Capítulo 2

ESTÁTICA DOS FLUIDOS

A ausência de movimento elimina os efeitos tangenciais e conseqüentemente a presença de tensões de cisalhamento. A presença exclusiva de efeitos normais faz com que o objetivo deste capítulo seja o estudo da pressão. Nesse caso são vistas suas propriedades num fluido em repouso, suas unidades, as escalas para a medida, alguns instrumentos básicos e a equação manométrica, de grande utilidade. Estuda-se o cálculo da resultante das pressões em superfícies submersas, o cálculo do empuxo, que também terá utilidade nos problemas do Capítulo 9, a determinação da estabilidade de flutuantes e o equilíbrio relativo. É importante ressaltar, em todas as aplicações, que o fluido está em repouso, para que o leitor não tente aplicar, indevidamente, alguns conceitos deste capítulo em fluidos em movimento. Para que não haja confusão, quando a pressão é indicada na escala efetiva ou relativa, não se escreve nada após a unidade, quando a escala for a absoluta, escreve-se (abs) após a unidade. Exercício 2.1

( )

N13510101035,1G

Pa1035,1205104,5

AA

pp

Pa104,5210

5,21072,21010500AA

ApApp

ApGApAp

Pa1072,22000.136hp

ApAApAp

45

55

IV

III34

553

HII

II2I13

V4

IV4III3

5Hg2

II2HII3I1

=×××=

×=××==

×=−

××−××=

−−

=

==

×=×=γ=

+−=

−

Exercício 2.2

kN10N000.10525400

DD

FF

4DF

4D

F

N4001,02,0200F

1,0F2,0F

2

21

22

BO22

21

BO

BO

BOAO

==⎟⎠⎞

⎜⎝⎛×=⎟

⎟⎠

⎞⎜⎜⎝

⎛=⇒

π=

π

=×=

×=×

Exercício 2.3

mm3681000000.136

5000.10h

hh

Hg

OHOHHgHg 22

=××

=

γ=γ

Exercício 2.4

)abs(mmHg3400)abs(cmkgf62,4)abs(MPa453,0)abs(

mkgf200.46)abs(atm47,4p

mca10atm97,0MPa098,0Pa108,9cmkgf1

mkgf000.1074,0600.13hp

mca2,36000.1200.36ph

bar55,398,0cmkgf62,310

mkgf200.36p

MPa355,0108,9mkgf200.3666,2600.13hp

mmHg26601

5,3760p

patm5,3mmHg760atm1

22abs

422HgHgatm

O2HO2H

24

2

62HgHg

=====

===×≅=≅×=γ=

==γ

=

=×=×=

=××=×=γ=

=×

=

→→

−

−

Exercício 2.5

kPa35,13Pa350.13025,0000.101,0000.136p

01,0025,0p

1

HgOH1 2

==×−×=

=×γ−×γ+

Exercício 2.6

kPa1,132Pa100.1321000.13625,0000.108,0000.8pp

p8,0125,0p

BA

BOHgO2HA

−=−=×−×−×=−

=×γ−×γ+×γ+

Exercício 2.7

kPa6,794,20100p

kPa4,20Pa400.2015,0000.13615,0pp100p

m

HgA

Am

=−=

==×=×γ=−=

Exercício 2.8

kPa55,36103,0500.834p

p3,0p)b

)abs(kPa13410034ppp

kPa100Pa000.10074,0000.136hpkPa34Pa000.348,0500.83,0000.136p

07,03,07,08,0p)a

3M

MOar

atmarabsar

HgHgatm

ar

O2HHgO2HOar

=××+=

=×γ+

=+=+=

≅≅×=γ===×−×=

=×γ−×γ−×γ+×γ+

−

)abs(kPa55,13610055,36ppp atmMabsM =+=+= Exercício 2.9

( )

