Solucionario Mecanica Vectorial para ingenieros Estatica - Edicion 9

CHAPTER 3

80 mm

PROBLEM 3.1

A foot valve for a pneumatic system is hinged at B. Knowingthat a - 28°, determine the moment of the 1 6-N force about

Point B by resolving the force into horizontal and vertical

components.

SOLUTION

Note that

and

=a- 20° = 28° -20° = 8°

Fx = (1 6 N)cos 8° = 1 5.8443 NFv =(16N)sin8° = 2.2268N

^£*. \©* — tL> Kl

* U^r^-^T^P,,/|t^^^u^d

^r^^^^-,Also x = (0. 1 7 m) cos 20° = 0. 1 59748 m

y = (0.17 m)sin 20° = 0.058143 m.

>n^y c

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

MB =xFy +yFx= (0.1 59748 m)(2.2268N)

+(0.058143 m)(l 5.8443 N)

= 1.277 N-m or MB =1.277N-m^)4

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153

PROBLEM 3.2

A foot valve for a pneumatic system is hinged at B. Knowingthat a = 28°, determine the moment of the 1 6-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

g = (16 N) sin 28°

= 7.5115 N

Then MB = rmQ°^7^

= (0.17m)(7.5115N)

= 1.277N-m or MB = 1.277 N-m^^

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154

200 in..,

25"

-100 mm- 200 mm

i'"> mm

PROBLEM 3.3

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

creates the same moment about D.

SOLUTION

(a)

O.2.

PC

0>T-**>

Fv =(300N)cos25°

= 27.1.89 N

Fy=(300 N) sin 25°

= 126.785 NF = (27 1 .89 N)i + (1 26.785 N)

j

r = ZM = -(0.1m)i-(0.2m)j

MD =rxF

MD = HO. 1 m)i - (0.2 m)j] x [(271 .89 N)i + (1.26.785 N)j]

= -(12.6785 N • m)k + (54.378 N • m)k

= (41.700 N-m)k

M =41.7N-nO^

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

DB at 45°^£L

MD =Q(DB)

41 .700 N m = (2(0.28284 m) Q = 147.4 N ^L 45° <

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155

200 mn.

25°

•-lOOnim-* -200 mm *.

125 il

•C

H

PROBLEM3.4

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.

SOLUTION

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

M =41.7N-m^HCl^n

0>\7.Siy\

c = at.

(b) Since C is horizontal C = Ci

r = DC = (0.2 m)i - (0. 125 m)j

M D =rxCi = C(0.l25m)k

4l.7N-m = (0.l25m)(C)

C = 333.60 N

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

C = 334N <

tan6^0.125 m0.2 m

a = 32.0°

M D = C(£>C); DC = V( -2 m)

2+ (°- 125 m)'

= 0.23585 m

41.70 Nm = C(0.23585m) C = 176.8 N^L 5HX)°<

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156

PROBLEM 3.5

An 8-1b force P is applied to a shift lever. Determine the moment of V about Bwhen a is equal to. 259.

SOLUTION

First note Px = (8 lb) cos 25°

= 7.2505 lb

/^ =(8 lb) sin 25°

= 3.3809 lb

Noting that the direction of the moment of each force component about B is

clockwise, have

= -(8in.)(3.3809 1b)

- (22 in.)(7.2505 lb)

= -186.6 lb -in.

i*22. •**.

or Mj =186.6 lb -in. J) ^

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157

PROBLEM 3.6

For the shift lever shown, determine the magnitude and the direction of the smallest

force P that has a 21 0-lb • in. clockwise moment about B.

22 in.

SOLUTION

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

a =e

, 8 in ^T^s*. K

22 in. /= 19.98° i

and MB =dP^n p ZZ i«.

Where d = rAIB fl$ u=^m.y+(22m.y JB w= 23.409 in.

Then_210Ib-in.

' min " 23.409 in.

-8.97 lb Pmin =8.97 lb^ 19.98° <

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158

PROBLEM 3.7

An 1 1 -lb force P is applied to a shift lever. The moment of P about B is clockwise

and has a magnitude of250 lb • in. Determine the value ofa.

22 in.

SOLUTION

By definition

where

and

also

Then

or

or

and

MB =rmPsm.6

= a + (9Q°-tf>)

_i 8 in.<p - tan" 19.9831'

22 in.

rf/fl =V(8in.)

2+(22in.)

2

= 23.409 in.

250lb-in = (23.409in.)(lllb)

xsin(tf + 90° -19.9831°)

sin (« + 70.01 69°) = 0.97088

a + 70.0 169° = 76. 1391°

a+ 70.0169° = 103.861° a = 6.12° 33.8° <


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159

PROBLEM 3.8

It is known that a vertical force of 200 lb 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 ~ .1 0°, (c) the

smallest force P that creates the same moment about B.

V—Ar 4 in.n

SOLUTION

(a) We have MB =raBFN(4 in.)(200 lb)

800 lb -in.

or MB =H00\b-m.)<

A- m.(/;) By definition MB ~rA/BPsin

6» = 10° + (180°~70°)

= 120°

Then 800 lb • in. = (18 in.) x Psin 1 20°

or P = 51.3 lb <

(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

or

or

A*W™.cl = fA IB

800 lb • in. = (18 in.)Pm

^i„=44.4 1b Pmitl

=44.41b^l20 ^

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160

PROBLEM 3.9

A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 1040 N andlength 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.

0.2 m

0.875 m

SOLUTION

(a) Slope of line

Then

EC =0.875 m

1.90 m + 0.2 m 12

*abx - ,~ (Tab)

12

13

960 N

(1040 N) 0,1«y

and *W=-0040N)

= 400 N

Then

(b) We have

MD - TABx (0.875 m)~TABy (0.2 m)

= (960 N)(0.875 m) - (400 N)(0.2 m)

= 760 N •

m

MD ~rm (y) + TABx (x)

= (960 N)(0) + (400 N)(I .90 m)

= 760 N •

m

or M./J=760N-m

>

)^

or M7>= 760N-m

v)<«

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161

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.

0.875 m

0.2 j.i»

SOLUTION

\*.»£^

*AB

'My

o. a? 5^i

z-acw OiZ^

Slope of line

Then

and

We have

EC =0.875 m 7

7'

2.80 m + 0.2 m 24

24My

r/ffl)>

25

125

T/)B

rAH

24 7960N-m=—7^(0)+—7^(2.80*)

7^= 1224 N or r^=1224N ^

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162

PROBLEM 3.11

It is known that a force with a moment of 960 N • m aboutD is required to straighten the fence post CD. Ifthe

capacity of winch puller AB is 2400 N, determine the minimum value of distance d to create the specified

moment about Point D.

fe

0.2 in

0.875 in

SOLUTION

o.mtn

czom

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

MD ={TABmK ),(d)

where

Now

MD =960N-m

(^flmax )y = TAIHmx sin & = (2400 N)sfo

. . 0.875msin #

960 N • m = 2400 N

^(t/ + 0.20)2+(0.875)

2m

0.875(d)

+ 0.20)2+(0.875)

2

or ^ + 0.20)2 + (0.875)

2 = 2. ! 875d

or (J + 0.20)2 + (0.875)

2 = 4.7852rf2

or 3 .7852</2 - 0.40c/- .8056 =

Using the quadratic equation, the minimum values ofd are 0.51719 m and -.41151 m.

Since only the positive value applies here, d — 0.5 1 7 1 9 mor d ~ 5 1 7 mm ^

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163

5.3 in. PROBLEM 3.12

12.0 in.

2.33 in.

mmM1 >iii. i

The tailgate of a car is supported by the hydraulic lift BC. If the

lift exerts a 125-lb force directed along its centerline on the ball

and socket at B, determine the moment of the force about A.

SOLUTION

First note

Then

and

Now

where

Then

dcB == 7(12.0 in.)2

= 12.224 lin.

+ (2.33 in.)2

cos 9 =12.0 in.

12.2241 in.

sin 9-2.33 in.

12.2241 in.

*cb = FCB cos 9\ -FCB $m9l1251b

12.2241 in.

[(12.0 in.)i- (2.33 in.) j]

M.A= VB/A X ^CB

XBIA = (15.3in.)i-- (12.0 in.+ 2.33 in.) j

M

= (15.3 in.) i--(14.33 in.) j

ru-nimn* 1251b

\«5»,a> im.

\Z.& >N.

2.33 ttvi.

— (12.01 — 2.331)12.2241 in.

(1393.87 lb in.)k

(116.156 lb -ft)k or M„ = 116.2 lb- ft *)<



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164

20.5 in.

h—

-

4.38 in

i

*Ti

7.62

1

t§|1.7.2 hi.

PROBLEM 3.13

The tailgate of a car is supported by the hydraulic lift EC. If the lift

exerts a 125-lb force directed along its centerline on the ball andsocket at B, determine the moment of the force about A.

SOLUTION

First note

Then

dCB := 7(17.2 in.)2

= 18.8123 in.

+ (7.62 in.)2

cos -17.2 in.

18.8123 in.

sin 6 -7.62 in.

18.8123 in.

Z.O.'S «w.

and

Now

where

Then

fcd = (Ft:v*cos 0)\ - (FCB sin 6>)j

=S(,7 '2i"-)i + (7 '62i^

r^ = (20.5 in.)i- (4.38 in.)j

M, = [(20.5 in.)i - (4.38 in.)j] xt

.

1251b(1 7.21 - 7.62j)

18.8 123 in.

n.z .«»».

4.?.e. >Ni.

XKvZ >KJ.

(1538.53 lb -in.)k

(128.2 lb -ft)k or M^ =128.2 lb- ft^H

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165

120mm

65 mm

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.

SOLUTION

We have M c =rwc xFj,

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

clockwise.

Where

and

Mc =xFBy +yFBx

x - 120 mm - 65 mm = 55 mm

y - 72 mm + 90 mm - 1 62 mm

4 j,

F65

lix

Fa

7(65)2+(72)

2

72

V(65)2 + (72)

3

-(485N) = 325N

.(485 N)- 360 N

iWc = (55 mm)(360 N)+ (1 62)(325 N)

= 72450 N- mm= 72.450 N-m or Mc =72.5 N-m J) <





166

By definition:

Now

and

PROBLEM 3.15

Form the vector products B x C and B' * C, where B = B\ and use the results

obtained to prove the identity

sin a cos/? = -sin (a + P) + -sin (a - p).

SOLUTION

N(>te: B = £(cos/?i + sin/?j)

B'= 5(cos/?i-sin/?j)

C - C(cosa i + sin a j)

|BxC| = flCsin(a-jff)

|B'xC| = 5Csin(flf + y?)

B xC = Z?(cos /?i + sin y9j)x C(cos tfi + sin orj)

= BC(cos /?sin « - sin /?cos «)k

B'xC - /?(cos /?i - sin /?j)x C(cos ai -f sin #j)

- £C(cos yffsin ar+ sin /?cos ar)k

Equating the magnitudes of BxC from Equations (I ) and (3) yields:

BCs'm(a-p)~ BC(cos ps'm a- sin pcos a)

Similarly, equating the magnitudes of B'xC from Equations (2) and (4) yields;

BCsm(a + p) = BC(cos ps'm a + sln pcos a)

Adding Equations (5) and (6) gives:

sin(a- p) + s'm(a+ p) - 2cos /?sin a

0,)

(2)

(3)

(4)

(5)

(6)

or sin « cos /? --sin(ar + /?) + -sin(flf - /?) ^

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167

PROBLEM 3.16

A line passes through the Points (20 m, 16 m) and (-1 m, ~4 m). Determine the perpendicular distance rffrom

the tine to the origin O of the system of coordinates.

SOLUTION

dAB = V[20m- (-1 m)]2 + [1 6 m - (-4 m)f

- 29 m

Assume that a force F, or magnitude F(N), acts at Point A and is

directed ixomA to B.

Then,,

Where

By definition

Where

Then

¥~FX m

'All*B~ rA

d.41!

= -—<21i + 20j)29

V

MQ = \rA xF\ = dF

r,=-(lm)i-(4m)j

F

E> (Zom, K-rti^

M =[-(-1 m)i-(4 m)j]x—-[(21 m)i + (20 m)j]29 m

a -(20)k + (84)k]

~F|k N-m29

-*x

Finally64

29F = </(F)

. 64rf-— m

29</ = 2.21m ^

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168

PROBLEM 3.17

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

SOLUTION

(a) We have

where

Then

(b) We have

where

v4 = |PxQf

P = -7i + 3j-3k

Q = 2i + 2j + 5k

PxQk

-3

> J

-7 3

2 2 5

= [(15 + 6)i + (-6 + 35)j + (-14-6)k]

= (21)1 + (29)j(-20)k

^ = V(20)2+(29)

2+(-20)

2

A = \PxQ\

P = 6i~5j-2k

Q = ~2i + 5in.j~lk

or 4 = 41.0 <4

Then PxQi fci

6 -5

-2 5

= [(5 + 1 0)i + (4 + 6)j + (30-1 0)k]

«(15)i + (10)j + (20)k

A=j(i5)-+(\oy + (2oy or 4 = 26.9 ^

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169

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 - 5k and 4i - 7j - 5k, (b) 3i - 3j + 2k and -2i + 6j - 4k.

SOLUTION

{a) We have

where

Then

and

(b) We have

where

Then

and

AxB

A

B

AxB

|AxB|

:li + 2j-5k

= 4i - 7 j- 5k

i J *

1 +2 -5

4 -7 -5

(-1 ~ 35)i + (20 + 5)j + (-7 - 8)k

15(31 -1J -Ik)

|AxB|

X--

X-

A

B

AxB

15V(-3)2 +H)2

+(-l)2-I5>/ri

15(-3f-lJ-lk)or X

15VH

AxB

(-3i-j~k) 4

|AxB|

3i-3j + 2k

:-2i + 6j-4k

I J k

3-3 2

-2 6 -4

(12-12)i + (-4+ l2)j + (18-6)k

(8j + 12k)

|AxB| 4^(2)2 +(37 = 4^13

4(2j + 3k)

4^13or X,

V^(2j + 3k) <





170

PROBLEM 3.19

Determine the moment about the origin O of the force F = 4i + 5j - 3k that acts at a Point A. Assume that the

positioji vector ofA is {a) r - 2i - 3j + 4k, (£) r - 2i + 2.5j - 1 .5k, (c) r - 2i + 5j + 6k.

SOLUTION

(a) M,

(h) M,

' J k

2-3 4

4 5 -3

(9-20)i + (16 + 6)j + (I0+ 12)k

i j k

2 2.5 -1.5

4 5 -:s

Ma = -lli + 22j + 22k ^

(-7.5 + 7.5)i + (-6+ 6)j + (10-1 0)k M, «

(<0 M,

' J

2 5

4 5

(-1 5 - 30)i + (24 + 6)j + (10- 20)k M, -45i + 30j-10k A

Note: The answer to Part b could have been anticipated since the elements ofthe last, two rows of the

determinant are proportional.

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171

PROBLEM 3.20

Determine the moment about the origin of the force F == -2i + 3j + 5k that acts at a Point A. Assume that the

position vector ofA is (a) r == i + j + k,(6)r == 2i + 3j- 5k, (c)r == -4i + 6j + 10k.

SOLUTION

i j k

(a) M(}= 1 1 1

-2 3 5

= (5-3)i + (-2-5)j + (3 + 2)k M = 2i-7j + 5k A

i j k

(h) M - 2 3 -f

-2 3 5

= (15 + 15)i + (10-10)j + (6+ 6)k Mo =30i + I2k A

1 J k

(c) M a - -4 6 10

-2 3 5

= (30 - 30)i + (-20 + 20)j + (-.12 + 1 2)k M = <

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

determinant are proportional.



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172

-VI

200 N

r

PROBLEM 3.21

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

SOLUTION

We have

where

Then

M, =%.i x *c

rCIA = (0.06 m)i + (0.075 m)j

*c = ~(200 N)cos 30°j + (200 N)sin 30°k

i j &

MA-.= 200 0.06 0.075

-cos 30° sin 30°

= 200[(0.075sin 30°)i - (0.06sin 30°)j ~ (0.06 cos 30°)k]

or M,, = (7.50 N m)i- (6.00 N-m)j-(l 0.39 N-m)k <4

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173

'/ PROBLEM 3.22

£^-^

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 BCare 555 N and 660 N, respectively, determine the moment about O

. .'vjB 4.25 in

of the resultant force exerted on the tree by the cables at B.

/ ,.-"r

^J""

0.75 in/\-^1 "I

.T

SOLUTION

We have

where

and

M,

1

ff/O

rl3/Q

X F#

(7m)j

Tdff + Tec

'yf/f ~ ,VBA' AB

l J!C

-(0.75m)i-(7m)J+ (6m)k

(.75)2+(7)

2+(6)

2 m

*-BC*BC

(555 N)

(4.25m)i-(7m)j + (lm)k

^(4.25)2+(7)

2+(l)

2 m-(660 N)

Fi?= [-(45.00 N)i - (420.0 N)j -f- (360.0 N)k]

+[(340.0 N)i - (560.0 N)j + (80.00 M)k]

= (295.0 N)i - (980.0 N)j + (440.0 N)k

M,

i J k

7

295 980 440

Nm

(3080 N • m)i - (2070 N • m)k or M,}= (3080 N-m)i- (2070 N-m)k <





174

fi in

PROBLEM 3.23

The 6-m boom ^5 has a fixed end /*. A steel cable is stretched fromthe 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 ofthe force exerted by the cable at B.

SOLUTION

First note

Then

We have

where

Then

^c-VK>)2+(2.4)

2+(-4)

:

= 7.6 m

V=-?4^(-« + 2.4j-4k)7.6

^A ~ r/i//l

XTBC

lB/A (6 m)i

M;4 ^(6m.)ix^4r^-(~6» + 2.4j-4k)

7.6

or M //== (7.89 kN-m)j + (4.74 kN-m)k <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

175

3{ji)i.. C- 48itii

:'6-iftl

PROBLEM 3.24

A wooden board AB, which is used as a temporary prop to support

a small roof, exerts at Point A of the roof a 57-lb force directed

along BA. Determine the moment about C of that force.

SOLUTION

We have

where

and

rA/c = (48 in.)i - (6 in.)j + (36 in.)k

\<BA— xM pM

-(5in.)i + (90in.)j-(30in.)k

V(5)2+(90)

2 + (30)2in.

= -(31b)i + (541b)j-(181b)k

(57 lb)

M c lb -in.

i J k

48 6 36

3 54 18

-(1 836 lb • in.)i + (756 lb • in.)j + (2574 lb • in.)

or Mc = -(1 53 .0 lb • ft)i + (63.0 lb • ft)j + (2 1 5 lb • ft)k A


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176

/.)«'

0.6 id

0.6 m "X

PROBLEM 3.25

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^riy,xTM

where **/•/,(= (2-3 m)j

Tw: =*"DE^f.)E

- (0.6m)l + (3.3in)J-(3m)k

V(0.6)2+(3.3)

2+(3)

2m= (1 08 N)i + (594 N)j - (540 N)k

i J k

M,,= 2.3

108 594 -540

Nm

= -(1 242 N - m)i - (248.4 N • m)k

or M/i=-(1242N-m)i-(248N-m)k <

(b) We have ^A=rG//lxTCG

where r(,,=(2.7m)i + (2.3m)j

T — 1 T*CO ~~ *"CG'CG

= -(.6m)i + (3.3m)j-(3m)k(810N)

V(.6)2+(3.3)

2+(3)

2m= -(108 N)i + (594 N)j - (540 N)k

i J k

M A= 2.7 2.3

-108 594 -540

N-m

= -(1 242 N • m)i + (1458 N • m)j + (1 852 N • m)k

or MA = -(1242 N • m)i + (1 458 N • m)j + (1 852 N • m)k A


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177

PROBLEM 3.26

A smaJI boat hangs from two davits, one of which is shown in the

figure. The tension in line ABAD is 82 lb. Determine the momentabout C of the resultant force R,, exerted, on the davit at A.

SOLUTION

We have

where

and

Thus

Also

Using Eq. (3.21):

R.=2R(/)

. + F,D

,„~-(82 1b)j

ADAD I AD— = (82 lb)

AD6i~7.75j-3k

10.25

F,/3=(48lb)i-(62 1b)j-(24 1b)k

RA = 2¥AB +FAD = (48 lb)i - (226 lb)j - (24 lb)k

l A/C

Mc

(7.75ft)j + (3ft)k

1 J

7.75

48 -226

k

3

-24

(492 lb • ft)i + (1 44 lb • ft)j - (372 lb • ft)k

Mc = (492 lb • ft)i + (1 44.0 lb ft)j - (372 lb • ft)k <





178

PROBLEM 3.27

In Problem 3.22, determine the perpendicular distance fromPoint O to cable AB.

