 3 in. 28 lb
13B
 = − + − + + +  M j i i i j j i i j× × × × 
( ) 221.246 lb in.B∴ = ⋅M k 
 
 
 PROBLEM 3.107 CONTINUED 
Since B B=M r R× 
( ) ( )221.246 lb in. 0 51.297 18.9461
18.9461 51.297 0
x y x y∴ ⋅ = = +
−
i j k
k k 
For 221.2460, 4.3130 in.
51.297
y x= = = or 4.31 in.x = W 
For 221.2460, 11.6776 in.
18.9461
x y= = = or 11.68 in.y = W 
 
 
 
 
 
PROBLEM 3.108 
As four holes are punched simultaneously in a piece of aluminum sheet 
metal, the punches exert on the piece the forces shown. Knowing that the 
forces are perpendicular to the surfaces of the piece, determine (a) the 
resultant of the applied forces when 45α = ° and the point of 
intersection of the line of action of that resultant with a line drawn 
through points A and B, (b) the value of α so that the line of action of the 
resultant passes through fold EF. 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
 
 
Position the origin for the coordinate system along the centerline of the 
sheet metal at the intersection with line EF. 
(a) Have Σ =F R 
( )2.6 5.25 10.5 cos 45 sin 45 3.2 kN = − − − ° + ° − R j j i j i 
( ) ( )10.6246 kN 15.2746 kN∴ = − −R i j 
( ) ( )2 22 2 10.6246 15.2746x yR R R= + = + 
 18.6064 kN= 
1 1 15.2746tan tan 55.179
10.6246
y
x
R
R
θ − −  − = = = °   −  
 
 or 18.61 kN=R 55.2°W 
 Have EF EFM M= Σ 
 where 
 ( )( ) ( )( )2.6 kN 90 mm 5.25 kN 40 mmEFM = + 
 ( )( ) ( ) ( )10.5 kN 20 mm 3.2 kN 40 mm sin 45 40 mm − − ° +  
15.4903 N mEFM∴ = ⋅ 
 To obtain distance d left of EF, 
 Have ( )15.2746 kNEF yM dR d= = − 
3
3
15.4903 N m 1.01412 10 m
15.2746 10 N
d −−
⋅∴ = = − ×− × 
 or 1.014 mm left of d EF= W 
 
 
 
PROBLEM 3.108 CONTINUED 
(b) Have 0EFM = 
( )( ) ( )( )0 2.6 kN 90 mm 5.25 kN 40 mmEFM = = + 
 ( )( )10.5 kN 20 mm− 
 ( ) ( )3.2 kN 40 mm sin 40 mmα − +  
( )128 N m sin 106 N mα∴ ⋅ = ⋅ 
 sin 0.828125α = 
 55.907α = ° 
 or 55.9α = °W 
 
 
 
 
 
PROBLEM 3.109 
As four holes are punched simultaneously in a piece of aluminum sheet 
metal, the punches exert on the piece the forces shown. Knowing that the 
forces are perpendicular to the surfaces of the piece, determine (a) the 
value of α so that the resultant of the applied forces is parallel to the 
10.5 N force, (b) the corresponding resultant of the applied forces and the 
point of intersection of its line of action with a line drawn through points 
A and B. 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
 (a) For the resultant force, R, to be parallel to the 10.5 kN force, 
α φ= 
tan tan y
x
R
R
α φ∴ = =where 
 ( )3.2 kN 10.5 kN sinxR α= − − 
( )2.6 kN 5.25 kN 10.5 kN cosyR α= − − − 
3.2 10.5 sintan
7.85 10.5cos
αα α
+∴ = + 
 and 3.2tan 0.40764
7.85
α = = 
 22.178α = ° or 22.2α = °W 
(b) From 22.178α = ° 
 ( )3.2 kN 10.5 kN sin 22.178 7.1636 kNxR = − − ° = − 
( )7.85 kN 10.5 kN cos 22.178 17.5732 kNyR = − − ° = − 
( ) ( )2 22 2 7.1636 17.5732 18.9770 kNx yR R R= + = + = 
 or 18.98 kN=R 67.8°W 
 Then 
EF EFM M= Σ 
 where 
( )( ) ( )( ) ( )( )2.6 kN 90 mm 5.25 kN 40 mm 10.5 kN 20 mmEFM = + − 
 ( ) ( )3.2 kN 40 mm sin 22.178 40 mm − ° +  
 57.682 N m= ⋅ 
 
PROBLEM 3.109 CONTINUED 
 To obtain distance d left of EF, 
 Have ( )17.5732EF yM dR d= = − 
3
3
57.682 N m 3.2824 10 m
17.5732 10 N
d −⋅∴ = = − ×− × 
 or 3.28 mm left of d EF= W 
 
 
 
 
 
PROBLEM 3.110 
A truss supports the loading shown. Determine the equivalent force 
acting on the truss and the point of intersection of its line of action with a 
line through points A and G. 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
 
 
Have = ΣR F 
( )( ) ( )240 N cos70 sin 70 160 N= ° − ° −R i j j 
 ( )( ) ( )300 N cos 40 sin 40 180 N+ − ° − ° −i j j 
( ) ( )147.728 N 758.36 N∴ = − −R i j 
( ) ( )2 22 2 147.728 758.36x yR R R= + = + 
 772.62 N= 
1 1 758.36tan tan 78.977
147.728
y
x
R
R
θ − −  − = = = °   −  
 
 or 773 N=R 79.0°W 
Have A yM dRΣ = 
where 
[ ]( ) [ ]( )240 N cos70 6 m 240 Nsin 70 4 mAMΣ = − ° − ° 
 ( )( ) [ ]( )160 N 12 m 300 N cos 40 6 m− + ° 
 [ ]( ) ( )( )300 Nsin 40 20 m 180 N 8 m− ° − 
 7232.5 N m= − ⋅ 
7232.5 N m 9.5370 m
758.36 N
d − ⋅∴ = =− 
 or 9.54 m to the right of d A= W 
 
 
 
 
 
 
PROBLEM 3.111 
Three forces and a couple act on crank ABC. For 5 lbP = and 40 ,α = ° 
(a) determine the resultant of the given system of forces, (b) locate the 
point where the line of action of the resultant intersects a line drawn 
through points B and C, (c) locate the point where the line of action of the 
resultant intersects a line drawn through points A and B. 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 (a) 5 lb, 40P α= = ° 
 Have = ΣR F 
 ( )( ) ( ) ( )5 lb cos 40 sin 40 3 lb 2 lb= ° + ° − −i j i j 
( ) ( )0.83022 lb 1.21394 lb∴ = +R i j 
( ) ( )2 22 2 0.83022 1.21394x yR R R= + = + 
1.47069 lb= 
1 1 1.21394tan tan 55.632
0.83022
y
x
R
R
θ − −   = = = °     
 
 or 1.471 lb=R 55.6°W 
(b) From B B yM M dR= Σ = 
 where 
( ) ( ) ( )5 lb cos 40 15 in. sin 50 5 lb sin 40BM      = − ° ° − °      
 ( ) ( ) ( )15 in. sin 50 3 lb 6 in. sin 50   × ° + °    
 ( )( )2 lb 6 in. 50 lb in.− + ⋅ 
 23.211 lb in.BM∴ = − ⋅ 
 and 23.211 lb in. 19.1205 in.
1.21394 lb
B
y
Md
R
− ⋅= = = − 
 or 19.12 in. to the left of d B= W 
 
 
 
 
 
PROBLEM 3.111 CONTINUED 
(c) From /B D B=M r R× 
 ( ) ( )1 123.211 lb in. cos50 sin 50d d− ⋅ = − ° + °k i j 
 ( ) ( )0.83022 lb 1.21394 lb × − + i j 
 ( ) ( )1 123.211 lb in. 0.78028 0.63599d d− ⋅ = − −k k 
1
23.211 16.3889 in.
1.41627
d∴ = = 
or 1 16.39 in. from along line d B AB= 
 or 1.389 in. above and to the left of AW 
 
 
 
 
 
PROBLEM 3.112 
Three forces and a couple act on crank ABC. Determine the value of d so 
that the given system of forces is equivalent to zero at (a) point B, (b) 
point D. 
 
SOLUTION 
 
 
 
 
 
 
Based on 0xFΣ = 
cos 3 lb 0P α − = 
 cos 3 lbP α∴ = (1) 
and 0yFΣ = 
sin 2 lb 0P α − = 
 sin 2 lbP α∴ = (2) 
Dividing Equation (2) by Equation (1), 
2tan
3
α = 
 33.690α∴ = ° 
Substituting into Equation (1), 
3 lb 3.6056 lb
cos33.690
P = =° 
or 3.61 lb=P 33.7° 
(a) Based on 0BMΣ = 
( ) ( )3.6056 lb cos33.690 6 in. sin 50d   − ° + °    
 ( ) ( )3.6056 lb sin 33.690 6 in. cos50d   − ° + °    
 ( ) ( ) ( )( )3 lb 6 in. sin 50 2 lb 6 in. 50 lb in. 0 + ° − + ⋅ =  
3.5838 30.286d− = − 
8.4509 in.d∴ = 
 or 8.45 in.d = W 
 
 
 
 
 
 
 
PROBLEM 3.112 CONTINUED 
(b) Based on 0DMΣ = 
 ( ) ( )3.6056 lb cos33.690 6 in. sin 50d   − ° + °    
 ( ) ( )3.6056 lb sin 33.690 6 in. cos50 6 in.d   − ° + ° +    
 ( ) ( )3 lb 6 in. sin 50 50 lb in. 0 + ° + ⋅ =  
3.5838 30.286d− = − 
8.4509 in.d∴ = 
 or 8.45 in.d = W 
 This result is expected, since 0=R and 0RB =M for 
 8.45 in.d = implies that 0 and 0= =R M at any other point for 
 the value of d found in part a. 
 
 
 
 
 
 
PROBLEM 3.113 
Pulleys A and B are mounted on bracket CDEF. The tension on each side 
of the two belts is as shown. Replace the four forces with a single 
equivalent force, and determine where its line of action intersects the 
bottom edge of the bracket. 
 
SOLUTION 
 
 
 
 
 
 
Equivalent force-couple at A due to belts on pulley A 
Have : 120 N 160 N ARΣ − − =F 
 280 NA∴ =R 
Have ( ): 40 N 0.02 mA AMΣ − =M 
 0.8 N mA∴ = ⋅M 
Equivalent force-couple at B due to belts on pulley B 
Have ( ): 210 N 150 NΣ +F 25 B° = R 
 360 NB∴ =R 25° 
Have ( ): 60 N 0.015 mB BMΣ − =M 
 0.9 N mB∴ = ⋅M 
Equivalent force-couple at F 
Have ( ) ( )( ): 280 N 360 N cos 25 sin 25FΣ = − + ° + °F R j i j 
 ( ) ( )326.27 N 127.857 N= −i j 
( ) ( )2 22 2 326.27 127.857 350.43 NF Fx FyR R R R= = + = + = 
1 1 127.857tan tan 21.399
326.27
Fy
Fx
R
R
θ − −  − = = = − °     
 
 or 350 NF = =R R 21.4°W 
 
 
 
 
 
 
 
PROBLEM 3.113 CONTINUED 
Have 
 ( )( ): 280 N 0.06 m 0.80 N mF FMΣ = − − ⋅M 
 ( ) ( )360 N cos 25 0.010 m − °  
 ( ) ( )360 N sin 25 0.120 m 0.90 N m + ° − ⋅  
( )3.5056 N mF = − ⋅M k 
To determine where a single resultant force will intersect line FE, 
F yM dR= 
3.5056 N m 0.027418 m 27.418 mm
127.857 N
F
y
Md
R
− ⋅∴ = = = =− 
 or 27.4 mmd = W 
 
 
 
 
 
 
 
PROBLEM 3.114 
As follower AB rolls along the surface of member C, it exerts a constant 
force F perpendicular to the surface. (a) Replace F with an equivalent 
force-couple system at the point D obtained by drawing the perpendicular 
from the point of contact to the x axis (b) For 1 ma = and 2 m,b = 
determine the value of x for which the moment of the equivalent force-
couple system at D is maximum. 
 
SOLUTION 
 
 
 
 
 
 
 (a) The slope of any tangent to the surface of member C is 
2
2 2
21dy d x bb x
dx dx a a
   −= − =      
 
 Since the force F is perpendicular to the surface, 
1 2 1tan
2
dy a
dx b x
α
−   = − =       
 For equivalence 
:FΣ =F R 
( )( ): cosD A DM F y MαΣ = 
 where 
( ) ( )
2
22 22
2cos , 1
2
A
bx xy b
aa bx
α  = = −   +
 
3
2
2
4 2 2
2
 
4
D
xxFb
a
M
a b x
 −   ∴ =
+
 
 Therefore, the equivalent force-couple system at D is 
 F=R 
2
1tan
2
a
bx
−     
W 
 
3
2
2
4 2 2
2
a 4
xFb x
a
b x
 −   =
+
M W 
 
 
 
 
 
 
 
 
 
 
PROBLEM 3.114 CONTINUED 
(b) To maximize M, the value of x must satisfy 0dM
dx
= 
 where, for 1 m, 2 ma b= = 
( )3
2
8
1 16
F x x
M
x
−=
+
 
( ) ( ) ( )( )
( )
1
2 2 3 2 2
2
11 16 1 3 32 1 16
2
 8 0
1 16
x x x x x x
dM F
dx x
− + − − − +   ∴ = =+
( )( ) ( )2 2 31 16 1 3 16 0x x x x x+ − − − = 
 or 4 232 3 1 0x x+ − = 
( )( )
( )2 2 2
3 9 4 32 1
0.136011 m and 0.22976 m
2 32
x
− ± − −= = − 
 Using the positive valueof 2,x 
0.36880 mx = 
 or 369 mmx = W 
 
 
 
 
 
 
PROBLEM 3.115 
As plastic bushings are inserted into a 3-in.-diameter cylindrical sheet 
metal container, the insertion tool exerts the forces shown on the 
enclosure. Each of the forces is parallel to one of the coordinate axes. 
Replace these forces with an equivalent force-couple system at C. 
 
SOLUTION 
 
For equivalence 
: A B C D CΣ + + + =F F F F F R 
( ) ( ) ( ) ( )5 lb 3 lb 4 lb 7 lbC = − − − −R j j k i 
 ( ) ( ) ( ) 7 lb 8 lb 4 lbC∴ = − − −R i j k W 
Also for equivalence 
/ / /:C A C A B C B D C D C′ ′ ′Σ + + =M r F r F r F M× × × 
or 
0 0 1.5 in. 1 in. 0 1.5 in. 0 1.5 in. 1.5 in.
0 5 lb 0 0 3 lb 0 7 lb 0 0
C = − + − +
− −
i j k i j k i j k
M 
 ( ) ( ) ( )7.50 lb in. 0 0 4.50 lb in. + 3.0 lb in. 0   = − ⋅ − + − ⋅ − ⋅ −   i i k 
 ( ) ( )10.5 lb in. 0 0 10.5 lb in. + ⋅ − + + ⋅ j k 
 ( ) ( ) ( )or 12.0 lb in. 10.5 lb in. 7.5 lb in.C = − ⋅ + ⋅ + ⋅M i j k W 
 
 
 
 
 
 
 
PROBLEM 3.116 
Two 300-mm-diameter pulleys are mounted on line shaft AD. The belts B 
and C lie in vertical planes parallel to the yz plane. Replace the belt forces 
shown with an equivalent force-couple system at A. 
 
SOLUTION 
 
 
 
 
 
 
 
 
Equivalent force-couple at each pulley 
Pulley B 
( )( )290 N cos 20 sin 20 430 NB = − ° + ° −R j k j 
 ( ) ( )702.51 N 99.186 N= − +j k 
 ( )( )430 N 290 N 0.15 mB = − −M i 
 ( )21 N m= − ⋅ i 
Pulley C 
( )( )310 N 480 N sin10 cos10C = + − ° − °R j k 
 ( ) ( )137.182 N 778.00 N= − −j k 
 ( )( )480 N 310 N 0.15 mC = −M i 
 ( )25.5 N m= ⋅ i 
Then ( ) ( )839.69 N 678.81 NB C= + = − −R R R j k 
 or ( ) ( )840 N 679 N= − −R j kW 
 / /A B C B A B C A C= + + +M M M r R r R× × 
 ( ) ( )21 N m 25.5 N m 0.45 0 0 N m
0 702.51 99.186
= − ⋅ + ⋅ + ⋅
−
i j k
i i 
 0.90 0 0 N m
0 137.182 778.00
+ ⋅
− −
i j k
 
 ( ) ( ) ( )4.5 N m 655.57 N m 439.59 N m= ⋅ + ⋅ − ⋅i j k 
 ( ) ( ) ( )or 4.50 N m + 656 N m 440 N mA = ⋅ ⋅ − ⋅M i j k W 
 
 
 
 
 
PROBLEM 3.117 
A mechanic uses a crowfoot wrench to loosen a bolt at C. The mechanic 
holds the socket wrench handle at points A and B and applies forces at 
these points. Knowing that these forces are equivalent to a force-couple 
system at C consisting of the force ( ) ( )40 N 20 N= − +C i k and the 
couple ( )40 N mC = ⋅M i , determine the forces applied at A and B when 
10zA = N. 
 
SOLUTION 
 
 
 
 
 
 
 
Have :Σ + =F A B C 
or : 40 Nx x xF A B+ = − 
 ( ) 40 Nx xB A∴ = − + (1) 
: 0y y yF A BΣ + = 
or y yA B= − (2) 
: 10 N 20 Nz zF BΣ + = 
or 10 NzB = (3) 
Have / /:C B C A C CΣ + =M r B r A M× × 
( ) 0.2 0 0.05 0.2 0 0.2 N m 40 N m
10 10x y x yB B A A
∴ − + ⋅ = ⋅
i j k i j k
i 
or ( ) ( )0.05 0.2 0.05 2 0.2 2y x x xB A B A− + − − + −i j 
 ( ) ( )0.2 0.2 40 N my yB A+ + = ⋅k i 
From - coefficient 0.05 0.2 40 N my yB A− = ⋅i (4) 
 - coefficient 0.05 0.2 4 N mx xB A− + = ⋅j (5) 
 - coefficient 0.2 0.2 0y yB A+ =k (6) 
 
 
 
PROBLEM 3.117 CONTINUED 
From Equations (2) and (4): ( )0.05 0.2 40y yB B− − = 
160 N, 160 Ny yB A= = − 
From Equations (1) and (5): ( )0.05 40 0.2 4x xA A− − − + = 
8 NxA = 
From Equation (1): ( )8 40 48 NxB = − + = − 
 ( ) ( ) ( ) 8 N 160 N 10 N∴ = − +A i j kW 
 ( ) ( ) ( )48 N 160 N 10 N= − + +B i j k W 
 
 
 
 
 
PROBLEM 3.118 
While using a pencil sharpener, a student applies the forces and couple 
shown. (a) Determine the forces exerted at B and C knowing that these 
forces and the couple are equivalent to a force-couple system at A 
consisting of the force ( ) ( )3.9 lb 1.1 lbyR= + −R i j k and the couple 
( ) ( )1.5 lb ft 1.1 lb ft .RA xM= + ⋅ − ⋅M i j k . (b) Find the corresponding 
values of yR and .xM 
 
SOLUTION 
Have :Σ + =F B C R 
 : 3.9 lb or 3.9 lbx x x x xF B C B CΣ + = = − (1) 
 :y y yF C RΣ = (2) 
 : 1.1 lbz zF CΣ = − (3) 
Have / /:
R
A B A C A B AΣ + + =M r B r C M M× × 
( ) ( ) ( )1 1 0 4.5 4 0 2.0 2 lb ft 1.5 lb ft 1.1 lb ft
12 12
0 0 1.1
x
x x y
x M
B C C
∴ + + ⋅ = + ⋅ − ⋅
−
i j k i j k
i i j k 
( ) ( ) ( )2 0.166667 0.375 0.166667 0.36667 0.33333y x x yC B C C− + + + +i j k 
 ( ) ( )1.5 1.1xM= + −i j k 
From - coefficient 2 0.166667 y xC M− =i (4) 
 - coefficient 0.375 0.166667 0.36667 1.5x xB C+ + =j (5) 
 - coefficient 0.33333 1.1 or 3.3 lby yC C= − = −k (6) 
(a) From Equations (1) and (5): 
( )0.375 3.9 0.166667 1.13333x xC C− + = 
0.32917 1.58000 lb
0.20833x
C = = 
 From Equation (1): 3.9 1.58000 2.32 lbxB = − = 
 ( ) 2.32 lb∴ =B iW 
 ( ) ( ) ( )1.580 lb 3.30 lb 1.1 lb= − −C i j kW 
(b) From Equation (2): 3.30 lby yR C= = − ( )or 3.30 lby = −R jW 
 From Equation (4): ( )0.166667 3.30 2.0 2.5500 lb ftxM = − − + = ⋅ 
 ( )or 2.55 lb ftx = ⋅M iW 
 
 
 
 
 
PROBLEM 3.119 
A portion of the flue for a furnace is attached to the ceiling at A. While 
supporting the free end of the flue at F, a worker pushes in at E and pulls 
out at F to align end E with the furnace. Knowing that the 10-lb force at 
F lies in a plane parallel to the yz plane, determine (a) the angle α the 
force at F should form with the horizontal if duct AB is not to tend to 
rotate about the vertical, (b) the force-couple system at B equivalent to 
the given force system when this condition is satisfied. 
 
SOLUTION 
(a) Duct AB will not have a tendency to rotate about the vertical or y-axis if: 
( )/ / 0R RBy B F B F E B EM = Σ = + =j M j r F r F⋅ ⋅ × × 
 where 
 ( ) ( ) ( )/ 45 in. 23 in. 28 in.F B = − +r i j k 
( ) ( ) ( )/ 54 in. 34 in. 28 in.E B = − +r i j k 
( ) ( )10 lb sin cosF α α = + F j k 
 ( )5 lbE = −F k 
 ( ) ( )( ) 10 lb 45 in. 23 in. 28 in. 5 lb 2 in. 27 17 14
0 sin cos 0 0 1
R
B
α α
∴ Σ = − + −
−
i j k i j k
M 
 ( ) ( ) ( )230cos 280sin 170 450cos 270 450sin lb in.α α α α = − − + − − + ⋅ i j k 
 Thus, 450cos 270 0RByM α= − + = 
cos 0.60α = 
53.130α = ° 
 or 53.1α = °W 
 
 
 
 PROBLEM 3.119 CONTINUED 
 (b) E F= +R F F 
 where 
 ( )5 lbE = −F k 
 ( )( ) ( ) ( )10 lb sin 53.130 cos53.130 8 lb 6 lbF = ° + ° = +F j k j k 
 ( ) ( ) 8 lb 1 lb∴ = +R j kW 
 and ( ) ( ) ( ) ( )230 0.6 280 0.8 170 450 0.6 270 450 0.8RB      = Σ = − + − − − +     M M i j k 
 ( ) ( ) ( )192 lb in. 0 360 lb in.= − ⋅ − + ⋅i j k 
 ( ) ( )or 192 lb in. 360 lb in.= − ⋅ + ⋅M i kW 
 
 
 
 
 
PROBLEM 3.120 
A portion of the flue for a furnace is attached to the ceiling at A. While 
supporting the free end of the flue at F, a worker pushes in at E and pulls 
out at F to align end E with the furnace. Knowing that the 10-lb force at 
F lies in a plane parallel to the yz plane and that α = 60°, (a) replace the 
given force system with an equivalent force-couple system at C, (b) 
determine whether duct CD will tend to rotate clockwise or 
counterclockwise relative to elbow C, as viewed from D to C. 
 
SOLUTION 
(a) Have F E= Σ = +R F F F 
 where ( ) ( ) ( ) ( )10 lb sin 60 cos60 8.6603 lb 5.0 lbF  = ° + ° = + F j k j k 
 ( )5 lbE = −F k 
 ( ) 8.6603 lb∴ =R j ( )or 8.66 lb=R jW 
 Have ( ) / /RC F C F E C E= Σ = +M r F r F r F× × × 
 where ( ) ( )/ 9 in. 2 in.F C = −r i j 
 ( ) ( )/ 18 in. 13 in.E C = −r i j 
 9 2 0 lb in. 18 13 0 lb in.
0 8.6603 5.0 0 0 5
R
C∴ = − ⋅ + − ⋅
−
i j k i j k
M 
 ( ) ( ) ( )55 lb in. 45 lb in. 77.942 lb in.= ⋅ + ⋅ + ⋅i j k 
 ( ) ( ) ( )or 55.0 lb in. 45.0 lb in. 77.9 lb in.RC = ⋅ + ⋅ + ⋅M i j kW 
(b) To determine which direction duct section CD has a tendency to turn, have 
R R
CD DC CM = Mλ ⋅ 
 where 
( )( ) ( )18 in. 4 in. 1 9 2
2 85 in. 85DC
− += = − +i j i jλ 
 Then ( ) ( )1 9 2 55 45 77.942 lb in.
85
R
CDM = − + + + ⋅i j i j k⋅ 
 ( )53.690 9.7619 lb in.= − + ⋅ 
 43.928 lb in.= − ⋅ 
 Since 0,RDC C <Mλ ⋅ duct DC tends to rotate clockwise relative to elbow C as viewed from D to C. W 
 
 
 
 
 
 
 
PROBLEM 3.121 
The head-and-motor assembly of a radial drill press was originally 
positioned with arm AB parallel to the z axis and the axis of the chuck 
and bit parallel to the y axis. The assembly was then rotated o25 about 
the y axis and o20 about the centerline of the horizontal arm AB, bringing 
it into the position shown. The drilling process was started by switching 
on the motor and rotating the handle to bring the bit into contact with the 
workpiece. Replace the force and couple exerted by the drill press with an 
equivalent force-couple system at the center O of the base of the vertical 
column. 
 
SOLUTION 
 
 
 
 
 
 
 
 
Have =R F 
 ( ) ( ) ( ) ( )44 N sin 20 cos 25 cos 20 sin 20 sin 25 = ° ° − ° − ° ° i j k 
 ( ) ( ) ( )13.6389 N 41.346 N 6.3599 N= − −i j k 
 ( ) ( ) ( )or 13.64 N 41.3 N 6.36 N= − −R i j k W 
Have /O B O C= +M r F M× 
where 
( ) ( ) ( )/ 0.280 m sin 25 0.300 m 0.280 m cos 25B O    = ° + + °   r i j k 
 ( ) ( ) ( )0.118333 m 0.300 m 0.25377 m= + +i j k 
( ) ( ) ( ) ( )7.2 N m sin 20 cos 25 cos 20 sin 20 sin 25C  = ⋅ ° ° − ° − ° ° M i j k 
 ( ) ( ) ( )2.2318 N m 6.7658 N m 1.04072 N m= ⋅ − ⋅ − ⋅i j k 
 0.118333 0.300 0.25377 N m
13.6389 41.346 6.3599
O∴ = ⋅
− −
i j k
M 
 ( )2.2318 6.7658 1.04072 N m+ − − ⋅i j k 
 ( ) ( ) ( )10.8162 N m 2.5521 N m 10.0250 N m= ⋅ − ⋅ − ⋅i j k 
 ( ) ( ) ( )or 10.82 N m 2.55 N m 10.03 N mO = ⋅ − ⋅ − ⋅M i j kW 
 
 
 
 
 
 
PROBLEM 3.122 
While a sagging porch is leveled and repaired, a screw jack is used to 
support the front of the porch. As the jack is expanded, it exerts on the 
porch the force-couple system shown, where 300 NR = and 
37.5 N m.M = ⋅ Replace this force-couple system with an equivalent 
force-couple system at C. 
 
SOLUTION 
From ( ) ( ) ( ) ( )0.2 m 1.4 m 0.5 m300 N 300 N
1.50 mC AB
 − + −= = =   
i j k
R R λ 
 ( ) ( ) ( )40.0 N 280 N 100 NC = − + −R i j kW 
From /C A C= +M r R M× 
where 
( ) ( )/ 2.6 m 0.5 mA C = +r i k 
( ) ( ) ( ) ( ) ( )0.2 m 1.4 m 0.5 m37.5 N m 37.5 N m
1.50 mBA
 − += ⋅ = ⋅   
i j k
M λ 
 ( ) ( ) ( )5.0 N m 35.0 N m 12.5 N m= ⋅ − ⋅ + ⋅i j k 
( ) ( ) ( ) ( ) 10 N m 2.6 0 0.5 5.0 N m 35.0 N m 12.5 N m
4 28 10
C∴ = ⋅ + ⋅ − ⋅ + ⋅
− −
i j k
M i j k 
 ( ) ( ) ( )140 5 N m 20 260 35 N m 728 12.5 N m     = − + ⋅ + − + − ⋅ + + ⋅     i j k 
 ( ) ( ) ( )or 135.0 N m 205 N m 741 N mC = − ⋅ + ⋅ + ⋅M i j k W 
 
 
 
 
 
 
 
PROBLEM 3.123 
Three children are standing on a 15 15-ft× raft. If the weights of the 
children at points A, B, and C are 85 lb, 60 lb, and 90 lb, respectively, 
determine the magnitude and the point of application of the resultant of 
the three weights. 
 
