Sistema de Forças resultantesestatica2015.weebly.com › uploads › 3 › 7 › 5 › 7 › 37571293 › aula06-… · Momento de uma força (Torque) • Quando uma força é aplicada

Sistema de Forças resultantes

Prof. Ettore Baldini-Neto

Application of forces to the handles of these wrenches will produce a tendency torotate each wrench about its end. It is important to know how to calculate this effectand, in some cases, to be able to simplify this system to its resultants.

A aplicação de forças nos cabos das chaves faz com que estas chaves tendam a rodar em suas extremidades. É importante conhecermos como calcular este efeito e em alguns casos estarmos aptos a simplificar este problema às suas forças resultantes

Momento de uma força (Torque)• Quando uma força é aplicada sobre um corpo ela vai produzir uma tendência de rotação em um ponto que não está em sua linha de ação.

• Esta tendência de rotação é conhecida como torque ou momento de uma força.

• O módulo do momento da força é proporcional `a intensidade da força e à distância perpendicular `a aplicação da força, ou braço do momento.

Force SystemResultants

CHAPTER OBJECTIVES

• To discuss the concept of the moment of a force and show how tocalculate it in two and three dimensions.

• To provide a method for finding the moment of a force about aspecified axis.

• To define the moment of a couple.

• To present methods for determining the resultants of nonconcurrentforce systems.

• To indicate how to reduce a simple distributed loading to a resultantforce having a specified location.

4.1 Moment of a Force—Scalar Formulation

When a force is applied to a body it will produce a tendency for the bodyto rotate about a point that is not on the line of action of the force. Thistendency to rotate is sometimes called a torque, but most often it is calledthe moment of a force or simply the moment. For example, consider awrench used to unscrew the bolt in Fig. 4–1a. If a force is applied to thehandle of the wrench it will tend to turn the bolt about point O (or the zaxis). The magnitude of the moment is directly proportional to themagnitude of F and the perpendicular distance or moment arm d. Thelarger the force or the longer the moment arm, the greater the moment orturning effect. Note that if the force F is applied at an angle ,Fig. 4–1b, then it will be more difficult to turn the bolt since the momentarm will be smaller than d. If F is applied along the wrench,Fig. 4–1c, its moment arm will be zero since the line of action of F willintersect point O (the z axis).As a result, the moment of F about O is alsozero and no turning can occur.

d¿ = d sinu

u Z 90°

4z

O d

F

(a)z

O

F

d¿ ! d sin u

(b)

u

d

z

O

(c)

F

Fig. 4–1

117

Note que: Se o ângulo entre F e d for diferente de 90o, a força aplicada para rodar o parafuso será maior.

Se, por sua vez a força for aplicada junto ao eixo da chave o momento será zero e não haverá rotação.


CHAPTER OBJECTIVES








d¿ = d sinu

u Z 90°

4z

O d

F

(a)z

O

F

d¿ ! d sin u

(b)

u

d

z

O

(c)

F

Fig. 4–1

117


CHAPTER OBJECTIVES








d¿ = d sinu

u Z 90°

4z

O d

F

(a)z

O

F

d¿ ! d sin u

(b)

u

d

z

O

(c)

F

Fig. 4–1

117

Generalizando a discussão

118 CHAPTER 4 FORCE SYSTEM RESULTANTS

4

We can generalize the above discussion and consider the force F andpoint O which lie in the shaded plane as shown in Fig. 4–2a. The moment

about point O, or about an axis passing through O and perpendicularto the plane, is a vector quantity since it has a specified magnitude anddirection.

Magnitude. The magnitude of is

(4–1)

where d is the moment arm or perpendicular distance from the axis atpoint O to the line of action of the force. Units of moment magnitudeconsist of force times distance, e.g., or

Direction. The direction of is defined by its moment axis, whichis perpendicular to the plane that contains the force F and its momentarm d. The right-hand rule is used to establish the sense of direction of

. According to this rule, the natural curl of the fingers of the righthand, as they are drawn towards the palm, represent the tendency forrotation caused by the moment. As this action is performed, the thumbof the right hand will give the directional sense of , Fig. 4–2a. Noticethat the moment vector is represented three-dimensionally by a curlaround an arrow. In two dimensions this vector is represented only bythe curl as in Fig. 4–2b. Since in this case the moment will tend to cause acounterclockwise rotation, the moment vector is actually directed out ofthe page.

