Statics Mechanics of Materials 4th Asia Edition Solutions © Pearson Education South Pte Ltd Hibbeler 2014. All rights reserved. ThisManual 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. Full Download: http://alibabadownload.com/product/statics-mechanics-of-materials-4th-edition-hibbeler-solutions-manual/
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2–2. If u = 60° and F = 450 N, determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.
y F u 15⬚ 700 N
SOLUTION The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of consines to Fig. b, FR = 27002 + 4502 - 2(700)(450) cos 45° = 497.01 N = 497 N
Ans.
This yields sin 45° sin a = 700 497.01
a = 95.19°
Thus, the direction of angle f of FR measured counterclockwise from the positive x axis, is f = a + 60° = 95.19° + 60° = 155°
Ans.
18
x
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2–3. y
If the magnitude of the resultant force is to be 500 N, directed along the positive y axis, determine the magnitude of force F and its direction u.
F u 15⬚ 700 N
SOLUTION 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, F = 25002 + 7002 - 2(500)(700) cos 105° = 959.78 N = 960 N
Ans.
Applying the law of sines to Fig. b, and using this result, yields sin (90° + u) sin 105° = 700 959.78 u = 45.2°
Ans.
19
x
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2–4. Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured clockwise from the positive u axis. 70 u
30 45 F2
SOLUTION FR = 2(300)2 + (500)2 - 2(300)(500) cos 95° = 605.1 = 605 N
Ans.
500 605.1 = sin 95° sin u u = 55.40° f = 55.40° + 30° = 85.4°
Ans.
20
500 N v
F1
300 N
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2–5. Resolve the force F1 into components acting along the u and v axes and determine the magnitudes of the components. 70 u
30 45 F2
SOLUTION F1u 300 = sin 40° sin 110° F1u = 205 N
Ans.
F1v 300 = sin 30° sin 110° F1v = 160 N
Ans.
21
500 N v
F1
300 N
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2–6. Resolve the force F2 into components acting along the u and v axes and determine the magnitudes of the components. 70⬚ 30⬚ 45⬚ F2 ⫽ 500 N v
SOLUTION F2u 500 = sin 45° sin 70° F2u = 376 N
Ans.
F2v 500 = sin 65° sin 70° F2v = 482 N
Ans.
22
u
F1 ⫽ 300 N
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2–7. If FB = 2 kN and the resultant force acts along the positive u axis, determine the magnitude of the resultant force and the angle u.
y
FA ⫽ 3 kN u 30⬚
B
u FB
23
x
A
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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.
y
FA ⫽ 3 kN u 30⬚
B
u FB
24
x
A
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2–9. Resolve F1 into components along the u and v axes and determine the magnitudes of these components.
v F1
F2
SOLUTION
150 N
30
30
Sine law:
105
F1v 250 = sin 30° sin 105°
F1v = 129 N
Ans.
F1u 250 = sin 45° sin 105°
F1u = 183 N
Ans.
25
250 N
u
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2–10. Resolve F2 into components along the u and v axes and determine the magnitudes of these components.
v F1
F2
SOLUTION
150 N
30
30
Sine law:
105
F2v 150 = sin 30° sin 75°
F2v = 77.6 N
Ans.
F2u 150 = sin 75° sin 75°
F2u = 150 N
Ans.
26
250 N
u
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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.
y¿
y 10⬚
x¿x¿ xx 60⬚ 360 N
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29
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2–14. 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.
y¿
y 10⬚
x¿ x 60⬚ 360 N
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30
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2–15. The plate is subjected to the two forces at A and B as shown. If u = 60°, determine the magnitude of the resultant of these two forces and its direction measured clockwise from the horizontal.
FA u
A
SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using law of cosines (Fig. b), we have FR = 282 + 62 - 2(8)(6) cos 100° 40
= 10.80 kN = 10.8 kN
Ans. B
The angle u can be determined using law of sines (Fig. b).
FB
sin 100° sin u = 6 10.80 sin u = 0.5470 u = 33.16° Thus, the direction f of FR measured from the x axis is f = 33.16° - 30° = 3.16°
Ans.
31
6 kN
8 kN
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2–16. Determine the angle of u 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?
FA u
A
SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using law of sines (Fig .b), we have sin (90° - u) sin 50° = 6 8
40 B
sin (90° - u) = 0.5745
FB
u = 54.93° = 54.9°
Ans.
From the triangle, f = 180° - (90° - 54.93°) - 50° = 94.93°. Thus, using law of cosines, the magnitude of FR is FR = 282 + 62 - 2(8)(6) cos 94.93° = 10.4 kN
Ans.
32
6 kN
8 kN
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33
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2–19. Determine the magnitude and direction of the resultant FR = F1 + F2 + F3 of the three forces by first finding the resultant F¿ = F1 + F2 and then forming FR = F¿ + F3.
y F1
30 N 5 3
F3
4
50 N x
20
SOLUTION
F2
F¿ = 2(20)2 + (30)2 - 2(20)(30) cos 73.13° = 30.85 N 30 30.85 = ; sin 73.13° sin (70° - u¿)
u¿ = 1.47°
FR = 2(30.85)2 + (50)2 - 2(30.85)(50) cos 1.47° = 19.18 = 19.2 N
Ans.
30.85 19.18 = ; sin 1.47° sin u
Ans.
u = 2.37°
35
20 N
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2–20. Determine the magnitude and direction of the resultant FR = F1 + F2 + F3 of the three forces by first finding the resultant F¿ = F2 + F3 and then forming FR = F¿ + F1.
y F1
30 N 5 3
F3
4
50 N x
20
SOLUTION
F2
F ¿ = 2(20)2 + (50)2 - 2(20)(50) cos 70° = 47.07 N 20 sin u
¿
=
47.07 ; sin 70°
u¿ = 23.53°
FR = 2(47.07)2 + (30)2 - 2(47.07)(30) cos 13.34° = 19.18 = 19.2 N 19.18 30 = ; sin 13.34° sin f
Ans.
f = 21.15°
u = 23.53° - 21.15° = 2.37°
Ans.
36
20 N
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2–21. Two forces act on the screw eye. If F1 = 400 N and F2 = 600 N, determine the angle u(0° … u … 180°) between them, so that the resultant force has a magnitude of FR = 800 N.
F1
u
SOLUTION The parallelogram law of addition and triangular rule are shown in Figs. a and b, respectively. Applying law of cosines to Fig. b, 800 = 2400 + 600 - 2(400)(600) cos (180° - u°) 2
2
8002 = 4002 + 6002 - 480000 cos (180° - u) cos (180° - u) = - 0.25 180° - u = 104.48 u = 75.52° = 75.5°
Ans.
37
F2
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2–22. Two forces F1 and F2 act on the screw eye. If their lines of action are at an angle u apart and the magnitude of each force is F1 = F2 = F, determine the magnitude of the resultant force FR and the angle between FR and F1.
F1
u
SOLUTION F F = sin f sin (u - f) sin (u - f) = sin f
F2
u - f = f f =
u 2
Ans.
FR = 2(F)2 + (F)2 - 2(F)(F) cos (180° - u) Since cos (180° - u) = -cos u FR = F A 22 B 21 + cos u u 1 + cos u Since cos a b = 2 A 2 Then u FR = 2F cosa b 2
Ans.
38
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2–23. Two forces act on the screw eye. If F = 600 N, determine the magnitude of the resultant force and the angle u if the resultant force is directed vertically upward.
y F 500 N
u
30⬚
x
SOLUTION The parallelogram law of addition and triangular rule are shown in Figs. a and b respectively. Applying law of sines to Fig. b, sin 30° sin u = ; sin u = 0.6 600 500
u = 36.87° = 36.9°
Ans.
Using the result of u, f = 180° - 30° - 36.87° = 113.13° Again, applying law of sines using the result of f, FR 500 = ; sin 113.13° sin 30°
FR = 919.61 N = 920 N
Ans.
39
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2–24. Two forces are applied at the end of a screw eye in order to remove the post. Determine the angle u 10° … u … 90°2 and the magnitude of force F so that the resultant force acting on the post is directed vertically upward and has a magnitude of 750 N.
y F 500 N
θ
30°
x
SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using law of sines (Fig. b), we have sin f sin 30° = 750 500 sin f = 0.750 f = 131.41° 1By observation, f 7 90°2 Thus, u = 180° - 30° - 131.41° = 18.59° = 18.6°
Ans.
F 500 = sin 18.59° sin 30° F = 319 N
Ans.
40
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2-25 . Determine the magnitude and direction of the resultant force F 1 . Express the result in terms of the magnitudes of the component F 2 and resultant F R and the angle T .
Solution: F1
2
F R F2 2FR F 2 cos T 2
2
Since cos 180 deg I F1
cos I ,
F R F2 2 F R F 2 cos T 2
2
Ans.
41
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2–26. The beam is to be hoisted using two chains. Determine 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°.
y FB
FA u
30
SOLUTION x
FA 600 = ; sin 45° sin 105°
FA = 439 N
Ans.
FB 600 = ; sin 30° sin 105°
FB = 311 N
Ans.
42
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2–27. The beam is to be hoisted using two chains. If the resultant force is to be 600 N directed along the positive y axis, determine the magnitudes of forces FA and FB acting on each chain and the angle u of FB so that the magnitude of FB is a minimum. FA acts at 30° from the y axis, as shown.
y FB
FA u
30
SOLUTION x
For minimum FB, require u = 60°
Ans.
FA = 600 cos 30° = 520 N
Ans.
FB = 600 sin 30° = 300 N
Ans.
43
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2–28. If the resultant force of the two tugboats is 3 kN, directed along the positive x axis, determine the required magnitude of force FB and its direction u.
y
A
FA ⫽ 2 kN 30⬚
x
u C
SOLUTION 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, FB = 222 + 32 - 2(2)(3)cos 30° = 1.615kN = 1.61 kN
Ans.
Using this result and applying the law of sines to Fig. b, yields sin u sin 30° = 2 1.615
u = 38.3°
Ans.
44
FB
B
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2–29. If FB = 3 kN and u = 45°, determine the magnitude of the resultant force of the two tugboats and its direction measured clockwise from the positive x axis.
y
A
FA ⫽ 2 kN 30⬚
x
u C
SOLUTION 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, FR = 222 + 32 - 2(2)(3) cos 105° = 4.013 kN = 4.01 kN
Ans.
Using this result and applying the law of sines to Fig. b, yields sin 105° sin a = 3 4.013
a = 46.22°
Thus, the direction angle f of FR, measured clockwise from the positive x axis, is f = a - 30° = 46.22° - 30° = 16.2°
Ans.
45
FB
B
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2–30. If the resultant force of the two tugboats is required to be directed towards the positive x axis, and FB is to be a minimum, determine the magnitude of FR and FB and the angle u.
y
A
FA ⫽ 2 kN 30⬚
x
u
SOLUTION
C
FB
For FB to be minimum, it has to be directed perpendicular to FR. Thus, u = 90°
Ans.
The parallelogram law of addition and triangular rule are shown in Figs. a and b, respectively. By applying simple trigonometry to Fig. b, FB = 2 sin 30° = 1 kN
Ans.
FR = 2 cos 30° = 1.73 kN
Ans.
46
B
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2-31. If the tension in the cable is F1, determine the magnitude and direction of the resultant force acting on the pulley. This angle defines the same angle T of line AB on the tailboard block. Given: F1
400 N
T1
30 deg
Solution: FR FR
F1 F 1 2F 1 F 1 cos 90 deg T 1 2
2
400 N
sin 90 deg T FR
Ans.
sin T 1 F1
§ FR
90 deg asin ¨
T
© F1
T
60 deg
·
sin T 1 ¸
¹ Ans.
47
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2-32 . Determine the magnitude of the resultant force and its direction, measured clockwise from the positive x axis. Given: F1
70 N
F2
50 N
F3
65 N
T
30 deg
I
45 deg
Solution: o
F Rx = 6Fx;
F RX
F 1 F 2 cos T F 3 cos I
F Ry = 6F y;
F RY
F2 sin T F 3 sin I
n
FR
T FR
T
2
FRX FRY
2
§ FRY · ¸ © FRX ¹
atan ¨
97.8 N
Ans.
46.5 deg
Ans.
48
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2–33. Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.
y F3
750 N 45
x
SOLUTION
3
5 4
+ F = ©F ; : Rx x
F Rx =
4 (850) - 625 sin 30° - 750 sin 45° = - 162.8 N 5
30 F2
+ c FRy = ©Fy ;
FRy
3 = - (850) - 625 cos 30° + 750 cos 45° = - 520.9 N 5
FR = 2 ( - 162.8)2 + (-520.9)2 = 546 N f = tan - 1 B
Ans.
