brunet capitulo 2

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7
•2–1. If and , determine the magnitude
of the resultant force acting on the eyebolt and its direction
measured clockwise from the positive x axis.
T = 6 kNu = 30°
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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.
8 kN
T
x
y
u
45
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The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively.
Applying the law of cosines to Fig. b,
Applying the law of sines to Fig. b and using this result, yields
Thus, the direction angle f of FR measured clockwise from the positive x axis is
f = a – 60° = 63.05° – 60° = 3.05° Ans
FR = 6
2 + 82 – 2(6)(8) cos 75°
 
= 8.669 kN = 8.67 kN
 Ans
sin a
8
sin 75°
8.669
= a = 63.05°
8
2–2. If and , determine the magnitude
of the resultant force acting on the eyebolt and its direction
measured clockwise from the positive x axis.
T = 5 kNu = 60°
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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.
8 kN
T
x
y
u
45
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Applying the law of sines to Fig. b and using this result, yields
Thus, the direction angle f of FR measured clockwise from the positive x axis is
f = a – 30° = 47.54° – 30° = 17.5° Ans
FR = 5
2 + 82 – 2(5)(8) cos 105°
 
= 10.47 kN = 10.5 kN
 Ans
sin a
8
sin 105°
10.47
= a = 47.54°
The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively.
Applying the law of cosines to Fig. b,
9
2–3. If the magnitude of the resultant force is to be 9 kN
directed along the positive x axis, determine the magnitude of
force T acting on the eyebolt and its angle .u
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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.
8 kN
T
x
y
u
45
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The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively.
Applying the law of cosines to Fig. b,
Applying the law of sines to Fig. b and using this result, yields
T = 82 + 92 – 2(8)(9) cos 45°
 
= 6.571 kN = 6.57 kN
 Ans
sin (90° – u)
8
sin 45°
6.571
=
u = 30.6° Ans
10
*2–4. Determine the magnitude of the resultant force
acting on the bracket and its direction measured
counterclockwise from the positive u 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.
u
F1 200 lb
F2 150 lbv
30
30
45
N
N
N
N
N
N
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NN
Applying the law of sines to Fig. b and using this result, yields
Thus, the direction angle f of FR measured clockwise from the positive x axis is
f = a – 60° = 63.05° – 60° = 3.05° Ans
FR = 200
2 + 1502 – 2(200)(150) cos 75°
 
= 216.72 N = 217 N
 Ans
sin a
200
sin 75°
216.72
= a = 63.05°
The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively.
Applying the law of cosines to Fig. b,
11
•2–5. Resolve F1 into components along the u and axes,
and determine the magnitudes of these components.
v
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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.
u
F1 200 lb
F2 150 lbv
30
30
45
N
N
N
N
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Fv
sin 45°
200
sin 30°
= Fv = 283 N
Fu
sin 105°
200
sin 30°
= Fu = 386 N
The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively.
Applying the law of sines to Fig. b, yields
Ans
Ans
12
2–6. Resolve F2 into components along the u and axes,
and determine the magnitudes of these components.
v
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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.
u
F1 200 lb
F2 150 lbv
30
30
45
N
N
N
N
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Fv
sin 120°
150
sin 30°
= Fv = 260 N
Fu
sin 30°
150
sin 30°
= Fu = 150 N Ans
Ans
The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively.
Applying the law of sines to Fig. b,
13
2–7. If and the resultant force acts along the
positive u axis, determine the magnitude of the resultant
force and the angle .u
FB = 2 kN
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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.
y
x
u
B
FA 3 kN
FB
A
u 30
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sin f
3
sin 30°
2
= f = 48.59°
The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively.
Applying the law of sines to Fig. b, yields
Thus,
With the result u = 78.590, applying the law of sines to Fig. b again, yields
u = 30° + f = 30° + 48.59° = 78.59° = 78.6° Ans
FR
sin (180° – 78.59°)
2
sin 30°
= FR = 3.92 kN Ans
14
*2–8. If the resultant force is required to act along the
positive u axis and have a magnitude of 5 kN, determine the
required magnitude of FB and its direction .u
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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.
y
x
u
B
FA 3 kN
FB
A
u 30
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sin u
5
sin 30°
2.832
= u = 62.0° Ans
The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively.
Applying the law of cosines to Fig. b,
Using this result and realizing that sin (180° – u) = sin u, the application of 
the sine law to Fig. b, yields
FB = 3
2 + 52 – 2(3)(5) cos 30°
 
= 2.832 kN = 2.83 kN
 Ans
15
•2–9. The plate is subjected to the two forces at A and B
as shown. If , determine the magnitude of the
resultant of these two forces and its direction measured
clockwise from the horizontal.
u = 60°
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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
B
FA 8 kN
FB 6 kN
40
u
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Parallelogram Law : The parallelogram law of addition is shown in
Fig. (a).
Trigonometry : Using law of cosines[Fig. (b)], we have
sin u
6
sin u = 0.5470
 u = 33.16° 
sin 100°
10.80
=
The angleu can be determined using law of sines [Fig. (b)].
Thus, the direction f of FR measured from the x axis is
f = 33.16° – 30° = 3.16° Ans
FR = 8
2 + 62 – 2(8)(6) cos 100°
 
= 10.80 kN = 10.8 kN
 Ans
16
2–10. Determine the angle of for connecting member A
to the plate so that the resultant force of FA and FB is
directed horizontally to the right.Also, what is the magnitude
of the resultant force?
u
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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
B
FA 8 kN
FB 6 kN
40
u
2–11. If the tension in the cable is 400 N, determine the
magnitude and direction of the resultant force acting on 
the pulley. This angle is the same angle of line AB on the
tailboard block.
u
400 N
30
y
A
B
x
400 N
u
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Parallelogram Law : The parallelogram law of addition is shown in
Fig. (a).
Trigonometry : Using law of sines [Fig. (b)], we have
sin (90° – u)
6
sin (90° – u) = 0.5745
 u = 54.93° = 54.9° Ans 
sin 50°
8
=
From the triangle, u = 180° – (90° – 54.93°) – 50° = 94.93°. Thus, using
law of cosines, the magnitude of FR is
FR = 8
2 + 62 – 2(8)(6) cos 94.93°
 
= 10.4 kN
 Ans
FR = (400)
2 + (400)2 – 2(400)(400) cos 60° = 400 N Ans
sin u
400
u = 60° Anssin 60°
400
= ;
17
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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.
*2–12. The device is used for surgical replacement of the
knee joint. If the force acting along the leg is 360 N,
determine its components along the x and y axes.¿
60
360 N
10
y
x
y¿
x¿
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–Fx
sin 20°
Fx = –125 N Ans
360
sin 100°
= ;
Fy
sin 60°
Fy = 317 N Ans
360
sin 100°
= ;
18
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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.
•2–13. The device is used for surgical replacement of the
knee joint. If the force acting along the leg is 360 N,
determine its components along the x and y axes.¿
60
360 N
10
y
x
y¿
x¿
02a Ch02a 7-56.indd 1802a Ch02a 7-56.indd 18 6/12/09 8:05:30 AM6/12/09 8:05:30 AM
–Fx
sin 30°
Fx = –183 N Ans
360
sin 80°
= ;
Fy
sin 70°
Fy = 344 N Ans
360
sin 80°
= ;
19
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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.
2–14. Determine the design angle for
strut AB so that the 400-lb horizontal force has a
component of 500 lb directed from A towards C.What is the
component of force acting along member AB? Take
.f = 40°
u (0° … u … 90°) A
C
B
400 lb
u
f
2–15. Determine the design angle 
between struts AB and AC so that the 400-lb horizontal
force has a component of 600 lb which acts up to the left, in
the same direction as from B towards A. Take .u = 30°
f (0° … f … 90°) A
C
B
400 lb
u
f
800 N
800 N
1000 N
800 N
1000 N
800 N
800 N
1200 N
1200 N
800 N
2–14. Determine the design angle (0° 90°) for 
strut AB so that the 800-N horizontal force has a 
component of 1000 N directed from A towards C. What is 
the component of force acting along member AB? Take 
 40°.
2–15. Determine the design angle (0° 90°) 
between struts AB and AC so that the 800 N horizontal 
force has a component of 1200 N which acts up to the 
left, in the same direction as from B towards A. Take 
 30°.
Parallelogram Law : The parallelogram law of addition is 
shown in Fig. (a).
Trigonometry : Using law of cosines [Fig. (b)], we have
FAC = 800 + 1200 – 2(800)(1200) cos 30°
2 2 
= 645.93 N
The angle can be determined using law of sines [Fig. (b)].
sin
800
 = 
sin 30°
645.93
 sin = 0.6193
 = 38.3° Ans
Parallelogram Law : The parallelogram law of addition is 
shown in Fig. (a).
Trigonometry : Using law of sines [Fig. (b)], we have
sin
1000
 = 
sin 40°
800
 sin = 0.8035
 = 53.46° = 53.5° Ans
Thus, = 180° – 40° – 53.46° = 86.54°
Using law of sines [Fig. (b)]
FAB
sin 86.54°
 = 
800
sin 40°
 FAC = 1242 N Ans
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20
*2–16. Resolve F1 into components along the u and axes
and determine the magnitudes of these components.
v
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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.
F1 250 N
F2 150 N u
v
30
30
105
•2–17. Resolve F2 into components along the u and axes
and determine the magnitudes of these components.
v
F1 250 N
F2 150 N u
v
30
30
105
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F1v
sin 30°
F1v = 129 N Ans
Sine law :
250
sin 105°
=
F1u
sin 45°
F1u = 183 N Ans
250
sin 105°
=
F2v
sin 30°
F2v = 77.6 N Ans
Sine law :
150
sin 75°
=
F2u
sin 75°
F2u = 150 N Ans
150
sin 75°
=
21
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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.
2–18. The truck is to be towed using two ropes. Determine
the magnitudes of forces FA and FB acting on each rope in
order to develop a resultant force of 950 N directed along
the positive x axis. Set .u = 50°
y
20°
x
A
B
FA
FB
u
2–19. The truck is to be towed using two ropes. If the
resultant force is to be 950 N, directed along the positive x
axis, determine the magnitudes of forces FA and FB acting
on each rope and the angle of FB so that the magnitude of
FB is a minimum. FA acts at 20° from the x axis as shown.
u
y
20°
x
A
B
FA
FB
u
02a Ch02a 7-56.indd 2102a Ch02a 7-56.indd 21 6/12/09 8:05:34 AM6/12/09 8:05:34 AM
Parallelogram Law : The parallelogram law of addition is 
shown in Fig. (a).
Trigonometry : Using law of sines [Fig. (b)], we have
FB
sin 20°
 = 
950
sin 110°
 FB = 346 N Ans
FA
sin 50°
 = 
950
sin 110°
 FA = 774 N Ans
Parallelogram Law : In order to produce a minimum force FB, FB has 
to act perpendicular to FA. The parallelogram law of addition is
shown in Fig. (a).
Trigonometry : Fig. (b).
The angle u is 
 u = 90° – 20° = 70.0° Ans
 FB = 950 sin 20° = 325 N Ans
 FA = 950 cos 20° = 893 N Ans
22
*2–20. If , , and the resultant force is 
6 kN directed along the positive y axis, determine the required
magnitude of F2 and its direction .u
F1 = 5 kNf = 45°
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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.
F1
F2
x
y
u
f
60
02a Ch02a 7-56.indd 2202a Ch02a 7-56.indd 22 6/12/09 8:05:35 AM6/12/09 8:05:35 AM
Using this result and applying the law of sines to Fig. b, yields
F
2
 = 62 + 52 – 2(6)(5) cos 45°
 
