CHAPTER 3CHAPTER 7CHAPTER 4 - · PDF fileCHAPTER 3CHAPTER 7CHAPTER 4. 3. ... or distributed...

CHAPTER 3CHAPTER 7CHAPTER 7CHAPTER 3CHAPTER 4CHAPTER 4

3

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

PROBLEM 4.1

A tractor of mass 950 kg is used to lift gravel weighing 4 kN. Determine the reaction at each of the two (a) rear wheels A, (b) front wheels B.

SOLUTION

(a) Rear wheels + S - -M AB = + =0 9319 5 1000 4000 1250 2 1500 0: ( . )( ) ( )( ) ( )N mm N mm mm

A = +1439 83. N A = 1440 N ≠

(b) Front wheels + SM BA: ( . )( ) ( )( ) ( )- - + =9319 5 500 4000 2750 2 1500 0N mm N mm mm

B = +5219 92. N B = 5220 N ≠

Check: + = + - - =SF A By 0 2 2 9320 4000: N N 0

2 1440 2 5220 9320 4000 0( ) ( )+ - - =

0 0= ( )Checks

4


PROBLEM 4.2

A gardener uses a 60-N wheelbarrow to transport a 250-N bag of fertilizer. What force must she exert on each handle?

SOLUTION

Free-Body Diagram:

+ SM FA = - - =0 2 1 60 0 15 250 0 3 0: ( )( ) ( )( . ) ( )( . )m N m N m

F = 42 0. N ≠

5


PROBLEM 4.3

The gardener of Problem 4.2 wishes to transport a second 250-N bag of fertilizer at the same time as the first one. Determine the maximum allowable horizontal distance from the axle A of the wheelbarrow to the center of gravity of the second bag if she can hold only 75 N with each arm.

PROBLEM 4.2 A gardener uses a 60-N wheelbarrow to transport a 250-N bag of fertilizer. What force must she exert on each handle?

SOLUTION

Free-Body Diagram:

+ SM

xA = -

- - =0 2 75 1 60 0 15

250 0 3 250 0

: ( )( ) ( )( . )

( )( . ) ( )

N m N m

N m N

x = 0 264. m

6


PROBLEM 4.4

For the beam and loading shown, determine (a) the reaction at A, (b) the tension in cable BC.

SOLUTION

Free-Body Diagram:

(a) Reaction at A SF Ax x= =0 0:

+ SMB = + ++

0 75 700 100 550 175 350

100 150

: ( )( ) ( )( ) ( )( )

( )(

N mm N mm N mm

N mmm mm) ( )- =Ay 150 0

Ay = +1225 N A = 1225 N ≠

(b) Tension in BC + SMA = + +-

0 75 550 100 400 175 200

75 150

: ( )( ) ( )( ) ( )( )

( )(

N mm N mm N mm

N mm)) ( )- =FBC 150 0mm

FBC = +700 N FBC = 700 N

Check: + SF A Fy BC= - - - - - + - =

- + - =

0 75 100 175 100 75 0

525 1225 700 0

: N N N N N

0 0= ( )Checks

7


PROBLEM 4.5

Two crates, each of mass 350 kg, are placed as shown in the bed of a 1400-kg pickup truck. Determine the reactions at each of the two (a) rear wheels A, (b) front wheels B.

SOLUTION

Free-Body Diagram:

W

Wt

= =

= =

( )( . ) .

( )( . ) .

350 9 81 3 434

1400 9 81 13 734

kg m/s kN

kg m/s kN

2

2

(a) Rear wheels + SM W W W AB t= + + + - =0 1 7 2 05 1 2 2 3 0: ( . ) ( . ) ( . ) ( )m 2.05 m m m m

( . )( . ) ( . )( . )

( . )( . ) ( )

3 434 3 75 3 434 2 05

13 734 1 2 2 3

kN m kN m

kN m m

++ - =A 00

A = +6 0663. kN A = 6 07. kN ≠

(b) Front wheels + SF W W W A By t= - - - + + =0 2 2 0:

- - - + + =3 434 3 434 13 734 0. . .kN kN kN 2(6.0663 kN) 2B

B = +4 2347. kN B = 4 23. kN ≠

8


PROBLEM 4.6

Solve Problem 4.5, assuming that crate D is removed and that the position of crate C is unchanged.

PROBLEM 4.5 Two crates, each of mass 350 kg, are placed as shown in the bed of a 1400-kg pickup truck. Determine the reactions at each of the two (a) rear wheels A, (b) front wheels B.

SOLUTION

Free-Body Diagram:

W

Wt

= =

= =

( )( . ) .

( )( . ) .

350 9 81 3 434

1400 9 81 13 734

kg m/s kN

kg m/s kN

2

2

(a) Rear wheels + SM W W AB t= + + - =0 1 7 1 2 2 3 0: ( . ) ( . ) ( )m 2.05 m m m

( . )( . ) ( . )( . ) ( )3 434 3 75 13 734 1 2 2 3 0kN m kN m m+ - =A

A = +4 893. kN A = 4 89. kN ≠

(b) Front wheels + SM W W A By t= - - + + =0 2 2 0:

- - + + =3 434 13 734 0. .kN kN 2(4.893 kN) 2B

B = +3 691. kN B = 3 69. kN ≠

9


PROBLEM 4.7

A T-shaped bracket supports the four loads shown. Determine the reactions at A and B (a) if a = 250 mm, (b) if a = 175 mm. Neglect the weight of the bracket.

SOLUTION

Free-Body Diagram:

+

SF Bx x= =0 0:

+ SM a a AB = - - + + =0 200 150 150 50 200 300 0: ( )( ) ( ) ( )( ) ( )N mm N N mm mm

Aa= -( )200 20000

300 (1)

+ SM a

aA = - - - +

-0 200 150 250 300 150 300

50

: ( )( ) ( )( ) ( )( )

( )(

N mm N mm N mm

N ++ + =500 300 0mm mm) ( )By

Ba

y = +( )175000 200

300

Since B Ba

x = = +0

175000 200

300

( ) (2)

(a) For mma = 250

Eq. (1): A = ¥ - = +( )200 250 20000

300100 N A = 100 0. N Ø

Eq. (2): B = + ¥ = +( )175000 200 250

300750 N B = 750 N Ø

(b) For mma = 175

Eq. (1): A = ¥ - = +( )200 175 20000

30050 N A = 50 0. N Ø

Eq. (2): B = + ¥ = +( )175000 200 175

300700 N B = 700 N ≠

10


PROBLEM 4.8

For the bracket and loading of Problem 4.7, determine the smallest distance a if the bracket is not to move.

PROBLEM 4.7 A T-shaped bracket supports the four loads shown. Determine the reactions at A and B (a) if a = 250 mm, (b) if a = 175 mm. Neglect the weight of the bracket.

SOLUTION

Free-Body Diagram:

For no motion, reaction at A must be downward or zero; smallest distance a for no motion corresponds to A = 0.

+ SM a a AB = - - + + =0 200 150 150 50 200 300 0: ( )( ) ( ) ( )( ) ( )N mm N N mm mm

Aa= -( )200 20000

300

A a= - =0 200 20000 0: ( ) a = 100 0. mm

11


PROBLEM 4.9

The maximum allowable value of each of the reactions is 180 N. Neglecting the weight of the beam, determine the range of the distance d for which the beam is safe.

SOLUTION

S F Bx x= =0 0:

B By=

+ SM d d d B dA = - - - - + - =0 50 100 0 45 150 0 9 0 9 0: ( ) ( )( . ) ( )( . ) ( . )N N m N m m

50 45 100 135 150 0 9d d d B Bd- + - + + -.

dB

A B=

◊ --

180 0 9

300

N m m( . ) (1)

+ SM A dB = - - + =0 50 0 9 0 9 100 0: ( )( . ) ( . ) (N m m N)(0.45 m)

45 0 9 45 0- + + =. A Ad

dA

A=

- ◊( . )0 9 90m N m (2)

Since B �180 N, Eq. (1) yields

d �180 0 9 180

300 180

18

1200 15

--

= =( . ). m d �150 0. mm v

Since A�180 N, Eq. (2) yields

d �( . )

.0 9 180 90

180

72

1800 40

- = = m d � 400 mm v

Range: 150 0. mm 400 mm� �d

12


PROBLEM 4.10

Solve Problem 4.9 if the 50-N load is replaced by an 80-N load.

PROBLEM 4.9 The maximum allowable value of each of the reactions is 180 N. Neglecting the weight of the beam, determine the range of the distance d for which the beam is safe.

SOLUTION

SF Bx x= =0 0:

B By=

+ SM d d d B dA = - - - - + - =0 80 100 0 45 150 0 9 0 9 0: ( ) ( )( . ) ( )( . ) ( . )N N m N m m

80 45 100 135 150 0 9 0d d d B Bd- + - + + - =.

dB

B=

◊ --

180 0 9

330

N m

N

. (1)

+ SM A dB = - - + =0 80 0 9 0 9 100 0: ( )( . ) ( . ) (N m m N)(0.45 m)

dA

A= -0 9 117.

(2)

Since B �180 N, Eq. (1) yields

d � ( . ) ( ) .180 0 9 180 330 18018

1500 12- ¥ - = =/ m d = 120 0. mm v

Since A�180 N, Eq. (2) yields

d � ( . ) .0 9 180 11245

1800 25¥ - = =/180 m d = 250 mm v

Range: 120 0. mm 250 mm� �d

13


PROBLEM 4.11

For the beam of Sample Problem 4.2, determine the range of values of P for which the beam will be safe, knowing that the maximum allowable value of each of the reactions is 135 kN and that the reaction at A must be directed upward.

SOLUTION

SF Bx x= =0 0:

B = By ≠

+ SM P BA = - + - - =0 0 9 2 7 27 3 3 27: ( . ) ( . ) ( )( . ) (m m kN m kN)(3.9 m) 0

P B= -3 216 kN (1)

+ SM A PB = - + - - =0 2 7 1 8 27 0 6 27 0: ( . ) ( . ) ( )( . ) (m m kN m kN)(1.2 m)

P A= +1 5 27. kN (2)

Since B �135 kN, Eq. (1) yields

P � ( )( )3 135 216- kN P �189 0. kN v

Since 0 135� �A kN, Eq. (2) yields

0 27 1 5 135 27+ +kN � �P ( . )( )

27 229 5kN kN� �P . v

Range of values of P for which beam will be safe:

27 0. kN 189.0 kN� �P

14


PROBLEM 4.12

The 10-m beam AB rests upon, but is not attached to, supports at C and D. Neglecting the weight of the beam, determine the range of values of P for which the beam will remain in equilibrium.

SOLUTION

Free-Body Diagram:

+ SM P DC = - - + =0 2 4 3 20 8 6: ( ) ( )( ) ( )( ) (m kN m kN m m) 0

P D= -86 3kN (1)

+ SM P CD = + - - =0 8 4 3 20 2 6 0: ( ) ( )( ) ( )( ) (m kN m kN m m)

P C= +3 5 0 75. .kN (2)

For no motion C � 0 and D � 0

For C � 0 from (2) P � 3 50. kN

For D � 0 from (1) P � 86 0. kN

Range of P for no motion:

3 50. kN 86.0 kN� �P

15


PROBLEM 4.13

The maximum allowable value of each of the reactions is 50 kN, and each reaction must be directed upward. Neglecting the weight of the beam, determine the range of values of P for which the beam is safe.

SOLUTION

Free-Body Diagram:

+ SM P DC = - - + =0 2 4 3 20 8 6: ( ) ( )( ) ( )( ) (m kN m kN m m) 0

P D= -86 3kN (1)

+ SM P CD = + - - =0 8 4 3 20 2 6 0: ( ) ( )( ) ( )( ) (m kN m kN m m)

P C= +3 5 0 75. .kN (2)

For C � 0, from (2): P � 3 50. kN v

For D � 0, from (1): P � 86 0. kN v

For C � 50 kN, from (2):

P

P

�

�

3 5 0 75 50

41 0

. . ( )

.

kN kN

kN

+ v

For D � 50 kN, from (1):

P

P

�

�

86 3 50

64 0

kN kN

kN

--

( )

. v

Comparing the four criteria, we find

3 50. kN 41.0 kN� �P

16


PROBLEM 4.14

For the beam and loading shown, determine the range of the distance a for which the reaction at B does not exceed 450-N downward or 900-N upward.

SOLUTION

Assume B is positive when directed ≠

Sketch showing distance from D to forces.

+ SM a a BD = - - - - + =0 1350 200 1350 50 225 100 400: ( )( ) ( )( ) ( )( ) ( )N mm N N mm 00

- + + =2700 315000 400 0a B

aB

=+( )315000 400

2700 (1)

For B = 450 N Ø = -450 N, Eq. (1) yields:

a �[ ( )]315000 400 450

2700

+ - a � 50 0. mm v

For B = 900 N ≠ = +900 N, Eq. (1) yields:

a �[ ( )]315000 400 900

2700

+ a � 250 mm v

Required range: 50 0 250. mm mm� �a

17


PROBLEM 4.15

Two links AB and DE are connected by a bell crank as shown. Knowing that the tension in link AB is 720 N, determine (a) the tension in link DE, (b) the reaction at C.

SOLUTION

Free-Body Diagram:

+ SM F FC AB DE= - =0 100 120 0: ( ) ( )mm mm

F FDE AB= 5

6 (1)

(a) For FAB = 720 N

FDE = 5

6720( )N FDE = 600 N

(b) +

SF Cx x= - + =03

5720 0: ( )N

Cx = +432 N

+SF C

C

y y

y

= - + - =

= +

04

5720 600

1176

: ( )N N 0

N

C == ∞

1252 84

69 829

.

.

N

a

C = 1253 N 69.8°

18


PROBLEM 4.16

Two links AB and DE are connected by a bell crank as shown. Determine the maximum force that may be safely exerted by link AB on the bell crank if the maximum allowable value for the reaction at C is 1600 N.

SOLUTION

See solution to Problem 4.15 for F. B. D. and derivation of Eq. (1)

F FDE AB= 5

6 (1)

+ SF F C C Fx AB x x AB= - + = =03

50

3

5:

+ SF F C Fy AB y DE= - + - =04

50:

- + - =

=

4

5

5

60

49

30

F C F

C F

AB y AB

y AB

C C C

F

C F

x y

AB

AB

= +

= +

=

2 2

2 21

3049 18

1 74005

( ) ( )

.

For C FAB= =1600 1 74005N, 1600 N . FAB = 920 N

19


PROBLEM 4.17

The required tension in cable AB is 900 N. Determine (a) the vertical force P that must be applied to the pedal, (b) the corresponding reaction at C.

SOLUTION

Free-Body Diagram:

BC = 175 mm

(a) + SM PC = - =0 375 900 151 5544 0: ( ) ( )( . )mm N mm

P = 363 731. N P = 364 N Ø

(b) + SF Cx x= - =0 900: N 0 Cx = 900 N Æ

+SF C P Cy y y= - = - =0 0 363 731: . N 0 Cy = 364 N ≠

a = ∞=

22 0

970 722

.

.C N

C = 971 N 22.0∞

20


PROBLEM 4.18

Determine the maximum tension that can be developed in cable AB if the maximum allowable value of the reaction at C is 1125 N.

SOLUTION

Free-Body Diagram:

BC = 175 mm

+ SM P T P TC = - = =0 375 151 5544 0 0 40415: ( ) ( . ) .mm mm

+ SF P C T Cy y y= - + = - + =0 0 0 40415: . 0

C Ty = 0 40415.

+ SF T Cx x= - + =0: 0 C Tx =

C C C T T

C T

x y= + = +

=

2 2 2 20 40415

1 0786

( . )

.

For C = 1125 N

1125 1 0786 1043 02N N= fi =. .T T T = 1043 N

21


PROBLEM 4.19

The bracket BCD is hinged at C and attached to a control cable at B. For the loading shown, determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION

At B: T

Ty

x

= 0 18

0 24

.

.

m

m

T Ty x= 3

4 (1)

(a) + SM TC x= - - =0 0 18 240 0 4 240 0 8 0: ( . ) ( )( . ) ( )( . )m N m N m

Tx = +1600 N

Eq. (1) Ty = =3

41600 1200( )N N

T T Tx y= + = + =2 2 2 21600 1200 2000 N T = 2 00. kN

(b) + SF C Tx x x= - =0: 0

C Cx x- = = +1600 1600N 0 N Cx = 1600 N Æ

+SF C Ty y y= - - - =0 240 240 0: N N

Cy - - =1200 480 0N N

Cy = +1680 N Cy = 1680 N ≠

a = ∞=

46 4

2320

.

C N C = 2 32. kN 46.4°

22


PROBLEM 4.20

Solve Problem 4.19, assuming that a = 0.32 m.

PROBLEM 4.19 The bracket BCD is hinged at C and attached to a control cable at B. For the loading shown, determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION

At B: T

T

T T

y

x

y x

=

=

0 32

0 24

4

3

.

.

m

m

+ SM TC x= - - =0 0 32 240 0 4 240 0 8 0: ( . ) ( )( . ) ( )( . )m N m N m

Tx = 900 N

Eq. (1) Ty = =4

3900 1200( )N N

T T Tx y= + = + =2 2 2 2900 1200 1500 N T = 1 500. kN

+ SF C Tx x x= - =0: 0

C Cx x- = = +900 900N 0 N Cx = 900 N Æ

+SF C Ty y y= - - - =0 240 240 0: N N

Cy - - =1200 480 0N N

Cy = +1680 N Cy = 1680 N ≠

a = ∞=

61 8

1906

.

C N C = 1 906. kN 61.8°

23


PROBLEM 4.21

Determine the reactions at A and B when (a) h = 0, (b) h = 200 mm.

SOLUTION

Free-Body Diagram:

+ SM B B hA = ∞ - ∞ - =0 60 0 5 60 150 0 25 0: ( cos )( . ) ( sin ) ( )( . )m N m

Bh

=-37 5

0 25 0 866

.

. . (1)

(a) When h = 0:

Eq. (1): B = =37 5

0 25150

.

.N B = 150 0. N 30.0°

+ SF A By x= - ∞ =0 60 0: sin

Ax = ∞ =( )sin .150 60 129 9 N A x = 129 9. N Æ +SF A By y= - + ∞ =0 150 60 0: cos

Ay = - ∞ =150 150 60 75( )cos N A y = 75 N ≠

a = ∞=

30

150 0A . N A = 150 0. N 30.0°

(b) When h = 200 mm = 0.2 m

Eq. (1): B =-

=37 5

0 25 0 866 0 2488 3

.

. . ( . ). N B = 488 N 30.0°

+ SF A Bx x= - ∞ =0 60 0: sin

Ax = ∞ =( . )sin .488 3 60 422 88 N A x = 422 88. N Æ

+SF A By y= - + ∞ =0 150 60 0: cos

Ay = - ∞ = -150 488 3 60 94 15( . )cos . N A y = 94 15. N Ø

a = ∞=

12 55

433 2

.

.A N A = 433 N 12.6°

24


PROBLEM 4.22

For the frame and loading shown, determine the reactions at A and E when (a) a = ∞30 , (b) a = ∞45 .

SOLUTION

Free-Body Diagram:

+ SM E EA = +- -

0 200 125

90 250 90 75

: ( sin )( ) ( cos )( )

( )( ) ( )(

a amm mm

N mm N mmm) = 0

E =+

29250

200 125sin cosa a

(a) When a = 30°: E =∞ + ∞

=29250

200 30 125 30140 454

sin cos. N E = 140 5. N 60.0°

+ SF Ax x= - + ∞ =0 90 30 0: sinN (140.454 N)

Ax = +19 773. N A x = 19 77. N Æ

+SF Ay y= - + ∞ =0 90 140 454 30 0: ( . )cosN N

Ay = -31 637. N A y = 31 7. N Ø

A = 37 3. N 58.0°

25


PROBLEM 4.22 (Continued)

(b) When a = ∞45 :

E =∞ +

=29250

200 45 125127 279

sin cos.

aN E = 127 3. N 45.0°

+ SF Ax x= - + ∞ =0 90 127 279 45 0: ( . )sinN N

Ax = 0 A x = 0

+SF Ay y= - + ∞ =0 90 127 279 45 0: ( . )cosN

Ay = 0 Ay = 0 A = 0

26


PROBLEM 4.23

For each of the plates and loadings shown, determine the reactions at A and B.

SOLUTION

(a) Free-Body Diagram:

+ SM BA = - - =0 500 250 100 200 250 0: ( ) ( )( ) ( )( )mm N mm N mm

B = +150 N B = 150 0. N ≠

+ SF Ax x= + =0 200 0: N

Ax = -200 N A x = 200 N ¨

+SF A By y= + - =0 250: N 0

Ay + - =150 250 0N N

Ay = +100 N A y = 100 0. N ≠

a = ∞=

26 56

223 607

.

.A N A = 224 N 26.6°

27



(b) Free-Body Diagram:

+ SM BA = ∞ - - =0 30 500 200 250 250 100 0: ( cos )( ) ( )( ) ( )( )mm N mm N mm

B = 173 205. N B = 173 2. N 60.0°

+ SF A Bx x= - ∞ +0 30 200: sin N

Ax - ∞ + =( . )sin173 205 30 200 0N N

Ax = -113 3975. N A x = 113 4. N ¨

+ SF A By y= + ∞ - =0 30 250: cos N 0

Ay + ∞ - =( . )cos173 205 30 250 0N

Ay = +100 N A y = 100 0. N ≠

a =

=41 4

151 192

.

.

∞A N A = 151 2. N 41.4°

28


PROBLEM 4.24

For each of the plates and loadings shown, determine the reactions at A and B.

SOLUTION

(a) Free-Body Diagram:

+ SM AB = - + - =0 500 250 400 200 250 0: ( ) ( )( ) ( )( )mm N mm N mm

A = +100 N A = 100 0. N ≠

+ SF Bx x= + =0 200 0: N

Bx = -200 N Bx = 200 N ¨

+SF A By y= + - =0 250: N 0

100 250 0N N+ - =By

By = +150 N B y = 150 0. N ≠

a = ∞=

36 87

250

.

B N B = 250 N 36.9°

29



(b)

+ SM AA = - ∞ - + =0 30 500 200 250 250 400 0: ( cos )( ) ( )( ) ( )( )mm N mm N mm

A = 115 47. N A = 115 5. N 60.0°

+ SF A Bx x= ∞ + + =0 30 200 0: sin N

( . )sin115 47 30 200 0N N∞ + + =Bx

Bx = -257 735. N Bx = 258 N ¨

+SF A By y= ∞ + - =0 30 250: cos N 0

( . )cos115 47 30 250N N 0∞ + - =By

By = +150 N B y = 150 0. N ≠

a = ∞=

30 2

298 207

.

.B N B = 298 N 30.2°

30


PROBLEM 4.25

Determine the reactions at A and B when (a) a = 0, (b) a = 90 ,∞ (c) a = 30 .∞

SOLUTION

(a) a = 0∞

+ SM BA = - ◊ =0 0 5 100 0: ( . )m N m

B = 200 N

+ SF Ax x= =0 0:

+SF Ay y= + =0 200 0:

Ay = -200 N

A = 200 N Ø B = 200 N ≠

(b) a = ∞90

+ SM BA = - ◊ =0 0 3 100: ( . )m N m 0

B = =1000

3333 33N N.

+ SF A AA x x= - = =0 333 33 0 333 33: . , .N N

+ SF Ay y= =0 0:

A = 333 N Æ B = 333 N ¨

(c) a = ∞30

+ SM B BA = ∞ + ∞ - ◊ =0 30 0 5 30 0 3 100: ( cos )( . ) ( sin )( . )m m N m 0

B = 171 523. N

+ SF Ax x= - ∞ =0 171 523 30 0: ( . )sinN

Ax = 85 762. N

+ SF A Ay y y= + ∞ = = -0 171 523 0 148 543: ( . cos .N) 30 N

A = = ∞171 523. N 60.0a

A = 171 5. N 60.0° B = 171 5. N 60.0°

31


PROBLEM 4.26

A rod AB hinged at A and attached at B to cable BD supports the loads shown. Knowing that d = 200 mm, determine (a) the tension in cable BD, (b) the reaction at A.

