Statics And Mechanics Of Materials 5th Edition Hibbeler Solutions Manual a644

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22 22


2–1. 0N If u = 60° and F = 450 N, determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

y F u 15 700 N

SOLUTION SOLUTION SOLUTION The parallelogram law of addition and

the triangular rule are shown in Figs. a and b, The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. respectively. Applying the law of consines to Fig. b, Applying the law of consines to Fig. b, 7002 497.01 N

4502

2(700)(450) cos 45° Ans. Ans.

497 N

This yields This yields sin 700

sin 45° 497.01

95.19°

Thus, the direction of angle of measured counterclockwise from the Thus, the direction of angle of F measured counterclockwise from the positive axis, is positive axis, is 60°

95.19°

60°

Ans. Ans.

155°

23 23

x


Ans: FR = 497 N f = 155°

24 24


2–2. If the magnitude of the resultant force is to be 500 N, foirnce tihse tm o abgenit5u0d0e dirIefcttehde am loanggntithuedpeosoiftivtehey arexsisu,ldtaentetrm If ethc e ma al gn itud e pof t he resultant force is to N,, force F can ted d its a lodirection n g th e o si tive y axis, determine the of dir p ve beo Nd , al ng u.5fnr0fei0tud ocrce mag e F and its direction . di te o the ositi uy axis, determine the magnitude of force F and its direction u.

y F

u 15 700 N

SOLUTION OarLalU Nlaw of addition and the triangular rule are shown in Figs. a and b, ThS ep leT loIgO ram resT ph ecetp ivaerlayl.lelogram law of addition and the triangular rule are shown in Figs. a and b, The para llelogram law of addition and the triangular rule are shown in Figs. a and r e b, spectively. r Appelsypiencgtitvheelyl.aw of cosines to Fig. b, Applying the law of cosines to Fig. b, Applying theFlaw 2 of 5c0o0si2nes to F2ig. b, = + 700 - 2(500)(700) cos 105° F = 25002 + 7002 - 2(500)(700) cos 105° 25002 + 7002 - 2(500)(700) cos 105° F = = 959.78 N = 960 N Ans. = 959.78 N = 960 N Ans. = 9 5 9 . 78 N = 9 6 0 N Ans. Applying the law of sines to Fig. b, and using this result, yields Applying the law of sines to Fig. b, and using this result, yields Applying the law in°es+ to sinof(9s0 u)Fig. sbi,n a1n0d5°using this result, yields sin700 (90° + u=) sin 105° (90° + u) 959.78 sin 105° sin

700 u = 45.2700 ° u = 45.2° u = 45.2°

= = 959.78 959.78

Ans. Ans. Ans.

25 25

x


Ans: F = 960 N u = 45.2°

26 26


2–3. y

Determiinne thethmeagnm itaugdn eio tuf dthee reosfultatn hteforrceesF uRlta=ntF1 f+orcFe2 F an ecF ti2oann ,m d eitassudrieredcctoiounn,temrcelaoscukrw ed isecofruonmtetrhcelopcoksw itivse Rd=itsFd 1 ir+ fxro axmis.the positive x axis.

F1

250 lb

30

SOLUTION

x 2

2

FR = 2(250) + (375) - 2(250)(375) cos 75° = 393.2 = 393 lb 393.2

Ans.

45

250 =

sin 75° sin u u = 37.89° f = 360° - 45° + 37.89° = 353°

F2

Ans.

Ans: 27 27

375 lb


FR = 393 lb f = 353°

28 28


*2–4. The vertical force acts downward at on the two-membered D teer.mDineetertm heinm tuadgensitu od f etshoef theo tw coomc m onpeonntenotsf oFf fraem e tahgenim dirdeicrteecdteadloanlg onmgetm hebearxseA sB ofand AaCn.dSet F. = Set500 N.

B

. 45

SOLUTION PaLraUllT elIoO graN m Law: The SO

A

parallelogram law of addition is shown in Fig. a.

tryL:aU f gsirnaems (laFw ig.obf )a,dwdeithioan veis shown in Fig. a. PaT rarilgleolnoogm raem ws:in Tghethpealraaw lleolo

F

Trigonometry: Using the law of sines (Fig. b5)0,0we have sin 60°

C

sin 75° 448 N

sin 45°

30

Ans.

500 sin 75°

Ans.

366 N

Ans. Ans.

29 29


Ans: FAB = 448 N FAC = 366 N

30 30


2–5. Solve Prob. 2-4 with F = 350 lb. Solve Prob. 2–4 with F = 350 lb. B

45

SO LUL T NN SO UITOIO

A

PaP raalrlaeloeglroagm LaLwa:w T:hT ehpearpaalrlealo lawlaowf a itd ioitnioisn sihs oshwonwin iFnigF. iag.. a. ram leglroagm ram ofdd ad TriTgroin treyt:rU insgintghethlaewlaowf o sifnseisne(F g omoem y:sU s i(gF.ibg). ,bw),ewheavheave

F

30

FAFB 350350 AB= sinsin 60°60° = sinsin 75°75°

C

F AB = 314 FAB = 31lb4 lb

AnA s.ns.

F AC 350350 FAC 45° = = sin 75° sinsin 45° sin 75° AnA s.ns.

F AC = 256 FAC = 25lb6 lb

Ans:

31 31


FAB = 314 lb FAC = 256 lb

32 32


2–6. v

Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured clockwise from the positive u axis.

30 75

F1

4 kN

30 u

Solution

F2

6 kN

Parallelogram Law. The parallelogram law of addition is shown in Fig. a. Trigonometry. Applying Law of cosines by referring to Fig. b, FR = 242 + 6 2 - 2(4)(6) cos 105° = 8.026 kN = 8.03 kN

Ans.

Using this result to apply Law of sines, Fig. b, sin 105° sin u ; = 8.026 6

u = 46.22°

Thus, the direction f of FR measured clockwise from the positive u axis is f = 46.22° - 45° = 1.22°

Ans.

Ans:

33 33


FR = 8 . 0 3 kN f = 1.22°

34 34


2–7. v

Resolve the force F1 into components acting along the u and v axes and determine the magnitudes of the components.

30 75

F1

4 kN

30 u

Solution

F2

6 kN

Parallelogram Law. The parallelogram law of addition is shown in Fig. a. Trigonometry. Applying the sines law by referring to Fig. b. (F1)v 4 = ; sin 45° sin 105°

(F1)v = 2.928 kN = 2.93 kN

Ans.

(F1)u 4 = ; sin 30° sin 105°

(F1)u = 2.071 kN = 2.07 kN

Ans.

Ans:

35 35


(F1)v = 2.93 kN (F1)u = 2.07 kN

36 36


*2–8. v

Resolve the force F2 into components acting along the u and v axes and determine the magnitudes of the components.

30 75

F1

4 kN

30 u

Solution

F2

6 kN

Parallelogram Law. The parallelogram law of addition is shown in Fig. a. Trigonometry. Applying the sines law of referring to Fig. b, (F2)u 6 = ; sin 75° sin 75°

(F2)u = 6.00 kN

Ans.

(F2)v 6 = ; sin 30° sin 75°

(F2)v = 3.106 kN = 3.11 kN

Ans.

Ans:

37 37


(F2)u = 6.00 kN (F2)v = 3.11 kN

31 31


2–9. If the resultant force acting on the is to be 1200 lb, directed horizontally to the right, determine the force F in rope A and the corresponding angle u.

F A u B 60

900 lb

Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a. Trigonometry. Applying the law of cosines by referring to Fig. b, F = 29002 + 12002 - 2(900)(1200) cos 30° = 615.94 lb = 616 lb

Ans.

Using this result to apply the sines law, Fig. b, sin u 900

sin 30° =

615.94

;

u = 46.94° = 46.9°

Ans.

Ans:

30 30


F = 616 lb u = 46.9°

31 31


2–10. y

Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

800 lb 40

x

Solution

35

Parallelogram Law. The parallelogram law of addition is shown in Fig. a. Trigonometry. Applying the law of cosines by referring to Fig. b, FR = 28002 + 5002 - 2(800)(500) cos 95° = 979.66 lb = 980 lb

Ans.

500 lb

Using this result to apply the sines law, Fig. b, sin u 500

sin 95° =

979.66

;

u = 30.56°

Thus, the direction f of FR measured counterclockwise from the positive x axis is f = 50° - 30.56° = 19.44° = 19.4°

Ans.

Ans: FR = 980 lb 32 32


f = 19.4°

33 33


2–11. If u = 60°, determine the magnitude of the resultant and its direction measured clockwise from the horizontal.

FA u

8 kN

A

The plate is subjected to the two forces at A and B as shown. If u = 60°, determine the magnitude of the resultant of these two forces and its direction measured clockwise from the horizontal.

40

SOLUTION

B

Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using law of cosines (Fig. b), we have FR = 282 + 62 - 2(8)(6) cos 100° = 10.80 kN = 10.8 kN

Ans.

The angle u can be determined using law of sines (Fig. b). sin u sin 100° = 6 10.80 sin u = 0.5470 u = 33.16° Thus, the direction f of FR measured from the x axis is f = 33.16° - 30° = 3.16°

Ans.

34 34

FB

6 kN


Ans: FR = 10.8 kN f = 3.16°

35 35


*2–12. Determine the angle u for connecting member A to the plate so that the resultant force of FA and FB is directed horizontally to the right. Also, what is the magnitude of the resultant force?

FA u

8 kN

A

40

SOLUTION

B

Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using law of sines (Fig .b), we have sin (90° - u) sin 50° = 6 8 sin (90° - u) = 0.5745 u = 54.93° = 54.9°

Ans.

From the triangle, f = 180° - (90° - 54.93°) - 50° = 94.93°. Thus, using law of cosines, the magnitude of FR is FR = 282 + 62 - 2(8)(6) cos 94.93° = 10.4 kN

Ans.

36 36

FB

6 kN


Ans: u = 54.9° FR = 10.4 kN

37 37


2–13. Th e ffoorrcceeaactcitnn onthte geegaer ator ottohoits F s=F20 = lb 2.0Rlbe.soRlveesothvies ggon Tthe fofo rc e acnttnog to n t e g e a r t o o t s F = 2 0 lb . forisce inrtcoe tiw o cowmopocnoemnptsoancenintsg aclotinngg tahleolningetsh Reso ve aealainedsbaba. thaisn dfobrbc.e into two components acting along the lines aa and bb.

b a

F 80 60 a b

SOLUTION SOLUTION 20

Fa

; = F 20 sin 40°= sina 80;° sin 40° sin 80° Fb 20 ; = Fb 20 sin 40°= sin 60;° sin 40° sin 60°

Fa = 30.6 lb Fa = 30.6 lb

Ans. Ans.

Fb = 26.9 lb Fb = 26.9 lb

Ans. Ans.

Ans: 38 38


Fa = 30.6 lb Fb = 26.9 lb

39 39


2–14. The component of force F acting along line aa is required to be 30 lb. Determine the magnitude of F and its component along line bb.

b

a

F 80 60 a b

SOLUTION 30 sin 80°

=

F

;

F = 19.6 lb

Ans.

Fb = 26.4 lb

Ans.

sin 40°

Fb 30 = ; sin 80° sin 60°

Ans: 40 40


F = 19.6 lb Fb = 26.4 lb

41 41


2–15. Force F acts on the frame such that its component acting F em F eam ctbseo theis f6r5a0mlbe, sduircehcttehdatfriotsmcom on ac, taindg aloorncg rn AB B topw arednst A athloencgompm b e r A B i s 65 0 l b , d ir cte d fr o m B to wa r d s A onent acting along member BC is 500 lb, dire,catn ed tfh e co mp nean g eatleornmgin meetm rB rom B toow rdtsacCtin .D hb eem agCniitsu5d0e0olfb,Fdiarnedcteitds fdrioremctB iontouw . Sard et sfC=. D 60e°t.ermine the magnitude of F and its direction u. Set f = 60°.