- 520.9 R = 72.64° - 162.8
u = 180° + 72.64° = 253°
Ans.
49
625 N
F1
850 N
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2–34.
y 60⬚
Resolve F1 and F2 into their x and y components.
30⬚ F1 ⫽ 400 N
45⬚
SOLUTION F1 = {400 sin 30°( +i) +400 cos 30°(+j)} N
F2 ⫽ 250 N
= {200i+346j} N
Ans.
F2 = {250 cos 45°(+i)+250 sin 45°( -j)} N = {177i -177j} N
Ans.
50
x
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2–35. Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis.
y 60⬚ 30⬚ F1 ⫽ 400 N
SOLUTION Rectangular Components: By referring to Fig. a, the x and y components of F1 and F2 can be written as (F1)x = 400 sin 30° = 200 N
(F1)y = 400 cos 30° = 346.41 N
(F2)x = 250 cos 45° = 176.78 N
(F2)y = 250 sin 45° = 176.78 N
F2 ⫽ 250 N
Resultant Force: Summing the force components algebraically along the x and y axes, we have + : ©(FR)x = ©Fx;
(FR)x = 200 + 176.78 = 376.78 N
+ c ©(FR)y = ©Fy;
(FR)y = 346.41 - 176.78 = 169.63 N c
The magnitude of the resultant force FR is FR = 2(FR)x2 + (FR)y2 = 2376.782 + 169.632 = 413 N
Ans.
The direction angle u of FR, Fig. b, measured counterclockwise from the positive axis, is u = tan-1 c
(FR)y 169.63 d = tan-1 a b = 24.2° (FR)x 376.78
Ans.
51
45⬚
x
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2–36. Resolve each force acting on the gusset plate into its x and y components, and express each force as a Cartesian vector.
y F3 ⫽ 650 N 3
F2 ⫽ 750 N
5 4
45⬚ x F1 ⫽ 900 N
F1 = {900( +i)} = {900i} N
Ans.
F2 = {750 cos 45°( +i) + 750 sin 45°( +j)} N = {530i + 530j} N
Ans.
F3 = e 650a
4 3 b(+i) + 650 a b(-j) f N 5 5
= {520 i - 390j)} N
Ans.
52
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2–37. Determine the magnitude of the resultant force acting on the plate and its direction, measured counterclockwise from the positive x axis.
y F3 ⫽ 650 N 3
F2 ⫽ 750 N
5 4
SOLUTION
45⬚
Rectangular Components: By referring to Fig. a, the x and y components of F1, F2, and F3 can be written as (F1)x = 900 N
(F1)y = 0
(F2)x = 750 cos 45° = 530.33 N 4 (F3)x = 650 a b = 520 N 5
(F2)y = 750 sin 45° = 530.33 N 3 (F3)y = 650a b = 390 N 5
Resultant Force: Summing the force components algebraically along the x and y axes, we have + : ©(FR)x = ©Fx;
(FR)x = 900 + 530.33 + 520 = 1950.33 N :
+ c ©(FR)y = ©Fy;
(FR)y = 530.33 - 390 = 140.33 N c
The magnitude of the resultant force FR is FR = 2(FR)x2 + (FR)y2 = 21950.332 + 140.332 = 1955 N = 1.96 kN Ans. The direction angle u of FR, measured clockwise from the positive x axis, is u = tan-1 c
(FR)y 140.33 d = tan-1 a b = 4.12° (FR)x 1950.33
Ans.
53
x F1 ⫽ 900 N
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.
54
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2–39. y
Resolve each force acting on the support into its x and y components, and express each force as a Cartesian vector.
F2 ⫽ 600 N
F1 ⫽ 800 N
45⬚ 60⬚ x 13 5
SOLUTION F1 = {800 cos 60°( +i) + 800 sin 60°( +j)} N = {400i + 693j} N
Ans.
F2 = {600 sin 45°( -i) + 600 cos 45°( +j)} N = {-424i + 424j} N F3 = e 650a
Ans.
12 5 b(+i) + 650a b(-j) f N 13 13
Ans.
= {600i - 250j} N
55
12
F3 ⫽ 650 N
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2–40. y
Determine the magnitude of the resultant force and its direction u, measured counterclockwise from the positive x axis.
F2 ⫽ 600 N
F1 ⫽ 800 N
45⬚ 60⬚ x 13 5
SOLUTION Rectangular Components: By referring to Fig. a, the x and y components of F1, F2, and F3 can be written as (F1)x = 800 cos 60° = 400 N
(F1)y = 800 sin 60° = 692.82 N
(F2)x = 600 sin 45° = 424.26 N
(F2)y = 600 cos 45° = 424.26 N
(F3)x = 650 a
12 b = 600 N 13
(F3)y = 650 a
5 b = 250 N 13
Resultant Force: Summing the force components algebraically along the x and y axes, we have + : ©(FR)x = ©Fx;
(FR)x = 400 - 424.26 + 600 = 575.74 N :
+ c ©(FR)y = ©Fy;
(FR)y = - 692.82 + 424.26 - 250 = 867.08 N c
The magnitude of the resultant force FR is FR = 2(FR)x2 + (FR)y2 = 2575.742 + 867.082 = 1041 N = 1.04 kN Ans. The direction angle u of FR , Fig. b, measured counterclockwise from the positive x axis, is u = tan-1 c
(FR)y 867.08 d = tan-1 a b = 56.4° (FR)x 575.74
Ans.
56
12
F3 ⫽ 650 N
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2-41. Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis. Units Used: 3
kN
10 N
Given: F1
30 kN
F2
26 kN
T
30 deg
c
5
d
12
Solution: F Rx = 6F x; o
F Rx
F 1 sin T
c § · ¨ 2 2 ¸ F2 © c d ¹
F Rx
25 kN
F Ry
F 1 cos T
§ d ·F ¨ 2 2¸ 2 © c d ¹
F Ry
2 kN
FR
FRx FRy
FR
25.1 kN
n
F Ry = 6F y;
2
2
§ FRy · ¸ © FRx ¹
I
atan ¨
I
4.5 deg
E
180 deg I
E
184.5 deg
57
Ans.
Ans.
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2–42. Determine the magnitude and orientation u of FB so that the resultant force is directed along the positive y axis and has a magnitude of 1500 N.
y FB
FA = 700 N
30° A
B θ
x
SOLUTION Scalar Notation: Summing the force components algebraically, we have + F = ©F ; : Rx x
0 = 700 sin 30° - FB cos u FB cos u = 350
+ c FRy = ©Fy ;
(1)
1500 = 700 cos 30° + FB sin u FB sin u = 893.8
(2)
Solving Eq. (1) and (2) yields u = 68.6° FB = 960 N
Ans.
58
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2–43. Determine the magnitude and orientation, measured counterclockwise from the positive y axis, of the resultant force acting on the bracket, if FB = 600 N and u = 20°.
y FB
FA
30
700 N
A
B u
x
SOLUTION Scalar Notation: Summing the force components algebraically, we have + F = ©F ; : Rx x
FRx = 700 sin 30° - 600 cos 20° = - 213.8 N = 213.8 N ;
+ c FRy = ©Fy ;
FRy = 700 cos 30° + 600 sin 20° = 811.4 N c
The magnitude of the resultant force FR is FR = 2F2Rx + F2Ry = 2213.82 + 811.42 = 839 N
Ans.
The direction angle u measured counterclockwise from the positive y axis is u = tan - 1
FRx FRy
= tan - 1 ¢
213.8 ≤ = 14.8° 811.4
Ans.
59
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2–44. The magnitude of the resultant force acting on the bracket is to be 400 N. Determine the magnitude of F1 if f = 30°.
y
u
F2 ⫽ 650 N 4
5 3
45⬚
F1 f x
45⬚
SOLUTION Rectangular Components: By referring to Fig. a, the x and y components of F1, F2, and F3 can be written as (F1)x = F1 cos 30° = 0.8660F1
(F1)y = F1 sin 30° = 0.5F1
3 (F2)x = 650a b = 390 N 5
4 (F2)y = 650 a b = 520 N 5
(F3)x = 500 cos 45° = 353.55 N
(F3)y = 500 sin 45° = 353.55 N
Resultant Force: Summing the force components algebraically along the x and y axes, we have + : ©(FR)x = ©Fx;
(FR)x = 0.8660F1 - 390 + 353.55 = 0.8660F1 - 36.45
+ c ©(FR)y = ©Fy;
(FR)y = 0.5F1 + 520 - 353.55 = 0.5F1 + 166.45
Since the magnitude of the resultant force is FR = 400 N, we can write FR = 2(FR)x2 + (FR)y2 400 = 2(0.8660F1 - 36.45)2 + (0.5F1 + 166.45)2 F12 + 103.32F1 - 130967.17 = 0
Ans.
Solving, F1 = 314 N
or
F1 = -417 N
Ans.
The negative sign indicates that F1 = 417 N must act in the opposite sense to that shown in the figure.
60
F3 ⫽ 500 N
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2–45. y
If the resultant force acting on the bracket is to be directed along the positive u axis, and the magnitude of F1 is required to be minimum, determine the magnitudes of the resultant force and F1.
u
F2 ⫽ 650 N 4
5 3
45⬚
F1 f x
45⬚
SOLUTION 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
3 (F2)x = 650a b = 390 N 5
4 (F2)y = 650 a b = 520 N 5
(F3)x = 500 cos 45° = 353.55 N
(F3)y = 500 sin 45° = 353.55 N
(FR)x = FR cos 45° = 0.7071FR
(FR)y = FR sin 45° = 0.7071FR
Resultant Force: Summing the force components algebraically along the x and y axes, we have + : ©(FR)x = ©Fx; + c ©(FR)y = ©Fy;
0.7071FR = F1 cos f - 390 + 353.55
(1)
0.7071FR = F1 sin f + 520 - 353.55
(2)
Eliminating FR from Eqs. (1) and (2), yields F1 =
202.89 cos f - sin f
(3)
The first derivative of Eq. (3) is sin f + cos f dF1 = df (cos f - sin f)2
(4)
The second derivative of Eq. (3) is d2F1 df2 For F1 to be minimum,
=
2(sin f + cos f)2 (cos f - sin f)3
+
1 cos f - sin f
(5)
dF1 = 0 . Thus, from Eq. (4) df sin f + cos f = 0 tan f = -1 f = -45°
Substituting f = - 45° into Eq. (5), yields d2F1 df2
= 0.7071 > 0
This shows that f = -45° indeed produces minimum F1. Thus, from Eq. (3) F1 =
202.89 = 143.47 N = 143 N cos (- 45°) - sin (-45°)
Ans.
Substituting f = - 45° and F1 = 143.47 N into either Eq. (1) or Eq. (2), yields FR = 91.9 N
Ans. 61
F3 ⫽ 500 N
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2–46. y
If the magnitude of the resultant force acting on the bracket is 600 N, directed along the positive u axis, determine the magnitude of F and its direction f.
u
F2 ⫽ 650 N 4
5 3
45⬚
F1 f x
45⬚
SOLUTION 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
3 (F2)x = 650a b = 390 N 5
4 (F2)y = 650 a b = 520 N 5
(F3)x = 500 cos 45° = 353.55 N
(F3)y = 500 cos 45° = 353.55 N
(FR)x = 600 cos 45° = 424.26 N
(FR)y = 600 sin 45° = 424.26 N
Resultant Force: Summing the force components algebraically along the x and y axes, we have + : ©(FR)x = ©Fx;
424.26 = F1 cos f - 390 + 353.55
(1)
F1 cos f = 460.71 + c ©(FR)y = ©Fy;
424.26 = F1 sin f + 520 - 353.55
(2)
F1 sin f = 257.82 Solving Eqs. (1) and (2), yields f = 29.2°
F1 = 528 N
Ans.
62
F3 ⫽ 500 N
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2–47. Determine the magnitude and direction u of the resultant force FR. Express the result in terms of the magnitudes of the components F1 and F2 and the angle f.
SOLUTION
F1
F 2R
=
F 21
+
F 22
f
Since cos (180° - f) = - cos f, FR =
2F 21
+
F 22
+ 2F1F2 cos f
Ans.
From the figure, tan u =
FR
- 2F1F2 cos (180° - f)
F1 sin f F2 + F1 cos f
u = tan –1 ¢
F1 sin f ≤ F2 + F1 cos f
Ans.
