Equilíbrio de um ponto material Forças em 3D

Prof. Ettore Baldini-Neto

Sistema de forças tridimensionais

X~F = 0

XFx

= 0X

Fy

= 0X

Fz

= 0Receita de bolo:

1) Diagrama de corpo livre

a) Desenhe os eixos x, y e z orientados adequadamente b) Identifique todas as forças

2) Equações de equilíbrio a) Aplique e resolva adequadamente. b) Se for complicada a geometria 3D, expresse cada força no DCL como um vetor cartesiano e substitua-os na equação de equilíbrio igualando a zero cada componente i, j e k. c) O sentido de uma força que tenha intensidade desconhecida pode ser assumido

Exemplo 1: Uma carga de 90lb é suspensa pelo gancho mostrado na figura. Se ela é apoiada por dois cabos e uma mola de constante elástica 500lb/ft, determine a força no cabos e a elongação da mola para manter o equilíbrio. O cabo AD está no plano xy e o cabo AC no plano xz.

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EXAMPLE 3.5

A 90-lb load is suspended from the hook shown in Fig. 3–10a. If theload is supported by two cables and a spring having a stiffness

, determine the force in the cables and the stretch of thespring for equilibrium. Cable AD lies in the x–y plane and cable AClies in the x–z plane.

SOLUTIONThe stretch of the spring can be determined once the force in the springis determined.

Free-Body Diagram. The connection at A is chosen for theequilibrium analysis since the cable forces are concurrent at thispoint. The free-body diagram is shown in Fig. 3–10b.

Equations of Equilibrium. By inspection, each force can easily beresolved into its x, y, z components, and therefore the three scalarequations of equilibrium can be used. Considering componentsdirected along each positive axis as “positive,” we have

(1)

(2)

(3)

Solving Eq. (3) for then Eq. (1) for and finally Eq. (2) for yields

Ans.

Ans.

Ans.

The stretch of the spring is therefore

Ans.

NOTE: Since the results for all the cable forces are positive, eachcable is in tension; that is, it pulls on point A as expected, Fig. 3–10b.

sAB = 0.416 ft

207.8 lb = (500 lb>ft)1sAB2FB = ksAB

FB = 207.8 lb

FD = 240 lb

FC = 150 lb

FB,FD,FC,

A35 BFC - 90 lb = 0©Fz = 0;

-FD cos 30° + FB = 0©Fy = 0;

FD sin 30° - A45 BFC = 0©Fx = 0;

k = 500 lb>ft

x

y

z

(a)

30!

C

90 lb

A

5 34k = 500 lb/ft

B

D

Fig. 3–10

y

x

z

(b)

30!

90 lb

A

5 3

4

FC

FB

FD

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EXAMPLE 3.5

A 90-lb load is suspended from the hook shown in Fig. 3–10a. If theload is supported by two cables and a spring having a stiffness

, determine the force in the cables and the stretch of thespring for equilibrium. Cable AD lies in the x–y plane and cable AClies in the x–z plane.

SOLUTIONThe stretch of the spring can be determined once the force in the springis determined.

Free-Body Diagram. The connection at A is chosen for theequilibrium analysis since the cable forces are concurrent at thispoint. The free-body diagram is shown in Fig. 3–10b.

Equations of Equilibrium. By inspection, each force can easily beresolved into its x, y, z components, and therefore the three scalarequations of equilibrium can be used. Considering componentsdirected along each positive axis as “positive,” we have

(1)

(2)

(3)

Solving Eq. (3) for then Eq. (1) for and finally Eq. (2) for yields

Ans.

Ans.

Ans.

The stretch of the spring is therefore

Ans.

NOTE: Since the results for all the cable forces are positive, eachcable is in tension; that is, it pulls on point A as expected, Fig. 3–10b.

sAB = 0.416 ft

207.8 lb = (500 lb>ft)1sAB2FB = ksAB

FB = 207.8 lb

FD = 240 lb

FC = 150 lb

FB,FD,FC,

A35 BFC - 90 lb = 0©Fz = 0;

-FD cos 30° + FB = 0©Fy = 0;

FD sin 30° - A45 BFC = 0©Fx = 0;

k = 500 lb>ft

x

y

z

(a)

30!