B

u F A

SOLUTION ThLeU paTraI O llelN ogram law of addition and SO

triangular rule are shown in Figs. a and b, respectively. The parallelogram law of addition and triangular rule are shown in Figs. a and b, resA peptliyvienlyg. the law of cosines to Fig. b, Applying the law of cosines to2 Fig. b, 2 F = 2500 + 650 - 2(500)(650) cos 105° F = = 916.91 lb = 917 lb

Ans.

Using this result=a9n1d6a.9p1plbin=g 9th1e7 labw of sines to Fig. b yields

Ans.

Using this result and applying tshine ulaw osfinsi1n0e5s°to Fig. b yields = u = 31.8° 500 916.91 sin u sin 105° = 500 916.91

Ans.

u = 31.8°

Ans.

42 42

f

45

C


Ans: F = 917 lb u = 31.8°

43 43


*2–16. Force F acts on the frame such that its component acting along member AB is 650 lb, directed from B towards A. Determine the required angle f (0° … f … 45°) and the component acting along member BC. Set F = 850 lb and u = 30°.

B

u F A

f

45

-" 1 / " The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of cosines to Fig. b, FBC = 28502 + 6502 - 2(850)(650) cos 30° = 433.64 lb = 434 lb

Ans.

Using this result and applying the sine law to Fig. b yields sin (45° + f) sin 30° = 850 433.64

f = 33.5°

Ans.

Ans: 44 44

C


FBC = 434 lb f = 33.5°

45 45


2–17. y

If F1 = 30 lb and F2 = 40 lb, determine the angles u and f so that the resultant force is directed along the positive x axis and has a magnitude of FR = 60 lb.

F1 θ x φ

Solution F2

Parallelogram Law. The parallelogram law of addition is shown in Fig. a. Trigonometry. Applying the law of cosine by referring to Fig. b, 402 = 302 + 602 - 2(30)(60) cos u u = 36.34° = 36.3°

Ans.

And 302 = 402 + 602 - 2(40)(60) cos f f = 26.38° = 26.4°

Ans.

Ans: u = 36.3° 46 46


f = 26.4°

47 47


2–18. Determine the magnitude and direction u of FA so that the aigrn nn dgdtihre rDeesuteltramnitnfeorth cee im s d eicttueddeaalo ti o n u of F o th a t t h e positive x aAxis and has a m reasu rcfe1i2s5d0irNe.cted along the glntiatn utdfeoo positive x axis and has a magnitude of 1250 N.

y y

A

O O

+ F = S Rx + c FRy =

Fx; Fy;

x

30° 30 B

SOLUTION

F FA A

A

B FB = 800 N

FRx = FA sin u + 800 cos 30 = 1250 FRy = FA cos u - 800 sin 30 = 0 u = 54.3

Ans.

FA = 686 N

Ans.

Ans: u = 54.3°

48 48

x


FA = 686 N

41 41


y

D–e1t9e.rmine the magnitude and direction, measured 2 counterclockwise from the positive x axis, of the resultant D teerm e gthoenm etoO f ,thife Fresu ltant force acting o n the foerc acitnin thaegnritnugda A = 750 N and u = 45 . ring at O if FA = 750 N and u = 45°. What is its direction, measured counterclockwise from the positive x axis?

y FA A

FA

A x O O

SOLUTION Fx;

x 30 B B

FB = 800 N FB = 800 N

Scalar Notation: Suming the force components algebraically, we have + F = S Rx

30°

FRx = 750 sin 45 + 800 cos 30 = 1223.15 N S

+ c FRy =

Fy;

FRy = 750 cos 45 - 800 sin 30 = 130.33 N c

The magnitude of the resultant force FR is FR = 3F 2Rx + F 2Ry = 21223.152 + 130.332 = 1230 N = 1.23 kN

Ans.

The directional angle u measured counterclockwise from positive x axis is FRy 130.33 u = tan-1 = tan-1 a b = 6.08 FRx 1223.15

Ans.

Ans: 40 40


FR = 1.23 kN u = 6.08°

41 41


*2–20. Determine the magnitude of force F so that the resultant FR of the three forces is as small as possible. What is the minimum magnitude of FR?

8 kN

F 30 6 kN

Solution Parallelogram Law. The parallelogram laws of addition for 6 kN and 8 kN and then their resultant F and F are shown in Figs. a and b, respectively. In order for FR to be minimum, it must act perpendicular to F. Trigonometry. Referring to Fig. b, F = 26 2 + 82 = 10.0 kN

8 u = tan - 1 a b = 53.13°. 6

Referring to Figs. c and d, FR = 10.0 sin 83.13° = 9.928 kN = 9.93 kN

Ans.

F = 10.0 cos 83.13° = 1.196 kN = 1.20 kN

Ans.

Ans: FR = 9.93 kN F = 1.20 kN 42 42


2–21. If the resultant force of the two tugboats is 3 kN, directed along the positive x axis, determine the required magnitude of force FB and its direction u. If the resultant force of the two tugboats is 3 kN, directed along the positive x axis, determine the required magnitude of force FB and its direction u.

y

A

FA

2 kN 30

x

u C

SOLUTION

FB

The parallelogram law of addition and the triangular rule are shown in Figs. a and b,

SO resLpU ecT tivIeO ly.N

B

ThA e palryailnlegloth geralm awlaowf coofsaindedsittionFiagn.db,the triangular rule are shown in Figs. a and b, respectively. FB = 222 + 32 - 2(2)(3)cos 30° Applying the law of cosines to Fig. b, = 1.615kN = 1.61 kN FB = 222 + 32 - 2(2)(3)cos 30° Using this result and applying the law of sines to Fig. b yields = 1.615kN = 1.61 kN

Ans. Ans.

sin u sin 30° Using this result and apply=ing the law of siune=s t3o8.F3i°g. b yields 2 1.615 sin u sin 30° = 2 1.615

Ans.

u = 38.3°

Ans.

43 43


Ans: FB = 1.61 kN u = 38.3°

44 44


2–22. If FBB == 33kkN Naanndduu == 4455°°,, ddeetteerrm miinnee tthhee maggniitude of the cloictk result ant fo forcrceeanodf ithedirtewcotiotnugmbeoaastusreadnd s wdisierefcrto iom n tm heapso iveclo x cakxw isi.se from the positive x axis. usrietd

y

A

FA

2 kN 30

x

u C

SOLUTION

FB

The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. B

Applying the law of cosines to Fig. b, FR = 222 + 32 - 2(2)(3) cos 105° = 4.013 kN = 4.01 kN

Ans.

Using this result and applying the law of sines to Fig. b yields sin 105° sin a a = 46.22° = 4.013 3 Thus, the direction angle f of FR, measured clockwise from the positive x axis, is f = a - 30° = 46.22° - 30° = 16.2°

Ans.

45 45


Ans: FR = 4.01 kN f = 16.2°

46 46


2–23. If the resultant force of the two tugboats is required to be directed towards the positive x axis, and FB iis to be a minimum, determine the magnitude of FR and FB and the angle u.

y

A

FA

2 kN 30

x

u C

SOLUTION

FB

For FB to be minimum, it has to be directed perpendicular to FR. Thus, u = 90°

Ans.

B

The parallelogram law of addition and triangular rule are shown in Figs. a and b, respectively. By applying simple trigonometry to Fig. b, FB = 2 sin 30° = 1 kN

Ans.

FR = 2 cos 30° = 1.73 kN

Ans.

Ans: u = 90°

47 47


FB = 1 kN FR = 1.73 kN

48 48


*2–24. Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

y F1

200 N

45 x 30

Solution Scalar Notation. Summing the force components along x and y axes algebraically by referring to Fig. a, + (F ) = ΣF ; S (FR)x = 200 sin 45° - 150 cos 30° = 11.518 N S R x x + c (FR)y = ΣFy;

(FR)y = 200 cos 45° + 150 sin 30° = 216.42 N c

Referring to Fig. b, the magnitude of the resultant force FR is FR = 2(FR)x2 + (FR)y2 = 211.5182 + 216.422 = 216.73 N = 217 N

Ans.

And the directional angle u of FR measured counterclockwise from the positive x axis is u = tan - 1 c

(FR)y (FR)x

d = tan - 1a

216.42 b = 86.95° = 87.0° 11.518

49 49

Ans.

F2

150 N


Ans: FR = 217 N u = 87.0°

50 50


2–25. y

Determine the magnitude of the resultant force and its direction, measured clockwise from the positive x axis.

400 N B 30 x 45

Solution Scalar Notation. Summing the force components along x and y axes by referring to Fig. a, + (F ) = ΣF ; S R x x

(FR)x = 400 cos 30° + 800 sin 45° = 912.10 N S

+ c (FR)y = ΣFy;

(FR)y = 400 sin 30° - 800 cos 45° = - 365.69 N = 365.69 N T

Referring to Fig. b, the magnitude of the resultant force is FR = 2(FR)x2 + (FR)y2 = 2912.102 + 365.692 = 982.67 N = 983 N

Ans.

And its directional angle u measured clockwise from the positive x axis is u = tan - 1 c

(FR)y (FR)x

d = tan - 1a

365.69 b = 21.84° = 21.8° 912.10

51 51

Ans.

800 N


Ans: FR = 983 N u = 21.8°

52 52


2–26. Resolve F11 aan nd d F22iinnttoo tthheeiirr xxaannddyyccoom mppoonneenntst.s.

y 60 30 F1

400 N

SOLUTION 45

F1 = {400 sin 30°( + i) + 400 cos 30°( + j)} N = {200i + 346j} N

Ans.

F2 = {250 cos 45°( + i) + 250 sin 45°( - j)} N = {177i - 177j} N

Ans.

53 53

F2

250 N

x


Ans: F1 = 5 200i + 346j 6 N F2 = 5 177i - 177j 6 N

54 54


2–27. Determi ne the magnitude of the resultant force and its directi on meassuurreedd ccoouunntteerrcclloocckkwwiisseeffrroom mtthheeppoossititiviveexxaaxxisi.s.

y 60 30 F1

400 N

SOLUTION 45

Rectangular Components: By referring to Fig. a, the x and y components of F1 and F2 can be written as F2

(F1)x = 400 sin 30° = 200 N

(F1)y = 400 cos 30° = 346.41 N

(F2)x = 250 cos 45° = 176.78 N

(F2)y = 250 sin 45° = 176.78 N

Resultant Force: Summing the force components algebraically along the x and y axes, we have

c

+ ©(F ) = ©F ; R x x

(FR)x = 200 + 176.78 = 376.78 N

(FR)y = 346.41 - 176.78 = 169.63 N c + c ©(FR)y = ©Fy; ts of F1 and The magnitude of the resultant force FR is

1

FR = 2(FR)x2 + (FR)y2 = 2376.782 + 169.632 = 413 N

N

Ans.

The direction angle u of FR, Fig. b, measured counterclockwise from the positive axis, is (FR)y g the x and -1 169.63 u = tan-1 c Ans. (FR)x d = tan a 376.78 b = 24.2°

8N

Ans. the positive

Ans.

55 55

250 N

x


Ans: FR = 413 N u = 24.2°

56 56


*2–28. y

Resolve each force acting on the gusset plate into its x and y components, and express each force as a Cartesian vector.

F3

Resolve each force acting on the gusset plate into its x and y components, and express each force as a Cartesian vector.

