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© 2011 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.
7 7
•2–1. If and , determine the magnitudeof the resultant force acting on the eyebolt and its directionmeasured clockwise from the positive x axis.
T = 6 kNu = 30°
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. 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
02a Ch02a 7-56.indd 7 6/12/09 8:05:02 AM
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 = 62 + 82 – 2(6)(8) cos 75° = 8.669 kN = 8.67 kN Ans
sin a8
sin 75°8.669
= a = 63.05°
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88
2–2. If and , determine the magnitudeof the resultant force acting on the eyebolt and its directionmeasured clockwise from the positive x axis.
T = 5 kNu = 60°
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. 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
02a Ch02a 7-56.indd 8 6/12/09 8:05:05 AM
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 = 52 + 82 – 2(5)(8) cos 105° = 10.47 kN = 10.5 kN Ans
sin a8
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,
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9 9
2–3. If the magnitude of the resultant force is to be 9 kNdirected along the positive x axis, determine the magnitude offorce T acting on the eyebolt and its angle .u
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8 kN
T
x
y
u
45
02a Ch02a 7-56.indd 9 6/12/09 8:05:11 AM
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
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1010
*2–4. Determine the magnitude of the resultant forceacting on the bracket and its direction measuredcounterclockwise from the positive u axis.
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u
F1 200 lb
F2 150 lbv
30
30
45
N
N
N
N
N
N
02a Ch02a 7-56.indd 10 6/12/09 8:05:16 AM
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 = 2002 + 1502 – 2(200)(150) cos 75° = 216.72 N = 217 N Ans
sin a200
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,
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1111
•2–5. Resolve F1 into components along the u and axes,and determine the magnitudes of these components.
v
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u
F1 200 lb
F2 150 lbv
30
30
45
N
N
N
N
02a Ch02a 7-56.indd 11 6/12/09 8:05:17 AM
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
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1212
2–6. Resolve F2 into components along the u and axes,and determine the magnitudes of these components.
v
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u
F1 200 lb
F2 150 lbv
30
30
45
N
N
N
N
02a Ch02a 7-56.indd 12 6/12/09 8:05:18 AM
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,
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1313
2–7. If and the resultant force acts along thepositive u axis, determine the magnitude of the resultantforce and the angle .u
FB = 2 kN
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y
x
uB
FA 3 kN
FB
Au 30
02a Ch02a 7-56.indd 13 6/12/09 8:05:21 AM
sin f3
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
13
2–7. If and the resultant force acts along thepositive u axis, determine the magnitude of the resultantforce and the angle .u
FB = 2 kN
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y
x
uB
FA � 3 kN
FB
Au 30�
02a Ch02a 7-56.indd 13 6/12/09 8:05:21 AM
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1414
*2–8. If the resultant force is required to act along thepositive u axis and have a magnitude of 5 kN, determine therequired magnitude of FB and its direction .u
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
x
uB
FA 3 kN
FB
Au 30
02a Ch02a 7-56.indd 14 6/12/09 8:05:23 AM
sin u5
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 = 32 + 52 – 2(3)(5) cos 30° = 2.832 kN = 2.83 kN Ans
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1515
•2–9. The plate is subjected to the two forces at A and Bas shown. If , determine the magnitude of theresultant of these two forces and its direction measuredclockwise from the horizontal.
u = 60°
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A
B
FA 8 kN
FB 6 kN
40
u
02a Ch02a 7-56.indd 15 6/12/09 8:05:26 AM
Parallelogram Law : The parallelogram law of addition is shown inFig. (a).
Trigonometry : Using law of cosines [Fig. (b)], we have
sin u6
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 = 82 + 62 – 2(8)(6) cos 100° = 10.80 kN = 10.8 kN Ans
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1616
2–10. Determine the angle of for connecting member Ato the plate so that the resultant force of FA and FB isdirected horizontally to the right.Also, what is the magnitudeof the resultant force?
u
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A
B
FA 8 kN
FB 6 kN
40
u
2–11. If the tension in the cable is 400 N, determine themagnitude and direction of the resultant force acting on the pulley. This angle is the same angle of line AB on thetailboard block.
u
400 N
30
y
A
Bx
400 N
u
02a Ch02a 7-56.indd 16 6/12/09 8:05:28 AM
Parallelogram Law : The parallelogram law of addition is shown inFig. (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, usinglaw of cosines, the magnitude of FR is
FR = 82 + 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 u400
u = 60° Anssin 60°400
= ;
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1717
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. 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 theknee 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 17 6/12/09 8:05:29 AM
–Fx
sin 20°Fx = –125 N Ans360
sin 100°= ;
Fy
sin 60°Fy = 317 N Ans360
sin 100°= ;
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1818
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. 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 theknee 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 18 6/12/09 8:05:30 AM
–Fx
sin 30°Fx = –183 N Ans360
sin 80°= ;
Fy
sin 70°Fy = 344 N Ans360
sin 80°= ;
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1919
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. 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 forstrut AB so that the 400-lb horizontal force has acomponent of 500 lb directed from A towards C.What is thecomponent 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 horizontalforce has a component of 600 lb which acts up to the left, inthe 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
02a Ch02a 7-56.indd 19 6/12/09 8:05:31 AM
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2020
*2–16. Resolve F1 into components along the u and axesand determine the magnitudes of these components.
v
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F1 250 N
F2 150 N u
v
30
30
105
•2–17. Resolve F2 into components along the u and axesand determine the magnitudes of these components.
v
F1 250 N
F2 150 N u
v
30
30
105
02a Ch02a 7-56.indd 20 6/12/09 8:05:32 AM
F1v
sin 30°F1v = 129 N Ans
Sine law :
250sin 105°
=
F1u
sin 45°F1u = 183 N Ans250
sin 105°=
F2v
sin 30°F2v = 77.6 N Ans
Sine law :
150sin 75°
=
F2u
sin 75°F2u = 150 N Ans150
sin 75°=
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2121
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2–18. The truck is to be towed using two ropes. Determinethe magnitudes of forces FA and FB acting on each rope inorder to develop a resultant force of 950 N directed alongthe 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 theresultant force is to be 950 N, directed along the positive xaxis, determine the magnitudes of forces FA and FB actingon each rope and the angle of FB so that the magnitude ofFB 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 21 6/12/09 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 isshown 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
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2222
*2–20. If , , and the resultant force is 6 kN directed along the positive y axis, determine the requiredmagnitude of F2 and its direction .u
F1 = 5 kNf = 45°
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F1
F2
x
y
u
f
60
02a Ch02a 7-56.indd 22 6/12/09 8:05:35 AM
Using this result and applying the law of sines to Fig. b, yields
F2 = 62 + 52 – 2(6)(5) cos 45° = 4.310 kN = 4.31 kN Ans
sin u5
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,
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2323
•2–21. If and the resultant force is to be 6 kNdirected along the positive y axis, determine the magnitudes ofF1 and F2 and the angle if F2 is required to be a minimum.u
f = 30°
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F1
F2
x
y
u
f
60
02a Ch02a 7-56.indd 23 6/12/09 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° Ansand
F1 = 6 cos 30° = 5.20 kN AnsF2 = 6 sin 30° = 3 kN Ans
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2424
2–22. If , , and the resultant force is tobe directed along the positive y axis, determine themagnitude of the resultant force if F2 is to be a minimum.Also, what is F2 and the angle ?u
F1 = 5 kNf = 30°
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F1
F2
x
y
u
f
60
02a Ch02a 7-56.indd 24 6/12/09 8:05:50 AM
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
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2525
2–23. If and , determine the magnitudeof the resultant force acting on the plate and its directionmeasured clockwise from the positive x axis.