C

90 lb

A

5 34k = 500 lb/ft

B

D

Fig. 3–10

y

x

z

(b)

30!

90 lb

A

5 3

4

FC

FB

FD

DCL

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EXAMPLE 3.7

y

x

z

(a)

8 ft

3 ft

4 ft

4 ft

C

B

D A

Fig. 3–12

y

x

z

W ! 40 lb

(b)

FB

A

FC

FD

Determine the force in each cable used to support the 40-lb crateshown in Fig. 3–12a.

SOLUTIONFree-Body Diagram. As shown in Fig. 3–12b, the free-body diagramof point A is considered in order to “expose” the three unknown forcesin the cables.

Equations of Equilibrium. First we will express each force inCartesian vector form. Since the coordinates of points B and C are

and C( 4 ft, 8 ft), we have

Equilibrium requires

Equating the respective i, j, k components to zero yields

(1)

(2)

(3)

Equation (2) states that Thus, solving Eq. (3) for and and substituting the result into Eq. (1) to obtain we have

Ans.

Ans.FD = 15.0 lb

FB = FC = 23.6 lb

FD,FCFBFB = FC.

0.848FB + 0.848FC - 40 = 0©Fz = 0;

-0.424FB + 0.424FC = 0©Fy = 0;

-0.318FB - 0.318FC + FD = 0©Fx = 0;

- 0.318FCi + 0.424FCj + 0.848FCk + FDi - 40k = 0

-0.318FBi - 0.424FBj + 0.848FBk

FB + FC + FD + W = 0©F = 0;

W = 5-40k6 lbFD = FDi

= -0.318FC i + 0.424FC j + 0.848FCk

FC = FC c -3i + 4j + 8k

2 1-322 + 1422 + 1822 d= -0.318FBi - 0.424FBj + 0.848FBk

FB = FB c -3i - 4j + 8k

2 1-322 + 1-422 + 1822 d-3 ft,B1-3 ft, -4 ft, 8 ft2

Exemplo 2: Determine as forças em cada cabo utilizado para suspender o caixote de peso W=40lb da figura abaixo.

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EXAMPLE 3.7

y

x

z

(a)

8 ft

3 ft

4 ft

4 ft

C

B

D A

Fig. 3–12

y

x

z

W ! 40 lb

(b)

FB

A

FC

FD

Determine the force in each cable used to support the 40-lb crateshown in Fig. 3–12a.

SOLUTIONFree-Body Diagram. As shown in Fig. 3–12b, the free-body diagramof point A is considered in order to “expose” the three unknown forcesin the cables.

Equations of Equilibrium. First we will express each force inCartesian vector form. Since the coordinates of points B and C are

and C( 4 ft, 8 ft), we have

Equilibrium requires

Equating the respective i, j, k components to zero yields

(1)

(2)

(3)

Equation (2) states that Thus, solving Eq. (3) for and and substituting the result into Eq. (1) to obtain we have

Ans.

Ans.FD = 15.0 lb

FB = FC = 23.6 lb

FD,FCFBFB = FC.

0.848FB + 0.848FC - 40 = 0©Fz = 0;

-0.424FB + 0.424FC = 0©Fy = 0;

-0.318FB - 0.318FC + FD = 0©Fx = 0;

- 0.318FCi + 0.424FCj + 0.848FCk + FDi - 40k = 0

-0.318FBi - 0.424FBj + 0.848FBk

FB + FC + FD + W = 0©F = 0;

W = 5-40k6 lbFD = FDi

= -0.318FC i + 0.424FC j + 0.848FCk

FC = FC c -3i + 4j + 8k

2 1-322 + 1422 + 1822 d= -0.318FBi - 0.424FBj + 0.848FBk

FB = FB c -3i - 4j + 8k

2 1-322 + 1-422 + 1822 d-3 ft,B1-3 ft, -4 ft, 8 ft2

DCL

3.4 THREE-DIMENSIONAL FORCE SYSTEMS 107

3

EXAMPLE 3.8

y1 m2 m

z

60! 135!2 m

D

120!

x

(a)

B

A

k " 1.5 kN/m

C

Fig. 3–13

y

x

z

W " 981 N

A

FC

(b)

FD

FB

Determine the tension in each cord used to support the 100-kg crateshown in Fig. 3–13a.