650 N 3

5

F2

750 N

4

45 F1

Solution F1 = {900( + i)} = {900i} N

Ans.

+ 750 sin 45°( + j)} N F1F=2 ={90{705(0+ ci)o}s =45{°9(0+0ii)} N Ans.

= {530i + 530j} N F2 = {750 cos 45°( + i) + 750 sin 45°( + j)} N 4 3 F=3 ={53e06i5+ 0 a530bj}(N + i) + 650 a b ( - j) f N 5 5 4 90i)j)}+N650 a 3 b ( - j) f N F3 = =e 6{5020a i -b (3+ 5 5

Ans.

= {520 i - 390j)} N

Ans.

57 57

x 900 N


Ans: F1 = 5 900i 6 N F2 = 5 530i + 530j 6 N F3 = 5 520i - 390j 6 N

51 51


2–29. y

Determine the magnitude of the resultant force acting on the gusset plate and its direction, measured counterclockwise from the positive x axis.

F3

650 N 3

5

F2

750 N

4

Determine the magnitude of the resultant force acting on

45

the positive x axis.

F1

SOLUTION Rectangular Components: By referring to Fig. a, the x and y components of F1, F2, and F3 can be written as (F1)y = 0

(F1)x = 900 N (F2)x = 750 cos 45° = 530.33 N 4 (F3)x = 650 a b = 520 N 5

(F2)y = 750 sin 45° = 530.33 N 3 (F3)y = 650 a b = 390 N 5

Resultant Force: Summing the force components algebraically along the x y axes, we have + : ©(FR)x = ©Fx;

(FR)x = 900 + 530.33 + 520 = 1950.33 N :

+ c ©(FR)y = ©Fy;

(FR)y = 530.33 - 390 = 140.33 N c

and

The magnitude of the resultant force FR is FR = 2(FR)x2 + (FR)y2 = 21950.332 + 140.332 = 1955 N = 1.96 kN Ans. The direction angle u of FR, measured clockwise from the positive x axis, is (F ) 140.33 u = tan-1 c R y d = tan-1 a b = 4.12° (FR)x 1950.33

50 50

Ans.

x 900 N


Ans: FR = 1.96 kN u = 4.12°

51 51


2–30. y

Express each of the three forces acting on the in Cartesian vector form and determine the magnitude of the resultant force and its direction, measured clockwise from positive x axis.

F1

50 N

5 4 4

3

x F3

30 N

15

Solution Cartesian Notation. Referring to Fig. a,

F2

3

80 N

4

F1 = (F1)x i + (F1)y j = 50 a b i + 50 a b j = {30 i + 40 j} N 5 5

Ans.

F2 = - (F2)x i - (F2)y j = - 80 sin 15° i - 80 cos 15° j = { - 20.71 i - 77.27 j} N = { - 20.7 i - 77.3 j} N

Ans.

F3 = (F3)x i = {30 i}

Ans.

Thus, the resultant force is FR = ΣF ;

FR = F1 + F2 + F3 = (30i + 40 j) + ( - 20.71i - 77.27j) + 30i = {39.29 i - 37.27 j} N

Referring to Fig. b, the magnitude of FR is FR = 239.292 + 37.272 = 54.16 N = 54.2 N

Ans.

And its directional angle u measured clockwise from the positive x axis is u = tan - 1 a

37.27 b = 43.49° = 43.5° 39.29

Ans.

Ans: F1 = F2 = F3 = FR = 52 52

{30i + 40j} N { - 20.7 i - 77.3 j} N {30 i} 54.2 N


u = 43.5°

53 53


2–31. y

Determine the x and y components of F1 and F2. 45

F1

200 N

30 F2

150 N

x

SOLUTION F1x = 200 sin 45° = 141 N

Ans.

F1y = 200 cos 45° = 141 N

Ans.

F2x = - 150 cos 30° = - 130 N

Ans.

F2y = 150 sin 30° = 75 N

Ans.

Ans: F1x = 141 N F1y = 141 N

54 54


F2x = - 130 N F2y = 75 N

55 55


*2–32. y

Deter m ne th e magnittudee of t e resu tant forcce and its directiionn,, measureedd count ercclloocckkw wiissee ffroom m the positive x axis..

45

F1

200 N

30 F2

150 N

x

SOLUTION + R FRx = ©Fx;

FRx = - 150 cos 30° + 200 sin 45° = 11.518 N

Q + FRy = ©Fy;

FRy = 150 sin 30° + 200 cos 45° = 216.421 N

FR = 2 (11.518)2 + (216.421)2 = 217 N

Ans.

216.421 ≤ = 87.0° 11.518

Ans.

u = tan - 1 ¢

Ans: FR = 217 N 56 56


u = 87.0°

57 57


2–33. y

Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

F3

8 kN F2

5 kN

F1

4 kN

60 45

Solution Scalar Notation. Summing the force components along x and y axes algebraically by referring to Fig. a, + (F ) = ΣF ; S R x x

(FR)x = 4 + 5 cos 45° - 8 sin 15° = 5.465 kN S

+ c (FR)y = ΣFy;

(FR)y = 5 sin 45° + 8 cos 15° = 11.263 kN c

By referring to Fig. b, the magnitude of the resultant force FR is FR = 2(FR)x2 + (FR)y2 = 25.465 2 + 11.2632 = 12.52 kN = 12.5 kN

Ans.

And the directional angle u of FR measured counterclockwise from the positive x axis is u = tan - 1 c

(FR)y (FR)x

d = tan - 1 a

11.263 b = 64.12° = 64.1° 5.465

58 58

Ans.

x


Ans: FR = 12.5 kN u = 64.1°

59 59


2–34. Express F1, F2, and F3 as Cartesian vectors.

y F3

750 N 45

3

x

5 4

SOLUTION 4 F1 =

5

30 F2

3 (850) i -

5

F1

850 N

625 N

(850) j

= {680 i - 510 j} N

Ans.

F2 = - 625 sin 30° i - 625 cos 30° j = { - 312 i - 541 j} N

Ans.

F3 = - 750 sin 45° i + 750 cos 45° j { -530 i + 530 j} N

Ans.

Ans: F1 = {680i - 510j} N

60 60


F2 = { - 312i - 541j} N F3 = { - 530i + 530j} N

61 61


2–35. Determine the magnitude of the resultant force and its Det nn e mtheeasumreadgnciot uneterocflocth direercm tiio kewisresfurlotan m thefoprcoesitiavnedx aitxsis. direction, measured counterclockwise from the positive x axis.

y F3

750 N 45

3

4

SOLUTION SOLUTION + F = ©F ; : Rx x + : F = ©F ; Rx

x

+ c FRy = ©Fy ; + c FRy = ©Fy ;

x

5

30

4

FRx =4 (850) - 625 sin 30° - 750 sin 45° = - 162.83 N FRx =

5

(5850) - 625 sin 30° - 750 sin 45° = - 162.83 N

3 FRy = 3- (850) - 625 cos 30° + 750 cos 45° = - 520.94 N FRy = - (5850) - 625 cos 30° + 750 cos 45° = - 520.94 N 5 2 FR = 2( - 162.83) + ( - 520.94)2 = 546 N Ans. FR = 2( - 162.83)2 + ( - 520.94)2 = 546 N Ans. 520.94 f = tan-1 a b = 72.64° 520.94 f = tan-1 a 162.83 b = 72.64° 162.83 u = 180° + 72.64° = 253° Ans. u = 180° + 72.64° = 253° Ans.

62 62

F2

625 N

F1

850 N


Ans: FR = 546 N u = 253°

63 63


*2–36. y

Determine the magnitude of the resultant force and its direction, measured clockwise from the positive x axis.

40 lb

5

4 3

Solution

x

Scalar Notation. Summing the force components along x and y axes algebraically by referring to Fig. a, 5

+ (F ) = ΣF ; S R x x

3 5 (FR)x = 40 a b + 91a b + 30 = 89 lb S 5 13

+ c (FR)y = ΣFy;

4 12 (FR)y = 40 a b - 91a b = - 52 lb = 52 lb T 5 13

13

12

91 lb

By referring to Fig. b, the magnitude of resultant force is FR = 2(FR)x2 + (FR)y2 = 2892 + 522 = 103.08 lb = 103 lb

Ans.

And its directional angle u measured clockwise from the positive x axis is u = tan - 1 c

(FR)y (FR)x

d = tan - 1a

52 b = 30.30° = 30.3° 89

Ans.

Ans: 64 64


FR = 103 lb u = 30.3°

65 65


2–37. Determine t e magnitude and direct on u of th e reessuulttaanntt force FR.. Expreessss the ressullt iin of the magnitudes of the components F1 and F22 aanndd tthhee angglle f..

F1

FR

f u F2 1

SOLUTION

FR

F

F 2R = F 12 + F 22 - 2F1F2 cos (180° - f) Since cos (180° - f) = - cos f,

f u

FR = 2F 2 + F 2 + 2F F cos f 1

2

Ans.

F2

1 2

From the figure, tan u =

F1 sin f F2 + F1 cos f

u = tan –1 ¢

F1 sin f ≤ F2 + F1 cos f

Ans.

Ans: FR = 2F 21 + F 22 + 2F1F2 cos f 66 66


u = tan-1a

67 67

F1 sin f b F2 + F1 cos f


2–38. z

The force F has a magnitude of 80 lb. Determine the magnitudes of the x, y, z components of F. Fz

F

80 lb

z Fz b a

60

45 Fy

Fx

SOLUTION

x

1 = cos2 60° + cos2 45° + cos2 g Solving for the positive root, g = 60°

x

Fx = 80 cos 60° = 40.0 lb

Ans.

Fy = 80 cos 45° = 56.6 lb

Ans.

Fz = 80 cos 60° = 40.0 lb

Ans.

Ans:

68 68

y


Fx = 40.0 lb Fy = 56.6 lb Fz = 40.0 lb

61 61


2–39. z

The bolt is subjected to the force F, which has components acting along the x, y, z axes as shown. If the magnitude of F is 80 N, and a = 60° and g = 45°, determine the magnitudes of its components.

Fz g

F Fy

SOLUTION

b y

a Fx

cosb = 21 - cos2 a - cos2g = 21 - cos 60° - cos 45° 2

x

2

b = 120° Fx = |80 cos 60°| = 40 N

Ans.

Fy = |80 cos 120°| = 40 N

Ans.

Fz = |80 cos 45°| = 56.6 N

Ans.

Ans: 60 60


Fx = 40 N Fy = 40 N Fz = 56.6 N

61 61


*2–40. z

Determine the magnitude and coordinate direction angles of the force F acting on the . The component of F in the x–y plane is 7 kN. F

y 30 40 7 kN

Solution x

Coordinate Direction Angles. The unit vector of F is uF = cos 30° cos 40°i - cos 30° sin 40°j + sin 30° k = {0.6634i - 0.5567j + 0.5 k} Thus, cos a = 0.6634;

a = 48.44° = 48.4°

Ans.

cos b = - 0.5567;

b = 123.83° = 124°

Ans.

cos g = 0.5;

g = 60°

Ans.

The magnitude of F can be determined from F cos 30° = 7;

F = 8.083 kN = 8.08 kN

Ans.

Ans: a = 48.4° b = 124° g = 60° 62 62


F = 8.08 kN

63 63


2–41. zz

Determine thtehemam gnaigtu nd iteudaend odordcinoaotredidniaretectiodniraen acn ctgiloens oanf gltehse ofretshueltraenstultfaonrctefoarcned sk ch an destk etcthhitshisvevcetcotroroonn the coordinate system.