SOLUTION 
 
Have : A B CΣ + + =F F F F R 
( ) ( ) ( )85 lb 60 lb 90 lb− − − =j j j R 
 ( )235 lb− =j R or 235 lbR = W 
Have ( ) ( ) ( ) ( ):x A A B B C C DM F z F z F z R zΣ + + = 
( )( ) ( )( ) ( )( ) ( )( )85 lb 9 ft 60 lb 1.5 ft 90 lb 14.25 ft 235 lb Dz+ + = 
 9.0957 ftDz∴ = or 9.10 ftDz = W 
Have ( ) ( ) ( ) ( ):z A A B B C C DM F x F x F x R xΣ + + = 
( )( ) ( )( ) ( )( ) ( )( )85 lb 3 ft 60 lb 4.5 ft 90 lb 14.25 ft 235 lb Dx+ + = 
 7.6915 ftDx∴ = or 7.69 ftDx = W 
 
 
 
 
 
 
 
 
PROBLEM 3.124 
Three children are standing on a 15 15-ft× raft. The weights of the 
children at points A, B, and C are 85 lb, 60 lb, and 90 lb, respectively. If a 
fourth child of weight 95 lb climbs onto the raft, determine where she 
should stand if the other children remain in the positions shown and the 
line of action of the resultant of the four weights is to pass through the 
center of the raft. 
 
SOLUTION 
 
 
Have : A B C DΣ + + + =F F F F F R 
( ) ( ) ( ) ( )85 lb 60 lb 90 lb 95 lb− − − − =j j j j R 
 ( ) 330 lb∴ = −R j 
Have ( ) ( ) ( ) ( ) ( ):x A A B B C C D D HM F z F z F z F z R zΣ + + + = 
( )( ) ( )( ) ( )( ) ( )( ) ( )( )85 lb 9 ft 60 lb 1.5 ft 90 lb 14.25 ft 95 lb 330 lb 7.5 ftDz+ + + = 
 3.5523 ftDz∴ = or 3.55 ftDz = W 
Have ( ) ( ) ( ) ( ) ( ):z A A B B C C D D HM F x F x F x F x R xΣ + + + = 
( )( ) ( )( ) ( )( ) ( )( ) ( )( )85 lb 3 ft 60 lb 4.5 ft 90 lb 14.25 ft 95 lb 330 lb 7.5 ftDx+ + + = 
 7.0263 ftDx∴ = or 7.03 ftDx = W 
 
 
 
 
 
 
 
 
PROBLEM 3.125 
The forces shown are the resultant downward loads on sections of the flat 
roof of a building because of accumulated snow. Determine the 
magnitude and the point of application of the resultant of these four 
loads. 
 
SOLUTION 
 
 
Have : A B C DΣ + + + =F F F F F R 
( ) ( ) ( ) ( )580 kN 2350 kN 330 kN 140 kN− − − − =j j j j R 
 ( ) 3400 kN∴ = −R j 3400 kNR = W 
Have ( ) ( ) ( ) ( ) ( ):x A A B B C C D D EM F z F z F z F z R zΣ + + + = 
( )( ) ( )( ) ( )( ) ( )( ) ( )( )580 kN 8 m 2350 kN 16 m 330 kN 6 m 140 kN 33.5 m 3400 kN Ez+ + + = 
 14.3853 mEz∴ = or 14.39 mEz = W 
Have ( ) ( ) ( ) ( ) ( ):z A A B B C C D D EM F x F x F x F x R xΣ + + + = 
( )( ) ( )( ) ( )( ) ( )( ) ( )( )580 kN 10 m 2350 kN 32 m 330 kN 54 m 140 kN 32 m 3400 kN Ex+ + + = 
 30.382 mEx∴ = or 30.4 mEx = W 
 
 
 
 
 
 
 
 
PROBLEM 3.126 
The forces shown are the resultant downward loads on sections of the flat 
roof of a building because of accumulated snow. If the snow represented 
by the 580-kN force is shoveled so that the this load acts at E, determine 
a and b knowing that the point of application of the resultant of the four 
loads is then at B. 
 
SOLUTION 
 
 
Have : B C D EΣ + + + =F F F F F R 
( ) ( ) ( ) ( )2350 kN 330 kN 140 kN 580 kN− − − − =j j j j R 
 ( ) 3400 kN∴ = −R j 
Have ( ) ( ) ( ) ( ) ( ):x B B C C D D E E BM F z F z F z F z R zΣ + + + = 
( )( ) ( )( ) ( )( ) ( )( ) ( )( )2350 kN 16 m 330 kN 6 m 140 kN 33.5 m 580 kN 3400 kN 16 mb+ + + = 
 17.4655 mb∴ = or 17.47 mb = W 
Have ( ) ( ) ( ) ( ) ( ):z B B C C D D E E BM F x F x F x F x R xΣ + + + = 
( )( ) ( )( ) ( )( ) ( )( ) ( )( )2350 kN 32 m 330 kN 54 m 140 kN 32 m 580 kN 3400 kN 32 ma+ + + = 
 19.4828 ma∴ = or 19.48 ma = W 
 
 
 
 
 
 
PROBLEM 3.127 
A group of students loads a 2 4-m× flatbed trailer with two 
0.6 0.6 0.6-m× × boxes and one 0.6 0.6 1.2-m× × box. Each of the 
boxes at the rear of the trailer is positioned so that it is aligned with both 
the back and a side of the trailer. Determine the smallest load the students 
should place in a second 0.6 0.6 1.2-m× × box and where on the trailer 
they should secure it, without any part of the box overhanging the sides 
of the trailer, if each box is uniformly loaded and the line of action of the 
resultant of the weights of the four boxes is to pass through the point of 
intersection of the centerlines of the trailer and the axle. (Hint: Keep in 
mind that the box may be placed either on its side or on its end.) 
 
SOLUTION 
 
For the smallest weight on the trailer so that the resultant force of the four weights acts over the axle at the 
intersection with the center line of the trailer, the added 0.6 0.6 1.2-m× × box should be placed adjacent to 
one of the edges of the trailer with the 0.6 0.6-m× side on the bottom. The edges to be considered are based 
on the location of the resultant for the three given weights. 
Have ( ) ( ) ( ): 200 N 400 N 180 NΣ − − − =F j j j R 
( ) 780 N∴ = −R j 
Have ( )( ) ( )( ) ( )( ) ( )( ): 200 N 0.3 m 400 N 1.7 m 180 N 1.7 m 780 NzM xΣ + + = 
 1.34103 mx∴= 
Have ( )( ) ( )( ) ( )( ) ( )( ): 200 N 0.3 m 400 N 0.6 m 180 N 2.4 m 780 NxM zΣ + + = 
0.93846 mz∴ = 
From the statement of the problem, it is known that the resultant of R from the original loading and the 
lightest load W passes through G, the point of intersection of the two center lines. Thus, 0.GΣ =M 
Further, since the lightest load W is to be as small as possible, the fourth box should be placed as far from G 
as possible without the box overhanging the trailer. These two requirements imply 
( ) ( )0.3 m 1 m 1.8 m 3.7 mx z≤ ≤ ≤ ≤ 
 
 
 
 PROBLEM 3.127 CONTINUED 
Let 0.3 m,x = ( )( ) ( )( ) ( )( ) ( ): 200 N 0.7 m 400 N 0.7 m 180 N 0.7 m 0.7 m 0GzM WΣ − − + = 
 380 NW∴ = 
( )( ) ( )( ) ( )( ) ( )( ): 200 N 1.5 m 400 N 1.2 m 180 N 0.6 m 380 N 1.8 m 0GxM zΣ − − + + − = 
 3.5684 m 3.7 m acceptablez∴ = < ∴ 
Let 3.7 m,z = ( )( ) ( )( ) ( )( ) ( ): 200 N 1.5 m 400 N 1.2 m 180 N 0.6 m 1.7 m 0GxM WΣ − − + + = 
 395.29 N 380 NW∴ = > 
Since the weight W found for 0.3 mx = is less than W found for 3.7 m, 0.3 mz x= = results in the 
smallest weight W. 
 ( )or 380 N at 0.3 m, 0, 3.57 mW = W 
 
 
 
 
 
 
PROBLEM 3.128 
Solve Problem 3.127 if the students want to place as much weight as 
possible in the fourth box and that at least one side of the box must 
coincide with a side of the trailer. 
Problem 3.127: A group of students loads a 2 4-m× flatbed trailer with 
two 0.6 0.6 0.6-m× × boxes and one 0.6 0.6 1.2-m× × box. Each of the 
boxes at the rear of the trailer is positioned so that it is aligned with both 
the back and a side of the trailer. Determine the smallest load the students 
should place in a second 0.6 0.6 1.2-m× × box and where on the trailer 
they should secure it, without any part of the box overhanging the sides 
of the trailer, if each box is uniformly loaded and the line of action of the 
resultant of the weights of the four boxes is to pass through the point of 
intersection of the centerlines of the trailer and the axle. (Hint: Keep in 
mind that the box may be placed either on its side or on its end.) 
 
SOLUTION 
 
 
For the largest additional weight on the trailer with the box having at least one side coinsiding with the side of 
the trailer, the box must be as close as possible to point G. For 0.6 m,x = with a small side of the box 
touching the z-axis, satisfies this condition. 
Let 0.6 m,x = ( )( ) ( )( ) ( )( ) ( ): 200 N 0.7 m 400 N 0.7 m 180 N 0.7 m 0.4 m 0GzM WΣ − − + = 
 665 NW∴ = 
and ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ): 200 N 1.5 m 400 N 1.2 m 180 N 0.6 m 665 N 1.8 m 0GXM zΣ − − + + − = 
( ) 2.8105 m 2 m 4 m acceptablez z∴ = < < ∴ 
 ( )or 665 N at 0.6 m, 0, 2.81 mW = W 
 
 
 
 
 
 
 
PROBLEM 3.129 
A block of wood is acted upon by three forces of the same magnitude P 
and having the directions shown. Replace the three forces with an 
equivalent wrench and determine (a) the magnitude and direction of the 
resultant R, (b) the pitch of the wrench, (c) the point where the axis of the 
wrench intersects the xy plane. 
 
SOLUTION 
 
First, reduce the given force system to a force-couple at the origin. 
Have : P P PΣ − − =F i i k R 
 P∴ = −R k 
Have ( ) ( ) ( ): 3 3 3 RO OP a P a P a aΣ − − + − + =M k j i j M 
( ) 3RO Pa∴ = − −M i k 
Then let vectors ( )1, R M represent the components of the wrench, where their directions are the same. 
(a) P= −R k or Magnitude of P=R W 
 Direction of : 90 , 90 , 180x y zθ θ θ= ° = ° = − °R W 
(b) Have 1
R
R OM = Mλ ⋅ 
 ( )3Pa = − − − k i k⋅ 
 3Pa= 
 and pitch 1 3 3M Pap a
R P
= = = or 3p a= W 
 
 
 PROBLEM 3.129 CONTINUED 
(c) Have 1 2
R
O = +M M M 
 ( ) ( )2 1 3 3RO Pa Pa Pa∴ = − = − − − − = −M M M i k k i 
 Require 2 /Q O=M r R× 
( ) ( )Pa x y P Px Py− = + − = −i i j k j i× 
 From : or Pa Py y a− = − =i 
 : 0x =j 
 ∴ The axis of the wrench is parallel to the z-axis and intersects the xy plane at 0, x y a= = W 
 
 
 
 
 
 
 
PROBLEM 3.130 
A piece of sheet metal is bent into the shape shown and is acted upon by 
three forces. Replace the three forces with an equivalent wrench and 
determine (a) the magnitude and direction of the resultant R, (b) the pitch 
of the wrench, (c) the point where the axis of the wrench intersects the 
yz plane. 
 
SOLUTION 
 
First, reduce the given force system to a force-couple system at the origin. 
Have ( ) ( ) ( ): 2P P PΣ − + =F i j j R 
( ) 2P∴ =R i 
Have ( ): RO O OΣ Σ =M r F M× 
( )2 2 2.5 0 0 4 1.5 5 6
2 1 0 0 1 0
R
O Pa Pa= + = − + −
−
i j k i j k
M i j k 
(a) 2P=R i or Magnitude of 2P=R W 
 Direction of : 0 , 90 , 90x y zθ θ θ= ° = − ° = °R W 
(b) Have 1 
R
R O RM R
= = RMλ ⋅ λ 
 ( )1.5 5 6Pa Pa Pa= − + −i i j k⋅ 
 1.5Pa= − 
 and pitch 1 1.5 0.75
2
M Pap a
R P
−= = = − or 0.75p a= − W 
 
 
 PROBLEM 3.130 CONTINUED 
(c) Have 1 2
R
O = +M M M 
 ( ) ( )2 1 5 6RO Pa Pa∴ = − = −M M M j k 
 Require 2 /Q O=M r R× 
 ( ) ( ) ( ) ( ) ( ) ( )5 6 2 2 2Pa Pa y z P Py Pz− = + = − +j k j k i k j× 
 From : 5 2Pa Pz=i 
 2.5z a∴ = 
 From : 6 2Pa Py− = −k 
 3y a∴ = 
 ∴ The axis of the wrench is parallel to the x-axis and intersects the yz-plane at 3 , 2.5y a z a= = W 
 
 
 
 
 
 
 
 
PROBLEM 3.131 
The forces and couples shown are applied to two screws as a piece of 
sheet metal is fastened to a block of wood. Reduce the forces and the 
couples to an equivalent wrench and determine (a) the resultant force R, 
(b) the pitch of the wrench, (c) the point where the axis of the wrench 
intersects the xz plane. 
 
SOLUTION 
 
First, reduce the given force system to a force-couple at the origin. 
Have ( ) ( ): 10 N 11 NΣ − − =F j j R 
( ) 21 N∴ = −R j 
Have ( ): RO O C OΣ Σ + Σ =M r F M M× 
( )0 0 0.5 N m 0 0 0.375 N m 12 N m
0 10 0 0 11 0
R
O = ⋅ + − ⋅ − ⋅
− −
i j k i j k
M j 
 ( ) ( )0.875 N m 12 N m= ⋅ − ⋅i j 
(a) ( )21 N= −R j ( )or 21 N= −R jW 
(b) Have 1 
R
R O RM R
= = RMλ ⋅ λ 
 ( ) ( ) ( )0.875 N m 12 N m = − − j i j⋅ ⋅ ⋅ 
 ( )112 N m and 12 N m= ⋅ = − ⋅M j 
 and pitch 1 12 N m 0.57143 m
21 N
Mp
R
⋅= = = or 0.571 mp = W 
 
 
 PROBLEM 3.131 CONTINUED 
(c) Have 1 2
R
O = +M M M 
( )2 1 0.875 N mRO∴ = − = ⋅M M M i 
 Require 2 /Q O=M r R× 
( ) ( ) ( ) 0.875 N m 21 Nx z  ∴ ⋅ = + − i i k j× 
 ( ) ( )0.875 21 21x z= − +i k i 
 From i: 0.875 21z= 
 0.041667 mz∴ = 
 From k: 0 21x= − 
 0z∴ = 
 ∴ The axis of the wrench is parallel to the y-axis and intersects the xz-plane at 0, 41.7 mmx z= = W 
 
 
 
 
 
 
 
PROBLEM 3.132 
The forces and couples shown are applied to two screws as a piece of 
sheet metal is fastened to a block of wood. Reduce the forces and the 
couples to an equivalent wrench and determine (a) the resultant force R, 
(b) the pitch of the wrench, (c) the point where the axis of the wrench 
intersects the xz plane. 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
First, reduce the given force system to a force-couple system. 
Have ( ) ( ): 6 lb 4.5 lb 7.5 lbRΣ − − = =F i j R 
Have ( ): RO O C OΣ ∑ + ∑ =M r F M M× 
( ) ( ) ( )6 lb 8 in. 160 lb in. 72 lb in.RO = − − ⋅ − ⋅M j i j 
 ( ) ( )160 lb in. 120 lb in.= − ⋅ − ⋅i j 
200 lb in.ROM = ⋅ 
(a) ( ) ( )6 lb 4.5 lb= − −R i jW 
(b) Have 1 
R
R OM R
= = RMλ ⋅ λ 
 ( ) ( ) ( )0.8 0.6 160 lb in. 120 lb in. = − − − ⋅ − ⋅ i j i j⋅ 
 200 lb in.= ⋅ 
 and ( )1 200 lb in. 0.8 0.6= ⋅ − −M i j 
 Pitch 1 200 lb in. 26.667 in.
7.50 lb
Mp
R
⋅= = = 
 or 26.7 in.p = W 
(c) From above note that1
R
O=M M 
 Therefore, the axis of the wrench goes through the origin. The line 
 of action of the wrench lies in the xy plane with a slope of 
 3
4
dy
dx
= W 
 
 
 
 
 
 
 
PROBLEM 3.133 
Two bolts A and B are tightened by applying the forces and couple 
shown. Replace the two wrenches with a single equivalent wrench and 
determine (a) the resultant R, (b) the pitch of the single equivalent 
wrench, (c) the point where the axis of the wrench intersects the xz plane. 
 
SOLUTION 
 
First, reduce the given force system to a force-couple at the origin. 
Have ( ) ( ) ( ) ( ): 20 lb 21 lb 21 lb 20 lb 29 lbRΣ − − = − − = =F k j j k R 
and ( ): RO O C OΣ ∑ + ∑ =M r F M M× 
( ) ( ) ( )20 lb 4 in. 4 3 0 21 lb 4 in. 6 0 1 300 320 lb in.
0 0 1 0 1 0
R
O+ + − − ⋅ =
− −
i j k i j k
j k M 
( ) ( ) ( ) 156 lb in. 20 lb in. 824 lb in.RO∴ = − ⋅ + ⋅ − ⋅M i j k 
(a) ( ) ( )21 lb 20 lb= − −R j kW 
(b) Have 1 
R
R O RM R
= = RMλ ⋅ λ 
 ( ) ( ) ( )21 20 156 lb in. 20 lb in. 824 lb in.
29
− −  = − − ⋅ + ⋅ − ⋅ j k i j k⋅ 
 553.80 lb in.= ⋅ 
 
 
 PROBLEM 3.133 CONTINUED 
 and ( ) ( )1 1 401.03 lb in. 381.93 lb in.RM= = − ⋅ − ⋅M j kλ 
 Then pitch 1 553.80 lb in. 19.0964 in.
29 lb
Mp
R
⋅= = = or 19.10 in.p = W 
(c) Have 1 2
R
O = +M M M 
( ) ( )2 1 156 20 824 401.03 381.93 lb in.RO  ∴ = − = − + − − − − ⋅ M M M i j k j k 
 ( ) ( ) ( )156.0 lb in. 421.03 lb in. 442.07 lb in.= − ⋅ + ⋅ − ⋅i j k 
 Require 2 /Q O=M r R× 
( ) ( ) ( )156 421.03 442.07 21 20x z− + − = + − −i j k i k j k× 
 ( ) ( ) ( )21 20 21z x x= + −i j k 
 From i: 156 21z− = 
 7.4286 in.z∴ = − 
 or 7.43 in.z = − 
 From k: 442.07 21x− = − 
 21.051 in.x∴ = 
 or 21.1 in.x = 
 ∴ The axis of the wrench intersects the xz-plane at 
 21.1 in., 7.43 in.x z= = − W 
 
 
 
 
 
 
PROBLEM 3.134 
Two bolts A and B are tightened by applying the forces and couple 
shown. Replace the two wrenches with a single equivalent wrench and 
determine (a) the resultant R, (b) the pitch of the single equivalent 
wrench, (c) the point where the axis of the wrench intersects the xz plane. 
 
SOLUTION 
 
First reduce the given force system to a force-couple at the origin at B. 
(a) Have ( ) ( ) 8 15: 79.2 lb 51 lb
17 17
 Σ − − + =  F k i j R 
 ( ) ( ) ( ) 24.0 lb 45.0 lb 79.2 lb∴ = − − −R i j k W 
 and 94.2 lbR = 
 Have /:
R
B A B A A B BΣ + + =M r F M M M× 
( )8 150 20 0 660 714 1584 660 42 8 15
17 17
0 0 79.2
R
B
 = − − − + = − − +  −
i j k
M k i j i k i j 
( ) ( ) ( ) 1248 lb in. 630 lb in. 660 lb in.RB∴ = ⋅ − ⋅ − ⋅M i j k 
(b) Have 1 
R
R O RM R
= = RMλ ⋅ λ 
 ( ) ( ) ( )24.0 45.0 79.2 1248 lb in. 630 lb in. 660 lb in.
94.2
− − −  = ⋅ − ⋅ − ⋅ i j k i j k⋅ 
 537.89 lb in.= ⋅ 
 
 
 PROBLEM 3.134 CONTINUED 
 and 1 1 RM=M λ 
 ( ) ( ) ( )137.044 lb in. 256.96 lb in. 452.24 lb in.= − ⋅ − ⋅ − ⋅i j k 
 Then pitch 1 537.89 lb in. 5.7101 in.
94.2 lb
Mp
R
⋅= = = or 5.71 in.p = W 
(c) Have 1 2
R
B = +M M M 
( ) ( )2 1 1248 630 660 137.044 256.96 452.24RB∴ = − = − − − − − −M M M i j k i j k 
 ( ) ( ) ( )1385.04 lb in. 373.04 lb in. 207.76 lb in.= ⋅ − ⋅ − ⋅i j k 
 Require 2 /Q B=M r R× 
 1385.04 373.04 207.76 0
24 45 79.2
x z− − =
− − −
i j k
i j k 
 ( ) ( ) ( ) ( )45 24 79.2 45z z x x= − + −i j j k 
From i: 1385.04 45 30.779 in.z z= ∴ = 
From k: 207.76 45 4.6169 in.x x− = − ∴ = 
 ∴ The axis of the wrench intersects the xz-plane at 
 4.62 in., 30.8 in.x z= = W 
 
 
 
 
 
 
 
PROBLEM 3.135 
A flagpole is guyed by three cables. If the tensions in the cables have the 
same magnitude P, replace the forces exerted on the pole with an 
equivalent wrench and determine (a) the resultant force R, (b) the pitch of 
the wrench, (c) the point where the axis of the wrench intersects the xz 
plane. 
 
SOLUTION 
 
(a) First reduce the given force system to a force-couple at the origin. 
 Have : BA DC DEP P PΣ + + =F Rλ λ λ 
4 3 3 4 9 4 12
5 5 5 5 25 5 25
P  −     = − + − + − +            R j k i j i j k 
 ( )3 2 20
25
P∴ = − −R i j k W 
( ) ( ) ( )2 2 23 27 52 20 1
25 25
PR P= + + = 
 Have ( ): RO OPΣ Σ =M r M× 
( ) ( ) ( )4 3 3 4 9 4 1224 20 20
5 5 5 5 25 5 25
R
O
P P P P P P Pa a a− −     − + − + − + =          j j k j i j j i j k M× × × 
( )24 
5
R
O
Pa∴ = − −M i k 
(b) Have 1
R
R OM = Mλ ⋅ 
 where ( ) ( )3 25 12 20 2 20
25 27 5 9 5R
P
R P
= = − − = − −R i j k i j kλ 
 
 
 PROBLEM 3.135 CONTINUED 
 Then ( ) ( )1 1 24 82 20 59 5 15 5
Pa PaM −= − − − − =i j k i k⋅ 
 and pitch 1 8 25 8
8115 5 27 5
M Pa ap
R P
− − = = =   or 0.0988p a= − W 
(c) ( ) ( )1 1 8 1 82 20 2 2067515 5 9 5R
Pa PaM −  = = − − = − + +  M i j k i j kλ 
 Then ( ) ( ) ( )2 1 24 8 82 20 403 20 4065 675 675RO
Pa Pa Pa= − = − − − − + + = − − −M M M i k i j k i j k 
 Require 2 /Q O=M r R× 
( ) ( ) ( )8 3403 20 406 2 20
675 25
Pa Px z   − − − = + − −      i j k i k i j k× 
 ( )3 20 2 20
25
P z x z x   = + + −     i j k 
 From i: ( ) 38 403 20
675 25
Pa Pz  − =    1.99012z a∴ = − 
 From k: ( ) 38 406 20 2.0049
675 25
Pa Px x a − = − ∴ =   
 ∴ The axis of the wrench intersects the xz-plane at 
 2.00 , 1.990x a z a= = − W 
 
 
 
 
 
 
 
PROBLEM 3.136 
Determine whether the force-and-couple system shown can be reduced to 
a single equivalent force R. If it can, determine R and the point where the 
line of action of R intersects the yz plane. If it cannot be so reduced, 
replace the given system with an equivalent wrench and determine its 
resultant, its pitch, and the point where its axis intersects the yz plane. 
 
SOLUTION 
 
First, reduce the given force system to a force-couple at D. 
Have : DA ED DA DA ED EDF FΣ + = + =F F F Rλ λ 
where ( ) ( ) ( )0.300 m 0.225 m 0.200 m136 N
0.425 mDA
 − + +=   
i j k
F 
 ( ) ( ) ( )96 N 72 N 64 N= − + +i j k 
( ) ( ) ( ) ( )0.150 m 0.200 m120 N 72 N 96 N
0.250 mED
 − −= = − −  
i k
F i k 
 ( ) ( ) ( )168 N 72 N 32 N∴ = − + −R i j k W 
Have : RD A DΣ =M M M 
or ( ) ( ) ( ) ( ) ( )0.150 m 0.150 m 0.450 m 16 N m16 N m 3
0.150 11 m 11
R
D
 − − + ⋅= ⋅ = − − +  
i j k
M i j k 
 
 
 PROBLEM 3.136 CONTINUED 
The force-couple at D can be replaced by a single force if R is perpendicular to .RDM To be perpendicular, 
0.RD =R M⋅ 
Have ( ) ( )16168 72 32 3
11
R
D = − + − − − +R M i j k i j k⋅ ⋅ 
 ( )128 21 9 12
11
= − − 
 0= 
 ∴ Force-couple can be reduced to a single equivalent force.W 
To determine the coordinates where the equivalent single force intersects the yz-plane, /
R
D Q D=M r R× 
where ( ) ( ) ( )/ 0 0.300 m 0.075 m 0 mQ D y z     = − + − + −     r i j k 
( ) ( ) ( )16 N m 3 8 N 0.3 0.075 m
11 21 9 4
y z⋅∴ − − + = − −
− −
i j k
i j k 
or 
( ) ( ) ( ) ( ) ( ){ }16 N m 3 8 N 4 0.075 9 21 1.2 2.7 21 0.075 m
11
y z z y⋅    − − + = − − − + − − + − + −   i j k i j k 
From j: ( )16 8 21 1.2
11
z− = − − 0.028427 m 28.4 mmz∴ = − = − 
From k: ( )48 8 2.7 21 0.075 0.28972 m 290 mm
11
y y = − + − ∴ = =  
 ∴ line of action of R intersects the yz-plane at 
 290 mm, 28.4 mmy z= = − W 
 
 
 
 
 
 
 
PROBLEM 3.137 
Determine whether the force-and-couple system shown can be reduced to 
a single equivalent force R. If it can, determine R and the point where the 
line of action of R intersects the yz plane. If it cannot be so reduced, 
replace the given system with an equivalentwrench and determine its 
resultant, its pitch, and the point where its axis intersects the yz plane. 
 
SOLUTION 
 
First, reduce the given force system to a force-couple at the origin. 
Have : A GΣ + =F F F R 
( ) ( ) ( ) ( ) ( ) ( ) ( )4 in. 6 in. 12 in. 10 lb 14 lb 4 lb 6 lb 2 lb
14 in.
 + −∴ = + = + −  
i j k
R k i j kW 
and 56 lbR = 
Have ( ): RO O C OΣ ∑ + ∑ =M r F M M× 
 ( ) ( ) ( ) ( ) ( ) ( ){ }12 in. 10 lb 16 in. 4 lb 6 lb 12 lbRO    = + + −   M j k i i j k× × 
 ( ) ( ) ( ) ( ) ( ) ( ) ( )16 in. 12 in. 4 in. 12 in. 6 in.84 lb in. 120 lb in.
20 in. 14 in.
   − − ++ ⋅ + ⋅      
i j i j k
 
( ) ( ) ( )0 221.49 lb in. 38.743 lb in. 147.429 lb in.R∴ = ⋅ + ⋅ + ⋅M i j k 
 ( ) ( ) ( )18.4572 lb ft 3.2286 lb ft 12.2858 lb ft= ⋅ + ⋅ + ⋅i j k 
 
 
 PROBLEM 3.137 CONTINUED 
The force-couple at O can be replaced by a single force if the direction of R is perpendicular to .ROM 
To be perpendicular 0RO =R M⋅ 
Have ( ) ( )4 6 2 18.4572 3.2286 12.2858 0?RO = + − + + =R M i j k i j k⋅ ⋅ 
 73.829 19.3716 24.572= + − 
 0≠ 
∴ System cannot be reduced to a single equivalent force. 
To reduce to an equivalent wrench, the moment component along the line of action of P is found. 
 1 
R
R O RM R
= = RMλ ⋅ λ 
 ( ) ( )4 6 2 18.4572 3.2286 12.2858
56
 + −= + +  
i j k
i j k⋅ 
 9.1709 lb ft= ⋅ 
and ( )( )1 1 9.1709 lb ft 0.53452 0.80178 0.26726RM= = ⋅ + −M i j kλ 
And pitch 1 9.1709 lb ft 1.22551 ft
56 lb
Mp
R
⋅= = = 
 or 1.226 ftp = W 
Have 
( ) ( )( )2 1 18.4572 3.2286 12.2858 9.1709 0.53452 0.80178 0.26726RO= − = + + − + −M M M i j k i j k 
 ( ) ( ) ( )13.5552 lb ft 4.1244 lb ft 14.7368 lb ft= ⋅ − ⋅ + ⋅i j k 
Require 2 /Q O=M r R× 
 ( ) ( ) ( )13.5552 4.1244 14.7368 4 6 2y z− + = + + −i j k j k i j k× 
 ( ) ( ) ( )2 6 4 4y z z y= − + + −i j k 
From j: 4.1244 4z− = or 1.0311 ftz = − 
From k: 14.7368 4 or 3.6842 fty y= − = − 
 ∴ line of action of the wrench intersects the yz plane at 
 3.68 ft, 1.031 fty z= − = W 
 
 
 
 
 
 
PROBLEM 3.138 
Replace the wrench shown with an equivalent system consisting of two 
forces perpendicular to the y axis and applied respectively at A and B. 
 
