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CHAPTER 3CHAPTER 3CHAPTER 3

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 3.1

A foot valve for a pneumatic system is hinged at B. Knowing that a = ∞28 , determine the moment of the 16-N force about Point B by resolving the force into horizontal and vertical components.

A

BC

a=28 °20 °

20 °

θ ° 16N

0.17m

x

yFy

FxSOLutiOn

Note that q a= - ∞ = ∞ - ∞ = ∞20 28 20 8

and F

Fx

y

= ∞ == ∞ =

( )cos .

( )sin .

16 8 15 8443

16 8 2 2268

N N

N N

Also x

y

= ∞ == ∞ =

( . )cos .

( . )sin .

0 17 20 0 159748

0 17 20 0 058143

m m

m mNoting that the direction of the moment of each force component about B is counterclockwise,

M xF yFB y x= +

=+

=

( . )( . )

( . )( . )

.

0 159748 2 2268

0 058143 15 8443

1 27

m N

m N

77 N m◊ or MB = ◊1 277. N m

4


PROBLEM 3.2

A foot valve for a pneumatic system is hinged at B. Knowing that a = ∞28 , determine the moment of the 16-N force about Point B by resolving the force into components along ABC and in a direction perpendicular to ABC.

SOLutiOn

First resolve the 4-lb force into components P and Q, where

Q = ∞=

( )sin

.

16 28

7 5115

N

N

Then M r QB A B==

/

m N( . )( . )0 17 7 5115

= ◊1 277. N m or MB = ◊1 277. N m

B

A

0.17 mC

16 NQ

P

α = 28°

5


PROBLEM 3.3

A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the smallest force applied at B that creates the same moment about D.

D

r

A25°

0.1 m300 N

0.2 m

Q

B

45°

D

0.2 m

0.2 m

0.28284 m

SOLutiOn

(a) F

F

x

y

= ∞== ∞

==

( )cos

.

( )sin

.

( .

300 25

271 89

300 25

126 785

271

N

N

N

N

F 889 126 785 N N) ( . )i j+

r i j

M r F

M i j

= = - -= ¥= - - ¥

DA

D

D

u ruu( . ) ( . )

[ ( . ) ( . ) ] [(

0 1 0 2

0 1 0 2 271

m m

m m .. ) ( . ) ]

( . ) ( . )

( .

89 126 785

12 6785 54 378

41 7

N N

N m N m

i j

k k

+= - ◊ + ◊= 000 N m◊ )k

MD = ◊41 7. N m

(b) The smallest force Q at B must be perpendicular to DBu ruu

at 45°

MD Q DB

Q

=◊ =

( )

. ( . )

u ruu

41 700 0 28284 N m m Q = 147 4. N 45°

6


PROBLEM 3.4

A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the magnitude and sense of the horizontal force applied at C that creates the same moment about D, (c) the smallest force applied at C that creates the same moment about D.

0.2 m

C

D

0.125 m aa

SOLutiOn

(a) See Problem 3.3 for the figure and analysis leading to the determination of MD

MD = ◊41 7. N m

0.2 m

0.125 m

D

C

r

C= Ci

(b) Since C is horizontal C i= C

r i j

M r i k

= = -= ¥ =

◊ =

DC

C CD

u ruu( . ) ( . )

( . )

. (

0 2 0 125

0 125

41 7 0

m m

m

N m .. )( )

.

125

333 60

m

N

C

C = C = 334 N (c) The smallest force C must be perpendicular to DC; thus, it forms a with the vertical

tan.

..

( ); ( . ) ( . )

.

a

a

=

= ∞

= = +=

0 125

0 232 0

0 2 0 125

0

2 2

m

m

m mMD C DC DC

223585 m

41 70 0 23585. ( . ) N m m◊ = C C = 176 8. N 58.0°

7


PROBLEM 3.5

A 150-N force acts on the end of the 1 m lever as shown. Determine the moment of the force about O.

SOLutiOn

First note P

P

x

y

= ∞== ∞

=

150 30

129 9

150 30

75

cos

.

sin

N

N

Noting that the direction of the moment of Px about O is clockwise and P

y about O

is counterclockwise,

M xP yPB y x= -

= -= - ◊

( . )( ) ( . )( . )

.

0 643 75 0 766 129 9

51 3 N m

or MB = 51.3 N · m

30°20°

P = 150 NPx

Py

o

1 m

x = 1 cos 50° = 0.643 m

y = 1 sin 50° = 0.766 m

50°

8


PROBLEM 3.6

For the shift lever shown, determine the magnitude and the direction of the smallest force P that has a 25 N · m clockwise moment about B.

A

B

0.6 m

0.2 m

Pα

θ

SOLutiOn

For P to be minimum, it must be perpendicular to the line joining Points A and B. Thus,

a q=

=

= ∞

-tan.

..

1 0 2

0 618 43

m

m

and M dPB = min

where d rA B=

= +=

/

m m

m

( . ) ( . )

.

0 2 0 6

0 632

2 2

Then Pmin .

.

=◊

=

25

0 632

39 6

N m

m

N Pmin .= 39 6 N 18.43°

9


PROBLEM 3.7

A 55-N force P is applied to a shift lever. The moment of P about B is clockwise and has a magnitude of 34 N · m. Determine the value of a.

A

θ

rA/B

B

0.2 m

0.6 m

P= 55 N

a

af

fSOLutiOn

By definition M r PB A B= / sinq

where q a f= + ∞ -( )90

and f = = ∞-tan.

..1 0 2

0 618 43

m

m

also rA B/ m m

m

= +=

( . ) ( . )

.

0 2 0 6

0 632

2 2

Then N m m N

or

34 0 632 55 90 18 43

71 57 0

◊ = + ∞ - ∞+ ∞ =

( . )( )sin( . )

sin( . ) .

aa 9978

71 57 77 96

71 57 102 04

aa

+ ∞ = ∞+ ∞ = ∞

. .

. .or a = ∞ ∞6 39 30 5. , .

10


PROBLEM 3.8

It is known that a vertical force of 1000 N is required to remove the nail at C from the board. As the nail first starts moving, determine (a) the moment about B of the force exerted on the nail, (b) the magnitude of the force P that creates the same moment about B if a = ∞10 , (c) the smallest force P that creates the same moment about B.

P

A

450 mm

FN=1000 N

BC

100 mm

70°

a

70°

q

Pa = 10°

rA/B

P

A a=20°

BC

70°

SOLutiOn

(a) We have M r FB C B N=

= = ◊= ◊

/

mm N N mm

N m

( )( )100 1000 10

100

5

or MB = ◊100 N m

(b) By definition M r PB A B== ∞ + ∞ - ∞= ∞

/ sin

( )

qq 10 180 70

120 Then 100 0 45 120N m m◊ = ¥ ∞( . ) sinP or P = 257 N

(c) For P to be minimum, it must be perpendicular to the line joining Points A and B. Thus, P must be directed as shown.

Thus M dP

d rB

A B

==

min

/

or 100 0 45N m m min◊ = ( . )P

or Pmin = 222 N Pmin = 222 N 20°

11


PROBLEM 3.9

A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 1040 N and length d is 1.90 m, determine the moment about D of the force exerted by the cable at C by resolving that force into horizontal and vertical components applied (a) at Point C, (b) at Point E.

1.90 m(b)

E

C

D12

13 5

TAByTAB

TABx

0.2 m(a)

1.90 m

0.875 mTABy

TABx

TAB= 1040 NC

E D

13

125

SOLutiOn

(a) Slope of line EC =+

=0 875

1 90 0 2

5

12

.

. .

m

m m

Then T TABx AB=

=

=

12

1312

131040

960

( )

( )N

N

and TABy =

=

5

131040

400

( )N

N

Then M T TD ABx ABy= -

= -

( . ) ( . )

( )( . ) ( )( . )

0 875 0 2

960 0 875 400 0 2

m m

N m N m

= ◊760 N m or MD = ◊760 N m

(b) We have M T y T xD ABx ABx= += += ◊

( ) ( )

( )( ) ( )( . )960 0 400 1 90

760

N N m

N m

or MD = ◊760 N m

12


PROBLEM 3.10

It is known that a force with a moment of 960 N · m about D is required to straighten the fence post CD. If d = 2.80 m, determine the tension that must be developed in the cable of winch puller AB to create the required moment about Point D.

SOLutiOn

E D

C

0.2 m

0.875 m

2.80 m

257

TAB

TABx

TABy

Slope of line EC =+

=0 875

2 80 0 2

7

24

.

. .

m

m m

Then T TABx AB= 24

25

and T TABy AB= 7

25

We have M T y T xD ABx ABy= +( ) ( )

96024

250

7

252 80

1224

N m m

N

◊ = +

=

T T

T

AB AB

AB

( ) ( . )

or TAB = 1224 N

13


PROBLEM 3.11

It is known that a force with a moment of 960 N ? m about D is required to straighten the fence post CD. If the capacity of winch puller AB is 2400 N, determine the minimum value of distance d to create the specified moment about Point D.

SOLutiOnC

q DE(TABmax)x

(TABmax)yTABmax

d

0.875 m

0.20 m

The minimum value of d can be found based on the equation relating the moment of the force TAB about D:

M T dD AB y= ( ) ( )max

where MD = ◊960 N m

( ) sin ( )sinmax maxT TAB y AB= =q q2400 N

Now sin.

( . ) ( . )

.

( . ) (

q =+ +

◊ =+ +

0 875

0 20 0 875

960 24000 875

0 20

2 2

2

m

m

N m N

d

d 00 875 2. )( )

È

ÎÍÍ

˘

˚˙˙

d

or ( . ) ( . ) .d d+ + =0 20 0 875 2 18752 2

or ( . ) ( . ) .d d+ + =0 20 0 875 4 78522 2 2

or 3 7852 0 40 8056 02. . .d d- - =Using the quadratic equation, the minimum values of d are 0.51719 m and - . .41151 m

Since only the positive value applies here, d = 0 51719. m

or d = 517 mm

14


PROBLEM 3.12

The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 625-N force directed along its centerline on the ball and socket at B, determine the moment of the force about A.

300 mm

300 mm

380 mm

58 mmB

C +

A

FCB

yx

θ

SOLutiOn

First note dCB = +=

( ) ( )

.

300 58

305 6

2 2mm mm

mm

Then cos.

sin

q

q

=

=

300

305 6

58

mm

mm

mm

305.6 mm

and FCB CB CBF F= -

= -

cos sin

.[( ) ( ) ]

q qi j

i j625

305 6300 58

N

mmmm mm

Now M r FA B A CB= ¥/

where r i j

i jB A/ mm mm mm

mm mm

= - += -

( ) ( )

( ) ( )

380 300 58

380 358

Then M i j i jA = - ¥ -

= - ◊

[( ) ( ) ].

( )380 358625

305 6300 58

45075

mm mmN

mmmm mm

N mmm N mm

174575 N mm

N m

i + ◊= ◊= ◊

219650

175

k

k

k or MA = ◊175 N m

15


PROBLEM 3.13

The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 625-N force directed along its centerline on the ball and socket at B, determine the moment of the force about A.

512 mm

110 mm

190 mm

430 mm

AB

FCBy

x

C +

SOLutiOn

First note dCB = +=

( ) ( )

.

430 190

470 1

2 2mm mm

mm

Then cos.

sin.

q

q

=

=

430

470 1

190

470 1

mm

mm

mm

mm

and F i j

i j

CB CB CBF F= -

= +

( cos ) ( sin )

.[( ) ( ) ]

q q625

470 1430 190

N

mmmm mm

Now M r FA B A CB= ¥/

where r i jB A/ mm) mm= -( ( )512 110

Then M i j i jA = - ¥ +

=

[( ) ( ) ]( .

( )512 110625

470 1430 190

129334

mm mmN

mm)mm mm

NN mm N mm

N mm

N m

◊ + ◊= ◊= ◊

k k

k

k

62886

192220

192 or MA = 192 N.m

16


120 mm

C

90 mm

72 mm

65 mm

FBx

FBy

SOLutiOn

We have M r FC B C B= ¥/

Noting the direction of the moment of each force component about C is clockwise.

MC By BxxF yF= ¥

where x = 120 mm - 65 mm = 55 mm

y = 72 mm + 90 mm = 162 mm

and F

F

Bx

By

=+

=+

65

65 72

72

65 72

2 2

2 2

( ) ( )

( ) ( )

(485N)=325N

(485N)=360 N

MC === ◊= ◊

(55mm)(360 N)+(162)(325N)

72450

72 450

N mm

N m.

or MC = ◊72 5. N m

PROBLEM 3.14

A mechanic uses a piece of pipe AB as a lever when tightening an alternator belt. When he pushes down at A, a force of 485 N is exerted on the alternator at B. Determine the moment of that force about bolt C if its line of action passes through O.

17


PROBLEM 3.15

Form the vector products B 3 C and B9 3 C, where B B9, and use the results obtained to prove the identity

sin cos sin ( ) sin ( ).a b a b a b= + + -1

2

1

2

SOLutiOn

Note: B i j

B i j

C i j

= +¢ = -

= +

B

B

C

(cos sin )

(cos sin )

(cos sin )

b bb ba a

By definition: | |B C¥ = -BC sin ( )a b (1)

| |¢ ¥ = +B C BC sin ( )a b (2)

Now B C i j i j¥ = + ¥ +B C(cos sin ) (cos sin )b b a a

= -BC(cos sin sin cos )b a b a k (3)

and ¢ ¥ = - ¥ +B C i j i jB C(cos sin ) (cos sin )b b a a

= +BC(cos sin sin cos )b a b a k (4)

Equating the magnitudes of B C¥ from Equations (1) and (3) yields:

BC BCsin( ) (cos sin sin cos )a b b a b a- = - (5)

Similarly, equating the magnitudes of ¢ ¥B C from Equations (2) and (4) yields:

BC BCsin( ) (cos sin sin cos )a b b a b a+ = + (6)

Adding Equations (5) and (6) gives:

sin( ) sin( ) cos sina b a b b a- + + = 2

or sin cos sin( ) sin( )a b a b a b= + + -1

2

1

2

18


PROBLEM 3.16

A line passes through the Points (20 m, 16 m) and (21 m, 24 m). Determine the perpendicular distance d from the line to the origin O of the system of coordinates.

SOLutiOn

dAB = - - + - -=

[ ( )] [ ( )]20 1 16 4

29

2 2 m m m m

m

Assume that a force F, or magnitude F(N), acts at Point A and is directed from A to B.

Then, F = F ABl

where lABB A

ABd=

-

= +

r r

1

2921 20( )i j

By definition M r FO A dF= ¥ =| |

where r i jA = - -( ) ( )1 4 m m

Then M i j i j

k

OF

F

= - - - ¥ +

= - +

[ ( ) ( ) ] [( ) ( ) ]

[ ( ) (

1 429

21 20

2920 84

m m m

m m

)) ]k

k= ÊËÁ

ˆ¯̃

◊64

29F N m

Finally 64

29F d FÊ

ËÁˆ¯̃

= ( )

d = 64

29 m d = 2 21. m

+

x

yB (20 m, 16 m)

A (–1 m, –4 m)d

F

rB

rA

0

+

lAB

19


PROBLEM 3.17

The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when (a) P 2 7i 3j 2 3k and Q 2i 2j 5k, (b) P 6i 2 5j 2 2k and Q 2 2i 5j 2 k.

SOLutiOn

(a) We have A = ¥| |P Q

where P i j k= - + -7 3 3

Q i j k= + +2 2 5

Then P Q

i j k

i j k

i j

¥ = - -

= + + - + + - -= + -

7 3 3

2 2 5

15 6 6 35 14 6

21 29 2

[( ) ( ) ( ) ]

( ) ( ) ( 00)k

A = + + -( ) ( ) ( )20 29 202 2 2 or A = 41 0.

(b) We have A = ¥| |P Q

where P i j k= - -6 5 2

Q i j k= - + -2 5 1in.

Then P Q

i j k

i j k

i j

¥ = - -- -

= + + + + -= + +

6 5 2

2 5 1

5 10 4 6 30 10

15 10 2

[( ) ( ) ( ) ]

( ) ( ) ( 00)k

A = + +( ) ( ) ( )15 10 202 2 2 or A = 26 9.

20


PROBLEM 3.18

A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal to, respectively, (a) i 2j 2 5k and 4i 2 7j 2 5k, (b) 3i 2 3j 2k and 2 2i 6j 2 4k.

SOLutiOn

(a) We have l = ¥¥

A BA B| |

where A i j k= + -1 2 5

B i j k= - -4 7 5

Then A B

i j k

i j k

i j k

¥ = + -- -

= - - + + + - -= - -

1 2 5

4 7 5

10 35 20 5 7 8

15 3 1 1

( ) ( ) ( )

( )

and | | ( ) ( ) ( )A B¥ = - + - + - =15 3 1 1 15 112 2 2

l = - - -15 3 1 1

15 11

( )i j k or l = - - -1

113( )i j k

(b) We have l = ¥¥

A BA B| |

where A i j k= - +3 3 2

B i j k= - + -2 6 4

Then A B

i j k

i j k

j k

¥ = -- -

= - + - + + -= +

3 3 2

2 6 4

12 12 4 12 18 6

8 12

( ) ( ) ( )

( )

and | |A B¥ = + =4 2 3 4 132 2( ) ( )

l = +4 2 3

4 13

( )j k or l = +1

132 3( )j k

21


PROBLEM 3.19

Determine the moment about the origin O of the force F 4i 5j 2 3k that acts at a Point A. Assume that the position vector of A is (a) r 2i 2 3j 4k, (b) r 2i 2.5j 2 1.5k, (c) r 2i 5j 6k. The components of F and r are in N and m respectively.

SOLutiOn

(a) M

i j k

O = --

2 3 4

4 5 3

= - + + + +( ) ( ) ( )9 20 16 6 10 12i j k M i j k O = - + +11 22 22 N m◊

(b) M

i j k

O = --

2 2 5 1 5

4 5 3

. .

= - + + - + + -( . . ) ( ) ( )7 5 7 5 6 6 10 10i j k MO = 0

(c) M

i j k

O =-

2 5 6

4 5 3

= - - + + + -( ) ( ) ( )15 30 24 6 10 20i j k M i j k O = - + -45 30 10 N m◊

Note: The answer to Part b could have been anticipated since the elements of the last two rows of the determinant are proportional.

22


PROBLEM 3.20

Determine the moment about the origin O of the force F 2 2i 3j 5k that acts at a Point A. Assume that the position vector of A is (a) r i j k, (b) r 2i 3j 2 5k, (c) r 2 4i 6j 10k. The components of F and r are in N and m respectively.

SOLutiOn

(a) M

i j k

O =-1 1 1

2 3 5

= - + - - + +( ) ( ) ( )5 3 2 5 3 2i j k M i j kO = 2 7 + 5 N m- ◊

(b) M

i j k

O = --2 3 5

2 3 5

= + + - + +( ) ( ) ( )15 15 10 10 6 6i j k M i kO = 30 +12 N m◊

(c) M

i j k

O = --4 6 10

2 3 5

= - + - + + - +( ) ( ) ( )30 30 20 20 12 12i j k MO = 0

Note: The answer to Part c could have been anticipated since the elements of the last two rows of the determinant are proportional.

23


PROBLEM 3.21

A 200-N force is applied as shown to the bracket ABC. Determine the moment of the force about A.

SOLutiOn

We have M r FA C A C= ¥/

where r i j

F j kC A

C

/ m m

N N

= += - ∞ + ∞

( . ) ( . )

( )cos ( )sin

0 06 0 075

200 30 200 30

Then M

i j k

i

A =- ∞ ∞

= ∞ -

200 0 06 0 075 0

0 30 30

200 0 075 30 0 06

. .

cos sin

[( . sin ) ( . ssin ) ( . cos ) ]30 0 06 30∞ - ∞j k

or M i j kA = ◊ - ◊ - ◊( . ) ( . ) ( . )7 50 6 00 10 39 N m N m N m

24


PROBLEM 3.22

The 6-m boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a Point C located on the vertical wall. If the tension in the cable is 2.5 kN, determine the moment about A of the force exerted by the cable at B.

SOLutiOn

First note dBC = - + + -=

( ) ( . ) ( )

.

6 2 4 4

7 6

2 2 2

m

Then T i j kBC = - + -2 5

7 66 2 4 4

.

.( . )

kN

We have M r TA B A BC= ¥/

where r iB A/ m= ( )6

Then M i i j kA = ¥ - + -( ).

.( . )6

2 5

7 66 2 4 4 m

kN

or M j kA = ◊ + ◊( . ) .7 89 4 74 kN m ( kN m)

25


PROBLEM 3.23

A wooden board AB, which is used as a temporary prop to support a small roof, exerts at Point A of the roof a 300-N force directed along BA. Determine the moment about C of that force.

SOLutiOn

We have M r FC A C BA= ¥/

where r i j kA C/ mm mm mm= - +( ) ( ) ( )1200 150 900

and F

i j k

BA BA BAF=

=- + -

+ +

l

( ) ( ) ( )

( ) ( ) ( )

125 2250 750

125 2250 302 2

mm mm mm22

300

16 64 299 5 99 84

È

ÎÍÍ

˘

˚˙˙

= - + -

( )

( . ) ( . ) ( . )

N

N N Ni j k

M

i j k

i

C = -- -

= - +

1200 150 900

16 64 299 5 99 84

254574 1048

. . .

( ) (

N m

N mm

◊

◊ 332 356904N mm N mm◊ ◊) ( )j k+

or M i j kC = - + +( ) ( . ) ( )255 104 8 357N m N m N m◊ ◊ ◊

1200 mm

150 mm

900 mm

A

x

y

z

FBA

rA/C D

C

26


PROBLEM 3.24

The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 810 N. Determine the moment about A of the force exerted by (a) the cable at D, (b) the cable at C.

SOLutiOn

(a) We have M r TA E A DE= ¥/

where r j

T

i j k

E A

DE DE DET/ m

m m m

==

=+ -

+ +

( . )

( . ) ( . ) ( )

( . ) ( . )

2 3

0 6 3 3 3

0 6 3 32 2

l

(( )( )

( ) ( ) ( )

3810

108 594 540

2 mN

N N N= + -i j k

M

i j k

i k

A =-

◊

= - ◊ - ◊

0 2 3 0

108 594 540

1242 248 4

.

( ) ( . )

N m

N m N m

or M i kA = - ◊ - ◊( ) ( )1242 248N m N m

(b) We have M r TA G A CG= ¥/

where r i j

T

i j k

G A

CG CG CGT/ m m

m m m

= +=

=- + -

( . ) ( . )

( . ) ( . ) ( )

( . )

2 7 2 3

0 6 3 3 3

0 6

l

22 2 23 3 3810

108 594 540

+ += - + -

( . ) ( )( )

( ) ( ) ( )

mN

N N Ni j k

M

i j k

i j

A =- -

◊

= - ◊ + ◊ + ◊

2 7 2 3 0

108 594 540

1242 1458 1852

. .

( ) ( ) (

N m

N m N m N mm)k

or M i j kA = - ◊ + ◊ + ◊( ) ( ) ( )1242 1458 1852N m N m N m

27


PROBLEM 3.25

A small boat hangs from two davits, one of which is shown in the figure. The tension in line ABAD is 410 N. Determine the moment about C of the resultant force R

A exerted on the davit at A.

SOLutiOn

We have R F FA AB AD= +2

where F jAB = -( )410 N

and F Fi j k

F

AD AD

AD

AD

AD= = - -

+ +=

u ruu( )

.

( ) ( . ) ( )

(

4102 5

2 2 25 1

258

2 2 2N

(2 )

.. ) ( . ) ( . )5 323 1 129 25N N Ni j k- -

Thus R F F i j kA AB AD= + = - -2 258 5 1143 1 129 25( . ) ( . ) ( . )N N N

Also r j kA C/ m m= +( . ) ( )2 5 1

Using Eq. (3.21): M

i j k

i j

C =- -

= + -

0 2 5 1

258 5 1143 1 129 25

820 259 646

.

. . .

( ) ( ) (N m N m N m◊ ◊ ◊ ))k

M i j kC = + -( ) ( ) (820 259 646N m N m N m)◊ ◊ ◊

28


PROBLEM 3.26

In Sample Problem 3.5, determine the perpendicular distance from Point O to cable AB.

SAMPLE PROBLEM 3.5 Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tensions in cables AB and BC are 555 N and 660 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B.

SOLutiOn

We have | |MO BAT d=

where d O AB= perpendicular distancefrom to line .

Now M r TO B O BA= ¥/

and r jB O/ m= ( )7

T

i j kBA BA ABT=

=- - +

+ += -

l( . ) ( ) ( )

( . ) ( ) ( )( )

0 75 7 6

0 75 7 6555

2 2 2

m m m

mN

(( . ) ( ) ( )

( .

45 0 420 360

0 7 0

45 420 360

2520 0

N N N

N m

N m

i j k

M

i j k

- +

=- -

◊

= ◊

O

)) ( . )i k+ ◊315 00 N m

and | | ( . ) ( . )

.

MO = +

= ◊

2520 0 315 00

2539 6

2 2

N m

2539 6. N m (555 N)◊ = d

or d = 4 5759. m or d = 4 58. m

7 mrB/O TBA

6 m

0.75 m

B

A

y

x

z

d0

29


PROBLEM 3.27

In Sample Problem 3.5, determine the perpendicular distance from Point O to cable BC.

SAMPLE PROBLEM 3.5 Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tensions in cables AB and BC are 555 N and 660 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B.

SOLutiOn

We have | |MO BCT d=

where d O BC= perpendicular distance from to line .

