Equilíbrio de uma partícula. Prof. Ettore Baldini-Neto

Equilíbrio de uma partícula

Condição de equilíbrio estático de uma

partícula

• Uma partícula está em equilíbrio quando:

• Está em repouso

• Move-se com velocidade constante

• No equilíbrio estático, o objeto está em repouso.

X

~

F = 0

Esta é uma condição necessária e suficiente para se determinar o

equilíbrio estático.

Diagrama de corpo livre

• As equações de equilíbrio devem levar em conta todas as forças

(conhecidas e desconhecidas) que atuam sobre a partícula.

• Para isto, vamos aprender a traçar o Diagrama de Corpo Livre da

partícula.

• Conexões encontradas em problemas de equilíbrio:

•

Cabos e Polias:

Todos os cabos, salvo dito o contrário, tem massa

desprezível e são indeformáveis. Um cabo aqui, poderá suportar

apenas uma força de tração que atua sempre na direção do cabo.

• Molas:

Vale a lei de Hooke.

86 CH A P T E R 3 EQ U I L I B R I U M O F A PA R T I C L E

3.2

The Free-Body Diagram

To apply the equation of equilibrium, we must account for all the known and unknown forces which act on the particle. The best way to do this is to think of the particle as isolated and “free” from its surroundings. A drawing that shows the particle with all the forces that act on it is called a free-body diagram (FBD).

Before presenting a formal procedure as to how to draw a free-body diagram, we will first consider two types of connections often encountered in particle equilibrium problems.

Springs.

If a linearly elastic spring (or cord) of undeformed length l_o is used to support a particle, the length of the spring will change in direct proportion to the force F acting on it, Fig. 3–1. A characteristic that

defines the “elasticity” of a spring is the spring constant or stiffness k.

The magnitude of force exerted on a linearly elastic spring which has a stiffness k and is deformed (elongated or compressed) a distance

, measured from its unloaded position, is

(3–2)

If s is positive, causing an elongation, then F must pull on the spring;

whereas if s is negative, causing a shortening, then F must push on it. For

example, if the spring in Fig. 3–1 has an unstretched length of 0.8 m and a stiffness and it is stretched to a length of 1 m, so

that then a force

is needed.

Cables and Pulleys.

Unless otherwise stated, throughout this book, except in Sec. 7.4, all cables (or cords) will be assumed to have negligible weight and they cannot stretch. Also, a cable can support only a tension or “pulling” force, and this force always acts in the direction of the cable. In Chapter 5, it will be shown that the tension force developed in a continuous cable which passes over a frictionless pulley must have a

constant magnitude to keep the cable in equilibrium. Hence, for any angle shown in Fig. 3–2, the cable is subjected to a constant tension T

throughout its length. u, 500 N_{>m(0.2 m) = 100 N}s = l - lo = 1 m - 0.8 m = 0.2 m, F = ks = k = 500 N>m F = ks s = l - l_o 1©F2 3 F !s o Fig. 3–1 T T Cable is in tension u Fig. 3–2

86

C

H A P T E R

3

E

Q U I L I B R I U M O F A

P

A R T I C L E

3.2

The Free-Body Diagram

To apply the equation of equilibrium, we must account for all the known

and unknown forces

which act on the particle. The best way to do

this is to think of the particle as isolated and “free” from its surroundings.

A drawing that shows the particle with all the forces that act on it is called

a free-body diagram (FBD).

Before presenting a formal procedure as to how to draw a free-body

diagram, we will first consider two types of connections often

encountered in particle equilibrium problems.

Springs.

If a linearly elastic spring (or cord) of undeformed length l

_o

is used to support a particle, the length of the spring will change in direct

proportion to the force

F acting on it, Fig. 3–1. A characteristic that

defines the “elasticity” of a spring is the spring constant or stiffness k.

