Equilíbrio Do Ponto Materia_aula2

7/24/2019 Equilbrio Do Ponto Materia_aula2

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zx

y

Esttica da Partcula

no espao

Fx = 0 Fy = 0 Fz= 0Fx = 0 Fy = 0

no plano

x

y

Sir Isaac Newton

England, 4 January

1643

31 March 1727


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Equilbrio da Partcula no plano

Condies necessrias de equilbrio no plano

0FFFFFR 4321

0)()(

jFiF yx

00 yx FF

01000500150060cos200060cos10001500F oox 0173286686660sen200060sen1000866F ooy

F1A

F2

F3

F4

30o

30o

F1= 1500 NF2= 866 N

F3= 1000 N

F4= 2000 N

x

y


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Diagrama de Corpo Livre

Picture from chapter 2, BEER F. P. e

JOHNSTON E. R. Vectors Mechanics for

Engineers

Statics, sixth edition, Mc

Graw-Hill, NC, USA, 1996.


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Engineers



AA

T

T

mg

x

y

A

TT

mg

x

y

A

2T

mg

x

y


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Engineers



A

T

T

mg

x

y

A

TT

mg

x

y

A

2T

mg

x

y

A



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Engineers



x A

3T

mg

x

y

A A

TT

mg

x

y

T

A

T

T

mg

y

T



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Engineers



A

3T

mg

x

y

A A

TT

mg

x

y

T

xA

T

T

mg

y

T



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Engineers

Statics, sixth edition, McGraw-Hill, NC, USA, 1996.

A

ATT

mg

x

y

TT

A

TT

mg

x

y

T

T

A

mg

x

y

4T


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Equilbrio da Partcula no plano e no espao

Mtodo de Soluo de Problemas

1

DCL: Diagrama de Corpo Livre

2 Foras dadas por suas componentes retangulares

3 Equaes de equilbrio da partcula

4 Resoluo

5 - Verificao


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Equilbrio da Partcula no plano Exemplo 1

0F0F yx

01000500F60cos200060cos1000FF 1oo

1x 01732866F60sen200060sen1000FF 2

oo2y

F1

A

F2

F3

F4

30o

30o

F1= ?

F2= ?F3= 1000 N

F4= 2000 N

x

yDeterminar as intensidades das foras

F1 e F2 para que o ponto material A

permanea em repouso.

N1500F1 N866F2


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050cosT30cosTF oABo

ACx

073650senT30senTF oABoACy

Determinar as traes nas cordas, AB

e AC para equilibrar a carga de 75 kg.

N647TAB

N480TAC

A

TAB

TAC

mg

30o50o

x

y

mg = (75kg)x(9,81m/s2) = 736N

0F0F yx


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Determinar as traes nas cordas, AB

e AC para equilibrar a carga de 75 kg.

N647TAB N480TAC

A

TAB

TAC

mg60

o

50o

B

C

A

TAB

TAC

736N

60o

40o

B

C

80o

Lei dos senos

oo

AC

o

AB

80sen

N736

40sen

T

60sen

T

Tringulo ABC

mg = (75kg)x(9,81m/s2) = 736N


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0F0F yx

0cosP20Psen2F ox 01570Psen20cosP2F oy

Determinar a trao P na corda e a inclinao

para equilibrar a carga m de 160 kg. b= 20o

m

A

P

P

P

mg

b

x

y

ou

o47

N601P

o

47

N1365P

mg = (160kg)x(9,81m/s2) = 1570N


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Determinar a trao P na corda e a inclinao