F2 = 6 kNu = 30°
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y
x
F3 5 kN
F1 4 kN
F2
u
02a Ch02a 7-56.indd 25 6/12/09 8:05:55 AM
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 = 62 + 6.4032 – 2(6)(6.403) cos (51.34° + 30°)
= 8.09 kN Ans
sin b6
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 forcessuccessively, 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 determinedas follows.
54
F� = 42 + 52 = 6.403 kN
tan a = a = 51.34°
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2626
*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 themagnitudes of FR and F2 and the angle .u … 90°
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. 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
02a Ch02a 7-56.indd 26 6/12/09 8:05:59 AM
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,
54
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2727
•2–25. Two forces F1 and F2 act on the screw eye. If theirlines of action are at an angle apart and the magnitude of each force is determine the magnitude ofthe resultant force FR and the angle between FR and F1.
F1 = F2 = F,u
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F2
F1
u
02a Ch02a 7-56.indd 27 6/12/09 8:06:00 AM
=
=
FR = (F)2 + (F)2 – 2(F)(F) cos (180° – u)
FR = F( 2) 1 + cos u
FR = 2F cos ( )
Fsin f
Fsin(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 u2
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2833
*2–26. Determine the magnitude of the resultant forceacting on the pin and its direction measured clockwise fromthe positive x axis.
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x
yF1 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 y2 = (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
FR y
R x
)
)
⎡
⎣⎢⎢
⎤
⎦⎥⎥
= tan–1 66.43
331.61
⎛
⎝⎜
⎞
⎠⎟ = 11.3° Ans
02a Ch02a 7-56.indd 33 6/12/09 8:06:09 AM
2–26.
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2934
•2–33. If and , determine themagnitude of the resultant force acting on the eyebolt andits direction measured clockwise from the positive x axis.
f = 30°F1 = 600 N
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y
x
345
F2 500 N
F1
F3 450 N
f
60
02a Ch02a 7-56.indd 34 6/12/09 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 y2 = 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
FR y
R x
)
)
⎡
⎣⎢⎢
⎤
⎦⎥⎥
= tan–1 493.01
499.62
⎛
⎝⎜
⎞
⎠⎟ = 44.6° Ans
5
3⎛
⎝⎜
⎞
⎠⎟
5
4⎛
⎝⎜
⎞
⎠⎟
•2–27.
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3035
2–28. If the magnitude of the resultant force acting on the eyebolt is 600 N and its direction measured clockwisefrom the positive x axis is , determine the magni-tude of F1 and the angle .f
u = 30°
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y
x
345
F2 500 N
F1
F3 450 N
f
60
02a Ch02a 7-56.indd 35 6/12/09 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 N5
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)
*2–28.
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3136
2–29. The contact point between the femur and tibiabones of the leg is at A. If a vertical force of 875 N is appliedat this point, determine the components along the x and yaxes. Note that the y component represents the normalforce on the load-bearing region of the bones. Both the xand y components of this force cause synovial fluid to besqueezed out of the bearing space.
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x
A
875 N
125
13
y
875 N
875 N
Fx = 8755
13 = 336.5 N Ans
Fy = –87512
13 = –807.7 N Ans
02a Ch02a 7-56.indd 36 6/12/09 8:06:14 AM
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3237
*2–36. If and , determine the magnitudeof the resultant force acting on the plate and its direction measured clockwise from the positive x axis.
u
F2 = 3 kNf = 30°
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x
y
F2
5
4
3
F1 4 kN
F3 5 kN
f
30
02a Ch02a 7-56.indd 37 6/12/09 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 y2 = 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
FR y
R x
)
)
⎡
⎣
⎤
⎦ = tan–1 4.964
4.598
= 47.2° Ans
5
4
5
3
2–30.
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3338
•2–31. If the magnitude for the resultant force acting onthe plate is required to be 6 kN and its direction measuredclockwise from the positive x axis is , determine themagnitude of F2 and its direction .f
u = 30°
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x
y
F2
5
4
3
F1 4 kN
F3 5 kN
f
30
02a Ch02a 7-56.indd 38 6/12/09 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 kN5
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)
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3439
2–32. If and the resultant force acting on thegusset plate is directed along the positive x axis, determinethe magnitudes of F2 and the resultant force.
f = 30°
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x
y
F2
5
4
3
F1 4 kN
F3 5 kN
f
30
02a Ch02a 7-56.indd 39 6/12/09 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 kN5
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
*2–32.
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3540
2–33. Determine the magnitude of F1 and its direction so that the resultant force is directed vertically upward andhas a magnitude of 800 N.
u
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Ax
y
F1
400 N600 N
34
5
30
u
02a Ch02a 7-56.indd 40 6/12/09 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;
53
, = 800 = F1 cos u + 400 sin 30° + 600 F1 cos u = 240 [2]
53
x
FRx
FRy
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3641
*2–34. Determine the magnitude and direction measuredcounterclockwise from the positive x axis of the resultantforce of the three forces acting on the ring A. Take
and .u = 20°F1 = 500 N
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Ax
y
F1
400 N600 N
34
5
30
u
02a Ch02a 7-56.indd 41 6/12/09 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 + FRx 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–1F
F
Ry
Rx
= tan–1 1029.8
37.42
⎛
⎝⎜
⎞
⎠⎟ = 87.9° Ans
FRx
FRy
2–34.
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3742
•2–41. Determine the magnitude and direction of FB sothat the resultant force is directed along the positive y axisand has a magnitude of 1500 N.
u
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FB
x
y
B A
30FA 700 N
u
02a Ch02a 7-56.indd 42 6/12/09 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
•2–35.
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3843
2–36. Determine the magnitude and angle measuredcounterclockwise from the positive y axis of the resultantforce acting on the bracket if and .u = 20°FB = 600 N
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FB
x
y
B A
30FA 700 N
u
02a Ch02a 7-56.indd 43 6/12/09 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 + FRx Ry
2 = 213.82 + 811.42 = 839 N Ans
The direction angle u measured counterclockwise from positive y axis is
f = tan –1F
F
Rx
Ry
= tan–1 213.8
811.4
⎛
⎝⎜
⎞
⎠⎟ = 14.8° Ans
*2–36.
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3944
2–43. If and , determine themagnitude of the resultant force acting on the bracket andits direction measured clockwise from the positive x axis.
F1 = 250 lbf = 30°
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F3 260 lb
F2 300 lb5
1213
34
5
x
yF1
f
1.3 kN1.5 kN
1.25 kN
1.783 kN
1.475 kN
1.5 kN
1.3 kN
2–37. 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 = 1.54
5
= 1.2 kN (F2)y = 1.53
5
= 0.9 kN
(F3)x = 1.35
13
= 0.5 kN (F3)y = 1.312
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 y2 = 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
FR y
R x
)
)
⎡
⎣⎢⎢
⎤
⎦⎥⎥
= tan–1 1.475
1.783
= 39.6° Ans
02a Ch02a 7-56.indd 44 6/12/09 8:06:23 AM
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4045
*2 If the magnitude of the resultant force acting onthe bracket is 400 lb directed along the positive x axis,determine the magnitude of F1 and its direction .f
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F3 260 lb
F2 300 lb5
1213
34
5
x
yF1
f
1.5 kN
1.3 kN
1.5 kN
1.3 kN
2 kN
*2–34. 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 = 1.54
5
= 1.2 kN (F2)y = 1.53
5
= 0.9 kN
(F3)x = 1.35
13
= 0.5 kN (F3)y = 1.312
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 45 6/12/09 8:06:26 AM
2–38.