M r T

r j

T

i j k

O B O BC

B O

BC BC BCT

= ¥==

=- +

+

/

/ m

m m m

7

4 25 7 1

4 25 2

l( . ) ( ) ( )

( . ) (( ) ( )( )

( ) ( ) ( )

7 1660

340 560 80

0 7 0

340 560 80

2 2+= - +

=-

mN

N N Ni j k

M

i j k

O

== ◊ - ◊( ) ( )560 2380N m N mi k

and | | ( ) ( )

.

MO = += ◊

560 2380

2445 0

2 2

N m

2445 0

3 7045

.

.

N m (660 N)

m

◊ ==

d

d or d = 3 70. m

d

C

B

x

y

z1 m

4.25 m

7 m

0

TBC

rB/O

30


PROBLEM 3.28

In Problem 3.23, determine the perpendicular distance from Point D to a line drawn through Points A and B.

PROBLEM 3.23 A wooden board AB, which is used as a temporary prop to support a small roof, exerts at Point A of the roof a 300 N force directed along BA. Determine the moment about C of that force.

SOLutiOn

We have | |MD BAF d=

where d D AB= perpendicular distance from to line .

M r F

r j k

F

i

D A D BA

A D

BA BA BAF

= ¥= - +=

=- +

/

/ mm mm

mm

( ) ( )

( ) (

150 900

125 22

l

550 750

125 2250 30300

16 64

2 2 2

mm mmN

) ( )

( ) ( ) ( )( )

( .

j k-

+ +

È

ÎÍÍ

˘

˚˙˙

= - NN N N) ( . ) ( . )i j k+ -299 5 99 84

M

i j k

i

D = -- -

= - -

0 150 900

16 64 299 5 99 84

254574 14976

. . .

( ) (

N m

N mm N

◊

◊ ◊ mmm N mm) ( )j k- 2496 ◊

and | | ( ) ( ) ( )Muru

D = + +=

254574 14976 2496

255026

2 2 2

N mm◊

255026

850

N mm (300 N)

mm

◊ ==

d

d or d = 850 mm

1200 mm

150 mm

900 mm

A

x

y

z

FBA

rA/D D

C

31


PROBLEM 3.29

In Problem 3.23, determine the perpendicular distance from Point C to a line drawn through Points A and B.

PROBLEM 3.23 A wooden board AB, which is used as a temporary prop to support a small roof, exerts at Point A of the roof a 300 N force directed along BA. Determine the moment about C of that force.

SOLutiOn

We have | |MC BAF d=

where d C AB= perpendicular distance from to line .

M r F

r i j kC A BA

A C

= ¥= - +

/

/ mm mm mmC

( ) ( ) ( )1200 150 900

F

i j k

BA BA BAF=

=- + -

+ +

l

( ) ( ) ( )

( ) ( ) ( )

125 2250 750

125 2250 302 2

mm mm mm22

300

16 64 299 5 99 84

È

ÎÍÍ

˘

˚˙˙

= - + -

( )

( . ) ( . ) ( . )

N

N N Ni j k

M

i j k

i

C = -- -

= - +

1200 150 900

16 64 299 5 99 84

254574 1048

. . .

( ) (

N m

N mm

◊

◊ 332 356904N mm N mm◊ ◊) ( )j k+

and | | ( ) ( ) ( )Muru

C = + +=

254574 104832 356904

450753

2 2 2

N mm◊

450753 300

1503

==

( )N d

mmd or d = 1 503. m

1200 mm

150 mm

900 mm

A

x

y

z

FBA

rA/C D

C

32


PROBLEM 3.30

In Problem 3.24, determine the perpendicular distance from Point A to portion DE of cable DEF.

PROBLEM 3.24 The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 810 N. Determine the moment about A of the force exerted by (a) the cable at D, (b) the cable at C.

SOLutiOn

We have | |MA DET d=

where d = perpendicular distance from A to line DE.

M

jA E A DE

E A

DE DE DE

= ¥==

=+ -

r T

r

T T

i j k

/

/ m

m m m

( . )

( . ) ( . ) ( )

( .

2 3

0 6 3 3 3

0

l

66 3 3 3810

108 594 540

0 2 3 0

2 2 2) ( . ) ( )( )

( ) ( ) ( )

.

+ += + -

=

mN

N Ni j k

M

i j k

N

A

1108 594 540

1242 00 248 00

N m

N m N m

◊

= - ◊ - ◊( . ) ( . )i k

and | |

N m

MA = += ◊

( . ) ( . )

.

1242 00 248 00

1266 52

2 2

1266 52 810

1 56360

. ( )

.

N m N

m

◊ ==

d

d or d = 1 564. m

y

E

A

D1 m x

TDE

rE/A

3 m0.6 m

d

2.3 m

33


PROBLEM 3.31

In Problem 3.24, determine the perpendicular distance from Point A to a line drawn through Points C and G.

PROBLEM 3.24 The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 810 N. Determine the moment about A of the force exerted by (a) the cable at D, (b) the cable at C.

SOLutiOn

We have | |MA CGT d=

where d = perpendicular distance from A to line CG.

M r TA G A CG= ¥/

r i j

T

i j k

G A

CG CG CGT/ m m

m m m

= +=

=- + -

( . ) ( . )

( . ) ( . ) ( )

( . )

2 7 2 3

0 6 3 3 3

0 6

l

22 2 23 3 3810

108 594 540

2 7 2

+ += - + -

=

( . ) ( )( )

( ) ( ) ( )

. .

m N

N N Ni j k

M

i j k

A 33 0

108 594 540

1242 00 1458 00 1852 00

- -◊

= - ◊ + ◊ + ◊

N m

N m N m N m( . ) ( . ) ( .i j ))k

and | |

N m

MA = + += ◊

( . ) ( . ) ( . )

.

1242 00 1458 00 1852 00

2664 3

2 2 2

2664 3 810

3 2893

. ( )

.

N m N

m

◊ ==

d

d or d = 3 29. m

3 mC

TCG

rG/A

2.7 m

Az

1 m x

d

B

Ey

6

0.6 m

23 m

34


PROBLEM 3.32

In Problem 3.25, determine the perpendicular distance from Point C to portion AD of the line ABAD.

PROBLEM 3.25 A small boat hangs from two davits, one of which is shown in the figure. The tension in line ABAD is 410 N. Determine the moment about C of the resultant force R

A exerted on the davit

at A.

SOLutiOn

First compute the moment about C of the force FDA exerted by the line on D:

From Problem 3.25: F F

i j k

M r F

i

DA AD

C D C DA

= -

= - + +

= ¥

= +

( . ) ( . ) ( . )

( )

258 5 323 1 129 25

2

N N N

m

/

¥¥ - + +

= - +

[ ( . ) ( . ) ( . ) ]

( . ) ( .

258 5 323 1 129 25

258 5 646 2

N N N

N m N m

i j k

j◊ ◊ ))

( . ) ( . )

.

k

MC = +

=

258 5 646 2

695 99

2 2

N m◊

Then M FC DAd=

Since FDA = 410 N

dM

FC

DA

=

=695 99

410

. N m

N

◊ d = 1 698. m

35


PROBLEM 3.33

Determine the value of a that minimises the per pendicular distance from Point C to a section of pipeline that passes through Points A and B.

SOLutiOn

Assuming a force F acts along AB,

| | | |M r FC A C F d= ¥ =/ ( )

where d = perpendicular distance from C to line AB

F

i j k

=

=+ -

+ +

=

l ABF

F

( ) ( . ) ( . )

( ) ( . ) ( . )

( .

8 7 32 9 05

8 7 32 9 05

0 566

2 2 2

m m m

44 0 5183 0 6407

1 3 05 3 05

1

i j k

r i j k

M

i j k

+ -

= - - -

= -

. . )

( ) ( . ) ( . )A C

C

a/ m m m

33 05 3 05

0 5664 0 5183 0 6407

0 37332 0 5183 2 368

. .

. . .

[( . . ) ( .

--

= + +

a F

a i 222 0 5664 2 24582- +. ) ( . ) ]a Fj k

Since | | | | or | |/ /M r F r FC A C A C dF= ¥ ¥ =2 2 2( )

( . . ) ( . . ) ( . )0 3733 0 5183 2 368 0 5664 2 2462 2 2 2+ + - + =a a d

Setting dda d( )2 0= to find a to minimize d

[ ( . )( . . ) ( . )( . . )]2 0 5183 0 3733 0 5183 2 0 5664 2 368 0 5664 0+ + - - =a a

Solving a = 1 947. m or a = 1 947. m

AC

By

F

rA/C

α

36


PROBLEM 3.34

Given the vectors P i j k= - +3 2 , Q i j k= + -4 5 3 , and S i j k= - + -2 3 , compute the scalar products P ? Q, P ? S, and Q ? S.

SOLutiOn

P Q i j k i j k◊ = - + ◊ - -= + - - + -=

( ) ( )

( )( ) ( )( ) ( )( )

3 1 2 4 5 3

3 4 1 5 2 3

1 or P Q◊ = 1

P S i j k i j k◊ = - + ◊ - + -= - + - + -= -

( ) ( )

( )( ) ( )( ) ( )( )

3 1 2 2 3 1

3 2 1 3 2 1

11 or P S◊ = -11

Q S i j k i j k◊ = - - ◊ - + -= - + + - -=

( ) ( )

( )( ) ( )( ) ( )( )

4 5 3 2 3 1

4 2 5 3 3 1

10 or Q S◊ = 10

37


PROBLEM 3.35

Form the scalar products B C◊ and ¢B C× , where B B= ¢, and use the results obtained to prove the identity

cos cos cos( ) cos( ).a b a b a b= + + -1

2

1

2

x

yC

B

B

α – β

α + β

SOLutiOn

By definition B C◊ = -BC cos( )a b

where B

C i j

= += +

B

C

[(cos ) (sin ) ]

[(cos ) (sin ) ]

b ba a

i j

( cos )( cos ) ( sin )( sin ) cos( )B C B C BCb a b a a b+ = -

or cos cos sin sin cos( )b a b a a b+ = - (1)

By definition ¢ = +B C◊ BC cos( )a b

where ¢ = -B i j[(cos ) (sin ) ]b b

( cos )( cos ) ( sin )( sin ) cos( )B C B C BCb a b a a b+ - = +

or cos cos sin sin cos( )b a b a a b- = + (2)

Adding Equations (1) and (2),

2cos cos cos( ) cos( )b a a b a b= - + +

or cos cos cos( ) cos( )a b a b a b= + + -1

2

1

2

38


PROBLEM 3.36

Section AB of a pipeline lies in the yz plane and forms an angle of 37° with the z axis. Branch lines CD and EF join AB as shown. Determine the angle formed by pipes AB and CD.

SOLutiOn

First note AB AB

CD

u ruu= ∞ - ∞= - ∞ ∞ + ∞ -

(sin cos )

( cos cos sin cos

37 37

40 55 40

j k

CD j j 440 55∞ ∞sin )k

C

D

40°

Y

XZ

(CD)xz

(CD)y

(CD)z55°

(CD)x

(CD)

Now AB CD AB CDu ruu u ruu

◊ = ( )( )cos q

or AB CD(sin cos ) ( cos cos sin cos sin37 37 40 55 40 40 55∞ - ∞ ◊ - ∞ ∞ + ∞ - ∞ ∞j k i j kk)

= ( )( )cosAB CD q

or cos (sin )(sin ) ( cos )( cos sin )

.

q = ∞ ∞ + - ∞ - ∞ ∞=

37 40 37 40 55

0 88799 or q = ∞27 4.

39


PROBLEM 3.37

Section AB of a pipeline lies in the yz plane and forms an angle of 37° with the z axis. Branch lines CD and EF join AB as shown. Determine the angle formed by pipes AB and EF.

SOLutiOn

First note AB AB

EF EF

u ruuu ruu

= ∞ - ∞

= ∞ ∞ + ∞ -

(sin cos )

(cos cos sin

37 37

32 45 32

j k

i j ccos sin )32 45∞ ∞k

(EF)x

(EF)y(EF)xz(EF)z

EF

y

E

zx45°

32°

F

Now AB EF AB EFu ruu u ruu

◊ = ( )( )cos q

or AB EF(sin cos ) (cos cos sin cos sin37 37 32 45 32 32 45∞ - ∞ ◊ ∞ ∞ + ∞ - ∞ ∞j k j j k))

= ( )( )cosAB EF q

or cos (sin )(sin ) ( cos )( cos sin )

.

q = ∞ ∞ + - ∞ - ∞ ∞=

37 32 37 32 45

0 79782 or q = ∞37 1.

40


PROBLEM 3.38

Consider the volleyball net shown. Determine the angle formed by guy wires AB and AC.

SOLutiOn

First note AB

AC

= - + - +=

= + - +=

( . ) ( . ) ( . )

.

( ) ( . ) ( )

.

1 98 2 44 0 61

3 2

0 2 44 2

3 1

2 2 2

2 2 2

m

555m

and AB

AC

u ruuu ruu

= - - +

= - +

( . ( . ) ( . )

( . ) ( )

1 98 2 44 0 61

2 44 2

m) m m

m m

i j k

j k

By definition AB AC AB ACu ruu u ruu

◊ = ( )( )cosq

or [ ( . ) ( . ) ( . ) [ ( . ) ( ) ( . )( . )cos- - + ◊ - + =1 98 2 44 0 61 2 44 2 3 2 3 155i j k j k] ] qqq[( . ) ( . )] [( . )( )] . cos- ¥ - + =2 44 2 44 0 61 2 10 096

or cos .q = 0 711 or q = ∞44 7.

41


PROBLEM 3.39

Consider the volleyball net shown. Determine the angle formed by guy wires AC and AD.

SOLutiOn

First note AC

AD

= + - +=

= + - +=

( ) ( . ) ( )

.

( . ) ( . ) ( . )

.

0 2 44 2

3 155

1 22 2 44 0 3

2 7

2 2 2

2 2 2

m

444 mand AC

AD

u ruuu ruu

= - +

= - +

( . ( )

( . ) ( . ) ( . )

2 44 2

1 22 2 44 0 3

m) m

m m m

j k

i j k

By definition AC AD AC ADu ruu u ruu

◊ = ( )( )cosq

or ( . ) ( . . . ) ( . )( . )cos- + ◊ - + =2 44 2 1 22 2 44 0 3 3 155 2 744j k i j k q

( . )( . ) ( . ) . cos- - + =2 44 2 44 2 0 3 8 6573 q

or cos .q = 0 757 or q = ∞40 8.

42


PROBLEM 3.40

Knowing that the tension in cable AC is 1260 N, determine (a) the angle between cable AC and the boom AB, (b) the projection on AB of the force exerted by cable AC at Point A.

SOLutiOn

(a) First note AC

AB

= - + +=

= - + - +=

( . ) ( . ) ( . )

.

( . ) ( . ) ( )

.

2 4 0 8 1 2

2 8

2 4 1 8 0

3 0

2 2 2

2 2 2

m

mm

and AC

AB

u ruuu ruu

= - + +

= - -

( . ( . ) ( . )

( . ) ( . )

2 4 0 8 1 2

2 4 1 8

m) m m

m m

i j k

i jj

By definition AC AB AC ABu ruu u ruu

◊ = ( )( )cosq

or ( . . . ) ( . . ) ( . )( ) cos- + + ◊ - - = ¥2 4 0 8 1 2 2 4 1 8 2 8 30i j k i j q

or ( . )( . ) ( . )( . ) ( . )( ) . cos- - + - + =2 4 2 4 0 8 1 8 1 2 0 8 4 q

or cos .q = 0 51429 or q = ∞59 0.

(b) We have ( )

cos

( )( . )

T

TAC AB AC AB

AC

= ◊==

T lq

1260 0 51429 N or ( )TAC AB = 648 N

43


PROBLEM 3.41

Knowing that the tension in cable AD is 405 N, determine (a) the angle between cable AD and the boom AB, (b) the projection on AB of the force exerted by cable AD at Point A.

SOLutiOn

(a) First note AD

AB

= - + + -=

= - + - +=

( . ) ( . ) ( . )

.

( . ) ( . ) ( )

.

2 4 1 2 2 4

3 6

2 4 1 8 0

3 0

2 2 2

2 2 2

m

m

and &AD i j k

AB i j

= - + -= - -

( . ( . ) ( . )

( . ) ( . )

2 4 1 2 2 4

2 4 1 8

m) m m

m m

By definition, AD AB& ◊ = ( )( )cosAD AB q

( . . . ) ( . . ) ( . )( . )cos

( . )( . )

- + - ◊ - - =- - +

2 4 1 2 2 4 2 4 1 8 3 6 3 0

2 4 2 4

i j k i j q(( . )( . ) ( . )( ) . cos1 2 1 8 2 4 0 10 8- + - = q

cosq = 1

3 q = ∞70 5.

(b) ( )

cos

T

TAD AB AD AB

AD

= ◊=

T lq

= ÊËÁ

ˆ¯̃

( )4051

3 N ( ) .TAD AB = 135 0 N

44


PROBLEM 3.42

Slider P can move along rod OA. An elastic cord PC is attached to the slider and to the vertical member BC. Knowing that the distance from O to P is 150 mm and that the tension in the cord is 15 N, determine (a) the angle between the elastic cord and the rod OA, (b) the projection on OA of the force exerted by cord PC at Point P.

SOLutiOn

First note OA = + + -=

( ) ( ) ( )300 300 150

450

2 2 2

mm

Then lOAOA

OA= = + -

= + -

u ruu1

450300 300 150

1

32 2

( )

( )

i j k

i j k

Now OP OP OA= fi =1501

3mm. ( )

The coordinates of Point P are (100 mm, 100 mm, 2 50 mm)

so that PCu ruu

= + +( ) ( ) ( )125 275 350mm mm mmi j k

and PC = + + =( ) ( ) ( ) .125 275 350 462 332 2 2 mm

(a) We have PC PCOA

u ruu◊ =l ( )cosq

or ( ) ( ) . cos125 275 3501

32 2 462 33i j k i j k+ + ◊ + - = q

or cos .q = 0 3244 or q = ∞71 1.

(b) We have ( )

( )

cos

( )

T

T

TPC

PC

T

PC PC OA

PC PC OA

PC OA

PC

OA= ◊

= ◊

= ◊

=

=

T l

l l

lu ruu

q

15N (( . )0 32444 or ( ) .TPC OA = 4 87 N

45


PROBLEM 3.43

Slider P can move along rod OA. An elastic cord PC is attached to the slider and to the vertical member BC. Determine the distance from O to P for which cord PC and rod OA are perpendicular.

SOLutiOn

First note OA = + + -=

( ) ( ) ( )300 300 150

450

2 2 2

mm

Then lOAOA

OA= = + -

= + -

u ruu1

450300 300 150

1

32 2

( )

( )

i j k

i j k

Let the coordinates of Point P be (x mm, y mm, z mm). Then

PC x y zu ruu

= - + - + -[( ) ( ) [( )225 375 300i j k]

Also, OP dd

OP OAOP

u ruu= = + -l

32 2( )i j k

and OP x y z

x d y d z dOP OP OP

u ruu= + +

= = =

i j k

2

3

2

3

1

3

The requirement that OA and PC be perpendicular implies that

lOA PC◊ =u ruu

0

or 1

32 2 225 375 300 0( ) [( ) ( ) ( ) ]i j k i j k+ - ◊ - + - + - =x y z

or ( ) ( ) ( )2 2252

32 375

2

31 300

1

3-Ê

ËÁˆ¯̃

+ -ÊËÁ

ˆ¯̃

+ - - -ÊËÁ

ˆ¯̃

d d dOP OP OPÈÈÎÍ

˘˚̇

= 0

or dOP = 300 mm

46


PROBLEM 3.44

Determine the volume of the parallelepiped of Fig. 3.25 when (a) P 4i 2 3j 2k, Q 2 2i 2 5j k, and S 7i j 2 k, (b) P 5i 2 j 6k, Q 2i 3j k, and S 2 3i 2 2j 4k.

SOLutiOn

Volume of a parallelepiped is found using the mixed triple product.

(a) Vol = ◊ ¥

=-

- --

= - - + + -=

P Q S( )

( )

4 3 2

2 5 1

7 1 1

20 21 4 70 6 4

67 or Volume = 67 0.

(b) Vol = ◊ ¥

=-

- -= + - + + +=

P Q S( )

( )

5 1 6

2 3 1

3 2 4

60 3 24 54 8 10

111 or Volume = 111 0.

47


PROBLEM 3.45

Given the vectors P i j k Q i j k= - + = + -4 2 3 2 4 5, , and S i j k= - +Sx 2 , determine the value of Sx for which the three vectors are coplanar.

SOLutiOn

If P, Q, and S are coplanar, then P must be perpendicular to ( ).Q S¥

P Q S◊ ¥ =( ) 0

(or, the volume of a parallelepiped defined by P, Q, and S is zero).

Then

4 2 3

2 4 5

1 2

0

--

-=

Sx

or 32 10 6 20 8 12 0+ - - + - =S Sx x Sx = 7

48


PROBLEM 3.46

The 0 61 1 00. .¥ -m lid ABCD of a storage bin is hinged along side AB and is held open by looping cord DEC over a frictionless hook at E. If the tension in the cord is 66 N, determine the moment about each of the coordinate axes of the force exerted by the cord at D.

0.61 m

Az

0.11 mD

SOLutiOn

First note z = -=

( . ) ( . )

.

0 61 0 11

0 60

2 2

m

Then dDE = + + -=

( . ) ( . ) ( . )

.

0 3 0 6 0 6

0 9

2 2 2

m

and TDE = + -

= + -

66

0 90 3 0 6 0 6

22 1 2 2

N

N N N.

( . . . )

[( ) ( ) ( ) ]

i j k

i j k

Now M r TA D A DE= ¥/

where r j kD A/ m m= +( . ) ( . )0 11 0 60

Then M

i j k

i j k

A =-

= - - + -= -

22 0 0 11 0 60

1 2 2

22 0 22 1 20 0 60 0 11

31 2

. .

[( . . ) . . ]

( . 44 13 20 2 42 N m N m N m◊ + ◊ - ◊) ( . ) ( . )i j k

M M Mx y z= - ◊ = ◊ = - ◊31 2 13 20 2 42. , . . N m N m, N m

49


PROBLEM 3.47

The 0 61 1 00. .¥ -m lid ABCD of a storage bin is hinged along-side AB and is held open by looping cord DEC over a frictionless hook at E. If the tension in the cord is 66 N, determine the moment about each of the coordinate axes of the force exerted by the cord at C.

zB

C

0.11 m

0.61 m

SOLutiOn

First note z = -=

( . ) ( . )

.

0 61 0 11

0 60

2 2

m

Then dCE = - + + -=

( . ) ( . ) ( . )

.

0 7 0 6 0 6

1 1

2 2 2

m

and TCE = - + -

= - + -

66

1 10 7 0 6 0 6

6 7 6 6

N

N N N.

( . . . )

[ ( ) ( ) ( ) ]

i j k

i j k

Now M r TA E A CE= ¥/

where r i jE A/ m m= +( . ) ( . )0 3 0 71

Then M

i j k

i j k

A =- -

= - + + += - ◊

6 0 3 0 71 0

7 6 6

6 4 26 1 8 1 8 4 97

25 56

. .

[ . . ( . . ) ]

( . N mm N m N m) ( . ) ( . )i j k+ ◊ + ◊10 80 40 62

M M Mx y z= - ◊ = ◊ = ◊25 6 10 80 40 6. , . . N m N m, N m

50


PROBLEM 3.48

To lift a heavy crate, a man uses a block and tackle attached to the bot tom of an I-beam at hook B. Knowing that the moments about the y and the z axes of the force exerted at B by portion AB of the rope are, respectively, 120 N ? m and 2460 N ? m, determine the distance a.

SOLutiOn

First note BA au ruu

= - -( . ) ( . ) ( )2 2 3 2 m m mi j k

Now M r TD A D BA= ¥/

where r i j

T i j k

A D

BABA

BA

T

da

/ m m

N

= +

= - -

( . ) ( . )

( . . ) ( )

2 2 1 6

2 2 3 2

Then M

i j k

i j

DBA

BA

BA

BA

T

da

T

da a

=- -

= - + + -

2 2 1 6 0

2 2 3 2

1 6 2 2 2 2 3

. .

. .

{ . . [( . )( .. ) ( . )( . )] }2 1 6 2 2- k

Thus MT

da

MT

d

yBA

BA

zBA

BA

= ◊

= - ◊

2 2

10 56

. ( )

. ( )

N m

N m

Then forming the ratio M

My

z

120

460

2 2

10 56

N m

N m

N m

N m

◊- ◊

=◊

- ◊

. ( )

. ( )

Td

Td

BA

BA

BA

BA

or a = 1 252. m

51


PROBLEM 3.49

To lift a heavy crate, a man uses a block and tackle attached to the bot tom of an I-beam at hook B. Knowing that the man applies a 195-N force to end A of the rope and that the moment of that force about the y axis is 132 N ? m, determine the distance a.

SOLutiOn

First note d a

a

BA = + - + -

= +

( . ) ( . ) ( )

.

2 2 3 2

15 08

2 2 2

2 m

and T i j kBABAd

a= - -1952 2 3 2

N( . . )

Now M y A D BA= ◊ ¥j r T( )/

where r i jA/ m m0 2 2 1 6= +( . ) ( . )

Then Md

a

da

yBA

BA

=- -

= ◊

1950 1 0

2 2 1 6 0

2 2 3 2

1952 2

. .

. .