( ) )abs(mca12,17000.10000.171p

h

)abs(Pa200.171200.95000.76pppPa200.95000.1367,0p

Pa000.76p000.574

pp

000.57pp000.30p000.27p000.27pppap

000.30pp

p4p4AA

AA

AA

ApApAApApAp

2AA

kPa30pp

OH

absBOH

atmBB

atm

BB

B

ABAB

BCBC

AC

ABH

2

H

1

1

2

HB2AH1B1B2A

1

2

AC

22

efabs

==γ

=

=+=+==×=

=→=−

=−→=−−−=→=γ+

=−

=→==×

=→−−=

=

=−

Exercício 2.10

)abs(kPa991001pppkPa1Pa000.12,010500ghp

mkg500

2,01,0000.1

hhhh0ghp

0ghp

atm0abs0

AA0

3A

BBABBAABB0

AA0

=+−=+=−=−=××−=ρ−=

=×=ρ=ρ⇒ρ=ρ⇒=ρ+

=ρ+

Exercício 2.11

( ) ( )

( ) ( )

33243

o

OH

OHo

OHo

cm833.47m107833,41043,06

45,0xA6DV)c

m45,03,05,0000.8

6,04,0000.10x5,0x2y

D

m3,02

4,012

yyxyyx2

x2yx5,0D)b

m4,0000.10

5,0000.8y

y5,0)a

2

2

2

=×=××+×π

=+π

=

=−−+

=−−γ

+γ=

=−

=−′

=→′=+

+γ=++γ

=×

=

×γ=×γ

−−

Exercício 2.12

( )( ) ( )

m105

5,11sen5,4

1000.8

10

senDd

pL0LsenDdLp

DdLH

4DH

4dL

Pa10001,010001,0p

0LsenHp

3

o22

x2

x

222

4O2Hx

x

−×=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎠

⎞⎜⎝

⎛

−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡α+⎟

⎠⎞

⎜⎝⎛γ

−=⇒=

⎥⎥⎦

⎤

⎢⎢⎣

⎡α+⎟

⎠⎞

⎜⎝⎛γ+

⎟⎠⎞

⎜⎝⎛=⇒

π=

π

−=−×=−×γ=

=α+γ+

Exercício 2.13

( )( )

( )

( )

( )

mca7,3000.10000.37p

Pa000.37000.17000.20000.17pp)b

absmmHg831684147p

mmHg147m147,0000.136000.20Pa000.20p

000.17p10331p104:)1(nadoSubstituin

p000.17p

p4,0000.104,0000.5005,0000.102p

m05,04,717,35

24,0

Dd

2hh

4d

2h

4Dh

phhh2p

1p10331p104

0357,00714,04

p3140714,0p

dD4

pF4Dp)a

2

12

abs1

1

1

21

21

2221

21

21

21

ar

arar

ar

ar

ar3

ar3

arar

arar

2222

arOHmOHar

ar3

ar3

22ar

2ar

22ar

2ar

==

=+=+=

=+=

====

+×=+×

=+

=×−×+×××

=⎟⎠

⎞⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛=Δ→

π=

πΔ

=γ−γ+Δγ+

×=+×

−π

=+×π

−π

=+π

−−

−−

Exercício 2.14

( )