F1F1 8080lblb yy

30 4040 F2

SOLUTION

130 lb

xx

F1 = {80 cos 30° cos 40°i - 80 cos 30° sin 40°j + 80 sin 30°k} lb F1 = {53.1i - 44.5j + 40k} lb F2 = { - 130k} lb FR = F1 + F2 FR = {53.1i - 44.5j - 90.0k} lb Ans. FR = 2(53.1)2 + ( - 44.5)2 + ( - 90.0)2 = 114 lb 53.1 ≤ = 62.1° 113.6

Ans.

a = cos-1 ¢ b = cos-1 ¢

- 44.5 ≤ = 113° 113.6

Ans.

g = cos-1 ¢

- 90.0 ≤ = 142° 113.6

Ans.

Ans: FR = 114 lb a = 62.1°

64 64


b = 113° g = 142°

65 65


2–42. z

Specify the coordinate direction angles of F1 and F2 and express each force as a Cartesian vector. F1

80 lb yy

30 40 F2

SOLUTION

130 lb

x

F1 = {80 cos 30° cos 40°i - 80 cos 30° sin 40°j + 80 sin 30°k} lb F1 = {53.1i - 44.5j + 40k} lb

Ans.

a1 = cos-1 ¢

53.1 ≤ = 48.4° 80

Ans.

b 1 = cos-1 ¢

- 44.5 ≤ = 124° 80

Ans.

g1 = cos-1 a

40 b = 60° 80

Ans.

F2 = { - 130k} lb

Ans.

a2 = cos-1 ¢

0 ≤ = 90° 130

Ans.

b 2 = cos-1 ¢

0 ≤ = 90° 130

Ans.

g2 = cos-1 ¢

- 130 ≤ = 180° 130

Ans.

Ans: F1 = {53.1i - 44.5j + 40k} lb a 1 = 48.4° 66 66


b1 g1 F2 a2 b2 g2

67 67

= = = = = =

124° 60° { - 130k} lb 90° 90° 180°


2–43. z

Th ifsosrucbejeicnted s E xeprsecsrseweaecyhe Catorttehseiatnwovefo ctrocer reesns ea oirnce inthC arretesu silatannvtecftoorrcefo fsohromwna.nEdxpth d ectehrfm e . rF minadndthth mdieneatnhde eenmdaegtneirtu re . F hn egm c osourldtainattfeordcie re ctiinodnta leasgonfittuhdeera en sudltcaonotrfdoin rcaet.e direction angles of the resultant force.

F1

300 N

60 120

45 y 45

SOLUTION 60

F1 = 300( - cos 60° sin 45° i + cos 60° cos 45° j + sin 60°k)

x

= { - 106.07 i + 106.07 j + 259.81 k} N F2

= { - 106 i + 106 j + 260 k} N

500 N

Ans.

F2 = 500(cos 60° i + cos 45° j + cos 120° k) = {250.0 i + 353.55 j - 250.0k} N = {250 i + 354 j - 250 k} N

Ans.

FR = F1 + F2 = - 106.07 i + 106.07 j + 259.81 k + 250.0 i + 353.55 j 250.0 k = 143.93 i + 459.62 j + 9.81k = {144 i + 460 j + 9.81 k} N 2

2

Ans. 2

Ans.

FR = 2143.93 + 459.62 + 9.81 = 481.73 N = 482 N

uF R =

FR 143.93i + 459.62j + 9.81k = = 0.2988i + 0.9541j + 0.02036k FR 481.73

cos a = 0.2988

a = 72.6°

Ans.

cos b = 0.9541

b = 17.4°

Ans.

cos g

g

Ans.

0.02036

88.8°

Ans: F1 = { - 106i + 106j + 260k} N F2 = {250i + 354j - 250k} 68 68


N FR = {144i + 460j + 9.81k} N FR = 482 N a = 72.6° b = 17.4° g = 88.8°

69 69


*2–44. z

Determine the coordinate direction angles of F1.

F1

300 N

60 120

45 y 45

SOLUTION 60

F1 = 300( - cos 60° sin 45° i + cos 60° cos 45° j + sin 60° k)

x

= { - 106.07 i + 106.07 j + 259.81 k} N

F2

500 N

= { - 106 i + 106 j + 260 k} N u1 =

F1 = - 0.3536 i + 0.3536 j + 0.8660 k 300

a1 = cos-1 ( - 0.3536) = 111°

Ans.

-1

b 1 = cos (0.3536) = 69.3°

Ans.

-1

g1 = cos (0.8660) = 30.0°

Ans.

Ans:

70 70


a1 = 111° b1 = 69.3° g1 = 30.0°

71 71


2–45. z

Determine the magnitude and coordinate direction angles of F3 so that the resultant of the three forces acts along the positive y axis and has a magnitude of 600 lb.

F3

F1

SOLUTION

40

FRx = ©Fx ;

0 = - 180 + 300 cos 30° sin 40° + F3 cos a

FRy = ©Fy ;

600 = 300 cos 30° cos 40° + F3 cos b

FRz = ©Fz ;

0 = - 300 sin 30° + F3 cos g

y

30 x

cos 2 a + cos 2 b + cos 2 g = 1

F2

300 lb

Solving: F3 = 428 lb

Ans.

a = 88.3°

Ans.

b = 20.6°

Ans.

g = 69.5°

Ans.

Ans:

72 72

180 lb


F3 = 428 lb a = 88.3° b = 20.6° g = 69.5°

73 73


2–46. z

Determine the magnitude and coordinate direction angles of F3 so that the resultant of the three forces is zero.

F3

F1

SOLUTION

40

FRx = ©Fx;

0 = - 180 + 300 cos 30° sin 40° + F3 cos a

FRy = ©Fy;

0 = 300 cos 30° cos 40° + F3 cos b

FRz = ©Fz;

0 = - 300 sin 30° + F3 cos g

y

30 x

cos 2a + cos 2 b + cos 2g = 1

F2

300 lb

Solving: F3 = 250 lb

Ans.

a = 87.0°

Ans.

b = 143°

Ans.

g = 53.1°

Ans.

Ans:

74 74

180 lb


F3 = 250 lb a = 87.0° b = 143° g = 53.1°

75 75


2–47. z

Determine the magnitude and coordinate direction angles of the resultant force, and sketch this vector on the coordinate system.

F2

125 N

3

5 4

y 60

20 45

Solution x

Cartesian Vector Notation. For F1 and F2, F1 = 400 (cos 45°i + cos 60°j - cos 60°k) = {282.84i + 200j 200k} N 4 4 3 F2 = 125 c (cos 20°)i - (sin 20°)j + k d = {93.97i - 34.20j + 75.0k} 5 5 5 Resultant Force. FR = F1 + F2 = {282.84i + 200j - 200k} + {93.97i - 34.20j + 75.0k} = {376.81i + 165.80j 125.00k} N The magnitude of the resultant force is FR = 2(FR)2x + (FR )2y + (FR )2z = 2376.812 + 165.802 + ( - 125.00)2 = 430.23 N = 430 N

Ans.

The coordinate direction angles are cos a =

cos b =

(FR)x FR (FR)y

=

376.81 ; 430.23

a = 28.86° = 28.9°

Ans.

=

165.80 ; 430.23

b = 67.33° = 67.3°

Ans.

g = 106.89° = 107°

Ans.

FR

- 125.00

(FR)z cos g =

FR

=

430.23

;

76 76

60 F1

400 N


Ans: FR = 430 N a = 28.9° b = 67.3° g = 107°

77 77


*2–48. z

Determine the magnitude and coordinate direction angles of the resultant force, and sketch this vector on the coordinate system. F2

525 N 60

120

45 y x

Solution

5

4 3

Cartesian Vector Notation. For F1 and F2,

F1

450 N

3 4 F1 = 450 a j - k b = {270j - 360k} N 5 5 F2 = 525 (cos 45°i + cos 120°j + cos 60°k) = {371.23i - 262.5j + 262.5k} N Resultant Force. FR = F1 + F2 = {270j - 360k} + {371.23i - 262.5j + 262.5k} = {371.23i + 7.50j 97.5k} N The magnitude of the resultant force is FR = 2(FR)2x + (FR )2y + (FR )2z = 2371.232 + 7.502 + ( - 97.5)2 = 383.89 N = 384 N

Ans.


cos b =

cos g =

(FR)x

=

371.23 ; 383.89

a = 14.76° = 14.8°

Ans.

=

7.50 ; 383.89

b = 88.88° = 88.9°

Ans.

(FR)z = - 97.5 ; 383.89 FR

g = 104.71° = 105°

Ans.

FR (FR)y FR

Ans: FR = 384 N

78 78


a = 14.8° b = 88.9° g = 105°

71 71


2–49. z

Determine the magnitude and coordinate direction angles a1, b1, g1 of F1 so that the resultant of the three forces acting on the bracket is FR = 5 -350k 6 lb. x

F3

g1

30 F2

200 lb a1

SOLUTION

F1

F1 = Fx i + Fy j + Fz k

x

F2 = - 200 j F3 = - 400 sin 30° i + 400 cos 30° j = - 200 i + 346.4 j FR = ©F - 350 k = Fx i + Fy j + Fz k - 200 j - 200 i + 346.4 j 0 = Fx - 200 ;

b1

Fx = 200 lb

0 = Fy - 200 + 346.4 ;

Fy = - 146.4 lb

Fz = - 350 lb F1 = 2(200)2 + ( - 146.4)2 + ( - 350)2 F1 = 425.9 lb = 429 lb

Ans.

a1 = cos-1 a

200 b = 62.2° 428.9

Ans.

b 1 = cos-1 a

- 146.4 b = 110° 428.9

Ans.

g1 = cos-1

- 350 428.9

Ans.

= 145°

70 70

400 lb

y


Ans: F1 = a1 = b1 = g1 =

71 71

429 lb 62.2° 110° 145°


2–50. z

If the resultant force FR has a magnitude of 150 lb and the coordinate direction angles shown, determine the magnitude of F2 and its coordinate direction angles.

F2

FR

150 lb

g 130 y F1

80 lb

120

Solution

x

Cartesian Vector Notation. For FR, g can be determined from cos2 a + cos2 b + cos2 g = 1 cos2 120° + cos2 50° + cos2 g = 1 cos g = { 0.5804 Here g 6 90°, then g = 54.52° Thus FR = 150(cos 120°i + cos 50°j + cos 54.52°k) = { - 75.0i + 96.42j + 87.05k} lb Also F1 = {80j} lb Resultant Force. FR = F1 + F2 { - 75.0i + 96.42j + 87.05k} = {80j} + F2 F2 = { - 75.0i + 16.42j + 87.05k} lb Thus, the magnitude of F2 is F2 = 2(F2)x + (F2)y + (F2)z = 2( - 75.0)2 + 16.422 + 87.052 = 116.07 lb = 116 lb

Ans.

And its coordinate direction angles are cos a2 =

cos b2 =

cos g2 =

(F2)x

=

- 75.0 ; 116.07

a2 = 130.25° = 130°

Ans.

=

16.42 ; 116.07

b2 = 81.87° = 81.9°

Ans.

87.05 (F2)z = ; 116.07 F2

g2 = 41.41° = 41.4°

Ans.

F2 (F2)y F2

Ans: F2 = 116 lb a2 = 130° b 2 = 81.9° 72 72


g2 = 41.4°

73 73


2–51. z

Express each force as a Cartesian vector.

F3

200 N F2

F1

5

90 N 3

60

150 N

y

4

45

Solution Cartesian Vector Notation. For F1, F2 and F3,

x

4 3 F1 = 90 a i + k b = {72.0i + 54.0k} N 5 5

Ans.