( . ) ( )N m

Substituting for My and d

BA

132195

15 082 2

2 N m◊ =

+.( . )

aa

or 0 30769 15 08 2. . + =a a

Squaring both sides of the equation

0 094675 15 08 2 2. ( . )+ =a a

or a = 1 256. m

52


PROBLEM 3.50

A small boat hangs from two davits, one of which is shown in the figure. It is known that the moment about the z axis of the resultant force RA exerted on the davit at A must not exceed 470 N · m in absolute value. Determine the largest allowable tension in line ABAD when x = 2m.

SOLutiOn

First note R T TA AB AD= +2

Also note that only TAD

will contribute to the moment about the z axis.

Now AD = + - + -=

( ) ( . ) ( )

.

2 2 5 1

3 354

2 2 2

m

Then, T

i j k

AD TAD

ADT

=

= - -

u ruu

3 3542 2 5

.( . )

Now Mz A C AD= ◊ ¥k r T( )/

where r j kA C/ m m= +( . ) ( )2 5 1

Then for Tmax, 4703 354

0 0 1

0 2 5 1

2 2 5 1

3 3542 2 25

=- -

=

T

T

max

max

..

.

.( )( . )

or Tmax = 350 N

CD2.5 m

1 m

A

TAD

2TAB

y

x

xz

53


PROBLEM 3.51

For the davit of Problem 3.50, determine the largest allowable distance x when the tension in line ABAD is 300 N.

SOLutiOn

From the solution of Problem 3.50, TAD is now

T TAD

AD

x

x

AD =

= - -

+ - + -

300 2 5

2 5 12 2 2

N( . )

( . ) ( )

i j k

Then Mz A C AD= ◊ ¥k r T( )/ becomes

470300

2 5 1

0 0 1

0 2 5 1

2 5 1

470300

7 252 5

2 2 2

2

=+ - + - - -

=+

-

x x

x

( . ) ( ).

.

.| ( . ))( ) |

.

. .

x

x x

x x

470 7 25 750

0 627 7 25

2

2

+ =

+ =

Squaring both sides: 0 393 2 8502 2. .x x+ =

x2 4 695= . x = 2 17. m

54


PROBLEM 3.52

To loosen a frozen valve, a force F of magnitude 350 N is applied to the handle of the valve. Knowing that q = ∞25 , M

x = -102 N m◊ and Mz = -72 N m◊ , determine f and d.

SOLutiOn

We have SM r F MO A O O: / ¥ =

where r i j k

F i j k

A O d

F

/ mm mm= - + -

= - +

( ) ( ) ( )

(cos cos sin cos sin )

100 275

q f q q f

For F = = ∞350 25N, q

F i j k

M

i

= - +

=

( )[( . cos ) . ( . sin ) ]

( )

350 0 90631 0 42262 0 90631

350

N

N

f f

O

jj k

- --

=

100 275

0 90631 0 42262 0 90631

350 249 2

d

. cos . . sin

( )[( .

f fmm

N ssin . ) ( . cos . sin )

( . .

f f f- + - +

+ -

0 42262 0 90631 90 631

42 262 249 2

d di j

ccos ) ]f k mm

and M dx = - = - ¥( )( . sin . ) ( ) /350 249 2 0 4226 102 3N mm N m 10 mm mf ◊ (1)

Mz = - = -( )( . . cos ) ( /350 42 262 249 2 72N mm N m)(1000 mm m)f ◊ (2)

From Equation (2) f = ÊËÁ

ˆ¯̃

= ∞-cos.

..1 247 98

249 25 68 or f = ∞5 68.

From Equation (1) d = ÊËÁ

ˆ¯̃

=316 09

0 4226747 97

.

.. mm or d = 748mm

55


PROBLEM 3.53

When a force F is applied to the handle of the valve shown, its moments about the x and z axes are, respectively, Mx = -130 N m◊ and Mz = -135N m◊ . For d = 675mm,determine the moment M

y of F about the y axis.

SOLutiOn

We have SM r F MO A O O: / ¥ =

Where r i j k

F i jA O

F/ mm mm mm= - + -

= - +( ) ( ) ( )

(cos cos sin cos si

100 275 675

q f q q nn )fk

M

i j k

O F

F

= - --

=

100 275 675

275

cos cos sin cos sin

[( cos sin

q f q q f

q f

N mm◊

-- + - +

+ -

675 675 100

100 275

sin ) ( cos cos cos sin )

( sin cos co

q q f q f

q q

i j

ss ) ]f k N mm◊

and M Fx = -( cos sin sin )(275 675q f q N mm)◊ (1)

M Fy = - +( cos cos cos sin ) ( )675 100q f q f N mm◊ (2)

M Fz = -( sin cos cos ) ( )100 275q q f N mm◊ (3)

Now, Equation (1) cos sin sinq f q= +ÊËÁ

ˆ¯̃

1

275675

M

Fx (4)

and Equation (3) cos cos sinq f q= -ÊËÁ

ˆ¯̃

1

275100

M

Fz (5)

Substituting Equations (4) and (5) into Equation (2),

M FM

F

M

Fyz x= - -Ê

ËÁˆ¯̃

ÈÎÍ

˘˚̇

+ +ÊËÁ

ˆ675

1

275100 100

1

275675sin sinq q

¯̃̄ÈÎÍ

˘˚˙

ÏÌÓ

¸˝˛

or M M My z x= +1

275675 100( )

56


PROBLEM 3.53 (continued)

Noting that the ratios 675275

and 100275 are the ratios of lengths, have

M y = - + -

= -= -

675

275135

100

275130

331 36 47 273

378

( ) ( )

( . ) ( . )

(

N m N m◊ ◊

.. )63 N m◊ or M y = -379 N m◊

57


PROBLEM 3.54

The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable.

SOLutiOn

MAD AD B A BH= ◊ ¥l ( )/r T

where l AD

B A

= -

=

1

54 3

0 5

( )

( . )/

i k

r i m

and dBH = + + -=

( . ) ( . ) ( . )

.

0 375 0 75 0 75

1 125

2 2 2

m

Then T i j k

i j

BH = + -

= + -

450

1 1250 375 0 75 0 75

150 300 300

N

N N .

( . . . )

( ) ( ) ( NN)k

Finally MAD =-

-

= -

1

5

4 0 3

0 5 0 0

150 300 300

1

53 0 5 300

.

[( )( . )( )]

or M AD = - ◊90 0. N m

58


PROBLEM 3.55

In Problem 3.54, determine the moment about the diagonal AD of the force exerted on the frame by portion BG of the cable.

SOLutiOn

MAD AD B A BG= ◊ ¥l ( )r T/

where l AD

B A

= -

=

1

54 3

0 5

( )

( . )/

i k

r j m

and BG = - + + -=

( . ) ( . ) ( . )

.

0 5 0 925 0 4

1 125

2 2 2

m

Then T BGuv

= - + -

= - + -

450

1 1250 5 0 925 0 4

200 370 16

N

N N.

( . . . )

( ) ( ) (

i j k

i j 00 N)k

Finally MAD =

-

- -

15

4 0 3

0 5 0 0

200 370 160

.

= -1

53 0 5 370[( )( . )( )] M AD = - ◊111 0. N m

59


PROBLEM 3.56

The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 55 N, determine the moment of that force about the line joining Points D and B.

SOLutiOn

First note dAE = + - + =( . ) ( . ) ( . ) .0 9 0 6 0 2 1 12 2 2 m

Then T i j k

i j k

AE = - +

= - +

55

1 10 9 0 6 0 2

5 9 6 2

N

N N N.

( . . . )

[( ) ( ) ( ) ]

Also DB = + - +=

( . ) ( . ) ( )

.

1 2 0 35 0

1 25

2 2 2

m

Then lDBDB

DB=

= -

= -

u ruu

1

1 251 2 0 35

1

2524 7

.( . . )

( )

i j

i j

Now MDB DB A D AE= ◊ ¥l ( )r T/

where T j kDA = - +( . ) ( . )0 1 0 2 m m

Then MDB =-

--

= - - +

1

255

24 7 0

0 0 1 0 2

9 6 2

1

54 8 12 6 28 8

( ) . .

( . . . )

or MDB = ◊2 28. N m

60


PROBLEM 3.57

The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 33 N, determine the moment of that force about the line joining Points D and B.

SOLutiOn

First note dCF = + + =( . ) ( . ) ( . ) .0 6 0 9 0 2 1 12 2 2- - m

Then T i j k

i j k

CF = - +

= - -

33

1 10 6 0 9 0 2

3 6 9 2

N

N N N.

( . . . )

[( ) ( ) ( ) ]

Also DB = + - +=

( . ) ( . ) ( )

.

1 2 0 35 0

1 25

2 2 2

m

Then lDBDB

DB=

= -

= -

u ruu

1

1 251 2 0 35

1

2524 7

.( . . )

( )

i j

i j

Now MDB DB C D CF= ◊ ¥l ( )/r T

where r j kC D/ ( . ) ( . )= -0 2 0 4 m m

Then MDB =-

-- -

= - + -

1

253

24 7 0

0 0 2 0 4

6 9 2

3

259 6 16 8 86 4

( ) . .

( . . . )

or MDB = - ◊9 50. N m

61


SOLutiOn

We have MOA OA C O= ◊ ¥l ( )r P/

where

From triangle OBC ( )

( ) ( ) tan

OAa

OA OAa a

x

z x

=

= ∞ = ÊËÁ

ˆ¯̃

=

2

302

1

3 2 3

Since ( ) ( ) ( ) ( )OA OA OA OAx y z2 2 2 2= + +

or aa

OAa

OA aa a

a

y

y

22

22

22 2

2 2 3

4 12

2

3

= ÊËÁ

ˆ¯̃

+ + ÊËÁ

ˆ¯̃

= - - =

( )

( )

Then r i j kA Oa

aa

/ = + +2

2

3 2 3

and lOA = + +1

2

2

3

1

2 3i j k

P

i ki k

r i

= = ∞ - ∞ = -

=

lBC

C O

Pa a

aP

P

a

( sin ) ( cos )( ) ( )

30 30

23

/

M aP

aP aP

OA =

-

ÊËÁ

ˆ¯̃

= -Ê

ËÁˆ

¯̃- =

1

2

2

3

1

2 31 0 0

1 0 3

2

2

2

31 3

2

( )

( )( ) MaP

OA =2

B

Cx

z

O(OA)z

(OA)x

30° 60°

A

Cx

BP

rC/O

lOA

z

O

y

PROBLEM 3.58

A regular tetrahedron has six edges of length a. A force P is directed as shown along edge BC. Determine the moment of P about edge OA.

62


PROBLEM 3.59

A regular tetrahedron has six edges of length a. (a) Show that two opposite edges, such as OA and BC, are perpendicular to each other. (b) Use this property and the result obtained in Problem 3.59 to determine the perpendicular distance between edges OA and BC.

SOLutiOn

(a) For edge OA to be perpendicular to edge BC,

OA BCu ruu u ruu

◊ = 0

where

From triangle OBC ( )

( ) ( ) tan

(

OAa

OA OAa a

OAa

x

z x

=

= ∞ = ÊËÁ

ˆ¯̃

=

= ÊËÁ

ˆ¯̃

+

2

302

1

3 2 3

2

u ruui OOA

ay) j k+ Ê

ËÁˆ¯̃2 3

and BC a a

a a a

u ruu= ∞ - ∞

= - = -

( sin ) ( cos )

( )

30 30

2

3

2 23

i k

i k i k

Then a

OAa a

y2 2 33

20i j k i k+ + Ê

ËÁˆ¯̃

È

ÎÍ

˘

˚˙ ◊ - =( ) ( )

or a

OAa

OA BC

y

2 2

40

40

0

+ - =

◊ =

( ) ( )u ruu u ruu

so that OAu ruu

is perpendicular to BCu ruu

(b) Have M PdOA = , with P acting along BC and d the perpendicular distance from OAu ruu

to BCu ruu

.

From the results of Problem 3.56

MPa

PaPd

OA =

=

2

2 or d

a=2

(OA)x

(OA)z

C xO

B

z

30°60°

y

A

B

Cd

O

OA

BCx

z

63


PROBLEM 3.60

A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EF at E is 230 N, determine the moment of that force about the line joining points A and D.

SOLutiOn

First note that BC = + =( ) ( )1200 900 15002 2 mm and that BEBC = =1125

150034 . The coordinates of Point E are then

34

341200 2400 900 900 2400 675¥ ¥( ), , ( )or mm, mm, mm . Then

dEF = - + - +=

( ) ( ) ( )15 110 30

2875

2 2 2

mm

Then T

i j k

EF = - + - +

= - - +

= -

( ) ( ) ( )

( )

[

375 2750 750

230

2875375 2750 750

2

2 2 2

N

(( ) ( ) ( ) ]15 110 30N N Ni j k- +

Also AD = + - +=

( ) ( ) ( )

.

1200 300 900

1529 71

2 2 2

mm

Then l ADAD

AD=

= - +

= - +

u ruu

1

1529 711200 300 900

0 7845 0 1961 0.

( )

( . . .

i j k

i j 55883k)

64



Now MAD AD E A EF= ◊ ¥l ( )r T/

where r i j kE A/ mm mm mm= + +( ) ( ) ( )900 2400 675

Then M AD =

- +

- -

= + -

0 7845 0 1961 0 5883

900 2400 675

30 220 60

229466 14560 7

. . .

44126

169900= N mm◊

MAD = 169 9. N m◊

65


SOLutiOn

First note that BC = + =( ) ( )1200 900 15002 2 mm and that BEBC = =1125

150034 . The coordinates of Point E are

then 34

341200 2400 900 900 2400 675¥ ¥( ), , ( )or mm, mm, mm Then

dEG = + - + -=

( ) ( ) ( )275 2200 1100

2475

2 2 2

mm

Then T i j k

i j k

EG = - -

= - -

270

2475275 2200 1100

30 240 120

N

N N N

( )

( ) ( ) ( )

Also AD = + - +=

( ) ( ) ( )

.

1200 300 900

1529 71

2 2 2

mm

Then l ADAD

AD=

= - +

= - +

u ruu

1

1529 711200 300 900

0 7845 0 1961 0.

( )

( . . .

i j k

i j 55883k)

PROBLEM 3.61

A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EG at E is 270 N, determine the moment of that force about the line joining points A and D.

66



Now MAD AD E A EF= ◊ ¥l ( )r T/

where r i j kE A/ mm mm mm= + +( ) ( ) ( )900 2400 675

Then MAD =- +

- -

= - - -

0 7845 0 1961 0 5883

900 2400 675

30 240 120

98847 25150

. . .

1169430

293427

N mm

N mm

◊

= - ◊

or MAD = - ◊293 N m

67


PROBLEM 3.62

Two forces F1 and F2 in space have the same magnitude F. Prove that the moment of F1 about the line of action of F2 is equal to the moment of F2 about the line of action of F1 .

SOLutiOn

F1

A A

B B

F1

rB/ArA/B

F2F2

First note that F F1 1 1 2 2= =F Fl land 2

Let M1 = moment of 2F about the line of action of M1 and M2 = moment of F1 about the line of action of M2

Now, by definition M

F

M

B A

B A

A B

A B

1 1 2

1 2 2

2 2 1

2 1

= ◊ ¥= ◊ ¥

= ◊ ¥= ◊ ¥

ll l

ll l

( )

( )

( )

( )

/

/

/

/

r F

r

r F

r FF1

Since F F F

M F

M F

A B B A

B A

B A

1 2

1 1 2

2 2 1

= = = -= ◊ ¥

= ◊ - ¥

and / /

/

/

r r

r

r

l l

l l

( )

( )

Using Equation (3.38) l l l l1 2 2 1◊ ¥ = ◊ - ¥( ) ( )r rB A B A/ /

so that M FB A2 1 2= ◊ ¥l l( )r / M M12 21=

68


PROBLEM 3.63

In Problem 3.54, determine the perpendicular distance between portion BH of the cable and the diagonal AD.

PROBLEM 3.54 The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable.

SOLutiOn

From the solution to Problem 3.54: TBH

BH

== + -

450

150 300 300

N

N N NT i j k( ) ( ) ( )

| | N mM AD = ◊90 0.

l AD = -1

54 3( )i k

Based on the discussion of Section 3.11, it follows that only the perpendicular component of TBH

will contributeto the moment of T

BH about line AD

u ruu.

Now ( )

( ) ( )

[( )(

TBH BH ADparallel = ◊

= + - ◊ -

=

T

i j k i k

l

150 300 3001

54 3

1

5150 44 300 3

300

) ( )( )]+ - -

= N

Also T T TBH BH BH= +( ) ( )parallel perpendicular

so that ( ) ( ) ( ) .TBH perpendicular N= - =450 300 335 412 2

Since l AD and ( )TBH perpendicular are perpendicular, it follows that

M d TAD BH= ( )perpendicular

or 90 0 335 41. ( . ) N m N◊ = d

d = 0 26833. m d = 0 268. m

69


PROBLEM 3.64

In Problem 3.55, determine the perpendicular distance between portion BG of the cable and the diagonal AD.

PROBLEM 3.55 In Problem 3.54, determine the moment about the diagonal AD of the force exerted on the frame by portion BG of the cable.

SOLutiOn

From the solution to Problem 3.55: T

T i j kBG

BG

== - + -

450

200 370 160

N

N N N( ) ( ) ( )

| | N mM AD = ◊111

lAD = -1

54 3( )i k

Based on the discussion of Section 3.11, it follows that only the perpendicular component of TBG

will contribute to the moment of T

BG about line AD

u vuu.

Now ( )

( ) ( )

[(

T T

i j k i k

BG BG ADparallel = ◊

= - + - ◊ -

= -

l

200 370 1601

54 3

1

5200))( ) ( )( )]4 160 3

64

+ - -

= - NAlso T T TBG BG BG= +( ) ( )parallel perpendicular

so that ( ) ( ) ( ) .TBG perpendicular N= - - =450 64 445 432 2

Since l AD and ( )TBG perpendicular are perpendicular, it follows that

M d TAD BG= ( )perpendicular

or 111 445 43 N m N◊ = d( . )

d = 0 24920. m d = 0 249. m

70


PROBLEM 3.65

In Problem 3.56, determine the perpendicular distance between cable AE and the line joining Points D and B.

PROBLEM 3.56 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 55 N, determine the moment of that force about the line joining Points D and B.

SOLutiOn

From the solution to Problem 3.56 T

T i j kAE

AE

== - +

55

5 9 6 2

N

N N N[( ) ( ) ( ) ]

| | N mMDB = ◊2 28.

lDB = -1

2524 7( )i j

Based on the discussion of Section 3.11, it follows that only the perpendicular component of TAE


AE about line DB

u ruu.

Now ( )

( ) ( )

[( )( ) (

T T

i j k i j

AE AE DBparallel = ◊

= - + ◊ -

= +

l

5 9 6 21

2524 7

1

59 24 -- -

=

6 7

51 6

)( )]

. N

Also T T TAE AE AE= +( ) ( )parallel perpendicular

so that ( ) ( ) ( . ) .TAE perpendicular N= + =55 51 6 19 03792 2

Since lDB and ( )TAE perpendicular are perpendicular, it follows that

M d TDB AE= ( )perpendicular

or 2 28 19 0379. ( . ) N m N◊ = d

d = 0 119761. d = 0 1198. m

71


PROBLEM 3.66

In Problem 3.57, determine the perpendicular distance between cable CF and the line joining Points D and B.

PROBLEM 3.57 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 33 N, determine the moment of that force about the line joining Points D and B.

SOLutiOn

From the solution to Problem 3.57 T

T i j kCF

CF

== - -

33

3 6 9 2

N

N N N[( ) ( ) ( ) ]

| | N mMDB = ◊9 50.

lDB = -1

2524 7( )i j

Based on the discussion of Section 3.11, it follows that only the perpendicular component of TCF


CF about line DB

u ruu.

Now ( )

( ) ( )

[( )( )

TCF CF DBparallel = ◊

= - - ◊ -

= +

T

i j k i j

l

3 6 9 21

2524 7

3

256 24 (( )( )]

.

- -

=

9 7

24 84 N

Also T T TCF CF CF= +( ) ( )parallel perpendicular

so that ( ) ( ) ( . )

.

TCF perpendicular

N

= -

=

33 24 84

21 725

2 2

Since lDB and ( )TCF perpendicular are perpendicular, it follows that

| | perpendicularM d TDB CF= ( )

or 9 50 21 725. . N m N◊ = ¥d

or d = 0 437. m

72


PROBLEM 3.67

In Problem 3.60, determine the perpendicular distance between cable EF and the line joining Points A and D.

PROBLEM 3.60 A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EF at E is 230 N, determine the moment of that force about the line joining Points A and D.

SOLutiOn

From the solution to Problem 3.61 TEF

EF

== - - +

230

30 220 60

N

N N NT i j k( ) ( ) ( ) | | N mM AD = ◊169 9.

lAD = - +( . . . )0 7845 0 1961 0 5883i j k

Based on the discussion of Section 3.11, it follows that only the perpendicular component of TEF


EF about line AD

u ruu.

Now ( )

( ) ( . . .

TEF EF ADparallel = ◊

= - - + ◊ - +

T

i j k i j

l

30 220 60 0 7845 0 1961 0 58883

54 905

k)

.= N

Also T T TEF EF EF= +( ) ( )parallel perpendicular

so that ( ) ( ) ( . ) .TEF perpendicular N= - =230 54 905 223 352 2

Since l AD and ( )TEF perpendicular are perpendicular, it follows that

M d TAD EF= ( )perpendicular

or 169 9 223 35. . N m N◊ = ¥d or md = 0 761.

73


PROBLEM 3.68

In Problem 3.61, determine the perpendicular distance between cable EG and the line joining Points A and D.

PROBLEM 3.61 A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EG at E is 270 N, determine the moment of that force about the line joining Points A and D.

SOLutiOn

From the solution to Problem 3.62 TEG

EG

== - -

270

30 240 120

N

N N NT i j k( ) ( ) ( )

| | N mM AD = ◊293

l AD = - +( . . . )0 7845 0 1961 0 5883i j k

Based on the discussion of Section 3.11, it follows that only the perpendicular component of TEG


EG about line AD

u ruu.

Now ( )

( ) ( . . .

TEG EG ADparallel = ◊

= - - ◊ - +

T

i j k i j

l

30 240 120 0 7845 0 1961 0 58883

0

k)

=Thus, ( )T TEG EGperpendicular N= = 270

Since l AD and ( )TEG perpendicular are perpendicular, it follows that

| | perpendicularM d TAD EG= ( )

or 293 270 N m N◊ = ¥d

or d = 1 085. m

74


PROBLEM 3.69

Two parallel 60-N forces are applied to a lever as shown. Determine the moment of the couple formed by the two forces (a) by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples, (b) by using the perpendicular distance between the two forces, (c) by summing the moments of the two forces about Point A.

SOLutiOn

(a) We have SM MB x yd C d C: - + =1 2

where d

d

1 0 360

0 360 55

0 20649

= ∞== ∞=

( . )sin

( . )sin

.

m 55

0.29489 m

m

m2

C

C

x

y

= ∞== ∞

=

( )cos

.

( sin

.

60 20

56 382

60 20

20 521

N

N

N)

N

M k k= - += -

( . ( .0 29489 0 20649 m)(56.382 N) m)(20.521 N)

(12.3893 NN m)◊ k or M = ◊12 39. N m

(b) We have M k

k

= -= ∞ - ∞ -

Fd( )

)sin( )]( )60 55 20 N[(0.360 m

= - ◊( .12 3893 N m)k or M = ◊12 39. N m

(c) We have S S ¥M r F r F r F MA A B A B C A C: / /( )¥ = ¥ + =

M = ∞ ∞- ∞ - ∞

+

( . cos sin

cos sin

( .

0 520 55 55 0

20 20 0

0 800

m)(60 N)

m)

i j k

((60 N)

N m N

i j k

cos sin

cos sin

( . .

55 55 0

20 20 0

17 8956 30 285

∞ ∞∞ ∞

= ◊ - ◊◊= - ◊

m)

N m)

k

k( .12 3892

or M = ◊12 39. N m

C

AA

20°

20°

55°55°

d1

d2

Bx By

Cx

Cy

BB

d

60 N60 N

75


PROBLEM 3.70

A plate in the shape of a parallelogram is acted upon by two couples. Determine (a) the moment of the couple formed by the two 105 N forces, (b) the perpendicular distance between the 60 N forces if the resultant of the two couples is zero, (c) the value of a if the resultant couple is 9 N m◊ clockwise and d is 1050 mm.

SOLutiOn

(a) We have M d F1 1 1=

where d

F1

1

400

105

==

mm

N

M1 400

42000

== ◊

( mm)(105 N)

N mm or M1 42= ◊ N m

(b) We have M M1 2 0+ =

or 42 60 02 N m N)◊ - =d ( d2 0 7 700= =. m mm

(c) We have M M Mtotal = +1 2

or - ◊ = ◊ - ÊËÁ

ˆ¯̃

91050

100060 N m 42 N m m N)(sin )(a

sin .a = 0 80952

and a = ∞54 049. or a = ∞54 0.

D

D

A

A

B

B

d

d sina

a

C

C

60 N

60 N

60 N

60 N

d2

d2

76


PROBLEM 3.71

A couple M of magnitude 18 N m◊ is applied to the handle of a screwdriver to tighten a screw into a block of wood. Determine the magnitudes of the two smallest horizontal forces that are equivalent to M if they are applied (a) at corners A and D, (b) at corners B and C, (c) anywhere on the block.

SOLutiOn

(a) We have M Pd=

or 18 24 N m m)◊ = P(.

P = 75 0. N or Pmin .= 75 0 N

(b) d BE ECBC = +

= +=

( ) ( )

(.

.