PROBLEM 3.22 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

where

Now

and

IM TBAd

d = perpendicular distance from O to line AB.

M, rB/0 X *BA

rBIO ={lm)\

BA' AB

(0.75m)i-(7m)j + (6m)k

*BA ~~ '"BA 1 AB

M,

and

or

(555 N)'(0.75)

2+(7)

2+(6)

2m-(45.0 N)i - (420 N)j + (360 N)k

i j k

7 N-m-45 -420 360

(2520.0 N • m)i + (315.00 N • m)k

|M1= V(252o-°)

2 + (31 5.00);

= 2539.6 N-m

2539.6 N-m = (555 N)rf

d = 4.5759 m or d = 4.58 m <

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 limiteddistribution to teachers andeducators permittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.

179

PROBLEM 3,28

In Problem 3.22, determine the perpendicular distance from

Point O to cable BC.

PROBLEM 3.22 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.#.

SOLUTION

We have

where

[M, Tscd

d — perpendicular distance from O to line BC.

M,

lB/0

rB/0 X ^liC

7mj

P — 1 T'BC ™ *"BC* HC

(4.25m)i-(7m)j + (lm)k(660 N)

M,

and

V(4.25)2+(7)

2+(l)

2m

(340 N)i - (560 N)j + (80 N)k

I J k

7

340 -560 80

(560 N • m)i - (2380 N • m)k

|M!= 7(560)

2+(2380)

= 2445.0 N-m

2445.0 N-m = (660 N)rf

d = 3.7045 m or d~ 3.70 m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, inc. All rights reserved. No part of this Manual may be displayed,




180

A."(i in.

•IS in.

Cj-iii.

.90 in,:

()(> in.

~~*S B ~~^~£

PROBLEM 3.29

In Problem 3.24, determine the perpendicular distance from Point Dto a line drawn through Points A and B.

PROBLEM 3.24 A wooden board AB, which is used as a temporary

prop to support a small roof, exerts at Point A of the roof a 57-lb

force directed along BA. Determine the moment about C of that

force.

SOLUTION

We have

where

and

IMJ^rf

d — perpendicular distance from D to line AB.

M.l}^rAlii x¥]iA

r,/D =-(6in.)j + (36in.)k

^liA ~ ^BA^BA

(-(5in.)i + (90in.)j-(30 in.)k)

V(5)2+(90)

2 + (30)2in.

-(31b)i + (541b)j-(181b)k

(57 lb)

M,;)

I J k

-6 36 lb -in.

-3 54 -18

-(1 836.00 lb • in.)i - (1 08.000 lb • in.)j- (1 8.0000 lb • in.)k

IM 836.00)2+(108.000)

2 + (18.0000)'

= 1839.26 lb -in.

1839.26 lb -in =(57 lb)e/

</ = 32.268 in. or </ = 32.3in. <

PROPRIETARY MATERIAL. & 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers andeducatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

181

(if) in.

B^-<

PROBLEM 3.30

In Problem 3.24, determine the perpendicular distance from Point C to a

line drawn through Points A and B.

PROBLEM 3.24 A wooden board AB, which is used as a temporary

prop to support a small roof, exerts at Point A of the roof a 57-3b force

directed along BA. Determine the moment about C of that force.

SOLUTION4

We have

where

\Mc \~FBAd

d - perpendicular distance from C to line AB.

3fc.. .

M.c ^rA/c x¥BA ^%f M / >lD ^-.

rAIC - (48 in.)i - (6 in,)j + (36 in.)k ^^\ \ //S^fylA ~ ^BA^IIA f*

_ (~(5 in.)i + (90 in.)j - (30 in.)k) fe*

7(5)2+(90)

2+(30)

2in.

= ™(3 lb)i + (54 lb)j - (1 8 lb)k

i J k

Mc = 48 -6 36

-3 54 -18

lb -in.

and

= -(1 836.001b -i

|M

n.)i - (756.00 lb • in.)j + (2574.0 lb • in.)k

2

c |= V(l 836.00)

2 + (756.00)2 + (2574.0)

-3250.8 lb -in.

3250.8 lb in. = 57 lb

d- 57.032 in. or d = 57.0 in. ^

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. ;Vt> />flrf o//Afc Manual may be displayed,




182

0.6 m

PROBLEM 3.31

In Problem 3.25, determine the perpendicular distance fromPoint A to portion DE of cable DEF.

PROBLEM 3.25 The ramp ABCD is supported by cables at

comers 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

where

IM TDEd

and

d

—

perpendicular distance from A to line DE,

Mr„HIA (2.3 m)j

'DE ~ A"DE*-DE

(0.6m)i + (3.3m)j-(3m)k(8I0N)

M

V(0.6)2 + (3.3)

2+(3)

2 m(108 N)i + (594 N)j - (540 N)k

i J b

2.3 N-m108 594 540

~(1 242.00 N • m)i - (248.00 N • m)k

|MJ =-N/(1242.00)

2+(248.00)

:

= 1266.52 N-m

1266.52Nm = (8 10 N)d

d = 1 .56360 m

O'fcn,

or d = 1.564 m

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 limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it withoutpermission.

183

PROBLEM 3.32

In Problem 3.25, determine the perpendicular distance from

Point A to a line drawn through Points C and G.

PROBLEM 3.25 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, (h) the cable at C.

0.6 m

SOLUTION

We have

where d - perpendicular distance from A to line CG.

rG/A X IceM

IVu/A

lCG

M

(810 N)

(2.7m)i + (2.3m)j

-(0.6.m)i + (3.3m)j-(3 m)k

A/(0.6)2+(3.3)

2 + (3)2m

-(108N)i + (594N)j-(540N)k

i J k

2.7 2.3 N-m-108 594 -540

-(1242.00 N • m)i + (1458.00 N • m)j +(1 852.00 N m)k

and

= 2664.3 N-m

2664.3 N-m = (8 ION)d

d = 3.2893 m or d = 3.29 m 4



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184

PROBLEM 3.33

.In Problem 3.26, determine the perpendicular distance fromPoint C to portion AD of the line ABAD.

PROBLEM 3.26 A small boat hangs from two davits, one ofwhich is shown in the figure. The tension in line ABAD is

82 lb. Determine the moment about C of the resultant force

R4 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 3.26:

= -(48 lb)i+ (62 lb)j + (24 lb)k

Mc =rm; xF^= +(6 ft)i x[-(48 lb)i + (62 lb)j + (24 lb)k]

= -(1441b-ft)j + (372 1b-ft)k

Mc =V044>2+(372)

2

= 398.90 lb • ft

Then Mc = VMd

Since F,w = 82 1b

398.90 lb -ft

82 1bd = 4.86 ft <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,yon are using it without permission.

185

*-* PROBLEM 3.34

Determine the value of a that minimizes the

i

16 ft perpendicular 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.cMrA/c xF\^F(d)

Where d - perpendicular distance from C to line AB

%ABF

(24ft)i + (24ft)j-(28)k

F =W7(24)

2+(24)

2+(18)

2ft

•F

(6)i + (6)j-(7)k

AK; = (3 ft)i - (1 ft)j- {a - 1 ft)k

i J k

3 -10 10a

6 6-7Mc

.[(10+ 6fl)i + (81-6.ai)j + 78k]11

Since

12!

^/c xr or r^xF2

|=W

(1 + 6a)2 +(81- 6a)

1 + (78)2 = d*

d / j2Setting -j-{d )- to find a to minimize c/

1

[2(6)(l + 6a) + 2(-6)(8 1 - 6a)] =

Solving

121

a = 5.92 ft or o - 5.92 ft ^

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Afo par/ o//Aw Minna/ »m>' fc displayed




186

PROBLEM 3.35

Given the vectors P =3i -

P S, and Q • S.

-J + 2k,Q == 4i + 5j-3k, and S = -2i + 3j - k, compute the scalar products P • Q,

SOLUTION

P-Q = (31-1j + 2k)-(4i-5J-3k)

= (3)(4) + (-l)(~5) + (2)(-3)

= 1 or P Q = I AP • S = (3i - lj + 2k) • (-2i + 3j - Ik)

= (3)(-2) + H)(3) + (2)H)= -U. or P-S = -U <

Q-S-(4i-5j~3k)-(-2i + 3j-l.k)

= (4K-2) + (5)(3)+ (-3X-l)

= 10 or Q-S = 10 <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers andeducatorspermittedby McGraw-Hillfor their individual coursepreparation, ifyou are a student using this Manual,you are using it without permission.

187

PROBLEM 3.36

Form the scalar products B • C and B' • C, where B = B', and use the

results obtained to prove the identity

cos a cosP ~— cos (a + /?)+— cos (a - ft).

SOLUTION

y By definition

where

(1)

B-C = £Ccos(a~/?)

B = /?[(cos/?)i + (sin/?)j]

C = C[(cos a)\ + (sin a)j]

(B cos /?)(Ccos a) + (B sin /?)(Csin a) - BCcos(a~ (3)

or cos /?cos a+ sin /?sin a = cos(a - j3)

By definition B' • C - BC cos (or + /?)

where B' = [(cos fi)i ~ (sin /?)j]

(B cos p)(C cos or) + (~B sin /?)(C sin or) = BCcos (a + /?)

or cos /?cos « - sin ft sin a = cos{a + /?)

Adding Equations (1) and (2),

2 cos /?cos a = cos (a-fi)+ cos (a+ /?)

or cosacosfi = ~-cos(a + fi) + ~cos(a~ ft) A

(2)

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188

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 CD.

SOLUTION

First note AB = AB{$m 37°j - cos 37°k)

CD = CD(- cos 40° cos 55°j + sin 40°j- cos 40° sin 55°k)

fc.b\

Now

or

or

AB- CD = (AB)(CD) cos

AB(sm 37°j - cos 37°k) • CD(-cos 40° cos 55°i + sin 40°j - cos 40°sin 55°k)

~(AB)(CD) cos

cose? = (sin 37°)(sin 40°) + (~cos 37°)(--cos 4()°sin 55°)

= 0.88799

or (9 = 27.4° A

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 limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

189

PROBLEM 3.38

Section AB of a pipeline lies in theyz 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 = /f£?(sm37 j-cos37 k)__.

= £F(cos 32° cos 45°i + sin. 32°j - cos 32° sin 45°k)

1

£

F

l/^ (€t\ *W A

Now AB-EF--= (AB)(EF)cosO

or AB(sm

:

17°j - cos 37°k) - £F(cos 32° cos 45°j + sin 32°j - cos 32° sin 45°k)

= (AB)(EF)cos0

or cos = (sin 37°)(s.in 32°)+ (-cos 37°)(-cos 32° sin 45°)

= 0.79782

or (9 = 37.1° ^

PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No parr of this Manual may be displayed,




1 90

PROBLEM 3.39

Consider the volleyball net shown.

Determine the angle formed by guywires AB and AC,

SOLUTION

First note

and

By definition

or

or

AB - V(-6.5)2 + (-8)

2 + (2)2 = 1 0.5 ft

AC^yj(0f+(~S)2 + (6)2 =]0 ft

AB = -(6.5 ft)i - (8 ft)j + (2 ft)k

7c = -(8ft)j + (6ft)k

AB-AC = (ABXAC)cos0

(-6,51 - 8j + 2k) • (-8j + 6k) = (1 0.5)0 0) cos -

(-6.5)(0) + (-8)(-8)+ (2)(6) - 1 05 cos 6

cos = 0.72381 or = 43.6° <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it withoutpermission.

191

sit

PROBLEM 3.40

Consider the volleyball net shown.

Determine the angle formed by guy

wiresAC and AD.

SOLUTION

First note

and

By definition

or

or

^C = V(0)2+(~8)

2+(6)

2

-10 ft

AD = J(4)2+(-&)

2 + (l?

= 9 ft

IC = ~(8ft)j + (6ft)k

75 = (4ft)j-(8ft)j + (lft)k

AC AD = (AC)(AD)cos0

(-8j+ 6k) • (4i - 8j + k) = (1 0)(9) cos t

(0)(4) + (-8)(-8) + (6)(1) = 90cos 6

cos0 = 0.77778 or # = 38.9° <

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192

1.2 in

2.-1

PROBLEM 3.41

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 ofthe force exerted by cableAC at Points.

-f

2.6 in

2.4 m

SOLUTION

(a) First note

and

By definition

or

or

or

(b) We have

AC~-=^j(~2A)2 + (0.8)

2+(1.2)

2

= 2.8m

AB = yj(-2A)2+(-].&? +(Q)

2

= 3.0m

^C = -(2.4 m)i + (0.8 m)j + (1 .2 m)k

I/? = -(2.4m)i-(K8m)j

AC -AB = (AC){AB) cos

(-2.41 + 0.8j + 1 .2k) • (-2.4i - 1 .8j) = (2.8)(30) x cos

(-24X-2.4) + (0.8X-1 .8) + (1 .2X0) - 8.4cos

cos# = 0.51429

= 7^c cos<9

= (1260N)(0.51429)

or = 59.OC

or C^cU=648N

PROPRIETARY MATERIAL. © 20 JO The McGraw-Hill Companies, Inc. All rights reserved. ;Vo />«/•/ o//Afe Muiua/ /w/y to? displayed,reproduced or distributed in any form or by any weans, without (he prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

193

2.4 m

PROBLEM 3.42

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 = >/(-2.4)2 + (i ,2)

2 + (-2.4)2

= 3.6m

AB = V(-2.4)2 + (-1 .8)

2 + (0)2

-3.0m

and AD = -(2.4 m)i + (1 .2 m)j - (2.4 m)k

AB = -(2.4m)i~(1.8m)j

By definition, ADAB = {AD)(AB)cos0

(-2.4i + 1 .2j - 2.4k) (-2.4i - .1 .8j) = (3.6)(3.0)cos#

(-2.4)(-2.4) 4- (1 .2)(-I.8) + (-2.4X0) = 10.8 cos

cos $ — —3

^^70.5° <

(b) (*AD'AB ~ *AD ' ^Ali

= TAD cos&

=(405N)^ Pad)ab = 135.0 N <

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194

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. Knowingthat the distance from O to P is 6 in. and that the tension in thecord is 3 lb, determine (a) the angle between the elastic cordand the rod OA, (b) the projection on OA of the force exertedby cord PC at Point P.

SOLUTION

First note

Then

^ = >/(12)2+(12)

2+(-6)

2=18in.

» OA 1 ,

= i(2i + 2j~k)

Now OP = 6 in, => OP = -(04)

The coordinates ofPoint P are (4 in., 4 in., -2 in.)

PC = (5 in.)i + (1 .1 in..)j + (14 in.)k

PC = 7(5)2+ (1 1)

2 + (14)2 = V342 in.

~PCXOA ^(PC)cosO

so that

and

(a) We have

or

or

(5i + llj + 14k)--(2i + 2(i-k) = 7342cos^

cos]

3V342

0.32444

[(5)(2) + (ll)(2) + (14)(~l)j

(b) We have

= {Tpc'K]>c)-XOA

PCftpc OA

— Tpr COS&

= (3 1b)(0.32444)

or = 71.1° ^

or (Tpc)oa =0-973 lb ^

=^^S~S5£=SS£=f£S195

PROBLEM 3.44

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

Then

(M^/(12)2+(12)

2 +(-6)- =18 in.

OA 1

XCH ~ — (12i + 12j-6k)OA OA 18

(2i + 2j-k)

Let the coordinates of Point P be (x in., j> in., z in.). Then

PC = [(9 - x)in.]i + (15- y)in.]j + [(12- z)in.]k

Also,

and

OP = rfo^o, = -^(21 + 2J-k)

OP ~ (x in.)i + (>' in.)j + (z in.)k

2 2dOP

The requirement that CM and PC" be perpendicular implies that

^PC-0

or -(2j + 2j-k)-[(9-.v)i + (15-y)j + (12-2)k] =

or (2)|9--^>| + (2) 'l5-|rfw |+ H) 12 -rf0P

or Jnp = 12.00 in. ^W

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196

PROBLEM 3.45

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

(/;) P = 5i - j + 6k, Q = 2i + 3j + k5 and S = -3i - 2j + 4k.

SOLUTION

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

O) VoI = P-(QxS)

4-3 2

-2 -5 1 in.3

7 1 -1

(20-21.-4 + 70 + 6-4)

67

or Volume=67.0 4(b) Vol=P-(QxS)

5 -1 6

2 3 I in.3

-3 -2 4

(60 + 3-24 + 54 + 8+10)

111

or Volunie= 111.0 A

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197

PROBLEM 3.46

Given the vectors P - 4i - 2j + 3k,Q = 2i + 4j - 5k, and S = SJ - j + 2k, 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 (Qx S).

P-(QxS) = ()

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

4-2 3

Then 2 4 -5-0

Sx -1 2

or 32 + lO.S: -6-20 + 8-125;. =0 S =7 <

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198

0.1 1 in...

PROBLEM 3.47

The 0.61x].00-m lid ABCD of a storage 'bin is hingedalong 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 ofthe force exerted by the cord at D.

SOLUTION

First note

Then

and

Now

where

Then

z = Vi°-61)2-(0.Jl)

2

0.60 m0.11m

dm = V(0.3)2+(0.6)

2

+(-0.6r

= 0.9 m

1 l)E

66 N

0.9(0.3i + 0.6j-0.6k)

= 22[(lN)i + (2N)j-(2N)k]

MA = rD/A xTDf:

r/)//(

= (0.ll m)j + (0.60m)k

i J k

M/(= 220 0.11 0.60

1 2 -2

= 22[(-0.22 ~ 1 .20)i + 0.60j - 0. 1 1k

]

= - (3 1 .24 N • m)i + (1 3.20 N • m)j - (2.42 N • m)k

-3 1 .2 N • m, Mv= 1 3.20 N • m, A/

2= -2.42 N - m A

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//™ Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individualcoursepreparation. Ifyou are. a student using this Manual,you are using it without permission.

199

PROBLEM 3.48

The 0.61xl.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.

SOLUTION

First note

Then

and

Now

where

Then

x6\y-(o.uy

0.60 m

= l.lm

lCE66 N1.1

(-0.7i + 0.6j-0.6k)

= 6[-(7N)i + (6N)j-(6N)k]

M A ^vm x\E

r£M = (0.3m)I + (0.71m)J

i J k

MA -6 0.3 0.71

-7 6 -6

= 6[-4.26i + 1 .8j + (1 .8 + 4.97)k]

= - (25.56 N • m)i + (1 0.80 N • m)j + (40.62 N • m)k

Mx- -25.6 N • m, M

y= 1 0.80 N • m, Afz = 40.6 N •m -4

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200

PROBLEM 3,49

To lift a heavy crate, a man uses a block and tackle attached to thebottom of an I-beam at hook B. Knowing that the moments about the yand the z axes of the force exerted at B by portion AB of the rope are,

respectively, 120 N • m and -460 N • m, determine the distance a.

SOLUTION

First note

Now

where

Then

Thus

^ = (2.2m)i-(3.2 m)j-(am)k

™*D ~ VAID X '«/j

'AID (2.2m}i + (1.6m)j

TT,BA

I!Ad

ISA

M,T,BA

dBA

(2.2i-3.2j-ak)(N)

i J k

2.2 1.6

2.2 -3.2 -a

T,BA

dt .

{- 1 .6a \ + 22a\ + [(2.2)(~3 .2) - (1 .6)(2.2)]k}

Mv =2.2-^-a

BA

Then forming the ratio

M,

itM.

-I0.56-7*4

d

(N • m)

(N • m)HA

120 N-m 2 -27Z"(N-m)dB,t

-460 N-m -10.56-^- (N-m)or a = 1 .252 m ^j

PROPRI&IARl MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of ,his Manual may be displayedreproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manualyou are using it without permission.

201

PROBLEM 3.50

To lift a heavy crate, a man uses a block and tackle attached to the

bottom 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 they

axis is 132 N • m, determine the distance a.