= 4.310 kN = 4.31 kN
 Ans
sin u
5
sin 45°
4.310
= u= 55.1° Ans
The parallelogram law of addition and the triangular rule are shown in 
Figs. a and b, respectively.
Applying the law of cosines to Fig. b,
23
•2–21. If and the resultant force is to be 6 kN
directed along the positive y axis, determine the magnitudes of
F1 and F2 and the angle if F2 is required to be a minimum.u
f = 30°
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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.
F1
F2
x
y
u
f
60
02a Ch02a 7-56.indd 2302a Ch02a 7-56.indd 23 6/12/09 8:05:43 AM6/12/09 8:05:43 AM
For F2 to be minimum, it has to be directed perpendicular to FR.
The parallelogram law of addition and triangular rule are shown in Figs. a and b, respectively.
By applying simple trigonometry to Fig. b,
u = 90° – 30° = 60° Ans
and
F1 = 6 cos 30° = 5.20 kN Ans
F2 = 6 sin 30° = 3 kN Ans
24
2–22. If , , and the resultant force is to
be directed along the positive y axis, determine the
magnitude of the resultant force if F2 is to be a minimum.
Also, what is F2 and the angle ?u
F
1
= 5 kNf = 30°
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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.
F1
F2
x
y
u
f
60
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Parallelogram Law and Triangular Rule: The parallelogram law of addition and 
triangular rule are shown in Figs. a and b, respectively.
For F2 to be minimum, it must be directed perpendicular to the resultant force. Thus,
u = 90° Ans
By applying simple trigonometry to Fig. b,
F2 = 5 sin 30° = 2.50 kN Ans
FR = 5 cos 30° = 4.33 kN Ans
25
2–23. If and , determine the magnitude
of the resultant force acting on the plate and its direction
measured clockwise from the positive x axis.
F2 = 6 kNu = 30°
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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.
y
x
F3 5 kN
F1 4 kN
 F2
u
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Using the results for F� and a and referring to Fig. c, FR and b can be determined.
Thus, the direction angle f of FR , measured clockwise from the positive x axis, is
FR = 6
2 + 6.4032 – 2(6)(6.403) cos (51.34° + 30°)
 = 8.09 kN
 Ans
sin b
6
sin (51.34° + 30°)
8.089
= b = 47.16°
f = a + b = 51.34° + 47.16° = 98.5° Ans
Parallelogram law and Triangular Rule: This problem can be solved by adding the forces
successively, using the parallelogram law of addition, shown in Fig. a. Two triangular 
force diagrams, shown in Figs. b and c, can be derived from the parallelogram.
Determination of Unknowns: Referring to Fig. b, F� and a can be determined
as follows.
5
4
F� = 42 + 52 = 6.403 kN
tan a = a = 51.34°
26
*2–24. If the resultant force FR is directed along a 
line measured 75° clockwise from the positive x axis and 
the magnitude of F2 is to be a minimum, determine the
magnitudes of FR and F2 and the angle .u … 90°
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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.
y
x
F3 5 kN
F1 4 kN
 F2
u
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Using the results for u, a, and F�, FR and F2 can be determined by referring to Fig. c.
Referring to Fig. b, F� and a can be determined.
u = 90° – 75° = 15° Ans
FR = 6.403 sin (15° + 51.43°) = 5.86 kN Ans
F2 = 6.403 cos (15° + 51.43°) = 2.57 kN Ans
F� = 42 + 52 = 6.403 kN
tan a = a = 51.34°
This problem can be solved by adding the forces successively, using the parallelogram 
law of addition, shown in Fig. a. Two triangular force diagrams, shown in Figs. b 
and c, can be derived from the parallelograms.
For F1 to be minimum, it must be perpendicular to the resultant force’s line 
of action. Thus,
5
4
27
•2–25. Two forces F1 and F2 act on the screw eye. If their
lines of action are at an angle apart and the magnitude 
of each force is determine the magnitude of
the resultant force FR and the angle between FR and F1.
F1 = F2 = F,
u
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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.
F2
F1
u
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=
=
FR = (F)
2 + (F)2 – 2(F)(F) cos (180° – u)
FR = F( 2) 1 + cos u
FR = 2F cos ( )
F
sin f
F
sin(u – f)
sin(u – f) = sin f
Since cos (180° – u) = –cos u
Since cos ( ) = 
Then
u – f = f
u
2
Ans
Ans
f
u
2
u
2
1 + cos u
2
28
2–26. The log is being towed by two tractors A and B.
Determine the magnitudes of the two towing forces FA and
FB if it is required that the resultant force have a magnitude
and be directed along the x axis. Set .u = 15°FR = 10 kN
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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.
x
y
B
A
30
FA
FB
u
2–27. The resultant FR of the two forces acting on the log is
to be directed along the positive x axis and have a magnitude
of 10 kN, determine the angle of the cable, attached to B such
that the magnitude of force FB in this cable is a minimum.
What is the magnitude of the force in each cable for this
situation?
u
x
y
B
A
30
FA
FB
u
02a Ch02a 7-56.indd 2802a Ch02a 7-56.indd 28 6/12/09 8:06:01 AM6/12/09 8:06:01 AM
Parallelogram Law : The parallelogram law of addition is shown in 
Fig. (a). 
Trigonometry : Using law of sines [Fig. (b)], we have
FA
sin 15°
 = 
10
sin 135°
FA = 3.66 kN Ans
FB
sin 30°
 = 
10
sin 135°
FB = 7.07 kN Ans
Parallelogram Law : In order to produce a minimum force FB, FB has
to act perpendicular to FA. The parallelogram law of addition is shown in 
Fig. (a). 
Trigonometry : Fig. (b).
The angle u is
FB = 10 sin 30° = 5.00 kN Ans
FA = 10 cos 30° = 8.66 kN Ans
u = 90° – 30° = 60.0° Ans
29
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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.
*2–28. The beam is to be hoisted using two chains. Deter-
mine the magnitudes of forces FA and FB acting on each chain
in order to develop a resultant force of 600 N directed along
the positive y axis. Set .u = 45°
FB FA
y
x
30
u
02a Ch02a 7-56.indd 2902a Ch02a 7-56.indd 29 6/12/09 8:06:03 AM6/12/09 8:06:03 AM
= ;FB
sin 30°
FB = 311 N Ans
600
sin 105°
= ;FA
sin 45°
FA = 439 N Ans
600
sin 105°
30
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•2–29. The beam is to be hoisted using two chains. If the
resultant force is to be 600 N directed along the positive y
axis, determinethe magnitudes of forces FA and FB acting on
each chain and the angle of FB so that the magnitude of FB
is a minimum. FA acts at 30° from the y axis, as shown.
u
FB FA
y
x
30
u
02a Ch02a 7-56.indd 3002a Ch02a 7-56.indd 30 6/12/09 8:06:05 AM6/12/09 8:06:05 AM
For minimum FB, require
u = 60° Ans
FA = 600 cos 30° = 520 N Ans
FB = 600 sin 30° = 300 N Ans
31
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–30. Three chains act on the bracket such that they create
a resultant force having a magnitude of 500 lb. If two of the
chains are subjected to known forces, as shown, determine
the angle of the third chain measured clockwise from the
positive x axis, so that the magnitude of force F in this chain
is a minimum. All forces lie in the x–y plane. What is the
magnitude of F? Hint: First find the resultant of the two
known forces. Force F acts in this direction.
u
300 lb
200 lb
x
y
F
30
u
2–30. Three chains act on the bracket such that they create 
a resultant force having a magnitude of 1000 N. If two of the
Cosine law:
FR1 = 600 + 400 – 2(600)(400) cos 60°
2 2
= 529.15 N
Sine law:
sin (30° + )
400
 = 
sin 60°
529.15
 = 10.9° Ans
When F is directed along FR1, F will be minimum to create 
the resultant force.
 FR = FR1 + F
1000 = 529.15 + Fmin
 Fmin = 471 N Ans
600 N
400 N
800 N
600 N
400 N
600 N
1000 N
529.15 N
400 N
02a Ch02a 7-56.indd 3102a Ch02a 7-56.indd 31 6/12/09 8:06:06 AM6/12/09 8:06:06 AM
32
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–31. Three cables pull on the pipe such that they create a
resultant force having a magnitude of 900 lb. If two of the
cables are subjected to known forces, as shown in the figure,
determine the angle of the third cable so that the
magnitude of force F in this cable is a minimum. All forces
lie in the x–y plane. What is the magnitude of F? Hint: First
find the resultant of the two known forces.
u
45
30
y
x
400 lb
600 lb
F
u
1200 N
800 N
800 N
1200 N
1800 N
1605.3 N
1200 N
800 N
2–31. Three cables pull on the pipe such that they create a 
resultant force having a magnitude of 1800 N. If two of the
F ʹ = (1200) + (800) – 2(1200)(800) cos 105°2 2
= 1605.3 N
F = 1800 – 1605.3 = 194.7 N Ans
sin
1200
 = 
sin 105°
1605.3
: = 46.22°
 = 46.22° – 30° = 16.2° Ans
02a Ch02a 7-56.indd 3202a Ch02a 7-56.indd 32 6/12/09 8:06:07 AM6/12/09 8:06:07 AM
33
*2–32. Determine the magnitude of the resultant force
acting on the pin and its direction measured clockwise from
the positive x 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.
x
y
F1 30 lb
F2 40 lb
F3 25 lb
15
15
45
150 N
200 N
125 N
66.43 N
331.61 N
150 N
200 N
125 N
Rectangular components: By referring to Fig. a, the x and y components of F1, F2, and F3 
can be written as
(F1)x = 150 cos 45° = 106.07 N (F1)y = 150 sin 45° = 106.07 N
(F2)x = 200 cos 15° = 193.19 N (F2)y = 200 sin 15° = 51.76 N
(F3)x = 125 sin 15° = 32.35 N (F3)y = 125 cos 15° = 120.74 N
Resultant Force: Summing the force components algebraically along the x and y axes,
+
→Σ(FR)x = ΣFx; (FR)x = 106.07 + 193.19 + 32.35 = 331.61 N →
+↑Σ(FR)y = ΣFy; (FR)y = 106.07 – 51.76 – 120.74 = –13.29 N = –66.43 N ↓
The magnitude of the resultant force FR is
FR = ( ) + ( )
2F FR x R y
2 = (331.61) + (–66.43)2 2 = 338.2 N Ans
The direction angle of FR, measured clockwise from the positive x axis, is
 = tan–1
(
(
F
F
R y
R x
)
)
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 = tan–1
66.43
331.61
⎛
⎝
⎜
⎞
⎠
⎟ = 11.3° Ans
02a Ch02a 7-56.indd 3302a Ch02a 7-56.indd 33 6/12/09 8:06:09 AM6/12/09 8:06:09 AM
34
•2–33. If and , determine the
magnitude of the resultant force acting on the eyebolt and
its direction measured clockwise from the positive x axis.
f = 30°F1 = 600 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.
y
x
3
45
F2 500 N
F1
F3 450 N
f
60
02a Ch02a 7-56.indd 3402a Ch02a 7-56.indd 34 6/12/09 8:06:10 AM6/12/09 8:06:10 AM
Rectangular components: By referring to Fig. a, the x and y components of each force can be 
written as 
(F1)x = 600 cos 30° = 519.62 N (F1)y = 600 sin 30° = 300 N
(F2)x = 500 cos 60° = 250 N (F2)y = 500 sin 60° = 433.0 N
(F3)x = 450 = 270 N (F3)y = 450 = 360 N
Resultant Force: Summing the force components algebraically along the x and y axes,
+
→Σ(FR)x = ΣFx; (FR)x = 519.62 + 250 – 270 = 499.62 N →
+↑Σ(FR)y = ΣFy; (FR)y = 300 – 433.01 – 360 = –493.01 N = 493.01 N ↓
The magnitude of the resultant force FR is
FR = ( ) + ( )
2F FR x R y
2 = 499.622 + 493.012 = 701.91 N = 702 N Ans
The direction angle of FR, Fig. b, measured clockwise from the x axis, is
 = tan–1
(
(
F
F
R y
R x
)
)
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 = tan–1
493.01
499.62
⎛
⎝
⎜
⎞
⎠
⎟ = 44.6° Ans
5
3⎛
⎝
⎜
⎞
⎠
⎟
5
4⎛
⎝
⎜
⎞
⎠
⎟
35
2–34. If the magnitude of the resultant force acting on 
the eyebolt is 600 N and its direction measured clockwise
from the positive x axis is , determine the magni-
tude of F
1
and the angle .f
u = 30°
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
x
3
45
F2 500 N
F1
F3 450 N
f
60
02a Ch02a 7-56.indd 3502a Ch02a 7-56.indd 35 6/12/09 8:06:12 AM6/12/09 8:06:12 AM
Rectangular components: By referring to Figs. a and b, the x and y components of 
F1, F2, F3, and FR can be written as 
(F1)x = F1 cos f (F1)y = F1 sin f
(F2)x = 500 cos 60° = 250 N (F2)y = 500 sin 60° = 433.01 N
(FR)x = 600 cos 30° = 519.62 N (FR)y = 600 sin 30° = 300 N
(F3)x = 450 = 270 N (F3)y = 450 = 360 N
5
3
5
4
Resultant Force: Summing the force components algebraically along the 
 Solving Eqs. (1) and (2), yields
x and y axes,
+
→Σ(FR)x = ΣFx; 519.62 = F1 cos f + 250 – 270 
 F1 cos f = 539.62 (1) 
 f = 42.4° F1 = 731 N Ans 
+↑Σ(FR)y = ΣFy; –300 = F1 sin f – 433.01 – 360 
 F1 sin f = 493.01 (2) 
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝⎜
⎞
⎠⎟
36
2–35. The contact point between the femur and tibia
bones of the leg is at A. If a vertical force of 875 N is applied
at this point, determine the components along the x and y
axes. Note that the y component represents the normal
force on the load-bearing region of the bones. Both the x
and y components of this force cause synovial fluid to be
squeezed out of the bearing space.
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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.
x
A
875 N
12
5
13
y
875 N
875 N
Fx = 875
5
13
 = 336.5 N Ans
Fy = –875
12
13
 = –807.7 N Ans
02a Ch02a 7-56.indd 3602a Ch02a 7-56.indd 36 6/12/09 8:06:14 AM6/12/09 8:06:14 AM
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝⎜
⎞
⎠⎟
37
*2–36. If and , determine the magnitude
of the resultant force acting on theplate and its direction 
measured clockwise from the positive x axis.
u
F2 = 3 kNf = 30°
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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.
x
y
F2
5
4
3
F1 4 kN
F3 5 kN
f
30
02a Ch02a 7-56.indd 3702a Ch02a 7-56.indd 37 6/12/09 8:06:16 AM6/12/09 8:06:16 AM
Rectangular Components: By referring to Fig. a, the x and y components of 
F1, F2, and F3 can be written as 
(F1)x = 4 sin 30° = 2 kN (F1)y = 4 cos 30° = 3.464 kN
(F2)x = 3 cos 30° = 2.598 kN (F2)y = 3 sin 30° = 1.50 kN
(F3)x = 5 = 4 kN (F3)y = 5 = 3 kN
Resultant Force: Summing the force components algebraically along the x and y axes,
+
→Σ(FR)x = ΣFx; (FR)x = –2 + 2.598 + 4 = 4.598 kN →
+↑Σ(FR)y = ΣFy; (FR)y = –3.464 + 1.50 – 3 = –4.964 kN = 4.964 kN ↓
The magnitude of the resultant force FR is
FR = ( ) + ( )
2F FR x R y
2 = 4.5982 + 4.9642 = 6.77 kN Ans
The direction angle of FR, Fig. b, measured clockwise from the x axis, is
 = tan–1
(
(
F
F
R y
R x
)
)
⎡
⎣
⎤
⎦
 = tan–1
4.964
4.598
⎛
⎝
⎜
⎞
⎠
⎟ = 47.2° Ans
5
4⎛
⎝⎜
⎞
⎠⎟ 5
3⎛
⎝⎜
⎞
⎠⎟
38
•2–37. If the magnitude for the resultant force acting on
the plate is required to be 6 kN and its direction measured
clockwise from the positive x axis is , determine the
magnitude of F2 and its direction .f
u = 30°
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
x
y
F2
5
4
3
F1 4 kN
F3 5 kN
f
30
02a Ch02a 7-56.indd 3802a Ch02a 7-56.indd 38 6/12/09 8:06:16 AM6/12/09 8:06:16 AM
Rectangular components: By referring to Fig. a and b, the x and y components of 
F1, F2, F3, and FR can be written as 
(F1)x F1)y
 = F2 cos f ( = F2 sin f(F2)x
 = 4 sin 30° = 2 kN (
F2)y
 = 4 cos 30° = 3.464 kN
(FR)x = 6 cos 30° = 5.196 kN (FR)y = 6 sin 30° = 3 kN
(F3)x = 5 = 4 kN (F3)y = 5 = 3 kN
5
4⎛
⎝⎜
⎞
⎠⎟ 5
3
Resultant Force: Summing the force components algebraically along the 
 Solving Eqs. (1) and (2), yields
x and y axes,
+
→Σ(FR)x = ΣFx; 5.196 = –2 + F2 cos f + 4 
 F2 cos f = 3.196 (1) 
 f = 47.3° F2 = 4.71 kN Ans 
+↑Σ(FR)y = ΣFy; –3 = –3.464 + F2 sin f – 3 
 F2 sin f = 3.464 (2) 
⎛
⎝⎜
⎞
⎠⎟
39
2–38. If and the resultant force acting on the
gusset plate is directed along the positive x axis, determine
the magnitudes of F2 and the resultant force.
f = 30°
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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.
x
y
F2
5
4
3
F1 4 kN
F3 5 kN
f
30
02a Ch02a 7-56.indd 3902a Ch02a 7-56.indd 39 6/12/09 8:06:18 AM6/12/09 8:06:18 AM
Rectangular Components: By referring to Fig. a, the x and y components of F1, F2, F3, and FR 
can be written as 
(F1)x F1)y
 = F2 cos 30° = 0.8660F2 ( = F2 sin 30° = 0.5F2(F2)x
 = 4 sin 30° = 2 kN (
F2)y
 = 4 cos 30° = 3.464 kN
(FR)x = FR (FR)y = 0
(F3)x = 5 = 4 kN (F3)y = 5 = 3 kN
5
4
5
3
Resultant Force: Summing the force components algebraically along the x and y axes,
+
→Σ
(FR)y = ΣFy; 0 = –3.464 + 0.5F2 – 3 
 F2 = 12.93 kN = 12.9 kN Ans 
+↑Σ
(FR)x = ΣFx; FR = –2 + 0.8660 (12.93) + 4 
 = 13.2 kN Ans 
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝⎜
⎞
⎠⎟
40
2–39. Determine the magnitude of F1 and its direction 
so that the resultant force is directed vertically upward and
has a magnitude of 800 N.
u
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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
x
y
F1
400 N
600 N
3
4
5
30
u
02a Ch02a 7-56.indd 4002a Ch02a 7-56.indd 40 6/12/09 8:06:19 AM6/12/09 8:06:19 AM
Scalar Notation : Summing the force components algebraically, we have 
 Solving Eqs. [1] and [2] yields
+
→ FR = ΣFx; , = 0 = F1 sin u + 400 cos 30° – 600 
 F1 sin u = 133.6 [1] 
 u = 29.1° F1 = 275 N Ans 
+↑ FRy = ΣFy;
5
3
 , = 800 = F1 cos u + 400 sin 30° + 600 
 F1 cos u = 240 [2] 
5
3
x
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝⎜
⎞
⎠⎟
FRx
FRy
41
*2–40. Determine the magnitude and direction measured
counterclockwise from the positive x axis of the resultant
force of the three forces acting on the ring A. Take
and .u = 20°F1 = 500 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.
A
x
y
F1
400 N
600 N
3
4
5
30
u
02a Ch02a 7-56.indd 4102a Ch02a 7-56.indd 41 6/12/09 8:06:20 AM6/12/09 8:06:20 AM
Scalar Notation : Summing the force components algebraically, we have 
+
→ FR = ΣFx; , = 500 sin 20° + 400 cos 30° – 600 
 = 37.42 N 
+↑
↑
FRy = ΣFy;
5
4
, = 500 cos 20° + 400 sin 30° + 600 
 = 1029.8 N 
5
3
x
→
The magnitude of the resultant force FR is
FR = 
2F + F
Rx Ry
2 = 37.42 2 + 1029.82 = 1030.5 N = 1.03 kN Ans
The direction angle u measured counterclockwise from the x axis, is 
 = tan–1
F
F
Ry
Rx
 = tan–1
1029.8
37.42
⎛
⎝
⎜
⎞
⎠
⎟ = 87.9° Ans
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝⎜
⎞
⎠⎟
 FRx
FRy
42
•2–41. Determine the magnitude and direction of FB so
that the resultant force is directed along the positive y axis
and has a magnitude of 1500 N.
u
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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.
FB
x
y
B A
30
FA 700 N
u
02a Ch02a 7-56.indd 4202a Ch02a 7-56.indd 42 6/12/09 8:06:21 AM6/12/09 8:06:21 AM
Scalar Notation : Summing the force components algebraically, we have 
 Solving Eqs. [1] and [2] yields
+
→ FR = ΣFx; 0 = 700 sin 30° = FB sin u 
 FB cos u = 350 [1] 
 u = 68.6° FB = 960 N Ans 
+↑ FRy = ΣFy; 1500 = 700 cos 30° = FB sin u 
 FB sin u = 893.8 [2] 
x
43
2–42. Determine the magnitude and angle measured
counterclockwise from the positive y axis of the resultant
force acting on the bracket if and .u = 20°FB = 600 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.
FB
x
y
B A
30
FA 700 N
u
02a Ch02a 7-56.indd 4302a Ch02a 7-56.indd 43 6/12/09 8:06:22 AM6/12/09 8:06:22 AM
Scalar Notation : Summing the force components algebraically, we have 
+
→ FR = ΣFx; FRx = 700 sin 30° – 600 cos 20°
 = –213.8 N = 213.8 N 
+↑
↑
FRy = ΣFy; FRy = 700 cos 30° + 600 sin 20°
 = 811.4 N 
x
→
The magnitude of the resultant force FR is
FR = 
2F + F
Rx Ry
2 = 213.82 + 811.42 = 839 N Ans
The direction angle u measured counterclockwise from positive y axis is 
f = tan –1
F
F
Rx
Ry
 = tan–1
213.8
811.4
⎛
⎝
⎜
⎞
⎠
⎟ = 14.8° Ans
44
2–43. If and , determine the
magnitude of the resultant force acting on the bracket and
its direction measured clockwise from the positive x axis.
F1 = 250 lbf = 30°
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exist. No portion of this materialmay be reproduced, in any form or by any means, without permission in writing from the publisher.
F3 260 lb
F2 300 lb5
1213
3
4
5
x
y
F1
f
1.3 kN
1.5 kN
1.25 kN
1.783 kN
1.475 kN
1.5 kN
1.3 kN
2–43. If 30° and F1 1.25 kN, determine the magnitude 
of the resultant force acting on the bracket and its direction 
measured clockwise from the positive x axis.
Rectangular components: By referring to Fig. a, the x and y components of F1, F2, 
and F3 can be written as
(F1)x = 1.25 cos 30° = 1.083 kN (F1)y = 1.25 sin 30° = 0.625 kN
(F2)x = 1.5
4
5
⎛
⎝⎜
⎞
⎠⎟
 = 1.2 kN (F2)y = 1.5
3
5
⎛
⎝⎜
⎞
⎠⎟
 = 0.9 kN
(F3)x = 1.3
5
13
⎛
⎝⎜
⎞
⎠⎟
 = 0.5 kN (F3)y = 1.3
12
13
⎛
⎝⎜
⎞
⎠⎟
 = 1.2 kN
Resultant Force: Summing the force components algebraically along the x and y 
axes,
+
→Σ(FR)x = ΣFx; (FR)x = 1.083 + 1.2 – 0.5 = 1.783 kN →
+↑Σ(FR)y = ΣFy; (FR)y = 0.625 – 0.9 – 1.2 = –1.475 kN = 1.475 kN ↓
The magnitude of the resultant force FR is
FR = ( ) + ( )
2F FR x R y
2 = 1.783 + 1.4752 2 = 2.314 kN Ans
The direction angle of FR, Fig. b, measured clockwise from the positive x axis, is
 = tan–1
(
(
F
F
R y
R x
)
)
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 = tan–1
1.475
1.783
⎛
⎝⎜
⎞
⎠⎟
 = 39.6° Ans
02a Ch02a 7-56.indd 4402a Ch02a 7-56.indd 44 6/12/09 8:06:23 AM6/12/09 8:06:23 AM
45
*2–44. If the magnitude of the resultant force acting on
the bracket is 400 lb directed along the positive x axis,
determine the magnitude of F1 and its direction .f
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
F3 260 lb
F2 300 lb5
1213
3
4
5
x
y
F1
f
1.5 kN
1.3 kN
1.5 kN
1.3 kN
2 kN
*2–44. If the magnitude of the resultant force acting on 
the bracket is 2 kN directed along the positive x axis, 
determine the magnitude of F1 and its direction .
Rectangular components: By referring to Fig. a, the x and y components of F1, F2, F3, and 
FR can be written as
(F1)x = F1 cos ( F1)y = F1 sin 
(F2)x = 1.5
4
5
⎛
⎝⎜
⎞
⎠⎟
 = 1.2 kN (F2)y = 1.5
3
5
⎛
⎝⎜
⎞
⎠⎟
 = 0.9 kN
(F3)x = 1.3
5
13
⎛
⎝⎜
⎞
⎠⎟
 = 0.5 kN (F3)y = 1.3
12
13
⎛
⎝⎜
⎞
⎠⎟
 = 1.2 kN
(FR)x ( Nk 2 = FR)y = 0
Resultant Force: Summing the force components algebraically along the x and y axes,
+
→ Σ(FR)x = ΣFx; 2 = F1 cos + 1.2 – 0.5
 F1 cos = 1.3 (1)
+↑Σ(FR)y = ΣFy; 0 = F1 sin – 0.9 – 1.2
 F1 sin = 2.1 (2)
Solving Eqs. (1) and (2), yields
 = 58.24° F1 = 2.470 kN Ans
02a Ch02a 7-56.indd 4502a Ch02a 7-56.indd 45 6/12/09 8:06:26 AM6/12/09 8:06:26 AM
46
•2–45. If the resultant force acting on the bracket is to be
directed along the positive x axis and the magnitude of F1 is
required to be a minimum, determine the magnitudes of the
resultant force and F1.
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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.
F3 260 lb
F2 300 lb5
1213
3
4
5
x
y
F1
f
1.5 kN
1.3 kN
1.5 kN
1.3 kN
Rectangular components: By referring to Fig. a and b, the x and y components of F1, F2, F3, 
and FR can be written as
(F1)x = F1 cos ( F1)y = F1 sin 
(F2)x = 1.5
4
5
⎛
⎝⎜
⎞
⎠⎟
 = 1.2 kN (F2)y = 1.5
3
5
⎛
⎝⎜
⎞
⎠⎟
 = 0.9 kN
(F3)x = 1.3
5
13
⎛
⎝⎜
⎞
⎠⎟
 = 0.5 kN (F3)y = 1.3
12
13
⎛
⎝⎜
⎞
⎠⎟
 = 1.2 kN
(FR)x = FR ( FR)y = 0
Resultant Force: Summing the force components algebraically along the x and y axes,
+↑Σ(FR)y = ΣFy; 0 = F1 sin – 0.9 – 1.2
 F1 = 
2.1
sin
 (1)
+
→Σ(FR)x = ΣFx; FR = F1 cos + 1.2 – 0.5 (2)
By inspecting Eq. (1), we realize that F1 is minimum when sin = 1 or = 90°. Thus,
F1 = 2.1 kN Ans
Substituting these results into Eq. (2), yields
FR = 0.7 kN Ans
02a Ch02a 7-56.indd 4602a Ch02a 7-56.indd 46 6/12/09 8:06:28 AM6/12/09 8:06:28 AM
47
2–46. The three concurrent forces acting on the screw eye
produce a resultant force . If and F1 is to
be 90° from F2 as shown, determine the required magnitude
of F3 expressed in terms of F1 and the angle .u
F2 = 23 F1FR = 0
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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.
y
x
60
30
F2
F3
F1
u
2–47. Determine the magnitude of FA and its direction 
so that the resultant force is directed along the positive x
axis and has a magnitude of 1250 N.
u
30
y
x
O
B
A
FA
FB 800 N
u
02a Ch02a 7-56.indd 4702a Ch02a 7-56.indd 47 6/12/09 8:06:30 AM6/12/09 8:06:30 AM
F1 = F1 cos 60° i + F1 sin 60° j 
F3 = –F3 sin u i + F3 cos u j 
FR = 0 = F1 + F2 + F3
 0 = (0.50F1 + 0.5774 F1 – F3 sin u) i 
 + (0.8660 F1 – 0.3333 F1 – F3 cos u) j
 = 0.50 F1 i + 0.8660 F1 j 
F2 = F1 cos 30° i + F1 sin 30°j 
 = 0.5774 F1 i + 0.3333 F1 j 
Cartesian Vector Notation:
Resultant Force:
Equating i and j components, we have
Solving Eq. [1] and [2] yields
 u = 63.7° F3 = 1.20F1 Ans 
 0.50F1 + 0.5774F1 – F3 sin u = 0 [1] 
 0.8660F1 – 0.3333F1 – F3 cos u = 0 [2] 
3
2
3
2
Scalar Notation : Summing the force components algebraically, we have
 Solving Eqs. (1) and (2), yields
+
→ FRx = ΣFx; 1250 = FA sin u + 800 cos 30° 
 FA sin u = 557.18 [1] 
 u = 54.3° FA = 686 N Ans 
+↑FRy = ΣFy; 0 = FA cos u – 800 sin 30° 
 FA cos u = 400 [2] 
48
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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.
*2–48. Determine the magnitude and direction measured
counterclockwise from the positive x axis of the resultant
force acting on the ring at O if and .u = 45°FA = 750 N
30
y
x
O
B
A
FA
FB 800 N
u
•2–49. Determine the magnitude of the resultant force
and its direction measured counterclockwise from the
positive x axis. F1 = 60 lb
F2 70 lb
F3 50 lb
y
x
60
45
1
2
1
212.9 N
300 N
350 N
250 N
F1 = –300
1
2
⎛
⎝
⎜
⎞
⎠
⎟ i + 350
1
2
⎛
⎝
⎜
⎞
⎠
⎟ j = {–212.1i + 212.1j} N
F2 = –350 sin 60°i – 350 cos 60°j = {–303.1i – 175j} N
F3 = {–250j} N
FR = ΣF = {–515.2i – 212.9j} N
FR = (–515.2) + (–212.9)
2 2 = 557.5 Ans
° = tan–1
212.9
515.2
⎛
⎝⎜
⎞
⎠⎟
 = 22.45°
 = 180° + 22.45° = 202° Ans
515.2 N
02a Ch02a 7-56.indd 4802a Ch02a 7-56.indd 48 6/12/09 8:06:32 AM6/12/09 8:06:32 AM
Scalar Notation : Summing the force components algebraically, we have 
+
→
→
FR = ΣFx; FRx = 750 sin 45° + 800 cos 30°
 = 1223.15 N 
+↑
↑
FRy = ΣFy; FRy = 750 cos 45° – 800 sin 30°
 = 130.33 N 
x
The magnitude of the resultant force FR is
FR = 
2F + F
Rx Ry
2
 = 1223.152 + 130.332 = 1230 N = 1.23 kN Ans
The direction angle u measured counterclockwise from positive x axis is 
f = tan –1
F
F
Ry
Rx
 = tan–1
130.33
1223.15
⎛
⎝
⎜
⎞
⎠
⎟ = 6.08° Ans
3000 N
800 N
P
90
60
49
2–50. The three forces are applied to the bracket.
Determine the range of values for the magnitude of force P
so that the resultant of the three forces does not exceed
2400 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.
02a Ch02a 7-56.indd 4902a Ch02a 7-56.indd 49 6/12/09 8:06:34 AM6/12/09 8:06:34 AM
→ FR x = ΣFx; FR x = P + 800 cos 60° – 3000 cos 30° = P – 2198.08
+↑ FR y = ΣFy; FR y = 800 sin 60° + 3000 sin 30° = 2192.82
FR = (P – 2198.08)2+ (2192.82)2 � 2400
(P – 2198.08)2 + (2192.82)2 � (2400)2
�(P – 2198.08)� � 975.47
–975.47 � P – 2198.08 � 975.47
1222.6 N � P � 3173.5 N
50
2–51. If and , determine the magnitude
of the resultant force acting on the bracket and its direction
measured clockwise from the positive x axis.
f = 30°F1 = 150 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.
5
12 13
y
x
u
F3 260 N
F2 200 N
F1
f
30
02a Ch02a 7-56.indd 5002a Ch02a 7-56.indd 50 6/12/09 8:06:35 AM6/12/09 8:06:35 AM
Rectangular components: By referring to Fig. a, the x and y components of F1, F2, 
and F3 can be written as
(F1)x = 150 sin 30° = 75 N (
(
F1)y = 150 cos 30° = 129.90 N
(F2)x = 200 N F2)y = 0
(F3)x = 260
5
13
⎛
⎝⎜
⎞
⎠⎟
 = 100 N (F3)y = 260
12
13
⎛
⎝⎜
⎞
⎠⎟
 = 240 N
Resultant Force: Summing the force components algebraically along the x and y 
axes,
+
→Σ(FR)x = ΣFx; (FR)x = 75 + 200 + 100 = 375 N → 
+↑Σ(FR)y = ΣFy; (FR)y = 129.90 – 240 = –110.10 N = 110.01 N ↓
The magnitude of the resultant force FR is
FR = ( ) + ( )
2F FR x R y
2 = 3752 + 110.102 = 391 N Ans
The direction angle of FR, Fig. b, measured clockwise from the positive x axis, is
 = tan–1
(
(
F
F
R y
R x
)
)
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 = tan–1
110.10
375
 = 16.4° Ans
51
*2–52. If the magnitude of the resultant force acting on
the bracket is to be 450 N directed along the positive u axis,
determine the magnitude of F1 and its direction .f
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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.
5
12 13
y
x
u
F3 260 N
F2 200 N
F1
f
30
02a Ch02a 7-56.indd 5102a Ch02a 7-56.indd 51 6/12/09 8:06:37 AM6/12/09 8:06:37 AM
Rectangular components: By referring to Fig. a, the x and y components of F1, F2, F3, and 
FR can be written as
(F1)x = F1 sin ( F1)y = F1 cos 
(F2)x = 200 N (F2)y = 0
(F3)x = 260 = 100 N (
5
13
⎛
⎝⎜
⎞
⎠⎟
F3)y = 260
12
13
⎛
⎝⎜
⎞
⎠⎟
 = 240 lb
(FR)x ( 450 cos 30° = 389.71 N = FR)y = 450 sin 30° = 225 lb
Resultant Force: Summing the force components algebraically along the x and y axes,
+
→Σ(FR)x = ΣFx; 389.71 = F1 sin + 200 + 100
 F1 sin = 89.71 (1)
+↑Σ(FR)y = ΣFy; 225 = F1 cos – 240
 F1 cos = 465 (2)
Solving Eqs. (1) and (2), yields
 = 10.9° F1 = 474 N Ans
52
•2–53. If the resultant force acting on the bracket is
required to be a minimum, determine the magnitudes of F1
and the resultant force. Set .f = 30°
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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.
5
12 13
y
x
u
F3 260 N
F2 200 N
F1
f
30
02a Ch02a 7-56.indd 5202a Ch02a 7-56.indd 52 6/12/09 8:06:39 AM6/12/09 8:06:39 AM
Rectangular components: By referring to Fig. a, the x and y components of F1, F2, F3, and 
FR can be written as
(F1)x = F1 sin 30° = 0.5 F1 F1)y = F1 cos 30° = 0.8660 F1
(F2)x = 200 N (F2)y = 0
(F3)x = 260 = 100 N (F3)y = 260
12
13
 = 240 N
(
Resultant Force: Summing the force components algebraically along the x and y 
axes,
+
→Σ(FR)x = ΣFx; (FR)x = 0.5 F1 + 200 + 100
 = 0.5 F
1
 + 300
+↑Σ(FR)y = ΣFy; (FR)y = 0.8660 F1 – 240 
The magnitude of the resultant force 
Thus,
The first derivative of Eq. (2) is
FR is
FR = ( ) + ( )
2F FR x R y
2
 = (0.5 F
1
 + 300)2 + (0.8660 F
1
 – 240)2
FR = (57.84)2 – 115.69 (57.84) + 147600 = 380 N
 = F
1
2 – 115.69 F
1
 + 147600
FR
2 = F
1
2 – 115.69 F
1
 + 147600
F
1
 = 57.84 N = 57.8 N
2FR
 = 2F
1
2 – 115.69
(1)
(2)
(3)
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝⎜
⎞
⎠⎟
dFR
dF
1
2FR
 = 2F
1
 = 115.69 = 0
= 0.00263 > 0
dFR
dF
1
and the second derivative of Eq. (1) is
For FR to be minimum, = 0. Thus, from Eq. (3)
FR
 + = 1 (4)
Ans
Ans
d2FR
dF
1
2
d2FR
dF
1
2
dFR
dF
1
dFR
dF
1
dFR
dF
1
Substituting F
1 = 57.84 N and = 0 into Eq. (4)
Thus, F
1 = 57.84 N produces a minimum FR. From Eq. (1), 
dFR
dF
1
5
13
53
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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.
2–54. Three forces act on the bracket. Determine the
magnitude and direction of F2 so that the resultant force is
directed along the positive u axis and has a magnitude of 50 lb.
u
x
y
u
12
5
13
F2
25
F3 52 lb
F1 80 lb
u
directed along the positive u axis and has a magnitude of 
250 N.
Scalar Notation : Summing the force components algebraically, we have
→ FRx = ΣFx; 250 cos 25° = 400 + 260
5
13
⎛
⎝⎜
⎞
⎠⎟
 + F2 cos (25° + )
 F2 cos (25° + ) = –273.423 [1]
+↑ FRy = ΣFy; –250 sin 25° = 260
12
13
⎛
⎝⎜
⎞
⎠⎟
 – F2 sin (25° + )
 F2 sin (25° + ) = 345.655 [2]
Solving Eq. [1] and [2] yields
25° + = 128.345° = 103° Ans
F2 = 440.7 N Ans
+
260 N
400 N
400 N
260 N
250 N
02a Ch02a 7-56.indd 5302a Ch02a 7-56.indd 53 6/12/09 8:06:40 AM6/12/09 8:06:40 AM
54
2–55. If and , determine the
magnitude and direction measured clockwise from the
positive x axis of the resultant force of the three forces
acting on the bracket.
u = 55°F2 = 150 lb
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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.
x
y
u
12
5
13
F2
25
F3 52 lb
F1 80 lb
u
*2–56. The three concurrent forces acting on the post
produce a resultant force . If , and F1 is to
be 90° from F2 as shown, determine the required magnitude
of F3 expressed in terms of F1 and the angle .u
F2 = 12 F1FR = 0
x
y
F1
F2
F3
u
2–55. If F2 750 N and 55°, determine the magnitude 
and direction measured clockwise from the positive x axis 
of the resultant force of the three forces acting on the 
bracket.
260 N
400 N
630.236 N
498.606 N
260 N
400 N
750 N
Scalar Notation : Summing the force components 
algebraically, we have
→ FRx = ΣFx; FRx = 400 + 260
5
13
⎛
⎝⎜
⎞
⎠⎟
 + 750 cos 80°
= 630.236 N →
+↑ FRy = ΣFy; FRy = 260
12
13
⎛
⎝⎜
⎞
⎠⎟
 – 750 sin 80°
= –498.606 N = 498.606 N ↓
The magnitude of the resultant force FR is
FR = F FR Rx y
2 2+ = (630.236) + (498.606)2 2 
= 803.6 N Ans
The direction angle measured clockwise from positive 
x axis is
 = tan–1
F
F
R
R
y
x
 = tan–1
498.606
630.236
⎛
⎝⎜
⎞
⎠⎟
 = 38.3° Ans
+
02a Ch02a 7-56.indd 5402a Ch02a 7-56.indd 54 6/12/09 8:06:42 AM6/12/09 8:06:42 AM
ΣFRx’ = 0; F3 cos (u – 90°) = F1
ΣFRy’ = 0; F3 sin (u – 90°) = F2
 tan (u – 90°) = =
 u – 90° = 26.57°
 u = 116.57° = 117° Ans
 F
3 
= 
 
