Aula 03

Exercícios gerais vetores

1.Expresse a força F mostrada abaixo como um vetor cartesiano

2.6 ADDITION OF CARTESIAN VECTORS 47

2

Important Points

• Cartesian vector analysis is often used to solve problems in threedimensions.

• The positive directions of the x, y, z axes are defined by theCartesian unit vectors i, j, k, respectively.

• The magnitude of a Cartesian vector is .

• The direction of a Cartesian vector is specified using coordinatedirection angles which the tail of the vector makes with thepositive x, y, z axes, respectively. The components of the unitvector represent the direction cosines of . Onlytwo of the angles have to be specified. The third angle isdetermined from the relationship .

• Sometimes the direction of a vector is defined using the twoangles q and as in Fig. 2–28. In this case the vector componentsare obtained by vector resolution using trigonometry.

• To find the resultant of a concurrent force system, express eachforce as a Cartesian vector and add the i, j, k components of allthe forces in the system.

f

cos2 a + cos2 b + cos2 g = 1a, b, g

a, b, guA = A>Aa, b, g

A = 2A2x + A2

y + A2z

EXAMPLE 2.8

Express the force F shown in Fig. 2–30 as a Cartesian vector.

SOLUTIONSince only two coordinate direction angles are specified, the third angleα must be determined from Eq. 2–8; i.e.,

Hence, two possibilities exist, namely,

By inspection it is necessary that , since Fx must be in the +xdirection.

Using Eq. 2–9, with , we have

Ans.

Show that indeed the magnitude of .F = 200 N

= 5100.0i + 100.0j + 141.4k6 N= (200 cos 60° N)i + (200 cos 60° N)j + (200 cos 45° N)k

F = F cos ai + F cos bj + F cos gk

F = 200 N

a = 60°

a = cos-1(0.5) = 60° or a = cos-1(-0.5) = 120°

cos a = 21 - (0.5)2 - (0.707)2 = ;0.5 cos2 a + cos2 60° + cos2 45° = 1

cos2 a + cos2 b + cos2 g = 1

z

y

x

45!

F " 200 N

60!a

Fig. 2–30

The resultant force acting on the bow theship can be determined by firstrepresenting each rope force as a Cartesianvector and then summing the i, j, and kcomponents.

2. Duas forças atuam sobre o gancho mostrado abaixo. Calcule a intensidade de F2 e seus ângulos de direção coordenados, de modo que a força resultante FR atue junto ao eixo y positivo e tenha módulo de 800N.

50 CH A P T E R 2 FO R C E VE C T O R S

2

EXAMPLE 2.11

Two forces act on the hook shown in Fig. 2–32a. Specify the magnitudeof and its coordinate direction angles of that the resultant forceFR acts along the positive y axis and has a magnitude of 800 N.

SOLUTIONTo solve this problem, the resultant force FR and its two components,F1 and F2, will each be expressed in Cartesian vector form. Then, asshown in Fig. 2–33a, it is necessary that .

Applying Eq. 2–9,

Since FR has a magnitude of 800 N and acts in the +j direction,

We require

To satisfy this equation the i, j, k components of FR must be equal tothe corresponding i, j, k components of (F1 + F2). Hence,

The magnitude of F2 is thus

Ans.

We can use Eq. 2–9 to determine 2, 2, 2.

Ans.

Ans.

Ans.

These results are shown in Fig. 2–32b.

g2 = 77.6° cos g2 = 150700

;

b2 = 21.8° cos b2 = 650700

;

a2 = 108° cos a2 = -212.1700

;

gba

= 700 N

F2 = 2(-212.1 N)2 + (650 N)2 + (150 N)2

0 = 212.1 + F2x

800 = 150 + F2y 0 = -150 + F2z

F2x = -212.1 NF2y = 650 NF2z = 150 N

800j = (212.1 + F2x)i + (150 + F2y)j + (-150 + F2z)k 800j = 212.1i + 150j - 150k + F2x i + F2yj + F2zk

FR = F1 + F2

FR = (800 N)(+ j) = 5800j6 NF2 = F2xi + F2y j + F2zk

= 5212.1i + 150j - 150k6N= 300 cos 45° i + 300 cos 60° j + 300 cos 120° k

F1 = F1 cos a1i + F1 cos b1j + F1 cos g1k

FR = F1 + F2

F2F2

z

F2

F1 ! 300 N

(a)

x

y60"

45"

120"

z

(b)

F1 ! 300 N

F2 ! 700 N

FR ! 800 N

x

y

g2 ! 77.6"

b2 ! 21.8"

a2 ! 108"

Fig. 2–33

3. Uma tira de borracha está presa entre dois pontos A e B conforme mostrado abaixo. Determine seu comprimento e sua direção medidos no sentido de A para B.