SOLUTION

Free-Body Diagram:

(a) Move T along BD until it acts at Point D.

+ SM TA = ∞ + + =0 45 0 2 90 0 1 90 0 2 0: ( sin )( . ) ( )( . ) ( )( . )m N m N m

T = 190 92. N T = 190 9. N

(b) + SF Ax x= - ∞ =0 190 92 45 0: ( . )cosN

Ax = +135 N A x = 135 0. N Æ

+ SF Ay y= - - + ∞ =0 90 90 0: sinN N (190.92 N) 45

Ay = +45 N A y = 45 0. N ≠

A = 142 3. N 18.43°

32


PROBLEM 4.27

A rod AB hinged at A and attached at B to cable BD supports the loads shown. Knowing that d = 150 mm, determine (a) the tension in cable BD, (b) the reaction at A.

SOLUTION

Free-Body Diagram:

tan ;a a= = ∞10

1533.69

(a) Move T along BD until it acts at Point D.

+ SM TA = ∞ - - =0 33 69 0 15 90 0 1 90 0 2 0: ( sin . )( . ) ( )( . ) ( )( . )m N m N m

T = 324 5. N T = 324 N

(b) + SF Ax x= - ∞ =0 324 99 33 69 0: ( . )cos .N

Ax = +270 N A x = 270 N Æ

+SF Ay y= - - + ∞ =0 90 90 33 69 0: sin .N N (324.5 N)

Ay = 0 A y = 0 A = 270 N Æ

33


PROBLEM 4.28

A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 500-N horizontal force at B, determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION

Triangle ACD is isosceles with SC = ∞ + ∞ = ∞90 30 120 S SA D= = ∞ - ∞ = ∞1

2180 120 30( )

Thus DA forms angle of 60° with horizontal.

(a) We resolve FAD into components along AB and perpendicular to AB.

+ SM FC AD= ∞ - =0 30 250 500 100 0: ( sin )( ) ( )( )mm N mm FAD = 400 N

(b) + SF Cx x= - ∞ + - =0 400 60 500 0: ( )cosN N Cx = +300 N

+ SF Cy y= - ∞ + =0 400 0: ( sinN) 60 Cy = +346 4. N

C = 458 N 49.1°

34


PROBLEM 4.29

A force P of magnitude 1260 N is applied to member ABCD, which is supported by a frictionless pin at A and by the cable CED. Since the cable passes over a small pulley at E, the tension may be assumed to be the same in portions CE and ED of the cable. For the case when a = 75 mm, determine (a) the tension in the cable, (b) the reaction at A.

SOLUTION

Free-Body Diagram:

(a) + SM

T T

T

A = -

-

- =

0 1260 75

3007

25300

24

25200 0

: ( )( )

( ) ( )

( )

N mm

mm mm

mm

24 94500

3937 5

T

T

== . N T = 3940 N

(b) + SF Ax x= + + =07

253937 5 3937 5 0: ( . ) .

Ax = -5040 N A x = 5040 N ¨

+ SF Ay y= - - =0 126024

253937 5 0: ( . )N N

Ay = +5040 N A y = 5040 N ≠

A = 7128 N 45.0°

35


PROBLEM 4.30

Neglecting friction, determine the tension in cable ABD and the reaction at support C.

SOLUTION

Free-Body Diagram:

+ SM T TC = - - =0 0 25 0 1 120 0 1 0: ( . ) ( . ) ( )( . )m m N m T = 80 0. N

+ SF C Cx x x= - = = +0 80 80: N 0 N Cx = 80 0. N Æ

+ SF C Cy y y= - + = = +0 120 40: N 80 N 0 N Cy = 40 0. N ≠

C = 89 4. N 26.6°

36


PROBLEM 4.31

Rod ABC is bent in the shape of an arc of circle of radius R. Knowing the q = 30°, determine the reaction (a) at B, (b) at C.

SOLUTION

Free-Body Diagram: + SM C R P RD x= - =0 0: ( ) ( )

C Px = +

+ SF C Bx x= - =0 0: sinq

P B

B P

- ==

sin

/sin

qq

0

B = P

sinq q

+SF C B Py y= + - =0 0: cosq

C P P

C P

y

y

+ - =

= -ÊËÁ

ˆ¯̃

( /sin )cos

tan

q q

q

0

11

For q = ∞30 :

(a) B P P= ∞ =/sin30 2 B = 2P 60.0°

(b) C P C Px x= + = Æ

C P Py = - ∞ = -( /tan ) . /1 1 30 0 732 Cy P= 0 7321. Ø

C = 1 239. P 36.2°

37


PROBLEM 4.32

Rod ABC is bent in the shape of an arc of circle of radius R. Knowing the q = 60°, determine the reaction (a) at B, (b) at C.

SOLUTION

See the solution to Problem 4.31 for the free-body diagram and analysis leading to the following expressions:

C P

C P

BP

x

y

= +

= -ÊËÁ

ˆ¯̃

=

11

tan

sin

q

q

For q = ∞60 :

(a) B P P= ∞ =/sin .60 1 1547 B = 1 155. P 30.0°

(b) C P C Px x= + = Æ

C P Py = - ∞ = +( /tan ) .1 1 60 0 4226 Cy P= 0 4226. Ø

C = 1 086. P 22.9°

38


PROBLEM 4.33

Neglecting friction, determine the tension in cable ABD and the reaction at C when q = 60°.

SOLUTION

+ SM T a a Ta PaC = + - + =0 2 0: ( cos )q

TP=

+1 cosq (1)

+ SF C Tx x= - =0 0: sinq

C TP

x = =+

sinsin

cosq q

q1

+ SF C T T Py y= + + - =0 0: cosq

C P T P P

C

y

y

= - + = - ++

=

( cos )cos

cos1

1

10

q qq

Since C C Cy x= =0, C =+

Psin

cos

qq1

Æ (2)

For q = ∞60 :

Eq. (1): TP P=

+ ∞=

+1 60 1 12cos

T P= 2

3

Eq. (2): C P P= ∞+ ∞

=+

sin

cos

.60

1 60

0 866

1 12

C = 0 577. P Æ

39


PROBLEM 4.34

Neglecting friction, determine the tension in cable ABD and the reaction at C when q = 45°.

SOLUTION

Free-Body Diagram:

Equilibrium for bracket:

+ SM T a P a T a

T a aC = - - + ∞ ∞

+ ∞ + ∞0 45 2 45

45 2 45

: ( ) ( ) ( sin )( sin )

( cos )( cos ) == 0

T = 0 58579. or T P= 0 586.

+ SF C Px x= + ∞ =0 0 58579 45 0: ( . )sin

C Px = 0 41422.

+SF C P P Py y= + - + ∞ =0 0 58579 0 58579 45 0: . ( . )cos

Cy = 0 or C = 0 414. P Æ

40


PROBLEM 4.35

A light rod AD is supported by frictionless pegs at B and C and rests against a frictionless wall at A. A vertical 600 N force is applied at D. Determine the reactions at A, B, and C.

SOLUTION

Free-Body Diagram:

SF Ax = ∞ - ∞ =0 30 600 60 0: cos ( )cosN

A = 346 4102. N A = 346 N Æ

+ SM CB = - ∞+

0 200 600 400 30

346 4102 200 30

: ( ) ( )( )cos

( . )( )sin

mm N mm

N mm ∞∞ = 0

C = 866 0253. N C = 866 N 60.0°

+ SM BC = - ∞+

0 200 600 200 30

346 4102 400 30

: ( ) ( )( )cos

( . )( )sin

mm N mm

N mm ∞∞ = 0

B = 173 205. N B = 173 2. N 60.0°

41


PROBLEM 4.36

A light bar AD is suspended from a cable BE and supports a 250-N block at C. The ends A and D of the bar are in contact with frictionless vertical walls. Determine the tension in cable BE and the reactions at A and D.

SOLUTION

Free-Body Diagram:

SF A Dx = =0:

SFy = 0: TBE = 250 N

We note that the forces shown form two couples.

+ SM A= - =0 200 250 75 0: ( ) ( )( )mm N mm

A = 93 75. N

A = 93 75. N Æ D = 93 75. N ¨

42


PROBLEM 4.37

Bar AC supports two 400-N loads as shown. Rollers at A and C rest against frictionless surfaces and a cable BD is attached at B. Determine (a) the tension in cable BD, (b) the reaction at A, (c) the reaction at C.

SOLUTION

Similar triangles: ABE and ACD

AE

AD

BE

CD

BEBE= = =;

.

. .; .

0 15

0 5 0 250 075

m

m mm

(a) + SM T TA x x= - ÊËÁ

ˆ¯̃

- -0 0 250 075

0 350 5 400 0 1 400: ( . )

.

.( . ) ( )( . ) (m m N m N))( . )0 4 0m =

Tx = 1400 N

Ty =

=

0 075

0 351400

300

.

.( )N

N T = 1432 lb

43



(b) + SF Ay = - - - =0 300 400 400 0: N N N

A = +1100 N A = 1100 N ≠

(c) + SF Cx = - + =0 1400 0: N

C = + 1400 N C = 1400 N ¨

44


PROBLEM 4.38

Determine the tension in each cable and the reaction at D.

SOLUTION

tan.

.

tan.

.

a

a

b

b

=

= ∞

=

= ∞

0 08

21 80

0 08

38 66

m

0.2 m

m

0.1 m

+ SM TB CF= - ∞ =0 600 0 1 38 66 0 1 0: ( )( . ) ( sin . )( . )N m m

TCF = 960 47. N TCF = 96 0. N

+ SM TC BE= - ∞ =0 600 0 2 21 80 0 1 0: ( )( . ) ( sin . )( . )N m m

TBE = 3231 1. N TBE = 3230 N

+ SF T T Dy BE CF= + - =0 0: cos cosa b

( . )cos . ( . )cos .3231 1 21 80 960 47 38 66 0N N∞ + ∞ - =D

D = 3750 03. N D = 3750 N ¨

45


PROBLEM 4.39

Two slots have been cut in plate DEF, and the plate has been placed so that the slots fit two fixed, frictionless pins A and B. Knowing that P = 75 N, determine (a) the force each pin exerts on the plate, (b) the reaction at F.

SOLUTION

Free-Body Diagram:

+ SF Bx = - ∞ =0 75 30 0: sinN B = 150 N 60.0°

+ SM B B FA = - + ∞ + ∞ -0 150 100 30 75 30 275 325: ( )( ) sin ( ) cos ( ) ( )N mm mm mm mm == 0

- + + - =15000 5625 35723 55 325 0. ( )F

F = +81 0725. N F = 81 1. N Ø

+ SF A B Fy = - + ∞ - =0 150 30 0: cosN

A - + ∞ - =150 150 30 81 0725 0N N N( )cos .

A = +101 169. N A = 101 2. N ≠

46


PROBLEM 4.40

For the plate of Problem 4.39 the reaction at F must be directed downward, and its maximum allowable value is 100 N. Neglecting friction at the pins, determine the required range of values of P.

PROBLEM 4.39 Two slots have been cut in plate DEF, and the plate has been placed so that the slots fit two fixed, frictionless pins A and B. Knowing that P = 75 N, determine (a) the force each pin exerts on the plate, (b) the reaction at F.

SOLUTION

Free-Body Diagram:

+ SF P Bx = - ∞ =0 30 0: sin B = 2P 60°

+ SM B B FA = - + ∞ + ∞ -0 150 100 30 75 30 275 325: ( )( ) sin ( ) cos ( ) ( )N mm mm mm mm == 0

- ◊ ∞ + ∞ - =-15000 2 30 75 2 30 275 325 0

15000

N mm+ mm mm mmP P Fsin ( ) cos ( ) ( )

++ + - =75 476 314 325 0P P F.

PF= +325 15000

551 314. (1)

For F = 0: P = =15000

551 31427 208

.. N

For P = 100 N: P = + =32500 15000

551 31486 158

.. N

For 0 100� �F N 27 2. N 86.2 N� �P

47


PROBLEM 4.41

Bar AD is attached at A and C to collars that can move freely on the rods shown. If the cord BE is vertical (a = 0), determine the tension in the cord and the reactions at A and C.

SOLUTION

Free-Body Diagram:

+ SF Ty = - ∞ + ∞ =0 30 80 30 0: cos ( )cosN

T = 80 N T = 80 0. N

+ SM AC = ∞ - - =0 30 0 4 80 0 2 80 0 2 0: ( sin )( . ) ( )( . ) ( )( . )m N m N m

A = +160 N A = 160 0. N 30.0°

+ SM CA = - + ∞ =0 80 0 2 80 0 6 30 0 4 0: ( )( . ) ( )( . ) ( sin )( . )N m N m m

C = +160 N C = 160 0. N 30.0°

48


PROBLEM 4.42

Solve Problem 4.41 if the cord BE is parallel to the rods (a = 30°).

PROBLEM 4.41 Bar AD is attached at A and C to collars that can move freely on the rods shown. If the cord BE is vertical (a = 0), determine the tension in the cord and the reactions at A and C.

SOLUTION

Free-Body Diagram:

+ SF Ty = - + ∞ =0 80 30 0: ( )cosN

T = 69 282. N T = 69 3. N

+ SM

AC = - ∞

- + ∞ =0 69 282 30 0 2

80 0 2 30 0 4 0

: ( . )cos ( . )

( )( . ) ( sin )( . )

N m

N m m

A = +140 000. N A = 140 0. N 30.0°

+ SM

CA = + ∞

- + ∞ =0 69 282 30 0 2

80 0 6 30 0 4 0

: ( . )cos ( . )

( )( . ) ( sin )( . )

N m

N m m

C = +180 000. N C = 180 0. N 30.0°

49


PROBLEM 4.43

An 8-kg mass can be supported in the three different ways shown. Knowing that the pulleys have a 100-mm radius, determine the reaction at A in each case.

SOLUTION

W mg= = =( )( . ) .8 9 81 78 48kg m/s N2

(a) SF Ax x= =0 0:

+SF A Wy y= - =0 0: A y = 78 48. N ≠

+ SM M WA A= - =0 1 6 0: ( . )m

M A = + ( . )( . )78 48 1 6N m MA = ◊125 56. N m

A = 78 5. N ≠ MA = ◊125 6. N m

(b) + SF A Wx x= - =0 0: A x = 78 48. ≠

+ SF A Wy y= - =0 0: A y = 78 48. ¨

A = =( . ) .78 48 2 110 99N N 45°

+ SM M WA A= - =0 1 6 0: ( . )m

M A A= + = ◊( . )( . ) .78 48 1 6 125 56N m N mM

A = 111 0. N 45° MA = ◊125 6. N m

(c) SF Ax x= =0 0:

+ SF A Wy y= - =0 2 0:

A Wy = = =2 2 78 48 156 96( . ) .N N ≠

+ SM M WA A= - =0 2 1 6 0: ( . )m

M A = +2 78 48 1 6( . )( . )N m MA = ◊125 1. N m

A = 157 0. N ≠ MA = ◊125 N m

50


PROBLEM 4.44

A tension of 22.5 N is maintained in a tape as it passes through the support system shown. Knowing that the radius of each pulley is 10 mm, determine the reaction at C.

SOLUTION

From f.b.d. of system

+ SF Cx x= + =0 22 5 0: ( . )N

Cx = -22 5. N

+ SF Cy y= - =0 22 5 0: ( . )N

Cy = 22 5. N

Then C C Cx y= +

= +=

( ) ( )

( . ) ( . )

.

2 2

2 222 5 22 5

31 8198 N

and q = +-

ÊËÁ

ˆ¯̃

= - ∞-tan.

.1 22 5

22 545 or C = 31 8. N 45.0∞

+ SM MC C= + + =0 22 5 160 22 5 60 0: ( . )( ) ( . )( )N mm N mm

MC = - ◊4950 N mm or MC = ◊4950 N mm

51


PROBLEM 4.45

Solve Problem 4.44, assuming that 15 mm radius pulleys are used.

PROBLEM 4.44 A tension of 22.5 N is maintained in a tape as it passes through the support system shown. Knowing that the radius of each pulley is 10 mm, determine the reaction at C.

SOLUTION

From f.b.d. of system

+ SF Cx x= + =0 22 5 0: N( . )

Cx = -22 5. N

+ SF Cy y= - =0 22 5 0: N( . )

Cy = 22 5. N

Then C C Cx y= +

= +=

( ) ( )

( . ) ( . )

.

2 2

2 222 5 22 5

31 8198 N

and q =-

ÊËÁ

ˆ¯̃

= - ∞-tan.

..1 22 5

22 545 0 or C = 31 8. N 45 0. ∞

+ SM MC C= + + =0 22 5 165 22 5 65 0: N mm N mm( . )( ) ( . )( )

MC = - ◊5175 N mm or MC = ◊5180 N mm

52


PROBLEM 4.46

The frame shown supports part of the roof of a small building. Knowing that the tension in the cable is 150 kN, determine the reaction at the fixed end E.

SOLUTION

Free-Body Diagram:

4.5m

1.8m

20kN

D

C

6m

150kNME

Ex

My

E F

BA

20kN20kN20kN

1.8m1.8m1.8m

Equilibrium Equations DF = + =( . ( .4 5 6 7 52 2m) m) m,

+ SF Ex x= + =04 5

7 5150 0:

.

.( )kN

Ex = -90 0. kN Ex = 90 0. kN ¬

+ SF Ey y= - - =0 4 206

7 5150 0: (

.( )kN) kN

Ey = +200 kN Ey = 200 kN

+ SME = + +

+

0 20 20 5 4 20 3 6

20

: ( ( )( . ) ( )( . )

(

kN)(7.2 m) kN m kN m

kN)(1.8m) -- 6

7 5150 0

.( kN)(4.5m)+ ME =

ME = + ◊180 0. kN m ME = ◊180 0. kN m

53


PROBLEM 4.47

Beam AD carries the two 200-N loads shown. The beam is held by a fixed support at D and by the cable BE that is attached to the counterweight W. Determine the reaction at D when (a) W = 500 N, (b) W = 450 N.

SOLUTION

(a) W = 500 N

From f.b.d. of beam AD

+ SF Dx x= =0 0:

+ SF Dy y= - - + =0 200 200 500: N N N 0

Dy = -100 N or D = 100 0. N Ø

+ SM MD D= - ++ =

0 500 1 5 200 2 4

200

: ( )( . ) ( )( . )

(

N m N m

N)(1.2 m) 0

MD = ◊30 N m or MD = ◊30 0. N m

(b) W = 450 N

From f.b.d. of beam AD

+ SF Dx x= =0 0:

+ SF Dy y= + - - =0 450 200 200: N N N 0

Dy = -50 N or D = 50 0. N Ø

+ SM MD D= - ++ =

0 450 1 5 200 2 4

200 1 2 0

: ( )( . ) ( )( . )

( )( . )

N m N m

N m

MD = - ◊45 N m or MD = - ◊45 0. N m

54


PROBLEM 4.48

For the beam and loading shown, determine the range of values of W for which the magnitude of the couple at D does not exceed 60 N · m.

SOLUTION

For Wmin , MD = - ◊60 N m

From f.b.d. of beam AD + SM WD = -+ - ◊ =

0 200 2 4 1 5

200 1 2 60 0

: ( )( . ) ( . )

( )( . )minN m m

N m N m

Wmin N= 440

For Wmax, MD = ◊60 N m

From f.b.d. of beam AD + SM WD = -+ + ◊ =

0 200 2 4 1 5

200 1 2 60 0

: ( )( . ) ( . )

( )( . )maxN m m

N m N m

Wmax = 520 N or N 520 N440 � �W

55


PROBLEM 4.49

Knowing that the tension in wire BD is 1300 N, determine the reaction at the fixed support C of the frame shown.

SOLUTION

T

T T

T T

x

y

=

=

=

=

=

1300

5

13500

12

131200

N

N

N

+ SM C Cx x x= - + = = -0 450 500 50: N N 0 N Cx = 50 N ¨

+ SF C Cy y y= - - = = +0 750 120 1950: N 0 N 0 N Cy = 1950 N ≠

C = 1951 N 88.5°

+ SM MC C= + +- =

0 750 0 5 4 50 0 4

1200 0 4 0

: ( )( . ) ( . )( . )

( )( . )

N m N m

N m

MC = - ◊75 0. N m MC = ◊75 0. N m

56


PROBLEM 4.50

Determine the range of allowable values of the tension in wire BD if the magnitude of the couple at the fixed support C is not to exceed 100 N m.◊

SOLUTION

+ SM T

T

C = + - ÊËÁ

ˆ¯̃

-ÊËÁ

0 750 450 0 45

130 6

12

13

: ( ( )( . ) ( . )N)(0.5 m) N m m

ˆ̂¯̃

+ =( .0 15 0m) MC

375 1804 8

130N m N m m◊ + ◊ - Ê

ËÁˆ¯̃

+ =.T MC

T MC= +13

4 8555

.( )

For M TC = - ◊ = - =10013

4 8555 100 1232N m: N

.( )

For M TC = + ◊ = + =10013

4 8555 100 1774N m: N

.( )

For | | :MC �100 N m◊ 1 232. kN 1.774 kN� �T

57


PROBLEM 4.51

A vertical load P is applied at end B of rod BC. (a) Neglecting the weight of the rod, express the angle q corresponding to the equilibrium position in terms of P, l, and the counterweight W. (b) Determine the value of q corresponding to equilibrium if P W= 2 .

SOLUTION

(a) Triangle ABC is isosceles.

We have CD BC l= =( )cos cosq q2 2

+ SM W l P lC = ÊËÁ

ˆ¯̃

- =02

0: cos ( sin )q q

Setting sin sin cos : cos sin cosq q q q q q= - =22 2 2

22 2

0Wl Pl

W P- =22

0sinq

q = ÊËÁ

ˆ¯̃

-22

1sinW

P b

(b) For P W= 2 : sin .q2 2 4

0 25= = =W

P

W

W

q2

14 5= ∞. q = ∞29 0. b

or q q2

165 5 331= ∞ = ∞. ( )discard

58


PROBLEM 4.52

A slender rod AB, of weight W, is attached to blocks A and B, which move freely in the guides shown. The blocks are connected by an elastic cord that passes over a pulley at C. (a) Express the tension in the cord in terms of W and q. (b) Determine the value of q for which the tension in the cord is equal to 3W.

SOLUTION

(a) From f.b.d. of rod AB

+ SM T l W T lC = + ÊËÁ

ˆ¯̃

ÈÎÍ

˘˚̇

- =01

20: ( sin ) cos ( cos )q q q

TW=

-cos

(cos sin )

qq q2

Dividing both numerator and denominator by cos q,

TW=

-ÊËÁ

ˆ¯̃2

1

1 tanq or T

W

=( )

-2

1( tan )q b

(b) For T W= 3 , 31

11

6

2WW

=( )

-

- =

( tan )

tan

q

q

or q = ÊËÁ

ˆ¯̃

= ∞-tan .1 5

639 806 or q = ∞39 8. b

59


PROBLEM 4.53

Rod AB is acted upon by a couple M and two forces, each of magnitude P. (a) Derive an equation in q, P, M, and l that must be satisfied when the rod is in equilibrium.(b) Determine the value of q corresponding to equilibrium when M = 1500 N m,� P = 200 N, and l = 600 mm.

SOLUTION

Free-Body Diagram: (a) From free-body diagram of rod AB

+ SM P l P l MC = + - =0 0: ( cos ) ( sin )q q

or sin cosq q+ = M

Pl b

(b) For M P l= ◊ = =

+ =◊

=

1500 600

1500

N m, 200 N, and mm

N m

(200 N)(600 mm)sin cosq q 55

41 25= .