SOLUTION

First note

and

Now

where

Then

Substituting for My and dtHA

dBA - J(22f+(-3.2)2 + (-af

= Vl5.08 + «2 m

195 NTs,=^^-!-(2.2i-3.2j-ok)

My~\ -(rA/D xTw )

Vf/0

M,

(2.2m)i + (1.6m)j

195

195

d

1

2.2 1.6

2.2 -3.2 -a

(2.2a) (N • m)HA

132 N-m195

J\Jm+a*(2.2a)

or 0.30769^1 5.08 + a* - a

Squaring both sides of the equation

0.094675(15.08 + a2 ) = a2

or a = 1.256 m A

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202

PROBLEM 3.51

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 Rtexerted on the davit at A must not exceed

279 lb • ft in absolute value. Determine the largest allowable

tension in line ABAD when x ~ 6 ft.

SOLUTION

First note R 21^ +TADAlso note that only TAD will contribute to the moment about the z axis.

Now

Then,

Now

where

Then for 71.

= 10.25 ft

'AD

M..

yAIC

279

TADADr

(6i-7.75j-3k)10.25

(7.75ft)j + (3ft)k

7\

10.25

1

7,75 3

6 -7.75 -3

~^H-(1)(7.75)(6)|10.25'

A ;I

or 7" =61.5 lb <

PROPRIETARY MATERIAL. CO 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.

203

PROBLEM 3.52

For the davit of Problem 3.51, determine the largest allowable

distance x when the tension in line ABAD is 60 lb.

SOLUTION

From the solution ofProblem 3.51, TAD is now

l AI)

ADAD

60 lb

yjX2 +(-lJS)

2+(-lf

(xi-7.75j-3k)

Then M = k • (rA/c xT,(0) becomes

279

279.

60

V*2+(-7.75)

2 +(-3)'

60

1

7.75 3

jc -7.75 -3

six2 +69.0625

(D(7.75)(x)

279>/x2 + 69^0625 = 465.x

0.6Vjc2 + 69.0625 =x

Squaring both sides: 0.36x2 + 24.8625 = r

x1 =38.848 x = 6.23 ft 4

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204

PROBLEM 3,53

To loosen a frozen valve, a force F of magnitude 70 lb is

applied to the handle of the valve. Knowing that 6 = 25°,

Mx - -6.1 lb • ft, and Mz~ -43 lb • ft, determine (j) and d.

SOLUTION

We have

where

Pol-

and

From Equation (3)

From Equation (1)

rivi(

VAIO'

F

F

ML:

M.

My

M.

^ = cos

-(4in.)i + (llin.)j-(rf)k

F(cos #eos fi- sin $\ + cos #sin <J)k)

70 1b, 9 = 25°

- (70 lb)[(0.9063 1 cos 0)i - 0.42262j + (0.9063 1 sin 0)k]

ij k

(701b) -4 11 -J in.

-0.90631 cos -0.42262 0.90631 sin

(70 lb)[(9.9694sin - 0.42262^)1 + (-0.9063 Wcos + 3.6252sin <p) j

+ (1.69048 -9.9694 cos 0)k] in.

(70 lb)(9.9694sin - 0.42262d)m. = -(61 lb • ft)(I2 in./ft) (1)

(70 lb)(-0.9063 Irfcos (j) + 3.6252 sin<f>)

in. (2)

(70 lb)(l .69048 - 9.9694cos^ in. - -43 lb • ft(l 2 in./ft) (3)

U 634.33

697.8624.636'

d1022.90

29.583= 34.577 in.

or (p = 24.6° <

or c/ = 34.6in. ^

PROPRIETARY MATERIAL. €5 2010 The McGraw-Hill Companies, inc. All righis reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers andeducators permittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.

205

i I in.

PROBLEM 3.54

When a force F is applied to the handle of the valve shown, its

moments about the x and z axes are, respectively, Mx ~ -11 lb • ft

and M.z- -81 lb • ft. For d~21 in., determine the moment My of

F about they axis.

SOLUTION

We have

Where

and

ZM : rw xF =M(

r^=-(4in.)i + (lim.)j-(27in.)k

F = F(cos 0cos <p\ - sin. $\ + cos sin flk)

i J k

-4 1 .1 -27

cos 0cos (p -sin 6 cos #sin

F [(11 cos sin 0-27 sin 0)i

+ (-27 cos Bcos <p+ 4 cos #sin 0)j

+ (4sin - 1 1 cos 6*cos ^)k](lb • in.)

Ma =F lb in.

Mx= F(l. 1 cos (9sin0-27 sin 0)(lb • in.)

MJ?

= F(-21 cos 0cos0+ 4cos 0sin 0) (lb • in.)

Mz= F(4 sin <9 - 1 .1 cos cos (j)) (lb • in.)

Now, Equation (1) cos #sin <f>-11

MF^+ 27sin#

cos 0cos 0-— 4sin —

—

-

n{ Fand Equation (3)

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

0)

(2)

(3)

(4)

(5)

M, =/N-27 4sin0-F

+ 41 (M111 F

*- + 27sin0

or M.11

(27MZ+4.M

V )

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206

PROBLEM 3.54 (Continued)

Noting that the ratiosyp and± are the ratios of lengths, have

Mv=— (-81 lb -ft)+—(-77 lb -ft)

- 11 11

= 226.82 lb • ft or Mv-= -227 lb - ft A

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207

0.35 in

0.75 m

0.75 in*

PROBLEM 3.55

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 ofthe cable.

SOLUTION

MAD ~ "'AD ' \VBIA X *Bll)

Where X„,D ---(4i-3k)

'a/a (0.5 m)i

and

Then

dm - vv0.375)2+(0.75)

2+(-0.75)

2

= 1.125 m

T.SH

450 N—(0.375i + 0.751 - 0,75k)

1.1.25J

(1 50 N)i + (300 N)j - (300 N)k

Finally Hu> =

4 -3|

0.5!

150 300 -300;

[(-3X0.5X300)]

or MAD 90.0 N-m

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208

0.35 m

0.75 m

PROBLEM 3.56

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

SOLUTION

Where

and

Then

Finally

mad=^ad-(*biaXTbg)

M/) =-(4i-3k)

r»MA (0.5 m)j

BG - V(-°-5) + (0.925)2+ (-0.4)

2

= 1.125 ra

450 NT liC!=- -(-0.5i + 0.925j-0.4k)

1.125J '

= -(200 N)i + (370 N)j - (1 60 N)k

-3

u»'i

4

0.5

-200 370 -160

[(-3X0.5X370)] MAD -Ul.ON-m

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

209

0.7 m

.O.G.Ti]

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 AE at A is 55 N, determine the

moment of that force about the line joining Points D and B.

SOLUTION

First note

Then

Also

Then

Now

where

Then

4E = V(°-9) + (~0-6) + (0-2)2=1.1 m

«=-yp(0.9i-0.6J + 0.2k)

= 5[(9N)i-(6N)j + (2N)k]

DB - y[(\ 2f + (-0.35)2 + (0)

2

X

1.25m

DBOB DB

1.25

1

(1.2I-0.35J)

25(241 - 7j)

MDB ~ '"DB"*DB '\VAID X T Ui)

TO/}=-(0.1m)j + (0.2m)k

!M»-^(5)24 -7

-0.1 0.2

9 -6 2

H1.8- 12.6 + 28.8)

or MDfl=2.28N-m ^

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210

PROBLEM 3.58

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 -- 0.2)2 = 1.1 m= V(0.6)

2+(-0.9)

2 + (-

Then \:f'-

33 N-_ (0.6i-0.9j + 0.2k)

Also DB :

= 3[(6N)i~(9N)j-(2N)k]

= V0-2)2+(-0.35)

2+(0)

2

~ 1 .25 m

Then ^DB~~

_ DB' DB

=— (I.2i- 0.35j)1.25

J

= _L(24i-7j)25

V J;

Now Mm -A'OB '(rC/D X *cp)

where yan :-(0.2m)j-(0.4m)k

24 -7

Then Mm = -(3)25

0.2 -0.4

6 -9 -2

= —(-9.6 + 16.8-86.4)25

or Mm ^~9.50N-m <


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211

l>

PROBLEM 3.59

A regular tetrahedron has six edges of length a. A force P is

directed as shown along edge BC. Determine the moment of Pabout edge OA.

>-

SOLUTION

We have

where

From triangle OBC

Since

or

Then

and

MoA=*OA<*aoX*)

{OA)x2

f i \

(OA)z=(OA)x tanW

n/3, 2>/3vv-1 /

(OA)2=(OA)l +(OA)l +{OAzf

2 [a

2+ {OA)l +

( Ya

a , 12 . a4,0

2 h 2>/3

J. 2. 1 ,

x- =l,+Vl

J+ivr

k

p = = (a sin30°)i-(a coS 30°)k(/)) =

P(|_^

« 2

rc/0 =oi

M(A!

2 V3

I

1

2V3

Kf)]

-73"

2

' Tv - 3

y

0X->/3) =aP

Moa =aP_

72"

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212

PROBLEM 3.60

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-BC = Q

where

From triangle OBC (OA)x

(OA)z =(OA)x tm\30° = -

04 = I-|I + (CM)J +

( i>

\

V3 J 2^3

ka

\2Sj

and BC = (asm 30°) i - (a cos 30°)k

Then

or

so that

a .

2

2

i+(.oa)vi+

2

c \a

^+ (O^)v(0)-~ =

4 } 4

OA-BC^0

(i-V3k)~ =

<9/i is perpendicular to #C.

(6) Have A^fW = Pt/, with P acting along BC and d the perpendicular distance from OA to 5C.

From the results ofProblem 3.57

ft?M0A

Pa

4i

4i

~Pd or d =—T A72

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213

y PROBLEM 3.61

y^ji \ 45 in. A sign erected on uneven ground is

4 V guyed by cables EF and EG. If the

11 _^#g& force exerted by cable £7** at E is 46 lb,

si /,: .^-^.,-f?;"" .--.. determine the moment of that force

96 ii r. 8»v jm^M^^^W% about the line joining Points A and D.;|*:

=

i|

r\ .

1 fPIMl

'/

L#j >' M^:

^îÎbr>>V - 47 in .

l:-;iy';!t'^ " ':>-^ :'::

:

'"'

' <*><. -

.--•'-•

. •• -'• -/••- ,"-~-J

" "^v "^

\ .--^ &%8 iftô^

i 1^Xl7in. >

• v-ii:—;T*V ->''' T. •

-'*'"- I'-i i"»?»*" ^' fT /^ "''

i

"x*'' '\

SOLUTION

First note that EC ~-V(48)

2+ (36)

2 - 60 in. and that J§ = -g = |. The coordinates of Point E are

then(fx48,96, J> 36) or (36 in., 96 in,, 27 in,). Then

^,=>/H5)2 + H10)

2+(30)

2

= 115 in.

Then T^==^(-l5I-110J + 30k)

Also

= 2[-(3 lb)i - (22 lb)j + (6 lb)k]

AD = 7C48)2 + (-12)

2 + (36)2

= 3.2726 in.

Then

_

=—!—(48i-12j + 36k)12^26

=ir(4M+3k)

Now MAD =-Kiiy(-rEI,4 XTEF)

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214


where

Then M

rm = (36 in.)i + (96 in.)j + (27 in.)k

1

2

(2)

4 -I 3

36 96 27

-3 -22 6

(2304 + 81-2376 + 864 + 216 + 2376)

or M^, =1359 lb -in. <«

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215

>J PROBLEM 3.62

% in.

\-1 > in. A sign erected on uneven ground is

guyed by cables EF and EG. If the

^0^ force exerted by cable EG at E is 54 lb,

,, ,,^'y^^ determine the moment of that force

Is. ^M^M^M:-}-^:-^ about the line joining Points A and D.

"&ftK?'?^^T?^i*K;

: V^lS ]ff ;1 - :i' p":

>^' *

*-"' "\ ^^^x in.XJ7in-

?|V ; 36iri7 \^

\| 12 1 ®^,

SOLUTION

First note that BC = - 60 in. and that ~ ~~ — ~. The coordinates of Point E areBC 60 4=V(48>

2+ (36)

2=

then (Jx 48, 96, f <36) or (36 in , 96 in., 27 in.). Then

4*;=Vai)2+(-88)

2+(-44)

2

= 99 in.

Then T*;= ^(lli-88j -44k)

Also

= 6[(llb)i-(8 1b)j-(4 1.b)k]

^/> - s/(48)2 + (~1 2)

2 + (36)2

= 12>/26 in.

Then AD AD

=—!=(48i~I2j + 36k)12^26

J

= ~=-(4i-j + 3k)V26

Now MAD ^kAD <rm xTEG )

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21fi

where

Then


rBA = (36 in.)i + (96 in.)j + (27 in.)k

(6)

(-1536 -27 »864 -.288 -144 + 864)

MAD26

_6_

V26

4 -I 3

36 96 27

I -8 -4

or M,n =~23501b-in. <

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217

PROBLEM 3.63

Two forces F| and F, in space have the same magnitude F, Prove that the moment of F, about the line of

action of F2 is equal to the moment of F2about the line of action of F,

,

SOLUTION

First note that ¥x~f\Ay and F2 ~F2

a\2

Let Mj = moment of F2 about the line of action of M, and M2= moment of F, about the line of

action of M,

Now, by definition

Since

Using Equation (3.39)

so that

itfi=4-(rBM xF2 )

= A\-( i:wa x A2)f2

Fi=F2 =F and xmMi=^r(»*^xi2

)F

M2 =X2-{-*

BIA xAx)F

h i?BIA XX2 ) = ^2 HfiM X ^1

)

M2= A

}-(rm xZ

2 )F

-r»BM

M[2 =M2i A

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218

0.35 m

^m

0.925 m

0.875 m

//

0.75 in

L _y_

0.75 m0.5 n>*K-^^ 1C,<

0.5 m "A,

PROBLEM 3.64

In Problem 3.55, determine the perpendicular distance betweenportion BH of the cable and the diagonal AD.

PROBLEM 3.55 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 ADofthe force exerted on the frame by portionBH of the cable.

SOLUTION

From the solution to Problem 3.55: 7)W =450N

TBH = (1 50 N)i + (300 N)j - (300 N)k

|A/^! = 90,0N-m

X,D =-(4i.-3k)

Based on the discussion of Section 3.1 1, it follows that only the perpendicular component of TBH will

contribute to the moment ofTbh about line AD.

Now\'lill )para!lei

~~ *IIH '"'AD

= (1 501 + 300j - 300k) • I(4i - 3k)

= ^l'050)(4) + (-300)(-3)]

= 300 N

Also

so that

T*liH

v BH 'perpendicular

=V */>•// /parallel

+ ( '«// ^perpendicular

= V(450)2 - (300)

2 = 335.41 N

Since %AD and (TM )pcrpcndicuiQr are perpendicular, it follows that

MAD ~ "('BH Jpcipendiculftr

or 90.0 N-m

d

= rf(335.41N)

= 0.26833 m d = 0.268 m A

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219

0.35 in

(;:v.l 0.875 m

PROBLEM 3.65

In Problem 3.56, determine the perpendicular distance between

portionBG of the cable and the diagonal AD,

PROBLEM 3.56 In Problem 3.55, 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 3.56: T}iG- 450 N

Tbg =-(200M)i + (370N)j-(160N)k

\MAD \= \WH-m

^D--(4i-3k)

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

contribute to the moment ofTbg about line AD.

Now (Tbg )parallel ~ * BO ' ^ 41)

Also

so that

- (-2001 + 370j - 1 60k) - ~(4i - 3k)

= |[(-200X4) + (-I60X-3)]

= -64 N

*BG ~ '*/J6" /parallel

"*~V *BG )perpendicular

(TBG )pcrpenc.icuiar= V(450)

2 - (~64)2= 445.43 N

Since XAD and (Tso )pC1.

p,ndicularare perpendicular, it follows that

™AD ~ "('ag)perpendicular

llJN-m = J(445.43N)

</ = 0.24920 m

or

d = 0.249 m A





220

PROBLEM 3.66

In Problem 3.57, determine the perpendicular distance betweencable AE 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 AEat A is 55 N, determine the moment of that force about the line

joining Points D and B,

SOLUTION

From the solution to Problem 3.57 T„? =55N

T^=5[(9N)i-(6N)j + (2N)k]

|MD/i |= 2.28N-m

X/«-'—(24i-7j)

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

contribute to the moment ofTAe about line DB.

Now (*AE )parallel ~ AE * ^ Dli

5(9i-6j + 2k)-—(24i~7j)

[(9)(24) + (-6)(-7)]

Also

so that

5

= 51.6N

*AE ~ ( *Ae)parallel+ ( *AEJperpendicular

(''./: ),,, neodfcul* - J(5S)2+(51.6)

2 - 1 9.0379 N' perpendicuiar

Since A,ra and (T(/i )pcrpendicular are perpendicular, it follows that

MDB ^"(T^Operpendieiilar

or 2.28 N-m = </(!. 9.0379 N)

d~ 0.1 19761 J = 0.1198 m ^

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Ill

PROBLEM 3.67

In Problem 3.58, determine the perpendicular distance

between cable CF and the line joining Points D and B.

PROBLEM 3.58 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.

:0.3iJi

SOLUTION

From the solution to Problem 3.58 Tcr = 33N

Tc/,=3[(6N)i-(9N)j-(2N)k]

jA*D«| = 9.50N-m

Xm =25

(241 ~ 7,)

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

contribute to the moment ofTcr about line DB,

Now ( i-CF )parallcl~" *CF " * o«

= 3(61-9j-2k)~ (241 - 7 j)

= ^[(6)(24) + (-9)(-7)]

- 24.84 N

AlSO Tc y;- — (TcF )jwral le!

+ ( *-C/' )perpendicular

s° that (Ta.Wendicuiar = V(33)2 - (24.84)

2

= 2.1 .725 N

Since XDB and (TCF )pcrpendicillar are perpendicular, it follows that

\MDB I "V'CY'Vperpcndicperpendicular

or 9,50N-m = </x21.725N

or d~ 0.437 m A

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Ill

% ill

PROBLEM 3.68

In Problem 3.61, determine the perpendicular

distance between cable EF 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 EF at E is 46 lb,

determine the moment of that force about

the line joining Points A. and D.

SOLUTION

From the solution to Problem 3.61 7V„=46 1bEl-

\M

X

T^-2H3 1b)i-(22 1b)j + (6 1b)k]

lb in.

(4i-j + 3k)

1Di -1359 1b-in.

1

AD

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

contribute to the moment ofTEF about line AD.

Now (TeF /parallel ~ ^EF ' ^, AD

2(-3i-22j + 6k)

2

(4i-j + 3k)

Also

so that

[(~3)(4)+ (-22)H) + (6)(3)j'26

= 10.9825 lb

iEF - ( lEF )para||c]+ (lEF )perpemiicular

(^perpendicular = V(46)2 - (10.9825)

2 = 44.670 lb

Since \AD and OW ^^are perpendicular, it follows that

or

MAD - d{TEl, )pcrpeiidicular

1359 lb • in. = rfx 44.670 lb or t/ = 30.4in. <

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223

PROBLEM 3.69

In Problem 3.62, determine the perpendicular

distance between cable EG and the line joining

Points A and D.

PROBLEM 3.62 A sign erected on uneven

ground is guyed by cables EF and EG. If the

force exerted by cable EG at E is 54 lb,

determine the moment of that force about, the

line joining Points A and D.

SOLUTION

From the solution to Problem 3.62 TB0 = 54 lb

TfiC =6[(llb)i-(81b)i-(41b)k]

|M/JD |= 23501b-in.

^ =4^(41 -j + 3k)V26

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

contribute to the moment of Teg about line AD.

N»w (TEG )para„e,= TEG XAD

- 6(i - 8j - 4k) •-fL(4i - j + 3k)

V26

= ~™L[(I)(4) + (-8X-D + (-4X3)] -

Thus, (TBG )perpendiciliar=T£G = 54 lb

Since %A0 and (T£6 pe[pcildicutarare perpendicular, it follows that

1™ AD \~ "\*EG /perpendicular

or 2350 lb • in. = dx 54 lb

or rf = 43.5 in. 4

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224

PROBLEM 3.70

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 componentsand adding the moments of the two resulting couples, (b) by using

the perpendicular distance between the two forces, (c) by summingthe moments of the two forces about Points.

