Vetores -...

11
Vetores Prof. Ettore Baldini-Neto

Transcript of Vetores -...

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VetoresProf. Ettore Baldini-Neto

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Notação de um vetor em termos dos vetores unitários

• Vetores unitários: Um vetor unitário é adimensional e têm módulo igual a 1.

• No sistema de referência cartesiano dextrógiro, os vetores unitários são usualmente representados pela tríade (ı, |, k)

~A = Ax

ı+Ay

|+Az

k

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

2

Cartesian Vector Representation. Since the three componentsof A in Eq. 2–2 act in the positive i, j, and k directions, Fig. 2–24, we canwrite A in Cartesian vector form as

(2–3)

There is a distinct advantage to writing vectors in this manner.Separating the magnitude and direction of each component vector willsimplify the operations of vector algebra, particularly in threedimensions.

Magnitude of a Cartesian Vector. It is always possible toobtain the magnitude of A provided it is expressed in Cartesian vectorform. As shown in Fig. 2–25, from the blue right triangle,

, and from the gray right triangle, .Combining these equations to eliminate , yields

(2–4)

Hence, the magnitude of A is equal to the positive square root of the sumof the squares of its components.

Direction of a Cartesian Vector. We will define the directionof A by the coordinate direction angles a (alpha), b (beta), and g (gamma), measured between the tail of A and the positive x, y, z axesprovided they are located at the tail of A, Fig. 2–26. Note that regardlessof where A is directed, each of these angles will be between 0° and 180°.

To determine a, b, and g , consider the projection of A onto the x, y, zaxes, Fig. 2–27. Referring to the blue colored right triangles shown ineach figure, we have

(2–5)

These numbers are known as the direction cosines of A. Once theyhave been obtained, the coordinate direction angles a, b, g can then bedetermined from the inverse cosines.

cos a =Ax

Acos b =

Ay

Acos g =

Az

A

A = 2A2x + A2

y + A2z

A¿A¿ = 2A2

x + Ay2A = 2A¿2 + A2

z

A = Axi + Ay j + AzkA

Ax i

z

y

x

Ay j

Az k

k

i

j

Fig. 2–24

A

Axi

z

y

x

Ayj

Azk

A

A¿

Ay

Ax

Az

Fig. 2–25

| ~A| = A =p

~A · ~A

=qA2

x

+A2y

+A2z

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Direção do vetor cartesiano

An easy way of obtaining these direction cosines is to form a unitvector uA in the direction of A, Fig. 2–26. If A is expressed in Cartesianvector form, , then uA will have a magnitude ofone and be dimensionless provided A is divided by its magnitude, i.e.,

(2–6)

where . By comparison with Eqs. 2–7, it is seen thatthe i, j, k components of uA represent the direction cosines of A, i.e.,

(2–7)

Since the magnitude of a vector is equal to the positive square root ofthe sum of the squares of the magnitudes of its components, and uA has amagnitude of one, then from the above equation an important relationbetween the direction cosines can be formulated as

(2–8)

Here we can see that if only two of the coordinate angles are known,the third angle can be found using this equation.

Finally, if the magnitude and coordinate direction angles of A areknown, then A may be expressed in Cartesian vector form as

(2–9)= Axi + Ayj + Azk= A cos a i + A cos b j + A cos gk

A = AuA

cos2 a + cos2 b + cos2 g = 1

uA = cos a i + cos b j + cos gk

A = 2Ax2 + A2

y + A2z

uA = AA

=Ax

Ai + Ay

Aj +

Az

Ak

A = Axi + Ay j + Azk

2.5 CARTESIAN VECTORS 45

2

z

y

x

90!

A

Ax

az

y

x

90!

A

Ay

b

z

y

x

Az

90!

Ag

Fig 2–27

A

Axi

z

y

x

Ayj

Azk

uA

g

ab

Fig. 2–26

...é determinada pelos ângulos alfa, beta e gama medidos entre a cauda do vetor e os eixos x, y e z positivos desde que estejam junto a cauda de A.

