capitulo 2 mec flu resolucao

Prévia do material em texto

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
Pa1035,1
20
5104,5
A
A
pp
Pa104,5
210
5,21072,21010500
AA
ApAp
p
ApG
ApAp
Pa1072,22000.136hp
ApAApAp
45
55
IV
III
34
5
53
HII
II2I1
3
V4
IV4III3
5
Hg2
II2HII3I1
=×××=
×=××==
×=−
××−××=−
−=
=
=
×=×=γ=
+−=
−
 
 
Exercício 2.2 
 
kN10N000.10
5
25400
D
D
FF
4
D
F
4
D
F
N400
1,0
2,0200F
1,0F2,0F
2
2
1
2
2
BO2
2
2
1
BO
BO
BOAO
==⎟⎠
⎞⎜⎝
⎛×=⎟⎟⎠
⎞
⎜⎜⎝
⎛=⇒π=π
=×=
×=×
 
 
 
Exercício 2.3 
 
mm3681000
000.136
5000.10h
hh
Hg
OHOHHgHg 22
=××=
γ=γ
 
 
Exercício 2.4 
)abs(mmHg3400)abs(
cm
kgf62,4)abs(MPa453,0)abs(
m
kgf200.46)abs(atm47,4p
mca10atm97,0MPa098,0Pa108,9
cm
kgf1
m
kgf000.1074,0600.13hp
mca2,36
000.1
200.36ph
bar55,398,0
cm
kgf62,310
m
kgf200.36p
MPa355,0108,9
m
kgf200.3666,2600.13hp
mmHg2660
1
5,3760p
patm5,3
mmHg760atm1
22abs
4
22HgHgatm
O2H
O2H
2
4
2
6
2HgHg
=====
===×≅=≅×=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
kPa4,20Pa400.2015,0000.13615,0p
p100p
m
HgA
Am
=−=
==×=×γ=
−=
 
 
Exercício 2.8 
 
kPa55,36103,0500.834p
p3,0p)b
)abs(kPa13410034ppp
kPa100Pa000.10074,0000.136hp
kPa34Pa000.348,0500.83,0000.136p
07,03,07,08,0p)a
3
M
MOar
atmarabsar
HgHgatm
ar
O2HHgO2HOar
=××+=
=×γ+
=+=+=
≅≅×=γ=
==×−×=
=×γ−×γ−×γ+×γ+
−
 
)abs(kPa55,13610055,36ppp atmMabsM =+=+= 
 
 
Exercício 2.9 
 
( )
( ) )abs(mca12,17
000.10
000.171ph
)abs(Pa200.171200.95000.76ppp
Pa200.95000.1367,0p
Pa000.76p000.57
4
p
p
000.57pp000.30p000.27p
000.27pppap
000.30pp
p4p4
A
A
A
A
A
A
ApApAApApAp
2
A
A
kPa30pp
OH
absB
OH
atmBB
atm
B
B
B
ABAB
BCBC
AC
AB
H
2
H
1
1
2
HB2AH1B1B2A
1
2
AC
2
2
efabs
==γ=
=+=+=
=×=
=→=−
=−→=−−
−=→=γ+
=−
=→==×
=→−−=
=
=−
 
 
Exercício 2.10 
 
)abs(kPa991001ppp
kPa1Pa000.12,010500ghp
m
kg500
2,0
1,0000.1
h
hhh0ghp
0ghp
atm0abs0
AA0
3
A
B
BABBAABB0
AA0
=+−=+=
−=−=××−=ρ−=
=×=ρ=ρ⇒ρ=ρ⇒=ρ+
=ρ+
 
 
Exercício 2.11 
 
( ) ( )
( ) ( )
3324
3
o
OH
OHo
OHo
cm833.47m107833,41043,0
6
45,0xA
6
DV)c
m45,03,05,0
000.8
6,04,0000.10x5,0
x2y
D
m3,0
2
4,01
2
yyxyyx2
x2yx5,0D)b
m4,0
000.10
5,0000.8y
y5,0)a
2
2
2
=×=××+×π=+π=
=−−+=−−γ
+γ=
=−=−′=→′=+
+γ=++γ
=×=
×γ=×γ
−−
 