63
u F2
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2–48. If F1 = 600 N and f = 30°, determine the magnitude of the resultant force acting on the eyebolt and its direction measured clockwise from the positive x axis.
y
F1 f x
SOLUTION Rectangular Components: By referring to Fig. a, the x and y components of each force can be written as
60
(F1)x = 600 cos 30° = 519.62 N (F1)y = 600 sin 30° = 300 N (F2)x = 500 cos 60° = 250 N
5
(F2)y = 500 sin 60° = 433.01 N
3 (F3)x = 450 a b = 270 N 5
F3
4 (F3)y = 450a b = 360 N 5
Resultant Force: Summing the force components algebraically along the x and y axes, + ©(F ) = ©F ; : R x x + c ©(FR)y = ©Fy ;
(FR)x = 519.62 + 250 - 270 = 499.62 N
:
(FR)y = 300 - 433.01 - 360 = -493.01 N = 493.01 N T
The magnitude of the resultant force FR is FR = 2(FR)x2 + (FR)y2 = 2499.62 2 + 493.012 = 701.91 N = 702 N
Ans.
The direction angle u of FR, Fig. b, measured clockwise from the x axis, is u = tan - 1 B
(FR)y (FR)x
R = tan - 1 ¢
4
3
493.01 ≤ = 44.6° 499.62
Ans.
64
F2 450 N
500 N
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2–49. 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 u = 30°, determine the magnitude of F1 and the angle f.
y
F1 f x
SOLUTION 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
3 (F3)x = 450a b = 270 N 5
4 (F3)y = 450a b = 360 N 5
(FR)x = 600 cos 30° = 519.62 N
(FR)y = 600 sin 30° = 300 N
5
F3
519.62 = F1 cos f + 250 - 270 F1 cos f = 539.62
+ c ©(FR)y = ©Fy ;
(1)
- 300 = F1 sin f - 433.01 - 360 F1 sin f = 493.01
(2)
Solving Eqs. (1) and (2), yields f = 42.4°
4
3
Resultant Force: Summing the force components algebraically along the x and y axes, + ©(F ) = ©F ; : R x x
60
F1 = 731 N
Ans.
65
F2 450 N
500 N
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2–50. Determine the magnitude of F1 and its direction u so that the resultant force is directed vertically upward and has a magnitude of 800 N.
y u
600 N 3
400 N
5 4
30
SOLUTION
x A
Scalar Notation: Summing the force components algebraically, we have + F = ©F ; : Rx x
4 FRx = 0 = F1 sin u + 400 cos 30° - 600a b 5 F1 sin u = 133.6
+ c FRy = ©Fy ;
F1
(1)
3 FRy = 800 = F1 cos u + 400 sin 30° + 600 a b 5 F1 cos u = 240
(2)
Solving Eqs. (1) and (2) yields u = 29.1°
F1 = 275 N
Ans.
66
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2–51. 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 F1 = 500 N and u = 20°.
y u
600 N 3
400 N
5 4
30
SOLUTION
x A
Scalar Notation: Summing the force components algebraically, we have + F = ©F ; : Rx x
4 FRx = 500 sin 20° + 400 cos 30° - 600 a b 5 = 37.42 N :
+ c FRy = ©Fy ;
F1
3 FRy = 500 cos 20° + 400 sin 30° + 600 a b 5 = 1029.8 N c
The magnitude of the resultant force FR is FR = 2F 2Rx + F 2Ry = 237.42 2 + 1029.82 = 1030.5 N = 1.03 kN
Ans.
The direction angle u measured counterclockwise from positive x axis is u = tan - 1
FRy FRx
= tan - 1 a
1029.8 b = 87.9° 37.42
Ans.
67
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2–52. Determine the magnitude of force F so that the resultant FR of the three forces is as small as possible. What is the minimum magnitude of FR?
5 kN
SOLUTION
30 F
Scalar Notation: Summing the force components algebraically, we have + F = ©F ; : Rx x
4 kN
FRx = 5 - F sin 30° = 5 - 0.50F :
+ c FRy = ©Fy ;
FRy = F cos 30° - 4 = 0.8660F - 4 c
The magnitude of the resultant force FR is FR = 2FR2 + F2Ry x
= 2(5 - 0.50F)2 + (0.8660F - 4)2 = 2F2 - 11.93F + 41
(1)
FR2 = F2 - 11.93F + 41 2FR
dFR = 2F - 11.93 dF
¢ FR
d2FR 2
dF
+
dFR dFR * ≤ = 1 dF dF
In order to obtain the minimum resultant force FR, 2FR
(2)
(3) dFR = 0. From Eq. (2) dF
dFR = 2F - 11.93 = 0 dF
F = 5.964 kN = 5.96 kN
Ans.
Substituting F = 5.964 kN into Eq. (1), we have FR = 25.9642 - 11.93(5.964) + 41 = 2.330 kN = 2.33 kN Substituting FR = 2.330 kN with
dFR = 0 into Eq. (3), we have dF
B (2.330) d2FR dF2
Ans.
d2FR dF2
+ 0R = 1
= 0.429 7 0
Hence, F = 5.96 kN is indeed producing a minimum resultant force.
68
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2–53. 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 the resultant force?
F
14 kN
30⬚
45⬚ 8 kN
SOLUTION + : FRx = ©Fx ;
FRz = 8 - F cos 45° - 14 cos 30° = -4.1244 - F cos 45°
+ c FRy = ©Fy ;
FRy = -F sin 45° + 14 sin 30° = 7 - F sin 45° FR2 = (-4.1244 - F cos 45°)2 + (7 - F sin 45°)2 2FR
From Eq. (1);
(1)
dFR = 2(-4.1244 - F cos 45°)(- cos 45°) + 2(7 - F sin 45°)( -sin 45°) = 0 dF F = 2.03 kN
Ans.
FR = 7.87 kN
Ans.
Also, from the figure require (FR)x¿ = 0 = ©Fx¿;
F + 14 sin 15° - 8 cos 45° = 0 F = 2.03 kN
(FR)y¿ = ©Fy¿;
Ans.
FR = 14 cos 15° - 8 sin 45° FR = 7.87 kN
Ans.
69
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2–54. Three forces act on the bracket. Determine the magnitude and direction u of F1 so that the resultant force is directed along the positive x¿ axis and has a magnitude of 1 kN.
y F2 45
SOLUTION
F3
200 N x
30
+ F = ©F ; : Rx x
1000 cos 30° = 200 + 450 cos 45° + F1 cos(u + 30°)
+ c FRy = ©Fy ;
- 1000 sin 30° = 450 sin 45° - F1 sin(u + 30°)
u x¿ F1
F1 sin(u + 30°) = 818.198 F1 cos(u + 30°) = 347.827 u + 30° = 66.97°,
450 N
u = 37.0°
Ans.
F1 = 889 N
Ans.
70
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2–55. If F1 = 300 N and u = 20°, determine the magnitude and direction, measured counterclockwise from the x¿ axis, of the resultant force of the three forces acting on the bracket.
y F2 ⫽ 450 N 45⬚
SOLUTION
F3 ⫽ 200 N x
30⬚
+ F = ©F ; : Rx x
FRx = 300 cos 50° + 200 + 450 cos 45° = 711.03 N
+ c FRy = ©Fy ;
FRy = -300 sin 50° + 450 sin 45° = 88.38 N
F1
FR = 2 (711.03)2 + (88.38)2 = 717 N f¿ (angle from x axis) = tan - 1 B
u
Ans.
88.38 R 711.03
f¿ = 7.10° f (angle from x¿ axis) = 30° + 7.10° f = 37.1°
Ans.
71
x¿
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. .
72
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.
.
73
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2–58. 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.
y u
F1
f 30 x
SOLUTION
F2
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 f
(F1)y = F1 cos f
(F2)x = 200 N
( F2)y = 0
(F3)x = 260 ¢
5 ≤ = 100 N 13
(F3)y = 260 ¢
(FR)x = 450 cos 30° = 389.71 N
(FR)y = 450 sin 30° = 225 N
Resultant Force: Summing the force components algebraically along the x and y axes, + ©(F ) = ©F ; : R x x
389.71 = F1 sin f + 200 + 100 F1 sin f = 89.71
+ c ©(FR)y = ©Fy;
(1)
225 = F1 cos f - 240 F1 cos f = 465
(2)
Solving Eqs. (1) and (2), yields f = 10.9°
F1 = 474 N
Ans.
74
13
12 5
F3
12 ≤ = 240 N 13
200 N
260 N
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2–59. 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°.
y u
F1
f 30 x
SOLUTION
F2
Rectangular Components: By referring to Fig. a, the x and y components of F1, F2, and F3 can be written as (F1)x = F1 sin 30° = 0.5F1
(F1)y = F1 cos 30° = 0.8660F1
(F2)x = 200 N
(F2)y = 0
(F3)x = 260a
5 b = 100 N 13
(F3)y = 260a
Resultant Force: Summing the force components algebraically along the x and y axes, + ©(F ) = ©F ; : R x x
(FR)x = 0.5F1 + 200 + 100 = 0.5F1 + 300
+ c ©(FR)y = ©Fy;
(FR)y = 0.8660F1 - 240
The magnitude of the resultant force FR is FR = 2(FR)x2 + (FR)y2 = 2(0.5F1 + 300)2 + (0.8660F1 - 240)2 = 2F 21 - 115.69F1 + 147 600
(1)
Thus, FR2 = F 21 - 115.69F1 + 147 600
(2)
The first derivative of Eq. (2) is 2FR For FR to be minimum,
dFR = 2F1 - 115.69 dF1
(3)
dFR = 0. Thus, from Eq. (3) dF1
2FR
dFR = 2F1 - 115.69 = 0 dF1
F1 = 57.846 N = 57.8 N
Ans.
from Eq. (1), FR = 2(57.846)2 - 115.69(57.846) + 147 600 = 380 N
Ans.
75
13
12 5
F3
12 b = 240 N 13
200 N
260 N
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2–60. The stock mounted on the lathe is subjected to a force of 60 N. Determine the coordinate direction angle b and express the force as a Cartesian vector.
z 60 N
45⬚
SOLUTION
b
1 = 2cos a + cos b + cos g 2
2
2
60⬚
1 = cos 60° + cos b + cos 45° 2
2
2
y x
cos b = ; 0.5 b = 60°, 120° Use b = 120°
Ans.
F = 60 N(cos 60°i + cos 120°j + cos 45°k) = {30i - 30j + 42.4k} N
Ans.
76
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2–61. Determine the coordinate angle g for F2 and then express each force acting on the bracket as a Cartesian vector.
z
F1 ⫽ 450 N
45⬚ 30⬚ 45⬚
60⬚
x F2 ⫽ 600 N
77
y
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2–62. Determine the magnitude and coordinate direction angles of the resultant force acting on the bracket.
z
F1 ⫽ 450 N
45⬚ 30⬚ 45⬚
60⬚
x F2 ⫽ 600 N
78
y
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2–63. z
The bolt is subjected to the force F, which has components acting along the x, y, z axes as shown. If the magnitude of F is 80 N, and a = 60° and g = 45°, determine the magnitudes of its components.
Fz g
F Fy
a
SOLUTION Fx
cosb = 21 - cos2 a - cos2g = 21 - cos2 60° - cos2 45°
x
b = 120° Fx = |80 cos 60°| = 40 N
Ans.
Fy = |80 cos 120°| = 40 N
Ans.
Fz = |80 cos 45°| = 56.6 N
Ans.
79
b y
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2-64 . Determine the magnitude and coordinate direction angles of the force F acting on the stake. Given: Fh
40 N
T
70 deg
c
3
d
4
Solution:
§ c2 d2 · ¨ ¸ © d ¹
F
Fh
F
50 N
Ans.
Fx
F h cos T
Fy
F h sin T
Fz
c § · ¨ 2 2 ¸F © c d ¹
Fx
13.7 N
Fy
37.6 N
Fz
30 N
D
acos ¨
E
acos ¨
J
acos ¨
D
74.1 deg
E
41.3 deg
J
53.1 deg
§ Fx · ¸ ©F¹
§ Fy · ¸ ©F¹
§ Fz · ¸ ©F¹
80
Ans.
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2-6 5 . Determine the magnitude and coordinate direction angles of the force F acting on the stake. Given: F h := 40N θ := 50deg c := 3 d := 4 Solution:
2
c +d
F := F h⋅
2
d
F = 50 N
Ans.
c
F x := Fh⋅ cos ( θ )
F y := Fh⋅ sin ( θ )
F z :=
F x = 25.7 N
F y = 30.6 N
F z = 30 N
⎛ Fx ⎞ ⎟ ⎝F⎠
⎛ Fy ⎞ ⎟ ⎝F⎠
α := acos ⎜
α = 59 .1 deg
β = 52.2 deg
c +d
⋅F 2
⎛ Fz ⎞ ⎟ ⎝F⎠
β := acos ⎜ Ans.
