SM CH02 - solution manuals CH7 PDF

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u = 30° T = 6 kN of the resultant force acting on the eyebolt and its direction measured clockwise from the positive x axis.

y

T u

x 45

8 kN

The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of cosines to Fig. b, FR =

62 + 82 – 2(6)(8) cos 75°

= 8.669 kN = 8.67 kN

Ans

Applying the law of sines to Fig. b and using this result, yields sin a sin 75° = 8 8.669

a = 63.05°

Thus, the direction angle f of FR measured clockwise from the positive x axis is f = a – 60° = 63.05° – 60° = 3.05°

Ans

7

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2–2. If u = 60° and T = 5 kN, determine the magnitude of the resultant force acting on the eyebolt and its direction measured clockwise from the positive x axis.

y

T u

x 45

8 kN

The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of cosines to Fig. b, FR =

52 + 82 – 2(5)(8) cos 105°

= 10.47 kN = 10.5 kN

Ans

Applying the law of sines to Fig. b and using this result, yields sin a sin 105° = 8 10.47

a = 47.54°

Thus, the direction angle f of FR measured clockwise from the positive x axis is f = a – 30° = 47.54° – 30° = 17.5°

Ans

8

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2–3. If the magnitude of the resultant force is to be 9 kN directed along the positive x axis, determine the magnitude of force T acting on the eyebolt and its angle u .

y

T u

x 45

8 kN

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, T=

82 + 92 – 2(8)(9) cos 45°

= 6.571 kN = 6.57 kN

Ans

Applying the law of sines to Fig. b and using this result, yields sin (90° – u) sin 45° = 8 6.571 u = 30.6°

Ans

9

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*2–4. Determine the magnitude of the resultant force acting on the bracket and its direction measured counterclockwise from the positive u axis.

F2

v

150 N

30 u

30 45

F1

200 N

The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of cosines to Fig. b, FR =

2002 + 1502 – 2(200)(150) cos 75°

= 216.72 N = 217 N

Ans

Applying the law of sines to Fig. b and using this result, yields sin a sin 75° = 200 216.72

a = 63.05°

Thus, the direction angle f of FR measured clockwise from the positive x axis is f = a – 60° = 63.05° – 60° = 3.05°

Ans

N N

N

10

N

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v

F2

v

150 N

and determine the magnitudes of these components. 30 30 u 45 F1

200 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 Fu 200 = sin 105° sin 30°

F u = 386 N

Ans

Fv 200 = sin 45° sin 30°

F v = 283 N

Ans

N N

11

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

F2

v

150 N

30 30

u

45

F1

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, Fu 150 = sin 30° sin 30° Fv 150 = sin 120° sin 30°

F u = 150 N

Ans

F v = 260 N

Ans

12

200 N

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2–7. If FB = 2 kN and the resultant force acts along the positive u axis, determine the magnitude of the resultant force and the angle u.

y

FA u 30

B

Applying the law of sines to Fig. b, yields

sin f sin 30° = 3 2

f = 48.59°

Thus, u = 30° + f = 30° + 48.59° = 78.59° = 78.6°

Ans

With the result u = 78.590, applying the law of sines to Fig. b again, yields FR sin (180° – 78.59°)

=

2 sin 30°

F R = 3.92 kN

13

Ans

x

A

u FB

The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively.

3 kN

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*2–8. If the resultant force is required to act along the positive u axis and have a magnitude of 5 kN, determine the required magnitude of FB and its direction u.

y

FA

3 kN x

u 30

B

u FB

The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of cosines to Fig. b,

FB =

32 + 52 – 2(3)(5) cos 30°

= 2.832 kN = 2.83 kN

Ans

Using this result and realizing that sin (180° – u) = sin u, the application of the sine law to Fig. b, yields sin u sin 30° = 5 2.832

u = 62.0°

Ans

14

A

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FA

as shown. If u = 60° , determine the magnitude of the resultant of these two forces and its direction measured clockwise from the horizontal.

u

A

40 B FB

Parallelogram Law : The parallelogram law of addition is shown in Fig. (a). Trigonometry : Using law of cosines [Fig. (b)], we have FR =

82 + 62 – 2(8)(6) cos 100°

= 10.80 kN = 10.8 kN

Ans

The angleu can be determined using law of sines [Fig. (b)]. sin u sin 100° = 6 10.80 sin u = 0.5470 u = 33.16° Thus, the direction f of FR measured from the x axis is f = 33.16° – 30° = 3.16°

Ans

15

6 kN

8 kN

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2–10. Determine the angle of u for connecting member A to the plate so that the resultant force of FA and FB is directed horizontally to the right.Also, what is the magnitude of the resultant force?