2 2

24

0 25298

m) (.08 m)

m

2 2

We have M Pd=

18 0 25298

71 152

N m m)

N

◊ ==

P

P

( .

. or P = 71 2. N

(c) d AD DCAC = +

= +=

( ) ( )

( . ( .

.

2 2

0 24 0 32

0 4

m) m)

m

2 2

We have M Pd

PAC=

◊ =18 0 4 N m m)( .

P = 45 0. N or P = 45 0. N

A

DP

P

M

.24 m

A B

D C

P

P

M

.24 m

.08 m

dBC

E

A

CD

P

P.24 m

.32 m

M

77


PROBLEM 3.72

Four 25 mm-diameter pegs are attached to a board as shown. Two strings are passed around the pegs and pulled with the forces indicated. (a) Determine the resultant couple acting on the board. (b) If only one string is used, around which pegs should it pass and in what directions should it be pulled to create the same couple with the minimum tension in the string? (c) What is the value of that minimum tension?

SOLutiOn

(a) M = += ◊ + ◊

( )( ) ( )( )175 175 125 225

30625 28125

N mm N mm

N mm N mm M = ◊58750 N mm

(b) With only one string, pegs A and D, or B and C should be used. We have

tan . .q q q= = ∞ ∞ - = ∞150

20036 9 90 53 1

Direction of forces:

With pegs A and D: q = ∞53 1.

With pegs B and C: q = ∞53 1.

(c) The distance between the centers of the two pegs is

( ) ( )200 150 2502 2+ = mm

Therefore, the perpendicular distance d between the forces is

d = + Ê

ËÁˆ¯̃

=

250 225

2

275

mm mm

mm

We must have M Fd

F

= = ◊=

58750

275

N mm

mm( ) F = 213 6. N 214 N =

225 mm

175 N

125 N

175 mm

200 mmD

BF

C

–F

–FdA

F

150 mm q

q q

q

78


PROBLEM 3.73

Four pegs of the same diameter are attached to a board as shown. Two strings are passed around the pegs and pulled with the forces indicated. Determine the diameter of the pegs knowing that the resultant couple applied to the board is 60.25 N m◊ counterclockwise.

SOLutiOn

M d F d F

M

d

AD AD BC BC= +=

◊ = +60 25

60250 150

.

[( ) (

N m

N mm mm]

◊ = 60250 N ◊ mm

1175 200 125N mm] N) [( ) ( )]+ + d

d = 30 mm

79


PROBLEM 3.74

The shafts of an angle drive are acted upon by the two couples shown. Replace the two couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLutiOn

Based on M M M= +1 2

where M j

M k

M j k

1

9

12 9

= - ◊= - ◊= - ◊ - ◊

(12 N m)

N m)

N m) N m)2 (

( (

and | | N mM = + = ◊( ) ( )12 9 152 2 or M = ◊15 N m

l =

=- ◊ - ◊

◊= - -

MM

j k

j k

| |

N m) N m)

N m

( (

. .

12 9

15

0 8 0 6

or M M j k= = ◊ - -| | N m) )l ( ( . .15 0 8 0 6

cos

cos .

cos .

.

.

qq

q

qq

q

x

y

z

x

y

z

== -

= -

= ∞= ∞

= ∞

0

0 8

0 6

90

143 130

126 870

or q q qx y z= ∞ = ∞ = ∞90 0 143 1 126 9. . .

143.1°

y

M

z

126.9°

80


PROBLEM 3.75

If P = 0, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLutiOn

We have M M M= +1 2

where M r F1 1= ¥G C/

r i

F k

M i k

j

G C/ m

N

m N

N m

= -== - ¥= ◊

( . )

( )

( . ) ( )

( . )

0 3

18

0 3 18

5 4

1

1

Also, M r F

r i j

F

i j

2 2

2 2

15 08

15 08

= ¥= - +=

=+ +

D F

D F

EDF

/

/ m m

m m

(. ) (. )

(. ) (. ) (.

l117

15 08 1734

141 421

2 2 2

m

mN

N m(.15 .08 .17

)

(. ) (. ) (. )( )

.

k

i j

+ += ◊ + + kk)

M

i j k

i j

2 141 421

141 421 0136 0 0255

= ◊ --

= +

.

. (. .

N m .15 .08 0

.15 .08 .17

))N m◊

y

xz

.16 m

A

E

F F1

F1F2

F2

C

B

G

D

rD/F

rG/C

.17 m.15 m

.15 m

81



and M j i j

i

= ◊ + + ◊= ◊

[( . ] [ . (. . ) ]

( . )

5 4 141 421 0136 0255

1 92333

N m) N m

N m ++ ◊( .9 0062 N m)j

| | ( ) ( )

( . ) ( . )

.

M = +

= += ◊

M Mx y2 2

2 21 92333 9 0062

9 2093 N m or M = ◊9 21. N m

l = = ◊ + ◊◊

= +

MM

i j| |

( . ) ( . )

.

. .

1 92333 9 0062

9 2093

0 20885 0 97

N m N m

N m

7795

0 20885

77 945

cos .

.

qq

x

x

== ∞ or qx = ∞77 9.

cos .

.

q

qy

y

=

= ∞

0 97795

12 054 or q y = ∞12 05.

cos .qq

z

z

== ∞

0 0

90 or q z = ∞90 0.

82


PROBLEM 3.76

If P = 0, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLutiOn

M M M

M r F i j1

= + = == ¥ = ¥ - = -

1 2 1

1

80 200

750 80 60000

;

( ) ( ) (

F F

C

N, N

mm N 2

NN mm

mm mm/ /

◊= ¥ = -

= + +

)

; ( ) ( )

( ) ( )

k

M r F r i j2 E B E B

DEd

2

2 2

375 125

0 125 (( )

( )

( ( )

250 125 5

125 250

40 5 1 2

2

2

2

=

= -

= -( )

=

mm

200 N

125 5

N N

F j k

)j k

M 440 5 375 125 0

0 1 2

40 5 250 750 375

i j k

i j

--

= ◊ + ◊ + ◊[( ( ) ( ) N mm) N mm N mm kk

M i j k

i

]

(

( . ) (

= + += ◊ + ◊

22361 67082 33541

22 361 67082

) N.mm

N mm N mm)) ( )j k- ◊26459 N mm

M = + + -= ◊

( ) ( ) ( )22361 67082 26459

75499

2 2 2

N mm M = ◊75 5. N m

laxis = = + -

==

Mi j k

M

x

y

0 29617 0 88852 0 35045

0 29617

0 88

. . .

cos .

cos .

qq 8852

0 35045cos q z = - .

q q qx y= ∞ = ∞ = ∞72 8 27 3 110 5. . .z

83


PROBLEM 3.77

If P = 100 N, replace the three couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLutiOn

From the solution to Problem. 3.76

80 N force: M1 60000= - ◊ N mm k

200 N force: M2 8 5 22361 67082 33541= ◊ + ◊ + ◊[( ) ( ) ( ) ] N mm N mm N mmi j k

P = 100 N M PC3

750 100

75000

= ¥= ¥ -= ◊

r

i k)

j

( ) ( )(

( )

mm N

N mm

M M M M

i j k

= + += + -

=

1 2 3

22361 142082 26459

223

( ) ( ) ( )

(

N mm N mm N mm◊ ◊ ◊

M 661 142082 26459

146244

2 2 2) ( ) ( )+ + -= N mm◊

M = ◊146 2. N m

laxis = = + -

==

Mi j k

M

x

y

0 15290 0 97154 0 18092

0 15290

0 97

. . .

cos .

cos .

qq 1154

0 180921cos .q z = - q q qx y z= ∞ = ∞ = ∞81 2 13 70 100 4. . .

84


PROBLEM 3.78

If P = 20 N, replace the three couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLutiOn

We have M M M M= + +1 2 3

where

M r F

i j k

j

M r F

i j k

1 1

2 2

0 3 0 0

0 0 18

5 4

15 08

= ¥ = - ◊ = ◊

= ¥ = -

G C

D F

/

/

N m N m). ( .

. . 00

15 08 17

141 421 0136 0255

-◊

= + ◊. . .

. (. . )

141.421 N m

N mi j (See Solution to Problem 3.75)

M r F

i j k

i k

M

3 3 0 3 0 0 17

0 20 0

3 4 6

1 9233

= ¥ = ◊

= - ◊ + ◊=

C A/ N m

N m N m

. .

( . ) ( )

[( . 33 3 4 5 4 3 6062 6

1 47667 9 0062

- + + + ◊= - ◊ + ◊

. ) ( . . ) ( ) ]

( . ) ( .

i j k

i

N m

N m N mm N m) ( )j + ◊6

| |M = + +M M Mx y z2 2 2

= + += ◊

( . ) ( . ) ( )

.

1 47667 9 0062 6

10 9221

2

N m or M 10.92 N . m

85



l = = - + +

= - + +

MM

i

| |

1 47667 9 0062 6

10 9221

0 135200 0 82459 0 5493

. .

.

. . .j 44k

cos . .q qx x= - =0 135200 97 770 or qx = ∞97 8.

cos . .q qy y= =0 82459 34 453 or q y = ∞34 5.

cos . .q qz z= =0 54934 56 678 or q z = ∞56 7.

86


PROBLEM 3.79

Shafts A and B connect the gear box to the wheel assemblies of a tractor, and shaft C connects it to the engine. Shafts A and B lie in the vertical yz plane, while shaft C is directed along the x axis. Replace the couples applied to the shafts with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLutiOn

Represent the given couples by the following couple vectors:

M j k

j kA = - ∞ + ∞

= - ◊ + ◊1600 20 1600 20

547 232 1503 51

sin cos

( . ) ( . ) N m N m

MM j k

j k

M

B = ∞ + ∞= ◊ + ◊

1200 20 1200 20

410 424 1127 63

sin cos

( . ) ( . ) N m N m

CC = - ◊( )1120 N m i

The single equivalent couple is

M M M M

i j k

= + += - ◊ - ◊ + ◊

=

A B C

M

( ) ( . ) ( . )

(

1120 136 808 2631 1

112

N m N m N m

00 136 808 2631 1

2862 8

1120

2862 8

2 2 2) ( . ) ( . )

.

cos.

cos

+ += ◊

= - N m

q

q

x

y == -

=

136 808

2862 82631 1

2862 8

.

.

cos.

.q z

M x y z= ◊ = ∞ = ∞ = ∞2860 113 0 92 7 23 2 N m q q q. . .

87


PROBLEM 3.80

The tension in the cable attached to the end C of an adjustable boom ABC is 2.8 k N. Replace the force exerted by the cable at C with an equivalent force-couple system (a) at A, (b) at B.

SOLutiOn

(a) Based on SF F TA: = = 2800 N

or FA = 2800 N 20°

SM M T dA A A: ( sin )= ∞50 ( )

= ∞= ◊

( )sin ( . )2800 50 5 44

11668

N m

N m

or MA = ◊11 67. kN m

(b) Based on SF F TB: ( )= = 2800 N

or FB = 2800 N 20°

SM M T dB B B: ( sin )( )= ∞50

= ∞= ◊

( sin )2800 50

6435

N)( (3 m)

N m or MB = ◊6 43. kN m

Tcos 50°

Tsin 50°

B

A 20°

CdA

MA

MB

FA

FB

dB

C

B

A

20°

3 m

2.44 m

A

B30°

30°20°

40° T

C

88


PROBLEM 3.81

A 800 N force P is applied at Point A of a structural member. Replace P with (a) an equivalent force-couple system at C, (b) an equivalent system consisting of a vertical force at B and a second force at D.

SOLutiOn

(a) Based on SF P PC: = = 800 N or PC = 800 N 60° SM M P d P dC C x cy y Cx: = - +

where P

P

d

d

x

y

Cx

Cy

= ∞== ∞

===

( )cos

( )sin

.

.

.

800 60

400

800 60

692 82

1 2

0

N

N

N

N

m

883 m

MC = += ◊

( )( . ) ( . )( . )400 0 83 692 82 1 2

1163

N m N m

.4 N m or MC = ◊1163 N m

(b) Based on SF P Px Dx: cos= ∞60

= ∞=

( )cos800 60

400

N

N

SM P d P dD DA B DB: ( cos )( ) ( )60∞ =[( )cos ]( . ) ( . )800 60 0 45 1 8

100

N m m

N

∞ ==

P

PB

B or PB = ≠100 N

dCx

dCy

PC

60˚

MCC

A

PDx

dDA

PDX

PD

PB

dDB

D

B

A

q

89



SF P P Py B Dy: sin60∞ = +

( )sin

.

( ) ( )

( )

800 60

592 82

400

2 2

N 100 N

N

∞ = +

=

= +

=

P

P

P P P

Dy

Dy

D Dx Dy

22 2592 82

715 15

+=

( . )

. N

q =ÊËÁ

ˆ¯̃

= ÊËÁ

ˆ¯̃

= ∞

-

-

tan

tan.

.

1

1 592 82

400

55 991

P

PDy

Dx

or PD = 715 N 56.0°

90


PROBLEM 3.82

The 80-N horizontal force P acts on a bell crank as shown. (a) Replace P with an equivalent force-couple system at B. (b) Find the two vertical forces at C and D that are equivalent to the couple found in Part a.

SOLutiOn

(a) Based on SF F FB: = = 80 N or FB = 80 0. N ¨

SM M FdB B: =

== ◊

80

4 0000

N (.05 m)

N m.

or MB = ◊4 00. N m

(b) If the two vertical forces are to be equivalent to MB, they must be

a couple. Further, the sense of the moment of this couple must be counterclockwise.

Then, with FC and FD acting as shown,

SM M F dD C: =

4 0000 04

100 000

. (. )

.

N m m

N

◊ ==

F

FC

C or FC = Ø100 0. N

SF F Fy D C: 0 = -

FD = 100 000. N or NFD = ≠100 0.

A

A

D

D

C

CB

d

MBdB

FB

FB

FC FD

91


PROBLEM 3.83

A dirigible is tethered by a cable attached to its cabin at B. If the tension in the cable is 1040 N, replace the force exerted by the cable at B with an equivalent system formed by two parallel forces applied at A and C.

SOLutiOn

Require the equivalent forces acting at A and C be parallel and at an angle of a with the vertical.

Then for equivalence,

SF F Fx A B: ( )sin sin sin1040 30 N ∞ = +a a (1)

SF F Fy A B: ( )cos cos cos- ∞ = - -1040 30 N a a (2)

Dividing Equation (1) by Equation (2),

( )sin

( )cos

( )sin

( )cos

1040 30

1040 30

N

N

∞- ∞

=+

- +F F

F FA B

A B

aa

Simplifying yields a = ∞30

Based on

SM FC A: [( )cos ]( ) ( cos )( . )1040 30 4 30 10 7N m m∞ = ∞

FA = 388 79. N

or FA = 389 N 60°

Based on

SM FA C: [( )cos ]( . ) ( cos )( . )- ∞ = ∞1040 30 6 7 30 10 7N m m

FC = 651 21. N

or FC = 651 N 60°

A

A

B

B

C

C

FC

FA

30˚

30˚30˚

1040 N

4 m6.7 m

92


PROBLEM 3.84

The force P has a magnitude of 250 N and is applied at the end C of a 500-mm rod AC attached to a bracket at A and B. Assuming a = 30∞ and 60 ,b = ∞ replace P with (a) an equivalent force-couple system at B, (b) an equivalent system formed by two parallel forces applied at A and B.

SOLutiOn

(a) Equivalence requires SF F P F: = =or N250 60°

SMB M: ( . )( )= - = - ◊0 3 250 75 m N N m

The equivalent force-couple system at B is

F = 250 N 60° M = ◊75 0. N m

(b) Require

A

B

C

30°

60°

250 N

A

B

y x C

FB

FAφ

φ

Equivalence then requires

SF F Fx A B: cos cos0 = +f f

F FA B= - =or cosf 0

SF F Fy A B: sin sin- = - -250 f f

Now if F FA B= - fi - =250 0 reject

cos f = 0

or f = ∞90

and F FA B+ = 250

Also S M FB A: ( . )( ) ( . )- =0 3 250 0 2 m N m

or FA = -375 N

and FB = 625 N

FA = 375 N 60∞ FB = 625 N 60∞

93


PROBLEM 3.85

Solve Problem 3.84, assuming a b= = ∞ 25 .

SOLutiOn

A

B

C

25˚

25˚25˚

0.3 m

P = 250 N

(a) Equivalence requires

SF F P F: B B= =or N250 25 0. ∞

SMB BM: ( . )[( )sin ] .= - ∞ = - ◊0 3 250 50 57 453 m N N m

The equivalent force-couple system at B is

FB = 250 N 25 0. ∞ MB = ◊57 5. N m

(b) Require

25˚25˚

50˚

0.2 m250 N

A

MB

B

C C

250 N

BE

– Q

Q

A

Equivalence requires

M d Q

Q

Q

B AE= ∞= ∞=

( . )[( )sin ]

[( . )sin ]

0 3 250 50

0 2 50

375

m N

m

N

Adding the forces at B: FA = 375 N 25 0. ∞ FB = 625 N 25 0. ∞

94


PROBLEM 3.86

A force and a couple are applied as shown to the end of a cantilever beam. (a) Replace this system with a single force F applied at Point C, and determine the distance d from C to a line drawn through Points D and E. (b) Solve Part a if the directions of the two 360-N forces are reversed.

SOLutiOn

(a) We have SF F j j k: ( ) ( ) ( )= - -360 360 600N N N

or F k= -( )600 N

and SM dD : ( )( . ) ( )( )360 0 15 600N m N=

d = 0 09. m

or d ED= 90 0. mm below

(b) We have from Part a F k= -( )600 N

and SM dD : ( )( . ) ( )( )- = -360 0 15 600N m N

d 0.09 m

or d ED= 90 0. mm above

360 N

360 N

450 mm

150 mm

450 mm

600 N

600 N

360 N

360 N

B

B

D

y

y

x

x

A B

A B

E D

DD

F

F

+C

+C

C

E

E E

d d

d

A

A

z

z

+

95


PROBLEM 3.87

The shearing forces exerted on the cross section of a steel channel can be represented by a 900-N vertical force and two 250-N horizontal forces as shown. Replace this force and couple with a single force F applied at Point C, and determine the distance x from C to line BD. (Point C is defined as the shear center of the section.)

SOLutiOn

Replace the 250-N forces with a couple and move the 900-N force to Point C such that its moment about H is equal to the moment of the couple

250 N

H

H H

X

C

D

MH

900 N900 N 900 N

250 N

0.18 m

MH == ◊

( . )( )0 18 250

45

N

N m

Then M xH = ( )900 N

or 45 900

0 05

N m N

m

◊ ==

x

x

( )

. F = 900 N Ø x = 50 0. mm

96


PROBLEM 3.88

While tapping a hole, a machinist applies the horizontal forces shown to the handle of the tap wrench. Show that these forces are equivalent to a single force, and specify, if possible, the point of application of the single force on the handle.

SOLutiOn

Since the forces at A and B are parallel, the force at B can be replaced with the sum of two forces with one of the forces equal in magnitude to the force at A except with an opposite sense, resulting in a force-couple.

Have FB = - =15 13 2N N N, where the 13 N force be part of the couple. Combining the two parallel forces,

Mcouple N mm mm

N mm

= + ∞

= ◊

( )[( )cos ]

.

13 80 70 25

1767 3

and Mcouple N mm= ◊1767 3.25°

B

2 N

2 N

25°

1767.3 N·mm

B

A

A

z

z

x

–a

x

A single equivalent force will be located in the negative z-direction

Based on SM aB : . cos- ◊ = ∞1767 3 25N mm [(2 N) ]( )

a = 975 mm

F9 = ∞ + ∞( )(cos sin )2 25 25N i k

F9 = +( . ) ( . )1 813 0 845N Ni k and is applied on an extension of handle BD at a distance of 975 mm to the right of B

97


PROBLEM 3.89

Three control rods attached to a lever ABC exert on it the forces shown. (a) Replace the three forces with an equivalent force-couple system at B. (b) Determine the single force that is equivalent to the force-couple system obtained in Part a, and specify its point of application on the lever.

SOLutiOn

(a) First note that the two 100-N forces form A couple. Then

F = 240 N u

where q = ∞ - ∞ + ∞ = ∞180 60 55 65( )

and M MB== ∞ - ∞= -

S( )( )cos ( )( )cos750 240 55 1750 100 20

61202

mm N mm N

NN mm◊ The equivalent force-couple system at B is

F = 240 N 65° M = ◊61 2. N m

(b) The single equivalent force ¢F is equal to F. Further, since the sense of M is clockwise, ¢F must be applied between A and B. For equivalence.

SM M aFB : cos= - ¢ ∞55

where a is the distance from B to the point of application of F9. Then

- ◊ = - ∞61202 240 55 N mm Na( )cos

or a = 444 6. mm ¢ =F 240 N 65.0°

and is applied to the lever 445 mm

To the left of pin B

A

60°

55°

θ

98


PROBLEM 3.90

A hexagonal plate is acted upon by the force P and the couple shown. Determine the magnitude and the direction of the smallest force P for which this system can be replaced with a single force at E.

SOLutiOn

From the statement of the problem, it follows that SME = 0 for the given force-couple system. Further, for P

min, must require that P be perpendicular to rB E/ . Then

SM

P

E : ( . sin . )

( . )sin

( . ) min

0 2 30 0 2

0 2 30 300

0 4

∞ + ¥+ ∞ ¥-

m 300 N

m N

m == 0

or Pmin = 300 N

Pmin = 300 N 30.0°

0.2 m 0.2 m

300 N300 N

B

E

Pmin

60°

30°

a =30°

99


PROBLEM 3.91

A rectangular plate is acted upon by the force and couple shown. This system is to be replaced with a single equivalent force. (a) For a = 40∞, specify the magnitude and the line of action of the equivalent force. (b) Specify the value of a if the line of action of the equivalent force is to intersect line CD 300 mm to the right of D.

SOLutiOn

(a) The given force-couple system (F, M) at B is

F = Ø48 N

and M MB= = ∞ + ∞S ( . )( )cos ( . )( )sin0 4 15 40 0 24 15 40 m N m N

or M = ◊6 9103. N m

The single equivalent force F9 is equal to F. Further for equivalence

SM M dFB : = ¢

or 6 9103 48. N m N◊ = ¥d

or d = 0 14396. m ¢ =F 48 N

and the line of action of F9 intersects line AB 144 mm to the right of A.

(b) Following the solution to Part a but with d = 0 1. m and a unknown, have

SMB : ( . )( )cos ( . )( )sin0 4 15 0 24 15 m N m Na + a

= ( . )( )0 1 48 m N

or 5 3 4cos sina a+ =

Rearranging and squaring 25 4 32 2cos ( sin )a a= -

Using cos sin2 21a a= - and expanding

25 1 16 24 92 2( sin ) sin sin- = - +a a a

or 34 24 9 02sin sina a- - =

Then sin( ) ( )( )

( )

sin . sin .

a

a a

=± - - -

= = -

24 24 4 34 9

2 34

0 97686 0 27098

2

or

a a= ∞ = - ∞77 7 15 72. .or

M

F

B

dF′

100


PROBLEM 3.92

An eccentric, compressive 1220-N force P is applied to the end of a cantilever beam. Replace P with an equivalent force-couple system at G.

y

y

M

F

0.1 m

0.06 mII

z

rN/G

1220 N x

x

G

G

z

SOLutiOn

We have

SF i F: ( )- =1220 N

F i= - ( )1220 N

Also, we have

SM r P MG AG: / ¥ =

1220 0 1 06

1 0 0

i j k

M- --

◊ =. . N m

M j k= ◊ - - - - -( )[( . )( ) ( . )( ) ]1220 0 06 1 0 1 1N m

or M j k= ◊ - ◊( . ) ( )73 2 122N m N m

101


PROBLEM 3.93

To keep a door closed, a wooden stick is wedged between the floor and the doorknob. The stick exerts at B a 175-N force directed along line AB. Replace that force with an equivalent force-couple system at C.

SOLutiOn

We have

SF P F: AB C=

where

P

i j kAB AB ABP=

=+ -

l( ) ( ) ( )

.( )

33 990 594

1155 00175

mm mm mm

mmN

or F i j kC = + -( . ) ( ) ( . )5 00 150 90 0N N N

We have SM r P MC B C AB C: / ¥ =

M

i j k

i

C = --

◊

= - - - -

5 0 683 0 860 0

1 30 18

0 860 18 0 683 1

. .

. ( . )(

N m

(5){( )( ) 88

0 683 30 0 860 1

)

[( . )( ) ( . )( )] }

j

k+ -

or M i j kC = ◊ + ◊ + ◊( . ) ( . ) ( . )77 4 61 5 106 8N m N m N m

750 mm

67 mm

1850 mm

C

B

O

x

A

rB/C

PAB

100 mm 594 mm

990 mm

z

y

102


PROBLEM 3.94

An antenna is guyed by three cables as shown. Knowing that the tension in cable AB is 1.44 kN, replace the force exerted at A by cable AB with an equivalent force-couple system at the center O of the base of the antenna.

SOLutiOn

We have dAB = - + - + =( ) ( ) ( ) .19 38 5 42 7782 2 2 m

Then T i j k

i j

AB = - - +

= - - +

1440

42 77819 38 5

639 58 1279 16 16

N

N N.