SOLUTION

(a) We have

where

(b) We have

(c) We have

SM 8 : -t/,Cv +rf2C,,=M

</, = (0.360 m) sin 55°

= 0.29489 md2 =(0.360 m) sin 55°

= 0.20649 mC, =(60 N) cos 20°

- 56.382 NC,, =(60 N) sin 20°

= 20.521 N

M = -(0.29489 m)(56.382 N)k + (0.20649 m)(20.52 1 N)k

--=-(12.3893 N-m)k

M = Fd(-k)

= 60 N [(0.360 m)sin(55° - 20°)](-k)

= -(12.3893 N-m)k

£M A : £(rtx F) = rm x FB + rCIA xFc =M

i J k

M = (0.520 m)(60 N) cos 55° sin 55°

-cos20° -sin 20°

I J k

cos 55° sin 55°

cos 20° sin 20°

(1 7.8956 N •m - 30.285 N • m)k

-(12.3892 N-m)k

or M = 12.39 N-mJ^

or M = 12.39 N-m

+(0.800 m)(60N)

or M = 12.39 N-m

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225

1211./

A"1 ft

21. II >

1

1' lb 16 in.

PROBLEM 3.71

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 21 -lb forces, (/?) the perpendicular distance between the 12-lb

forces if the resultant of the two couples is zero, (c) the value of aifthe resultant couple is 72 lb- in. clockwise and dfc 42 in.

SOLUTION

\Z»V?

(a) We have Mx=d^

c where dx

~ 1 6 in.

Fj=211b

M, =(16in.)(21 lb)

-336 lb -in.

(/?) We have M,+M.2=0

1 2. Ho

or M, -336 lb -in. H

</, = 28.0 in. <or 336 lb • in. -^(1 2 lb) =

(c) We have Mlolll=M

1+M2

or -72 lb -in. = 336 lb -in. - (42 in.)(sin a){\ 2 lb)

sina~ 0.80952

and a = 54.049° or a = 54.0° ^

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226

100 imn KiOi

140 mm160 mm

2-10 mil

V^

SOLUTION

P D

1 nP A

(a) We have

or

(A)

We have

(c)

,32- .tv,

o

r%^ We have

PROBLEM 3,72

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.

M = Pd

18N-m = P(.24m)

P = 75.0N

dBC ^yj(BE)2 +{ECf

= V(-24m)2+(.08m)2

= 0.25298 m

M = Pd

18N-m = P(0.25298m)

P-71J52N

dAC =yl(ADf+(DCf

or Pmin =75.0N <

or P = 71.2N <

).24m)2+ (0.32 m)2

0.4 m

M = PdAC18N-m = />(0.4m)

P = 45.0N or P = 45.0N A

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers andeducatorspermittedby McGraw-Hillfar their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

227

.',:.) I!>

6 in.

25 lii

:25 ih

L Sin.-

PROBLEM 3.73

Four 1 -in.-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.

(/;) 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 ofthat minimum, tension?

SOLUTION

35"&

sty*

(a) +)M = (35 lb)(7 in.) + (25 lb)(9 in.)

-245 lb -in. + 225 lb -in.

M = 470 lb- in. ^)<

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

6tan

8

= 36.9< 90°- = 53.1°

Direction of forces:

With pegs /I andD:

With pegs/? and C:

(c) The distance between, the centers ofthe two pegs is

53.1° -4

53.1° <

F A s

<J&+62 =10 in.

Therefore, the perpendicular distance d between the forces is

We must have

c/ = 10in. + 2|-in.

= 1 1 in.

M = Fd 4701b-in. = F(llin.) F = 42.7 lb <

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228

25 lb

PROBLEM 3.74

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 knowingthat the resultant couple applied to the board is 485 lb in.

counterclockwise.

SOLUTION

M ^ dADpAD+dBCFBC485 lb • in. = [(6+ rf)in.](35 lb) + [(8 + rf)in.j(25 lb) rf = 1.250 in. <

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229

SllH'i

PROBLEM 3.75

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.

filb.fl

SOLUTION

Based on

where

I2M

M == M,+M 2

M,== -(81b-ft)j

M 2== -(61b-ft)k

M == -(8tb-ft)j-(61b-ft)k

|M| == >/(8)2+(6)

2 =10 lb -ft

3l =M

"|m:|

-(8 lb • ft)j - (6 lb • fl)k

or M = 10.00 lb- ft <

or M

10ib-ft

-0.8j-0.6k

jMjl = (101b-ft)(~0.8j-0.6k)

cos0v=O #

v=90°

cos<9v=-0.8 6>

(

.=143.130°

cos0z --O,6 Z=126.870°

or 9X - 90.0° 6 = 143. 1° O

z=\ 26.9° <

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230

170 mm

160 mm

18 N

150 mm

.1.50 mm

!8N

3-3 N

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

We have

where

Also,

M = M, + M,

M,

lGIC

rt?/c

x *i

-(0.3 m)i

(18N)k

M,=-(0.3m)ix(18N)k

= (5.4N-m)J

M 2 =rD/r xF2

'D/F -(.15m)i + (.08m)j

(.15m)i + (.08m)j + (.17m)k

M

(.15)2+(.08)

2+(.1.7)

2 m

= 141.421 N-m(.15i + .08j + .l7k)

i J k

= 141,421 N-m -.15 .08

-.15 .08 .17

- 141 .42 1(.01 361 + 0.0255j)N • m

(34 N)

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231


and M = [(5.4 N -m)j] + [14 1.42.1(.01 36i + .0255j) N • m]

= (1 .92333 N • m)i + (9.0062 N • m)j

|m:|-^(m,)2 +(m;v )

2

= V(l-92333)2+(9.0062)

2

= 9.2093 N • m or M = 9,2 1 N • m <

M = (1 .92333 N • m)i + (9.0062 N • m)j

~|Mj~ 9.2093 N-m= 0.20885 + 0.97795

cos 6>v=0.20885

0,= 77.945° or 9^11.9° <

cosy=0.97795

0, =12.054° or 0, =12.05°*

cos Z=0.0

=90° or 6' =90.0° <

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232

PROBLEM 3.77

If P~0, replace the two remaining couples with a single

equivalent couple, specifying its magnitude and the direction

of its axis.

SOLUTION

dOE

E,

M = M, +M 2 ; /'J= 16 lb, F

2= 40 lb

M, = rc x F, - (30 in.)* xH) 6 lb)jj = -(480 lb • in.)k

M 2 = rm x F2 ; rm = (1 5 in.)i - (5 in.)j

>/(0)2+ (5)

2+(10)

2 =5^ in.

= 8>/5[(llb)j-(2ib)k]

i j k

M 2 -8n/5 15 ~5

1 -2

- 8>/5[(10 lb • in.)i + (30 lb • in.)j + (1 5 lb • in.)k]

-(480 lb in.)k + 8V5[(1 lb • in.)i + (30 lb • in.)j + (1 5 lb • in.)k]

(178.885 lb • in.)i + (536.66 lb • in.)j - (21 1 .67 lb • in.)k

^/(178.885)2+(536.66)

2 +(-21 1.67)2

M

M

K

603.99 lb in

MA/ = 604 lb -in. <

cos0v

COS

COS 0..

M0.29617

0.88852

-0.35045

0.29617! + 0.88852j - 0.35045k

72.8° 0=27.3° 0,,=110.5° A

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233

PROBLEM 3.78

If P = 20 lb, replace the three couples with a single

equivalent couple, specifying its magnitude and the

direction of its axis.

SOLUTION

From the solution to Problem. 3.77

M, = -(480 lb • in.)k

M2= 8>/5 [(10 lb • in.)i + (30 lb • in.)j + (1 5 lb • in.)k]

16 lb force:

40 lb force:

F - 20 lb M3= rc x P

= (30in.)ix(20lb)k

= (600 lb - in.)j

M-M, +M 2 +M 3

= ~(480)k + 8V5 (1 Oi + 30j + 1 5k) + 600j

-(178.885 lb -in)i + (1136.66 lb -in.)j -(211.67 lb-in.)k

M = ./{178.885)2 + (1

1

3.66)2+(211 .67)

2

= 1169.96 lb in M = 1170 lb- in. <

X„MM

0. 1 52898i + 0.97 154j - 0. 1 8092 1k

cos ex = 0.152898

cos^ = 0.97 154

cos 0.= -0.1 8092.1 9. =81.2° 0=13.70° 6'=100.4° <





234

160 nun,- \

18 N

PROBLEM 3.79

If P = 20 N, replace the three couples with a single

equivalent couple, specifying its magnitude and the


SOLUTION

We have

where

M-M, +M2 +M 3

M^r^xF,i J k

0.3

18

N-m = (5.4N-m)j

M3=r /fXF2

i J k

,15 .08 141.421 N-m.15 .08 .17

141.421(.0136i + .0255j)N-m

(See Solution to Problem 3.76.)

M3= r(X4 xF3= 0.3 0.17 N-m

20

= -(3.4N-m)i + (6N-m)k

M=[(1.92333-3.4)i + (5.4 + 3.6062)j + (6)k]N.m

= -(1 .47667 N • m)i+ (9.0062 N • m)j + (6 N in)

llVff— 1**2 , nj2 , xj2M; +ML

V+ m;

V(l .47667) + (9.0062) + (6)2

10.9221 N-m or M = 10.92N-m A

PROPRIETARY MATERIAL. CO 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

235


XM -1.47667 + 9.0062 + 6

jM| 10.9221

= -0. 1 35200i + 0.82459J + 0.54934k

cos(?T= -0.135200 6X =97.770 or <9

V-97.8° <

cosV= 0.82459 #,,=34.453 or

y= 34.5° ^

cos#. = 0.54934Z=56.678 or <9

2= 56.7° ^

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236

z A } 1600 N-m

] 200 N-iii

1 1.20 N«m

PROBLEM 3.80

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 andB 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


SOLUTION

Represent the given couples by the following couple vectors:

M^ = -1 600sin 20°j + 1 600cos20°k

M:

Mc

-(547.232 N m)j + (1 503.5 1 N • m)k

1200sin 20°j + 1200cos20°k

(41 0.424 N • m)j + (11 27.63 N • m)k

-(1120N-m)i

The single equivalent couple is

M^M,, +M 5 +MC

= -(1 120 N • m)i - (1 36.808 N • m)j + (263 1.1 N- m)k

M = yj(\ 1 20)2 + (1 36.808)

2 + (263 1 . I)

2

= 2862.8 N-m-1120

cos 6,

COS tfy ^

cos 0„

2862.8

-136.808

2862.8

2631.1

2862.8

M = 2860N-m 0=113.0° B' =92.7* 23.2° <

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237

Aim W

20°

A*'

PROBLEM 3.81

The tension in the cabJe attached to the end C of an

adjustable boom ABC is 560 lb. 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 ZF: F,,=r = 560lb

mK^*^t^^ or ¥A = 560 lb ^ 20° ^

X^^S TscnSo ZMA : M^^sinSO )^,)

= (560lb)sin50°(18ft)^A

(b)

or

Based on

= 7721.7 lb -ft

M/(=7720.1b-ftjH

XF: 7^ = 7* = 560 lb

Ffi= 560 lb ^C 20° <

ZMB : MB =(Tsm50°)(dB )

I & or

«eA = (560 lb) sin 50° (10 ft)

3*6 \

^MB \ c

or

= 4289.8 lb ft

Mg =42901b-ftjH

\£&<!2th c*** F*

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238

PROBLEM 3.82

A 1 60-lb 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 Band a second force at D.

SOLUTION

(a) Based on

where

(b) Based on

2.F: Pc ^P = \60\b

XMC : Mc —Pxdcy + PydCx

Px =(160 lb) cos 60°

= 80 lb

/;=(1601b)sin60°

= 138.564 lb

rftt =4ft

JCl ,= 2.75 ft

Mc = (80 lb)(2.75 ft) + (1 38.564 lb)(4 ft)

= 220 lb -ft + 554.26 lb- ft

= 334.26 lb -ft

EFV : PDx =Pcos 60°

or P, =160 lb ^T.60°^

or Mc =334 lb • ft )<

= (160lb)cos60c

= 80 lb

TM, (Pcos60°)(clDA ) = PB(dDB )

[(160 lb)cos60°](1.5 ft) = PB (fi ft)

PB = 20.0 lb or P„=20.01bH

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239


ZFV

: Psin60o = PB + PI)},

(160 lb) sin60° = 20.0 lb + PDy

PDy =118.564 lb

2

^=>/('k)2 +('y

= 7(8°)2 +0 18 -564)

2

= 143.029 lb

# = tan_1(p \

p

J 118.564^- tan] — —I 80 )

= 55.991° or PD ,= 143.0 lb ^56.0°^





240

-*-r-HS A PROBLEM 3.83

50 nun F,- i

«i

100 mm

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

JvJ:

A(a) Based on IF: F

/?=F = 80N

XM: M n =Fdr

80 N (.05 m)

4.0000 N • m

or FB = 80.0 N --•••- ^

or M«-4,00N-m }^

5«*

^

1

(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,

XM: MD =Fcd

4.0000 N-m = Fc (.04m)

Fc =100.000 N or Fc =100.0nH

2Fy

: = FD -Fc

Fft =l 00.000 N or F„ =100.0nH

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241

PROBLEM 3.84

tsiSsif^S&&^-fift^iisKSS ......' A dirigible is tethered by a cable attached to its cabin at B. If

the tension in the cable is 1 040 N, replace the force exerted by

the cable at B with an equivalent system formed by twoA B C parallel forces applied at A and C,

6.7 m 4 m \

60°/\......,lA.^ ,

SOLUTION

Require the equivalent forces acting at A and C be parallel and at an

angle ofa with the vertical.

Then for equivalence,

XFX : (1 040 N) sin 30° = FA sin a +FB sina

ZFy

: -(1 040 N) cos 30° = -FA cosa - FB cosa

Dividing liquation (1) by Equation (2),

(1040 N) sin 30° __{FA +FB)s\na

-(.1040 N) cos 30°~ -{FA + FB )cosa

Simplifying yields a ~ 30°

Based on

XMC : [(1040 N)cos30°](4 m) = {FA cos30°)(l 0.7 m)

FA = 388.79 N

(1)

(2)

or

Based on

or

F,, = 389 N "^ 60° M

XM.A :- [(1 040 N) cos 30°](6.7 m) = (Fc cos 30°)(l 0.7 m)

Fc =651.21 M

FC =651N ^60°^

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242

:,-l

A\ 200 mm

300 mm

PROBLEM 3.85

The force P has a magnitude of 250 N and is applied at the end Cof a 500-mm rod AC attached to a bracket at A and B. Assuming

a ~ 30° and j3 — 60°, 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 £F: F = P or F = 250N^60°

1M B : M = -(0.3 m)(250 N) = -75 N m

The equivalent force-couple system at B is

F = 250N "^60° M = 75.0 N • m ) <

(b) Require

%/f/ MO NJ ™ 5^1*

9c

Equivalence then requi :es

£FV

: = F,, cos + FA> cos

F^ —~FR or cos0 = O

2)Fy

: --25 = -F4sin -Fe sin

Now if F.f= -Fn => -250 = reject

cos =

or = 90°

and F, + FB = 250

Also £.A/B : - (0.3 m)(250 N) = (0.2m)F/4

or F.,=-375N

and Fi?=625N

F^=375N ^60° FB = 625 N ^ 60° ^

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243

1'

'A\ 200 mm

300 mm

PROBLEM 3.86

Jk\ Solve Problem 3.85, assuming a-fi~ 25°.

SOLUTION

P= 2SO N\

id) Equivalence requires

£F: ¥B = P or Ffi= 250 N X' 25.0°

XMB : Mfi= ~-(0.3 m)[(250 N)sin 50°] = -57.453 N • m

The equivalent force-couple system at. B is

F« = 250 N ^ 25.0C M„^57.5N-m )<

(b) Require

Z5"o n! 25"0 M0-2. w>

Equivalence requires

Adding the forces at B:

MB = dAEQ (0.3 m)[(250 N) sin 50°]

= [(0.2m)sin50°]£

= 375N

F,, = 375 N ^ 25.0° F»=625N X: 25.0° <

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244

450 mm

rA

PROBLEM 3.87

A force and a couple are applied as shown to the end of a

cantilever beam, (a) Replace this system with a single force Fapplied at Point C, and determine the distance d from C to a

line drawn through Points D and E. (h) Solve Part a if the

directions of the two 360-N forces are reversed.

pi.v

i 150 mm

SOLUTION

A3fcON

A

3(>ON

1

E COON E

—*-J

w

—

p

Tc r

150 win

3&0N

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

or F = -(600N)k A

and SMD : (360N)(0.]5m) = (600N)(rf)

d - 0.09 m

or d - 90.0 mm below ED A

(b) We have from Part a F = -(600N)k 4

and 1MD : -<360N)(0.15m) = -(600N)(rf)

d = 0.09 m

or d = 90.0 mm above ED A

-H---'r '""fe

d

D



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245

|*-]2()inm-*)*i+

i '' [—~—-"'-^^i^;-; ; --::;:-: ;j

I

250 N y90 mm M

C

1 900 N90 ram

PROBLEM 3.88

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 Fapplied at Point C, and determine the distance x from C to line BD.

(Point C is defined as the shear center of the section.)

250 N

SOLUTION

Replace the 250-N forces with a couple and move the 900-N force to Point C such that its moment about 11 is

equal to the moment of the couple

25"o M k^

O.tftwi ?JPh

iOoM•ZSoM

T^OO MD

Then

or

Mu = (0.18)(250 N)

= 45N-m

Mlf= x(900 N)

45N-m = x(900N)

x- 0.05 m

F = 900NJ x = 50.0 mm A





246

X3.2 in'

2.05 I!)

2.8 in.

1)¥

PROBLEM 3,89

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 oftwo forces with one of

the forces equal in magnitude to the force atA except with an opposite sense, resulting in a force-couple.

Have FB - 2.9 lb- 2.65 lb = 0.25 lb, where the 2.65 lb force be part of the couple. Combining the two

parallel forces,

Couple = (2.65 lb)[(3.2in. + 2.8 in.)cos25°]

= 14.4103 lb -in.

and Mcouple

14.41 03 lb -in.

H 1A 6

*

A

f is^h

-Q

X

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

Based on ZMn-1 4.4 1 03 lb • in. = [(.25 lb) cos 25°](a)

a = 63.600 in.

F'= (.25 lb)(cos 25°i + sin 25°k)

F'= (0.227 lb)i + (0. 1 057 lb)k and is applied on an extension of handle BD at a

distance of 63.6 in. to the right ofB

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247

48 lb

2o:20

40 in.

20 ib

PROBLEM 3.90

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.

SOLUTIONIK

(a) First note that the two 20-lb forces form A couple. Then a\6

where

F = 48 Ib ALd

= 180° -(60° + 55°) = 65°

*E>S°

and M = 2,MB

= (30in.)(481b)cos55°- (70 in.)(20 lb)cos20c

= -489.62 lb • in

The equivalent force-couple system at B is

F = 48.0 lb ^65° M == 490 lb- in.jH

0>) The single

between Aequivalent force F' is equa

and B. For equivalence.

I to F. Further, since the sense ofM is clockwise,F' must be applied

EM,: M = -aF' cos 55°

where a is the distance from B to the

-489.62 lb •

point of application of F'

in. = -tf(48.01b)cos55°

. Then

oi- o = 17.78 in. F' = 48.0 lb ^L 65.0° <

and is applied to the lever 17.78 in.

To the left of pin B <

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248

300 N

300 N

PROBLEM 3,91

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 is.

SOLUTION

From the statement ofthe problem, it follows that XME = for the given force-couple system. Further,

for Pmi„, must require that P be perpendicular to rB/E . Then

XME : (0.2 sin 30°+ 0.2)mx300N

+ (0.2m)sin30°x300N

~(0-4m)P- =0

3ooM

or R,, -300N

3<x>tv\

P., 300 N 30.0° <

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249

PROBLEM 3,92

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 or if the line

of action of the equivalent force is to intersect line CD 300 mmto the right ofZX

SOLUTION

(a) The given force-couple system (fi\ M) at B is

F = 48Nj

and M =ZMB = (0.4 m)(l 5 N)cos 40° + (0.24 m)(1 5 N)sin 40

or M =6.91.03 N -m

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

~ZMB : M = dF'

or 6.9.1.03 N -m = dx 48 N

or d~ 0.14396 m

and the line of action of/"7intersects line AB 144 mm to the right ofA.