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4146
•2–39. If the resultant force acting on the bracket is to bedirected along the positive x axis and the magnitude of F1 isrequired to be a minimum, determine the magnitudes of theresultant force and F1.
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F3 260 lb
F2 300 lb5
1213
34
5
x
yF1
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 = 1.54
5
= 1.2 kN (F2)y = 1.53
5
= 0.9 kN
(F3)x = 1.35
13
= 0.5 kN (F3)y = 1.312
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 46 6/12/09 8:06:28 AM
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4247
2–40. The three concurrent forces acting on the screw eyeproduce a resultant force . If and F1 is tobe 90° from F2 as shown, determine the required magnitudeof F3 expressed in terms of F1 and the angle .u
F2 = 23 F1FR = 0
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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 xaxis and has a magnitude of 1250 N.
u
30
y
xO
B
A
FA
FB 800 N
u
02a Ch02a 7-56.indd 47 6/12/09 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]
*2–40.
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4357
2–41. Determine the coordinate angle for F2 and thenexpress each force acting on the bracket as a Cartesianvector.
g
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y
z
F2 600 N
F1 450 N
45
30
4560
x
02b Ch02b 57-106.indd 57 6/12/09 8:21:51 AM
Rectangular Components: Since cos2 a2 + cos2 b2 + cos2 g2 = 1, then cos g2z = ± 1 – cos2 45° – cos2 60° = ±0.5.
However, it is required that g2 > 90°, thus, g2 = cos–1 (–0.5) = 120°. 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 expressed in Cartesian vector form as
F1 = 450 cos 45° sin 30° (–i) + 450 cos 45° cos 30° (+j) + 450 sin 45° (+k) = {–159i + 276j + 318k} N Ans
F2 = 600 cos 45°i + 600 cos 60°j + 600 cos 120°k = {424i + 300j – 300k} N Ans
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4458
*2–42. Determine the magnitude and coordinate directionangles of the resultant force acting on the bracket.
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y
z
F2 600 N
F1 450 N
45
30
4560
x
02b Ch02b 57-106.indd 58 6/12/09 8:21:51 AM
Rectangular Components: Since cos2 a2 + cos2 b2 + cos2 g2 = 1, then cos g2z = ± 1 – cos2 45° – cos2 60° = ±0.5.
However, it is required that a2 > 90°, thus, g2 = cos–1 (–0.5) = 120°. 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 expressed in Cartesian vector form, as
Resultant Force: By adding F1 and F2 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)x2 + (FR)y
2 + (FR)z2
= 265.162 + 575.572 + 18.202 = 633.97N = 634N Ans
F1 = 450 cos 45° sin 30° (–i) + 450 cos 45° cos 30° (+j) + 450 sin 45° (+k) = {–159.10i + 275.57j + 318.20k} N AnsF2 = 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
265.16633.97
b = cos–1 = cos–1 = 24.8° Ans(FR)y
FR
575.57633.97
g = cos–1 = cos–1 = 88.4° Ans(FR)z
FR
18.20633.97
•2–42.
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4559
•2–43. Express each force acting on the pipe assembly inCartesian vector form.
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z
y
x
5
34
F2 400 lb
F1 600 lb 120
60
3 kN
2 kN
3 kN
3 kN
F1 = 34
5
(+i) + 0j + 33
5
(+k)
= [2.4i + 1.8k] kN AnsF2 = 2 cos 60°i + 2 cos 45°j + 2 cos 120°k
= [i + 1.414j – k] kN Ans
02b Ch02b 57-106.indd 59 6/12/09 8:21:52 AM
Rectangular Components: Since cos2 a2 + cos2 b2 + cos2 g2 = 1, then cos b2 = ± 1 – cos2 60° – cos2 120° = ±0.7071.
However, it is required that b2 > 90°, thus, b2 = cos–1 (–0.7071) = 45°. 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 expressed in Cartesian vector form, as
© 2011 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.
4660
2–44. Determine the magnitude and direction of theresultant force acting on the pipe assembly.
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z
y
x
5
34
F2 400 lb
F1 600 lb 120
60
2 kN
3 kN
F1 = 34
5
(+i) + 0j + 33
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 z2 2
= (3.4) + (1.4142) + (0.8)2 2 2 = 3.768 kN Ans
The coordinate direction angles of FR are
= cos–1 (F
FR x
R
)
= cos–1 3.4
3.768
= 25.5° Ans
= cos–1(F
FR y
R
)
= cos–1 1.4142
3.768
= 1.4142° Ans
= cos–1(F
FR z
R
)
= cos–1 0.8
3.768
= 77.7° Ans
02b Ch02b 57-106.indd 60 6/12/09 8:21:54 AM
Force Vectors: Since cos2 a2 + cos2 b2 + cos2 g2 = 1, then cos g2 = ± 1 – cos2 60° – cos2 120° = ±0.7071.
However, it is required that g2 < 90°, thus, g2 = cos–1 (–0.7071) = 45°. 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 expressed in Cartesian vector form, as
*2–44.
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4761
2–45. The force F acts on the bracket within the octantshown. If , , and , determine thex, y, z components of F.
g = 45°b = 60°F = 400 N
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F
y
z
x
a
b
g
02b Ch02b 57-106.indd 61 6/12/09 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
cos2 a + 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
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4862
*2–46. The force F acts on the bracket within the octantshown. If the magnitudes of the x and z components of Fare 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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F
y
z
x
a
b
g
02b Ch02b 57-106.indd 62 6/12/09 8:21:56 AM
Rectangular Components: The magnitude of F is given by
F = Fx2 + Fy
2 + Fz2
F = 3002 + Fy2 + 6002
F2 = Fy2 + 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
F600
774.60
2–46.
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4963
•2–65. The two forces F1 and F2 acting at A have aresultant force of . Determine themagnitude and coordinate direction angles of F2.
FR = 5-100k6 lb
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y
x
F2
A
30
50
F1 60 lb
z
B
60 N
N
02b Ch02b 57-106.indd 63 6/12/09 8:21:57 AM
Cartesian Vector Notation:
FR = {–100k} N
F1 = 60 {–cos 50° cos 30°i + cos 50° sin 30°j – sin 50°k} N
= {–33.40i + 19.28j – 45.96k} N
F2 = {F2xi + F2y
j + F2zk} N
Equating i, j and k components, we have
The coordinate direction angles for F2 are
F2x – 33.40 = 0 F2x
= 33.40 N
F2y + 19.28 = 0 F2y
= –19.28 N
F2z – 45.96 = –100 F2z
= –54.04 N
Resultant Force:
FR = F1 + F2
–100k = {(F2x – 33.40)i + (F2y
+ 19.28)j + (F2z – 45.96)k}
The magnitude of force F2 is
F2 = F 22x
+ F 22y
+ F 22z
= 33.402 + (–19.28)2 + (–54.04)2
= 66.39 lb = 66.4 N Ans
cos a = = a = 59.8° AnsF2x
F2
33.4066.39
cos b = = b = 107° AnsF2y
F2
–19.2866.39
cos g = = g = 144° AnsF2z
F2
–54.0466.39
•2–47.
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5064
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–49. The spur gear is subjected to the two forces causedby contact with other gears. Express each force as aCartesian vector.
135
F1 50 lb
F2 180 lb
24
7
25
60
60
z
y
x
*2–50. The spur gear is subjected to the two forces causedby contact with other gears. Determine the resultant of thetwo forces and express the result as a Cartesian vector.
135
F1 50 lb
F2 180 lb
24
7
25
60
60
z
y
x
2–48. Determine the coordinate direction angles of theforce 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 64 6/12/09 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 obtainedabove, we have
cos a = –0.5567 a = 124° Anscos b = 0.3214 b = 71.3° Anscos g = –0.7660 g = 140° Ans
*2–48.