SOLUTION 
 
 
 
 
 
Express the forces at A and B as 
x zA A= +A i k 
x zB B= +B i k 
Then, for equivalence to the given force system 
 : 0x x xF A BΣ + = (1) 
 :z z zF A B RΣ + = (2) 
 ( ) ( ): 0x z zM A a B a bΣ + + = (3) 
 ( ) ( ):z x xM A a B a b MΣ − − + = (4) 
From Equation (1), x xB A= − 
Substitute into Equation (4) 
( ) ( )x xA a A a b M− + + = 
 and x x
M MA B
b b
∴ = = − 
From Equation (2), z zB R A= − 
and Equation (3), ( )( ) 0z zA a R A a b+ − + = 
 1z
aA R
b
 ∴ = +   
 
 
 
 
 
 
 
PROBLEM 3.138 CONTINUED 
and 1z
aB R R
b
 = − +   
 z
aB R
b
∴ = − 
Then 1M aR
b b
   = + +      A i kW 
 M a R
b b
   = − −      B i kW 
 
 
 
 PROBLEM 3.139 
Show that, in general, a wrench can be replaced with two forces chosen in 
such a way that one force passes through a given point while the other 
force lies in a given plane. 
 
SOLUTION 
 
First, choose a coordinate system so that the xy plane coincides with the given plane. Also, position the 
coordinate system so that the line of action of the wrench passes through the origin as shown in Figure a. 
Since the orientation of the plane and the components (R, M) of the wrench are known, it follows that the 
scalar components of R and M are known relative to the shown coordinate system. 
A force system to be shown as equivalent is illustrated in Figure b. Let A be the force passing through the 
given point P and B be the force that lies in the given plane. Let b be the x-axis intercept of B. 
The known components of the wrench can be expressed as 
 and x y z x y zR R R M M M M= + + = + +R i j k i j k 
while the unknown forces A and B can be expressed as 
 and x y z x zA A A B B= + + = +A i j k B i k 
Since the position vector of point P is given, it follows that the scalar components (x, y, z) of the position 
vector Pr are also known. 
Then, for equivalence of the two systems 
 :x x x xF R A BΣ = + (1) 
 :y y yF R AΣ = (2) 
 :z z z zF R A BΣ = + (3) 
 :x x z yM M yA zAΣ = − (4) 
 :y y x z zM M zA xA bBΣ = − − (5) 
 :z z y xM M xA yAΣ = − (6) 
 
 
 PROBLEM 3.139 CONTINUED 
Based on the above six independent equations for the six unknowns ( ), , , , , ,x y z x zA A A B B b there exists a 
unique solution for A and B. 
From Equation (2) y yA R= W 
 Equation (6) ( )1x y zA xR My = −   W 
 Equation (1) ( )1x x y zB R xR My = − −   W 
 Equation (4) ( )1z x yA M zRy = +   W 
 Equation (3) ( )1z z x yB R M zRy = − +   W 
 Equation (5) ( )( )x y zx z y
xM yM zM
b
M yR zR
+ += − + W 
 
 
 PROBLEM 3.140 
Show that a wrench can be replaced with two perpendicular forces, one 
of which is applied at a given point. 
 
SOLUTION 
 
First, observe that it is always possible to construct a line perpendicular to a given line so that the constructed 
line also passes through a given point. Thus, it is possible to align one of the coordinate axes of a rectangular 
coordinate system with the axis of the wrench while one of the other axes passes through the given point. 
See Figures a and b. 
Have and and are known.R M= =R j M j 
The unknown forces A and B can be expressed as 
 and x y z x y zA A A B B B= + + = + +A i j k B i j k 
The distance a is known. It is assumed that force B intersects the xz plane at (x, 0, z). Then for equivalence 
 : 0x x xF A B∑ = + (1) 
 :y y yF R A B∑ = + (2) 
 : 0z z zF A B∑ = + (3) 
 : 0x yM zB∑ = − (4) 
 :y z z xM M aA xB zB∑ = − − + (5) 
 : 0z y yM aA xB∑ = + (6) 
Since A and B are made perpendicular, 
 0 or 0x x y y z zA B A B A B= + + =A B⋅ (7) 
There are eight unknowns: , , , , , , , x y z x y zA A A B B B x z 
But only seven independent equations. Therefore, there exists an infinite number of solutions. 
 
 
 PROBLEM 3.140 CONTINUED 
Next consider Equation (4): 0 yzB= − 
If 0,yB = Equation (7) becomes 0x x z zA B A B+ = 
Using Equations (1) and (3) this equation becomes 2 2 0x zA A+ = 
Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that 
0,yB ≠ so that from Equation (4), 0.z = 
To obtain one possible solution, arbitrarily let 0.xA = 
(Note: Setting , ,y zA A or zB equal to zero results in unacceptable solutions.) 
The defining equations then become. 
 0 xB= (1)′ 
 y yR A B= + (2) 
 0 z zA B= + (3) 
 z zM aA xB= − − (5)′ 
 0 y yaA xB= + (6) 
 0y y z zA B A B+ = (7)′ 
Then Equation (2) can be written y yA R B= − 
 Equation (3) can be written z zB A= − 
 Equation (6) can be written y
y
aA
x
B
= − 
Substituting into Equation (5)′, 
( )yz z
y
R B
M aA a A
B
 −= − − − −   
 
or z y
MA B
aR
= − (8) 
Substituting into Equation (7)′, 
( ) 0y y y yM MR B B B BaR aR  − + − =     
 
 
 PROBLEM 3.140 CONTINUED 
or 
2 3
2 2 2y
a RB
a R M
= + 
Then from Equations (2), (8), and (3) 
 
2 3 2
2 2 2 2 2 2y
a R RMA R
a R M a R M
= − =+ + 
2 3 2
2 2 2 2 2 2z
M a R aR MA
aR a R M a R M
 = − = −  + + 
 
 
2
2 2 2z
aR MB
a R M
= + 
In summary 
 ( )2 2 2RM M aRa R M= −+A j k 
( )22 2 2aR aR Ma R M= ++B j k 
Which shows that it is possible to replace a wrench with twoperpendicular forces, one of which is applied at a 
given point. 
Lastly, if 0 and 0,R M> > it follows from the equations found for A and B that 0 and 0.y yA B> > 
From Equation (6), 0 (assuming 0).x a< > Then, as a consequence of letting 0,xA = force A lies in a plane 
parallel to the yz plane and to the right of the origin, while force B lies in a plane parallel to the yz plane but to 
the left of the origin, as shown in the figure below. 
 
 
 
 PROBLEM 3.141 
Show that a wrench can be replaced with two forces, one of which has a 
prescribed line of action. 
 
SOLUTION 
 
First, choose a rectangular coordinate system where one axis coincides with the axis of the wrench and 
another axis intersects the prescribed line of action ( ).AA′ Note that it has been assumed that the line of 
action of force B intersects the xz plane at point ( ), 0, .P x z Denoting the known direction of line AA′ by 
A x y zλ λ λ= + +i j kλ 
it follows that force A can be expressed as 
( )A x y zA A λ λ λ= = + +i j kλA 
Force B can be expressed as 
x y zB B B= + +B i j k 
Next, observe that since the axis of the wrench and the prescribed line of action AA′ are known, it follows 
that the distance a can be determined. In the following solution, it is assumed that a is known. 
Then, for equivalence 
 : 0x x xF A BλΣ = + (1) 
 :y y yF R A BλΣ = + (2) 
 : 0z z zF A BλΣ = + (3) 
 : 0x yM zBΣ = − (4) 
 :y z x zM M aA zB xBλΣ = − + − (5) 
 : 0z y yM aA xBλΣ = + (6) 
Since there are six unknowns ( ), , , , , x y zA B B B x z and six independent equations, it will be possible to 
obtain a solution. 
 
 
 PROBLEM 3.141 CONTINUED 
Case 1: Let 0z = to satisfy Equation (4) 
Now Equation (2) y yA R Bλ = − 
 Equation (3) z zB Aλ= − 
 Equation (6) ( )y y
y y
aA ax R B
B B
λ  = − = − −   
 
Substitution into Equation (5) 
( )( )z y z
y
aM aA R B A
B
λ λ   = − − − − −     
 
1 y
z
MA B
aRλ
 ∴ = −    
Substitution into Equation (2) 
1
y y y
z
MR B B
aR
λλ
 = − +   
2
 zy
z y
aRB
aR M
λ
λ λ∴ = − 
Then 
z y y z
MR RA aRaR M
M
λ λ λ λ
= − =− −
 
x
x x
z y
MRB A
aR M
λλ λ λ= − = − 
z
z z
z y
MRB A
aR M
λλ λ λ= − = − 
In summary 
 A
y z
P
aR
M
λ λ
=
−
A λ W 
 ( )x z z
z y
R M aR M
aR M
λ λ λλ λ= + +−B i j k W 
and z 21 1
y
y z
aR MRx a a R
B aR
λ λ
λ
   − = − = −           
 
 or y
z
Mx
R
λ
λ= W 
Note that for this case, the lines of action of both A and B intersect the x axis. 
 
 
 PROBLEM 3.141 CONTINUED 
Case 2: Let 0yB = to satisfy Equation (4) 
Now Equation (2) 
y
RA λ= 
 Equation (1) xx
y
B R λλ
 = −    
 
 Equation (3) zz
y
B R λλ
 = −    
 
 Equation (6) 0 which requires 0yaA aλ = = 
Substitution into Equation (5) 
 or x z z x y
y y
MM z R x R x z
R
λ λ λ λ λλ λ
           = − − − − =                   
 
This last expression is the equation for the line of action of force B. 
In summary 
A
y
R
λ
 =    
A λ 
( )x z
y
R λ λλ
 = − −   
B i k 
Assuming that , , 0,x y zλ λ λ > the equivalent force system is as shown below. 
 
Note that the component of A in the xz plane is parallel to B. 
 
 
 
 
 
 
PROBLEM 3.142 
A worker tries to move a rock by applying a 360-N force to a steel bar as 
shown. (a) Replace that force with an equivalent force-couple system at 
D. (b) Two workers attempt to move the same rock by applying a vertical 
force at A and another force at D. Determine these two forces if they are 
to be equivalent to the single force of part a. 
 
SOLUTION 
 
(a) Have ( ) ( ) ( ): 360 N sin 40 cos40 231.40 N 275.78 NΣ − ° − ° = − − =F i j i j F 
 or 360 N=F 50°W 
 Have /:D B DΣ =M r R M× 
 where ( ) ( )/ 0.65 m cos30 0.65 m sin 30B D    = − ° + °   r i j 
 ( ) ( )0.56292 m 0.32500 m= − +i j 
( ) 0.56292 0.32500 0 N m 155.240 75.206 N m
231.40 275.78 0
 ∴ = − ⋅ = + ⋅ 
− −
i j k
M k 
 ( )230.45 N m= ⋅ k or 230 N m= ⋅M W 
(b) Have /:D A D AΣ =M M r F× 
 where ( ) ( )/ 1.05 m cos30 1.05 m sin 30A D    = − ° + °   r i j 
 ( ) ( )0.90933 m 0.52500 m= − +i j 
 
 
 PROBLEM 3.142 CONTINUED 
[ ] 0.90933 0.52500 0 N m 230.45 N m
0 1 0
AF∴ − ⋅ = ⋅
−
i j k
k 
 or ( )0.90933 230.45AF =k k 
 253.42 NAF∴ = or 253 NA =F W 
 Have : A DΣ = +F F F F 
( ) ( ) ( ) ( )231.40 N 275.78 N 253.42 N cos sinDF θ θ− − = − + − −i j j i j 
 From : 231.40 N cosDF θ=i (1) 
 : 22.36 N sinDF θ=j (2) 
Equation (2) divided by Equation (1) 
tan 0.096629θ = 
 5.5193θ∴ = ° or 5.52θ = ° 
Substitution into Equation (1) 
231.40 232.48 N
cos5.5193D
F = =° 
 or 232 ND =F 5.52°W 
 
 
 
 
 
 
PROBLEM 3.143 
A worker tries to move a rock by applying a 360-N force to a steel bar as 
shown. If two workers attempt to move the same rock by applying a force 
at A and a parallel force at C, determine these two forces so that they will 
be equivalent to the single 360-N force shown in the figure. 
 
SOLUTION 
 
Have : A CΣ = +F R F F 
( ) ( ) ( ) ( )360 N sin 40 360 N cos 40 sin cosA C A CF F F Fθ θ      − ° − ° = − + − +       i j i j 
From ( ) ( ): 360 N sin 40 sinA CF F θ° = +i (1) 
 ( ) ( ): 360 N cos 40 cosA CF F θ° = +j (2) 
Dividing Equation (1) by Equation (2), 
tan 40 tanθ° = 
 40θ∴ = ° 
Substituting 40θ = ° into Equation (1), 
 360 NA CF F+ = (3) 
Have / /:C B C A C AΣ =M r R r F× × 
where ( )( ) ( ) ( )/ 0.35 m cos30 sin 30 0.30311 m 0.175 mB C = − ° + ° = − +r i j i j 
 
 
 PROBLEM 3.143 CONTINUED 
( )( ) ( ) ( )360 N sin40 cos 40 231.40 N 275.78 N= − ° − ° = − −R i j i j 
( )( ) ( ) ( )/ 0.75 m cos30 sin 30 0.64952 m 0.375 mA C = − ° + = − +r i j i j 
( ) ( )sin 40 cos40 0.64279 0.76604A A AF F= − ° − ° = − −F i j i j 
 0.30311 0.175 0 N m 0.64952 0.375 0 N m
231.40 275.78 0 0.64279 0.76604 0
AF∴ − ⋅ = − ⋅
− − − −
i j k i j k
 
( )83.592 40.495 0.49756 0.24105 AF+ = + 
 168.002 NAF∴ = or 168.0 NAF = 
Substituting into Equation (3), 
360 168.002 191.998 NCF = − = or 192.0 NCF = 
 or 168.0 NA =F 50°W 
 192.0 NC =F 50°W 
 
 
 
 
 
 
PROBLEM 3.144 
A force and a couple are applied as shown to the end of a cantilever 
beam. (a) Replace this system with a single force F applied at point C, 
and determine the distance d from C to a line drawn through points D and 
E. (b) Solve part a if the directions of the two 360-N forces are reversed. 
 
SOLUTION 
(a) 
(b) 
 
 (a) Have ( ) ( ) ( ): 360 N 360 N 600 NΣ = − −F F j j k 
 or ( )600 N= −F k W 
 and ( )( ) ( )( ): 360 N 0.15 m 600 NDM dΣ = 
 0.09 md∴ = 
 or 90.0 mm below d ED= W 
(b) Have from part a ( )600 N= −F k W 
 and ( )( ) ( )( ): 360 N 0.15 m 600 NDM dΣ − = − 
 0.09 md∴ = 
 or 90.0 mm above d ED= W 
 
 
 
 
 
 
PROBLEM 3.145 
A crate of mass 80 kg is held in the position shown. Determine (a) the 
moment produced by the weight W of the crate about E, (b) the smallest 
force applied at B which creates a moment of equal magnitude and 
opposite sense about E. 
 
SOLUTION 
 
 
 
 
 (a) By definition ( )280 kg 9.81 m/s 784.8 NW mg= = = 
 Have ( )( ): 784.8 N 0.25 mE EM MΣ = 
 196.2 N mE∴ = ⋅M W 
(b) For the force at B to be the smallest, resulting in a moment ( )EM 
 about E, the line of action of force BF must be perpendicular to the 
 line connecting E to B. The sense of BF must be such that the force 
 produces a counterclockwisemoment about E. 
 Note: ( ) ( )2 20.85 m 0.5 m 0.98615 md = + = 
 Have ( ): 196.2 N m 0.98615 mE BM FΣ ⋅ = 
 198.954 NBF∴ = 
 and 1 0.85 mtan 59.534
0.5 m
θ −  = = °   
 or 199.0 NB =F 59.5°W 
 
 
 
 
 
 
PROBLEM 3.146 
A crate of mass 80 kg is held in the position shown. Determine (a) the 
moment produced by the weight W of the crate about E, (b) the smallest 
force applied at A which creates a moment of equal magnitude and 
opposite sense about E, (c) the magnitude, sense, and point of application 
on the bottom of the crate of the smallest vertical force which creates a 
moment of equal magnitude and opposite sense about E. 
 
SOLUTION 
 
 
 
 
 (a) By definition ( )280 kg 9.81 m/s 784.8 NW mg= = = 
 Have ( )( ): 784.8 N 0.25 mE EM MΣ = 
 196.2 N mE∴ = ⋅M W 
(b) For the force at A to be the smallest, resulting in a moment about E, 
 the line of action of force AF must be perpendicular to the line 
 connecting E to A. The sense of AF must be such that the force 
 produces a counterclockwise moment about E. 
 Note: ( ) ( )2 20.35 m 0.5 m 0.61033 md = + = 
 Have ( ): 196.2 N m 0.61033 mE AM FΣ ⋅ = 
 321.47 NAF∴ = 
 and 1 0.35 mtan 34.992
0.5 m
θ −  = = °   
 or 321 NA =F 35.0°W 
(c) The smallest force acting on the bottom of the crate resulting in a 
 moment about E will be located at the point on the bottom of the 
 crate farthest from E and acting perpendicular to line CED. The 
 sense of the force will be such as to produce a counterclockwise 
 moment about E. A force acting vertically upward at D satisfies 
 these conditions. 
 
 
 
 
 
 
PROBLEM 3.146 CONTINUED 
 Have /:E E D E DΣ =M M r F× 
( ) ( ) ( )196.2 N m 0.85 m DF⋅ =k i j× 
( ) ( )196.2 N m 0.85 DF⋅ =k k 
 230.82 NDF∴ = 
 or 231 ND =F W 
 
 
 
 
 
 
 
PROBLEM 3.147 
A farmer uses cables and winch pullers B and E to plumb one side of a 
small barn. Knowing that the sum of the moments about the x axis of the 
forces exerted by the cables on the barn at points A and D is equal to 
4728 lb ft,⋅ determine the magnitude of DET when 255 lb.AB =T 
 
SOLUTION 
 
 
 
 
The moment about the x-axis due to the two cable forces can be found 
using the z-components of each force acting at their intersection with the 
xy-plane (A and D). The x-components of the forces are parallel to the x-
axis, and the y-components of the forces intersect the x-axis. Therefore, 
neither the x or y components produce a moment about the x-axis. 
Have ( ) ( ) ( ) ( ):x AB A DE D xz zM T y T y MΣ + = 
where ( ) ( )AB AB AB ABzT T= =k T k⋅ ⋅ λ 
 12 12255 lb 180 lb
17
 − − + = =    
i j kk ⋅ 
 ( ) ( )DE DE DE DEzT T= =k T k⋅ ⋅ λ 
 1.5 14 12 0.64865
18.5DE DE
T T − + = =    
i j kk ⋅ 
12 ftAy = 
14 ftDy = 
4728 lb ftxM = ⋅ 
( )( ) ( )( ) 180 lb 12 ft 0.64865 14 ft 4728 lb ftDET∴ + = ⋅ 
and 282.79 lbDET = 
 or 283 lbDET = W 
 
 
 
 
 
 
PROBLEM 3.148 
Solve Problem 3.147 when the tension in cable AB is 306 lb. 
Problem 3.147: A farmer uses cables and winch pullers B and E to 
plumb one side of a small barn. Knowing that the sum of the moments 
about the x axis of the forces exerted by the cables on the barn at points A 
and D is equal to 4728 lb ft,⋅ determine the magnitude of DET when 
255 lb.AB =T 
 
SOLUTION 
 
 
 
 
The moment about the x-axis due to the two cable forces can be found 
using the z components of each force acting at the intersection with the xy 
plane (A and D). The x components of the forces are parallel to the x axis, 
and the y components of the forces intersect the x axis. Therefore, neither 
the x or y components produce a moment about the x axis. 
Have ( ) ( ) ( ) ( ):x AB A DE D xz zM T y T y MΣ + = 
where ( ) ( )AB AB AB ABzT T= =k T k⋅ ⋅ λ 
 12 12306 lb 216 lb
17
 − − + = =    
i j kk ⋅ 
 ( ) ( )DE DE DE DEzT T= =k T k⋅ ⋅ λ 
 1.5 14 12 0.64865
18.5DE DE
T T − + = =    
i j kk ⋅ 
12 ftAy = 
14 ftDy = 
4728 lb ftxM = ⋅ 
( )( ) ( )( ) 216 lb 12 ft 0.64865 14 ft 4728 lb ftDET∴ + = ⋅ 
and 235.21 lbDET = 
 or 235 lbDET = W 
 
 
 
 
 
 
PROBLEM 3.149 
As an adjustable brace BC is used to bring a wall into plumb, the force-
couple system shown is exerted on the wall. Replace this force-couple 
system with an equivalent force-couple system at A knowing that 
21.2 lbR = and 13.25 lb ft.M = ⋅ 
 
SOLUTION 
 
 
 
Have : A BCΣ = =F R R Rλ 
where ( ) ( ) ( )42 in. 96 in. 16 in.
106 in.BC
− −= i j kλ 
( )21.2 lb 42 96 16
106A
∴ = − −R i j k 
 ( ) ( ) ( )or 8.40 lb 19.20 lb 3.20 lbA = − −R i j k W 
Have /:A C A AΣ + =M r R M M× 
where ( ) ( ) ( )/ 142 in. 48 in. 42 48 ft12C A = + = +r i k i k 
 ( ) ( )3.5 ft 4.0 ft= +i k 
( ) ( ) ( )8.40 lb 19.20 lb 3.20 lb= − −R i j k 
 BCM= −M λ 
 ( )42 96 16 13.25 lb ft
106
− + += ⋅i j k 
 ( ) ( ) ( )5.25 lb ft 12 lb ft 2 lb ft= − ⋅ + ⋅ + ⋅i j k 
 
 
 
 
 
PROBLEM 3.149 CONTINUED 
Then ( )3.5 0 4.0 lb ft 5.25 12 2 lb ft
8.40 19.20 3.20
A⋅ + − + + ⋅ =
− −
i j k
i j k M 
( ) ( ) ( ) 71.55 lb ft 56.80 lb ft 65.20 lb ftA∴ = ⋅ + ⋅ − ⋅M i j k 
 ( ) ( ) ( )or 71.6 lb ft 56.8 lb ft 65.2 lb ftA = ⋅ + ⋅ − ⋅M i j k W 
 
 
 
 
 
 
 
PROBLEM 3.150 
Two parallel 60-N forces are applied to a lever as shown. Determine the 
moment of the couple formed by the two forces (a) by resolving each 
force into horizontal and vertical components and adding the moments of 
the two resulting couples, (b) by using the perpendicular distance 
between the two forces, (c) by summing the moments of the two forces 
about point A. 
 
SOLUTION 
 
(a) Have 1 2:B x yd C d CΣ − + =M M 
 where ( )1 0.360 m sin 55 0.29489 md = ° = 
 ( )2 0.360 m cos55 0.20649 md = ° = 
 ( )60 N cos 20 56.382 NxC = ° = 
 ( )60 N sin 20 20.521 NyC = ° = 
( )( ) ( )( ) ( ) 0.29489 m 56.382 N 0.20649 m 20.521 N 12.3893 N m∴ = − + = − ⋅M k k k 
 or 12.39 N m= ⋅M W 
(b) Have ( ) ( ) ( ) ( )60 N 0.360 m sin 55 20Fd  = − = ° − ° − M k k 
 ( )12.3893 N m= − ⋅ k 
 or 12.39 N m= ⋅M W 
 
 
 PROBLEM 3.150 CONTINUED 
(c) Have ( ) / /:A A B A B C A CΣ Σ = + =M r F r F r F M× × × 
( )( ) ( )( ) 0.520 m 60 N cos55 sin 55 0 0.880 m 60 N cos55 sin 55 0
cos 20 sin 20 0 cos 20 sin 20 0
M∴ = ° ° + ° °
− ° − ° ° °
i j k i j k
 
 ( ) ( )17.8956 N m 30.285 N m 12.3892 N m= ⋅ − ⋅ = − ⋅k k 
 or 12.39 N m= ⋅M W 
 
 
 
 
 
 
 
PROBLEM 3.151 
A 32-lb motor is mounted on the floor. Find the resultant of the weight 
and the forces exerted on the belt, and determine where the line of action 
of the resultant intersects the floor. 
 
SOLUTION 
 
Have ( ) ( ) ( )( ): 60 lb 32 lb 140 lb cos30 sin 30Σ − + ° + ° =F i j i j R 
( ) ( ) 181.244 lb 38.0 lb∴ = +R i j 
 or 185.2 lb=R 11.84°W 
Have :O O yM M xRΣ Σ = 
( ) ( ) ( ) ( ) 140 lb cos 30 4 2 cos 30 in. 140 lb sin 30 2 in. sin 30       ∴ − ° + ° − ° °        
 ( )( ) ( )60 lb 2 in. 38.0 lbx− = 
( )1 694.97 70.0 120 in.
38.0
x = − − − 
and 23.289 in.x = − 
 Or, resultant intersects the base (x axis) 23.3 in. to the left of the vertical centerline (y axis) of the motor. W 
 
 
 
 
 
 
PROBLEM 3.152 
To loosen a frozen valve, a force F of magnitude 70 lb is applied to the 
handle of the valve. Knowing that 25 ,θ = ° 61 lb ft,xM = − ⋅ and 
43 lb ft,zM = − ⋅ determine θ and d. 
 
SOLUTION 
Have /:O A O OΣ =M r F M× 
where ( ) ( ) ( )/ 4 in. 11 in.A O d= − + −r i j k 
( )cos cos sin cos sinF θ φ θ θ φ= − +F i j k 
For 70 lb, 25F θ= = ° 
( ) ( ) ( )70 lb 0.90631cos 0.42262 0.90631sinφ φ = − + F i j k( ) 70 lb 4 11 in.
0.90631cos 0.42262 0.90631sin
O d
φ φ
∴ = − −
− −
i j k
M 
 ( ) ( ) ( )70 lb 9.9694sin 0.42262 0.90631 cos 3.6252sind dφ φ φ= − + − + i j 
 ( )1.69048 9.9694cos in.φ + − k 
and ( )( ) ( )( )70 lb 9.9694sin 0.42262 in. 61 lb ft 12 in./ftxM dφ= − = − ⋅ (1) 
 ( )( )70 lb 0.90631 cos 3.6252sin in.yM d φ φ= − + (2) 
 ( )( ) ( )70 lb 1.69048 9.9694cos in. 43 lb ft 12 in./ftzM φ= − = − ⋅ (3) 
 
 
 PROBLEM 3.152 CONTINUED 
From Equation (3) 
1 634.33cos 24.636
697.86
φ −  = = °   
 or 24.6φ = °W 
From Equation (1) 
1022.90 34.577 in.
29.583
d  = =   
 or 34.6 in.d = W 
 
 
 
 
 
 
 
PROBLEM 3.153 
When a force F is applied to the handle of the valve shown, its moments 
about the x and z axes are, respectively, 77 lb ftxM = − ⋅ and 
81 lb ft.zM = − ⋅ For 27 in.,d = determine the moment yM of F about 
the y axis. 
 