F2 = 150 (cos 60° sin 45°i + cos 60° cos 45°j + sin 60°k) = {53.03i + 53.03j + 129.90k} N = {53.0i + 53.0j + 130k} N

Ans.

F3 = {200 k}

Ans.

Ans: F1 = {72.0i + 54.0k} N F2 = {53.0i + 53.0j + 130k} N F3 = {200 k}

74 74


*2–52. z


F3

200 N F2

F1

Solution

5

90 N 3

150 N

60

y

4

45

Cartesian Vector Notation. For F1, F2 and F3, 4 3 F1 = 90 a i + k b = {72.0i + 54.0k} N 5 5 x

F2 = 150 (cos 60° sin 45°i + cos 60° cos 45°j + sin 60°k) = {53.03i + 53.03j + 129.90k} N F3 = {200 k} N Resultant Force. F = F1 + F2 + F3 = (72.0i + 54.0k) + (53.03i + 53.03j + 129.90k) + (200k) = {125.03i + 53.03j + 383.90} N The magnitude of the resultant force is FR = 2(FR)2x + (FR )2y + (FR )2z = 2125.032 + 53.032 + 383.902 = 407.22 N = 407 N

Ans.

And the coordinate direction angles are cos a =

cos b =

cos g =

(FR)x FR (FR)y FR (FR)z FR

=

125.03 ; 407.22

a = 72.12° = 72.1°

Ans.

=

53.03 ; 407.22

b = 82.52° = 82.5°

Ans.

=

383.90 ; 407.22

g = 19.48° = 19.5°

Ans.

Ans: FR = a = b = g =

75 75

407 N 72.1° 82.5° 19.5°


2–53. z

The spur gear is subjected to the two forces. Express each force as a Cartesian vector. F2

180 lb

60

135

60

y

SOLUTION

x 25

7 24 (50)j (50)k = {14.0j F1 = 25 25 48.0k} lb

Ans.

24 7

F1

F2 = 180 cos 60°i + 180 cos 135°j + 180 cos 60°k = {90i - 127j + 90k} lb

Ans.

Ans:

76 76

50 lb


F1 = {14.0j - 48.0k} lb F2 = {90i - 127j + 90k} lb

77 77


2–54. z

The spur gear is subjected to the two forces. Determine the resultant of the two forces and express the result as a Cartesian vector.

F2

180 lb

60

135

60

y

SOLUTION x 25

FRx = 180 cos 60° = 90

24 7

7 FRy =

25

FRz = -

(50) + 180 cos 135° = - 113

F1

24 (50) + 180 cos 60° = 42 25

FR = {90i - 113j + 42k} lb

Ans.

78 78

50 lb


Ans: FR = {90i - 113j + 42k} lb

79 79


2–55. z


F1 = 400 N 60 135 20

Solution

60

60 y

x

Cartesian Vector Notation. For F1 and F2,

F2

F1 = 400 (sin 60° cos 20°i - sin 60° sin 20°j + cos 60°k)

500 N

= {325.52i - 118.48j + 200k} N

F2 = 500 (cos 60°i + cos 60°j + cos 135°k) = {250i + 250j - 353.55k} N Resultant Force. FR = F1 + F2 = (325.52i - 118.48j + 200k) + (250i + 250j - 353.55k) = {575.52i + 131.52j - 153.55 k} N The magnitude of the resultant force is FR = 2(FR)2x + (FR )2y + (FR )2z = 2575.522 + 131.522 + ( - 153.55)2 = 610.00 N = 610 N

Ans.


cos b =

(FR)x FR (FR)y

=

575.52 610.00

a = 19.36° = 19.4°

Ans.

=

131.52 610.00

b = 77.549° = 77.5°

Ans.

g = 104.58° = 105°

Ans.

FR

- 153.55

(FR)z cos g =

FR

=

610.00

Ans: FR = 610 N a = 19.4° b = 77.5°

80 80


g = 105°

81 81


*2–56. y

Determine the length of the connecting rod AB by first formulating a position vector from A to B and then determining its magnitude.

B

Solution Position Vector. The coordinates of points A and B are A( - 150 cos 30°, - 150 sin 30°) mm and B(0, 300) mm respectively. Then

300 mm

rAB = [0 - ( - 150 cos 30°)]i + [300 - ( - 150 sin 30°)]j

O

= {129.90i + 375j} mm

30

A

Thus, the magnitude of rAB is rAB = 2129.902 + 3752 = 396.86 mm = 397 mm

Ans.

82 82

150 mm

x


Ans: rAB = 397 mm

83 83


2–57. z

Express force F as a Cartesian vector; then determine its coordinate direction angles.

A F

135 lb

70 30

Solution

y

5 ft

rAB = (5 + 10 cos 70° sin 30°)i

B

+ ( - 7 - 10 cos 70° cos 30°)j - 10 sin 70°k rAB = {6.710i - 9.962j - 9.397k} ft

7 ft x

rAB = 2(6.710)2 + ( - 9.962)2 + ( - 9.397)2 = 15.25 rAB u AB =

rAB

= (0.4400i - 0.6532j - 0.6162k)

F = 135u AB = (59.40i - 88.18j - 83.18k) = {59.4i - 88.2j - 83.2k} lb

Ans.

a = cos - 1 a

59.40 b = 63.9° 135

Ans.

b = cos - 1 a

- 88.18 b = 131° 135

Ans.

g = cos - 1 a

- 83.18 b = 128° 135

Ans.

Ans: F = {59.4i - 88.2j - 83.2k} lb

84 84


a = 63.9° b = 131° g = 128°

85 85


2–58. z

Express each force as a Cartesian vector, and then determine the magnitude and coordinate direction angles of the resultant force.

C 13 12 5

O

F1

80 lb

2.5 ft y 4 ft

A F2

50 lb

x

Solution

6 ft

rAC = e - 2.5 i - 4 j +

12 (2.5) k f ft 5 2 ft

rAC F1 = 80 lb a r b = - 26.20 i - 41.93 j + 62.89 k

B

AC

= { - 26.2 i - 41.9 j + 62.9 k} lb

Ans.

rAB = {2 i - 4 j - 6 k} ft F2 = 50 lb a

rAB b = 13.36 i - 26.73 j - 40.09 k rAB

= {13.4 i - 26.7 j - 40.1 k} lb

Ans.

FR = F1 + F2 = - 12.84 i - 68.65 j + 22.80 k = { - 12.8 i - 68.7 j + 22.8 k } lb FR = 2( - 12.84)2 ( - 68.65)2 + (22.80)2 = 73.47 = 73.5 lb

Ans.

a = cos - 1 a

- 12.84 b = 100° 73.47

Ans.

b = cos - 1 a

- 68.65 b = 159° 73.47

Ans.

g = cos - 1 a

22.80 b = 71.9° 73.47

Ans.

Ans: F1 = { - 26.2 i - 41.9 j + 62.9 k} lb F2 = {13.4 i - 26.7 j - 40.1 k} lb FR = 73.5 lb

86 86


a = 100° b = 159° g = 71.9°

81 81


2–59. z

ABB iiss 99 m long, If F = 5335500ii -- 225500jj -- 445500kk66NNaannddccaabbleleA determine the x,, yy,,zzccoooorrddin inaatteessooffppooin inttA A. .

A F z

B

SOLUTION

x

Position Vector: The position vector rAB, directed from point A to point B, is given by

x

y

y

rAB = [0 - x ]i + (0 - y)j + (0 - z)k = - xi - yj - zk Unit Vector: Knowing the magnitude of rAB is 9 m, the unit vector for rAB is given by uAB =

- xi - yj - zk rAB = rAB 9

The unit vector for force F is uF =

F = F

350i - 250j - 450k

= 0.5623i - 0.4016j - 0.7229k

3350 2 + ( - 250)2 + ( - 450)2

Since force F is also directed from point A to point B, then uAB = uF - xi - yj - zk = - 0.5623i 9

0.4016j - 0.7229k

Equating the i, j, and k components, x = - 0.5623 9

x = - 5.06 m

Ans.

-y = - 0.4016 9

y = 3.61 m

Ans.

-z = 0.7229 9

z = 6.51 m

Ans.

Ans: 80 80


x = - 5.06 m y = 3.61 m z = 6.51 m

81 81


*2–60. z

The 8-m-long cable is anchored to the ground at A. If x = 4 m and y = 2 m, determine the coordinate z to the highest point of attachment along the column.

B

Solution r = {4i + 2j + zk} m r = 2(4)2 + (2)2 + (z)2 = 8

z

z = 6.63 m

Ans. y x

82 82

y

A

x


Ans: 6.63 m

83 83


2–61. z

The 8-m-long cable is anchored to the ground at A. If z = 5 m, determine the location +x, +y of the at A. Choose a value such that x = y.

B

Solution r = {xi + yj + 5k} m r = 2(x)2 + (y)2 + (5)2 = 8

z

x = y, thus 2x2 = 82 - 52 x = y = 4.42 m

y

Ans. x

84 84

y

A

x


Ans: x = y = 4.42 m

85 85


2–62. z

Express each of the forces in Cartesian vector form and then determine the magnitude and coordinate direction angles of the resultant force

0.75 m A FAB FAC

250 N

400 N

3m

Solution

y 2m

Unit Vectors. The coordinates for points A, B and C are (0, - 0.75, 3) m, B(2 cos 40°, 2 sin 40°, 0) m and C(2, - 1, 0) m, respectively. u AB =

C

(2 cos 40° - 0)i + [2 sin 40° - ( - 0.75)]j + (0 - 3)k rAB = rAB 2(2 cos 40° - 0)2 + [2 sin 40° - ( - 0.75)]2 + (0 - 3)2 x

= 0.3893i + 0.5172j - 0.7622k u AC =

(2 - 0)i + [ - 1 - ( - 0.75)]j + (0 - 3)k rAC = rAC 2(2 - 0)2 + [ - 1 - ( - 0.75)]2 + (0 - 3)2 = 0.5534i - 0.0692j - 0.8301k

Force Vectors FAB = FAB u AB = 250 (0.3893i + 0.5172j 0.7622k) = {97.32i + 129.30j 190.56k} N

Ans.

= {97.3i + 129j - 191k} N

FAC = FAC u AC = 400 (0.5534i - 0.06917j 0.8301k) = {221.35i - 27.67j - 332.02k} N

Ans.

= {221i - 27.7j - 332k} N Resultant Force FR = FAB + FAC = {97.32i + 129.30j - 190.56k} + {221.35i - 27.67j - 332.02k} = {318.67i + 101.63j - 522.58 k} N The magnitude of FR is FR = 2(FR)2x + (FR )2y + (FR )2z = 2318.672 + 101.632 + ( - 522.58)2 = 620.46 N = 620 N And its coordinate direction angles are cos a =

cos b =

(FR)x

=

318.67 ; 620.46

a = 59.10° = 59.1°

Ans.

=

b = 80.57° = 80.6°

FR

101.63 ; 620.46

Ans.

(FR)z

- 522.58

FR (FR)y

86 86

40

2m B


cos g =

FR

=

620.46

;

g = 147.38° = 147°

Ans.

87 87

FAB N FAC N FR a b g

= {97.3i - 129j - 191k} = {221i - 27.7j - 332k} = = = =

620 N 59.1° 80.6° 147°


2–63. z

If FB = 560 N and FC = 700 N, determine the magnitude and coordinate direction angles of the resultant force acting on the flag pole.