SOLUTIONFree-Body Diagram. The force in each of the cords can bedetermined by investigating the equilibrium of point A.The free-bodydiagram is shown in Fig. 3–13b. The weight of the crate is

Equations of Equilibrium. Each force on the free-body diagram isfirst expressed in Cartesian vector form. Using Eq. 2–9 for andnoting point for we have

Equilibrium requires

W = 5-981k6 N= -0.333FD i + 0.667FD j + 0.667FDk

FD = FD c -1i + 2j + 2k

2 1-122 + 1222 + 1222 d= -0.5FC i - 0.707FC j + 0.5FCk

FC = FC cos 120°i + FC cos 135°j + FC cos 60°k

FB = FB i

FD,D1-1 m, 2 m, 2 m2 FC

W = 10019.812 = 981 N.

Equating the respective i, j, k components to zero,(1)

(2)

(3)

Solving Eq. (2) for in terms of and substituting this into Eq. (3)yields is then determined from Eq. (2). Finally, substituting theresults into Eq. (1) gives Hence,

Ans.

Ans.

Ans.FB = 694 N

FD = 862 N

FC = 813 N

FB.FDFC.

FCFD

0.5FC + 0.667FD - 981 = 0©Fz = 0;

-0.707FC + 0.667FD = 0©Fy = 0;

FB - 0.5FC - 0.333FD = 0©Fx = 0;

- 0.333FD i + 0.667FD j + 0.667FDk - 981k = 0

FB i - 0.5FC i - 0.707FC j + 0.5FCk

FB + FC + FD + W = 0©F = 0;

Exemplo 3: Determine as trações em cada corda usada para suspender o caixote de massa 100kg da figura abaixo.

3.4 THREE-DIMENSIONAL FORCE SYSTEMS 107

3

EXAMPLE 3.8

y1 m2 m

z

60! 135!2 m

D

120!

x

(a)

B

A

k " 1.5 kN/m

C

Fig. 3–13

y

x

z

W " 981 N

A

FC

(b)

FD

FB

Determine the tension in each cord used to support the 100-kg crateshown in Fig. 3–13a.

SOLUTIONFree-Body Diagram. The force in each of the cords can bedetermined by investigating the equilibrium of point A.The free-bodydiagram is shown in Fig. 3–13b. The weight of the crate is

Equations of Equilibrium. Each force on the free-body diagram isfirst expressed in Cartesian vector form. Using Eq. 2–9 for andnoting point for we have

Equilibrium requires

W = 5-981k6 N= -0.333FD i + 0.667FD j + 0.667FDk

FD = FD c -1i + 2j + 2k

2 1-122 + 1222 + 1222 d= -0.5FC i - 0.707FC j + 0.5FCk

FC = FC cos 120°i + FC cos 135°j + FC cos 60°k

FB = FB i

FD,D1-1 m, 2 m, 2 m2 FC

W = 10019.812 = 981 N.

Equating the respective i, j, k components to zero,(1)

(2)

(3)

Solving Eq. (2) for in terms of and substituting this into Eq. (3)yields is then determined from Eq. (2). Finally, substituting theresults into Eq. (1) gives Hence,

Ans.

Ans.

Ans.FB = 694 N

FD = 862 N

FC = 813 N

FB.FDFC.

FCFD

0.5FC + 0.667FD - 981 = 0©Fz = 0;

-0.707FC + 0.667FD = 0©Fy = 0;

FB - 0.5FC - 0.333FD = 0©Fx = 0;

- 0.333FD i + 0.667FD j + 0.667FDk - 981k = 0

FB i - 0.5FC i - 0.707FC j + 0.5FCk

FB + FC + FD + W = 0©F = 0;

DCL

Exemplo 4: A lâmpada de massa 10kg é suspensa por três cordas de comprimentos iguais. Determine a menor distância vertical s a partir do teto se a força máxima que cada corda pode suportar não excede 50N.

3.4 THREE-DIMENSIONAL FORCE SYSTEMS 105

3

EXAMPLE 3.6

The 10-kg lamp in Fig. 3-11a is suspended from the three equal-lengthcords. Determine its smallest vertical distance s from the ceiling if theforce developed in any cord is not allowed to exceed 50 N.

xy

s

(a)

z

DA

B

C

600 mm120!