58 CH A P T E R 2 FO R C E VE C T O R S

2

EXAMPLE 2.12

An elastic rubber band is attached to points A and B as shown inFig. 2–37a. Determine its length and its direction measured from Atoward B.

SOLUTIONWe first establish a position vector from A to B, Fig. 2–37b. Inaccordance with Eq. 2–11, the coordinates of the tail A(1 m, 0, –3 m) aresubtracted from the coordinates of the head B(–2 m, 2 m, 3 m), whichyields

These components of r can also be determined directly by realizingthat they represent the direction and distance one must travel alongeach axis in order to move from A to B, i.e., along the x axis {–3i} m,along the y axis {2j} m, and finally along the z axis {6k} m.

The length of the rubber band is therefore

Ans.

Formulating a unit vector in the direction of r, we have

The components of this unit vector give the coordinate directionangles

Ans.

Ans.

Ans.

NOTE: These angles are measured from the positive axes of a localizedcoordinate system placed at the tail of r, as shown in Fig. 2–37c.

g = cos-1a67b = 31.0°

b = cos-1a27b = 73.4°

a = cos-1a- 37b = 115°

u = rr

= - 37

i + 27

j + 67

k

r = 2(-3 m)2 + (2 m)2 + (6 m)2 = 7 m

= 5-3i + 2j + 6k6 mr = [-2 m - 1 m]i + [2 m - 0] j + [3 m - (-3 m)]k

(a)

z

y

x 3 m

1 mA

B3 m

2 m

2 m

(b)

z

y

A

B

{6 k}

{2 j} m{!3 i} m

rx

(c)

A

B

z¿

y¿

x¿

r " 7 m

g " 31.0#

a " 115#b " 73.4#

Fig. 2–37

4. Um homem puxa uma corda com uma força de 350N conforme mostra a figura. Represente esta força que atua sobre um suporte A como um vetor cartesiano e determine sua intensidade. Conversão 1ft≈0,3048m

60 CH A P T E R 2 FO R C E VE C T O R S

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EXAMPLE 2.13

The man shown in Fig. 2–39a pulls on the cord with a force of 70 lb.Represent this force acting on the support A as a Cartesian vector anddetermine its direction.

SOLUTIONForce F is shown in Fig. 2–39b. The direction of this vector, u, isdetermined from the position vector r, which extends from A to B.Rather than using the coordinates of the end points of the cord, r canbe determined directly by noting in Fig. 2–39a that one must travel fromA {–24k} ft, then {–8j} ft, and finally {12i} ft to get to B. Thus,

The magnitude of r, which represents the length of cord AB, is

Forming the unit vector that defines the direction and sense of bothr and F, we have

Since F, has a magnitude of 70 lb and a direction specified by u, then

Ans.

The coordinate direction angles are measured between r (or F) andthe positive axes of a localized coordinate system with origin placed atA, Fig. 2–39b. From the components of the unit vector:

Ans.

Ans.

Ans.

NOTE: These results make sense when compared with the anglesidentified in Fig. 2–39b.

g = cos-1a -2428b = 149°

b = cos-1a -828b = 107°

a = cos-1a1228b = 64.6°

= 530i - 20j - 60k6 lb

F = Fu = 70 lba1228

i - 828

j - 2428

kbu = r

r= 12

28i - 8

28j - 24

28k

r = 2(12 ft)2 + (-8 ft)2 + (-24 ft)2 = 28 ft

r = 512i - 8j - 24k6 fty

x

z

A

30 ft

8 ft

6 ft

12 ft

B

(a)

F ! 70 lb

(b)

x¿

y¿

z¿

A

u

r

B

g

b

a

Fig. 2–39

5. Determine as intensidades das projeções da força F sobre os eixos u e v.

2.9 DOT PRODUCT 71

2

Important Points

• The dot product is used to determine the angle between twovectors or the projection of a vector in a specified direction.