The magnitude of force exerted on a linearly elastic spring which has a

stiffness k and is deformed (elongated or compressed) a distance

, measured from its unloaded position, is

(3–2)

If s is positive, causing an elongation, then

F must pull on the spring;

whereas if s is negative, causing a shortening, then

F must push on it. For

example, if the spring in Fig. 3–1 has an unstretched length of 0.8 m and a

stiffness

and it is stretched to a length of 1 m, so

that

then a force

is needed.

Cables and Pulleys.

Unless otherwise stated, throughout this

book, except in Sec. 7.4, all cables (or cords) will be assumed to have

negligible weight and they cannot stretch. Also, a cable can support only

a tension or “pulling” force, and this force always acts in the direction of

the cable. In Chapter 5, it will be shown that the tension force developed

in a continuous cable which passes over a frictionless pulley must have a

constant

magnitude to keep the cable in equilibrium. Hence, for any

angle shown in Fig. 3–2, the cable is subjected to a constant tension T

throughout its length.

u,

500 N

_{>m(0.2 m) = 100 N}

s

= l - l

o

= 1 m - 0.8 m = 0.2 m,

F

= ks =

k

= 500 N>m

F

= ks

s

= l - l

_o

1©F2

3

F !s o

Fig. 3–1

T T Cable is in tension u

Fig. 3–2

Como desenhar o diagrama de corpo livre.

• Desenhe o contorno da partícula estudada

• Desenhe todas as forças que atuam sobre ela

cuidadosamente.

3.2 THE FREE-BODY DIAGRAM 87

3

The bucket is held in equilibrium by the cable, and instinctively we know that the force in the cable must equal the weight of the bucket. By drawing a free-body diagram of the bucket we can understand why this is so. This diagram shows that there are only two forces acting on the bucket, namely, its weight W and the force T

of the cable. For equilibrium, the resultant of these forces must be equal to zero, and so T _{= W.}

W T

The spool has a weight W and is suspended from the crane boom. If we wish to obtain the forces in cables AB and AC, then we should consider the free-body diagram of the ring at A.Here the cables

AD exert a resultant force of W on the ring and

the condition of equilibrium is used to obtain

and T_C. TB T_C T_B W D C A A B

Procedure for Drawing a Free-Body Diagram

Since we must account for all the forces acting on the particle when applying the equations of equilibrium, the importance of first drawing a body diagram cannot be overemphasized. To construct a free-body diagram, the following three steps are necessary.

Draw Outlined Shape.

Imagine the particle to be isolated or cut “free” from its surroundings by drawing its outlined shape.

Show All Forces.

Indicate on this sketch all the forces that act on the particle. These forces can be active forces, which tend to set the particle in motion, or they can be reactive forces which are the result of the constraints or supports that tend to prevent motion. To account for all these forces, it may be helpful to trace around the particle’s boundary, carefully noting each force acting on it.

Identify Each Force.

The forces that are known should be labeled with their proper magnitudes and directions. Letters are used to represent the magnitudes and directions of forces that are unknown.

3.2 T

HE

F

REE

-B

ODY

D

IAGRAM

87

3

The bucket is held in equilibrium by

the cable, and instinctively we know

that the force in the cable must equal

the weight of the bucket. By drawing

a free-body diagram of the bucket we

can understand why this is so. This

diagram shows that there are only

two forces acting on the bucket,

namely, its weight

W and the force T

of the cable. For equilibrium, the

resultant of these forces must be

equal to zero, and so T

_{= W.}

W

T

The spool has a weight W and is suspended from

the crane boom. If we wish to obtain the forces in

cables AB and AC, then we should consider the

free-body diagram of the ring at A.Here the cables

AD

exert a resultant force of

W on the ring and

the condition of equilibrium is used to obtain

and

T

_C

.

T

B

T

_C

T

_B

W

D

C

A

A

B

Procedure for Drawing a Free-Body Diagram

Since we must account for all the forces acting on the particle when

applying the equations of equilibrium, the importance of first drawing

a body diagram cannot be overemphasized. To construct a

free-body diagram, the following three steps are necessary.