( )

( . ) ( . ) ( 88 31. )N kNow

M M r T

j i j k

= = ¥= ¥ - - +=

O A O AB/

[ ( . ) ( . ) ( . ) )]

(

38 639 58 1279 16 168 31

6396 N m N m◊ + ◊) ( )i k24304

The equivalent force-couple system at O is

F i j k= - - +( ) ( ) ( . )640 1279 168 3N N N

M i k= ◊ + ◊( . ) ( . )6 40 24 3 kN m kN m

103


PROBLEM 3.95

An antenna is guyed by three cables as shown. Knowing that the tension in cable AD is 1.35 kN, replace the force exerted at A by cable AD with an equivalent force-couple system at the center O of the base of the antenna.

SOLutiOn

We have dAD = - + - + -=

( ) ( ) ( )19 38 38

57

2 2 2

m

Then T i j k

i j k

AD = - - +

= - - -

1350

5719 38 38

450 2 2

N

N

( )

( )

Now M M r T

j i j k

i

= = ¥= ¥ - - -= - ◊ + ◊

O A O AD/

N m N m

128 450 2 2

115200 57600

( )

( ) ( ))k

The equivalent force-couple system at O is

F i j k= - - -( ) ( ) ( )450 900 900 N N N

M i k= - ◊ + ◊( . ) ( . )115 2 57 6 kN m kN m

104


PROBLEM 3.96

Replace the 150-N force with an equivalent force-couple system at A.

SOLutiOn

Equivalence requires SF F j k

j k

: ( )( cos sin )

( . ) ( . )

= - ∞ - ∞= - -

150 35 35

122 873 86 036

N

N N SM M r FA D A: = ¥/

where r i j kD A/ m m m= - +( . ) ( . ) ( . )0 18 0 12 0 1

Then M

i j k

= -- -

◊

= - - -

0 18 0 12 0 1

0 122 873 86 036

0 12 86 036 0

. . .

. .

[( . )( . ) (

N m

.. )( . )]

[ ( . )( . )]

[( . )( . )]

(

1 122 873

0 18 86 036

0 18 122 873

-+ - -+ -

=

i

j

k

222 6 15 49 22 1. ) ( . ) ( . ) N m N m N m◊ + ◊ - ◊i j k

The equivalent force-couple system at A is

F j k= - -( . ) ( . )122 9 86 0 N N

M i j k= ◊ + ◊ - ◊( . ) ( . ) ( . )22 6 15 49 22 1 N m N m N m

105


PROBLEM 3.97

A 77-N force F1 and a 31-N . m couple M

1 are

applied to corner E of the bent plate shown. If F

1 and M

1 are to be replaced with an equivalent

force-couple system (F2, M

2) at corner B and if

(M2)z = 0, determine (a) the distance d, (b) F

2

and M2.

SOLutiOn

(a) We have SM MBz z: 2 0=

k r F◊ ¥ + =( )/H B zM1 1 0 (1)

where r i jH B/ ( . ) ( . )= -0 31 0 0233m

F

i j k

i

1 1

0 06 0 06 0 07

0 1177

42 42

=

=+ -

= +

lEH F

( . ) ( . ) ( . )

.( )

( ) ( )

m m m

mN

N N jj k

k M

M

i j k

-= ◊=

=- + -

+

( )

( . ) ( . )

.(

49

0 03 0 07

0 0058

1 1

1 1

2

N

m m

m

M

M

d

d

z

EJl

331 N m◊ )

Then from Equation (1),

0 0 1

0 31 0 0233 0

42 42 49

0 07 31

0 00580

2. .

( . )( )

.-

-+

- ◊

+=

m N m

d

Solving for d, Equation (1) reduces to

( . . ).

.13 0200 0 9786

2 17

0 00580

2+ -

◊

+=

N m

d

From which d = 0 1350. m or d = 135 0. mm

106



(b) F F i j k2 1 42 42 49= = + -( )N or F i j k2 42 42 49= + -( ) ( ) ( )N N N

M r F M

i j ki j k

2 1 1

0 31 0 0233 0

42 42 49

0 1350 0 03 0 07

= ¥ +

= --

+ + -

H B/

. .( . ) . .

00 15500031

1 14170 15 1900 13 9986

.( )

( . . . )

(

N m

N m

27.000

◊

= + + ◊

+ -

i j k

i ++ - ◊

= - ◊ + ◊

6 0000 14 0000

25 858 21 1902

. .

( . ) ( . )

j k

M i j

) N m

N m N m

or M i j2 25 9 21 2= - ◊ + ◊( . ) ( . )N m N m

107


PROBLEM 3.98

A 230 N force F and a 265 N?m couple M are applied to corner A of the block shown. Replace the given force-couple system with an equivalent force-couple system at corner H.

SOLutiOn

We have dAJ = + - + - =( ) ( ) ( )450 350 75 5752 2 2 mm

Then F i j k

i j k

= - -

= - -

230

575450 350 75

180 140 30

N

N N N

( )

( ) ( ) ( )

Also dAC = - + + - =( ) ( ) ( )1125 0 700 13252 2 2 mm

Then M i k

i k

=◊

- -

= - ◊ - ◊

265

13251125 700

225 140

N m

N m N m

( )

( ) ( )

Now ¢ = + ¥M M r FA H/

where r i jA H/ mm mm= +( ) ( )1125 350

Then ¢ = - - +- -

= +M i k

i j k

i j( ) . . ( . ) ( )225 140 1 125 0 35 0

180 140 30

1 125 350m m

M ¢ -= - - - +

= - ◊ + ◊225 140 10 5 33 75

235 5 33 75

i k i j k

i j

. .

( . ) ( . )

220.5

N m N m -- ◊( . )360 5 N m k

The equivalent force-couple system at H is ¢ = - -F i j k( ) ( ) ( )180 140 30 N N N

¢ = - ◊ + ◊ - ◊M i j k( . ) ( . ) ( . )235 5 33 75 360 5 N m N m N m

108


PROBLEM 3.99

The handpiece for a miniature industrial grinder weighs 3 N, and its center of gravity is located on the y axis. The head of the handpiece is offset in the xz plane in such a way that line BC forms an angle of 25∞ with the x direction. Show that the weight of the handpiece and the two couples M

1 and M

2 can be replaced with a

single equivalent force. Further, assuming that M1 = 85 N?mm and

M2 = 81.25 N?mm, determine (a) the magnitude and the direction

of the equivalent force, (b) the point where its line of action intersects the xz plane.

SOLutiOn

First assume that the given force W and couples M1 and M

2 act at the origin.

Now W j= -W

and M M M

i k

= += - ∞ + - ∞

1 2

2 1 225 25( cos ) ( sin )M M M

Note that since W and M are perpendicular, it follows that they can be replaced

with a single equivalent force.

(a) We have F W= = - = -W F jjor N( )3 or F j= - (3N)

(b) Assume that the line of action of F passes through Point P(x, 0, z). Then for equivalence

M r F= ¥P /0

where r i kP x z/0 = +

- ∞ + - ∞( cos ) ( sin )M M M2 1 225 25i k

=-

= -i j k

i kx z

W

Wz Wx0

0 0

( ) ( )

Equating the i and k coefficients, zM

Wx

M M

Wz=

- ∞= -

- ∞ÊËÁ

ˆ¯̃

cos sin25 251 2and

(b) For W M M= = ◊ = ◊3 85 81 251 2N N mm N mm.

x = - - ∞-

ÊËÁ

ˆ¯̃

=0 85 81 25 25

316 887

. . sin. mm or x = 16 89. mm

z =- ∞

= -81 25 25

324 546

. cos. mm or z = -24 5. mm

y

zx

M1

M2W

0

25˚

y yF

zzx

x

rP/O

W

o oM

P(x, 0, z)

109


PROBLEM 3.100

A 4-m-long beam is subjected to a variety of loadings. (a) Replace each loading with an equivalent force-couple system at end A of the beam. (b) Which of the loadings are equivalent?

SOLutiOn

(a) (a) We have SF Ry a: - - =400 200N N

or Ra = Ø600 N

and SM MA a: ( )( )1800 200 4N m N m◊ - =

or Ma = ◊1000 N m

(b) We have SF Ry b: - =600 N

or Rb = Ø600 N

and SM MA b: - ◊ =900 N m

or Mb = ◊900 N m

(c) We have SF Ry c: 300 900N N- =

or Rc = Ø600 N

and SM MA c: ( )( )4500 900 4N m N m◊ - =

or Mc = ◊900 N m

A B

R

M4 m

110



(d) We have SF Ry d: - + =400 800N N

or Rd = ≠400 N

and SM MA d: ( ( )800 4 2300N) m N m- ◊ =

or Md = ◊900 N m

(e) We have SF Ry e: - - =400 200N N

or Re = Ø600 N

and SM MA e: (200 400 200N m N m N)(4 m)◊ + ◊ - =

or Me = ◊200 N m

(f) We have SF Ry f: - + =800 200N N

or R f = Ø600 N

and SM MA f: (- ◊ + ◊ + =300 300 200N m N m N)(4 m)

or M f = ◊800 N m

(g) We have SF Ry g: - - =200 800N N

or Rg = Ø1000 N

and SM MA g: (200 4000 800N m N m N)(4 m)◊ + ◊ - =

or Mg = ◊1000 N m

(h) We have SF Ry h: - - =300 300N N

or Rh = Ø600 N

and SM MA h: (2400 300 300N m N m N)(4 m)◊ - ◊ - =

or Mh = ◊900 N m

(b) Therefore, loadings (c) and (h) are equivalent.

111


PROBLEM 3.101

A 4-m-long beam is loaded as shown. Determine the loading of Problem 3.100 which is equivalent to this loading.

SOLutiOn

We have SFy R: - - =200 400N N or R = 600 N Ø

and SM MA: ( )( )- ◊ + ◊ - =400 2800 400 4N m N m N m

or M = ◊800 N m

Equivalent to case (f) of Problem 3.101

Problem 3.101 Equivalent force-couples at A

Case R M

(a) 600 N Ø 1000 N m◊

(b) 600 N Ø 900 N m◊

(c) 600 N Ø 900 N m◊

(d) 400 N ≠ 900 N m◊

(e) 600 N Ø 200 N m◊

(f) 600 N Ø 800 N m◊

(g) 1000 N Ø 1000 N m◊

(h) 600 N Ø 900 N m◊

R

A B

M 4 m

112


PROBLEM 3.102

Determine the single equivalent force and the distance from Point A to its line of action for the beam and loading of (a) Problem 3.101b, (b) Problem 3.101d, (c) Problem 3.101e.

PROBLEM 3.101 A 4-m-long beam is subjected to a variety of loadings. (a) Replace each loading with an equivalent force-couple system at end A of the beam. (b) Which of the loadings are equivalent?

SOLutiOn

(a) For equivalent single force at distance d from A

We have SF Ry : - =600 N

or R = 600 N Ø

and SM dC : ( )( )600 900 0N N m- ◊ =

or d = 1 500. m

(b) We have SF Ry : - + =400 800N N

or R = 400 N ≠

and SM d dC : ( )( ) ( )( )400 800 4

2300 0

N N

N m

+ -- ◊ =

or d = 2 25. m

(c) We have SF Ry : - - =400 200N N

or R = 600 N Ø

and SM d

dC : ( ( )

( )( )

200 400

200 4 400 0

N m N)

N N m

◊ +- - + ◊ =

or d = 0 333. m

600 N 900 N·m

B

B

B

A

A

A B

B

A

A

C

C

R

R

R

d

d

A

C

C

C

C

400 N 2300 N·m

200 N

400 N·m

400 N

200 N·m

800 N4 m

4 m

4 m

B

d

113


PROBLEM 3.103

Five separate force-couple systems act at the corners of a piece of sheet metal, which has been bent into the shape shown. Determine which of these systems is equivalent to a force F (50 N)i and a couple of moment M (22.5 N ? m)j (22.5 N ? m)k located at the origin.

SOLutiOn

First note that the force-couple system at F cannot be equivalent because of the direction of the force [The force of the other four systems is (10 N)i]. Next move each of the systems to the origin O; the forces remain unchanged.

A A O: ( . ) ( . ) ( . ) ( )M M j k k i= = ◊ + ◊ + ¥S 7 5 22 5 0 6 50 N m N m m N

= ◊ + ◊( . ) ( . )37 5 22 5 N m N mj k

D D O: ( . ) ( . )

[( . ) ( . ) ( .

M M j k

i j

= = - ◊ + ◊+ + +

S 7 5 37 5

1 35 0 3 0 6

N m N m

m m mm N

N m N m

) ] ( )

( . ) ( . )

k i

j k

¥ 50

22 5 22 5= ◊ + ◊

G

IG O

I I

: ( . ) ( . )

: ( . ) ( .

M M i j

M M j

= = ◊ + ◊= = ◊ -

SS

22 5 22 5

22 5 7 5

N m N m

N m NN m

m m N

◊+ + ¥

)

[( . ) ( . ) ] ( )

k

i j i1 35 0 3 50 = ◊ - ◊( . ) ( . )22 5 22 5 N m N mj k

The equivalent force-couple system is the system at corner D.

114


PROBLEM 3.104

The weights of two children sitting at ends A and B of a seesaw are 420 N and 320 N, respectively. Where should a third child sit so that the resultant of the weights of the three children will pass through C if she weighs (a) 300 N, (b) 260 N.

SOLutiOn

C

A B

420 N

1.8 m 1.8 m

320 Nd Wc

(a) For the resultant weight to act at C, SM WC C= =0 300 N

Then ( )( . ) ( ( ) ( ( . )420 1 8 300 320 1 8 0 N m N) N) m- - =d

d C= 0 6. m to the right of

(b) For the resultant weight to act at C, SM WC C= =0 260 N

Then ( )( . ) ( ( ) ( ( . )420 1 8 260 320 1 8 0 N m N) N) m- - =d

d C= 0 692. m to the right of

115


PROBLEM 3.105

Three stage lights are mounted on a pipe as shown. The lights at A and B each weigh 20.5 N, while the one at C weighs 17.5 N. (a) If d 625 mm, determine the distance from D to the line of action of the resultant of the weights of the three lights. (b) Determine the value of d so that the resultant of the weights passes through the midpoint of the pipe.

SOLutiOn

A

20.5 N 20.5 N 17.5 N

B C

E

L

D ED

R

For equivalence

SF Ry : or 58.5 N- - - = - = Ø20 5 20 5 17 5. . . R

M

dD : mm N mm N

mm N

- -- +

( )( . ) ( )( . )

[( ) ]( .

250 20 5 1100 20 5

1100 17 5 )) ( )( . )= - L mm N58 5

or 46925 17 5 58 5+ =. . ( ,d L d L in mm)

(a) d = 625 mm

We have 46925 17 5 625 58 5 989+ = =( . )( ) . L Lor mm

The resultant passes through a Point 989 mm to the right of D.

(b) L = 1050 mm

We have 46925 17 5 58 5 1050+ =( . ) ( . )( )d or d = 829 mm

116


PROBLEM 3.106

A beam supports three loads of given magnitude and a fourth load whose magnitude is a function of position. If b = 1.5 m and the loads are to be replaced with a single equivalent force, determine (a) the value of a so that the distance from support A to the line of action of the equivalent force is maximum, (b) the magnitude of the equivalent force and its point of application on the beam.

SOLutiOn

1300 N 400 N 600 N

B

L R

A BA400 Nab

For equivalence

SFa

bRy : - + - - = -1300 400 400 600

or Ra

b= -Ê

ËÁˆ¯̃

2300 400 N (1)

SM

a a

ba a b LRA:

2400 400 600Ê

ËÁˆ¯̃

- - + = -( ) ( )( )

or L

a bab

ab

=+ -

-

1000 600 200

2300 400

2

Then with b La a

a= =

+ -

-1 5

10 943

2383

2

. m (2)

where a, L are in m

(a) Find value of a to maximize L

dL

da

a a a a=

-ÊËÁ

ˆ¯̃ -Ê

ËÁˆ¯̃ - + -Ê

ËÁˆ¯̃ -Ê

ËÁˆ¯̃

-

1083

2383

10 943

83

23

2

883

2

aÊËÁ

ˆ¯̃

117



or 230184

3

80

3

64

9

80

324

32

902 2- - + + + - =a a a a a

or 16 276 1143 02a a- + =

Then a =± - -276 276 4 16 1143

2 16

2( ) ( )( )

( )

or a a= =10 3435 6 9065. . m and m

Since AB = 9 m, a must be less than 9 m a = 6 91. m

(b) Using Eq. (1) R = -2300 4006 9065

1 5

.

. or R = 458 N

and using Eq. (2) L =+ -

-=

10 6 9065 943

6 9065

2383

6 90653 16

2( . ) ( . )

( . ). m

R is applied 3.16 m to the right of A

118


43

2160°

15 m

33 m45° 4

21 m3

27 m 30 30 30

mO m m60 m

PROBLEM 3.107

Four tugboats are used to bring an ocean liner to its pier. Each tugboat exerts a 25 kN force in the direction shown. Determine (a) the equivalent force-couple system at the foremast O, (b) the point on the hull where a single, more powerful tugboat should push to produce the same effect as the original four tugboats.

SOLutiOn

(a) Force-Couple System at O. Each of the given forces is resolved into components in the diagram shown (kip units are used). The force-couple system at O equivalent to the given system of forces consists of a force R and a couple M

OR defined as follows:

R F== + + - +

S- 21.65 - 20 -17.68( . ) ( ) ( ) ( .12 5 15 25 17 68i j i j j i j)

= (45.188 )i j- 48.97

MOR = ¥

= - + ¥ -+ + ¥ -+

S( )

( ) ( . . )

( ) ( )

(

r F

i j i j

i j i j

27 15 12 5 21 65

30 21 15 20

1120 21 25

90 21 17 68 17 68

584 55 187 5

i j j

i j i j

+ ¥ -+ + ¥ -

= - -

) ( )

( ) ( . . )

( . . 6600 315 3000 1591 2 371 28

1555 47

- - + += -

. . )

.

k

k

The equivalent force-couple system at O is thus,

R i j M k= - - ◊( . ) ( . ) ( )45 2 49 0 1555kN kN kN mOR

or R = 666 kN 47.3° MOR = ◊1400 9. kN m

(b) Single Tugboat. The force exerted by a single tugboat must be equal to R, and its point of application A must be such that the moment of R about O is equal to M

OR. Observing that the position

vector of A is

r i j= +x 21

we write r R M¥ = OR

( ) ( . ) .

( . ) . .

x +

x

i j i j k

k k

21 45 18 4847 1555 47

48 97 948 78 1555

¥ - = -- = -- 447k x = 12.39 m

15 m

33 m

F1F2 F3

F4

17.68i

12.5i 15i

17.68j

– 25j– 20j– 21.65j

21 m27 m 30 30 30

mO m m60 m

40.65i

47.3°

44.1jR

MRO= – (1400.9 k)

O

40.65i

44.1j

21 mA

R

x

Oo

119


PROBLEM 3.108

A couple of magnitude M = 7 N?m and the three forces shown are applied to an angle bracket. (a) Find the resultant of this system of forces. (b) Locate the points where the line of action of the resultant intersects line AB and line BC.

R

E

C

c

Ba

D

A

SOLutiOn

(a) We have SF j i

j i

i

: ( ) ( cos )

sin ( )

( ) ( .

R = - + ∞+ ∞ + -

= - +

50 150 60

150 60 225

150 79 N 99 N)j

or R = 170 N 28.0°

(b) First reduce the given forces and couple to an equivalent force-couple system ( , )R M B at B.

We have SM MB B: ( ) ( )( ) ( )( )= ◊ + -= -

7000 300 50 200 225

23000

N mm mm N mm N

N ◊◊mm

Then with R at D SM aB : ( . )- ◊ =23000 79 9 N m N

or a = 288 m

and with R at E SM cB : ( )- ◊ =23000 150 N m N

or c = 153 3. mm

The line of action of R intersects line AB 288 mm to the left of B and intersects line BC 153.3 mm below B.

120


PROBLEM 3.109

A couple M and the three forces shown are applied to an angle bracket. Find the moment of the couple if the line of action of the resultant of the force system is to pass through (a) Point A, (b) Point B, (c) Point C.

SOLutiOn

In each case, must have MiR = 0

(a) + M M MAB

A= = + ∞ - =S ( )[( )sin ] ( )( )300 150 60 200 225 0 mm N mm N

M = + ◊6028 9. N mm M = ◊6 03. mN

(b) + M M MBR

B= = + - =S ( )( ) ( )( )300 50 200 225 0 mm N mm N

M = + ◊30000 N mm M = ◊30 N m

(c) + M M MCR

C= = + - ∞ =S ( )( ) ( )[( )cos ]300 50 200 150 60 0 mm N mm N

M = 0 M = 0

121


PROBLEM 3.110

Four forces act on a 700 3 375-mm plate as shown. (a) Find the resultant of these forces. (b) Locate the two points where the line of action of the resultant intersects the edge of the plate.

SOLutiOn

(a) R

N N 760 N

N N N

N)

== - + -

+ + += -

SF

i

j

i

( )

( )

(

400 160

600 300 300

1000 ++ ( )1200 N j

R = +

=

= -ÊËÁ

ˆ¯̃

= -

( ) ( )

.

tan

1000 1200

1562 09

1200

1

2 2 N N

N

N

1000 Nq

..

.

20000

50 194q = - ∞

R = 1562 N 50.2°

(b) M r F

i j

k

CR = ¥

= ¥ += ◊

S( . ) ( )

( )

0 5 300 300

300

m N N

N m

( ) ( )

.

(

300 1200

0 25000

250

1

N m N

m

mm

(300 N m)

◊ = ¥==

◊ = ¥ -

k i j

j

x

x

x

y 0000

0 30000

300

N)

m

mm

i

y

y

==

.

Intersection 250 mm to right of C and 300 mm above C

C

D

A B

E

600 N760 N

340 N500 N

3 174 8

5 15

y

C

x

– 1000 N1200 N

122


PROBLEM 3.111

Solve Problem 3.110, assuming that the 760-N force is directed to the right.

PROBLEM 3.110 Four forces act on a 700 3 375-mm plate as shown. (a) Find the resultant of these forces. (b) Locate the two points where the line of action of the resultant intersects the edge of the plate.

SOLutiOn

(a) R

i

j

i

== - + +

+ + += +

SF

N N 760 N

N N N

N

( )

( )

( ) (

400 160

600 300 300

520 12000 N)j

R = + =

= ÊËÁ

ˆ¯̃

=

( ) ( ) .

tan .

520 1200 1307 82

12002 307

2 2 N N N

N

520 Nq 77

66 5714q = ∞.

R = 1308 N 66.6°

(b) M r F

i j

k

CR = ¥

= ¥ += ◊

S( . ) ( )

( )

0 5 300 300

300

m N N

N m

( ) ( )

.

300 1200

0 25000

N m N

m

◊ = ¥=

k i jx

x or x = 0 250. mm

(300 N m) m N N◊ = ¢ + ¥ +

= ¢ -k i j i j[ ( . ) ] [( ) ( ) ]

(

x

x

0 375 520 1200

1200 1995)k ¢ =x 0 41250. m or ¢ =x 412 5. mm

Intersection 412 mm to the right of A and 250 mm to the right of C

A B

E 760N600N

500N 340N

DC

1517

8

3 5

4

A B

E

x

x´

1200N

1200 N

520N

0.375m

C

123


PROBLEM 3.112

A truss supports the loading shown. Determine the equivalent force acting on the truss and the point of intersection of its line of action with a line drawn through Points A and G.

SOLutiOn

We have R F

R i j j

i

== ∞ - ∞ -

+ - ∞ -

S( )(cos sin ) ( )

( )( cos si

1200 70 70 800

1500 40

N N

N nn ) ( )40 900∞ -j jN

R i j= - -

= +

= - +=

( . ) ( . )

( . ) ( . )

738 64 3791 8

738 6 3791 8

3863

2 2

2 2

N N

R R Rx y

NN

q =ÊËÁ

ˆ¯̃

= --

ÊËÁ

ˆ¯̃

= ∞

-

-

tan

tan.

.

.

1

1 3791 8

738 6

78 977

R

Ry

x

or R = 3860 N 79 0. ∞

We have SM dRA y=

where SM A = - ∞ - ∞- +[ cos ]( . ) [ sin ]( . )

( )( . )

1200 70 1 8 1200 70 1 2

800 3 6

N m N m

N m [[ cos ]( . )

[ sin ]( . ) ( )( . )

1500 40 1 8

1500 40 6 0 900 2 4

108

N m

N m N m

∞- ∞ -

= - 448 75. N m◊

d =- ◊

-=

10848 75

3791 8

2 861

.

.

.

N m

N

m

or d A= 2 86. m to the right of

A

y

B D F

E

G

C

1200 N800 N 1500 N

900 N

x

40˚70˚

y

A

B D F

C E

G x

dR

124


PROBLEM 3.113

Pulleys A and B are mounted on bracket CDEF. The tension on each side of the two belts is as shown. Replace the four forces with a single equivalent force, and determine where its line of action intersects the bottom edge of the bracket.

SOLutiOn

Equivalent force-couple at A due to belts on pulley A

We have SF: - - =600 800N N RA

R A = Ø1400 N

We have SMA: - =200 50N mm( ) M A

MA = ◊10000 N mm

Equivalent force-couple at B due to belts on pulley B

We have SF: N N( )1050 750+ 25∞ = RB

RB = 1800 N 25°

We have SMB BM: - =300 N(37.5 mm)

MB = ◊11250 N mm

Equivalent force-couple at F

We have SF R j i j: N NF = - + ∞ + ∞( ) ( )(cos sin )1400 1800 25 25

= -=

= +

= + -

( . ) ( . )

( . ) ( .

1631 35 639 285

1631 35 639 28

2 2

2

N Ni j

R R

R R

F

Fx Fy

55

1752 13

639 285

1631 35

2

1

1

)

.

tan

tan.

.