Resultant Moment. For two-dimensional problems, where all theforces lie within the x–y plane, Fig. 4–3, the resultant momentabout point O (the z axis) can be determined by finding the algebraic sumof the moments caused by all the forces in the system. As a convention,we will generally consider positive moments as counterclockwise sincethey are directed along the positive z axis (out of the page). Clockwisemoments will be negative. Doing this, the directional sense of eachmoment can be represented by a plus or minus sign. Using this signconvention, the resultant moment in Fig. 4–3 is therefore

a

If the numerical result of this sum is a positive scalar, will be acounterclockwise moment (out of the page); and if the result is negative,

will be a clockwise moment (into the page).(MR)O

(MR)O

+ (MR)O

= ©Fd; (MR)O

= F1d1 - F2d2 + F3d3

(MR)O

MO

MO

MO

lb # ft.N # m

MO = Fd

MO

MO

Sense of rotation

O

Moment axis

dF

MO

MO

F

d

O

(a)

(b)

Fig. 4–2

y

xO

F3

F2

F1

M3

M2 M1

d3

d2d1

Fig. 4–3


4




(4–1)





a



(MR)O

+ (MR)O

= ©Fd; (MR)O

= F1d1 - F2d2 + F3d3

(MR)O

MO

MO

MO

lb # ft.N # m

MO = Fd

MO

MO

Sense of rotation

O

Moment axis

dF

MO

MO

F

d

O

(a)

(b)

Fig. 4–2

y

xO

F3

F2

F1

M3

M2 M1

d3

d2d1

Fig. 4–3

M0 = Fd [N.m]

O Momento de uma força é uma grandeza vetorial

Direção:

Intensidade:

Momento Resultante


4




(4–1)





a



(MR)O

+ (MR)O

= ©Fd; (MR)O

= F1d1 - F2d2 + F3d3

(MR)O

MO

MO

MO

lb # ft.N # m

MO = Fd

MO

MO

Sense of rotation

O

Moment axis

dF

MO

MO

F

d

O

(a)

(b)

Fig. 4–2

y

xO

F3

F2

F1

M3

M2 M1

d3

d2d1

Fig. 4–3Em problemas bi-dimensionais, o momento resultante das forças que atuam no plano xy é simplesmente a soma algébrica dos vários momentos com a seguinte convenção: Quando as forças atuam em sentido anti-horário, o momento é positivo. Quando atuam no sentido horário o momento é negativo.

MR0 = F1d1 � F2d2 + F3d3

Exemplo 1: Para cada caso determine o momento da força em volta do ponto O.

4.1 MOMENT OF A FORCE—SCALAR FORMULATION 119

4

EXAMPLE 4.1

For each case illustrated in Fig. 4–4, determine the moment of theforce about point O.

SOLUTION (SCALAR ANALYSIS)The line of action of each force is extended as a dashed line in order toestablish the moment arm d.Also illustrated is the tendency of rotationof the member as caused by the force. Furthermore, the orbit of theforce about O is shown as a colored curl. Thus,

Fig. 4–4a b Ans.

Fig. 4–4b b Ans.

Fig. 4–4c b Ans.

Fig. 4–4d d Ans.

Fig. 4–4e d Ans.MO = 17 kN214 m - 1 m2 = 21.0 kN # mMO = 160 lb211 sin 45° ft2 = 42.4 lb # ftMO = 140 lb214 ft + 2 cos 30° ft2 = 229 lb # ftMO = 150 N210.75 m2 = 37.5 N # mMO = 1100 N212 m2 = 200 N # m

2 m

O

(a)

100 N

Fig. 4–4

2 m

O

(b)

50 N

0.75 m

(d)

O1 sin 45! ft

60 lb

3 ft

45!1 ft

2 m

O (e)

4 m

1 m7 kN

2 ft

(c)

O

4 ft2 cos 30! ft

40 lb30!


