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190
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•4–1. If A, B, and D are given vectors, prove the
distributive law for the vector cross product, i.e.,
.A : (B + D) = (A : B) + (A : D)
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Consider the three vectors; with A vertical.
Note obd is perpendicular to A.
od = �A × (B + D)� = �A�(�B + D�) sin u
3
 
ob = �A × B� = �A� �B� sin u
1
 
A × (B + D) = A × B + A × D (QED)
Also, these three cross products all lie in the plane
obd since they are all perpendicular to A. As noted
the magnitude of each cross product is proportional
to the length of each side of the triangle. 
bd = �A × D� = �A� �D� sin u
2
 
A = Axi + Ayj + Azk 
B = Bxi + Byj + Bzk 
D = Dxi + Dyj + Dzk 
Note also,
 i j k
 Ax Ay Az
 Bx + Dx By + Dy Bz + Dz
A × (B + D) = 
 i j k
 Ax Ay Az
 Bx By Bz
 i j k
 Ax Ay Az
 Dx Dy Dz
= [Ay (Bz + Dz) – Az (By + Dy)]i
 – [Ax (Bz + Dz) – Az (Bz + Dz)] j
 + [Ax (By + Dy) – Ay (Bz + Dz)]k
= [(AyBz – AzBy)i – (AzBz – AzBz)j + (AzBy – AyBx)k]
 + [(AyDz – AzDy)i – (AzDz – AzDz)j + (AzDy – AyDx)k]
= (A × B) + (A × D) (QED)
The three vector cross - products also form a closed
triangle o�b�d� which is similar to triangle obd. Thus 
from the figure, 
= + 
4–2. Prove the triple scalar product identity
.A # B : C = A : B # C
191
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As shown in the figure
Area = B(C sin u) = �B : C�
Thus,
Volume of parallelepiped is �B : C��h�
But,
�h� = �A # m(B : C)� = A # 
Thus,
Volume = �A # B : C�
Since �A : B # C � represents this same volume then
A # B : C = A : B # C (QED)
⎛
⎝⎜
⎞
⎠⎟
 i j k
 Bx By Bz
 Cx Cy Cz
 i j k
 Ax Ay Az
 Bx By Bz
Also, 
LHS = A # B : C
 = (Ax i + Ay j + Az k) # 
 
 = Ax (By Cz – Bz Cy) – Ay (Bx Cz – Bz Cx) + Az (Bx Cy – By Cz)
 = Ax By Cz – Ax Bz Cy – Ay Bx Cz + Ay Bz Cx + Az Bx Cy – AzBy Cx
RHS = A : B # C
 = # (Cx i + Cy j + Cz k)
 
 = Cx (Ay Bz – Az By) – Cy (Ax Bz – Az Bx) + Cz (Ax By – Ay Bx)
 = Az By Cz – Ax Bz Cy – Ay Bx Cz + Ay Bz Cx + Ax Bz Cy – AxBy Cz
Thus, LHS = RHS
A # B : C = A : B # C (QED)
�B : C�
B : C
192
4–3. Given the three nonzero vectors A, B, and C, show
that if , the three vectors must lie in the
same plane.
A # (B : C) = 0
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*4–4. Two men exert forces of and on
the ropes. Determine the moment of each force about A.
Which way will the pole rotate, clockwise or counterclockwise?
P = 50 lbF = 80 lb
A
P
F
B
C
6 ft
45
12 ft
3
4
5
•4–5. If the man at B exerts a force of on his
rope, determine the magnitude of the force F the man at C
must exert to prevent the pole from rotating, i.e., so the
resultant moment about A of both forces is zero.
P = 30 lb
A
P
F
B
C
6 ft
45
12 ft
3
4
5
1.8 m
3.6 m
1.8 m
3.6 m
*4–4. Two men exert forces of F = 400 N and P = 250 N on
•4–5. If the man at B exerts a force of P = 150 N on his
+ (MA)C = 400
4
5
⎛
⎝⎜
⎞
⎠⎟
(3.6) = 1152 N · m Ans
+ (MA)B = 250 (cos 45°) (5.4) = 954.6 N · m Ans
 Since (MA)C > (MA)B 
 Clockwise Ans
+ 150 (cos 45°) (5.4) = F
4
5
⎛
⎝⎜
⎞
⎠⎟
(3.6) = 0
 F = 198.9 N Ans
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Consider,
�A # (B : C)� = �A��B : C� cos u
 = (�A� cos u) �B : C� 
 = �h��B : C�
 = BC �h� sin f
 = volume of parallelepiped.
If A # (B : C) = 0, then the volume equals zero, so that A, B, and C are coplanar. 
4–6. If , determine the moment produced by the 
4-kN force about point A.
u = 45° 3 m
0.45 m
4 kN
A
u
193
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Resolving the 4-kN force into its horizontal and vertical components, Fig. a, and applying the 
principle of moments,
 + MA = 4 cos 45° (0.45) – 4 sin 45° (3)
 = –7.21 kN · m = 7.21 kN · m (clockwise) Ans
194
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4–7. If the moment produced by the 4-kN force about
point A is clockwise, determine the angle , where
.0° … u … 90°
u10 kN # m
3 m
0.45 m
4 kN
A
u
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Resolving the 4-kN force into its horizontal and vertical components, Fig. a, and 
applying the principle of moments,
 + MA = –10 = 4 cos u (0.45) – 4 sin u (3)
 12 sin u – 1.8 cos u = 10 (1)
Referring to the geometry of Fig. a,
cos f = sin f = (2)
Dividing Eq. (1) by 147.24 yields
 sin u – cos u = (3)
 
Substituting Eq. (2) into 3 yields
sin u cos f – cos u sin f = 
sin (u – f) = 
However, f = tan–1 = 8.531°. Thus,
sin (u – 8.531°) = 
u – 8.531° = 55.50°
u = 64.0° Ans
1.8
12
⎛
⎝⎜
⎞
⎠⎟
12
147.24
1.8
147.24
12
147.24
10
147.24
1.8
147.24
10
147.24
10
147.24
10
147.24
*4–8. The handle of the hammer is subjected to the force
of Determine the moment of this force about the
point A.
F = 20 lb.
F
B
A
18 in.
5 in.
30
195
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125 mm
450 mm
1000 N
(100 sin 30°) N
(100 cos 30°) N
125 mm
450 mm
of F = 1000 N. Determine the moment of this force about 
the point A.
+ MA = [–1000 cos 30° (450) – 1000 sin 30° (125)] (10
–3)
= –452.2 N · m = 452 N · m (clockwise) 
Resolving the 1000-N force into components parallel and perpendicular to the hammer, Fig. a, 
and applying the principle of moments,
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•4–9. In order to pull out the nail at B, the force F exerted
on the handle of the hammer must produce a clockwise
moment of about point A. Determine the
required magnitude of force F.
500 lb # in.
F
B
A
18 in.
5 in.
30
125 mm
450 mm
0.125 m
0.450 m
moment of 60 N · m about point A. Determine the required 
magnitude of force F.
+ MA = –60 = –F cos 30° (0.450) – F sin 30° (0.125)
F = 132.7 N Ans
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Resolving forceF into components parallel and perpendicular to the hammer, Fig. a, 
and applying the principle of moments,
4–10. The hub of the wheel can be attached to the axle
either with negative offset (left) or with positive offset
(right). If the tire is subjected to both a normal and radial
load as shown, determine the resultant moment of these
loads about point O on the axle for both cases.
4 kN
800 N 800 N
4 kN
Case 1 Case 2
0.4 m
0.05 m
0.05 m
0.4 m
O
O
197
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For case 1 with negative offset, we have
 + MO = 800(0.4) – 4000(0.05)
 = 120 N · m (Counterclockwise) Ans
For case 2 with positive offset, we have
 + MO = 800(0.4) + 4000(0.05)
 = 520 N · m (Counterclockwise) Ans
198
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4–11. The member is subjected to a force of . If
, determine the moment produced by F about
point A.
u = 45°
F = 6 kN
A
6 m
1.5 m
u
F 6 kN
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Resolving force F into horizontal and vertical components, Fig. a, and applying the principle of moments,
 + MA = –6 cos 45° (6) – 6 sin 45° (3)
 = –38.18 kN · m = 38.2 kN · m (clockwise) Ans
*4–12. Determine the angle of the
force F so that it produces a maximum moment and a
minimum moment about point A. Also, what are the
magnitudes of these maximum and minimum moments?
u (0° … u … 180°)
A
6 m
1.5 m
u
F 6 kN
199
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In order to produce the maximum moment about point A, force F must act perpendicular to line AB, 
Fig. a. From the geometry of the diagram,
f = tan–1 = 63.43°
u = 90° – f = 90° – 63.43° = 26.6° Ans
Also,
d = 62 + 32 = 45 m
The maximum moment of F about point A is given by
(MA)max = Fd = 6 45 = 40.2 kN · m 
The minimum moment of F about point A occurs when the line of action of F passes through point A. 
Referring to Fig. b,
u = 180° – f = 180° – 63.43° = 117° Ans
and
(MA)min = Fd = 6(0) = 0 Ans
6
3
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝
⎞
⎠
200
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•4–13. Determine the moment produced by the force F
about point A in terms of the angle . Plot the graph of 
versus , where .0° … u … 180°u
MAu
A
6 m
1.5 m
u
F 6 kN
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Moment Function: Resolving force F into horizontal and vertical components, Fig. a, and applying the 
principle of moments,
 + MA = –6 cos u (6) – 6 sin u (3)
 = –(36 cos u + 18 sin u) kN · m
 = (36 cos u + 18 sin u) kN · m (clockwise)
The maximum moment occurs when = 0.
 = –36 sin u + 18 cos u = 0.
u = 26.6°
The maximum moment of F about point A is given by
(MA)max = 36 cos 26.57° + 18 sin 26.57° = 40.2 kN · m
Also,
MA�u= 0° = 36 cos 0° + 18 sin 0° = 36 kN · m
MA�u= 90° = 36 cos 90° + 18 sin 90° = 18 kN · m
MA�u= 180° = 36 cos 180° + 18 sin 180° = –36 kN · m
When MA = 0,
0 = 36 cos u + 18 sin u u = 117°
The plot of MA versus u is shown in Fig. b. 
dMA
du
dMA
du
4–14. Serious neck injuries can occur when a football
player is struck in the face guard of his helmet in the
manner shown, giving rise to a guillotine mechanism.
Determine the moment of the knee force about
point A. What would be the magnitude of the neck force F
so that it gives the counterbalancing moment about A?
250 N
50 mm
100 mm
150 mm
30
60
P 250 N
F
A
201
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4–15. The Achilles tendon force of is
mobilized when the man tries to stand on his toes. As this is
done, each of his feet is subjected to a reactive force of
Determine the resultant moment of and
about the ankle joint A.
NF400 N.
650 N
100 mm
65 mm
200 mm
A
Nf 400 N
Ft
5
+ MA = 250 sin 60° (100) – 250 cos 60° (50) 
= 15400.6 N · mm = 15.4 N · m Ans
 15400.6 = F cos 30° (150) 
 F = 118.6 N Ans
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Referring to Fig. a,
 + (MR)A = ΣFd; (MR)A = 400 (0.1) = 650 (0.65) cos 5°
 = –2.09 N · m = 2.09 N · m (clockwise) Ans 
 