Draw Outlined Shape.

Imagine the particle to be isolated or cut “free” from its surroundings

by drawing its outlined shape.

Show All Forces.

Indicate on this sketch all the forces that act on the particle. These

forces can be active forces, which tend to set the particle in motion,

or they can be reactive forces which are the result of the constraints

or supports that tend to prevent motion. To account for all these

forces, it may be helpful to trace around the particle’s boundary,

carefully noting each force acting on it.

Identify Each Force.

The forces that are known should be labeled with their proper

magnitudes and directions. Letters are used to represent the

magnitudes and directions of forces that are unknown.

3.2 THE FREE-BODY DIAGRAM 87

3

The bucket is held in equilibrium by the cable, and instinctively we know that the force in the cable must equal the weight of the bucket. By drawing a free-body diagram of the bucket we can understand why this is so. This diagram shows that there are only two forces acting on the bucket, namely, its weight W and the force T

of the cable. For equilibrium, the resultant of these forces must be equal to zero, and so T _{= W.}

W T

The spool has a weight W and is suspended from the crane boom. If we wish to obtain the forces in cables AB and AC, then we should consider the free-body diagram of the ring at A.Here the cables

AD exert a resultant force of W on the ring and

the condition of equilibrium is used to obtain

andT_C. TB T_C T_B W D C A A B

Procedure for Drawing a Free-Body Diagram

Since we must account for all the forces acting on the particle when applying the equations of equilibrium, the importance of first drawing a body diagram cannot be overemphasized. To construct a free-body diagram, the following three steps are necessary.

Draw Outlined Shape.

Imagine the particle to be isolated or cut “free” from its surroundings by drawing its outlined shape.

Show All Forces.

Indicate on this sketch all the forces that act on the particle. These forces can be active forces, which tend to set the particle in motion, or they can be reactive forces which are the result of the constraints or supports that tend to prevent motion. To account for all these forces, it may be helpful to trace around the particle’s boundary, carefully noting each force acting on it.

Identify Each Force.

The forces that are known should be labeled with their proper magnitudes and directions. Letters are used to represent the magnitudes and directions of forces that are unknown.

3.2 T

HE

F

REE

-B

ODY

D

IAGRAM

87

3

The bucket is held in equilibrium by

the cable, and instinctively we know

that the force in the cable must equal

the weight of the bucket. By drawing

a free-body diagram of the bucket we

can understand why this is so. This

diagram shows that there are only

two forces acting on the bucket,

namely, its weight

W and the force T

of the cable. For equilibrium, the

resultant of these forces must be

equal to zero, and so T

_{= W.}

W

T

The spool has a weight W and is suspended from

the crane boom. If we wish to obtain the forces in

cables AB and AC, then we should consider the

free-body diagram of the ring at A.Here the cables

AD

exert a resultant force of

W on the ring and

the condition of equilibrium is used to obtain

and

T

_C

.

T

B

T

_C

T

_B

W

D

C

A

A

B

Procedure for Drawing a Free-Body Diagram

Since we must account for all the forces acting on the particle when

applying the equations of equilibrium, the importance of first drawing

a body diagram cannot be overemphasized.To construct a

free-body diagram, the following three steps are necessary.

Draw Outlined Shape.

Imagine the particle to be isolated or cut “free” from its surroundings

by drawing its outlined shape.

Show All Forces.

Indicate on this sketch all the forces that act on the particle. These

forces can be active forces, which tend to set the particle in motion,

or they can be reactive forces which are the result of the constraints

or supports that tend to prevent motion. To account for all these

forces, it may be helpful to trace around the particle’s boundary,

carefully noting each force acting on it.

Identify Each Force.

The forces that are known should be labeled with their proper

magnitudes and directions. Letters are used to represent the

magnitudes and directions of forces that are unknown.