4

EXAMPLE 4.1



Fig. 4–4a b Ans.

Fig. 4–4b b Ans.

Fig. 4–4c b Ans.

Fig. 4–4d d Ans.


2 m

O

(a)

100 N

Fig. 4–4

2 m

O

(b)

50 N

0.75 m

(d)

O1 sin 45! ft

60 lb

3 ft

45!1 ft

2 m

O (e)

4 m

1 m7 kN

2 ft

(c)

O

4 ft2 cos 30! ft

40 lb30!


4

EXAMPLE 4.1



Fig. 4–4a b Ans.

Fig. 4–4b b Ans.

Fig. 4–4c b Ans.

Fig. 4–4d d Ans.


2 m

O

(a)

100 N

Fig. 4–4

2 m

O

(b)

50 N

0.75 m

(d)

O1 sin 45! ft

60 lb

3 ft

45!1 ft

2 m

O (e)

4 m

1 m7 kN

2 ft

(c)

O

4 ft2 cos 30! ft

40 lb30!


4

EXAMPLE 4.1



Fig. 4–4a b Ans.

Fig. 4–4b b Ans.

Fig. 4–4c b Ans.

Fig. 4–4d d Ans.


2 m

O

(a)

100 N

Fig. 4–4

2 m

O

(b)

50 N

0.75 m

(d)

O1 sin 45! ft

60 lb

3 ft

45!1 ft

2 m

O (e)

4 m

1 m7 kN

2 ft

(c)

O

4 ft2 cos 30! ft

40 lb30!


4

EXAMPLE 4.1



Fig. 4–4a b Ans.

Fig. 4–4b b Ans.

Fig. 4–4c b Ans.

Fig. 4–4d d Ans.


2 m

O

(a)

100 N

Fig. 4–4

2 m

O

(b)

50 N

0.75 m

(d)

O1 sin 45! ft

60 lb

3 ft

45!1 ft

2 m

O (e)

4 m

1 m7 kN

2 ft

(c)

O

4 ft2 cos 30! ft

40 lb30!

Exemplo 2: Determine o momento resultante em relação ao ponto O das forças atuando no bastão. Assuma que os momentos positivos atuam na direção k (sentido anti-horário).


4

EXAMPLE 4.2

50 N

40 N

20 N3 m

2 m 2 m

Ox

y

60 N

30!

Fig. 4–5

As illustrated by the example problems, the moment of aforce does not always cause a rotation. For example, the forceF tends to rotate the beam clockwise about its support at Awith a moment The actual rotation would occurif the support at B were removed.

MA = FdA.

O

FN

FH

The ability to remove the nail will require the momentof about point O to be larger than the moment ofthe force about O that is needed to pull the nail out.FN

FH

MA " FdA

dA

F

A B

Determine the resultant moment of the four forces acting on the rodshown in Fig. 4–5 about point O.

SOLUTIONAssuming that positive moments act in the direction, i.e.,counterclockwise, we have

a

b Ans.

For this calculation, note how the moment-arm distances for the 20-Nand 40-N forces are established from the extended (dashed) lines ofaction of each of these forces.

MRO= -334 N # m = 334 N # m

-40 N14 m + 3 cos 30° m2MRO= -50 N12 m2 + 60 N102 + 20 N13 sin 30° m2+MRO= ©Fd;

+k

Revisão: Produto Vetorial

~A⇥ ~B = det

2

4ı̂ |̂ k̂A

x

Ay

Az

Bx

By

Bz

3

5

| ~A⇥ ~B| = ABsen(✓)

O produto vetorial de dois vetores A e B é definido como:

O módulo deste produto é dado por:

A direção do vetor resultante do produto vetorial de dois vetores é dada pela regra da mão direita

4.2 CROSS PRODUCT 121

4

4.2 Cross Product

The moment of a force will be formulated using Cartesian vectors in thenext section. Before doing this, however, it is first necessary to expand ourknowledge of vector algebra and introduce the cross-product method ofvector multiplication.