SOLUTION 
Have /:O A O OΣ =M r F M× 
where ( ) ( ) ( )/ 4 in. 11 in. 27 in.A O = − + −r i j k 
( )cos cos sin cos sinF θ φ θ θ φ= − +F i j k 
 4 11 27 lb in.
cos cos sin cos sin
O F
θ φ θ θ φ
∴ = − − ⋅
−
i j k
M 
 ( ) ( )11cos sin 27sin 27cos cos 4cos sinF θ φ θ θ φ θ φ= − + − + i j 
 ( ) ( )4sin 11cos cos lb in.θ θ φ + − ⋅k 
and ( )( )11cos sin 27sin lb in.xM F θ φ θ= − ⋅ (1) 
 ( )( )27cos cos 4cos sin lb in.yM F θ φ θ φ= − + ⋅ (2) 
 ( )( )4sin 11cos cos lb in.zM F θ θ φ= − ⋅ (3) 
Now, Equation (1) 1cos sin 27sin
11
xM
F
θ φ θ = +   (4) 
and Equation (3) 1cos cos 4sin
11
zM
F
θ φ θ = −   (5) 
Substituting Equations (4) and (5) into Equation (2), 
1 127 4sin 4 27sin
11 11
z x
y
M MM F
F F
θ θ        = − − + +               
 
or ( )1 27 4
11y z x
M M M= + 
 
 
 PROBLEM 3.153 CONTINUED 
Noting that the ratios 27
11
 and 4
11
 are the ratios of lengths, have 
( ) ( )27 481 lb ft 77 lb ft 226.82 lb ft
11 11y
M = − ⋅ + − ⋅ = ⋅ 
 or 227 lb ftyM = − ⋅ W 
 
 
 
 
PROBLEM 4.1 
The boom on a 4300-kg truck is used to unload a pallet of shingles of 
mass 1600 kg. Determine the reaction at each of the two (a) rear 
wheels B, (b) front wheels C. 
 
SOLUTION 
 
 ( )( )21600 kg 9.81 m/sA AW m g= = 
 15696 N= 
or 15.696 kNA =W 
 ( )( )24300 kg 9.81 m/sG GW m g= = 
 42 183 N= 
or 42.183 kNG =W 
(a) From f.b.d. of truck with boom 
 ( ) ( ) ( )0: 15.696 kN 0.5 0.4 6cos15 m 2 0.5 0.4 4.3 mC BM F   Σ = + + ° − + +    
 ( )( )42.183 kN 0.5 m 0+ = 
 126.1852 24.266 kN
5.2B
F∴ = = 
 or 12.13 kNB =F W 
(b) From f.b.d. of truck with boom 
 ( ) ( ) ( ) ( )0: 15.696 kN 6cos15 4.3 m 42.183 kN 4.3 0.4 mBM    Σ = ° − − +    
 ( )2 4.3 0.9 m 0CF  + + =  
174.7862 33.613 kN
5.2C
F∴ = = 
 or 16.81 kNC =F W 
 Check: ( )0: 33.613 42.183 24.266 15.696 kN 0?yFΣ = − + − = 
 ( )57.879 57.879 kN 0 ok− = 
 
 
 PROBLEM 4.2 
Two children are standing on a diving board of mass 65 kg. Knowing that 
the masses of the children at C and D are 28 kg and 40 kg, respectively, 
determine (a) the reaction at A, (b) the reaction at B. 
 
SOLUTION 
 
 ( )( )265 kg 9.81 m/s 637.65 NG GW m g= = = 
 ( )( )228 kg 9.81 m/s 274.68 NC CW m g= = = 
 ( )( )240 kg 9.81 m/s 392.4 ND DW m g= = = 
(a) From f.b.d. of diving board 
 ( ) ( )( ) ( )( ) ( )( )0: 1.2 m 637.65 N 0.48 m 274.68 N 1.08 m 392.4 N 2.08 m 0B yM AΣ = − − − − = 
1418.92 1182.43 N
1.2y
A∴ = − = − 
 or 1.182 kNy =A W 
(b) From f.b.d. of diving board 
 ( ) ( ) ( ) ( )0: 1.2 m 637.65 N 1.68 m 274.68 N 2.28 m 392.4 N 3.28 m 0A yM BΣ = − − − = 
2984.6 2487.2 N
1.2y
B∴ = = 
 or 2.49 kNy =B W 
 Check: ( )0: 1182.43 2487.2 637.65 274.68 392.4 N 0?yFΣ = − + − − − = 
 ( )2487.2 2487.2 N 0 ok− = 
 
 
 
 
 
PROBLEM 4.3 
Two crates, each weighing 250 lb, are placed as shown in the bed of a 
3000-lb pickup truck. Determine the reactions at each of the two 
(a) rear wheels A, (b) front wheels B. 
 
SOLUTION 
 
(a) From f.b.d. of truck 
 ( )( ) ( )( ) ( )( ) ( )( )0: 250 lb 12.1 ft 250 lb 6.5 ft 3000 lb 3.9 ft 2 9.8 ft 0B AM FΣ = + + − = 
163502 1668.37 lb
9.8A
F∴ = = 
 834 lbA∴ =F W 
(b) From f.b.d. of truck 
 ( )( ) ( )( ) ( )( ) ( )( )0: 2 9.8 ft 3000 lb 5.9 ft 250 lb 3.3 ft 250 lb 2.3 ft 0A BM FΣ = − − + = 
179502 1831.63 lb
9.8B
F∴ = = 
 916 lbB∴ =F W 
 Check: ( )0: 250 1668.37 250 3000 1831.63 lb 0?yFΣ = − + − − + = 
 ( )3500 3500 lb 0 ok− = 
 
 
 
 
PROBLEM 4.4 
Solve Problem 4.3 assuming that crate D is removed and that the position 
of crate C is unchanged. 
P4.3 The boom on a 4300-kg truck is used to unload a pallet of 
shingles of mass 1600 kg. Determine the reaction at each of the two 
(a) rear wheels B, (b) front wheels C 
 
 
SOLUTION 
 
(a) From f.b.d. of truck 
 ( )( ) ( )( ) ( )( )0: 3000 lb 3.9 ft 2 9.8 ft 250 lb 12.1 ft 0B AM FΣ = − + = 
14725 2 1502.55 lb
9.8A
F∴ = = 
 or 751 lbA =F W 
(b) From f.b.d. of truck 
 ( )( ) ( )( ) ( )( )0: 2 9.8 ft 3000 lb 5.9 ft 250 lb 2.3 ft 0A BM FΣ = − + = 
17125 2 1747.45 lb
9.8B
F∴ = = 
 or 874 lbB =F W 
 Check: ( )0: 2 751 874 3000 250 lb 0?yF  Σ = + − − =  
 ( )3250 3250 lb 0 ok− = 
 
 
 
 
PROBLEM 4.5 
A T-shaped bracket supports the four loads shown. Determine the reactions 
at A and B if (a) 100 mm,a = (b) 70 mm.a = 
 
SOLUTION 
 (a) 
 
 From f.b.d. of bracket 
 ( )( ) ( )( ) ( )( ) ( )0: 10 N 0.18 m 30 N 0.1 m 40 N 0.06 m 0.12 m 0BM AΣ = − − + + = 
 2.400 20 N
0.12
A∴ = = or 20.0 N=A W 
 ( ) ( )( ) ( )( ) ( )( ) ( )( )0: 0.12 m 40 N 0.06 m 50 N 0.12 m 30 N 0.22 m 10 N 0.3 m 0AM BΣ = − − − − = 
 18.000 150 N
0.12
B∴ = = or 150.0 N=B W 
(b) 
 
 From f.b.d. of bracket 
 ( )( ) ( )( ) ( )( ) ( )0: 10 N 0.15 m 30 N 0.07 m 40 N 0.06 m 0.12 m 0BM AΣ = − − + + = 
 1.200 10 N
0.12
A∴ = = or 10.00 N=A W 
 ( ) ( )( ) ( )( ) ( )( )0: 0.12 m 40 N 0.06 m 50 N 0.12 m 30 N 0.19 mAM BΣ = − − − 
 ( )( )10 N 0.27 m 0− = 
 16.800 140 N
0.12
B∴ = = or 140.0 N=B W 
 
 
 
 
 
PROBLEM 4.6 
For the bracket and loading of Problem 4.5, determine the smallest 
distance a if the bracket is not to move. 
P4.5 A T-shaped bracket supports the four loads shown. Determine the 
reactions at A and B if (a) 100 mm,a = (b) 70 mm.a = 
 
SOLUTION 
 
 
The mina value will be based on 0=A 
From f.b.d. of bracket 
 ( )( ) ( )( ) ( )( )0: 40 N 60 mm 30 N 10 N 80 mm 0BM a aΣ = − − + = 
1600 40 mm
40
a∴ = = 
 or min 40.0 mma = W 
 
 
 
 
 
PROBLEM 4.7 
A hand truck is used to move two barrels, each weighing 80 lb. 
Neglecting the weight of the hand truck, determine (a) the vertical force 
P which should be applied to the handle to maintain equilibrium when 
o35 ,α = (b) the corresponding reaction at each of the two wheels. 
 
SOLUTION 
 
 
 ( ) ( )1 20 in. sin 8 in. cosa α α= − 
( ) ( )2 32 in. cos 20 in. sina α α= − 
 ( )64 in. cosb α= 
From f.b.d. of hand truck 
 ( ) ( ) ( )2 10: 0BM P b W a W aΣ = − + = (1) 
 0: 2 2 0yF P w BΣ = − + = (2) 
For 35α = ° 
1 20sin 35 8cos35 4.9183 in.a = ° − ° = 
2 32cos35 20sin 35 14.7413 in.a = ° − ° = 
 64cos35 52.426 in.b = ° = 
(a) From Equation (1) 
( ) ( ) ( )52.426 in. 80 lb 14.7413 in. 80 lb 4.9183 in. 0P − + = 
 14.9896 lbP∴ = or 14.99 lb=P W 
(b) From Equation (2) 
( )14.9896 lb 2 80 lb 2 0B− + = 
 72.505 lbB∴ = or 72.5 lb=B W 
 
 
 
 
PROBLEM 4.8 
Solve Problem 4.7 when o40 .α = 
P4.7 A hand truck is used to move two barrels, each weighing 80 lb. 
Neglecting the weight of the hand truck, determine (a) the vertical force P 
which shouldbe applied to the handle to maintain equilibrium when 
o35 ,α = (b) the corresponding reaction at each of the two wheels. 
 
SOLUTION 
 
 
 ( ) ( )1 20 in. sin 8 in. cosa α α= − 
( ) ( )2 32 in. cos 20 in. sina α α= − 
 ( )64 in. cosb α= 
From f.b.d. of hand truck 
 ( ) ( ) ( )2 10: 0BM P b W a W aΣ = − + = (1) 
 0: 2 2 0yF P w BΣ = − + = (2) 
For 40α = ° 
1 20sin 40 8cos 40 6.7274 in.a = ° − ° = 
2 32cos 40 20sin 40 11.6577 in.a = ° − ° = 
 64cos 40 49.027 in.b = ° = 
(a) From Equation (1) 
( ) ( ) ( )49.027 in. 80 lb 11.6577 in. 80 lb 6.7274 in. 0P − + = 
 8.0450 lbP∴ = 
 or 8.05 lb=P W 
(b) From Equation (2) 
( )8.0450 lb 2 80 lb 2 0B− + = 
 75.9775 lbB∴ = 
 or 76.0 lb=B W 
 
 
 
 
PROBLEM 4.9 
Four boxes are placed on a uniform 14-kg wooden plank which rests 
on two sawhorses. Knowing that the masses of boxes B and D are 
4.5 kg and 45 kg, respectively, determine the range of values of the 
mass of box A so that the plank remains in equilibrium when box C is 
removed. 
 
SOLUTION 
 
 
 
 
 
45A A D DW m g W m g g= = = 
4.5 14B B G GW m g g W m g g= = = = 
For ( )min, 0Am E = 
 ( )( ) ( )( )0: 2.5 m 4.5 1.6 mF AM m g gΣ = + 
 ( )( ) ( )( )14 1 m 45 0.6 m 0g g+ − = 
2.32 kgAm∴ = 
For ( )max, 0:Am F = 
 ( ) ( )( ) ( )( )0: 0.5 m 4.5 0.4 m 14 1 mE AM m g g gΣ = − − 
 ( )( )45 2.6 m 0g− = 
265.6 kgAm∴ = 
 or 2.32 kg 266 kgAm≤ ≤ W 
 
 
 
 
 
PROBLEM 4.10 
A control rod is attached to a crank at A and cords are attached at B and 
C. For the given force in the rod, determine the range of values of the 
tension in the cord at C knowing that the cords must remain taut and that 
the maximum allowed tension in a cord is 180 N. 
 
SOLUTION 
 
 
 
For ( )max, 0C BT T = 
 ( ) ( ) ( )( )max0: 0.120 m 400 N 0.060 m 0O CM TΣ = − = 
 ( ) maxmax 200 N 180 NCT T= > = 
( )max 180.0 NCT∴ = 
For ( ) maxmin , 180 NC BT T T= = 
 ( ) ( ) ( )( )min0: 0.120 m 180 N 0.040 mO CM TΣ = + 
 ( )( )400 N 0.060 m 0− = 
( )min 140.0 NCT∴ = 
Therefore, 140.0 N 180.0 NCT≤ ≤ W 
 
 
 
 
 
 
PROBLEM 4.11 
The maximum allowable value of each of the reactions is 360 N. 
Neglecting the weight of the beam, determine the range of values of the 
distance d for which the beam is safe. 
 
SOLUTION 
 
 
 
From f.b.d. of beam 
 0: 0 so that x x yF B B BΣ = = = 
 ( )0: 100 200 300 N 0yF A BΣ = + − + + = 
or 600 NA B+ = 
Therefore, if either A or B has a magnitude of the maximum of 360 N, 
the other support reaction will be ( )360 N 600 N 360 N 240 N .< − = 
 ( )( ) ( )( ) ( )( )0: 100 N 200 N 0.9 300 N 1.8AM d d dΣ = − − − − 
 ( )1.8 0B d+ − = 
or 720 1.8
600
Bd
B
−= − 
Since 360 N,B ≤ 
( )720 1.8 360 0.300 m or 300 mm
600 360
d d
−= = ≥− 
 ( )( ) ( ) ( )( )0: 100 N 1.8 1.8 200 N 0.9 0BM A dΣ = − − + = 
or 1.8 360Ad
A
−= 
Since 360 N,A ≤ 
( )1.8 360 360 0.800 m or 800 mm
360
d d
−= = ≤ 
 or 300 mm 800 mmd≤ ≤ W 
 
 
 
 
 
 
PROBLEM 4.12 
Solve Problem 4.11 assuming that the 100-N load is replaced by a 160-N 
load. 
P4.11 The maximum allowable value of each of the reactions is 360 N. 
Neglecting the weight of the beam, determine the range of values of the 
distance d for which the beam is safe. 
 
SOLUTION 
 
 
 
From f.b.d of beam 
 0: 0 so that x x yF B B BΣ = = = 
 ( )0: 160 200 300 N 0yF A BΣ = + − + + = 
or 660 NA B+ = 
Therefore, if either A or B has a magnitude of the maximum of 360 N, 
the other support reaction will be ( )360 N 660 360 300 N .< − = 
 ( ) ( ) ( )0: 160 N 200 N 0.9 300 N 1.8AM d d dΣ = − − − − 
 ( )1.8 0B d+ − = 
or 720 1.8
660
Bd
B
−= − 
Since 360 N,B ≤ 
( )720 1.8 360 0.240 m or 240 mm
660 360
d d
−= = ≥− 
 ( ) ( ) ( )0: 160 N 1.8 1.8 200 N 0.9 0BM A dΣ = − − + = 
or 1.8 468Ad
A
−= 
Since 360 N,A ≤ 
( )1.8 360 468 0.500 m or 500 mm
360
d d
−= = ≥ 
or 240 mm 500 mmd≤ ≤ W 
 
 
 
 
 
 
PROBLEM 4.13 
For the beam of Sample Problem 4.2, determine the range of values of 
P for which the beam will be safe knowing that the maximum 
allowable value of each of the reactions is 45 kips and that the 
reaction at A must be directed upward. 
 
SOLUTION 
 
For the force of P to be a minimum, A = 0. 
With 0,A = 
 ( ) ( )( ) ( )( )min0: 6 ft 6 kips 2 ft 6 kips 4 ft 0BM PΣ = − − = 
min 6.00 kipsP∴ = 
For the force P to be a maximum, max 45 kips= =A A 
With 45 kips,A = 
 ( )( ) ( ) ( )( ) ( )( )max0: 45 kips 9 ft 6 ft 6 kips 2 ft 6 kips 4 ft 0BM PΣ = − + − − = 
max 73.5 kipsP∴ = 
A check must be made to verify the assumption that the maximum value of P is based on the reaction force at 
A. This is done by making sure the corresponding value of B is 45 kips.< 
 0: 45 kips 73.5 kips 6 kips 6 kips 0yF BΣ = − + − − = 
max40.5 kips 45 kips ok or 73.5 kipsB P∴ = < ∴ = 
 and 6.00 kips 73.5 kipsP≤ ≤ W 
 
 
 
 
 
 
PROBLEM 4.14 
For the beam and loading shown, determine the range of values of the 
distance a for which the reaction at B does not exceed 50 lb 
downward or 100 lb upward. 
 
SOLUTION 
 
 
 
 
 
 
To determine maxa the two 150-lb forces need to be as close to B without 
having the vertical upward force at B exceed 100 lb. 
From f.b.d. of beam with 100 lb=B 
 ( )( ) ( )( )max max0: 150 lb 4 in. 150 lb 1 in.DM a aΣ = − − − − 
 ( )( )25 lb 2 in.− ( )( )100 lb 8 in. 0+ = 
or max 5.00 in.a = 
To determine mina the two 150-lb forces need to be as close to A without 
having the vertical downward force at B exceed 50 lb. 
From f.b.d. of beam with 50 lb=B 
 ( )( ) ( )( )min min0: 150 lb 4 in. 150 lb 1 in.DM a aΣ = − − − 
 ( )( ) ( )( )25 lb 2 in. 50 lb 8 in. 0− − = 
or min 1.00 in.a = 
Therefore, or 1.00 in. 5.00 in.a≤ ≤ W 
 
 
 
 
 
 
PROBLEM 4.15 
A follower ABCD is held against a circular cam by a stretched spring, 
which exerts a force of 21 N for the position shown. Knowing that the 
tension in rod BE is 14 N, determine (a) the force exerted on the roller 
at A, (b) the reaction at bearing C. 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
Note: From f.b.d. of ABCD 
cos60
2x
AA A= ° = 
3sin 60
2y
A A A= ° = 
(a) From f.b.d. of ABCD 
 ( ) ( )0: 40 mm 21 N 40 mm2C
AM  Σ = −   
 ( )14 N 20 mm 0+ = 
28 NA∴ = 
or 28.0 N =A 60°W 
(b) From f.b.d. of ABCD 
 ( )0: 14 N 28 N cos60 0x xF CΣ = + + ° = 
28 N or 28.0 Nx xC∴ = − =C 
 0:yFΣ = ( )21 N 28 N sin 60 0yC − + ° = 
3.2487 N or 3.25 Ny yC∴ = − =C 
Then ( ) ( )2 22 2 28 3.2487 28.188 Nx yC C C= + = + = 
and 1 1 3.2487tan tan 6.6182
28
y
x
C
C
θ − −  − = = = °   −  
 
or 28.2 N =C 6.62°W 
 
 
 
 
 
 
PROBLEM 4.16 
A 6-m-long pole AB is placed in a hole and is guyed by three cables. 
Knowing that the tensions in cables BD and BE are 442 N and 322 N, 
respectively, determine (a) the tension in cable CD, (b) the reaction 
at A. 
 
SOLUTION 
 
 
 
Note: 
( ) ( )2 22.8 5.25 5.95 mDB = + = 
( ) ( )2 22.8 2.10 3.50 mDC = + = 
(a) From f.b.d. of pole 
 ( )( ) ( ) ( )2.8 m0: 322 N 6 m 442 N 6 m5.95 mAM
  Σ = − +      
 ( )2.8 m 2.85 m 0
3.50 m CD
T  + =     
300 NCDT∴ = 
or 300 NCDT = W 
(b) From f.b.d. of pole 
 ( )2.8 m0: 322 N 442 N
5.95 mx
F  Σ = −    
 ( )2.8 m 300 N 0
3.50 m x
A − + =   
126 N or 126 Nx xA∴ = =A 
 ( )5.25 m0: 442 N5.95 my yF A  Σ = −    ( )
2.10 m 300 N 0
3.50 m
 − =   
570 N or 570 Ny yA∴ = =A 
Then ( ) ( )2 22 2 126 570 583.76 Nx yA A A= + = + = 
and 1570 Ntan 77.535
126 N
θ −  = = °   
or 584 N=A 77.5°W 
 
 
 
 
 
 
PROBLEM 4.17 
Determine the reactions at A and C when (a) 0,α = (b) o30 .α = 
 
SOLUTION 
 (a) 
 
 
 
 
 
 
 
 
 
 
 (b) 
 
 
(a) 0α = ° 
 From f.b.d. of member ABC 
 ( )( ) ( )( ) ( )0: 80 lb 10 in. 80 lb 20 in. 40 in. 0CM AΣ = + − = 
 60 lbA∴ = 
or 60.0 lb=A W 
 0: 60 lb 0y yF CΣ = + = 
 60 lb or 60 lby yC∴ = − =C 
 0: 80 lb 80 lb 0x xF CΣ = + + = 
 160 lb or 160 lbx xC∴ = − =C 
Then ( ) ( )2 22 2 160 60 170.880 lbx yC C C= + = + = 
and 1 1 60tan tan 20.556
160
y
x
C
C
θ − −  − = = = °   −  
 
or 170.9 lb=C 20.6°W 
(b) 30α = ° 
 From f.b.d. of member ABC 
 ( )( ) ( )( )0: 80 lb 10 in. 80 lb 20 in.CMΣ = + ( )( )cos30 40 in.A− ° 
 ( )( )sin 30 20 in. 0A+ ° = 
97.399 lbA∴ = 
or 97.4 lb=A 60°W 
 
 
 
 
PROBLEM 4.17 CONTINUED 
 ( )0: 80 lb 80 lb 97.399 lb sin 30 0x xF CΣ = + + ° + = 
 208.70 lb or 209 lbx xC∴ = − =C 
 ( )0: 97.399 lb cos30 0y yF CΣ = + ° = 
 84.350 lb or 84.4 lby yC∴ = − =C 
Then ( ) ( )2 22 2 208.70 84.350 225.10 lbx yC C C= + = + = 
and 1 1 84.350tan tan 22.007
208.70
y
x
C
C
θ − −  − = = = °   −  
 
or 225 lb=C 22.0°W 
 
 
 
 
 
 
PROBLEM 4.18 
Determine the reactions at A and B when (a) 0,h = (b) 8 in.h = 
 
SOLUTION 
 (a) 
 
 
 
 
 
 
 
 
 
 
 (b) 
 
 
(a) 0h = 
 From f.b.d. of plate 
 ( )( ) ( )( )0: sin 30 20 in. 40 lb 10 in. 0AM BΣ = ° − = 
40 lbB∴ = 
or 40.0 lb=B 30°W 
 ( )0: 40 lb cos30 0x xF AΣ = − ° = 
 34.641 lbxA∴ = or 34.6 lbx =A 
 ( )0: 40 lb 40 lb sin 30 0y yF AΣ = − + ° = 
 20 lbyA∴ = or 20.0 lby =A 
Then ( ) ( )2 22 2 34.641 20 39.999 lbx yA A A= + = + = 
and 1 1 20tan tan 30.001
34.641
y
x
A
A
θ − −   = = = °     
 
or 40.0 lb=A 30°W 
(b) 8 in.h = 
 From f.b.d. of plate 
 ( )( ) ( )( )0: sin 30 20 in. cos30 8 in.AM B BΣ = ° − ° 
 ( )( )40 lb 10 in. 0− = 
130.217 lbB∴ = 
or 130.2 lb=B 30.0°W 
 
 
 
 
PROBLEM 4.18 CONTINUED 
 ( )0: 130.217 lb cos30 0x xF AΣ = − ° = 
 112.771 lb or 112.8 lbx xA∴ = =A 
 ( )0: 40 lb 130.217 lb sin 30 0y yF AΣ = − + ° = 
 25.108 lb or 25.1 lby yA∴ = − =A 
Then ( ) ( )2 22 2 112.771 25.108 115.532 lbx yA A A= + = + = 
and 1 1 25.108tan tan 12.5519
112.771
y
x
A
A
θ − −  − = = = − °     
 
or 115.5 lb=A 12.55°W 
 
 
 
 
 
 
PROBLEM 4.19 
The lever BCD is hinged at C and is attached to a control rod at B. If 
200 N,P = determine (a) the tension in rod AB, (b) the reaction at C. 
 
SOLUTION 
 
 
(a) From f.b.d. of lever BCD 
 ( ) ( )0: 50 mm 200 N 75 mm 0C ABM TΣ = − = 
 300 NABT∴ = W 
(b) From f.b.d. of lever BCD 
 ( )0: 200 N 0.6 300 N 0x xF CΣ = + + = 
 380 N or 380 Nx xC∴ = − =C 
 ( )0: 0.8 300 N 0y yF CΣ = + = 
 y240 N or 240 NyC∴ = − =C 
Then ( ) ( )2 22 2 380 240 449.44 Nx yC C C= + = + = 
and 1 1 240tan tan 32.276
380
y
x
C
C
θ − −  − = = = °   −  
 
or 449 N=C 32.3°W 
 
 
 
 
 
 
PROBLEM 4.20 
The lever BCD is hinged at C and is attached to a control rod at B. 
Determine the maximum force P which can be safely applied at D if the 
maximum allowable value of the reaction at C is 500 N. 
 
SOLUTION 
 
 
 
From f.b.d. of lever BCD 
 ( ) ( )0: 50 mm 75 mm 0C ABM T PΣ = − = 
 1.5ABT P∴ = (1) 
 0: 0.6 0x AB xF T P CΣ = + − = 
 0.6x ABC P T∴ = + (2) 
From Equation (1) ( )0.6 1.5 1.9xC P P P= + = 
 0: 0.8 0y AB yF T CΣ = − = 
 0.8y ABC T∴ = (3) 
From Equation (1) ( )0.8 1.5 1.2yC P P= = 
From Equations (2) and (3) 
( ) ( )2 22 2 1.9 1.2 2.2472x yC C C P P P= + = + = 
Since max 500 N,C = 
max500 N 2.2472P∴ = 
or max 222.49 lbP = 
 or 222 lb=P W 
 
 
 
 
 
PROBLEM 4.21 
The required tension in cable AB is 800 N. Determine (a) the vertical 
force P which must be applied to the pedal, (b) the corresponding 
reaction at C. 
 
SOLUTION 
 
 
 
 
 
(a) From f.b.d. of pedal 
 ( ) ( ) ( )0: 0.4 m 800 N 0.18 m sin 60 0CM P  Σ = − ° =  
 311.77 NP∴ = 
or 312 N=P W 
(b) From f.b.d. of pedal 
 0: 800 N 0x xF CΣ = − = 
 800 NxC∴ = 
 or 800 Nx =C 
 0: 311.77 N 0y yF CΣ = − = 
 311.77 NyC∴ = 
 or 311.77 Ny =C 
 Then ( ) ( )2 22 2 800 311.77 858.60 Nx yC C C= + = + = 
 and 1 1 311.77tan tan 21.291
800
y
x
C
C
θ − −   = = = °     
 
or 859 N=C 21.3°W 
 
 
 
 
 
PROBLEM 4.22 
Determine the maximum tension which can be developed in cable AB 
if the maximum allowable value of the reaction at C is 1000 N. 
 
SOLUTION 
 
 
 
 
 
 
Have max 1000 NC = 
Now 2 2 2x yC C C= + 
 ( )2 2 1000y xC C∴ = − (1) 
From f.b.d. of pedal 
 max0: 0x xF C TΣ = − = 
 max xC T∴ = (2) 
 ( ) ( )max0: 0.4 m 0.18 m sin 60 0D yM C T  Σ = − ° =  
 max 0.38971yC T∴ = (3) 
Equating the expressions for yC in Equations (1) and (3), with maxxC T= 
from Equation (2) 
( )2 2max max1000 0.389711T T− = 
2
max 868,150T∴ = 
and max 931.75 NT = 
maxor 932 NT = W 
 
 
 
 
 
PROBLEM 4.23 
A steel rod is bent to form a mounting bracket. For each of the mounting 
brackets and loadings shown, determine the reactions at A and B. 
 