202
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*4–16. The Achilles tendon force is mobilized when the
man tries to stand on his toes.As this is done, each of his feet
is subjected to a reactive force of If the resultant
moment produced by forces and about the ankle joint
A is required to be zero, determine the magnitude of .Ft
NtFt
Nt = 400 N.
Ft
100 mm
65 mm
200 mm
A
Nf 400 N
Ft
5
•4–17. The two boys push on the gate with forces of
and as shown. Determine the moment of each
force about C. Which way will the gate rotate, clockwise or
counterclockwise? Neglect the thickness of the gate.
FA = 30 lb
60
6 ft
C
B
A
3 ft
3
4
5
FB
FA
4–18. Two boys push on the gate as shown. If the boy at B
exerts a force of , determine the magnitude of
the force the boy at A must exert in order to prevent the
gate from turning. Neglect the thickness of the gate.
FA
FB = 30 lb
60
6 ft
C
B
A
3 ft
3
4
5
FB
FA
1.8 m 0.9 m
1.8 m 0.9 m
•4–17. The two boys push on the gate with forces of FA = 
150 N and FB = 250 N as shown. Determine the moment of 
each force about C. Which way will the gate rotate, clockwise 
or counterclockwise? Neglect the thickness of the gate.
4–18. Two boys push on the gate as shown. If the boy at B 
exerts a force of FB = 150 N determine the magnitude of 
the force FA the boy at A must exert in order to prevent the 
gate from turning. Neglect the thickness of the gate.
+ (MFA)C = –150
3
5
⎛
⎝⎜
⎞
⎠⎟
(2.7) 
= –243 N · m = 243 N · m (Clockwise) Ans
+ (MFB)C = 250 (sin 60°) (1.8) 
= 389.7 N · m (Counterclockwise) Ans
Since (MFB)C > (MFA)C, the gate will rotate Counterclockwise. Ans
In order to prevent the gate from turning, the resultant 
moment about point C must be equal to zero.
+ MRC = ΣFd; MRC = 0 = 150 sin 60° (1.8) – FA
3
5
⎛
⎝⎜
⎞
⎠⎟
(2.7)
 FA = 144.3 N Ans
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Referring to Fig. a,
 + (MR)A = ΣFd; 0 = 400 (0.1) = F cos 5° (0.065)
 F = 618 N Ans 
 