Exercício 2.12 
 ( )
( ) ( )
m105
5,11sen
5,4
1000.8
10
sen
D
d
pL0Lsen
D
dLp
D
dLH
4
DH
4
dL
Pa10001,010001,0p
0LsenHp
3
o
22
x
2
x
222
4
O2Hx
x
−×=
⎥⎥⎦
⎤
⎢⎢⎣
⎡ +⎟⎠
⎞⎜⎝
⎛
−=
⎥⎥⎦
⎤
⎢⎢⎣
⎡ α+⎟⎠
⎞⎜⎝
⎛γ
−=⇒=
⎥⎥⎦
⎤
⎢⎢⎣
⎡ α+⎟⎠
⎞⎜⎝
⎛γ+
⎟⎠
⎞⎜⎝
⎛=⇒π=π
−=−×=−×γ=
=α+γ+
Exercício 2.13 
 
( )
( )
( )
( )
( )
mca7,3
000.10
000.37p
Pa000.37000.17000.20000.17pp)b
absmmHg831684147p
mmHg147m147,0
000.136
000.20Pa000.20p
000.17p10331p104:)1(nadoSubstituin
p000.17p
p4,0000.104,0000.5005,0000.102p
m05,0
4,71
7,35
2
4,0
D
d
2
hh
4
d
2
h
4
Dh
phhh2p
1p10331p104
0357,00714,0
4
p31
4
0714,0p
dD
4
pF
4
Dp)a
2
12
abs1
1
1
21
21
2221
21
21
21
ar
arar
ar
ar
ar
3
ar
3
arar
arar
2222
arOHmOHar
ar
3
ar
3
22
ar
2
ar
22
ar
2
ar
==
=+=+=
=+=
====
+×=+×
=+
=×−×+×××
=⎟⎠
⎞⎜⎝
⎛=⎟⎠
⎞⎜⎝
⎛=Δ→π=πΔ
=γ−γ+Δγ+
×=+×
−π=+×π
−π=+π
−−
−−
 
 
Exercício 2.14 
( )
1
2
11
22
222
111
arar
21
ar
HgO2Har
T
T
Vp
VpmRTVp
mRTVp)c
Pa050.12p0000.1361,0000.10155,0p
cm5
1
105,0hA.hA.y)b
Pa200.25000.10000.1362,0p
02,02,0p)a
=⇒=
=
=′⇒=×−×+′
=×=Δ⇒Δ=Δ
=−=
=×γ−×γ+
 
C44K317
100
95
200.125
050.112373T
cm95105,01010V
050.112000.100050.12p
)abs(Pa200.125000.100200.25p
o
2
3
2
abs2
abs1
==××=
=×−×=
=+=
=+=
 
 
 
Exercício 2.15 
 
 
3A
A
A
atmAAabs
atm
OH
A
OH
A
2222
A
212A
m
kg12,1
293287
576.94
RT
p
)abs(Pa576.94200.95624ppp
Pa200.95000.1367,0p)b
mca0624,0
000.10
624ph
Pa6240015,02000.8600p
m0015,0
40
4
2
3,0
D
d
2
hh
4
d
2
h
4
Dh
h2000.83,0000.103,0000.8p
0hhh2p)a
2
2
=×==ρ
=+−=+=
=×=
−=−=γ=
−=××−−=
=⎟⎠
⎞⎜⎝
⎛=⎟⎠
⎞⎜⎝
⎛=Δ→π=πΔ
Δ×−×−×=
=γ−γ+Δγ+
 
 
Exercício 2.16 
 
3
1
2
2
1
12
1
2
11
22
absgásO2Hgás
O2Hgás
absgás
atm
gásO2HHggás
m16,2
293
333
100
952
T
T
p
pVV
T
T
Vp
Vp
)abs(kPa1001090pkPa10Pa000.101000.10z.p)c
m5,0
000.10
000.5zz.p)b
)abs(kPa95590p
kPa90Pa032.90662,0000.136p
Pa500016,0000.10025,0000.136p16,0025,0p)a
=××==⇒=
=+=′⇒==×=′γ=′
==⇒γ=
=+=
==×=
=×+×=⇒×γ+×γ=
 
 
 