Using identity sin cos

sin ( sin ) .

( sin ) . sin

2 2

2 1 2

2 1 2

1

1 1 25

1 1 25

1

q q

q q

q q

+ =

+ - =

- = -

/

/

-- = - +sin . . sin sin2 21 5625 2 5q q q

2 2 5 0 5625 02sin . sin .q q- + =

Using quadratic formula

sin( . ) ( ) ( )( . )

( )

. .

q =- - ± -

= ±

2 5 625 4 2 0 5625

2 2

2 5 1 75

4

or sin . sin .

. .

q qq q

= == ∞ = ∞

0 95572 0 29428

72 886 17 1144

and

and

or andq q= ∞ = ∞17 11 72 9. . b

60


PROBLEM 4.54

Rod AB is attached to a collar at A and rests against a small roller at C. (a) Neglecting the weight of rod AB, derive an equation in P, Q, a, l, and q that must be satisfied when the rod is in equilibrium. (b) Determine the value of q corresponding to equilibrium when P = 80 N, Q = 60 N, l = 500 mm, and a = 125 mm.

SOLUTION

Free-Body Diagram:

+SF C P Qy = - - =0 0: cosq

CP Q= +cosq

(a) + SM Ca

PlA = - =0 0:cos

cosq

q

P Q a

Pl+ ◊ - =

cos coscos

q qq 0 cos

( )3 q = +a P Q

Pl b

(b) For P Q l a= = = =80 60 500 125N, N, mm and mm,

cos

( )( )

( )( )

.

3 125 80 60

80 500

0 4375

q =+

=

mm N N

N mm

cos .q = 0 75915 q = ∞40 6. b

61


PROBLEM 4.55

A collar B of weight W can move freely along the vertical rod shown. The constant of the spring is k, and the spring is unstretched when q = 0. (a) Derive an equation in q, W, k, and l that must be satisfied when the collar is in equilibrium. (b) Knowing that W = 300 N, l = 500 mm, and k = 800 N/m, determine the value of q corresponding to equilibrium.

SOLUTION

First note T ks=

where k

s

ll

l

==

= -

= -

spring constant

elongation of spring

cos

cos( co

q

q1 ss )

cos( cos )

q

qq T

kl= -1

(a) From f.b.d. of collar B + Â = - =F T Wy 0 0: sinq

or kl

Wcos

( cos )sinq

q q1 0- - = or tan sinq q- = W

kl b

(b) For W

l

k

l

===

= =

- =

300

500

800

5000 5

300

N

mm

N m

mm

2400 N/mm

N

800

/

.

tan sin(

q qNN/m m)( . )

.0 5

0 75=

Solving numerically, q = ∞57 957. or q = ∞58.0 b

62


PROBLEM 4.56

A vertical load P is applied at end B of rod BC. The constant of the spring is k, and the spring is unstretched when q = ∞90 . (a) Neglecting the weight of the rod, express the angle q corresponding to equilibrium in terms of P, k, and l. (b) Determine the value of q corresponding to equilibrium when P kl= 1

4 .

SOLUTION

First note T ks= =tension in spring

where s

AB AB

l l

=

= -

= ÊËÁ

ˆ¯̃

-

= ∞

elongation of spring

( ) ( )

sin sin

q q

q90

22

2990

2

22

1

2

∞ÊËÁ

ˆ¯̃

= ÊËÁ

ˆ¯̃

- ÊËÁ

ˆ¯̃

È

ÎÍ

˘

˚˙l sin

q

T kl= ÊËÁ

ˆ¯̃

- ÊËÁ

ˆ¯̃

È

ÎÍ

˘

˚˙2

2

1

2sin

q

(1)

(a) From f.b.d. of rod BC + SM T l P lC = ÊËÁ

ˆ¯̃

ÈÎÍ

˘˚̇

- =02

0: cos ( sin )q q

Substituting T from Equation (1)

22

1

2 20kl l P lsin cos sin

q q qÊËÁ

ˆ¯̃

- ÊËÁ

ˆ¯̃

È

ÎÍ

˘

˚˙

ÊËÁ

ˆ¯̃

ÈÎÍ

˘˚̇

- ( ) =

222

1

2 22

22kl Plsin cos sin c

q q qÊËÁ

ˆ¯̃

- ÊËÁ

ˆ¯̃

È

ÎÍ

˘

˚˙

ÊËÁ

ˆ¯̃

- ÊËÁ

ˆ¯̃ oos

q2

0ÊËÁ

ˆ¯̃

ÈÎÍ

˘˚̇

=

Factoring out 22

l cos ,qÊ

ËÁˆ¯̃

leaves

kl Psin sinq q2

1

2 20Ê

ËÁˆ¯̃

- ÊËÁ

ˆ¯̃

È

ÎÍ

˘

˚˙ - Ê

ËÁˆ¯̃

=

or sinq2

1

2ÊËÁ

ˆ¯̃

=-

ÊËÁ

ˆ¯̃

kl

kl P q =

-È

ÎÍ

˘

˚˙-2

21sin

( )

kl

kl P b

63



(b) Pkl

kl

kl

kl

kl

kl

=

= -( )È

ÎÍ

˘

˚˙

= ÊËÁ

ˆ¯̃

ÈÎÍ

˘˚̇

=

-

-

4

22

22

4

3

2

1

4

1

q sin

sin

ssin- ÊËÁ

ˆ¯̃

1 4

3 2

== ∞

-2 0 94281

141 058

1sin ( . )

. or 141.1q = ∞ b

64


PROBLEM 4.57

Solve Sample Problem 4.5, assuming that the spring is unstretched when q = ∞90 .

SOLUTION

First note T ks= =tension in spring

where s

r

F kr

===

deformation of spring

bb

From f.b.d. of assembly + SM W l F r0 0 0= - =: ( cos ) ( )b

or Wl kr

kr

Wl

cos

cos

b b

b b

- =

=

2

2

0

For k

r

l

W

====

=

45

75

200

1800

45 75

1800 20

2

N mm

mm

mm

N

N/mm mm

N

/

cos( )( )

( )(b

00 mm)b

or cos .b b= 0 703125

Solving numerically, b = 0 89245. rad

or b = ∞51 134.

Then q = ∞ + ∞ = ∞90 51 134 141 134. . or 141.1q = ∞ b

65


PROBLEM 4.58

A slender rod AB, of weight W, is attached to blocks A and B that move freely in the guides shown. The constant of the spring is k, and the spring is unstretched when q = 0. (a) Neglecting the weight of the blocks, derive an equation in W, k, l, and q that must be satisfied when the rod is in equilibrium. (b) Determine the value of q when W = 375 N, l = 750 mm, and k = 600 N/m.

SOLUTION

Free-Body Diagram:

Spring force: F ks k l l kls = = - = -( cos ) ( cos )q q1

(a) + SM F l Wl

D s= - ÊËÁ

ˆ¯̃

=02

0: ( sin ) cosq q

kl lW

l( cos ) sin cos12

0- - =q q q

klW

( cos ) tan12

0- - =q q or ( cos ) tan12

- =q q W

kl b

(b) For given values of W

l

k

== ==

- = -

=

375

750 0 75

600

1

375

2 60

N

mm m

N/m

N

.

( cos ) tan tan sin

(

q q q q

00 0 75

0 41667

N/m m)( . )

.=

Solving numerically: q = ∞49 710. or q = ∞49 7. b

66


PROBLEM 4.59

Eight identical 500 750-mm¥ rectangular plates, each of mass m = 40 kg, are held in a vertical plane as shown. All connections consist of frictionless pins, rollers, or short links. In each case, determine whether (a) the plate is completely, partially, or improperly constrained, (b) the reactions are statically determinate or indeterminate, (c) the equilibrium of the plate is maintained in the position shown. Also, wherever possible, compute the reactions.

SOLUTION

1. Three non-concurrent, non-parallel reactions

(a) Plate: completely constrained

(b) Reactions: determinate

(c) Equilibrium maintained

A C= = 196 2. N ≠





B C D= = =0 196 2, . N ≠

3. Four non-concurrent, non-parallel reactions


(b) Reactions: indeterminate


A x = 294 N Æ, Dx = 294 N ¨

(A Dy y+ = 392 N ≠ )

4. Three concurrent reactions (through D)

(a) Plate: improperly constrained


(c) No equilibrium ( )SMD π 0

67



5. Two reactions

(a) Plate: partial constraint



C R= = 196 2. N ≠





B = 294 N Æ, D = 491 N 53.1°

7. Two reactions

(a) Plate: improperly constrained

(b) Reactions determined by dynamics

(c) No equilibrium ( )SFy π 0

8. For non-concurrent, non-parallel reactions




B D= =y 196 2. N ≠

( )C D+ =x 0

68


PROBLEM 4.60

The bracket ABC can be supported in the eight different ways shown. All connections consist of smooth pins, rollers, or short links. For each case, answer the questions listed in Problem 4.59, and, wherever possible, compute the reactions, assuming that the magnitude of the force P is 500 N.

SOLUTION


(a) Bracket: complete constraint



A = 601N 56.3°, B = 333 N ¨

2. Four concurrent, reactions (through A)

(a) Bracket: improper constraint


(c) No equilibrium ( )SM A π 0

3. Two reactions

(a) Bracket: partial constraint



A = 250 N ≠, C = 250 N ≠





A = 250 N ≠, B = 417 N 36.9°, C = 333N Æ

69






(c) Equilibrium maintained ( )SMC y= =0 250 NA ≠





A x = 333 N ®, B = 333 N ¨

(A By y+ = 500 N ≠ )





A C= = 250 N ≠

8. Three concurrent, reactions (through A)

(a) Bracket: improper constraint


(c) No equilibrium ( )SM A π 0

70


PROBLEM 4.61

Determine the reactions at A and B when a = 180 mm.

SOLUTION

Reaction at B must pass through D where A and 300-N load intersect. Free-Body Diagram: (Three-force member)

DBCD : tan b

b

=

= ∞

240

180

53 13.

Force triangle

A = ∞=

( ) tan .300 53 13

400

N

N A = 400 N ≠ b

B =

=

300

500

N

cos 53.13 N

∞ B = 500 N 53 1. ∞ b

71


PROBLEM 4.62

For the bracket and loading shown, determine the range of values of the distance a for which the magnitude of the reaction at B does not exceed 600 N.

SOLUTION

Reaction at B must pass through D where A and 300-N load intersect.

Free-Body Diagram:(Three-force member)

a = 240 mm

tan b (1)

Force Triangle

(with N)B = 600

cos .

.

b

b

= =

= ∞

3000 5

60 0

N

600 N

Eq. (1) a =

=

240

138 56

mm

tan 60.0 mm

∞.

For B � 600 138 6 N mma ≥ . b

72


PROBLEM 4.63

Using the method of Section 4.7, solve Problem 4.17.

PROBLEM 4.17 The required tension in cable AB is 900 N. Determine (a) the vertical force P that must be applied to the pedal, (b) the corresponding reaction at C.

SOLUTION

Free-Body Diagram: (Three-Force body)

Reaction at C must pass through E, where P and 900 N forceintersect.

tan.

.

b

b

=

=

151 554

375

22 006

m

mm

∞

Force triangle

(a) P = ( )900 N tan 22.006∞

P = 363 7. P = 364 N Ø b

(b) C =∞

=900

22 006970 7

N N

cos .. C = 971 N 22 0. ∞ b

73


PROBLEM 4.64


PROBLEM 4.18 Determine the maximum tension that can be developed in cable AB if the maximum allowable value of the reaction at C is 1125 N.

SOLUTION


Reaction at C must pass through E, where D and the force T intersect.

tan

.

.

b

b

=

= ∞

151 554

375

22 006

mm

mm

Force triangle

T

T

= ∞=

( )cos .

.

1125 22 006

1043 04

N

N T = 1043 N b

74


PROBLEM 4.65

The spanner shown is used to rotate a shaft. A pin fits in a hole at A, while a flat, frictionless surface rests against the shaft at B. If a 300-N force P is exerted on the spanner at D, find the reactions at A and B.

SOLUTION

Free-Body Diagram:

(Three-Force body)

The line of action of A must pass through D, where B and P intersect.

tansin

cos.

.

a

a

= ∞∞ +

== ∞

75 50

75 50 3750 135756

7 7310

Force triangle A

B

=

=

=

=

300

2230 11

300

2209 842

N

sin 7.7310 N

N

tan 7.7310 N

∞

∞

.

.

A = 2230 N 7 73. ∞ b

B = 2210 N Æ b

75


PROBLEM 4.66

Determine the reactions at B and D when b = 60 mm.

SOLUTION

Since CD is a two-force member, the line of action of reaction at D must pass through Points C and D.

Free-Body Diagram:

(Three-Force body)

Reaction at B must pass through E, where the reaction at D and 80 N force intersect.

tan

.

b

b

=

= ∞

220

41 348

mm

250 mm

Force triangle

Law of sines

80

45 131 348888 0

942 8

N

sin 3.652 N

N

∞ ∞ ∞= =

==

B D

B

D

sin sin ..

. B = 888 N 41 3. ∞ D = 943 N 45 0. ∞ b

76


PROBLEM 4.67

Determine the reactions at B and D when b = 120 mm.

SOLUTION

Since CD is a two-force member, line of action of reaction at D must pass through C and D.

Free-Body Diagram:

(Three-Force body)

Reaction at B must pass through E, where the reaction at D and 80-N force intersect.

tan

.

b

b

=

= ∞

280

48 24

mm

250 mm

Force triangle

Law of sines 80

135 41 76

N

sin 3.24∞=

∞=

∞B D

sin sin .

B

D

==

1000 9

942 8

.

.

N

N B = 1001 N 48 2 943. ∞ =D N 45 0. ∞ b

77


PROBLEM 4.68

Determine the reactions at B and C when a = 37 5. mm.

SOLUTION

Since CD is a two-force member, the force it exerts on member ABD is directed along DC.

Free-Body Diagram of ABD: (Three-Force member)

The reaction at B must pass through E, where D and the 250 N load intersect.

Triangle CFD: tan .

.

a

a

= =

= ∞

75

1250 6

30 964

Triangle EAD: AE

GE AE AG

= == - = - =

250 150

150 37 5 112 5

tan

. .

a mm

mm

Triangle EGB: tan.

.

b

b

= =

= ∞

GB

GE

75

112 533 690

Force triangle B D

sin . sin .120 964 33 690

250

∞=

∞=

∞ N

sin 25.346

B

D

==

500 773

323 943

.

.

N

N B = 501 N 56 3. ∞ b

C D= = 324 N 31 0. ∞ b

78


PROBLEM 4.69

A 50-kg crate is attached to the trolley-beam system shown. Knowing that a = 1 5. m, determine (a) the tension in cable CD, (b) the reaction at B.

SOLUTION Free-Body Diagram:Three-Force body: W and TCD intersect at E.

tan.

..

b

b

=

= ∞

0 7497

1 526 56

m

m

Force triangle 3 forces intersect at E.

W ==

( )

.

50

490 5

kg 9.81 m/s

N

2

Law of sines 490 5

63 44 55

498 9

456 9

.

sin . sin

.

.

N

sin 61.56

N

N

∞=

∞=

∞

=

=

T B

T

B

CD

CD

(a) TCD = 499 N b

(b) B = 457 N 26 6. ∞ b

79


PROBLEM 4.70

Solve Problem 4.69, assuming that a = 3 m.

PROBLEM 4.69 A 50 kg crate is attached to the trolley-beam system shown. Knowing that a = 1 5. m, determine (a) the tension in cable CD, (b) the reaction at B.

SOLUTION

W and TCD intersect at E

Free-Body Diagram:

Three-Force body:

tan.

.

b

b

= =

= ∞

AE

AB

0 301

5 722

m

3 m

Force Triangle (Three forces intersect at E.)

W ==

( )50

490

kg 9.81 m/s

.5 N

2

Law of sines 490 5

95 722 55997 99

821

.

sin . sin.

.

N

sin 29.278 N

∞=

∞=

∞==

T B

T

B

CD

CD

559 N

(a) TCD = 998 N b

(b) B = 822 N 5 72. ∞ b

80


PROBLEM 4.71

One end of rod AB rests in the corner A and the other end is attached to cord BD. If the rod supports a 200-N load at its midpoint C, find the reaction at A and the tension in the cord.

SOLUTION


The line of action of reaction at A must pass through E, where T and the 200 N load intersect.

tan

.

tan

.

a

a

b

b

= =

= ∞

= =

= ∞

EF

AF

EH

DH

575

300

62 447

125

300

22 620

Force triangle A T

sin . sin .67 380 27 553

200

∞=

∞=

∞ N

sin 85.067

A = 185 3. N 62 4. ∞ b

T = 92 9. N b

81


PROBLEM 4.72

Determine the reactions at A and D when b = ∞30 .

SOLUTION

From f.b.d. of frame ABCD

+ SM AD = - + ∞+ ∞

0 0 18 150 30 0 10

150 30 0 28

: ( . [( sin ]( .

[( cos ]( .

m) N) m)

N) m)) = 0

A = 243 74. N

or A = 244 N Æ b

+ SF Dx x= + ∞ + =0 243 74 150 30 0: ( . ( sinN) N)

Dx = -318 74. N

+ SF Dy y= - ∞ =0 150 30 0: ( cosN)

Dy = 129 904. N

Then D D Dx x= +

= +=

( )

( . ) ( . )

.

2 2

2 2318 74 129 904

344 19 N

and q =ÊËÁ

ˆ¯̃

=-

ÊËÁ

ˆ¯̃

-

-

tan

tan.

.

1

1 129 904

318 74

D

Dy

x

= - ∞22 174. or D = 344 N 22 2. ∞ b

82


PROBLEM 4.73

Determine the reactions at A and D when b = ∞60 .

SOLUTION


+ SM AD = - + ∞+ ∞

0 0 18 150 60 0 10

150 60 0 28

: ( . [( sin ]( .

[( cos ]( .

m) N) m)

N) m)) = 0

A = 188 835. N

or A = 188 8. N Æ b

+ SF Dx x= + ∞ + =0 188 835 150 60 0: ( . ) ( )sinN N

Dx = -318 74. N

+ SF Dy y= - ∞ =0 150 60 0: ( )cosN

Dy = 75 0. N

Then D D Dx y= +

= +=

( ) ( )

( . ) ( . )

.

2 2

2 2318 74 75 0

327 44 N

and q =ÊËÁ

ˆ¯̃

-tan 1 D

Dy

x

=-

ÊËÁ

ˆ¯̃

= - ∞

-tan.

.

.

1 75 0

318 74

13 2409 or D = 327 N 13 24. ∞ b

83


PROBLEM 4.74

A 20-kg roller, of diameter 200 mm, which is to be used on a tile floor, is resting directly on the subflooring as shown. Knowing that the thickness of each tile is 8 mm, determine the force P required to move the roller onto the tiles if the roller is (a) pushed to the left, (b) pulled to the right.

SOLUTION

(a) Based on the roller having impending motion to the left, the only contact between the roller and floor will be at the edge of the tile.

First note W mg= = =( )( . ) .20 9 81 196 2kg m/s N2

From the geometry of the three forces acting on the roller

a =ÊËÁ

ˆ¯̃

=-cos .1 92

10023 074

mm

mm∞

and q a= - - = - =90 30 60 23 074 36 926∞ ∞ ∞ ∞ ∞. .

Applying the law of sines to the force triangle,

W P

sin sinq a=

or 196 2

36 926 23 074

.

sin . sin .

N

∞=

∞P

\ =P 127 991. N

or NP = 128 0. 30∞ b

(b) Based on the roller having impending motion to the right, the only contact between the roller and floor will be at the edge of the tile.

First note W mg= = ( )( . )20 9 81kg m/s2 = 196 2. N

From the geometry of the three forces acting on the roller

a =ÊËÁ

ˆ¯̃

=-cos .1 92

10023 074

mm

mm∞

and q a= + + = - =90 30 120 23 074 96 926∞ ∞ ∞ ∞ ∞. .

Applying the law of sines to the force triangle,

W P

sin sinq a=

or 196 2

96 926 23 074

.

sin . sin .

N

∞= P

\ =P 77 460. N

or NP = 77 5. 30∞ b

Free-Body Diagram:

Free-Body Diagram:

100 mm

f.b.d

30°

92 mm

N

P

W

120°

WN

P

100 mm P

W

N

αG

30°

92 mm

NW

P

60°θ

α

84


PROBLEM 4.75

Member ABC is supported by a pin and bracket at B and by an inextensible cord attached at A and C and passing over a frictionless pulley at D. The tension may be assumed to be the same in portions AD and CD of the cord. For the loading shown and neglecting the size of the pulley, determine the tension in the cord and the reaction at B.

SOLUTION

Reaction at B must pass through D. Free-Body Diagram:

tan

.

tan

.

a

a

b

b

=

= ∞

=

= ∞

175

30 256

175

16 26

mm

300 mm

mm

600 mm

Force triangle

Law of sines T T B

T T

sin . sin . sin .

(sin . ) (

59 744

360

13 996 106 26

13 996

∞= -

∞=

∞ = -

N

∞

3360

0 24185 360 0 86378

N)(sin 59.744 )∞

T T( . ) ( )( . )= -

T = 500 N T = 500 N b

B =

=

(sin .

.

50059 744

555 695

N)sin 106.26

N

∞

B = 556 N 30 3. ∞ b

85


PROBLEM 4.76

Member ABC is supported by a pin and bracket at B and by an inextensible cord attached at A and C and passing over a frictionless pulley at D. The tension may be assumed to be the same in portions AD and CD of the cord. For the loading shown and neglecting the size of the pulley, determine the tension in the cord and the reaction at B.

SOLUTION

Free-Body Diagram: Force triangle

Reaction at B must pass through D.

tan ; .a a= = ∞120

16036 9

T T B

4

75

3 5=

-=

lb

3 4 300 300

5

4

5

4300 375

T T T

B T

= - =

= = =

;

( )

lb

lb lb B = 375 lb 36.9°

86


PROBLEM 4.77

Rod AB is supported by a pin and bracket at A and rests against a frictionless peg at C. Determine the reactions at A and C when a 170-N vertical force is applied at B.

SOLUTION Free-Body Diagram: (Three-Force body)The reaction at A must pass through D where C and 170-N force intersect.

tan

.

a

a

=

= ∞

160

28 07

mm

300 mm

We note that triangle ABD is isosceles (since AC = BC ) and, therefore

SCAD = = ∞a 28 07.

Also, since CD CB^ , reaction C forms angle a = ∞28 07. with horizontal.

Force triangle

We note that A forms angle 2a with vertical. Thus A and C form angle

180 90 2 90∞ - ∞ - - = ∞ -( )a a a

Force triangle is isosceles and we have

A

C

===

170

2 170

160 0

N

N

N

( )sin

.

a

A = 170 0. N 33.9° C = 160 0. N 28.1° b

87


PROBLEM 4.78

Solve Problem 4.77, assuming that the 170-N force applied at B is horizontal and directed to the left.

PROBLEM 4.77 Rod AB is supported by a pin and bracket at A and rests against a frictionless peg at C. Determine the reactions at A and C when a 170-N vertical force is applied at B.

SOLUTION


The reaction at A must pass through D, where C and the 170-N force intersect.

tan

.

a

a

=

= ∞

160

28 07

mm

300 mm

We note that triangle ADB is isosceles (since AC = BC). Therefore S SA B= = ∞ -90 a.

Also S ADB = 2a

Force triangle

The angle between A and C must be 2a a a- =

Thus, force triangle is isosceles and

a = ∞28 07.

A

C

== =

170 0

2 170 300

.