Exemplo1:

A esfera na figura abaixo tem massa de 6kg e

está apoiada como mostrado. Desenhe o diagrama de corpo

livre da esfera, da corda CE e do nó em C.

88 CH A P T E R 3 EQ U I L I B R I U M O F A PA R T I C L E 3 EXAMPLE 3.1 45! 60! C E B A (a) D k Fig. 3–3 F_CE (Force of cord CE acting on sphere)

58.9 N (Weight or gravity acting on sphere) (b)

F_CE (Force of sphere acting on cord CE)

F_EC (Force of knot acting on cord CE)

(c)

C

FCBA(Force of cord CBA acting on knot)

F_CD (Force of spring acting on knot)

F_CE (Force of cord CE acting on knot) 60!

(d)

The sphere in Fig. 3–3a has a mass of 6 kg and is supported as shown. Draw a free-body diagram of the sphere, the cord CE, and the knot at C.

SOLUTION

Sphere. By inspection, there are only two forces acting on the

sphere, namely, its weight, 6 kg , and the force of

cord CE. The free-body diagram is shown in Fig. 3–3b.

Cord CE. When the cord CE is isolated from its surroundings, its free-body diagram shows only two forces acting on it, namely, the force of the sphere and the force of the knot, Fig. 3–3c. Notice that shown here is equal but opposite to that shown in Fig. 3–3b, a

consequence of Newton’s third law of action–reaction. Also, and

pull on the cord and keep it in tension so that it doesn’t collapse. For equilibrium,

Knot. The knot at C is subjected to three forces, Fig. 3–3d. They are caused by the cords CBA and CE and the spring CD. As required, the free-body diagram shows all these forces labeled with their magnitudes and directions. It is important to recognize that the weight of the sphere does not directly act on the knot. Instead, the cord CE subjects the knot to this force.

F_{CE = FEC}.

F_EC FCE

F_CE

Sistemas de forças coplanares

• No plano as equações de equilíbrio continuam valendo, ou

seja.

X

~

F = 0

X

F

_x

=

0

X

F

_y

=

0

Receita de bolo:

1) Diagrama de corpo livre

a) Desenhe os eixos x e y orientados adequadamente

b) Identifique todas as forças

2) Equações de equilíbrio

Exemplo2: Determine a tração nos cabos BA e BC

necessária para sustentar o cilindro de 60kg na figura.

3.3 COPLANAR FORCE SYSTEMS 91

3

Determine the tension in cables BA and BC necessary to support the 60-kg cylinder in Fig. 3-6a.

SOLUTION

Free-Body Diagram. Due to equilibrium, the weight of the cylinder causes the tension in cable to be , Fig. 3-6b. The forces in cables and BC can be determined by investigating the equilibrium of ring . Its free-body diagram is shown in Fig. 3-6c.The magnitudes of and are unknown, but their directions are known.

Equations of Equilibrium. Applying the equations of equilibrium along the x and y axes, we have

(1) (2) Equation (1) can be written as . Substituting this into Eq. (2) yields

So that

Ans.

Substituting this result into either Eq. (1) or Eq. (2), we get

Ans.

NOTE: The accuracy of these results, of course, depends on the accuracy of the data, i.e., measurements of geometry and loads. For most engineering work involving a problem such as this, the data as measured to three significant figures would be sufficient.

T_{A = 420 N} T_{C = 475.66 N = 476 N} T_C sin 45° ₊ A3₅B(0.8839T_C) _{- 60(9.81) N = 0} T_{A = 0.8839TC} T_C sin 45° ₊ A3₅BTA - 60(9.81) N = 0 + c ©Fy = 0; T_C cos 45° _- A4₅BT_A = 0 :+ ©Fx = 0; T_C T_A B BA T_{BD = 60(9.81) N} BD EXAMPLE 3.2 (a) B 3 4 5 A D C 45! Fig. 3–6 60 (9.81) N T_BD" 60 (9.81) N (b) T_BD" 60 (9.81) N T_A T_C y x (c) B 3 4 5 45!