SOLUTION 
 
(a) 
 
 
 
 
 
 
 
 
 
(b) 
 
 
 
(a) From f.b.d. of mounting bracket 
 ( ) ( )( )0: 8 in. 80 lb in. 10 lb 6 in.EM AΣ = − ⋅ − 
 ( )( )20 lb 12 in. 0− = 
 47.5 lbA∴ = 
or 47.5 lb=A W 
 0: 10 lb 47.5 lb 0x xF BΣ = − + = 
 37.5 lbxB∴ = − 
 or 37.5 lbx =B 
 0: 20 lb 0y yF BΣ = − = 
 20 lbyB∴ = 
 or 20.0 lby =B 
 Then ( ) ( )2 22 2 37.5 20.0 42.5 lbx yB B B= + = + = 
 and 1 1 20tan tan 28.072
37.5
y
x
B
B
θ − −   = = = − °   −  
 
or 42.5 lb=B 28.1°W 
(b) From f.b.d. of mounting bracket 
 ( )( )0: cos 45 8 in. 80 lb in.BM AΣ = ° − ⋅ 
 ( )( ) ( )( )10 lb 6 in. 20 lb 12 in. 0− − = 
 67.175 lbA∴ = 
or 67.2 lb=A 45°W 
 0: 10 lb 67.175cos 45 0x xF BΣ = − + ° = 
 37.500 lbxB∴ = − 
 or 37.5 lbx =B 
 
 
 
 
 
 
PROBLEM 4.23 CONTINUED 
 0: 20 lb 67.175sin 45 0y yF BΣ = − + ° = 
 27.500 lbyB∴ = − 
 or 27.5 lby =B 
 Then ( ) ( )2 22 2 37.5 27.5 46.503 lbx yB B B= + = + = 
 and 1 1 27.5tan tan 36.254
37.5
y
x
B
B
θ − −  − = = = °   −  
 
or 46.5 lb=B 36.3°W 
 
 
 
 
 
PROBLEM 4.24 
A steel rod is bent to form a mounting bracket. For each of the 
mounting brackets and loadings shown, determine the reactions at A 
and B. 
 
SOLUTION 
(a) 
 
 
 
 
 
(b) 
 
 
 
(a) From f.b.d. of mounting bracket 
 ( ) ( )( )0: 8 in. 20 lb 12 in.AM BΣ = − − 
 ( )( )10 lb 2 in. 80 lb in. 0+ − ⋅ = 
 37.5 lbB∴ = − 
or 37.5 lb=B W 
 0: 37.5 lb 10 lb 0x xF AΣ = − − + = 
 47.5 lbxA∴ = 
 or 47.5 lbx =A 
 0: 20 lb 0y yF AΣ = − + = 
 20 lbyA∴ = 
 or 20.0 lby =A 
 Then ( ) ( )2 22 2 47.5 20 51.539 lbx yA A A= + = + = 
 and 1 1 20tan tan 22.834
47.5
y
x
A
A
θ − −   = = = °     
 
or 51.5 lb=A 22.8°W 
(b) From f.b.d. of mounting bracket 
 ( )( ) ( )( )0: cos 45 8 in. 20 lb 2 in.AM BΣ = − ° − 
 ( )( )80 lb in. 10 lb 2 in. 0− ⋅ + = 
 53.033 lbB∴ = − 
or 53.0 lb=B 45°W 
 ( )0: 53.033 lb cos 45 10 0x xF AΣ = + − ° − = 
 47.500 lbxA∴ = 
 or 47.5 lbx =A 
 
 
 
 
 
 
PROBLEM 4.24 CONTINUED 
 ( )0: 53.033 lb sin 45 20 0y yF AΣ = − ° − = 
 17.500 lbyA∴ = − 
 or 17.50lby =A 
 Then ( ) ( )2 22 2 47.5 17.5 50.621 lbx yA A A= + = + = 
 and 1 1 17.5tan tan 20.225
47.5
y
x
A
A
θ − −  − = = = − °     
 
or 50.6 lb=A 20.2°W 
 
 
 
 
 
 
PROBLEM 4.25 
A sign is hung by two chains from mast AB. The mast is hinged at A and 
is supported by cable BC. Knowing that the tensions in chains DE and 
FH are 50 lb and 30 lb, respectively, and that 1.3 ft,d = determine 
(a) the tension in cable BC, (b) the reaction at A. 
 
SOLUTION 
 
 
 
 
 
 
 
First note ( ) ( )2 28.4 1.3 8.5 ftBC = + = 
(a) From f.b.d. of mast AB 
 ( ) ( )( )8.40: 2.5 ft 30 lb 7.2 ft8.5A BCM T
  Σ = −     
 ( )50 lb 2.2 ft 0− = 
 131.952 lbBCT∴ = 
or 132.0 lbBCT = W 
(b) From f.b.d. of mast AB 
 ( )8.40: 131.952 lb 0
8.5x x
F A  Σ = − =   
 130.400 lbxA∴ = 
 or 130.4 lbx =A 
 ( )1.30: 131.952 lb 30 lb 50 lb 08.5y yF A  Σ = + − − =   
 59.819 lbyA∴ = 
 or 59.819 lby =A 
 Then ( ) ( )2 22 2 130.4 59.819 143.466 lbx yA A A= + = + = 
 and 1 1 59.819tan tan 24.643
130.4
y
x
A
A
θ − −   = = = °     
 
or 143.5 lb=A 24.6°W 
 
 
 
 
 
 
PROBLEM 4.26 
A sign is hung by two chains from mast AB. The mast is hinged at A 
and is supported by cable BC. Knowing that the tensions in chains DE 
and FH are 30 lb and 20 lb, respectively, and that 1.54 ft,d = 
determine (a) the tension in cable BC, (b) the reaction at A. 
 
SOLUTION 
 
 
 
 
 
 
First note ( ) ( )2 28.4 1.54 8.54 ftBC = + = 
(a) From f.b.d. of mast AB 
 ( ) ( )8.40: 2.5 ft 20 lb 7.2 ft8.54A BCM T
  Σ = −     
 ( )30 lb 2.2 ft 0− = 
 85.401 lbBCT∴ = 
or 85.4 lbBCT = W 
(b) From f.b.d. of mast AB 
 ( )8.40: 85.401 lb 0
8.54x x
F A  Σ = − =   
 84.001 lbxA∴ = 
 or 84.001 lbx =A 
 ( )1.540: 85.401 lb 20 lb 30 lb 08.54y yF A  Σ = + − − =   
 34.600 lbyA∴ = 
 or 34.600 lby =A 
 Then ( ) ( )2 22 2 84.001 34.600 90.848 lbx yA A A= + = + = 
 and 1 1 34.6tan tan 22.387
84.001
y
x
A
A
θ − −   = = = °     
 
or 90.8 lb=A 22.4°W 
 
 
 
 
 
PROBLEM 4.27 
For the frame and loading shown, determine the reactions at A and E 
when (a) o30 ,α = (b) o45 .α = 
 
SOLUTION 
 
(a) 
 
 
 
 
(a) Given 30α = ° 
 From f.b.d. of frame 
 ( )( ) ( )( )0: 90 N 0.2 m 90 N 0.06 mAMΣ = − − 
 ( )( ) ( )( )cos60 0.160 m sin 60 0.100 m 0E E+ ° + ° = 
 140.454 NE∴ = 
or 140.5 N=E 60°W 
 ( )0: 90 N 140.454 N cos60 0x xF AΣ = − + ° = 
 19.7730 NxA∴ = 
 or 19.7730 Nx =A 
 ( )0: 90 N 140.454 N sin 60 0y yF AΣ = − + ° = 
 31.637 NyA∴ = − 
 or 31.6 Ny =A 
 Then ( ) ( )2 22 2 19.7730 31.637x yA A A= + = + 
 37.308 lb= 
 and 1 1 31.637tan tan
19.7730
y
x
A
A
θ − −  − = =     
 
 57.995= − ° 
 or 37.3 N=A 58.0°W 
 
 
 
 
 
(b) 
 
PROBLEM 4.27 CONTINUED 
(b) Given 45α = ° 
 From f.b.d. of frame 
 ( )( ) ( )( )0: 90 N 0.2 m 90 N 0.06 mAMΣ = − − 
 ( )( ) ( )( )cos 45 0.160 m sin 45 0.100 m 0E E+ ° + ° = 
127.279 NE∴ = 
or 127.3 N=E 45°W 
 ( )0: 90 127.279 N cos 45 0x xF AΣ = − + ° = 
 0xA∴ = 
 ( )0: 90 127.279 N sin 45 0y yF AΣ = − + ° = 
 0yA∴ = 
or 0=A W 
 
 
 
 
 
 
PROBLEM 4.28 
A lever AB is hinged at C and is attached to a control cable at A. If the 
lever is subjected to a 300-N vertical force at B, determine 
(a) the tension in the cable, (b) the reaction at C. 
 
SOLUTION 
 
 
 
 
First 
( )0.200 m cos 20 0.187 939 mACx = ° = 
( )0.200 m sin 20 0.068 404 mACy = ° = 
Then 
0.240 mDA ACy y= − 
 0.240 m 0.068404 m= − 
 0.171596 m= 
and 0.171 596tan
0.187 939
DA
AC
y
x
α = = 
 42.397α∴ = ° 
and 90 20 42.397 27.603β = ° − ° − ° = ° 
(a) From f.b.d. of lever AB 
 ( )0: cos 27.603 0.2 mCM TΣ = ° 
 ( )300 N 0.3 m cos 20 0 − ° =  
 477.17 NT∴ = or 477 NT = W 
(b) From f.b.d. of lever AB 
 ( )0: 477.17 N cos 42.397 0x xF CΣ = + ° = 
 352.39 NxC∴ = − 
 or 352.39 Nx =C 
 ( )0: 300 N 477.17 N sin 42.397 0y yF CΣ = − − ° = 
 621.74 NyC∴ = 
 or 621.74 Ny =C 
 
 
 
 PROBLEM 4.28 CONTINUED 
 Then ( ) ( )2 22 2 352.39 621.74 714.66 Nx yC C C= + = + = 
 and 1 1 621.74tan tan 60.456
352.39
y
x
C
C
θ − −   = = = − °   −  
 
 or 715 N=C 60.5°W 
 
 
 
 
 
 
PROBLEM 4.29 
Neglecting friction and the radius of the pulley, determine the tension 
in cable BCD and the reaction at support A when 80 mm.d = 
 
SOLUTION 
 
 
 
 
 
First 
1 60tan 12.0948
280
α −  = = °   
1 60tan 36.870
80
β −  = = °   
From f.b.d. of object BAD 
 ( )( ) ( )( )0: 40 N 0.18 m cos 0.08 mAM T αΣ = + 
 ( )( ) ( )( )sin 0.18 m cos 0.08 mT Tα β+ − 
 ( )( )sin 0.18 m 0T β− = 
7.2 N m 128.433 N
0.056061
T ⋅ ∴ = =   
or 128.4 NT = W 
 ( )( )0: 128.433 N cos cos 0x xF Aβ αΣ = − + = 
 22.836 NxA∴ = 
or 22.836 Nx =A 
 ( )( )0: 128.433 N sin sin 40 N 0y yF A β αΣ = + + + = 
 143.970 NyA∴ = − 
or 143.970 Ny =A 
Then ( ) ( )2 22 2 22.836 143.970 145.770 Nx yA A A= + = + = 
and 1 1 143.970tan tan 80.987
22.836
y
x
A
A
θ − −  − = = = − °     
 
or 145.8 N=A 81.0°W 
 
 
 
 
 
 
PROBLEM 4.30 
Neglecting friction and the radius of the pulley, determine the tension in 
cable BCD and the reaction at support A when 144 mm.d = 
 
SOLUTION 
 
 
 
 
 
First note 
1 60tan 15.5241
216
α −  = = °   
1 60tan 22.620
144
β −  = = °   
From f.b.d. of member BAD 
 ( )( ) ( )( )0: 40 N 0.18 m cos 0.08 mAM T αΣ = + 
 ( )( ) ( )( )sin 0.18 m cos 0.08 mT Tα β+ − 
 ( )( )sin 0.18 m 0T β− = 
7.2 N m 404.04 N
0.0178199m
T
 ⋅∴ = =  
 
or 404 NT = W 
 ( )( )0: 404.04 N cos cos 0x xF A β αΣ = + − = 
 16.3402 NxA∴ = 
or 16.3402 Nx =A 
 ( )( )0: 404.04 N sin sin 40 N 0y yF A β αΣ = + + + = 
 303.54 NyA∴ = − 
or 303.54 Ny =A 
Then ( ) ( )2 22 2 16.3402 303.54 303.98 Nx yA A A= + = + = 
and 1 1 303.54tan tan 86.919
16.3402
y
x
A
A
θ − −  − = = = − °     
 
or 304 N=A 86.9°W 
 
 
 
 
 
 
PROBLEM 4.31 
Neglecting friction, determine the tension in cable ABD and the reaction 
at support C. 
 
SOLUTION 
 
 
 
 
From f.b.d. of inverted T-member 
 ( ) ( ) ( )( )0: 25 in. 10 in. 30 lb 10 in. 0CM T TΣ = − − = 
20 lbT∴ = 
 or 20.0 lbT = W 
 0: 20 lb 0x xF CΣ = − = 
 20 lbxC∴ = 
or 20.0 lbx =C 
 0: 20 lb 30 lb 0y yF CΣ = + − = 
 10 lbyC∴ = 
or 10.00 lby =C 
Then ( ) ( )2 22 2 20 10 22.361 lbx yC C C= + = + = 
and 1 1 10tan tan 26.565
20
y
x
C
C
θ − −   = = = °     
 
or 22.4 lb=C 26.6°W 
 
 
 
 
 
 
PROBLEM 4.32 
Rod ABC is bent in the shape of a circular arc of radius R. Knowing 
that o35 ,θ = determine the reaction (a) at B, (b) at C. 
 
SOLUTION 
 
 
 
 
For 35θ = ° 
(a) From the f.b.d. of rod ABC 
 ( ) ( )0: 0D xM C R P RΣ = − = 
 xC P∴ = 
 or x P=C 
 0: sin 35 0xF P BΣ = − ° = 
 1.74345
sin 35
PB P∴ = =° 
 or 1.743P=B 55.0°W 
(b) From the f.b.d. of rod ABC 
 ( )0: 1.74345 cos35 0y yF C P PΣ = + ° − = 
 0.42815yC P∴ = − 
 or 0.42815y P=C 
 Then ( ) ( )2 22 2 0.42815 1.08780x yC C C P P P= + = + = 
 and 1 1 0.42815tan tan 23.178y
x
C P
C P
θ − −  − = = = − °     
 
 or 1.088P=C 23.2°W 
 
 
 
 
 
 
PROBLEM 4.33 
Rod ABC is bent in the shape of a circular arc of radius R. Knowing 
that o50 ,θ = determine the reaction (a) at B, (b) at C. 
 
SOLUTIONFor 50θ = ° 
(a) From the f.b.d. of rod ABC 
 ( ) ( )0: 0D xM C R P RΣ = − = 
 xC P∴ = 
 or x P=C 
 0: sin 50 0xF P BΣ = − ° = 
 1.30541
sin 50
PB P∴ = =° 
 or 1.305P=B 40.0°W 
(b) From the f.b.d. of rod ABC 
 ( )0: 1.30541 cos50 0y yF C P PΣ = − + ° = 
 0.160900yC P∴ = 
 or 0.1609y P=C 
 Then ( ) ( )2 22 2 0.1609 1.01286x yC C C P P P= + = + = 
 and 1 1 0.1609tan tan 9.1405y
x
C P
C P
θ − −   = = = °     
 
 or 1.013P=C 9.14°W 
 
 
 
 
 
 
PROBLEM 4.34 
Neglecting friction and the radius of the pulley, determine (a) the 
tension in cable ABD, (b) the reaction at C. 
 
SOLUTION 
 
 
 
 
First note 
1 15tan 22.620
36
α −  = = °   
1 15tan 36.870
20
β −  = = °   
(a) From f.b.d. of member ABC 
 ( )( ) ( )( )0: 30 lb 28 in. sin 22.620 36 in.CM TΣ = − ° 
 ( )( )sin 36.870 20 in. 0T− ° = 
 32.500 lbT∴ = 
 or 32.5 lbT = W 
(b) From f.b.d. of member ABC 
 ( )( )0: 32.500 lb cos 22.620 cos36.870 0x xF CΣ = + ° + ° = 
 56.000 lbxC∴ = − 
 or 56.000 lbx =C 
 ( )( )0: 30 lb 32.500 lb sin 22.620 sin 36.870 0y yF CΣ = − + ° + ° = 
 2.0001 lbyC∴ = − 
 or 2.0001 lby =C 
 Then ( ) ( )2 22 2 56.0 2.001 56.036 lbx yC C C= + = + = 
 and 1 1 2.0tan tan 2.0454
56.0
y
x
C
C
θ − −  − = = = °   −  
 
 or 56.0 lb=C 2.05°W 
 
 
 
 
 
PROBLEM 4.35 
Neglecting friction, determine the tension in cable ABD and the 
reaction at C when o60 .θ = 
 
SOLUTION 
 
 
 
 
From f.b.d. of bent ACD 
 ( )( ) ( )( )0: cos30 2 sin 60 sin 30 2 cos60CM T a T a aΣ = ° ° + ° + ° 
 ( ) ( ) 0T a P a− − = 
 
1.5
PT∴ = 
 2or 
3
PT = W 
 20: cos30 0
3x x
PF C  Σ = − ° =   
 3 0.57735
3x
C P P∴ = = 
or 0.577x P=C 
 
2 20: cos60 0
3 3y y
PF C P P  Σ = + − + ° =   
 0yC∴ = 
 or 0.577P=C W 
 
 
 
 
 
 
PROBLEM 4.36 
Neglecting friction, determine the tension in cable ABD and the 
reaction at C when o30 .θ = 
 
SOLUTION 
 
 
 
 
From f.b.d. of bent ACD 
 ( )( ) ( )0: cos60 2 sin 30 sin 60 2 cos30CM T a T a aΣ = ° ° + ° + ° 
 ( ) ( ) 0P a T a− − = 
 0.53590
1.86603
PT P∴ = = 
 or 0.536T P= W 
 ( )0: 0.53590 cos60 0x xF C PΣ = − ° = 
 0.26795xC P∴ = 
or 0.268x P=C 
 ( )0: 0.53590 0.53590 sin 60 0y yF C P P PΣ = + − + ° = 
 0yC∴ = 
 or 0.268P=C W 
 
 
 
 
 
 
PROBLEM 4.37 
Determine the tension in each cable and the reaction at D. 
 
SOLUTION 
 
First note 
( ) ( )2 220 8 in. 21.541 in.BE = + = 
( ) ( )2 210 8 in. 12.8062 in.CF = + = 
From f.b.d. of member ABCD 
 ( )( ) ( )80: 120 lb 20 in. 10 in. 021.541C BEM T
  Σ = − =     
 646.24 lbBET∴ = 
 or 646 lbBET = W 
 ( )8 80: 120 lb 646.24 lb 0
21.541 12.8062y CF
F T   Σ = − + − =       
 192.099 lbCFT∴ = 
 or 192.1 lbCFT = W 
 ( ) ( )20 100: 646.24 lb 192.099 lb 0
21.541 12.8062x
F D   Σ = + − =       
 750.01 lbD∴ = 
 or 750 lb=D W 
 
 
 
 
 
PROBLEM 4.38 
Rod ABCD is bent in the shape of a circular arc of radius 80 mm and 
rests against frictionless surfaces at A and D. Knowing that the collar 
at B can move freely on the rod and that o45 .θ = determine (a) the 
tension in cord OB, (b) the reactions at A and D. 
 
SOLUTION 
 
 
 
 
(a) From f.b.d. of rod ABCD 
 ( ) ( ) ( )( )0: 25 N cos60 cos 45 0E OE OEM d T dΣ = ° − ° = 
 17.6777 NT∴ = 
 or 17.68 NT = W 
(b) From f.b.d. of rod ABCD 
 ( ) ( )0: 17.6777 N cos 45 25 N cos60xFΣ = − ° + ° 
 cos 45 cos 45 0D AN N+ ° − ° = 
 0A DN N∴ − = 
 or D AN N= (1) 
 ( )0: sin 45 sin 45 17.6777 N sin 45y A DF N NΣ = ° + ° − ° 
 ( )25 N sin 60 0− ° = 
 48.296 NA DN N∴ + = (2) 
 Substituting Equation (1) into Equation (2), 
2 48.296 NAN = 
24.148 NAN = 
 or 24.1 NA =N 45.0°W 
 and 24.1 ND =N 45.0°W 
 
 
 
 
 
PROBLEM 4.39 
Rod ABCD is bent in the shape of a circular arc of radius 80 mm and 
rests against frictionless surfaces at A and D. Knowing that the collar 
at B can move freely on the rod, determine (a) the value of θ for 
which the tension in cord OB is as small as possible, (b) the 
corresponding value of the tension, (c) the reactions at A and D. 
 
SOLUTION 
 
 
 
 
(a) From f.b.d. of rod ABCD 
 ( ) ( ) ( )( )0: 25 N cos60 cos 0E OE OEM d T dθΣ = ° − = 
 or 12.5 N
cos
T θ= (1) 
 is minimum when cos is maximum,T θ∴ 
 or 0θ = °W 
(b) From Equation (1) 
12.5 N 12.5 N
cos0
T = = 
 minor 12.50 NT = W 
(c) 0: cos 45 cos 45 12.5 Nx A DF N NΣ = − ° + ° + 
 ( )25 N cos60 0− ° = 
 0D AN N∴ − = 
 or D AN N= (2) 
 ( )0: sin 45 sin 45 25 N sin 60 0y A DF N NΣ = ° + ° − ° = 
 30.619 ND AN N∴ + = (3) 
 Substituting Equation (2) into Equation (3), 
2 30.619AN = 
15.3095 NAN = 
 or 15.31 NA =N 45.0°W 
 and 15.31 ND =N 45.0°W 
 
 
 
 
 
PROBLEM 4.40 
Bar AC supports two 100-lb loads as shown. Rollers A and C rest against 
frictionless surfaces and a cable BD is attached at B. Determine (a) the 
tension in cable BD, (b) the reaction at A, (c) the reaction at C. 
 
SOLUTION 
 
 
 
 
First note that from similar triangles 
10 3 in.
6 20
DB
DB
y y= ∴ = 
and ( ) ( )2 23 14 in. 14.3178 in.BD = + = 
14 0.97780
14.3178x
T T T= = 
3 0.20953
14.3178y
T T T= = 
(a) From f.b.d. of bar AC 
 ( )( ) ( )( )0: 0.97780 7 in. 0.20953 6 in.EM T TΣ = − 
 ( )( ) ( )( )100 lb 16 in. 100 lb 4 in. 0− − = 
 357.95 lbT∴ = 
 or 358 lbT = W 
(b) From f.b.d. of bar AC 
 ( )0: 100 0.20953 357.95 100 0yF AΣ = − − − = 
 275.00 lbA∴ = 
 or 275 lb=A W 
(c) From f.b.d of bar AC 
 ( )0: 0.97780 357.95 0xF CΣ = − = 
 350.00 lbC∴ = 
 or 350 lb=C W 
 
 
 
 
PROBLEM 4.41 
A parabolic slot has been cut in plate AD, and the plate has been 
placed so that the slot fits two fixed, frictionless pins B and C. The 
equation of the slot is 2/100,y x= where x and y are expressed in mm. 
Knowing that the input force 4 N,P = determine (a) the force each 
pin exerts on the plate, (b) the output force Q. 
 
SOLUTION 
 
The equation of the slot is 
2
100
xy = 
Now slope of the slot at 
C
dy C
dx
  =   
 
60 mm
2 1.200
100 x
x
=
 = =   
( )1 tan 1.200 50.194α −∴ = = ° 
and 90 90 50.194 39.806θ α= ° − = ° − ° = ° 
Coordinates of C are 
( ) 26060 mm, 36 mm
100C C
x y= = = 
Also, the coordinates of D are 60 mmDx = 
( )46 mm 40 mm sinDy β= + 
where 1 120 66tan 12.6804
240
β − − = = °   
 ( ) 46 mm 40 mm tan12.6804Dy∴ = + ° 
 55.000 mm= 
 
 
PROBLEM 4.41 CONTINUED 
Also, 60 mm 60 mm
tan tan12.6804ED
y β= = ° 
 266.67 mm= 
From f.b.d. of plate AD 
 ( ) ( ) ( )( ) ( )( )0: cos sin 4 N 0E C ED D C C C ED DM N y y y N x y yθ θ Σ = − − + − − =  
( ) ( ) ( )( ) ( )( )cos39.806 266.67 55.0 36.0 mm sin 39.806 60 mm 4 N 266.67 55.0 mm 0C CN N ° − − + ° − − =  
3.7025 NCN∴ = 
or 3.70 NC =N 39.8° 
 0: 4 N cos sin 0x CF N Qθ βΣ = − + + = 
( )4 N 3.7025 N cos39.806 sin12.6804 0Q− + ° + ° = 
 5.2649 NQ∴ = 
or 5.26 N=Q 77.3° 
 0: sin cos 0y B CF N N Qθ βΣ = + − = 
( ) ( )3.7025 N sin 39.806 5.2649 N cos12.6804 0BN + ° − ° = 
 2.7662 NBN∴ = 
or 2.77 NB =N 
(a) 2.77 NB =N , 3.70 NC =N 39.8°W 
(b) 5.26 N=Q ( )77.3 output° W 
 
 
 
PROBLEM 4.42 
A parabolic slot has been cut in plate AD, and the plate has been placed 
so that the slot fits two fixed, frictionless pins B and C. The equation of 
the slot is 2/100,y x= where x and y are expressed inmm. Knowing that 
the maximum allowable force exerted on the roller at D is 8.5 N, 
determine (a) the corresponding magnitude of the input force P, (b) the 
force each pin exerts on the plate. 
 
SOLUTION 
 
 
 
The equation of the slot is, 
2
100
xy = 
Now slope of slot at 
C
dy C
dx
  =   
 
60 mm
2 1.200
100 x
x
=
 = =   
( )1 tan 1.200 50.194α −∴ = = ° 
and 90 90 50.194 39.806θ α= ° − = ° − ° = ° 
Coordinates of C are 
( )26060 mm, 36 mm
100C C
x y= = = 
Also, the coordinates of D are 
60 mmDx = 
( )46 mm 40 mm sinDy β= + 
where 1 120 66tan 12.6804
240
β − − = = °   
( ) 46 mm 40 mm tan12.6804 55.000 mmDy∴ = + ° = 
Note: 0Ex = 
( )60 mm tanE Cy y θ= + 
 ( )36 mm 60 mm tan 39.806= + ° 
 86.001 mm= 
(a) From f.b.d. of plate AD 
 ( ) ( ) ( )0: 8.5 N sinE E E DM P y y yβ Σ = − −  
 ( ) ( )8.5 N cos 60 mm 0β − =  
 
 
 
 
 
 
PROBLEM 4.42 CONITNIUED 
( ) ( ) ( )86.001 mm 8.5 N sin12.6804 31.001 mmP − °   
 ( ) ( )8.5 N cos12.6804 60 mm 0− ° =   
 6.4581 NP∴ = 
 or 6.46 NP = W 
(b) ( )0: 8.5 N sin cos 0x CF P Nβ θΣ = − − = 
( )( ) ( )6.458 N 8.5 N sin12.6804 cos39.806 0CN− ° − ° = 
 5.9778 NCN∴ = 
 or 5.98 NC =N 39.8°W 
 ( )0: sin 8.5 N cos 0y B CF N N θ βΣ = + − = 
( ) ( )5.9778 N sin 39.806 8.5 N cos12.6804 0BN + ° − ° = 
 4.4657 NBN∴ = 
 or 4.47 NB =N W 
 
 
 
 
PROBLEM 4.43 
A movable bracket is held at rest by a cable attached at E and by frictionless 
rollers. Knowing that the width of post FG is slightly less than the distance 
between the rollers, determine the force exerted on the post by each roller 
when o20 .α = 
 
SOLUTION 
 
 
 
 
From f.b.d. of bracket 
 0: sin 20 60 lb 0yF TΣ = ° − = 
 175.428 lbT∴ = 
( )175.428 lb cos 20 164.849 lbxT = ° = 
( )175.428 lb sin 20 60 lbyT = ° = 
Note: and 60 lbyT force form a couple of 
( )60 lb 10 in. 600 lb in.= ⋅ 
 ( ) ( )0: 164.849 lb 5 in. 600 lb in. 8 in. 0B CDM FΣ = − ⋅ + = 
 28.030 lbCDF∴ = − 
or 28.0 lbCD =F 
 0: 0x CD AB xF F F TΣ = + − = 
28.030 lb 164.849 lb 0ABF− + − = 
 192.879 lbABF∴ = 
or 192.9 lbAB =F 
Rollers A and C can only apply a horizontal force to the right onto the 
vertical post corresponding to the equal and opposite force to the left on 
the bracket. Since FAB is directed to the right onto the bracket, roller B 
will react FAB. Also, since FCD is acting to the left on the bracket, it will 
act to the right on the post at roller C. 
 