FA u

8 kN

A

40 B 6 kN

FB

Parallelogram Law : The parallelogram law of addition is shown in Fig. (a). Trigonometry : Using law of sines [Fig. (b)], we have sin (90° – u) sin 50° = 6 8 sin (90° – u) = 0.5745 u = 54.93° = 54.9°

Ans

From the triangle, u = 180° – (90° – 54.93°) – 50° = 94.93°. Thus, using law of cosines, the magnitude of FR is FR =

82 + 62 – 2(8)(6) cos 94.93°

= 10.4 kN

Ans

2–11. If the tension in the cable is 400 N, determine the magnitude and direction of the resultant force acting on the pulley. This angle is the same angle u of line AB on the tailboard block.

y

400 N 30

u B A

FR =

(400)2 + (400)2 – 2(400)(400) cos 60° = 400 N

sin 60° sin u = ; u = 60° 400 400

Ans

Ans

16

x 400 N

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*2–12. The device is used for surgical replacement of the knee joint. If the force acting along the leg is 360 N, determine its components along the x and y ¿ axes.

y¿

y 10

x¿ x 60 360 N

360 –F x = ; F x = –125 N sin 20° sin 100°

Ans

F y = 360 ; F = 317 N y sin 60° sin 100°

Ans

171 7

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y¿

knee joint. If the force acting along the leg is 360 N, determine its components along the x¿ and y axes.

y 10

x¿ x 60 360 N

360 –F x = ; F x = –183 N sin 80° sin 30°

Ans

Fy 360 = ; F y = 344 N sin 80° sin 70°

Ans

18

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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°.

800 N A u B

f

C

Parallelogram Law : The parallelogram law of addition is shown in Fig. (a). Trigonometry : Using law of sines [Fig. (b)], we have sin 40° sin = 800 1000 sin = 0.8035 = 53.46° = 53.5° Thus,

800 N

Ans

= 180° – 40° – 53.46° = 86.54° 1000 N

Using law of sines [Fig. (b)] 800 F AB = sin 40° sin 86.54°

1000 N

FAC = 1242 N

Ans 800 N

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°.

800 N A u B

f

C

Parallelogram Law : The parallelogram law of addition is shown in Fig. (a). Trigonometry : Using law of cosines [Fig. (b)], we have FAC =

1200 N

800 2 + 1200 2 – 2(800)(1200) cos 30°

= 645.93 N The angle

800 N

can be determined using law of sines [Fig. (b)].

sin

sin 30° = 645.93 800 sin = 0.6193 = 38.3°

1200 N

Ans 800 N

19

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v

*2–16. Resolve F1 into components along the u and v axes and determine the magnitudes of these components.

F1

F2

150 N

30

250 N

u

30 105

Sine law : F 1v = 250 sin 30° sin 105°

F 1v = 129 N

Ans

F 1u = 250 sin 45° sin 105°

F 1u = 183 N

Ans

v

v and determine the magnitudes of these components.

F1

F2

150 N

30

30 105

Sine law : F 2v = 150 sin 30° sin 75°

F 2v = 77.6 N

Ans

F 2u = 150 sin 75° sin 75°

F 2u = 150 N

Ans

20

250 N

u

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2–18. The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop a resultant force of 950 N directed along the positive x axis. Set u = 50° .

y

FA A

B

u FB

Parallelogram Law : The parallelogram law of addition is shown in Fig. (a). Trigonometry : Using law of sines [Fig. (b)], we have FA sin 50°

=

950 sin 110°

FA = 774 N

Ans

950 FB = sin 110° sin 20° FB = 346 N

Ans

2–19. The truck is to be towed using two ropes. If the resultant force is to be 950 N, directed along the positive x axis, determine the magnitudes of forces FA and FB acting on each rope and the angle u of FB so that the magnitude of FB is a minimum. FA acts at 20° from the x axis as shown.

y

FA

A

B

u FB

Parallelogram Law : In order to produce a minimum force FB, FB has to act perpendicular to FA. The parallelogram law of addition is shown in Fig. (a). Trigonometry : Fig. (b). FB = 950 sin 20° = 325 N

Ans

FA = 950 cos 20° = 893 N

Ans

u = 90° – 20° = 70.0°

Ans

The angle u is

21

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*2–20. If f = 45°, F1 = 5 kN , and the resultant force is 6 kN directed along the positive y axis, determine the required magnitude of F2 and its direction u .

F1 y f u F2

60

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,

F2 =

62 + 52 – 2(6)(5) cos 45°

= 4.310 kN = 4.31 kN

Ans

Using this result and applying the law of sines to Fig. b, yields sin u sin 45° = 5 4.310

u = 55.1°

Ans

22

x

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f = 30° directed along the positive y axis, determine the magnitudes of F1 and F2 and the angle u if F2 is required to be a minimum.

F1 y f u F2

60

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, F1 = 6 cos 30° = 5.20 kN F2 = 6 sin 30° = 3 kN

Ans Ans

u = 90° – 30° = 60°

Ans

and

23

x

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2–22. If f = 30° , F1 = 5 kN , and the resultant force is to be directed along the positive y axis, determine the magnitude of the resultant force if F2 is to be a minimum. Also, what is F2 and the angle u?

F1 y f u F2

60

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

24

x

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2–23. If u = 30° and F2 = 6 kN , determine the magnitude of the resultant force acting on the plate and its direction measured clockwise from the positive x axis.

y F3

5 kN F2

Parallelogram law and Triangular Rule: This problem can be solved by adding the forces successively, using the parallelogram law of addition, shown in Fig. a. Two triangular force diagrams, shown in Figs. b and c, can be derived from the parallelogram. u

as follows.

F1

tan a =

5 4

FR =

62 + 6.4032 – 2(6)(6.403) cos (51.34° + 30°)

a = 51.34°

= 8.09 kN sin b = sin (51.34° + 30°) 6 8.089

Ans b = 47.16°

Thus, the direction angle f of FR , measured clockwise from the positive x axis, is f = a + b = 51.34° + 47.16° = 98.5°

Ans

25

4 kN x

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