1

2

11

22222

111

arar

21

ar

HgO2Har

TT

VpVpmRTVp

mRTVp)cPa050.12p0000.1361,0000.10155,0p

cm51

105,0hA.hA.y)b

Pa200.25000.10000.1362,0p

02,02,0p)a

=⇒=

==′⇒=×−×+′

=×=Δ⇒Δ=Δ

=−=

=×γ−×γ+

C44K31710095

200.125050.112373T

cm95105,01010V

050.112000.100050.12p

)abs(Pa200.125000.100200.25p

o2

32

abs2

abs1

==××=

=×−×=

=+=

=+=

Exercício 2.15

3A

AA

atmAAabs

atm

OH

AOH

A

2222A

212A

m

kg12,1293287

576.94RTp

)abs(Pa576.94200.95624pppPa200.95000.1367,0p)b

mca0624,0000.10

624ph

Pa6240015,02000.8600p

m0015,0404

23,0

Dd

2hh

4d

2h

4Dh

h2000.83,0000.103,0000.8p0hhh2p)a

22

=×

==ρ

=+−=+==×=

−=−=γ

=

−=××−−=

=⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=Δ→

π=

πΔ

Δ×−×−×==γ−γ+Δγ+

Exercício 2.16

3

1

2

2

112

1

2

11

22

absgásO2Hgás

O2Hgás

absgás

atm

gásO2HHggás

m16,2293333

100952

TT

ppVV

TT

VpVp

)abs(kPa1001090pkPa10Pa000.101000.10z.p)c

m5,0000.10000.5zz.p)b

)abs(kPa95590pkPa90Pa032.90662,0000.136p

Pa500016,0000.10025,0000.136p16,0025,0p)a

=××==⇒=

=+=′⇒==×=′γ=′

==⇒γ=

=+===×=

=×+×=⇒×γ+×γ=

Exercício 2.17

( ) ( ) 23

222

21

23

322

212

21

1

32

21

3,0p1,05,0p5,0p4DpDD

4p

4Dp

000.22,0000.10pp000.10pp

×+−×=×→π

+−π

=π

=×=−=−

( )

( )kPa5,43Pa500.43p3480p08,0

180000.10p33,0p25,0180p33,0p25,0

000.2p09,0p24,0p25,0

11

11

21

221

==→=−−=

−=−+=

Exercício 2.18

3222

2

ct

c

tt

pGt

oG

p

22c

22p

22c

11p

mkg993.10

183,05,010950.34

LDgG4

L4Dg

GgVG)c

m183,05,0210

5,110005,0L

m0005,02

5,0501,02

DDDv

FLDLvF)b

N5,11FFFdesce196319755,0395030GsenF

cimaparaN196378549817F

N78544

5,0000.404DpF

N98174

5,0000.504DpF)a

=××π×

×=

π=

π==ρ

=×π××

×=

=−

=−

=ε

πμε

=⇒πε

μ=

=−=>=×==

=−=

=×π

×=π

=

=×π

×=π

=

−

Exercício 2.19

( ) ( )

( ) ( )

cm8,127m278,1278,01L

m278,0ym0278,0x0600.36x10098,1x000.9080008002

600.552

0200.735,0x15000.10x98,0800AF2

x10yy2,0x2

0200.7330ysen30sen1y000.10y25,0x55,0000.81,0AF2

mN200.73

30sen1

8,0000.101,0000.82600.55

30Lsen

8,01,0AF

030Lsen8,01,0AF

6

oo

3oo

21

3

o321

==+=′

=⇒=⇒=−×−+++×

=×+−×+++

=⇒=

=×+×+−×++−+×+

=×

×+×+=

×γ+×γ+=γ

=γ−×γ+×γ+

Exercício 2.20

( ) ( )

( ) ( )( ) ( )