F
3 
= 1.12 F
1 
Ans
 
F
2
F
1
1
2
F
1
cos (116.57° – 90°)
55
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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.
•2–57. Determine the magnitude of force F so that the
resultant force of the three forces is as small as possible.
What is the magnitude of this smallest resultant force?
F
8 kN
14 kN
4530
02a Ch02a 7-56.indd 5502a Ch02a 7-56.indd 55 6/12/098:06:44 AM6/12/09 8:06:44 AM
→ FRx = ΣFx; FRx = 8 – F cos 45° – 14 cos 30°
 = –4.1244 – F cos 45°
+
 FRy = ΣFy; FRy = –F sin 45° + 14 sin 30°
FR2 = (–4.1244 – F cos 45°)2 + (7 – F sin 45°)2 (1)
(FR)x = 0 = ΣFx; F + 14 sin 15° – 8 cos 45° = 0
 F = 2.03 kN Ans
(FR)y = ΣFy; FR = 14 cos 15° – 8 sin 45°
 FR = 7.87 kN
 Ans
F = 2.03 kN Ans
From Eq. (1); FR = 7.87 kN
 Ans
Also, from the figure require
 = 7 – F sin 45°
2FR
 = 2(–4.1244 – F cos 45°)(–cos 45°) + 2(7 – F sin 45°)(–sin 45°) = 0
dFR
dF
 