Exercício 2.17 
 
 
( ) ( ) 232222123322212211
32
21
3,0p1,05,0p5,0p
4
DpDD
4
p
4
Dp
000.22,0000.10pp
000.10pp
×+−×=×→π+−π=π
=×=−
=−
 
( )
( )
kPa5,43Pa500.43p3480p08,0
180000.10p33,0p25,0
180p33,0p25,0
000.2p09,0p24,0p25,0
11
11
21
221
==→=
−−=
−=
−+=
 
 
Exercício 2.18 
 
3222
2
ct
c
t
t
pGt
o
G
p
22
c
22p
22
c
11p
m
kg993.10
183,05,010
950.34
LDg
G4
L
4
Dg
G
gV
G)c
m183,0
5,0210
5,110005,0L
m0005,0
2
5,0501,0
2
DD
Dv
FLDLvF)b
N5,11FFF
desce196319755,0395030GsenF
cimaparaN196378549817F
N7854
4
5,0000.40
4
DpF
N9817
4
5,0000.50
4
DpF)a
=××π×
×=π=π==ρ
=×π××
×=
=−=−=ε
πμ
ε=⇒πεμ=
=−=
>=×==
=−=
=×π×=π=
=×π×=π=
−
 
 
Exercício 2.19 
( ) ( )
( ) ( )
cm8,127m278,1278,01L
m278,0ym0278,0x0600.36x10098,1x000.908000800
2
600.552
0200.735,0x15000.10x98,0800
A
F2
x10yy2,0x2
0200.7330ysen30sen1y000.10y25,0x55,0000.81,0
A
F2
m
N200.73
30sen1
8,0000.101,0000.8
2
600.55
30Lsen
8,01,0
A
F
030Lsen8,01,0
A
F
6
oo
3oo
21
3
o
321
==+=′
=⇒=⇒=−×−+++×
=×+−×+++
=⇒=
=×+×+−×++−+×+
=×
×+×+
=
×γ+×γ+
=γ
=γ−×γ+×γ+
 
Exercício 2.20 
 
( ) ( )
( ) ( )
( ) ( )
( )
kPa50109,39ppp)c
)abs(kPa1,60)abs(Pa100.6039908000.100pPa908.39
103,50
150102013,50000.10100
A
FAApGp
FApAApAApG
cm3,50
4
8
4
DA;cm201
4
16
4
DA)b
N15005,008,016,0
001,0
58,0DDvF
s
m.N8,0
10
000.810
g
)a
abm
absb
4
4
2
t12a
b
t2bH1aH2a
2
22
2
2
2
22
1
1
21t
2
3
−=−−=−=
==−+=
−=×
−×−×+=−−+=
++−=−+
=×π=π==×π=π=
=×+×π××=+πεμ=
=×=μγ=ν
−
−
−
l
 
 
 
Exercício 2.21 
 
2
3
p
p
p
p
p
p
2
p
p
pp2
12
m
s.N8,0
10
000.810
g
m001,0
2
998,01
2
DD
D
vL4pLv
4
D
p
LD
4
D
p
pistãonomédiapressãopondephp
000.10pp
=×=νγ=μ
=−=−=ε
ε
μ=→εμ=
τπ=π
==γ+
=−
−
 
 
 
 
 
 
 
 
 
Exercício 2.22 
 
N33933,0
4
2,1000.10b
4
RF
N160.23,02,16,0000.10AhF
22
y
x
=××π×=πγ=
=×××=γ=
 
 
 
kPa23,25Pa230.25000.10230.15000.10pp
m
N230.152000.85,769hpp
Pa5,769
998,0001,0
2,02,18,04p
21
2p2
p
−=−=−−=−=
=×−=γ−=
=×
×××=
Exercício 2.23 
 
m4,02,06,0b
m2,0
6
h
h
2
hAh
Ihh
N920.252,1
2
2,1000.30hhApF
m2,14,06,0
000.30
000.804,06,0h
6,0.4,0.h
2
12
4h
CG
cp
22
p
m
m
=−=
==
×
==−
=××=γ==
=−×=−×γ
γ=
γ=γ+γ
 