2
γ := acos ⎜ Ans.
81
γ = 53.1 deg
Ans.
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2–66. Determine the magnitude and coordinate direction angles of F1 = 560i - 50j + 40k6 N and F2 = 5-40i - 85j + 30k6 N. Sketch each force on an x, y, z reference.
SOLUTION F1 = 60 i - 50 j + 40 k F1 = 216022 + 1-5022 + 14022 = 87.7496 = 87.7 N
Ans.
a1 = cos-1 a
60 b = 46.9° 87.7496
Ans.
b 1 = cos-1 a
-50 b = 125° 87.7496
Ans.
g1 = cos-1 a
40 b = 62.9° 87.7496
Ans.
F2 = - 40 i - 85 j + 30 k F2 = 21-4022 + 1-8522 + 13022 = 98.615 = 98.6 N
Ans.
a2 = cos-1 a
-40 b = 114° 98.615
Ans.
b 2 = cos-1 a
-85 b = 150° 98.615
Ans.
g2 = cos-1
30 98.615
Ans.
= 72.3°
82
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2–67.
z
Express each force in Cartesian vector form.
F1 5 kN
60
( F1 ) x = 0 ( F2 ) x = 5cos 60D = 2.5 kN ( F1 ) y = 2 kN ( F2 ) y = 5cos 45D = 3.54 kN ( F1 ) z = 0 ( F2 ) z = 5cos 60D = 2.5 kN Thus, F1 = 0 ( i ) + 2 ( − j) + 0 ( k ) = {−2 j} N Ans. F2 = 2.5 ( i ) + 3.54 ( j) + 2.5 ( k ) = {2.5i + 3.54 j+ 2.5k} N Ans.
F2 2 kN
45 y 60
x
83
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2–68. z
Express each force as a Cartesian vector.
30⬚ 30⬚
x
SOLUTION
F1 ⫽ 300 N
45⬚
Rectangular Components: By referring to Figs. a and b, the x, y, and z components of F1 and F2 can be written as (F1)x = 300 cos 30° = 259.8 N
(F2)x = 500 cos 45° sin 30° = 176.78 N
(F1)y = 0
(F2)y = 500 cos 45° cos 30° = 306.19 N
(F1)t = 300 sin 30° = 150 N
(F2)z = 500 sin 45° = 353.55 N
F2 ⫽ 500 N
Thus, F1 and F2 can be written in Cartesian vector form as
F1 = 259.81(+i) + 0j + 150(-k) = {260i - 150k} N
Ans.
F2 = 176.78(+i) + 306.19(+j) + 353.55(-k) = 2{177i + 306j - 354k} N
Ans.
84
y
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2–69. z
Determine the magnitude and coordinate direction angles of the resultant force acting on the hook.
30⬚ y 30⬚
x
SOLUTION Force Vectors: By resolving F1 and F2 into their x, y, and z components, as shown in Figs. a and b, respectively, F1 and F2 can be expessed in Cartesian vector form as
F1 ⫽ 300 N
45⬚
F1 = 300 cos 30°(+i) + 0j + 300 sin 30°( -k) F2 ⫽ 500 N
= {259.81i - 150k} N F2 = 500 cos 45°sin 30°( +i) + 500 cos 45° cos 30°(+j) + 500 sin 45°( -k) = {176.78i - 306.19j - 353.55k} N Resultant Force: The resultant force acting on the hook can be obtained by vectorally adding F1 and F2. Thus,
FR = F1 + F2 = (259.81i - 150k) + (176.78i + 306.19j - 353.55k) = {436.58i) + 306.19j - 503.55k} N The magnitude of FR is FR = 2(FR)x2 + (FR)y 2(FR)z 2 = 2(436.58)2 + (306.19)2 + (-503.55)2 = 733.43 N = 733 N The coordinate direction angles of FR are (FR)x 436.58 d = cos-1 a b = 53.5° ux = cos-1 c FR 733.43 uy = cos-1 c uz = cos-1 c
(FR)y FR (FR)z FR
Ans.
Ans.
d = cos-1 a
306.19 b = 65.3° 733.43
Ans.
d = cos-1 a
-503.55 b = 133° 733.43
Ans.
85
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. . . .
86
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2–71. 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°.
z
g
F
500 N
SOLUTION Force Vectors: By resolving F1 and F into their x, y, and z components, as shown in Figs. a and b, respectively, F1 and F can be expressed in Cartesian vector form as
b a
F1 = 600 cos 30° sin 30°(+i) + 600 cos 30° cos 30°(+j) + 600 sin 30°( - k)
30
= {259.81i + 450j - 300k} N
y x
F = 500 cos ai + 500 cos bj + 500 cos gk
30
F1
Since the resultant force FR is directed towards the positive y axis, then
600 N
FR = FR j Resultant Force: FR = F1 + F FR j = (259.81i + 450j - 300k) + (500 cos ai + 500 cos bj + 500 cos gk) 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, 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.
However, since cos2 a + cos2 b + cos2 g = 1, a = 121.31°, and g = 53.13°, cos b = ; 21 - cos2 121.31° - cos2 53.13° = ;0.6083 If we substitute cos b = 0.6083 into Eq. (1), 2
FR = 450 + 500(0.6083) = 754 N
Ans. 2
and b = cos - 1 (0.6083) = 52.5°
Ans.
2
87
2
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2-72 . Specify the magnitude F 3 and directions D3, E3, and J3 of F 3 so that the resultant force of the three forces is FR. Units Used: kN
3
10 N
Given: F1
12 kN
c
5
F2
10 kN
d
12
T
30 deg
FR
§0· ¨ 9 ¸ kN ¨ ¸ ©0¹
Solution: Initial Guesses:
F 3x
Given
§ F3x · § 0 · ¨ ¸ ¨ ¸ ¨ F3y ¸ F1¨ cos T ¸ ¨F ¸ © sin T ¹ © 3z ¹
§ F3x · ¨ ¸ ¨ F3y ¸ ¨F ¸ © 3z ¹ § D3 · ¨ ¸ ¨ E3 ¸ ¨J ¸ © 3¹
FR
1 kN
Find F3x F 3y F3z
§ F3 · acos ¨ ¸ © F3 ¹
F 3y
F3
§ D3 · ¨ ¸ ¨ E3 ¸ ¨J ¸ © 3¹
1 kN
F 3z
1 kN
§ d · ¨0¸ ¸ 2 2¨ c d © c ¹ F2
§ F3x · ¨ ¸ ¨ F3y ¸ ¨F ¸ © 3z ¹ § 15.5 · ¨ 98.4 ¸ deg ¨ ¸ © 77.0 ¹
88
F3
§ 9.2 · ¨ 1.4 ¸ kN ¨ ¸ © 2.2 ¹
F3
9.6 kN
Ans.
Ans.
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2-73. Determine the magnitude and coordinate direction angles of F3 so that the resultant of the three forces acts along the positive y axis and has magnitude F. Given: F
600 1
F1
180 N
F2
300 N
D1
30 deg
D2
40 deg
Solution: Initial guesses:
D
40 deg
J
50 deg
E
50 deg
F3
45 N
Given F Rx = 6F x;
0
F1 F2 cos D 1 sin D 2 F 3 cos D
F Ry = 6F y;
F
F2 cos D 1 cos D 2 F3 cos E
F Rz = 6F z;
0
F2 sin D 1 F3 cos J
cos D cos E cos J 2
§ F3 · ¨ ¸ ¨D¸ ¨E ¸ ¨ ¸ ©J ¹
Find F 3 D E J
2
F3
2
1
428 N
89
§¨ D ·¸ ¨E ¸ ¨J ¸ © ¹
§ 88.3 · ¨ 20.6 ¸ deg ¨ ¸ © 69.5 ¹
Ans.
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2-74 . Determine the magnitude and coordinate direction angles of F3 so that the resultant of the three forces is zero. Given: F1
180 N
D1
30 deg
F2
300 N
D2
40 deg
Solution: Initial guesses:
D
40 deg
J
E
50 deg
F3
50 deg 45 N
Given F Rx = 6F x;
0
F1 F2 cos D 1 sin D 2 F 3 cos D
F Ry = 6F y;
0
F2 cos D 1 cos D 2 F3 cos E
F Rz = 6F z;
0
F2 sin D 1 F3 cos J
cos D cos E cos J 2
§ F3 · ¨ ¸ ¨D¸ ¨E ¸ ¨ ¸ ©J ¹
Find F 3 D E J
2
F3
250 N
90
2
1
§¨ D ·¸ ¨E ¸ ¨J ¸ © ¹
§ 87.0 · ¨ 142.9 ¸ deg ¨ ¸ © 53.1 ¹
Ans.
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2–75. z
Determine the coordinate direction angles of force F1. F1 ⫽ 600 N
F2 ⫽ 450 N
5 3
4
45⬚ 30⬚
SOLUTION
x
Rectangular Components: By referring to Figs. a, the x, y, and z components of F1 can be written as 4 (F1)x = 600 a b cos 30° N 5
4 (F1)y = 600 a b sin 30° N 5
3 (F1)z = 600a b N 5
Thus, F1 expressed in Cartesian vector form can be written as F1 = 600 e
4 4 3 cos 30°(+i) + sin 30°(-j) + (+k) f N 5 5 5
= 600[0.6928i - 0.4j + 0.6k] N Therefore, the unit vector for F1 is given by uF1 =
F1 600(0.6928i - 0.4j + 0.6k = = 0.6928i - 0.4j + 0.6k F1 600
The coordinate direction angles of F1 are a = cos-1(uF1)x = cos-1(0.6928) = 46.1°
Ans.
b = cos-1(uF1)y = cos-1(-0.4) = 114°
Ans.
g = cos-1(uF1)z = cos-1(0.6) = 53.1°
Ans.
91
y
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2–76. z
Determine the magnitude and coordinate direction angles of the resultant force acting on the eyebolt. F1 ⫽ 600 N
F2 ⫽ 450 N
5 3
4
45⬚ 30⬚
x
SOLUTION Force Vectors: By resolving F1 and F2 into their x, y, and z components, as shown in Figs. a and b, respectively, they are expressed in Cartesian vector form as 4 4 3 F1 = 600 a bcos 30°(+ i) + 600a bsin 30°( -j) + 600 a b(+k) 5 5 5 = 5415.69i - 240j + 360k6 N F2 = 0i + 450 cos 45°(+j) + 450 sin 45°( +k) = 5318.20j + 318.20k6 N Resultant Force: The resultant force acting on the eyebolt can be obtained by vectorally adding F1 and F2. Thus, FR = F1 + F2 = (415.69i - 240j + 360k) + (318.20j + 318.20k) = 5415.69i + 78.20j + 678.20k6 N The magnitude of FR is given by FR = 3(FR)x 2 + (FR)y 2 + (FR)z 2 = 3(415.69)2 + (78.20)2 + (678.20)2 = 799.29 N = 799 N
Ans.
The coordinate direction angles of FR are a = cos-1 c b = cos-1 c g = cos-1 c
(FR)x 415.69 d = cos-1 a b = 58.7° FR 799.29 (FR)y FR (FR)z FR
Ans.
d = cos-1 a
78.20 b = 84.4° 799.29
Ans.
d = cos-1 a
678.20 b = 32.0° 799.29
Ans.
92
y
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2–77. The cables attached to the screw eye are subjected to the three forces shown. Express each force in Cartesian vector form and determine the magnitude and coordinate direction angles of the resultant force.
z F1 = 350 N
F3 = 250 N
40°
60°
120°
45°
y 60°
SOLUTION Cartesian Vector Notation:
45° x
F1 = 3505sin 40°j + cos 40°k6 N
F2 = 100 N
= 5224.98j + 268.12k6 N = 5225j + 268k6 N
Ans.
F2 = 1005cos 45°i + cos 60°j + cos 120°k6 N = 570.71i + 50.0j - 50.0k6 N = 570.7i + 50.0j - 50.0k6 N
Ans.
F3 = 2505cos 60°i + cos 135°j + cos 60°k6 N = 5125.0i - 176.78j + 125.0k6 N = 5125i - 177j + 125k6 N
Ans.
Resultant Force: FR = F1 + F2 + F3 = 5170.71 + 125.02i + 1224.98 + 50.0 - 176.782j + 1268.12 - 50.0 + 125.02k6 N = 5195.71i + 98.20j + 343.12k6 N The magnitude of the resultant force is FR = 2F2Rx + F2Ry + F2Rz = 2195.712 + 98.202 + 343.122 = 407.03 N = 407 N
Ans.