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

J.MB : (0.4 m)(l 5 N) cos a+ (0.24 m)(l 5 N) sin a

= (0.!m)(48N)

or 5 cosa + 3 sin a - 4

Rearranging and squaring 25 cos2 a ~ (4 - 3 sin. a)

2

Using cos2 a~ 1 -sin

2 a and expanding

25(1 -sin2a) -16 -24 sin «r + 9 sin

2or

or 34 sin2or ~ 24 sin or - 9 =

CJ O

F' = 48N A

Then sin a2(34)

sin a = 0.97686 or sin a = -0.27098

a = 77.7° or a = -15.72°





250

100 mm

""60 mm

PROBLEM 3.93

An eccentric, compressive 1 220-N force P is applied to the end

of a cantilever beam. Replace P with an equivalent force-couple

system at G.

SOLUTION

0, trvn

We have

ZF: -(1220N)i=F

F = -(1220N)i <

Also, we have

1220

IMG : i>c xP =M

N-m = Mi J k

-.3 -.06

-1

M = (1 220 N m)[(~0.06)(-l)j- (~0. l)(-l)k]

or M=(73.2N-m)j-(122N-m)k <



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251

.67 mm

594 mm

PROBLEM 3.94

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.

1.00 mm

SOLUTION

We have Kt ^ (,!-*.«

where

IF: Plfi=Fc

* AB=

*"AB'AB

iat

I.3>(33 mm )i + (990mm)j-

1155.00 mm(594mm)k

()75N)

or Fc =(5.00N)i + (150N)j-(90.0N)k <

We have JM.C : rm:: xPAli=MC

» J k

Mc = 5 0.683 -0.860 N-m

1 30 -18

= (5){(-0.860)(-1 8)i - (0.683)(-l 8)j

+ [(0.683)(30)--(0.860)(l)]k}

or Mc = (77.4 N-m)i + (61.5 N •m)j + (106.8 N-m)k A





252

16 ft

PROBLEM 3.95

An antenna is guyed by three cables as shown. Knowingthat the tension in cable AB is 288 lb, 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

Then

Now

dAB =yl(r64f +(-12Sf +{16)2 = 144 ft

T^=~^(-64i-128j + 16k)

= (32 1b)(-4i-8j + k)

M =M =r }/0 xT,„,

= 128jx32(-4i-8j + k)

- (4096 lb • ft)i + (1 6,384 lb • ft)k

The equivalent force-couple system at O is

F = -(1 28.0 lb)i - (256 lb) j + (32,0 lb)k -4

M = (4. 1 kip • fl)i + (1 6.38 kip • ft)k <

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253

J.6ft

PROBLEM 3.96

An antenna is guyed by three cables as shown. Knowing

that the tension in cable AD is 270 lb, replace the force

exerted at A by cable AD with an equivalent force-couple

system at the center O ofthe base of the antenna.

SOLUTION

We have *ad = VC-*4)2+C-128)

2+(-128)

2

= 192 ft.

Then T^=~5Jb(~64i-128j + 128k)

= (90 1b)(-i-2j-2k)

Now M-M = r^xT^= 128jx90(-i~2j-2k)

= -(23,040 lb • ft)i + (1 1,520 lb • ft)k

The equivalent for ;e-couple system at O is

F = -(90.0 lb)i - (1 80.0 lb)j - (1 80.0 lb)k <

M = -(23 .0 kip • ft)i + (1 1 .52 kip • ft)k <

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254

PROBLEM 3.97

Replace the J50-N force with an equivalent force-couple

system at A.

120mm

Y\

20mm 60 mm

SOLUTION


where

LF: F = (150N)(~cos35 oj-sin35ok)

= -(122.873 N)j - (86.036 N)k

XMA : M=rm xF

rDIA = (0. 1 8 m)i - (0. 1 2 m)j + (0. 1 m)k

i J k

Then M=0.18 -0.12 0.1 N-m

-122.873 -86.036

- [(-0. 1 2)(-86.036) - (0. l)(-i 22,873)]i

+ [-(0.18)(-86.036)]j

+ [(0.18)(-122.873)jk

= (22.6 N • m)i + (1 5.49 N • m)j - (22. 1 N • m)k

The equivalent force-couple system atA is

F = -(1 22.9 N)j- (86.0 N)k <

M = (22.6 N • m)i + (1 5.49 N • m)j - (22. 1 N • m)k <4

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

255

83.3 mm

PROBLEM 3.98

A 77-N force Ffand a 31-N • m couple Mi are

applied to corner E of the bent plate shown. If

Fj and Mi are to be replaced with an equivalent

force-couple system (F2, M2) at corner B and if

(M2)7_ - 0, determine (a) the distance d, (b) F2

and M2 .

SOLUTION

(a) We have XM& : M2z=0

M^xFJ +M^O (1)

where rHfB = (0.3 lm)i- (0,0233)j

_ (0.06 m)i + (0.06 m)j - (0.07 m)k (nn Kn

0. 1 1 m= (42 N)i + (42 N)j - (49 N)k

Ml2=k-M.,

M, = KuM\

~d\ + (0.03 m) j- (0.07 m)k „ . .VE .- _ ^ 1*—

i

J— (3 ] |s| . m)

V^2+0.0058 m

Then from Eq nation dX

1

0.31 -0.0233(-0.07m)(31N-m) n+ - J -0

Vc/2+0.0058

42 42 -49

Solving for d, Equat on (I ) reduces to

(13.0200 + 0.978 , }

2.17N,m _^

4d2+0.0058

From which d = 0.1350 in or d ~ 135.0mm A





256


(/>) F2 = F.= (42i + 42j-- 49k)N or F

2= (42 N)i

M2- VWB x.F, + M,

i

0.31

42

J*

-0.0233

42

k

-49

+(0.1350)1 + 0.03j -0.07k

0.155000(3IN-m)

M,

(1. 14

1

70i + 1 5. 1900J + 1 3.9986k) Nm+ (-27.000i + 6.0000j - 14.0000k) N • m-(25.858 N • m)i + (21.190 N • m)j

or M* =- (25.9N-m)i + (21.2N-m)j <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior mitten permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

257

SOLUTION

We have

Then

Also

Then.

Now

where

Then

PROBLEM 3.99

A 46-lb force F and a 2120-lb • in. couple M are

applied to corner A of the block shown. Replace

the given force-couple system with an equivalent

force-couple system at comer //.

3 in.

dAJ

F

18)z+(-14)-+(-3r =23 in.

46 1b

23-(18i-14j-3k)

d 4C

M

(36 tb)i - (28 lb)j - (6 1b)k

: 7(-45)2+(0)

2+(-28)

2 = 53 in.

2120 lb in.

53(-451 - 28k)

-(1800 1b-in.)i-(l120 1b-in.)k

rt +r^xF

r/J/y/

=(45in.)i + (14in..)]

M'^M +r^xF

M' = (-1800i~I120k) +

i J k

45 14

36 -28 -6

- (-1 8001 - 1 1 20k) + {[(1 4)(-6)]i + [-(45)(-6)]j + [(45)(-28) - (1 4)(36)]k}

= (- 1 800 - 84)i + (270)j + (-1120-1 764)k

= -(1 884 lb • in.)i + (270 lb • in.)j - (2884 lb • in.)k

= -(1.57 lb • ft)i + (22.5 lb • ft)j - (240 lb • ft)k

The equivalent force-couple system atH is F' = (36.0 lb)i - (28.0 lb)j - (6.00 lb)k <

M' = -(] 57 lb • ft)i + (22.5 lb • ft)j - (240 lb • ft)k 4





258

PROBLEM 3.100

The handpiece for a miniature industrial grinder weighs 0.6 lb, andits 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 formsan angle of 25° with the x direction. Show that the weight of the

handpiece and the two couples M.j and M2 can be replaced with asingle equivalent force. Further, assuming that M

}

- 0.68 lb • in.

and M2 ~ 0.65 lb • in., determine (a) the magnitude and the direction

of the equivalent force, (/>) the point where its line of action

intersects the xz plane.

SOLUTION

First assume that the given forceW and couples M, and M2 act at the origin.

Now w = ™^

and M = M, +M2 M &

= ~(M2 cos 25°)i + (M, -M

2 sin 25°)k

Note that sinceW andM are perpendicular, it follows that they can be replaced

with a single equivalent force.

(a) We have F-W or V^-iV. =-(0.6 lb)j

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

M = r/vo xF

where rm = xi + zk

-(M2 cos 25°)i + (M, -M2 sin 25°)k

(Wz)\ -~ (Wx)k

Equating the I and k coefficients, z =~^ COs25°

and x =( M"'M* sin 25

'

ft

W M 2

or F = -(0.600 lb)j <

i J k

X z

-w

(/;) For

W { W

W = 0.61b Mx

= 0.68 lb • in. M2-0.65 lb -in.

0.68 -0.65 sin 25°

-0.6

•0.65 cos 25°

0.6

0.67550 in.

-0.98 1 83 in.

or x = 0.675 in, A

or 2 = -0.982 in. ^

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individualcoursepreparation. Ifyou are a student using this Manual,you are using it withoutpermission.

259

PROBLEM 3.101

A 4-m-Iong 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?

-100 *

\

/

V

=SCK> N

•on \

4 ni

ifl)

(s)

Mi X mo n

i';00N-

2300 N-nt

.-'

100 N

800 N

•'i.iN 200 N- 1.

1

?-'\m Nmii

300 N

MJON-iu

900 N-ni

(W

200 N

.

(e)

100 N-tn

,100 N

300 N- us

(ft)

"A"

000 N

.100 N I

800 N300 N-m

: OON-iii

;i00N-ri)

2(Hi \

(/)

SOLUTION

(o) (a) We have s/y -400N-20()N = /?o

and £JW/. 1 800 N-m - (200 N)(4 in.) =Ma

(/?) We have £7y -600N = tf,,

and L/V/^ -900N-m = A/ft

(c) We have I/y 300N~900N^#,.

and Z/V/^ 4500 N-m - (900 N)(4 m) = Mc

&

M

A

A /w

or Rtt=600NH

m) = M„

or Mfl= lOOON-m^)^

or R6 =600NH

or M, -900 N -m)<

or Rc= 600 N J <

1 m) = Mc

or M, = 900 N • m ") <

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260

or R,,=400NH

or M,,=900N-m)^


{d) We have £F„: -400 N + 800 N = Rd

and ZM,t: (800 N)(4m)- 2300 N-m = M

rf

(e) We have My - 400 N - 200 N = Re

or Re =600NJ^

and ZMA : 200 N • m + 400 N m - (200 N)(4 m) =Me

or Me =200N-m >)^

(,/') We have SFJt

: -800N + 200N = /?/

or R/ =600nJ^

and ZMA : - 300 N • m + 300 N • m + (200 N)(4 m) = M.f

or M/ =800N-m>

)^

(g) We have £F : -200 N- 800 N = tf,

or R^=1000n|^

and ZMA : 200 N •m + 4000 N • m - (800 N)(4 m) =M

or M£=1000N-m

>

)^(A) We have Z/y -300N-300N = ifA

or RA= 600 N

J <

and LMA : 2400 N • m - 300 N •m - (300 N)(4 m) = Mh

or M,,=900N-m^H(b) Therefore, loadings (c) and (h) are equivalent. -4

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reproduced or distributed m anyjorm or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers andeducatorspermittedby McGraw-Hillfor their individual coursepreparation, ifyou are a student using this Manualyou are using it without permission.

261

200 N 100 N

4 mPROBLEM 3.102

\

100 N-m

SOLUTION

We have

and

or

: > \ a 4-m-long beam is loaded as shown. Determine the loading of

-' Problem 3.101 which is equivalent to this loading.:'S00N'in

XFV

: -200N-400N = fi

ZMA : -400 N • m + 2800 N-m - (400 N)(4 m) = M

M=800N-nO

Problem 3.101 Equivalent force-couples atA

M

or R =600NHR

HFMm.

Equivalent to case (/) ofProblem 3.101 -4

Case R M

(a) 600 N| 1000 N-m)

(*) 600 N 900 N • m )

(c) 600 N|

900 N • m ^)

id) 400 N|

900 N • m. ^)

(*) 600 NJ

200 N-m}

if) 600 N|

800 N • m ")

(g) 1000 N|

1.000 Nm")

(A) 600 N|

900 N • m ^)

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262

PROBLEM 3.103

Determine the single equivalent force and the distance from Point A to its line of action for the beam andloading of (a) Problem 3 . 1 1 Z>, (/?) Problem 3.10k/, (c) Problem 3. 1 1 e.

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

400 N

-J0U X

./lit) \

200 N-ii

4 in -Hit) X w.\ \

1 KOI) N i>>

(«)

i')(H)N-m101) N

(W

. cA Av.xt n 200 V:..

W)

800 NMJON-ii)

i,.\

11,'

""

1000 N-c ''

(g> (A)

>' \-lll

200 \r

'/:

"Oil \

300 N- mi

,'

NO!) N

i: '< \ in

»iN

« : N-ni

i,.

:/)

900 N

".'

! .-'MN-in

:«iu X-

.., vj

SOLUTION

I GVCH <\OOH«r\

^~~ c

a£h--4

w

W~t ^400n 2.2>oorvm

8<30rt

\*Y\

[v~d -^t'

200fc<W i%

til!

lm

(a) For equivalent single force at distance d fromA

We have ZF„: -600 N^^

ZOOiM

and

(b) We have

and

(c) We have

and

rf»c sje

or R ~ 600 N ^

XMC : (600N)(rf)-900N-m =

or d^ 1.500 m -«

£/y -400N + 800N = #

or R = 400Nf^

ZMC : (400N)(rf) + (800N)(4~</)

-2300N-m =

or c/ = 2.25 m «

lFy

: -400N-200N = /f

or R = 600 NJ^

ZMC : 200N-m + (400N)(t/)

-(200N)(4-c/) + 400Nm =^0

or tf = 0.333 m <

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263

PROBLEM 3.104

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 = (10 lb)i and a couple of moment

M - (15 lb • ft)] + (15 lb • ft)k located at the origin.

lb.lt

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 lb)i]. Next move each of the systems to the origin 0\ the forces remain

unchanged.

A: M^ =IM =(5 lb-ft)j + (15 lb-ft)k + (2 ft)kx(10 Ib)i

-(25 1b-fl)j + (15 1b-ft)k

D: MD = 2M = -(5 lb • ft)] + (25 lb • ft)k

+ [(4.5 ft)j + (l ft)] + (2 ft)k"jxl.O lb)i

= (151b-ft)i + (151b-ft)k

G : M6. =XM - (1 5 lb • ft)i + (1 5 lb • ft)

j

/: M;=SM

/=(151b-ft)]-(51b-ft)k

+ [(4.5ft)i + (lft)jjx(101b)j

= (15 1b-ft)]-(15 1b-ft)k

The equivalent force-couple system is the system at corner D, ^





264

6 ft"

6 ft'

m B

fe& _ ,

,

PROBLEM 3,105

The weights oftwo children sitting at ends A and B of a seesaware 84 lb and 64 lb, 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) 60 lb, (/?) 52 lb.

SOLUTION

&\\b w„

K.^£± % w „

(a) For the resultant weight, to act at C, £Afc =0 Wc

. = 60 lb

Then (84 lb)(6 ft) - 60 \h{d) - 64 lb(6 ft) =

(b) For the resultant weight to act at C, ZMC = Wc - 52 lb

Then (84 lb)(6 ft) - 52 \h{d) - 64 lb(6 ft) =

feMlb

t

d = 2.00 ft to the right ofC

d = 2.31 ft to the right ofC

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. JVo part of this Manual may be displayedreproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.

265

kl ii

34 in.

P

PROBLEM 3.106

Three stage lights are mounted on a pipe as shown.

The lights at A and B each weigh 4.1 lb, while the one

at C weighs 3.5 lb. (a) If d = 25 in., 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

4.Wb LU \h 3.£

• A <b C-!

1

L K

3D E D

For equivalence

or

(b)

TFy

: -4.1-4.1-3.5 = -/? or R = 11.7 lb|

X/^: -(10 in.)(4.1 lb) -(44 in.)(4.1 lb)

-[(4.4 + rf)in.](3.5 lb)«-(£ in.)(11.7 lb)

375.4 + 3.5^ = 1 1.71 (rf.iinin.)

rf = 25 in.

We have 375.4+ 3.5(25) = ! 1.7 1 or I = 39.6 in.

The resultant passes through a Point 39.6 in. to the right ofD.

1 = 42 in.

We have 375.4 + 3.5</ = 11.7(42) or </ = 33.lin. A

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266

PROBLEM 3.107

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 ofa so that

the distance from support A to the line of action of the equivalent force is maximum, (/?) the magnitude of the

equivalent force and its point of application on the beam.

U00N

9 in

400 N 1 I 600 N

400 ~ N

SOLUTION

For equivalence

liooN

4ooMj i toco N

or

4oo| N B.

r L ~iR

2.FV : -1300 + 400- - 400 - 600 = -R

b

/?= 2300-400- N

ZMA : ^400^A2 1 b

0)

*(400) ~(a + b)(6()Q) = -LR

or

1000a + 6006 -200

2300-400-

Then with 6 = 1.5 m L =10a + 9—

a

2

3

Where a, L are in m(a) Find value ofa to maximize L

dL=

da

s V3 ,

r*823- a3

23--al-M0a+9—-o ;

(2)

'<n8

23— a

\ 2

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

267


or„_ 184 80 64 ,80 n . 32 >

,,230 a a+—a~+—a + 24 a~ -03 3 9 3 9

or

Then

16a2 -276a + 1143 =

276± J(-276)2 - 4(1 6)(.l 143)

a~ —2(16)

or a = 1 0.3435 m and a ~ 6,9065 m

Since ^42? - 9 tn, a must be less than 9 m a = 6.9 1 m -4

(b) Using Eq. (1) .# = 2300 4006 "9065

or K-458N <«

1.5

1 0(6.9065) + 9 -- (6.9065)2

,t = 3 = 3.16 m23— (6.9065)

and using Eq. (2)

R is applied 3.16 m to the right ofA. M





268

PROBLEM 3.108

Gear C is rigidly attached to arm AB. If the forces and couple

shown can be reduced to a single equivalent force at A,

determine the equivalent force and the magnitude of the

couple M.

SOLUTION

We have

For equivalence

or

or

Then

or

Also

YFX : -1 8 sin 30° + 25 cos40° = Rx

R =10.1511 lb

£F : -1 8 cos 30° - 40 - 25 sin 40° = R,.

ffv=-71.658 lb

R = yl(\QA5\\)2+(7l.65$y

= 72.416

71.658tan# =

10.1511

= 81.9° R = 72.4 lb "%" 81.9°^

IMA : M - (22 in.)(l 8 lb) sin 35° - (32 i.n.)(40 lb) cos 25°

~(48in.)(25 1b)sin65 = ()

M = 2474.8 lb -in. or A/ = 2061b-ft <

PROPRIETARY MATERIAL, © 2010 The McGraw-Hill Companies, Inc. Al! rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual coarsepreparation. Ifyou are a student using this Manual,you are using it without permission.

269

•12 in.

U"(Y

!| j 8in.

i I j I

--* I'Mh

**>»» PROBLEM 3,109

A couple of magnitude M = 54 lb • in. and the three forces shown are

applied to an angle bracket, (a) Find the resultant of this system of

forces. (/?) Locate the points where the line of action of the resultant

intersects line AB and line BC.

SOLUTION

(a) We have LF : R = (-1 Oj) + (30 cos 60°)i

+ 30sin60°j + (-45i)

= -(30 lb)i + (15.9808 lb)

j

or R = 34.0 lb ^ 28.0° <«

(b) First reduce the given forces and couple to an equivalent force-couple system (R, M/f )

at B.

We have 2Affl

: Mfi=(54 lb -in) + (12 in.)(10 lb) -(8 in.)(45 lb)

Then with R at D

or

|\ iv_l— —

cLM

f}: -186 lb • in = a(.l 5.9808 lb)

t>

S>-

1*

a = U .64 in. ~ ^

.

and with R at E LMtf

: -1 86 lb in = C(30 lb) «Jc

or C - 6.2 in.

The line of action

below B.

ofR intersects line AB 1 1 .64 in. to the left of# and intersects line BC 6.20 in.

<



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270

10 lb

M

-12 in.

30 II

,60'

Sin.

'-^ ,!.3li.

PROBLEM 3.110

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 ofthe resultant of the

force system is to pass through (a) Point A, (/?) Point B, (c) Point C.