N640.8
2,1
4,025920
h
bFFbFhF pp =×==→×=× 
 
Exercício 2.24 
 
N948.59100.115,42,1F
N668.7
2
100.11100.5100.52,16,0F
N755.285,46,0
2
100.11100.55,46,0
2
100.5FFF
Pa100.116,0000.10100.56,0pp
Pa100.56,0500.86,0p
f
B
21A
212
11
=××=
=⎟⎠
⎞⎜⎝
⎛ ++××=
=××++××=+=
=×+=×γ+=
=×=×γ=
 
 
Exercício2.25 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
N500.225,121500.7AhF
m0833,10833,01
m0833,0
5,124AhAh
Ihh
N102,15,124000.10AhApF
F2FF
2o2
1
12
325,1
1
12
3bh
1
CG
11CP
5
1O2H11
22B11
=×××=γ=
=+=
=××===−
×=×××=γ==
+×=
×
l
ll
 
m333,1333,01
m333,0
5,121Ah
hh
2
12
325,1
2
12
3bh
22CP
=+=
=××==−
×
l
N105F
333,1500.222F0833,1102,1
4
B
B
5
×=
×+×=××
F
Fp 
h hcp 
b 
h 
5m 
2 m 
A 
B 
1l 2l 
3 m 
F1 F2 
FB 
Exercício 2.26 
 
m736,0
634.7
680.42,1
F
FyxxFyF
N634.73,0
4
8,1000.10b
4
RF
m2,18,1
3
2R
3
2y
N860.43,0
2
8,1000.10bR
2
RF
y
x
CPCPCPyCPx
22
y
c
2
x
=×==⇒=
=××π×=πγ=
=×==
=×=••γ=
 
 
 
Exercício 2.27 
 
m65,230cos75,02h
AhApF
o =×+=
γ==
 
 
 
kN4,991075,365,2000.10F
m75,35,25,1A
3
2
=×××=
=×=
− 
 
Exercício 2.28 
 
( )
( )
( ) ( )
3
oO2H
2
O2H
22
oinfsup
2
2
O2Hinf
2
osup
m
N000.35
6,0
5,2000.86,05,3000.10
6,0
h6,0h
4
D6,0h6,0
4
D
4
DhFGF
6,0
4
DG
4
D6,0hF
4
DhF
=×−+×=γ−+
′γ=γ
π+′γ=×πγ+πγ⇒=+
×πγ=
π+′γ=
πγ=
 
 
 
Exercício 2.29 
 
 
 
 
 
 
 
 
xCG CG 
γ1
γ2 
R 
R 
O 
Fx1 F2 
Fy1 
21 ll = 
2
bRRb
2
RF
AhF
FxFF
2
1
11x
1111x
22CG1y11x
γ=γ=
γ=
=+ ll
 
 
6
R
Rb
2
RAh
Ihh 12
3bR
CG
11CP ===− 
3
1
22
3
R
2
bR
3
R4
4
bR
3
R
2
bR
b
4
RVF
2
bR
Rb
2
RAhF
3
R
6
R
2
R
2
12
1
1
2
2
2
1
2
1
2
11y
2
2
22222
21
1
=γ
γ→γ=γ+γ
×γ=π×
πγ+×γ
πγ=γ=
γ=γ=γ=
==−= ll
 
 
Exercício 2.30 
 
( )
( )
N3,465
1
579,0300.14583,0000.15
BA
brFbrF
FBAFMM
m579,0079,05,0br
m079,0
5,106,1
125,0
Ay
Iyy
m06,156,05,0y
m56,0
000.9
032.5ph
N300.145,11532.9ApFPa532.9
2
032.14032.5
2
pp
p
Pa032.141000.950321pp
Pa032.5037,0000.136037,0pp
m583,0083,05,0br
m083,0
5,11
125,0yy
m125,0
12
15,1
12
bI
Ay
Iyy)b
N000.155,11000.10ApFPa000.10
2
000.15000.5
2
pp
p
Pa000.155,1000.105,1p
Pa000.55,0000.105,0p)a
esqesqdirdir
BBesqdir
esq
esq
CG
esqCP
esq
o
ar
areq
esqesq
esqBesqA
esq
oesqAesqB
HgaresqA
dir
dirCP
4
33
CG
CG
CP
dirdir
dirBdirA
dir
O2HdirB
O2HdirA
=×−×=−=⇒×+=
=+=
=×==−
=+=
==γ=
≅××==⇒=+=+=
=×+=×γ+=
=×=×γ==
=+=
=×=−
=×==→=−
=××==⇒=+=+=
=×=×γ=
=×=×γ=
l
 
