equilíbrio de uma partícula · 2019-09-21 · condição de equilíbrio estático de uma...

Equilíbrio de uma partícula

Prof. Ettore Baldini-Neto

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 RT I C L E

3.2 The Free-Body Diagram

To apply the equation of equilibrium, we must account for all the knownand unknown forces which act on the particle. The best way to dothis 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 calleda free-body diagram (FBD).

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

Springs. If a linearly elastic spring (or cord) of undeformed length lois used to support a particle, the length of the spring will change in directproportion to the force F acting on it, Fig. 3–1. A characteristic thatdefines the “elasticity” of a spring is the spring constant or stiffness k.

The magnitude of force exerted on a linearly elastic spring which has astiffness 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. Forexample, if the spring in Fig. 3–1 has an unstretched length of 0.8 m and astiffness and it is stretched to a length of 1 m, sothat then a force

is needed.

Cables and Pulleys. Unless otherwise stated, throughout thisbook, except in Sec. 7.4, all cables (or cords) will be assumed to havenegligible weight and they cannot stretch. Also, a cable can support onlya tension or “pulling” force, and this force always acts in the direction ofthe cable. In Chapter 5, it will be shown that the tension force developedin a continuous cable which passes over a frictionless pulley must have aconstant magnitude to keep the cable in equilibrium. Hence, for anyangle 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 NF = ks =s = l - lo = 1 m - 0.8 m = 0.2 m,

k = 500 N>mF = ks

s = l - lo

1©F23

F

!s

o

Fig. 3–1

T

TCable is in tension

u

Fig. 3–2


3.2 The Free-Body Diagram

To apply the equation of equilibrium, we must account for all the knownand unknown forces which act on the particle. The best way to dothis 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 calleda free-body diagram (FBD).

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

Springs. If a linearly elastic spring (or cord) of undeformed length lois used to support a particle, the length of the spring will change in directproportion to the force F acting on it, Fig. 3–1. A characteristic thatdefines the “elasticity” of a spring is the spring constant or stiffness k.

The magnitude of force exerted on a linearly elastic spring which has astiffness 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. Forexample, if the spring in Fig. 3–1 has an unstretched length of 0.8 m and astiffness and it is stretched to a length of 1 m, sothat then a force

is needed.

Cables and Pulleys. Unless otherwise stated, throughout thisbook, except in Sec. 7.4, all cables (or cords) will be assumed to havenegligible weight and they cannot stretch. Also, a cable can support onlya tension or “pulling” force, and this force always acts in the direction ofthe cable. In Chapter 5, it will be shown that the tension force developedin a continuous cable which passes over a frictionless pulley must have aconstant magnitude to keep the cable in equilibrium. Hence, for anyangle 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 NF = ks =s = l - lo = 1 m - 0.8 m = 0.2 m,

k = 500 N>mF = ks

s = l - lo

1©F23

F

!s

o

Fig. 3–1

T

TCable is in tension

u

Fig. 3–2F = ks = k(l � l0)

Para qualquer ângulo a tração permanece constante

Como desenhar o diagrama de corpo livre.

• Desenhe o contorno da partícula estudada

• Desenhe todas as forças que atuam sobre ela cuidadosamente.

• Identifique cada força

3.2 THE FREE-BODY DIAGRAM 87

3

The bucket is held in equilibrium bythe cable, and instinctively we knowthat the force in the cable must equalthe weight of the bucket. By drawinga free-body diagram of the bucket wecan understand why this is so. Thisdiagram shows that there are onlytwo forces acting on the bucket,namely, its weight W and the force Tof the cable. For equilibrium, theresultant of these forces must beequal to zero, and so T = W.

W

T

The spool has a weight W and is suspended fromthe crane boom. If we wish to obtain the forces incables AB and AC, then we should consider thefree-body diagram of the ring at A.Here the cablesAD exert a resultant force of W on the ring andthe condition of equilibrium is used to obtain and TC.

TB

TCTB

WD

C

AA

B

Procedure for Drawing a Free-Body Diagram

Since we must account for all the forces acting on the particle whenapplying the equations of equilibrium, the importance of first drawinga free-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 surroundingsby drawing its outlined shape.

Show All Forces.Indicate on this sketch all the forces that act on the particle. Theseforces can be active forces, which tend to set the particle in motion,or they can be reactive forces which are the result of the constraintsor supports that tend to prevent motion. To account for all theseforces, 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 propermagnitudes and directions. Letters are used to represent themagnitudes and directions of forces that are unknown.