=

=ÊËÁ

ˆ¯̃

= -ÊËÁ

ˆ¯

-

-

N

qR

RFy

Fx

˜̃

= - ∞21 399.

or R RF = = 1752 N 21.4°

y

xF

AC

BE

D

MA = 10000 N· mm

RA= 1400 NRB= 1800 N

MB= 11250 N· mm

y

xF

C

E

DMF

RF= R

y

xF

C

E

D

Rd

125



We have SMF FM: N mm N mm= - - ◊( )( ) ,1400 150 10 000

- ∞+ ∞ - ◊

[( )cos ]( )

[( )sin ]( )

1800 25 25

1800 25 300 11250

N mm

N mm N mm M kF = - ◊( )43820 N mm

To determine where a single resultant force will intersect line FE,

M dR

dM

R

F y

F

y

=

=

= --

=

43820

639 28568 545

.. mm

or d = 68 6. mm

126


PROBLEM 3.114

A machine component is subjected to the forces and couples shown. The component is to be held in place by a single rivet that can resist a force but not a couple. For P = 0, determine the location of the rivet hole if it is to be located (a) on line FG, (b) on line GH.

SOLutiOn

We have120 N

80 N200 N

42 N·m

40 N·m P

Ry

Rx

RMG

+a

b

G H

J

++

+ F

I

First replace the applied forces and couples with an equivalent force-couple system at G.

Thus SF P Rx x: cos cos200 15 120 70∞ - ∞ + =

or R Px = +( . )152 142 N

SF Ry y: sin sin- ∞ - ∞ - =200 15 120 70 80

or Ry = -244 53. N SMG : ( . )( )cos ( . )( )sin

( . )(

- ∞ + ∞+

0 47 200 15 0 05 200 15

0 47

m N m N

m 1120 70 0 19 120 70

0 13 0 59

N m N

m N) m

)cos ( . )( )sin

( . )( ( . )

∞ - ∞- -P (( )80 42 N N m

40 N m

+ ◊+ ◊ = MG

or M PG = - + ◊( . . )55 544 0 13 N m (1)

Setting P = 0 in Eq. (1):Now with R at I SM aG : . ( .- ◊ = -55 544 244 53 N m N)

or a = 0 227. mand with R at J SM bG : . ( .- ◊ = -55 544 152 142 N m N)

or b = 0 365. m

(a) The rivet hole is 0.365 m above G.

(b) The rivet hole is 0.227 m to the right of G.

127


PROBLEM 3.115

Solve Problem 3.114, assuming that P = 60 N.

PROBLEM 3.115 A machine component is subjected to the forces and couples shown. The component is to be held in place by a single rivet that can resist a force but not a couple. For P = 0, determine the location of the rivet hole if it is to be located (a) on line FG, (b) on line GH.

SOLutiOn

See the solution to Problem 3.115 leading to the development of Equation (1)

and

M P

R PG

x

= - + ◊= +

( . . )

( . )

55 544 0 13

152 142

N m

N

For P = 60 N

We have R

M

x

G

= +== - += - ◊

( . )

.

[ . . ( )]

.

152 142 60

212 14

55 544 0 13 60

63 344

N

N m

Then with R at I SM aG : . ( . )- ◊ = -63 344 244 53 N m N

or a = 0 259. m

and with R at J SM bG : . ( .- ◊ = -63 344 212 14 N m N)

or b = 0 299. m

(a) The rivet hole is 0.299 m above G.

(b) The rivet hole is 0.259 m to the right of G.

128


PROBLEM 3.116

A 160 N motor is mounted on the floor. Find the resultant of the weight and the forces exerted on the belt, and determine where the line of action of the resultant intersects the floor.

SOLutiOn

We have

SF i j i j R: ( ) ( ) ( )(cos sin )300 160 700 30 30N N N- + ∞ + ∞ =

R i j= +( . ) ( )906 22 190N N

or R = 926 N 11.84°

We have S SM M xRO O y: =

- ∞ + ∞ - ∞[( )cos ][( cos ) ] [( )sin ]

[( )sin

700 30 100 50 30 700 30

50

N mm N

mm 330 300 50 190∞ - =] ( )( ) ( )N mm Nx

x = - - - ◊1

19086871 8750 15000( ) N mm

and x = -582 21. mm

Or, resultant intersects the base (x axis) 582 mm to the left of the vertical centerline (y axis) of the motor.

700 sin 30°

700 cos 30°

y

O x

50 mm 160 N 50 mm

50 mm300 N

129


PROBLEM 3.117

As follower AB rolls along the surface of member C, it exerts a constant force F perpendicular to the surface. (a) Replace F with an equivalent force-couple system at the Point D obtained by drawing the perpendicular from the point of contact to the x axis. (b) For a = 1 m and b = 2 m, determine the value of x for which the moment of the equivalent force-couple system at D is maximum.

SOLutiOn

(a) The slope of any tangent to the surface of member C is

dy

dx

d

dxb

x

a

b

ax= -

Ê

ËÁˆ

¯̃

È

ÎÍÍ

˘

˚˙˙

= -1

22

2 2

Since the force F is perpendicular to the surface,

tana = -ÊËÁ

ˆ¯̃

= ÊËÁ

ˆ¯̃

-dy

dx

a

b x

1 2

2

1

For equivalence

SF : F R=

SM F y MD A D: ( cos )( )a =

where

cos( ) ( )

a =+

= -Ê

ËÁˆ

¯̃

=-

ÊËÁ

ˆ¯̃

+

2

2

1

2

2 2 2

2

2

23

2

4

bx

a bx

y bx

a

M

Fb xx

a

a

A

D44 2 2b x

Therefore, the equivalent force-couple system at D is

R = F tan- Ê

ËÁˆ

¯̃1

2

2

a

bx

M =

-ÊËÁ

ˆ¯̃

+

2

4

23

2

4 2 2

Fb xx

a

a b x

2bx

a2

a

b

D

y

x

F

A

Cy = b 1 x2

a2−

130



(b) To maximize M, the value of x must satisfy dM

dx= 0

where, for a b= =1 2m, m

MF x x

x

dM

dxF

x x x x x x

= -

+

=+ - - - +

8

1 16

81 16 1 3

12

32 1 16

3

2

2 2 3 2

( )

( ) ( ) ( )( )--ÈÎÍ

˘˚̇

+=

1 2

21 160

/

( )x

( )( ) ( )1 16 1 3 16 02 2 3+ - - - =x x x x x

or 32 3 1 04 2x x+ - =

x2 3 9 4 32 1

2 320 136011 0 22976=

- ± - -= -

( )( )

( ). .m and m2 2

Using the positive value of x2 x = 0 36880. m or x = 369 mm

131


PROBLEM 3.118

Four forces are applied to the machine component ABDE as shown. Replace these forces by an equivalent force-couple system at A.

SOLutiOn

R j i i k

R i j

= - - - -= - - -

( ) ( ( ) ( )

( ) ( (

50 300 120 250

420 50

N N) N N

N N) 2250

0 2

0 2 0 16

0 2 0 1

N

m

m m

m m

)

( . )

( . ) ( . )

( . ) ( .

k

r i

r i k

r i

B

D

E

== += - )) ( . )j k+ 0 16 m

M r i j

r k r i

j

AR

B

D

= ¥ - -+ ¥ - + ¥ -

=

[ ( ) ( ) ]

(

300 50

250

N N

N) ( 120 N)

0.2

i k

m 0

0 N N

m m

N

m m

0

30 50 0

0 2 0 0 16

0 0 250

0 2 0 1 0

- -+

-

+ -

i j k

i j k

. .

. . .116

120 0 0

10 50 19 2 12

m

N

N m N m N m N m

-

= - ◊ + ◊ - ◊ - ◊( ) ( ) ( . ) ( )k j j k

Force-couple system at A is

R i j k M j k= - - - = ◊ - ◊( ) ( ) ( ) ( . ) ( )420 50 250 30 8 220 N N N N m N mAR

A

y

xD

E

B0.16 m

0.2 m

0.1m

- (300 N) i

- (50 N) j

- (250 N) k- (120 N) i

z

132


PROBLEM 3.119

Two 150-mm-diameter pulleys are mounted on line shaft AD. The belts at B and C lie in vertical planes parallel to the yz plane. Replace the belt forces shown with an equivalent force-couple system at A.

SOLutiOn

Equivalent force-couple at each pulley

Pulley B R j k j

j k

M

B

B

= - ∞ + ∞ -= - +

( )( cos sin )

( . ) ( . )

145 20 20 215

351 26 49 593

N N

N N

== - -= - ◊

( )( . )

( . )

215 145 0 075

5 25

N N m

N m

i

i

Pulley C R j k

j k

M

C

C

= + - ∞ - ∞= - -=

( )( sin cos )

( . ) ( . )

155 240 10 10

68 591 389 00

N N

N N

(( )( . )

( . )

240 155 0 075

6 3750

N N m

N m

-= ◊

i

i

Then R R R j k= + = - -B C ( . ) ( . )419 85 339 41N or R j k= -( ) ( )420 339N N M M M r R r R

i i

i j kA B C B A B C A C= + + ¥ + ¥

= - ◊ + ◊ +

/ /

( . ) ( . ) .5 25 6 3750 0 225N m N m 00 0

0 351 26 49 593

0 45 0 0

0 68 591 389 00

1 12500

-◊

- -◊

=

. .

.

. .

( .

N m

+ N m

N

i j k

◊◊ + ◊ - ◊m N m N m) ( . ) ( . )i j k163 892 109 899

or M i j kA = ◊ + ◊ - ◊( . ) ( . ) ( . )1 125 163 9 109 9N m N m N m

z

y

z

240 N

155 N 10°

y

z

y

145N 215N

20˚

10˚

10˚

B

C C

z

y

MB

MC

RB

RC

B

133


PROBLEM 3.120

While using a pencil sharpener, a student applies the forces and couple shown. (a) Determine the forces exerted at B and C knowing that these forces and the couple are equivalent to a force-couple system at A consisting of the force R i j= -(13N) + (3.5N)Ry k and the couple M i kA

RxM (1 5 N m) (1.08 N m) .= + ◊ ◊ - ◊j (b) Find the corresponding

values of Ry and Mx .

SOLutiOn

(a) From the statement of the problem, equivalence requires

Â + =F B C R:

or SF B Cx x x: + = 13 N (1)

SF C Ry y y: - = (2)

SF C Cz z z: . .- = - =3 5 3 5 N or N

and Â ¥ + + ¥ =M r B M r CA : ( )/ /B A B C A ARM

or Â ◊ + =Mx y xC M: ( ) ( )( )1500 45N mm mm (3)

SM B Cy x x: ( )( ) ( )( ) ( )( )95 45 85 3 5 1500mm mm mm mm+ + ◊ = ◊N N

or 95 45 1202 5B Cx x+ = .

Using Eq. (1) 95 45 13 1202 5B Bx x+ - =( ) .

or Bx = 12 35. N

and Cx = 0 65. N

SM Cz y: ( )- ( ) = - ◊85 1080mm N mm

or Cy = 12 706. N

B C= = - -( . ) ( . ) ( . ) ( . )12 35 0 65 12 71 3 5 N N N Ni i j k

(b) Eq. (2) fi Ry = -12 71. N

Using Eq. (3) 1500 45 12 706+ =( )( . ) Mx

2071.77 N . mm

or Mx = ◊2 07. N m

134


PROBLEM 3.121

A mechanic uses a crowfoot wrench to loosen a bolt at C. The mechanic holds the socket wrench handle at Points A and B and applies forces at these points. Knowing that these forces are equivalent to a force-couple system at C consisting of the force C i k= (40 N) + (20 N) and the couple MC = ◊(45 N m) ,idetermine the forces applied at A and at B when Az = 10 N.

SOLutiOn

We have SF: A B C+ =

or F A Bx x x: + = 40 N

B Ax x= - +( )40 N (1)

SF A By y y: + = 0

or A By y= - (2)

SF Bz z: 10 20N N+ =

or Bz = 10 N (3)

We have SM r B r A MC B C A C C: / /¥ + ¥ =

i j k i j k

i200 0 50

10

200 0 200

10

- + ◊ = ◊B B A Ax y x y

N mm 45000 N mm)(

or ( ) ( )50 200 50 2000 200 2000B A B Ay y x x- + - + -i j

+ + = ◊( ) ( )200 200 45000B Ay y k iN mm

From i-coefficient 50 200 45000B Ay y- = (4)

j-coefficient 50 200 4000B Ax x+ = (5)

k-coefficient 200 200 0B Ay y+ = (6)

40 N

20 N

Az=10N

Ay

By

BBx

Ax

45000 N·mm

y

C

A

xz Bz

135



From Equations (2) and (4): 50 200 45000B By y- - =( )

B Ay y= = -180 180N N

From Equations (1) and (5): 50 40 200 4000( )- - + =A Ax x

Ax = 40 N

From Equation (1): Bx = - + = -( )40 40 80 N

A i j k= - +( ( ) ( )40 180 10N) N N

B i j k= - + +( ( ) ( )80 180 10N) N N

136


PROBLEM 3.122

As an adjustable brace BC is used to bring a wall into plumb, the force-couple system shown is exerted on the wall. Replace this force-couple system with an equivalent force-couple system at A if R = 106 N and M = ◊20 N m.

SOLutiOn

We have SF R R R: = =A BCl

where lBC =- -( ) ( ) ( )1050 2400 400

2650

mm mm mmi j k

mm

R i j kA = - -106

26501050 2400 400

N( )

or R i j kA = - -( ) ( ) ( )42 96 16N N N

We have SM r R M MA C A A: / ¥ + =

where r i k

kC A/ mm

mj m

= += +

( ) ( )

. .

1050 1200

1 05 1 2

mm

R i j k

M

i j k

= - -= -

= - + + ◊

( ) ( ) ( )

( )

42 96 16

1050 2400 400

265020

N N N

N m

lBC M

== - ◊ + ◊ + ◊( . ) ( . ) ( . )7 92 18 113 3 019N m N m N mi j k

Then

i j k

i j k1 05 0 1 2

42 96 16

7 924 18 113 3 019. . ( . . . )

- -◊ + - + + ◊N m N m

M i j kA = ◊ + ◊ - ◊( . ) ( . ) ( . )107 28 85 313 97 781N m N m N m

or M i j kA = ◊ + ◊ - ◊( . ( . )107 3 97 8N m) (85.3 N m) N m

R

M

rC/A

B

A

C

x

y

z

137


PROBLEM 3.123

A mechanic replaces a car’s exhaust system by firmly clamping the catalytic converter FG to its mounting brackets H and I and then loosely assembling the mufflers and the exhaust pipes. To position the tailpipe AB, he pushes in and up at A while pulling down at B. (a) Replace the given force system with an equivalent force-couple system at D. (b) Determine whether pipe CD tends to rotate clockwise or counterclockwise relative to muffler DE, as viewed by the mechanic.

SOLutiOn


SF R A B: = +

= ∞ - ∞ -= - -

( )(cos sin ) ( )

( . ) ( )

100 30 30 115

28 4 50

N N

N N

j k j

j k and SM M r F r F/ /D D A D A B D B: = ¥ + ¥

where r i j k

r iA D

B D

/

/

m m m

m

= - - += - +

( . ) ( . ) ( . )

( . ) ( .

0 48 0 225 1 12

0 38 0 82 mm)k

Then

M

i j k i j k

D = - -∞ - ∞

+ --

100 0 48 0 225 1 12

0 30 30

115 0 38 0 0 82

0 1

. . .

cos sin

. .

00

= ∞ - ∞ + - ∞+ - ∞100 0 225 30 1 12 30 0 48 30

0 48 30

[( . sin . cos ) ( . sin )

( . cos

i j

)) ] [( . ) ( . ) ]

. . .

k i k

i j k

+ += - +

115 0 82 0 38

8 56 24 0 2 13

138



The equivalent force-couple system at D is

R j k= - -( . ) ( .28 4 50 0 N N)

MD = ◊ - ◊ + ◊( . ) ( . ) ( . )8 56 24 0 2 13 N m N m N mi j k

(b) Since ( )MD z is positive, pipe CD will tend to rotate counterclockwise relative to muffler DE.

139


PROBLEM 3.124

For the exhaust system of Problem 3.123, (a) replace the given force system with an equivalent force-couple system at F, where the exhaust pipe is connected to the catalytic converter, (b) determine whether pipe EF tends to rotate clockwise or counterclockwise, as viewed by the mechanic.

SOLutiOn


SF R A B: = +

= ∞ - ∞ -= - -

( )(cos sin ) ( )

( . ) ( )

100 30 30 115

28 4 50

N N

N N

j k j

j k and M M r A r B/ /F F A F B F: = ¥ + ¥

where r i j k

r iA F

B F

/

/

m m m

m

= - - += - -

( . ) ( . ) ( . )

( . ) ( .

0 48 0 345 2 10

0 38 0 12 mm m) ( . )j k+ 1 80

Then M

i j k i j k

F = - -∞ - ∞

+ -100 0 48 0 345 2 10

0 30 30

115 0 38 0 12 1 80. . .

cos sin

. . .

00 1 0-

M i jF = ∞ - ∞ + - ∞+ -

100 0 345 30 2 10 30 0 48 30

0 48 3

[( . sin . cos ) ( . sin )

( . cos 00 115 1 80 0 38

42 4 24 0 2 13

∞ + += - +

) ] [( . ) ( . ) ]

. . .

k i k

i j k

140



The equivalent force-couple system at F is

R j k= - -( . ) (28 4 50 N N)

MF = ◊ - ◊ + ◊( . ) ( . ) ( . )42 4 24 0 2 13 N m N m N mi j k

(b) Since ( )MF z is positive, pipe EF will tend to rotate counterclockwise relative to the mechanic.

141


Problem 3.125

The head-and-motor assembly of a radial drill press was originally positioned with arm AB parallel to the z axis and the axis of the chuck and bit parallel to the y axis. The assembly was then rotated 25∞ about the y axis and 20∞ about the centerline of the horizontal arm AB, bringing it into the position shown. The drilling process was started by switching on the motor and rotating the handle to bring the bit into contact with the workpiece. Replace the force and couple exerted by the drill press with an equivalent force-couple system at the center O of the base of the vertical column.

Solution

We have R F i j k= = ∞ ∞ - ∞ - ∞ ∞=

( sin cos ) (cos ) (sin sin ) ]

( .

55 20 25 20 20 25

17

N)[( ]

00485 51 683 7 9499 N N) N) ( . ( . )i j k- -

or R i j k= - -( . ) ( . ) ( . )17 05 51 7 7 95N N

We have M r F MO B O C= ¥ +/

wherer i j kB O/ mm mm= ∞ + + ∞

=

[( )sin ] ( ) [( )cos ]

( .

350 25 375 350 25

147 9175

mm

mmm mm) ( ) ( . )

[( . ) ( . ) ( .

i j k

r i j

+ +

= + +

375 317 208

0 1479 0 375 0

mm

m m/B O 33172

11 20 25 20 20 25

m)

( )[(sin cos ) (cos ) (sin sin

k

M i jC = ◊ ∞ ∞ - ∞ - ∞ ∞N m )) ]

( . ) ( . )

k

i j k= ◊ - ◊ - ◊(3.4097 N m) N m N m10 337 1 59

M

i j k

iO =- -

◊ -0 1479 0 375 0 3172

17 048 51 683 7 9499

. . .

. . .

N m + (3.0497 10..337 1.59) N m

= (16.462 (3.6165 (15.627

j

i j

- ◊

◊ - ◊ - ◊N m) N m N m) ))k

or M kO = ◊ - ◊ - ◊( . ( . ( .16 46 3 62 15 63N m) N m) N m)i j

Fy25°

20°

c

y

x

Fx

F=55 N

Mc=11N·m

Fz

z

142


Problem 3.126

Three children are standing on a 5 5-m ¥ raft. If the weights of the children at Points A, B, and C are 375 N, 260 N, and 400 N, respectively, determine the magnitude and the point of application of the resultant of the three weights.

Solution

G G

C

FA

AB

D

F

E E

xx

F

z

y y

RO O

FB

FC

xDz zD

We have SF F F F R: A B C+ + =

- - - =

- =( ) ( ) ( )

( )

375 260 400

1035

N N N

N

j j j R

j R or NR = 1035

We have SM F z F z F z R zx A A B B C C D: ( ) ( ) ( ) ( )+ + =

( )( ) ( )( . ) ( ( )375 3 260 0 5 400 1035N m N m N)(4.75 m) N)(+ + = zD

zD = 3 0483. m or mzD = 3 05.

We have SM F x F x F x R xz A A B B C C D: ( ) ( ) ( ) ( )+ + =

375 1 1035N( m) (260 N)(1.5 m) (400 N)(4.75 m) N+ + = ( )( )xD

xD = 2 5749. m or mxD = 2 57.

143


Problem 3.127

Three children are standing on a 5 5-m raft.¥ The weights of the children at Points A, B, and C are 375 N, 260 N, and 400 N, respectively. If a fourth child of weight 425 N climbs onto the raft, determine where she should stand if the other children remain in the positions shown and the line of action of the resultant of the four weights is to pass through the center of the raft.

Solution

y y

C

E EH

G

R

xx

F F

OO

2.5 m

2.5 m

FC

DAG

FA FB

FDxD

z

zD

z

We have SF F F F R: A B C+ + = - - - - =( ) ( ) ( ) ( )375 260 400 425N N N Nj j j j R

R j= -( )1460 N

We have SM F z F z F z F z R zx A A B B C C D D H: ( ) ( ) ( ) ( ) ( )+ + + =

( )( ) ( )( . ) (

) (

375 3 260 0 5 400

1460 2

N m N m N)(4.75 m)

(425 N)( N)(

+ ++ =zD .. )5 m

zD = 1 16471. m or mzD = 1 165.

We have SM F x F x F x F x R xz A A B B C C D D H: ( ) ( ) ( ) ( ) ( )+ + + =

( )(

( )(

375 1

1460 2

N)( m) (260 N)(1.5 m) (400 N 4.75 m)

(425 N)( ) N

+ ++ =xD .. )5 m

xD = 2 3235. m or mxD = 2 32.

144


Problem 3.128

Four signs are mounted on a frame spanning a highway, and the magnitudes of the horizontal wind forces acting on the signs are as shown. Determine the magnitude and the point of application of the resultant of the four wind forces when a 0.3 m= and b = 4 m.

Solution

We have

x

x

x

PR

yyy

450 N

250 N800 N

525 N

zz

Assume that the resultant R is applied at Point P whose coordinates are (x, y, 0).Equivalence then requires

SF Rz : - - - - = -525 450 800 250

or R = 2025 N

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

( . )(

1 5 525 0 3 450 0 9 800

1 65

m N) m N m N

m

- ++ 2250 2025 N N) ( )= - y

or y = -0 8815. m

SM y : ( . )( ( )( ) ( . )( )

( . )(

1 65 525 4 450 4 65 800

7 05

m N) m N m N

m

+ +

+ 2250 202 N 5 N) ( )= -x

or x = ◊4 02. N m

R acts 4.02 N . m to the right of member AB and 0.881 m below member BC.

145


PROBLEM 3.129

Four signs are mounted on a frame spanning a highway, and the magnitudes of the horizontal wind forces acting on the signs are as shown. Determine a and b so that the point of application of the resultant of the four forces is at G.

SOLutiOn

Since R acts at G, equivalence then requires that SMG of the applied system of forces also be zero. Then

at G M ax: : ( . ) ( ) ( . )( )

( . )( )

S - + ¥ ++ =

0 9 450 0 6 525

0 75 250 0

N m N

m N

m

or a = 0 217. m

SM by : ( )( ) ( . ) ( )

( . )( )

- - - ¥

+ =

3 525 4 65 450

2 4 250 0

m N m N

m N

or b = 6 82. m

146


PROBLEM 3.130*

A group of students loads a 2 3.3-m¥ flatbed trailer with two 0 66. ¥ ¥0.66 0.66-m boxes and one 0.66 0.66 1.2-m¥ ¥ box. Each of the boxes at the rear of the trailer is positioned so that it is aligned with both the back and a side of the trailer. Determine the smallest load the students should place in a second 0.66 0.66 1.2-m¥ ¥ box and where on the trailer they should secure it, without any part of the box overhanging the sides of the trailer, if each box is uniformly loaded and the line of action of the resultant of the weights of the four boxes is to pass through the point of intersection of the centerlines of the trailer and the axle. (Hint: Keep in mind that the box may be placed either on its side or on its end.)

SOLutiOn

zL

y

B

AC

D

xL

WL

224 N

176 N

392 N

0.33 m

0.33 m0.33 m

x0.6 m

1.4 mz

z

y

B

G

R

A

C

DxR

x

zR

1.8 m

1.5 m

1 m1 m

For the smallest weight on the trailer so that the resultant force of the four weights acts over the axle at the intersection with the centerline of the trailer, the added 0 66 0 66 1 2. . .¥ ¥ -m box should be placed adjacent to one of the edges of the trailer with the 0 66 0 66. .¥ -m side on the bottom. The edges to be considered are based on the location of the resultant for the three given weights.

We have SF j j j R: ( ) ( ) ( )- - - =224 392 176N N N

R j= -( )792 N

We have SMz : ( )( . ) ( )( . ) ( )( . ) ( )- - - = -224 0 33 392 1 67 176 1 67 792N m N m N m N (( )x

xR = 1 29101. m

We have SM zx : ( )( . ) ( )( . ) ( )( . ) ( )( )224 0 33 392 0 6 176 2 0 792N m N m N m N+ + =

zR = 0 83475. m

From the statement of the problem, it is known that the resultant of R from the original loading and the lightest load W passes through G, the point of intersection of the two centerlines. Thus, SMG = 0.