 
 
 
 
 PROBLEM 4.43 CONTINUED 
 0∴ = =A D 
192.9 lb=B 
 28.0 lb=C 
Forces exerted on the post are 
 0= =A D W 
 192.9 lb=B W 
 28.0 lb=C W 
 
 
 
 
 
PROBLEM 4.44 
Solve Problem 4.43 when o30 .α = 
P4.43 A movable bracket is held at rest by a cable attached at E and by 
frictionless rollers. Knowing that the width of post FG is slightly less 
than the distance between the rollers, determine the force exerted on the 
post by each roller when o20 .α = 
 
SOLUTION 
 
 
 
 
From f.b.d. of bracket 
 0: sin 30 60 lb 0yF TΣ = ° − = 
 120 lbT∴ = 
( )120 lb cos30 103.923 lbxT = ° = 
( )120 lb sin 30 60 lbyT = ° = 
Note: and 60 lbyT force form a couple of 
( )( )60 lb 10 in. 600 lb in.= ⋅ 
 ( )( ) ( )0: 103.923 lb 5 in. 600 lb in. 8 in. 0B CDM FΣ = − ⋅ + = 
 10.0481 lbCDF∴ = 
or 10.05 lbCD =F 
 0: 0x CD AB xF F F TΣ = + − = 
10.0481 lb 103.923 lb 0ABF+ − = 
 93.875 lbABF∴ = 
or 93.9 lbAB =F 
Rollers A and C can only apply a horizontal force to the right on the 
vertical post corresponding to the equal and opposite force to the left on 
the bracket. The opposite direction apply to roller B and D. Since both 
ABF and CDF act to the right on the bracket, rollers B and D will react 
these forces. 
 0∴ = =A C 
93.9 lb=B 
10.05 lb=D 
Forces exerted on the post are 
 0= =A C W 
 93.9 lb=B W 
 10.05 lb=D W 
 
 
 
 
 
PROBLEM 4.45 
A 20-lb weight can be supported in the three different ways shown. 
Knowing that the pulleys have a 4-in. radius, determine the reaction at A 
in each case. 
 
SOLUTION 
 
 
 
 
(a) From f.b.d. of AB 
 0: 0x xF AΣ = = 
 0: 20 lb 0y yF AΣ = − = 
 or 20.0 lbyA = 
 and 20.0 lb=A W 
 ( )( )0: 20 lb 1.5 ft 0A AM MΣ = − = 
30.0 lb ftAM∴ = ⋅ 
 or 30.0 lb ftA = ⋅M W 
(b) Note: 
1 ft4 in. 0.33333 ft
12 in.
  =   
 From f.b.d. of AB 
 0: 20 lb 0x xF AΣ = − = 
 or 20.0 lbxA = 
 0: 20 lb 0y yF AΣ = − = 
 or 20.0 lbyA = 
 Then ( ) ( )2 22 2 20.0 20.0 28.284 lbx yA A A= + = + = 
 28.3 lb∴ =A 45°W 
 ( )( )0: 20 lb 0.33333 ftA AM MΣ = + 
 ( )( )20 lb 1.5 ft 0.33333 ft 0− + = 
 30.0 lb ftAM∴ = ⋅ 
 or 30.0 lb ftA = ⋅M W 
 
 
 
 
 
PROBLEM 4.45 CONTINUED 
(c) From f.b.d. of AB 
 0: 0x xF AΣ = = 
 0: 20 lb 20 lb 0y yF AΣ = − − = 
 or 40.0 lbyA = 
 and 40.0 lb=A W 
 ( )( )0: 20 lb 1.5 ft 0.33333 ftA AM MΣ = − − 
 ( )( )20 lb 1.5 ft 0.33333 ft 0− + = 
 60.0 lb ftAM∴ = ⋅ 
 or 60.0 lb ftA = ⋅M W 
 
 
 
 
 
 
PROBLEM 4.46 
A belt passes over two 50-mm-diameter pulleys which are mounted on a 
bracket as shown. Knowing that 0M = and 24 N,i OT T= = determine 
the reaction at C. 
 
SOLUTION 
 
 
 
 
From f.b.d. of bracket 
 0: 24 N 0x xF CΣ = − = 
24 NxC∴ = 
 0: 24 N 0y yF CΣ = − = 
 24 NyC∴ = 
Then ( ) ( )2 22 2 24 24 33.941 Nx yC C C= + = + = 
 33.9 N∴ =C 45.0°W 
 ( ) ( )0: 24 N 45 25 mmC CM M  Σ = − −  
 ( ) ( )24 N 25 50 60 mm 0 + + − =  
 120 N mmCM∴ = ⋅ 
 or 0.120 N mC = ⋅M W 
 
 
 
 
 
PROBLEM 4.47 
A belt passes over two 50-mm-diameter pulleys which are mounted on a 
bracket as shown. Knowing that 0.40 N mM = ⋅ m and that iT and OT 
are equal to 32 N and 16 N, respectively, determine the reaction at C. 
 
SOLUTION 
 
 
 
 
From f.b.d. of bracket 
 0: 32 N 0x xF CΣ = − = 
 32 NxC∴ = 
 0: 16 N 0y yF CΣ = − = 
 16 NyC∴ = 
Then ( ) ( )2 22 2 32 16 35.777 Nx yC C C= + = + = 
and 1 1 16tan tan 26.565
32
y
x
C
C
θ − −   = = = °     
 
 or 35.8 N=C 26.6°W 
 ( )( )0: 32 N 45 mm 25 mmC CM MΣ = − − 
 ( )( )16 N 25 mm 50 mm 60 mm 400 N mm 0+ + − − ⋅ = 
 800 N mmCM∴ = ⋅ 
 or 0.800 N mC = ⋅M W 
 
 
 
 
 
PROBLEM 4.48 
A 350-lb utility pole is used to support at C the end of an electric wire. 
The tension in the wire is 120 lb, and the wire forms an angle of 15° 
with the horizontal at C. Determine the largest and smallest allowable 
tensions in the guy cable BD if the magnitude of the couple at A may not 
exceed 200 lb ft.⋅ 
 
SOLUTION 
 
 
 
 
First note 
 ( ) ( )2 24.5 10 10.9659 ftBDL = + = 
maxT : From f.b.d. of utility pole with 200 lb ftA = ⋅M 
 ( ) ( )0: 200 lb ft 120 lb cos15 14 ftAM  Σ = − ⋅ − °  
 ( )max4.5 10 ft 010.9659 T
  + =     
max 444.19 lbT∴ = 
 or max 444 lbT = W 
minT : From f.b.d. of utility pole with 200 lb ftA = ⋅M 
 ( ) ( )0: 200 lb ft 120 lb cos15 14 ftAM  Σ = ⋅ − °  
 ( )min4.5 10 ft 010.9659 T
  + =     
min 346.71 lbT∴ = 
 or min 347 lbT = W 
 
 
 
 
 
PROBLEM 4.49 
In a laboratory experiment, students hang the masses shown from a beam 
of negligible mass. (a) Determine the reaction at the fixed support A 
knowing that end D of the beam does not touch support E. (b) Determine 
the reaction at the fixed support A knowing that the adjustable support E 
exerts an upward force of 6 N on the beam. 
 
SOLUTION 
 
 
 
 
( )( )21 kg 9.81 m/s 9.81 NB BW m g= = = 
 ( )( )20.5 kg 9.81 m/s 4.905 NC CWm g= = = 
(a) From f.b.d. of beam ABCD 
 0: 0x xF AΣ = = 
 0: 0y y B CF A W WΣ = − − = 
9.81 N 4.905 N 0yA − − = 
 14.715 NyA∴ = 
 or 14.72 N=A W 
 ( ) ( )0: 0.2 m 0.3 m 0A A B CM M W WΣ = − − = 
 ( )( ) ( )( )9.81 N 0.2 m 4.905 N 0.3 m 0AM − − = 
 3.4335 N mAM∴ = ⋅ 
 or 3.43 N mA = ⋅M W 
(b) From f.b.d. of beam ABCD 
 0: 0x xF AΣ = = 
 0: 6 N 0y y B CF A W WΣ = − − + = 
9.81 N 4.905 N 6 N 0yA − − + = 
 8.715 NyA∴ = or 8.72 N=A W 
 ( ) ( ) ( )( )0: 0.2 m 0.3 m 6 N 0.4 m 0A A B CM M W WΣ = − − + = 
 ( )( ) ( )( ) ( )( )9.81 N 0.2 m 4.905 N 0.3 m 6 N 0.4 m 0AM − − + = 
 1.03350 N mAM∴ = ⋅ 
 or 1.034 N mA = ⋅M W 
 
 
 
 
 
PROBLEM 4.50 
In a laboratory experiment, students hang the masses shown from a beam 
of negligible mass. Determine the range of values of the force exerted on 
the beam by the adjustable support E for which the magnitude of the 
couple at A does not exceed 2.5 N m.⋅ 
 
SOLUTION 
 
 
 
 
 ( )21 kg 9.81 m/s 9.81 NB BW m g= = = 
( )20.5 kg 9.81 m/s 4.905 NC CW m g= = = 
Maximum AM value is 2.5 N m⋅ 
min :F From f.b.d. of beam ABCD with 2.5 N mA = ⋅M 
 ( ) ( )0: 2.5 N m 0.2 m 0.3 mA B CM W WΣ = ⋅ − − 
 ( )min 0.4 m 0F+ = 
( )( ) ( )( ) ( )min2.5 N m 9.81 N 0.2 m 4.905 N 0.3 m 0.4 m 0F⋅ − − + = 
min 2.3338 NF∴ = 
or min 2.33 NF = 
max :F From f.b.d. of beam ABCD with 2.5 N mA = ⋅M 
( ) ( )0: 2.5 N m 0.2 m 0.3 mA B CM W WΣ = − ⋅ − − 
 ( )max 0.4 m 0F+ = 
( )( ) ( )( ) ( )max2.5 N m 9.81 N 0.2 m 4.905 N 0.3 m 0.4 m 0F− ⋅ − − + = 
max 14.8338 NF∴ = 
or max 14.83 NF = 
 or 2.33 N 14.83 NEF≤ ≤ W 
 
 
 
 
 
 
PROBLEM 4.51 
Knowing that the tension in wire BD is 300 lb, determine the reaction at 
fixed support C for the frame shown. 
 
SOLUTION 
 
From f.b.d. of frame with 300 lbT = 
 50: 100 lb 300 lb 0
13x x
F C  Σ = − + =   
 15.3846 lbxC∴ = − or 15.3846 lbx =C 
 
120: 180 lb 300 lb 0
13y y
F C  Σ = − − =   
 456.92 lbyC∴ = or 456.92 lby =C 
Then ( ) ( )2 22 2 15.3846 456.92 457.18 lbx yC C C= + = + = 
and 
1 1 456.92tan tan 88.072
15.3846
y
x
C
C
θ − −   = = = − °   −  
 
 or 457 lb=C 88.1°W 
 ( )( ) ( )( ) ( )120: 180 lb 20 in. 100 lb 16 in. 300 lb 16 in. 013C CM M
  Σ = + + − =     
 769.23 lb in.CM∴ = − ⋅ 
 or 769 lb in.C = ⋅M W 
 
 
 
 
 
 
PROBLEM 4.52 
Determine the range of allowable values of the tension in wire BD if the 
magnitude of the couple at the fixed support C is not to exceed 75 lb ft.⋅ 
 
SOLUTION 
 
maxT From f.b.d. of frame with 75 lb ftC = ⋅M 900 lb in.= ⋅ 
 ( )( ) ( )( ) ( )max120: 900 lb in. 180 lb 20 in. 100 lb 16 in. 16 in. 013CM T
  Σ = ⋅ + + − =     
max 413.02 lbT∴ = 
minT From f.b.d. of frame with 75 lb ftC = ⋅M 900 lb in.= ⋅ 
 ( )( ) ( )( ) ( )min120: 900 lb in. 180 lb 20 in. 100 lb 16 in. 16 in. 013CM T
  Σ = − ⋅ + + − =     
min 291.15 lbT∴ = 
 291 lb 413 lbT∴ ≤ ≤ W 
 
 
 
 
 
 
PROBLEM 4.53 
Uniform rod AB of length l and weight W lies in a vertical plane and is 
acted upon by a couple M. The ends of the rod are connected to small 
rollers which rest against frictionless surfaces. (a) Express the angle θ 
corresponding to equilibrium in terms of M, W, and l. (b) Determine the 
value of θ corresponding to equilibrium when 1.5 lb ft,M = ⋅ 
4 lb,W = and 2 ft.l = 
 
SOLUTION 
 
 
(a) From f.b.d. of uniform rod AB 
 0: cos 45 cos 45 0xF A BΣ = − ° + ° = 
 0A B∴ − + = or B A= (1) 
 0: sin 45 sin 45 0yF A B WΣ = ° + ° − = 
 2A B W∴ + = (2) 
 From Equations (1) and (2) 
 2 2A W= 
 1
2
A W∴ = 
 From f.b.d. of uniform rod AB 
 0: cos2B
lM W Mθ  Σ = +     
 ( )1 cos 45 0
2
W l θ   − ° − =     (3) 
 From trigonometric identity 
( )cos cos cos sin sinα β α β α β− = + 
 Equation (3) becomes 
 ( )cos cos sin 0
2 2
Wl WlMθ θ θ   + − + =       
 
 
 
 
 
 
PROBLEM 4.53 CONTINUED 
 or cos cos sin 0
2 2 2
Wl Wl WlMθ θ θ     + − − =           
2 sin M
Wl
θ∴ = 
 or 1 2sin M
Wl
θ −  =   W 
(b) ( )( )( )1
2 1.5 lb ft
sin 22.024
4 lb 2 ft
θ −  ⋅= = °   
 
 or 22.0θ = °W 
 
 
 
 
 
 
PROBLEM 4.54 
A slender rod AB, of weight W, is attached to blocks A and B, which 
move freely in the guides shown. The blocks are connected by an elastic 
cord which passes over a pulley at C. (a) Express the tension in the cord 
in terms of W and .θ (b) Determine the value of θ for which the tension 
in the cord is equal to 3W. 
 
SOLUTION 
 
 
 
(a) From f.b.d. of rod AB 
 ( ) ( )0: sin cos cos 02C
lM T l W T lθ θ θ  Σ = + − =     
( )
cos
2 cos sin
WT θθ θ∴ = − 
 Dividing both numerator and denominator by cos ,θ 
 1
2 1 tan
WT θ
 =  −  
 or ( )
2
1 tan
W
T θ
   = − W 
(b) For 3 ,T W= 
 ( )
23
1 tan
W
W θ
   = − 
 1 1 tan
6
θ∴ − = 
 or 1 5tan 39.806
6
θ −  = = °   
 or 39.8θ = °W 
 
 
 
 
 
 
 
PROBLEM 4.55 
A thin, uniform ring of mass m and radius R is attached by a frictionless 
pin to a collar at A and rests against a small roller at B. The ring lies in a 
vertical plane, and the collar can move freely on a horizontal rod and is 
acted upon by a horizontal force P. (a) Express the angle θ 
corresponding to equilibrium in terms of m and P. (b) Determine the 
value of θ corresponding to equilibrium when 500 gm = and 5 N.P = 
 
SOLUTION 
 
 
 
(a) From f.b.d. of ring 
 ( ) ( )0: cos cos sin 0CM P R R W Rθ θ θΣ = + − = 
2 tanP W θ= where W mg= 
 2 tan P
mg
θ∴ = 
 or 1 2tan P
mg
θ −  =   W 
(b) Have 500 g 0.500 kg and 5 Nm P= = = 
 ( )( )( )1 2
2 5 N
 tan
0.500 kg 9.81 m/s
θ −
  ∴ =   
 
 63.872= ° 
 or 63.9θ = °W 
 
 
 
 
 
 
PROBLEM 4.56 
Rod AB is acted upon by a couple M and two forces, each of magnitude 
P. (a) Derive an equation inθ , P, M, and l which must be satisfied when 
the rod is in equilibrium. (b) Determine the value of θ corresponding to 
equilibrium when 150 lb in.,M = ⋅ 20 lb,P = and 6 in.l = 
 
SOLUTION 
 
 
 
(a) From f.b.d. of rod AB 
 ( ) ( )0: cos sin 0CM P l P l Mθ θΣ = + − = 
 or sin cos M
Pl
θ θ+ = W 
(b) For 150 lb in., 20 lb, and 6 in.M P l= ⋅ = = 
 ( )( )
150 lb in. 5sin cos 1.25
20 lb 6 in. 4
θ θ ⋅+ = = = 
 Using identity 2 2sin cos 1θ θ+ = 
 ( )122sin 1 sin 1.25θ θ+ − = 
 ( )1221 sin 1.25 sinθ θ− = − 
 2 21 sin 1.5625 2.5sin sinθ θ θ− = − + 
22sin 2.5sin 0.5625 0θ θ− + = 
 Using quadratic formula 
 
( ) ( ) ( )( )
( )
2.5 6.25 4 2 0.5625
sin
2 2
θ − − ± −= 
 2.5 1.75
4
±= 
 or sin 0.95572 and sin 0.29428θ θ= = 
 72.886 and 17.1144θ θ∴ = ° = ° 
 or 17.11 and 72.9θ θ= ° = °W 
 
 
 
 
 
 
PROBLEM 4.57 
A vertical load P is applied at end B of rod BC. The constant of the spring 
is k, and the spring is unstretched when o90 .θ = (a) Neglecting the 
weight of the rod, express the angle θ corresponding to equilibrium in 
terms of P, k, and l. (b) Determine the value of θ corresponding to 
equilibrium when 1 .
4
P kl= 
 
SOLUTION 
 
 
 
First note 
 tension in spring T ks= = 
where elongation of springs = 
( ) ( )
90
AB ABθ θ = °= − 
 902 sin 2 sin
2 2
l lθ °   = −       
 12 sin
2 2
l θ   = −         
 1 2 sin
2 2
T kl θ   ∴ = −         (1) 
(a) From f.b.d. of rodBC 
 ( )0: cos sin 02CM T l P l
θ θ  Σ = − =     
 Substituting T From Equation (1) 
( )12 sin cos sin 0
2 22
kl l P lθ θ θ       − − =              
2 12 sin cos 2sin cos 0
2 2 2 22
kl Plθ θ θ θ           − − =                      
 Factoring out 2 cos , leaves
2
l θ    
 
 
 
 
 
 
PROBLEM 4.57 CONTINUED 
 1sin sin 0
2 22
kl Pθ θ     − − =           
 or 
 1sin
2 2
kl
kl P
θ   =   −    
 ( )1 2sin 2
kl
kl P
θ −  ∴ =  −  
W 
(b) 
4
klP = 
( )1 1 14
4 42sin 2sin 2sin
3 kl2 3 22 kl
kl kl
kl
θ − − −       = = =    −       
 
 ( )12sin 0.94281−= 
 141.058= ° 
 or 141.1θ = °W 
 
 
 
 
 
 
 
PROBLEM 4.58 
Solve Sample Problem 4.5 assuming that the spring is unstretched when 
o90 .θ = 
 
SOLUTION 
 
 
 
First note 
 tension in springT ks= = 
where deformation of springs = 
rβ= 
 F krβ∴ = 
From f.b.d. of assembly 
 ( ) ( )0 0: cos 0M W l F rβΣ = − = 
or 2cos 0Wl krβ β− = 
2
 cos kr
Wl
β β∴ = 
For 250 lb/in., 3 in., 8 in., 400 lbk r l W= = = = 
 ( )( )( )( )
2250 lb/in. 3 in.
cos
400 lb 8 in.
β β= 
or cos 0.703125β β= 
Solving numerically, 
 0.89245 radβ = 
or 51.134β = ° 
Then 90 51.134 141.134θ = ° + ° = ° 
 or 141.1θ = °W 
 
 
 
 
 
 
PROBLEM 4.59 
A collar B of weight W can move freely along the vertical rod shown. The 
constant of the spring is k, and the spring is unstretched when 0.θ = 
(a) Derive an equation in θ , W, k, and l which must be satisfied when the 
collar is in equilibrium. (b) Knowing that 3 lb,W = 6 in.,l = and 
8 lb/ft,k = determine the value of θ corresponding to equilibrium. 
 
SOLUTION 
 
 
 
First note T ks= 
where spring constantk = 
elongation of springs = 
 ( )1 cos
cos cos
l ll θθ θ= − = − 
 ( ) 1 cos
cos
klT θθ∴ = − 
(a) From f.b.d. of collar B 
 0: sin 0yF T WθΣ = − = 
 or ( )1 cos sin 0
cos
kl Wθ θθ − − = 
 or tan sin W
kl
θ θ− = W 
(b) For 3 lb, 6 in., 8 lb/ftW l k= = = 
 6 in. 0.5 ft
12 in./ft
l = = 
( )( )
3 lbtan sin 0.75
8 lb/ft 0.5 ft
θ θ− = = 
 Solving Numerically, 
 57.957θ = ° 
 or 58.0θ = °W 
 
 
 
 
 
 
PROBLEM 4.60 
A slender rod AB, of mass m, is attached to blocks A and B which move 
freely in the guides shown. The constant of the spring is k, and the spring 
is unstretched when 0θ = . (a) Neglecting the mass of the blocks, derive 
an equation in m, g, k, l, and θ which must be satisfied when the rod is in 
equilibrium. (b) Determine the value of θ when 2 kg,m = 750l = 
mm, and 30 N/m.k = 
 
SOLUTION 
 
 
 
First note 
 spring forcesF ks= = 
where spring constantk = 
 spring deformations = 
 cosl l θ= − 
 ( )1 cosl θ= − 
 ( ) 1 cossF kl θ∴ = − 
(a) From f.b.d. of assembly 
 ( )0: sin cos 02D s
lM F l Wθ θ Σ = − =   
 ( )( )1 cos sin cos 0
2
lkl l Wθ θ θ − − =   
 ( )sin cos sin cos 0
2
Wkl θ θ θ θ − − =   
 Dividing by cosθ 
 ( )tan sin
2
Wkl θ θ− = 
 tan sin
2
W
kl
θ θ∴ − = 
 or tan sin
2
mg
kl
θ θ− = W 
(b) For 2 kg, 750 mm, 30 N/mm l k= = = 
 750 mm 0.750 ml = = 
 
 
 
 
 PROBLEM 4.60 CONTINUED 
 Then 
( )( )
( )( )
22 kg 9.81 m/s
tan sin 0.436
2 30 N/m 0.750 m
θ θ− = = 
 Solving Numerically, 
50.328θ = ° 
 or 50.3θ = °W 
 
 
 
 
PROBLEM 4.61 
The bracket ABC can be supported in the eight different ways shown. All 
connections consist of smooth pins, rollers, or short links. In each case, 
determine whether (a) the plate is completely, partially, or improperly 
constrained, (b) the reactions are statically determinate or indeterminate, 
(c) the equilibrium of the plate is maintained in the position shown. Also, 
wherever possible, compute the reactions assuming that the magnitude of 
the force P is 100 N. 
 
SOLUTION 
 
 
 
 
1. Three non-concurrent, non-parallel reactions 
 (a) Completely constrained W 
 (b) Determinate W 
 (c) Equilibrium W 
From f.b.d. of bracket: 
 ( ) ( )( )0: 1 m 100 N 0.6 m 0AM BΣ = − = 
 60.0 N∴ =B W 
 0: 60 N 0x xF AΣ = − = 
 60.0 Nx∴ =A 
 0: 100 N 0y yF AΣ = − = 
 100 Ny∴ =A 
Then ( ) ( )2 260.0 100 116.619 NA = + = 
and 1 100tan 59.036
60.0
θ −  = = °   
 116.6 N∴ =A 59.0°W 
2. Four concurrent reactions through A 
 (a) Improperly constrained W 
 (b) Indeterminate W 
 (c) No equilibrium W 
3. Two reactions 
 (a) Partially constrained W 
 (b) Determinate W 
 (c) Equilibrium W 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
PROBLEM 4.61 CONTINUED 
From f.b.d. of bracket 
 ( ) ( )( )0: 1.2 m 100 N 0.6 m 0AM CΣ = − = 
 50.0 N∴ =C W 
 0: 100 N 50 N 0yF AΣ = − + = 
 50.0 N∴ =A W 
4. Three non-concurrent, non-parallel reactions 
 (a) Completely constrained W 
 (b) Determinate W 
 (c) Equilibrium W 
From f.b.d. of bracket 
1 1.0tan 39.8
1.2
θ −  = = °   
( ) ( )2 21.2 1.0 1.56205 mBC = + = 
 ( ) ( )( )1.20: 1 m 100 N 0.6 m 01.56205AM B
  Σ = − =     
 78.1 N∴ =B 39.8°W 
 ( )0: 78.102 N cos39.806 0xF CΣ = − ° = 
 60.0 N∴ =C W 
 ( )0: 78.102 N sin 39.806 100 N 0yF AΣ = + ° − = 
 50.0 N∴ =A W 
5. Four non-concurrent, non-parallel reactions 
 (a) Completely constrained W 
 (b) Indeterminate W 
 (c) Equilibrium W 
From f.b.d. of bracket 
 ( )( ) ( )0: 100 N 0.6 m 1.2 m 0C yM AΣ = − = 
 50 NyA∴ = or 50.0 Ny =A W 
6. Four non-concurrent non-parallel reactions 
 (a) Completely constrained W 
 (b) Indeterminate W 
 (c) Equilibrium W 
 
 
 
 
 
 
 
PROBLEM 4.61 CONTINUED 
From f.b.d. of bracket 
 ( ) ( )( )0: 1 m 100 N 0.6 m 0A xM BΣ = − − = 
 60.0 NxB∴ = − 
 or 60.0 Nx =B W 
 0: 60 0x xF AΣ = − + = 
 60.0 NxA∴ = 
 or 60.0 Nx =A W 
7. Three non-concurrent, non-parallel reactions 
 (a) Completely constrained W 
 (b) Determinate W 
 (c) Equilibrium W 
From f.b.d. of bracket 
 0: 0x xF AΣ = = 
 ( ) ( )( )0: 1.2 m 100 N 0.6 m 0AM CΣ = − = 
 50.0 NC∴ = 
 or 50.0 N=C W 
 0: 100 N 50.0 N 0y yF AΣ = − + = 
 50.0 NyA∴ = 
 50.0 N∴ =A W 
8. Three concurrent, non-parallel reactions 
 (a) Improperly constrained W 
 (b) Indeterminate W 
 (c) No equilibrium W 
 
 
 
 
 
PROBLEM 4.62 
Eight identical 20 30-in.× rectangular plates, each weighing 50 lb, are 
held in a vertical plane as shown. All connections consist of frictionless 
pins, rollers, or short links. For each case, answer the questions listed in 
Problem 4.61, and, wherever possible, compute the reactions. 
P6.1 The bracket ABC can be supported in the eight different ways 
shown. All connections consist of smooth pins, rollers, or short links. In 
each case, determine whether (a) the plate is completely, partially, or 
improperly constrained, (b) the reactions are statically determinate or 
indeterminate, (c) the equilibrium of the plate is maintained in the 
position shown. Also, wherever possible, compute the reactions assuming 
that the magnitude of the force P is 100 N. 
 
SOLUTION 
 
 
 
 
1. Three non-concurrent, non-parallel reactions 
 (a) Completely constrained W 
 (b) Determinate W 
 (c) Equilibrium W 
From f.b.d. of plate 
 ( ) ( )0: 30 in. 50 lb 15 in. 0AM CΣ = − = 
 25.0 lb=C W 
 0: 0x xF AΣ = = 
 0: 50 lb 25 lb 0y yF AΣ = − + = 
 25 lbyA = 25.0 lb=A W 
2. Three non-current, non-parallel reactions 
 (a) Completely constrained W 
 (b) Determinate W 
 (c) Equilibrium WFrom f.b.d. of plate 
 0:xFΣ = 0=B W 
 ( )( ) ( )0: 50 lb 15 in. 30 in. 0BM DΣ = − = 
 25.0 lb=D W 
 0: 25.0 lb 50 lb 0yF CΣ = − + = 
 25.0 lb=C W 
 
 
 
 
 
 
PROBLEM 4.62 CONTINUED 
3. Four non-concurrent, non-parallel reactions 
 (a) Completely constrained W 
 (b) Indeterminate W 
 (c) Equilibrium W 
From f.b.d. of plate 
 ( ) ( )( )0: 20 in. 50 lb 15 in.D xM AΣ = − 
 37.5 lbx∴ =A W 
 0: 37.5 lb 0x xF DΣ = + = 
 37.5 lbx∴ =D W 
4. Three concurrent reactions 
 (a) Improperly constrained W 
 (b) Indeterminate W 
 (c) No equilibrium W 
5. Two parallel reactions 
 (a) Partial constraint W 
 (b) Determinate W 
 (c) Equilibrium W 
From f.b.d. of plate 
 ( ) ( )( )0: 30 in. 50 lb 15 in. 0DM CΣ = − = 
 25.0 lb=C W 
 0: 50 lb 25 lb 0yF DΣ = − + = 
 25.0 lb=D W 
6. Three non-concurrent, non-parallel reactions 
 (a) Completely constrained W 
 (b) Determinate W 
 (c) Equilibrium W 
From f.b.d. of plate 
 ( ) ( )( )0: 20 in. 50 lb 15 in. 0DM BΣ = − = 
 37.5 lb=B W 
 0: 37.5 lb 0 37.5 lbx x xF DΣ = + = =D 
 0: 50 lb 0 50.0 lby y yF DΣ = − = =D 
 or 62.5 lb=D 53.1°W 
 
 
 
 
 
 
 
PROBLEM 4.62 CONTINUED 
7. Two parallel reactions 
 (a) Improperly constrained W 
 (b) Reactions determined by dynamics W 
 (c) No equilibrium W 
8. Four non-concurrent, non-parallel reactions 
 (a) Completely constrained W 
 (b) Indeterminate W 
 (c) Equilibrium W 
From f.b.d. of plate 
 ( ) ( )( )0: 30 in. 50 lb 15 in. 0DM BΣ = − = 
 25.0 lb=B W 
 0: 50 lb 25.0 lb 0y yF DΣ = − + = 
 25.0 lby =D W 
 0: 0x xF D CΣ = + = 
 
 
 
 
 
 
PROBLEM 4.63 
Horizontal and vertical links are hinged to a wheel, and forces are applied 
to the links as shown. Knowing that 3.0 in.,a = determine the value of P 
and the reaction at A. 
 