( )cos

N

N Na A = 170 0. N 56.1° C = 300 N 28.1° b

88


PROBLEM 4.79


PROBLEM 4.21 Determine the reactions at A and B when(a) h = 0, (b) h = 200 mm.

SOLUTION

Free-Body Diagram: (a) h = 0

Reaction A must pass through C where 150-N weight and B interect.

Force triangle is equilateral

A = 150 0. N 30.0° b

B = 150 0. N 30.0° b

(b) h = 200 mm

tan.

.

b

b

=

= ∞

55 662

25012 552

Force triangle

Law of sines

150

60 102 552433 247

488 31

N

sin17.448N

N

∞=

∞=

∞==

A B

A

B

sin sin ..

.

A = 433 N 12.55° b

B = 488 N 30.0° b

89


PROBLEM 4.80


PROBLEM 4.28 A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 500-N horizontal force at B, determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION Free-Body Diagram: (Three-Force body)

Reaction at C must pass through E, where FAD and 500-N force intersect.

Since AC CD= = 250 mm, triangle ACD is isosceles.

We have SC = ∞ + ∞ = ∞90 30 120

and S SA D= = ∞ - ∞ = ∞1

2180 120 30( ) Dimensions in mm

On the other hand, from triangle BCF:

CF BC

FD CD CF

= ∞ = ∞ == - = - =

( )sin30 200 30 100

250 100 150

sin mm

mmFrom triangle EFD, and sinceSD = ∞30 :

EF FD= ∞ = ∞ =( ) tan tan .30 150 30 86 60 mm

From triangle EFC: tan

.

a

a

= =

= ∞

CF

EF

100

49 11

mm

86.60 mm

Force triangle

Law of sines F C

F C

AD

AD

sin . sin49 11 60

500

400 458

∞=

∞=

= =

N

sin 70.89

N, N

∞

(a) FAD = 400 N b

(b) C = 458 N 49.1° b

90


PROBLEM 4.81

Knowing that q = 30°, determine the reaction (a) at B, (b) at C.

SOLUTION Free-Body Diagram: (Three-Force body)Reaction at C must pass through D where force P and reaction at B intersect.

In DCDE: tan( )

.

b

b

= -

= -= ∞

3 1

3 1

36 2

R

R

Force triangle

Law of sines P B C

B P

C P

sin . sin . sin.

.

23 8 126 2 302 00

1 239

∞=

∞=

∞==

(a) B = 2P 60.0° b

(b) C = 1 239. P 36.2° b

91


PROBLEM 4.82

Knowing that q = 60°, determine the reaction (a) at B, (b) at C.

SOLUTION

Reaction at C must pass through D where force P and reaction at B intersect.

In DCDE: tan

.

b

b

=-

= -

= ∞

R

R

R3

11

322 9

Force triangle

Law of sines P B C

B P

C P

sin . sin . sin.

.

52 9 67 1 601 155

1 086

∞=

∞=

∞==

(a) B = 1 155. P 30.0° b

(b) C = 1 086. P 22.9° b

Free-Body Diagram:(Three-Force body)

92


PROBLEM 4.83

Rod AB is bent into the shape of an arc of circle and is lodged between two pegs D and E. It supports a load P at end B. Neglecting friction and the weight of the rod, determine the distance c corresponding to equilibrium when a = 20 mm and R = 100 mm.

SOLUTION

Since y x aED ED= = , Free-Body Diagram:

Slope of ED is 45∞

slope of isHC 45∞

Also DE a= 2

and DH HE DEa= = Ê

ËÁˆ¯̃

=1

2 2

For triangles DHC and EHC sin b = =a

R

a

R2

2Now c R= ∞ -sin( )45 b

For a R= =

=

== ∞

20 100

2 100

0 141421

8 1301

mm and mm

20 mm

mmsin

( )

.

.

b

b

and c = ∞ - ∞( )sin( . )100 mm 45 8 1301

= 60 00. mm or c = 60 0. mm b

93


PROBLEM 4.84

A slender rod of length L is attached to collars that can slide freely along the guides shown. Knowing that the rod is in equilibrium, derive an expression for the angle q in terms of the angle b.

SOLUTION

As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the forcegeometry

Free-Body Diagram:

tan b =x

yGB

AB

where

y LAB = cosq

and

x LGB = 1

2sinq

tansin

cos

tan

bqq

q

=

=

12

1

2

L

L

or tan tanq b= 2 b

94


PROBLEM 4.85

An 8 kg slender rod of length L is attached to collars that can slide freely along the guides shown. Knowing that the rod is in equilibrium and that b = 30°, determine (a) the angle q that the rod forms with the vertical, (b) the reactions at A and B.

SOLUTION

(a) As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the geometry of the forces

Free-Body Diagram:

tan b =x

yCB

BC

where

x LCB = 1

2sinq

and y LBC =

=

cos

tan tan

q

b q1

2

or tan tanq b= 2

For b = ∞30

tan tan

.

.

q

q

= ∞== ∞

2 30

1 15470

49 107 or q = ∞49 1. b

(b) W mg= = =( )( ) .8 78 480kg 9.81 m/s N2

From force triangle A W== ∞

tan

( . ) tan

b78 480 30N

= 45 310. N or A = 45 3. N ¬ b

and BW= =

∞=

cos

.

cos.

b78 480

3090 621

NN or B = 90 6. N 60 0. ∞ b

95


PROBLEM 4.86

A slender uniform rod of length L is held in equilibrium as shown, with one end against a frictionless wall and the other end attached to a cord of length S. Derive an expression for the distance h in terms of L and S. Show that this position of equilibrium does not exist if S > 2L.

SOLUTION

From the f.b.d. of the three-force member AB, forces must intersect at D. Since the force T intersects Point D, directly above G,

y hBE =

For triangle ACE: S AE h2 2 22= +( ) ( ) (1)

For triangle ABE: L AE h2 2 2= +( ) ( ) (2)

Subtracting Equation (2) from Equation (1)

S L h2 2 23- = (3)

or hS L= -2 2

3 b

As length S increases relative to length L, angle q increases until

rod AB is vertical. At this vertical position:

h L S h S L+ = = -or

Therefore, for all positions of AB

h S L� - (4)

or S L

S L2 2

3

- -�

or S L S L

S SL L

S SL L

2 2 2

2 2

2 2

3

3 2

3 6 3

- -

= - +

= - +

� ( )

( )

or 0 2 6 42 2� S SL L- +

and 0 3 2 22 2� S SL L S L S L- + = - -( )( )

96



For S L S L- = =0

Minimum value of S is L

For S L S L- = =2 0 2

Maximum value of S is 2L

Therefore, equilibrium does not exist if S L� 2 b

97


PROBLEM 4.87

A slender uniform rod of length L = 500 mm is held in equilibrium as shown, with one end against a frictionless wall and the other end attached to a cord of length S = 750 mm. Knowing that the mass of the rod is 4.5 kg, determine (a) the distance h, (b) the tension in the cord, (c) the reaction at B.

SOLUTION

From the f.b.d. of the three-force member AB, forces must intersect at D. Since the force T intersects Point D, directly above G,

y hBE =

For triangle ACE: S AE h2 2 22= +( ) ( ) (1)

For triangle ABE: L AE h2 2 2= +( ) ( ) (2)

Subtracting Equation (2) from Equation (1)

S L h2 2 23- =

or hS L= -2 2

3(a) For L S= =500 750mm and mm

h = -

=

( ) ( )

.

750 500

3322 7486

2 2

mm or h = 322 mm b

(b) We have W = ¥ =4 5 9 81 44 145. . . N

and q

q

= ÊËÁ

ˆ¯̃

= ÈÎÍ

˘˚̇

= ∞

-

-

sin

sin( . )

.

1

1

2

2 322 7486

750

59 391

h

s

98



From the force triangle

TW=

=∞

=

sin.

sin ..

q44 145

59 39151 2919

N

N or T = 51 3. N b

(c) BW=

=∞

tan.

tan .

q44 145

59 391

N

= 26 1166. N or B = 26 1. N ¬ b

99


PROBLEM 4.88

A uniform rod AB of length 2R rests inside a hemispherical bowl of radius R as shown. Neglecting friction, determine the angle q corresponding to equilibrium.

SOLUTION

Based on the f.b.d., the uniform rod AB is a three-force body. Point E is the point of intersection of the three forces. Since force A passes through O, the center of the circle, and since force C is perpendicular to the rod, triangle ACE is a right triangle inscribed in the circle. Thus, E is a point on the circle.

Note that the angle a of triangle DOA is the central angle corresponding to the inscribed angleq of triangle DCA.

a q= 2

The horizontal projections of AE xAE, ( ), and AG xAG, ( ), are equal.

x x xAE AG A= =

or ( )cos ( )cosAE AG2q q=

and ( )cos cos2 2R Rq q=

Now cos cos2 2 12q q= -

then 4 22cos cosq q- =

or 4 2 02cos cosq q- - =

Applying the quadratic equation

cos . cos .q q= = -0 84307 0 59307and

q q= ∞ = ∞32 534 126 375. . ( )and Discard or q = ∞32 5. b

100


PROBLEM 4.89

A slender rod of length L and weight W is attached to a collar at A and is fitted with a small wheel at B. Knowing that the wheel rolls freely along a cylindrical surface of radius R, and neglecting friction, derive an equation in q, L, and R that must be satisfied when the rod is in equilibrium.

SOLUTION

Free-Body Diagram (Three-Force body)

Reaction B must pass through D where B and W intersect.

Note that DABC and DBGD are similar.

AC AE L= = cosq

In DABC: ( ) ( ) ( )

( cos ) ( sin )

cos sin

CE BE BC

L L R

R

L

2 2 2

2 2 2

22

2

4

+ =

+ =

ÊËÁ

ˆ¯̃ = +

q q

q 22

22 2

22

4 1

3 1

q

q q

q

R

L

R

L

ÊËÁ

ˆ¯̃ = + -

ÊËÁ

ˆ¯̃ = +

cos cos

cos

cos221

31q = Ê

ËÁˆ¯̃ -

È

ÎÍ

˘

˚˙

R

L b

101


PROBLEM 4.90

Knowing that for the rod of Problem 4.89, L = 375 mm, R = 500 mm, and W = 45 N, determine (a) the angle q corresponding to equilibrium, (b) the reactions at A and B.

SOLUTION

See the solution to Problem 4.89 for free-body diagram and analysis leading to the following equation

cos221

31q = Ê

ËÁˆ¯̃ -

È

ÎÍ

˘

˚˙

R

LFor L R W= = =375 500 45mm, mm, and N

(a) cos ; .22

1

3

5001 59 39q q=

ÊËÁ

ˆ¯̃

-È

ÎÍÍ

˘

˚˙˙

= ∞mm

375 mm q = ∞59 4. b

In DABC: tansin

costan

tan tan . .

.

a qq

q

a

a

= = =

= ∞ =

= ∞

BE

CE

L

L2

1

21

259 39 0 8452

40 2

Force triangle

A W

BW

= = ∞ =

= =∞

=

tan ( ) tan . .

cos

( )

cos ..

a

a

45 40 2 38 028

45

40 258 916

N N

NN

(b) A = 38 0. N ® b

(c) B = 58 9. N 49.8° b

102


PROBLEM 4.91

A 1 2. m 2.4 m¥ sheet of plywood of mass 17 kg has been temporarily placed among three pipe supports. The lower edge of the sheet rests on small collars at A and B and its upper edge leans against pipe C. Neglecting friction at all surfaces, determine the reactions at A, B, and C.

SOLUTION

W = 17 × 9.81 = 166.77 N

r i j kG B/ = + +1 125

2

0 41758

20 3

. ..

We have 5 unknowns and six equations of equilibrium.

Plywood sheet is free to move in z direction, but equilibrium is maintained ( ).SFz = 0

SM r A A r C r wB A B x y C B G B= ¥ + + ¥ - + ¥ - =0 0: ( ) ( ) ( )/ / /i j i j

i j k i j k i j k

0 0 1 2

0

1 125 0 41758 0 6

0 0

0 5625 0 20879 0 3

0

. . . . . . .

A A Cx y

+-

+-1166 77 0

0

.

=

- + - + + - =1 2 1 2 0 6 0 41758 50 031 93 808125 0. . . . . .A A C Cy xi j j k i k

Equating coefficients of unit vectors to zero: i: - + =1 2 50 031 0. .Ay Ay = 41 6925. N

j: - + =0 6 1 2 0. .C Az A Cx = = =1

2

1

2224 64 112 32( . ) . N

k: 0 41758 93 808125 0. .C - = C = 224 64. N C = 225 N b

SFx = 0: A B Cx x+ - = 0: Bx = - =224 64 112 32 112 32. . . N

SFy = 0: A B Wy y+ - = 0: By = - =166 77 41 6925 125 08. . . N

A i j B i j C i= + = + = -( . ( . ) ( . ) ( . ) ( )112 3 41 7 112 3 125 1 225N) N N N N b

103


PROBLEM 4.92

Two tape spools are attached to an axle supported by bearings at A and D. The radius of spool B is 30 mm and the radius of spool C is 40 mm. Knowing that TB = 80 N and that the system rotates at a constant rate, determine the reactions at A and D. Assume that the bearing at A does not exert any axial thrust and neglect the weights of the spools and axle.

SOLUTION

Dimensions in mm

We have six unknowns and six equations of equilibrium.

SM T D D DA C x y= + ¥ - + + ¥ - + ¥ + +0 90 30 80 210 40 300: ( ) ( ) ( ) ( ) ( ) (i k j i j k i i j zzk) = 0

- + + - + - =7200 2400 210 40 300 300 0k i j i k jT T D DC C y z

Equate coefficients of unit vectors to zero:

i : 2400 40 0- =TC TC = 60 N

j: ( )( )210 300 0 210 60 300 0T D DC z z- = - = Dz = 42 N

k : - + =7200 300 0Dy Dy = 24 N

SFx = 0: Dx = 0

SF A Dy y y= + - =0 80: N 0 Ay = - =80 24 56 N

SF A Dz z z= + - =0 60: N 0 Az = - =60 42 18 N

A j k D j k= + = +( . ) ( . ) ( . ) ( . )56 0 18 00 24 0 42 0N N N N b

104


PROBLEM 4.93

Solve Problem 4.92, assuming that the spool C is replaced by a spool of radius 50 mm.

PROBLEM 4.92 Two tape spools are attached to an axle supported by bearings at A and D. The radius of spool B is 30 mm and the radius of spool C is 40 mm. Knowing that TB = 80 N and that the system rotates at a constant rate, determine the reactions at A and D. Assume that the bearing at A does not exert any axial thrust and neglect the weights of the spools and axle.

SOLUTION

Dimensions in mm


SM T D D DA C x y= + ¥ - + + ¥ - + ¥ + +0 90 30 80 210 50 300: ( ) ( ) ( ) ( ) ( ) (i k j i j k i i j zzk) = 0

- + + - + - =7200 2400 210 50 300 300 0k i j i k jT T D DC C y z


i : 2400 50 0- =TC TC = 48 N

j: ( )( )210 300 0 210 48 300 0T D DC z z- = - = Dz = 33 6. N

k : - + =7200 300 0Dy Dy = 24 N

SFx = 0: Dx = 0

SF A Dy y y= + - =0 80: N 0 Ay = - =80 24 56 N

SFz = 0: A Dz z+ - =48 0 Az = - =48 33 6 14 4. . N

A j k D j k= + = +( . ) ( . ) ( . ) ( . )56 0 14 40 24 0 33 6N N N N b

105


PROBLEM 4.94

Two transmission belts pass over sheaves welded to an axle supported by bearings at B and D. The sheave at A has a radius of 50 mm, and the sheave at C has a radius of 40 mm. Knowing that the system rotates at a constant rate, determine (a) the tension T, (b) the reactions at B and D. Assume that the bearing at D does not exert any axial thrust and neglect the weights of the sheaves and axle.

SOLUTION

Assume moment reactions at the bearing supports are zero. From f.b.d. of shaft

(a) SM Tx-axis : N N mm N mm= - + - =0 120 90 100 150 80 0( )( ) ( )( ) T = 187 5. N b

(b) SF Bx x= =0 0:

SM BD z y( ) ( )( ) ( )-axis : N 7.5 N mm mm= + - =0 150 18 150 300 0

By = 168 75. N

SM BD y z( ) ( )( ) ( )-axis : N N mm mm= + + =0 120 90 500 300 0

Bz = -350 N

or N NB j k= -( . ) ( )168 8 350 b

SM DB z y( ) ( )( ) ( )-axis : N 7.5 N mm mm= - + + =0 150 18 150 300 0

NDy = 168 75.

SM DB y z( ) ( )( ) ( )-axis : N N mm mm= + + =0 120 90 200 300 0

Dz = -140 N

or N ND j k= -( . ) ( . )168 8 140 0 b

106


PROBLEM 4.95

A 200-mm lever and a 240-mm-diameter pulley are welded to the axle BE that is supported by bearings at C and D. If a 720-N vertical load is applied at A when the lever is horizontal, determine (a) the tension in the cord, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust.

SOLUTION

Dimensions in mm

We have six unknowns and six equations of equilibrium—OK

SM k i j j k i k i jC x yD D T= - ¥ + + - ¥ + - ¥ - =0 120 120 160 80 200 720: ( ) ( ) ( ) ( ) ( ) 00

- + - - + ¥ + ¥ =120 120 120 160 57 6 10 144 10 03 3D D T Tx yj i k j i k.

Equating to zero the coefficients of the unit vectors:

k : - + ¥ =120 144 10 03T (a) T = 1200 N b

i : 120 57 6 10 0 4803D Dy y+ ¥ = = -. N

j: ( )- - =120 160 1200 0Dx N Dx = -1600 N

SFx = 0: C D Tx x+ + = 0 Cx = - =1600 1200 400 N

SFy = 0: C Dy y+ - =720 0 Cy = + =480 720 1200 N

SFz = 0: Cz = 0

(b) C i j D i j= + = - -( ) ( ) ( ) ( )400 1200 1600 480N N N N b

107


PROBLEM 4.96

Solve Problem 4.95, assuming that the axle has been rotated clockwise in its bearings by 30° and that the 720-N load remains vertical.

PROBLEM 4.95 A 200-mm lever and a 240-mm-diameter pulley are welded to the axle BE that is supported by bearings at C and D. If a 720-N vertical load is applied at A when the lever is horizontal, determine (a) the tension in the cord, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust.

SOLUTION

Dimensions in mm


SM k i j j k i k iC x yD D T= - ¥ + + - ¥ + - ¥ -0 120 120 160 80 173 21 720: ( ) ( ) ( ) ( . ) ( jj) = 0

- + - - + ¥ + ¥ =120 120 120 160 57 6 10 124 71 10 03 3D D T Tx yj i k j i k. .


k : . .- + ¥ = =120 124 71 10 0 1039 23T T N T = 1039 N b

i : .120 57 6 10 0 4803D Dy y+ ¥ = = - N

j: ( . )- -120 160 1039 2Dx Dx = -1385 6. N

SFx = 0: C D Tx x+ + = 0 Cx = - =1385 6 1039 2 346 4. . .

SFy = 0: C Dy y+ - =720 0 Cy = + =480 720 1200 N

SFz = 0: Cz = 0

(b) C i j D i j= + = - -( ) ( ) ( ) ( )346 1200 1386 480N N N N b

108


PROBLEM 4.97

An opening in a floor is covered by a 1 1.2-m¥ sheet of plywood of mass 18 kg. The sheet is hinged at A and B and is maintained in a position slightly above the floor by a small block C. Determine the vertical component of the reaction (a) at A, (b) at B, (c) at C.

SOLUTION

r i

r i k

r i k

B A

C A

G A

/

/

/

== += +

0 6

0 8 1 05

0 3 0 6

.

. .

. .

W mg

W

= ==

( ) .

.

18 9 81

176 58

kg

N

SM B C WA B A C A G A= ¥ + ¥ + ¥ - =0 0: ( )r j r j r j/ / /

( . ) ( . . ) ( . . ) ( )0 6 0 8 1 05 0 3 0 6 0i j i k j i k j¥ + + ¥ + + ¥ - =B C W

0 6 0 8 1 05 0 3 0 6 0. . . . .B C C W Wk k i k i+ - - + =

Equate coefficients of unit vectors of zero:

i : . ..

..1 05 0 6 0

0 6

1 05176 58C W C+ = = Ê

ËÁˆ¯̃

=N 100.90 N

k : . . .0 6 0 8 0 3 0B C W+ - =

0 6 0 8 100 90 0 3 176 58 0 46 24. . ( . . ( . ) .B B+ - = = -N) N N

SF A B C Wy = + + - =0 0:

A A- + + = =46 24 121 92. .N 100.90 N 176.58 N 0 N

( ) . ( ) . ( ) .a A b B c C= = - =121 9 46 2 100 9N N N b

109


PROBLEM 4.98

Solve Problem 4.97, assuming that the small block C is moved and placed under edge DE at a point 0.15 m from corner E.

PROBLEM 4.97 An opening in a floor is covered by a 1 1.2-m¥ sheet of plywood of mass 18 kg. The sheet is hinged at A and B and is maintained in a position slightly above the floor by a small block C. Determine the vertical component of the reaction (a) at A, (b) at B, (c) at C.

SOLUTION

r i

r i k

r i k

B A

C A

G A

/

/

/

== += +

0 6

0 65 1 2

0 3 0 6

.

. .

. .

W mg

W

= ==

( ) .

.

18 9 81

176 58

kg m/s

N

2

SM B C WA B A C A G A= ¥ + ¥ + ¥ - =0 0: ( )r j r j r j/ / /

0 6 0 65 1 2 0 3 0 6 0. ( . . ) ( . . ) ( )i j i k j i k j¥ + + ¥ + + ¥ - =B C W

0 6 0 65 1 2 0 3 0 6 0. . . . .B C C W Wk k i k i+ - - + =


i : . ..

..- + = = Ê

ËÁˆ¯̃

=1 2 0 6 00 6

1 2176 58C W C N 88.29 N

k : . . .0 6 0 65 0 3 0B C W+ - =

0 6 0 65 88 29 0 3 176 58 0 7 36. . ( . . ( . ) .B B+ - = = -N) N N

SF A B C Wy = + + - =0 0:

A A- + - = =7 36 95 648. .N 88.29 N 176.58 N 0 N

( ) . ( ) . ( ) .a A b c= -95 6 7 36 88 3N N N b

110


PROBLEM 4.99

The rectangular plate of mass 40 kg and is supported by three vertical wires. Determine the tension in each wire.

SOLUTION

Free-Body Diagram: W = 40 × 9.81 N = 392.4 N

SM r j r j jB A B A C B C G BT T r= ¥ + ¥ + ¥ - =0 392 4 0: ( . )/ / / N

( ) [( ( ) ] [( ) ( ) ]1500 1500 375 750 750mm mm) mm mm mmk j i k j i k¥ + + ¥ + +T TA C ¥¥ - =( . )392 4 0N j

- + - - + =1500 1500 375 294300 294300 0T T TA C Ci k i k i


k : 1500 294300 0 196 2T TC C- = fi = . N TC = 196 2. N b

i : - - + = fi =1500 375 196 2 294300 0 147 15T TA A( . ) . N TA = 147 2. N b

SFy = 0: T T TA B C+ + - =392 4 0. N

147 15 196 2 392 4 0 49 05. . . .N N+ + - = fi =T TB B TB = 49 1. N b

111


PROBLEM 4.100

The rectangular plate of mass 40 kg is supported by three vertical wires. Determine the weight and location of the lightest block that should be placed on the plate if the tensions in the three wires are to be equal.

SOLUTION

Free-Body Diagram: W = 40 × 9.81 = 392.4 N

Let -Wb j be the weight of the block and x and z the block’s coordinates.