120!

Fig. 3–11

SOLUTIONFree-Body Diagram. Due to symmetry, Fig. 3-11b, the distance

. It follows that from and, the tension T in each cord will be the same. Also, the angle

between each cord and the axis is .

Equation of Equilibrium. Applying the equilibrium equation alongthe axis, with , we have

From the shaded triangle shown in Fig. 3-11b,

Ans.s = 519 mm

tan 49.16° = 600 mms

g = cos-1 98.1150

= 49.16°

3[(50 N) cos g] - 10(9.81) N = 0gFz = 0;

T = 50 Nz

gzgFy = 0

gFx = 0DA = DB = DC = 600 mm

x

y

s

600 mm

D

z

(b)

A

B

C

10(9.81) N

TT

Tg

3.4 THREE-DIMENSIONAL FORCE SYSTEMS 105

3

EXAMPLE 3.6

The 10-kg lamp in Fig. 3-11a is suspended from the three equal-lengthcords. Determine its smallest vertical distance s from the ceiling if theforce developed in any cord is not allowed to exceed 50 N.

xy

s

(a)

z

DA

B

C

600 mm120!

120!

Fig. 3–11

SOLUTIONFree-Body Diagram. Due to symmetry, Fig. 3-11b, the distance

. It follows that from and, the tension T in each cord will be the same. Also, the angle

between each cord and the axis is .

Equation of Equilibrium. Applying the equilibrium equation alongthe axis, with , we have

From the shaded triangle shown in Fig. 3-11b,

Ans.s = 519 mm

tan 49.16° = 600 mms

g = cos-1 98.1150

= 49.16°

3[(50 N) cos g] - 10(9.81) N = 0gFz = 0;

T = 50 Nz

gzgFy = 0

gFx = 0DA = DB = DC = 600 mm

x

y

s

600 mm

D

z

(b)

A

B

C

10(9.81) N

TT

Tg

DCL

Exercício 1: Determine as intensidades das forças na figura abaixo de tal modo que a partícula esteja em equilíbrio.

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FUNDAMENTAL PROBLEMS

All problem solutions must include an FBD.

F3–7. Determine the magnitude of forces sothat the particle is held in equilibrium.

F3,F2,F1,

F3–10. Determine the tension developed in cables AB,AC, and AD.

F3–8. Determine the tension developed in cables AB, AC,and AD.

F3–9. Determine the tension developed in cables AB, AC,and AD.

F3–11. The 150-lb crate is supported by cables AB, AC,and AD. Determine the tension in these wires.

900 N

600 N

z

x y

4

44

3

3

3

5

5 F1

F2

F3

5

A

C

z

y

xB

D

33

4

45

5

900 lb

2 m1 m

2 mA

C

z

y

x B

D

600 N

30!

A

Cz

y

x

B

60º

300 lb

30!

45!

120!

60!

D

A

D

E

B

C

2 ft

3 ft

3 ft

2 ft

6 ft

F3–9 F3–11

F3–8

F3–7

F3–10

Exercício 2: Determine as intensidades das trações nos caos AB, AC e AD da figura.

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FUNDAMENTAL PROBLEMS

All problem solutions must include an FBD.

F3–7. Determine the magnitude of forces sothat the particle is held in equilibrium.

F3,F2,F1,

F3–10. Determine the tension developed in cables AB,AC, and AD.

F3–8. Determine the tension developed in cables AB, AC,and AD.

F3–9. Determine the tension developed in cables AB, AC,and AD.

F3–11. The 150-lb crate is supported by cables AB, AC,and AD. Determine the tension in these wires.

900 N

600 N

z

x y

4

44

3

3

3

5

5 F1

F2

F3

5

A

C

z

y

xB

D

33

4

45

5

900 lb

2 m1 m

2 mA

C

z

y

x B

D

600 N

30!

A

Cz

y

x

B

60º

300 lb

30!

45!

120!

60!

D

A

D

E

B

C

2 ft

3 ft

3 ft

2 ft

6 ft

F3–9 F3–11

F3–8

F3–7

F3–10