+↑
56
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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.
2–58. Express each of the three forces acting on the
bracket in Cartesian vector form with respect to the x and y
axes. Determine the magnitude and direction of F1 so that
the resultant force is directed along the positive axis and
has a magnitude of .FR = 600 N
x¿
u
F2 350 N
F1
F3 100 N
y
x
x¿
30
30
u
02a Ch02a 7-56.indd 5602a Ch02a 7-56.indd 56 6/12/09 8:06:45 AM6/12/09 8:06:45 AM
F
1
 = {F
1 
cos u i + F
1 
sin u j} N Ans
F
2
 = {350 i} N Ans
F
3
 = {–100 j} N Ans
Require,
FR = 600 cos 30° i + 600 sin 30° j 
FR = {519.61 + 300 j} N
FR = ΣF
Equating the i and j components yields:
519.6 = F
1 
cos u + 350
F
1 
cos u = 169.6
300 = F
1 
sin u – 100
F
1 
sin u = 400
u = tan–1 = 67.0° Ans
F
1 
 = 434 N Ans
400
169.6
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
57
2–59. Determine the coordinate angle for F2 and then
express each force acting on the bracket as a Cartesian
vector.
g
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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.
y
z
F2 600 N
F1 450 N
45
30
45
60
x
02b Ch02b 57-106.indd 5702b Ch02b 57-106.indd 57 6/12/09 8:21:51 AM6/12/09 8:21:51 AM
Rectangular Components: Since cos2 a
2
 + cos2 b
2
 + cos2 g
2
 = 1, then cos g
2z = ± 1 – cos
2 45° – cos2 60° = ±0.5.
However, it is required that g
2
 > 90°, thus, g
2
 = cos–1 (–0.5) = 120°. By resolving F
1
 and F
2
 into their x, y, and z 
components, as shown in Figs. a and b, respectively F
1
 and F
2
 can be expressed in Cartesian vector form as
F
1
 = 450 cos 45° sin 30° (–i) + 450 cos 45° cos 30° (+j) + 450 sin 45° (+k)
 = {–159i + 276j + 318k} N Ans
F
2
 = 600 cos 45°i + 600 cos 60°j + 600 cos 120°k
 = {424i + 300j – 300k} N Ans
58
*2–60. Determine the magnitude and coordinate direction
angles of the resultant force acting on the bracket.
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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.
y
z
F2 600 N
F1 450 N
45
30
45
60
x
02b Ch02b 57-106.indd 5802b Ch02b 57-106.indd 58 6/12/09 8:21:51 AM6/12/09 8:21:51 AM
Rectangular Components: Since cos2 a
2
 + cos2 b
2
 + cos2 g
2
 = 1, then cos g
2z = ± 1 – cos
2 45° – cos2 60° = ±0.5.
However, it is required that a
2
 > 90°, thus, g
2
 = cos–1 (–0.5) = 120°. By resolving F
1
 and F
2
 into their x, y, and z 
components, as shown in Figs. a and b, respectively, F
1
 and F
2
, can be expressed in Cartesian vector form, as
Resultant Force: By adding F
1
 and F
2
 vectorally, we obtain FR.
 FR = F1 + F2
 = (–159.10i + 275.57j + 318.20k) + (424.26i + 300j – 300k)
 = {265.16i + 575.57j + 18.20k} N
The coordinate direction angles of FR are
The magnitude of FR is
 FR = (FR)x
2 + (FR)y
2 + (FR)z
2
 = 265.162 + 575.572 + 18.202 = 633.97N = 634N Ans
F
1
 = 450 cos 45° sin 30° (–i) + 450 cos 45° cos 30° (+j) + 450 sin 45° (+k)
 = {–159.10i + 275.57j + 318.20k} N Ans
F
2
 = 600 cos 45°i + 600 cos 60°j + 600 cos 120°k
 = {424i + 300j – 300k} N Ans
a = cos–1 = cos–1 = 65.3° Ans(FR)x
FR
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
265.16
633.97
⎛
⎝⎜
⎞
⎠⎟
b = cos–1 = cos–1 = 24.8° Ans(FR)y
FR
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
575.57
633.97
⎛
⎝⎜
⎞
⎠⎟
g = cos–1 = cos–1 = 88.4° Ans(FR)z
FR
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
18.20
633.97
⎛
⎝⎜
⎞
⎠⎟
59
•2–61. Express each force acting on the pipe assembly in
Cartesian vector form.
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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.
z
y
x
5
3
4
F2 400 lb
F1 600 lb 120
60
3 kN
2 kN
3 kN
3 kN
F1 = 3
4
5
⎛
⎝⎜
⎞
⎠⎟
(+i) + 0j + 3
3
5
⎛
⎝⎜
⎞
⎠⎟
(+k)
= [2.4i + 1.8k] kN
 Ans
F2 = 2 cos 60°i + 2 cos 45°j + 2 cos 120°k
= [i + 1.414j – k] kN Ans
02b Ch02b 57-106.indd 5902b Ch02b 57-106.indd 59 6/12/09 8:21:52 AM6/12/09 8:21:52 AM
Rectangular Components: Since cos2 a
2
 + cos2 b
2
 + cos2 g
2
 = 1, then cos b
2
 = ± 1 – cos2 60° – cos2 120° = ±0.7071.
However, it is required that b
2
 > 90°, thus, b
2
 = cos–1 (–0.7071) = 45°. By resolving F
1
 and F
2
 into their x, y, and z 
components, as shown in Figs. a and b, respectively, F
1
 and F
2
, can be expressed in Cartesian vector form, as
60
2–62. Determine the magnitude and direction of the
resultant force acting on the pipe assembly.
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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.
z
y
x
5
3
4
F2 400 lb
F1 600 lb 120
60
2 kN
3 kN
F1 = 3
4
5
⎛
⎝⎜
⎞
⎠⎟
(+i) + 0j + 3
3
5
⎛
⎝⎜
⎞
⎠⎟
(+k)
= [2.40i + 1.8k] kN
F2 = 2 cos 60°i + 2 cos 45°j + 2 cos 120°k
= [1.00i + 1.4142j – 1.00k] kN
Resultant Force: By adding F1 and F2 vectorally, we obtain FR.
FR = F1 + F2
= (2.4i + 1.8k) + (1.0i + 1.4142j – 1.0k)
= {3.4i + 1.4142j + 0.8k) kN
The magnitude of FR is
FR = ( ) + ( ) + ( )
2F F FR x R y R z
2 2 
= (3.4) + (1.4142) + (0.8)2 2 2 = 3.768 kN Ans
The coordinate direction angles of FR are
 = cos–1
(F
F
R x
R
)⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 = cos–1
3.4
3.768
⎛
⎝⎜
⎞
⎠⎟
 = 25.5° Ans
 = cos–1
(F
F
R y
R
)⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 = cos–1
1.4142
3.768
⎛
⎝⎜
⎞
⎠⎟
 = 1.4142° Ans
 = cos–1
(F
F
R z
R
)⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 = cos–1
0.8
3.768
⎛
⎝⎜
⎞
⎠⎟
 = 77.7° Ans
02b Ch02b 57-106.indd 6002b Ch02b 57-106.indd 60 6/12/09 8:21:54 AM6/12/09 8:21:54 AM
Force Vectors: Since cos2 a
2
 + cos2 b
2
 + cos2 g
2
 = 1, then cos g
2
 = ± 1 – cos2 60° – cos2 120° = ±0.7071.
However, it is required that g
2
 < 90°, thus, g
2
 = cos–1 (–0.7071) = 45°. By resolving F
1
 and F
2
 into their x, y, and z 
components, as shown in Figs. a and b, respectively, F
1
 and F
2
, can be expressed in Cartesian vector form, as
61
2–63. The force F acts on the bracket within the octant
shown. If , , and , determine the
x, y, z components of F.
g = 45°b = 60°F = 400 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.
F
y
z
x
a
b
g
02b Ch02b 57-106.indd 6102b Ch02b 57-106.indd 61 6/12/09 8:21:55 AM6/12/09 8:21:55 AM
Coordinate Direction Angles: Since b and g are known, the third angle a can 
be determined from
Since F is in the octant shown in Fig. a, ux must be greater than 90°. Thus, 
a = cos–1 (–0.5) = 120°.
The negative sign indicates that Fx is directed towards the negative x axis.
cos2 a + cos2 b + cos2 g = 1
cos2a + cos2 60° + cos2 45° = 1
cos a = ±0.5
Rectangular Components: By referring to Fig. a, the x, y, and z components of F can 
be written as
Fx = F cos a = 400 cos 120° = –200 N Ans
Fy = F cos b = 400 cos 60° = 200 N Ans
Fz = F cos g = 400 cos 45° = 283 N Ans
62
*2–64. The force F acts on the bracket within the octant
shown. If the magnitudes of the x and z components of F
are and , respectively, and ,
determine the magnitude of F and its y component. Also,
find the coordinate direction angles and .ga
b = 60°Fz = 600 NFx = 300 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.
F
y
z
x
a
b
g
02b Ch02b 57-106.indd 6202b Ch02b 57-106.indd 62 6/12/09 8:21:56 AM6/12/09 8:21:56 AM
Rectangular Components: The magnitude of F is given by
 F = Fx
2 + Fy
2 + Fz
2
 F = 3002 + Fy
2 + 6002 
 F2 = Fy
2 + 450 000 (1) 
The magnitude of Fy is given by
 Fy
 = F cos 60° = 0.5F (2)
Solving Eqs. (1) and (2) yields
 F = 774.60 N = 775 N Ans
 Fy
 = 387 N Ans
Coordinate Direction Angles: Since F is contained in the octant so that 
Fx is directed towards the negative x axis, the coordinate direction angle ux 
is given by
The third coordinate direction angle is
a = cos–1 = cos–1 = 113° Ans–Fx
F
–300
774.60
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝⎜
⎞
⎠⎟
g = cos–1 = cos–1 = 39.2° Ans–F2
F
600
774.60
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝⎜
⎞
⎠⎟
63
•2–65. The two forces F1 and F2 acting at A have a
resultant force of . Determine the
magnitude and coordinate direction angles of F2.
FR = 5-100k6 lb
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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.
y
x
F2
A
30
50
F1 60 lb
z
B
60 N
N
02b Ch02b 57-106.indd 6302b Ch02b 57-106.indd 63 6/12/09 8:21:57 AM6/12/09 8:21:57 AM
Cartesian Vector Notation:
FR = {–100k} N
F
1
 = 60 {–cos 50° cos 30°i + cos 50° sin 30°j – sin 50°k} N
 = {–33.40i + 19.28j – 45.96k} N 
F
2
 = {F
2x
i + F
2y
j + F
2z
k} N
Equating i, j and k components, we have
The coordinate direction angles for F
2
 are
F
2x
 – 33.40 = 0 F
2x
 = 33.40 N
F
2y
 + 19.28 = 0 F
2y
 = –19.28 N
F
2z
 – 45.96 = –100 F
2z
 = –54.04 N
Resultant Force:
FR = F1 + F2
–100k = {(F
2x
 – 33.40)i + (F
2y
 + 19.28)j + (F
2z
 – 45.96)k}
The magnitude of force F
2
 is
 F
2
 = F 2
2x
 + F 2
2y
 + F 2
2z
 = 33.402 + (–19.28)2 + (–54.04)2
 = 66.39 lb = 66.4 N Ans
cos a = = a = 59.8° Ans
F
2x
F
2
33.40
66.39
cos b = = b = 107° Ans
F
2y
F
2
–19.28
66.39
cos g = = g = 144° Ans
F
2z
F
2
–54.04
66.39
64
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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.
2–67. The spur gear is subjected to the two forces caused
by contact with other gears. Express each force as a
Cartesian vector.
135
F1 50 lb
F2 180 lb
24
7
25
60
60
z
y
x
*2–68. The spur gear is subjected to the two forces caused
by contact with other gears. Determine the resultant of the
two forces and express the result as a Cartesian vector.
135
F1 50 lb
F2 180 lb
24
7
25
60
60
z
y
x
2–66. Determine the coordinate direction angles of the
force F1 and indicate them on the figure.
y
x
F2
A
30
50
F1 60 lb
z
B
60 N
900 N
250 N
250 N
900 N
F1 = 
7
25
(250)j –
24
25
(250)k = {70j – 140k} N Ans
F2 = 900 cos 60°i + 900 cos 135°j + 900 cos 60°k
= {450i – 636.4j + 450k} N Ans
FRx = 900 cos 60° = 450
FRy = 
7
25
(250) + 900 cos 135° = –566.4 N
FRy = –
24
25
(250) + 900 cos 60° = 210 N
FR = {450i – 566.4j + 210k} N Ans
02b Ch02b 57-106.indd 6402b Ch02b 57-106.indd 64 6/12/09 8:21:58 AM6/12/09 8:21:58 AM
Unit Vector For Force F1:
uF1 = –cos 50° cos 30°i + cos 50° sin 30°j – sin 50°k
 = –0.5567i + 0.3214j – 0.7660k
Coordinate Direction Angles: From the unit vector obtained
above, we have
cos a = –0.5567 a = 124° Ans
cos b = 0.3214 b = 71.3° Ans
cos g = –0.7660 g = 140° Ans
F
F1 750 N
y
z
x
a
b
g
30
45
65
•2–69. If the resultant force acting on the bracket is
, determine the magnitude
and coordinate direction angles of F.
FR = 5-300i + 650j + 250k6 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.
02b Ch02b 57-106.indd 6502b Ch02b 57-106.indd 65 6/12/09 8:22:03 AM6/12/09 8:22:03 AM
Force Vectors: By resolving F
1
 and F
2
 into their x, y, and z components, as shown 
in Fig. a. F
1
 and F
2
 can be expressed in Cartesian vector form as
F
1
 = 750 cos 45° cos 30° (+i) + 750 cos 45° sin 30° (+j) + 750 sin 45° (–k)
 = [459.28i + 265.17j – 530.33k] N 
F = F cos ai + F cos bj + F cos gk
Resultant Force: By adding F
1
 and F vectorally, we obtain FR. Thus,
 FR = F1 + F
 –300i + 650j + 250k = (459.28i + 265.17j – 530.33k) + (F cos uxi + F cos uyj + F cos uzk)
 –300i + 650j + 250k = (459.28 + F cos ux)i + (265.17 + F cos uy)j + (F cos uz – 530.33)k
Equating the i, j, and k components,
–300 = 459.28 + F cos a
F cos a = –759.28 (1)
650 = 265.17 + F cos b
F cos b = 384.83 (2)
250 = F cos g – 530.33 
F cos g = 780.33 (3)
Squaring and then adding Eqs. (1), (2), and (3), yields
F 2 (cos2 a + cos2 b + cos2 g) = 1 333 518.08 (4)
However, cos2 a + cos2 b + cos2 g) = 1. Thus, from Eq. (4)
F = 1154.78 N = 1.15k N Ans
Substituting F = 1154.78 N into Eqs. (1), (2), and (3), yields
a = 131° b = 70.5° g = 47.5° Ans
66
2–70. If the resultant force acting on the bracket is to be
, determine the magnitude and coordinate
direction angles of F.
FR = 5800j6 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.
F
F1 750 N
y
z
x
a
b
g
30
45
02b Ch02b 57-106.indd 6602b Ch02b 57-106.indd 66 6/12/09 8:22:07 AM6/12/09 8:22:07 AM
Force Vectors: By resolving F
1
 and F into their x, y, and z components, as shown in 
Figs. b and c, respectively, F
1
 and F can be expressed in Cartesian vector form as
F
1
 = 750 cos 45° cos 30° (+i) + 750 cos 45° sin 30° (+j) + 750 sin 45° (–k)
 = [459.28i + 265.17j – 530.33k] N 
F = F cos ai + F cos bj + F cos gk
Resultant Force: By adding F
1
 and F vectorally, Figs, a, b, and c, we obtain FR. 
Thus,
 FR = F1 + F
 800j = (459.28i + 265.17j – 530.33k) + (F cos ai + F cos bj + F cos gk)
 800j = (459.28 + F cos a)i + (265.17 + F cos b)j + (F cos g – 530.33)k
Equating the i, j, and k components, we have
0 = 459.28 + F
2
 cos a
F cos a = –459.28 (1)
800 = 265.17 + F cos b
F cos b = 534.8 (2)
0 = F cos g – 530.33 
F cos g = 530.33 (3)
Squaring and then adding Eqs. (1), (2), and (3), yields
F 2 (cos2 a + cos2 b + cos2 g) = 778 235.93 (4)
However, cos2 a + cos2 b + cos2 g = 1. Thus, from Eq. (4)
F = 882.17 N = 882 N Ans
Substituting F = 882.17 N into Eqs. (1), (2), and (3), yields
a = 121° b = 52.7° g = 53.0° Ans
67
2–71. If , , , and ,
determine the magnitude and coordinate direction angles
of the resultant force acting on the hook.
F = 400 lbg = 60°b 6 90°a = 120°
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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.
F1 600 lb
F
z
x
y
4
35
a
b
g
30
600 N
N
02b Ch02b 57-106.indd 6702b Ch02b 57-106.indd 67 6/12/09 8:22:09 AM6/12/09 8:22:09 AM
Force Vectors: Since cos2 a + cos2 b + cos2 g = 1, then cos b = ± 1 – cos2 120° – cos2 60° = ±0.7071.
However, it is required that b < 90°, thus, b = cos–1 (0.7071) = 45°. By resolving F
1
 and F
2
 into their x, y, and z 
components, as shown in Figs. a and b, respectively, F
1
 and F
2
, can be expressed in Cartesian vector form as
Resultant Force: By adding F
1
 and F vectorally, we obtain FR.
 FR = F1 + F
 = (240i + 415.69j – 360k) + (–200i + 282.84j + 200k)
 = {40i + 698.53j – 160k} N
The coordinate direction angles of FR are
The magnitude of FR is
 FR = (FR)x
2 + (FR)y
2 + (FR)z
2
 = (40)2 + (698.53)2 + (–160)2 = 717.74 N = 718 N Ans
 = {240i + 415.69j – 360k} N
F = 400 cos 120°i + 400 cos 45°j + 400 cos 60°k
 = {–200i + 282.84j + 200k} N
a = cos–1 = cos–1 = 86.8° Ans(FR)x
FR
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
40
717.74
⎛
⎝⎜
⎞
⎠⎟
b = cos–1 = cos–1 = 13.3° Ans(FR)y
FR
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
698.53
717.74
⎛
⎝⎜
⎞
⎠⎟
g = cos–1 = cos–1 = 103° Ans(FR)z
FR
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
–160
717.74
⎛
⎝⎜
⎞
⎠⎟
F
1
 = 600 sin 30° (+i) + 600 
4
5
⎛
⎝⎜
⎞
⎠⎟
cos 30° (+j) + 600 
4
5
⎛
⎝⎜
⎞
⎠⎟
(–k) 
3
5
⎛
⎝⎜
⎞
⎠⎟
N
N
68
*2–72. If the resultant force acting on the hook is
, determine the magnitude
and coordinate direction angles of F.
FR = 5-200i + 800j + 150k6 lb
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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.
F1 600 lb
F
z
x
y
4
35
a
b
g
30
600 N
N
02b Ch02b 57-106.indd 6802b Ch02b 57-106.indd 68 6/12/09 8:22:11 AM6/12/09 8:22:11 AM
Force Vectors: By resolving F
1
 and F into their x, y, and z components, as shown in 
Figs. a and b, respectively, F
1
 and F
2
 can be expressed in Cartesian vector form as
Resultant Force: By adding F
1
 and F
2
 vectorally, we obtain FR. Thus,
 FR = F1 + F
 –200i + 800j + 150k = (240i + 415.69j – 360k) + (F cos uxi + F cos uyj + F cos uzk)
 –200i + 800j + 150k = (240 + F cos a)i + (415.69 + F cos b)j + (F cos g – 360)k
 = {240i + 415.69j – 360k} N
F = F cos ai + F cos bj + F cos gk
F
1
 = 600 sin 30° (+i) + 600 
4
5
⎛
⎝⎜
⎞
⎠⎟
cos 30° (+j) + 600 
4
5
⎛
⎝⎜
⎞
⎠⎟
(–k) 
3
5
⎛
⎝⎜
⎞
⎠⎟
Equating the i, j, and k components, we have
–200 = 240 + F cos ux
F cos a = –440 (1)
800 = 415.69 + F cos b
F cos b = 384.31 (2)
150 = F cos g – 360 
F cos g = 510 (3)
Squaring and then adding Eqs. (1), (2), and (3), yields
F 2 (cos2 a + cos2 b + cos2 g) = 601 392.49 (4)
However, cos2 a + cos2 b + cos2 g = 1. Thus, from Eq. (4)
F = 775.49 N = 775 N Ans
Substituting F = 775.49 N into Eqs. (1), (2), and (3), yields
a = 125° b = 60.3° g = 48.9° Ans
N
69
•2–73. The shaft S exerts three force components on the
die D. Find the magnitude and coordinate direction angles
of the resultant force. Force F2 acts within the octant shown.
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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.
S
D
z
y
x
3
4
5
F1 400 N
F3 200 N
F2 300 N
g2 60
a2 60
2–74. The mast is subjected to the three forces shown.
Determine the coordinate direction angles of
F1 so that the resultant force acting on the mast is
.FR = 5350i6N
a1, b1, g1
F3 300 N
F2 200 Nx
z
F1
y
b1
a1
g1
02b Ch02b 57-106.indd 6902b Ch02b 57-106.indd 69 6/12/09 8:22:12 AM6/12/09 8:22:12 AM
F
1
 = 400i
Since cos2 60° + cos2 b
2
 + cos2 60° = 1 
Solving for the positive root, b
2
 = 45°
F
2
 = 300 cos 60°i + 300 cos 45°j + 300 cos 60°k
 = 150i + 212.1j + 150k
F
1
 = 500 cos a
1
i + 500 cos b
1
j + 500 cos g
1
k
FR = F1 + (–300j) + (–200k) 
350i = 500 cos a
1
i + (500 cos b
1
 – 300)j + (500 cos g
1
 – 200)k
350 = 500 cos a
1
; a
1
 = 45.6° Ans
0 = 500 cos b
1
 – 300; b
1
 = 53.1° Ans
0 = 500 cos g
1
 – 200; g
1
 = 66.4° Ans
Then
FR = F1 + F2 + F3 = 550i + 52.1j + 270k 
 = –160j + 120k
4
5
3
5
F
3
 = –200 j + 200 
⎛
⎝⎜
⎞
⎠⎟
k 
⎛
⎝⎜
⎞
⎠⎟
FR = (550)
2 + (52.1)2 + (270)2 = 614.9 N = 615 N Ans 
550
614.9
a = cos–1 = 26.6° 
⎛
⎝⎜
⎞
⎠⎟
Ans
52.1
614.9
b = cos–1 = 85.1° 
⎛
⎝⎜
⎞
⎠⎟
Ans
270
614.9
g = cos–1 = 64.0° 
⎛
⎝⎜
⎞
⎠⎟
Ans
70
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–75. The mast is subjected to the three forces shown.
Determine the coordinate direction angles of
F1 so that the resultant force acting on the mast is zero.
a1, b1, g1
F3 300 N
F2 200 Nx
z
F1
y
b1
a1
g1
*2–76. Determine the magnitude and coordinate
direction angles of F2 so that the resultant of the two forces
acts along the positive x axis and has a magnitude of 500 N.
y
x
z
F1 180 N
F2
60
15
b2
a2
g2
02b Ch02b 57-106.indd 7002b Ch02b 57-106.indd 70 6/12/09 8:22:13 AM6/12/09 8:22:13 AM
F
1
 = {500 cos a
1
i + 500 cos b
1
j + 500 cos g
1
k} N
F
2
 = {–200k} N 
F
3
 = {–300j} N 
FR = F1 + F2 + F3 = 0 
500 cos a
1
 = 0; a
1
 = 90° Ans
500 cos b
1
 = 300; b
1
 = 53.1° Ans
500 cos g
1
 = 200; g
1
 = 66.4° Ans
F
1
 = (180 cos 15°) sin 60°i + (180 cos 15°) cos 60°j – 180 sin 15°k
 = 150.57i + 86.93j – 46.59k
F
2
 = F
2
 cos a
2
i + F
2
 cos b
2
j + F
2
 cos g
2
k 
FR = {500i} N 
FR = F1 + F2 
i components:
500 = 150.57 + F
2
 cos a
2
F
2x = F2 cos a2 = 349.43 
j components:
0 = 86.93 + F
2
 cos b
2
F
2y = F2 cos b2 = –86.93 
Thus,
F
2
 = F 2
2x + F
2
2y + F
2
2z = (349.43)
2 + (–86.93)2 + (46.59)2
F
2
 = 363 N Ans
a
2
 = 15.8° Ans
b
2
 = 104° Ans
g
2
 = 82.6° Ans
k components:
0 = –46.59 + F
2
 cos g
2
F
2z = F2 cos g2 = 46.59 
71
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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.
•2–77. Determine the magnitude and coordinate direction
angles of F2 so that the resultant of the two forces is zero.
y
x
z
F1 180 N
F2
60
15
b2
a2
g2
02b Ch02b 57-106.indd 7102b Ch02b 57-106.indd 71 6/12/09 8:22:14 AM6/12/09 8:22:14 AM
F
1
 = (180 cos 15°) sin 60°i + (180 cos 15°) cos 60°j – 180 sin 15°k
 = 150.57i + 86.93j – 46.59k
F
2
 = F
2
 cos a
2
i + F
2
 cos b
2
j + F
2
 cos g
2
k 
FR = 0 
i components: k components:
F
2
 cos a
2
 = –150.57 
F
2
 cos b
2
 = –86.93
F
2
 cos g
2
 = 46.59 
0 = 150.57 + F
2
 cos a
2
0 = 86.93 + F
2
 cos b
2
0 = –46.59 + F
2
 cos g
2
j components:
F
2
 = (–150.57)2 + (–86.93)2 + (46.59)2
Solving,
F
2
 = 180 N Ans
a
2
 = 147° Ans
b
2
 = 119° Ans
g
2
 = 75.0° Ans
72
2–78. If the resultant force acting on the bracket is directed
along the positive y axis, determine the magnitude of the
resultant force and the coordinate direction angles of F so
that .b 6 90°
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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.
x
y
z
F 500 N
F1 600 N
a
b
g
30
30
02b Ch02b 57-106.indd 7202b Ch02b 57-106.indd 72 6/12/098:22:15 AM6/12/09 8:22:15 AM
Force Vectors: By resolving F
1
 and F into their x, y, and z components, as shown in Figs. a and b, 
respectively, F
1
 and F can be expressed in Cartesian vector form as
F
1
 = 600 cos 30° sin 30° (+i) + 600 cos 30° cos 30° (+j) + 600 sin 30° (–k) 
 = {259.81i + 450j – 300k} N
F = 500 cos ai + 500 cos bj + 500 cos gk
Since the resultant force FR is directed towards the positive y axis, then
FR = FR j
Resultant Force:
FR = F1 + F
FR j = (259.81i + 450j – 300k) + (500 cos uxi + 500 cos uy j + 500 cos uzk)
FR j = (259.81 + 500 cos a)i + (450 + 500 cos b)j + (500 cos g – 300)k 
Equating the i, j, and k components,
However, since cos2 a + cos2 b + cos2 g = 1, a = 121.31°, and g = 53.13°,
0 = 259.81 + 500 cos a
a = 121.31° = 121° Ans
FR = 450 + 500 cos b (1)
0 = 500 cos g – 300
g = 53.13° = 53.1° Ans
cos b = ± 1 – cos2 121.31° – cos2 53.13° = ±0.6083
If we substitute cos b = 0.6083 into Eq. (1),
FR = 450 + 500(0.6083) = 754 N Ans 
and
b = cos–1(0.6083) = 52.5° Ans 
73
2–79. Specify the magnitude of F3 and its coordinate
direction angles so that the resultant force
.FR = 59j6 kN
a3, b3, g3
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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.
x
z
5
12
13
y
F3
30
F2 10 kN
F1 12 kN
g3
b3
a3
02b Ch02b 57-106.indd 7302b Ch02b 57-106.indd 73 6/12/09 8:22:16 AM6/12/09 8:22:16 AM
F
1
 = 12 cos 30°j – 12 sin 30°k = 10.392j – 6k
9j = 10.392j – 6k – 9.231i + 3.846k + F
3
F
3
 = 9.231i – 1.392j + 2.154k
F
3
 = (9.231)2 + (–1.392)2 + (2.154)2
F
3
 = 9.581 kN = 9.58 kN Ans
FR = F1 + F2 + F3
Require
Hence,
F
2
 = – (10)i + (10)k = –9.231i + 3.846k
12
13
5
13
a
3
 = cos–1 = 15.5° Ans9.231
9.581
⎛
⎝⎜
⎞
⎠⎟
b
3
 = cos–1 = 98.4° Ans–1.392
9.581
⎛
⎝⎜
⎞
⎠⎟
g
3
 = cos–1 = 77.0° Ans2.154
9.581
⎛
⎝⎜
⎞
⎠⎟
74
*2–80. If , , and = 45°, determine the
magnitude and coordinate direction angles of the resultant
force acting on the ball-and-socket joint.
fu = 30°F3 = 9 kN
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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
3
5
F3
F2 8 kN
F1 10 kN
z
y
x
u
f
30
60
02b Ch02b 57-106.indd 7402b Ch02b 57-106.indd 74 6/12/09 8:22:18 AM6/12/09 8:22:18 AM
Force Vectors: By resolving F
1
, F
2
 and F
3
 into their x, y, and z components, as shown in Figs. 
respectively, F
1
, F
2
 and F
3
 can be expressed in Cartesian vector form as
FR = F1 + F2 + F3
 = (–2.5i + 4.330j – 8.660k) + (–4.8i + 6.4k) + (3.182i – 5.511j – 6.364k)
 = {–4.118i – 1.181j – 8.624k} kN
Resultant Force: By adding F
1
, F
2
 and F
3
 vectorally, we obtain FR. Thus,
The coordinate direction angles of FR are
The magnitude of FR is
 = {–4.8i + 6.4k} kN
F
3
 = 9 cos 45° sin 30° (+i) + 9 cos 45° cos 30°(–j) + 9 sin 45° (–k)
 = {3.182i – 5.511j – 6.364k} kN
F
1
 = 10 cos 60° sin 30° (–i) + 10 cos 60° cos 30° (+j) + 10 sin 60° (–k)
 = {–2.5i + 4.330j – 8.660k} kN
FR = (FR)x
2 + (FR)y
2 + (FR)z
2
 = (–4.118)2 + (–1.181)2 + (–8.624)2 = 9.630 kN = 9.63 kN Ans
a = cos–1 = cos–1 = 115° Ans(FR)x
FR
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
–4.118
9.630
⎛
⎝⎜
⎞
⎠⎟
b = cos–1 = cos–1 = 97.0° Ans(FR)y
FR
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
–1.181
9.630
⎛
⎝⎜
⎞
⎠⎟
g = cos–1 = cos–1 = 154° Ans(FR)z
FR
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
–8.624
9.630
⎛
⎝⎜
⎞
⎠⎟
F
2
 = 8 (–i) + 0j + 8
3
5
⎛
⎝⎜
⎞
⎠⎟
(+k)
4
5
⎛
⎝⎜
⎞
⎠⎟
75
•2–81. The pole is subjected to the force F, which has
components acting along the x, y, z axes as shown. If the
magnitude of F is 3 kN, , and , determine
the magnitudes of its three components.
g = 75°b = 30°
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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.
z
Fz
Fy
Fx
F
y
x
a
b
g
2–82. The pole is subjected to the force F which has
components and . If ,
determine the magnitudes of F and Fy.
b = 75°Fz = 1.25 kNFx = 1.5 kN
z
Fz
Fy
Fx
F
y
x
a
b
g
02b Ch02b 57-106.indd 7502b Ch02b 57-106.indd 75 6/12/09 8:22:19 AM6/12/09 8:22:19 AM
cos2 a + cos2 b + cos2 g = 1 
cos2 a + cos2 30° + cos2 75° = 1 
a = 64.67°
Fx = 3 cos 64.67° = 1.28 kN Ans
Fy = 3 cos 30° = 2.60 kN Ans
Fz = 3 cos 75° = 0.776 kN Ans
F = 2.02 kN Ans
Fy = 2.02 cos 75° = 0.523 kN Ans
cos2 a + cos2 b + cos2 g = 1 
+ cos2 75° + = 11.5
F
⎛
⎝⎜
⎞
⎠⎟
2 1.25
F
⎛
⎝⎜
⎞
⎠⎟
2
76
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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.
2–83. Three forces act on the ring. If the resultant force FR
has a magnitude and direction as shown, determine the
magnitude and the coordinate direction angles of force F3.
x
y
z
3 4
5
F3
45
30
F1 80 N
F2 110 N
FR 120 N
*2–84. Determine the coordinate direction angles of F1
and FR.
x
y
z
3 4
5
F3
45
30
F1 80 N
F2 110 N
FR 120 N
02b Ch02b 57-106.indd 7602b Ch02b 57-106.indd 76 6/12/09 8:22:20 AM6/12/09 8:22:20 AM
Cartesian Vector Notation:
F
2
 = {–110k} N
F
3
 = {F
3x
i + F
3y
j + F
3z
k} N
FR = 120 {cos 45° sin 30°i + cos 45° cos 30°j + sin 45°k} N
 = {42.43i + 73.48j + 84.85k} N
64.0 + F
3x
 = 42.43 F
3x
 = –21.57 N
 F
3y
 = 73.48 N
48.0 – 110 + F
3z
 = 84.85 F
3z
 = 146.85 N
Resultant Force:
Equating i, j and k components, we have
FR = F1 + F2 + F3
{42.43i + 73.48j + 84.85k}
 = {(64.0 + F
3x
)i + F
3y
j + (48.0 – 110 + F
3x
)k}
F
1
 = 80 i + N = {64.0i + 48.0k} N
4
5
k
3
5
Unit Vector of F
1
 and FR:
uR = cos 45° sin 30°i + cos 45° cos 30°j + sin 45°k
 = 0.3536i + 0.6124j + 0.7071k
uF1 = i +
4
5
k = 0.8i + 0.6k
3
5
The magnitude of F
3
 is
The coordinate direction angles for F
3
 are
F
3
 = F
3x
2 + F
3y
2 + F
3z
2
 = (–21.57)2 + 73.482 + 146.852 
 = 165.62 N = 166 N Ans
cos a = = a = 97.5° Ans
F
3x
F
3
–21.57
165.62
Thus, the coordinate direction angles F
1
 and FR are
cos b = = b = 63.7° Ans
F
3y
F
3
73.48
165.62
cos g = = g = 27.5° Ans
F
3z
F
3
146.85
165.62
cos aF1
 = 0.8 aF1
 = 36.9° Ans
cos bF1
 = 0 bF1
 = 90.0° Ans
cos gF1
 = 0.6 gF1
 = 53.1° Ans
cos aR = 0.3536 aR = 69.3° Ans
cos bR = 0.6124 bR = 52.2° Ans
cos gR = 0.7071 gR = 45.0° Ans
F2
80
110
x
y
z
g
F1 20 lb
FR 50 lb
77
•2–85. Two forces F1 and F2 act on the bolt. If the resultant
force FR has a magnitude of 50 lb and coordinate direction
angles and , as shown, determine the
magnitude of F2 and its coordinate direction angles.
b = 80°a = 110°
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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.
2–86. Determine the position vector r directed from point
A to point B and the length of cord AB. Take .z = 4 m
3 m
2 m
6 m
z
y
z
B
x
A
20 N
50 N
N
02b Ch02b 57-106.indd 7702b Ch02b 57-106.indd 77 6/12/09 8:22:22 AM6/12/09 8:22:22 AM
(1)2 = cos2 110° + cos2 80° + cos2 g
g = 157.44°
FR = F1 + F2
50 cos 110° = (F
2
)x
50 cos 80° = (F
2
)y
50 cos 157.44° = (F
2
)z – 20
(F
2
)x = –17.10
(F
2
)y = 8.68
(F
2
)z = –26.17
F
2
 = (–17.10)2 + (8.68)2 + (–26.17)2 = 32.4 N Ans
a
2
 = cos–1 = 122° Ans–17.10
32.4
⎛
⎝⎜
⎞
⎠⎟
b
2
 = cos–1 = 74.5° Ans8.68
32.4
⎛
⎝⎜
⎞
⎠⎟
g
2
 = cos–1 = 144° Ans–26.1732.4
⎛
⎝⎜
⎞
⎠⎟
rAB = (0 –3)i + (6 – 0)j + (4 – 2)k
 = {–3i + 6j + 2k} m Ans
Position Vector: The coordinates for points A and B are A(3, 0, 2) m and 
B(0, 6, 4) m, respectively. Thus,
rAB = (–3)
2 + 62 + 22 = 7 m Ans
The length of cord AB is
78
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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.
2–87. If the cord AB is 7.5 m long, determine the
coordinate position +z of point B
3 m
2 m
6 m
y
B
x
A
*2–88. Determine the distance between the end points A
and B on the wire by first formulating a position vector
from A to B and then determining its magnitude.
x
B
A
y
25 mm
75 mm
200 mm
50 mm
30
60
rAB = (200 sin 60° – (–75 sin 30°))i + (200 cos 60° – 75 cos 30°)j + (–50 – 25)k
rAB = (210.71i + 35.048j – 75k) mm
rAB = (210.71) + (35.048) + (–75)
2 2 2 = 226.4 mm Ans
02b Ch02b 57-106.indd 7802b Ch02b 57-106.indd 78 6/12/09 8:22:24 AM6/12/09 8:22:24 AM
rAB = (0 –3)i + (6 – 0)j + (z – 2)k
 = {–3i + 6j + (z – 2)k} m
Position Vector: The coordinates for points A and B are A(3, 0, 2) m and 
B(0, 6, z) m, respectively. Thus,
rAB = 7.5 = (–3)
2 + 62 + (z – 2)2
56.25 = 45 + (z – 2)2
z – 2 = ±3.354
z = 5.35 m Ans
Since the length of cord is equal to the magnitude of rAB, then
79
•2–89. Determine the magnitude and coordinate
direction angles of the resultant force acting at A.
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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.
0.6 m
1.2 m
0.9 m
0.9 m
1.2 m
0.75 m
B
A
x
C
FC 3.75 kN
FB 3 kN
Unit Vectors: The coordinate points A, B, and C are shown in Fig. a. Thus,
uB = 
rB
Br
 = 
(0.9 – 0) + (–0.9 – 0) + (0.75 – 1.2)
(0.
i j k
99 – 0) + (–0.9 – 0) + (0.75 – 1.2)2 2 2
= 
2
3
i – 
2
3
j – 
1
3
k
uC = 
rC
Cr
 = 
(0.6 – 0) + (1.2 – 0) + (0 – 1.2)
(0.6 –
i j k
00) + (1.2 – 0) + (0 – 1.2)2 2 2
= 
1
3
i + 
2
3
j – 
2
3
k
Force Vectors: Multiplying the magnitude of the force with its unit vector, we have
FB = FBuB = 3
2
3
–
2
3
–
1
3
i j k
⎛
⎝⎜
⎞
⎠⎟
 = {2i – 2j – 1k} kN Ans
FC = FCuC = 3.75
1
3
+
2
3
–
2
3
i j k
⎛
⎝⎜
⎞
⎠⎟
 = {1.25i + 2.5j – 2.5k} kN Ans
FR = FB + FC = 2i – 2j – 1k + 1.25i + 2.5j – 2.5k
FR = {3.25i + 0.5j – 3.5k} kN
FR = (3.25) + (0.5) + (–3.5)2 2 2 = 4.80 kN Ans
 = cos–1
(F
F
R x
R
)⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 = cos–1
3.25
4.80
⎛
⎝⎜
⎞
⎠⎟
 = 47.4° Ans
 = cos–1
(F
F
R y
R
)⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 = cos–1
0.5
4.80
⎛
⎝⎜
⎞
⎠⎟
 = 84.0° Ans
 = cos–1
(F
F
R z
R
)⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 = cos–1
–3.5
4.80
⎛
⎝⎜
⎞
⎠⎟
 = 137° Ans
A(0, 0, 1.2) m
C(0.6, 1.2, 0) m
B(0.9, –0.9, 0.75) m
02b Ch02b 57-106.indd 7902b Ch02b 57-106.indd 79 6/12/09 8:22:27 AM6/12/09 8:22:27 AM
80
2–90. Determine the magnitude and coordinate direction
angles of the resultant force.
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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.
x
z
y
C
B
A
600 N 500 N
8 m
4 m
4 m
2 m
02b Ch02b 57-106.indd 8002b Ch02b 57-106.indd 80 6/12/09 8:22:29 AM6/12/09 8:22:29 AM
rAB = (–2j – 4k) m; rA-B = 4.472 m
uAB = = –0.447j – 0.894k
rAB
rAB
⎛
⎝⎜
⎞
⎠⎟
uAC = = 0.485i + 0.728j – 0.485k
rAC
rAC
⎛
⎝⎜
⎞
⎠⎟
FAB = 600 uAB = {–268.33j – 536.66k} N
rAC = {4i + 6j – 4k}m; rAC = 8.246 m
FAC = 500 uAC = {242.54i + 363.80j – 242.54k} N
FR = FAB + FAC
FR = {242.54i + 95.47j – 779.20k}
FR = (242.54)
2 + (95.47)2 + (–779.20)2 = 821.64 = 822 N Ans
a = cos–1 = 72.8° Ans242.54
821.64
⎛
⎝⎜
⎞
⎠⎟
b = cos–1 = 83.3° Ans95.47
821.64
⎛
⎝⎜
⎞
⎠⎟
g = cos–1 = 162° Ans–779.20
821.64
⎛
⎝⎜
⎞
⎠⎟
81
2–91. Determine the magnitude and coordinate direction
angles of the resultant force acting at A.
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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.
y
x
B
C
A
6 m
3 m
45
4.5 m
6 m
FB 900 N
FC 600 N
z
02b Ch02b 57-106.indd 8102b Ch02b 57-106.indd 81 6/12/09 8:22:31 AM6/12/09 8:22:31 AM
Force Vectors: The unit vectors uB and uC of FB and FC must be determined first. From Fig. a
Thus, the force vectors FB and FC are given by
uB = 
rB
Br
 = 
(4.5 sin 45° – 0)2 + (–4.5 cos 45° – 0)2 + (0 – 6)2
= 0.4243i – 0.4243j – 0.8k
uC = 
rC
Cr
 = 
(–3 – 0)i + (–6 – 0)j + (0 – 6)k
(–3 – 0)2 + (–6 – 0)2 + (0 – 6)2
= –
1
3
i – 
2
3
j – 
2
3
k
(4.5 sin 45° – 0)i + (–4.5 cos 45° – 0)j + (0 – 6)k
FB = FBuB = 900(0.4243i – 0.4243j – 0.8k) = {381.84i – 381.84j – 720k} N
Resultant Force:
The magnitude of FR is
The coordinate direction angles of FR are
FR = FB + FC = (381.84i – 381.84j – 720k) + (–200i – 400j – 400k)
 = {181.84i – 781.84j – 1120k} N
3
–
2
3
–
2
3
⎛
⎝⎜
⎞
⎠⎟
= {–200i – 400j – 400k} NFC = FCuC = 600 – i j k
1
FR = (FR)x
2 + (FR)y
2 + (FR)z
2
 = (181.84)2 + (–781.84)2 + (–1120)2 = 1377.95 N = 1.38 kN Ans
a = cos–1 = cos–1 = 82.4° Ans(FR)x
FR
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
181.84
1377.95
⎛
⎝⎜
⎞
⎠⎟
b = cos–1 = cos–1 = 125° Ans(FR)y
FR
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
–781.84
1377.95
⎛
⎝⎜
⎞
⎠⎟
g = cos–1 = cos–1 = 144° Ans(FR)z
FR
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
–1120
1377.95
⎛
⎝⎜
⎞
⎠⎟
82
*2–92. Determine the magnitude and coordinate direction
angles of the resultant force.
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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
C
B
1.2 m
2.1 m
0.9 m
x
y
F2 405 N
F1 500 N
40
1.2 m
•2–93. The chandelier is supported by three chains which
are concurrent at point O. If the force in each chain has a
magnitude of 300 N, express each force as a Cartesian vector
and determine the magnitude and coordinate direction
angles of the resultant force.
120
y
120 1.2 m
A
B
C
1.8 m
O
FA
FB
FC
x
120
F1 = –500
3
5
⎛
⎝⎜
⎞
⎠⎟
 sin 40°i + 500
3
5
⎛
⎝⎜
⎞
⎠⎟
 cos 40°j – 500
4
5
⎛
⎝⎜
⎞
⎠⎟
k
= {–192.84i + 229.81j – 400k} N
F2 = 405 N
4
9
–
7
9
–
4
9
i j k
⎛
⎝⎜
⎞
⎠⎟
= {180i – 315j – 180k} N
FR = F1 + F2 = {–12.84i – 85.19j – 580k} N
FR = (12.84) + (85.19) + (580)2 2 2 = 586.36 N = 586 N Ans
 = cos–1
–12.84
586.36
⎛
⎝⎜
⎞
⎠⎟
 = 91.3° Ans
 = cos–1
–85.19
586.36
⎛
⎝⎜
⎞
⎠⎟
 = 98.4° Ans
 = cos–1
–580
586.36
⎛
⎝⎜
⎞
⎠⎟
 = 172° Ans
FA = 300
(1.2 cos 30° – 1.2 sin 30° – 1.8 )
(1.2 co
i j k
ss 30°) + (–1.2 sin 30°) + (–1.8)222
= {144.1i – 83.2j – 249.6k} N Ans
FB = 300
(–1.2 cos 30° – 1.2 sin 30° – 1.8 )
(–1.2
i j k
ccos 30°) + (–1.2 sin 30°) + (–1.8)222
= {–144.1i – 83.2j – 249.6k} N Ans
FC = 300
(1.2 – 1.8 )
(1.2) + (–1.8)2 2
j k
= {166.4j – 249.6k} N Ans
FR = FA + FB + FC = {–748.8k} N
FR = 748.8 N Ans
 = 90° Ans
 = 90° Ans
 = 180° Ans
02b Ch02b 57-106.indd 8202b Ch02b 57-106.indd 82 6/12/09 8:22:32 AM6/12/09 8:22:32 AM
83
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2–94. The chandelier is supported by three chains which
are concurrent at point O. If the resultant force at O has a
magnitude of 650 N and is directed along the negative z axis,
determinethe force in each chain.
120
y
120 1.2 m
A
B
C
1.8 m
O
FA
FB
FC
x
120
2–95. Express force F as a Cartesian vector; then
determine its coordinate direction angles.
y
x
B
A
3 m
70
30
2.1 m
1.5 m
F 675 N
FC = F
(1.2 – 1.8 )
(1.2) + (–1.8)2 2
j k
 = 0.5547 F j – 0.8321 F k
FA = FB = FC
FRz = ΣFz; 650 = 3(0.8321 F)
F = 260.4 N Ans
Unit Vector : The coordinates of point A are
A (–3 cos 70° sin 30°, 3 cos 70° cos 30°, 3 sin 70°) m
= A (–0.513, 0.8886, 2.819) m
Then
rAB = {[1.5 – (–0.513)]i + (–2.1 – 0.8886)j + (0 – 2.819)k} m
= {2.013i – 2.9886j – 2.819k} m
rAB = 2.013 + (–2.9886) + (–2.819)
2 2 2 = 4.575 m
uAB = 
rAB
ABr
 = 
2.013 – 2.9886 – 2.819
4.575
i j k
= 0.4400i – 0.6532j – 0.6162k
Force Vector :
F = FuAB = 675{0.4400i – 0.6532j – 0.6162k} N
= {297i – 440.9j – 415.9k} N Ans
Coordinate Direction Angles : From the unit vector uAB obtained 
above, we have
cos = 0.4400 = 63.9° Ans
cos = –0.6532 = 131° Ans
cos = –0.6162 = 128° Ans
02b Ch02b 57-106.indd 8302b Ch02b 57-106.indd 83 6/12/09 8:22:34 AM6/12/09 8:22:34 AM
84
*2–96. The tower is held in place by three cables. If the
force of each cable acting on the tower is shown, determine
the magnitude and coordinate direction angles of
the resultant force. Take , .y = 15 mx = 20 m
a, b, g
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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.
x
z
yxy
6 m
4 m
18 m
C
A
D
400 N
800 N
600 N
24 m
O
16 m
B
02b Ch02b 57-106.indd 8402b Ch02b 57-106.indd 84 6/12/09 8:22:35 AM6/12/09 8:22:35 AM
FDA = 400 Ni + 
20
34.66
j – 
15
34.66
k 
24
34.66
⎛
⎝⎜
⎞
⎠⎟
FDB = 800 Ni + 
–6
25.06
j – 
4
25.06
k 
24
25.06
⎛
⎝⎜
⎞
⎠⎟
FDC = 600 Ni – 
16
34
j – 
18
34
k
24
34
⎛
⎝⎜
⎞
⎠⎟
FR = FDA + FDB + FDC
 = {321.66i – 16.82j – 1466.71k} N
FR = (321.66)
2 + (–16.82)2 + (–1466.71)2
 = 1501.66 N = 1.50 kN Ans
a = cos–1 = 77.6° Ans321.66
1501.66
⎛
⎝⎜
⎞
⎠⎟
b = cos–1 = 90.6° Ans–16.82
1501.66
⎛
⎝⎜
⎞
⎠⎟
g = cos–1 = 168° Ans–1466.71
1501.66
⎛
⎝⎜
⎞
⎠⎟
85
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•2–97. The door is held opened by means of two chains. If
the tension in AB and CD is and ,
respectively, express each of these forces in Cartesian 
vector form.
FC = 250 NFA = 300 N
x
y
z
2.5 m
1.5 m
0.5 m1 m
30
A
C
B
D
FA 300 N
FC 250 N
2–98. The guy wires are used to support the telephone
pole. Represent the force in each wire in Cartesian vector
form. Neglect the diameter of the pole.
y
B
C
D
A
x
z
4 m
4 m
1.5 m
1 m
3 m
2 m
FA 250 N
FB 175 N
02b Ch02b 57-106.indd 8502b Ch02b 57-106.indd 85 6/12/09 8:22:36 AM6/12/09 8:22:36 AM
Unit Vector: First determine the position vector rAB and rCD. The coordinates of 
points A and C are
Then
A[0, –(1 + 1.5 cos 30°), 1.5 sin 30°] m = A[0, –2.299, 0.750) m
C[–2.50, –(1 + 1.5 cos 30°), 1.5 sin 300°] m = C[–2.50, –2.299, 0.750) m
rAB = {(0 – 0)i + [0 – (–2.299)]j + (0 – 0.750)k} m
 = {2.299j – 0.750k} m
rAB = 2.299
2 + (–0.750)2 = 2.418 m
Unit Vector:
rAC = {(–1 – 0)i + (4 – 0)j + (0 – 4)k} m = {–1i + 4j – 4k} m
rAC = (–1)
2 + 42 + (–4)2 = 5.745 m
uAB = = 0.9507j – 0.3101k=
rAB
rAB
2.299j – 0.750k
2.418
uAC = = –0.1741i + 0.6963j – 0.6963k=
rAC
rAC
–1i + 4j – 4k
5.745
rCD = {[–0.5 – (–2.5)]i + [0 – (–2.299)]j + (0 – 0.750)k} m
 = {2.00i + 2.299j – 0.750k} m
rCD = 2.00
2 + 2.2992 + (–0.750)2 = 3.138 m
uCD = = 0.6373i + 0.7326j – 0.2390k=
rCD
rCD
2.00i + 2.299j – 0.750k
3.138
Force Vector:
FA = FAuAB = 300{0.9507j – 0.3101k} N
 = {285.21j – 93.04k} N
 = {285j – 93.0k} N Ans
FC = FCuCD = 250{0.6373i + 0.7326j – 0.2390k} N
 = {159.33i + 183.15j – 59.75k} N
 = {159i + 183j – 59.7k} N Ans
Force Vector:
FA = FAuAC = 250{–0.1741i + 0.6963j – 0.6963k} N
 = {–43.52i + 174.08j – 174.08k} N
 = {–43.5i + 174j – 174k} N Ans
FB = FBuBD = 175{0.3041i – 0.4562j – 0.8363k} N
 = {53.22i – 79.83j – 146.36k} N
 = {53.2i – 79.8j – 146k} N Ans
rBD = {(2 – 0)i + (–3 – 0)j + (0 – 5.5)k} m = {2i – 3j – 5.5k} m
rBD = 2
2 + (–3)2 + (–5.5)2 = 6.576 m
uBD = = 0.3041i – 0.4562j – 0.8363k=
rBD
rBD
2i – 3j – 5.5k
6.576
86
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2–99. Two cables are used to secure the overhang boom in
position and support the 1500-N load. If the resultant force
is directed along the boom from point A towards O,
determine the magnitudes of the resultant force and forces
FB and FC. Set and .z = 2 mx = 3 m
z
A
x y
6 m
1500 N
3 m
FB
FC
B
C
2 mx
z
02b Ch02b 57-106.indd 8602b Ch02b 57-106.indd 86 6/12/09 8:22:38 AM6/12/09 8:22:38 AM
Force Vectors: The unit vectors uB and uC must be determined first. From Fig. a,
 = 
(–2 – 0)2 + (0 – 6)2 + (3 – 0)2
(–2 – 0)i + (0 – 6)j + (3 – 0)kuB =
rB
rB
= – i – 2
7
j + 6
7
k 3
7
Thus, the force vectors FB and FC are given by
FB = FBuB = – FBi – 
2
7
FB j + 
6
7
FBk 
3
7
0 = – (1)FB + 
2
7
FC
3
7
0 = (3)FB + 
3
7
FC – 1500
2
7
–FR = – (2)FB – 
6
7
FC
6
7
Resultant Force: The vector addition of FB, FC, and W is equal to FR. Thus,
Equating the i, j, and k components,
Solving Eqs. (1), (2), and (3) yields
FR = FB + FC + W
Since the resultant force FR is directed along the negative y axis, and the 
load W is directed along the z axis, these two forces can be written as
FR = – FR j and W = [–1500k] N 
FC = 1615.38 N = 1.62 kN Ans
FB = 2423.08 N = 2.42 kN Ans
FR = 3461.53 N = 3.46 kN Ans 
FC = FCuC = – FCi – 
3
7
FC j + 
6
7
FCk 
2
7
 = 
(3 – 0)2 + (0 – 6)2 + (2 – 0)2
(3 – 0)i + (0 – 6)j + (2 – 0)kuC =
rC
rC
= i – 3
7
j + 6
7
k 2
7
–FR j = – + + (–1500k)FBi – 
2
7
FB j + 
6
7
FBk 
3
7
⎛
⎝⎜
⎞
⎠⎟
FCi – 
3
7
FC j + 
6
7
FCk 
2
7
⎛
⎝⎜
⎞
⎠⎟
–FR j = – i + j + kFB + 
2
7
FC 
3
7
⎛
⎝⎜
⎞
⎠⎟
– FB – 
6
7
FC
6
7
FB + 
3
7
FC – 1500
2
7
⎛
⎝⎜
⎛
⎝⎜
⎞
⎠⎟
⎞
⎠⎟
87
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*2–100. Two cables are used to secure the overhang boom
in position and support the 1500-N load. If the resultant
force is directed along the boom from point A towards O,
determine the values of x and z for the coordinates of point
C and the magnitude of the resultant force. Set
and .FC = 2400 NFB = 1610 N
z
A
x y
6 m
1500 N
3 m
FB
FC
B
C
2 mx
z
02b Ch02b 57-106.indd 8702b Ch02b 57-106.indd 87 6/12/09 8:22:41 AM6/12/09 8:22:41 AM
Force Vectors: From Fig. a,
Thus,
 = 
(–2 – 0)2 + (0 – 6)2 + (3 – 0)2
(–2 – 0)i + (0 – 6)j + (3 – 0)kuB =
rB
rB
= – i – 2
7
j + 6
7
k 3
7
 = 
(x – 0)2 + (0 – 6)2 + (z – 0)2
(x – 0)i + (0 – 6)j + (z – 0)k
x2 + z2 + 36
xuC =
rC
rC
= i – 
x2 + z2 + 36
z k 
x2 + z2 + 36
6 j + 
FB = FBuB = 1610 – = [–460i – 1380j + 690k] Ni – 
2
7
j + 6
7
k 3
7
⎛
⎝⎜
⎞
⎠⎟
x2 + z2 + 36
x
FC = FCuC = 2400 i – 
x2 + z2 + 36
z k 
x2 + z2 + 36
6 j + 
x2 + z2 + 36
2400x
= i – 
x2 + z2 + 36
2400z k 
x2 + z2 + 36
14 400 j + 
⎛
⎝⎜
⎞
⎠⎟
–FR j = (–460i – 1380j + 690k) + + (–1500k)
x2 + z2 + 36
2400x i – 
x2 + z2 + 36
2400z k 
x2 + z2 + 36
14 400 j + ⎛
⎝⎜
⎞
⎠⎟
–FR j = 
x2 + z2 + 36
2400x
–460 i – ⎛
⎝⎜
⎞
⎠⎟ x2 + z2 + 36
14 400
+1380j + 690 +⎛
⎝⎜
⎞
⎠⎟ x2 + z2 + 36
2400z
–1500 k⎛
⎝⎜
⎞
⎠⎟
0 = 
x2 + z2 + 36
2400x
–460
x2 + z2 + 36
2400x
= 460 (1)
0 = 690 + 
x2 + z2 + 36
2400z
–1500
x2 + z2 + 36
2400z
= 810 (3)
Since the resultant force FR is directed along the negative y axis, and the load 
is directed along the z axis, these two forces can be written as
FR = – FR j and W = [–1500k] N 
Resultant Force:
Equating the i, j, and k components,
Dividing Eq. (1) by Eq. (3), yields
x = 0.5679z (4)
Substituting Eq. (4) into Eq. (1), and solving
z = 2.197 m = 2.20 m Ans
FR = FB + FC + W
–FR = – FR =
x2 + z2 + 36
14 400
+1380
x2 + z2 + 36
14 400
+1380 (2)⎛
⎝⎜
⎞
⎠⎟
Substituting z = 2.197 m into Eq. (4), yields
x = 1.248 m = 1.25 m Ans
Substituting x = 1.248 m and z = 2.197 m into Eq. (2), yields
FR = 3591.85 N = 3.59 kN Ans
88
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•2–101. The cable AO exerts a force on the top of the pole
 fo htgnel a sah elbac eht fI. fo
34 ft, determine the height z of the pole and the location 
(x, y) of its base.
F = 5-120i - 90j - 80k6 lb
y
z
A
z
x
F
x
y
O
•2–101. The cable AO exerts a force on the top of the pole 
of F = {–120i – 90j – 80k} N. If the cable has a length of 
1.02 m, determine the height z of the pole and the location 
(x, y) of its base.
F = (–120) + (–90) + (–80)2 2 2 = 170 N
u = 
F
F
 = –
120
170
i – 
90
170
j – 
80
170
k
r = 1.02u = {–0.72i – 0.54j – 0.48k} m
Thus,
x = 0.72 m Ans
y = 0.54 m Ans
z = 0.48 m Ans
02b Ch02b 57-106.indd 8802b Ch02b 57-106.indd 88 6/12/09 8:22:46 AM6/12/09 8:22:46 AM
89
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2–102. If the force in each chain has a magnitude of 2.25 kN,
determine the magnitude and coordinate direction angles
of the resultant force.
120
120
0.9 m
2.1 m
120
FAFBFC
C
A
D
B
y
x
D(0, 0, 2.1) m
A(–0.9 sin 30°, 0.9 cos 30°, 0) m
B(–0.9 sin 30°, –0.9 cos 30°, 0) m
C(0.9, 0, 0) m
Force Vectors: The unit vectors uA, uB, and uC must be determined first. From Fig. a,
uA = 
(–0.9 sin 30° – 0) + (0.9 cos 30° – 0) +i j ((0 – 2.1)
(–0.9 sin 30° – 0) + (0.9 cos 32
k
00° – 0) + (0 – 2.1)2 2
 = –0.1970i + 0.3411j – 0.9191k
uB = 
(–0.9 sin 30° – 0) + (–0.9 cos 30° – 0) +i j (0 – 2.1)
(–0.9 sin 30° – 0) + (–0.9 cos2
k
30° – 0) + (0 – 2.1)2 2
 