Since tensions in wires are equal, let

T T T TA B C= = =

SM T T T W x z WA B C G b0 0 0= ¥ + ¥ + ¥ + ¥ - + + ¥ - =: ( ) ( ) ( ) ( ) ( ) ( )r j r j r j r j i k j

or, ( ) ( ) ( ) ( ) ( ) (1875 375 1500 750 750 1125k j k j i k j i k j¥ + ¥ + + ¥ + + ¥ - +T T T W xii k j+ ¥ - =z Wb) ( ) 0

or, - - + - - + - + =1875 375 1500 750 750 1125 0T T T T W W xW zWb bi i k i k i k i


i : - + + =3000 1125 0T W zWB (1)

k : 1500 750 0T W xWB- - = (2)

Also, SFy = 0: 3 0T W Wb- - = (3)

Eq. (1) + 1000 Eq. (3): 125 1000 0W z Wb+ - =( ) (4)

Eq. (2) – 500 Eq. (3): - - - =250 500 0W x Wb( ) (5)

112



Solving (4) and (5) for W Wb / and recalling of 0 1500� �x mm, 0 2250� �z mm,

Eq.(4): W

W zb =

- -=125

1000

125

1000 00 125� .

Eq.(5): W

W xb =

- -=250

500

250

500 00 5� .

Thus, ( ) . . ( . ) .minW Wb = = =0 5 0 5 392 4 196 2N N ( ) .minWb = 196 2 N b

Making W Wb = 0 5. in (4) and (5):

125 1000 0 5 0W z W+ - =( )( . ) z = 750 mm b

- - - =250 500 0 5 0W x W( )( . ) x = 0 mm b

113


PROBLEM 4.101

Two steel pipes AB and BC, each having a mass per unit length of 8 kg/m, are welded together at B and supported by three wires. Knowing that a = 0.4 m, determine the tension in each wire.

SOLUTION

W m g

W m g1

2

0 6

1 2

= ¢= ¢

.

.

SM T W W TD A D A E D F D C D C= ¥ + ¥ - + ¥ - + ¥ =0 01 2: ( ) ( )r j r j r j r j/ / / /

( . . ) ( . . ) ( ) . ( ) .- + ¥ + - + ¥ - + ¥ - + ¥ =0 4 0 6 0 4 0 3 0 2 0 81 2i k j i k j i j i jT W W TA C 00

- - + + - + =0 4 0 6 0 4 0 3 0 2 0 8 01 1 2. . . . . .T T W W W TA A Ck i k i k k


i : . . ; . .- + = = = ¢ = ¢0 6 0 3 01

2

1

20 6 0 31 1T W T W m g m gA A

k : . . . .- + - + =0 4 0 4 0 2 0 8 01 2T W W TA C

- ¢ + ¢ - ¢ + =0 4 0 3 0 4 0 6 0 2 1 2 0. ( . ) . ( . ) . ( .m g m g m g TC) 0.8

Tm g

m gC = - - ¢ = ¢( . . . )

.0 12 0 24 0 24

0 150.8

SF T T T W Wy A C D= + + - - =0 01 2:

0 3 0 15 0 6 1 2 0

1 35

. . . .

.

¢ + ¢ + - ¢ - ¢ == ¢

m g m g T m g m g

T m gD

D ¢ = =m g (8 kg/m)(9.81m/s ) N/m2 78 48.

T m gA = ¢ = ¥0 3 0 3 78 45. . . TA = 23 5. N b

T m gB = ¢ = ¥0 15 0 15 78 45. . . TB = 11 77. N b

T m gC = ¢ = ¥1 35 1 35 78 45. . . TC = 105 9. N b

114


PROBLEM 4.102

For the pipe assembly of Problem 4.101, determine (a) the largest permissible value of a if the assembly is not to tip, (b) the corresponding tension in each wire.

SOLUTION

W m g

W m g1

2

0 6

1 2

= ¢= ¢

.

.

SM T W W TD A D A E D F D C D C= ¥ + ¥ - + ¥ - + ¥ =0 01 2: ( ) ( )r j r j r j r j/ / / /

( . ) ( . ) ( ) ( . ) ( ) ( . )- + ¥ + - + ¥ - + - ¥ - + - ¥a T a W a W aAi k j i k j i j i0 6 0 3 0 6 1 21 2 TTC j = 0

- - + + - - + - =T a T W a W W a T aA A Ck i k i k k0 6 0 3 0 6 1 2 01 1 2. . ( . ) ( . )


i : . . ; . .- + = = = ¢ = ¢0 6 0 3 01

2

1

20 6 0 31 1T W T W m g m gA A

k : ( . ) ( . )- + - - + - =T a W a W a T aA C1 2 0 6 1 2 0

- ¢ + ¢ - ¢ - + - =0 3 0 6 1 2 1 2 0. . . ( . )m ga m ga m g a T aC(0.6 )

Ta a a

aC = - + --

0 3 0 6 1 2 0 6

1 2

. . . ( . )

.For Max a and no tipping, TC = 0

(a) - + - =- + - =0 3 1 2 0 6 0

0 3 0 72 1 2 0

. . ( . )

. . .

a a

a a

1 5 0 72. .a = a = 0 480. m b

115



(b) Reactions: ¢ = =m g ( ) . .8 9 81 78 48kg/m m/s N/m2

T m gA = ¢ = ¥ =0 3 0 3 78 48 23 544. . . . N TA = 23 5. N b

SF T T T W Wy A C D= + + - - =0 01 2:

T T m g m gA D+ + - ¢ - ¢ =0 0 6 1 2 0. .

T m g TD A= ¢ - = ¥ - =1 8 1 8 78 48 23 544 117 72. . . . . TD = 117 7. N b

116


SOLUTION

r i k

r i k

r i k

B A

C A

G A

a

a/

/

/

= += += +

750

750

375 375

By symmetry: B C=

SM B C WA B A C G A= ¥ + ¥ + ¥ - =0 0: ( )r j r j r j/ /

( ) ( ) ( ) ( )a B a B Wi k j i k j i k j+ ¥ + + ¥ + + ¥ - =750 750 375 375 0

Ba B B Ba W Wk i k i k i- + - - + =750 750 375 375 0

Equate coefficient of unit vector i to zero:

i : - - + =750 375 0B Ba W

BW

aC B

W

a=

+= =

+375

750

375

750 (1)

SF A B C Wy = + + - =0 0:

AW

aW A

aW

a+

+ÈÎÍ

˘˚̇

- = =+

2375

7500

750; (2)

PROBLEM 4.103

The 120 N square plate shown is supported by three vertical wires. Determine (a) the tension in each wire when a = 250 mm, (b) the value of a for which the tension in each wire is 40 N.

117



(a) For a = 250 mm, W = 120 N

Eq. (1) C B= = =375 120

100045

¥N

Eq. (2) A = =250 120

100030

( )N A B C= = =30 0 45 0. .N N b

(b) For tension in each wire = 40 N

Eq. (1) 40375 120

750N =

+¥

( )a

750 1125mm mm+ =a a = 375 mm b

118


PROBLEM 4.104

The table shown weighs 150 N and has a diameter of 1.2 m. It is supported by three legs equally spaced around the edge. A vertical load P of magnitude 500 N is applied to the top of the table at D. Determine the maximum value of a if the table is not to tip over. Show, on a sketch, the area of the table over which P can act without tipping the table.

SOLUTION

r b r= = ∞ =0 6 30 0 3. sin .m m

We shall sum moments about AB.

( ) ( )b r C a b P bW+ + - - = 0

( . . ) ( . ) ( . )0 3 0 6 0 3 500 0 3 150 0+ + - - =C a

C a= - -1

0 945 0 3 500

.[ ( . ) ]

If table is not to tip, C � 0

[ ( . ) ]

( . )

45 0 3 500 0

45 0 3 500

- --

a

a

�

�

a a a- =0 3 0 09 0 39 0 39. . . .� � m m

Only ^ distance from P to AB matters. Same condition must be satisfied for each leg. P must be located in shaded area for no tipping

119


PROBLEM 4.105

A 3-m boom is acted upon by the 3.4-kN force shown. Determine the tension in each cable and the reaction at the ball-and-socket joint at A.

SOLUTION

We have five unknowns and six equations of equilibrium but equilibrium is maintained ( ).SMx = 0

Free-Body Diagram:

BD BD

BE

u ruu

= - + + == - + +

( . ) ( . ) ( . ) .

( . ) ( . )

1 8 2 1 1 8 3 3

1 8 2 1

m m m m

m m

i j k

i j (( . ) .

.( . . . )

1 8 3 3

3 31 8 2 1 1 8

m mk

i j k

BE

T TBD

BD

T TBD BD

BD BD

=

= = - + + =u ruu

1116 7 6

3 31 8 2 1 1 8

1

( )

.( . . . )

- + +

= = - + - =

i j k

i j kT TBE

BE

T TBE BE

BE BE

u ruu

116 7 6( )- + -i j k

SM T TA B BD B BE C= ¥ + ¥ + ¥ - =0 3 4 0: ( . )r r r j

1 811

6 7 6 1 811

6 7 6 3 3 4 0. ( ) . ( ) ( . )i i j k i i j k i j¥ - + + + ¥ - + - + ¥ - =T TBD BE

12 6

11

10 8

11

12 6

11

10 8

1110 2 0

. . . ..T T T TBD BD BE BEk j k j k- + + - =

Equate coefficients of unit vectors to zero.

j:. .- + = =10 8

11

10 8

110T T T TBD BE BE BD

k :. .

.12 6

11

12 6

1110 2 0T TBD BE+ - =

212 6

1110 2

..TBD

ÊËÁ

ˆ¯̃

= TBD = 4 45. kN b

TBD

= 4.4524 kN TBE = 4 45. kN b

120



SF Ax x= - - =06

114 4524

6

114 4524 0: ( . ) ( . )kN kN

Ax = 4 8572. kN

SF Ay y= + + - =07

114 4524

7

114 4524 3 4 0: ( . ) ( . ) .kN kN kN

Ay = -2 2667. kN

SF Az z= + - =06

114 4524

6

114 4524 0: ( . ) ( . )kN kN

Az = 0 A i j= -( . ) ( . )4 86 2 27kN kN b

121


PROBLEM 4.106

A 2.4-m boom is held by a ball-and-socket joint at C and by two cables AD and AE. Determine the tension in each cable and the reaction at C.

SOLUTION

Free-Body Diagram: Five Unknowns and six equations of equilibrium. Equilibrium is maintained ( ).SMAC = 0

r k

r kB

A

==

1 2

2 4

.

.

AD AD

AE AE

u ruu

u ruu

= - + - =

= + - =

0 8 0 6 2 4 2 6

0 8 1 2 2 4 2 8

. . . .

. . . .

i j k

i j k

m

m

T

AD

AD

T

TAE

AE

T

ADAD

AEAE

= = - + -

= =

u ruu

u ruu

2 60 8 0 6 2 4

2 80

.( . . . )

.( .

i j k

88 1 2 2 4i j k+ -. . )

SMC A AD A AE B= ¥ + ¥ + ¥ - =0 3 0: ( )r T r T r jkN

i j k i j k

k0 0 2 4

0 8 0 6 2 42 6

0 0 2 4

0 8 1 2 2 42 8

1 2.

. . ..

.

. . ..

. (

- -+

-+ ¥ -

T TAD AE 33 6 0. )kN j =


i

j

: . . .

: . .

- - + =- + =

0 55385 1 02857 4 32 0

0 73846 0 68671 0

T T

T TAD AE

AD AE

(1)

T TAD AE= 0 92857. (2)

122



Eq. (1): - - + =0 55385 0 92857 1 02857 4 32 0. ( . ) . .T TAE AE

1 54286 4 32

2 800

. .

.

T

TAE

AE

== kN TAE = 2 80. kN b

Eq. (2): TAD = =0 92857 2 80 2 600. ( . ) . kN TAD = 2 60. kN b

S

S

F C C

F C

x x x

y y

= - + = =

= +

00 8

2 62 6

0 8

2 82 8 0 0

00 6

2 6

:.

.( . )

.

.( . )

:.

.(

kN kN

22 61 2

2 82 8 3 6 0 1 800

02 4

2 62

. ).

.( . ) ( . ) .

:.

.(

kN kN kN kN+ - = =

= -

C

F C

y

z zS .. ).

.( . ) .6

2 4

2 82 8 0 4 80kN kN kN- = =Cz

C j k= +( . ) ( . )1 800 4 80kN kN b

123


PROBLEM 4.107

Solve Problem 4.106, assuming that the 3.6-kN load is applied at Point A.

PROBLEM 4.106 A 2.4-m boom is held by a ball-and-socket joint at C and by two cables AD and AE. Determine the tension in each cable and the reaction at C.

SOLUTION

Free-Body Diagram: Five unknowns and six equations of equilibrium. Equilibrium is maintained ( ).S MAC = 0

AD AD

AE AE

u ruu

u ruu

= - + - =

= + - =

0 8 0 6 2 4 2 6

0 8 1 2 2 4 2 8

. . . .

. . . .

i j k

i j k

m

m

T

AD

AD

T

TAE

AE

T

ADAD

AEAE

= = - + -

= =

u ruu

u ruu

2 60 8 0 6 2 4

2 80

.( . . . )

.( .

i j k

88 1 2 2 4i j k+ -. . )

SMC A AD A AE A= ¥ + ¥ + ¥ -0 3 6: ( . )r T r T r jkN

Factor rA: r T T jA AD AE¥ + -( ( . ) )3 6 kN

or: T T jAD AE+ - =( )3 0kN (Forces concurrent at A)

Coefficient of i: - + =T TAD AE

2 60 8

2 80 8 0

.( . )

.( . )

T TAD AE= 2 6

2 8

.

. (1)

Coefficient of j: T T

T

AD AE

AE

2 60 6

2 81 2 3 6 0

2 6

2 8

0 6

2 6

1 2

2

.( . )

.( . ) .

.

.

.

.

.

.

+ - =

ÊËÁ

ˆ¯̃

+

kN

883 6 0

0 6 1 2

2 83 6

T

T

AE

AE

- =

+ÊËÁ

ˆ¯̃

=

.

. .

..

kN

kN

TAE = 5 600. kN TAE = 5 60. kN b

124



Eq. (1): TAD = =2 6

2 85 6 5 200

.

.( . ) . kN TAD = 5 20. kN b

S

S

F C C

F C

x x x

y y

= - + = =

= +

00 8

2 65 2

0 8

2 85 6 0 0

00 6

2 6

:.

.( . )

.

.( . ) ;

:.

.

kN kN

(( . ).

.( . ) .

:.

.( . )

5 21 2

2 85 6 3 6 0 0

02 4

2 65 2

2

kN kN kN

kN

+ - = =

= - -

C

F C

y

z zS ..

.( . ) .

4

2 85 6 0 9 60kN kN= =Cz

C k= ( . )9 60 kN b

Note: Since forces and reaction are concurrent at A, we could have used the methods of Chapter 2.

125


PROBLEM 4.108

A 300-kg crate hangs from a cable that passes over a pulley B and is attached to a support at H. The 100-kg boom AB is supported by a ball-and-socket joint at A and by two cables DE and DF. The center of gravity of the boom is located at G. Determine (a) the tension in cables DE and DF, (b) the reaction at A.

SOLUTION

Free-Body Diagram:

W

WC

G

= ¥ == ¥ =

300 9 81 2943

100 9 81 981

.

.

N

N

We have five unknowns ( , , , , )T T A A ADE DF x y z and five equilibrium equations. The boom is free to spin about the AB axis, but equilibrium is maintained, since SM AB = 0.

We have BH BH

DE

u vuu

u vuu

= - =

= -

( . ) ( ) .

( . ).

.(

9 15 7 11 5205

4 152 65

3 657

m m m

m

i j

i mm m

m m m m

) ( )

( . ) ( . ) ( ) .

( .

j k

i j k

+

= - + =

=

2

4 15 5 0822 2 6 8594

4 15

DE

DFu vuu

mm m m m) ( . ) ( ) .i j k- - =5 0822 2 6 8594DF

126



Thus: T Ti j

iBH BHBH

BH= = - = -

u vuu

( ).

.( . ) ( .2943

9 15 7

11 52052337 438 1788N N 2204

6 85944 15 5 0822 2

N)

.( . . )

j

T T i j k

T T

DE DEDE

DF DF

DE

DE

T

D

= = - +

=

u vuu

FF

DF

TDF

u vuu

= - -6 8594

4 15 5 0822 2.

( . . )i j k

(a) SM r W r W r T r T r T

iA J C K G H BH E DE F DF= ¥ + ¥ + ¥ + ¥ + ¥ =

-0 0

3 65

: ( ) ( ) ( ) ( ) ( )

( . ) ¥¥ - - ¥ - + ¥ -( ) ( . ) ( ) ( . ) ( . . )2943 1 825 981 5 5 2337 438 1788 204j i j i i j

+-

+ --

T TDE DF

6 85941 5 0 2

4 15 5 0822 26 8594

1 5 0 2

4 15 5 08.

.

. ..

.

. .

i j k i j k

222 2

0

-=

or, 10741 95 1790 325 9835 1222 5 0822

6 8594. . . ( )

.

.k k k i+ - + -

¥( )T TDE DF

+ - - + =5 3

6 8594

7 6233

6 85940

.

.( )

.

.( )T T T TDE DF DE DFj k


i or j: T T T TDE DF DE DF- = =0 *

k : . . ..

.( )10741 95 1790 325 9835 122

7 6233

6 85942 0+ - - =TDE TDE = 1213 441. N

T TDE DF= = 1213 N b

(b) SF A Ax x x= + + ÊËÁ

ˆ¯̃

= = -0 2337 438 24 15

6 85941213 441 0 3805 72: .

.

.( . ) . NN

SF Ay y= - - - - ÊËÁ

ˆ¯̃

0 2943 981 1788 204 25 0822

6 85941213 441: .

.

.( . ) == =0 7510 31Ay . N

SF Az z= =0 0: A i j= - +( ) ( )3810 7510N N b

*Remark: The fact that T TDE DF= could have been noted at the outset from the symmetry of structure with respect to xy plane.

127


SOLUTION

Free-Body Diagram:

By symmetry with xy plane

T T T

CD

CD

CD CE= =

= - + +=

u ruu

3 1 5 1 2

3 562

i j k. .

. m

T T

CD

CDT

T T

CD

CE

= = - + +

= - + -

u vuu

3 1 5 1 2

3 5623 1 5 1 2

3 562

i j k

i j k

. .

.. .

.

r i r iB A C A/ /= =2 3

SMA C A CD C A CE B A= ¥ + ¥ + ¥ - =0 5 0: ( )r T r T r j/ / / kN

i j k i j k i j k

3 0 0

3 1 5 1 23 562

3 0 0

3 1 5 1 23 562

2 0 0

0 5 0

0

-+

- -+

-=

. ..

. ..

T T

Coefficient of k: 2 3 1 53 562

10 0 3 958¥ ¥ÈÎÍ

˘˚̇

- = =..

.T

T kN

SF A T TCD CE= + + - =0 5 0: j

PROBLEM 4.109

A 3-m pole is supported by a ball-and-socket joint at A and by the cables CD and CE. Knowing that the 5-kN force acts vertically downward ( ),f = 0 determine (a) the tension in cables CD and CE, (b) the reaction at A.

128



Coefficient of k: Az = 0

Coefficient of i: A Ax x- ¥ = =2 3 958 3 3 562 0 6 67[ . . ] ./ kN

Coefficient of j: A Ay y+ ¥ - = =2 3 958 1 5 3 562 5 0 1 667[ . . . ] ./ kN

(a) T TCD CE= = 3 96. kN

(b) A i j= +( . ) ( . )6 67 1 667kN kN b

129


PROBLEM 4.110

A 3-m pole is supported by a ball-and-socket joint at A and by the cables CD and CE. Knowing that the line of action of the 5-kN force forms an angle f = ∞30 with the vertical xy plane, determine (a) the tension in cables CD and CE, (b) the reaction at A.

SOLUTION

Free-Body Diagram:

Five unknowns and six Eqs. of equilibrium but equilibrium ismaintained ( )SM AC = 0

r i

r iB A

C A

/

/

==

2

3Load at B. = - +

= - +( cos ) ( sin )

. .

5 30 5 30

4 33 2 5

j k

j k

CD CD

TCD

CD

TCD CD

u ruu

u ruu

= - + + =

= = - +

3 1 5 1 2 3 562

3 5623 1

i j k

T i

. . .

.( .

m

55 1 2j k+ . )

Similarly, T i j kCET= - + -

3 5623 1 5 1 2

.( . . )

SMA C A CD C A CE B A= ¥ + ¥ + ¥ - + =0 4 33 2 5 0: ( . . )r T r T r j k/ / /

i j k i j k i j k

3 0 0

3 1 5 1 23 562

3 0 0

3 1 5 1 23 562

2 0 0

0 4 3-+

- -+

-. ..

. ..

.

T TCD CE

33 2 5

0

.

=


j: ..

..

- + - =3 63 562

3 63 562

5 0T TCD CE

- + - =3 6 3 6 17 810 0. . .T TCD CE (1)

130



k : ..

..

.4 53 562

4 53 562

8 66 0T TCD CE+ - =

4 5 4 5 30 846. . .T TCD CE+ = (2)

( ) . ( ):2 1 25 1+ 9 53 11 0 5 901T TCE CE- = =. . kN

Eq. (1): - + - =3 6 3 6 5 901 17 810 0. . ( . ) .TCD

TCD = 0 954. kN

SF CD CE= + + - + =0 4 33 2 5 0: . .A T T j k

i :.

.( )

.

.( )Ax + - + - =0 954

3 5623

5 901

3 5623 0

Ax = 5 77. kN

j:.

.( . )

.

.( . ) .Ay + + - =0 954

3 5621 5

5 901

3 5621 5 4 33 0

Ay = 1 443. kN

k :.

.( . )

.

.( . ) .Az + + - + =0 954

3 5621 2

5 901

3 5621 2 2 5 0

Az = -0 833. kN

Answers:

(a) TCD = 0 954. kN TCE = 5 90. kN b

(b) A i j k= + -( . ) ( . ) ( . )5 77 1 443 0 833kN kN kN b

131


PROBLEM 4.111

A 1200 mm boom is held by a ball-and-socket joint at C and by two cables BF and DAE; cable DAE passes around a frictionless pulley at A. For the loading shown, determine the tension in each cable and the reaction at C.

SOLUTION

Free-Body Diagram:

Five unknowns and six equations of equilibrium but equilibrium is maintained ( ).S MAC = 0

T = Tension in both parts of cable DAE.

r k

r kB

A

==

750

1200

mm

mm

AD AD

AE AE

BF

u ruu

u ruu

u ru

= - - =

= - =

500 1200 1300

500 1200 1300

i k

j k

mm

mmuu

= - =400 750 850i k mmBF

T i k i k

T

AD

AE

TAD

AD

T T

TAE

AE

= = - - = - -

=

u ruu

u ruu

1300500 1200

135 12( ) ( )

== - = -

= =

T T

TBF

BF

TBF BF

BF

1300500 1200

135 12

850400

( ) ( )

(

j k j k

T i

u ruu

-- = -75017

8 15k i k) ( )TBF

SM r T r T r T r jC A AD A AE B BF B= ¥ + ¥ + ¥ + ¥ - =0 1600 0: ( )N

i j k i j k i j k

0 0 1200

5 0 1213

0 0 1200

0 5 1213

0 0 750

8 0 1517

7

- -+

-+

-+T T TBF ( 550 1600 0k j) ( )¥ - =

Coefficient of i: - + = =6000

131 200 000 0 2600T T, , N

132



Coefficient of j: - + =6000

13

6000

170T TBD

T T TBF BF= = =17

13

17

132600 3400( ) N

SF T T T j= + + - + =0 1600 0: AD AE BF C

Coefficient of i: - + + =5

132600

8

173400 0( ) ( ) Cx

Cx = -600 N

Coefficient of j: 5

132600 1600 0( ) - + =Cy

Cy = +600 N

Coefficient of k: - - - + =12

132600

12

132600

15

173400 0( ) ( ) ( ) Cz

Cz = 7800 N

Answers: T TDAE = TDAE = 2600 N b

TBD = 3400 N b

C i j k= - + +( ) ( ) ( )600 600 7800N N N b

133


PROBLEM 4.112

Solve Problem 4.111, assuming that the 1600 N load is applied at A.