The cross product of two vectors A and B yields the vector C, which iswritten

(4–2)

and is read “C equals A cross B.”

Magnitude. The magnitude of C is defined as the product of themagnitudes of A and B and the sine of the angle between their tails

Thus,

Direction. Vector C has a direction that is perpendicular to the planecontaining A and B such that C is specified by the right-hand rule; i.e.,curling the fingers of the right hand from vector A (cross) to vector B,the thumb points in the direction of C, as shown in Fig. 4–6.

Knowing both the magnitude and direction of C, we can write

(4–3)

where the scalar defines the magnitude of C and the unit vectordefines the direction of C. The terms of Eq. 4–3 are illustrated

graphically in Fig. 4–6.uC

AB sin u

C = A * B = 1AB sin u2uC

C = AB sin u.10° … u … 180°2. u

C = A * B

C ! A " B

A

B

u

uC

Fig. 4–6

Propriedades

B

A

! C " B # A

C " A # B

B

A

Fig. 4–7


4

Laws of Operation.● The commutative law is not valid; i.e., . Rather,

This is shown in Fig. 4–7 by using the right-hand rule. The crossproduct yields a vector that has the same magnitude but actsin the opposite direction to C; i.e.,

● If the cross product is multiplied by a scalar a, it obeys the assoc-iative law;

This property is easily shown since the magnitude of the resultantvector and its direction are the same in each case.

● The vector cross product also obeys the distributive law of addition,

● The proof of this identity is left as an exercise (see Prob. 4–1). It isimportant to note that proper order of the cross products must bemaintained, since they are not commutative.

Cartesian Vector Formulation. Equation 4–3 may be usedto find the cross product of any pair of Cartesian unit vectors. Forexample, to find the magnitude of the resultant vector is

and its direction is determined usingthe right-hand rule. As shown in Fig. 4–8, the resultant vector points inthe direction. Thus, In a similar manner,

These results should not be memorized; rather, it should be clearlyunderstood how each is obtained by using the right-hand rule and thedefinition of the cross product. A simple scheme shown in Fig. 4–9 ishelpful for obtaining the same results when the need arises. If the circle isconstructed as shown, then “crossing” two unit vectors in acounterclockwise fashion around the circle yields the positive third unitvector; e.g., “Crossing” clockwise, a negative unit vector isobtained; e.g., i * k = - j.

k * i = j.

k * i = j k * j = - i k * k = 0j * k = i j * i = -k j * j = 0i * j = k i * k = - j i * i = 0

i * j = 112k.+k

1i21j21sin 90°2 = 112112112 = 1,i * j,

A * 1B + D2 = 1A * B2 + 1A * D21 ƒa ƒAB sin u2

a1A * B2 = 1aA2 * B = A * 1aB2 = 1A * B2aB * A = -C.

B * A

A * B = -B * A

A * B Z B * A

y

x

z

k " i # j

j

i

Fig. 4–8

$

!

i

j k

Fig. 4–9

B

A

! C " B # A

C " A # B

B

A

Fig. 4–7


4










k * i = j.


i * j = 112k.+k

1i21j21sin 90°2 = 112112112 = 1,i * j,



B * A

A * B = -B * A

A * B Z B * A

y

x

z

k " i # j

j

i

Fig. 4–8

$

!

i

j k

Fig. 4–9

B

A

! C " B # A

C " A # B

B

A

Fig. 4–7


4










k * i = j.


i * j = 112k.+k

1i21j21sin 90°2 = 112112112 = 1,i * j,



B * A

A * B = -B * A

A * B Z B * A

y

x

z

k " i # j

j

i

Fig. 4–8

$

!

i

j k

Fig. 4–9

Em termos dos vetores unitário i,j k temos:

B

A

! C " B # A

C " A # B

B

A

Fig. 4–7


4










k * i = j.


i * j = 112k.+k

1i21j21sin 90°2 = 112112112 = 1,i * j,



B * A

A * B = -B * A

A * B Z B * A

y

x

z

k " i # j

j

i

Fig. 4–8

$

!