SOLUTION 
 
 
 
 
As shown on the f.b.d., the wheel is a three-force body. Let point D be 
the intersection of the three forces. 
From force triangle 
21 lb
5 4 3
A P= = 
( )4 21 lb 28 lb
3
P∴ = = 
 or 28.0 lbP = W 
and ( )5 21 lb 35 lb
3
A = = 
1 3tan 36.870
4
θ −  = = °   
 35.0 lb∴ =A 36.9°W 
 
 
 
 
 
 
PROBLEM 4.64 
Horizontal and vertical links are hinged to a wheel, and forces are applied 
to the links as shown. Determine the range of values of the distance a for 
which the magnitude of the reaction at A does not exceed 42 lb. 
 
SOLUTION 
 
 
 
 
 
Let D be the intersection of the three forces acting on the wheel. 
From the force triangle 
2
21 lb
16
A
a a
=
+
 
or 2
1621 1A
a
= + 
For 42 lbA = 
2
21 lb 42 lb
16a a
=
+
 
or 
2
2 16
4
aa += 
or 16 2.3094 in.
3
a = = 
 or 2.31 in.a ≥ W 
Since 2
1621 1A
a
= + 
as a increases, A decreases 
 
 
 
 
 
 
PROBLEM 4.65 
Using the method of Section 4.7, solve Problem 4.21. 
P4.21 The required tension in cable AB is 800 N. Determine (a) the 
vertical force P which must be applied to the pedal, (b) the corresponding 
reaction at C. 
 
SOLUTION 
 
 
 
 
 
Let E be the intersection of the three forces acting on the pedal device. 
First note 
( )1 180 mm sin 60tan 21.291
400 mm
α −  °= = °  
 
From force triangle 
(a) ( )800 N tan 21.291P = ° 
 311.76 N= 
 or 312 N=P W 
(b) 800 N
cos 21.291
C = ° 
 858.60 N= 
 or 859 N=C 21.3°W 
 
 
 
 
 
 
PROBLEM 4.66 
Using the method of Section 4.7, solve Problem 4.22. 
P4.22 Determine the maximum tension which can be developed in cable 
AB if the maximum allowable value of the reaction at C is 1000 N. 
 
SOLUTION 
 
 
 
Let E be the intersection of the three forces acting on the pedal device. 
First note 
( )1 180 mm sin 60tan 21.291
400 mm
α −  °= = °  
 
From force triangle 
( )max 1000 N cos 21.291T = ° 
 931.75 N= 
 maxor 932 NT = W 
 
 
 
 
 
 
PROBLEM 4.67 
To remove a nail, a small block of wood is placed under a crowbar, and a 
horizontal force P is applied as shown. Knowing that 3.5 in.l = and 
30 lb,P = determine the vertical force exerted on the nail and the 
reaction at B. 
 
SOLUTION 
 
 
 
Let D be the intersection of the three forces acting on the crowbar. 
First note 
( )1 36 in. sin 50tan 82.767
3.5 in.
θ −  °= = °  
 
From force triangle 
( )tan 30 lb tan82.767NF P θ= = ° 
 236.381 lb= 
 on nail 236 lbN∴ =F W 
30 lb 238.28 lb
cos cos82.767B
PR θ= = =° 
 or 238 lbB =R 82.8°W 
 
 
 
 
 
 
PROBLEM 4.68 
To remove a nail, a small block of wood is placed under a crowbar, and a 
horizontal force P is applied as shown. Knowing that the maximum 
vertical force needed to extract the nail is 600 lb and that the horizontal 
force P is not to exceed 65 lb, determine the largest acceptable value of 
distance l. 
 
SOLUTION 
 
 
 
Let D be the intersection of the three forces acting on the crowbar. 
From force diagram 
600 lbtan 9.2308
65 lb
NF
P
θ = = = 
 83.817θ∴ = ° 
From f.b.d. 
( )36 in. sin 50tan
l
θ °= 
( )36 in. sin 50 2.9876 in.
tan83.817
l
°∴ = =° 
 or 2.99 in.l = W 
 
 
 
 
 
 
PROBLEM 4.69 
For the frame and loading shown, determine the reactions at C and D. 
 
SOLUTION 
 
Since member BD is acted upon by two forces, B and D, they must be colinear, have the same magnitude, and 
be opposite in direction for BD to be in equilibrium. The force B acting at B of member ABC will be equal in 
magnitude but opposite in direction to force B acting on member BD. Member ABC is a three-force body with 
member forces intersecting at E. The f.b.d.’s of members ABC and BD illustrate the above conditions. The 
force triangle for member ABC is also shown. The angles and α β are found from the member dimensions: 
1 0.5 mtan 26.565
1.0 m
α −  = = °   
1 1.5 mtan 56.310
1.0 m
β −  = = °   
Applying the law of sines to the force triangle for member ABC, 
( ) ( ) ( )
150 N
sin sin 90 sin 90
C B
β α α β= =− ° + ° − 
or 150 N
sin 29.745 sin116.565 sin 33.690
C B= =° ° ° 
( )150 N sin116.565 270.42 N
sin 29.745
C
°∴ = =° 
 or 270 N=C 56.3°W 
and ( )150 N sin 33.690 167.704 N
sin 29.745
D B
°= = =° 
 or 167.7 N=D 26.6°W 
 
 
 
 
 
 
PROBLEM 4.70 
For the frame and loading shown, determine the reactions at A and C. 
 
SOLUTION 
 
Since member AB is acted upon by two forces, A and B, they must be colinear, have the same magnitude, and 
be opposite in direction for AB to be in equilibrium. The force B acting at B of member BCD will be equal in 
magnitude but opposite in direction to force B acting on member AB. Member BCD is a three-force body with 
member forces intersecting at E. The f.b.d.’s of members AB and BCD illustrate the above conditions. The 
force triangle for member BCD is also shown. The angle β is found from the member dimensions: 
1 60 mtan 30.964
100 m
β −  = = °   
Applying of the law of sines to the force triangle for member BCD, 
( )
130 N
sin 45 sin sin135
B C
β β= =° − ° 
or 130 N
sin14.036 sin 30.964 sin135
B C= =° ° ° 
( )130 N sin 30.964 275.78 N
sin14.036
A B
°∴ = = =° 
 or 276 N=A 45.0°W 
and ( )130 N sin135 379.02 N
sin14.036
C
°= =° 
 or 379 N=C 59.0°W 
 
 
 
 
 
PROBLEM 4.71 
To remove the lid from a 5-gallon pail, the tool shown is used to apply an 
upward and radially outward force to the bottom inside rim of the lid. 
Assuming that the rim rests against the tool at A and that a 100-N force is 
applied as indicated to thehandle, determine the force acting on the rim. 
 
SOLUTION 
 
 
 
 
The three-force member ABC has forces that intersect at D, where 
1 90 mmtan
45 mmDC BCy y
α −  =  − − 
 
and 
( )360 mm cos35
tan 20 tan 20
BC
DC
xy
°= =° ° 
 810.22 mm= 
 ( )360 mm sin 35BCy = ° 
 206.49 mm= 
1 90tan 9.1506
558.73
α −  ∴ = = °   
Based on the force triangle, the law of sines gives 
100 N
sin sin 20
A
α = ° 
( )100 N sin 20 215.07 N
sin 9.1506
A
°∴ = =° 
or 215 N=A 80.8 on tool° 
and 215 N=A 80.8 on rim of can° W 
 
 
 
 
 
 
PROBLEM 4.72 
To remove the lid from a 5-gallon pail, the tool shown is used to apply 
an upward and radially outward force to the bottom inside rim of the 
lid. Assuming that the top and the rim of the lid rest against the tool at 
A and B, respectively, and that a 60-N force is applied as indicated to 
the handle, determine the force acting on the rim. 
 
SOLUTION 
 
 
 
The three-force member ABC has forces that intersect at point D, where, 
from the law of sines ( )CDE∆ 
( )150 mm 19 mm tan 35
sin 95 sin 30
L + °=° ° 
325.37 mmL∴ = 
Then 
1 45 mmtan
BDy
α −  =    
where 
22 mmBD AEy L y= − − 
 19 mm325.37 mm 22 mm
cos35
= − −° 
 280.18 mm= 
1 45 mmtan 9.1246
280.18 mm
α −  ∴ = = °   
Applying the law of sines to the force triangle, 
60 N
sin150 sin 9.1246
B =° ° 
189.177 NB∴ = 
Or, on member 189.2 N=B 80.9° 
and, on lid 189.2 N=B 80.9°W 
 
 
 
 
 
PROBLEM 4.73 
A 200-lb crate is attached to the trolley-beam system shown. Knowing 
that 1.5 ft,a = determine (a) the tension in cable CD, (b) the reaction 
at B. 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
 
From geometry of forces 
1tan
1.5 ft
BEyβ −  =    
where 
2.0BE DEy y= − 
 2.0 1.5 tan 35= − ° 
 0.94969 ft= 
1 0.94969tan 32.339
1.5
β −  ∴ = = °   
and 90 90 32.339 57.661α β= ° − = ° − ° = ° 
 35 32.339 35 67.339θ β= + ° = ° + ° = ° 
Applying the law of sines to the force triangle, 
200 lb
sin sin sin 55
T B
θ α= = ° 
or ( )200 lb
sin 67.339 sin 57.661 sin 55
T B= =° ° ° 
(a) ( )( )200 lb sin 57.661 183.116 lb
sin 67.339
T
°= =° 
or 183.1 lbT = W 
(b) ( )( )200 lb sin 55 177.536 lb
sin 67.339
B
°= =° 
or 177.5 lb=B 32.3°W 
 
 
 
 
 
PROBLEM 4.74 
Solve Problem 4.73 assuming that 3 ft.a = 
P4.73 A 200-lb crate is attached to the trolley-beam system shown. 
Knowing that 1.5a = ft, determine (a) the tension in cable CD, (b) the 
reaction 
at B. 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
 
From geometry of forces 
1tan
3 ft
BEyβ −  =    
where 
2.0 ftBE DEy y= − 
 3 tan 35 2.0= ° − 
 0.100623 ft= 
1 0.100623tan 1.92103
3
β −  ∴ = = °   
and 90 90 1.92103 91.921α β= ° + = ° + ° = ° 
 35 35 1.92103 33.079θ β= ° − = ° − ° = ° 
Applying the law of sines to the force triangle, 
200 lb
sin sin sin 55
T B
θ α= = ° 
or 200 lb B
sin 33.079 sin 91.921 sin 55
T= =° ° ° 
(a) ( )( )200 lb sin 91.921 366.23 lb
sin 33.079
T
°= =° 
or 366 lbT = W 
(b) ( )( )200 lb sin 55 300.17 lb
sin 33.079
B
°= =° 
or 300 lb=B 1.921°W 
 
 
 
 
 
PROBLEM 4.75 
A 20-kg roller, of diameter 200 mm, which is to be used on a tile floor, is 
resting directly on the subflooring as shown. Knowing that the thickness 
of each tile is 8 mm, determine the force P required to move the roller 
onto the tiles if the roller is pushed to the left. 
 
SOLUTION 
 
 
 
 
 
Based on the roller having impending motion to the left, the only contact 
between the roller and floor will be at the edge of the tile. 
First note ( )( )220 kg 9.81 m/s 196.2 NW mg= = = 
From the geometry of the three forces acting on the roller 
1 92 mmcos 23.074
100 mm
α −  = = °   
and 90 30θ α= ° − ° − 
 60 23.074= ° − 
 36.926= ° 
Applying the law of sines to the force triangle, 
sin sin
W P
θ α= 
or 196.2 N
sin 36.926 sin 23.074
P=° ° 
127.991 NP∴ = 
or 128.0 N=P 30°W 
 
 
 
 
 
 
PROBLEM 4.76 
A 20-kg roller, of diameter 200 mm, which is to be used on a tile floor, is 
resting directly on the subflooring as shown. Knowing that the thickness 
of each tile is 8 mm, determine the force P required to move the roller 
onto the tiles if the roller is pulled to the right. 
 
SOLUTION 
 
 
 
 
 
 
 
 
Based on the roller having impending motion to the right, the only 
contact between the roller and floor will be at the edge of the tile. 
First note ( )( )220 kg 9.81 m/sW mg= = 
 196.2 N= 
From the geometry of the three forces acting on the roller 
1 92 mmcos 23.074
100 mm
α −  = = °   
and 90 30θ α= ° + ° − 
 120 23.074= ° − ° 
 96.926= ° 
Applying the law of sines to the force triangle, 
sin sin
W P
θ α= 
or 196.2 N
sin 96.926 sin 23.074
P=° 
77.460 NP∴ = 
or 77.5 N=P 30°W 
 
 
 
 
 
 
PROBLEM 4.77 
A small hoist is mounted on the back of a pickup truck and is used to lift 
a 120-kg crate. Determine (a) the force exerted on the hoist by the 
hydraulic cylinder BC, (b) the reaction at A. 
 
SOLUTION 
 
 
 
First note ( )( )2120 kg 9.81 m/s 1177.2 NW mg= = = 
From the geometry of the three forces acting on the small hoist 
 ( )1.2 m cos30 1.03923 mADx = ° = 
 ( )1.2 m sin 30 0.6 mADy = ° = 
and ( )tan 75 1.03923 m tan 75 3.8785 mBE ADy x= ° = ° = 
Then 1 10.4 m 3.4785tan tan 73.366
1.03923
BE
AD
y
x
α − − −  = = = °      
 75 75 73.366 1.63412β α= ° − = ° − ° = ° 
 180 15 165 1.63412 163.366θ β= ° − ° − = ° − ° = ° 
Applying the law of sines to the force triangle, 
sin sin sin15
W B A
β θ= = ° 
or 1177.2 N
sin1.63412 sin163.366 sin15
B A= =° ° ° 
(a) 11 816.9 NB = 
 or 11.82 kN=B 75.0°W 
(b) 10 684.2 NA = 
 or 10.68 kN=A 73.4°W 
 
 
 
 
 
 
PROBLEM 4.78 
The clamp shown is used to hold the rough workpiece C. Knowing that 
the maximum allowable compressive force on the workpiece is 200 N 
and neglecting the effect of friction at A, determine the corresponding 
(a) reaction at B, (b) reaction at A, (c) tension in the bolt. 
 
SOLUTION 
 
 
 
 
 
From the geometry of the three forces acting on the clamp 
 ( )105 mm tan 78 493.99 mmADy = ° = 
 ( )70 mm 493.99 70 mm 423.99 mmBD ADy y= − = − = 
Then 1 1 423.99tan tan 65.301
195 mm 195
BDyθ − −   = = = °       
90 12 78 65.301 12.6987α θ= ° − − ° = ° − ° = ° 
(a) Based on the maximum allowable compressive force on the 
 workpiece of 200 N, 
( ) 200 NB yR = 
 or sin 200 NBR θ = 
200 N 220.14 N
sin 65.301B
R∴ = =° 
or 220 NB =R 65.3°W 
 Applying the law of sines to the force triangle, 
( )sin12 sin sin 90B A
R N T
α θ= =° ° + 
 or 220.14 N
sin12 sin12.6987 sin155.301
AN T= =° ° ° 
(b) 232.75 NAN = 
or 233 NA =N W 
(c) 442.43 NT = 
or 442 NT = W 
 
 
 
 
 
 
PROBLEM 4.79 
A modified peavey is used to lift a 0.2-m-diameter log of mass 36 kg. 
Knowing that 45θ = ° and that the force exerted at C by the worker is 
perpendicular to the handle of the peavey, determine (a) the force exerted 
at C, (b) the reaction at A. 
 
SOLUTION 
 
 
 
First note ( )( )236 kg 9.81 m/s 353.16 NW mg= = = 
From the geometry of the three forces acting on the modified peavey 
 1 1.1 mtan 40.236
1.1 m 0.2 m
β −  = = ° +  
 45 45 40.236 4.7636α β= ° − = ° − ° = ° 
Applying the law of sines to the force triangle, 
sin sin sin135
W C A
βα= = ° 
or 353.16 N
sin 40.236 sin 4.7636 sin135
C A= =° ° 
(a) 45.404 NC = 
or 45.4 N=C 45.0°W 
(b) 386.60 NA = 
or 387 N=A 85.2°W 
 
 
 
 
 
 
PROBLEM 4.80 
A modified peavey is used to lift a 0.2-m-diameter log of mass 36 kg. 
Knowing that 60θ = ° and that the force exerted at C by the worker is 
perpendicular to the handle of the peavey, determine (a) the force exerted 
at C, (b) the reaction at A. 
 
SOLUTION 
 
 
 
 
 
 
First note ( )( )236 kg 9.81 m/s 353.16 NW mg= = = 
From the geometry of the three forces acting on the modified peavey 
1 1.1 mtan
0.2 mDC
β −  =  +  
where ( )1.1 m tan 30DC a= + ° 
 
tan 30
Ra R = − °  
 0.1 m 0.1 m
tan 30
 = − °  
 0.073205 m= 
( )1.173205 tan 30DC∴ = ° 
 0.67735 m= 
and 1 1.1tan 51.424
0.87735
β −  = = °   
60 60 51.424 8.5756α β= ° − = ° − ° = ° 
Applying the law of sines to the force triangle, 
sin sin sin120
W C A
β α= = ° 
or 353.16 N
sin 51.424 sin8.5756 sin120
C A= =° ° ° 
(a) 67.360 NC = 
or 67.4 N=C 30°W 
(b) 391.22 NA = 
or 391 N=A 81.4°W 
 
 
 
 
 
 
 
PROBLEM 4.81 
Member ABC is supported by a pin and bracket at B and by an 
inextensible cord at A and C and passing over a frictionless pulley at D. 
The tension may be assumed to be the same in portion AD and CD of the 
cord. For the loading shown and neglecting the size of the pulley, 
determine the tension in the cord and the reaction at B. 
 
SOLUTION 
 
 
 
 
From the f.b.d. of member ABC, it is seen that the member can be treated 
as a three-force body. 
From the force triangle 
300 3
4
T
T
− = 
 3 4 1200T T= − 
 1200 lbT∴ = W 
Also, 5
4
B
T
= 
 ( )5 5 1200 lb 1500 lb
4 4
B T∴ = = = 
 1 3tan 36.870
4
θ −  = = °   
 and 1500 lb=B 36.9°W 
 
 
 
 
 PROBLEM 4.82 
Member ABCD is supported by a pin and bracket at C and by an 
inextensible cord attached at A and D and passing over frictionless 
pulleys at B and E. Neglecting the size of the pulleys, determine the 
tension in the cord and the reaction at C. 
 
SOLUTION 
 
 
 
From the geometry of the forces acting on member ABCD 
1 200tan 33.690
300
β −  = = °   
 1 375tan 61.928
200
α −  = = °   
 61.928 33.690 28.237α β− = ° − ° = ° 
 180 180 61.928 118.072α° − = ° − ° = ° 
Applying the law of sines to the force triangle, 
 ( ) ( )
80 N
sin sin sin 180
T T C
α β β α
− = =− ° − 
or 80 N
sin 28.237 sin 33.690 sin118.072
T T C− = =° ° ° 
Then ( )80 N sin 33.690 sin 28.237T T− ° = ° 
 543.96 NT∴ = 
 or T 544 N= W 
and ( )543.96 N sin118.072 sin 33.690C= ° 
 865.27 NC∴ = 
 or 865 N=C 33.7°W 
 
 
 
 
 
 
PROBLEM 4.83 
Using the method of Section 4.7, solve Problem 4.18. 
P4.18 Determine the reactions at A and B when (a) 0h = , (b) 8h = in. 
 
SOLUTION 
 
 
 
 
 
 (a) Based on symmetry 
30α = ° 
 From force triangle 
40 lbA B= = 
 or 40.0 lb=A 30°W 
 and 40.0 lb=B 30°W 
 
 
 
 
(b) From geometry of forces 
 ( )1 8 in. 10 in. tan 30tan 12.5521
10 in.
α −  − °= = °  
 
 Also, 
30 30 12.5521 17.4479α° − = ° − ° = ° 
 90 90 12.5521 102.5521α° + = ° + ° = ° 
 Applying law of sines to the force triangle, 
( ) ( )
40 lb
sin 30 sin 60 sin 90
A B
α α= =° − ° ° + 
 or 40 lb
sin17.4479 sin 60 sin102.5521
A B= =° ° 
115.533 lbA = 
 or 115.5 lb=A 12.55°W 
 130.217 lbB = 
 or 130.2 lb=B 30.0°W 
 
 
 
 
 
 
 
PROBLEM 4.84 
Using the method of Section 4.7, solve Problem 4.28. 
P4.28 A lever is hinged at C and is attached to a control cable at A. If the 
lever is subjected to a 300-N vertical force at B, determine 
(a) the tension in the cable, (b) the reaction at C. 
 
SOLUTION 
 
 
 
 
 
From geometry of forces acting on lever 
1tan DA
DA
y
x
α −  =    
where 
 ( )0.24 m 0.24 m 0.2 m sin 20DA ACy y= − = − ° 
 0.171596 m= 
 ( )0.2 m cos 20DAx = ° 
 0.187939 m= 
 1 0.171596 tan 42.397
0.187939
α −  ∴ = = °   
 190 tan AC EA
CE
y y
x
β −  += ° −   
 
where ( )0.3 m cos 20 0.28191 mCEx = ° = 
 ( )0.2 m sin 20 0.068404 mACy = ° = 
 ( ) tanEA DA CEy x x α= + 
 ( )0.187939 0.28191 tan 42.397= + ° 
 0.42898 m= 
 1 0.49739 90 tan 29.544
0.28191
β −  ∴ = ° − = °   
Also, ( )90 90 71.941 18.0593α β° − + = ° − ° = ° 
 90 90 42.397 132.397α° + = ° + ° = ° 
 
 
 
 
 
 
PROBLEM 4.84 CONTINUED 
Applying the law of sines to the force triangle, 
( ) ( )
300 N
sin sin 90sin 90
T C
β αα β = =  ° +° − + 
 
or 300 N
sin18.0593 sin 29.544 sin132.397
T C= =° ° ° 
(a) 477.18 NT = or 477 NT = W 
(b) 714.67 NC = or 715 N=C 60.5°W 
 
 
 
 
 
 
 
PROBLEM 4.85 
Knowing that o35 ,θ = determine the reaction (a) at B, (b) at C. 
 
SOLUTION 
 
 
 
From the geometry of the three forces applied to the member ABC 
1tan CDy
R
α −  =    
where 
tan 55 0.42815CDy R R R= ° − = 
 ( )1 tan 0.42815 23.178α −∴ = = ° 
Then 55 55 23.178 31.822α° − = ° − ° = ° 
 90 90 23.178 113.178α° + = ° + ° = ° 
Applying the law of sines to the force triangle, 
 ( ) ( )sin 55 sin 90 sin 35
P B C
α α= =° − ° + ° 
or 
sin 31.822 sin113.178 sin 35
P B C= =° ° ° 
(a) 1.74344B P= 
 or 1.743P=B 55.0°W 
(b) 1.08780C P= 
 or 1.088P=C 23.2°W 
 
 
 
 
 
 
 
PROBLEM 4.86 
Knowing that o50 ,θ = determine the reaction (a) at B, (b) at C. 
 
 
SOLUTION 
 
 
 
 
From the geometry of the three forces acting on member ABC 
1tan DCy
R
α −  =    
where 
( )1 tan 90 50DC ADy R y R  = − = − ° − °  
 0.160900R= 
( )1 tan 0.160900 9.1406α −∴ = = ° 
Then 90 90 9.1406 80.859α° − = ° − ° = ° 
 40 40 9.1406 49.141α° + = ° + ° = ° 
Applying the law of sines to the force triangle, 
 ( ) ( )sin 40 sin 90 sin 50
P B C
α α= =° + ° − ° 
or ( )sin 49.141 sin 80.859 sin 50
P B C= =° ° ° 
(a) 1.30540B P= 
 or 1.305P=B 40.0°W 
(b) 1.01286C P= 
 or 1.013P=C 9.14°W 
 
 
 
 
 
 
PROBLEM 4.87 
A slender rod of length L and weight W is held in equilibrium as shown, 
with one end against a frictionless wall and the other end attached to a 
cord of length S. Derive an expression for the distance h in terms of L and 
S. Show that this position of equilibrium does not exist if 2 .S L> 
 
SOLUTION 
 
 
 
 
 
 
 
From the f.b.d of the three-force member AB, forces must intersect at D. 
Since the force T intersects point D, directly above G, 
 BEy h= 
For triangle ACE: 
 ( ) ( )2 22 2S AE h= + (1) 
For triangle ABE: 
 ( ) ( )2 22L AE h= + (2) 
Subtracting Equation (2) from Equation (1) 
 2 2 23S L h− = (3) 
 or 
2 2
3
S Lh −= W 
As length S increases relative to length L, angle θ increases until rod AB 
is vertical. At this vertical position: 
orh L S h S L+ = = − 
Therefore, for all positions of AB h S L≥ − (4) 
or 
2 2
3
S L S L− ≥ − 
or ( ) ( )22 2 2 2 2 23 3 2 3 6 3S L S L S SL L S SL L− ≥ − = − + = − + 
or 2 20 2 6 4S SL L≥ − + 
and ( )( )2 20 3 2 2S SL L S L S L≥ − + = − − 
For 0S L S L− = = 
 Minimum value of is S L∴ 
For 2 0 2S L S L− = = 
 Maximum value of is 2S L∴ 
Therefore, equilibrium does not exist if 2S L> W 
 
 
 
 
 
 
PROBLEM 4.88 
A slender rod of length 200 mmL = is held in equilibrium as shown, 
with one end against a frictionless wall and the other end attached to a 
cord of length 300 mm.S = Knowing that the mass of the rod is 1.5 kg, 
determine (a)the distance h, (b) the tension in the cord, (c) the reaction 
at B. 
 
 
SOLUTION 
 
 
 
 
 
From the f.b.d of the three-force member AB, forces must intersect at D. 
Since the force T intersects point D, directly above G, 
 BEy h= 
For triangle ACE: 
 ( ) ( )2 22 2S AE h= + (1) 
For triangle ABE: 
 ( ) ( )2 22L AE h= + (2) 
Subtracting Equation (2) from Equation (1) 
 2 2 23S L h− = 
 or 
2 2
3
S Lh −= 
(a) For 200 mm and 300 mmL S= = 
 ( ) ( )2 2300 200 129.099 mm
3
h
−= = 
 or 129.1 mmh = W 
(b) Have W mg= ( )( )21.5 kg 9.81 m/s= 14.715 N= 
 and ( )1 1 2 129.0992sin sin
300
h
s
θ − −   = =      
 
 59.391θ = ° 
 From the force triangle 
 14.715 N 17.0973 N
sin sin 59.391
WT θ= = =° 
 or 17.10 NT = W 
(c) 14.715 N 8.7055 N
tan tan 59.391
WB θ= = =° 
 or 8.71 N=B W 
 
 
 
 
 
 
PROBLEM 4.89 
A slender rod of length L and weight W is attached to collars which can 
slide freely along the guides shown. Knowing that the rod is in 
equilibrium, derive an expression for the angle θ in terms of the 
angle β . 
 
SOLUTION 
 
 
 
As shown in the f.b.d of the slender rod AB, the three forces intersect at 
C. From the force geometry 
 tan GB
AB
x
y
β = 
where 
 cosABy L θ= 
and 1 sin
2GB
x L θ= 
 
1
2 sin 1 tan tan
cos 2
L
L
θβ θθ∴ = = 
 or tan 2 tanθ β= W 
 
 
 
 
 
 
PROBLEM 4.90 
A 10-kg slender rod of length L is attached to collars which can slide 
freely along the guides shown. Knowing that the rod is in equilibrium and 
that 25 ,β = ° determine (a) the angle θ that the rod forms with the 
vertical, (b) the reactions at A and B. 
 
SOLUTION 
 
 
 
 (a) As shown in the f.b.d. of the slender rod AB, the three forces 
 intersect at C. From the geometry of the forces 
 tan CB
BC
x
y
β = 
 where 
 1 sin
2CB
x L θ= 
 and cosBCy L θ= 
 1 tan tan
2
β θ∴ = 
 or tan 2 tanθ β= 
 For 25β = ° 
 tan 2 tan 25 0.93262θ = ° = 
 43.003θ∴ = ° 
 or 43.0θ = °W 
(b) W mg= ( )( )210 kg 9.81 m/s= 98.1 N= 
 From force triangle 
 tanA W β= 
 ( )98.1 N tan 25= ° 
 45.745 N= 
 or 45.7 N=A W 
 and 98.1 N 108.241 N
cos cos 25
WB β= = =° 
 or 108.2 N=B 65.0°W 
 
 
 
 
 
 
 
 
PROBLEM 4.91 
A uniform slender rod of mass 5 g and length 250 mm is balanced on a 
glass of inner diameter 70 mm. Neglecting friction, determine the angle θ 
corresponding to equilibrium. 
 