4–19. The tongs are used to grip the ends of the drilling
pipe P. Determine the torque (moment) that the
applied force 750 N exerts on the pipe about point P
as a function of . Plot this moment versusfor
.0 … u … 90°
uMPu
F =
MP
43 in.
6 in.
F
P
MP
u
203
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*4–20. The tongs are used to grip the ends of the drilling
pipe P. If a torque (moment) of is needed
at P to turn the pipe, determine the cable force F
F
F F
F
F
that must
be applied to the tongs. Set .u = 30°
MP
43 in.
6 in.
F
P
MP
u
150 mm
1075 mm
150 mm
1075 mm
806.25
112.5
814.1
Mp(N·m)
*
 = 1200 N · m
MP = 750 cos (1.075) + 750 sin (0.150)
= (806.25 cos + 112.5 sin ) N · m Ans
dM
d
P = –806.25 sin + 112.5 cos = 0
tan = 
112.5
806.25
 = 7.943°
At = 7.943°, MP is maximum.
(MP)max = 806.25 cos 7.943° + 112.5 sin 7.943° = 814.1 N · m
Also (MP)max = 750 ((1.075)
2 + (0.150)2)
1
2 = 814.1 N · m
MP = cos 30° (1.075) + sin 30° (0.150)
Set MP = 1200 N · m
1200 = cos 30° (1.075) + sin 30° (0.150)
 = 1193 N Ans
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204
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•4–21. Determine the direction for of the
force F so that it produces the maximum moment about
point A. Calculate this moment.
0° … u … 180°u F 400 N
3 m
2 m
A
u
04a Ch04a 190-253.indd 20404a Ch04a 190-253.indd 204 6/12/09 8:33:16 AM6/12/09 8:33:16 AM
 + MA = 400 (3)
2 + (2)2 = 1442 N · m
 MA = 1.44 kN · m Ans
 f = tan–1 = 33.69°
 u = 90° – 33.69° = 56.3° Ans
2
3
⎛
⎝⎜
⎞
⎠⎟
205
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4–22. Determine the moment of the force F about point A
as a function of . Plot the results of M (ordinate) versus 
(abscissa) for .0° … u … 180°
uu
F 400 N
3 m
2 m
A
u
4–23. Determine the minimum moment produced by 
the force F about point A. Specify the angle 
.u … 180°)
u (0° …
F 400 N
3 m
2 m
A
u
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Mmin = 400 (0) = 0 Ans
umin = 90° + 56.3° = 146° Ans
 + MA = 400 sin u (3) + 400 cos u (2)
 = 1200 sin u + 800 cos u Ans
 = 1200 cos u – 800 sin u = 0
 u = tan–1 = 56.3°
(MA)max = 1200 sin 56.3° + 800 cos 56.3° 
 = 1442 N · m
dMA
du
1200
800
⎛
⎝⎜
⎞
⎠⎟
206
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*4–24. In order to raise the lamp post from the position
shown, force F is applied to the cable. If 
determine the moment produced by F about point A.
F = 200 lb,
F
75C
A
B
10 ft
20 ft6 m
3 m
6 m
3 m
1000 N
1000 sin 
1000 cos 
shown, force is applied to the cable. If F = 1000 N,
Geometry: Applying the law of cosines to Fig. a,
BC2 = 32 + 62 – 2(3)(6) cos 105°
BC = 7.370 m
Then, applying the law of sines,
sin
3
 = 
sin 105°
7.37°
 = 23.15°
Moment About Point A: By resolving force F into components parallel and perpendicular to the lamp pole, 
Fig. a, and applying the principle of moments,
+ (MR)A = ΣFd; MA = 1000 sin 23.15° (6) + 1000 cos 23.15° (0)
= 2358.8 N · m = 2.36 kN · m (counterclockwise) 
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207
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•4–25. In order to raise the lamp post from the position
shown, the force F on the cable must create a counterclockwise
moment of about point A. Determine the
magnitude of F that must be applied to the cable.
1500 lb # ft
F
75C
A
B
10 ft
20 ft6 m
3 m
3 m
6 m
moment of 2250 N · m about point A. Determine the
Geometry: Applying the law of cosines to Fig. a,
BC2 = 32 + 62 – 2(3)(6) cos 105°
BC = 7.370 m
Then, applying the law of sines,
sin
3
 = 
sin 105°
7.37°
 = 23.15°
Moment About Point A: By resolving force F into components parallel and perpendicular to the lamp pole, 
Fig. a, and applying the principle of moments,
+ (MR)A = ΣFd; 2250 = F sin 23.15° (6)
 F = 953.9 N Ans
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208
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4–26. The foot segment is subjected to the pull of the two
plantarflexor muscles. Determine the moment of each force
about the point of contact A on the ground.
60
30
100 mm
A
25 mm
87.5 mm
70
F2 150 N
F1 100 N
4–27. The 70-N force acts on the end of the pipe at B.
Determine (a) the moment of this force about point A, and
(b) the magnitude and direction of a horizontal force, applied
at C, which produces the same moment.Take 60°.
A
C
0.3 m 0.7 m
0.9 m
B
70 N
(MA)1 = 100 cos 30° (0.1125) + 100 sin 30° (0.100) = 14.74 N · m Ans
(MA)2 = 150 cos 70° (0.100) + 150 sin 70° (0.0875) = 17.46 N · m Ans
04a Ch04a 190-253.indd 20804a Ch04a 190-253.indd 208 6/12/09 8:33:25 AM6/12/09 8:33:25 AM
(a) + MA = 70 sin 60° (0.7) + 70 cos 60° (0.9)
 MA = 73.94 = 73.9 N · m Ans
(b) FC (0.9) = 73.94
 FC = 82.2 N d Ans
209
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*4–28. The 70-N force acts on the end of the pipe at B.
Determine the angles of the force that
will produce maximum and minimum moments about
point A. What are the magnitudes of these moments?
u (0° … u … 180°)
A
C
0.3 m 0.7 m
0.9 m
B
70 N
u
•4–29. Determine the moment of each force about the
bolt located at A. Take FB = 40 lb, FC = 50 lb. 225 mm
750 mm
bolt located at A. Take FB = 200 N, FC = 250 N.
+ MB = 200 cos 25° (0.750) = 135.9 N · m Ans
+ MC = 250 cos 30° (0.975) = 211.1 N · m Ans
04a Ch04a 190-253.indd 20904a Ch04a 190-253.indd 209 6/12/09 8:33:29 AM6/12/09 8:33:29 AM
 + MA = 70 sin u (0.7) + 70 cos u (0.9)
 MA = 49 sin u + 63 cos u
For maximum moment = 0
 = 0; 49 cos u – 63 sin u = 0
 u = tan–1 = 37.9° Ans
(MA)max = 49 sin 37.9° + 63 cos 37.9° 
 = 79.8 N · m Ans
For minimum moment MA = 0
 MA = 0; 49 sin u + 63 cos u = 0
 u = 180° + tan–1 = 128° Ans
 (MA)min = 49 sin 128° + 63 cos 128° = 0 Ans 
dMA
du
dMA
du
49
63
⎛
⎝⎜
⎞
⎠⎟
–63
49
⎛
⎝⎜
⎞
⎠⎟
210
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4–30. If and determine the resultant
moment about the bolt located at A.
FC = 45 lb,FB = 30 lb
4–31. The rod on the power control mechanism for a
business jet is subjected toa force of 80 N. Determine the
moment of this force about the bearing at A.
20
60
A
80 N
150 mm
*4–32. The towline exerts a force of at the end
of the 20-m-long crane boom. If determine the
placement x of the hook at A so that this force creates a
maximum moment about point O. What is this moment?
u = 30°,
P = 4 kN
1.5 m
O
20 m
A
B
P 4 kN
x
u
225 mm
750 mm
4–30. If FB = 150 N and FC = 225 N, determine the resultant
+ MA = 150 cos 25° (0.750) + 225 cos 30° (0.975)
= 291.9 N · m Ans
04a Ch04a 190-253.indd 21004a Ch04a 190-253.indd 210 6/12/09 8:33:32 AM6/12/09 8:33:32 AM
+ MA = 80 cos 20° (0.15 sin 60°) – 80 sin 20° (0.15 cos 60°) 
 = 7.71 N · m Ans 
Maximum moment, OB › BA
 + (MO)max = 4 kN (20) = 80 kN · m Ans
4 kN sin 60° (x) – 4 kN cos 60° (1.5) = 80 kN · m
x = 24.0 m Ans
•4–33. The towline exerts a force of at the end
of the 20-m-long crane boom. If determine the
position of the boom so that this force creates a maximum
moment about point O. What is this moment?
u
x = 25 m,
P = 4 kN
1.5 m
O
20 m
A
B
P 4 kN
x
u
211
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04a Ch04a 190-253.indd 21104a Ch04a 190-253.indd 211 6/12/09 8:33:37 AM6/12/09 8:33:37 AM
2.259
2.712
⎛
⎝⎜
⎞
⎠⎟
20 + y
z
25 + z
y
Maximum moment, OB › BA
 + (MO)max = 4000 (20) = 80 000 N · m = 80.0 kN · m Ans
4000 sin f (25) – 4000 cos f (1.5) = 80 000
25 sin f – 1.5 cos f = 20
f = 56.43°
u = 90° – 56.43° = 33.6° Ans
Also, 
(1.5)2 + z2 = y2
2.25 + z2 = y2
Similar triangles
 = 
20y + y2 = 25z + z2
20( 2.25 + z2) + 2.25 + z2 = 25z + z2
z = 2.259 m
y = 2.712 m
u = cos–1 = 33.6° Ans
212
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–34. In order to hold the wheelbarrow in the position
shown, force F must produce a counterclockwise moment
of about the axle at A. Determine the required
magnitude of force F.
200 N # m B
0.65 m
0.5 m
1.2 m
30 
0.3 m
F
G
A
04a Ch04a 190-253.indd 21204a Ch04a 190-253.indd 212 6/12/09 8:33:41 AM6/12/09 8:33:41 AM
Resolving force F into horizontal and vertical components, Fig. a, 
and applying the principle of moments,
 + MA = 200 = F sin 30° (1.5) + F cos 30° (1.5)
 F = 115 N Ans
213
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4–35. The wheelbarrow and its contents have a mass of
50 kg and a center of mass at G. If the resultant moment
produced by force F and the weight about point A is to be
zero, determine the required magnitude of force F.
B
0.65 m
0.5 m
1.2 m
30 
0.3 m
F
G
A
04a Ch04a 190-253.indd 21304a Ch04a 190-253.indd 213 6/12/09 8:33:44 AM6/12/09 8:33:44 AM
Resolving force F into its horizontal and vertical components, Fig. a, 
and applying the principle of moments,
 + (MR)A = ΣFd; 0 = F sin 30° (1.5) + F cos 30° (1.5) – 50 (9.81) (0.3)
 F = 84.3 N Ans 
214
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*4–36. The wheelbarrow and its contents have a center of
mass at G. If and the resultant moment produced
by force F and the weight about the axle at A is zero,
determine the mass of the wheelbarrow and its contents.
F = 100 N
B
0.65 m
0.5 m
1.2 m
30 
0.3 m
F
G
A
04a Ch04a 190-253.indd 21404a Ch04a 190-253.indd 214 6/12/09 8:33:47 AM6/12/09 8:33:47 AM
Resolving force F into its horizontal and vertical components, Fig. a, 
and applying the principle of moments,
 + (MR)A = ΣFd; 0 = 100 cos 30° (1.15) + 100 sin 30° (1.5) – M (9.81) (0.3)
 M = 59.3 kg Ans 
215
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•4–37. Determine the moment produced by about
point O. Express the result as a Cartesian vector.
F1
y
x
z
1 ft
2 ft
2 ft
A
O
3 ft
F2 { 10i 30j 50k} lb
F1 { 20i 10j 30k} lb
0.4 m
0.2 m
0.6 m
0.4 m
N
N
A(0.6, 0.6, –0.4) m
Position Vector: The position vector rOA, Fig. a, must be determined first.
rOA = (0.6 – 0)i + (0.6 – 0)j + (–0.4 – 0)k = [0.6i + 0.6j – 0.4k] m
Vector Cross Product: The moment of F1 about point O is
MO = rOA × F1 = 
i j k
0.6 0.6 – .
–
0 4
20 10 30
 = [22i – 10j + 18k] N · m Ans
04a Ch04a 190-253.indd 21504a Ch04a 190-253.indd 215 6/12/09 8:33:50 AM6/12/09 8:33:50 AM
216
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4–38. Determine the moment produced by about
point O. Express the result as a Cartesian vector.
F2
y
x
z
1 ft
2 ft
2 ft
A
O
3 ft
F2 { 10i 30j 50k} lb
F1 { 20i 10j 30k} lb
0.4 m
0.2 m
0.6 m
0.4 m
N
N
A(0.6, 0.6, –0.4) m
Position Vector: The position vector rOA, Fig. a, must be determined first.
rOA = (0.6 – 0)i + (0.6 – 0)j + (–0.4 – 0)k = [0.6i + 0.6j – 0.4k] m
Vector Cross Product: The moment of F2 about point O is
MO = rOA × F2 = 
i j k
0.6 0.6 – .
– –
0 4
10 30 50
 = [18i – 26j – 12k] N · m Ans
04a Ch04a 190-253.indd 21604a Ch04a 190-253.indd 216 6/12/09 8:33:53 AM6/12/09 8:33:53 AM
217
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–39. Determine the resultant moment produced by the two
forces about point O. Express the result as a Cartesian vector.
y
x
z
1 ft
2 ft
2 ft
A
O
3 ft
F2 { 10i 30j 50k} lb
F1 { 20i 10j 30k} lb
0.4 m
0.2 m
0.6 m
0.4 m
N
N
A(0.6, 0.6, –0.4) m
Position Vector: The position vector rOA, Fig. a, must be determined first.
rOA = (0.6 – 0)i + (0.6 – 0)j + (–0.4 – 0)k = [0.6i + 0.6j – 0.4k] m
Resultant Moment: The resultant moment of F1 and F2 about point O can be determined by
(MR)O = rOA × F1 + rOA × F2
= 
i j k
0.6 0.6 – .
–
0 4
20 10 30
 + 
i j k
0.6 0.6 – .
– –
0 4
10 30 50
= [40i – 36j + 6k] N · m 
Or we can apply the principle of moments which gives
(MR)O = rOA × (F1 + F2)
= 
i j k
0.6 0.6 – .
– –
0 4
30 20 80
 
= [40i – 36j + 6k] N · m 
04a Ch04a 190-253.indd 21704a Ch04a 190-253.indd 217 6/12/09 8:33:55 AM6/12/09 8:33:55 AM
218
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*4–40. Determine the moment produced by force 
about point O. Express the result as a Cartesian vector.
FB
y
x
z
C
O
B
A
6 m
3 m
2 m
2.5 m
FC 420 N
FB 780 N
04a Ch04a 190-253.indd 21804a Ch04a 190-253.indd 218 6/12/09 8:33:58 AM6/12/09 8:33:58 AM
Position Vectorand Force Vectors: Either position vector rOA or rOB can be used
to determine the moment of FB about point O.
rOA = [6k] m rOB = [2.5j] m
The force vector FB is given by
FB = FBuFB = 780 = [300j – 720k] N
Vector Cross Product: The moment of FB about point O is given by
MO = rOA × FB = = [–1800i] N · m = [–1.80i] kN · m Ans
or
MO = rOB × FB = = [–1800i] N · m = [–1.80i] kN · m Ans
 i j k
 0 0 6
 0 300 –720
 i j k
 0 2.5 0
 0 300 –720
(0 – 0)i + (2.5 – 0)j + (0 – 6)k
(0 – 0)2 + (2.5 – 0)2 + (0 – 6)2 � �
219
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•4–41. Determine the moment produced by force 
about point O. Express the result as a Cartesian vector.
FC
y
x
z
C
O
B
A
6 m
3 m
2 m
2.5 m
FC 420 N
FB 780 N
04a Ch04a 190-253.indd 21904a Ch04a 190-253.indd 219 6/12/09 8:34:01 AM6/12/09 8:34:01 AM
Position Vector and Force Vectors: Either position vector rOA or rOC can be used
to determine the moment of FC about point O.
rOA = {6k} m
rOC = (2 – 0)i + (–3 – 0)j + (0 – 0)k = [2i – 3j] m
The force vector FC is given by
FC = FCuFC = 420
 