i

j k

Fig. 4–9

B

A

! C " B # A

C " A # B

B

A

Fig. 4–7


4










k * i = j.


i * j = 112k.+k

1i21j21sin 90°2 = 112112112 = 1,i * j,



B * A

A * B = -B * A

A * B Z B * A

y

x

z

k " i # j

j

i

Fig. 4–8

$

!

i

j k

Fig. 4–9

Momento de uma força -Formulação Vetorial-

r1r3 r2

O

F

MO ! r1 " F ! r2 " F ! r3 " F

Line of action

Fig. 4–11


4

4.3 Moment of a Force—VectorFormulation

The moment of a force F about point O, or actually about the moment axispassing through O and perpendicular to the plane containing O and F,Fig. 4–10a, can be expressed using the vector cross product, namely,

(4–6)

Here r represents a position vector directed from O to any point on theline of action of F. We will now show that indeed the moment whendetermined by this cross product, has the proper magnitude and direction.

Magnitude. The magnitude of the cross product is defined fromEq. 4–3 as where the angle is measured between thetails of r and F. To establish this angle, r must be treated as a slidingvector so that can be constructed properly, Fig. 4–10b. Since themoment arm then

which agrees with Eq. 4–1.

Direction. The direction and sense of in Eq. 4–6 are determinedby the right-hand rule as it applies to the cross product. Thus, sliding r tothe dashed position and curling the right-hand fingers from r toward F, “rcross F,” the thumb is directed upward or perpendicular to the planecontaining r and F and this is in the same direction as the moment of the force about point O, Fig. 4–10b. Note that the “curl” of the fingers,like the curl around the moment vector, indicates the sense of rotationcaused by the force. Since the cross product does not obey thecommutative law, the order of must be maintained to produce the correct sense of direction for .

Principle of Transmissibility. The cross product operation isoften used in three dimensions since the perpendicular distance ormoment arm from point O to the line of action of the force is notneeded. In other words, we can use any position vector r measured frompoint O to any point on the line of action of the force F, Fig. 4–11. Thus,

Since F can be applied at any point along its line of action and still createthis same moment about point O, then F can be considered a slidingvector. This property is called the principle of transmissibility of a force.

MO = r1 * F = r2 * F = r3 * F

MO

r * F

MO,

MO

MO = rF sin u = F1r sin u2 = Fd

d = r sin u,u

uMO = rF sin u,

MO,

MO = r * F

O

Moment axis

MO

rA

F

(a)

Fig. 4–10

O

Moment axis

d

MO

rAr

F

(b)

u

u

~MO = ~r ⇥ ~F

Módulo: MO = rFsen✓

r1r3 r2

O

F

MO ! r1 " F ! r2 " F ! r3 " F

Line of action

Fig. 4–11


4



(4–6)







MO = r1 * F = r2 * F = r3 * F

MO

r * F

MO,

MO


d = r sin u,u

uMO = rF sin u,

MO,

MO = r * F

O

Moment axis

MO

rA

F

(a)

Fig. 4–10

O

Moment axis

d

MO

rAr

F

(b)

u

uDireção: Regra da mão direita do produto vetorial. MO é sempre perpendicular ao plano gerado por r e por F.

Princípio da Transmissibilidade

Pode-se usar qualquer vetor r medido a partir do ponto O até qualquer ponto sobre a linha de ação da força F desde que o braço do momento do ponto O até a linha de ação da força não é necessário. Isto implica que:

~MO = ~r1 ⇥ ~F = ~r2 ⇥ ~F = ~r3 ⇥ ~F

r1r3 r2

O

F

MO ! r1 " F ! r2 " F ! r3 " F

Line of action

Fig. 4–11


4



(4–6)







MO = r1 * F = r2 * F = r3 * F

MO

r * F

MO,

MO


d = r sin u,u

uMO = rF sin u,

MO,

MO = r * F

O

Moment axis

MO

rA

F

(a)

Fig. 4–10

O

Moment axis

d

MO

rAr

F

(b)

u

u

F é chamado de vetor deslizante, pois pode ser aplicado em qualquer ponto ao longo de sua linha de ação e criar o mesmo momento MO

Calculando o Momento da força

~

MO = ~r ⇥ ~

F = det

2

4ı̂ |̂ k̂

x y z

Fx Fy Fz

3

5

Onde x, y e z são as componentes do vetor posição r definido a partir do ponto O até qualquer ponto sobre a linha de ação da força F e Fx, Fy e Fz são as componentes da força F.