An easy way of obtaining these direction cosines is to form a unitvector uA in the direction of A, Fig. 2–26. If A is expressed in Cartesianvector form, , then uA will have a magnitude ofone and be dimensionless provided A is divided by its magnitude, i.e.,

(2–6)

where . By comparison with Eqs. 2–7, it is seen thatthe i, j, k components of uA represent the direction cosines of A, i.e.,

(2–7)

Since the magnitude of a vector is equal to the positive square root ofthe sum of the squares of the magnitudes of its components, and uA has amagnitude of one, then from the above equation an important relationbetween the direction cosines can be formulated as

(2–8)

Here we can see that if only two of the coordinate angles are known,the third angle can be found using this equation.

Finally, if the magnitude and coordinate direction angles of A areknown, then A may be expressed in Cartesian vector form as

(2–9)= Axi + Ayj + Azk= A cos a i + A cos b j + A cos gk

A = AuA

cos2 a + cos2 b + cos2 g = 1

uA = cos a i + cos b j + cos gk

A = 2Ax2 + A2

y + A2z

uA = AA

=Ax

Ai + Ay

Aj +

Az

Ak

A = Axi + Ay j + Azk

2.5 CARTESIAN VECTORS 45

2

z

y

x

90!

A

Ax

az

y

x

90!

A

Ay

b

z

y

x

Az

90!

Ag

Fig 2–27

A

Axi

z

y

x

Ayj

Azk

uA

g

ab

Fig. 2–26

cos(↵) =A

x

A

An easy way of obtaining these direction cosines is to form a unitvector uA in the direction of A, Fig. 2–26. If A is expressed in Cartesianvector form, , then uA will have a magnitude ofone and be dimensionless provided A is divided by its magnitude, i.e.,

(2–6)

where . By comparison with Eqs. 2–7, it is seen thatthe i, j, k components of uA represent the direction cosines of A, i.e.,

(2–7)

Since the magnitude of a vector is equal to the positive square root ofthe sum of the squares of the magnitudes of its components, and uA has amagnitude of one, then from the above equation an important relationbetween the direction cosines can be formulated as

(2–8)

Here we can see that if only two of the coordinate angles are known,the third angle can be found using this equation.

Finally, if the magnitude and coordinate direction angles of A areknown, then A may be expressed in Cartesian vector form as

(2–9)= Axi + Ayj + Azk= A cos a i + A cos b j + A cos gk

A = AuA

cos2 a + cos2 b + cos2 g = 1

uA = cos a i + cos b j + cos gk

A = 2Ax2 + A2

y + A2z

uA = AA

=Ax

Ai + Ay

Aj +

Az

Ak

A = Axi + Ay j + Azk

2.5 CARTESIAN VECTORS 45

2

z

y

x

90!

A

Ax

az

y

x

90!

A

Ay

b

z

y

x

Az

90!

Ag

Fig 2–27

A

Axi

z

y

x

Ayj

Azk

uA

g

ab

Fig. 2–26 cos(�) =Ay

A

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O mesmo vale para o ângulo gama. Portanto:

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

2

Cartesian Vector Representation. Since the three componentsof A in Eq. 2–2 act in the positive i, j, and k directions, Fig. 2–24, we canwrite A in Cartesian vector form as

(2–3)

There is a distinct advantage to writing vectors in this manner.Separating the magnitude and direction of each component vector willsimplify the operations of vector algebra, particularly in threedimensions.

Magnitude of a Cartesian Vector. It is always possible toobtain the magnitude of A provided it is expressed in Cartesian vectorform. As shown in Fig. 2–25, from the blue right triangle,

, and from the gray right triangle, .Combining these equations to eliminate , yields

(2–4)

Hence, the magnitude of A is equal to the positive square root of the sumof the squares of its components.