( )kPa50109,39ppp)c

)abs(kPa1,60)abs(Pa100.6039908000.100p

Pa908.39103,50

150102013,50000.10100A

FAApGp

FApAApAApG

cm3,5048

4DA;cm201

416

4DA)b

N15005,008,016,0001,058,0DDvF

sm.N8,0

10000.810

g)a

abm

absb

4

4

2

t12ab

t2bH1aH2a

222

22

222

11

21t

2

3

−=−−=−=

==−+=

−=×

−×−×+=

−−+=

++−=−+

=×π

=π

==×π

=π

=

=×+×π××=+πε

μ=

=×

=μγ

=ν

−

−

−

l

Exercício 2.21

2

3

p

pp

pp

p

2p

p

pp2

12

ms.N8,0

10000.810

g

m001,02998,01

2DD

DvL4pLv

4D

p

LD4D

p

pistãonomédiapressãopondephp000.10pp

=×

=νγ

=μ

=−

=−

=ε

εμ

=→ε

μ=

τπ=π

==γ+=−

−

Exercício 2.22

N33933,04

2,1000.10b4RF

N160.23,02,16,0000.10AhF22

y

x

=××π

×=π

γ=

=×××=γ=

kPa23,25Pa230.25000.10230.15000.10ppmN230.152000.85,769hpp

Pa5,769998,0001,0

2,02,18,04p

21

2p2

p

−=−=−−=−=

=×−=γ−=

=×

×××=

Exercício 2.23

m4,02,06,0b

m2,06h

h2hAh

Ihh

N920.252,122,1000.30hhApF

m2,14,06,0000.30000.804,06,0h

6,0.4,0.h

2

124h

CGcp

22p

m

m

=−=

==×

==−

=××=γ==

=−×=−×γγ

=

γ=γ+γ

N640.82,14,025920

hbFFbFhF pp =×==→×=×

Exercício 2.24

N948.59100.115,42,1F

N668.72

100.11100.5100.52,16,0F

N755.285,46,02

100.11100.55,46,02100.5FFF

Pa100.116,0000.10100.56,0ppPa100.56,0500.86,0p

f

B

21A

212

11

=××=

=⎟⎠⎞

⎜⎝⎛ ++

××=

=××+

+××=+=

=×+=×γ+==×=×γ=

Exercício2.25

N500.225,121500.7AhFm0833,10833,01

m0833,05,124AhAh

Ihh

N102,15,124000.10AhApF

F2FF

2o2

1

12325,1

1

123bh

1

CG11CP

51O2H11

22B11

=×××=γ=

=+=

=××

===−

×=×××=γ==

+×=

×

l

ll

m333,1333,01

m333,05,121Ah

hh

2

12325,1

2

123bh

22CP

=+=

=××

==−×

l

N105F

333,1500.222F0833,1102,14

B

B5

×=

×+×=××

F

Fp

h hcp

b

h

5m 2 m

A

B

1l 2l

3 m

F1 F2

FB

Exercício 2.26

m736,0634.7680.42,1

FFyxxFyF

N634.73,04

8,1000.10b4RF

m2,18,132R

32y

N860.43,028,1000.10bR

2RF

y

xCPCPCPyCPx

22

y

c

2

x

=×==⇒=

=××π

×=π

γ=

=×==

=×=••γ=

Exercício 2.27

m65,230cos75,02h

AhApFo =×+=

γ==

kN4,991075,365,2000.10F

m75,35,25,1A3

2

=×××=

=×=−

Exercício 2.28

( )

( )

( ) ( )3

oO2H

2

O2H

22

oinfsup

2

2

O2Hinf

2

osup

mN000.35

6,05,2000.86,05,3000.10

6,0

h6,0h4D6,0h6,0

4D

4DhFGF

6,04DG

4D6,0hF

4DhF

=×−+×

=γ−+′γ

=γ

π+′γ=×

πγ+

πγ⇒=+

×π

γ=

π+′γ=

πγ=

Exercício 2.29

xCG CG

γ1

γ2

R R

O

Fx1 F2

Fy1

21 ll =

2bRRb

2RF

AhF

FxFF

21

11x

1111x

22CG1y11x

γ=γ=

γ=

=+ ll

6R

Rb2RAh

Ihh 123bR

CG11CP ===−

31

22

3R

2bR

3R4

4bR

3R

2bR

b4RVF

2bR

Rb2RAhF

3R

6R

2R

2

121

1

22

21

21

211y

22

22222

21

1

=γγ

→γ

=γ+γ

×γ

=π

×πγ

+×γ

πγ=γ=

γ=γ=γ=

==−= ll

Exercício 2.30

( )

( )