PROBLEM 4.111 A 1200 mm boom is held by a ball-and-socket joint at C and by two cables BF and DAE; cable DAE passes around a frictionless pulley at A. For the loading shown, determine the tension in each cable and the reaction at C.

SOLUTION

Free-Body Diagram:

Five unknowns and six equations of equilibrium but equilibrium ismaintained ( ).SMAC = 0

T = tension in both parts of cable DAE.

r k

r kB

A

==

750

1200

mm

mm

AD AD

AE AE

BF

u ruu

u ruu

u ru

= - - =

= - =

500 1200 1300

500 1200 1300

i k

j k

mm

mmuu

= - =400 750 850i k BF mm

T i k i k

T

AD

AE

TAD

AD

T T

TAE

AE

= = - - = - -

=

u ruu

u ruu

1300500 1200

135 12( ) ( )

== - = -

= =

T T

TBF

BF

TBF BF

BF

1300500 1200

135 12

850400

( ) ( )

(

j k j k

T i

u ruu

-- = -75017

8 15k i k) ( )TBF

SMC A AD A AE B BF A= ¥ + ¥ + ¥ + ¥ - =0 1600 0: ( )r T r T r T r jN

i j k i j k i j k

0 0 1200

5 0 1213

0 0 1200

0 5 1213

0 0 750

8 0 1517

12

- -+

-+

-+T T TBF 000 1600 0k j¥ - =( )

Coefficient of i: - + = =6000

131 920 000 0 4160T T, , N

134



Coefficient of j: - + =6000

13

6000

170T TBD

T T TBD BD= = =17

13

17

134160 5440( ) N

SF T T T j C= + + - + =0 1600 0: AD AE BF

Coefficient of i: - + + =5

134160

8

175440 0( ) ( ) Cx

Cx = -960 N

Coefficient of j: 5

134160 1600 0( ) - + =Cy

Cy = 0

Coefficient of k: - - - + =12

134160

12

134160

15

175440 0( ) ( ) ( ) Cz

Cz = 12480 N

Answers: T TDAE = TDAE = 4160 N b

TBD = 5440 N b

C i k= - +( ) ( )960 12480N N b

135


PROBLEM 4.113

A 20-kg cover for a roof opening is hinged at corners A and B. The roof forms an angle of 30° with the horizontal, and the cover is maintained in a horizontal position by the brace CE. Determine (a) the magnitude of the force exerted by the brace, (b) the reactions at the hinges. Assume that the hinge at A does not exert any axial thrust.

SOLUTION

Force exerted by CD

F i j

F i j

= ∞ + ∞= +

= =

F F

F

W mg

(sin ) (cos )

( . . )

( .

75 75

0 2588 0 9659

20 9 81kg m//s N2

/

/

/

) .

.

. .

. .

( .

=== += +=

196 2

0 6

0 9 0 6

0 45 0 3

0 25

r k

r i k

r i k

F

A B

C B

G B

F 888 0 9659i j+ . )

SM r j r F r AB G B C B A B= ¥ - + ¥ + ¥ =0 196 2 0: ( . )/ / /

i j k i j k i j k

0 45 0 0 3

0 196 2 0

0 9 0 0 6

0 2588 0 9659 0

0 0 0 6. .

.

. .

. .

.

-+

++F

A Ax y 00

0=

Coefficient of i: + - - =58 86 0 5796 0 6 0. . .F Ay (1)

Coefficient of j: + + =0 1553 0 6 0. .F Ax (2)

Coefficient of k: - + = =88 29 0 8693 0 101 56. . : .F F N

Eq. (2): + - - = =58 86 0 5796 101 56 0 6 0 0. . ( . ) . A Ay y

Eq. (3): + + = = -0 1553 101 56 0 6 0 26 29. ( . ) . .A Ax x N

F = 101 6. N A i= -( . )26 3 N b

SF A B F j: + + - =W 0

Coefficient of i: 26 29 0 2588 101 56 0 0. . ( . )+ + = =B Bx x

Coefficient of j: B By y+ - = =0 9659 101 56 196 2 0 98 1. ( . ) . . N

Coefficient of k: Bz = 0 B j= ( . )98 1 N b

136


SOLUTION

Free-Body Diagram:

W = −mgj = −(30 kg)(9.81 m/s2)j = −(294 N)j

DC DC

TDC

DCT

u ruu

u ruu

= - + - =

= = -

( ( (480 240 160 560

67

mm) mm) mm) mmi j k

T ii j k+ -37

27T T

Equilibrium Equations:

SF i j k i j T j

i

= + + + + + - =

+ - + + +

0 294 0

67

37

: ( )

( ) (

A A A B B

A B T A B

x y z x y

x x y y

N

TT A Tz- + - =294 027N) ( )j k (1)

S SM r FB = ¥ =( :) 0

2 2 29467

37

27r A A A r r T T T r rx y zk i j k i k i j k i k¥ + + + + ¥ - + - + + ¥ -( ) ( ) ( ) ( ) ( NN

N) N

)

( ( ) ( )

j

i j k

=

- - + + - + - =

0

2 294 2 294 037

27

67A T r A T r T ry x (2)

Setting the coefficients of the unit vectors equal to zero in Eq. (2)

Ax = +49.0 N A

y = +73.5 N T

= 343 N b

Setting the coefficients of the unit vectors equal to zero in Eq. (1), we obtain three more scalar equations. After substituting the values of T, A

x and A

y into these equations we obtain

Az = +98.0 N B

x = +245 N B

y = +73.5 N

The reactions at A and B therefore

A = +(49.0 N)i + (73.5 N)j + (98.0 N)k b

B = +(245 N)i + (73.5 N)j b

PROBLEM 4.114

A uniform pipe cover of radius r = 240 mm and mass 30 kg is held in a horizontal position by the cable CD. Assuming that the bearing at B does not exert any axial thrust, determine the tension in the cable and the reaction at A and B.

r = 240 mm

r = 240 mm160 mm

80 mm

240 mm

r = 240 mm

W = – (294 N) j

T

D

A

C

Axi

Azkz

Bxi

Ayj

Byj

x

B

y

137


PROBLEM 4.115

A 100-kg uniform rectangular plate is supported in the position shown by hinges A and B and by cable DCE that passes over a frictionless hook at C. Assuming that the tension is the same in both parts of the cable, determine (a) the tension in the cable, (b) the reactions at A and B. Assume that the hinge at B does not exert any axial thrust.

SOLUTION Dimensions in mm

r i i

r i k

i k

r

B A

G A

C A

/

/

/

( )960 180 780

960

290

450

2

390 225

- =

= -ÊËÁ

ˆ¯̃

+

= +== +600 450i k

T = Tension in cable DCE

CD CD

CE CE

u ruu

u ruu

= - + - =

= + - =

690 675 450 1065

270 675 450 855

i j k

i j k

mm

mmm

T i j k

T i j k

W i

CD

CE

T

T

mg

= - + -

= + -

= - = -

1065690 675 450

855270 675 450

( )

( )

(( )( . ) ( )100 9 81 981kg m/s N2 j j= -

SM r T r T r j r BA C A CD C A CE G A B AW= ¥ + ¥ + ¥ - + ¥ =0 0: ( )/ / / /

i j k i j k

i j k

600 0 450

690 675 4501065

600 0 450

270 675 450855

39

- -+

-+

+

T T

00 0 225

0 981 0

780 0 0

0

0

-+ =

i j k

B By z

138



Coefficient of i: - - + ¥ =( )( ) ( )( ) .450 6751065

450 675855

220 725 10 03T T

T = 344 6. N T = 345 N b

Coefficient of j: ( ).

( ).- ¥ + ¥ + ¥ + ¥ -690 450 600 450

344 6

1065270 450 600 450

344 6

855780Bzz = 0

Bz = 185 49. N

Coefficient of k: ( )( ).

( )( ).

.600 675344 6

1065600 675

344 6

855382 59 10 7803+ - ¥ + =B By y 1113 2. N

B j k= +( . ) ( . )113 2 185 5N N b

SF A B T T W= + + + + =0 0: CD CE

Coefficient of i: Ax - + =690

1065344 6

270

855344 6 0( . ) ( . ) Ax = 114 4. N

Coefficient of j: A Ay y+ + + - = =113 2675

1065344 6

675

855344 6 981 0 377. ( . ) ( . ) N

Coefficient of k: Az + - - =185 5450

1065344 6

450

855344 6 0. ( . ) ( . ) Az = 141 5. N

A i j k= + +( . ) ( ) ( . )114 4 377 144 5N N N b

139


PROBLEM 4.116

Solve Problem 4.115, assuming that cable DCE is replaced by a cable attached to Point E and hook C.

PROBLEM 4.115 A 100-kg uniform rectangular plate is supported in the position shown by hinges A and B and by cable DCE that passes over a frictionless hook at C. Assuming that the tension is the same in both parts of the cable, determine (a) the tension in the cable, (b) the reactions at A and B. Assume that the hinge at B does not exert any axial thrust.

SOLUTION

See solution to Problem 4.115 for free-body diagram and analysis leading to the following:

CD

CE

==

1065

855

mm

mm

Now: T i j k

T i j k

W i

CD

CE

T

T

mg

= - + -

= + -

= - = -

1065690 675 450

855270 675 450

( )

( )

(( ) (100 981kg)(9.81 m/s N)2 j j= -

SM r r j rA C A CE G A B AT W B= ¥ + ¥ - + ¥ =0 0: / / /( )

i j k i j k i j k

600 0 450

270 675 450855

390 0 225

0 981 0

780 0 0

0

0

-+

-+ =T

B By z

Coefficient of i: - + ¥ =( )( ) .450 675855

220 725 10 03T

T = 621 3. N T = 621 N b


.270 450 600 450621 3

855980 0 364 7¥ + ¥ - = =B Bz z N

Coefficient of k: ( )( ).

. .600 675621 3

855382 59 10 780 0 113 23- ¥ + = =B By y N

B j k= +( . (113 2 365 N) N) b

140



SF A B T W= + + + =0 0: CE

Coefficient of i: A Ax x+ = = -270

855621 3 0 196 2( . ) . N

Coefficient of j: A Ay y+ + - = =113 2675

855621 3 981 0 377 3. ( . ) . N

Coefficient of k: Az + - =364 7450

855621 3 0. ( . ) Az = -37 7. N

A i j k= - + -( . ( ( .196 2 377 37 7 N) N) N) b

141


PROBLEM 4.117

The rectangular plate shown weighs 40 kg and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B.

SOLUTION

Weight of plate = 392.4 N

r i i

r i k

i k

r

B A

E A

G A

/

/

/

= - == - += +

=

( )

( )

1000 200 800

750 100 500

650 500

10000

2250

500 250

250 625 500

838 525

i k

i k

i j k

T

+

= +

= + -=

=

EF

EF

TA

u ruu

. mm

FF

AFT

TA E A G A

u ruu

( . . . )

(

0 29814 0 74536 0 59629

0 3

i j k

M r r

+ -

= ¥ + ¥ -S : / / 992 4 0. )j r+ ¥ =B A B/

i j k i j k i j k

650 0 500

0 29814 0 74536 0 59629

500 0 250

0 392 4 0

80

. . . .-+

-+T 00 0 0

0

0

B By z

=

Coefficient of i: - + = fi =372 68 98100 0 263 23. .T T N T = 263 N b

Coefficient of j: ( . . )( . ) :387 5885 149 07 263 23 800+ - =Bz 176.581 N

Coefficient of k: ( . )( . ) :484 484 263 23 196200 800 0- + =By By = 85 837. kN

B j k= +( . ( .85 8 176 6 N) N) b

142



SF A B T j= + + - =0 392 4: N) 0( .

Coefficient of i: A Ax x+ = = -0 29814 263 23 0 78 479. ( . ) . N

Coefficient of j: A Ay y+ + - = =85 837 0 74536 263 23 392 4 0 110 362. . ( . ) . . N

Coefficient of k: A Az z+ - = = -176 581 0 59629 263 23 0 19 6196. . ( . ) . N

A i j k= - + -( . ( . ( .78 5 110 4 19 62 N) N) N) b

143


PROBLEM 4.118

Solve Problem 4.117, assuming that cable EF is replaced by a cable attached at points E and H.

PROBLEM 4.117 The rectangular plate shown weighs 40 kg and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B.

SOLUTION

r i i

r i kB A

E A

/

/

= - == +

( )1000 200 800

650 500

r i kG A/ = +500 250

EH

EH

u ruu

= - + -=

750 300 500

950

i j k

mm

T i j k= = - + -TEH

EH

Tu ruu

950750 300 500( )

SM r T r r BA E A G A B A= ¥ + ¥ - + ¥ =0 392 4 0: / / /( . )

i j k i j k i j k

650 0 500

750 300 500950

500 0 250

0 392 4 0

800 0 0

0- -+

-+T

B By z.

== 0

Coefficient of i: - + =( )( ) ( . )( )300 500950

392 4 250 0T

T = 621.3 N T = 621N b


325000 375000621 3

950800 0-

( )- =Bz Bz = - 40 875. N

Coefficient of k: ( )( )( . )

( )( . )650 300621 3

950500 392 4 800 0- + =By By = 85 8375. N

B j k= -( . ( .85 8 40 9N) N) b

144



SF = 0: A B T j+ + - =( .392 4 N) 0

Coefficient of i: A Ax x- = =750

950621 3 0 490 5( . ) . N

Coefficient of j: A Ay y+ + - = =85 8375300

950621 3 392 4 0 110 3625. ( . ) . . N

Coefficient of k: A Az z- - = =40 875500

950621 3 0 367 875. ( . ) . N

A i j k= + +( ( . (491 110 4 368 N) N) N) b

145


PROBLEM 4.119

Solve Problem 4.114, assuming that the bearing at D is removed and that the bearing at C can exert couples about axes parallel to the y and z axes.

PROBLEM 4.114 The bent rod ABEF is supported by bearings at C and D and by wire AH. Knowing that portion AB of the rod is 250 mm long, determine (a) the tension in wire AH, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust.

SOLUTION

Free-Body Diagram: DABH is Equilateral

Dimensions in mm

r i j

r i k

T j k j

H C

F C

T T T

/

/

= - += += ∞ - ∞ =

50 250

350 250

30 30 0 5(sin ) (cos ) ( . -- 0 866. )k

SM r j r j kC F C H C C y C zT M M= ¥ - + ¥ + + =0 400 0: / /( ) ( ) ( )

i j k i j k

j k350 0 250

0 400 0

50 250 0

0 0 5 0 866

0

-+ -

-+ + =

. .

( ) ( )T M MC y C z

Coefficient of i: + ¥ - = =100 10 216 5 0 461 93 . .T T N T = 462 N b

Coefficient of j: - + =43 3 461 9 0. ( . ) ( )MC y

( )

) .

M

M

C y

C y

= ¥ ◊

= ◊

20 10

20 0

3 N mm

( N m

146



Coefficient of k: - ¥ - + =140 10 25 461 9 03 ( . ) ( )MC z

( ) .

) .

M

MC z

C z

= ¥ ◊= ◊

151 54 10

151 5

3 N mm

( N m

SF C T= + - =0 400 0: j

M j kC = ◊ + ◊( . ( .20 0 151 5 N m) N m) b

Coefficient of i: Cx = 0

Coefficient of j: C Cy y+ - = =0 5 461 9 400 0 169 1. ( . ) . N

Coefficient of k: C Cz z- = =0 866 461 9 0 400. ( . ) N

C j k= +( . (169 1 400 N) N) b

147


SOLUTION

Weight of plate = 40 × 9.81 = 392.4 N

r i k

r i k

i j k

E A

G A

EF

EF

/

/

= += +

= + -=

650 500

500 250

250 625 500

838 52

u ruu

. 55

0 29814 0 75436 0 59629

mm

T TEF

EFT= = + -

u ruu

( . . . )i j k

SM r T r j j kA E A G A A y A zM M= ¥ + ¥ - + + =0 75 0: / / ( ) ( ) ( )

i j k i j k

650 0 500

0 29814 0 75430 0 59629

500 0 250

0 392 4 0. . . .

( )

-+

-+T M A y jj k+ =( )M A z 0

Coefficient of i: - + = =372 68 98100 0 263 23. .T T N T = 263N b

Coefficient of j: ( . . )( . ) ( ) ( ) .387 5885 149 07 263 23 0 141264 6+ + = = ◊M MA y A y N mm

Coefficient of k: ( . )( . ) ( ) ( ) .484 484 263 23 196200 0 68669 3- + = = ◊M MA z A z N mm

SF A T= + - =0 392 44 0: . j

M j kA = - ◊ + ◊( (141300 68700 N mm) N mm) b

Coefficient of i: A Ax x+ = = -0 29814 263 23 0 78 479. ( . ) . N

Coefficient of j: A Ay y+ - = =( . )( . ) . .0 74536 263 23 392 4 0 196 199 N

Coefficient of k: Az – . ( . )0 59629 263 23 Az = 156 961. N

A i j k= - + +( . ( . ( .78 5 196 2 157 0N) N) N) b

PROBLEM 4.120

Solve Problem 4.117, assuming that the hinge at B is removed and that the hinge at A can exert couples about axes parallel to the y and z axes.

PROBLEM 4.117 The rectangular plate shown weighs 40 kg and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B.

148


PROBLEM 4.121

The assembly shown is used to control the tension T in a tape that passes around a frictionless spool at E. Collar C is welded to rods ABC and CDE. It can rotate about shaft FG but its motion along the shaft is prevented by a washer S. For the loading shown, determine (a) the tension T in the tape, (b) the reaction at C.

SOLUTION

Free-Body Diagram:

r j k

r i jA C

E C

/

/

= += -

105 50

40 60

SM T M MC A C E C C y C z= ¥ - + ¥ + + + =0 3 0: ( ) ( ) ( ) ( )r j r i k j k/ /

( ) ( ) ( ( ) ( )105 50 3 40 60 0 ) ( )j k j i j i k j k+ ¥ - + - ¥ + + + =T M MC y C z

Coefficient of i: 150 60 0- =T T = 2 5. N b

Coefficient of j: - + = = ◊40 2 5 0 100( . ) ( ) N) ( N mmM MC y C y

Coefficient of k: 60 2 5 0 150( . ( ) ( ) N) N mm+ = = - ◊M MC z C z

M j kC = ◊ - ◊( . ( .100 0 150 0 N mm) N mm) b

SF C C Cx y z= + + - + + =0 3 2 5: N) N) (2.5 N) 0i j k j i k( ( .


C C Cx y z= - = = -2 5 3 2 5. . N N N b

C i j k= - + -( . ( . ( .2 50 3 00 2 50 N) N) N) b

149


SOLUTION

Free-Body Diagram:

First note Ti j

CF CF CF CF

CF

T T

T

= = - +

+= -

l ( . ( . )

( . ) ( . )

(

0 08 0 06

0 08 0 06

0

2 2

m) m

m

.. . )

( . ( . )

( . ) ( . )

8 0 6

0 12 0 09

0 12 0 092 2

i j

Tj k

+

= = -

+DE DE DET Tl m) m

mDDE

DET= -( . . )0 8 0 6j k

(a) From f.b.d. of assembly

SF T Ty CF DE= + - =0 0 8 480: 0.6 N 0.

or 0 6 0 8 480. .T TCF DE+ = N (1)

SM T Ty CF DE= - + =0 0 8 0 135 0 6 0 08: m) m) 0( . )( . ( . )( .

or T TDE CF= 2 25. (2)

Substituting Equation (2) into Equation (1)

0 6 0 8 2 25 480. . [( . ) ]T TCF CF+ = N

TCF = 200 00. N

or TCF = 200 N b

and from Equation (2) TDE = =2 25 200 00. ( . N) 450.00

or TDE = 450 N b

PROBLEM 4.122

The assembly is welded to collar A that fits on the vertical pin shown. The pin can exert couples about the x and z axes but does not prevent motion about or along the y axis. For the loading shown, determine the tension in each cable and the reaction at A.

150



(b) From f.b.d. of assembly

SF A Az z z= - = =0 0 6 450 00 0 270 00: N) N( . )( . .

SF A Ax x x= - = =0 0 8 200 00 160 000: N) 0 N( . )( . .

or A i k= +( . (160 0 270 N) N) b

SM Mx Ax= + -

-

0 480 200 00

45

: N)(0.135 m) N)(0.6)](0.135 m)( [( .

[( 00 N)(0.8)](0.09 m) 0=

M Ax= - ◊16 2000. N m

SM Mz Az

= - +

+

0 480 200 00: N)(0.08 m) N)(0.6)](0.08 m

[(450

( [( . )

NN 0.8)](0.08 m 0)( ) = M Az

= 0

or M iA = - ◊( .16 20 N m) b

151


PROBLEM 4.123

The rigid L-shaped member ABC is supported by a ball-and-socket joint at A and by three cables. If a 2 kN load is applied at F, determine the tension in each cable.

SOLUTION

Free-Body Diagram:

In this problem: a = 500 mm

We have CD CD

BD

u ruu

u ruu

= - =

= - +

( (

( (

600 800 1000

1000 60

mm) mm) mm

mm)

j k

i 00 800 1414 21

1000 800

mm) mm) mm

mm) mm)

j k

i k

- =

= -

( .

( (

BD

BE Bu ruu

EE = 1280 63. mm

Thus T TCD

CDT

T TBD

BDT

CD CD CD

BD BD BD

= = -

= = -

u ruu

u ruu

( . . )

( .

0 6 0 8

0 7071

j k

i ++ -

= = -

0 4243 0 5657

0 7809 0 6247

. . )

( . . )

j k

i kT TBE

BETBE BE BE

u ruu

SM r T r T r T r WA C CD B BD B BE F= ¥ + ¥ + ¥ + ¥ =0 0: ( ) ( ) ( ) ( )

152



Noting that r i k

r k

r i k

C

B

F a

= - +== - +

( (

(

(

1000 800

800

800

mm) mm)

mm)

mm)

and using determinants, we write

i j k i j k

i j

--

+- -

+

1000 0 800

0 0 6 0 8

0 0 800

0 7071 0 4243 0 5657. . . . .

T TCD BD

kk i j k

0 0 800

0 7809 0 0 6247

0 800

0 2 0

0

. .-+ -

-=T aBE


i: - - + =480T TCD BD339 44 1600 0. (1)

j: - - + =800 565 68 624 72 0T T TCD BD BE. . (2)

k: - + =600 2 0T aCD (3)

Recalling that a = 500 mm, Eq. (3) yields

TCD = =2 500

600

5

3

( )kN TCD = 1 667. kN b

From (1): - ÊËÁ

ˆ¯̃

- + =4805

3339 44 1600 0. TBD

TBD = 2 35682. kN TBD = 2 36. kN b

From (2): - ÊËÁ

ˆ¯̃

- + =8005

3565 68 2 35682 624 72 0. ( . ) . TBE

TBE = 4 2684. kN TBE = 4 27. kN b

153


PROBLEM 4.124

Solve Problem 4.123, assuming that the 2 kN load is applied at C.

PROBLEM 4.123 The rigid L-shaped member ABC is supported by a ball-and-socket joint at A and by three cables. If a 2 kN load is applied at F, determine the tension in each cable.