Further, since the lightest load W is to be as small as possible, the fourth box should be placed as far from G as possible without the box overhanging the trailer. These two requirements imply

( . )( . )0 33 1 1 5m m m x z 2.97 m

147


PROBLEM 3.130* (continued)

With xL = 0 33. m

at G M WZ L: : ( . ) ( . ) ( )S 1 0 33 1 29101 1 792 0- ¥ - - ¥ = m m N

or WL = 344 00. N

Now must check if this is physically possible,

at G M Zx L: : ( . ) ( ) ( . . ) ( )S - ¥ - - ¥ =1 5 344 1 5 0 83475 792 0m N m N

or ZL = 3 032. m

which is not acceptable.

With ZL = 2 97. m:

at G M Wx L: : ( . . ) ( . . ) ( )S 2 97 1 5 1 5 0 83475 792 0- ¥ - - ¥ =m m N

or WL = 358 42. N

Now check if this is physically possible

at G M Xz L: : ( ) ( . ) ( . ) ( )S 1 358 42 1 29101 1 792 0- ¥ - - ¥ =m N m N

or X L = 0 357. m ok!

The minimum weight of the fourth box is WL = 358 N

And it is placed on end (A 0 66 0 66. .¥ -m side down) along side AB with the center of the box 0.357 m from side AD.

148


PROBLEM 3.131*

Solve Problem 3.131 if the students want to place as much weight as possible in the fourth box and at least one side of the box must coincide with a side of the trailer.

PROBLEM 3.130* A group of students loads a 2 3.3-m¥ flatbed trailer with two 0 66. ¥ ¥0.66 0.66-m boxes and one 0 66. ¥ ¥0.66 1.2-m box. Each of the boxes at the rear of the trailer is positioned so that it is aligned with both the back and a side of the trailer. Determine the smallest load the students should place in a second 0.66 0.66 1.2-m¥ ¥ box and where on the trailer they should secure it, without any part of the box overhanging the sides of the trailer, if each box is uniformly loaded and the line of action of the resultant of the weights of the four boxes is to pass through the point of intersection of the centerlines of the trailer and the axle. (Hint: Keep in mind that the box may be placed either on its side or on its end.)

SOLutiOn

First replace the three known loads with a single equivalent force R applied at coordinate ( , , )X ZR R 0

Equivalence requires

SF Ry : - - - = -224 392 176

or R = Ø792 N

SM

zx

R

: ( . )(

( )( ) ( )

0 33 224 392

2 176 792

m N) (0.6 m)( N)

m N N

++ =

or zR = 0 83475. m

SM

xz

R

: ( . )(

( . )( ) (

- -- =

0 33 224 392

1 67 176 79

m N) (1.67 m)( N)

m N 22 N)

or xR = 1 29101. m

From the statement of the problem, it is known that the resultant of R and the heaviest loads WH passes

through G, the point of intersection of the two centerlines. Thus,

SMG = 0

Further, since WH is to be as large as possible, the fourth box should be placed as close to G as possible while

keeping one of the sides of the box coincident with a side of the trailer. Thus, the two limiting cases are

x zH H= =0 6 2 7. . m or m

wH

xR

xHA

B

C

AXLE

D

R

Gz

zL

zH

cL

y

x

149



Now consider these two possibilities

With xH = 0 6. : m

at G M WZ H: : ( . ) ( . ) ( )S 1 0 6 1 29101 1 792 0- ¥ - - ¥ =m m N

or WH = 576 20. N

Checking if this is physically possible

at G M zx H: : ( . ) ( . ) ( . . ) ( )S - ¥ - - ¥ =1 5 576 20 1 5 0 83475 792 0m N m N

or zH = 2 414. m

which is acceptable.

With zH = 2 7. m

at G M Wx H: : ( . . ) ( . . ) ( )S 2 7 1 5 1 5 0 83475 792 0- - - ¥ = m N

or WH = 439 N

Since this is less than the first case, the maximum weight of the fourth box is

WH = 576 N

and it is placed with a 0 66 1 2. .¥ -m side down, a 0.66-m edge alongside AD, and the center 2.41 m from side DC.

150


PROBLEM 3.132

Three forces of the same magnitude P act on a cube of side a as shown. Replace the three forces by an equivalent wrench and determine (a) the magnitude and direction of the resultant force R, (b) the pitch of the wrench, (c) the axis of the wrench.

SOLutiOn

Force-couple system at O:

R i j k i j k= + + = + +P P P P( )

M j i k j i k

k i jOR a P a P a P

Pa Pa Pa

= ¥ + ¥ + ¥= - - -

M i j kOR Pa= - + +( )

Since R and MOR have the same direction, they form a wrench with M M1 = O

R . Thus, the axis of the wrench is the diagonal OA. We note that

cos cos cosq q qx y za

a= = = =

3

1

3

R P

M M Pa

x y z

OR

= = = = ∞

= = -

3 54 7

31

q q q .

Pitch = = = - = -pM

R

Pa

Pa1 3

3

(a) R P x y z= = = = ∞3 54 7q q q .

(b) - a

(c) Axis of the wrench is diagonal OA

z

x

y

a

O

A

3F = Pk

2F = P j 1F = P i

a

a

z

x

y

RA

O

M1 = MRO

151


PROBLEM 3.133*

A piece of sheet metal is bent into the shape shown and is acted upon by three forces. If the forces have the same magnitude P, replace them with an equivalent wrench and determine (a) the magnitude and the direction of the resultant force R, (b) the pitch of the wrench, (c) the axis of the wrench.

SOLutiOn

First reduce the given forces to an equivalent force-couple system R M, OR( ) at the origin.

We have

SF j j k: - + + =P P P R

or R k= P

SM j i kO ORaP aP aP M: ( ) ( )- + - + Ê

ËÁˆ¯̃

ÈÎÍ

˘˚̇

=5

2

or M i j kOR aP= - - +Ê

ËÁˆ¯̃

5

2(a) Then for the wrench

R P=

and laxis = =Rk

R

cos cos cosq q qx y z= = =0 0 1

or q q qx y z= ∞ = ∞ = ∞90 90 0

(b) Now

M

aP

aP

OR

1

5

2

5

2

= ◊

= ◊ - - +ÊËÁ

ˆ¯̃

=

laxis M

k i j k

Then PR

aP

P= =

M152 or P a= 5

2

F1R

F3

F2

y

zx xz

y

o

MRO

152



(c) The components of the wrench are ( , ),R M 1 where M1 1= M axisl , and the axis of the wrench is assumed to intersect the xy plane at Point Q whose coordinates are (x, y, 0). Thus require

M r Rz Q R= ¥

Where M M Mz O= ¥ 1

Then

aP aP x y P- - +ÊËÁ

ˆ¯̃

- = + +i j k k5

2

5

2( )i j k

Equating coefficientsi

j

:

:

- = = -- = - =

aP yP y a

aP xP x a

or

or

The axis of the wrench is parallel to the z axis and intersects the xy plane at x a y a= = -, .

yx

x

QrQ

z

yO

M1

R

153


PROBLEM 3.134*

The forces and couples shown are applied to two screws as a piece of sheet metal is fastened to a block of wood. Reduce the forces and the couples to an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz plane.

SOLutiOn

First, reduce the given force system to a force-couple system.

We have SF i j R: ( ) ( )- - = =20 15 25N N NR

We have S S SM r F M MO O C OR: ( )¥ + =

M j i j

i jOR = - - ◊ - ◊

= - ◊ - ◊20 4 1

4 3

N(0.1 m) N m N m)

N m N m

( ) (

( ) ( )(a) R i j= - -( . ) ( .20 0 15 0N N)

(b) We have MRR O

R1

0 8 0 6 4 3

= ◊ =

= - - ◊ - ◊ - ◊= ◊

l lMR

i j i j( . . ) [ ( ] (N m) N m) ]

5 N m

Pitch pM

R= =

◊=1 5

0 200N m

25 Nm.

or p = 0 200. m

(c) From above note that

M M1 = OR

Therefore, the axis of the wrench goes through the origin. The line of action of the wrench lies in the xy plane with a slope of

y x= 3

4

M1

O

43

y

R

x

154


PROBLEM 3.135*

The forces and couples shown are applied to two screws as a piece of sheet metal is fastened to a block of wood. Reduce the forces and the couples to an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz plane.

SOLutiOn

o oo

Q

y yy

x x

x

O

R RR

MR M1M1

M2

z z

z

z

First, reduce the given force system to a force-couple at the origin.

We have SF j j R: ( ) ( )- - =50 55N N

R j= - ( )105 N


M

i j k i j k

jOR =

-◊ + -

-◊ - ¥ ◊

=

0 0 500

0 50 0

0 0 375

0 55 0

1 5 1000N mm N mm N mm( . )

(44375 1500N mm N mm◊ - ◊) ( )i j

(a) R j= - ( )105 N or R j= - ( )105 N

(b) We have MRR O

RR1

4375 1500

1500

= ◊ =

= - ◊ ◊ - ◊= ◊

l lMR

j i j

M

( ) [( ) ( ) ]N mm N mm

N mm and 11 1500= - ◊( )N mm j

and pitch pM

R= =

◊=1 1500

14 2857N mm

105 Nmm. or p = 14 29. mm

155



(c) We have M M M

M M M i

OR

OR

= +

= - = ◊1 2

2 1 4375( )N mm

Require M r R

i i k j

i k

2

4375 105

4375 105 10

= ¥

◊ = + ¥ -

= - +

Q O

x z

x

/

( ) ( ) [ ( ) ]

( ) (

N mm N

55z)i

From i: 4375 105

41 667

==

z

z . mm. From k: 0 105

0

= -=

x

x

The axis of the wrench is parallel to the y axis and intersects the xz plane at x z= =0 41 7, . mm

156


PROBLEM 3.136*

Two bolts at A and B are tightened by applying the forces and couples shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz plane.

SOLutiOnMR

O

RR

R

Qxx

M1

xx

OO

o

y

z zz

z

M2

M1

y y


We have SF j k R: ( ) ( )- - = =84 80 116N N NR

and S S SM r F M MO O C OR: ( )¥ + =

i j k i j k

j k M0 6 0 0 1

0 84 0

0 4 0 3 0

0 0 80

30 32. . . . ( )+ + - - ◊ =N m OR

M i j kOR = - ◊ + ◊ - ◊( . ) ( ) ( . )15 6 2 82 4N m N m N m

(a) R j k= - -( . ) ( . )84 0 80 0N N

(b) We have MRR O

RR1

84 80

11615 6 2 82 4

= ◊ =

= - - - ◊ - ◊ + ◊ - ◊

l lMR

j ki j[ ( . ( ) ( . )N m) N m N m kk]

.= ◊55 379 N m

and M j k1 1 40 102 38 192= = - ◊ - ◊M Rl ( . ) ( . )N m N m

Then pitch pM

R= =

◊=1 55 379

0 47741.

.N m

116 Nm or p = 0 477. m

157



(c) We have M M M

M M M i j k j k

OR

OR

= +

= - = - + - - - ◊1 2

2 1 15 6 2 82 4 40 102 38 192[( . . ) ( . . )] N mm

N m N m N m= - ◊ + ◊ - ◊( . ) ( . ) ( . )15 6 42 102 44 208i j k

Require M r R

i j k i k j k

i

2

15 6 42 102 44 208 84 80

84

= ¥

- + - = + ¥ -=

Q O

x z

z

/

( . . . ) ( ) ( )

( ) ++ -( ) ( )80 84x xj k

From i: - == -

15 6 84

0 185714

.

.

z

z m

or z = -0 1857. m

From k: - = -=

44 208 84

0 52629

.

.

x

x m

or x = 0 526. m

The axis of the wrench intersects the xz plane at

x y z= = = -0 526 0 0 1857. .m m

158


PROBLEM 3.137*

Two bolts at A and B are tightened by applying the forces and couples shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz plane.

SOLutiOny

y

B

Rx

x

x

Q

B

z z

z

MRB

First, reduce the given force system to a force-couple at the origin at B.

(a) We have SF k i j R: ( ) ( )- - +ÊËÁ

ˆ¯̃

=132 85200

425

375

425N N

R i j k= - - -( ) ( ) ( )40 75 132N N N

and R = 157 N

We have SM r F M M MB A B A A B BR: / ¥ + + =

M

i j k

k i jBR = -

-- - +Ê

ËÁˆ¯̃

=

0 250 0

0 0 132

27500 30000200

425

375

425

330000 27500 14117 6 26470 6

18882 4 26470 6

i k i j

M i

- - -

= ◊ - ◊

. .

( . ) ( .BR N mm N mmm) N mm

N m N m) N m

j k

M i j k

- ◊

= = ◊ - ◊ - ◊

( )

( . ) ( . ( . )

27500

18 88 26 5 27 5BR

159



(b) We have MRR O

RR1

40 75 132

15718 88 26 5 2

= ◊ =

= - - - ◊ ◊ - ◊ -

l lMR

i j ki j[( . ( . ) (N m) N m 77 5

4 81 12 66 23 1 30 95

. ) ]

. . . .

N m◊

= - + + = ◊

k

N m

and M i j k1 1 7 885 14 785 26 021= = - ◊ - ◊ - ◊M Rl ( . ) ( . ) ( . )N m N m N m

Then pitch pM

R= =

◊1 30 95. N m

157 N or p = 197 1. mm

(c) We have M M M

M M M i j k i j

BR

BR

= +

= - = - - - - - -1 2

2 1 18 88 26 5 27 5 7 885 14 785 2( . . . ) ( . . 66 021

26 765 11 715 1 479

. )

( . ) ( . ) ( . . )

k

i j k= ◊ - ◊ -N m N m N m

Require M r R2 = ¥Q B/

26 765 11 715 1 479 0

40 75 132

75 40 132

. . .

( ) ( ) (

i j k

i j k

i j

- - =- - -

= - +

x z

z z xx x) ( )j k- 75

From i: 26 765 75 0 357. .= =z z m = 357 mm

From k: - = - = =1 479 75 0 01972 19 72. . .x x m mm

The axis of the wrench intersects the xz plane at x y z= = =19 72 0 357. mm mm

160


PROBLEM 3.138*

Two ropes attached at A and B are used to move the trunk of a fallen tree. Replace the forces exerted by the ropes with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the yz plane.

SOLutiOn

(a) First replace the given forces with an equivalent force-couple system R M, OR( ) at the origin.

We have

d

d

AC

BD

= + + =

= + + =

( ) ( ) ( )

( ) ( ) ( )

6 2 9 11

14 2 5 15

2 2 2

2 2 2

m

m

Then

TAC = = + +

= + +

16506 2 9

900 300 1350

N

11 N N N

( )

( ) ( ) ( )

i j k

i j k and

TBD = = + +

= + +

150014 2 5

1400 200 500

N

15 N N N

( )

( ) ( ) ( )

i j k

i j k

Equivalence then requires

SF R T T

i j k

i j k

i

:

( )

( )

( )

= += + +

+ + +=

AC BD

900 300 1350

1400 200 500

2300 N ++ +( ) ( )500 1850 N Nj k

SM M r T r TO OR

A AC B BD: = ¥ + ¥

= ¥ + ++ ¥ +( ) [( ( ( ) ]

( [( (

12 900 300 1350

9 1400

m N) N) N

m) N)

k i j k

i i 2200 500

3600 10800 4500 1800

3600

N) N

N

j k

i j k

+= - + - += -

( ) ]

( ) ( ) ( )

( ◊◊ + ◊ + ◊m N m) (1800 N m)) (i j k6300

The components of the wrench are ( , ),R M 1 where

R i j k= + +( ) ( ) ( )2300 500 1850 N N N

161



(b) We have

R = + + =100 23 5 18 5 2993 72 2 2( ) ( ) ( . ) . N

Let

laxis = = + +Ri j k

R

1

29 93723 5 18 5

.( . )

Then M OR

1

1

29 93723 5 18 5 3600 6300 1800

1

0

= ◊

= + + ◊ - + +

=

laxis M

.( . ) ( )i j k i j k

..[( )( ) ( )( ) ( . )( )]

.29937

23 36 5 63 18 5 18

601 26

- + +

= - ◊ N m

Finally PM

R= = - ◊1 601 26. N m

2993.7 N or P = - 0 201. m (c) We have M M1 1

601 261

29 93723 5 18 5

=

= - ◊ ¥ + +

N m

axisl

( . ).

( . )i j k

or M i j k1 461 93 100 421 371 56= - ◊ - ◊ - ◊( . ) ( . ) ( . N m N m N m)

Now M M M

i j k

i j k

2 1

3600 6300 1800

461 93 100 421 371 56

= -= - + +

- - - -

OR

( )

( . . . )

== - ◊ + ◊ + ◊( . ) ( . ( . )3138 1 6400 4 2171 6 N m N m) N mi j k

For equivalence

y

y

y

xxO

OR

RPM1

M1

M2

z

zz

162



Thus require M r R r j k2 = ¥ = +P y z( )

Substituting

- + + =3138 1 6400 4 2171 6 0

2300 500 1850

. . .i j k

i j k

y z

Equating coefficientsj

k

: . .

: . .

6400 4 2300 2 78

2171 6 2300 0 944

= == - = -

z z

y y

or m

or m

The axis of the wrench intersects the yz plane at y z= - =0 944 2 78. . m m

163


PROBLEM 3.139*

A flagpole is guyed by three cables. If the tensions in the cables have the same magnitude P, replace the forces exerted on the pole with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz plane.

SOLutiOn

R R

R

y y y

o

oo

Q

M2

M1

M1

OMR

x x xx

zz

z

z

(a) First reduce the given force system to a force-couple at the origin.

We have SF R: P P PBA DC DEl l l+ + =

R j k i j i j k= -ÊËÁ

ˆ¯̃

+ -ÊËÁ

ˆ¯̃

+ - - +ÊËÁ

ˆ¯̃

ÈÎÍ

˘˚̇

P4

5

3

5

3

5

4

5

9

25

4

5

12

25

R i j k= - -3

252 20

P( )

RP

P= + + =3

252 20 1

27 5

252 2 2( ) ( ) ( )

We have S SM r M: ( )O ORP¥ =

( ) ( ) ( )244

5

3

520

3

5

4

520

9

2a

P Pa

P Pa

Pj j k j i j j¥ - -Ê

ËÁˆ¯̃

+ ¥ -ÊËÁ

ˆ¯̃

+ ¥ -55

4

5

12

25i j k M- +Ê

ËÁˆ¯̃

=P POR

M i kOR Pa= - -24

5( )

164



(b) We have M R OR

1 = ◊l M

where lR R

P

P= = - - = - -R

i j k i j k3

252 20

25

27 5

1

9 52 20( ) ( )

Then MPa Pa

11

9 52 20

24

5

8

15 5= - - ◊ - - = -

( ) ( )i j k i k

and pitch pM

R

Pa

P

a= = - ÊËÁ

ˆ¯̃

= -1 8

15 5

25

27 5

8

81 or p a= -0 0988.

(c) M i j k i j k1 18

15 5

1

9 52 20

8

6752 20= = - Ê

ËÁˆ¯̃

- - = - + +MPa Pa

Rl ( ) ( )

Then M M M i k i j k i j2 124

5

8

6752 20

8

675430 20 4= - = - - - - + + = - - -O

R Pa Pa Pa( ) ( ) ( 006k)

Require M r R2 = ¥Q O/

8

675403 20 406

3

252 20

Pax z

PÊËÁ

ˆ¯̃

- - - = + ¥ ÊËÁ

ˆ¯̃

- -

=

( ) ( ) ( )i j k i k i j k

33

2520 2 20

Pz x z xÊ

ËÁˆ¯̃

+ + -[ ( ) ]i j k

From i: 8 403675

203

251 99012( ) .- = Ê

ËÁˆ¯̃

= -Paz

Pz a

From k: 8 406675

203

252 0049( ) .- = - Ê

ËÁˆ¯̃

=Pax

Px a

The axis of the wrench intersects the xz plane at

x a z a= = -2 00 1 990. , .

165


PROBLEM 3.140*

Determine whether the force-and-couple system shown can be reduced to a single equivalent force R. If it can, determine R and the point where the line of action of R intersects the yz plane. If it cannot be so reduced, replace the given system with an equivalent wrench and determine its resultant, its pitch, and the point where its axis intersects the yz plane.

SOLutiOn


We have SF F F R: A G+ =

R k

i j k

i

= ++ -È

ÎÍ

˘

˚˙

= +

( )( ) ( )

( ) (

50 7060 120

20 30

N N(40 mm) mm mm

140 mm

N NN N) ( )j k- 10

and R = 37 417. N


M j k i i j kOR = ¥ + ¥ + -

+

[( . ) ( ) ] {( . ) [( ) ( ) ( ) ]}

(

0 12 50 0 16 20 30 60

1

m N m N N N

00160 120

200

1440 120

N mmm mm

mm

N mmm mm

◊-È

ÎÍ

˘

˚˙

+ ◊-

)( ) ( )

( )( ) ( )

i j

i j ++È

ÎÍ

˘

˚˙

= ◊ - ◊ + ◊

( )

( ) ( . ) ( . )

60

140

18 8 4 10 8

mm

mm

N m N m N m

k

M i j kOR

To be able to reduce the original forces and couples to a single equivalent force, R and M must be

perpendicular. Thus, R M◊ = 0.

Substituting( ) ( . . )

?20 30 10 18 8 4 10 8 0i j k i j k+ - ◊ - + =

or ( )( ) ( )( . ) ( )( . )?

20 18 30 8 4 10 10 8 0+ - + - =

or 0 =?

0

R and M are perpendicular so that the given system can be reduced to the single equivalent force

R i j k= + -( . ) ( . ) ( . )20 0 30 0 10 00 N N N

166



Then for equivalence

R

x x

R

P

O

y y

yO

MR

O

z z

z

Thus require M r R r j kOR

p p y z= ¥ = +

Substituting

18 8 4 10 8 0

20 30 10

i j k

i j k

- + =-

. . y z

Equating coefficients

j

k

: . .

: . .

- = = -= - = -

8 4 20 0 42

10 8 20 0 54

z z

y y

or m

or m

The line of action of R intersects the yz plane at x y z= = - = -0 0 540 0 420 m m. .

167


PROBLEM 3.141*

Determine whether the force-and-couple system shown can be reduced to a single equivalent force R. If it can, determine R and the point where the line of action of R intersects the yz plane. If it cannot be so reduced, replace the given system with an equivalent wrench and determine its resultant, its pitch, and the point where its axis intersects the yz plane.

SOLutiOn

First determine the resultant of the forces at D. We have

d

d

DA

ED

= - + + =

= - + + - =

( ) ( ) ( )

( ) ( ) ( )

300 225 200 425

150 0 200 2

2 2 2

2 2 2

mm

550 mmThen

F i j k

i j k

DA = = - + +

= - + +

170300 225 200

120 90 80

N

425N N N

[( ) ( ) ( ) ]

( ) ( ) ( )and

F i k

i k

ED = = - -

= - -

150150 200

90 120

N

250N N

( )

( ) ( )Then

SF R F F: = +DA ED = - + -( ( )210 40N) (90 N) Ni j k

For the applied couple

dAK = - + - + =( ) ( ) ( )150 150 450 150 112 2 2 mm

Then

M i j k

i j k)

=◊

- - +

= - - + ◊

20000

150 11150 150 450

20000

11

N mm

N mm

( )

( 3

To be able to reduce the original forces and couple to a single equivalent force, R and M must beperpendicular. Thus

R M◊ =?

0

168



Substituting

( ) ( )?- + - ◊ - - + =210 90 40

20000

113 0i j k i j k

or 20000

11210 1 90 1 40 3 0[( )( ) ( )( ) ( )( )]

?- - + - + - =

or 0 0=?

R and M are perpendicular so that the given system can be reduced to the single equivalent force

R i j k= - + -( ) ( ) ( )210 90 40N N N

Then for equivalence

M

DR R

P OO

y y

yx x

zz

z

=

Thus require M r R= ¥P D/

where r i j kP D y z/ mm )mm mm)= - + - +( ) [( ] (300 75

Substituting

20000

113 300 75

210 90 40

75 40 90

( ) ( )

[( )( ) (

- - + = - -- -

= - - -

i j k

i j k

y z

y z ))]

[( )( ) ( )( )]

[( )( ) ( )( )]

i

j

k

+ - - - -+ - - - -

z

y

210 300 40

300 90 75 210Equating coefficients

j

k

: .

: (

- = - - = -

= - + -

20000

11210 12000 28 4

60000

1127000 210 75

z z

y

or mm

)) or mmy = 290

The line of action of R intersects the yz plane at x y z= = = -0 290 28 4mm mm.

169


PROBLEM 3.142*

Replace the wrench shown with an equivalent system consisting of two forces perpendicular to the y axis and applied respectively at A and B.

SOLutiOn

Express the forces at A and B as

A i k

B i k

= += +

A A

B Bx z

x z

Then, for equivalence to the given force system

SF A Bx x x: + = 0 (1)

SF A B Rz z z: + = (2)

SM A a B a bx z z: ( ) ( )+ + = 0 (3)

SM A a B a b Mz x x: ( ) ( )- - + = (4)

From Equation (1), B Ax x= -

Substitute into Equation (4)

- + + =A a A a b Mx x( ) ( )

AM

bB

M

bx x= = -and

From Equation (2), B R Az z= -

and Equation (3), A a R A a bz z+ - + =( )( ) 0

A Ra

bz = +ÊËÁ

ˆ¯̃

1

and B R Ra

bz = - +ÊËÁ

ˆ¯̃

1

Ba

bRz = -

Then A = ÊËÁ

ˆ¯̃

+ +ÊËÁ

ˆ¯̃

M

bR

a

bi k1

B i k= -ÊËÁ

ˆ¯̃

- ÊËÁ

ˆ¯̃

M

b

a

bR

z

Bz

y

b

a

Bx

Ax

x

OAz

170


PROBLEM 3.143*

Show that, in general, a wrench can be replaced with two forces chosen in such a way that one force passes through a given point while the other force lies in a given plane.