= [120i – 180j – 360k] N
Vector Cross Product: The moment of FC about point O is given by
MO = rOA × FC = = [1080i + 720j] N · m Ans
or
MO = rOC × FC = = [1080i + 720j] N · m Ans
(2 – 0)i + (–3 – 0)j + (0 – 6)k
(2 – 0)2 + (–3 – 0)2 + (0 – 6)2 � �
 i j k
 0 0 6
 120 –180 –360
 i j k
 2 –3 0
 120 –180 –360
220
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4–42. Determine the resultant moment produced by
forces and about point O. Express the result as a
Cartesian vector.
FCFB
y
x
z
C
O
B
A
6 m
3 m
2 m
2.5 m
FC 420 N
FB 780 N
04a Ch04a 190-253.indd 22004a Ch04a 190-253.indd 220 6/12/09 8:34:02 AM6/12/09 8:34:02 AM
Position Vector and Force Vectors: The position vector rOA and force vectors 
FB and FC, Fig. a, must be determined first.
rOA = {6k} m
FB = FBuFB = 780 = [300j – 720k] N
FC = FCuFC = 420 = [120i – 180j – 360k] N
Resultant Moment: The resultant moment of FB and FC about point O is given by
MO = rOA × FB + rOA × FC
M = + 
 = [–720i + 720j] N · m Ans
(2 – 0)i + (–3 – 0)j + (0 – 6)k
(2 – 0)2 + (–3 – 0)2 + (0 – 6)2 � �
 i j k
 0 0 6
 120 –180 –360
(0 – 0)i + (2.5 – 0)j + (0 – 6)k
(0 – 0)2 + (2.5 – 0)2 + (0 – 6)2 � �
 i j k
 0 0 6
 0 300 –720
221
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–43. Determine the moment produced by each force
about point O located on the drill bit. Express the results as
Cartesian vectors.
x
z
A BO
y
150 mm
600 mm
300 mm
150 mm
FA { 40i 100j 60k} N
FB { 50i 120j 60k} N
04a Ch04a 190-253.indd 22104a Ch04a 190-253.indd 221 6/12/09 8:34:05 AM6/12/09 8:34:05 AM
Position Vector: The position vectors rOA and rOB, Fig. a, must be determined first.
rOA = (0.15 – 0)i + (0.3 – 0)j + (0 – 0)k = [0.15i + 0.3j] m
rOB = (0 – 0)i + (0.6 – 0)j + (–0.15 – 0)k = [0.6j – 0.15k] m
Vector Cross Product: The moment of FA about point O is
(MR)O = rOA × FA
 = 
 = [–18i + 9j – 3k] N · m Ans
The moment of FB about point O is
(MR)O = rOB × FB
 = 
 = [18i + 7.5j + 30k] N · m Ans
 i j k
 0.15 0.3 0
 –40 –100 –60
 i j k
 0 0.6 –0.15
 –50 –120 60
222
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*4–44. a secudorp fo ecrof A
nigiro eht tuoba fo tnemom
of coordinates, point O. If the force acts at a point having an
x coordinate of determine the y and z coordinates.x = 1 m,
MO = 54i + 5j - 14k6 kN # m
F = 66i - 2j + 1k6 kN
•4–45. The pipe assembly is subjected to the 80-N force.
Determine the moment of this force about point A.
400 mm
y300 mm
200 mm
250 mm
x
z
30
40
F 80 N
B
C
A
04a Ch04a 190-253.indd 22204a Ch04a 190-253.indd 222 6/12/09 8:34:07 AM6/12/09 8:34:07 AM
 MRO = = [4i + 5j – 14k] kN · m
 y + 2z = 4
 –1 + 6z = 5
 –2 – 6y = –14
 y = 2 m Ans
 z = 1 m Ans
 i j k
 l y z
 6 –2 1
Position Vector And Force Vector:
 rAC = {(0.55 – 0)i + (0.4 – 0)j + (–0.2 – 0)k} m
 = {0.55i + 0.4j – 0.2k} m
 F = 80 (cos 30° sin 40°i + cos 30° cos 40°j – sin 30°k) N
 = {44.53i + 53.07j – 40.0k} N
Moment of Force F About Point A: Applying Eq. 4–7, we have
 MA = rAC × FB 
 = 
 = {–5.39i + 13.1j + 11.4k} N · m Ans 
 i j k
 0.55 0.4 –0.2
 44.53 53.07 –40.0
223
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4–47. a setaerc ecrof ehT
moment about point O . fo
If the force passes through a point having an x coordinate of
1 m, determine the y and z coordinates of the point. Also,
realizing that , determine the perpendicular
distance d from point O to the line of action of F.
MO = Fd
MO = 5-14i + 8j + 2k6 N # m
F = 56i + 8j + 10k6 N
d
z
x
y
O
y
1 m
z
P
F
MO
4–46. The pipe assembly is subjected to the 80-N force.
Determine the moment of this force about point B.
400 mm
y300 mm
200 mm
250 mm
x
z
30
40
F 80 N
B
C
A
04a Ch04a 190-253.indd 22304a Ch04a 190-253.indd 223 6/12/09 8:34:10 AM6/12/09 8:34:10 AM
Position Vector And Force Vector:
 rBC = {(0.55 – 0)i + (0.4 – 0.4)j + (–0.2 – 0)k} m
 = {0.55i – 0.2k} m
 F = 80 (cos 30° sin 40°i + cos 30° cos 40°j – sin 30°k) N
 = {44.53i + 53.07j – 40.0k} N
Moment of Force F About Point B: Applying Eq. 4–7, we have
 MB = rBC × F 
 
 = 
 = {10.6i + 13.1j + 29.2k} N · m Ans 
 i j k
 0.55 0 –0.2
 44.53 53.07 –40.0
–14i + 8j + 2k = 
 –14 = 10y – 8z
 8 = –10 + 6z
 2 = 8 – 6y
 y = 1 m Ans
 z = 3 m Ans
 MO = (–14)
2 + (8)2 + (2)2 = 16.25 N · m
 F = (6)2 + (8)2 + (10)2 = 14.14 N
 d = = 1.15 m Ans
 i j k
 l y z
 6 8 10
16.25
14.14
224
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*4–48. Force F acts perpendicular to the inclined plane.
Determine the moment produced by F about point A.
Express the result as a Cartesian vector.
z
x y
3 m
3 m
4 m
A
B C
F 400 N
04a Ch04a 190-253.indd 22404a Ch04a 190-253.indd 224 6/12/09 8:34:12 AM6/12/09 8:34:12 AM
12i + 9j + 12k
122 + 92 + 122 
b
b
 i j k
 0 4 –3
 –3 4 0
 i j k
 0 4 –3
 249.88 187.41 249.88
Force Vector: Since force F is perpendicular to the inclined plane, its unit vector uF 
is equal to the unit vector of the cross product, b = rAC × rBC, Fig. a. Here
 rAC = (0 – 0)i + (4 – 0)j + (0 – 3)k = [4j – 3k] m
 rAB = (0 – 3)i + (4 – 0)j + (0 – 0)k = [–3i + 4j] m
Thus,
 b = rCA × rCB = 
 = [12i + 9j + 12k] m2
Then,
 uF = = = 0.6247i + 0.4685j + 0.6247k 
And finally,
 F = FuF = 400 (0.6247i + 0.4685j + 0.6247k)
 = [249.88i + 187.41j + 249.88k] N
Vector Cross Product: The moment of F about point A is
 MA = rAC × F =
 = [–1.56i – 0.750j – 1k] kN · m Ans225
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•4–49. Force F acts perpendicular to the inclined plane.
Determine the moment produced by F about point B.
Express the result as a Cartesian vector.
z
x y
3 m
3 m
4 m
A
B C
F 400 N
04a Ch04a 190-253.indd 22504a Ch04a 190-253.indd 225 6/12/09 8:34:13 AM6/12/09 8:34:13 AM
12i + 9j + 12k
122 + 92 + 122 
b
b
 i j k
 0 4 –3
 –3 4 0
 i j k
 –3 4 0
 249.88 187.41 249.88
Force Vector: Since force F is perpendicular to the inclined plane, its unit vector uF 
is equal to the unit vector of the cross product, b = rAC × rBC, Fig. a. Here
 rAC = (0 – 0)i + (4 – 0)j + (0 – 3)k = [4j – 3k] m
 rBC = (0 – 3)i + (4 – 0)j + (0 – 0)k = [–3i + 4j] m
Thus,
 b = rCA × rCB = 
 = [12i + 9j + 12k] m2
Then,
 uF = = = 0.6247i + 0.4685j + 0.6247k 
And finally,
 F = FuF = 400 (0.6247i + 0.4685j + 0.6247k)
 = [249.88i + 187.41j + 249.88k] N
Vector Cross Product: The moment of F about point B is
 MB = rBC × F =
 = [–i + 0.750j –1.56k] kN · m Ans
 
226
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–50. A 20-N horizontal force is applied perpendicular to
the handle of the socket wrench. Determine the magnitude
and the coordinate direction angles of the moment created
by this force about point O.
15
200 mm
75 mm
20 N
A
O
x
y
z
04a Ch04a 190-253.indd 22604a Ch04a 190-253.indd 226 6/12/09 8:34:14 AM6/12/09 8:34:14 AM
rA = 0.2 sin 15°i + 0.2 cos 15°j + 0.075k
 = 0.05176i + 0.1932j + 0.075j
F = –20 cos 15°i + 20 sin 15°j
 = –19.32i + 5.176j
MO = rA × FB = 
 = {–0.3882i – 1.449j + 4.00k} N · m
MO = 4.272 = 4.27 N · m Ans
a = cos–1 = 95.2° Ans
 