Direction of a Cartesian Vector. We will define the directionof A by the coordinate direction angles a (alpha), b (beta), and g (gamma), measured between the tail of A and the positive x, y, z axesprovided they are located at the tail of A, Fig. 2–26. Note that regardlessof where A is directed, each of these angles will be between 0° and 180°.

To determine a, b, and g , consider the projection of A onto the x, y, zaxes, Fig. 2–27. Referring to the blue colored right triangles shown ineach figure, we have

(2–5)

These numbers are known as the direction cosines of A. Once theyhave been obtained, the coordinate direction angles a, b, g can then bedetermined from the inverse cosines.

cos a =Ax

Acos b =

Ay

Acos g =

Az

A

A = 2A2x + A2

y + A2z

A¿A¿ = 2A2

x + Ay2A = 2A¿2 + A2

z

A = Axi + Ay j + AzkA

Ax i

z

y

x

Ay j

Az k

k

i

j

Fig. 2–24

A

Axi

z

y

x

Ayj

Azk

A

A¿

Ay

Ax

Az

Fig. 2–25

Vetor unitário na direção de A.

uA =~A

| ~A|

An easy way of obtaining these direction cosines is to form a unitvector uA in the direction of A, Fig. 2–26. If A is expressed in Cartesianvector form, , then uA will have a magnitude ofone and be dimensionless provided A is divided by its magnitude, i.e.,

(2–6)

where . By comparison with Eqs. 2–7, it is seen thatthe i, j, k components of uA represent the direction cosines of A, i.e.,

(2–7)

Since the magnitude of a vector is equal to the positive square root ofthe sum of the squares of the magnitudes of its components, and uA has amagnitude of one, then from the above equation an important relationbetween the direction cosines can be formulated as

(2–8)

Here we can see that if only two of the coordinate angles are known,the third angle can be found using this equation.

Finally, if the magnitude and coordinate direction angles of A areknown, then A may be expressed in Cartesian vector form as

(2–9)= Axi + Ayj + Azk= A cos a i + A cos b j + A cos gk

A = AuA

cos2 a + cos2 b + cos2 g = 1

uA = cos a i + cos b j + cos gk

A = 2Ax2 + A2

y + A2z

uA = AA

=Ax

Ai + Ay

Aj +

Az

Ak

A = Axi + Ay j + Azk

2.5 CARTESIAN VECTORS 45

2

z

y

x

90!

A

Ax

az

y

x

90!

A

Ay

b

z

y

x

Az

90!

Ag

Fig 2–27

A

Axi

z

y

x

Ayj

Azk

uA

g

ab

Fig. 2–26

uA = cos(↵)ı+ cos(�)|+ cos(�)k

(1)

(2)

|uA|2 = 1 ! cos

2(↵) + cos

2(�) + cos

2(�) = 1

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18 | C H A P T E R 1 Measurement and Vectors

vector is called resolving the vector into its components. The components of a vec-tor along the x, y, and z directions, illustrated in Figure 1-13 for a vector in the xyplane, are called the rectangular (or Cartesian) components. Note that the compo-nents of a vector do depend on the coordinate system used, although the vector it-self does not.

We can use right-triangle geometry to find the rectangular components of a vec-tor. If is the angle measured counterclockwise* from the direction to the di-rection of (see Figure 1-13), then

1-2x COMPONENT OF A VECTOR

and

1-3y COMPONENT OF A VECTOR

where A is the magnitude of If we know and we can find the angle from

1-4

and the magnitude A from the Pythagorean theorem:

1-5a

In three dimensions,

1-5b

Components can be positive or negative. The x component of a vector is posi-tive if the x coordinate of an ant as it walks from the tail to the head of the vectorincreases. Thus, if points in the positive x direction, then is positive, and if points in the negative x direction, then is negative.

It is important to note that in Equation 1-4, the inverse tangent function is mul-tiple valued. This issue is clarified in Example 1-9.