2–50.
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51
F
F1 750 N
y
z
x
a
b
g
30 45
65
•2–51. 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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02b Ch02b 57-106.indd 65 6/12/09 8:22:03 AM
Force Vectors: By resolving F1 and F2 into their x, y, and z components, as shown in Fig. a. F1 and F2 can be expressed in Cartesian vector form as
F1 = 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 F1 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 aF cos a = –759.28 (1)
650 = 265.17 + F cos bF 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
2–51.
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5266
2–52. 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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F
F1 750 N
y
z
x
a
b
g
30 45
02b Ch02b 57-106.indd 66 6/12/09 8:22:07 AM
Force Vectors: By resolving F1 and F into their x, y, and z components, as shown in Figs. b and c, respectively, F1 and F can be expressed in Cartesian vector form as
F1 = 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 F1 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 + F2 cos aF cos a = –459.28 (1)
800 = 265.17 + F cos bF 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
*2–52.
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5367
2–71. If , , , and ,determine the magnitude and coordinate direction anglesof the resultant force acting on the hook.
F = 400 lbg = 60°b 6 90°a = 120°
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F1 600 lb
F
z
xy
4
35
a
b
g
30
600 N
N
02b Ch02b 57-106.indd 67 6/12/09 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 F1 and F2 into their x, y, and z components, as shown in Figs. a and b, respectively, F1 and F2, can be expressed in Cartesian vector form as
Resultant Force: By adding F1 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)x2 + (FR)y
2 + (FR)z2
= (40)2 + (698.53)2 + (–160)2 = 717.74 N = 718 N Ans
= {240i + 415.69j – 360k} NF = 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
40717.74
b = cos–1 = cos–1 = 13.3° Ans(FR)y
FR
698.53717.74
g = cos–1 = cos–1 = 103° Ans(FR)z
FR
–160717.74
F1 = 600 sin 30° (+i) + 600 45
cos 30° (+j) + 600 45
(–k) 35
N
N
2–53.
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5468
*2–54. 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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F1 600 lb
F
z
xy
4
35
a
b
g
30
600 N
N
02b Ch02b 57-106.indd 68 6/12/09 8:22:11 AM
Force Vectors: By resolving F1 and F into their x, y, and z components, as shown in Figs. a and b, respectively, F1 and F2 can be expressed in Cartesian vector form as
Resultant Force: By adding F1 and F2 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} NF = F cos ai + F cos bj + F cos gk
F1 = 600 sin 30° (+i) + 600 45
cos 30° (+j) + 600 45
(–k) 35
Equating the i, j, and k components, we have
–200 = 240 + F cos ux
F cos a = –440 (1)
800 = 415.69 + F cos bF 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
2–54.
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5569
•2–73. The shaft S exerts three force components on thedie D. Find the magnitude and coordinate direction anglesof the resultant force. Force F2 acts within the octant shown.
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. 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 NF2 300 N
g2 60
a2 60
2–56. The mast is subjected to the three forces shown.Determine the coordinate direction angles ofF1 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 69 6/12/09 8:22:12 AM
F1 = 400i
Since cos2 60° + cos2 b2 + cos2 60° = 1
Solving for the positive root, b2 = 45°
F2 = 300 cos 60°i + 300 cos 45°j + 300 cos 60°k = 150i + 212.1j + 150k
F1 = 500 cos a1i + 500 cos b1j + 500 cos g1k
FR = F1 + (–300j) + (–200k)
350i = 500 cos a1i + (500 cos b1 – 300)j + (500 cos g1 – 200)k
350 = 500 cos a1; a1 = 45.6° Ans
0 = 500 cos b1 – 300; b1 = 53.1° Ans
0 = 500 cos g1 – 200; g1 = 66.4° Ans
Then
FR = F1 + F2 + F3 = 550i + 52.1j + 270k
= –160j + 120k
45
35
F3 = –200 j + 200
k
FR = (550)2 + (52.1)2 + (270)2 = 614.9 N = 615 N Ans
550614.9
a = cos–1 = 26.6°
Ans
52.1614.9
b = cos–1 = 85.1°
Ans
270614.9
g = cos–1 = 64.0°
Ans
•2–55.
*2–56.
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5670
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–57. The mast is subjected to the three forces shown.Determine the coordinate direction angles ofF1 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
*2–58. Determine the magnitude and coordinatedirection angles of F2 so that the resultant of the two forcesacts 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 70 6/12/09 8:22:13 AM
F1 = {500 cos a1i + 500 cos b1j + 500 cos g1k} N
F2 = {–200k} N
F3 = {–300j} N
FR = F1 + F2 + F3 = 0
500 cos a1 = 0; a1 = 90° Ans
500 cos b1 = 300; b1 = 53.1° Ans
500 cos g1 = 200; g1 = 66.4° Ans
F1 = (180 cos 15°) sin 60°i + (180 cos 15°) cos 60°j – 180 sin 15°k
= 150.57i + 86.93j – 46.59k
F2 = F2 cos a2i + F2 cos b2 j + F2 cos g2k
FR = {500i} N
FR = F1 + F2
i components:
500 = 150.57 + F2 cos a2
F2x = F2 cos a2 = 349.43
j components:
0 = 86.93 + F2 cos b2
F2y = F2 cos b2 = –86.93
Thus,
F2 = F 22x + F 2
2y + F 22z = (349.43)2 + (–86.93)2 + (46.59)2
F2 = 363 N Ans
a2 = 15.8° Ans
b2 = 104° Ans
g2 = 82.6° Ans
k components:
0 = –46.59 + F2 cos g2
F2z = F2 cos g2 = 46.59
2–58.
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5771
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•2–59. Determine the magnitude and coordinate directionangles 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 71 6/12/09 8:22:14 AM
F1 = (180 cos 15°) sin 60°i + (180 cos 15°) cos 60°j – 180 sin 15°k
= 150.57i + 86.93j – 46.59k
F2 = F2 cos a2i + F2 cos b2 j + F2 cos g2k
FR = 0
i components: k components:
F2 cos a2 = –150.57
F2 cos b2 = –86.93
F2 cos g2 = 46.59
0 = 150.57 + F2 cos a2
0 = 86.93 + F2 cos b2
0 = –46.59 + F2 cos g2
j components:F2 = (–150.57)2 + (–86.93)2 + (46.59)2
Solving,
F2 = 180 N Ans
a2 = 147° Ans
b2 = 119° Ans
g2 = 75.0° Ans
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5872
2–60. If the resultant force acting on the bracket is directedalong the positive y axis, determine the magnitude of theresultant force and the coordinate direction angles of F sothat .b 6 90°
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xy
z
F 500 N
F1 600 N
a
b
g
30
30
02b Ch02b 57-106.indd 72 6/12/09 8:22:15 AM
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
F1 = 600 cos 30° sin 30° (+i) + 600 cos 30° cos 30° (+j) + 600 sin 30° (–k) = {259.81i + 450j – 300k} NF = 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 + FFR 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 aa = 121.31° = 121° AnsFR = 450 + 500 cos b (1)
0 = 500 cos g – 300g = 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
*2–60.