b = cos–1 = 110° Ans 
g = cos–1 = 20.6° Ans 
 i j k
 0.05176 0.1932 0.075
 –19.32 5.176 0
–0.3882
4.272
⎛
⎝
⎜
⎞
⎠
⎟
–1.449
4.272
⎛
⎝
⎜
⎞
⎠
⎟
4
4.272
⎛
⎝
⎜
⎞
⎠
⎟
227
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–51. Determine the moment produced by force F about
the diagonal AF of the rectangular block. Express the result
as a Cartesian vector.
3 m
1.5 m
3 mx
C
A B
G
F
y
z
O
D
F { 6i 3j 10k} N
04a Ch04a 190-253.indd 22704a Ch04a 190-253.indd 227 6/12/09 8:34:16 AM6/12/09 8:34:16 AM
Moment About Diagonal AF: Either position vector rAB or rFB, Fig a, can be used to 
find the moment of F about diagonal AF.
 rAB = (0 – 0)i + (3 – 0)j + (1.5 – 1.5)k = [3j] m
 rFB = (0 – 3)i + (3 – 3)j + (1.5 – 0)k = [–3i + 1.5k] m
The unit vector uAF, Fig a, that specifies the direction of diagonal AF is given by 
 uAF = = i + j – k 
The magnitude of the moment of F about diagonal AF axis is 
 MAF = uAF · rAB × F = 
 = [3(10) – (3)(0)] – [0(10) – (–6)(0)] + – [0(3) – (–6)(3)]
 = 14 N · m
or
 MAF = uAF · rFB × F =
 = [0(10) – (3)(1.5)] – [(–3)(10) – (–6)(1.5)] + – [(–3)(3) – (–6)(0)]
 = 14 N · m
Thus, MAF can be expressed in Cartesian vector form as
MAF = MAF uAF = 14 i + j – k = [9.33i + 9.33j – 4.67k] N · m Ans
(3 – 0)i + (3 – 0)j + (0 – 1.5)k
(3 – 0)2 + (3 – 0)2 + (0 – 1.5)2 
2
3
2
3
1
3
 2 2 1
 3 3 3
 –3 0 1.5
 –6 3 10
 2 2 1
 3 3 3
 0 3 0
 –6 3 10
2
3
1
3
2
3
⎛
⎝
⎜
⎞
⎠
⎟
2
3
2
3
1
3
⎛
⎝
⎜
⎞
⎠
⎟
2
3
2
3
2
3
⎛
⎝
⎜
⎞
⎠
⎟
228
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*4–52. Determine the moment produced by force F about
the diagonal OD of the rectangular block. Express the
result as a Cartesian vector.
3 m
1.5 m
3 mx
C
A B
G
F
y
z
O
D
F { 6i 3j 10k} N
04a Ch04a 190-253.indd 22804a Ch04a 190-253.indd 228 6/12/09 8:34:17 AM6/12/09 8:34:17 AM
Moment About Diagonal OD: Either position vector rOB or rDB, Fig a, can be used to 
find the moment of F about diagonal OD.
 rOB = (0 – 0)i + (3 – 0)j + (1.5 – 0)k = [3j + 1.5j] m
 rDB = (0 – 3)i + (3 – 3)j + (1.5 – 1.5)k = [–3i] m
The unit vector uOD, Fig a, that specifies the direction of diagonal OD is given by 
 uAF = = i + j – k 
The magnitude of the moment of F about diagonal OD is 
 MOD = uOD · rOB × F = – 
 = [3(10) – (3)(1.5)] – [0(10) – (–6)(1.5)] + [0(3) – (–6)(3)]
 = 17 N · m
or
 MOD = uOD · rDB × F = 
–
 = [0(10) – (3)(0)] – [(–3)(10) – (–6)(0)] + [(–3)(3) – (–6)(0)]
 = 17 N · m
Thus, MOD can be expressed in Cartesian vector form as
MOD = MOD uOD = 17 i + j + k = [11.3i + 11.3j + 5.67k] N · m Ans
(3 – 0)i + (3 – 0)j + (1.5 – 0)k
(3 – 0)2 + (3 – 0)2 + (0 – 1.5)2 
 2 2 1
 3 3 3
 –3 0 0
 –6 3 10
 2 2 1
 3 3 3
 0 3 1.5
 –6 3 10
2
3
1
3
2
3
1
3
2
3
2
3
1
3
2
3
2
3
2
3
1
3
2
3
⎛
⎝
⎜
⎞
⎠
⎟
229
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•4–53. The tool is used to shut off gas valves that are
difficult to access. If the force F is applied to the handle,
determine the component of the moment created about the
z axis of the valve.
x
y
0.4 m
F { 60i 20j 15k} N
30
z
0.25 m
4–54. Determine the magnitude of the moments of the
force F about the x, y, and z axes. Solve the problem (a) using
a Cartesian vector approach and (b) using a scalar approach.
4 ft
3 ft
2 ft
y
z
C
A
B
F {4i 12j 3k} lb
x
kN
4 m
3 m
2 m
m
kN · m
kN · m
kN · m
kN · m
kN · m
kN · m
04a Ch04a 190-253.indd 22904a Ch04a 190-253.indd 229 6/12/09 8:34:20 AM6/12/09 8:34:20 AM
 0 0 1
 0.125 0.2165 0
 –60 20 15
 u = k
 r = 0.25 sin 30°i + 0.25 cos 30°j
 = 0.125i + 0.2165j
 Mz = = 15.5 N · m Ans
a) Vector Analysis
Position Vector:
 rAB = {(4 – 0)i + (3 – 0)j + (–2 – 0)k} ft = {4i + 3j – 2k} m
Moment of Force F About x, y and z. Axes: The unit vectors 
along x, y and z axes are i, j and k respectively. Applying Eq. 
4–7, we have
 Mx = i · (rAB × F)
 
 = 
 = 1[3(–3) – (12)(–2)] – 0 + 0 = 15.0 kN · m Ans
 Mx = j · (rAB × F)
 
 = 
 = 0 – 1[4(–3) – (4)(–2)] + 0 = 4.00 kN · m Ans
 Mx = k · (rAB × F)
 
 = 
 = 0 – 0 + 1[4(12) – 4(3)] = 36.0 kN · m Ans
 1 0 0
 4 3 –2
 4 12 –3
 0 1 0
 4 3 –2
 4 12 –3
 0 0 1
 4 3 –2
 4 12 –3
b) Scalar Analysis
 Mx = ΣMx; Mx = 12(2) – 3(3) = 15.0 kN · m Ans
My = ΣMy; My = –4(2) + 3(4) = 4.00 kN · m Ans
Mz = ΣMz; Mz = –4(3) + 12(4) = 36.0 kN · m Ans
230
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4–55. Determine the moment of the force F about an axis
extending between A and C. Express the result as a
Cartesian vector.
4 ft
3 ft
2 ft
y
z
C
A
B
F {4i 12j 3k} lb
x
kN
4 m
3 m
2 mm
mm
kN.
kN · m
kN · m
kN · m
04a Ch04a 190-253.indd 23004a Ch04a 190-253.indd 230 6/12/09 8:34:21 AM6/12/09 8:34:21 AM
Position Vector:
rCB = {–2k} m
rAB = {(4 – 0)i + (3 – 0)j + (–2 – 0)k} m = (4i + 3j – 2k} m
Unit Vector Along AC Axis:
uAC =
 (4 – 0)i + (3 – 0)j 
= 0.8i + 0.6j
 (4 – 0)2 + (3 – 0)2
Moment of Force F About AC Axis: With F = {4i + 12j – 3k) kN 
applying Eq. 4–7, we have
MAC = uAC · (rCB : F)
 0.8 0.6 0
 = 0 0 –2
 4 12 –3
 = 0.8[(0)(–3) – 12(–2)] – 0.6[0(–3) – 4(–2)] + 0
 = 14.4 kN · m
Or
MAC = uAC · (rCB : F)
 0.8 0.6 0
 = 4 3 –2
 4 12 –3
 = 0.8[(3)(–3) – 12(–2)] – 0.6[4(–3) – 4(–2)] + 0
 = 14.4 kN · m
Expressing MAC as a Cartesian vector yields
MAC = MAC uAC
 = 14.4(0.8i + 0.6j)
 = {11.5i + 8.64j} kN · m Ans
231
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*4–56. Determine the moment produced by force F about
segment AB of the pipe assembly. Express the result as a
Cartesian vector.
y
x
z
4 m
4 m
3 m
A
B
C
F { 20i 10j 15k} N
04a Ch04a 190-253.indd 23104a Ch04a 190-253.indd 231 6/12/09 8:34:24 AM6/12/09 8:34:24 AM
Moment About Line AB: Either position vector rAC or rBC can be 
conveniently used to determine the moment of F about line AB.
rAC = (3 – 0)i + (4 – 0)j + (4 – 0)k = [3i + 4j + 4k] m
rBC = (3 – 3)i + (4 – 4)j + (4 – 0)k = [4k] m
The unit vector uAB, Fig. a, that specifies the direction of line AB is 
given by
uAB =
 (3 – 0)i + (4 – 0)j + (0 – 0)k 
=
 3 
i + 
4 
j 
(3 – 0)2 + (4 – 0)2 + (0 – 0)2
 5 5
Thus, the magnitude of the moment of F about line AB is given by
 3 4 
0
MAB = uAB · rAC : F =
 5 5
 3 4 4
 –20 10 15
 
=
 3 
[4(15) – 10(4)] –
 4 
[3(15) – (–20)(4)] + 0
 5 5
 = –88 N · m
or
 3 4 
0
MAB = uAB · rAC : F =
 5 5
 0 0 4
 –20 10 15
 