SOLUTION

See solution of Problem 4.123 for free-body diagram and derivation of Eqs (1), (2), and (3):

- - + =480 339 44 1600 0T TCD BD. (1)

- - + =800 565 68 624 72 0T T TCD BD BE. . (2)

- + =600 2 0T aCD (3)

In this problem, the 2kN load is applied at C and we have a = 1000 mm. Carrying into (3) and solving for TCD ,

TCD = 10

3kN TCD = 3 33. kN b

From (1): - ÊËÁ

ˆ¯̃

- + =48010

3339 44 1600 0. TBD TBD = 0 b

From (2): - ÊËÁ

ˆ¯̃

- + = =80010

30 624 72 0. T TBE BEfi 4.2685 kN TBE = 4 27. kN b

154


PROBLEM 4.125

Frame ABCD is supported by a ball-and-socket joint at A and by three cables. For a = 150 mm, determine the tension in each cable and the reaction at A.

SOLUTION

First note Ti j

DG DG DG DGT T= =- +

+l

( . ( .

( . ) ( . )

0 48 0 14

0 48 0 142 2

m) m)

m

= - +

= +

0 48 0 14

0 50

2524 7

. .

.

( )

i j

i j

T

T

DG

DG

Ti k

BE BE BE BET T= =- +

+l

( . ( .

( . ) ( . )

0 48 0 2

0 48 0 22 2

m) m)

m

= - +

= - +

0 48 0 2

0 52

1312 5

. .

.

( )

i k

j k

T

T

BE

BE )


SM Tx DG= ÊËÁ

ˆ¯̃

- =07

250 3 350 0 15 0: m) N)( m)( . ( .

or TDG = 625 N b

SM Ty BE= ¥ÊËÁ

ˆ¯̃ - Ê

ËÁˆ¯̃

=024

25625 0 3

5

130 48 0: N m) m)( . ( .

or TBE = 975 N b

SM Tz CF= + ¥ÊËÁ

ˆ¯̃

- =0 0 147

25625 0 48 350 0 48 0: m) N m) N)( m)( . ( . ( .

or TCF = 600 N b

155



SF A T T Tx x CF BE x DG x= + + + =0 0: ( ) ( )

Ax - - ¥ÊËÁ

ˆ¯̃

- ¥ÊËÁ

ˆ¯̃

=600 97524

25625 0N

12

13N N

Ax = 2100 N

SF A Ty y DG y= + - =0 350 0: N( )

Ay + ¥ÊËÁ

ˆ¯̃

- =7

25625 350 0N N

Ay = 175 0. N

SF A Tz z BE z= + =0 0: ( )

Az + ¥ÊËÁ

ˆ¯̃

=5

13975 0N

Az = -375 N

Therefore A i j k= + -( ) ( . ) ( )2100 175 0 375 N N N b

156


PROBLEM 4.126

Frame ABCD is supported by a ball-and-socket joint at A and by three cables. Knowing that the 350-N load is applied at D (a = 300 mm), determine the tension in each cable and the reaction at A.

SOLUTION

First note Ti j

DG DG DG DGT T= =- +

+l

( . ( .

( . ) ( . )

0 48 0 14

0 48 0 142 2

m) m)

m

= - +

= +

0 48 0 14

0 50

2524 7

. .

.

( )

i j

i j

T

T

DG

DG

Ti k

BE BE BE BET T= =- +

+l

( . ( .

( . ) ( . )

0 48 0 2

0 48 0 22 2

m) m)

m

= - +

= - +

0 48 0 2

0 52

1312 5

. .

.

( )

i k

i k

T

T

BE

BE


SM Tx DG= ÊËÁ

ˆ¯̃

- =07

250 3 350 0 3 0: m) N)( m)( . ( .

or TDG = 1250 N b

SM Ty BE= ¥ÊËÁ

ˆ¯̃ - Ê

ËÁˆ¯̃

=024

251250 0 3

5

130 48 0: N m m)( . ) ( .

or TBE = 1950 N b

SM Tz CF= + ¥ÊËÁ

ˆ¯̃

- =0 0 147

251250 0 48 350 0 48 0: m) N m) N)( m)( . ( . ( .

or TCF = 0 b

157



SF A T T Tx x CF BE x DG x= + + + =0 0: ( ) ( )

Ax + - ¥ÊËÁ

ˆ¯̃

- ¥ÊËÁ

ˆ¯̃

=0 195024

251250 0

12

13N N

Ax = 3000 N

SF A Ty y DG y= + - =0 350 0: N( )

Ay + ¥ÊËÁ

ˆ¯̃

- =7

251250 350 0N N

Ay = 0

SF A Tz z BE z= + =0 0: ( )

Az + ¥ÊËÁ

ˆ¯̃

=5

131950 0N

Az = -750 N

Therefore A i k= -( (3000 750N) N) b

158


SOLUTION

From f.b.d. of weldment

SM r A r B r CO A O B O C O= ¥ + ¥ + ¥ =0 0: / / /

i j k i j k i j k

300 0 0

0

0 200 0

0

0 0 250

0

0

A A B B C Cy z x z x y

+ + =

( ) ( ) ( )- + + - + - + =300 300 200 200 250 250 0A A B B C Cz y z x y xj k i k i j

From i-coefficient 200 250 0B Cz y- =

or B Cz y= 1 25. (1)

j-coefficient - + =300 250 0A Cz x

or C Ax z= 1 2. (2)

k-coefficient 300 200 0A By x- =

or B Ax y= 1 5. (3)

SF A B C= + + - =0 0: P

or ( ) ( . ) ( )B C A C A Bx x y y z z+ + + - + + =i j k1 2 0kN

From i-coefficient B Cx x+ = 0

or C Bx x= - (4)

j-coefficient A Cy y+ - =1 2 0. kN

or A Cy y+ = 1 2. kN (5)

PROBLEM 4.127

Three rods are welded together to form a “corner” that is supported by three eyebolts. Neglecting friction, determine the reactions at A, B, and C when P = 1 2. kN , a b= =300 200mm, mm and c = 250 mm.

159



k-coefficient A Bz z+ = 0

or A Bz z= - (6)

Substituting Cx from Equation (4) into Equation (2)

- =B Az z1 2. (7)

Using Equations (1), (6), and (7)

CB A B B

yz z x x= =

-= Ê

ËÁˆ¯̃

=1 25 1 25

1

1 25 1 2 1 5. . . . . (8)

From Equations (3) and (8)

CA

C Ayy

y y= =1 5

1 5

.

.or

and substituting into Equation (5)

2 1 2Ay = . kN

A Cy y= = 600 N (9)

Using Equation (1) and Equation (9)

Bz = =1 25 600 750. ( )N N

Using Equation (3) and Equation (9)

Bx = =1 5 600 900. ( )N N

From Equation (4) Cx = -900 N

From Equation (6) Az = -750 N

Therefore A j k= -( ) ( )600 750N N b

B i k= +( ) ( )900 750N N b

C i j= - +( ) ( )900 600N N b

160


PROBLEM 4.128

Solve Problem 4.127, assuming that the force P is removed and is replaced by a couple M j= + ◊(75 N mm) acting at B.

PROBLEM 4.127 Three rods are welded together to form a “corner” that is supported by three eyebolts. Neglecting friction, determine the reactions at A, B, and C when P = 1 2. kN, a b= =300 200mm, mm, and c = 250 mm.

SOLUTION

From f.b.d. of weldment

SM r A r B r C MO A O B O C O= ¥ + ¥ + ¥ + =0 0: / / /

i j k i j k i j k

j300 0 0

0

0 200 0

0

0 0 250

0

75000 0

A A B B C Cy z x z x y

+ + + ◊ =( )N mm

( ) ( ) ( ) (- + + - + - + + ◊300 300 200 200 250 250 75000A A B B C Cz y z x y xj k j k i j N mmm)j = 0

From i-coefficient 200 250 0B Cz y- =

or C By z= 0 8. (1)

j-coefficient - + + =300 250 75 000 0A Cz x ,

or C Ax z= -1 2 300. (2)

k-coefficient 300 200 0A By x- =

or B Ax y= 1 5. (3)

SF A B C= + + =0 0:

( ) ( ) ( )B C A C A Bx x y y z z+ + + + + =i j k 0

From i-coefficient C Bx x= - (4)

j-coefficient C Ay y= - (5)

k-coefficient A Bz z= - (6)

161



Substituting Cx from Equation (4) into Equation (2)

AB

zx= - Ê

ËÁˆ¯̃

2501 2.

(7)

Using Equations (1), (6), and (7)

C B A By z z x= = - = ÊËÁ

ˆ¯̃

-0 8 0 82

3200. . (8)

From Equations (3) and (8)

C Ay y= - 200

Substituting into Equation (5) 2 200Ay =

Ay = 100 0. N

From Equation (5) Cy = -100 0. N

Equation (1) Bz = -125 0. N

Equation (3) Bx = 150 0. N

Equation (4) Cx = -150 0. N

Equation (6) Az = 125 0. N

Therefore A j k= +( . ) ( . )100 0 125 0N N b

B i k= -( . ) ( . )150 0 125 0N N b

C i j= - -( . ) ( . )150 0 100 0N N b

162


SOLUTION

From f.b.d. of pipe assembly ABCD

SF Bx x= =0 0:

SM BD x z( ) : ( )( . ) ( )-axis N m m= - =0 48 2 5 2 0

Bz = 60 0. N

and B k= ( . )60 0 N b

SM CD z y( ) : ( )-axis m N m= - ◊ =0 3 90 0

Cy = 30 0. N

SM CD y axis z( ) : ( ) ( . )( ) ( )( )- = - - + =0 3 60 0 4 48 4 0m N m N m

Cz = -16 00. N

and C j k= -( . ) ( . )30 0 16 00N N b

SF Dy y= + =0 30 0 0: .

Dy = -30 0. N

SF Dz z= - + - =0 16 00 60 0 48 0: . .N N N

Dz = 4 00. N

and D j k= - +( . ) ( . )30 0 4 00N N b

PROBLEM 4.129

In order to clean the clogged drainpipe AE, a plumber has disconnected both ends of the pipe and inserted a power snake through the opening at A. The cutting head of the snake is connected by a heavy cable to an electric motor that rotates at a constant speed as the plumber forces the cable into the pipe. The forces exerted by the plumber and the motor on the end of the cable can be represented by the wrench F k M k= - = - ◊( , ( .48 90N) N m) Determine the additional reactions at B, C, and D caused by the cleaning operation. Assume that the reaction at each support consists of two force components perpendicular to the pipe.

163


PROBLEM 4.130

Solve Problem 4.129, assuming that the plumber exerts a force F k= -(48 N) and that the motor is turned off ( ).M = 0

PROBLEM 4.129 In order to clean the clogged drainpipe AE, a plumber has disconnected both ends of the pipe and inserted a power snake through the opening at A. The cutting head of the snake is connected by a heavy cable to an electric motor that rotates at a constant speed as the plumber forces the cable into the pipe. The forces exerted by the plumber and the motor on the end of the cable can be represented by the wrench F k M= - = - ◊( , ( ) .48 90N) N m k Determine the additional reactions at B, C, and D caused by the cleaning operation. Assume that the reaction at each support consists of two force components perpendicular to the pipe.

SOLUTION

From f.b.d. of pipe assembly ABCD

SF Bx x= =0 0:

SM BD x z( ) : ( )( . ) ( )-axis N m m= - =0 48 2 5 2 0

Bz = 60 0. N and B k= ( . )60 0 N b

SM C BD z y x( ) : ( ) ( )-axis m 2 m= - =0 3 0

Cy = 0

SM CD y z( ) : ( ) ( . )( ) ( )( )-axis m N m N m= - + =0 3 60 0 4 48 4 0

Cz = -16 00. N and C k= - ( . )16 00 N b

SF D Cy y y= + =0 0:

Dy = 0

SF D B C Fz z z z= + + - =0 0:

Dz + - - =60 0 16 00 48. .N N N 0

Dz = 4 00. N and D k= ( . )4 00 N b

164


SOLUTION

r i j

T i j r j k

T j k r

E F

B B B F

C C C F

T

T

/

/

/

( ) /

( ) /

= - +

= - = -

= - + =

200 80

2 80 200

2 2000 80

2 80 200

i j

T i j r j k

+

= - + = +D D D ET ( ) / /

SM T PF B F B C F C D F D E F= ¥ + ¥ + ¥ + ¥ - =0 0: ( )/ / / /r T r T r r j

i j k i j k i j k i j k

0 80 200

1 1 02

200 80 0

0 1 12

0 80 200

1 1 02

2--

+-

+- -

+ -T T TB c D 000 80 0

0 0

0

-=

P

Equate coefficients of unit vectors to zero and multiply each equation by 2.

i : - + + =200 80 200 0T T TB C D (1)

j: - - - =200 200 200 0T T TB C D (2)

k : - - + + =80 200 80 200 2 0T T T PB C D (3)

80

2002 80 80 80 0( ): - - - =T T TB C D (4)

Eqs. ( ) ( ):3 4+ - - + =160 280 200 2 0T T PB C (5)

Eqs. ( ) ( ):1 2+ - - =400 120 0T TB C

T T TB C C= - -120

4000 3. (6)

PROBLEM 4.131

The assembly shown consists of an 80-mm rod AF that is welded to a cross consisting of four 200 mm arms. The assembly is supported by a ball-and-socket joint at F and by three short links, each of which forms an angle of 45∞ with the vertical. For the loading shown, determine (a) the tension in each link, (b) the reaction at F.

165



Eqs. ( )6 Æ ( ):5 - - - + =160 0 3 280 200 2 0( . )T T PC C

- + =232 200 2 0T PC

T PC = 1 2191. T PC = 1 219. b

From Eq. (6): T P PB = - = - =0 3 1 2191 0 36574. ( . ) . T PB = -0 366. b

From Eq. (2): - - - - =200 0 3657 200 1 2191 200 0( . ) ( . )P P T Dq

T PD = -0 8534. T PD = -0 853. b

SF = 0: F T T T j+ + + - =B C D P 0

i :( . ) ( . )

FP P

x +-

--

=0 36574

2

0 8534

20

F P F Px x= - = -0 3448 0 345. .

j:( . ) ( . ) ( . )

FP P P

y --

- --

- =0 36574

2

1 2191

2

0 8534

2200 0

F P F Py y= =

k :( . )

FP

z + =1 2191

20

F P F Pz z= - = -0 8620 0 862. . F i j k= - + -0 345 0 862. .P P P b

166


SOLUTION

Free-Body Diagram:

Five unknowns and six equations of equilibrium. But equilibrium is maintained ( )SMAB = 0

W mg

W

=

==

( ) .

.

10 9 81

98 1

kg m/s

N

2

GC GC

TGC

GC

T

u ruu

u ruu

= - + - =

= = - + -

300 200 225 425

425300 200

i j k

T i j

mm

( 2225

600 400 150

300 200 75

k

r i j

r i j

)

/

/

B A

G A

= - + +

= - + +

mm

mm

SM WA B A G A G A= ¥ + ¥ + ¥ - =0 0: ( )/ / /r B r T r j

i j k i j k i j k

- + -- -

+ -600 400 150

0 0

300 200 75

300 200 225425

300 200 75

0B

T

--98 1 0.

Coefficient of i : ( . . ) .- - + =105 88 35 29 7357 5 0T

T = 52 12. N T = 52 1. N b

PROBLEM 4.132

The uniform 10-kg rod AB is supported by a ball-and-socket joint at A and by the cord CG that is attached to the midpoint G of the rod. Knowing that the rod leans against a frictionless vertical wall at B, determine (a) the tension in the cord, (b) the reactions at A and B.

167



Coefficient of j : ( ).

150 300 75 300 22552 12

4250B - ¥ + ¥ =

B = 73 58. N B i= ( . )73 6 N b

SF A B T j= + + - =0 0: W

Coefficient of i : . .Ax + - =73 58 52 15300

4250 Ax = 36 8. N b

Coefficient of j: . .Ay + - =52 15200

42598 1 0 Ay = 73 6. N b

Coefficient of k : .Az - =52 15225

4250 Az = 27 6. N b

168


SOLUTION

Free-Body Diagram:

DF DF

TDE

DF

T

u ruu

u ruu

= - + - =

= = - + -

400 275 200 525

525400 275

i j k

T i j

mm

( 2200

400

400 350

175 600

625

k

r i

r i k

i k

)

/

/

D E

C E

EAEA

EA

=

= -

= = -lu ruu

S l lMEA EA D E EA C E= ◊ ¥ + ◊ ◊ - =0 300 0: ( ) ( ( ))/ /r T r j

175 0 600

400 0 0

400 275 200525 625

175 0 600

400 0 350

0 300 0

1

62

-

- -¥

+--

-

T

550

600 400 275

525 625

175 300 350 600 400 300

6250

880

=

- ¥ ¥¥

+ - ¥ ¥ + ¥ ¥ =

-

T

0000

753625000 0

426 5625

T

T

+ =

= . N T = 427 N b

PROBLEM 4.133

The bent rod ABDE is supported by ball-and-socket joints at A and E and by the cable DF. If a 300-N load is applied at C as shown, determine the tension in the cable.

169


PROBLEM 4.134

Solve Problem 4.133, assuming that cable DF is replaced by a cable connecting B and F.

PROBLEM 4.133 The bent rod ABDE is supported by ball-and-socket joints at A and E and by the cable DF. If a 300-N load is applied at C as shown, determine the tension in the cable.

SOLUTION

Free-Body Diagram:

r i

r i kB A

C A

/

/

=

= +

225

225 250

BF BF

TBF

BF

T

u ruu

u ruu

= - + + =

= = = -

400 275 400 628 987

25 1616

i j k

T i

.

.(

mm

++ +

= = -

11 16

7 24

j k

i k25

)

l AEAE

AE

u ruu

S lMAE AF B A AE C A= ◊ ¥ + ◊ ◊ - =0 300 0: ( ) ( ( ))/ /r T r jl

7 0 24

225 0 0

16 11 1625 25 16

7 0 24

225 0 250

0 300 0

1

250

-

-¥

+-

-=T

.

- ¥ ¥¥

+ ¥ ¥ + ¥ ¥24 225 11

25 25 16

24 225 300 7 250 300

25.T

2360 89 2 145 000 0. , ,T - = T = 909 N b

170


SOLUTION

Free-Body Diagram:

W mg

W

CE

CE

= ==

= - + -=

( )( . )

.

50 9 81

490 50

240 600 400

760

kg m/s

N

2

u ruu

i j k

mmm

T i j k

i j

= = - + -

= = -

TCE

CE

T

AB

ABAB

u ruu

u ruu

760240 600 400

480 200

( )

l5520

1

1312 5= -( )i j

S l lM r r jAB AB E A AB G AT W= ◊ ¥ + ◊ ¥ - =0 0: ( ) ( )/ /

r i j r i j kE A G A/ /;= + = - +240 400 240 100 200

12 5 0

240 400 0

240 600 40013 20

12 5 0

240 100 200

0 0

1

130

12

-

- -¥

+-

--

=

-

T

W

( ¥¥ ¥ - ¥ ¥ + ¥ =400 400 5 240 400760

12 200 0)T

W

T W= =0 76 0 76 490 50. . ( . )N T = 373 N b

PROBLEM 4.135

The 50-kg plate ABCD is supported by hinges along edge AB and by wire CE. Knowing that the plate is uniform, determine the tension in the wire.

171


PROBLEM 4.136

Solve Problem 4.135, assuming that wire CE is replaced by a wire connecting E and D.

PROBLEM 4.135 The 50-kg plate ABCD is supported by hinges along edge AB and by wire CE. Knowing that the plate is uniform, determine the tension in the wire.

SOLUTION

Free-Body Diagram:

Dimensions in mm

W mg

W

DE

DE

= ==

= - + -=

( )( . )

.

50 9 81

490 50

240 400 400

614

kg m/s

N

2

u ruu

i j k

..

.( )

5

614 5240 400 400

480 2

mm

T i j k

i

= = + -

= = -

TDE

DE

T

AB

ABAB

u ruu

u ruu

l 000

520

1

1312 5

ji j= -( )

r i j r i j kE A G A/ /;= + = - +240 400 240 100 200

12 5 0

240 400 0

240 400 40013 614 5

12 5 0

240 100 200

0 0

1

130

1

-

-¥

+ --

=

-

T

W.

( 22 400 400 5 240 400614 5

12 200 0¥ ¥ - ¥ ¥ + ¥ ¥ =).

TW

T W= =0 6145 0 6145 490 50. . ( . )N

T = 301 N b

172


SOLUTION

First note lBD =- - +

+ +

=

( ) ( ) ( )

( ) ( ) ( )

(

150 225 300

150 225 300

1

29

2 2 2

mm mm mm

mm

i j k

-- - +

= -

==

=

2 3 4

150

400

200

i j k

r i

P k

r i

C j

)

( )

( )

( )

( )

/

/

A B

C D

C

mm

N

mm

From the f.b.d. of the plates

S l lMBD BD A B BD C D= ◊ ¥ + ◊ ¥ =0 0: ( ) ( )/ /r rP C

- -- È

ÎÍ˘˚̇

+- -

ÈÎÍ

˘˚̇

=2 3 4

1 0 0

0 0 1

150 400

29

2 3 4

1 0 0

0 1 0

200

290

( ) ( )

(

C

-- + =3 150 400 4 200 0)( )( ) ( )( )C

C = 225 N or NC j= ( )225 b

PROBLEM 4.137

Two rectangular plates are welded together to form the assembly shown. The assembly is supported by ball-and-socket joints at B and D and by a ball on a horizontal surface at C. For the loading shown, determine the reaction at C.

173


SOLUTION

Free-Body Diagram:

AF AF

AF

G A

u ruu

= - - =

= - -

= -

1 2 0 6 1 2 1 8

1

32 2

0 6 0 31

. . . .

( )

. .

i j k

i j k

r i j

m

/

l

rr i j k

r iG A

B A

21 2 0 3 0 6

1 2

/

/

= - -

=

. . .

.

S l l lM TAF AF G A AF G A AF B A= ◊ ¥ - + ◊ ¥ - + ◊ ¥ =0 60 60 02: / / /( ( ) ( ( )) ( )r j r j r

2 1 2

0 6 0 3 0

0 60 0

1

3

2 1 2

1 2 0 3 0 6

0 60 0

1

3

- ---

+- -

- --

+ ◊ ¥ =. . . . . ( )l AF B Ar T/ 00

( . ) ( . . ) ( )2 0 6 601

32 0 6 60 2 1 2 60

1

30¥ ¥ + - ¥ ¥ + ¥ ¥ + ◊ ¥ =l AF B Ar T/

l lAF B A A F B A◊ ¥ = - ◊ ¥ = -( ) ( )r T T r/ / /or48 48 (1)

PROBLEM 4.138

Two 0 6 1 2. .¥ m plywood panels, each of weight 60 N, are nailed together as shown. The panels are supported by ball-and-socket joints at A and F and by the wire BH. Determine (a) the location of H in the xy plane if the tension in the wire is to be minimum, (b) the corresponding minimum tension.

174



Projection of T on ( )l AF B A¥ r / is constant. Thus, Tmin is parallel to

l AF B A¥ = - - ¥ = - +r i j k i j k/1

32 2 1 2

1

32 4 1 2( ) . ( . . )

Corresponding unit vector is 15

2( )- +j k

T Tmin ( )= - +21

5j k (2)

Eq. (1): T

T5

21

32 2 1 2 48

52

1

32 4 1 2

( ) ( ) .

( ) ( . .

- + ◊ - - ¥ÈÎÍ

˘˚̇

= -

- + ◊ - +

j k i j k i

j k j kk) = -48

T

T3 5

4 8 1 2 483 5 48

624 5( . . )

( )+ = - = - =

T = 53 6656. N

Eq. (2) T Tmin

min

( )

( )

( (

= - +

= - +

= - +

21

5

24 5 21

548 24

j k

j k

T j k N) N )

Since Tmin has no i component, wire BH is parallel to the yz plane, and x = 1 2. m

(a) x y= =1 200 2 40. . m; m b

(b) Tmin .= 53 7 N b

175


SOLUTION

Free-Body Diagram:

AF AF

AF

G A

u ruu

= - - =

= - -

= -

1 2 0 6 1 2 1 8

1

32 2

0 6 0 31

. . . .