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5973
2–79. Specify the magnitude of F3 and its coordinatedirection angles so that the resultant force
.FR = 59j6 kNa3, b3, g3
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x
z
5
12
13
y
F3
30
F2 10 kN
F1 12 kN
g3b3
a3
02b Ch02b 57-106.indd 73 6/12/09 8:22:16 AM
F1 = 12 cos 30°j – 12 sin 30°k = 10.392j – 6k
9j = 10.392j – 6k – 9.231i + 3.846k + F3
F3 = 9.231i – 1.392j + 2.154k
F3 = (9.231)2 + (–1.392)2 + (2.154)2
F3 = 9.581 kN = 9.58 kN Ans
FR = F1 + F2 + F3
Require
Hence,
F2 = – (10)i + (10)k = –9.231i + 3.846k1213
513
a3 = cos–1 = 15.5° Ans9.2319.581
b3 = cos–1 = 98.4° Ans–1.3929.581
g3 = cos–1 = 77.0° Ans2.1549.581
2–61.
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60
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 resultantforce FR has a magnitude of 50 lb and coordinate directionangles and , as shown, determine themagnitude of F2 and its coordinate direction angles.
b = 80°a = 110°
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2–62. Determine the position vector r directed from pointA 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 77 6/12/09 8:22:22 AM
(1)2 = cos2 110° + cos2 80° + cos2 g
g = 157.44°
FR = F1 + F2
50 cos 110° = (F2)x
50 cos 80° = (F2)y
50 cos 157.44° = (F2)z – 20
(F2)x = –17.10
(F2)y = 8.68
(F2)z = –26.17
F2 = (–17.10)2 + (8.68)2 + (–26.17)2 = 32.4 N Ans
a2 = cos–1 = 122° Ans–17.1032.4
b2 = cos–1 = 74.5° Ans8.6832.4
g2 = 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
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6178
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2–63. If the cord AB is 7.5 m long, determine thecoordinate position +z of point B
3 m
2 m
6 m
y
B
x
A
*2–64. Determine the distance between the end points Aand B on the wire by first formulating a position vectorfrom A to B and then determining its magnitude.
xB
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 78 6/12/09 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
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6279
•2–65. Determine the magnitude and coordinatedirection angles of the resultant force acting at A.
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0.6 m
1.2 m
0.9 m
0.9 m
1.2 m
0.75 m
B
A
xC
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
3i –
2
3j –
1
3k
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
3i +
2
3j –
2
3k
Force Vectors: Multiplying the magnitude of the force with its unit vector, we have
FB = FBuB = 32
3–
2
3–
1
3i j k
= {2i – 2j – 1k} kN Ans
FC = FCuC = 3.751
3+
2
3–
2
3i 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
FR x
R
)
= cos–1 3.25
4.80
= 47.4° Ans
= cos–1(F
FR y
R
)
= cos–1 0.5
4.80
= 84.0° Ans
= cos–1(F
FR 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 79 6/12/09 8:22:27 AM
•2–65. Determine the magnitude and coordinate
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6380
2–66. Determine the magnitude and coordinate directionangles of the resultant force.
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x
z
y
C
B
A
600 N 500 N
8 m
4 m
4 m
2 m
02b Ch02b 57-106.indd 80 6/12/09 8:22:29 AM
rAB = (–2j – 4k) m; rA-B = 4.472 m
uAB = = –0.447j – 0.894krABrAB
uAC = = 0.485i + 0.728j – 0.485krACrAC
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° Ans242.54821.64
b = cos–1 = 83.3° Ans95.47821.64
g = cos–1 = 162° Ans–779.20821.64
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6481
2–67. Determine the magnitude and coordinate directionangles of the resultant force acting at A.
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y
x
B
C
A
6 m
3 m
454.5 m
6 m
FB 900 N
FC 600 N
z
02b Ch02b 57-106.indd 81 6/12/09 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
3i –
2
3j –
2
3k
(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 k1
FR = (FR)x2 + (FR)y
2 + (FR)z2
= (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
181.841377.95
b = cos–1 = cos–1 = 125° Ans(FR)y
FR
–781.841377.95
g = cos–1 = cos–1 = 144° Ans(FR)z
FR
–11201377.95
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6582
*2–68. Determine the magnitude and coordinate directionangles of the resultant force.
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A
C
B
1.2 m2.1 m
0.9 m
xy
F2 405 NF1 500 N
40
1.2 m
•2–69. The chandelier is supported by three chains whichare concurrent at point O. If the force in each chain has amagnitude of 300 N, express each force as a Cartesian vectorand determine the magnitude and coordinate directionangles of the resultant force.
120y
120 1.2 m
A
B
C
1.8 m
O
FA
FBFC
x
120
F1 = –5003
5
sin 40°i + 5003
5
cos 40°j – 5004
5
k
= {–192.84i + 229.81j – 400k} N
F2 = 405 N4
9–
7
9–
4
9i 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 82 6/12/09 8:22:32 AM
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6683
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2–70. The chandelier is supported by three chains whichare concurrent at point O. If the resultant force at O has amagnitude of 650 N and is directed along the negative z axis,determine the force in each chain.
120y
120 1.2 m
A
B
C
1.8 m
O
FA
FBFC
x
120
2–71. Express force F as a Cartesian vector; thendetermine 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
ThenrAB = {[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 83 6/12/09 8:22:34 AM
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6792
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•2–72. If the force in each cable tied to the bin is 350 N,determine the magnitude and coordinate direction anglesof the resultant force.
B
C
E
D
A
xy
1.8 m
0.9 m
0.9 m
0.6 m0.6 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
7i –
2
7j –
6
7k
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
7i +
2
7j –
6
7k
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
7i +
2
7j –
6
7k
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
7i –
2
7j –
6
7k
Thus, the force vectors FA, FB, FC and FD are given by
FA = FAuA = 3503
7–
2
7–
6
7i j k
= [150i – 100j – 300k] N
FB = FBuB = 3503
7+
2
7–
6
7i j k
= [150i + 100j – 300k] N
FC = FCuC = 350 –3
7+
2
7–
6
7i j k
= [–150i + 100j – 300k] N
FD = FDuD = 350 –3
7–
2
7–
6
7i 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 z2 2
= 0 + 0 + (–1200)2 = 1200 N Ans
The coordinate direction angles of FR are
= cos–1 (F
FR x
R
)
= cos–1 0
1200
= 90° Ans
= cos–1(F
FR y
R
)
= cos–1 0
1200
= 90° Ans
= cos–1(F
FR z
R
)
= cos–1 –1200
1200
= 180° Ans
02b Ch02b 57-106.indd 92 6/12/09 8:22:52 AM
*2–72.
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6893
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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 cablesis the same.
F 1.8 k kN
B
C
E
D
A
xy
1.8 m
0.9 m
0.9 m
0.6 m0.6 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
7i –
2
7j –
6
7k
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
7i +
2
7j –
6
7k
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
7i +
2
7j –
6
7k
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
7i –
2
7j –
6
7k
Since the magnitudes of FA, FB, FC and FD are the same and denoted as F, they can be written as
FA = FAuA = F3
7–
2
7–
6
7i j k
FB = FBuB = F3
7+
2
7–
6
7i j k
FC = FCuC = F –3
7+
2
7–
6
7i j k
FD = FDuD = F –3
7–
2
7–
6
7i 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} = F3
7–
2–
6
7i j k
7
+ F3
7+
2–
6
7i j k
7
+ F –3
7+
2–
6
7i j k
7
+ F –3
7–
2–
6
7i j k
7
–1.8k = –24
7k
Thus,
1.8 = 24
7F F = 0.525 kN Ans
02b Ch02b 57-106.indd 93 6/12/09 8:22:54 AM
2–73.
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6985
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•2–74. The door is held opened by means of two chains. Ifthe tension in AB and CD is and ,respectively, express each of these forces in Cartesian vector form.