=
 3 
[0(15) – 10(4)] –
 4 
[0(15) – (–20)(4)] + 0
 5 5
 = –88 N · m
Thus, MAB can be expressed in Cartesian vector form as
 MAB = MAB uAB = –88
 3 i + 4 j = [–52.8i – 70.4j] N · m Ans
 5 5
⎛
⎝⎜
⎞
⎠⎟
232
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•4–57. Determine the magnitude of the moment that the
force F exerts about the y axis of the shaft. Solve the
problem using a Cartesian vector approach and using a
scalar approach.
200 mm
250 mm
45
B
x
yz
A
O
30
50 mm
 16 NF
04a Ch04a 190-253.indd 23204a Ch04a 190-253.indd 232 6/12/09 8:34:26 AM6/12/09 8:34:26 AM
a) Vector Analysis
Position Vector and Force Vector:
 rOB = {0.2 cos 45°i – 0.2 sin 45°k} m = {0.1414i – 0.1414k} m
 F = 16{–cos 30°i + sin 30°k} N = {–13.856i + 8.00k} N
Moment of Force F About y Axis: The unit vector along the y 
axis is j. Applying Eq. 4–11, we have
 My = j · (rOB : F)
 0 1 0
 = 0.1414 0 –0.1414
 –13.856 0 8
 = 0 – 1[0.1414(8) – (–13.856)(–0.1414)] + 0
 = 0.828 N · m Ans
b) Scalar Analysis
 My = ΣMy; My = 16 cos 30° (0.2 sin 45°) 
 – 16 sin 30° (0.2 cos 45°)
 = 0.828 N · m Ans
233
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–58. If , determine the magnitude of the
moment produced by this force about the x axis.
F = 450 N
300 mmx
y
z
A B
60
60
45
F
100 mm
150 mm
04a Ch04a 190-253.indd 23304a Ch04a 190-253.indd 233 6/12/09 8:34:28 AM6/12/09 8:34:28 AM
Moment About the x axis: Either position vector rAB or rCB can be used to 
determine the moment of F about the x axis.
rAB = (–0.15 – 0)i + (0.3 – 0)j + (0.1 – 0)k = [–0.15i + 0.3j + 0.1k] m
rCB = [(–1.5 – (–0.15)]i + (0.3 – 0)j + (0.1 – 0)k = [0.3j + 0.1k] m
The force vector F is given by
F = 45(– cos 60°i + cos 60°j + cos 45°k) = [–225i + 225j + 318.20k] N
Knowing that the unit vector of the x axis is i, the magnitude of the moment of F 
about the x axis is given by
 1 0 0
Mx = i · rAB : F = –0.15 0.3 0.1
 –225 225 318.20
 = 1[0.3(318.20) – (225)(0.1)] + 0 + 0 = 73.0 N · m Ans
or
 1 0 0
Mx = i · rCB : F = 0 0.3 0.1
 –225 225 318.20
 = 1[0.3(318.20) – (225)(0.1)] + 0 + 0 = 73.0 N · m Ans
234
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–59. The friction at sleeve A can provide a maximum
resisting moment of about the x axis. Determine
the largest magnitude of force F that can be applied to the
bracket so that the bracket will not turn.
125 N # m
300 mmx
y
z
A B
60
60
45
F
100 mm
150 mm
04a Ch04a 190-253.indd 23404a Ch04a 190-253.indd 234 6/12/09 8:34:30 AM6/12/09 8:34:30 AM
Moment About the x axis: The position vector rAB, Fig. a, will be used to determine the 
moment of F about the x axis.
rAB = (–0.15 – 0)i + (0.3 – 0)j + (0.1 – 0)k = [–0.15i + 0.3j + 0.1k] m
The force vector F is given by
F = F(–cos 60°i + cos 60°j + cos 45°k) = –0.5Fi + 0.5Fj + 0.7071Fk
Knowing that the unit vector of the x axis is i, the magnitude of the moment of F about 
the x axis is given by
 1 0 0
Mx = i · rAB : F = –0.15 0.3 0.1
 –0.5 F 0.5F 0.7071F
 = 1[0.3(0.7071F) – 0.5F(0.1)] + 0 + 0 = 0.1621F
Since the friction at sleeve A can resist a moment of Mx = 125N · m, the maximum 
allowable magnitude of F is given by
 125 = 0.1621F
 F = 771 N Ans 
235
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*4–60. Determine the magnitude of the moment
produced by the force of about the hinged axis
(the x axis) of the door.
F = 200 N
y
x
z
15
A
B
2.5 m
2 m
F 200 N
0.5 m
1 m
04a Ch04a 190-253.indd 23504a Ch04a 190-253.indd 235 6/12/09 8:34:33 AM6/12/09 8:34:33 AM
Moment About the x axis: Either position vector rOB or rCA can be 
used to determine the moment of F about the x axis.
rCA = (2.5 – 2.5)i + (0.9659 – 0)j + (0.2588 – 0)k = [0.9659j + 0.2588k] m
rOB = (0.5 – 0)i + (0 – 0)j + (2 – 0)k = [0.5i + 2k] m
The force vector F is given by
F = FuAB = 200
 (0.5 – 2.5)i + (0 – 0.9659)j + (2 – 0.2588)k 
= [–141.73i – 68.45j + 123.39k] N
 (0.5 – 2.5)2 + (0 – 0.9659)2 + (2 – 0.2588)2
Knowing that the unit vector of the x axis is i, the magnitude of the moment of F about the x axis is 
given by
 1 0 0
Mx = i · rCA : F = 0 0.9659 0.2588 = 137 N · m Ans
 –141.73 –68.45 123.39
or
 1 0 0
Mx = i · rOB : F = 0.5 0 2 = 137 N · m Ans
 –141.73 –68.45 123.39
� �
236
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•4–61. If the tension in the cable is , determine
the magnitude of the moment produced by this force about
the hinged axis, CD, of the panel.
F = 140 lb
6 ft
4 ft
4 ft
6 ft
y
z
A
C
F
D
B
6 ft
x
1.2 m
1.2 m
1.8 m
1.8 m
1.8 m
B(0, 1.2, 3.6) m
D(0, 2.4, 1.8) m
A(1.8, 0, 0) m
•4–61. If the tension in the cable is F = 700 N, determine
Moment About the CD axis: Either position vector rCA or rDB, Fig. a, can be used 
to determine the moment of F about the CD axis.rCA = (1.8 – 0)i + (0 – 0)j + (0 – 0)k = [1.8i] m
rDB = (0 – 0)i + (1.2 – 2.4)j + (3.6 – 1.8)k = [–1.2j + 1.8k] m
Referring to Fig. a, the force vector F can be written as
F = FuAB = 700
(0 – 1.8) + (1.2 – 0) + (3.6 – 0)
(0 – 1.
i j k
88) + (1.2 – 0) + (3.6 – 0)2 2 2
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 = [–300i + 200j + 600k] N
The unit vector uCD, Fig. a, that specifies the direction of the CD axis is given by
uCD = 
(0 – 0) + (2.4 – 0) + (1.8 – 0)
(0 – 0)2
i j k
++ (2.4 – 0) + (1.8 – 0)2 2
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 = 
4
5
j + 
3
5
k
Thus, the magnitude of the moment of F about the CD axis is given by
MCD = uCD · rCA × F = 
0
1.8 0 0
–300 200 600
4
5
3
5
= 0 – 
4
5
[1.8(600) – (–300)(0)] + 
3
5
[1.8(200) – (–300)(0)]
= –648 N · m Ans
or
MCD = uCD · rDB × F = 
0
0 –1.2 1.8
–300 200 600
4
5
3
5
= 0 – 
4
5
[0(600) – (–300)(1.8)] + 
3
5
[0(200) – (–300)(–1.2)]
= –648 N · m Ans
The negative sign indicates that MCD acts in the opposite sense to that of uCD.
04a Ch04a 190-253.indd 23604a Ch04a 190-253.indd 236 6/12/09 8:34:35 AM6/12/09 8:34:35 AM
237
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–62. Determine the magnitude of force F in cable AB in
order to produce a moment of about the hinged
axis CD, which is needed to hold the panel in the position
shown.
500 lb # ft
6 ft
4 ft
4 ft
6 ft
y
z
A
C
F
D
B
6 ft
x
1.2 m
1.2 m
1.8 m
1.8 m
1.8 m
B(0, 1.2, 3.6) m
D(0, 2.4, 1.8) m
A(1.8, 0, 0) m
order to produce a moment of 750 N · m about the hinged
Moment About the CD axis: Either position vector rCA or rCB, Fig. a, can be used 