( )

. .

i j k

i j k

r i j

m

/

l

rr i j k

r iG A

B A

21 2 0 3 0 6

1 2

/

/

= - -

=

. . .

.

S l l lM TAF AF G A AF G A AF B A= ◊ ¥ - + ◊ ¥ - + ◊ ¥ =0 60 60 02: / / /( ( ) ( ( )) ( )r j r j r

2 1 2

0 6 0 3 0

0 60 0

1

3

2 1 2

1 2 0 3 0 6

0 60 0

1

3

- ---

+- -

- --

+ ◊ ¥ =. . . . . ( )l AF B Ar T/ 00

( . ) ( . . ) ( )2 0 6 601

32 0 6 60 2 1 2 60

1

30¥ ¥ + - ¥ ¥ + ¥ ¥ + ◊ ¥ =l AF B Ar T/

l AF B A◊ ¥ = -( )r T/ 48 (1)

BH y BH yu ruu

= - + - = +1 2 1 2 2 88 2 1 2. . ( . )i j k /

Ti j k= = - + -

+T

BH

BHT

y

y

u ruu

1 2 1 2

32 2 1 2

. .

( ) /

PROBLEM 4.139

Solve Problem 4.138, subject to the restriction that H must lie on the y axis.

PROBLEM 4.138 Two 0 6 1 2. .¥ m plywood panels, each of weight 60 N, are nailed together as shown. The panels are supported by ball-and-socket joints at A and F and by the wire BH. Determine (a) the location of H in the xy plane if the tension in the wire is to be minimum, (b) the corresponding minimum tension.

176



Eq. (1): l AF B A T

y

T

y◊ ¥ =

- -

- - += -( ) .

. .( . )

r /

2 1 2

1 2 0 0

1 2 1 2

1

3 2 8848

2

( . . ) ( . )( . )

( . .- - = - ¥ + = +

+1 44 2 4 3 48 2 88 144

2 88

1 44 2 42 1 2

2 1 2

y T y Ty

y/

/

))

( . )= +

+ÊËÁ

ˆ¯̃

1002 88

153

2 12y

y (2)

dT

dy

yy y y

=

ÊËÁ

ˆ¯̃ + + + Ê

ËÁ-

0

53

12

2 88 2 2 8853

2 1 2 2 1 2

: 100

1+ / /( . ) ( ) ( . ) ˆ̂¯̃

ÏÌÓ

¸˝˛

+ÊËÁ

ˆ¯̃1

53

2y

Numerator = 0: 15

32 88

5

32+Ê

ËÁˆ¯̃

= +yy y( . )

5

3

24

5

5

32 2y y y+ = + y = 4 80. m b

Eq. (2): T = +

+ ¥ÊËÁ

ˆ¯̃

=100 2 88 4 8

15 4 8

3

56 5692 1 2( . . )

..

/

N Tmin .= 56 6 N b

177


PROBLEM 4.140

The pipe ACDE is supported by ball-and-socket joints at A and E and by the wire DF. Determine the tension in the wire when a 640-N load is applied at B as shown.

SOLUTION

Free-Body Diagram:

Dimensions in mm

AE

AE

AE

AEAE

u ruu

u ruu

= + -=

= = + -

480 160 240

560

480 160 240

5

i j k

i j k

mm

l660

6 2 3

7200

480 160

l AE

B A

D A

= + -

== +

i j k

r i

r i j/

/

DF DF

TDF

DFTDF DF DF

u ruu

u ruu

= - + - =

= = - +

480 330 240 630

480 3

i j k

Ti

; mm

330 240

630

16 11 8

21

j k i j k- = - + -TDF

S lM AE AE D A DF AE B A= - ¥ + - ¥ - =( ) ( ( ))r T r j/ /l 600 0

6 2 3

480 160 0

16 11 821 7

6 2 3

200 0 0

0 640 0

1

70

-

- -¥

+-

-=

TDE

- ¥ ¥ + ¥ ¥ - ¥ ¥ - ¥ ¥

¥+ ¥ ¥ =6 160 8 2 480 8 3 480 11 3 160 16

21 7

3 200 640

70TDF

- + ¥ =1120 384 10 03TDF

TDF = 342 86. N TDF = 343 N b

178


PROBLEM 4.141

Solve Problem 4.140, assuming that wire DF is replaced by a wire connecting C and F.

PROBLEM 4.140 The pipe ACDE is supported by ball-and-socket joints at A and E and by the wire DF. Determine the tension in the wire when a 640-N load is applied at B as shown.

SOLUTION

Free-Body Diagram:

Dimensions in mm

AE

AE

AE

AEAE

u ruu

u ruu

= + -=

= = + -

480 160 240

560

480 160 240

5

i j k

i j k

mm

l660

6 2 3

7200

480

480 490 240

l AE

B A

C A

CF C

= + -

==

= - + -

i j k

r i

r i

i j k

/

/u ruu

; FF

TCE

CFCF CF

=

= = - + -

726 70

480 490 240

726 70

.

.

mm

Ti j k

u ruu

S l lMAE AE C A CF AE B A= ◊ ¥ + ◊ ¥ - =0 600 0: ( ) ( ( ))r T r j/ /

6 2 3

480 0 0

480 490 240726 7 7

6 2 3

200 0 0

0 640 0

1

70

-

- + -¥

+-

-=

TCF

.

2 480 240 3 480 490

726 7 7

3 200 640

70

¥ ¥ - ¥ ¥¥

+ ¥ ¥ =.

TCF

- + ¥ =653 91 384 10 03. TCF TCF = 587 N b

179


SOLUTION

Free-Body Diagram:

W mg

a

= = == -

( ) .

( (

40 392 40

300 801

kg)(9.81 m/s N

mm)sin mm)cos

2

a aaa aa

a

b2 430 300

930

= -=

( (

(

mm)cos mm)sin

mm)cos

From free-body diagram of hand truck

Dimensions in mm

+ SM P b W a W aB = - + =0 02 1: ( ) ( ) ( ) (1)

+SF P W By = - + =0 2 2 0: (2)

For a = ∞35

a

a1 300 35 80 35 106 541

430 35 300 35 180

= ∞ - ∞ == ∞ - ∞ =

sin cos .

cos sin

mm

2 ..

cos .

162

930 35 761 81

mm

mmb = ∞ =(a) From Equation (1)

P( . . .761 81 392 40 392 40 mm) N(180.162 mm) N(106.54 mm) 0- + =

P = 37 921. N or P = 37 9. N ≠ b

(b) From Equation (2)

37 921 2 0. N 2(392.40 N)- + =B or B = 373 N ≠ b

PROBLEM 4.142

A hand truck is used to move two kegs, each of mass 40 kg. Neglecting the mass of the hand truck, determine (a) the vertical force P that should be applied to the handle to maintain equilibrium when a = ∞35 , (b) the corresponding reaction at each of the two wheels.

180


SOLUTION

(a) a = ∞0

From f.b.d. of member ABC

+ SM AC = + - =0 300 300 0 8 0: ( ( ( . N)(0.2 m) N)(0.4 m) m)

A = 225 N or A = 225 N ≠ b

+SF Cy y= + =0 225 0: N

Cy y= - =225 225N or NC Ø

+ SF Cx x= + + =0 300 300 0: N N

Cx x= - =600 600 N or NC ¨

Then C C Cx y= + = + =2 2 2 2600 225 640 80( ) ( ) . N

and q =ÊËÁ

ˆ¯̃

= --

ÊËÁ

ˆ¯̃

= ∞- -tan tan .1 1 225

60020 556

C

Cy

x

or C = 641 N 20 6. ∞ b

(b) a = ∞30

From f.b.d. of member ABC

+ SM A

AC = + - ∞

+0 300 300 30 0 8: ( ( ( cos )( .

( sin

N)(0.2 m) N)(0.4 m) m)

330 20 0∞ =)( )in.

A = 365 24. N or A = 365 N 60 0. ∞ b

+ SF Cx x= + + ∞ + =0 300 300 365 24 30 0: ( . sin N N N)

Cx = -782 62.

PROBLEM 4.143

Determine the reactions at A and C when (a) a = 0, (b) a = ∞30 .

181



+SF Cy y= + ∞ =0 365 24 30 0: ( . cos N)

Cy y= - =316 31 316. N or NC Ø

Then C C Cx y= + = + =2 2 2 2782 62 316 31 884 12( . ) ( . ) . N

and q =ÊËÁ

ˆ¯̃

= --

ÊËÁ

ˆ¯̃

= ∞- -tan tan.

..1 1 316 31

782 6222 007

C

Cy

x

or C = 884 N 22 0. ∞ b

182


SOLUTION

Free-Body Diagram:

Geometry:

xAC = ∞ =

= ∞ =( cos .

( sin .

250 20 234 923

250 20 85 505

mm) mm

mm) mmyAC

fi = - =yDA 300 85 505 214 495 mm mm mm. .

a =ÊËÁ

ˆ¯̃

= ÊËÁ

ˆ¯̃

= ∞- -tan tan.

..1 1 214 495

234 92342 397

y

xDA

AC

b = ∞ - ∞ - ∞ = ∞90 20 42 397 27 603. .

Equilibrium for lever:

(a) + SM TC AD= ∞ - ∞ =0 27 603 250 350 0: cos . ( (mm) N)[(375 mm)cos 20 ]

TAD = 556 703. N TAD = 557 N b

(b) + SF Cx x= + ∞ =0 556 703 42 397 0: ( . cos . N)

Cx = -411 12. N

+SF Cy y= - - ∞ =0 350 42 397 0: sin .N N)(556.703

Cy = 725 364. N

Thus: C C Cx y= + = - + =2 2 2 2411 12 725 364 833 77( . ) ( . ) . N

and q = = = ∞- -tan tan.

..1 1 725 364

411 1260 456

C

Cy

x

C = 834 N 60 5. ∞ b

PROBLEM 4.144

A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 350 N vertical force at B, determine (a) the tension in the cable, (b) the reaction at C.

183


SOLUTION

Free-Body Diagram:

Dimensions in mm

Geometry:

Distance AD = + =( . ) ( . ) .0 36 0 150 0 392 2 m

Distance BD = + =( . ) ( . ) .0 2 0 15 0 252 2 m

Equilibrium for beam:

(a) + SM T TC = - ÊËÁ

ˆ¯̃

- ÊËÁ

0 1200 15

0 390 36

0 15

0 25: (

.

.( .

.

. N)(0.28 m) m) ˆ̂

¯̃=( .0 2 0m)

T = 130 000. N

or T = 130 0. N b

(b) + SF Cx x= + ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

00 36

0 39130 000

0 2

0 25130 000:

.

.( . )

.

.( . N N)) = 0

Cx = -224 00. N

+SF Cy y= + ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

00 15

0 39130 00

0 15

0 25130 00:

.

.( .

.

.( . N) N) -- =120 0N

Cy = -8 0000. N

Thus: C C Cx y= + = - + -( ) =2 2 2 2224 8 224 14( ) . N

and q = = = ∞- -tan tan .1 1 8

2242 0454

C

Cy

x

C = 224 N 2 05. ∞ b

PROBLEM 4.145

Neglecting friction and the radius of the pulley, determine (a) the tension in cable ADB, (b) the reaction at C.

184


SOLUTION

Free-Body Diagram:

+SF Ey = ∞ - - =0 30 100 200 0: cos

E =∞

=300346 41

N

cos 30 N. E = 346 N 60 0. ∞ b

+ SM

C ED = -

- + ∞0 200

75 30 75

: (100 N)(100 mm) N)(100 mm)

mm) m

(

( sin ( mm) 0=

C = 39 8717. N C = 39 9. N Æ b

SF E C Dx = ∞ + - =0 30 0: sin

( . )( . ) .346 41 0 5 39 8717 0 N N+ - =D

D = 213 077. N D = 213 N ¨ b

PROBLEM 4.146

The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C, D, and E when q = ∞30 .

185


SOLUTION

Free-Body Diagram:

+SF Ey = - - =0 100 200 0: cosq

E = 300

cosq (1)

+ SM CD = - -0 200 75: (100 N)(100 mm) N)(100 mm) mm)

+300

cos

( (

sinq

qÊÊËÁ

ˆ¯̃

=75 0mm

C = - +400

3300 tanq

(a) For C = 0, 300400

3tanq =

tan .q q= = ∞4

923 962 q = ∞24 0. b

Eq. (1) E =∞

=300

23 962328 294

cos ..

+ SF D C Ex = - + + =0 0: sinq

D = =( . )sin . .328 294 23 962 133 33 N

(b) C D= =0 133 33. N ¨ E = 328 N 66 0. ∞ b

PROBLEM 4.147

The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine (a) the smallest value of u for which the equilibrium of the bracket is maintained, (b) the corresponding reactions at C, D, and E.

186


PROBLEM 4.148

For the frame and loading shown, determine the reactions at A and C.

SOLUTION

Since member AB is acted upon by two forces, A and B, they must be colinear, have the same magnitude, and be opposite in direction for AB to be in equilibrium. The force B acting at B of member BCD will be equal in magnitude but opposite in direction to force B acting on member AB. Member BCD is a three-force body with member forces intersecting at E. The f.b.d.’s of members AB and BCD illustrate the above conditions. The force triangle for member BCD is also shown. The angle b is found from the member dimensions:

b = ÊËÁ

ˆ¯̃

= ∞-tan .1 15030 964

mm

250 mm

Applying of the law of sines to the force triangle for member BCD,

150

45 135

N

sin( ) sin sin∞ -= =

∞b bB C

or 150

14 036 30 964 135

N

sin . sin . sin∞=

∞=

∞B C

A B= =∞

∞=

( )sin .

sin ..

150 30 964

14 036318 205

N N

or A = 318 2. N 45 0. ∞ b

and C = ∞∞

=( )sin

sin ..

150 135

14 036437 33

N N

or C = 437 3. N 59 0. ∞ b

187


SOLUTION

Free-Body Diagram: (Three-force body)

Reaction A must pass through Point D where 100-N forceand B intersect

In right D BCD

a = ∞ - ∞ = ∞

= ∞ =90 75 15

250 75 933 01BD tan . mm

Dimensions in mm

In right D ABD

tan

.

.

g

g

= =

=

AB

BD

150

933 01

9 13

mm

mm

Force Triangle

Law of sines

100

15 155 87163 1 257 6

N

9.13N; N

sin sin sin .. .

∞=

∞=

∞= =

A B

A B

A = 163 1. N 74.1° B = 258 N 65.0° b

PROBLEM 4.149

Determine the reactions at A and B when b = ∞50 .

188


SOLUTION

Free-Body Diagram:

Five unknowns and six equations of equilibrium, but equilibrium is maintained

( )SM AC = 0

r j

r jB

C

==

3

6

CF CF

BD BD

BE

u ruu

u ruu

u ruu

= - - + =

= - - =

= -

3 6 2 7

1 5 3 3 4 5

1 5 3

i j k

i j k

i

m

m. .

. jj k+ =3 4 5BE . m

P i j k

T i j k

= = - - +

= = - -

PCF

CE

P

TBD

BD

TBD BD

BD

u ruu

u ruu

73 6 2

4 51 5 3 3

( )

.( . )) ( )

( )

= - -

= = = - +

T

TBE

BE

T

BD

BE BEBD

32 2

32 2

i j k

T i j k

u ruu

SM

T T

A B BD B BE C

BD BE

= ¥ + ¥ + ¥ =

- -+

-+

0 0

0 3 0

1 2 23

0 3 0

1 2 23

: r T r T r P

i j k i j k i jj k

0 6 0

3 6 27

0

- -=P

Coefficient of i: - + + =2 212

70T T PBD BE (1)

Coefficient of k: - - + =T T PBD BF18

70 (2)

PROBLEM 4.150

The 6-m pole ABC is acted upon by a 455-N force as shown. The pole is held by a ball-and-socket joint at A and by two cables BD and BE. For a = 3 m, determine the tension in each cable and the reaction at A.

189



Eq. (1) + 2 Eq. (2): - + = =448

70

12

7T P T PBD BD

Eq. (2): - - + = =12

7

18

70

6

7P T P T PBE BE

Since P TBD= =44512

7455N ( ) TBD = 780 N b

TBE = 6

7455( ) TBE = 390 N b

SF T T P A= + + + =0 0: BD BE

Coefficient of i: 780

3

390

3

455

73 0+ - + =( ) Ax

260 130 195 0 195 0+ - + = =A Ax x . N

Coefficient of j: - - - + =780

32

390

32

455

76 0( ) ( ) ( ) Ay

- - - + = =520 260 390 0 1170A Ay y N

Coefficient of k: - + + + =780

32

390

32

455

72 0( ) ( ) ( ) Az

- + + + = = +520 260 130 0 130 0A Az z . N

A i j k= - + +( . ) ( ) ( . )195 0 1170 130 0N N N b

190


PROBLEM 4.151

Solve Problem 4.150 for a = 1.5 m.

PROBLEM 4.150 The 6-m pole ABC is acted upon by a 455-N force as shown. The pole is held by a ball-and-socket joint at A and by two cables BD and BE. For a m,= 3 determine the tension in each cable and the reaction at A.

SOLUTION

Free-Body Diagram:

Five unknowns and six equations of equilibrium but equilibrium is maintained

( )SM AC = 0

r j

r jB

C

==

3

6

CF CF

BD BD

BE

u ruu

u ruu

u ruu

= - - + =

= - - =

=

1 5 6 2 6 5

1 5 3 3 4 5

1

. .

. .

.

i j k

i j k

m

m

55 3 3 4 5i j k- + =BE . m

P i j k i j k

T

= = - - + = - - +

=

PCF

CE

P P

TBD

BD BD

u ruu

u ruu

6 51 5 6 2

133 12 4

.( . ) ( )

BBD

T T

TBE

BE

T

BD BD

BE BEBD

= - - = - -

= = =

4 51 5 3 3

32 2

3

.( . ) ( )

(

i j k i j k

T

u ruu

ii j k- +2 2 )

SM

T T

A B BD B BE C

BD BE

= ¥ + ¥ + ¥ =

- -+

-+

0 0

0 3 0

1 2 23

0 3 0

1 2 23

: r T r T r P

i j k i j k i jj k

0 6 0

3 12 413

0

- - +=P

Coefficient of i: - + + =2 224

130T T PBD BE (1)

Coefficient of k: - - + =T T PBD BE18

130 (2)

191



Eq. (1) + 2 Eq. (2): - + = =460

130

15

13T P T PBD BD

Eq (2): - - + = =15

13

18

130

3

13P T P T PBE BE

Since P TBD= =44515

13455N ( ) TBD = 525 N b

TBE = 3

13455( ) TBE = 105 0. N b

SF T T P A= + + + =0 0: BD BE

Coefficient of i: 525

3

105

3

455

133 0+ - + =( ) Ax

175 35 105 0 105 0+ - + = =A Ax x . N

Coefficient of j: - - - + =525

32

105

32

455

1312 0( ) ( ) ( ) Ay

- - - + = =350 70 420 0 840A Ay y N

Coefficient of k: - + + + =525

32

105

32

455

134 0( ) ( ) ( ) Az

- + + + = =350 70 140 0 140 0A Az z . N

A i j k= - + +( . ) ( ) ( . )105 0 840 140 0N N N b

192


PROBLEM 4.152

The rigid L-shaped member ABF is supported by a ball-and-socket joint at A and by three cables. For the loading shown, determine the tension in each cable and the reaction at A.

SOLUTION

Free-Body Diagram:

r i

r i k

r i k

r i k

r

B A

F A

D A

E A

F A

/

/

/

/

/

== -= -= -=

300

300 200

300 400

300 600

3000 800i k-

BG

BG

BG

u ruu

= - +== - +

300 225

375

0 8 0 6

i k

i k

mm

l . .

DH DH

FJ

DH

u ruuu

u ruu

= - + = = - +

= - +

300 400 500 0 6 0 8

300 400

i k i j

i

; ; . .mm l

kk i j; . .FJ FJ= = - +500 0 6 0 8mm; l

SM r T r T r T

r j rA B A BG BG DA DH DH F A FJ FJ

F A E A

= ¥ + ¥ + ¥+ ¥ - +

0

120

:

( )/ /

/ /

l l l¥¥ - =

-+ -

-+

( )

. . . .

120 0

300 0 0

0 8 0 0 6

300 0 400

0 6 0 8 0

3

j

i j k i j k i j k

T TBG DH 000 0 800

0 6 0 8 0

300 0 200

0 120 0

300 0 600

0 120 0

0

--

+ --

+ --

=

. .

TFJ

i j k i j k

Coefficient of i: + + - - =320 640 24000 72000 0T TDH FJ (1)

Coefficient of k: + + - - =240 240 36000 36000 0T TDH FJ (2)

34 Eq. (1) − Eq. (2):

( )

( )

1 2 300

2 3000

fi + =fi + =

¸˝˛

fi =T T

T TTDH FJ

DH FJFJ TFJ = 0 b

193



Eq. (1): TDH = 300 N TDH = 300 N b

Coefficient of j: - + ¥ =180 240 300 0TBG ( ) TBG = 400 N b

SF A j j= + + + - - =0 24 24 0: T T TBG BG DH DH FJl l

Coefficient of i: A Ax x+ - + - = =( )( . ) ( )( . )400 0 8 300 0 6 0 500 N

Coefficient of j: Ay + - - =( )( . )300 0 8 120 120 0 Ay = 0

Coefficient of k: Az + + =( )( . )400 0 6 0 Az = -240 N

A i j= -( ) ( )500 240N N b

Note: The value Ay = 0

Can be confirmed by considering SMBF = 0

194


PROBLEM 4.153

A force P is applied to a bent rod ABC, which may be supported in four different ways as shown. In each case, if possible, determine the reactions at the supports.

SOLUTION

(a) + SM P C a aA a= - + ∞ + ∞ =0 45 2 45 0: ( sin ) (cos )

32

2

3

CP C P= = C = 0 471. P 45° b

+ SF A P AP

x x x= -Ê

ËÁˆ

¯̃=0

2

3

1

2 3: Æ

+SF A P P AP

y y y= - +Ê

ËÁˆ

¯̃=0

2

3

1

2

2

3: ≠

A = 0 745. P 63.4° b

(b) + SM Pa A a A aC = + - ∞ + ∞ =0 30 2 30 0: ( cos ) ( sin )

A P A P( . . ) .1 732 0 5 0 812- = =

A = 0 812. P 60.0° b

+ SF P C C Px x x= ∞ + = = -0 0 812 30 0 0 406: ( . )sin . +SF P P C C Py y y= ∞ - + = = -0 0 812 30 0 0 297: ( . )cos .

C = 0 503. P 36.2° b

195



(c) + SM Pa A a A aC = + - ∞ + ∞ =0 30 2 30 0: ( cos ) ( sin )

A P A P( . . ) .1 732 0 5 0 448+ = =

A P= 0 448. 60.0° b

+ SF P C C Px x x= - ∞ + = =0 0 448 30 0 0 224: ( . )sin . Æ

+SF P P C C Py y y= ∞ - + = =0 0 448 30 0 0 612: ( . )cos . ≠

C = 0 652. P 69.9° b

(d) Force T exerted by wire and reactions A and C all intersect at Point D.

+ SM PD a= =0 0:

Equilibrium not maintained

Rod is improperly constrained b

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Transcript of CHAPTER 3CHAPTER 7CHAPTER 4 - · PDF fileCHAPTER 3CHAPTER 7CHAPTER 4. 3. ... or distributed...