= –0.1970i – 0.3411j – 0.9191k
uC = 
(0.9 – 0) + (0 – 0) + (0 – 2.1)
(0.9 – 0)
i j k
22 2 2+ (0 – 0) + (0 – 2.1)
 = 0.3939i – 0.9191k
Thus, the force vectors FA, FB, and FC are given by
FA = FAuA = 2.25(–0.1970i + 0.3411j – 0.9191k) = {–0.443i + 0.767j – 2.068k} kN
FB = FBuB = 2.25(–0.1970i – 0.3411j – 0.9191k) = {–0.443i – 0.767j – 2.068k} kN
FC = FCuC = 2.25(0.3939i – 0.9191k) = {0.886i – 2.068k} kN
Resultant Force:
FR = FA + FB + FC = (–0.443i + 0.767j – 2.068k) + (–0.443i – 0.767j – 2.068k) + (0.886i – 2.068k)
= {–6.204k} kN
The magnitude of FR is
FR = ( ) + ( ) + ( )
2F F FR x R y R z
2 2 
= 0 + 0 + (–6.204)2 2 2 = 6.204 kN Ans
The coordinate direction angles of FR are
 = cos–1
(F
F
R x
R
)⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 = cos–1
0
6.204
⎛
⎝⎜
⎞
⎠⎟
 = 90° Ans
 = cos–1
(F
F
R y
R
)⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 = cos–1
0
6.204
⎛
⎝⎜
⎞
⎠⎟
 = 90° Ans
 = cos–1
(F
F
R z
R
)⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 = cos–1
–6.204
6.204
⎛
⎝⎜
⎞
⎠⎟
 = 180° Ans
02b Ch02b 57-106.indd 8902b Ch02b 57-106.indd 89 6/12/09 8:22:47 AM6/12/09 8:22:47 AM
90
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2–103. If the resultant of the three forces is
, determine the magnitude of the force in
each chain.
F 4.5 k kN
120
120
0.9 m
2.1 m
120
FAFBFC
C
A
D
B
y
x
D(0, 0, 2.1) m
A(–0.9 sin 30°, 0.9 cos 30°, 0) m
B(–0.9 sin 30°, –0.9 cos 30°, 0) m
Force Vectors: The unit vectors uA, uB, and uC must be determined first. From Fig. a,
uA = 
(–0.9 sin 30° – 0) + (0.9 cos 30° – 0) +i j ((0 – 2.1)
(–0.9 sin 30° – 0) + (0.9 cos 32
k
00° – 0) + (0 – 2.1)2 2
 = –0.1970i + 0.3411j – 0.9191k
uB = 
(–0.9 sin 30° – 0) + (–0.9 cos 30° – 0) +i j (0 – 2.1)
(–0.9 sin 30° – 0) + (–0.9 cos2
k
30° – 0) + (0 – 2.1)2 2
 