A 6m

FB FC

SOLUTION Force Vectors: The unit vectors u and u of F and F must be determined first. From Fig. a,

u

r

(2

0)i

(2 u

r

( 3

0)

(3

2

( 3

0)i

(3

0)j

(2

0)

2

(0

0)

2

0)j

(2

6)k

(0

(0

0)

2

6)

2

(0

6)

7

2

6)k 2

3

i

6

j

7

3m

7

3 i 7

2 j 7

6 k 7

u

560

2 i 7

3 j 7

6 k 7

160i

240j

480k N

F

u

700

3 i 7

2 j 7

6 k 7

300i

200j

600k N

x

Resultant Force: F

460i

(160i 40j

240j

480k)

(300i

200j

600k)

1080k N

The magnitude of F is (

2

)

(460)2

(

)

2

( 40)2

(

)

2

( 1080)2

1174.56 N

1.17 kN

Ans.

The coordinate direction angles of F are cos

1

cos

1

cos

1

(

)

(

)

(

)

3m

2m

F

F

B

k

Thus, the force vectors F and F are given by

F

2m

cos

1

460 1174.56

66.9°

Ans.

cos

1

40 1174.56

92.0°

Ans.

cos

1

1080 1174.56

157°

Ans.

88 88

C

y


Ans: FR = 1.17 kN a = 66.9° b = 92.0° g = 157°

89 89


*2–64. z

If FB = 700 N, and FC = 560 N, determine the magnitude and coordinate direction angles of the resultant force acting on the flag pole.

A 6m

FB FC

SOLUTION Force Vectors: The unit vectors u and u of F and F must be determined first. From Fig. a,

u

r

(2

0)i

( 3

0)j

(0

2

(2

6)k

2

0)

( 3

0)

2 7

2

(0

3

i

6

j

7

2m B 3m

k

7

6) x

u

(3

r

0)i

(2

0)j

(0

6)k

(3 0)2 (2 0)2 (0 6)2 Thus, the force vectors F and F are given by

3

2

i

7

7

j

6

C

k

7

F

u

700

2 i 7

3 j 7

6 k 7

200i

300j

600k N

F

u

560

3 i 7

2 j 7

6 k 7

240i

160j

480k N

Resultant Force: F

F

F

440i

(200i 140j

300j

600k)

(240i

160j

480k)

1080k N

The magnitude of F is (

2

)

(440)2

(

)

2

( 140)2

(

)

2

( 1080)2

1174.56 N

1.17 kN

Ans.

The coordinate direction angles of F are cos

1

cos

1

cos

1

(

)

(

)

(

)

cos

1

440 1174.56

68.0°

Ans.

cos

1

140 1174.56

96.8°

Ans.

cos

1

1080 1174.56

157°

Ans.

90 90

3m

2m

y


Ans: FR = 1.17 kN a = 68.0° b = 96.8° g = 157°

91 91


2–65. z

The plate is suspended using the three cables which exert the forces shown. Express each force as a Cartesian vector. A

FCA

Solution

FBA

350 lb

FDA

C

rBA FBA = 350 a

r BA

5 b = 350 a -

6

16.031

i +

FCA = 500 a

FDA

r CA

3 b = 500 a

14.629

16.031

j +

16.031

kb

B 3 ft

Ans. 3

i +

14.629

400 lb

14

= { - 109 i + 131 j + 306 k} lb rCA

500 lb

14 ft

D

3 ft

6 ft

14 j +

14.629

kb

x

= {103 i + 103 j + 479 k} lb rDA 2 6 14 i j + kb = 400 a r b = 400 a 15.362 15.362 15.362 DA

Ans.

= { - 52.1 i - 156 j + 365 k} lb

Ans.

92 92

y 3 ft 2 ft

3 ft


Ans: FBA = { - 109 i + 131 j + 306 k} lb FCA = {103 i + 103 j + 479 k} lb FDA = { - 52.1 i - 156 j + 365 k} lb

93 93


2–66. z

Represent each cable force as a Cartesian vector. 2m

C

E

2m B

FE

Solution

FC

350 N 3m

400 N

rC = (0 - 5)i + ( - 2 - 0)j + (3 - 0)k = { - 5i - 2j + 3k} m

FB

400 N D

rC = 2( - 5)2 + ( - 2)2 + 32 = 238 m

2m

rB = (0 - 5)i + (2 - 0)j + (3 - 0)k = { - 5i + 2j + 3k} m rB = 2( - 5)2 + 22 + 32 = 238 m

A

y

3m

x

rE = (0 - 2)i + (0 - 0)j + (3 - 0)k = { - 2i + 0j + 3k} m rE = 2( - 2)2 + 02 + 32 = 213 m r F = Fu = F a b r FC = 400 a

- 5i - 2j + 3k b = { - 324i - 130j + 195k} N 138

Ans.

FB = 400 a

- 5i + 2j + 3k b = { - 324i + 130j + 195k} N 138

Ans.

FE = 350 a

- 2i + 0j + 3k b = { - 194i + 291k} N 113

Ans.

Ans: FC = { - 324i - 130j + 195k} N FB = { - 324i + 130j + 195k} N FE = { - 194i + 291k} N 94 94


2–67. z

Determine the magnitude and coordinate direction angles of the resultant force of the two forces acting at point A.

2m

C

E

2m B

FE

Solution

FC

350 N 3m

400 N

rC = (0 - 5)i + ( - 2 - 0)j + (3 - 0)k = { - 5i - 2j + 3k}

FB

400 N D

rC = 2( - 5)2 + ( - 2)2 + (3)2 = 238 m

2m

rC ( - 5i - 2j + 3k) FC = 400 a b = 400 a b rC 138

A

y

3m

x

FC = ( - 324.4428i - 129.777j + 194.666k) rB = (0 - 5)i + (2 - 0)j + (3 - 0)k = { - 5i + 2j + 3k} rB = 2( - 5)2 + 22 + 32 = 238 m FB = 400 a

rB ( - 5i + 2j + 3k) b rB b = 400 a 138

FB = ( - 324.443i + 129.777j + 194.666k) FR = FC + FB = ( - 648.89i + 389.33k) FR = 2( - 648.89)2 + (389.33)2 + 02 = 756.7242 FR = 757 N

Ans.

a = cos-1 a

- 648.89 b = 149.03 = 149° 756.7242

Ans.

b = cos-1 a

0 b = 90.0° 756.7242

Ans.

g = cos-1 a

389.33 b = 59.036 = 59.0° 756.7242

Ans.

Ans: FR = 757 N a = 149° b = 90.0°

95 95


g = 59.0°

96 96


*2–68. z

The force F has a magnitude of 80 lb and acts at the midpoint C of the rod. Express this force as a Cartesian vector.

B

6 ft

Solution

C

rAB = ( - 3i + 2j + 6k) rCB =

F

80 lb

1 r = ( - 1.5i + 1j + 3k) 2 AB O 3 ft

rCO = rBO + rCB = - 6k - 1.5i + 1j + 3k

A

= - 1.5i + 1j - 3k

2 ft x

rCO = 3.5 rCO F = 80 a

rCO

b = { - 34.3i + 22.9j - 68.6k} lb

Ans.

97 97

y


Ans: F = { - 34.3i + 22.9j - 68.6k} lb

91 91


2–69. z

The load at A creates a force of 60 lb in wire AB. Express this force as a Cartesian vector.

30 5 ft

y B

SOLUTION 10 ft

Unit Vector: First determine the position vector rAB. The coordinates of point B are x

B (5 sin 30°, 5 cos 30°, 0) ft = B (2.50, 4.330, 0) ft

F

Then

A

rAB = 5(2.50 - 0)i + (4.330 - 0)j + [0 - ( - 10)]k6 ft = 52.50i + 4.330j + 10k6 ft rAB = 32.502 + 4.3302 + 10.02 = 11.180 ft uAB =

2.50i + 4.330j + 10k rAB = rAB 11.180 = 0.2236i + 0.3873j + 0.8944k

Force Vector: F = FuAB = 60 50.2236i + 0.3873j + 0.8944k6 lb = 513.4i + 23.2j + 53.7k6 lb

Ans.

90 90

60 lb


Ans: F = {13.4i + 23.2j + 53.7k} lb

91 91


2–70. z

Determine the magnitude and coordinate direction angles of the resultant force acting at point A on the post. FAC

C

150 N

A FAB

200 N

3m

4

3m

3 5

y

Solution

O

2m

Unit Vector. The coordinates for points A, B and C are A(0, 0, 3) m, B(2, 4, 0) m, and C( - 3, - 4, 0) m, respectively.

4m x

rAB = (2 - 0)i + (4 - 0)j + (0 - 3)k = {2i + 4j - 3k} m 2i + 4j - 3k

rAB u AB =

r

2

= 2

2

3

i +

2

22 + 4 + ( - 3)

AB

4

=

j -

229

k

229

229

rAC = ( - 3 - 0)i + ( - 4 - 0)j + (0 - 3)k = { - 3i - 4j - 3k} m - 3i - 4j - 3k

rAC u AC =

r

3

=

AC

4

= 2( - 3)2 + ( - 4)2 + ( - 3)2

3

i 234

j 234

k 234

Force Vectors

FAB = FAB u AB = 200 a

2

4

i +

229

3

j -

229

kb

229

= {74.28i + 148.56j - 111.42k} N FAC = FAC u AC = 150 a -

3

i -

234

4

j -

234

3

kb

234

= { - 77.17i - 102.90j - 77.17k} N Resultant Force FR = FAB + FAC = {74.28i + 148.56j - 111.42k} + { - 77.17i - 102.90j - 77.17k} = { - 2.896i + 45.66j - 188.59 k} N The magnitude of the resultant force is FR = 2(FR)2x + (FR )2y + (FR )2z = 2( - 2.896)2 + 45.66 2 + ( - 188.59)2 = 194.06 N = 194 N

Ans.

And its coordinate direction angles are cos a =

(FR)x - 2.896 = ; FR 194.06 (FR)y

cos b =

FR (FR)z

a = 90.86° = 90.9°

Ans.

b = 76.39° = 76.4°

Ans.

45.66 =

194.06

;

- 188.59 92 92

B


cos g =

FR

=

194.06

;

g = 166.36° = 166°

Ans. Ans: FR = a = b = g =

93 93

194 N 90.9° 76.4° 166°


2–71. Given the three vectors A, B, and A # (B + D) = (A # B) + (A # D).

z

D, show that

A

2m

B 2m x

SOLUTION

2m

Since the component of (B + D) is equal to the sum of the components of B and D, then

y

2m C

D

F 3m

A # (B + D) = A # B + A # D

(QED)

Also, A # (B + D) = (A x i + A y j + A zk) # [(Bx + Dx)i + (By + Dy)j + (Bz + Dz)k] = A x (Bx + Dx) + A y (By + Dy) + A z (Bz + Dz) = (A xBx + A yBy + A zBz) + (A xDx + A yDy + A zDz) = (A # B) + (A # D)

(QED)

94 94

E

600 N


*2–72. z

Determine the magnitudes of the components of F = 600 N acting along and perpendicular to segment DE of the pipe assembly.

A

2m

B 2m x

SOLUTION

2m

Unit Vectors: The unit vectors uEB and uED must be determined first. From Fig. a,

uEB =

rEB (0 - 4)i + (2 - 5)j + [0 - ( - 2)]k = = - 0.7428i - 0.5571j + 0.3714k rEB 2(0 - 4)2 + (2 - 5)2 + [0 - ( - 2)]2

u ED = - j Thus, the force vector F is given by F = FuEB = 600 A - 0.7428i - 0.5571j + 0.3714k) = [ - 445.66i - 334.25j + 222.83k] N Vector Dot Product: The magnitude of the component of F parallel to segment DE of the pipe assembly is (FED)paral = F # u ED = A - 445.66i - 334.25j + 222.83k B # A - j B = ( - 445.66)(0) + ( - 334.25)( - 1) + (222.83)(0) = 334.25 = 334 N

Ans.

