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Solutions Manual

Engineering Mechanics: Statics 2nd Edition

Michael E. Plesha University of Wisconsin–Madison

Gary L. Gray The Pennsylvania State University

Francesco Costanzo The Pennsylvania State University With the assistance of: Chris Punshon Andrew J. Miller Justin High Chris O’Brien Chandan Kumar Joseph Wyne Jonathan Fleischmann

Version: May 11, 2012

The McGraw-Hill Companies, Inc.

Copyright © 2002–2012 Michael E. Plesha, Gary L. Gray, and Francesco Costanzo

This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

3

Statics 2e

Important Information about this Solutions Manual We encourage you to occasionally visit http://www.mhhe.com/pgc2e to obtain the most up-to-date version of this solutions manual.

Contact the Authors If you find any errors and/or have questions concerning a solution, please do not hesitate to contact the authors and editors via email at: [email protected], [email protected]

and

We welcome your input.

This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

May 11, 2012

4

Solutions Manual

Accuracy of Numbers in Calculations Throughout this solutions manual, we will generally assume that the data given for problems is accurate to 3 significant digits. When calculations are performed, all intermediate numerical results are reported to 4 significant digits. Final answers are usually reported with 3 or 4 significant digits. If you verify the calculations in this solutions manual using the rounded intermediate numerical results that are reported, you should obtain the final answers that are reported to 3 significant digits.

This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

May 11, 2012

37

Statics 2e

Chapter 2 Solutions Problem 2.1 For each vector, write two expressions using polar vector representations, one using a positive value of & and the other a negative value, where & is measured counterclockwise from the right-hand horizontal direction.

Solution Part (a) ı

rED 12 in: @ 90

or

ı

rE D 12 in: @ —270

:

(1)

:

(2)

:

(3)

Part (b) E

F D 23 N @ 135

ı

E

ı

or

F D 23 N @ —225

or

vE D 15 m=s @ —120

Part (c) ı

vE D 15 m=s @ 240

ı

This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

May 11, 2012

Solutions Manual

38

Problem 2.2 E Add the two vectors shown to form a resultant vector R, and report your result using polar vector representation.

Solution Part (a) The vector polygon shown at the right corresponds to the addition of the two position vectors to obtain a resultant position vector R. Note that ˛ is ı ı ı given by ˛ D 180 —55 D 125 . Knowing this angle, the law of cosines may be used to determine R q 2 2 ı R D .101 mm/ C .183 mm/ — 2.101 mm/.183 mm/ cos 125 D 254:7 mm: (1)

R

E

β 101 mm

Next, the law of sines may be used to determine the angle ˇ: . Σ —1 183 mm ı ı 183 mm R ˇsin sin 125 36:05 : )D sin ˛ D sin ˇ 254:7 mm E Using these results, we may report the vector R using polar vector representation as E

ı

R D 255 mm @ 36:0

183 mm

α 55

◦

(2)

(3)

:

Part (b) The vector polygon shown at the right corresponds to the addition of the two force vectors to obtain a resultant force vector R. The law of cosines may be used to determine R q 2 2 ı R D .1:23 kip/ C .1:55 kip/ — 2.1:23 kip/.1:55 kip/ cos 45 D 1:104 kip: (4) E

Using the law of sines, we find that R

1:23 kip ı D

sin 45

sin ˇ

—1

) ˇ D sin

.

1:23 kip

ı

Σ

ı

sin 45

D 51:97 :

(5)

1:104 kip

E

ı

The direction o f R measured from the right-hand horizontal direction i s —90

E these results, we may report R using polar vector representation as E ı R D 1:10 kip @ —142

— 51:97 ı D —142ı. Using

:

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May 11, 2012

39

Statics 2e

Problem 2.3 E Add the two vectors shown to form a resultant vector R, and report your result using polar vector representation.

Solution 1.8 m

Part (a) The vector polygon shown at the right corresponds to the addition of the two E position vectors to obtain a resultant position vector R. The law of cosines may be used to determine R as q 2 2 ı R D .1:8 m/ C .2:3 m/ — 2.1:8 m/.2:3 m/ cos 65 (1) D 2:243 m:

65° R

2.3 m

The law of sines may be used to determine the angle ˛ as R

—1 2:3 m sin 65ı D 68:34ı: ) ˛ D sin

D 2:3 m

ı sin 65 sin ˛ Using polar vector representation, the resultant is

(2)

2:243 m

E

ı

R D 2:243 m @ — 68:34

(3)

: ı

ı

ı

If desired, this resultant may be stated using a positive angle, where 360 — 68:34 D 291:7 , as

E

R D 2:243 m @ 291:7

ı

:

Part (b) The vector polygon shown at the right corresponds to the addition of the two force vectors to obtain a resultant force vector R. The law of cosines may be used to determine R as q 2 2 ı R D .6 kN/ C .8:2 kN/ — 2.6 kN/.8:2 kN/ cos 20 D 3:282 kN:

(4)

E

(5)

Noting that ˇ appears to be an obtuse angle (see the Common Pitfall margin note in the text), we will use the law of sines to determine ˛ as . Σ 6 kNR 6 kN —1 ı ı ) ˛ D sin sin 20 D 38:70 : (6) D ı sin ˛ sin 20 3:282 kN

This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the i i fM G Hill i hibit d

May 11, 2012

Solutions Manual

40 Angle ˇ is obtained using ı

ı

(7)

20 C ˛ C ˇ D 180 ; ı

ı

ı

ı

(8)

E ı ı R D 3:282 kN @ —.180 — 121:3 /

(9)

ˇ D 180 — 20 — 38:70 D 121:3 : Using polar vector representation, the resultant is

E

ı

R D 3:282 kN @ —58:70

(10)

: ı

ı

ı

If desired, the resultant may be stated using a positive angle, where 360 — 58:70 D 301:3 , as

E

R D 3:282 kN @ 301:3

ı

:

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May 11, 2012

41

Statics 2e

Problem 2.4 E Add the two vectors shown to form a resultant vector R, and report your result using polar vector representation.

Solution 30°

Part (a) The vector polygon shown at the right corresponds to the addition of the two force vectors to obtain a resultant force vector R. Since the 54 N force is vertical, the angle ı ı ı ˛ may be obtained by inspection as ˛ D 90 C 30 D 120 . The law of cosines may be used to determine R as q 2 2 ı R D .48 N/ C .54 N/ — 2.48 N/.54 N/ cos 120 (1) D 88:39 N:

48 N

E

R 54 N

The law of sines may be used to determine the angle ˇ as R 54 N —1 54 N ı ı ) ˇ D sin sin 120 D 31:95 : D sin ˇ sin ˛ 88:39 N

(2)

Using polar vector representation, the resultant is E ı ı R D 88:39 N @ —.180 — 30 — ˇ/ E

R D 88:39 N @ —118:1

ı

(3) (4)

: ı

ı

ı

If desired, this resultant may be stated using a positive angle, where 360 — 118:1 D 241:9 , as

E

R D 88:39 N @ 241:9

ı

(5)

:

Part (b) The vector polygon shown at the right corresponds to the addition E 20° of the two position vectors to obtain a resultant position vector R. Given the ı ı 20 and 30 angles provided in the problem statement, we determine the angle 100 mm ı ı ı opposite R to be 70 C30 100D . The law of cosines may be used to determine R as q 2 2 ı R D .100 mm/ C .80 mm/ — 2.100 mm/.80 mm/ cos 100 D 138:5 mm:

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R 70°

30°

80 mm

20°

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May 11, 2012

Solutions Manual

42 The law of sines may be used to determine the angle ˛ as —1 80 mm ı ı 80 mm ˛ D sin sin 100 D 34:67 : R D ) ı sin 100 sin ˛ 138:5 mm Using polar vector representation, the resultant is E ı ı R D 138:5 mm @ 34:67 — 20

E

R D 138:5 mm @ 14:67

ı

:

This solutions manual, in any print or electronic form, remains the property of McGr aw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

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(8) (9)

May 11, 2012

43

Statics 2e

Problem 2.5 E Add the two vectors shown to form a resultant vector R, and report your result using polar vector representation.

Solution 30°

Part (a) The vector polygon shown at the right corresponds to the addition of the two 3 ft position vectors to obtain the resultant position vector R. The law of cosines may be used 30° to determine R as q R 2 2 ı R D .3 ft/ C .4 ft/ — 2.3 ft/.4 ft/ cos 120 4 ft (1) D 6:083 ft: E

The law of sines may be used to determine the angle ˛ as —1 4 ft ı ı 4 ft ) ˛ D sin sin 120 D 34:72 : R D ı sin 120 sin ˛ 6:083 ft Using polar vector representation, the resultant is E ı ı R D 6:083 ft @ —.180 — 30 — ˛/ E ı : R D 6:083 ft @ —115:3 ı

ı

(2)

(3) (4) ı

If desired, this resultant may be stated using a positive angle, where 360 — 115:3 D 244:7 , as

E

R D 6:083 ft @ 244:7

ı

:

(5)

Part (b) The vector polygon shown at the right corresponds to the addition of E ı and the two force vectors to obtain the resultant force vector R. Given the 10 ı 20 angles provided in the problem statement, we determine the angle opposite ı ı ı ı R to be 10 C 90 — 20 D 80 . The law of cosines may be used to determine R as

q 2 2 ı R D .300 lb/ C .400 lb/ — 2.300 lb/.400 lb/ cos 80 D 456:4 lb: The law of sines may be used to determine the angle ˛ as 300 lb ı ı 300 lb D R —1 sin 80 D 40:34 : )˛ D sin ı sin 80 sin ˛ 456:4 lb Using polar vector representation, the resultant is E ı ı R D 456:4 lb @ 180 C 20 — ˛ E ı : R D 456:4 lb @ 159:7

This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

(6)

(7)

(8) (9)

May 11, 2012

Solutions Manual

44

Problem 2.6 E Add the two vectors shown to form a resultant vector R, and report your result using polar vector representation.