to determine the moment of F about the CD axis.
rCA = (1.8 – 0)i + (0 – 0)j + (0 – 0)k = [1.8i] m
rCB = (0 – 0)i + (1.2 – 0)j + (3.6 – 0)k = [1.2j + 3.6k] m
Referring to Fig. a, the force vector F can be written as
F = FuAB = F
(0 – 1.8) + (1.2 – 0) + (3.6 – 0)
(0 – 1.
i j k
88) + (1.2 – 0) + (3.6 – 0)2 2 2
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 = –
3
7
Fi + 
2
7
Fj + 
6
7
Fk
The unit vector uCD, Fig. a, that specifies the direction of the CD axis is given by
uCD = 
(0 – 0) + (2.4 – 0) + (1.8 – 0)
(0 – 0)2
i j k
++ (2.4 – 0) + (1.8 – 0)2 2
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 = 
4
5
j + 
3
5
k
Thus, the magnitude of the moment of F about the CD axis is required to be 
MCD = ⎪500⎪ N · m. Thus,
MCD = uCD · rCA × F 
⎪750⎪ = 
0
1.8 0 0
–
4
5
3
5
3
7
2
7
6
7
F F F
–750 = 0 – 
4
5
1.8
6
7
– –
3
7
(0)F F
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 + 
3
5
1.8
2
7
– –
3
7
(0)F F
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
F = 810.2 N Ans
or
MCD = uCD · rDB × F 
⎪750⎪ = 
0
0 1.2 3.6
–
4
5
3
5
3
7
2
7
6
7
F F F
–750 = 0 – 
4
5
0
6
7
– –
3
7
(3.6)F F
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 + 
3
5
0
2
7
– –
3
7
(1.2)F F
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
F = 810.2 N Ans
04a Ch04a 190-253.indd 23704a Ch04a 190-253.indd 237 6/12/09 8:34:39 AM6/12/09 8:34:39 AM
238
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–63. The A-frame is being hoisted into an upright
position by the vertical force of . Determine the
moment of this force about the axis passing through
points A and B when the frame is in the position shown.
y¿
F = 80 lb
30
15
6 ft
y
y¿
x¿
C
A
B
F
x
z
3 ft
3 ft
2 m
1 m
1 m
2 cos 15° m
1 m
position by the vertical force of F = 400 N. Determine the
Scalar analysis:
My = 400 (2 cos 15°) = 772.7 N · m Ans
Vector analysis :
uAB = sin 30°i + cos 30° j
Coordinates of point C :
x = 1 sin 30° – 2 cos 15° cos 30° = –1.173 m
y = 1 cos 30° + 2 cos 15° sin 30° = 1.832 m
z = 2 sin 15° = 0.518 m
rAC = –1.173i + 1.832j + 0.518k
F = 400k
My = 
sin 30° cos 30° 0
1 173 1 832 0 518
0 0 400
– . . .
Mz = 772.7 N · m Ans
04a Ch04a 190-253.indd 23804a Ch04a 190-253.indd 238 6/12/09 8:34:42 AM6/12/09 8:34:42 AM
239
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*4–64. The A-frame is being hoisted into an upright
position by the vertical force of . Determine the
moment of this force about the x axis when the frame is in
the position shown.
F = 80 lb
30
15
6 ft
y
y¿
x¿
C
A
B
F
x
z
3 ft
3 ft
2 m
1 m
1 m
2 cos 15° m
1 m
position by the vertical force of F = 400 N. Determine the
Using x′, y′, z :
ux = cos 30°i′ + sin 30°j′
rAC = –2 cos 15°i′ + 1j′ + 2 sin 15°k
F = 400k
Mz = 
cos 30° sin 30°
–2 cos 15° 1 2 sin 15°
0
0 0 400
 = 346.41 + 386.37 + 0
Mz = 732.8 N · m Ans
Also, using x, y, z.
Coordinates of point C :
x = 1 sin 30° – 2 cos 15° cos 30° = –1.173 m
y = 1 cos 30° + 2 cos 15° sin 30° = 1.832 m
z = 2 sin 15° = 0.518 m
rAC = –1.173i + 1.832j + 0.518k
F = 80k
Mz = 
1 0
–1.173 1.832 0.518
0
0 0 400
 = 732.8 N · m Ans
04a Ch04a 190-253.indd 23904a Ch04a 190-253.indd 239 6/12/09 8:34:44 AM6/12/09 8:34:44 AM
240
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•4–65. The A-frame is being hoisted into an upright
position by the vertical force of . Determine the
moment of this force about the y axis when the frame is in
the position shown.
F = 80 lb
30
15
6 ft
y
y¿
x¿
C
A
B
F
x
z
3 ft
3 ft
2 m
1 m
1 m
2 cos 15° m
1 m
position by the vertical force of F = 400 N. Determine the
Using x′, y′, z :
uy = –sin 30°i′ + cos 30°j′
rAC = –2 cos 15°i′ + 1j′ + 2 sin 15°k
F = 400k
My = 
–sin 30° cos 30°
–2 cos 15° 1 2 sin 15°
0
0 0 400
 = –200 + 669.2 + 0
My = 469.2 N · m Ans
Also, using x, y, z.
Coordinates of point C :
x = 1 sin 30° – 2 cos 15° cos 30° = –1.173 m
y = 1 cos 30° + 2 cos 15° sin 30° = 1.832 m
z = 2 sin 15° = 0.518 m
rAC = –1.173i + 1.832j + 0.518k
F = 400k
My = 
0 1
–1.173 1.832 0.518
0
0 0 400
 = 469.2 N · m Ans
04a Ch04a 190-253.indd 24004a Ch04a 190-253.indd 240 6/12/09 8:34:46 AM6/12/09 8:34:46 AM
241
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–66. The flex-headed ratchet wrench is subjected to a
force of applied perpendicular to the handle as
shown. Determine the moment or torque this imparts along
the vertical axis of the bolt at A.
30 N,
60
A
250 mm
20 mm
P
4–67. If a torque or moment of is required to
loosen the bolt at A, determine the force P that must be
applied perpendicular to the handle of the flex-headed ratchet
wrench.
10 N m
60
A
250 mm
20 mm
P
M = 80(20 + 250 sin 60°)
1
1000
⎛
⎝⎜
⎞
⎠⎟
M = 18.92 N · m Ans
80 = P(20 + 250 sin 60°)
1
1000
⎛
⎝⎜
⎞
⎠⎟
P = 42.28 N Ans
04a Ch04a 190-253.indd 24104a Ch04a 190-253.indd 241 6/12/09 8:34:50 AM6/12/09 8:34:50 AM
242
*4–68. The pipe assembly is secured on the wall by the
two brackets. If the flower pot has a weight of 50 lb,
determine the magnitude of the moment produced by the
weight about the OA axis.
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exist. No portion of this material may be reproduced,in any form or by any means, without permission in writing from the publisher.
A
O
z
x y
4 ft
3 ft
3 ft
4 ft60
30
B
1.2 m
0.9 m
0.9 m
1.2 m
A(0, 1.2, 0.9) m
W = 250 N
B(0.9897, 1.742, 0.45) m
two brackets. If the flower pot has a weight of 250 N,
Moment About the OA axis: The coordinates of point B are [(1.2 + 0.9 cos 30°) cos 60°, 
(1.2 + 0.9 cos 30°) sin 60°, 0.9 sin 30°] m = (0.9897, 1.7142, 0.45) m. Either position vector 
rOB or rOC can be used to determine the moment of W about the OA axis.
rOB = (0.9897 – 0)i + (1.7142 – 0)j + (0.45 – 0)k = [0.9897i + 1.7142j + 0.45k] m
rAB = (0.9897 – 0)i + (1.7142 – 1.2)j + (0.45 – 0.9)k = [0.9897i + 0.5142j – 0.45k] m
Since W is directed towards the negative z axis, we can write
W = {–250k} N
The unit vector uOA, Fig. a, that specifies the direction of the OA axis is given by
uOA = 
(0 – 0) + (1.2 – 0) + (0.9 – 0)
(0 – 0)2
i j k
++ (1.2 – 0) + (0.9 – 0)2 2
 = 
4
5
j + 
3
5
k
The magnitude of the moment of W about the OA axis is given by
MOA = uOA · rOB × W = 
0
0.9897 1.742 0.45
0 0 –250
4
5
3
5
= 0 – 
4
5
[0.9897(–250) – 0(0.45)] + 
3
5
[0.9897(0) – 0(1.742)]
= 197.9 N · m Ans
or
 