FC = 250 NFA = 300 N
xy
z
2.5 m1.5 m
0.5 m1 m
30
A
C
B
D
FA 300 N
FC 250 N
2–75. The guy wires are used to support the telephonepole. Represent the force in each wire in Cartesian vectorform. Neglect the diameter of the pole.
y
B
CD
A
x
z
4 m4 m
1.5 m
1 m3 m
2 m
FA 250 NFB 175 N
02b Ch02b 57-106.indd 85 6/12/09 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 = 2.2992 + (–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=rABrAB
2.299j – 0.750k2.418
uAC = = –0.1741i + 0.6963j – 0.6963k=rACrAC
–1i + 4j – 4k5.745
rCD = {[–0.5 – (–2.5)]i + [0 – (–2.299)]j + (0 – 0.750)k} m
= {2.00i + 2.299j – 0.750k} m
rCD = 2.002 + 2.2992 + (–0.750)2 = 3.138 m
uCD = = 0.6373i + 0.7326j – 0.2390k=rCDrCD
2.00i + 2.299j – 0.750k3.138
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 = 22 + (–3)2 + (–5.5)2 = 6.576 m
uBD = = 0.3041i – 0.4562j – 0.8363k=rBDrBD
2i – 3j – 5.5k6.576
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7086
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2–76. Two cables are used to secure the overhang boom inposition and support the 1500-N load. If the resultant forceis directed along the boom from point A towards O,determine the magnitudes of the resultant force and forcesFB and FC. Set and .z = 2 mx = 3 m
z
A
x y6 m
1500 N
3 m
FB
FC
B
C
2 mx
z
02b Ch02b 57-106.indd 86 6/12/09 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 = rBrB
= – i – 27
j + 67
k 37
Thus, the force vectors FB and FC are given by
FB = FBuB = – FBi – 27
FB j + 67
FBk 37
0 = – (1)FB + 27
FC
37
0 = (3)FB + 37
FC – 150027
–FR = – (2)FB – 67
FC
67
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 – 37
FC j + 67
FCk 27
= (3 – 0)2 + (0 – 6)2 + (2 – 0)2
(3 – 0)i + (0 – 6)j + (2 – 0)kuC = rCrC
= i – 37
j + 67
k 27
–FR j = – + + (–1500k)FBi – 27
FB j + 67
FBk 37
FCi – 37
FC j + 67
FCk 27
–FR j = – i + j + kFB + 27
FC 37
– FB – 67
FC
67
FB + 37
FC – 150027
*2–76.
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7186
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2–76. Two cables are used to secure the overhang boom inposition and support the 1500-N load. If the resultant forceis directed along the boom from point A towards O,determine the magnitudes of the resultant force and forcesFB and FC. Set and .z = 2 mx = 3 m
z
A
x y6 m
1500 N
3 m
FB
FC
B
C
2 mx
z
02b Ch02b 57-106.indd 86 6/12/09 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 = rBrB
= – i – 27
j + 67
k 37
Thus, the force vectors FB and FC are given by
FB = FBuB = – FBi – 27
FB j + 67
FBk 37
0 = – (1)FB + 27
FC
37
0 = (3)FB + 37
FC – 150027
–FR = – (2)FB – 67
FC
67
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 – 37
FC j + 67
FCk 27
= (3 – 0)2 + (0 – 6)2 + (2 – 0)2
(3 – 0)i + (0 – 6)j + (2 – 0)kuC = rCrC
= i – 37
j + 67
k 27
–FR j = – + + (–1500k)FBi – 27
FB j + 67
FBk 37
FCi – 37
FC j + 67
FCk 27
–FR j = – i + j + kFB + 27
FC 37
– FB – 67
FC
67
FB + 37
FC – 150027
87
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*2–77. Two cables are used to secure the overhang boomin position and support the 1500-N load. If the resultantforce is directed along the boom from point A towards O,determine the values of x and z for the coordinates of pointC and the magnitude of the resultant force. Set
and .FC = 2400 NFB = 1610 N
z
A
x y6 m
1500 N
3 m
FB
FC
B
C
2 mx
z
02b Ch02b 57-106.indd 87 6/12/09 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 = rBrB
= – i – 27
j + 67
k 37
= (x – 0)2 + (0 – 6)2 + (z – 0)2
(x – 0)i + (0 – 6)j + (z – 0)k
x2 + z2 + 36
xuC = rCrC
= i – x2 + z2 + 36
z k x2 + z2 + 36
6 j +
FB = FBuB = 1610 – = [–460i – 1380j + 690k] Ni – 27
j + 67
k 37
x2 + z2 + 36
xFC = 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 +1380 j + 690 +
x2 + z2 + 36
2400z –1500 k
0 = x2 + z2 + 36
2400x –460x2 + z2 + 36
2400x = 460 (1)
0 = 690 + x2 + z2 + 36
2400z –1500x2 + 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), yieldsx = 0.5679z (4)
Substituting Eq. (4) into Eq. (1), and solvingz = 2.197 m = 2.20 m Ans
FR = FB + FC + W
–FR = – FR =x2 + z2 + 36
14 400 +1380x2 + z2 + 36
14 400 +1380 (2)
Substituting z = 2.197 m into Eq. (4), yieldsx = 1.248 m = 1.25 m Ans
Substituting x = 1.248 m and z = 2.197 m into Eq. (2), yieldsFR = 3591.85 N = 3.59 kN Ans
2–77.
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72
96
2–110. The cable attached to the shear-leg derrick exerts aforce on the derrick of . Express this force as aCartesian vector.
1.75 kN
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30
15 m
10.5 m
x
y
A
B
F 1.75 kN
2–78. 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
ThenrAB = {[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 96 6/12/09 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 component of (B + D) is equal to the sum of the components of B and D, then
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7397
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*2–79. Determine the projected component of the forceacting along cable AC. Express the result as a
Cartesian vector.FAB = 560 N
z
xy
CB
A
3 m
1.5 m1 m
3 m FAB 560 N
1.5 m
02b Ch02b 57-106.indd 97 6/12/09 8:22:59 AM
= (–1.5 – 0)2 + (0 – 3)2 + (1 – 0)2
(–1.5 – 0)i + (0 – 3)j + (1 – 0)kuAB = rABrAB
= – i – 37
j + 67
k 27
= (1.5 – 0)2 + (0 – 3)2 + (1 – 0)2
(1.5 – 0)i + (0 – 3)j + (3 – 0)kuAC = rACrAC
= i – 13
j + 23
k 23
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 – 37
j + 67
k 27
(FAB)AC = FAB · uAC = (–240i – 480j + 160k) ·
i – 13
j + 23
k 23
= (–240)
= 346.67 N
+ (–480) –
13
+ 160
23
23
(FAB)AC = (FAB)ACuAC = 346.67
i – 13
j + 23
k 23
= [116i – 231j + 231k] N Ans
2–79.
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7498
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•2–80. Determine the magnitudes of the components offorce acting along and perpendicular to line AO.F = 56 N
y
x
z
CO
D
A
B3 m
1.5 m
1 m
1 m F 56 N
02b Ch02b 57-106.indd 98 6/12/09 8:23:01 AM
= [0–(–1.5)]2 + (0 – 3)2 + (0 – 1)2
[0–(–1.5)]i + (0 – 3)j + (0 – 1)kuAO = rAOrAO
= i – 37
j – 67
k 27
= [0–(–1.5)]2 + (0 – 3)2 + (2 – 1)2
[0–(–1.5)]i + (0 – 3)j + (2 – 1)kuAD = rADrAD
= i – 37
j + 67
k 27
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 – 37
j + 67
k 27
(FAO)paral = F · uAO = (24i – 48j + 16k) ·
The component of F perpendicular to line AO is
i – 37
j – 67
k 27
= (24)
= 46.86 N = 46.9 N Ans
+ (–48) –
37
+ (16) –
27
67
(FAO)per = F 2 – (F AO)paral
= 562 – 46.862
= 30.7 N Ans
99
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2–114. Determine the length of side BC of the triangularplate. Solve the problem by finding the magnitude of rBC;then check the result by first finding q , rAB, and rAC andthen 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 99 6/12/09 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–1rAC · rAB
rAC rAB
5(5.0990)(3.6056)
*2–80.