N3,4651

579,0300.14583,0000.15BA

brFbrFFBAFMM

m579,0079,05,0br

m079,05,106,1

125,0Ay

Iyy

m06,156,05,0y

m56,0000.9032.5ph

N300.145,11532.9ApFPa532.92

032.14032.52

ppp

Pa032.141000.950321pp

Pa032.5037,0000.136037,0ppm583,0083,05,0br

m083,05,11

125,0yy

m125,012

15,112bI

AyIyy)b

N000.155,11000.10ApFPa000.102

000.15000.52

ppp

Pa000.155,1000.105,1p

Pa000.55,0000.105,0p)a

esqesqdirdirBBesqdir

esq

esq

CGesqCP

esq

o

arareq

esqesqesqBesqA

esq

oesqAesqB

HgaresqA

dir

dirCP

433

CGCG

CP

dirdirdirBdirA

dir

O2HdirB

O2HdirA

=×−×

=−

=⇒×+=

=+=

=×

==−

=+=

==γ

=

≅××==⇒=+

=+

=

=×+=×γ+=

=×=×γ===+=

=×

=−

=×

==→=−

=××==⇒=+

=+

=

=×=×γ=

=×=×γ=

l

Exercício 2.31

( ) ( ) N6363,06,04

3,0000.103,0D4

hApF

N107,14

6,06,0000.104D

hApF

2222MMMMM

322

FFFFF

=−π

××=−π

γ==

×=×π

××=π

γ==

Exercício 2.32

N230.762

083,1000.1205,0000.45F083,1F5,0F2F

m083,0412

2y12by

12/bAy

Iyy

N000.1205,12000.40ApFPa000.402

000.50000.30p

Pa000.505000.105p

m3000.10000.30ph

N000.455,11000.30ApF

Pa000.304,0000.1025,0000.1364,025,0p

BCAB

223CG

CP

BCBCBCBC

O2HC

O2H

AB

ABABAB

O2HHgAB

=×+×

=⇒×+×=×

=×

====−

=××=×=⇒=+

=

=×=×γ=

==γ

=

=××==

=×−×=×γ−×γ=

l

l

l

Exercício 2.33

Exercício 2.34

m1CBMM2CBbCB3M

33b3

23M

BCAB

BCAB

=⇒=

γ=→γ=

F1 F2

1l 2l ( )

( ) ( )

( ) ( )

m27,6z5,1108,225,6z5,2

5,115,2z

08,25,25,2z

5,2106,45,2z

08,25,25,2z10

m5,2N106,4251046pAF

5,2z08,25,2

55

2

532

1

==+−

=⎥⎦

⎤⎢⎣

⎡−

+−

××=⎥⎦

⎤⎢⎣

⎡−

+−

=×=×××==

−+=

l

l

Exercício 2.35

21

hxh

3x6h

3x

2xhxb

3xb

2x

2x

hxbF3x

xb2xAhF

FF

2

1

22

1

2

22

1

1111

2211

=→=→=γγ

×γ=×γ

=

γ=

=

γ=γ=

=

l

l

ll

Exercício 2.36

kN204H880.218015H

m.kN1805,1120MkN120000.1

134000.10V

m.kN880.2000.1

41126000.10M

V

x

=⇒+=×

=×=⇒=×××

=

=××××

=

Exercício 2.37 O ferro estará totalmente submerso.

N2183,04

3,0300.10h4DVE

22flfl =×

×π×=

πγ=γ=

A madeira ficará imersa na posição em que o peso seja igual ao empuxo.

sub2

fl

22mad

h4DE

N1593,04

3,0500.7h4DGE

πγ=

=××π

×=π

γ==

m218,03,0300.10

1594

D

E4h22

flsub =

×π×

×=

πγ=

Exercício 2.38

N625023,0000.25500VGG conconcil =×+=γ+=

F1

F2

1l

2l

( )

m3,02,05,0h

m5,01

23,0000.10

62504

DV/G4H

H4DVGEG

22con

2

con

=−=

=×π

⎟⎠⎞

⎜⎝⎛ −×

=π−γ

=

⎟⎟⎠

⎞⎜⎜⎝

⎛×

π+γ=⇒=

Exercício 2.39

( ) m7,29,08,1BAx:Logo

m9,0270

6,0080.13,0350.1F

GE

m6,038,1

3BA

m3,039,0

3IH

N270080.1350.1GEF:Logo

N080.112

6,08,1000.2b2

CBBAVG

N350.112

9,03,0000.10b2

IHCHVE

2BAIH

FGEEGF

2F

213

2

1

ccc

OHsubOH

321

22

=−−=−=

−=×−×

=−

=

===

===

=−=−=

=××

×=××

γ=γ=

=××

×=××

γ=γ=

=

+==+

l

lll

l

l

lll

A força deverá ser aplicada à direita do ponto B, fora da plataforma AB. Exercício 2.40