MOA = uOA · rAB × W = 
0
0.9897 0.5142 –0.45
0 0 –250
4
5
3
5
= 0 – 
4
5
[0.9897(–250) – 0(–0.45)] + 
3
5
[0.9897(0) – 0(0.5142)]
= 247.4 N · m Ans
04a Ch04a 190-253.indd 24204a Ch04a 190-253.indd 242 6/12/09 8:34:53 AM6/12/09 8:34:53 AM
243
•4–69. The pipe assembly is secured on the wall by the two
brackets. If the frictional force of both brackets can resist a
maximum moment of , determine the largest
weight of the flower pot that can be supported by the
assembly without causing it to rotate about the OA axis.
150 lb # ft
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
A
O
z
x y
4 ft
3 ft
3 ft
4 ft60
30
B
1.2 m
0.9 m
0.9 m
1.2 m
A(0, 1.2, 0.9) m
W
B(0.9897, 1.7142, 0.45) m
C(0.9897, 1.7142, 0) m
maximum moment of 225 N · m, determine the largest
Moment About the OA axis: The coordinates of point B are [(1.2 + 0.9 cos 30°) cos 60°, 
(1.2 + 0.9 cos 30°) sin 60°, 0.9 sin 30°] m = (0.9897, 1.7142, 0.45) m. Either position vector 
rOB or rOC can be used to determine the moment of W about the OA axis.
rOB = (0.9897 – 0)i + (1.7142 – 0)j + (0.45 – 0)k = [0.9897i + 1.7142j + 0.45k] m
rAB = (0.9897 – 0)i + (1.7142 – 1.2)j + (0.45 – 0.9)k = [0.9897i + 0.5142j – 0.45k] m
Since W is directed towards the negative z axis, we can write
W = –Wk
The unit vector uOA, Fig. a, that specifies the direction of the OA axis is given by
uOA = 
(0 – 0) + (1.2 – 0) + (0.9 – 0)
(0 – 0)2
i j k
++ (1.2 – 0) + (0.9 – 0)2 2
 = 
4
5
j + 
3
5
k
Since it is required that the magnitude of the moment of W about the OA axis not 
exceed 150 N · m, we can write
MOA = uOA · rOB × W 
⎪225⎪ = 
0
0.9897 1.7142 0.45
0 0 –
4
5
3
5
W
225 = 0 – 
4
5
[0.9897(–W) – 0(0.45)] + 
3
5
[0.9897(0) – 0(1.7142)]
W = 284.2 N Ans
or
MOA = uOA · rAB × W 
⎪225⎪ = 
0
0.9897 0.5142 –0.45
0 0 –
4
5
3
5
W
225 = 0 – 
4
5
[0.9897(–W) – 0(–0.45)] + 
3
5
[0.9897(0) – 0(0.5142)]
W = 284.2 N Ans
04a Ch04a 190-253.indd 24304a Ch04a 190-253.indd 243 6/12/09 8:34:57 AM6/12/09 8:34:57 AM
244
4–70. A vertical force of is applied to the
handle of the pipe wrench. Determine the moment that this
force exerts along the axis AB (x axis) of the pipe assembly.
Both the wrench and pipe assembly ABC lie in the 
plane. Suggestion: Use a scalar analysis.
x-y
F = 60 N
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
45
z
y
A
C
B
500 mm
200 mm
150 mm
F
x
04a Ch04a 190-253.indd 24404a Ch04a 190-253.indd 244 6/12/09 8:35:00 AM6/12/09 8:35:00 AM
Scalar Analysis: From the geometry, the perpendicular distance from
x axis to force F is d = 0.15 sin 45° + 0.2 sin 45° = 0.2475 m.
 Mx = ΣMx; Mx = –Fd = –60(0.2475) = –14.8 N · m
Negative sign indicates that Mx is directed toward negative x axis.
 Mx = –14.8 N · m Ans 
245
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–71. Determine the magnitude of the vertical force F
acting on the handle of the wrench so that this force
produces a component of moment along the AB axis (x axis)
of the pipe assembly of . Both the pipe
assembly ABC and the wrench lie in the plane.
Suggestion: Use a scalar analysis.
x-y
(MA)x = 5-5i6 N # m
45
z
y
A
C
B
500 mm
200 mm
150 mm
F
x
*4–72. The frictional effects of the air on the blades of the
standing fan creates a couple moment of on
the blades. Determine the magnitude of the couple forces
at the base of the fan so that the resultant couple moment
on the fan is zero.
MO = 6 N # m
0.15 m 0.15 m
FF
MO
04a Ch04a 190-253.indd 24504a Ch04a 190-253.indd 245 6/12/09 8:35:04 AM6/12/09 8:35:04 AM
Scalar Analysis: From the geometry, the perpendicular distance from 
x axis to F is d = 0.15 sin 45° + 0.2 sin 45° = 0.2475 m.
Mx = ΣMx; –5 = –F(0.2475)
 F = 20.2 N Ans
Couple Moment: The couple moment of F produces a counterclockwise moment of MC = F(0.15 + 0.15) = 0.3F. Since the 
resultant couple moment about the axis perpendicular to the page is required to be zero.
 +(MC) = ΣM; 0 = 0.3F – 6 F = 20 N Ans
�
246
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•4–73. Determine the required magnitude of the couple
moments and so that the resultant couple moment
is zero.
M3M2
M3
M2
45
M1 300 N m
04a Ch04a 190-253.indd 24604a Ch04a 190-253.indd 246 6/12/09 8:35:07 AM6/12/09 8:35:07 AM
Since the couple moment is the free vector, it can act at any point without altering its 
effect. Thus, the couple moments M
1
, M
2
, and M
3
 can be simplified as shown in Fig. a. 
Since the resultant of M
1
, M
2
, and M
3
 is required to be zero,
(MR)y = ΣMy; 0 = M2 sin 45° – 300
 M
2
 = 424.26 N · m = 424 N · m Ans
(MR)x = ΣMx; 0 = 424.26 cos 45° – M3
 M
3
 = 300 N · m Ans
247
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–74. The caster wheel is subjected to the two couples.
Determine the forces F that the bearings exert on the shaft
so that the resultant couple moment on the caster is zero.
40 mm
45 mm
100 mm
500 N
500 N
50 mm
F
F
A
B
04a Ch04a 190-253.indd 24704a Ch04a 190-253.indd 247 6/12/09 8:35:09 AM6/12/09 8:35:09 AM
 +ΣMA = 0; 500(50) – F (40) = 0
 F = 625 N Ans
248
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–75. If , determine the resultant couple
moment.
F = 200 lb
A
B
F
F
2 ft
2 ft
2 ft
2 ft
150 lb
150 lb
3
3
4
4
5
5
2 ft
30
30
0.2 m 0.2 m
0.2 m
0.2 m
0.2 m
1500 N
1500 N
1500 N
0.2 m
0.2 m0.2 m
2000(4/5) N
2000(3/5) N
2000 N
1500 sin 30°
1500 cos 30° N
0.2 m 0.2 m
2000 N
2000(4/5) N
1500 
sin 30° N
1500 N
1500 
cos 30° N
4–75. If F = 2000 N, determine the resultant couple 
moment.
a) By resolving the 1500-N and 2000-N couples into their x and y components, 
Fig. a, the couple moments (MC)1 and (MC)2 produced by the 1500-N and 
2000-N couples, respectively, are given by
+ (MC)1 = –1500 cos 30° (0.4) – 1500 sin 30° (0.4) 
= –819.62 N · m = 819.62 N · m
+ (MC)2 = 2000
4
5
⎛
⎝⎜
⎞
⎠⎟
(0.2) + 2000
3
5
⎛
⎝⎜
⎞
⎠⎟
(0.2) = 560 N · m 
 Thus, the resultant couple moment can be determined from
+ (MC)R = (MC)1 + (MC)2
= –819.62 + 560 = –259.62 N · m = 260 N · m (clockwise)
b) By resolving the 1500-N and 2000-N couples into their x and y components, Fig. a, and summing the 
moments of these force components algebraically about point A,
+ (MC)R = ΣMA; (MC)R = –1500 sin 30° (0.4) – 1500 cos 30° (0.6) + 2000
4
5
⎛
⎝⎜
⎞
⎠⎟
(0.2) + 2000
3
5
⎛
⎝⎜
⎞
⎠⎟
(0.6)
–2000
3
5
⎛
⎝⎜
⎞
⎠⎟
(0.4) + 2000
4
5
⎛
⎝⎜
⎞
⎠⎟
(0) + 1500 cos 30° (0.2) + 1500 sin 30° (0)
= –259.62 N · m = 260 N · m (clockwise) Ans
04a Ch04a 190-253.indd 24804a Ch04a 190-253.indd 248 6/12/09 8:35:11 AM6/12/09 8:35:11 AM
249
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*4–76. Determine the required magnitude of force F if the
resultant couple moment on the frame is ,
clockwise.
200 lb # ft
A
B
F
F
2 ft
2 ft
2 ft
2 ft
150 lb
150 lb
3
3
4
4
5
5
2 ft
30
30
0.2 m 0.2 m
0.2 m
0.2 m
0.2 m
1500 N
1500 N
1500 N
0.2 m
1500 sin 30° N
1500 cos 30° N
0.2 m 0.2 m
1500 N0.2 m
0.2 m
*4–76. Determine the required magnitude of force F if 
the resultant couple moment on the frame is 200 N · m, 
clockwise.
By resolving F and the 150-N couple into their x and y components, Fig. a, the couple moments (MC)1 and (MC)2 produced 
by F and the 5-kN couple, respectively, are given by
+ (MC)1 = F
4
5
⎛
⎝⎜
⎞
⎠⎟
(0.2) + F
3
5
⎛
⎝⎜
⎞
⎠⎟
(0.2) = 0.28F
+ (MC)2 = –1500 cos 30° (0.4) – 1500 sin 30° (0.4) = –819.62 N · m = 819.62 N · m 
The resultant couple moment acting on the beam is required to be 200 N · m, clockwise. Thus, 
+ (MC)R = (MC)1 + (MC)2
–200 = 0.28F – 819.62
F = 2213 N Ans
04a Ch04a 190-253.indd 24904a Ch04a 190-253.indd 249 6/12/09 8:35:15 AM6/12/09 8:35:15 AM
250
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–78. If , determine the magnitude of force F so that
the resultant couple moment is , clockwise.100 N # m
u = 30°
30
15
15
F
F
300 N
300 N
300 mm
30
u
u
•4–77. The floor causes a couple moment of
and on the brushes of the
polishing machine. Determine the magnitude of the couple
forces that must be developed by the operator on the
handles so that the resultant couple moment on the polisher
is zero. What is the magnitude of these forces if the brush 
at B suddenly stops so that MB = 0?
MB = 30 N # mMA = 40 N # m
0.3 m
MB
MA
F
F
A
B
04a Ch04a 190-253.indd 25004a Ch04a 190-253.indd 250 6/12/09 8:35:19 AM6/12/09 8:35:19 AM
By resolving F and the 300 N couple into their radial and tangential components, 
Fig. a, and summing the moment of these two force components about point O. 
 +(MC)R = ΣMO; –100 = F sin 45° (0.3) + F cos 15° (0.3) – 2(300 cos 30°)(0.3)
 F = 111 N Ans
Note: Since the line of action of the radial component of the forces pass through point 
O, no moment is produced about this point.
 +MB = 40 – 30 – F (0.3) = 0
 F = 33.3 N Ans
 +MB = 40 – F (0.3) = 0
 F = 133 N Ans
30
15
15
F
F
300 N
300 N
300 mm
30
251
4–79. If , determine the required angle so that
the resultant couple moment is zero.
200 N
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*4–80. Two couples act on the beam. Determine the
magnitude of F so that the resultant couple moment is
counterclockwise. Where on the beam does the
resultant couple moment act?
450 N m,
2000 N
2000 N
0.2 m
0.15 m 0.125 m
30
30
F
F
+ MR = ΣM; 450 = 2000 (0.15) + F cos 30° (0.125)
 F = 1386 N Ans
The resultant couple moment is a free vector. It can act at any point on the beam.
04a Ch04a 190-253.indd 25104a Ch04a 190-253.indd 251 6/12/09 8:35:24 AM6/12/09 8:35:24 AM
By resolving the 300 N and 200 N couples into their radial and tangential components, 
Fig. a, and summing the moment of these two force components about point O. 
 +(MC)R = ΣMO; 0 = 200 sin 45° (0.3) + 200 cos 15° (0.3) – 300 cos u (0.3) – 300 cos u (0.3)
 u = 56.1° Ans
Note: Since the line of action of the radial component of the forces pass through point O, no moment 
is produced about this point.
252
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–82. The cord passing over the two small pegs A and B of
the board is subjected to a tension of 100 N. Determine the
minimum tension P and the orientation of the cord
passing over pegs C and D, so that the resultant couple
moment produced by the two cords is , clockwise.20 N # m
u
100 N
100 N
P
P
BC
30
300 mm
300 mm
30
DA
45
u
u
•4–81. The cord passing over the two small pegs A and B of
the square board is subjected to a tension of 100 N.
Determine the required tension P acting on the cord that
passes over pegs C and D so that the resultant couple
produced by the two couples is acting clockwise.
Take .u = 15°
15 N # m
100 N
100 N
P
P
BC
30
300 mm
300 mm
30
DA
45
u
u
4–83. A device called a rolamite is used in various ways to
replace slipping motion with rolling motion. If the belt,
which wraps between the rollers, is subjected to a tension of
15 N, determine the reactive forces N of the top and bottom
plates on the rollers so that the resultant couple acting on
the rollers is equal to zero.
N
N
30
25 mm
A
B
25 mm
T 15 N
T 15 N
04a Ch04a 190-253.indd 25204a Ch04a 190-253.indd 252 6/12/09 8:35:27 AM6/12/09 8:35:27 AM
 +MB = 100 cos 30° (0.3) + 100 sin 30° (0.3) – P sin 15° (0.3) – P cos 15° (0.3) = 15
 P = 70.7 N Ans
For minimum P require u = 45° Ans
 +MB = 100 cos 30° (0.3) + 100 sin 30° (0.3) – P = 20
 P = 49.5 N Ans
0.3
cos 45°
⎛
⎝
⎜
⎞
⎠
⎟
 +ΣMA = 0; 15(50 + 50 sin 30°) – N (50 cos 30°) = 0
 N = 26.0 N Ans
253
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*4–84. Two couples act on the beam as shown. Determine
the magnitude of F so that the resultant couple moment is
counterclockwise. Where on the beam does the
resultant couple act?
300 lb # ft
200 lb
200 lb
1.5 ft
•4–85. Determine the resultant couple moment acting on
the beam. Solve the problem two ways: (a) sum moments
about point O; and (b) sum moments about point A.
1.5 m 1.8 m
45
45
30
30
A
2 kN
2 kN
8 kN
B
0.3 m
8 kN
O
2000 N
2000 N
0.4 m
0.15 m
300 N · m counterclockwise. Where on the beam does the 
resultant couple act?
+ (MC)R