PRACTICE PROBLEM 1-6A car travels 20.0 km in a direction north of west. Let east be the direction andnorth be the direction, as in Figure 1-14. Find the x and y components of the dis-placement vector of the car.

Once we have resolved a vector into its components, we can manipulate the in-dividual components. Consider two vectors and that lie in the plane. The rectangular components of each vector and those of the sum areshown in Figure 1-15. We see that the rectangular components of each vector andthose of the sum are equivalent to the two component equations

1-6aand

1-6b

In other words, the sum of the x components equals the x component of the re-sultant, and the sum of the y components equals the y component of the resultant.The angle and magnitude of the resultant vector can be found using Equations 1-4and 1-5a, respectively.

Cy ! Ay " By

Cx ! Ax " Bx

CS

! AS

" BS

CS

! AS

" BS

xyBS

AS

"y"x30.0°

Ax

AS

AxAS

A ! 3Ax2 " Ay2 " Az

2

A ! 3Ax2 " Ay2

tan u !AyAx

u ! tan#1AyAx

uAyAx

AS

.

Ay ! A sinu

Ax ! A cosu

AS

"xu

N

S

EW30.0

2.0 km

°

A

F I G U R E 1 - 1 4

y

x

Ax

Ay

Bx

By

Cx

CyA

B

C

F I G U R E 1 - 1 5

* This assumes the direction is counterclockwise from the direction."x90°"y

y

xAx

θ

AyAy = A sin θ

Ax = A cos θ

A

F I G U R E 1 - 1 3 The rectangularcomponents of a vector. is the angle betweenthe direction of the vector and the direction. The angle is positive if it ismeasured counterclockwise from the direction, as shown.

"x

"xu

~A = Ax

ı+Ay

|

A =qA2

x

+A2y

Lembrando ainda queA

x

= Acos(✓)

A

y

= Asen(✓)

Adição e Subtração Algébrica de Vetores no plano

~A = Ax

ı+Ay

|

~B = Bx

ı+By

|

Adição

Subtração

~A+ ~B = (Ax

+Bx

)ı+ (Ay

+By

)|

~A� ~B = (Ax

�Bx

)ı+ (Ay

�By

)|

Operações com vetores coplanares

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Multiplicação de um vetor por um escalar.

Note que alfa é um número real!

Produto Escalar de dois Vetores

Cuidado, o produto escalar de dois vetores não é um vetor e sim um número!

~A = Ax

ı+Ay

|+Az

k ~B = ↵. ~A = ↵Ax

ı+ ↵Ay

|+ ↵Az

k

~A = Ax

ı+Ay

|+Az

k

~B = Bx

ı+By

|+Bz

k

~A · ~B = Ax

Bx

+Ay

By

+Az

Bz

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~A

~B

~

A · ~B = ABcos(✓)

~A · ~B = Ax

Bx

+Ay

By

+Az

Bz

ABcos(✓) = A

x

B

x

+A

y

B

y

+A

z

B

z

cos(✓) =A

x

B

x

+A

y

B

y

+A

z

B

z

AB

✓ = arccos

~

A · ~BAB

!

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Produto Vetorial

• O produto vetorial de dois vetores é também um vetor e definido como:

~A⇥ ~B = det

2

4ı | kA

x

Ay

Az

Bx

By

Bz

3

5

~A⇥ ~B = � ~B ⇥ ~A

• Ainda podemos escrever o módulo do produto vetorial entre dois vetores como:

| ~A⇥ ~B| = ABsen(✓)

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Resultante de forças coplanares

2.4 ADDITION OF A SYSTEM OF COPLANAR FORCES 33

2

Cartesian Vector Notation. It is also possible to represent the xand y components of a force in terms of Cartesian unit vectors i and j.Each of these unit vectors has a dimensionless magnitude of one, and sothey can be used to designate the directions of the x and y axes,respectively, Fig. 2–16. *