SOLUTION

In each case, must have Mf -

{a) +)M*=2MA =M + (1 2 in.)[(30 lb)sin 60°] - (8 in.)(45 lb) =

M--= 4-48.231 lb -in. M = 48.21b -in.) <

(b) +)M*=!MB:= A/ + (12 in.)(10 lb) -(8 in.)(45 lb) =

M--= +240 lb in. M = 240 lb in.) <

(c) +)m£=z,mc =M + (12 in.)(l lb) - (8 in.)[{30 lb) cos 60°] =

M = M = <





271

3 10 N500 N

X A u/I

375 mm

:>/.,

-500 mm

GOO N

200 mm

TfiON

PROBLEM 3.111

Four forces act on a 700 x 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

(t>)

tan<9

= (-400 N + 160 N - 760 N)i

+(600N + 300N + 300N)j

= ~(1000N)i + (1200N)j

R = ^(1000 N)2+(1200 N)2

= 1562.09 N

1 200 N ^

1000N

-1.20000

-50.194°

M£=£rxF= (0.5m)ix(300N + 300N)j

= (300 N • m)k

(300N-m)k=xix(1200N)j

x = 0.25000 mx = 250 mm

(300 N m) = y\ x (-1 000 N)i

v = 0.30000 my - 300 mm

OOOM~V*Ot-\

R = 1562N ^50.2°^

-lOOO N

Intersection 250 mm to right ofC and 300 mm above C -4

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Ill

500 N;Vfo N

-i

/

/ s

/ j

/ I 375 mm;/':

C»

600 N

— 500 mm7<j0 N

200 mm

PROBLEM 3.112

Solve Problem 3.1 1 1, assuming that the 760-N force is directed

to the right.

PROBLEM 3.111 Four forces act on a 700 x 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

(b)

= (-400N + J60N + 760N)i

+(600N + 300N + 300N)j

= (520N)i + (1200N)j

tan#

R = 7(520 N)2+(1200 N)2 = 1 307.82

/l200N

N

..520 N

= 66.5714'

2.3077

M «

or

c

v - - -

/ fc>

>U n/\H V J&5»h\ /sw\

Ti^OwC?0O M

R = 1308N ^66.6C

or

IrxF

= (0.5m)ix(300N + 300N)j

- (300 N • m)k

(300N-m)k = xix(1200N)j

x = 0.25000 m

x = 0.250 mm

(300 N • m)k - [/i + (0.375 m)j]x[(520 N)i + (1200 N)j]

= (1200x'~l95)k

/ = 0,41250 m

x' ~ 412.5 mm

*~ S20 N

Intersection. 41.2 mm to the right ofA and 250 mm to the right ofC

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distribution to teachers and educators permittedby McGraw-Hillfor their individual coursepreparation, ifyou are a student using this Manual,

you are using ii without permission.

273

4 If

240 lb\

70A&-~

8ft 8 ft-

j

1.00

\\

\\

8 It—*\*— 8 ft

180 ih

8 ft

300 II)

76ft

PROBLEM 3.113

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 y4 and G.

SOLUTION

We have R = IF

R = (240 lb)(cos70°i - sin 70°j) - (1 60 lb)j

+ (300 lb)(- cos 40°i - sin 40°j)- (1 80 lb)

j

R = -(147.728 lb)i - (758.36 Ib)j

= 7(147.728)2+(758.36)

2

= 772.62 lb

v«*o lb

$ = tan

tan"

rA

^y-758.36

-147.728

= 78.977 cor R = 773 lb ^ 79.0° <

We have

where

YMA = dRv

ZM

d

-[240 lb cos 70°](6 ft) - [240 lb sin 70°](4 ft)

-(160 lb)(12 ft) + [300 lbcos40°](6 ft)

-[300 lb sin 40°](20 ft) - (1 80 lb)(8 ft)

-7232.5 lb -ft

-7232.5 lb -ft

-758.36 lb

9.5370 ft or rf = 9.54 ft to the right ofA <

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274

I] ill

- (> in

21011)

r 150 lb

PROBLEM 3.114

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 atA due to belts on pulleyA

We have IF: -120 lb -160 lb = J"?

,

R/(=280 lb

We have IMA ; -40 lb(2 in.) =MA

M„=801b-.in.J)

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

We have £F: (2 1.0 lb + 1. 50 lb)^L 25° = R„

R„=3601b^l!l25 <

We have 1M B : -601b(1.5in.) = MB

M g =90Ib-nO

Equivalent force-couple atF

R = 280 lb

AL«*HW<P^? ,b

A

We have LF: RF = (- 280 lb)] + (360 lb)(cos 25°i+ sin 25°j)

= (326.27 lb)i- (127.857 lb)j

R = RF

= 7(326.27)2+(127.857)

2

= 350.43 lb

r d \

= tan-i

tan"-127.857

326.27

f 1 in octN

-21.399'

Rr=f?

or R,, =R = 3501b^ 2.1.4° «

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275


We have TMF : MF = ~(280 lb)(6 in.) - 80 lb in.

-[(360lb)cos25°](1.0in.)

+[(360 lb) sin 25°](l 2 in.) - 90 lb • in.

M F =-(350.56 lb -in.)k

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

MF = clRy

M,

-127 -857 lb

2.74 18 in. or d = 2.1A in.

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distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation, ifyou are a student using this Manual,


276

420 N240 nun

•loo n h :

50 mm '- \ i • >

520 mm

fv

?u.\-m. :

50 nun

640 nun

SI! v

ISO mm

50mm

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 GIL

SOLUTION

We have

ten-™

tZO N

ftaKl

4oW«n^ .R,b

+ F

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

ZFX : 200cos 15° - 1 20cos 70° + P = Rx

tfv= (152.1.42 + .P)N

ZFr

: - 200sin 1 5°- 1 20sin 70° - 80 = R

Thus

or

or R. -244.53 N

or

Setting P-0 inEq.(l):

Now with R at /

or

and with R at J

or

ZMC : - (0.47 m)(200 N) cos 1 5° + (0.05 m)(200 N) sin .1.

5°

+ (0.47 m)(!20 N) cos 70° -(0.1 9 m)(120 N) sin 70°

- (0. 1 3 m)(P N) - (0.59 m)(80 N) + 42 N - m+40N-m = MG

MG =-(55.544 + 0.1 3P)N-m

ZMG : - 55.544 N • m = -a(244.53 N)

a- 0.227 m

SMC : - 55 .544 N m = -b(\ 52. 142 N)

b = 0.365 m

(I)

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

(b) The rivet hole is 0.227 m to the right of G. 4


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Ill

200

50.

520 mir

50 mm

240 mm'120 N

^

^.O-Nvm. ^kd

80 N

mm

640 mm

I

180 mm

- .50mm

PROBLEM 3.116

Solve Problem 3. 1 1.5, 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 GIL

SOLUTION

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

MG = -(55.544+ 0. 1 3P) N • mand /?, =(152.142 + /*) N

For P = 60 N

We have Rx=(152.142 + 60)

= 2.12.14N

A/c =-[55.544+ 0.13(60)]

= -63.344 N-m

Then with R at / ZMG : -63.344 N-m = -o(244.53N)

or a = 0.259 m

and with R at J XMG : -63.344 N-m = -£(212.14 N)

or b = 0.299 m

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

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



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278

)

PROBLEM 3.117

A 32-lb 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.

f

2 in.

i-vi

r2 in.

eJpM'-- -Si

SOLUTION

l4-OSth30°

*> I4-Ocos30°

We have

£F: (60 lb)i - (32 lb)j+ (1 40 lb)(cos30°i + sin 30°j) = R

R = (181.244 lb)i + (38.0 lb)j

or R = ]85.2Ib^1J.84°^

We have ZM : XM =xRy

-[(140 lb)cos30°][(4 + 2cos30°)in.] -[(140 lb)sin30°][(2 in.)sin 30°]

~(60 lb)(2 in.) = x(38.0 lb)

1

and

38.0

x = -23.289 in.

(-694.97 -70.0 -120) in.

Or, resultant intersects the base (x axis) 23.3 in. to the left of

the vertical, centerline (y axis) of the motor.

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279

»-»('-$)PROBLEM 3.118

As followerAB rolls along the surface ofmember C, it exerts

a constant force F perpendicular to the surface, (a) Replace Fwith an equivalent force-couple system at the Point Dobtained 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 ofx for which the moment of the equivalent force-

couple system at D is maximum.

SOLUTION

(a) The slope ofany tangent to the surface ofmember C is

~2bdy __ d

dx dxv * ;

Since the force F is perpendicular to the surface,

2

tana2b\x

For equivalence

where

IF: F = R

ZMD : (F cosa)(yA ) =MD

cos cr

:

2bx

^l(a2

)

2 +(2bxy

yA ~b'<4'

2Fb<

M,

.3\

Va4 +4*2jc2

Therefore, the equivalent force-couple system at £> is

Wb(.-s)

ZbX

R =F7 tan"

'«2 ^

2.F//

M

v2to

y

„3>

V 2 ..2^+4//x

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280


(/?) To maximize M, the value ofx must satisfy —— =dx

where, for

M 8F(jk-*3

)

a = 1 m, /? = 2 m

dx

/i + 16jc2(1-3jc

2)-(x-;c

3)

= 8F-

1

(32x)(l + 16x2 )" 1/2

or

(l + .16;r)

(1 + 1 6,y2)(1 - 3x

2) -

1

6x(x - x3) =

32x4 +3x2 -l=0

-3±V9-4(32)(-l)

2(32)

Using the positive value of.r

0.13601 ] m? and -0.22976 nV

x = 0.36880 m or x = 369mm -^

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Ail rights reserved. No part of this Manual may be displayed,




281

\i

, ::%, 200 mm

PROBLEM 3.119

Four forces are applied to the machine component

ABDE as shown. Replace these forces by an equivalent

force-couple system atA'

20 mm fijli 300 N

25! X \^f7-.^MM0'^ 1 60 i i i m'

j!

f :?;"| 100 mm

A

SOLUTION

R = -(50 N)j - (300 N)i - (1 20 N)i - (250 N)k

R = -(420 N)i - (50 N)j - (250 N)k

rw =(0.2m)i

rD =(0.2m)i + (0.16m)k

rB = (0.2 m)i - (0. 1 m)j+ (0. 1 6 m)k

M*=r;,x[-(300N)i-(50N)j]

+ rD x (-250 N)k + r x ( - 1 20 N)i

-(3oon)i

-(ztorji

i J k

0.2 m

-300 N -50 N

i J k

0.2 m 0.16m

-250 N

k» J

+ 0.2 m -0.1m 0.16 m

-120 N

= -(1.0 N • m)k + (50 N m)j - (1 9.2 N • m)j - (1 2 N m)k

Force-couple system atA is

R = -(420 N)i - (50 N)j - (250 N)k M* = (30.8 N • m)j -- (220 N • m)k <





2S2

225 i»hi

'<.Qj

i(>" ^r i!>* s

'i[}(^r ii0N

:/|5 \

145 N

ISO mm

PROBLEM 3.120

Two ] 50-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

Pulley C

Then

R/f= (i45 N)(-cos20°j + sin 20°k) - 2 1 5 Nj

= -(35 1 .26 N)j + (49.593 N)k

MB = -(21 5 N-.I45 NX0.075 m)i

= -(5.25N-m)i

Rc = (1 55 N + 240 N)(-sin 1 0°j - cos 1 0°k)

= -(68.591 N)j- (389.00 N)k

Mc = (240 N - 1 55 N)(0.075 m)i

= (6.3750 N • m)i

R = R/;+ Rc = - (4.1 9.85 N)j - (339.4 l)k

M.A =MB +Mc + xm x Rs + xcu xRc

- -(5.25 N • m)i + (6.3750 N m)i +

i

0.225

2lS>4

tS5e* (toi

2HOfi

&c

or R = (420N)j-(339N)k <

J k

-351.26 49.593

N-m

* J

0.45

-68.59:

k

-389.00

N-m

= (1.1 2500 N • m)i + (1 63.892 N • m)j - (1.09.899 N • m)k

or MA = (.1 . 125 N m)i + (1 63.9 N • m)j - (1 09.9 N • m)k <

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283

PROBLEM 3.121

..1! 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 = (2.61b)i +./?>,j~(0.71b)k

and the couple MRA =MX\ +(1.0 lb • ft)j - (0.72 lb • ft)k. (b) Find

the corresponding values of R and Mx .

SOLUTION

(a) From the statement ofthe problem, equivalence requires

B +C = R

# + C, = 2.6 lbor

and

or

or

Using Eq. (1)

or

and

or

(b) Eq.(2)=>

Using Eq. (3)

IF

ZFX

Y,Fy

XFZ

IMA :

LMV :

ZMy

:

ZM„

-C\,=Ry

-Cz=-0.7 1b or Cz

=0.7 1b

(rB/A xB + M.B ) + raA xC=M«

'1.75

(1)

(2)

(llb-ft) +

'3.75

12

12

ft (/?,) +

ft (CV )= M,

1.75

(3)

tt|(Cv H-|— (l

12 ' I 12(0.7 lb) = 1 lb • ft

1 +

3.75£T+ I.75C

V=9.55

3.755, + 1.75(2.65,) = 9.55

Bx = 2.5 lb

Cx =0.1. lb

'35 ^— ft (C,.) --0.721b- ft

12 )}

Cy= 2.4686 lb

B = (2.5 lb)i C = (0. 1000 lb)i - (2.47 lb)j- (0.700 lb)k <

Ry=*-2Al\b <

' .75^

12(2.4686) =Mx

or A/, =1.360 lb -ft ^





284

PROBLEM 3,122

A mechanic uses a crowfoot wrench to loosen a bolt at C Themechanic holds the socket wrench handle at Points A and Band applies forces at these points. Knowing that these forces

are equivalent to a force-couple system at C consisting of the

force C - (8 lb)i + (4 lb)k and the couple Mc = (360 lb • in.)i,

determine the forces applied at A and at B when A — 2 lb.

SOLUTION

We have

or

0ibXCm.

or

From i-coefficient

j-coefficient

k-coefficient

£F: A +B==C

F- A +B =8 lb

Bx =-(Ax +8)b)

IF- A. + # =0

or Ay=-B

y

LFZ

: 2lb + ^=4lb

B = 2 lb

SMC : r8/c xB + r,/c xA = M c

0)

(2)

(3)

i J k

8 2 +

Bx *, 2

i j k

8 8

4 4- 2

lb -in. = (360 lb -in.)!

(25y- SA

y)i + (25, -16 + %AX - 1 6)

j

+(85J,+8^)k = (360lb-in.)i

2Z?j;-8^, =360 lb -in.

-2.fi,+8/fv=32UVin.

8.^+8^=0

(4)

(5)

(6)

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285


From Equations (2) and (4): 2By-8(--5,) = 360

Z?v=361b A

r=36 lb

From Equations (1) and (5): 2("4~8) + 8,4t=32

4 = 1.6 lb

From Equation (1): Bx =-(1.6 + 8) = -9.6 lb

A = (1 .600 lb)i - (36.0 lb)] + (2.00 lb)k <

B = -(9.60 Ib)i + (36.0 lb)j + (2.00 Ib)k <4




you arc using it without permission.

286

64 in.

lfl} ? 96 in.

PROBLEM 3.123

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 Aif R = 2 1 .2 lb and M = 1 3.25 lb • ft.

SOLUTION5

We have 2F: R - R^ - RXBC

where , (42 in.)i - (96 in.)j - (1 6 in.)k

106 in.

A

r = (42i - 96j - 1 6k)aj06

J /

v<> \\ j

S* "v, \\ j

c

or R^, = (8.40 lb)i - (1 9.20 lb)j - (3.20 lb)k <

We have £MV,: rc//JxR +M=M

-4

where xCIA = (42 in.)i + (48 in..)k =— (421 + 48k)ft

= (3.5ft)i + (4.0ft)k

R = (8.40 lb)i - (19.50 lb)j - (3.20 lb)k

=-42i + 96j + 16k

106

= -(5.25 lb • ft)i + (1 2 lb • ft)j + (2 lb • ft)k

i j k

Then 3.5 4.0

8.40 -19.20 -3.20

lb-ft + (-5.25i + 12j-f 2k)lb-ft =Mi4

M.A = (71.55 lb • ft)i + (56.80 lb ft)j - (65.20 lb ft)k

or M ^ - (7 1 .6 lb • ft)i + (56.8 lb • ft)j - (65.2 lb - ft)k <

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287

PROBLEM 3,124

A mechanic replaces a car's exhaust system by firmly clamping the catalytic converter FG to its mounting

brackets H and ./ 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.

« .<f

0.14 in0.33

:-:i('

SOLUTION

(a) Equivalence requires

SF: R^A + B

(100 N)(cos 30°j - sin 30° k)- (1 1 5 N)j

-(28.4N)j-(50N)k

and

where

Then

SM.D : MD

AID

'aidk ^a +ybid x ^b

l B/D

-(0.48 m)i - (0.225 m)j + (1 . 1.2 m)k

-(0.38 m)i + (0.82 m)k

M n =100

i j k i J k

0.48 -0.225 1.1.2 + 115 -0.38 0.82

cos30° -sin 30° „1

1 00[(0.225 sin 30° - 1 . 1 2 cos 30°)i + (-0.48 sin 30°)

j

+ (-0.48 cos 30°)k] + 1 1 5[(0.82)i + (0.38)k]

8.56i-24.0j + 2.13k


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288


The equivalent force-couple system atD is

R = -(28.4 N)j - (50.0 N)k <

MD = (8.56 N • m)i - (24.0 N • m)j + (2. 1 3 N • m)k <

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

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2 SO

PROBLEM 3.125

For the exhaust system of Problem 3.124, (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 EFtends to rotate clockwise or counterclockwise, as viewed by the mechanic.

0..M m0.33 m

0.30 m

0.10 m

0.56 in

o \

SOLUTION

(a) Equivalence requires

and

where

Then

M

SF: R = A + B

= (100 N)(cos 30°j - sin 30° k) - (1 1 5 N)j

= -(28.4N)j-(50N)k

M^r^xA +r^xB

'A/F

y BIF

-(0.48 m)i - (0.345 m)j + (2.10 m)k

-(0.38 m)I - (0. 1 2 m)j + (1 .80 m)k

i J k i J k

0.48 -0.345 2.3.0 + 115 -0.38 0.12 1.80

cos 30° -sin 30° -1

M;, =100

MF = 1 00[(0.345 sin 30° - 2. 1 cos 30°)i + (-0.48 sin 30°)j

+ (-0.48 cos 30°)k] + 1 1 5[(1 .80)1 + (0.38)k]

= 42.4i-24.0j+ 2.13k

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290


The equivalent force-couple system atF is

R = -(28.4N)j~(50N)k A

M ,, = (42.4 N • m)i - (24.0 N • m)j + (2. 1 3 N • m)k <

(b) Since (MF )Z is positive, pipe EF will tend to rotate counterclockwise relative to the mechanic. -4

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291

PROBLEM 3.126

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 they 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

or

We have

where

R = F = (l 1 lb)[( sin 20° cos 25°)]i - (cos 20°)j - (sin 20° sin 25°)k]

= (3.4097 lb)i - (1 0.3366 lb)j - (1 .58998 lb)k

R = (3.4 1 lb)i - (1 0.34 lb)j - (1 .590 Ib)k

M =rB/0 x¥xM.c

rm) = [(l4in.)sin25°]i + (15 in.)j + [(14in.)cos25°]k

- (5.9167 in.)i + (15 in.)j + (12.6883 in.)k

Mc = (90 lb • in.)[(sin 20° cos 25°)i - (cos 20°)j - (sin 20° sin 25°)k]

= (27.898 lb • in.)i - (84.572 lb • in.)j - (13.0090 ib • in.)k

M.

i J k

5.9167 15 12.6883

3.4097 -10.3366 1.58998

lb -in.

p= ll tb

+(27.898 - 84.572 - 13.0090) lb - in.

= (1 35.202 lb • in.)* - (3 1 .90 1 lb • in.)j - (1 25.3 1 3 lb in.)k

or M = (1 35.2 lb in.)i - (3 1 .9 lb in.)j - (125.3 lb - in.)k

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292

PROBLEM 3.127

Three children are standing on a 5x5-tn raft. If the weights

of the children at Points A, B, and C are 375 N, 260 N, and400 N, respectively, determine the magnitude and the point

of application of the resultant of the three weights.