SOLutiOn

y y

x x

z z

R

M

Given plane

Given point

(a)

oo

P P

B

A

rp

b

(b)

First, choose a coordinate system so that the xy plane coincides with the given plane. Also, position the coordinate system so that the line of action of the wrench passes through the origin as shown in Figure a. Since the orientation of the plane and the components (R, M) of the wrench are known, it follows that the scalar components of R and M are known relative to the shown coordinate system.

A force system to be shown as equivalent is illustrated in Figure b. Let A be the force passing through the given Point P and B be the force that lies in the given plane. Let b be the x-axis intercept of B.

The known components of the wrench can be expressed as

R i j k i j k= + + = + +R R R M M M Mx y z x y zand

while the unknown forces A and B can be expressed as

A i j k B i k= + + = +A A A B Bx y z x zand

Since the position vector of Point P is given, it follows that the scalar components (x, y, z) of the

position vector rP are also known.

Then, for equivalence of the two systems SF R A Bx x x x: = + (1)

SF R Ay y y: = (2)

SF R A Bz z z z: = + (3)

SM M yA zAx x z y: = - (4)

SM M zA xA bBy y x z z: = - - (5)

SM M xA yAz z y x: = - (6)

Based on the above six independent equations for the six unknowns ( , , , , , , ),A A A B B B bx y z x y z there exists a unique solution for A and B.

From Equation (2) A Ry y=

171



Equation (6) Ay

xR Mx y z=ÊËÁ

ˆ¯̃

-1( )

Equation (1) B Ry

xR Mx x y z= -ÊËÁ

ˆ¯̃

-1( )

Equation (4) Ay

M zRz x y=ÊËÁ

ˆ¯̃

+1( )

Equation (3) B Ry

M zRz z x y= -ÊËÁ

ˆ¯̃

+1( )

Equation (5) bxM yM zM

M yR zRx y z

x z y

=+ +- +

( )

( )

172


PROBLEM 3.144*

Show that a wrench can be replaced with two perpendicular forces, one of which is applied at a given point.

SOLutiOn

M

y y

P

(a) (b)

P

A

BGiven pointx

x

o

a aR

zz

z

First, observe that it is always possible to construct a line perpendicular to a given line so that the constructed line also passes through a given point. Thus, it is possible to align one of the coordinate axes of a rectangular coordinate system with the axis of the wrench while one of the other axes passes through the given point.

See Figures a and b.

We have R j M j= =R Mand and are known.

The unknown forces A and B can be expressed as

A i j k B i j k= + + = + +A A A B B Bx y z x y zand

The distance a is known. It is assumed that force B intersects the xz plane at (x, 0, z). Then for equivalence

SF A Bx x x: 0 = + (1)

SF R A By y y: = + (2)

SF A Bz z z: 0 = + (3)

SM zBx y: 0 = - (4)

SM M aA xB zBy z z x: = - - + (5)

SM aA zBz y y: 0 = + (6)

Since A and B are made perpendicular,

A B◊ = + + =0 0or A B A B A Bx x y y z z (7)

There are eight unknowns: A A A B B B x zx y z x y z, , , , , , ,

But only seven independent equations. Therefore, there exists an infinite number of solutions.

Next consider Equation (4): 0 = - zBy

If By = 0, Equation (7) becomes A B A Bx x z z+ = 0

Using Equations (1) and (3) this equation becomes A Ax z2 2 0+ =

173



Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that By π 0, so that from Equation (4), z = 0.

To obtain one possible solution, arbitrarily let Ax = 0.

(Note: Setting A A By z z, , or equal to zero results in unacceptable solutions.)

The defining equations then become

0 = Bx (1)9

R A By y= + (2)

0 = +A Bz z (3)

M aA xBz z= - - (5)9

0 = +aA xBy y (6)

A B A By y z z+ = 0 (7)9

Then Equation (2) can be written A R By y= -

Equation (3) can be written B Az z= -

Equation (6) can be written xaA

By

y

= -

Substituting into Equation (5)9,

M aA aR B

BAz

y

yz= - - -

-Ê

ËÁ

ˆ

¯˜ -( )

or AM

aRBz y= - (8)

Substituting into Equation (7)9,

( )R B BM

aRB

M

aRBy y y y- + -Ê

ËÁˆ¯̃

ÊËÁ

ˆ¯̃

= 0

or Ba R

a R My =

+

2 3

2 2 2

Then from Equations (2), (8), and (3)

A Ra R

a R M

RM

a R M

AM

aR

a R

a R M

aR M

a

y

z

= -+

=+

= -+

Ê

ËÁˆ

¯̃= -

2 2

2 2 2

2

2 2 2

2 3

2 2 2

2

22 2 2

2

2 2 2

R M

BaR M

a R Mz

+

=+

174



In summary

A j k=+

-RM

a R MM aR

2 2 2( )

B j k=+

+aR

a R MaR M

2

2 2 2( )

Which shows that it is possible to replace a wrench with two perpendicular forces, one of which is applied at a given point.

Lastly, if R 0 and M 0, it follows from the equations found for A and B that Ay 0 and By 0.

From Equation (6), x 0 (assuming a 0). Then, as a consequence of letting Ax = 0, force A lies in a plane parallel to the yz plane and to the right of the origin, while force B lies in a plane parallel to the yz plane but to the left to the origin, as shown in the figure below.

B

y

O

x a

z

xP

A

175


PROBLEM 3.145*

Show that a wrench can be replaced with two forces, one of which has a prescribed line of action.

SOLutiOn

M

y y

P

A

AA

A′

A′

Bx

x

xO O

a aR

zz

z

First, choose a rectangular coordinate system where one axis coincides with the axis of the wrench and another axis intersects the prescribed line of action (AA9). Note that it has been assumed that the line of action of force B intersects the xz plane at Point P(x, 0, z). Denoting the known direction of line AA9 by

l l l lA x y z= + +i j k

it follows that force A can be expressed as

A i j k= = + +A AA x y zl l l l( )

Force B can be expressed as

B i j k= + +B B Bx y z

Next, observe that since the axis of the wrench and the prescribed line of action AA9 are known, it follows that the distance a can be determined. In the following solution, it is assumed that a is known.

Then, for equivalence

SF A Bx x x: 0 = +l (1)

SF R A By y y: = +l (2)

SF A Bz z z: 0 = +l (3)

SM zBx y: 0 = - (4)

SM M aA zB xBy z x z: = - + -l (5)

SM aA xBx y y: 0 = - +l (6)

Since there are six unknowns (A, Bx, B

y, B

z, x, z) and six independent equations, it will be possible to obtain a

solution.

176



Case 1: Let z = 0 to satisfy Equation (4)

Now Equation (2) A R By yl = -

Equation (3) B Az z= - l

Equation (6) xaA

B

a

BR By

y yy= - = -

Ê

ËÁ

ˆ

¯˜ -

l( )

Substitution into Equation (5)

M aAa

BR B A

AM

aRB

zy

y z

zy

= - - -Ê

ËÁ

ˆ

¯˜ - -

È

ÎÍÍ

˘

˚˙˙

= - ÊËÁ

ˆ¯̃

l l

l

( )( )

1


RM

aRB B

BaR

aR M

zy y y

yz

z y

= - ÊËÁ

ˆ¯̃

+

=-

1

2

ll

ll l

Then AMR

aR M

RaRM

B AMR

aR M

B AMR

a

z yy z

x xx

z y

z zz

z

= --

=-

= - =-

= - =

l l l l

ll

l l

ll

l RR My- l

In summary A =-

PaRMy z

A

l ll

B i j k=-

+ +R

aR MM aR M

z yx z zl l

l l l( )

and x aR

B

a RaR M

aR

y

z y

z

= -Ê

ËÁ

ˆ

¯˜

= --Ê

ËÁˆ

¯̃

È

ÎÍÍ

˘

˚˙˙

1

12

l l

l

or xM

Ry

z

=ll

Note that for this case, the lines of action of both A and B intersect the x axis.

177



Case 2: Let By = 0 to satisfy Equation (4)

Now Equation (2) AR

y

=l

Equation (1) B Rxx

y

= -Ê

ËÁ

ˆ

¯˜

ll

Equation (3) B Rzz

y

= -Ê

ËÁ

ˆ

¯˜

ll

Equation (6) aA yl = 0 which requires a = 0


M z R x R x zM

Rx

y

z

yz x= -

Ê

ËÁ

ˆ

¯˜

È

ÎÍÍ

˘

˚˙˙

- -Ê

ËÁ

ˆ

¯˜

È

ÎÍÍ

˘

˚˙˙

- = Êll

ll

l lorËËÁ

ˆ¯̃

ly

This last expression is the equation for the line of action of force B.

In summary

A =Ê

ËÁ

ˆ

¯˜

R

yAl

l

B i k=Ê

ËÁ

ˆ

¯˜ - -R

yx xl

l l( )

Assuming that l l lx y z, , , 0 the equivalent force system is as shown below.

z

y

MR

A

B

x

λyλx

MR

λyλz

Note that the component of A in the xz plane is parallel to B.

178


Problem 3.146

A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at B that creates a moment of equal magnitude and opposite sense about E.

Solution

(a) By definition W mg= = =80 784 8kg(9.81 m/s ) N2 .

We have SM ME E: ( . )( . )= 784 8 0 25N m

ME = ◊196 2. N m

(b) For the force at B to be the smallest, resulting in a moment (ME) about E, the line of action of force F

B

must be perpendicular to the line connecting E to B. The sense of FB must be such that the force produces

a counterclockwise moment about E.

Note: d = + =( . ) ( . ) .0 85 0 5 0 986152 2m m m

We have SM FE B: . ( . )196 2 0 98615N m m◊ =

FB = 198 954. N

and q =ÊËÁ

ˆ¯̃

= ∞-tan.

.1 0 8559 534

m

0.5 m or FB = 199 0. N 59.5°

A

A

C

C

E

E

D

D

d

q

B

B

FB

W0.25 m

0.5 m

0.85 m

179


PROBLEM 3.147

It is known that the connecting rod AB exerts on the crank BC a 1.5-kN force directed down and to the left along the centerline of AB. Determine the moment of the force about C.

SOLutiOn

Using (a)

M y F x FC AB x AB y= +

= ¥ÊËÁ

ˆ¯̃

+

1 1

0 028 1500 0 02124

( ) ( )

( . ( . ) m)7

25 N m

2251500

42

¥ÊËÁ

ˆ¯̃

= ◊

N

N m

or MC = ◊42 0. N m

Using (b)

M y FC AB x=

= ¥ÊËÁ

ˆ¯̃

= ◊

2

0 1 1500 42

( )

( . m)7

25 N N m

or MC = ◊42 0. N m

247(FAB)x

(FAB)yFAB

FABA

B By1

y2

c cx1

(a)(b)

180


PROBLEM 3.148

A 1.8 m-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line is 30 N. Determine the moment about A of the force exerted by the line at B.

SOLutiOn

We have Txz = ∞ =( )cos .30 8 29 708N N

Then T T

T T

T T

x xz

y BC

z xz

= ∞ == - ∞ = -

= - ∞ = -

sin .

sin .

cos

30 14 854

8 4 1752

30 25

N

N

..728 N

Now M r TA B A BC= ¥/

where r j k

j k

B A/

m

2

= ∞ - ∞

= -

( . sin ) ( . cos )

.( )

1 8 45 1 8 45

1 8

Then

M

i j k

i

A = -- -

= - - -

1 8

20 1 1

14 854 4 1752 25 728

1 8

225 728 4 1752

1

.

. . .

.( . . )

..( . )

.( . )

8

214 854

1 8

214 854j k-

or M iA = - ◊ - ◊ - ◊( . ) ( . ) ( . )38 1 18 91 18 91N m N m N mj k

yB

A

C

z

x

Ty

Tx

TBC = 30 N

Txz

Tz8˚

181


PROBLEM 3.149

Ropes AB and BC are two of the ropes used to support a tent. The two ropes are attached to a stake at B. If the tension in rope AB is 540 N, determine (a) the angle between rope AB and the stake, (b) the projection on the stake of the force exerted by rope AB at Point B.

SOLutiOn

First note BA

BD

= - + + - =

= - + + =

( ) ( ) ( . ) .

( . ) ( . ) ( . ) .

3 3 1 5 4 5

0 08 0 38 0 16 0 4

2 2 2

2 2 2

m

22 m

Then T i j k

i j k

BABA

BA

BD

T

T

BD

BD

= - + -

= - + -

= = -

4 53 3 1 5

32 2

1

0 420 08

.( . )

( )

.( .l ii j k

i j k

+ +

= - + +

0 38 0 16

1

214 19 8

. . )

( )

(a) We have TBA BD BAT◊ l q= cos

or TTBA

BA32 2

1

214 19 8( ) ( ) cos- + - - - + + =i j k i j k q

or cos [( )( ) ( )( ) ( )( )]

.

q = - - + + -

=

1

632 4 2 19 1 8

0 60317

or q = ∞52 9.

(b) We have ( )

cos

( )( . )

T

TBA BD BA BD

BA

= ◊==

T lq

540 0 603N 17 or ( )TBA BD = 326 N

182


PROBLEM 3.150

A farmer uses cables and winch pullers B and E to plumb one side of a small barn. If it is known that the sum of the moments about the x axis of the forces exerted by the cables on the barn at Points A and D is equal to 7092 N.m, determine the magnitude of T

DE when T

AB 1.275 kN.

SOLutiOn

The moment about the x axis due to the two cable forces can be found using the z components of each force acting at their intersection with the xy-plane (A and D). The x components of the forces are parallel to the x axis, and the y components of the forces intersect the x axis. Therefore, neither the x nor y components produce a moment about the x axis.

We have SM T y T y Mx AB z A DE z D x: ( ) ( ) ( ) ( )+ =

where ( )

( ) .

. .

.

T

TAB z AB

AB AB

= ◊= ◊ -

= ◊- - +Ê

ËÁˆ

k T

k

ki j k

l 0 3

12750 3 3 6 4

5 390

i

N¯̃̄

ÈÎÍ

˘˚̇

= 946 2. N

( )

( )

. .

.

T

T

T

DE z DE

DE DE

DE

= ◊= ◊

= ◊ - +ÊËÁ

ˆ¯̃

ÈÎÍ

˘˚

k T

k

ki j k

l

0 45 4 2 4

5 817 ˙̇

==== ◊

0 6876

3 6

4 2

7092

.

.

.

T

y

y

M

DE

A

D

x

m

m

N m

( . )( . ) ( . )( . )946 2 3 6 0 6876 4 2 7092N m m N m+ = ◊TDE

and TDE = 1276 24. N or TDE = 1 276. kN

y

AD

OB

C

TAB

TDE

E

F

xz

183


PROBLEM 3.151

Solve Problem 3.150 when the tension in cable AB is 1.53 kN.

PROBLEM 3.150 A farmer uses cables and winch pullers B and E to plumb one side of a small barn. If it is known that the sum of the moments about the x axis of the forces exerted by the cables on the barn at Points A and D is equal to 7092 N·m, determine the magnitude of T

DE when T

AB 1.275 kN

SOLutiOn

The moment about the x axis due to the two cable forces can be found using the z components of each force acting at their intersection with the xy plane (A and D). The x components of the forces are parallel to the x axis, and the y components of the forces intersect the x axis. Therefore, neither the x or y components produce a moment about the x axis.

We have SM T y T y Mx AB z A DE z D x: ( ) ( ) ( ) ( )+ =

where ( )

( )

. .

.

T

TAB z AB

AB AB

= ◊= ◊

= ◊- - +Ê

ËÁˆ¯̃

ÈÎÍ

k T

k

ki j k

l

15300 3 3 6 4

5 390N

˘̆˚̇

= 1135 44. N

( )

( )

. .

.

T

T

T

DE z DE

DE DE

DE

= ◊= ◊

= ◊ - +ÊËÁ

ˆ¯̃

ÈÎÍ

˘˚

k T

k

ki j k

l

0 45 4 2 4

5 817 ˙̇

==== ◊

0 6876

3 6

4 2

7092

.

.

.

T

y

y

M

DE

A

D

x

m

m

N m

( . )( . ) ( . )( . )1135 44 3 6 0 6876 4 2 7092N m m N m+ = ◊TDE

and TDE = 1040 34. N or TDE = 1 040. kN

y

AD

OB

C

TAB

TDE

E

F

xz

184


PROBLEM 3.152

A wiring harness is made by routing either two or three wires around 50 mm-diameter pegs mounted on a sheet of plywood. If the force in each wire is 15 N, determine the resultant couple acting on the plywood when a 450 mm and (a) only wires AB and CD are in place, (b) all three wires are in place.

SOLutiOn

In general, M dF= S , where d is the perpendicular distance between the lines of action of the two forces acting on a given wire.

(a)25 mm

25 mm

25 mm

25 mm

25 mm (450+250) mm

25 mm

600 mm

FAB

FAB

2418

FCD

FCD

3 4

450+250 mm

FEF

FEF

We have M d F d FAB AB CD CD= +

= + ¥ + ¥ÊËÁ

ˆ¯̃

¥( )50 600 15 700 15 mm N 50 +4

5Nmm

or M = ◊18900 N mm

(b)

25 mm25 mm

25 mm

25 mm

25 mm (450+250) mm

25 mm

600 mm

FAB

FAB

2418

FCD

FCD

3 4

450+250 mm

FEF

FEF

We have M d F d F d FAB AB CD CD EF EF= + +[ ]

= ◊ - ¥18900 N mm 700 mm 15N or M = ◊8 400. N mm

= 8.40 N . m

185


PROBLEM 3.153

A worker tries to move a rock by applying a 360-N force to a steel bar as shown. (a) Replace that force with an equivalent force-couple system at D. (b) Two workers attempt to move the same rock by applying a vertical force at A and another force at D. Determine these two forces if they are to be equivalent to the single force of Part a.

SOLutiOn

FA

FA

FO

D D

q1.05 m

M

B D

40°

30° 30° 30°0.65 m

360 N

(a) (b)

A A

(a) We have SF i j i j F: sin cos ( . ) ( . )360 40 231 40 275 78N( 40 ) N N- ∞ - ∞ = - - =

or NF = 360 50°

We have SM r R MD B D: / ¥ =

where r i j

iB D/ m m

m

= - ∞ + ∞= - +

[( . )cos ] [( . )sin ]

( . ) ( .

0 65 30 0 65 30

0 56292 0 325500

0 56292 0 32500 0

231 40 275 78 0

155 240 75

m

N m

)

. .

. .

[ .

j

M

i j k

= -- -

◊

= + .. ) ]206 N m◊ k

= ◊( . )230 45 N m k or N mM = ◊230

(b) We have SM M r FD A D A: = ¥/

where r i j

iB D/ m m

m

= - ∞ + ∞= - +

[( . )cos ] [( . )sin ]

( . ) ( .

1 05 30 1 05 30

0 90933 0 525500

0 90933 0 52500 0

0 1 0

230 45

m

N m

N m

)

. .

[ . ]

j

i j k

k

FA = --

◊

= ◊

186



or ( . ) .0 90933 230 45FA k k=

FA = 253 42. N or NFA = Ø253

We have SF F F F: = +A D

- - = - + - -( . ) ( . ) ( . ) ( cos sin )231 40 275 78 253 42N N Ni j j i jFD q q

From i : . cos231 40 N = FD q (1)

j: 22 36. sinN = FD q (2)

Equation (2) divided by Equation (1)

tan .

. .

qq q

== ∞ = ∞

0 096629

5 5193 5 52or Substitution into Equation (1)

FD =∞

=231 40

5 5193232 48

.

cos .. N or NFD = 232 5.52°

187


PROBLEM 3.154

A 110-N force acting in a vertical plane parallel to the yz plane is applied to the 220-mm long horizontal handle AB of a socket wrench. Replace the force with an equivalent force-couple system at the origin O of the coordinate system.

SOLutiOn

we have SF P F: B =

where P j k

j kB = - ∞ + ∞

= - +110 15 15

28 470 106 252

N[

N N

(sin ) (cos ) ]

( . ) ( . )

or N NF j k= - +( . ) ( . )28 5 106 3

We have SMO B O B O: r P M/ ¥ =

where r i j k

iB O/ m

m

= ∞ + - ∞= +

[( . cos ) ( . ) ( . sin ) ]

( . ) (

0 22 35 0 15 0 22 35

0 180213 00 15 0 126187. ) ( . )m mj k-

i j k

M0 180213 0 15 0 126187

0 28 5 106 3

. . .

. .-◊ =N m O

M i j kO = - - ◊[( . ) ( . ) ( . ) ]12 3487 19 1566 5 1361 N m

or N m N m) N mM i j kO = ◊ - ◊ - ◊( . ) ( . ( . )12 35 19 16 5 13

35°

y

y

B

o

o

x

F

x

15°

15°

110 N

150 mm

MO

rB/O

z

z

188


PROBLEM 3.155

Four ropes are attached to a crate and exert the forces shown. If the forces are to be replaced with a single equivalent force applied at a point on line AB, determine (a) the equivalent force and the distance from A to the point of application of the force when a = ∞30 , (b) the value of a so that the single equivalent force is applied at Point B.

SOLutiOn

We have

500 N A A B800 N

450 N

2000 N

Rdθ

(a) For equivalence SF Rx x: cos cos cos- ∞ + ∞ + ∞ =500 30 2000 65 450 65

or Rx = 602 4. N

SF Ry y: sin sin sin500 800 2000 65 450 65a + + ∞ + ∞ =

or Ry = +( . sin )3020 45 500 a N (1)

With a = ∞30 Ry = 3270 45. N

Then R = + =

= = ∞

( . ) ( . ) tan.

.. .

602 4 3270 453270 45

602 43325 5 79 6

2 2 q

qN or

Also SM A : ( . )( ) ( . )( ) sin ( . )(1 150 800 1 650 2000 65 0 650 2000m N m + ∞ +N m N)coos 65

m m) (450 N cos 65

∞+ ∞ + ∞ =( . )( ) ( . ) ( .1 650 450 65 0 9 3270N sin d 445)

or SM dA = ◊ =53043 1 622N m and m. R = 3330 N 79.6°

and R is applied 1.622 m. To the right of A.

(b) We have d = 1650 mm

Then SM RA y: . ( . ) ( )5304 3 1 65N m◊ = m

or Ry = 3214 7. N

Using Eq. (1) 3214 7 3020 45 500. . sin= + a or a = ∞22 9.

189


PROBLEM 3.156

A blade held in a brace is used to tighten a screw at A. (a) Determine the forces exerted at B and C, knowing that these forces are equivalent to a force-couple system at A consisting of R i j k= -(30 N) + + R Ry z and M iA

R = - ◊(12 N m) . (b) Find the corresponding values of Ry and Rz . (c) What is the orientation of the slot in the head of the screw for which the blade is least likely to slip when the brace is in the position shown?

SOLutiOn

(a) Equivalence requires SF R B C: = +

or - + + = - + - + +( ) ( )30 N i j k k i j kR R B C C Cy z x y z

Equating the i coefficients i : - = - =30 30 N or NC Cx x

Also SM M r B r CA AR

B A C A: = ¥ + ¥/ /

or - ◊ = + ¥ -+ ¥ - +

( ) [( . ) ( . ] ( )

( . ) [ ( )

12 0 2 0 15

0 4 30

N m m m)

m N

i i j k

i i

B

Cyy zCj k+ ]

Equating coefficients i

k

j

: ( . )

: ( . )

: ( . )

- ◊ = - == =

=

12 0 15 80

0 0 4 0

0 0 2

N m m or N

m or

m

B B

C Cy y

(( ) ( . )80 0 4 40 N m or N- =C Cz z

B k C i k= - = - +( . ) ( . ) ( . )80 0 30 0 40 0 N N N

(b) Now we have for the equivalence of forces

- + + = - + - +( ) ( ) [( ) ( ) ]30 80 30 40 N N N Ni j k k i kR Ry z

Equating coefficients j: Ry = 0 Ry = 0

k : Rz = - +80 40 or NRz = -40 0.

(c) First note that R i k= - -( ) ( ) .30 40 N N Thus, the screw is best able to resist the lateral force Rz when the slot in the head of the screw is vertical.

190


PROBLEM 3.157

A concrete foundation mat in the shape of a regular hexagon of side 12 m supports four column loads as shown. Determine the magnitudes of the additional loads that must be applied at B and F if the resultant of all six loads is to pass through the center of the mat.

SOLutiOn

From the statement of the problem it can be concluded that the six applied loads are equivalent to the resultant R at O. It then follows that

S S SMO x zM M= = =0 0 0or

For the applied loads.

A

B

F E

D

C

O

z

x

2 m

4 m

2√3 m

Then SMx B

F

F

F

= + -

- =

0 2 3 2 3 50 2 3 100

2 3 0

: ( ) ( )( ) ( )( )

(

m m kN m kN

m)

or F FB F- = 50 (1)

SMz BF= + -

- -0 4 75 2 2 50

4 150 2

: ( ( ) ( ) ( )( )

( )( ) ( )

m) kN m m kN

m kN m ((100 0kN) (2 m)+ =FF

or F FB F+ = 300 (2)

Then ( ) ( )1 2+ fi FB = Ø175 kN

and FF = Ø125 kN

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