 
 
 
Exercício 2.31 
 
( ) ( ) N6363,06,0
4
3,0000.103,0D
4
hApF
N107,1
4
6,06,0000.10
4
D
hApF
2222
MMMMM
3
22
F
FFFF
=−π××=−πγ==
×=×π××=πγ==
 
 
Exercício 2.32 
 
N230.76
2
083,1000.1205,0000.45F083,1F5,0F2F
m083,0
412
2
y12by
12/b
Ay
Iyy
N000.1205,12000.40ApFPa000.40
2
000.50000.30p
Pa000.505000.105p
m3
000.10
000.30ph
N000.455,11000.30ApF
Pa000.304,0000.1025,0000.1364,025,0p
BCAB
223
CG
CP
BCBCBCBC
O2HC
O2H
AB
ABABAB
O2HHgAB
=×+×=⇒×+×=×
=×====−
=××=×=⇒=+=
=×=×γ=
==γ=
=××==
=×−×=×γ−×γ=
l
l
l
 
 
 
Exercício 2.33 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Exercício 2.34 
 
m1CBMM
2
CBbCB3M
3
3b3
2
3M
BCAB
BCAB
=⇒=
γ=→γ=
 
 
 
F1 F2 
1l 2l 
( )
( ) ( )
( ) ( )
m27,6z
5,1108,225,6z5,2
5,11
5,2z
08,25,25,2z
5,2106,4
5,2z
08,25,25,2z10
m5,2
N106,4251046pAF
5,2z
08,25,2
55
2
53
2
1
=
=+−
=⎥⎦
⎤⎢⎣
⎡
−+−
××=⎥⎦
⎤⎢⎣
⎡
−+−
=
×=×××==
−+=
l
l
Exercício 2.35 
 
 
2
1
h
xh
3
x6h
3
x
2
xhxb
3
xb
2
x
2
x
hxbF
3
x
xb
2
xAhF
FF
2
1
2
2
1
2
22
1
1111
2211
=→=→=γ
γ
×γ=×γ
=
γ=
=
γ=γ=
=
l
l
ll
 
 
Exercício 2.36 
 
kN204H880.218015H
m.kN1805,1120MkN120
000.1
134000.10V
m.kN880.2
000.1
41126000.10M
V
x
=⇒+=×
=×=⇒=×××=
=××××=
 
 
Exercício 2.37 
 
O ferro estará totalmente submerso. 
N2183,0
4
3,0300.10h
4
DVE
22
flfl =××π×=πγ=γ= 
A madeira ficará imersa na posição em que o peso seja igual ao empuxo. 
 
sub
2
fl
22
mad
h
4
DE
N1593,0
4
3,0500.7h
4
DGE
πγ=
=××π×=πγ==
 
 
 
m218,0
3,0300.10
1594
D
E4h
22
fl
sub =×π×
×=
πγ
= 
 
Exercício 2.38 
 
N625023,0000.25500VGG conconcil =×+=γ+= 
F1 
F2 
1l 
2l 
( )
m3,02,05,0h
m5,0
1
23,0
000.10
62504
D
V/G4H
H
4
DVGEG
22
con
2
con
=−=
=×π
⎟⎠
⎞⎜⎝
⎛ −×
=π
−γ=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ ×π+γ=⇒=
 
 
 
Exercício 2.39 
 
 
 
 
 
 
 
 
 
 
 
( ) m7,29,08,1BAx:Logo
m9,0
270
6,0080.13,0350.1
F
GE
m6,0
3
8,1
3
BA
m3,0
3
9,0
3
IH
N270080.1350.1GEF:Logo
N080.11
2
6,08,1000.2b
2
CBBAVG
N350.11
2
9,03,0000.10b
2
IHCHVE
2
BAIH
FGE
EGF
2F
21
3
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 
 