Since the magnitude of each component of F is always a positivequantity, which is represented by the (positive) scalars Fx and Fy, then wecan express F as a Cartesian vector,

Coplanar Force Resultants. We can use either of the twomethods just described to determine the resultant of several coplanarforces. To do this, each force is first resolved into its x and y components,and then the respective components are added using scalar algebra sincethey are collinear. The resultant force is then formed by adding theresultant components using the parallelogram law. For example, considerthe three concurrent forces in Fig. 2–17a, which have x and y componentsshown in Fig. 2–17b. Using Cartesian vector notation, each force is firstrepresented as a Cartesian vector, i.e.,

The vector resultant is therefore

If scalar notation is used, then we have

These are the same results as the i and j components of FR determinedabove.

(+ c) FRy = F1y + F2y - F3y

(:+ ) FRx = F1x - F2x + F3x

= (FRx)i + (FRy)j= (F1x - F2x + F3x) i + (F1y + F2y - F3y) j= F1xi + F1y j - F2x i + F2y j + F3x i-F3y j

FR = F1 + F2 + F3

F1 = F1x i + F1y jF2 = -F2x i + F2y jF3 = F3x i - F3y j

F = Fx i + Fy j

F

Fx

Fy

y

xi

j

Fig. 2–16

F3

F1F2

(a)

x

y

(b)

x

y

F2x

F2yF1y

F1x

F3x

F3y

Fig. 2–17

*For handwritten work, unit vectors are usually indicated using a circumflex, e.g., and. These vectors have a dimensionless magnitude of unity, and their sense (or arrowhead)

will be described analytically by a plus or minus sign, depending on whether they arepointing along the positive or negative x or y axis.

j¿ i

¿

~FR

= (F1x � F2x + F3x)ı+ (F1y + F2y � F3y)|

~FR

= (FRx

)ı+ (FRy

)|

Em notação escalar

2.4 ADDITION OF A SYSTEM OF COPLANAR FORCES 33

2

Cartesian Vector Notation. It is also possible to represent the xand y components of a force in terms of Cartesian unit vectors i and j.Each of these unit vectors has a dimensionless magnitude of one, and sothey can be used to designate the directions of the x and y axes,respectively, Fig. 2–16. *

Since the magnitude of each component of F is always a positivequantity, which is represented by the (positive) scalars Fx and Fy, then wecan express F as a Cartesian vector,

Coplanar Force Resultants. We can use either of the twomethods just described to determine the resultant of several coplanarforces. To do this, each force is first resolved into its x and y components,and then the respective components are added using scalar algebra sincethey are collinear. The resultant force is then formed by adding theresultant components using the parallelogram law. For example, considerthe three concurrent forces in Fig. 2–17a, which have x and y componentsshown in Fig. 2–17b. Using Cartesian vector notation, each force is firstrepresented as a Cartesian vector, i.e.,

The vector resultant is therefore

If scalar notation is used, then we have

These are the same results as the i and j components of FR determinedabove.

(+ c) FRy = F1y + F2y - F3y

(:+ ) FRx = F1x - F2x + F3x

= (FRx)i + (FRy)j= (F1x - F2x + F3x) i + (F1y + F2y - F3y) j= F1xi + F1y j - F2x i + F2y j + F3x i-F3y j

FR = F1 + F2 + F3

F1 = F1x i + F1y jF2 = -F2x i + F2y jF3 = F3x i - F3y j

F = Fx i + Fy j

F

Fx

Fy

y

xi

j

Fig. 2–16

F3

F1F2

(a)

x

y

(b)

x

y

F2x

F2yF1y

F1x

F3x

F3y

Fig. 2–17

*For handwritten work, unit vectors are usually indicated using a circumflex, e.g., and. These vectors have a dimensionless magnitude of unity, and their sense (or arrowhead)

will be described analytically by a plus or minus sign, depending on whether they arepointing along the positive or negative x or y axis.

j¿ i

¿

FRx

= (F1x � F2x + F3x)