The coordinate direction angles are cos a = cos b = cos g =
60°
FRx FR FRy FR FRz FR
=
195.71 407.03
a = 61.3°
Ans.
=
98.20 407.03
b = 76.0°
Ans.
=
343.12 407.03
g = 32.5°
Ans.
93
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2–78. 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.
z F3 FR
110 N
F2
120 N
SOLUTION Cartesian Vector Notation:
F1
= {42.43i + 73.48j + 84.85k} N 3 4 F1 = 80 b i + k r N = {64.0i + 48.0k} N 5 5
x
F2 = { - 110k} N F3 = {F3x i + F3y j + F3z k} N Resultant Force: FR = F1 + F2 + F3
E A 64.0 + F3 x B i + F3 y j + A 48.0 - 110 + F3 z B k F
Equating i, j and k components, we have 64.0 + F3 x = 42.43
F3x = -21.57 N F3 y = 73.48 N
48.0 - 110 + F3 z = 84.85
F3 z = 146.85 N
The magnitude of force F3 is F3 = 2F 23 x + F 23 y + F 23 z = 2(- 21.57)2 + 73.482 + 146.852 = 165.62 N = 166 N
Ans.
The coordinate direction angles for F3 are cos a = cos b =
F3 x F3 F3 y F3
cos g = =
=
- 21.57 165.62
a = 97.5°
Ans.
=
73.48 165.62
b = 63.7°
Ans.
g = 27.5°
Ans.
F3 z F3
=
146.85 165.62
5 3
FR = 120{cos 45°sin 30°i + cos 45°cos 30°j + sin 45°k} N
{42.43i + 73.48j + 84.85k} =
80 N
94
4
45
y 30
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2–79. Determine the coordinate direction angles of F1 and FR.
z F3 FR
110 N
F2
120 N
SOLUTION Unit Vector of F1 and FR : u F1 =
F1
80 N
5 3
3 4 i + k = 0.8i + 0.6k 5 5
uR = cos 45° sin 30°i + cos 45° cos 30°j + sin 45°k = 0.3536i + 0.6124j + 0.7071k
x
Thus, the coordinate direction angles F1 and FR are cos aF1 = 0.8
aF1 = 36.9°
Ans.
cos b F1 = 0
b F1 = 90.0°
Ans.
cos gF1 = 0.6
gF1 = 53.1°
Ans.
cos aR = 0.3536
aR = 69.3°
Ans.
cos b R = 0.6124
b R = 52.2°
Ans.
cos gR = 0.7071
gR = 45.0°
Ans.
95
4
45
y 30
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2–80. The mast is subjected to the three forces shown. Determine the coordinate direction angles a1, b 1, g1 of F1 so that the resultant force acting on the mast is FR = 5350i6 N.
z F1 g1 a1
b1
F3 ⫽ 300 N y x
. . .
96
F2 ⫽ 200 N
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2–81. The mast is subjected to the three forces shown. Determine the coordinate direction angles a1, b 1, g1 of F1 so that the resultant force acting on the mast is zero.
z F1 g1 a1
b1
F3 ⫽ 300 N y x
. . .
97
F2 ⫽ 200 N
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2–82. 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.
z
F2 g2 b2 a2
y 60⬚ 15⬚
x F1 ⫽ 180 N
. . . .
98
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2–83. Determine the magnitude and coordinate direction angles of F2 so that the resultant of the two forces is zero.
z
F2 g2 b2 a2
y 60⬚ 15⬚
x F1 ⫽ 180 N
. . . .
99
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2–84. z
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, b = 30°, and g = 75°, determine the magnitudes of its three components.
Fz
F
g
SOLUTION cos2 a + cos2 b + cos2 g = 1
b a
cos2 a + cos2 30° + cos2 75° = 1
Fx
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.
100
x
Fy
y
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2–85. z
The pole is subjected to the force F which has components Fx = 1.5 kN and Fz = 1.25 kN. If b = 75°, determine the magnitudes of F and Fy.
Fz
F
g
SOLUTION cos2 a + cos2 b + cos2 g = 1 a
b a
1.25 2 1.5 2 b + cos2 75° + a b = 1 F F
Fx
F = 2.02 kN
Ans.
Fy = 2.02 cos 75° = 0.523 kN
Ans.
101
x
Fy
y
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.
102
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2–87. Determine the lengths of wires AD, BD, and CD. The ring at D is midway between A and B.
z C
D
A
SOLUTION Da
2m
0.5 m
2 + 0 0 + 2 1.5 + 0.5 , , b m = D(1, 1, 1) m 2 2 2
2m 1.5 m
rAD = (1 - 2)i + (1 - 0)j + (1 - 1.5)k = -1i + 1j - 0.5k
x
rBD = (1 - 0)i + (1 - 2)j + (1 - 0.5)k = 1i - 1j + 0.5k rCD = (1 - 0)i + (1 - 0)j + (1 - 2)k = 1i + 1j - 1k rAD = 2(-1)2 + 12 + (- 0.5)2 = 1.50 m
Ans.
rBD = 212 + ( - 1)2 + 0.52 = 1.50 m
Ans.
rCD = 212 + 12 + (- 1)2 = 1.73 m
Ans.
103
B 0.5 m y
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2–88. y
Determine the length of member AB of the truss by first establishing a Cartesian position vector from A to B and then determining its magnitude.
B A
1.5 m 1.2 m
SOLUTION rAB
40⬚
1.5 = (1.1) = - 0.80)i + (1.5 - 1.2)j tan 40°
x O
0.8 m 0.3 m
rAB = {2.09i + 0.3j} m rAB = 2(2.09)2 + (0.3)2 = 2.11 m
C
Ans.
104
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2–89. If F = 5350i - 250j - 450k6 N and cable AB is 9 m long, determine the x, y, z coordinates of point A.
z A F z
B
SOLUTION
x
Position Vector: The position vector rAB, directed from point A to point B, is given by rAB = [0 - ( -x)]i + (0 - y)j + (0 - z)k = xi - yj - zk Unit Vector: Knowing the magnitude of rAB is 9 m, the unit vector for rAB is given by uAB =
xi - yj - zk rAB = rAB 9
The unit vector for force F is uF =
350i - 250j - 450k F = = 0.5623i - 0.4016j - 0.7229k F 3 3502 + (-250)2 + (-450)2
Since force F is also directed from point A to point B, then uAB = uF xi - yj - zk = 0.5623i - 0.4016j - 0.7229k 9 Equating the i, j, and k components, x = 0.5623 9
x = 5.06 m
Ans.
-y = -0.4016 9
y = 3.61 m
Ans.
-z = 0.7229 9
z = 6.51 m
Ans.
105
x
y
y
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2–90. z
Express FB and FC in Cartesian vector form.
C
B
2m FB ⫽ 600 N
SOLUTION
0.5 m A
Force Vectors: The unit vectors uB and uC of FB and FC must be determined first. From Fig. a
uC =
1.5 m 1.5 m x
2 1 2 i - j + k 3 3 3
(-1.5 - 0.5)i + [0.5 - (-1.5)]j + (3.5 - 0)k rC = rC 3(-1.5 - 0.5)2 + [0.5 - ( -1.5)]2 + (3.5 - 0)2 = -
4 4 7 i + j + k 9 9 9
Thus, the force vectors FB and FC are given by 2 1 2 FB = FB uB = 600a - i - j + kb = 5 -400i - 200j + 400k6 N 3 3 3
Ans.
4 4 7 FC = FC uC = 450 a - i + j + kb = 5-200i + 200j + 350k6 N 9 9 9
Ans.
106
FC ⫽ 450 N
1m
(-1.5 - 0.5)i + [-2.5 - (-1.5)]j + (2 - 0)k rB uB = = rB 3(-1.5 - 0.5)2 + [-2.5 - (-1.5)]2 + (2 - 0)2 = -
3.5 m
0.5 m y
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2–91. z
Determine the magnitude and coordinate direction angles of the resultant force acting at A.
C
B
2m FB ⫽ 600 N
SOLUTION
0.5 m A
Force Vectors: The unit vectors uB and uC of FB and FC must be determined first. From Fig. a
uC =
x
2 1 2 i - j + k 3 3 3
4 4 7 i + j + k 9 9 9
Thus, the force vectors FB and FC are given by 2 1 2 FB = FB uB = 600a - i - j + kb = 5-400i - 200j + 400k6 N 3 3 3 4 4 7 FC = FC uC = 450 a - i + j + kb = 5 -200i + 200j + 350k6 N 9 9 9 Resultant Force: FR = FB + FC = (-400i - 200j + 400k) + (-200i + 200j + 350k) = 5-600i + 750k6 N The magnitude of FR is FR = 3(FR)x 2 + (FR)y 2 + (FR)z 2 = 3(-600)2 + 02 + 7502 = 960.47 N = 960 N The coordinate direction angles of FR are a = cos-1 c b = cos-1 c g = cos-1 c
(FR)x -600 d = cos-1 a b = 129° FR 960.47 (FR)y FR (FR)z FR
d = cos-1 a
Ans.
0 b = 90° 960.47
Ans.
760 b = 38.7° 960.47
Ans.
d = cos-1 a
1.5 m 1.5 m
(-1.5 - 0.5)i + [0.5 - ( -1.5)]j + (3.5 - 0)k rC = rC 3(-1.5 - 0.5)2 + [0.5 - ( -1.5)]2 + (3.5 - 0)2 = -
107
FC ⫽ 450 N
1m
(-1.5 - 0.5)i + [-2.5 - (-1.5)]j + (2 - 0)k rB uB = = rB 3(-1.5 - 0.5)2 + [-2.5 - (-1.5)]2 + (2 - 0)2 = -
3.5 m
0.5 m y
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2–92. If FB = 560 N and FC = 700 N, determine the magnitude and coordinate direction angles of the resultant force acting on the flag pole.
z
A 6m
FB FC
SOLUTION Force Vectors: The unit vectors uB and uC of FB and FC must be determined first. From Fig. a (2 - 0)i + ( -3 - 0)j + (0 - 6)k rB 2 3 6 uB = = = i - j - k rB 7 7 7 2 2 2 3(2 - 0) + (-3 - 0) + (0 - 6) (3 - 0)i + (2 - 0)j + (0 - 6)k rC 3 2 6 uC = = = i + j - k rC 7 7 7 3(3 - 0)2 + (2 - 0)2 + (0 - 6)2
x
2 3 6 FB = FB uB = 560a i - j - kb = 5160i - 240j - 480k6 N 7 7 7 3 2 6 FC = FC uC = 700 a i + j - k b = 5300i + 200j - 600k6 N 7 7 7 Resultant Force: FR = FB + FC = (160i - 240j - 480k) + (300i + 200j - 600k) = 5460i - 40j + 1080k6 N The magnitude of FR is FR = 3(FR)x 2 + (FR)y 2 + (FR)z 2 Ans.
The coordinate direction angles of FR are a = cos-1 c b = cos-1 c g = cos-1 c
(FR)x 460 d = cos-1 a b = 66.9° FR 1174.56 (FR)y FR (FR)z FR
d = cos-1 a
Ans.
-40 b = 92.0° 1174.56
Ans.
-1080 b = 157° 1174.56
Ans.
d = cos-1 a
108
B 3m
Thus, the force vectors FB and FC are given by
= 3(460)2 + (-40)2 + (-1080)2 = 1174.56 N = 1.17 kN
2m
3m
2m C
y
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2–93. If FB = 700 N, and FC = 560 N, determine the magnitude and coordinate direction angles of the resultant force acting on the flag pole.
z
A 6m
FB FC
SOLUTION Force Vectors: The unit vectors uB and uC of FB and FC must be determined first. From Fig. a uB =
2m B
(2 - 0)i + ( -3 - 0)j + (0 - 6)k rB 2 3 6 = = i - j - k rB 7 7 7 2 2 2 3(2 - 0) + (-3 - 0) + (0 - 6)
3m
(3 - 0)i + (2 - 0)j + (0 - 6)k rC 3 2 6 uC = = = i + j - k rC 7 7 7 3(3 - 0)2 + (2 - 0)2 + (0 - 6)2
x
Thus, the force vectors FB and FC are given by 2 3 6 FB = FB uB = 700a i - j - kb = 5200i - 300j - 600k6 N 7 7 7 3 2 6 FC = FC uC = 560 a i + j - k b = 5240i + 160j - 480k6 N 7 7 7 Resultant Force: FR = FB + FC = (200i - 300j - 600k) + (240i + 160j - 480k) = 5440i - 140j - 1080k6 N The magnitude of FR is FR = 3(FR)x 2 + (FR)y 2 + (FR)z 2 = 3(440)2 + (-140)2 + (-1080)2 = 1174.56 N = 1.17 kN
Ans.