( )( )
( )( ) 22dd444
3
odo
3
m1036,3A02,0A3,031055103,002,010
12
6,0
AARhGRA
26
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: 
3c
ccc
3
fl
fl
ap
m
N670.26
03,0
800
V
GVG
m03,0
000.10
300EVVE
N300500800EEGG
===γ→γ=
==γ=→γ=
=−=→+=
 
 
Exercício 2.42 
 
mm2,7m102,7
005,0
104,14
d
V4hh
4
dV
m104,11068,21082,2V
m1068,2
200.8
102,2GVVEG
m1082,2
800.7
102,2GVVEG
3
2
7
2
2
3766
36
2
2
2222
36
2
1
1111
=×=×π
××=π
Δ=Δ⇒Δ×π=Δ
×=×−×=Δ
×=×=γ=⇒γ==
×=×=γ=⇒γ==
−−
−−−
−−−−
 
 
Exercício 2.43 
 ( )
( )
( )
( )
m8,0hh000.16000.40h000.6000.32
h5,2000.16h000.6000.32
h5,14hp
m
N000.324000.8p4AApGAp
2Situação
m
N000.1622A4A
EG1Situação
ooo
oo
ooobase
2basebasecbasebasebasebase
3cbbc
=→−+=
−+=
−−γ+γ=
=×=→×γ=→=
=γ→γ=γ→×γ=×γ
=→
l
lll
 
 
Exercício 2.44 
 
m6
000.61009,2
2105,4x
N1009,2
12
210
26
DE
N105,4135,110AhF
GE
2FxxE3
3
2FxG
4
4
4
3
4
3
44
=−×
××=
×=×π×=×
πγ=
×=×××=γ=
−
×=⇒•=××+•
 
 
 
E 
G F 
Exercício 2.45 
 ( )
( )
( )
3B
B
BAbase
2b
bc
b
base
bbase
3cAbAbc
m
N000.25
4,02,0000.15000.13
2,06,02,0p
m
N000.13
1
000.1016,0000.5
A
FA6,0
A
FGp
FGAp
2Situação
m
N000.15000.5332,0A6,0AEG
1Situação
=γ
×γ+×=
−×γ+×γ=
=+××=+××γ=+=
+=
=×=γ=γ→×γ=×γ→=
 
 
Exercício 2.46 
 
( ) ( ) N171.10
6
121085,7132,110
6
DgG
1085,7
293400.41
200.95
TR
p
m
kg132,1
293287
200.95
TR
p
Pa200.957,0000.1367,0p
3
3
3
2Har
3
2H
2H
2H
3
ar
ar
ar
Hgatm
=×π××−×=πρ−ρ=
×=×==ρ
=×==ρ
=×=×γ=
−
− 
 
Exercício 2.47 
79,0x
21,0x
62
16466x:Raízes
01x6x6
0
2
x
2
1
x12
1xFazendo0
22
1
12
0
2
b
2
b
b
2
b
2
b0
V
I
r
bhbhbEG
2
2
cc
c
c
3
c
12
b
c
c
y
c
sub
2
sub
3
c
4
=′′
=′→×
××−±=
>+−
>+−→=γ
γ→>γ
γ+−γ
γ
>⎟⎟⎠
⎞
⎜⎜⎝
⎛
γ
γ−−
γ
γ
γ
γ−=→>−γ
γ=
γ
γ=→γ=γ→=
ll
l
l
l
l
l
l
l
ll
179,021,00 cc <γ
γ<<γ
γ<
ll 
Exercício 2.48 
 
 
estável0m037,00467,0
5,2
103,083.2000.10r
cm3,083.2
12
1025
12
bLI0
G
I
r
cm67,433,05cm5yCG
cm33,05,0
3
2yCC
cm5,0
10
5,2
L
Vh
hL
2
bh2V
m105,2
000.10
5,2GV
GVEG
8
4
33
y
yf
im
2
im
im
im
34
f
im
imf
⇒>=−××=
=×==→>−γ=
=−=⇒=→
=×=→
===
==
×==γ=
=γ⇒=
−
−
l
l
l
 