( )( )

( )( ) 22dd

4443

odo

3

m1036,3A02,0A3,031055103,002,01012

6,0

AARhGRA26

D

−×=⇒−+×+=××−××π

−+γ+=γ−γ××π

A B

C

I H

E

G

F

1l 2l

3l

Exercício 2.41 Supondo o empuxo do ar desprezível:

3cccc

3

flfl

ap

m

N670.2603,0

800VGVG

m03,0000.10

300EVVE

N300500800EEGG

===γ→γ=

==γ

=→γ=

=−=→+=

Exercício 2.42

mm2,7m102,7005,0

104,14dV4hh

4dV

m104,11068,21082,2V

m1068,2200.8102,2GVVEG

m1082,2800.7102,2GVVEG

32

7

2

2

3766

362

22222

362

11111

=×=×π

××=

πΔ

=Δ⇒Δ×π

=Δ

×=×−×=Δ

×=×

=γ

=⇒γ==

×=×

=γ

=⇒γ==

−−

−−−

−−

−−

Exercício 2.43

( )

( )

( )( )

m8,0hh000.16000.40h000.6000.32h5,2000.16h000.6000.32

h5,14hpm

N000.324000.8p4AApGAp

2Situaçãom

N000.1622A4A

EG1Situação

ooo

oo

ooobase

2basebasecbasebasebasebase

3cbbc

=→−+=−+=

−−γ+γ=

=×=→×γ=→=

=γ→γ=γ→×γ=×γ

=→

l

lll

Exercício 2.44

m6000.61009,22105,4x

N1009,212

21026

DE

N105,4135,110AhFGE2FxxE3

32FxG

4

4

43

43

44

=−×××

=

×=×π

×=×π

γ=

×=×××=γ=−×

=⇒•=××+•

E

G F

Exercício 2.45

( )

( )

( )

3B

B

BAbase

2b

bc

bbase

bbase

3cAbAbc

m

N000.25

4,02,0000.15000.132,06,02,0p

m

N000.131

000.1016,0000.5A

FA6,0A

FGp

FGAp2Situação

m

N000.15000.5332,0A6,0AEG

1Situação

=γ

×γ+×=−×γ+×γ=

=+××

=+××γ

=+

=

+=

=×=γ=γ→×γ=×γ→=

Exercício 2.46

( ) ( ) N171.106121085,7132,110

6DgG

1085,7293400.41

200.95TR

pmkg132,1

293287200.95

TRp

Pa200.957,0000.1367,0p

33

3

2Har

3

2H

2H2H

3ar

arar

Hgatm

=×π

××−×=π

ρ−ρ=

×=×

==ρ

=×

==ρ

=×=×γ=

−

−

Exercício 2.47

79,0x21,0x

6216466x:Raízes

01x6x6

02x

21

x121xFazendo0

221

12

02b

2b

b

2b

2b0

V

Ir

bhbhbEG

2

2

cc

c

c3

c

12b

c

c

y

csub

2sub

3c

4

=′′=′

→×

××−±=

>+−

>+−→=γγ

→>γγ

+−γγ

>⎟⎟⎠

⎞⎜⎜⎝

⎛γγ

−−γ

γ

γγ

−=→>−γ

γ=

γγ

=→γ=γ→=

ll

l

l

l

l

l

ll

ll

179,021,00 cc <γγ

<<γγ

<ll

Exercício 2.48

estável0m037,00467,05,2

103,083.2000.10r

cm3,083.212

102512bLI0

GI

r

cm67,433,05cm5yCG

cm33,05,032yCC

cm5,010

5,2L

Vh

hL2

bh2V

m105,2000.105,2GV

GVEG

8

433

yyf

im

2im

imim

34

fim

imf

⇒>=−××

=

=×

==→>−γ

=

=−=⇒=→

=×=→

===

==

×==γ

=

=γ⇒=

−

−

l

l

l

Exercício 2.49

( )