The coordinate direction angles of FR are a = cos-1 c b = cos-1 c g = cos-1 c
(FR)x 440 b = 68.0° d = cos-1 a FR 1174.56 (FR)y FR (FR)z FR
-140 b = 96.8° 1174.56
Ans.
-1080 b = 157° 1174.56
Ans.
d = cos-1 a d = cos-1 a
Ans.
109
3m
2m C
y
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2–94. 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 a, b, g of the resultant force. Take x = 15 m , y = 20 m .
z D 600 N 400 N
800 N
SOLUTION FDA = 400 a
24 20 15 i + j kb N 34.66 34.66 34.66
FDB = 800 a
4 24 -6 i + j kb N 25.06 25.06 25.06
FDC = 600a
24 m
16 m O
C
18 24 16 i j kb N 34 34 34
18 m
6m x
y A
FR = FDA + FDB + FDC
x
= {263.92i + 40.86j - 1466.71k} N FR = 2(263.92)2 + ( 40.86)2 + ( -1466.71)2 = 1490.83 N = 1.49 kN
Ans.
a = cos-1 a
263.92 b = 79.8° 1490.83
Ans.
b = cos-1 a
40.86 b = 88.4° 1490.83
Ans.
g = cos-1 a
-1466.71 b = 169.7° 1490.83
Ans.
110
4m
B
y
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2–95. z
At a given instant, the position of a plane at A and a train at B are measured relative to a radar antenna at O. Determine the distance d between A and B at this instant. To solve the problem, formulate a position vector, directed from A to B, and then determine its magnitude.
A
5 km 60⬚ 35⬚
SOLUTION
O
Position Vector: The coordinates of points A and B are
x
A(-5 cos 60° cos 35°, -5 cos 60° sin 35°, 5 sin 60°) km
y
40⬚ 25⬚ 2 km B
= A(-2.048, -1.434, 4.330) km B(2 cos 25° sin 40°, 2 cos 25° cos 40°, -2 sin 25°) km = B(1.165, 1.389, -0.845) km The position vector rAB can be established from the coordinates of points A and B. rAB = {[1.165 - (-2.048)]i + [1.389 - (-1.434)]j + ( -0.845 - 4.330)k} km = {3.213i + 2.822j - 5.175)k} km The distance between points A and B is d = rAB = 23.2132 + 2.8222 + (-5.175)2 = 6.71 km
Ans.
111
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2–96. 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 x = 3 m and z = 2 m.
z x
2m B 3m
C
z
x
FB 6 m FC
A y
1500 N
112
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2–97. 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 FB = 1610 N and FC = 2400 N.
z x
2m B 3m
C
z
x
FB 6 m FC
A y
1500 N
113
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.
.
114
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2–99. Determine the magnitude and coordinate direction angles of the resultant force acting at point A.
z 1.5 m A F1 F2
200 N
4m
SOLUTION rAC = {3i - 0.5j - 4k} m
F2 = 200a
2m
3i - 0.5j - 4k b = (119.4044i - 19.9007j - 159.2059k) 5.02494
x
rAB = (1.5i + 4.0981j + 4k) |rAB| = 2(1.5)2 + (4.0981)2 + (-4)2 = 5.9198 1.5i + 4.0981j - 4k b = (38.0079i + 103.8396j - 101.3545k) 5.9198
FR = F1 + F2 = (157.4124i + 83.9389j - 260.5607k) FR = 2(157.4124)2 + (83.9389)2 + (- 260.5604)2 = 315.7786 = 316 N
Ans.
a = cos-1 a
157.4124 b = 60.100° = 60.1° 315.7786
Ans.
b = cos-1 a
83.9389 b = 74.585° = 74.6° 315.7786
Ans.
g = cos-1 a
-260.5607 b = 145.60° = 146° 315.7786
Ans.
115
y
3m
B
C
rAB = (3 cos 60°i + (1.5 + 3 sin 60°) j - 4k)
F1 = 150a
60
3m
|rAC| = 232 + (- 0.5)2 + ( -4)2 = 225.25 = 5.02494
150 N
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2–100. 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.
z
SOLUTION
B
Unit Vector:
FB
175 N
rAC = {( -1 - 0)i + (4 - 0)j + (0 - 4)k} m = { -1i + 4j - 4k} m 2m
rAC = 2( -1)2 + 4 2 + (-4)2 = 5.745 m uAC
x
rBD = 22 2 + (-3)2 + (-5.5)2 = 6.576 m 2i - 3j - 5.5k rBD = 0.3041i - 0.4562j - 0.8363k = rBD 6.576
Force Vector: FA = FA uAC = 250{ -0.1741i + 0.6963j - 0.6963k} N = { -43.52i + 174.08j - 174.08k} 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.
116
250 N
4m D 3m
rBD = {(2 - 0) i + (- 3 - 0)j + (0 - 5.5)k} m = {2i - 3j - 5.5k} m
= { -43.5i + 174j - 174k} N
FA
4m C
- 1i + 4j - 4k rAC = -0.1741i + 0.6963j - 0.6963k = = rAC 5.745
uBD =
1.5 m A
1m
y
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2 –101. The force acting on the man, caused by his pulling on the anchor cord, is F. If the length of the cord is L, determine the coordinates A(x, y, - z) of the anchor. Given:
F
§ 40 · ¨ 20 ¸ N ¨ ¸ © 50 ¹
L
25 m
Solution: F
r
L
r
§ 14.9 · ¨ 7.5 ¸ m ¨ ¸ © 18.6 ¹
F
Ans.
117
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2–102. z
Each of the four forces acting at E has a magnitude of 28 kN. Express each force as a Cartesian vector and determine the resultant force. E
FEC
FEA FEB
FED D
SOLUTION FEA = 28a
6 4 12 i j kb 14 14 14
12 m A
FEA = {12i - 8j - 24k} kN FEB
Ans.
x
FEB = {12i + 8j - 24k} kN FEC = 28a
Ans.
-6 4 12 i + j kb 14 14 14
FEC = {-12i + 8j - 24k } kN FED = 28a
6m
4m 4m
6 4 12 = 28a i + j kb 14 14 14
Ans.
-6 4 12 i j kb 14 14 14
FED = {-12i - 8j - 24k} kN
Ans.
FR = FEA + FEB + FEC + FED = {-96k } kN
Ans.
118
C
B
6m y
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2–103. The cord exerts a force F on the hook. If the cord is length L, determine the location x,y of the point of attachment B, and the height z of the hook.
Units Used: Given:
§ 12 · F ¨ 9 ¸ kN ¨ ¸ © 8 ¹ L 4m a 1m
Solution : Initial guesses
Given
x 1m
§x a· ¨ y ¸ = L F ¨ ¸ F © z ¹
y 1m
z 1m
§x· ¨ y ¸ Find ( x y z) ¨ ¸ ©z¹
119
§x· ¨y¸ ¨ ¸ ©z¹
§ 3.82 · ¨ 2.12 ¸ m ¨ ¸ © 1.88 ¹
Ans.
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2–104. The cord exerts a force of magnitude F on the hook. If the cord length L, the distance z, and the x component of the force F x are given, determine the location x, y of the point of attachment B of the cord to the ground. Given: F 30kN L 4m z 2m Fx 25kN a 1m Solution : Guesses x 1m y 1m Given Fx =
xa F L
§x· ¨ ¸ Find ( x y) ©y¹
2
2
2
2
L = ( x a) y z
§x· ¨ ¸ ©y¹
§ 4.33 · m ¨ ¸ © 0.94 ¹
120
Ans.
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. . . .
121
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2–106 . The chandelier is supported by three chains which are concurrent at point O. If the force in each chain has magnitude F, express each force as a Cartesian vector and determine the magnitude and coordinate direction angles of the resultant force. Given: F 300N a 1.8m b 1.2m T 1 120deg T 2 120deg Solution:
T 3 360deg T 1 T 2
§ b sin T 1 · ¨ ¸ rOA ¨ b cos T 1 ¸ ¨ ¸ © a ¹ § b sin T 1 T 2 · ¨ ¸ rOB ¨ b cos T 1 T 2 ¸ ¨ ¸ a © ¹ §0· ¨b ¸ rOC ¨ ¸ © a ¹
FA F
FB F
FC F
rOA rOA
rOB rOB
rOC rOC
FR FA FB FC
FA
§ 144.12 · ¨ 83.21 ¸ N ¨ ¸ © 249.62 ¹
Ans.
FB
§ 144.12 · ¨ 83.21 ¸ N ¨ ¸ © 249.62 ¹
Ans.
FC
§ 0.00 · ¨ 166.41 ¸ N ¨ ¸ © 249.62 ¹
Ans.
748.85 N
Ans.
FR
122
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§¨ D ·¸ § FR · ¨ E ¸ acos ¨ ¸ FR ¨J ¸ © ¹ © ¹
§¨ D ·¸ ¨E ¸ ¨J ¸ © ¹
123
§ 90.00 · ¨ 90.00 ¸ deg ¨ ¸ © 180.00 ¹
Ans.
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2–107. The chandelier is supported by three chains which are concurrent at point O. If the resultant force at O has magnitude FR and is directed along the negative z axis, determine the force in each chain assuming F A = FB = F C . Given: a 1.8m b 1.2m F R 650N Solution: 2
F
F
a b 3a
260.4 N
2
FR
Ans.
124
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2–108. z
Determine the magnitude and coordinate direction angles of the resultant force. Set FB = 630 N, FC = 520 N and FD = 750 N, and x = 3 m and z = 3.5 m.
3m
B 4.5 m
2m C FC
FB A
SOLUTION
4m
Force Vectors: The unit vectors uB, uC, and uD of FB, FC, and FD must be determined first. From Fig. a,
x
(-3 - 0)i + (0 - 6)j + (4.5 - 2.5)k rB 3 6 2 = = - i j + k uB = rB 7 7 7 2 2 2 3(-3 - 0) + (0 - 6) + (4.5 - 2.5) uC =
(2 - 0)i + (0 - 6)j + (4 - 2.5)k rC 4 12 3 = = i j + k rC 13 13 13 2 2 2 3(2 - 0) + (0 - 6) + (4 - 2.5)
uD =
(3 - 0)i + (0 - 6)j + (-3.5 - 2.5)k rD 1 2 2 = = i j - k rD 3 3 3 2 2 2 3(0 - 3) + (0 - 6) + (-3.5 - 2.5)
FC = FC uC = 520 a
3 6 2 i - j + kb = 5-270i - 540j + 180k6 N 7 7 7
4 12 3 i j + k b = 5160i - 480j + 120k6 N 13 13 13
1 2 2 FD = FD uD = 750 a i - j - kb = 5250i - 500j - 500k6 N 3 3 3 Resultant Force: FR = FB + FC + FD = (-270i - 540j + 180k) + (160i - 480j + 120k) + (250i - 500j - 500k) = [140i - 1520j - 200k] N The magnitude of FR is FR = 3(FR)x 2 + (FR)y 2 + (FR)z 2 = 31402 + (-1520)2 + (-200)2 = 1539.48 N = 1.54 kN
Ans.
The coordinate direction angles of FR are a = cos-1 c b = cos-1 c g = cos-1 c
(FR)x 140 d = cos-1 a b = 84.8° FR 1539.48 (FR)y FR (FR)z FR
d = cos-1 a
Ans.
-1520 b = 171° 1539.48
Ans.
-200 b = 97.5° 1539.48
Ans.
d = cos-1 a
125
FD
z
6m D
Thus, the force vectors FB, FC, and FD are given by FB = FB uB = 630a -
O
x
2.5 m y
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2–109. z
If the magnitude of the resultant force is 1300 N and acts along the axis of the strut, directed from point A towards O, determine the magnitudes of the three forces acting on the strut. Set x = 0 and z = 5.5 m.