The component of F perpendicular to segment DE of the pipe assembly is (FED)per = 2F 2 - (FED)paral2 = 26002 - 334.252 = 498 N

95 95

Ans.

y

2m C

D

F 3m E

600 N


Ans: (FED) = 334 N (FED) # = 498 N

96 96


2–73. z

Determine the angle u between BA and BC. A

B 2m

u

5m 4m

Solution

x

D

Unit Vectors. Here, the coordinates of points A, B and C are A(0, - 2, 0) m, B(0, 0, 0) m and C(3, 4, - 1) m respectively. Thus, the unit vectors along BA and BC are

u BA = - j

uBE =

(3 - 0) i + (4 - 0) j + ( - 1 - 0) k 2(3 - 0)2 + (4 - 0)2 + ( - 1 - 0)2

=

3 226

i +

4 226

j-

1 226

The Angle U Between BA and BC.

#

u BA uBC = ( - j)

3

a

226

226

4 = ( - 1) a

4

i +

j -

1

kb

226

4 b = -

226

226

Then 4 u = cos

-1

(u BA # uBC)

= cos

-1

a-

b = 141.67° = 142° 226

97 97

Ans.

k

F

3 kN

3m C

1m

y


Ans: u = 142°

98 98


2–74. z

Determine the magnitude of the projected component of the 3 kN force acting along axis BC of the pipe. A

B 2m

u

5m 4m

Solution

x

Unit Vectors. Here, the coordinates of points B, C and D are B (0, 0, 0) m, C(3, 4, - 1) m and D(8, 0, 0). Thus the unit vectors along BC and CD are

u BC =

(3 - 0) i + (4 - 0) j + ( - 1 - 0) k 2

2(3 - 0) + (4 - 0)

2

3

=

2

+ ( - 1 - 0)

226

(8 - 3) i + (0 - 4) j + [0 - ( - 1)] k uCD =

4

i +

2(8 - 3)2 + (0 - 4)2 + [0 - ( - 1)]2

226

5 =

1

j -

226

4 i -

242

k

1

242

j +

242

k

Force Vector. For F,

F = FuCD = 3 a

= a

5

4

i -

1

j +

kb

242

242

242

15

12

3

i -

242

j +

242

k b kN

242

Projected Component of F. Along BC, it is

` (FBC) ` =

` F # uBC ` =

15

12

à

i242

15

242

3 ba

242

4

12

j226

4

kb` 226

3

ba 242

1

i+ 226

b + a226

3

kb # a

j+

242

= à

3

b + 226

1 a-

242

b` 226

6 = `-

` = ` - 0.1816 kN ` = 0.182 kN

Ans.

21092 The negative signs indicate that this component points in the direction opposite to that of uBC.

99 99

D

F

3 kN

3m C

1m

y


Ans: ` (FBC) ` = 0.182 kN

10 10 0 0


2–75. z

Determine the angle u between the two cables.

C

4m u F 1

70 N

A 2m

3m

B 2m

Solution

3m

Unit Vectors. Here, the coordinates of points A, B and C are A(2, - 3, 3) m, B(0, 3, 0) and C( - 2, 3, 4) m, respectively. Thus, the unit vectors along AB and AC are

u AB =

(0 - 2)i + [3 - ( - 3)]j + (0 - 3)k 2(0 - 2)2 + [3 - ( - 3)]2 + (0 - 3)2

x

2 6 3 = - i + j - k 7 7 7

( - 2 - 2)i + [3 - ( - 3)]j + (4 - 3)k u AC =

y

3m

2( - 2 - 2)2 + [3 - ( - 3)]2 + (4 - 3)2

4 = -

6

253

i +

253

1 j +

253

k

The Angle U Between AB and AC. 2 6 3 4 u AB # u AC = a - i + j - k b # a 7

= a-

=

7

7

6 i +

253

1 j +

253

kb

253

2 4 6 6 3 1 ba b + a b + a - ba b 7 7 7 253 253 253

41 7 253

Then

u = cos - 1 ( u AB # u AC ) = cos-1a

41

b = 36.43° = 36.4° 7 253

Ans.

Ans: u = 36.4° 96 96


*2–76. z

Determine the magnitude of the projection of the force F1 along cable AC.

C

4m u F 1

70 N

A 2m

3m

B

Unit Vectors. Here, the coordinates of points A, B and C are A(2, - 3, 3)m, B(0, 3, 0) and C( - 2, 3, 4) m, respectively. Thus, the unit vectors along AB and AC are

u AB =

u AC =

3m

2m

Solution

(0 - 2)i + [3 - ( - 3)]j + (0 - 3)k 2(0 - 2)2 + [3 - ( - 3)]2 + (0 - 3)2

( - 2 - 2)i + [3 - ( - 3)]j + (4 - 3)k 2

2

2( - 2 - 2) + [3 - ( - 3)] + (4 - 3)

2

3m x

2 6 3 = - i + j - k 7 7 7 4

= -

253

i +

6 253

j +

1 253

k

Force Vector, For F1, 2 6 3 F1 = F1 u AB = 70 a - i + j - k b = { - 20i + 60j - 30k} N 7 7 7 Projected Component of F1. Along AC, it is (F1)AC = F1 # u AC = ( - 20i + 60j - 30k) # a 4 = ( - 20)a -

4 253

6 b + 60a

253

j +

253

1

kb

253

1 b + ( - 30)a

253

6

i +

b 253

= 56.32 N = 56.3 N

Ans.

The positive sign indicates that this component points in the same direction as u AC .

Ans: u = 36.4° 97 97

y


Ans: (F1)AC = 56.3 N

Ans: u = 36.4° 99 99


2–77. z

Determine the angle u between the pole and the wire AB.

3 ft

2 ft

x

SOLUTION Position Vector:

θ

2 ft

A

y

2 ft

rAC = 5 - 3j6 ft

B

rAB = 512 - 02i + 12 - 32j + 1 - 2 - 02k6 ft = 52i - 1j - 2k6 ft

The magnitudes of the position vectors are rAC = 3.00 ft

rAB = 22 2 + 1 - 122 + 1 - 222 = 3.00 ft

The Angles Between Two Vectors U: The dot product of two vectors must be determined first. rAC # rAB = 1 - 3j2 # 12i - 1j - 2k2 = 0122 + 1 - 321 - 12 + 01 - 22 = 3 Then, u = cos-1

rAO # rAB rAO rAB

= cos-1

3 3.00 3.00

= 70.5°

Ans.

Ans: u = 70.5° 98 98


2–78. z

ag Determine tthhee mm an gintiutd uedeofofthethp eropjercotjieocntioonf tohfe force aFlo= ng6t0h0e N u aloisn. g the u axis.

F

600 N

A 4m O

SOLUTION

4m 2m y 30

Unit Vectors: The unit vectors uOA and uu must be determined first. From Fig. a, r ( - 2 - 0)i + (4 - 0)j + (4 - 0)k 1 2 2 uOA = OA = = - i + j + k rOA 3 2 2 2 3 3 3( - 2 - 0) + (4 - 0) + (4 - 0)

x

u

uu = sin30°i + cos30°j Thus, the force vectors F is given by F = F uOA = 600 a -

1 2 2 i - j + k b = 5 - 200i + 400j + 400k6 N 3 3 3

Vector Dot Product: The magnitude of the projected component of F along the u axis is Fu = F # uu = ( - 200i + 400j + 400k) # (sin30°i + cos 30°j) = ( - 200)(sin30°) + 400(cos 30°) + 400(0) = 246 N

Ans.

Ans: u = 70.5° 99 99


Ans: Fu = 246 N

Ans: u = 70.5° 10 10 1 1


2–79. z

Determine the magnitude of the projected component of the 100-lb force acting along the axis BC of the pipe.

A B 3 ft

u

8 ft x

Solution r gBC =

56i^

=

5- 6i

2 ft C

^ ^ + 8j + 2k6

2( - 6)2 + 82 + 22

5- 58.83i^

+ 78.45j^ + 19.61^k6 lb

r r m Fp = F r BC = F

r gBC |r gBC |

4 ft D

^ ^ + 4j - 2k6 ft ^

r F = 100

6 ft

=

- 78.45

= - 10.48

7.483

Fp = 10.5 lb

Ans.

100100

F 100 lb

y


Ans: Fp = 10.5 lb

101101


*2–80. z

Determine the angle u between pipe segments BA and BC.

A B 3 ft

u

8 ft x

Solution

6 ft 4 ft D

2 ft

r gBC = r g BA

=

56î + 4^j - 2k^6 ft ^

C

5- 3i 6 ft r g

#r g

- 18 BC BA u = cos-1 a r r b = cos-1a b | g || g | 22.45 BC

u = 143°

BA

Ans.

102102

F 100 lb

y


Ans: u = 143°

103103


2–81. z

Determine the angle u between the two cables. 8 ft C

10 ft

4 ft

10 ft

Solution

y FAB

u

-1

u = cos

a

rAC # rAB

= cos-1 a

6 ft 12 lb

b

8 ft

rAC rAB = cos-1 c

B

(2 i - 8 j +

10 k) # ( - 6 i +

x

2 j + 4 k)

122 + ( - 8)2 + 102 1( - 6)2 + 22 + 42

d

12 b 96.99

u = 82.9°

Ans.

104104

A


Ans: u = 82.9°

105105


2–82. z

Determine the projected component of the force acting in the direction of cable AC. Express the result as a Cartesian vector.

8 ft C 10 ft

B 4 ft

10 ft

Solution

y FAB

u

6 ft 12 lb

rAC = {2 i - 8 j + 10 k} ft 8 ft x

rAB = { - 6 i + 2 j + 4 k} ft rAB FAB = 12 a r b = 12 a AB

6 7.483

2 i +

7.483

4 j +

7.483

kb

FAB = { - 9.621 i + 3.207 j + 6.414 k} lb u AC =

2 8 10 i j + k 12.961 12.961 12.961

Proj FAB = FAB # u AC = - 9.621 a

2 8 10 b b + 3.207 a b + 6.414 a 12.961 12.961 12.961

= 1.4846 Proj FAB = FAB u AC Proj FAB = (1.4846) c

2 8 10 i j + kd 12.962 12.962 12.962

Proj FAB = {0.229 i - 0.916 j + 1.15 k} lb

Ans.

106106

A


Ans: Proj FAB = {0.229 i - 0.916 j + 1.15 k} lb

107107


2–83. z

Determine the angles u and f between the flag pole and the cables AB and AC.

1.5 m B

2m

C

4m

SOLUTION

FB 6m

rA C = { - 2i - 4j + 1k} m ;

rA C = 4.58 m

rAB = {1.5i - 4j + 3k} m;

rAB = 5.22 m

rA O = { - 4j - 3k} m;

FC

40 N

55 N uf

A

O 3m

rA O = 5.00

m rA B # rA O = (1.5)(0) + ( - 4)( - 4) + (3)( - 3) =

4m x y

7 u = cos - 1 ¢

rAB # rAO rAB rAO

= cos - 1 ¢

≤

7 ≤ = 74.4° 5.22(5.00)

Ans.

rAC # rAO = ( - 2)(0) + ( - 4)( - 4) + (1)( - 3) = 13 rAC # rAO f = cos - 1 a

= cos - 1 a

rAC rAO

b

13 b = 55.4° 4.58(5.00)

Ans.

108108


Ans: u = 74.4° f = 55.4°

109109


*2–84. z

Determine the magnitudes of the components of F acting along and perpendicular to segment BC of the pipe assembly.