3


W

T


TB

TCTB

WD

C

AA

B







3


W

T


TB

TCTB

WD

C

AA

B







3


W

T


TB

TCTB

WD

C

AA

B






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.


3

EXAMPLE 3.1

45!

60!

C

E

B

A

(a)

D

k

Fig. 3–3

FCE (Force of cord CE acting on sphere)

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

FCE (Force of sphere acting on cord CE)

FEC (Force of knot acting on cord CE)

(c)

C

FCBA (Force of cord CBA acting on knot)

FCD (Force of spring acting on knot)

FCE (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 thesphere, namely, its weight, 6 kg , and the force ofcord CE. The free-body diagram is shown in Fig. 3–3b.

Cord CE. When the cord CE is isolated from its surroundings, itsfree-body diagram shows only two forces acting on it, namely, theforce 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, aconsequence 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 arecaused by the cords CBA and CE and the spring CD. As required,the free-body diagram shows all these forces labeled with theirmagnitudes and directions. It is important to recognize that the weightof the sphere does not directly act on the knot. Instead, the cord CEsubjects the knot to this force.

FCE = FEC.FEC

FCE

FCE

19.81 m>s22 = 58.9 N

Sistemas de forças coplanares

• No plano as equações de equilíbrio continuam valendo, ou seja.

X~F = 0

XFx

= 0X

Fy

= 0Receita 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

Aplique e resolva adequadamente.

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 the60-kg cylinder in Fig. 3-6a.

SOLUTIONFree-Body Diagram. Due to equilibrium, the weight of the cylindercauses the tension in cable to be , Fig. 3-6b. Theforces in cables and BC can be determined by investigatingthe equilibrium of ring . Its free-body diagram is shown in Fig. 3-6c.Themagnitudes of and are unknown, but their directions are known.

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

(1)

(2)

Equation (1) can be written as . Substituting this intoEq. (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 theaccuracy of the data, i.e., measurements of geometry and loads. Formost engineering work involving a problem such as this, the data asmeasured to three significant figures would be sufficient.

TA = 420 N

TC = 475.66 N = 476 N

TC sin 45° + A35 B(0.8839TC) - 60(9.81) N = 0

TA = 0.8839TC

TC sin 45° + A35 BTA - 60(9.81) N = 0+ c©Fy = 0;

TC cos 45° - A45 BTA = 0:+ ©Fx = 0;

TCTA

BBA

TBD = 60(9.81) NBD

EXAMPLE 3.2

(a)

B

3

4

5

A

D

C

45!

Fig. 3–6

60 (9.81) N

TBD " 60 (9.81) N

(b)

TBD " 60 (9.81) N

TA TC

y

x

(c)

B

3

4

545!

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


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 whichthe crate can be suspended before one of the ropes breaks.

uAB10 kN

ACAB

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

Equations of Equilibrium. Applying the equations of equilibriumalong 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 thevalues for and into Eq. (1).

FB = 9.81 kN

10(103) N =FB

cos11.31°

FCuAB

u = sin-1(0.1962) = 11.31° = 11.3°

[10(103)N] sin u - 1962 N = 0

FC = 10 kNAB10 kNAC

cos u … 1FBFC

FC sin u - 1962 N = 0+ c©Fy = 0;

FC =FB

cos u- FC cos u + FB = 0:+ ©Fx = 0;

FD = 200 (9.81) N = 1962 N 6 10 kNFD

A

(a)

D

A B

C u

Fig. 3–7

FD ! 1962 N

y

x

(b)

A

FC

FBu

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 kAB=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 undeformedlength of spring AB is and the spring has a stiffness ofkAB = 300 N>m.

l¿AB = 0.4 m,

(a)

A B

! 300 N/m30"

2 m

C

kAB

Fig. 3–8

SOLUTIONIf the force in spring AB is known, the stretch of the spring can befound using From the problem geometry, it is then possible tocalculate 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.lAC = 1.32 m

2 m = lAC cos 30° + 0.853 m

lAB = 0.4 m + 0.453 m = 0.853 m

lAB = l¿AB + sAB

sAB = 0.453 m

135.9 N = 300 N>m1sAB2TAB = kABsAB;

TAB = 135.9 N

TAC = 157.0 N

TAC sin 30° - 78.5 N = 0+ c©Fy = 0;TAB - TAC cos 30° = 0:+ ©Fx = 0;

W = 819.812 = 78.5 N

F = ks.

y

x

W ! 78.5 N

A

(b)

30"

TAC

TAB

equilíbrio de uma partícula · 2019-09-21 · condição de equilíbrio estático de uma...

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