para equilibrar a carga m de 160 kg. b= 20o

m

o

47

N601P AP

2Pmg20

o

90B

C

Lei dos senos

)]9020(180[senmg

)90(senP2

20senP

ooooo

Tringulo ABC

A

P

P

Pmg

b

B

C

mg = (160kg)x(9,81m/s2) = 1570N

C


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Foras no espao

x

z

B

C

Oqx

Fy

FxFz

F A

D

E

y y

B

C

O

qy

Fy

FxFz

F A

E

z

x

Dx

B

C

O

qz

Fy

FxFz

F AD

z

E

y

xx cosFF q yy cosFF q zz cosFF q

kFjFiFF zyx

2z

2

y

2

x FFFF

Fvetordodireoaindicame, zyx

qqq

ExemploF = 1000 N

qx= 30

qy= 60

qz= 120o

Fx= (1000N)xcos30o= 866 N

Fy= (1000N)xcos60o= 500 N

Fz= (1000N)xcos120o

= -500 N

Componentes retangulares

C t t l


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x

z

Fy

FxFz

F

y

lx

ly

lz

l

lqqq

F)kcosjcosi(cosFkFjFiFFzyxzyx

kji)kcosjcosi(cos1 zyxzyx

lllqqql

l

FF

F

Fcos

F

Fcos

F

Fcos

zzz

y

yy

xxx

ql

ql

ql

1coscoscos z2

y

2

x

2

qqq

Foras no espao Componentes retangulares

F C t t l


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500,0

1000

500

F

Fcos

500,01000

500

F

Fcos

866,0

1000

866

F

Fcos

zzz

y

yy

xxx

ql

ql

ql

Exemplo

F = ?

qx= ?

qy= ?

qz= ?

Fx= 866 N

Fy= 500 N

Fz= -500 N

N1000FFFF 2

z

2

y

2

x

x = 30

y= 60

z= 120o

kFjFiFF zyx

k500,0j500,0i866,0

l

x

z

Fy

Fx

Fz

F

y

l

qz

qx

qy

l

FF


F C t t l


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Fora definida pela sua intensidade e por dois

pontos da sua linha de ao dMN

d

dcos

d

dcos

d

dcos

zzz

y

yy

xxx

ql

ql

ql

kdjdidd

1kji zyxzyx

llll

kdjdiddF

FF zyx

l

d

dFcosFF

d

dFcosFF

d

dFcosFF

zzz

y

yy

xxx

q

q

qx

z

F

y

l

M(x1,y1,z1)

N(x2,y2, z2)

dx=x2x1

dz=z2z1

dy=y2y1

O

2z

2y

2x dddd


EXEMPLO 1

y


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k)m30(j)m80(i)m40(m3,94

N2500FF

l

EXEMPLO 1

B

A

40 m 30 m

80 m

A trao no cabo AB que equilibra a

torre da figura abaixo de 2500 N.

Determinar ento as componentes

Fx, Fy e Fz desta fora agindo noponto A e os ngulos qx, qy e qz.

dmdmd zyx )3(0;80080;40400

m3,94ddddAB 2

z2

y2

x

k)m30(j)m80(i)m40(m3,94

1

AB

BA

l

A(40,0,-30)m

B(0,80,0)m

i

j

k

x

z

y

40 m 30 m

F

l80 m

2500N

EXEMPLO 1


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EXEMPLO 1

k)m30(j)m80(i)m40(m3,941

ABBA

l

k)m30(j)m80(i)m40(m3,94

N2500FF

l

A(40,0,-30)m

B(0,80,0)m

i

j

k

x

z

y

40 m 30 m

F

l

80 m

2500N

k)N795(j)N2120(i)N1060(F

N795FN2120FN1060F zyx

Direo da fora

Fora

2500795

FF

ddcos

2500

2120

F

F

d

dcos

25001060

FF

ddcos

zzzz

yy

yy

xxxx

ql

ql

ql

oz

oy

ox 7232115 qqq

EXEMPLO 1


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k)N795(j)N2120(i)N1060(F

oz

oy

ox 7232115 qqq

i

j

k

x

z

y

40 m 30 m

F

80 m qx

qy

qz

2500N

EXEMPLO 1

E ilb i d P t l


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Equilbrio da Partcula no espao

0FF...FFFR N321

0k)F(j)F(i)F(R zyx

0FR0FR0FR zzyyxx

A

F2

F1F3

x

y

z

O

F4

Eq ilbrio da partc la no espao E emplo 1


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Equilbrio da partcula no espao Exemplo 1

030000300FR

0300100100100FR

001000100FR

zz

yy

xx

k)N300(j)N300(i0F

k0j)N100(i)N100(F

k0j)N100(i0F

k)N300(j)N100(i)N100(F

4

3

2

1

F4

z

A

F2

F1F3

x

y

O

Exemplo


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Trs cabos esto conectados em A para

sustentar o peso P. etermine o valor da

trao em cada cabosabendo-se que o pesoP igual a 960 N.