FRy

= (F1y + F2y � F3y)

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

2

We can represent the components of the resultant force of any numberof coplanar forces symbolically by the algebraic sum of the x and ycomponents of all the forces, i.e.,

(2–1)

Once these components are determined, they may be sketched alongthe x and y axes with their proper sense of direction, and the resultantforce can be determined from vector addition, as shown in Fig. 2–17.From this sketch, the magnitude of FR is then found from thePythagorean theorem; that is,

Also, the angle , which specifies the direction of the resultant force, isdetermined from trigonometry:

The above concepts are illustrated numerically in the examples whichfollow.

u = tan-1 2 FRy

FRx

2u

FR = 2F2Rx + F2

Ry

FRx = ©Fx

FRy = ©Fy

The resultant force of the four cable forcesacting on the supporting bracket can bedetermined by adding algebraically theseparate x and y components of each cableforce. This resultant FR produces the samepulling effect on the bracket as all four cables.

Important Points

• The resultant of several coplanar forces can easily be determinedif an x, y coordinate system is established and the forces areresolved along the axes.

• The direction of each force is specified by the angle its line ofaction makes with one of the axes, or by a sloped triangle.

• The orientation of the x and y axes is arbitrary, and their positivedirection can be specified by the Cartesian unit vectors i and j.

• The x and y components of the resultant force are simply thealgebraic addition of the components of all the coplanar forces.

• The magnitude of the resultant force is determined from thePythagorean theorem, and when the components are sketchedon the x and y axes, the direction can be determined fromtrigonometry.

F1

F2

F3

F4

x

y

(c)

x

y

FRFRy

FRx

u

Fig. 2–17

~F1 = F1x ı+ F1y |

~F2 = �F2x ı+ F2y |

~F3 = F3x ı� F3y |

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Desde que as resultantes em x e y foram determinadas, elas podem ser desenhadas no plano obedecendo suas direções e calculamos uma soma simples de vetores

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

2

We can represent the components of the resultant force of any numberof coplanar forces symbolically by the algebraic sum of the x and ycomponents of all the forces, i.e.,

(2–1)

Once these components are determined, they may be sketched alongthe x and y axes with their proper sense of direction, and the resultantforce can be determined from vector addition, as shown in Fig. 2–17.From this sketch, the magnitude of FR is then found from thePythagorean theorem; that is,

Also, the angle , which specifies the direction of the resultant force, isdetermined from trigonometry:

The above concepts are illustrated numerically in the examples whichfollow.

u = tan-1 2 FRy

FRx

2u

FR = 2F2Rx + F2

Ry

FRx = ©Fx

FRy = ©Fy

The resultant force of the four cable forcesacting on the supporting bracket can bedetermined by adding algebraically theseparate x and y components of each cableforce. This resultant FR produces the samepulling effect on the bracket as all four cables.

Important Points

• The resultant of several coplanar forces can easily be determinedif an x, y coordinate system is established and the forces areresolved along the axes.

• The direction of each force is specified by the angle its line ofaction makes with one of the axes, or by a sloped triangle.

• The orientation of the x and y axes is arbitrary, and their positivedirection can be specified by the Cartesian unit vectors i and j.

• The x and y components of the resultant force are simply thealgebraic addition of the components of all the coplanar forces.

• The magnitude of the resultant force is determined from thePythagorean theorem, and when the components are sketchedon the x and y axes, the direction can be determined fromtrigonometry.

F1

F2

F3

F4

x

y

(c)

x

y

FRFRy

FRx

u

Fig. 2–17

FR

=qF 2Rx

+ F 2Ry

tan(✓) =FRy

FRx

Exemplo 2.5 (Hibbeler):

2.4 ADDITION OF A SYSTEM OF COPLANAR FORCES 35

2

EXAMPLE 2.5

Determine the x and y components of F1 and F2 acting on the boomshown in Fig. 2–18a. Express each force as a Cartesian vector.