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7598
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•2–80. Determine the magnitudes of the components offorce acting along and perpendicular to line AO.F = 56 N
y
x
z
CO
D
A
B3 m
1.5 m
1 m
1 m F 56 N
02b Ch02b 57-106.indd 98 6/12/09 8:23:01 AM
= [0–(–1.5)]2 + (0 – 3)2 + (0 – 1)2
[0–(–1.5)]i + (0 – 3)j + (0 – 1)kuAO = rAOrAO
= i – 37
j – 67
k 27
= [0–(–1.5)]2 + (0 – 3)2 + (2 – 1)2
[0–(–1.5)]i + (0 – 3)j + (2 – 1)kuAD = rADrAD
= i – 37
j + 67
k 27
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 – 37
j + 67
k 27
(FAO)paral = F · uAO = (24i – 48j + 16k) ·
The component of F perpendicular to line AO is
i – 37
j – 67
k 27
= (24)
= 46.86 N = 46.9 N Ans
+ (–48) –
37
+ (16) –
27
67
(FAO)per = F 2 – (F AO)paral
= 562 – 46.862
= 30.7 N Ans
99
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2–114. Determine the length of side BC of the triangularplate. Solve the problem by finding the magnitude of rBC;then check the result by first finding q , rAB, and rAC andthen 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 99 6/12/09 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–1rAC · rAB
rAC rAB
5(5.0990)(3.6056)
2–81.
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76100
2–82. Determine the magnitudes of the components ofacting along and perpendicular to segment DE
of the pipe assembly.F = 600 N
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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 100 6/12/09 8:23:03 AM
= (0 – 4)2 + (2 – 5)2 + [0 – (–2)]2
(0 – 4)i + (2 – 5)j + [0 – (–2)kuEB = rEBrEB
= –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)paral2 = 6002 – 334.252 = 498 N Ans
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77100
2–82. Determine the magnitudes of the components ofacting along and perpendicular to segment DE
of the pipe assembly.F = 600 N
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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 100 6/12/09 8:23:03 AM
= (0 – 4)2 + (2 – 5)2 + [0 – (–2)]2
(0 – 4)i + (2 – 5)j + [0 – (–2)kuEB = rEBrEB
= –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)paral2 = 6002 – 334.252 = 498 N Ans
101
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*2–83. Two forces act on the hook. Determine the anglebetween them.Also, what are the projections of F1 and F2
along the y axis?u
x
z
y
45
60120
F1 600 N
F2 {120i + 90j – 80k}N
u
•2–84. Two forces act on the hook. Determine themagnitude of the projection of F2 along F1.
x
z
y
45
60120
F1 600 N
F2 {120i + 90j – 80k}N
u
02b Ch02b 57-106.indd 101 6/12/09 8:23:04 AM
F2 = 120i + 90j – 80k; F2 = 170 N
u = j
So,
F1 · F2 = (–300)(120) + (300)(90) + (424.3)(–80) = –42 944
F1y = F1 · j = (300)(1) = 300 N Ans
F2y = F2 · j = (90)(1) = 90 N Ans
u1 = cos 120°i + cos 60°j + cos 45°k
Proj F2 = F2 · u1 = (120)(cos 120°) + (90)(cos 60°) + (–80)(cos 45°)
|Proj F2| = 71.6 N Ans
F1 = 600 cos 120°i + 600 cos 60°j + 600 cos 45°k
= –300i + 300j + 424.3k; F1 = 600 N
u = cos–1 = 115° Ans–42 944(170)(600)
2–83.
*2–84.
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78102
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2–85. Determine the projection of force alongline BC. Express the result as a Cartesian vector.
F = 80 N
F 80 N
A
E
B
y
FC
x
D
z
2 m
2 m
1.5 m1.5 m
2 m
2 m
02b Ch02b 57-106.indd 102 6/12/09 8:23:05 AM
= (4 – 2)2 + (0 – 2)2 + (0 – 0)2
(4 – 2)i + (0 – 2)j + (0 – 0)kuFC = rFCrFC
= 0.7071i – 0.7071j
= (2 – 2)2 + (0 – 2)2 + (1.5 – 0)2
(2 – 2)i + (0 – 2)j + (1.5 – 0)kuFD = rFDrFD
= – j + 45
k 35
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 + 45
k 35
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
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79102
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2–85. Determine the projection of force alongline BC. Express the result as a Cartesian vector.
F = 80 N
F 80 N
A
E
B
y
FC
x
D
z
2 m
2 m
1.5 m1.5 m
2 m
2 m
02b Ch02b 57-106.indd 102 6/12/09 8:23:05 AM
= (4 – 2)2 + (0 – 2)2 + (0 – 0)2
(4 – 2)i + (0 – 2)j + (0 – 0)kuFC = rFCrFC
= 0.7071i – 0.7071j
= (2 – 2)2 + (0 – 2)2 + (1.5 – 0)2
(2 – 2)i + (0 – 2)j + (1.5 – 0)kuFD = rFDrFD
= – j + 45
k 35
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 + 45
k 35
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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2–86. The clamp is used on a jig. If the vertical forceacting on the bolt is , determine themagnitudes of its components F1 and F2 which act along theOA axis and perpendicular to it.
F = {-500k} Nz
O
x
y
40 mm
40 mm
20 mm
F { 500 k} N
A
02b Ch02b 57-106.indd 103 6/12/09 8:23:06 AM
(0 – 20)2 + (0 – 40)2 + (0 – 40)2
(0 – 20)i + (0 – 40)j + (0 – 40)kuAO = = – i – 13
j – 23
k 23
Unit Vector: The unit vector along AO axis is
Projected Component of F Along OA Axis:
F1 = F · uAO = (–500k) ·
i – – 13
j – 23
k 23
= (0) –
= 333.33 N = 333 N Ans
+ (0) –
13
+ (–500) –
23
23
Component of F Perpendicular to OA Axis: Since the magnitude of force F is F = 500 N so that
F 2 = F 2 – F 12) = 5002 – 333.332 = 373 N Ans
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80104
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*2–87. Determine the magnitude of the projectedcomponent 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
7i –
2
7j –
6
7k
Thus, the force vector FAB is given by
FAB = FABuAB = 3.53
7–
2
7–
6
7i 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 104 6/12/09 8:23:07 AM
2–87.
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81104
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*2–87. Determine the magnitude of the projectedcomponent 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
7i –
2
7j –
6
7k
Thus, the force vector FAB is given by
FAB = FABuAB = 3.53
7–
2
7–
6
7i 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 104 6/12/09 8:23:07 AM
105
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•2–88. Determine the magnitude of the projectedcomponent 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 105 6/12/09 8:23:10 AM
*2–88.
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82106
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2–89. Determine the projection of force acting along line AC of the pipe assembly. Express the resultas a Cartesian vector.