A(960, 240, 0)mmB(0, 0, 380)mm

C(0, 0, -320)mm

D(0, 960, -220)mm AB

C

D

P

96

0

24

0

22

0

32

038

0

x

y

z

Exemplo


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A(960, 240, 0)mm

B(0, 0, 380)mm

C(0, 0, -320)mm

D(0, 960, -220)mm

AB

C

D

P960

960

240

220

320

380

x

y

z

TAC

TAD

TAB

k0j0iPP

)k380j240i960(1060

TT ABAB

)k320j240i960(1040

TT ACAC

)k220j720i960(1220

TT ADAD


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k0j0iPP

kT358,0jT226,0iT906,0T ABABABAB

kT308,0jT231,0iT923,0T ACACACAC

kT180,0jT590,0iT787,0T ADADADAD

A

P

y

z

TAC

TAD

TAB

x

0T180,0T308,0T358,0FR

0T590,0T231,0T226,0FR

0T787,0T923,0T906,0PFR

ADACABzz

ADACAByy

ADACABxx

0FR0FR0FR zzyyxx


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0T180,0T308,0T358,00T590,0T231,0T226,0

PT787,0T923,0T906,0

ADACAB

ADACAB

ADACAB

0F0F0F zyx

P = 960 N

0T180,0T308,0T358,0

0T590,0T231,0T226,0

960T787,0T923,0T906,0

ADACAB

ADACAB

ADACAB

N305T

N341T

N447T

AD

AC

AB

A

P

y

z

TACxTADy

TABz

x

TADz

TADx

TACz

TACy

TABy

TABx



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Mtodo de Soluo de Problemas

1

DCL: Diagrama de Corpo Livre



4

Resoluo

5 - Verificao

Exemplo


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Trs cabos esto conectados em A para

sustentar o peso P. etermine o valor da

trao em cada cabosabendo-se que o peso

P igual a 960 N.

A(960, 240, 0)mmB(0, 0, 380)mm

C(0, 0, -320)mm

D(0, 960, -220)mmA

B

C

D

P

960

240

220

320

380

x

y

z

Exemplo


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AB

C

D

P960

960

240

220

320

380

x

y

z

TAC

TAD

TAB

1DCL: Diagrama de Corpo Livre



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k0j0iPP

kT358,0jT226,0iT906,0T ABABABAB

kT308,0jT231,0iT923,0T ACACACAC

kT180,0jT590,0iT787,0T ADADADAD

A(960, 240, 0)mm

B(0, 0, 380)mm

C(0, 0, -320)mm

D(0, 960, -220)mm


l

FF

AB

C

D

P960

960

240

220

320

380

x

y

z

TAC

TAD

TAB


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0F0F0F zyx

0T180,0T308,0T358,00T590,0T231,0T226,0

960T787,0T923,0T906,0

ADACAB

ADACAB

ADACAB



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N305T

N341T

N447T

AD

AC

AB

4 Resoluo

0T180,0T308,0T358,0

0T590,0T231,0T226,0

960T787,0T923,0T906,0

ADACAB

ADACAB

ADACAB


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5 - Verificao

N305T

N341T

N447T

AD

AC

AB

0T180,0T308,0T358,0

0T590,0T231,0T226,0

960T787,0T923,0T906,0

ADACAB

ADACAB

ADACAB

00

00

960960

Documents

Equilíbrio Do Ponto Materia_aula2