SOLUTION 
 
 
 
From the geometry of the forces acting on the three-force member AB 
Triangle ACF 
 tanCFy d θ= 
Triangle CEF 2tan tanFE CFx y dθ θ= = 
Triangle AGE 
 
2tancos
2 2
FEd x d d
L L
θθ + += =         
 
 ( )22 1 tandL θ= + 
Now ( )2 2 11 tan sec and sec cosθ θ θ θ+ = = 
Then 2 2
2 2 1cos sec
cos
d d
L L
θ θ θ
 = =    
 3 2 cos d
L
θ∴ = 
For 70 mm and 250 mmd L= = 
 ( )3 2 70cos 0.56
250
θ = = 
 cos 0.82426θ∴ = 
and 34.487θ = ° 
 or 34.5θ = °W 
 
 
 
 
 
 
 
 
PROBLEM 4.92 
Rod AB is bent into the shape of a circular arc and is lodged between two 
pegs D and E. It supports a load P at end B. Neglecting friction and the 
weight of the rod, determine the distance c corresponding to equilibrium 
when 1a = in. and 5R = in. 
 
SOLUTION 
 
 
 
Since ,ED EDy x a= = 
Slope of ED is 45° 
 slope of isHC∴ 45° 
Also 2DE a= 
and 1
2 2
aDH HE DE = = =   
For triangles DHC and EHC 
 / 2sin
2
a a
R R
β = = 
Now ( )sin 45c R β= ° − 
For 1 in. and 5 in.a R= = 
 ( )
1 in.sin 0.141421
2 5 in.
β = = 
 8.1301β∴ = ° or 8.13β = °W 
and ( ) ( )5 in. sin 45 8.1301 3.00 in.c = ° − ° = 
 or 3.00 in.c = W 
 
 
 
 
 
 
 
PROBLEM 4.93 
A uniform rod AB of weight W and length 2R rests inside a hemispherical 
bowl of radius R as shown. Neglecting friction determine the angle θ 
corresponding to equilibrium. 
 
 
SOLUTION 
 
 
Based on the f.b.d., the uniform rod AB is a three-force body. Point E is 
the point of intersection of the three forces. Since force A passes through 
O, the center of the circle, and since force C is perpendicular to the rod, 
triangle ACE is a right triangle inscribed in the circle. Thus, E is a point 
on the circle. 
Note that the angle α of triangle DOA is the central angle corresponding 
to the inscribed angleθ of triangle DCA. 
2α θ∴ = 
The horizontal projections of ( ), ,AEAE x and ( ), ,AGAG x are equal. 
 AE AG Ax x x∴ = = 
or ( ) ( )cos 2 cosAE AGθ θ= 
and ( )2 cos 2 cosR Rθ θ= 
Now 2cos 2 2cos 1θ θ= − 
then 24cos 2 cosθ θ− = 
or 24cos cos 2 0θ θ− − = 
Applying the quadratic equation 
 cos 0.84307 and cos 0.59307θ θ= = − 
 32.534 and 126.375 (Discard)θ θ∴ = ° = ° 
 or 32.5θ = °W 
 
 
 
 
 
 
 
PROBLEM 4.94 
A uniform slender rod of mass m and length 4r rests on the surface shown 
and is held in the given equilibrium position by the force P. Neglecting 
the effect of friction at A and C, (a) determine the angle θ, (b) derive an 
expression for P in terms of m. 
 
 
SOLUTION 
 
 
 
 
The forces acting on the three-force member intersect at D. 
(a) From triangle ACO 
 1 1 1tan tan 18.4349
3 3
r
r
θ − −   = = = °       or 18.43θ = °W 
(b) From triangle DCG tan r
DC
θ = 
 3
tan tan18.4349
r rDC rθ∴ = = =° 
 and 3 4DO DC r r r r= + = + = 
 1tan DO
AG
y
x
α −  =   
 
 where ( ) ( )cos 4 cos18.4349DOy DO rθ= = ° 
 3.4947r= 
 and ( ) ( )2 cos 2 cos18.4349AGx r rθ= = ° 
 1.89737r= 
 1 3.4947 tan 63.435
1.89737
r
r
α −  ∴ = = °   
 where ( )90 90 45 135.00α θ° + − = ° + ° = ° 
 Applying the law of sines to the force triangle, 
( ) sinsin 90 A
mg R
θα θ = ° + − 
 
 ( ) 0.44721AR mg∴ = 
 Finally, cosAP R α= 
 ( )0.44721 cos63.435mg= ° 
 0.20000mg= or 
5
mgP = W 
 
 
 
 
 
 
PROBLEM 4.95 
A uniform slender rod of length 2L and mass m rests against a roller at D 
and is held in the equilibrium position shown by a cord of length a. 
Knowing that L = 200 mm, determine (a) the angle θ, (b) the length a. 
 
SOLUTION 
 
 
 (a) The forces acting on the three-force member AB intersect at E. Since 
 triangle DBC is isosceles, .DB a= 
 From triangle BDE 
tan 2 tan 2ED DB aθ θ= = 
 From triangle GED 
( )
tan
L a
ED θ
−= 
 ( ) tan 2 or tan tan 2 1
tan
L aa a Lθ θ θθ
−∴ = + = (1) 
 From triangle BCD 
( )1
2 1.25 or 1.6cos
cos
L La
a
θθ= = (2) 
 Substituting Equation (2) into Equation (1) yields 
 1.6cos 1 tan tan 2θ θ θ= + 
 Now sin sin 2tan tan 2
cos cos 2
θ θθ θ θ θ= 
 2
sin 2sin cos
cos 2cos 1
θ θ θ
θ θ= − 
 
( )2
2
2 1 cos
2cos 1
θ
θ
−= − 
 Then 
( )2
2
2 1 cos1.6cos 1
2cos 1
θθ θ
−= + − 
 or 33.2cos 1.6cos 1 0θ θ− − = 
 Solving numerically 23.515 or 23.5θ θ= ° = °W 
(b) From Equation (2) for 200 mm and 23.5L θ= = ° 
 ( )200 mm5 136.321 mm
8 cos 23.515
a = =° 
 or 136.3 mma = W 
 
 
 
 
 
 
 
PROBLEM 4.96 
Gears A and B are attached to a shaft supported by bearings at C and D. 
The diameters of gears A and B are 150 mm and 75 mm, respectively, and 
the tangential and radial forces acting on the gears are as shown. 
Knowing that the system rotates at a constant rate, determine the 
reactions at C and D. Assume that the bearing at C does not exert any 
axial force, and neglect the weights of the gears and the shaft. 
 
 
SOLUTION 
 
 
 
Assume moment reactions at the bearing supports are zero.From f.b.d. of 
shaft 
0: 0x xF DΣ = ∴ = 
 ( ) ( ) ( )( )-axis 0: 175 mm 482 N 75 mmyD zM CΣ = − + 
 ( )( )2650 N 50 mm 0+ = 
 963.71 NyC∴ = 
or ( )964 Ny =C j 
 ( ) ( ) ( )( )-axis 0: 175 mm 1325 N 75 mmzD yM CΣ = + 
 ( )( )964 N 50 mm 0+ = 
 843.29 NzC∴ = − 
or ( )843 Nz =C k 
 and ( ) ( )964 N 843 N= −C j kW 
 ( ) ( )( ) ( )-axis 0: 482 N 100 mm 175 mmyC zM DΣ = − + 
 ( )( )2650 N 225 mm 0+ = 
 3131.7 NyD∴ = − 
or ( )3130 Ny = −D j 
 ( ) ( )( ) ( )-axis 0: 1325 N 100 mm 175 mmzC yM DΣ = − − 
 ( )( )964 N 225 mm 0+ = 
 482.29 NzD∴ = 
or ( )482 Nz =D k 
 and ( ) ( )3130 N 482 N= − +D j kW 
 
 
 
 
 
PROBLEM 4.97 
Solve Problem 4.96 assuming that for gear A the tangential and radial 
forces are acting at E, so that FA = (1325 N)j + (482 N)k. 
P4.96 Gears A and B are attached to a shaft supported by bearings at C and 
D. The diameters of gears A and B are 150 mm and 75 mm, respectively, 
and the tangential and radial forces acting on the gears are as shown. 
Knowing that the system rotates at a constant rate, determine the reactions at 
C and D. Assume that the bearing at C does not exert any axial force, and 
neglect the weights of the gears and the shaft. 
 
 
SOLUTION 
 
 
Assume moment reactions at the bearing supports are zero. From f.b.d. of 
shaft 
0: 0x xF DΣ = ∴ = 
( ) ( ) ( )( )- 0: 175 mm 1325 N 75 mmyD z axisM CΣ = − − 
 ( )( )2650 N 50 mm 0+ = 
 189.286 NyC∴ = 
or ( )189.3 Ny =C j 
 ( ) ( ) ( )( )-axis 0: 175 mm 482 N 75 mmzD yM CΣ = + 
 ( )( )964 N 50 mm 0+ = 
 482.00 NzC∴ = − 
or ( )482 Nz = −C k 
 and ( ) ( )189.3 N 482 N= −C j k W 
 ( ) ( )( ) ( )-axis 0: 1325 N 100 mm 175 mmyC zM DΣ = + 
 ( )( )2650 N 225 mm 0+ = 
 4164.3 NyD∴ = − 
or ( )4160 Ny = −D j 
 ( ) ( )( ) ( )-axis 0: 482 N 100 mm 175 mmzC yM DΣ = − − 
 ( )( )964 N 225 mm 0+ = 
 964.00 NzD∴ = 
or ( )964 Nz =D k 
 and ( ) ( )4160 N 964 N= − +D j kW 
 
 
 
 
 
 
 
PROBLEM 4.98 
Two transmission belts pass over sheaves welded to an axle supported by 
bearings at B and D. The sheave at A has a radius of 50 mm, and the 
sheave at C has a radius of 40 mm. Knowing that the system rotates with 
a constant rate, determine (a) the tension T, (b) the reactions at B and D. 
Assume that the bearing at D does not exert any axial thrust and neglect 
the weights of the sheaves and the axle. 
 
 
SOLUTION 
 
Assume moment reactions at the bearing supports are zero. From f.b.d. of shaft 
(a) ( )( ) ( )( )-axis 0: 240 N 180 N 50 mm 300 N 40 mm 0xM TΣ = − + − = 
 375 NT∴ = W 
(b) 0: 0x xF BΣ = = 
 ( ) ( )( ) ( )-axis 0: 300 N 375 N 120 mm 240 mm 0yD zM BΣ = + − = 
 337.5 NyB∴ = 
 ( ) ( )( ) ( )-axis 0: 240 N 180 N 400 mm 240 mm 0zD yM BΣ = + + = 
 700 NzB∴ = − 
 or ( ) ( )338 N 700 N= −B j k W 
 ( ) ( )( ) ( )-axis 0: 300 N 375 N 120 mm 240 mm 0yB zM DΣ = − + + = 
 337.5 NyD∴ = 
 ( ) ( )( ) ( )-axis 0: 240 N 180 N 160 mm 240 mm 0zB yM DΣ = + + = 
280 NzD∴ = − 
 or ( ) ( )338 N 280 N= −D j kW 
 
 
 
 
 
PROBLEM 4.99 
For the portion of a machine shown, the 4-in.-diameter pulley A and 
wheel B are fixed to a shaft supported by bearings at C and D. The spring 
of constant 2 lb/in. is unstretched when θ = 0, and the bearing at C does 
not exert any axial force. Knowing that θ = 180° and that the machine is 
at rest and in equilibrium, determine (a) the tension T, (b) the reactions at 
C and D. Neglect the weights of the shaft, pulley, and wheel. 
 
SOLUTION 
 
First, determine the spring force, ,EF at 180 .θ = ° 
 E sF k x= 
where 2 lb/in.sk = 
 ( ) ( ) ( ) ( )final initial 12 in. 3.5 in. 12 in. 3.5 in. 7.0 in.E Ex y y= − = + − − = 
 ( )( ) 2 lb/in. 7.0 in. 14.0 lbEF∴ = = 
(a) From f.b.d. of machine part 
 ( )( ) ( )0: 34 lb 2 in. 2 in. 0xM TΣ = − = 
 34 lbT∴ = or 34.0 lbT = W 
(b) ( ) ( ) ( )-axis 0: 10 in. 2 in. 1 in. 0D y EzM C FΣ = − − + = 
 ( ) ( )10 in. 14.0 lb 3 in. 0yC− − = 
 4.2 lbyC∴ = − or ( )4.20 lby = −C j 
( ) ( ) ( ) ( )-axis 0: 10 in. 34 lb 4 in. 34 lb 4 in. 0zD yM CΣ = + + = 
 27.2 lbzC∴ = − or ( )27.2 lbz = −C k 
 and ( ) ( )4.20 lb 27.2 lb= − −C j k W 
 
 
 PROBLEM 4.99 CONTINUED 
 0: 0x xF DΣ = = 
 ( ) ( ) ( )-axis 0: 10 in. 12 in. 1 in. 0y EC zM D FΣ = − + = 
 or ( ) ( )10 in. 14.0 13 in. 0yD − = 
18.2 lbyD∴ = or ( )18.20 lby =D j 
 ( ) ( )( ) ( )-axis 0: 2 34 lb 6 in. 10 in. 0zC yM DΣ = − − = 
 40.8 lbzD∴ = − or ( )40.8 lbz = −D k 
 and ( ) ( )18.20 lb 40.8 lb= −D j k W 
 
 
 
 
 
 
PROBLEM 4.100 
Solve Problem 4.99 for θ = 90°. 
P4.99 For the portion of a machine shown, the 4-in.-diameter pulley A and 
wheel B are fixed to a shaft supported by bearings at C and D. The spring of 
constant 2 lb/in. is unstretched when θ = 0, and the bearing at C does not 
exert any axial force. Knowing that θ = 180° and that the machine is at rest 
and in equilibrium, determine (a) the tension T, (b) the reactions at C and D. 
Neglect the weights of the shaft, pulley, and wheel. 
 
 
SOLUTION 
 
First, determine the spring force, ,EF at 90 .θ = ° 
 E sF k x= 
where 2 lb/in.sk = 
and ( ) ( ) ( )2 2final initial 3.5 12 12 3.5 12.5 8.5 4.0 in.x L L  = − = + − − = − =   
 ( )( )2 lb/in. 4.0 in. 8.0 lbEF∴ = = 
Then ( ) ( ) ( ) ( )12.0 3.58.0 lb 8.0 lb 7.68 lb 2.24 lb
12.5 12.5E
−= + = − +F j k j k 
(a) From f.b.d. of machine part 
 ( )( ) ( ) ( )( )0: 34 lb 2 in. 2 in. 7.68 lb 3.5 in. 0xM TΣ = − − = 
 20.56 lbT∴ = or 20.6 lbT = W 
(b) ( ) ( ) ( )( )-axis 0: 10 in. 7.68 lb 3.0 in. 0D yzM CΣ = − − = 
 2.304 lbyC∴ = − or ( )2.30 lby = −C j 
( ) ( ) ( )( ) ( )( ) ( )( )-axis 0: 10 in. 34 lb 4.0 in. 20.56 lb 4.0 in. 2.24 lb 3 in. 0D zyM CΣ = + + − = 
 21.152 lbzC∴ = − or ( )21.2 lbz = −C k 
 and ( ) ( )2.30 lb 21.2 lb= − −C j k W 
 
 
 PROBLEM 4.100 CONTINUED 
 0: 0x xF DΣ = = 
 ( ) ( ) ( )( )-axis 0: 10 in. 7.68 lb 13 in. 0yC zM DΣ = − = 
 9.984 lbyD∴ = or ( )9.98 lby =D j 
( ) ( )( ) ( )( ) ( ) ( )( )-axis 0: 34 lb 6 in. 20.56 lb 6 in. 10 in. 2.24 lb 13 in. 0zC yM DΣ = − − − − = 
 35.648 lbzD∴ = − or ( )35.6 lbz = −D k 
 and ( ) ( )9.98 lb 35.6 lb= −D j kW 
 
 
 
 
 
 
 
PROBLEM 4.101 
A 1.2 2.4-m× sheet of plywood having a mass of 17 kg has been 
temporarily placed among three pipe supports. The lower edge of the 
sheet rests on small collars A and B and its upper edge leans against pipe 
C. Neglecting friction at all surfaces, determine the reactions at A, B, 
and C. 
 
SOLUTION 
 
First note ( )( )217 kg 9.81 m/s 166.77 NW mg= = = 
 ( ) ( )2 21.2 1.125 0.41758 mh = − = 
From f.b.d. of plywood sheet 
 ( ) ( )1.125 m0: 0
2z
M C h W
 Σ = − =  
 
 ( ) ( )( )0.41758 m 166.77 N 0.5625 m 0C − = 
 224.65 NC∴ = ( )or 225 N= −C i 
( ) ( )( ) ( )-axis 0: 224.65 N 0.6 m 1.2 m 0xB yM AΣ = − + = 
 112.324 NxA∴ = ( )or 112.3 Nx =A i 
 ( ) ( )( ) ( )-axis 0: 166.77 N 0.3 m 1.2 m 0yB xM AΣ = − = 
 41.693 NyA∴ = ( )or 41.7 Ny =A j 
 ( ) ( )( ) ( )-axis 0: 224.65 N 0.6 m 1.2 m 0xA yM BΣ = − = 
 112.325 NxB∴ = ( )or 112.3 Nx =B i 
 
 
 PROBLEM 4.101 CONTINUED 
 ( ) ( ) ( )( )-axis 0: 1.2 m 166.77 N 0.9 m 0yA xM BΣ = − = 
 125.078 NyB∴ = ( )or 125.1 Ny =B j 
 ( ) ( ) 112.3 N 41.7 N∴ = +A i jW 
 ( ) ( )112.3 N 125.1 N= +B i jW 
 ( )225 N= −C iW 
 
 
 
 
 
 
PROBLEM 4.102 
The 200 200-mm× square plate shown has a mass of 25 kg and is 
supported by three vertical wires. Determine the tensionin each wire. 
 
SOLUTION 
 
First note ( )( )225 kg 9.81 m/s 245.25 NW mg= = = 
From f.b.d. of plate 
( )( ) ( ) ( )0: 245.25 N 100 mm 100 mm 200 mm 0x A CM T TΣ = − − = 
 2 245.25 NA CT T∴ + = (1) 
( ) ( ) ( )( )0: 160 mm 160 mm 245.25 N 100 mm 0z B CM T TΣ = + − = 
 153.281 NB CT T∴ + = (2) 
 0: 245.25 N 0y A B CF T T TΣ = + + − = 
 245.25B C AT T T∴ + = − (3) 
Equating Equations (2) and (3) yields 
 245.25 N 153.281 N 91.969 NAT = − = (4) 
or 92.0 NAT = 
Substituting the value of TA into Equation (1) 
 ( )245.25 N 91.969 N 76.641 N
2C
T
−= = (5) 
or 76.6 NCT = 
Substituting the value of TC into Equation (2) 
153.281 N 76.641 N 76.639 NBT = − = or 76.6 NBT = 
 92.0 NAT = W 
 76.6 NBT = W 
 76.6 NCT = W 
 
 
 
 
 
 
PROBLEM 4.103 
The 200 200-mm× square plate shown has a mass of 25 kg and is 
supported by three vertical wires. Determine the mass and location of the 
lightest block which should be placed on the plate if the tensions in the 
three cables are to be equal. 
 
SOLUTION 
 
First note ( )( )21 25 kg 9.81 m/s 245.25 NG pW m g= = = 
( ) ( )21 9.81 m/s 9.81 NW mg m m= = = 
From f.b.d. of plate 
 10: 3 0y GF T W WΣ = − − = (1) 
( ) ( ) ( ) ( )10: 100 mm 100 mm 200 mm 0x GM W W z T TΣ = + − − = 
 1or 300 100 0GT W W z− + + = (2) 
 ( ) ( ) ( )10: 2 160 mm 100 mm 0z GM T W W xΣ = − − = 
 1or 320 100 0GT W W x− − = (3) 
( ) ( )Eliminate by forming 100 Eq. 1 Eq. 2 T  × +  
1 1100 0W W z− + = 
 100 mm 0 200 mm, okayz z∴ = ≤ ≤ ∴ 
( ) ( )Now, 3 Eq. 3 320 Eq. 1 yields   × − ×    
( ) ( ) ( )1 13 320 3 100 3 320 3 320 320 0G GT W W x T W W− − − + + = 
 
 
 PROBLEM 4.103 CONTINUED 
or ( ) 120 320 3 0GW x W+ − = 
or ( )1
20
3 320G
W
W x
= − 
The smallest value of 1
G
W
W
 will result in the smallest value of 1W since WG is given. 
max Use 200 mmx x∴ = = 
and then ( )1
20 1
3 200 320 14G
W
W
= =− 
( )1 245.25 N 17.5179 N minimum14 14G
WW∴ = = = 
and 1 2
17.5179 N 1.78571 kg
9.81 m/s
Wm
g
= = = 
 or 1.786 kgm = W 
 at 200 mm, 100 mmx z= = W 
 
 
 
 
 
 
PROBLEM 4.104 
A camera of mass 240 g is mounted on a small tripod of mass 200 g. 
Assuming that the mass of the camera is uniformly distributed and that 
the line of action of the weight of the tripod passes through D, determine 
(a) the vertical components of the reactions at A, B, and C when θ = 0, 
(b) the maximum value of θ if the tripod is not to tip over. 
 
SOLUTION 
 
First note ( )( )20.24 kg 9.81 m/s 2.3544 NC CW m g= = = 
 ( )( )2tp tp 0.20 kg 9.81 m/s 1.9620 NW m g= = = 
For 0θ = ( )60 mm 24 mm 36 mmCx = − − = − 
 0Cz = 
(a) From f.b.d. of camera and tripod as projected onto plane ABCD 
 tp0: 0y y y y CF A B C W WΣ = + + − − = 
 2.3544 N 1.9620 N 4.3164 Ny y yA B C∴ + + = + = (1) 
 ( ) ( )0: 38 mm 38 mm 0 x y y y yM C B C BΣ = − = ∴ = (2) 
( ) ( ) ( )( ) ( )0: 35 mm 35 mm 2.3544 N 36 mm 45 mm 0z y y yM B C AΣ = + + − = 
 9 7 7 16.9517y y yA B C∴ − − = (3) 
 Substitute yC with yB from Equation (2) into Equations (1) and (3), and solve by elimination 
( )7 2 4.3164y yA B+ = 
 9 14 16.9517y yA B− = 
 16 yA 47.166= 
 
 
 PROBLEM 4.104 CONTINUED 
 2.9479 NyA∴ = 
or 2.95 Ny =A W 
 Substituting 2.9479 NyA = into Equation (1) 
2.9479 N 2 4.3164yB+ = 
 0.68425 NyB∴ = 
0.68425 NyC = 
 or 0.684 Ny y= =B C W 
(b) 0yB = for impending tipping 
 
 From f.b.d. of camera and tripod as projected onto plane ABCD 
 tp0: 0y y y CF A C W WΣ = + − − = 
 4.3164 Ny yA C∴ + = (1) 
 ( ) ( ) ( )0: 38 mm 2.3544 N 36 mm sin 0x yM C θ Σ = − =  
 2.2305sinyC θ∴ = (2) 
( ) ( ) ( ) ( )0: 35 mm 45 mm 2.3544 N 36 mm cos 0z y yM C A θ Σ = − + =  
 ( ) 9 7 16.9517 N cosy yA C θ∴ − = (3) 
 ( ) ( )Forming 7 Eq. 1 Eq. 3 yields   × +    
 ( )16 30.215 N 16.9517 N cosyA θ= + (4) 
 
 
 PROBLEM 4.104 CONTINUED 
 Substituting Equation (2) into Equation (3) 
 ( ) ( )9 15.6134 N sin 16.9517 N cosyA θ θ− = (5) 
 ( ) ( )Forming 9 Eq. 4 16 Eq. 5 yields   × − ×    
( ) ( )249.81 N sin 271.93 N 118.662 N cosθ θ= − 
 or ( ) 22cos 2.2916 N 2.1053 N sinθ θ = −  
 Now 2 2cos 1 sinθ θ= − 
2 5.4323sin 9.6490sin 4.2514 0θ θ∴ − + = 
 Using quadratic formula to solve, 
sin 0.80981 and sin 0.96641θ θ= = 
 54.078 and 75.108θ θ∴ = ° = ° 
 maxor 54.1 before tippingθ = ° W 
 
 
 
 
 
PROBLEM 4.105 
Two steel pipes AB and BC, each having a weight per unit length of 
5 lb/ft, are welded together at B and are supported by three wires. 
Knowing that 1.25a = ft, determine the tension in each wire. 
 
SOLUTION 
 
First note ( )( )5 lb/ft 2 ft 10 lbABW = = 
( )( )5 lb/ft 4 ft 20 lbBCW = = 
30 lbAB BCW W W= + = 
To locate the equivalent force of the pipe assembly weight 
( ) ( ) ( )/G B AB BCG AB G BC= Σ = +i ir W r W r W r W× × × × 
or ( ) ( ) ( ) ( ) ( ) ( )30 lb 1 ft 10 lb 2 ft 20 lbG Gx z+ − = − + −i k j k j i j× × × 
( ) ( ) ( ) ( ) 30 lb 30 lb 10 lb ft 40 lb ftG Gx z∴ − + = ⋅ − ⋅k i i k 
From i-coefficient 10 lb ft 1 ft
30 lb 3G
z ⋅= = 
 k-coefficient 40 lb ft 11 ft
30 lb 3G
x ⋅= = 
From f.b.d. of piping 
 ( ) ( )0: 2 ft 0x G AM W z TΣ = − = 
1 1 ft 30 lb ft 5 lb
2 3A
T    ∴ = =       or 5.00 lbAT = 
0: 5 lb 30 lb 0y D CF T TΣ = + + − = 
 25 lbD CT T∴ + = (1) 
 
 
 PROBLEM 4.105 CONTINUED 
( ) ( ) 40: 1.25 ft 4 ft 30 lb ft 0
3z D C
M T T  Σ = + − =   
 1.25 4 40 lb ftD CT T∴ + = ⋅ (2) 
( )4 Equation 1 −   4 4 100D CT T− − = − (3) 
Equation (2) + Equation (3) 2.75 60DT− = − 
 21.818 lbDT∴ = or 21.8 lbDT = 
From Equation (1) 25 21.818 3.1818 lbCT = − = or 3.18 lbCT = 
Results: 5.00 lbAT = W 
 3.18 lbCT = W 
 21.8 lbDT = W 
 
 
 
 
 
 
 
PROBLEM 4.106 
For the pile assembly of Problem 4.105, determine (a) the largest 
permissible value of a if the assembly is not to tip, (b) the corresponding 
tension in each wire. 
P4.105 Two steel pipes AB and BC, each having a weight per unit length 
of 5 lb/ft, are welded together at B and are supported by three wires. 
Knowing that 1.25a = ft, determine the tension in each wire. 
 
 
SOLUTION 
 
First note ( )( )5 lb/ft 2 ft 10 lbABW = = 
( )( )5 lb/ft 4 ft 20 lbBCW = = 
From f.b.d. of pipe assembly 
 0: 10 lb 20 lb 0y A C DF T T TΣ = + + − − = 
 30 lbA C DT T T∴ + + = (1) 
 ( )( ) ( )0: 10 lb 1 ft 2 ft 0x AM TΣ = − = 
or 5.00 lbAT = (2) 
From Equations (1) and (2) 25 lbC DT T+ = (3) 
( ) ( ) ( )max0: 4 ft 20 lb 2 ft 0z C DM T T aΣ = + − = 
or ( ) max4 ft 40 lb ftC DT T a+ = ⋅ (4) 
 
 
 PROBLEM 4.106 CONTINUED 
Using Equation (3) to eliminate TC 
( ) max4 25 40D DT T a− + = 
or max
604
D
a
T
= − 
By observation, a is maximum when TD is maximum. From Equation (3), ( )maxDT occurs when 0.CT = 
Therefore, ( )max 25 lbDT = and 
max
604
25
 1.600 ft
a = −
=
 
Results: (a) max 1.600 fta = W 
 (b) 5.00 lbAT = W 
 0CT = W 
 25.0 lbDT = W 
 
 
 
 
 
 
 
 
PROBLEM 4.107 
A uniform aluminum rod of weight W is bent into a circular ring of radius 
R and is supported by three wires as shown. Determine the tension in 
each wire. 
 
SOLUTION 
 
 
 
 
From f.b.d. of ring 
 0: 0y A B CF T T T WΣ = + + − = 
 A B CT T T W∴ + + = (1) 
 ( ) ( )0: sin 30 0x A CM T R T RΣ = − ° = 
 0.5A CT T∴ = (2) 
( ) ( )0: cos30 0z C BM T R T RΣ = ° − = 
 0.86603B CT T∴ = (3) 
Substituting AT and BT from Equations (2) and (3) into Equation (1) 
0.5 0.86603C C CT T T W+ + = 
 0.42265CT W∴ = 
From Equation