Calculando o determinante temos:

~

MO = ı̂(yFz � zFy) + |̂(zFx � xFz) + k̂(xFy � yFx)

Momento resultante de um sistema de forças

~MOR =X

(~r ⇥ ~F )

4.3 MOMENT OF A FORCE—VECTOR FORMULATION 125

4

Cartesian Vector Formulation. If we establish x, y, z coordinateaxes, then the position vector r and force F can be expressed as Cartesianvectors, Fig. 4–12a. Applying Eq. 4–5 we have

(4–7)

whererepresent the x, y, z components of the positionvector drawn from point O to any point on theline of action of the force

represent the x, y, z components of the force vector

If the determinant is expanded, then like Eq. 4–4 we have

(4–8)

The physical meaning of these three moment components becomesevident by studying Fig. 4–12b. For example, the i component of can be determined from the moments of and about the x axis.The component does not create a moment or tendency to causeturning about the x axis since this force is parallel to the x axis. The lineof action of passes through point B, and so the magnitude of themoment of about point A on the x axis is . By the right-handrule this component acts in the negative i direction. Likewise, passesthrough point C and so it contributes a moment component of about the axis. Thus, as shown in Eq. 4–8. As anexercise, establish the j and k components of in this manner andshow that indeed the expanded form of the determinant, Eq. 4–8,represents the moment of F about point O. Once is determined,realize that it will always be perpendicular to the shaded planecontaining vectors r and F, Fig. 4–12a.

Resultant Moment of a System of Forces. If a body is actedupon by a system of forces, Fig. 4–13, the resultant moment of the forcesabout point O can be determined by vector addition of the moment ofeach force. This resultant can be written symbolically as

(4–9)MRO= ©1r * F2

MO

MO

1MO2x = 1ryFz - rzFy2 ryFziFz

rzFyFy

Fy

Fx

FzFy,Fx,MO

MO = 1ryFz - rzFy2i - 1rxFz - rzFx2j + 1rxFy - ryFx2kFx, Fy, Fz

rx, ry, rz

MO = r * F = 3 i j krx ry rz

Fx Fy Fz

3z

C

y

Fy

Fx

rz

r ry

rx

xA

B

O

F

(b)

Fz

Fig. 4–12

z

MO

Momentaxis

x

yO

F

(a)

r

z

x

yO

r2

r1r3

F3 F1

F2

MRO

Fig. 4–13

Exemplo 1: Determine o momento produzido pela força F na figura abaixo em relação ao ponto O. Expresse seu resultado como um vetor cartesiano.


4

EXAMPLE 4.3

Determine the moment produced by the force F in Fig. 4–14a aboutpoint O. Express the result as a Cartesian vector.

SOLUTIONAs shown in Fig. 4–14a, either or can be used to determine themoment about point O. These position vectors are

Force F expressed as a Cartesian vector is

Thus

Ans.

or

Ans.

NOTE: As shown in Fig. 4–14b, acts perpendicular to the planethat contains . Had this problem been worked using

notice the difficulty that would arise in obtaining themoment arm d.MO = Fd,

F, rA, and rB

MO

= 5-16.5i + 5.51j6 kN–m+ [4(1.376) - 12(0.4588)]k

= [12(-1.376) - 0(1.376)]i - [4(-1.376) - 0(0.4588)] j

MO = rB * F = 3 i j k4 12 0

0.4588 1.376 -1.376

3= 5-16.5i + 5.51j6 kN–m

+ [0(1.376) - 0(0.4588)]k

= [0(-1.376) - 12(1.376)]i - [0(-1.376) - 12(0.4588)] j

MO = rA * F = 3 i j k0 0 12

0.4588 1.376 -1.376

3= 50.4588i + 1.376j - 1.376k6 kN

F = FuAB = 2 kN c 54i + 12j - 12k6 m214 m22 + 112 m22 + 1-12 m22 drA = 512k6 m and rB = 54i + 12j6 m

rBrA

12 m

4 m

12 mA

B

O

x

y

z

(a)