= –0.1970i – 0.3411j – 0.9191k
uC = 
(0.9 – 0) + (0 – 0) + (0 – 2.1)
(0.9 – 0)
i j k
22 2 2+ (0 – 0) + (0 – 2.1)
 = 0.3939i – 0.9191k
C(0.9, 0, 0) m
02b Ch02b 57-106.indd 9002b Ch02b 57-106.indd 90 6/12/09 8:22:49 AM6/12/09 8:22:49 AM
FA = FAuA = –0.1970FAi + 0.3411FAj – 0.9191FAk
FB = FBuB = –0.1970FBi – 0.3411FB j – 0.9191FBk
FC = FCuC = 0.3939FCi – 0.9191FCk
Thus, the force vectors FA, FB and FC are given by
FR = FA + FB + FC
–4.5k = (–0.1970FAi + 0.3411FAj – 0.9191FAk) + (–0.1970FBi – 0.3411FB j – 0.9191FBk) + (0.3939FCi – 0.9191FCk)
–4.5k = (–0.1970FA – 0.1970FB + 0.3939FC)i + (0.3411FA – 0.3411FB)j + (–0.9191FA – 0.9191FB – 0.9191FC)k
Resultant Force: The vector addition of FA, FB and FC is equal to FR. Thus,
0 = –0.1970FA – 0.1970FB + 0.3939FC (1)
0 = 0.3411FA – 0.3411FB (2)
–4.5k = –0.9191FA – 0.9191FB – 0.9191FC (3)
Equating the i, j, and k components,
FA = FB = FC = 1.63 kN Ans
Solving Eqs. (1), (2), and (3), yields
91
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*2–104. The antenna tower is supported by three cables. If
the forces of these cables acting on the antenna are
, , and , determine the
magnitude and coordinate direction angles of the resultant
force acting at A.
FD = 560 NFC = 680 NFB = 520 N
24 m
10 m
18 m
8 m
16 m
12 m
18 m
z
x
y
A
O
C
DB
FB
FC
FD
02b Ch02b 57-106.indd 9102b Ch02b 57-106.indd 91 6/12/09 8:22:50 AM6/12/09 8:22:50 AM
FD = 560 = 560 – i + = –240i + 160j – 480k
rAD
rAD
⎛
⎝⎜
12
28
j –8
28
k24
28
⎛
⎝⎜
⎞
⎠⎟
⎞
⎠⎟
FC = 680 = 680 i + = 320i + 360j – 480k
rAC
rAC
⎛
⎝⎜
16
34
j –18
34
k24
34
⎛
⎝⎜
⎞
⎠⎟
⎞
⎠⎟
FB = 520 = 520 – j – = –200j – 480k
rAB
rAB
⎛
⎝⎜
10
26
k24
26
⎛
⎝⎜
⎞
⎠⎟
⎞
⎠⎟
FR = ©F = {80i + 320j – 1440k} N
FR = (80)
2 + (320)2 + (–1440)2 = 1477.3 = 1.48 kN Ans
a = cos–1 = 86.9° Ans80
1477.3
⎛
⎝⎜
⎞
⎠⎟
b = cos–1 = 77.5° Ans320
1477.3
⎛
⎝⎜
⎞
⎠⎟
g = cos–1 = 167° Ans–1440
1477.3
⎛
⎝⎜
⎞
⎠⎟
92
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•2–105. If the force in each cable tied to the bin is 350 N,
determine the magnitude and coordinate direction angles
of the resultant force.
B
C
E
D
A
x
y
1.8 m
0.9 m
0.9 m
0.6 m
0.6 m
FC
FD
FA
FB
E(0, 0, 1.8) m
D(–0.9, –0.6, 0) m
C(–0.9, 0.6, 0) m
B(0.9, 0.6, 0) m
A(0.9, –0.6, 0) m
Force Vectors: The unit vectors uA, uB, uC and uD of FA, FB, FC and FD must be determined first. 
From Fig. a,
uA = 
rA
Ar
 = 
(0.9 – 0) + (–0.6 – 0) + (0 – 1.8)
(0.9 –
i j k
0) + (–0.6 – 0) + (0 – 1.8)2 2 2
 = 
3
7
i – 
2
7
j – 
6
7
k
uB = 
rB
Br
 = 
(0.9 – 0) + (0.6 – 0) + (0 – 1.8)
(0.9 –
i j k
00) + (0.6 – 0) + (0 – 1.8)2 2 2
 
= 
3
7
i + 
2
7
j – 
6
7
k
uC = 
rC
Cr
 = 
(–0.9 – 0) + (0.6 – 0) + (0 – 1.8)
(–0.9
i j k
–– 0) + (0.6 – 0)+ (0 – 1.8)2 2 2
 = –
3
7
i + 
2
7
j – 
6
7
k
uD = 
rD
Dr
 = 
(–0.9 – 0) + (–0.6 – 0) + (0 – 1.8)
(–0.9
i j k
– 0) + (–0.6 – 0) + (0 – 1.8)2 2 2
 
= –
3
7
i – 
2
7
j – 
6
7
k
Thus, the force vectors FA, FB, FC and FD are given by
FA = FAuA = 350
3
7
–
2
7
–
6
7
i j k
⎛
⎝⎜
⎞
⎠⎟
 = [150i – 100j – 300k] N
FB = FBuB = 350
3
7
+
2
7
–
6
7
i j k
⎛
⎝⎜
⎞
⎠⎟
 = [150i + 100j – 300k] N
FC = FCuC = 350 –
3
7
+
2
7
–
6
7
i j k
⎛
⎝⎜
⎞
⎠⎟
 = [–150i + 100j – 300k] N
FD = FDuD = 350 –
3
7
–
2
7
–
6
7
i j k
⎛
⎝⎜
⎞
⎠⎟
 = [–150i – 100j – 300k] N
Resultant Force:
FR = FA + FB + FC + FD = (150i – 100j – 300k) + (150i + 100j – 300k) + (–150i + 100j – 300k) + (–150i – 100j – 300k)
= {–1200k} N
The magnitude of FR is
FR = ( ) + ( ) + ( )
2F F FR x R y R z
2 2 
= 0 + 0 + (–1200)2 = 1200 N Ans
The coordinate direction angles of FR are
 = cos–1
(F
F
R x
R
)⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 = cos–1
0
1200
⎛
⎝⎜
⎞
⎠⎟
 = 90° Ans
 = cos–1
(F
F
R y
R
)⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 = cos–1
0
1200
⎛
⎝⎜
⎞
⎠⎟
 = 90° Ans
 = cos–1
(F
F
R z
R
)⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 = cos–1
–1200
1200
⎛
⎝⎜
⎞
⎠⎟
 = 180° Ans
02b Ch02b 57-106.indd 9202b Ch02b 57-106.indd 92 6/12/09 8:22:52 AM6/12/09 8:22:52 AM
93
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2–106. If the resultant of the four forces is
, determine the tension developed in
each cable. Due to symmetry, the tension in the four cables
is the same.
F 1.8 k kN
B
C
E
D
A
x
y
1.8 m
0.9 m
0.9 m
0.6 m
0.6 m
FC
FD
FA
FB
E(0, 0, 1.8) m
D(–0.9, –0.6, 0) m
C(–0.9, 0.6, 0) m
B(0.9, 0.6, 0) m
A(0.9, –0.6, 0) m
Force Vectors: The unit vectors uA, uB, uC and uD of FA, FB, FC and FD must be determined first. 
From Fig. a,
uA = 
rA
Ar
 = 
(0.9 – 0) + (–0.6 – 0) + (0 – 1.8)
(0.9 –
i j k
0) + (–0.6 – 0) + (0 – 1.8)2 2 2
 = 
3
7
i – 
2
7
j – 
6
7
k
uB = 
rB
Br
 = 
(0.9 – 0) + (0.6 – 0) + (0 – 1.8)
(0.9 –
i j k
00) + (0.6 – 0) + (0 – 1.8)2 2 2
 
= 
3
7
i + 
2
7
j – 
6
7
k
uC = 
rC
Cr
 = 
(–0.9 – 0) + (0.6 – 0) + (0 – 1.8)
(–0.9
i j k
–– 0) + (0.6 – 0) + (0 – 1.8)2 2 2
 = –
3
7
i + 
2
7
j – 
6
7
k
uD = 
rD
Dr
 = 
(–0.9 – 0) + (–0.6 – 0) + (0 – 1.8)
(–0.9
i j k
– 0) + (–0.6 – 0) + (0 – 1.8)2 2 2
 