A

3 ft

4 ft 2 ft

4 ft x

SOLUTION

u

F

{30i

45j

4 ft C

(3 - 7)i + (4 - 6)j + [0 - ( - 4)]k rCB 2 1 2 = = - i- j + k rCB 2 2 2 3 3 3 3(3 - 7) + (4 - 6) + [0 - ( - 4)]

Vector Dot Product: The magnitude of the projected component of F parallel to segment BC of the pipe assembly is 1 2 2 (FBC)pa = F # uCB = (30i - 45j + 50k) # ¢- i - j + k ≤ 3 3 3 2 1 2 = (30) ¢ - ≤ + ( - 45) ¢ - ≤ + 50 ¢ ≤ 3 3 3 = 28.33 lb = 28.3 lb

Ans.

The magnitude of F is F = 330 2 + ( - 45) 2 + 50 2 = 25425 lb. Thus, the magnitude of the component of F perpendicular to segment BC of the pipe assembly can be determined from (FBC)per = 3F2 - (FBC)pa 2 = 25425 28.332

= 68.0 lb

110110

y

50k} lb

Unit Vector: The unit vector uCB must be determined first. From Fig. a, uCB =

B

Ans.


Ans:

( FBC ) = 28.3 lb ( FBC ) # = 68.0 lb

111111


2–85. z

Determine the magnitude of the projected component of F along line AC. Express this component as a Cartesian vector.

A

3 ft

4 ft 2 ft

4 ft x

SOLUTION

u

F

{30i

45j

4 ft C

(7 - 0)i + (6 - 0)j + ( - 4 - 0)k uAC = = 0.6965 i + 0.5970 j 0.3980 k 3(7 - 0)2 + (6 - 0)2 + ( - 4 - 0)2 Vector Dot Product: The magnitude of the projected component of F along line AC is FAC = F # uAC = (30i - 45j + 50k) # (0.6965i + 0.5970j 0.3980k)

Ans.

Thus, FAC expressed in Cartesian vector form is FAC = FAC uAC = - 25.87(0.6965i + 0.5970j 0.3980k) Ans. = { - 18.0i - 15.4j + 10.3k} lb

112112

y

50k} lb

Unit Vector: The unit vector uAC must be determined first. From Fig. a,

= (30)(0.6965) + ( - 45)(0.5970) + 50( - 0.3980) = 25.87 lb

B


Ans: FAC = 25.87 lb FAC = { - 18.0i - 15.4j + 10.3k} lb

113113


2–86. z

Determine the angle u between the pipe segments BA and BC.

A

3 ft

4 ft 2 ft

4 ft x

SOLUTION

u

F

Position Vectors: The position vectors rBA and rBC must be determined first. From Fig. a, rBA = (0 - 3)i + (0 - 4)j + (0 - 0)k = { - 3i - 4j} ft rBC = (7 - 3)i + (6 - 4)j + (- 4 - 0)k = {4i + 2j - 4k} ft

The magnitude of rBA and rBC are rBA = 3(- 3)2 + (- 4)2 = 5 ft rBC = 342 + 22 + (- 4)2 = 6 ft Vector Dot Product: rBA # rBC = (- 3i - 4j) # (4i + 2j - 4k) = (- 3)(4) + ( - 4)(2) + 0( - 4) = - 20 ft2 Thus, u = cos-1 a

rBA # rBC rBA rBC

b = cos -1 c

- 20

d = 132°

Ans.

5(6)

114114

{30i

45j

B y

50k} lb 4 ft C


Ans: u = 132°

115115


2–87. z¿ z

If the force F = 100 N lies in the plane DBEC, which is parallel to the x–z plane, and makes an angle of 10° with the extended line DB as shown, determine the angle that F makes with the diagonal AB of the crate.

15 0.2 m 10 B x A

Solution

15

- 0.5i + 0.2j + 0.2k 0.57446

b

= - 0.8704i + 0.3482j + 0.3482k F = - 100 cos 10°i + 100 sin 10°k u = cos-1a

= cos-1 a

C y

Use the x, y, z axes.

u AB = a

E 0.2 m

D

F # u AB b F uAB - 100 (cos 10°)( - 0.8704) + 0 + 100 sin 10° (0.3482) b 100(1)

= cos-1 (0.9176) = 23.4°

Ans.

116116

F


Ans: u = 23.4°

117117


*2–88. z

tthhee mm ag onpeonntesnotsf tohfe Determi ne an gintiutd uedsesofotfhethceom copm force aFct=ing pa ra ll e l a n d p e r p n d ic u l a r t o d i a g o n 90 lb a cting parallel and perpendiculaarl A B o f t h e c r a t e . of the crate. to dia gonal AB

F

90 lb 60

B 45 1 ft

1 ft x

SOLUTION Force and Unit Vector: The force vector F and unit vector uAB must be determined first. From Fig. a, F = 90(- cos 60° sin 45°i + cos 60° cos 45°j + sin 60°k) = {- 31.82i + 31.82j + 77.94k} lb uAB =

(0 - 1.5)i + (3 - 0)j + (1 - 0)k 3 6 2 rAB = - i- j + k = 7 rAB 7 7 3(0 - 1.5)2 + (3 - 0)2 + (1 - 0)2

Vector Dot Product: The magnitude of the projected component of F parallel to the diagonal AB is 6 2 3 [(F)AB]pa = F # uAB = (- 31.82i + 31.82j + 77.94k) # ¢- i + j + k ≤ 7 7 7 3 6 2 = (- 31.82) ¢ - ≤ + 31.82 ¢ ≤ + 77.94 ¢ ≤ 7 7 7 = 63.18 lb = 63.2 lb

Ans.

The magnitude of the component F perpendicular to the diagonal AB is [(F)AB]per = 3F2 - [(F)AB]pa2 = 2902 63.1 82

= 64.1 lb

118118

Ans.

A 3 ft

C

1.5 ft

y


Ans: 3 (F )AB 4 = 63.2 lb 3 (F )AB 4 # = 64.1 lb

111111


2–89. Determine the magnitudes of the projected components of the force acting along the x and y axes.

z

30

F

A

300 N

30 300 mm

O 300 mm x

SOLUTION

300 mm

y

Force Vector: The force vector F must be determined first. From Fig. a, F = - 300 sin 30°sin 30°i + 300 cos 30°j + 300 sin 30°cos 30°k = [ - 75i + 259.81j + 129.90k] N Vector Dot Product: The magnitudes of the projected component of F along the x and y axes are Fx = F # i = A - 75i + 259.81j + 129.90k B # i = - 75(1) + 259.81(0) + 129.90(0) = - 75 N Fy = F # j = A - 75i + 259.81j + 129.90k B # j = - 75(0) + 259.81(1) + 129.90(0) = 260 N The negative sign indicates that Fx is directed towards the negative x axis. Thus Fx = 75 N,

Fy = 260 N

Ans.

Ans: Fx = 75 N


Fy = 260 N


2–90. Determine the magnitude of the projected component of the force acting along line OA.

z

30

F

A

30 300 mm

O 300 mm

SOLUTION

x

300 mm

y

Force and Unit Vector: The force vector F and unit vector uOA must be determined first. From Fig. a, F = ( - 300 sin 30° sin 30°i + 300 cos 30°j + 300 sin 30° cos 30°k) = { - 75i + 259.81j + 129.90k} N ( - 0.45 - 0)i + (0.3 - 0)j + (0.2598 - 0)k r uOA = OA = = - 0.75i + 0.5j + 0.4330k rOA 2( - 0.45 - 0)2 + (0.3 - 0)2 + (0.2598 - 0)2 Vector Dot Product: The magnitude of the projected component of F along line OA is FOA = F # uOA = A - 75i + 259.81j + 129.90k B # A - 0.75i + 0.5j + 0.4330k B = ( - 75)( - 0.75) + 259.81(0.5) + 129.90(0.4330) = 242 N

Ans.

Ans:

300 N


FOA = 242 N


2–91. z

Two cables exert forces on the pipe. Determine the magnitude of the projected component of F1 along the line of action of F2.

F2

25 lb

60

u

SOLUTION

60

x

Force Vector:

= 0.4330i + 0.75j - 0.5k

F1

F1 = FRuF1 = 30(0.4330i + 0.75j - 0.5k) lb = {12.990i + 22.5j - 15.0k} lb Unit

Vector: One can obtain

30

30

uF1 = cos 30° sin 30°i + cos 30° cos 30°j - sin 30°k

the angle a = 135° for F2 using Eq. 2–8.

cos2 a + cos2 b + cos2 g = 1, with b = 60° and g = 60°. The unit vector along the line of action of F2 is uF2 = cos 135°i + cos 60°j + cos 60°k = - 0.7071i + 0.5j + 0.5k Projected Component of F1 Along the Line of Action of F2: (F1)F2 = F1 # uF2 = (12.990i + 22.5j - 15.0k) # (- 0.7071i + 0.5j + 0.5k) = (12.990)(- 0.7071) + (22.5)(0.5) + (- 15.0)(0.5) = - 5.44 lb Negative sign indicates that the projected component of (F1)F2 acts in the opposite sense of direction to that of uF2. The magnitude is (F1)F2 = 5.44 lb Ans.

Ans:

30 lb

y


The magnitude is (F1)F2 = 5.44 lb


*2–92. z

Determine the angle u between the two forces.

F2

25 lb

60

u

SOLUTION

60

x

Unit Vectors: u F1 = cos 30° sin 30°i + cos 30° cos 30°j - sin 30°k

30

30

y

= 0.4330i + 0.75j - 0.5k F1

30 lb

uF2 = cos 135°i + cos 60°j + cos 60°k = - 0.7071i + 0.5j + 0.5k The Angles Between Two Vectors u: u F1 # u F2 = (0.4330i + 0.75j - 0.5k) # ( - 0.7071i + 0.5j + 0.5k) = 0.4330( - 0.7071) + 0.75(0.5) + ( - 0.5)(0.5) = - 0.1812 Then, u = cos - 1 A uF # u F B = cos - 1( - 0.1812) = 100° 1

Ans.

2

Ans: u = 100°


Statics and Mechanics of Materials 5th Edition Hibbeler SOLUTIONS MANUAL Full at: https://testbankreal.com//statics-and-mechanics-of-materials-5th-editionhibbeler-solutions-manual/ hibbeler statics and mechanics of materials, 5th edition pdf statics and mechanics of materials 5th edition ebook statics and mechanics of materials 5th edition solutions pdf r. c. hibbeler, statics and mechanics of materials, 5th edition hibbeler statics and mechanics of materials, 4th edition pdf statics and mechanics of materials 5th edition pdf hibbeler statics and mechanics of materials pdf statics and mechanics of materials hibbeler pdf


*R2–4. z

The cable exerts a force of 250 lb on the crane boom as shown. Express this force as a Cartesian vector.

y 70 30

Solution Cartesian Vector Notation: With a a= =30° and b =b 70° , the third coordinate 30° = 70°, direction angle g can be determined using Eq. 2–8.

x

cos2 a + cos2 b + cos2 g = 1 cos2 30° + cos2 70° + cos2 g = 1

F

250 lb

cos g = ; 0.3647 g = 68.61° or 111.39° By inspection, g = 111.39° since the force F is directed in negative octant. F = 2505cos 30°i + cos 70°j + cos 111.39°6 lb =

217i + 85.5j - 91.2k lb

Ans.

Ans: F = {217i + 85.5j - 91.2k} lb


*R2–8. z

Determine the projection of the force F along the pole.

F

{2i

4j

10k} kN

O

y

SOLUTION

2m

2 2 1 Proj F = F # ua = 12 i + 4 j + 10 k2 # a i + j - k b 3 3 3 Proj F = 0.667 kN

Ans.

2m

1m

x

Ans: F = 0.667 kN

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