**Vector Mechanics for engineers Statics 7th - Cap 03**

Solucionário do cap 3 da 7ªed.

(Parte **1** de 3)

CHAPTER 3 CHAPTER 3

PROBLEM 3.1 A foot valve for a pneumatic system is hinged at B. Knowing that a - 28°, determine the moment of the 1 6-N force aboutPoint B by resolving the force into horizontal and verticalcomponents.

>ny c

Noting that the direction ofthe moment ofeach force component about B is counterclockwise,MB =xFy +yFx

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PROBLEM 3.2

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

First resolve the 4-lb force into components P and Q, whereg = (16 N)sin 28°

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-100 m- 200 m

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

(a) O.2.

PC 0>T-**>

Fv =(300N)cos25°

(b) The smallest force Q at B must be perpendicular toDB at 45°£LMD =Q(DB) 41 .700 N m = (2(0.28284 m) Q = 147.4 N L 45° <

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•-lOOnim-* -200 m *.

125 il

PROBLEM3.4

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

(a) See Problem 3.3 for the figure and analysis leading to the determination ofMd M =41.7N-mHCl^n

0>\7.Siy\ c = at. (b) Since C is horizontal C = Ci r = DC = (0.2 m)i -(0. 125 m)jMD =rxCi = C(0.l25m)k4l.7N-m = (0.l25m)(C)C = 3.60 N (c) The smallest force C must be perpendicular to DC; thus, it forms a with the vertical

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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°

Noting that the direction ofthe moment ofeach force component about B isclockwise, have

= -(8in.)(3.3809 1b) - (2 in.)(7.2505 lb) = -186.6 lb -in.

i*2. •**.

or Mj =186.6 lb -in.J) ^

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PROBLEM 3.6 For the shift lever shown, determine the magnitude and the direction of the smallestforce P that has a 21 0-lb • in. clockwise moment about B.

2 in.

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

2 in. /= 19.98° iand MB =dPn p Z i«.

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PROBLEM 3.7 An 1 1-lb force P is applied to a shift lever. The moment of P about B is clockwiseand has a magnitude of250 lb • in. Determine the value ofa.

2 in.

By definition where and also

Then

or and

MB =rmPsm.6 = a + (9Q°-tf>)

_i 8 in.<p - tan" 19.9831'2 in.

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PROBLEM 3.8

It is known that a vertical force of 200 lb is required to remove the nailat C from the board. As the nail first starts moving, determine (a) themoment 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 4in. n SOLUTION

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

800 lb -in. or MB =H00\b-m.)<

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

or or

A*W™. cl = fAIB

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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 thatforce into horizontal and vertical components applied (a) at Point C, (b) at Point E.

(a) Slope oflineThen

Then

(b) We have or M./J =760N-m>)^

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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. Ifd- 2,80 m, determine the tension that must be developed in the cable of winch puller AB to create therequired moment about Point D.

0.2 j.i» SOLUTION o. a?5^i

z-acw OiZ^

Slope of line

Then and We have

24My

T/)B rAH

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PROBLEM 3.1 It is known that a force with a moment of960 N • m about D is required to straighten the fence post CD. Ifthecapacity of winch puller AB is 2400 N, determine the minimum value of distance d to create the specifiedmoment about Point D.

0.2 in

0.875 in o.mtn czom The minimum value ofd can be found based on the equation relating the moment ofthe force TAB about D:MD ={TABmK),(d)

whereNow

0.875msin # |

or 3.7852</2 - 0.40c/ - .8056 = Using the quadratic equation, the minimum values ofdare 0.51719 m and -.41151 m.

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5.3in. PROBLEM 3.12 12.0 in.

2.33in.mmM 1 >i. 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 balland socket at B, determine the moment ofthe force about A.

SOLUTION First note

Then and

Now where

Then

12.2241 in.

sin 9- 2.3 in. 12.2241 in.

*cb = FCB cos 9\ -FCB $m9l 1251b

12.2241 in. [(12.0 in.)i- (2.3 in.) j]

M = (15.3 in.)i--(14.3 in.) j ru-nimn* 1251b

\«5»,a> im.

\Z.& >N. 2.3 ttvi.

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

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t§| 1.7.2 hi.

PROBLEM 3.13 The tailgate of a car is supported by the hydraulic lift EC. If the liftexerts a 125-lb force directed along its centerline on the ball andsocket at B, determine the moment of the force about A.

First noteThen cos - 17.2 in.

18.8123 in.

18.8123 in.

Z.O.'S «w.

and

Now where

Then r = (20.5 in.)i- (4.38 in.)j 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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120mm

PROBLEM 3.14

A mechanic uses a piece ofpipe 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 ifits line of action passes through O.

We have Mc =rwc xFj,

Noting the direction ofthe moment ofeach force component about C is clockwise.

Where and

Mc =xFBy +yFBx

4 j,

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By definition: Now and

PROBLEM 3.15

Form the vector products B x C and B' * C, where B =B\ and use the resultsobtained to prove the identity sinacos/? =-sin(a+ P) +-sin (a - p).

SOLUTION N(>te: B = £(cos/?i + sin/?j)

B'= 5(cos/?i-sin/?j)C - C(cosai +sin aj)|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 «)kB'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

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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 ofcoordinates.

Assume that a force F, or magnitude F(N), acts at Point A and isdirected ixomA to B. Then,,

Where

By definition Where

Then

¥~FXm

MQ = \rA xF\ = dFr,=-(lm)i-(4m)j F

E> (Zom, K-rti^

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PROBLEM 3.17 The vectors P and Q are two adjacent sides ofa parallelogram. Determine the area ofthe parallelogram when(tf)P--7i + 3j-3kandQ = 2i + 2j + 5k,(6)P = 6i-5j-2kandQ = -2i + 5j~k.

(a) We have where

Then

(b) We have where v4 = |PxQf

Then PxQ i fci 6 -5

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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.

{a) We have where

Then

and

(b) We have where

Then

and

AxB

AxB

|AxB| :li + 2j-5k

|AxB|

AxB

15V(-3)2 +H)2 +(-l)2 -I5>/ri

15(-3f-lJ-lk) or X15VHAxB (-3i-j~k) 4

|AxB| 3i-3j + 2k :-2i + 6j-4k

(12-12)i + (-4+ l2)j + (18-6)k (8j + 12k)

4(2j +3k)413 or X, V (2j + 3k) <

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(Parte **1** de 3)