= –
3
7
i – 
2
7
j – 
6
7
k
Since the magnitudes of FA, FB, FC and FD are the same and denoted as F, they can be written as
FA = FAuA = F
3
7
–
2
7
–
6
7
i j k
⎛
⎝⎜
⎞
⎠⎟
FB = FBuB = F
3
7
+
2
7
–
6
7
i j k
⎛
⎝⎜
⎞
⎠⎟
FC = FCuC = F –
3
7
+
2
7
–
6
7
i j k
⎛
⎝⎜
⎞
⎠⎟
FD = FDuD = F –
3
7
–
2
7
–
6
7
i j k
⎛
⎝⎜
⎞
⎠⎟
Resultant Force: The vectors addition of FA, FB, FC and FD is equal to FR. Thus,
FR = FA + FB + FC + FD 
{–1.8k} = F
3
7
–
2
–
6
7
i j k
7
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 + F
3
7
+
2
–
6
7
i j k
7
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 + F –
3
7
+
2
–
6
7
i j k
7
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
 + F –
3
7
–
2
–
6
7
i j k
7
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
–1.8k = –
24
7
k
Thus,
1.8 = 
24
7
F F = 0.525 kN Ans
02b Ch02b 57-106.indd 9302b Ch02b 57-106.indd 93 6/12/09 8:22:54 AM6/12/09 8:22:54 AM
94
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2–107. The pipe is supported at its end by a cord AB. If the
cord exerts a force of on the pipe at A, express
this force as a Cartesian vector.
60 N
0.9 m
20
yx
A
B
1.5 m
1.8 m
F 60 N
*2–108. The load at A creates a force of 200 N in wire AB.
Express this force as a Cartesian vector, acting on A and
directed towards B.
2 m
1 m
30
120
120
B
A
y
x
F 200 N
Unit Vector : The coordinates of point A are
A (1.5, 0.9 cos 20°, –0.9 sin 20°) m = A (1.5, 0.8457, –0.3078) m
Then
rAB = {[0 – 1.5]i + (0 – 0.8457)j + (–0.3078)k} m
= {–1.5i – 0.8457j + 2.1078k} m
rAB = (–1.5) + (–0.8457) + 2.1078
2 2 2 = 2.7218 m
uAB = 
rAB
ABr
 = 
–1.5 – 0.8457 + 2.1078
2.7218
i j k
= –0.5511i – 0.3107j – 0.7744k
Force Vector :
F = FuAB = 60{–0.5511i – 0.3107j – 0.7744k} N
= {–33.1i – 18.6j + 46.5k} N Ans
02b Ch02b 57-106.indd 9402b Ch02b 57-106.indd 94 6/12/09 8:22:56 AM6/12/09 8:22:56 AM
rAB = (1 sin 30° – 0)i + (1 cos 30° – 0)j + (2 – 0)k
 = {0.5i + 0.866j + 2k} m
F = 200uAB = {44.7i + 77.5j + 179k} N Ans
rAB = (0.5)
2 + (0.866)2 + (2)2 = 2.236 m
uAB = = 0.2236i + 0.3873j + 0.8944k
rAB
rAB
⎛
⎝⎜
⎞
⎠⎟
x
z
y
D
C
A
B
3 m
30
0.75 m
45
FB 8 kN
FC 5 kN
FA 6 kN
95
•2–109. The cylindrical plate is subjected to the three cable
forces which are concurrent at point D. Express each force
which the cables exert on the plate as a Cartesian vector,
and determine the magnitude and coordinate direction
angles of the resultant force.
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02b Ch02b 57-106.indd 9502b Ch02b 57-106.indd 95 6/12/09 8:22:57 AM6/12/09 8:22:57 AM
rA = (0 – 0.75)i + (0 – 0)j + (3 – 0)k = {–0.75i + 0j + 3k} m
rB = [0 – (–0.75 sin 30°)]i + (0 – 0.75 cos 30°)j + (3 – 0)k
 = {0.375i – 0.6495j + 3k} m
rC = [0 – (–0.75 sin 45°)]i + [0 – (–0.75 cos 45°)]j + (3 – 0)k
 = {0.5303i + 0.5303j + 3k} m
 = {–1.4552i + 5.8209k} kN
 = {–1.46i + 5.82k} kN Ans
rA = (–0.75)
2 + 02 + 32 = 3.0923 m
rC = (0.5303)
2 + (0.5303)2 + 32 = 3.0923 m
rB = (0.375)
2 + (–0.6495)2 + 32 = 3.0923 m
FA = FA = 6
rA
rA
⎛
⎝⎜
⎞
⎠⎟
– 0.75i + 3k
3.0923
⎛
⎝⎜
⎞
⎠⎟
 = {0.8575i + 0.8575j + 4.8507k} kN
 = {0.857i + 0.857j + 4.85k} kN Ans
FC = FC = 5
rC
rC
⎛
⎝⎜
⎞
⎠⎟
0.5303i + 0.5303j + 3k 
3.0923
⎛
⎝⎜
⎞
⎠⎟
 = {0.9701i – 1.6803j + 7.7611k} kN
 = {0.970i – 1.68j + 7.76k} kN Ans
FB = FB = 8
rB
rB
⎛
⎝⎜
⎞
⎠⎟
0.375i – 0.6495j + 3k 
3.0923
⎛
⎝⎜
⎞
⎠⎟
FR = FA + FB + FC
 = {–1.4552i + 5.8209k} + {0.9701i – 1.6803j + 7.7611k}
 + {0.8575i + 0.8575j + 4.8507k}
 = {0.3724i – 0.8228j + 18.4327k} kN
FR = (0.3724)
2 + (–0.8228)2 + (18.4327)2
 = 18.4548 kN = 18.5 kN Ans
 = 0.02018i – 0.04459j + 0.9988k
uR = =
FR
FR
0.3724i – 0.8228j + 18.4327k 
18.4548
cos a = 0.02018 a = 88.8° Ans
cos b = –0.04458 b = 92.6° Ans
cos g = 0.9988 g = 2.81° Ans
96
2–110. The cable attached to the shear-leg derrick exerts a
force on the derrick of . Express this force as a
Cartesian vector.
1.75 kN
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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.
30
15 m
10.5 m
x
y
A
B
F 1.75 kN
2–111. Given the three vectors A, B, and D, show that
.A (B D) (A B) (A D)
A (B D) A B A D (QED)
Unit Vector : The coordinates of point B are
B (15 sin 30°, 15 cos 30°, 0) m = B (7.5, 12.99, 0) m
Then
rAB = {[7.5 – 0]i + (12.99 – 0)j + (0 – 10.5)k} m
= {7.5i – 12.99j + 10.5k} m
rAB = 7.5 + 12.99 + (–10.5)
2 2 2 = 18.3096 m
uAB = 
rAB
ABr
 = 
7.5 – 12.99 + 10.5
18.3096
i j k
= 0.4096i – 0.7094j – 0.5735k
Force Vector :
F = FuAB = 1.75{0.4096i + 0.7094j – 0.5735k} N
= {0.717i + 1.24j – 1.00k} kN Ans
02b Ch02b 57-106.indd 9602b Ch02b 57-106.indd 96 6/12/09 8:22:58 AM6/12/09 8:22:58 AM
A · (B + D) = (Axi + Ay j + Azk) · [(Bx + Dx)i + (By + Dy)j + (Bz + Dz)k]
 = Ax(Bx + Dx) + Ay(By + Dy) + Az(Bz + Dz)
 = (AxBx + AyBy + AzBz) + (AxDx + AyDy + AzDz)
 = (A · B) + (A · D) (QED) 
Also,
Since the componentof (B + D) is equal to the sum of the components of B and D, then
97
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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.
*2–112. Determine the projected component of the force
acting along cable AC. Express the result as a
Cartesian vector.
FAB = 560 N
z
x
y
C
B
A
3 m
1.5 m
1 m
3 m FAB 560 N
1.5 m
02b Ch02b 57-106.indd 9702b Ch02b 57-106.indd 97 6/12/09 8:22:59 AM6/12/09 8:22:59 AM
 = 
(–1.5 – 0)2 + (0 – 3)2 + (1 – 0)2
(–1.5 – 0)i + (0 – 3)j + (1 – 0)kuAB =
rAB
rAB
= – i – 3
7
j + 6
7
k 2
7
 = 
(1.5 – 0)2 + (0 – 3)2 + (1 – 0)2
(1.5 – 0)i + (0 – 3)j + (3 – 0)kuAC =
rAC
rAC
= i – 1
3
j + 2
3
k 2
3
Force Vectors: The unit vectors uAB and uAC must be determined first. From Fig. a,
Thus, the force vector FAB is given by
Vector Dot Product: The magnitude of the projected component of FAB is
Thus, (FAB)AC expressed in Cartesian vector form is
FAB = FABuAB = 560 = [–240i – 480j + 160k] N
⎛
⎝⎜
⎞
⎠⎟
– i – 3
7
j + 6
7
k 2
7
(FAB)AC = FAB · uAC = (–240i – 480j + 160k) ·
⎛
⎝⎜
⎞
⎠⎟
i – 1
3
j + 2
3
k 2
3
= (–240)
= 346.67 N
+ (–480) –
⎛
⎝⎜
⎞
⎠⎟
1
3
+ 160
⎛
⎝⎜
⎞
⎠⎟
2
3
⎛
⎝⎜
⎞
⎠⎟
2
3
(FAB)AC = (FAB)ACuAC = 346.67
⎛
⎝⎜
⎞
⎠⎟
i – 1
3
j + 2
3
k 2
3
= [116i – 231j + 231k] N Ans
98
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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.
•2–113. Determine the magnitudes of the components of
force acting along and perpendicular to line AO.F = 56 N
y
x
z
C
O
D
A
B
3 m
1.5 m
1 m
1 m F 56 N
02b Ch02b 57-106.indd 9802b Ch02b 57-106.indd 98 6/12/09 8:23:01 AM6/12/09 8:23:01 AM
 = 
[0–(–1.5)]2 + (0 – 3)2 + (0 – 1)2
[0–(–1.5)]i + (0 – 3)j + (0 – 1)kuAO =
rAO
rAO
= i – 3
7
j – 6
7
k 2
7
 = 
[0–(–1.5)]2 + (0 – 3)2 + (2 – 1)2
[0–(–1.5)]i + (0 – 3)j + (2 – 1)kuAD =
rAD
rAD
= i – 3
7
j + 6
7
k 2
7
Unit Vectors: The unit vectors uAD and uAO must be determined first. From Fig. a,
Thus, the force vector F is given by
Vector Dot Product: The magnitude of the projected component of F parallel to line AO is
F = FuAD = 56 = [24i – 48j + 16k] N
⎛
⎝⎜
⎞
⎠⎟
i – 3
7
j + 6
7
k 2
7
(FAO)paral = F · uAO = (24i – 48j + 16k) ·
The component of F perpendicular to line AO is
⎛
⎝⎜
⎞
⎠⎟
i – 3
7
j – 6
7
k 2
7
= (24)
= 46.86 N = 46.9 N Ans
+ (–48) –
⎛
⎝⎜
⎞
⎠⎟
3
7
+ (16) –
⎛
⎝⎜
⎞
⎠⎟
2
7
⎛
⎝⎜
⎞
⎠⎟
6
7
(FAO)per = F
2 – (F AO)paral
 = 562 – 46.862
 = 30.7 N Ans
99
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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.
2–114. Determine the length of side BC of the triangular
plate. Solve the problem by finding the magnitude of rBC;
then check the result by first finding q , rAB, and rAC and
then using the cosine law.
y
x
A
C
B
z
1 m
4 m
3 m
3 m
1 m
5 m
u
02b Ch02b 57-106.indd 9902b Ch02b 57-106.indd 99 6/12/09 8:23:02 AM6/12/09 8:23:02 AM
rBC = {3i + 2j – 4k} m
rBC = (3)
2 + (2)2 + (–4)2 = 5.39 m Ans
rAC = {3i + 4j – 1k} m
rAC = (3)
2 + (4)2 + (–1)2 = 5.0990 m
rAB = {2j + 3k} m
rAB = (2)
2 + (3)2 = 3.6056 m
rAC · rAB = 0 + 4(2) + (–1)(3) = 5
u = 74.219°
rBC = (5.0990)
2 + (3.6056)2 – 2(5.0990)(3.6056) cos 74.219°
rBC = 5.39 m Ans
Also,
u = cos–1 = cos–1
rAC · rAB
rAC rAB
⎛
⎝⎜
5
(5.0990)(3.6056)
⎞
⎠⎟
100
2–115. Determine the magnitudes of the components of
acting along and perpendicular to segment DE
of the pipe assembly.
F = 600 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.
x y
E
D
C
B
A
z
2 m
2 m
2 m
2 m
3 m
F 600 N
02b Ch02b 57-106.indd 10002b Ch02b 57-106.indd 100 6/12/09 8:23:03 AM6/12/09 8:23:03 AM
 = 
(0 – 4)2 + (2 – 5)2 + [0 – (–2)]2
(0 – 4)i + (2 – 5)j + [0 – (–2)kuEB =
rEB
rEB
= –0.7428i – 0.5571j + 0.3714k 
F = FuEB = 600(–0.7428i – 0.5571j + 0.3714k) = [–445.66i – 334.25j + 222.83k] N 
uED = –j
Unit Vectors: The unit vectors uEB and uED must be determined first. From Fig. a,
Thus, the force vector F is given by
Vector Dot Product: The magnitude of the component of F parallel to segment DE of 
the pipe assembly is
The component of F perpendicular to segment DE of the pipe assembly is
(FED)paral = F · uED = (–445.66i – 334.25j + 222.83k) · (–j)
 = (–445.66)(0) + (–334.25)(–1) + (222.83)(0)
 = 334.25 = 334 N Ans
(F ED)per = F
2 – (F ED)paral
2 = 6002 – 334.252 = 498 N Ans
101
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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.
*2–116. Two forces act on the hook. Determine the angle
between them.Also, what are the projections of F1 and F2
along the y axis?
u
x
z
y
45
60
120
F1 600 N
F2 {120i + 90j – 80k}N
u
•2–117. Two forces act on the hook. Determine the
magnitude of the projection of F2 along F1.
x
z
y
45
60
120
F1 600 N
F2 {120i + 90j – 80k}N
u
02b Ch02b 57-106.indd 10102b Ch02b 57-106.indd 101 6/12/09 8:23:04 AM6/12/09 8:23:04 AM
F
2
 = 120i + 90j – 80k; F
2
 = 170 N
u = j
So,
F
1
 · F
2
 = (–300)(120) + (300)(90) + (424.3)(–80) = –42 944 
F
1y = F1 · j = (300)(1) = 300 N Ans
F
2y = F2 · j = (90)(1) = 90 N Ans
u
1
 = cos 120°i + cos 60°j + cos 45°k
Proj F
2
 = F
2
 · u
1
 = (120)(cos 120°) + (90)(cos 60°) + (–80)(cos 45°)
|Proj F
2
| = 71.6 N Ans
F
1
 = 600 cos 120°i + 600 cos 60°j + 600 cos 45°k
 = –300i + 300j + 424.3k; F
1
 = 600 N
u = cos–1 = 115° Ans–42 944
(170)(600)
⎛
⎝⎜
⎞
⎠⎟
102
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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.
2–118. Determine the projection of force along
line BC. Express the result as a Cartesian vector.
F = 80 N
F 80 N
A
E
B
y
F
C
x
D
z
2 m
2 m
1.5 m
1.5 m
2 m
2 m
02b Ch02b 57-106.indd 10202b Ch02b 57-106.indd 102 6/12/09 8:23:05 AM6/12/09 8:23:05 AM
 = 
(4 – 2)2 + (0 – 2)2 + (0 – 0)2
(4 – 2)i + (0 – 2)j + (0 – 0)kuFC =
rFC
rFC
= 0.7071i – 0.7071j
 = 
(2 – 2)2 + (0 – 2)2 + (1.5 – 0)2
(2 – 2)i + (0 – 2)j + (1.5 – 0)kuFD =
rFD
rFD
= – j + 4
5
k 3
5
Unit Vectors: The unit vectors uFD and uFC must be determined first. From Fig. a,
Thus, the force vector F is given by
F = FuFD = 80 = [–64j + 48k] N
⎛
⎝⎜
⎞
⎠⎟
– j + 4
5
k 3
5
Vector Dot Product: The magnitude of the projected component of F along line BC is
FBC = F · uFC = (–64j + 48k) · (0.7071i – 0.7071j)
 = (0)(0.7071) + (–64)(–0.7071) + 48(0)
 = 45.25 = 45.2 N Ans
The component of FBC can be expressed in Cartesian vector form as
FBC = FBC(uFC) = 45.25(0.7071i – 0.7071j)
 = {32i – 32j} N Ans
103
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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.
2–119. The clamp is used on a jig. If the vertical force
acting on the bolt is , determine the
magnitudesof its components F1 and F2 which act along the
OA axis and perpendicular to it.
F = {-500k} N
z
O
x
y
40 mm
40 mm
20 mm
F { 500 k} N
A
02b Ch02b 57-106.indd 10302b Ch02b 57-106.indd 103 6/12/09 8:23:06 AM6/12/09 8:23:06 AM
(0 – 20)2 + (0 – 40)2 + (0 – 40)2
(0 – 20)i + (0 – 40)j + (0 – 40)kuAO = = – i – 
1
3
j – 2
3
k 2
3
Unit Vector: The unit vector along AO axis is
Projected Component of F Along OA Axis:
F
1
 = F · uAO = (–500k) ·
⎛
⎝⎜
⎞
⎠⎟
i – – 1
3
j – 2
3
k 2
3
= (0) –
= 333.33 N = 333 N Ans
+ (0) –
⎛
⎝⎜
⎞
⎠⎟
1
3
+ (–500) –
⎛
⎝⎜
⎞
⎠⎟
2
3
⎛
⎝⎜
⎞
⎠⎟
2
3
Component of F Perpendicular to OA Axis: Since the magnitude of 
force F is F = 500 N so that
F
2
 = F 2 – F
1
2) = 5002 – 333.332 = 373 N Ans
104
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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.
*2–120. Determine the magnitude of the projected
component of force FAB acting along the z axis.
3 m
4.5 m
3 m
x
B
D
C
A
O
y
3 m
9 m
FAB 3.5 kN
FAC 3 kN
30
A(0, 0, 9) m
B(4.5, –3, 0) m
Unit Vector: The unit vector uAB must be determined first. From Fig. a,
uAB = 
rAB
ABr
 = 
(4.5 – 0) + (–3 – 0) + (0 – 9)
(4.5 – 0)2
i j k
+ (–3 – 0) + (0 – 9)2 2
 = 
3
7
i – 
2
7
j – 
6
7
k
Thus, the force vector FAB is given by
FAB = FABuAB = 3.5
3
7
–
2
7
–
6
7
i j k
⎛
⎝⎜
⎞
⎠⎟
 = {1.5i – 1j – 3k} kN
Vector Dot Product: The projected component of FAB along the z axis is
(FAB)z = FAB · k = (1.5i – 1j – 3k) · k
= –3 kN
The negative sign indicates that (FAB)z is directed towards the negative z axis. Thus
(FAB)z = 3 kN Ans
02b Ch02b 57-106.indd 10402b Ch02b 57-106.indd 104 6/12/09 8:23:07 AM6/12/09 8:23:07 AM
105
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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.
•2–121. Determine the magnitude of the projected
component of force FAC acting along the z axis.
3 m
4.5 m
3 m
x
B
D
C
A
O
y
3 m
9 m
FAB 3.5 kN
FAC 3 kN
30
A(0, 0, 9) m
C(3 sin 30°, 3 cos 30°, 0) m
Unit Vector: The unit vector uAC must be determined first. From Fig. a,
uAC = 
rAC
ACr
 = 
(3 sin 30° – 0) + (3 cos 30° – 0) + (0 –i j 99)
(3 sin 30° – 0) + (3 cos 30° – 0) + (22
k
00 – 9)2
 = 0.1581i + 0.2739j – 0.9487k
Thus, the force vector FAC is given by
FAC = FACuAC = 3(0.1581i + 0.2739j – 0.9487k) = {0.4743i + 0.8217j – 2.8461k} kN
Vector Dot Product: The projected component of FAC along the z axis is
(FAC)z = FAC · k = (0.4743i + 0.8217j – 2.8461k) · k
= –2.8461 kN
The negative sign indicates that (FAC)z is directed towards the negative z axis. Thus
(FAC)z = 2.846 kN Ans
02b Ch02b 57-106.indd 10502b Ch02b 57-106.indd 105 6/12/09 8:23:10 AM6/12/09 8:23:10 AM
106
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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.
2–122. Determine the projection of force 
acting along line AC of the pipe assembly. Express the result
as a Cartesian vector.
F = 400 N
x
A
B
C
y
z
4 m
3 m
F 400 N
30
45
02b Ch02b 57-106.indd 10602b Ch02b 57-106.indd 106 6/12/09 8:23:13 AM6/12/09 8:23:13 AM
 = 
(0 – 0)2 + (4 – 0)2 + (3 – 0)2
(0 – 0)i + (4 – 0)j + (3 – 0)kuAC =
rAC
rAC
= j + 4
5
k 3
5
F = 400(– cos 45° sin 30°i + cos 45° cos 30°j + sin 45°k)
 = {–141.42i + 244.95j + 282.84k}
Force and unit Vector: The force vector F and unit vector uAC must be determined first. 
From Fig. a,
Vector Dot Product: The magnitude of the projected component of F along line AC is
FAC = F · uAC = (–141.42i + 244.95j + 282.84k) · j + 
4
5
k 3
5
⎛
⎝⎜
⎞
⎠⎟
Thus, FAC written in Cartesian vector form is
FAC = FACuAC = 365.66 = {293j + 219k} N Ansj + 
4
5
k 3
5
⎛
⎝⎜
⎞
⎠⎟
= (–141.42)(0) + 244.95
= 365.66 N Ans
+ 282.84
4
5
⎛
⎝⎜
⎞
⎠⎟
3
5
⎛
⎝⎜
⎞
⎠⎟
107
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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.
2–123. Determine the magnitudes of the components of
force acting parallel and perpendicular to
segment BC of the pipe assembly.
F = 400 N
x
A
B
C
y
z
4 m
3 m
F 400 N
30
45
02c Ch02c 107-120.indd 10702c Ch02c 107-120.indd 107 6/12/09 8:25:47 AM6/12/09 8:25:47 AM
Force Vector: The force vector F must be determined first. From Fig. a, 
 F = 400 (–cos 45° sin 30°i + cos 45° cos 30°j + sin 45°k)
 = {–141.42i + 244.95j + 282.84k} N
Vector Dot Product: By inspecting Fig. (a) we notice that uBC = j. Thus, the magnitude of the
component of F parallel to segment BC of the pipe assembly is
 (FBC)paral = F · j = (–141.42i + 244.95j + 282.84k) · j
 = –141.42(0) + 244.95(1) + 282.84(0)
 = 244.951b = 245 N Ans
The magnitude of the component of F perpendicular to segment BC of the pipe assembly can be
determined from
F2 – (FBC) paral = 400
2 – 244.952 = 316 N Ans (FBC)per =
108
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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.
*2–124. Cable OA is used to support column OB.
Determine the angle it makes with beam OC.u
z
x
C
O
D
y
4 m
30
8 m
8 m
A
u
f
•2–125. Cable OA is used to support column OB.
Determine the angle it makes with beam OD.f
z
x
C
B
O
D
y
4 m
30
8 m
8 m
A
u
f
02c Ch02c 107-120.indd 10
8
02c Ch02c 107-120.indd 108 6/12/09 8:25:48 AM6/12/09 8:25:48 AM
Unit Vector :
uOC
uOC
 = 1i 
 = 
(4 – 0)i + (8 – 0)j + (–8 – 0)k
(4 – 0)2 + (8 – 0)2 + (–8 – 0)2
 = 
1
3
i + 
2
3
j – 
2
3
k
The Angles Between Two Vectors u : B
 uOC · uOA = (1i) ·
1
3
i + 
2
3
j – 
2
3
k
⎛
⎝⎜
⎛
⎝⎜
⎞
⎠⎟
2
3
1
3
⎞
⎠⎟
1
3
⎛
⎝⎜
⎞
⎠⎟
2
3
⎛
⎝⎜
⎞
⎠⎟
= 1 + (0) + 0 – =
Then,
1
3
u = cos–1 (uOC · uOA) = cos
–1 = 70.5° Ans
Unit Vector :
uOD
uOA
 = –sin 30°i + cos 30°j = –0.5i + 0.8660j 
 = 
(4 – 0)i + (8 – 0)j + (–8 – 0)k
(4 – 0)2 + (8 – 0)2 + (–8 – 0)2
 = 
1
3
i + 
2
3
j – 
2
3
k
The Angles Between Two Vectors f :
 uOD · uOA = (–0.5i + 0.8660j) ·
 = (–0.5) + (0.8660) + 0 –
 = 0.4107
Then,
f = cos–1 (uOD · uOA) = cos
–1 0.4107 = 65.8° Ans
1
3
i + 
2
3
j – 
2
3
k
⎛
⎝⎜
⎞
⎠⎟
1
3
⎛
⎝⎜
⎞
⎠⎟
2
3
⎛
⎝⎜
⎞
⎠⎟
2
3
⎛
⎝⎜
⎞
⎠⎟
109
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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.
2–126. The cables each exert a force of 400 N on the post.
Determine the magnitude of the projected component of F1
along the line of action of F2.
x
z
y
20
35
45
60
120
F1 400 N
F2 400 N
u
2–127. Determine the angle between the two cables
attached to the post.
u
x
z
y
20
35
45
60
120
F1 400 N
F2 400 N
u
02c Ch02c 107-120.indd 10902c Ch02c 107-120.indd 109 6/12/09 8:25:49 AM6/12/09 8:25:49 AM
Force Vector :
Unit Vector : The unit vector along the line of action of F
2
 is
Projected Component of F1 Along Line of Action of F2 : 
uF2 = cos 45°i + cos 60°j + cos 120°k
 = 0.7071i + 0.5j – 0.5kuF1 = sin 35° cos 20°i – sin 35° sin 20°j + cos 35°k
 = 0.5390i – 0.1962j + 0.8192k
 F
1 = F1 uF1 = 400(0.5390i – 0.1962j + 0.8192k) N
 = {215.59i – 78.47j + 327.66k} N
Negative sign indicates that the force component (F
1
)F2
 acts in the opposite sense
of direction to that of uF2
.
thus the magnitude (F
1
)F2 
= 50.6 N Ans 
 