• If vectors A and B are expressed in Cartesian vector form, thedot product is determined by multiplying the respective x, y, zscalar components and algebraically adding the results, i.e.,

.

• From the definition of the dot product, the angle formed betweenthe tails of vectors A and B is .

• The magnitude of the projection of vector A along a line whose direction is specified by ua is determined from the dotproduct .Aa = A # ua

aa

u = cos-1(A # B>AB)

A # B = AxBx + AyBy + AzBz

EXAMPLE 2.16

Determine the magnitudes of the projection of the force F in Fig. 2–44onto the u and axes.v

SOLUTIONProjections of Force. The graphical representation of the projectionsis shown in Fig. 2–44. From this figure, the magnitudes of the projectionsof F onto the u and v axes can be obtained by trigonometry:

Ans.

Ans.

NOTE: These projections are not equal to the magnitudes of thecomponents of force F along the u and axes found from theparallelogram law. They will only be equal if the u and axes areperpendicular to one another.

vv

(Fv)proj = (100 N)cos 15° = 96.6 N

(Fu)proj = (100 N)cos 45° = 70.7 N

F ! 100 N

u

(Fu)proj

v

15"

45"

(F )projv

Fig. 2–44

6. A estrutura mostrada abaixo está submetida a uma força horizontal F=300j. Determine a intensidade das componentes desta força que são paralelas e perpendiculares ao braço AB.

72 CH A P T E R 2 FO R C E VE C T O R S

2

EXAMPLE 2.17

(a)

z

y

x

6 m

2 m

3 mA

B F ! {300 j} N

(b)

F

F

FAB

z

y

x

A

BuB

Fig 2–45

The frame shown in Fig. 2–45a is subjected to a horizontal force F = {300j}. Determine the magnitude of the components of this force parallel and perpendicular to member AB.

SOLUTIONThe magnitude of the component of F along AB is equal to the dotproduct of F and the unit vector uB, which defines the direction of AB,Fig. 2–44b. Since

then

Ans.Since the result is a positive scalar, FAB has the same sense of directionas uB, Fig. 2–45b.

Expressing FAB in Cartesian vector form, we have

Ans.

The perpendicular component, Fig. 2–45b, is therefore

Its magnitude can be determined either from this vector or by usingthe Pythagorean theorem, Fig. 2–45b:

Ans.= 155 N

= 2(300 N)2 - (257.1 N)2F! = 2F2 - F2AB

= 5-73.5i + 80j - 110k6 NF! = F - FAB = 300j - (73.5i + 220j + 110k)

= 573.5i + 220j + 110k6N= (257.1 N)(0.286i + 0.857j + 0.429k)FAB = FABuB

= 257.1 N= (0)(0.286) + (300)(0.857) + (0)(0.429)

FAB = F cos u = F # uB = (300j) # (0.286i + 0.857j + 0.429k)

uB =rB

rB=

2i + 6j + 3k2(2)2 + (6)2 + (3)2= 0.286 i + 0.857 j + 0.429k

7. O tubo da figura abaixo está sujeito à uma força F de intensidade 800N. Determine o ângulo theta entre F e o segumento AB do tubo e a projeção de F ao longo deste segmento. (1ft=0,3048 m, 1lb=4,4482 N)

2.9 DOT PRODUCT 73

2

EXAMPLE 2.18

F ! 80 lb

2 ft

2 ft1 ft

B

1 ft

y

x

z

(a)

C

A

u

(c)

x F ! 80 lb

F

z

yA

B

FBAu

B

y

x

z

(b)

C

A

urBC

rBA

Fig. 2–46

The pipe in Fig. 2–46a is subjected to the force of F = 80 lb. Determinethe angle between F and the pipe segment BA and the projection ofF along this segment.

u

SOLUTIONAngle . First we will establish position vectors from B to A and Bto C; Fig. 2–46b. Then we will determine the angle between the tailsof these two vectors.