SOLUTION

We have

We have

We have

F f

XF: F,+Fa +FC =R

-(375 N)j - (260 N)j - (400 N)j =R-(1035N)j-R

or /? = 1035N <

^.v :FaM + F*(zb) + fc(zc) = R(zD )

(375 N)(3 m) + (260 N)(0.5 m) + (400 N)(4.75 m) = (1 035 N)(zD )

zD = 3.0483 m or Zf) = 3.05 m <

Z,MZ : PA (xA ) + FB (xB ) +Fc {xc ) = R(xD )

375 N(l m) + (260 N)( 1 .5 m) + (400 N)(4.75 m) = (1035 N)(.^ )

xD = 2.5749 m r xn = 2.57 m <

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293

0.5 in

0.25 ill" P ' 0.25 m

PROBLEM 3.128

Three children are standing on a 5x5-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 Nclimbs 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

We have

We have

We have

L.F: FA +FB +$C =R

-(375 N)j - (260 N)j - (400 N)j - (425 N)j - R

R=-(1460N)j

1MX : FA (zA ) +FB (zB ) + Fc (zc ) + FD {zD ) = R{zH )

(375 N)(3 m) + (260 N)(0.5 m) + (400 N)(4.75 m)

+(425 U)(zD ) = (1460 N)(2.5 m)

zD =1.16471 m

SMZ: FA (xA ) + FB (xa ) + Fc (xc ) + FD (xD ) = R{xH )

(375 N)(l m) + (260 M)(l .5 m) + (400 N)(4.75 m)

+(425 N)(x/} )

= (1460 N)(2.5 m)

xD = 2,3235 m

or Zt, =1.165 m M

or xn = 2.32 m A

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294

l'i><m f:-.i

feii )}•

8 If"

5 ft

<)ii

5.5 ft

"!Ufi

ir 2.5 ft

on.'ij

1:

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 the magnitude and the

point of application of the resultant of the four wind forces

when a -\ ft and 6 = 1.2 ft.

SOLUTION

We have

.» 1

Soft,

Assume that the resultant R is applied at PointP whose coordinates are (x,y, 0).

Equivalence then requires

E/<; : - 1 05 - 90 - 1 60 - 50 = -R

XMX : (5 ft)(1 05 lb) - (1 ft)(90 lb) + (3 it)(l 60 lb)

+ (5.5 ft)(50 lb) = -j/(405 lb)

or y = -2.94 ft

ZMy : (5.5 ft)(105 lb) + (12 ft)(90 lb) + (14.5 ft)(160 lb)

+ (22.5 ft)(50 lb) = -j(405 lb)

or x~ 12.60 ft

R acts 1 2.60 ft to the right ofmemberAB and 2.94 ft below member BC.

or R - 405 lb 4

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295

5 ft

'>>><:.. p. i

5.5 it

9 ft

S

H).5ft

PROBLEM 3.130

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 ofthe resultant of the four forces is at G.

SOLUTION

Since R acts at G, equivalence then i equires that 1M of the applied system of forces also be zero . Then

at G :EMV : -(a + 3) ft x (90 lb) + (2 ft)(l 05 lb)

+ (2.5 ft)(50 lb) =

or a~ 0.722 ft <

XMy

: -(9ft)(105ft)-

+ (8ft)(501b) =

-(14.5-6) ftx (90 lb)

Ol b-= 20.6 ft A

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2%

224 X

.1,5 m

PROBLEM 3.131*

A group of students loads a 2x3.3-m flatbed trailer with two

0.66 x 0.66x 0.66-m boxes and one 0.66x 0.66x 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 0.66x0.66x1.2-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

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 center line of the trailer, the added 0.66x0.66x1.2-m box should be placed adjacent to

one of the edges of the trailer with the 0.66x 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

We have

We have

EF : - (224 N)j - (392 N)j~(l 76 N)j = RR = -(792N)j

£MZ : -(224N)(0.33 m)-(392 N)(1.67 m) - (176 N)( 1.67 m) = (-792 N)(j)

xR =1.29101 m

XMX : (224 N)(0.33 m) + (392 N)(0.6 m) + (1 76 N)(2.0 m) = (792 N)(z)

z„= 0.83475 mFrom 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 center lines. Thus, IMa — 0.

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

(0.33 m < x< 1 m)(l .5 m < z < 2.97 m)



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297

PROBLEM 3.131* (Continued)

With XL = 0.33 m

at G: ZMZ : o- 0.33) mxWL -(1.29101-1) mx(792N) =

or wL~ 344.00 N

Now must check if this is physically possible,

at G: IMV : (Zz- 1 .5)mx344 N) - (1 .5 - 0.83475)mx (792 N) =

or ^ -3,032 m

which is not acceptable.

With ^ = 2.97 m:

at G: I,MX

(2.97 - 1 .5)m xWL- (1 .5 - 0.83475)m x (792 N) = )

or K = 358.42 N

Now check if thi s is physically possible

at G: XMZ

- (1- ^)mx(358.42 N)-(l .29101 -l)mx(792 N) =

or ^ -0.357 m ok!

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

And it is placed on end (A 0.66x0.66-m

from side AD.

side down) along side j42? with the center of the box 0.357 m<

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298

224 N

392 N

1..S m

PROBLEM 3.132*

Solve Problem 3. 1 31 if the students want to place as muchweight as possible in the fourth box and at least one side of

the box must coincide with a side of the trailer.

PROBLEM 3.131* A group of students loads a 2x3.3-m

flatbed trailer with two 0.66 x().66x().66-m boxes and one

0.66 x 0.66 x 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 0.66x0.66x1 .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 (XR , 0, Z/( )


224-392-176 = -/?£/y

or R = 792 N

Y

or

or

IMV

: (0.33 m)(224 N) + (0.6 m)(392 N)

+ (2m)(176N) = Zfl(792N)

zR -0.83475 m

ZMZ : -(0.33 m)(224 N)-(l .67 m)(392 N)

- (1 .67 m)(176 N) = xs (792 N)

xD =1.29101 m

A*»-tr

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

through G, the point of intersection of the two center lines. Thus,

£MC =

Further, since W/y 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

0.6 m or 2.7 m



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299


Now consider these two possibilities

With x„ -0.6 m:

at G: X.M,: (\~0,6)mxWH - (1.29 101 -l)mx (792 N) =

or W„ = 576.20 N

Checking if this is physically possible

at G: SMV

: {zH - 1.5)mx (576.20 N)-(1.5-0.83475)mx(792 N) =

or zfI=2.414 m

which is acceptable.

With zn = 2.7 m

at G: TMX : (2.7 - 1 .5) Wn - (1 .5 - 0.83475)m x (792 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 0.66x1 .2-m side down, a 0.66-m edge along side AD, and the center 2.41 mfrom side DC. <

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you are. using it without permission.

300

PROBLEM 3.133

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, (/>) the pitch of the wrench, (c) the axis of the wrench.

SOLUTION

Force-couple system at O:

r =n + pj + Pk = P(i + j + k)

Mj = c/j x Pi + ak x Pj + c/i x Pk

i *^ H

!

<?;- -

m:

Pak-Pai-Paj

-Pa(i + j + k)

y

Since R and M^ have the same direction, they form a wrench with M, =M^. Thus, the axis of the wrench

is the diagonal OA. We note that

cos X- cos - cos Z

——t=t = ~y=r

/? = Pfi X= #,, = ez = 54.7

C

M}= Mo = -PoV3

Pitch = pM, -PaS

-—a

(6) -a

(c) Axis ofthe wrench is diagonal 0yf

tf PV3

r^pS ex ^eY~ez

^5A.T

4





301

\Qj.\

PROBLEM 3.134*

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, (h) the pitch

of the wrench, (c) the axis of the wrench.

SOLUTION

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

We have

£F: -P\ + I>j + Pk = /?

or R = Pk /K|FA = J^K1M : -i*P)\ + -(*/>)! +

(-«/> k = "2 * 1 ^ r x

or M.g =qp(-\- j +V(a) Then for the wrench

R = P <

and /vaxis

~ ~ K

cos 6>v= cos 6

y= cos 9Z

= 1

or V^9O° 0,, =90° g =O° <

(b) Now

= k -opf-l-j+|kj

= *,/>

2

Then ^- Mi-!*P m. r>~ ,-, -4

rt /> 2

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302


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

xXaxi&>

and the axis of the wrench is

assumed to intersect ihexy plane at Point Q whose coordinates are (x,y, 0). Thus require

Where

Then

M z =rfixR*

M2 =M xM,

aP\ -i-j+ ~k -~aPk=;(xl + y$) + Fk

Equating coefficients

i : - aP — yP or y — -a

j: ~aP~-xP or x~a

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

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303

PROBLEM 3.135*

The forces and couples shown are applied to two screws as a piece

of sheet metal is fastened to a block ofwood. 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.

100 mm

SOLUTION

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

We have

We have

XF: -(20N)i-(15N)j = R # = 25N

IMa : S(r xF) +XMC = Mg

MJ = -20 N(0. 1 m)j - (4 N • m)i -(IN- m)j

= -(4N-m)i~(3N-m)j

(b) We have

R = -(20.0 N)i-(1 5.0 N)j <

M, =VM5 /U

(-0.81 - 0.6j) • [~(4 N • m)]i - (3 N m)j]

5N-m

Pitch

(c) From above note that

p =M, 5 N

m

R 25 N- 0.200 m

or p = 0.200 m <

M, =M /e

Therefore, the axis of the wrench goes through the origin. The line of action of the wrench lies in thcxy

plane with a slope of

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304

filh'in.

I .11 lit

PROBLEM 3,136*

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 at the origin.

We have

We have

(b) We have

IF: -(101b)J-(111b)j = R

R=-(211b)j

IM : Z(r x F) +ZMC =MJ

i J k

M£ = 20 lb- in. +

-10

= (351b-in.)i~(12lb-in.)j

R = -(21lb)j

R

and pitch

ij k

-.15

-11

lb in. - (12 lb- in)

j

or R = -(21.0 lb)j

M,=VM» kR =R

= (-j)-[(351b-in.)i-(121b-in.)j]

= 1 2 lb • in. and M , = -(1 2 lb in .)j

p=«L =H^i = 0.57I43i„.R 211b

or p~ 0.571 in.





305


(c) We have

Require

From i:

From k:

(35 lb

Mg*=M,+M2

M 2=Mj-M, =(35lb-in.)i

M 2 =re/0 xR

in.)i = (jri + zk)x[-(2Ilb)j]

35i = -(21x)k + (21z)i

35 = 21z

z = 1.66667 in.

= -21x

z =

The axis of the wrench is parallel to the y axis and intersects the xz plane at x — 0, z~= 1 .667 in. ^





306

IJ

0.1 m

0.6 m

30 N'rtt

M \

jum

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.

SOLUTION

3 M?

\RJ2 y

*


We have EF: -(84N)j~(80N)k = R rt = ll6N

and £M : I(r x F) +SMC =MJ

i J k

0.4 0.3

80

i J k

0.6 .1 +

84

+ (-30j-32k)N-m =M;

Mi = -(1 5.6 N • m)i + (2 N m)j - (82.4 N m)k

(a)

(b) We have

R = -(84.0 N)j- (80.0 N)k ^

.M,=^-Mg *,,=

-84j-80k

116

55.379 N • m

[-(15.6 N • m)i + (2 N • m)j-(82.4 N • m)k]

and

Then pitch

MX=M

XAR --(40.102 N-m)j-(38.192N-m)k

M, 55.379 N-mP

R 116 N0.47741m or p = 0.477 m <



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\


307

PROBLEM 3,137* (Continued)

(c) We have mJ=Mj+m 2

M 2 = JVf£~M., = [(-J.5.6i + 2j-82.4k)-(40.102j-38.I92k)]N-m

= -(15.6 N • m)i + (42.102 N • m)j- (44.208 N -m)k

Require M 2 = rQIO X R

(- 1 5.6i + 42. 1 02j - 44.208k) = (xi. + 2k)x (84j - 80k)

= (84z)i + (80x)j-(84x)k

From i: -15.6 = 842

z = -0.185714 m

or 2 = -0.1 857 m

From k: -44.208 = -84x

x = 0.52629 m.

or jc = 0.526 m

The axis <if the wrench intersects the xz plane at

x = 0.526 m y = 2 = -0.1857 m <



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308

PROBLEM 3.138*

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.

SOLUTION

E^b^&i

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

(a) We have ZF: -(26.4 lb)k -(17 lb)| —i+— j |

= R

and 7? = 31.41b

R = -(8.00 lb)i - (1 5.00 lb) j - (26.4 lb)k <

We have XM B : rAIB x FA +M A +M B =M %

(/;) We have

i\r

i J k

-10

™26.4

220k -2381 — i + -^ji

j7 ]7j= 2641 - 220k - 1 4(8i + J 5j)

M J = (152 lb • in.)i - (21 lb • in.)j - (220 lb • in.)k

M^X.-M.^ XRRR

_-8.()Oi-15.00j- 26.4k

31.4

= 246.56 lb- in.

[(1 52 ib • in.)i -(210 lb • in.)j - (220 ib • in.)k]

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309


and M, =MXXR = -(62.8 1 8 lb • in.)i - (1 1 7.783 lb • in.)j - (207.30 lb • in.)k

Then pitchM, 246.56 lb -in. „ ocoo . noc , ^p-—l- — = 7.8522 in. or » = 7.85 in. AR 31.41b '

(c) We have MJ=M,+M2

M2=Ml ~M, = (1 52i - 21 Oj -220k)™ (-62.8.181-1 17.783j- 207.30k)

= (214.82 lb in.)i - (92.21. 7 lb • in.)j - (12.7000 lb • in.)k

Require M2 =rp xR

i J *

214.821- 92.2 17J-12.7000k = x z

™8 -15 -26.4

= (15z)i-(8z)j + (26.4*)j-(15*)k

From i: 214.82 = 152 z = 14.3213 in.

From k: -12.7000 = -15* x = 0.84667 in.

The axis of the wrench intersects the xz plane at x~ 0.847 in. y - z~ 14.32 in, A.





310

PROBLEM 3.139*

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,

(/>) 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 f R, ftlf, ) at the origin.

We have

4,c=V(6)2+(2)

2+ (9)

2 ==llm

Then

TAC1650 NU

= (6i + 2j + 9k)

: (900 N)i + (300 N)j + (1 350 N)k

and

T,BO500 N15

(14i + 2j + 5k)

(1 400 N)i + (200 N)j + (500 N)k

Equivalence then requires

LF: R = T/JC+T/W

= (900i + 300j + 1350k)

+(I400i + 200j + 500k)

= (2300 N)i + (500 N)j + (1 850 N)k

£M : M.g-r,xT,c +rfixTfiD

- (12 m)kx[(900 N)i + (300 N)j + (1350 N)k]

+ (9 m)ix[(1400 N)i + (200 N)j + (500 N)k]

= -(3600)i + (1 0800 - 4500)j + (1 800)k

= -(3600 N • m)i + (6300 N • m)j + (1.800 N • m)k

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

R = (2300 N)i + (500 N)j + (1 850 N)k



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311


(b) We have

Let

Then

Finally

(c) We have

or

Now

For equivalence

tf = 1007(23)2 + (5)

2 + (I8.5)2 = 2993.7 N

K R !

R 29.937(23i + 5j + 18,5k)

^i=^„i.-MS

29.937

1

(23i + 5j + 1 8.5k) • (-36001 + 6300j + 1 800k)

[(23X-36) + (5X63) + (18.5X1 8)]0.29937

-601.26 N-m

M, -601.26 N-mR 2993.7 N

or P = -0.201 m <«

M,=M}Xaxis

= (-601.26 N-m) x

—

: (23i + 5j + 18.5k)29.937

M. = -(461.93 N • m)i - (1 00.42 1 N • m)j - (37 1 .56 N • m)k

M 2 -Mj-M,= (-36001 + 6300J + 1800k)

-(-46

1

,93i - 1 00.42 1j - 37 1 .56k)

= -(3138.1 N m)i + (6400.4 N • m)j + (2 1 7 1 .6 N • m)k

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312


Thus require M3=r,>xR r - (y\ + zk)

Substituting

i j k

-3 1 38. li + 6400.4j + 2 1 7 1.6k - y z

2300 500 1850


j: 6400.4 = 2300 z or z = 2.78 mk: 2171.6 = -2300^ or y = -0.944 m

The axis of the wrench intersects the yz plane at y ~ -0.944 m z - 2.78 in <

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313

PROBLEM 3.140*

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, (/>) the pitch of the wrench, (c) the point

where the axis of the wrench intersects the xz plane.

SOLUTION

,iM,

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

We have ZF: PXBA + PkDC + P

k

DE = R

R = P4. 3

+3, 4.^ (~9. 4. 12

i— il+ l— i— 1 +— k

5 5 J U5 5 25

R =—(2i-20j-k) A25

«=—V(2) +(2°) +(!)

We have

(-AP 3P x

EM: I(r xP) =M

3P 4P+ (20a)jx|—j-_j| + (20fl)jx

-9P. AP . YIP.1 1 +

25 5 25m;

m:24/J«

(-i-k)



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314


(b) We have

where

Then

and pitch

(c)

MX=X8 >M*

X„

M,

P

(2I-20J-k)25

nfip 9V5R ~25

—M2i~20}~k) —(~i~k) = r9V5 5 15V5

(2i-20j-k)

My -SPa ( 25 ^

R \5yf5 .21Sp

M, = MX,-%Pa

-8a

81or p = ~ 0.0988a -*

15^5 ,9>/5

1 8 Pa

v.-'v-'/

(2i-20j-k) =—(-2l + 20j + k.)

675

Then M2=M*-M, =-?^H-k)-—(-21 + 20J + k)=—(-4301- 20j- 406k)

5 675 675

Require

( %Pa

I 675

M2~ r

e/(?xR

\ f

(-4031 - 20j - 406k) = (xi + 2k)x3P^l-(2l-20J-k)

^-Wi25j

[20zi + (x+ 2z)j-20xk]

From i:

Pa8(~403) = 20z

675

Pa

rw \

V 25y

8(_4()6)^l ^ _20jc675

From k:

The axis of the wrench intersects the jez plane at

3P

25

z = -1,990! 2a

x = 2.0049a

x = 2.00a, z~ -1.990a <

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315

60 mm

/" [M

.1 I4N-IH

40 mm

8M£ v" V .1

10 N-m ^^Pq160mm /^¥ ^

40 mm

PROBLEM 3.141*

Determine whether the force-and-couple system shown can be

reduced to a single equivalent force R. If it can, determine Rand 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 theyz plane.

SOLUTION


We have XF: F^+FG =R"(40 mm)i + (60 mm)j - (120 mm)k

and

We have

R = (50N)k + 70N

= (20N)i + (30N)j-(10N)k

/? = 37.417 N

140 mm

1M : £(r xF)+2M(

Mi

M.g = [((). 12 m)j x (50 N)k] + {(0. 16 m)i x[(20 N)i + (30 N)j - (60 N)k]}

+ (I0N-m)

+ (14N-m)

(160mm)i-(120 mm)j

200mm

(40 mm)i - (1 20 mm) j + (60 mm)k

1 40 mmm; (1 8 N • m)i - (8.4N • m)j + (.1 0.8 N • m)k

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

(20i + 30j - 1 0k) - (1 8i - 8.4j + 1 0.8k) =

(20)(1 8) + (30)(-8.4) + (-10)(10.8) =or

or 0^0

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

R = (20.0 N)i + (30.0 N)j-(1 0.00 N)k <



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316


Then for equivalence

Thus require

Substituting

MS=r_xR r=yl + zk

18i-8.4j + 10.8k

i j k

y z

20 30 -10


j: -8.4 = 20z or 2 = -0.42 mk: 10.8 = -20^ or y = -0.54 m

The line of action ofR intersects the yz plane at x = y = -0.540 m z = -0.420 m

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317

PROBLEM 3.142*

Determine whether the force-and-coupie system shown can be

reduced to a single equivalent force R. If it can, determine Rand 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 force

dDA

is at D. We have

= VH2)2+(9)

2+(8)

2=17in.

dm :^y](-~6)2+(Q)

2+(-%)

2 =\0m.