SOLUTIONScalar Notation. By the parallelogram law, F1 is resolved into x andy components, Fig. 2–18b. Since F1x acts in the –x direction, and F1y actsin the +y direction, we have

Ans.

Ans.

The force F2 is resolved into its x and y components as shown inFig. 2–17c. Here the slope of the line of action for the force isindicated. From this “slope triangle” we could obtain the angle , e.g.,

, and then proceed to determine the magnitudes of thecomponents in the same manner as for F1. The easier method, how-ever, consists of using proportional parts of similar triangles, i.e.,

Similarly,

Notice how the magnitude of the horizontal component, F2x, wasobtained by multiplying the force magnitude by the ratio of thehorizontal leg of the slope triangle divided by the hypotenuse;whereas the magnitude of the vertical component, F2y, was obtainedby multiplying the force magnitude by the ratio of the vertical legdivided by the hypotenuse. Hence,

Ans.

Ans.

Cartesian Vector Notation. Having determined the magnitudesand directions of the components of each force, we can express eachforce as a Cartesian vector.

Ans.

Ans.F2 = 5240i - 100j6N

F1 = 5-100i + 173j6N

F2y = -100 N = 100 NT

F2x = 240 N = 240 N:

F2y = 260 Na 513b = 100 N

F2x = 260 Na 1213b = 240 N

F2x

260 N= 12

13

u = tan-1( 512)

u

F1y = 200 cos 30° N = 173 N = 173 Nc

F1x = -200 sin 30° N = -100 N = 100 N;

y

x

F1 ! 200 N

F2 ! 260 N

30"

(a)

512

13

y

x

F1 ! 200 N

F1x ! 200 sin 30" N

30"

F1y ! 200 cos 30" N

(b)

y

x

F2 ! 260 N

(c)

512

13

F2x ! 260 12—13( (N

F2y ! 260 5—13( (N

Fig. 2–18

Determine as componentes x e y de F1 e F2 que atuam no bastão.

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

2

EXAMPLE 2.6

The link in Fig. 2–19a is subjected to two forces F1 and F2. Determinethe magnitude and direction of the resultant force.

SOLUTION IScalar Notation. First we resolve each force into its x and ycomponents, Fig. 2–19b, then we sum these components algebraically.

The resultant force, shown in Fig. 2–18c, has a magnitude of

Ans.From the vector addition,

Ans.

SOLUTION IICartesian Vector Notation. From Fig. 2–19b, each force is firstexpressed as a Cartesian vector.

Then,

The magnitude and direction of FR are determined in the samemanner as before.

NOTE: Comparing the two methods of solution, notice that the use ofscalar notation is more efficient since the components can be founddirectly, without first having to express each force as a Cartesian vectorbefore adding the components. Later, however, we will show thatCartesian vector analysis is very beneficial for solving three-dimensionalproblems.

= 5236.8i + 582.8j6 N+ (600 sin 30° N + 400 cos 45° N)j

FR = F1 + F2 = (600 cos 30° N - 400 sin 45° N)i

F2 = 5-400 sin 45°i + 400 cos 45°j6 N

F1 = 5600 cos 30°i + 600 sin 30°j6N

u = tan-1a582.8 N236.8 N

b = 67.9°

= 629 N

FR = 2(236.8 N)2 + (582.8 N)2

= 582.8 Nc+ cFRy = ©Fy; FRy = 600 sin 30° N + 400 cos 45° N

= 236.8 N::+ FRx = ©Fx; FRx = 600 cos 30° N - 400 sin 45° N

y

F1 ! 600 N

x

F2 ! 400 N

45"

30"

(a)

y

F1 ! 600 N

x

F2 ! 400 N

30"

(b)

45"

y

FR

x

(c)

582.8 N

236.8 N

u

Fig. 2–19

Exemplo 2.6 (Hibbeler): Determine o módulo e a direção da força resultante na figura abaixo.