F = 400 N
x
A
BC
y
z
4 m
3 m
F 400 N
30
45
02b Ch02b 57-106.indd 106 6/12/09 8:23:13 AM
= (0 – 0)2 + (4 – 0)2 + (3 – 0)2
(0 – 0)i + (4 – 0)j + (3 – 0)kuAC = rACrAC
= j + 45
k 35
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 + 45
k 35
Thus, FAC written in Cartesian vector form is
FAC = FACuAC = 365.66 = {293j + 219k} N Ansj + 45
k 35
= (–141.42)(0) + 244.95
= 365.66 N Ans
+ 282.8445
35
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83106
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2–89. Determine the projection of force acting along line AC of the pipe assembly. Express the resultas a Cartesian vector.
F = 400 N
x
A
BC
y
z
4 m
3 m
F 400 N
30
45
02b Ch02b 57-106.indd 106 6/12/09 8:23:13 AM
= (0 – 0)2 + (4 – 0)2 + (3 – 0)2
(0 – 0)i + (4 – 0)j + (3 – 0)kuAC = rACrAC
= j + 45
k 35
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 + 45
k 35
Thus, FAC written in Cartesian vector form is
FAC = FACuAC = 365.66 = {293j + 219k} N Ansj + 45
k 35
= (–141.42)(0) + 244.95
= 365.66 N Ans
+ 282.8445
35
107
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2–90. Determine the magnitudes of the components offorce acting parallel and perpendicular tosegment BC of the pipe assembly.
F = 400 N
x
A
BC
y
z
4 m
3 m
F 400 N
30
45
02c Ch02c 107-120.indd 107 6/12/09 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 thecomponent 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 bedetermined from
F2 – (FBC) paral = 4002 – 244.952 = 316 N Ans (FBC)per =
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84108
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*2–91. Cable OA is used to support column OB.Determine the angle it makes with beam OC.u
z
x
C
O
D
y4 m
30
8 m
8 m
A
u
f
•2–92. Cable OA is used to support column OB.Determine the angle it makes with beam OD.f
z
x
C
B
O
D
y4 m
30
8 m
8 m
A
u
f
02c Ch02c 107-120.indd 108 6/12/09 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
3i +
2
3j –
2
3k
The Angles Between Two Vectors u : B
uOC · uOA = (1i) ·1
3i +
2
3j –
2
3k
2
3
1
3
1
3
2
3
= 1 + (0) + 0 – =
Then,1
3u = 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
3i +
2
3j –
2
3k
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
3i +
2
3j –
2
3k
1
3
2
3
2
3
2–91.
*2–92.
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85108
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*2–91. Cable OA is used to support column OB.Determine the angle it makes with beam OC.u
z
x
C
O
D
y4 m
30
8 m
8 m
A
u
f
•2–92. Cable OA is used to support column OB.Determine the angle it makes with beam OD.f
z
x
C
B
O
D
y4 m
30
8 m
8 m
A
u
f
02c Ch02c 107-120.indd 108 6/12/09 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
3i +
2
3j –
2
3k
The Angles Between Two Vectors u : B
uOC · uOA = (1i) ·1
3i +
2
3j –
2
3k
2
3
1
3
1
3
2
3
= 1 + (0) + 0 – =
Then,1
3u = 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
3i +
2
3j –
2
3k
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
3i +
2
3j –
2
3k
1
3
2
3
2
3
109
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2–93. The cables each exert a force of 400 N on the post.Determine the magnitude of the projected component of F1along the line of action of F2.
x
z
y
20
35
4560
120
F1 400 N
F2 400 N
u
2–94. Determine the angle between the two cablesattached to the post.
u
x
z
y
20
35
4560
120
F1 400 N
F2 400 N
u
02c Ch02c 107-120.indd 109 6/12/09 8:25:49 AM
Force Vector :
Unit Vector : The unit vector along the line of action of F2 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 – 0.5k
uF1 = sin 35° cos 20°i – sin 35° sin 20°j + cos 35°k
= 0.5390i – 0.1962j + 0.8192k
F1 = F1 uF1 = 400(0.5390i – 0.1962j + 0.8192k) N
= {215.59i – 78.47j + 327.66k} N
Negative sign indicates that the force component (F1)F2 acts in the opposite sense
of direction to that of uF2.
thus the magnitude (F1)F2 = 50.6 N Ans
(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
Unit Vector :
The Angles Between Two Vector u : The dot product of two unitvectors 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
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86110
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*2–128. A force of is applied to the handle ofthe wrench. Determine the angle between the tail of theforce 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
CB
Fu
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–1r rAB AC
AB ACr r
·
= cos–1 26.73
5.1788 (6.2426)
= 34.2° Ans
02c Ch02c 107-120.indd 110 6/12/09 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
2–95.
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87114
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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-lbforce.
y
x
700 lb
30
60
700 N
700 N
700 N
N
02c Ch02c 107-120.indd 114 6/12/09 8:25:59 AM
*2–96.
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88115
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*2–97. Determine the magnitude of the projectedcomponent of the 100-lb force acting along the axis BC ofthe pipe.
yC
BA
D
z
8 ft
3 ft
6 ft
2 ft
4 ftx
F 100 lb
u
•2–98. Determine the angle between pipe segmentsBA and BC.
u
yC
BA
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–1r rBA BC
BA BCr r
·
= cos–1 –1.62
0.9 (2.245)
= 143° Ans
500-N
02c Ch02c 107-120.indd 115 6/12/09 8:26:00 AM
2–97.
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89116
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x
y
30 30
45
F1 80 NF2 75 N
F3 50 N
2–99. Determine the magnitude and direction of theresultant of the three forces by firstfinding 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 116 6/12/09 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 f80
sin 105°104.7
= ; f = 47.54°
sin b104.7
sin 162.46°177.7
= ; b = 10.23°
FR = 177.7 = 178 N Ans
u = 75° + 10.23° = 85.2° Ans
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90
2–139. Determine the design angle ( < 90°) betweenthe two struts so that the 500-lb horizontal force has acomponent of 600 lb directed from A toward C. What is thecomponent of force acting along member BA?
uu
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500 lb
20
A
B
C
u
N
N
N
N
500 N
NN
02c Ch02c 107-120.indd 117 6/12/09 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 u500
sin 20°214.91
=
The parallelogram law of addition and the triangular rule are shown in Figs. a and b.
*2–100.
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91118
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*2–101. Determine the magnitude and direction of thesmallest force F3 so that the resultant force of all threeforces has a magnitude of 20 lb.
F2 10 lb
F34
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 + 503
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 118 6/12/09 8:26:03 AM
2–101.
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92119
•2–102. Resolve the 250-N force into components actingalong the u and axes and determine the magnitudes ofthese components.
v
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u
v
40
20
250 N
2–103. Cable AB exerts a force of 80 N on the end of the3-m-long boom OA. Determine the magnitude of theprojection of this force along the boom.
O
A
80 N
3 m
B
z
y
x
4 m
60
02c Ch02c 107-120.indd 119 6/12/09 8:26:04 AM
Vector Analysis:
Scalar Analysis:
F = 80rAB
rAB
3 cos 60°
5
= 80 i – – j +
3 sin 60°
54 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°
43
Proj F = 80 cos 53.13° = 48.0 N Ans
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93120
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2–104. The three supporting cables exert the forces shownon 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 NFC 400 N
FE 350 N
02c Ch02c 107-120.indd 120 6/12/09 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 + 0.4867krAB
rAB
–5i + 2j + 3k6.164
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 + 0.4867krAC
rAC
–5i – 2j + 3k6.164
rDE = {(0 – 2)i + (0 – 0)j + (3 – 0)k} m ={–2i + 3k} m
rDE = (–2)2 + 32 = 3.605 m
uDE = = = –0.5547i + 0.8321krDE
rDE
–2i + 3k3.605
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
*2–104.