Thus,

Ans.

Components of F. The component of F along BA is shown in Fig. 2–46b.We must first formulate the unit vector along BA and forceF as Cartesian vectors.

Thus,

Ans.= 59.0 lb

= 0 a-23b + (-75.89)a- 2

3b + (25.30) a1

3b

FBA = F # uBA = (-75.89j + 25.30k) # a - 23

i - 23

j + 13

kbF = 80 lba rBC

rBCb = 80a -3j + 1k210

b = -75.89j + 25.30k

uBA =rBA

rBA=

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

= - 23

i - 23

j + 13

k

u = 42.5°

= 0.7379 cos u =rBA

# rBC

rBArBC=

(-2)(0) + (-2)(-3) + (1)(1)

3210

rBC = 5-3j + 1k6 ft, rBC = 210ft

rBA = 5-2i - 2j + 1k6 ft, rBA = 3 ft

uu

NOTE: Since is known, then also, .FBA = F cos u = 80 lb cos 42.5º = 59.0 lbu

8. Determine o ângulo theta entre a força e a linha AO.

74 CH A P T E R 2 FO R C E VE C T O R S

2

2 m

2 m

1 m

z

y

A

O

x

F ! {"6 i # 9 j # 3 k} kN

u

F2–25

FUNDAMENTAL PROBLEMS

F2–26. Determine the angle between the force and theline .AB

u

F2–27. Determine the angle between the force and the line .

F2–28. Determine the component of projection of theforce along the line .OA

OAu

F2–30. Determine the components of the force actingparallel and perpendicular to the axis of the pole.

F2–25. Determine the angle between the force and the line AO.

u F2–29. Find the magnitude of the projected component ofthe force along the pipe.

F ! 650 N

x

A

O

y

13

125u

F2–27/28

O

z

yx

4 m

6 m

5 m B

A

F ! 400 N

4 m

F2–29

yx

z

AF ! 600 N

C

B

4 m

4 m

3 m

u

F2–26

z

x

y

A

F ! 600 lb

60$

30$

4 ft

2 ft

4 ft

O

F2–30

9. Se a corda AB tem 7.5m de comprimento, determine a coordenada z do ponto B na figura.

64 CH A P T E R 2 FO R C E VE C T O R S

2

2–86. Determine the position vector r directed from pointA to point B and the length of cord AB. Take .

2–87. If the cord AB is 7.5 m long, determine thecoordinate position +z of point B

z = 4 m

*2–88. Determine the distance between the end points Aand B on the wire by first formulating a position vectorfrom A to B and then determining its magnitude.

•2–89. Determine the magnitude and coordinatedirection angles of the resultant force acting at A.

2–90. Determine the magnitude and coordinate directionangles of the resultant force.

PROBLEMS

2 ft

4 ft

3 ft

3 ft

4 ft

2.5 ft

B

A

xC

z

FC ! 750 lb

FB ! 600 lb

Prob. 2–89

z

xB

A

y

1 in.3 in.

8 in.

2 in.

30"

60"

Prob. 2–88

3 m

2 m

6 m

z

y

z

B

x

A

Probs. 2–86/87

x

z

y

C

B

A

600 N 500 N

8 m

4 m

4 m

2 m

Prob. 2–90

2.6 ADDITION OF CARTESIAN VECTORS 51

2

F2–14. Express the force as a Cartesian vector.

F2–15. Express the force as a Cartesian vector.

F2–17. Express the force as a Cartesian vector.

F2–18. Determine the resultant force acting on the hook.

FUNDAMENTAL PROBLEMS

F ! 500 Nz

yx

60"

60"

F2–14

F ! 500 N

z

y

x

45"

60"

F2–15

y

z

x 30"

F ! 75 lb

45"

F2–13

z

yx

345

F ! 50 lb

45"

F2–16

F ! 750 Nz

y

x

45"

60"

F2–17

F2 ! 800 lb

F1 ! 500 lb

34

5

y

z

x30"

45"

F2–18

F2–13. Determine its coordinate direction angles of theforce.

F2–16. Express the force as a Cartesian vector.

10. Expresse a força da figura como um vetor cartesiano.