Exemplo 3: A caixa de 200kg é suspensa usando as cordas AB e

AC. Cada corda pode suportar uma força máxima de 10kN antes

de romper. Se AB sempre está na horizontal, determine

o menor ângulo para o qual a caixa pode ser suspensa

antes que uma das cordas se rompa.

EXAMPLE 3.3

92 CH A P T E R 3 EQ U I L I B R I U M O F A PA R T I C L E

3

The 200-kg crate in Fig. 3-7a is suspended using the ropes and . Each rope can withstand a maximum force of before it breaks. If always remains horizontal, determine the smallest angle to which the crate can be suspended before one of the ropes breaks.

u AB

10 kN AB AC

SOLUTION

Free-Body Diagram. We will study the equilibrium of ring . There are three forces acting on it, Fig. 3-7b. The magnitude of is equal to the weight of the crate, i.e., .

Equations of Equilibrium. Applying the equations of equilibrium along the x and y axes,

; (1)

(2) From Eq. (1), is always greater than since . Therefore, rope will reach the maximum tensile force of

before rope . Substituting into Eq. (2), we get

Ans.

The force developed in rope can be obtained by substituting the values for and into Eq. (1).

F_{B = 9.81 kN} 10(103) N ₌ FB cos 11.31° F_C u AB u = sin-1(0.1962) = 11.31° = 11.3° [10(103) N] sin u _{- 1962 N = 0} F_{C = 10 kN} AB 10 kN AC cos u _{… 1} FB FC FC sin u - 1962 N = 0 + c ©Fy = 0; FC = F_B cos u - FC cos u + FB = 0 :+ ©Fx = 0; F_{D = 200 (9.81) N = 1962 N 6 10 kN} F_D A (a) D A B C _u Fig. 3–7 F_D! 1962 N y x (b) A F_C F_B u

Exemplo4: Determine o comprimento da corda AC na figura, de

modo que a luminária de 8kg seja suspensa na posição mostrada.

O comprimento não deformado da mola AB é l’

AB

=0,4m e a mola

tem uma rigidez k

AB

=300N/m.

EXAMPLE 3.4

3.3 COPLANAR FORCE SYSTEMS 93

3

Determine the required length of cord AC in Fig. 3–8a so that the 8-kg lamp can be suspended in the position shown. The undeformed

length of spring AB is and the spring has a stiffness of

k_{AB = 300 N>m.} l¿_AB = 0.4 m, (a) A B ! 300 N/m 30" 2 m C k_AB Fig. 3–8 SOLUTION

If the force in spring AB is known, the stretch of the spring can be

found using From the problem geometry, it is then possible to

calculate the required length of AC.

Free-Body Diagram. The lamp has a weight

and so the free-body diagram of the ring at A is shown in Fig. 3–8b.

Equations of Equilibrium. Using the x, y axes,

Solving, we obtain

The stretch of spring AB is therefore

so the stretched length is

The horizontal distance from C to B, Fig. 3–8a, requires

Ans. l_{AC = 1.32 m} 2 m _{= lAC} cos 30° _{+ 0.853 m} l_{AB = 0.4 m + 0.453 m = 0.853 m} l_{AB = l¿AB + sAB} s_AB = 0.453 m 135.9 N _{= 300 N>m1sAB2} T_{AB = kAB}s_AB; T_{AB = 135.9 N} T_{AC = 157.0 N} T_AC sin 30° _{- 78.5 N = 0} + c ©Fy = 0; T_AB - T_AC cos 30° _{= 0} :+ ©Fx = 0; W _{= 819.812 = 78.5 N} F _{= ks.} y x W ! 78.5 N A (b) 30" T_AC T_AB