Then

*DA==34lb

-( 12i + 9j + 8k)17

= -(241b)i + (18Ib)j + (16 1b)k

and

*w10

= -{18 1b)i-(241b)k

Then

If : R = FZM +FaD= (-241 + 1 8j+

1

6k + (-1 8i - 24k)

= -(42 1b)i + (18lb)j-(8!b)k

For the applied couple

dAK = V(-6)2+ ("6)

2 + o 8)2= 6^ in -

Then

M 1601b-in., .. ,. 10I ,_

—

(-61-61 + 18k)671

1

-1^ [ (lib- in,)" - lb • i«-)j + (3 lb in.)k]

To be able to rixiuce the original ibrces and couple to a single equivalent force R and Mmust be perpendicular. Thus

R-M=0

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318

Substituting


(-42i + 18j-8k)-^E(-i-j + 3k) =

or

or

160[(-42)H) + (18)(~l)+ (-8)(3)] =

0^0

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

R = -(42.0 Ib)i + (18.00 lb)j -(8.00 lb)k

Then for equivalence

o

T5"

Thus require

where

Substituting

M - rp/D xR

vPID = -(12 in.)i + [{y - 3)in.Jj + (z in.)k

160i j k

H-J + 3k)= -12 (y-3) z

-42 1 8 -8

= [(>>~3X-8)-(z)(.l.8)]i

+ [(z)(-42)-(-12)(-8)]j

+ [(-12)(1.8)-(.y-3)(-42)]k


k:

160

480

-42z-96 or z = -1.137 in.

-216 + 42(^-3) or y = 11.59 in.

The line of action ofR intersects theyz plane at x — y - 1 1 .59 in. z ~ - 1 . 1 37 in.


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319

u PROBLEM 3.143*

A*

1.,

1-

Replace the wrench shown with an equivalent system consisting of two

forces perpendicular to they axis and applied respectively at A and B.

b

\ a

^J"S"s\

* X

SOLUTIONy

Express the forces at A and B asB^^

A =

B =

Ax i + Azk

Bxi + Bzk

Then, for equivalence to the given force system £^*

ZFX

Ax + Bx =0 0)

ZFZ

AZ+ B

Z=R (2)

ZMX Az(a) + Bz (a+ b) = (3)

ZMZ : -Ax(a)-Bx (a + b) = M (4)

From Equation (I ), BX ^~AX

Substitute into Equation (4)

-Ax(a) + Ax (a + b) = M

AM A D M

xb * b

From Equation (2), K~--R~AZ

and Equation (3), Aza+ (R~Az)(a + b) = Q

^K)and Bz

= -- R - RK b)

Bz=

b

Then A =[bj

i +Rn+-)k 4

~(2Mf>«



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320

PROBLEM 3.144*

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

SOLUTION

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 ofR 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 = Rxi + Ryj + Rzk and M = MJ + MJ+

M

zk

while the unknown forces A and B can be expressed as

A = Ax\ + Ay $ + Azk and B - Bx i + Bzk

Since the position vector of Point P is given, it follows that the scalar components (\%y, z) of theposition vector i> are also known.

Then, for equivalence ofthe two systems

(1)

(2)

(3)

(4)

bBz (5)

(6)

Based on the above six independent equations for the six unknowns (Ax ,Ayy Az , Bx , B , B2 , b), there

exists a unique solution for A and B.

From Equation (2) a. - $ -4

£/V ** == AX +BX

ZFy

: V= Ar

X/y /?,=-• A2 + Bz

£MT : Mx

= yAz -zAy

ZMy

; My~ zA

x- xAz

2M_: Mz= xA

v-yAx

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 writ/en permission of the publisher, or used beyond the limiteddistribution to teachers andeducatorspermittedby McGraw-Hillfor their individualcoursepreparation. Ifyou are a student using this Manual,you are using it without permission.

321


r oEquation (6) 4 =

,yj(xR

y~Mz ) A

(OEquation (1) Bx

--= *,-,yj

(xRy-M2 ) 4

Equation (4) Az= flW+zR,) <

( i>

Equation (3) Bz~~-K-

,yj

(Mx + zRy ) <

Equation (5) b(xM\ +yM

y +zMJ(Mx -yRz +zRy )





322

PROBLEM 3.145*

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

SOLUTION

M

*- 6iv«*> point

u-Co) (b) y

First, observe that it is always possible to construct a line perpendicular to a given line so that the constructedline also passes through a given point. Thus, it is possible to align one of the coordinate axes of a rectangularcoordinate 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 and M = JWj and are known.

The unknown forces A and B can be expressed as

A = AJ + AJ + AM and B = BJ + BJ+ Bsk

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

tt?x : = Ax +Bx

XFy

: R = Ay+B

y

-LBV. 0^Az+B

z

-zB..

-aAz ~xBz +zB x

= aAy+ xB

XMV

:

2Af„: MXMZ :

Since A and B are made perpendicular,

A-B = or AXBX+ AVB V + AR

There are eight unknowns: Ax , Ay , A., Bx , B > Bz , x, z

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

Next consider Equation (4): 0~-zB

If Bv- 0, Equation (7) becomes A.B. + A,B ~0

(I)

(2)

(3)

(4)

(5)

(6)

(7)

Using Equations (I.) and (3) this equation becomes At + At=0

PROPRIETARY MATERIAL. © 20 1 The McGraw-Hill Companies, Inc. All rights reserved. No pari of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers andeducatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

323


Since the components ofA must be real, a nontrivial solution is not possible. Thus, it is required that By * 0,

so that from Equation (4), z = 0.

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

(Note: Setting Ay , Az , or Bz equal to zero results in unacceptable solutions.)

The defining equations then become

= 5.

Then Equation (2) can be written

Equation (3) can be written

Equation (6) can be written

Substituting into Equation (5)',

R = Ay+B

y

Q~AZ+ B

Z

M = ~aA~xB,

0-aAy+xB

y

AvB

¥+ AzBz

=0

Ay=R~B

y

or

aAy

M=~aAz

aR >

R-By

B(~AZ )

y J

Substituting into Equation (7)',

(R-By)B

y+

aR } [aR }

or B sa R

ya2R2 ±M 2

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

A„=R-a R rm:

a2R2 +M 2

a2R2 +M :

A..

B.

aR

2»3 Aa R

a2 R 2 +M 7

aR2M

aRlMJ2. r>2cfR£ + M"

a2R2 +M2

or

(2)

(3)

(5/

(6)

(7)'

(8)

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324


In summary

A

B

RM„2 n 2a'R

A + AT

aRl

a2R2 +M :

-(Mj-aRk)

-(aRj + Mk)

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 > and M > 0, it follows from the equations found for A and B that A > and B, > 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 theyz plane and to the right of the origin, while force B lies in a plane parallel to theyz planebut to the left to the origin, as shown in the figure below.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

325

PROBLEM 3.146*

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

SOLUTION

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 (AA'). Mote 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 AA' by

it follows that force A can be expressed as

Force B can be expressed as

B = Bxi + Byj+ Bzk

Next, observe that since the axis of the wrench and the prescribed line of action AA' 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

IFX : 0=AAx +B (

ZIy. R = AAy+ B

y

IF' = AA, + B„

SM.,.: zB.

JMV : M = -aAX. + zBx- xB.

EMV: Q = -aAZ

y+ xB

y

Since there are six unknowns (A, Bx> By, Bz, x, z) and six independent equations, it will be possible to

obtain a solution.

(!)

(2)

(3)

(4)

(5)

(6)





32(»


Case /: Let z - to satisfy Equation (4)

Now Equation (2)

Equation (3)

Equation (6)

Substitution into Equation (5)


Then

In summary

and

AXy=R-B

y

Bz= ~AX

Z

aAX, ( \a

M

A

-aAX7-

J ( M

v^v(R-B

y )

a

KB>>

(R-BY-AX)

# = -

X2\aR

I ( M

B.

X.z V

)bi+bvaR) y y y

B.XaR<

1

XzaR~X

vM

MRA =—

XzaR~X

yM ^ aR

M X,

BX =-AXX

B=~AX,

XXMRXzaR-Xv

MXZMR

XzaR-XyM

BR

A = P

, aR ,"A

Av~ A.

y M

XA <

XaR-XM(XxM + X

zaR] + XzMk) <

x~ay J

\~R(

X

zaR~X

yM

XmR 2

KM .

or x =™— A.X,R

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

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AH riglils reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

327


Case 2: Let B ~ to satisfy Equation (4)

Now Equation (2)

Equation (1)

Equation (3)

Equation (6)


A

B=-RfX.^

B=-R( X^

\. y J

aAXy~ which requires a -

M=z R

r x?

K. y J

R(X^xxv

or X^x — Xr ihThis last expression is the equation for the line of action of force B.

In summary

A

B

r R^

v*yj

f R^

<. y J

K

M,i-4k)

Assuming that Xx , Xy , Xz > 0, the equivalent force system is as shown below.

3 A.

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

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328

0.6 m -*U_ 0.6 in

0.5 in

C\V' \\

O.V> m

PROBLEM 3.147

A crate of mass 80 kg is held in the position shown. Determine(a) the moment produced by the weightW of the crate about E,

(b) the smallest force applied at B that creates a moment ofequal magnitude and opposite sense about E.

SOLUTION

(a) By definition

We have

o.2Sm

W = mg = 80 kg(9.8 1 mis2) = 784.8 N

ZME : ME = (784.8 N)(0.25 m)

^r

W

M£ = 196.2N-m^(/?) For the force at B to be the smallest, resulting in a moment. (M£) about .£, the line of action of force Ffl

must be perpendicular to the line connecting E to B. The sense of F« must be such that the forceproduces a counterclockwise moment about E.

Note:

We have

and

or

d = ^(0.85 m)2 + (0.5 m)2 = 0.9861 5 m

ZME : 1 96.2 N • m = FB (0.986 1 5 m)

Fs =198.954 N

A

0,5 in

= tan_1 0.85 m

0.5 m59.534c

C £

1 99.0N^ 59,5° <

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329

PROBLEM 3.148

iXin (m.Ba™r

\%

21 mm

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.

28 nm

SOLUTION

Using (a)

(&> MC =yl(FAB )x +X\(FAB)y

(0.028 m)

42N-m

— X1500N25

24+ (0.021 m)|—X1500N

1

25

or Mc =42.0N-m)*«

Using (b)

MC =y2(FAB)x

= (0.1 m)25

X1500N = 42N-m

or Mc = 42.0 N • m *) <

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330

PROBLEM 3.149

A 6-ft-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 6 lb. Determine the moment aboutA ofthe force exerted by the line at B.

F^^^S^^RSilSi^R®^/i ;->"- v --~-.r^t;~~-r~-^^i

^-~ -------

SOLUTION

We have

Then

Now

where

Then

or

TX2= (6 lb) cos 8° = 5.9416 lb

Tx = Txz sin 30° = 2.9708 lb

Tv= Tsc sin 8° = -0.83504 lb

T, = r cos 30<

™*A ~" rBfA X TffC

[/?//»

M

M

-5.14561b

(6sin45°)j-(6cos45°)k

__ 6 ft

6

6

(j-k)

i J k

1 -1

2.9708 -0.83504 -5.1456

._ (-5.1456~0.83504)i -~~ (2.9708)j-~s-(2.9708)k

~(25.4 lb-ft)i-(I2.60 lb-ft)j-(12.60 lb-ft)k

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, lac. AH rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual course,preparation- Ifyou are a student using this Manual,yon are using if withoutpermission.

331

3 m

x 0.16 mDetail of the stake at B

PROBLEM 3.150

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 = V("3)2 + Of + (-1 -5)

2 - 4.5 m

BD = ,J(rO.QS)2 +(0M)2

+(0.16)2 =0.42 m

Then TiM =^-(-3I + 3J-1.5k)

=Zk(_2i + 2j~k)

Xlin=:

BD=

]

( 0.081 + 0.38J + 0.1 6k)m BD 0.42

=—(-41 + 191 + 8k)21

J

(a) We have %li^BD= TBA COS0

or ^L(~2l + 2j - k) •—(-41 + 1 9j + 8k) = TBA cos 6

orcos - J-K-2X-4) + (2)(19) + (-1X8)]

63

-0,60317

or (9 = 52.9° ^

(b) We have

= TBA cos6

= (540N)(0.60317)

or (7,

JM )JD =326N ^





332

PROBLEM 3.151

A former uses cables and winch pullers B and E to plumbone 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 4728 lb • ft,

determine the magnitude ofTDE when TAti - 255 lb.

SOLUTION

The moment about the x axis due to the two cable forces can be found using the z components of eachforce 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 oi-y componentsproduce a moment about the x axis.

We have

where

£M,: (TM )M (yA ) + (Tm )z (yD ) = Mx

\'ab)z ~*- '*AB

= k-{T4BAAB )

k 2551b-i-12j-H2k

17

801b

(TD/i )z ~k-%

= k

DE

1.5i-14j + 12k N

T,DE8.5

0.64865r,DE

yA

y»

12 ft

14 ft

4728 lb -ft

and

(1 80 Ib)(12 ft)+ (0.648657^ )(1 4 ft) = 4728 lb • ft

TDE = 282.79 lb or TnF -2831b <DE

PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may he displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

3M

/;,;:,..,,..,.. PROBLEM 3.152

Solve Problem 3.151 when the tension in cable AB is 306 lb.

H /' -{ Hft/hb - i"- ^

PROBLEM 3.151 A farmer uses cables and winch pullers Band 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 4728 lb • ft,

determine the magnitude ofT;^ when Tab ~ 255 lb.

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 the 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

Where

£M,: iTAB )z {yA ) + {Tm )z {yD ) =Mx

(TAB )x =k-rAB

M^sV)k 306 lb

-i-12j + 12k

17

= 21.6 lb

(TDE ) -k T

0.5i-14j + 12k^k T,DE

8.5

0.648657 DE

>>A

yu

12 ft

14 ft

and

MA.=47281b-ft

(2 1 6 lb)(12 ft) + (0.648657^ )(14 It) = 4728 lb - ft

TDE = 235.211b or 7V„, = 235 lb ^/j/;

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334

'3 Hi

HMt^fiffi^v

PROBLEM 3.153

A wiring harness is made by routing either two or

three wires around 2-in.-diameter pegs mounted on a

sheet of plywood. If the force in each wire is 3 lb,

determine the resultant couple acting on the plywood

when a ~ 1 8 in. and (a) only wires AB and CD are in

place, (//) all three wires are in place.

SOLUTION

In general, M - "ZdF, where d is the perpendicular distance between the lines of action of the two forces

acting on a given wire.

(a)

A»

2.4 »*•

We have M = dABFAB+da>FCD

(2 + 24)in.x3 1b + |2 + -x28 in.x3 1b or M = 151.2 lb- in. ") <

(b)

tlM

Af

We have M ^idab fab+ dCDFCD \ + dEFFEF

= 151.2 lb-in.-28in.x3 lb or M = 67.2 lb -in. )<

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Alt rights reserved. No part of this Manual may be displayedreproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual coursepreparation, ifyou are a student using this Manual,you are using it without permission.

335

I

PROBLEM 3.154

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

(a) We have

We have

where

LF: 360 N(-sin40°i - cos40°j) = -(23 1.40 N)i - (275.78 N)j = F

or F = 360N ^50°^

JMD : r/;//)xR =M

re/D = -[(0.65 m) cos 30°]i + [(0.65 m)sin 30°].)

= -(0.56292 m)i + (0.32500 m)j

i J k

-0.56292 0.32500

-231.40 -275.78

= [1 55.240 + 75.206)N-m]k

- (230.45 N m)k

M N-m

or M = 230N-mX

) A

(b) We have

where

£MD : M = r,/D xF,

'hid=40 -05 m)cos30°]i + [(1 .05 m)sin30°]j

= -(0.90933 m)i + (0.52500 m)j

i J k

FA = -0.90933 0.52500 N-m

-I

= [230.45 N-m]k





336


or (0.90933F, )k = 230.45k

FA = 253.42 N or F^=253NH

We have XF: F^F^+F^

-(231.40 N)i -(275.78 N)j = -(253.42 N^ + FoC-cos^i-sin^j)

From i: 23 1 .40 N = FD cos (9 (D

j: 22.36 fi = FD sm0 (2)

Equation (2) divided by Equation (1)

tan = 0.096629

(9 = 5.5193° or #-5.52°


FD =mM

= 232.48 N° cos 5.5 193°

or FD = 232N ^5.52°^

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distribution to teachers and educatorspermitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

337

1.10 N

PROBLEM 3.155

A 1 10-N force acting in a vertical plane parallel to theyz 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

where

We have

where

SF: PS =F

yB = 11 N[~ (sin 1 5°)j + (cos 1 5°)k]

= -(28.470 N)j + (1 06.252 N)k

YM \ %,xPB

or F = -(28.5 N)j + (1 06.3 N)k <

hio = [(°-22 cos 35 °)» + (°- ] 5)J - (0.22sin 35°)k]m

= (0. 1.802 1 3 m)i + (0. 1 5 m)j - (0. 126 1 87 m)k

i j k

0.180213 0.15 0.126187

-28.5 106.3

N • m = ML

M o [(12.3487)i-(19.1566)j-(5.1361)k]Nm

or M = (1 2.35 N • m)i - (1 9. 1 6 N • m)j - (5 . 1 3 N • m)k <

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338

fiO Ht PROBLEM 3.156

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

(a) For equivalence

tOOIfes

\U3 >t>

* 1 i_ .. Ka "1^&

f-

XFV

: -100 cos 30° + 400coi 65° + 90 cos K° = RX

or Rx = 120.480 lb

LFy

: 1 00 sin a + 1 60 + 400 sin 65°+ 90 sin 65° - Ry

or Ry= (604.09 + 100 sin a) lb 0)

With a -30°

Then R = 4

54.09 lb

„ 654.09tan =

120.480J20.480)

2+(654.09)

2

= 665 lb or = 79.6°

Also 1,MA : (46 in.)(l 60 lb) + (66 in.)(400 lb) sin 65°

+(26 in.)(400 lb) cos 65° + (66 in.)(90 lb)sin65°

+(36 in.)(90 lb) cos 65° = </(654.09 lb)

or LMA =42,435 lb -in. and c/ = 64.9 in. R~= 665 lb ^79.6°^

and R is applied 64.9 in. To the right ofA. <(b) We have d — 66 in.

Then LMA : 42, 435 lb • in = (66 m.)Ry

or Ry= 642.95 lb

Using Eq. (I) 642,95 = 604.09 + 100sina or a = 22.9° ^

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339

200 mm

H 200 mm

PROBLEM 3.157

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 = -(30 N)i + Rv \

+ Rzk

and Mj =-(l2N m)i. (b) Find the corresponding

values of Ryand 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

(") Equivalence requires IF; R = B +C

or -(30 N)i + Ryj+ Rzk = -j?k + (-Cri + C

y j+ C2k)

Equating the i coefficients i: -30N = -CV

or CV=30N

Also JMA : M*=r^xB + rc7,xC

-(l 2 N • m)i = [(0.2 m)i + (0. 1 5 m)j] x (~B)k

+(0.4 m)ix[-(30 N)i + C,,j + Czk]

Equating coefficients i: -12 N-m = -(0.15 m)/? or 5 = SON

k: = (0.4m)Cy

or Cy=0

j: = (0.2m)(80N)-(0.4m)C, or C,=40N

B = -(80.0N)k C = -(30.0 N)i + (40.0 N)k <

(b) Now we have for the equivalence of forces

-(30 N)i + RJ + Rzk = -(80 N)k + [(-30 N)i + (40 N)k]

Equating coefficients j: Ry=0 R

y^0 <

k : Rz= -80 + 40 or R

z= -40.0 N <

(c) First note that R = -(30 N)i -- (40 N)k. Thus, the screw is best able to resist the lateral force R.z

when the slot in the head of the screw is vertical. 4





340

PROBLEM 3.158

A concrete foundation mat in the shape of a regular hexagon of side

12 ft 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 ofthe mat.

SOLUTION

From the statement ofthe problem it can be concluded that the six applied loads

R at 0. It then follows that

are equivalent to the resultant

£Mo =0 or 1MX= ZM

Z^Q

For the applied loads.

* J

f—E

/ \ 1K

\7t> x

—17. «L—

1

\

Then SM , = 0: (6>/3 &)FS + (6>/3 ft)(10 kips) - (6>/3 ft)(20 kips)

-(673 ft)/>=0

or FB -FF =10 (1)

SM, = 0: (1.2 ft)(15 kips) + (6 ft)FB -(6 ft)(10 kips)

-(12 ft)(30 kips) - (6 ft)(20 kips) + (6 ft)FF =

or Ffl+ 7^=60 (2)

Then (l) + (2)=> FB = 35.0 kips j 4

and Ff = 25.0 kips J -^



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341

Solucionario Mecanica Vectorial para ingenieros Estatica - Edicion 9

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