( )⎟⎟⎠

⎞⎜⎜⎝

⎛γγ

−γγ

<→−

<

<−−→>+−

=γγ

>γγ

+−γγ

→>γγ

+−γ

γ→>⎟⎟

⎠

⎞⎜⎜⎝

⎛γγ

−−γπ

πγ

⎟⎟⎠

⎞⎜⎜⎝

⎛γγ

−=−=π

=γπ=

>−γ

=

γγ

=

γπ=πγ

=

ll

l

l

l

l

l

l

l

l

l

l

l

l

l

l

12

1RH

x1x21

RH

01x2x2R

H0R

H2.xR

H2x1

:RportudodividindoexFazendo

0H2H2R02H

2H

H4R

0HH21

HR4

R

HH21

2h

2H

4RIHRG

0G

Ir

Hh

HRhR

GE

22

2

2

2

2

2

2

2222

2

4

sub4

y2

y

sub

2sub

2

CG

CC 0,5cm

Exercício 2.50

z6gg51z

ga

1zp yz Δγ=⎟⎟

⎠

⎞⎜⎜⎝

⎛+Δγ=⎟⎟

⎠

⎞⎜⎜⎝

⎛±Δγ=Δ

Exercício 2.51

hkm2,646,3

sm83,17557,3tav)b

s

m57,320tg8,9a20tggag

axz)a

x

2o

xo

xx

=×=×==

=×=→=→=ΔΔ

Exercício 2.52

ooo

x 4130tg30cos8,9

45,2tgcosgatg =θ⇒+

×=α+

α=θ

Exercício 2.53

( )

2x

3x

3

Hg

s

m72,15,1

257,010xzga

m257,0000.136

10140175zg

axz)b

m29,1000.13610175ph)a

=×=ΔΔ

=

=×−

=Δ→=ΔΔ

=×

=γ

=

Exercício 2.54

)abs(kPa10610

6,010000.1100ghpp

)abs(kPa7,12510

6,010000.17,119ghpp

)abs(kPa7,119100106,025,10000.1p

srd5,10

601002n2pr

2p

3atmC

3AB

322

A

atm2

2

A

=××

+=ρ+=

=××

+=ρ+=

=+×⎟⎟⎠

⎞⎜⎜⎝

⎛××=

=×π×=π=ω→+Δω

ρ=

−

Exercício 2.55

2xx

sm78,2

106,3

100

tva

gatg)a ===→=α

140

175 Pa

zΔ

( ) ( )( ) ( ) Pa600.314,05,0000.10h5,0p

Pa400.614,05,0000.10h5,0p

m14,0278,05,0h5,0htg)b

5,15278,01078,2tg

O2HB

O2HA

o

=−×=Δ−γ=

=+×=Δ+γ=

=×=Δ→Δ

=α

=α→==α

Exercício 2.56

2o

xxo

ooo

4

3dir

dir

4

3esq

esq

sm8,530tg10a

ga30tg

m73,130tg1

30tghL

Lh30tg

m11011hm1110

10110ph

m1010

10100ph

=×=⇒=

==Δ

=⇒Δ

=

=−=Δ⇒×

=γ

=

=×

=γ

=

Exercício 2.57

s54

6,372

avt

tva

s

m45,02,010a

ga

tg

xx

2x

x

===→=

=×=

=α

Exercício 2.58

( ) kN6,13N600.131010006,31000GmaFmaGFsm6,31

000.10200.27600.13g1

zppa

ga

1zpp

Pa600.131,0000.1361,0p

Pa200.272,0000.1362,0p

212

yy

12

Hg2

Hg1

−=−=×−−×=−=⇒=+

−=⎟⎠⎞

⎜⎝⎛ +

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛+

Δγ−

=⇒⎟⎟⎠

⎞⎜⎜⎝

⎛+Δγ=−

=×=×γ=

=×=×γ=