3m
B 4.5 m
2m C FC
FB A
SOLUTION
4m
Force Vectors: The unit vectors uB, uC, uD, and uFR of FB, FC, FD, and FR must be determined first. From Fig. a,
x
O z
(-3 - 0)i + (0 - 6)j + (4.5 - 2.5)k rB 3 6 2 = = - i j + k uB = rB 7 7 7 3(-3 - 0)2 + (0 - 6)2 + (4.5 - 2.5)2 uC =
(2 - 0)i + (0 - 6)j + (4 - 2.5)k rC 4 12 3 = = i j + k rC 13 13 13 3(2 - 0)2 + (0 - 6)2 + (4 - 2.5)2
uD =
(0 - 0)i + (0 - 6)j + (-5.5 - 2.5)k rD 3 4 = = - j + k rD 5 5 2 2 2 3(0 - 0) + (0 - 6) + (-5.5 - 2.5)
uFR =
(0 - 0)i + (0 - 6)j + (0 - 2.5)k rAO 12 5 = = j + k rAO 13 13 2 2 2 3(0 - 0) + (0 - 6) + (0 - 2.5)
3 6 2 FB = FB uB = - FB i - FB j + FB k 7 7 7 4 12 3 F i F j + F k 13 C 13 C 13 C
3 4 FD = FD uD = - FD j - FD k 5 5 FR = FR uR = 1300 a -
12 5 j kb = [- 1200j-500k] N 13 13
Resultant Force: FR = FB + FC + FD 3 6 2 4 12 3 4 3 -1200j - 500k = a - FBi - FBj + FBkb + a FCi F j + F kb + a - FDj - FDkb 7 7 7 13 13 C 13 C 5 5 3 4 6 12 3 2 3 4 -1200j - 500k = a - FB + F bi + a - FB F - FDjb + a FB + F - FD bk 7 13 C 7 13 C 5 7 13 C 5 Equating the i, j, and k components, 4 3 F 0 = - FB + 7 13 C
(1)
6 12 3 -1200 = - FB - FC - FD j 7 13 5 3 4 2 F - F -500 = FB + 7 13 C 5 D
(2) (3)
Solving Eqs. (1), (2), and (3), yields FC = 442 N
FB = 318 N
FD = 866 N
Ans.
126
6m D
Thus, the force vectors FB, FC, FD, and F R are given by
FC = FC uC =
FD x
2.5 m y
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2 –110 . The positions of point A on the building and point B on the antenna have been measured relative to the electronic distance meter (EDM) at O. Determine the distance between A and B. Hint: Formulate a position vector directed from A to B; then determine its magnitude. Given: a
460 m
b
653 m
D
60 deg
E
55 deg
T
30 deg
I
40 deg
Solution:
rOA
§¨ a cos I sin T ¸· ¨ a cos I cos T ¸ ¨ ¸ a sin I © ¹
rOB
§¨ b cos E sin D ¸· ¨ b cos E cos D ¸ ¨ ¸ b sin E © ¹
rAB
rOB rOA
rAB
§ 148.2 · ¨ 492.4 ¸ m ¨ ¸ © 239.2 ¹
127
rAB
567.2 m
Ans.
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2–111. 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.
z D
FC ⫽ 5 kN
3m FB ⫽ 8 kN
C
B
45⬚
30⬚ y
A
0.75 m
FA ⫽ 6 kN
x
.
.
. .
.
.
.
128
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2–112. Given the three vectors A, B, and D, show that A # (B + D) = (A # B) + (A # D).
SOLUTION Since the component of (B + D) is equal to the sum of the components of B and D, then A # (B + D) = A # B + A # D
(QED)
Also, A # (B + D) = (A x i + A y j + A zk) # [(Bx + Dx)i + (By + Dy)j + (Bz + Dz)k] = A x (Bx + Dx) + A y (By + Dy) + A z (Bz + Dz) = (A xBx + A yBy + A zBz) + (A xDx + A yDy + A zDz) = (A # B) + (A # D)
(QED)
129
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2–113. Determine the angle u between the edges of the sheetmetal bracket.
z 400 mm
250 mm
x u
SOLUTION r1 = {400i + 250k} mm ;
r1 = 471.70 mm
r2 = {50i + 300j} mm ;
r2 = 304.14 mm
300 mm 50 mm
r1 # r2 = (400) (50) + 0(300) + 250(0) = 20 000 u = cos-1 ¢ = cos-1 ¢
y
r1 # r2 ≤ r1 r2 20 000 ≤ = 82.0° (471.70) (304.14)
Ans.
130
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2–114. Determine the angle u between the sides of the triangular plate.
z
3m B
4m
SOLUTION
A 1m
rAC = 53 i + 4 j - 1 k6 m
rAB = 52 j + 3 k6 m
5m x
rAB = 21222 + 1322 = 3.6056 m rAC # rAB = 0 + 4122 + 1-12132 = 5 rAC # rAB rACrAB
y 1m
rAC = 21322 + 1422 + 1-122 = 5.0990 m
u = cos-1
= cos-1
θ
5 15.0990213.60562
u = 74.219° = 74.2°
Ans.
131
3m C
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2–115. 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 , rAB, and rAC and then using the cosine law.
z
3m
B
SOLUTION 4m
rBC = {3 i + 2 j - 4 k} m
A
rBC = 2(3)2 + (2)2 + ( - 4)2 = 5.39 m
1m
Ans.
u y
1m
Also, rAC = {3 i + 4 j - 1 k} m
5m
rAC = 2(3)2 + (4)2 + (- 1)2 = 5.0990 m
x
rAB = {2 j + 3 k} m rAB = 2(2)2 + (3)2 = 3.6056 m rAC # rAB = 0 + 4(2) + (- 1)(3) = 5 u = cos-1 a
rAC # rAB 5 b = cos-1 rAC rAB (5.0990)(3.6056)
u = 74.219° rBC = 2(5.0990)2 + (3.6056)2 - 2(5.0990)(3.6056) cos 74.219° rBC = 5.39 m
Ans.
132
3m C
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2–116. Determine the angle T between the tails of the two vectors. Given: r1
9m
r2
6m
D
60 deg
E
45 deg
J
120 deg
I
30 deg
H
40 deg
Solution: Determine the two position vectors and use the dot product to find the angle
r1v
T
§¨ sin H cos I ¸· r1 ¨ sin H sin I ¸ ¨ cos H ¸ © ¹ § r1v r2v · ¸ © r1v r2v ¹
acos ¨
r2v
T
§¨ cos D ¸· r2 ¨ cos E ¸ ¨ cos J ¸ © ¹
109.4 deg
133
Ans.
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2 –117. Determine the magnitude of the projected component of r1 along r2, and the projection of r2 along r1. Given: r1
9m
r2
6m
D
60 deg
E
45 deg
J
120 deg
I
30 deg
H
40 deg
Solution: Write the vectors and unit vectors
r1v
§¨ sin H cos I ¸· r1 ¨ sin H sin I ¸ ¨ cos H ¸ © ¹
r2v
§¨ cos D ¸· r2 ¨ cos E ¸ ¨ cos J ¸ © ¹
u1
r1v r1v
u2
r2v r2v
r1v
§ 5.01 · ¨ 2.89 ¸ m ¨ ¸ © 6.89 ¹
r2v
§ 3 · ¨ 4.24 ¸ m ¨ ¸ © 3 ¹
u1
§ 0.557 · ¨ 0.321 ¸ ¨ ¸ © 0.766 ¹
u2
§ 0.5 · ¨ 0.707 ¸ ¨ ¸ © 0.5 ¹
The magnitude of the projection of r1 along r2.
r1v u2
2.99 m
Ans.
The magnitude of the projection of r2 along r1.
r2v u1
1.99 m
Ans.
134
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2–118. Determine the projection of the force F along the pole.
z
F = {2i + 4j + 10k} kN
O
y
SOLUTION Proj F
2 = F # ua = 12 i + 4 j + 10 k2 # a i +
Proj F = 0.667 kN
3
2m
2 1 j - kb 3 3 Ans.
2m x
135
1m
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2 –119. Determine the angle T between the two cords. Given: a
3m
b
2m
c
6m
d
3m
e
4m
Solution:
rAC
§b· ¨a¸ m ¨ ¸ ©c¹
rAB
§0· ¨ d ¸ m ¨ ¸ ©e ¹
T
136
§ rAC rAB · ¸ © rAC rAB ¹
acos ¨
T
64.6 deg
Ans.
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2–120. Two forces act on the hook. Determine the angle u between them. Also, what are the projections of F1 and F2 along the y axis?
z
F1 ⫽ 600 N 45⬚
60⬚ 120⬚ u
.
y x
. .
137
F2 ⫽ {120i + 90j – 80k}N
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2–121. Two forces act on the hook. Determine the magnitude of the projection of F2 along F1.
z
F1 ⫽ 600 N 45⬚
60⬚ 120⬚ u y x
.
138
F2 ⫽ {120i + 90j – 80k}N
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2–122. Determine the magnitude of the projected component of force FAB acting along the z axis.
Z A FAC FAB
3 kN
9m
3.5 kN D
4.5 m O
Unit Vector: The unit vector uAB must be determined first. From Fig. a, uAB =
rAB rAB
=
(4.5 – 0)i + (–3 – 0)j + (0 – 9)k 2
2
(4.5 – 0) + (–3 – 0) + (0 – 9)
= 2
3 2 6 i– j– k 7 7 7
3m x
Thus, the force vector FAB is given by ⎛3 2 6 ⎞ FAB = FABuAB = 3.5 ⎜ i – j – k⎟ = {1.5i – 1j – 3k} kN 7 7 ⎠ ⎝7 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.
A(0, 0, 9) m
B(4.5, –3, 0) m
139
3m
B 3m C
30
y
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2–123. Determine the magnitude of the projected component of force FAC acting along the z axis.
Z A FAC FAB
3 kN
9m
3.5 kN D
4.5 m O
3m
B 3m x
Unit Vector: The unit vector uAC must be determined first. From Fig. a, uAC =
rAC rAC
=
(3 sin 30° – 0)i + (3 cos 30° – 0)j + (0 – 9)k
= 0.1581i + 0.2739j – 0.9487k
(3 sin 30° – 0) 2 + (3 cos 30° – 0) 2 + (0 – 9) 2
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.
A(0, 0, 9) m
C(3 sin 30°, 3 cos 30°, 0) m
140
3m C
30
y
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2–124. Determine the projection of force F = 400 N acting along line AC of the pipe assembly. Express the result as a Cartesian vector.
z F ⫽ 400 N B
45⬚
C 30⬚ A 3m x
4m y
N
N
141
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2–125. Determine the magnitudes of the components of force F = 400 N acting parallel and perpendicular to segment BC of the pipe assembly.
z F ⫽ 400 N B
45⬚
C 30⬚ A 3m x
4m y
N
142
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2–126. Cable OA is used to support column OB . Determine the angle u it makes with beam OC.
z D 30⬚
O f C
4m
u 8m
x 8m
B
.
143
A
y
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2–127. Cable OA is used to support column OB. Determine the angle f it makes with beam OD.
z D 30⬚
O f C
4m
u 8m
x 8m
B
.
144
A
y
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2–128. Determine the angles T and I between the axis OA of the pole and each cable, AB and AC. Given: F1
50 N
F2
35 N
a
1m
b
3m
c
2m
d
5m
e
4m
f
6m
g
4m
Solution:
rAO
§0· ¨ g ¸ ¨ ¸ ©f ¹
rAB
§e· ¨a¸ ¨ ¸ ©f¹
rAC
§ c · ¨a b¸ ¨ ¸ © f ¹
§ rAO rAB · ¸ © rAO rAB ¹
T
52.4 deg
Ans.
§ rAO rAC · ¸ © rAO rAC ¹
I
68.2 deg
Ans.
T
acos ¨
I
acos ¨
145
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2 –129. The two cables exert the forces shown on the pole. Determine the magnitude of the projected component of each force acting along the axis OA of the pole. Given: F1
50 N
F2
35 N
a
1m
b
3m
c
2m
d
5m
e
4m
f
6m
g
4m
Solution:
rAB
§e· ¨a¸ ¨ ¸ ©f¹
F 1v
F1
F 2v
F2
rAC
rAC rAC rAB rAB
§ c · ¨a b¸ ¨ ¸ © f ¹
rAO
§0· ¨ g ¸ ¨ ¸ ©f ¹
uAO
rAO rAO
F 1AO
F 1v uAO
F 1AO
18.5 N
Ans.
F 2AO
F 2v uAO
F 2AO
21.3 N
Ans.