 (F
1
)F2 
= F
1
 · uF2 = (215.59i – 78.47j + 327.66k) · (0.7071i + 0.5j – 0.5k)
 = (215.59)(0.7071) + (–78.47)(0.5) + (327.66)(–0.5)
 = –50.6 N 
 
Unit Vector :
The Angles Between Two Vector u : The dot product of two unit
vectors must be determined first.
uF1 = sin 35° cos 20°i – sin 35° sin 20°j + cos 35°k
 = 0.5390i – 0.1962j + 0.8192k
uF1 · uF2 = (0.5390i – 0.1962j + 0.8192k) · (0.7071i + 0.5j – 0.5k)
 = 0.5390(0.7071) + (–0.1962)(0.5) + 0.8192(–0.5)
 = –0.1265
uF2 = cos 45°i + cos 60°j + cos 120°k
 = 0.7071i +0.5j – 0.5k
Then,
u = cos–1 (uF1 · uF2) = cos
–1 (–0.1265) = 97.3° Ans
110
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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.
*2–128. A force of is applied to the handle of
the wrench. Determine the angle between the tail of the
force and the handle AB.
u
F = 80 N
x
z
B
A
y
300 mm
500 mm
F 80 N
30
45
u
•2–129. Determine the angle between cables AB and AC.u
y
z
x
8 ft
3 ft
12 ft
8 ft
15 ft
A
C
B
F
u
2.4 m
2.4 m
4.5 m
3.6 m
0.9 m
Position Vector :
rAB = {(0 – 4.5)i + (0.9 – 0)j + (2.4 – 0)k} m
= {–4.5i + 0.9j + 2.4k} m
rAC = {(0 – 4.5)i + (–0.9 – 0)j + (3.6 – 0)k} m
= {–4.5i + 0.9j + 3.6k} m
The magnitudes of the position vectors are
rAB = (–4.5) + (0.9) + (2.4)
2 2 2 = 5.1788 m
rAC = (–4.5) + (–2.4) + (3.6)
2 2 2 = 6.2426 m
The Angles Between Two Vectors :
rAB · rAC = (–4.5i + 0.9j + 2.4k) · (–4.5i – 2.4j + 3.6k)
= (–4.5)(–4.5) + (0.9)(–2.4) + (2.4)(3.6)
= 26.73 m2
Then,
 = cos–1
r rAB AC
AB ACr r
·⎛
⎝
⎜
⎞
⎠
⎟ = cos–1
26.73
5.1788 (6.2426)
⎡
⎣
⎢
⎤
⎦
⎥ = 34.2° Ans
02c Ch02c 107-120.indd 11002c Ch02c 107-120.indd 110 6/12/09 8:25:50 AM6/12/09 8:25:50 AM
uF = – cos 30° sin 45°i + cos 30° cos 45°j + sin 30°k
cos u = uF · uAB = (–0.6124i + 0.6124j + 0.5k) · (–j)
 = –0.6124
 u = 128° Ans
uAB = – j
 = – 0.6124i + 0.6124j + 0.5k
111
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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.
2–130. If F has a magnitude of 55 lb, determine the
magnitude of its projected components acting along the x
axis and along cable AC.
y
z
x
8 ft
3 ft
12 ft
8 ft
15 ft
A
C
B
F
u
2–131. Determine the magnitudes of the projected
components of the force acting along the x and
y axes.
F = 300 N
z
A
O
x y
300 mm
300 mm
300 mm
F 300 N
30
30
2.4 m
2.4 m
4.5 m
3.6 m
0.9 m
Force Vector :
uAB = 
(0 – 4.5) + (0.9 – 0) + (2.4 – 0)
(0 – 4.
i j k
55) + (0.9 – 0) + (2.4 – 0)2 2 2
= –0.8689i + 0.1738j + 0.4634k
F = FuAB = 250(–0.8689i + 0.1738j + 0.4634k) N
= {–217.225i + 43.45j + 115.85k} N
Unit Vector : The unit vector along negative x axis and AC are
ux = –1i
uAC = 
(0 – 4.5) + (–2.4 – 0) + (3.6 – 0)
(0 – 4
i j k
)0–6.3(+)0–4.2–(+)5.. 2 2 2
= –0.7209i – 0.3845j + 0.5767k
Projected Component of F :
Fx = F · ux = (–217.225i + 43.45j + 115.85k) · (–1i)
= (–217.225)(–1) + 43.45(0) + 115.85(0)
= 217.2 N Ans
FAC = F · uAC = (–217.225i + 43.45j + 115.85k) · (–0.7209i – 0.3845j + 0.5767k)
= (–217.225)(–0.7209) + (43.45)(–0.3845) + (115.85)(0.5767)
= 206.7 N Ans
250 N,
02c Ch02c 107-120.indd 11102c Ch02c 107-120.indd 111 6/12/09 8:25:53 AM6/12/09 8:25:53 AM
Force Vector: The force vector F must be determined first. From Fig. a, 
 F = –300 sin 30° sin 30°i + 300 cos 30°j + 300 sin 30° cos 30°k
 = [–75i + 259.81j + 129.90k] N
The negative sign indicates that Fx is directed towards the negative x axis. Thus
Fx = 75 N, Fy = 260 N Ans
 Fy = F · j = (–75i + 259.81j + 129.90k) · j
 = –75(0) + 259.81(1) + 129.90(0)
 = 260 N
Vector Dot Product: The magnitudes of the projected component of F along the x and y axes are
 Fx = F · i = (–75i + 259.81j + 129.90k) · i
 = –75(1) + 259.81(0) + 129.90(0)
 = –75 N
112
*2–132. Determine the magnitude of the projected
component of the force acting along line OA.F = 300 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.
z
A
O
x y
300 mm
300 mm
300 mm
F 300 N
30
30
02c Ch02c 107-120.indd 11202c Ch02c 107-120.indd 112 6/12/09 8:25:55 AM6/12/09 8:25:55 AM
Force and Unit Vector: The force vector F and unit vector uOA must be determined first. 
From Fig. a, 
 F = –300 sin 30° sin 30°i + 300 cos 30°j + 300 sin 30° cos 30°k)
 = {–75i + 259.81j + 129.90k} N
Vector Dot Product: The magnitude of the projected component of F along line OA is
 FOA = F · uOA = (–75i + 259.81j + 129.90k) · (–0.75i + 0.5j + 0.4330k) 
 = (–75)(–0.75) + 259.81(0.5) + 129.90(0.4330)
 = 242 N Ans
uOA = 
rOA
OAr
 = 
(–0.45 – 0)i + (0.3 – 0)j + (0.2598 – 0)k
(–0.45 – 0)2 + (0.3 – 0)2 + (0.2598 – 0)2 
 = –0.75i + 0.5j + 0.4330k 
113
•2–133. Two cables exert forces on the pipe. Determine
the magnitude of the projected component of F1 along the
line of action of F2.
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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.
60
y
z
60
30
30
x
F2 25 lb
F1 30 lb
u
2–134. Determine the angle between the two cables
attached to the pipe.
u
60
y
z
60
30
30
x
F2 25 lb
F1 30 lb
u
25 N
30 N
25 N
30 N
02c Ch02c 107-120.indd 11302c Ch02c 107-120.indd 113 6/12/09 8:25:57 AM6/12/09 8:25:57 AM
Force Vector :
F
1
 = FFuF1 = 30(0.4330i + 0.75j – 0.5k) N
 = {12.990i + 22.5j – 15.0k} N
uF2 = cos 135°i + cos 60°j + cos 60°k = –0.7071i + 0.5j + 0.5k
uF1 = cos 30° sin 30°i + cos 30° cos 30°j – sin 30°k
 = 0.4330i + 0.75j – 0.5k
Unit Vector : One can obtain the angle a = 135° for F
2
 using Eq. 2–8
Projected Component of F1 Along the Line of Action of F2 :
cos2 a + cos2 b + cos2 g = 1, with b = 60° and g = 60°. The unit vector along the
line of action of F
2
 is
 
(F
1
)F2
 = F
1 
· uF2 = (12.990i + 22.5j – 15.0k) · (–0.7071i + 0.5j + 0.5k)
 = (12.990)(–0.7071) + (22.5)(0.5) + (–15.0)(0.5)
 = –5.44 N 
The Angles Between Two Vectors u :
uF1 · uF2 = (0.4330i + 0.75j – 0.5k) · (–0.7071i + 0.5j + 0.5k)
 = 0.4330(–0.7071) + 0.75(0.5) + (–0.5)(0.5)
 = –0.1812
Negative sign indicates that the projected component (F
1
)F2
 acts in the opposite
sense of direction to that of uF2.
The magnitude is (F
1
)F2
 = 5.44 N. Ans
Then,
u = cos–1 (uF1 · uF2) = cos
–1 (–0.1812) = 100° Ans
Unit Vector :
uF2 = cos 135°i + cos 60°j + cos 60°k
 = –0.7071i + 0.5j + 0.5k
uF1 = cos 30° sin 30°i + cos 30° cos 30°j – sin 30°k
 = 0.4330i + 0.75j – 0.5k
114
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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.
Fx = –700 cos 30° = –606 N Ans
Fy = 700 sin 30° = 350 N Ans
2–135. Determine the x and y components of the 700-lb
force.
y
x
700 lb
30
60
700 N
700 N
700 N
N
02c Ch02c 107-120.indd 11402cCh02c 107-120.indd 114 6/12/09 8:25:59 AM6/12/09 8:25:59 AM
115
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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.
*2–136. Determine the magnitude of the projected
component of the 100-lb force acting along the axis BC of
the pipe.
yC
B
A
D
z
8 ft
3 ft
6 ft
2 ft
4 ftx
F 100 lb
u
•2–137. Determine the angle between pipe segments
BA and BC.
u
yC
B
A
D
z
8 ft
3 ft
6 ft
2 ft
4 ftx
F 100 lb
u
2.4 m
500 N
1.8 m
0.9 m
0.6 m
1.2 m
2.4 m
500 N
1.8 m
0.9 m
0.6 m
1.2 m
Force Vector :
uCD = 
(0 – 1.8) + (3.6 – 1.2) + [0 – (–0.6)]
(0
i j k
– 1.8) + (3.6 – 1.2) + [0 – (–0.6)]2 2 2
= –0.5883i + 0.7845j + 0.1961k
F = FuCD = 500(–0.5883i + 0.7845j + 0.1961k)
= {–294.17i + 392.23j + 98.058k} N
Unit Vector : The unit vector along CB is
uCB = 
(0 – 1.8) + (0 – 1.2) + [0 – (–0.6)]
(0 –
i j k
1.8) + (0 – 1.2) + [0 – (–0.6)]2 2 2
= –0.8018i – 0.5345j + 0.2673k
Projected Component of F Along CB:
FCB = F · uCB = (–294.17i + 392.23j + 98.058k) · (–0.8018i – 0.5345j + 0.2673k)
= (–294.17)(–0.8018) + (392.23)(–0.5345) + (98.058)(0.2673)
= 52.43 N Ans
Position Vector :
rBA = {–0.9i} m
rBC = {(1.8 – 0)i + (1.2 – 0)j + (–0.6 – 0)k} m
= {1.8i + 1.2j + 0.6k} m
The magnitudes of the position vectors are
rBA = 0.9 m
rBC = 1.8 + 1.2 + (–0.6)
2 2 2 = 2.245 m
The Angles Between Two Vectors :
rBA · rBC = (–0.9i)(1.8i + 1.2j – 0.6k)
= (–0.9)(1.8) + (0)(1.2) + (0)(–0.6)
= –1.62 m2
Then,
 = cos–1
r rBA BC
BA BCr r
·⎛
⎝
⎜
⎞
⎠
⎟ = cos–1
–1.62
0.9 (2.245)
⎡
⎣
⎢
⎤
⎦
⎥ = 143° Ans
500-N
02c Ch02c 107-120.indd 11502c Ch02c 107-120.indd 115 6/12/09 8:26:00 AM6/12/09 8:26:00 AM
116
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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.
x
y
30
30
45
F1 80 N
F2 75 N
F3 50 N
2–138. Determine the magnitude and direction of the
resultant of the three forces by first
finding the resultant and then forming
. Specify its direction measured counter-
clockwise from the positive x axis.
FR = F¿ + F2
F¿ = F1 + F3
FR = F1 + F2 + F3
02c Ch02c 107-120.indd 11602c Ch02c 107-120.indd 116 6/12/09 8:26:01 AM6/12/09 8:26:01 AM
FR = (104.7)
2 + (75)2 – 2(104.7)(75) cos 162.46°
F ¿ = (80)2 + (50)2 – 2(80)(50) cos 105° = 104.7 N
sin f
80
sin 105°
104.7
= ; f = 47.54°
sin b
104.7
sin 162.46°
177.7
= ; b = 10.23°
FR = 177.7 = 178 N Ans
u = 75° + 10.23° = 85.2° Ans
2–139. Determine the design angle ( < 90°) between
the two struts so that the 500-lb horizontal force has a
component of 600 lb directed from A toward C. What is the
component of force acting along member BA?
uu
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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.
500 lb
20
A
B
C
u
N
N
N
N
500 N
N
N
02c Ch02c 107-120.indd 11702c Ch02c 107-120.indd 117 6/12/09 8:26:02 AM6/12/09 8:26:02 AM
117
Applying the law of cosines to Fig. b,
Applying the law of sines to Fig. b and using this result yields
5002 + 6002 – 2(500)(600) cos 20°FR =
214.91 N = 215 N
u = 52.7°
 = Ans
Anssin u
500
sin 20°
214.91
=
The parallelogram law of addition and the triangular rule are shown in Figs. a and b.
118
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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.
*2–140. Determine the magnitude and direction of the
smallest force F3 so that the resultant force of all three
forces has a magnitude of 20 lb.
F2 10 lb
F3
4
3
5
F1 5 lb
u
50 N
25 N
50 N
25 N
68.007 N
100 N
100 N.
F3 is minimum :
F3 = 100 – ⎪F1 + F2⎪
F1 + F2 = 25 + 50
3
5
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝
⎜
⎞
⎠
⎟ i + 50
4
5
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝
⎜
⎞
⎠
⎟ j = 55i + 40j
⎪F1 + F2⎪ = 55 + 402 2 = 68.007 N
 = tan–1
40
55
⎛
⎝⎜
⎞
⎠⎟
 = 36.0° Ans
Thus
(F3)min = 100 – 68.007 = 31.993 N Ans
02c Ch02c 107-120.indd 11802c Ch02c 107-120.indd 118 6/12/09 8:26:03 AM6/12/09 8:26:03 AM
119
•2–141. Resolve the 250-N force into components acting
along the u and axes and determine the magnitudes of
these components.
v
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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.
u
v
40
20
250 N
2–142. Cable AB exerts a force of 80 N on the end of the
3-m-long boom OA. Determine the magnitude of the
projection of this force along the boom.
O
A
80 N
3 m
B
z
y
x
4 m
60
02c Ch02c 107-120.indd 11902c Ch02c 107-120.indd 119 6/12/09 8:26:04 AM6/12/09 8:26:04 AM
Vector Analysis:
Scalar Analysis:
 F = 80
rAB
rAB
⎛
⎝⎜
⎛
⎝⎜
3 cos 60°
5
⎞
⎠⎟
= 80 i – – j + 
⎞
⎠⎟
3 sin 60°
5
4 
5
k
– 24 i – 41.57 j + 64 k
uAO = – cos 60° i – sin 60° j = – 0.5 i – 0.866 j
Proj F = F · uAO (–24)(–0.5) + (–41.57)(–0.866) + (64)0 = 48.0 N Ans
 =
250 
sin 120°
Fu
sin 40°
= ; Fu = 186 N Ans
250 
sin 120°
Fu
sin 20°
= ; Fu = 98.7 N Ans
 Angle OAB = tan–1 = 53.13°
⎛
⎝⎜
⎞
⎠⎟
4
3
 Proj F = 80 cos 53.13° = 48.0 N Ans
120
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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.
2–143. The three supporting cables exert the forces shown
on the sign. Represent each force as a Cartesian vector.
2 m
z
C
2 m
y
x
A
D
E
B
3 m
3 m
2 m
FB 400 N
FC 400 N
FE 350 N
02c Ch02c 107-120.indd 12002c Ch02c 107-120.indd 120 6/12/09 8:26:05 AM6/12/09 8:26:05 AM
Unit Vector :
Force Vector :
rAB = {(0 – 5)i + (2 – 0)j + (3 – 0)k} m ={–5i + 2j + 3k} m
rAB = (–5)
2 + 22 + 32 = 6.164 m
uAB = = = –0.8111i + 0.3244j + 0.4867k
rAB
rAB
–5i + 2j + 3k
6.164
rAC = {(0 – 5)i + (–2 – 0)j + (3 – 0)k} m ={–5i – 2j + 3k} m
rAC = (–5)
2 + (–2)2 + 32 = 6.164 m
uAC = = = –0.8111i – 0.3244j + 0.4867k
rAC
rAC
–5i – 2j + 3k
6.164
rDE = {(0 – 2)i + (0 – 0)j + (3 – 0)k} m ={–2i + 3k} m
rDE = (–2)
2 + 32 = 3.605 m
uDE = = = –0.5547i + 0.8321k
rDE
rDE
–2i + 3k
3.605
FE = FEuDE = 350{–0.5547i + 0.8321k} N
 = {–194.15i + 291.22k} N
 = {–194i + 291k} N Ans
FC = FCuAB = 400{–0.8111i – 0.3244j + 0.4867k} N
 = {–324.44i – 129.78j + 194.67k} N
 = {–324i – 130j + 195k} N Ans
FB = FBuAB = 400{–0.8111i + 0.3244j + 0.4867k} N
 = {–324.44i = 129.78j = 194.67k} N
 = {–324i + 130j + 195k} N Ans
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