Solution Part (a) The vector polygon shown at the right corresponds to the addition of the E two force vectors to obtain a resultant force vector R. Since the two forces being added are perpendicular, basic trigonometry may be used to obtain R and ˛ as

139 lb

R

200 lb

q RD

2 2 .139 lb/ C .200 lb/ D 243:6 lb;

˛ D tan—1

200 lb

139 lb

(1)

D 55:20ı:

(2)

Using polar vector representation, the resultant is E ı ı R D 243:6 lb @ —.90 — 55:20 / E

ı

R D 243:6 lb @ —34:80

(3) (4)

: ı

ı

ı

If desired, this resultant may be stated using a positive angle, where 360 — 34:80 D 325:2 , as

E

R D 243:6 lb @ 325:2

ı

(5)

:

Part (b) The vector polygon shown at the right corresponds to the addition of the two position vectors to obtain a resultant position vector R. The law of cosines may be used to determine R as q 2 2 ı R D .6 in./ C .8 in./ — 2.6 in./.8 in./ cos 80 D 9:129 in.(6) E

The law of sines may be used to determine the angle ˛ as 8 in.

D

R

—1 ) ˛ D sin

ı sin ˛ sin 80 Using polar vector representation, the resultant is

8 in.

ı

ı

sin 80 D 59:66 :

(7)

9:129 in.

E ı ı R D 9:129 in. @ 180 — 20 — ˛

E R D 9:129 in. @ 100:3

ı

(8)

:

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May 11, 2012

45

Statics 2e

Problem 2.7 E Add the two vectors shown to form a resultant vector R, and report your result using polar vector representation.

Solution Part (a) The vector polygon shown at the right corresponds to the addition of the two E force vectors to obtain a resultant force vector R. Using this vector polygon, we determine R and ˇ as

35 kN

!

q 2 2 R D .35 kN/ C .18 kN/ D 39:36 kN;

—1 35 kN ˇ D tan 18 kN

E

β ı

D 62:78 : ı

The direction for R measured from the right-hand horizontal direction is 180

E the polar vector representation for R is

E

ı

R D 39:4 kN @ 117

R

(1)

18 kN

— 62:78 ı D 117:2 ı . Therefore,

(2)

: R

Part (b) The vector polygon shown at the right corresponds to the addition of the 45◦β two position vectors to obtain a resultant position vector R. We observe from this α 1.89 ft 1.23 ft ı ı ı ı ◦ ◦ vector polygon that ˛ D 180 — 60 — 45 D 75 . Using the law of cosines 45 60 q 2 2 ı R D .1:23 ft/ C .1:89 ft/ — 2.1:23 ft/.1:89 ft/ cos 75 D 1:970 ft: (3) E

Next, use the law of sines to find ˇ, such that 1:89 ft

R

sin ˇ D sin ˛

. Σ —1 1:89 ft ı ı ) ˇ D sin D 67:91 : 1:970 ft sin 75

E

ı

The direction of R measured from the right-hand horizontal direction is 67:91

E the polar vector representation of R is

E

R D 1:97 ft @ 22:9

ı

(4)

— 45 ı D 22:91 ı . Therefore,

:

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(5)

May 11, 2012

46

Solutions Manual

Problem 2.8 E

E

ı

D 6 m @ 90 ı . Sketch the vector polygons and evaluate R for the following, reporting your answer using polar vector representation. E E E (a) R D A C B, E E E (b) R D 2A — B, E E E E E (c) R D jAj B C jBj A, E E A B E

E

Let A D 2 m @ 0and B

(d) R D

E

jAj

C

E

.

jBj

Solution

Part (a) The vector polygon is shown to the right. The magnitude R of vector R is given by

E

q

p 2 2 2 2 R D A C B D .2 m/ C .6 m/ D 6:325 m: Referring to the figure again, we find ˇ in the following manner: ! 2m 1 ı R cos ˇ D A ) ˇ D cos— 6:325 m D 71:57 :

B (1)

R β A

(2)

E The polar vector representation of R is

E

R D 6:32 m @ 71:6

ı

(3)

: 2A

Part (b) Referring to the vector polygon shown at the right, we determine the values for R and ˇ as ! q 6m 2 2 —1 ı .2 · 2 m/ C .6 m/ D 7:211 m; ˇ D sin D 56:31 : (4) RD 7:211 m

β −B

R

E The polar vector representation of R is E R D 7:21 m @ —56:3

ı

:

(5) 12 m

2 EE E E ; since they are Part (c) Each vector jAjB and jBjA has a magnitude of 12 m ı ı perpendicular to one another, it follows that ˇ D 45 and & D 45 . The magnitude of R is given by q 22 22...