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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°


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°


8 kN

T

x

y

u

45

02a Ch02a 7-56.indd 8 6/12/09 8:05:05 AM



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°




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


8 kN

T

x

y

u

45

02a Ch02a 7-56.indd 9 6/12/09 8:05:11 AM




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


1010

*2–4. Determine the magnitude of the resultant forceacting on the bracket and its direction measuredcounterclockwise from the positive u axis.


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



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°




1111

•2–5. Resolve F1 into components along the u and axes,and determine the magnitudes of these components.

v


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


Applying the law of sines to Fig. b, yields

Ans

Ans


1212

2–6. Resolve F2 into components along the u and axes,and determine the magnitudes of these components.

v


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


Applying the law of sines to Fig. b,


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


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°


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


y

x

uB

FA � 3 kN

FB

Au 30�

02a Ch02a 7-56.indd 13 6/12/09 8:05:21 AM


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


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



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


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°


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


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


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

= ;


1717


*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°= ;


1818


•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°= ;


1919


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



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


2020

*2–16. Resolve F1 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

•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°=


2121


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



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


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°


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.



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°


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


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°


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


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°


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°


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°


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


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


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


2833

*2–26. Determine the magnitude of the resultant forceacting on the pin and its direction measured clockwise fromthe positive x axis.


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.


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


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


+→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 ↓


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.


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°


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.


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.


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


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°


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


+→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 ↓


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.


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°


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


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)


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°


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


+→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.


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


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


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


Ax

y

F1

400 N600 N

34

5

30

u

02a Ch02a 7-56.indd 41 6/12/09 8:06:20 AM


+→ 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

→


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.


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


FB

x

y

B A

30FA 700 N

u

02a Ch02a 7-56.indd 42 6/12/09 8:06:21 AM


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.


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


FB

x

y

B A

30FA 700 N

u

02a Ch02a 7-56.indd 43 6/12/09 8:06:22 AM


+→ 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

→


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.


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°


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


+→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 ↓


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


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


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


+→ 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)


= 58.24° F1 = 2.470 kN Ans

02a Ch02a 7-56.indd 45 6/12/09 8:06:26 AM

2–38.


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.


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


+↑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


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


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



+→ 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.


4357

2–41. Determine the coordinate angle for F2 and thenexpress each force acting on the bracket as a Cartesianvector.

g


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


4458

*2–42. Determine the magnitude and coordinate directionangles of the resultant force acting on the bracket.


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.


4559

•2–43. Express each force acting on the pipe assembly inCartesian vector form.


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


4660

2–44. Determine the magnitude and direction of theresultant force acting on the pipe assembly.


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


FR = ( ) + ( ) + ( )2F F FR x R y R z2 2

= (3.4) + (1.4142) + (0.8)2 2 2 = 3.768 kN Ans


= 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.


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


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


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


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.


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


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.


5064


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.


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


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.


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


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)


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


a = 121° b = 52.7° g = 53.0° Ans

*2–52.


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°


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



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.


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


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)


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


a = 125° b = 60.3° g = 48.9° Ans

N

2–54.


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.


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.


5670


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.


5771


•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


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°


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


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.


5973

2–79. Specify the magnitude of F3 and its coordinatedirection angles so that the resultant force

.FR = 59j6 kNa3, b3, g3


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.


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°


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


6178


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


6279

•2–65. Determine the magnitude and coordinatedirection angles of the resultant force acting at A.


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


6380

2–66. Determine the magnitude and coordinate directionangles of the resultant force.


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


6481

2–67. Determine the magnitude and coordinate directionangles of the resultant force acting at A.


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:



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


6582

*2–68. Determine the magnitude and coordinate directionangles of the resultant force.


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


6683


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


6792


•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


FR = ( ) + ( ) + ( )2F F FR x R y R z2 2

= 0 + 0 + (–1200)2 = 1200 N Ans


= 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.


6893


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.


6985


•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


7086


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


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,


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.


7186


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


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,


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


*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:


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.


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


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


7397


*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.


7498


•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


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.


7598


•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,


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


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.


76100

2–82. Determine the magnitudes of the components ofacting along and perpendicular to segment DE

of the pipe assembly.F = 600 N


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,


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


77100

2–82. Determine the magnitudes of the components ofacting along and perpendicular to segment DE

of the pipe assembly.F = 600 N


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,


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


*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.


78102


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,


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


79102


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,


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


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


80104


*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


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.


81104


*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


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


•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.


82106


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


83106


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


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 =


84108


*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.


85108


*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


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


86110


*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.


87114


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.


88115


*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.


89116


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


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


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 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.


91118


*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.


92119

•2–102. Resolve the 250-N force into components actingalong the u and axes and determine the magnitudes ofthese components.

v


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


93120


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.

Documents

8 kN - Faculty Personal Homepage- KFUPMfaculty.kfupm.edu.sa/ce/saghamdi/hmpage/CE202/S_Manual/SM_CH02.pdfs = ) = + • *