F ! 2 kN

uAB

Fig. 4–14

(b)

A

B

O

x

y

z

F

rB

rA

MO


4

EXAMPLE 4.3

Determine the moment produced by the force F in Fig. 4–14a aboutpoint O. Express the result as a Cartesian vector.

SOLUTIONAs shown in Fig. 4–14a, either or can be used to determine themoment about point O. These position vectors are

Force F expressed as a Cartesian vector is

Thus

Ans.

or

Ans.

NOTE: As shown in Fig. 4–14b, acts perpendicular to the planethat contains . Had this problem been worked using

notice the difficulty that would arise in obtaining themoment arm d.MO = Fd,

F, rA, and rB

MO

= 5-16.5i + 5.51j6 kN–m+ [4(1.376) - 12(0.4588)]k

= [12(-1.376) - 0(1.376)]i - [4(-1.376) - 0(0.4588)] j

MO = rB * F = 3 i j k4 12 0

0.4588 1.376 -1.376

3= 5-16.5i + 5.51j6 kN–m

+ [0(1.376) - 0(0.4588)]k

= [0(-1.376) - 12(1.376)]i - [0(-1.376) - 12(0.4588)] j

MO = rA * F = 3 i j k0 0 12

0.4588 1.376 -1.376

3= 50.4588i + 1.376j - 1.376k6 kN

F = FuAB = 2 kN c 54i + 12j - 12k6 m214 m22 + 112 m22 + 1-12 m22 drA = 512k6 m and rB = 54i + 12j6 m

rBrA

12 m

4 m

12 mA

B

O

x

y

z

(a)

F ! 2 kN

uAB

Fig. 4–14

(b)

A

B

O

x

y

z

F

rB

rA

MO

Exemplo 2: Duas forças agem sobre a barra na figura abaixo. Determine o momento resultante que elas criam em relação ao ponto O. Expresse o resultado como um vetor cartesiano.

4.3 MOMENT OF A FORCE—VECTOR FORMULATION 127

4

EXAMPLE 4.4

x

z

O

5 ft

4 ft

2 ft

A

B

F2 ! {80i " 40j # 30k} lb

F1 ! {#60i " 40j " 20k} lb

(a)

y

Fig. 4–15

x

y

z

O A

B

(b)

rA

rB

F1

F2

x

y

z

O

!39.8$

!67.4$

!121$

MRO! {30i # 40j " 60k} lb · ft

(c)

a

gb

SOLUTIONPosition vectors are directed from point O to each force as shown inFig. 4–15b. These vectors are

The resultant moment about O is therefore

rB = 54i + 5j - 2k6 ftrA = 55j6 ft

NOTE: This result is shown in Fig. 4–15c. The coordinate directionangles were determined from the unit vector for Realize that thetwo forces tend to cause the rod to rotate about the moment axis inthe manner shown by the curl indicated on the moment vector.

MRO.

Ans.= 530i - 40j + 60k6 lb # ft+ [51-302 - 1-221402]i - [41-302 - (-2)1802]j + [41402 - 51802]k= [51202 - 01402]i - [0] j + [01402 - (5)1-602]k= 3 i j k

0 5 0-60 40 20

3 + 3 i j k4 5 -280 40 -30

3= rA * F1 + rB * F3

MRO= ©1r * F2

Two forces act on the rod shown in Fig. 4–15a. Determine theresultant moment they create about the flange at O. Express the resultas a Cartesian vector.

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Sistema de Forças resultantesestatica2015.weebly.com › uploads › 3 › 7 › 5 › 7 › 37571293 › aula06-… · Momento de uma força (Torque) • Quando uma força é aplicada