 
Exercício 2.49 
( )
( )
⎟⎟⎠
⎞
⎜⎜⎝
⎛
γ
γ−γ
γ<→−<
<−−→>+−
=γ
γ
>γ
γ+−γ
γ→>γ
γ+−γ
γ→>⎟⎟⎠
⎞
⎜⎜⎝
⎛
γ
γ−−
γπ
πγ
⎟⎟⎠
⎞
⎜⎜⎝
⎛
γ
γ−=−=π=γπ=
>−γ=
γ
γ=
γπ=πγ
=
ll
l
l
l
l
l
l
l
l
l
l
l
l
l
l
12
1
R
H
x1x2
1
R
H
01x2x2
R
H0
R
H2.x
R
H2
x
1
:RportudodividindoexFazendo
0H2H2R0
2
H
2
H
H4
R
0HH
2
1
HR4
R
HH
2
1
2
h
2
H
4
RIHRG
0
G
I
r
Hh
HRhR
GE
2
2
2
2
2
2
2
2
222
2
2
4
sub
4
y
2
y
sub
2
sub
2
 
 
 
CG 
CC 0,5cm 
Exercício 2.50 
 
z6
g
g51z
g
a
1zp yz Δγ=⎟⎟⎠
⎞
⎜⎜⎝
⎛ +Δγ=⎟⎟⎠
⎞
⎜⎜⎝
⎛ ±Δγ=Δ 
 
 
Exercício 2.51 
 
h
km2,646,3
s
m83,17557,3tav)b
s
m57,320tg8,9a20tgga
g
a
x
z)a
x
2
o
x
o
x
x
=×=×==
=×=→=→=Δ
Δ
 
 
Exercício 2.52 
 
oo
o
x 4130tg
30cos8,9
45,2tg
cosg
atg =θ⇒+×=α+α=θ 
 
 
Exercício 2.53 
 
( )
2x
3
x
3
Hg
s
m72,1
5,1
257,010
x
zga
m257,0
000.136
10140175z
g
a
x
z)b
m29,1
000.136
10175ph)a
=×=Δ
Δ=
=×−=Δ→=Δ
Δ
=×=γ=
 
 
Exercício 2.54 
 
)abs(kPa106
10
6,010000.1100ghpp
)abs(kPa7,125
10
6,010000.17,119ghpp
)abs(kPa7,119100106,0
2
5,10000.1p
s
rd5,10
60
1002n2pr
2
p
3atmC
3AB
32
2
A
atm
2
2
A
=××+=ρ+=
=××+=ρ+=
=+×⎟⎟⎠
⎞
⎜⎜⎝
⎛ ××=
=×π×=π=ω→+Δωρ=
−
 
 
Exercício 2.55 
 
 2x
x
s
m78,2
10
6,3
100
t
va
g
atg)a ===→=α 
140 
175 Pa 
zΔ
( ) ( )
( ) ( ) Pa600.314,05,0000.10h5,0p
Pa400.614,05,0000.10h5,0p
m14,0278,05,0h
5,0
htg)b
5,15278,0
10
78,2tg
O2HB
O2HA
o
=−×=Δ−γ=
=+×=Δ+γ=
=×=Δ→Δ=α
=α→==α
 
 
Exercício 2.56 
 
2
o
x
xo
oo
o
4
3
dir
dir
4
3
esq
esq
s
m8,530tg10a
g
a30tg
m73,1
30tg
1
30tg
hL
L
h30tg
m11011hm11
10
10110ph
m10
10
10100ph
=×=⇒=
==Δ=⇒Δ=
=−=Δ⇒×=γ=
=×=γ=
 
 
Exercício 2.57 
 
s5
4
6,3
72
a
vt
t
va
s
m4
5,0
2,010a
g
a
tg
x
x
2x
x
===→=
=×=
=α
 
 
Exercício 2.58 
 
( ) kN6,13N600.131010006,31000GmaFmaGF
s
m6,31
000.10
200.27600.13g1
z
ppa
g
a
1zpp
Pa600.131,0000.1361,0p
Pa200.272,0000.1362,0p
2
12
y
y
12
Hg2
Hg1
−=−=×−−×=−=⇒=+
−=⎟⎠
⎞⎜⎝
⎛ +−=⎟⎟⎠
⎞
⎜⎜⎝
⎛ +Δγ
−=⇒⎟⎟⎠
⎞
⎜⎜⎝
⎛ +Δγ=−
=×=×γ=
=×=×γ=