146
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2–130. Determine the angle u between the pipe segments BA and BC.
z
1.5 m
A
2m 1m
2m B
x
SOLUTION
y
F ⫽ {30i ⫺ 45j ⫹ 50k} N
Position Vectors: The position vectors rBA and rBC must be determined first. From Fig. a,
2m C
rBA = (0 - 1.5)i + (0 - 2)j + (0 - 0)k = { - 1.5i - 2j} m rBC = (3.5 - 1.5)i + (3 - 2)j + (- 2 - 0)k = {2i + 1j - 2k} m The magnitude of rBA and rBC are rBA = 3(- 1.5 )2 + (- 2)2 = 2.5 m
B(1.5, 2, 0) m
rBC = 32 + 1 + (- 2) = 3 m 2
2
2
Vector Dot Product: rBA # rBC = (- 1.5 i - 2j) # (2i + 1j - 2k)
C(3.5, 3, –2) m
= (- 1.5)(2) + (- 2)(1) + 0(- 2) = - 5 m2 Thus, u = cos-1 a
rBA # rBC -5 d = 132° b = cos-1 c rBA rBC 2.5(3)
Ans.
147
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2–131. Determine the angles u and f made between the axes OA of the flag pole and AB and AC, respectively, of each cable.
z 1.5 m 2m
B
C
4m
SOLUTION
FC
rA C = { - 2i - 4j + 1k} m ;
rA C = 4.58 m
rAB = {1.5i - 4j + 3k} m;
rAB = 5.22 m
rA O = { - 4j - 3k} m;
rA O = 5.00 m
FB
u
6m
4m x y
7 = cos - 1 ¢ ≤ = 74.4° 5.22(5.00)
Ans.
rAC # rAO = (-2)(0) + (- 4)(- 4) + (1)( -3) = 13
= cos - 1 a
A
3m
rAB # rAO ≤ rAB rAO
f = cos - 1 a
f
O
rA B # rA O = (1.5)(0) + ( -4)( -4) + (3)(- 3) = 7 u = cos - 1 ¢
40 N
55 N
rAC # rAO b rAC rAO 13 b = 55.4° 4.58(5.00)
Ans.
148
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2–132. 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.
z F1
400 N
35
SOLUTION 120
Force Vector: uF1 = sin 35° cos 20°i - sin 35° sin 20°j + cos 35°k 20
= 0.5390i - 0.1962j + 0.8192k F1 = F1uF1 = 400(0.5390i - 0.1962j + 0.8192k) N
x
= {215.59i - 78.47j + 327.66k} N Unit Vector: The unit vector along the line of action of F2 is uF2 = cos 45°i + cos 60°j + cos 120°k = 0.7071i + 0.5j - 0.5k Projected Component of F1 Along Line of Action of F2: (F1)F2 = F1 # 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 Negative sign indicates that the force component (F1)F2 acts in the opposite sense of direction to that of uF2. Thus the magnitude is (F1 )F2 = 50.6 N
Ans.
149
u 45
y
60
F2
400 N
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2–133. Determine the angle u between the two cables attached to the post.
z F1
400 N
35
SOLUTION 120
Unit Vector: uF1 = sin 35° cos 20°i - sin 35° sin 20°j + cos 35°k 20
= 0.5390i - 0.1962j + 0.8192k uF2 = cos 45°i + cos 60°j + cos 120°k
x
= 0.7071i + 0.5j - 0.5k The Angle Between Two Vectors u: The dot product of two unit vectors must be determined first. 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 Then, u = cos-1 A uF1 # uF2 B = cos-1(- 0.1265) = 97.3°
Ans.
150
u 45
y
60
F2
400 N
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2 –134 . Force F is applied to the handle of the wrench. Determine the angle T between the tail of the force and the handle AB. Given: a
300 mm
b
500 mm
F
80 N
T1
30 deg
T2
45 deg
Solution:
Fv
T
§ cos T 1 sin T 2 · ¨ ¸ F ¨ cos T 1 cos T 2 ¸ ¨ ¸ sin T 1 © ¹ § Fv uab · ¸ © F ¹
acos ¨
uab
T
§0· ¨ 1 ¸ ¨ ¸ ©0¹ 127.8 deg
151
Ans.
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2–135. Determine the projected component of the force F acting along the axis AB of the pipe. Given: F
80 N
a
4m
b
3m
c
12 m
d
2m
e
6m
Solution: Find the force and the unit vector
rA
rAB
§ e · ¨ a b ¸ ¨ ¸ © dc ¹ § e · ¨ b ¸ ¨ ¸ ©d¹
rA
§ 6 · ¨ 7 ¸ m ¨ ¸ © 10 ¹
rAB
§ 6 · ¨ 3 ¸ m ¨ ¸ ©2¹
F
Fv
rA rA
rAB
uAB
rAB
Now find the projection using the Dot product. F AB
F v uAB
F AB
31.1 N
Ans.
152
Fv
uAB
§ 35.3 · ¨ 41.2 ¸ N ¨ ¸ © 58.8 ¹ § 0.9 · ¨ 0.4 ¸ ¨ ¸ © 0.3 ¹
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2–136. Determine the components of F that act along rod AC and perpendicular to it. Point B is located at the midpoint of the rod.
z A
B
4m
O
SOLUTION rAC = (- 3i + 4j - 4k), rAB =
rAC = 2(- 3) + 4 + ( - 4) = 241 m 2
2
D
rBD = rAD - rAB = (4i + 6j - 4k) - ( - 1.5i + 2j - 2k) = {5.5i + 4j - 2k} m rBD = 2(5.5)2 + (4)2 + (- 2)2 = 7.0887 m rBD b = 465.528i + 338.5659j - 169.2829k rBD
Component of F along rAC is F| | (465.528i + 338.5659j - 169.2829k) # ( -3i + 4j - 4k) F # rAC = rAC 241
F| | = 99.1408 = 99.1 N
Ans.
Component of F perpendicular to rAC is F F2 + F2|| = F2 = 6002 F2 = 6002 - 99.14082 F
3m
6m
rAD = rAB + rBD
F| | =
4m C
600 N
x 2
- 3i + 4j + 4k rAC = = -1.5i + 2j - 2k 2 2
F = 600a
F
= 591.75 = 592 N
Ans.
153
4m
y
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2–137. Determine the components of F that act along rod AC and perpendicular to it. Point B is located 3 m along the rod from end C.
z A
B
4m
O
SOLUTION
D
3 (r ) = 1.40556i - 1.874085j + 1.874085k = 6.403124 CA
rOB = rOC + rCB = -3i + 4j + r CB = -1.59444i + 2.1259j + 1.874085k rOD = rOB + rBD rBD = rOD - rOB = (4i + 6j) - rOB = 5.5944i + 3.8741j - 1.874085k rBD = 2(5.5944)2 + (3.8741)2 + ( -1.874085)2 = 7.0582 rBD ) = 475.568i + 329.326j - 159.311k rBD
rAC = (- 3i + 4j - 4k),
rAC = 241
Component of F along rAC is F| F| | =
|
(475.568i + 329.326j - 159.311k) # (- 3i + 4j - 4k) F # rAC = rAC 241
F| | = 82.4351 = 82.4 N
Ans.
Component of F perpendicular to rAC is F F2 + F2|| = F2 = 6002 F2 = 6002 - 82.43512 F
3m
6m
rCA = 6.403124
F = 600(
4m C
600 N
x
rCA = 3i - 4j + 4k
rCB
F
= 594 N
Ans.
154
4m
y
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2–138. Determine the magnitudes of the projected components of the force F = 300 N acting along the x and y axes.
30
F
z A
30 300 mm
SOLUTION
O 300 mm
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
x
= [-75i + 259.81j + 129.90k] N Vector Dot Product: The magnitudes of the projected component of F along the x and y axes are Fx = F # i =
A -75i + 259.81j + 129.90k B # i
= -75(1) + 259.81(0) + 129.90(0) = -75 N Fy = F # j =
A - 75i + 259.81j + 129.90k B # j
= -75(0) + 259.81(1) + 129.90(0) = 260 N The negative sign indicates that Fx is directed towards the negative x axis. Thus Fx = 75 N,
Fy = 260 N
Ans.
155
300 mm
y
300 N
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2–139. Determine the magnitude of the projected component of the force F = 300 N acting along line OA.
30
F
z A
30 300 mm
O 300 mm
SOLUTION
x
300 mm
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 uOA =
(-0.45 - 0)i + (0.3 - 0)j + (0.2598 - 0)k rOA = = -0.75i + 0.5j + 0.4330k rOA 2( -0.45 - 0)2 + (0.3 - 0)2 + (0.2598 - 0)2
Vector Dot Product: The magnitude of the projected component of F along line OA is FOA = F # uOA =
A -75i + 259.81j + 129.90k B # A - 0.75i + 0.5j + 0.4330k B
= (- 75)( -0.75) + 259.81(0.5) + 129.90(0.4330) = 242 N
Ans.
156
y
300 N
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2–140. Determine the length of the connecting rod AB by first formulating a Cartesian position vector from A to B and then determining its magnitude.
y
25
B
A
SOLUTION
125 mm
x O
rAB = [400 - (- 125 sin 30°)]i + (0 - 125 cos 30°) j 400 mm
= {462.5 i - 108.25 j} mm rAB = 2(462.5)2 + (108.25)2 = 475 mm
Ans.
157
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2–141. y
Determine the x and y components of F1 and F2.
45⬚ F1 ⫽ 200 N
30⬚
SOLUTION F1x = 200 sin 45° = 141 N
Ans.
F1y = 200 cos 45° = 141 N
Ans.
F2x = -150 cos 30° = -130 N
Ans.
F2y = 150 sin 30° = 75 N
Ans.
F2 ⫽ 150 N x
158
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2–142. y
Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis. 45⬚
F1 ⫽ 200 N
30⬚
SOLUTION +R FRx = ©Fx;
FRx = -150 cos 30° + 200 sin 45° = 11.518 N
Q+FRy = ©Fy;
FRy = 150 sin 30° + 200 cos 45° = 216.421 N
F2 ⫽ 150 N x
FR = 2 (11.518)2 + (216.421)2 = 217 N
Ans.
216.421 ≤ = 87.0° 11.518
Ans.
u = tan - 1 ¢
159
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2–143. Resolve the 250-N force into components acting along the u and v axes and determine the magnitudes of these components. u 20⬚ 250 N 40⬚
v
. .
160
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2–144 . Express F1 and F2 as Cartesian vectors.
y
F2 = 26 kN 13
12 5
x
SOLUTION F1 = -30 sin 30° i - 30 cos 30° j = 5 -15.0 i - 26.0 j6 kN F2 = =
30°
Ans.
12 5 1262 i + 1262 j 13 13
F1 = 30 kN
-10.0 i + 24.0 j kN
Ans.
161
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2–145. Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis.
y
F2 = 26 kN 13
12 5
x
SOLUTION + F = ©F ; : Rx x + c FRy = ©Fy ;
FRx = - 30 sin 30° FRy
5 1262 = -25 kN 13
30°
12 = - 30 cos 30° + 1262 = -1.981 kN 13
F1 = 30 kN
FR = 21-2522 + 1-1.98122 = 25.1 kN f = tan-1 a
Ans.
1.981 b = 4.53° 25
u = 180° + 4.53° = 185°
Ans.
162
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2–146. 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.
z B
4m
O
y 60⬚
.
163
3m A
x
.
80 N
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2–147. y
Determine the magnitude and direction of the resultant FR = F1 + F2 + F3 of the three forces by first finding the resultant F¿ = F1 + F3 and then forming FR = F¿ + F2. Specify its direction measured counterclockwise from the positive x axis.
F2 ⫽ 75 N
F1 ⫽ 80 N
F3 ⫽ 50 N 30⬚
SOLUTION
45⬚ x
F¿ = 2(80)2 + (50)2 - 2(80)(50) cos 105° = 104.7 N sin f sin 105° = ; 80 104.7
f = 47.54°
FR = 2(104.7)2 + (75)2 - 2(104.7)(75) cos 162.46° FR = 177.7 = 178 N sin b sin 162.46° = ; 104.7 177.7
30⬚
Ans. b = 10.23°
u = 75° + 10.23° = 85.2°
Ans.
164
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2–148. If u = 60° and F = 20 kN, determine the magnitude of the resultant force and its direction measured clockwise from the positive x axis.
y 50 kN
5
3
4
x u
SOLUTION
2
1
F
1 4 = 50a b + (40) - 20 cos 60° = 58.28 kN 5 22
+ F = ©F ; : Rx x
FRx
+ c FRy = ©Fy ;
1 3 FRy = 50a b (40) - 20 sin 60° = -15.60 kN 5 22 FR = 2(58.28)2 + (- 15.60)2 = 60.3 kN f = tan - 1 B
40 kN
Ans.
15.60 R = 15.0° 58.28
Ans.
165
1
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2–14 9. Determine the design angle u (0° … u … 90°) for strut AB so that the 400-N horizontal force has a component of 500 N directed from A towards C. What is the component of force acting along member AB? Take f = 40°.
400 N A u f
B
C
N
.
N
N N
. N
166
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