Diese Präsentation wurde erfolgreich gemeldet.
Wir verwenden Ihre LinkedIn Profilangaben und Informationen zu Ihren Aktivitäten, um Anzeigen zu personalisieren und Ihnen relevantere Inhalte anzuzeigen. Sie können Ihre Anzeigeneinstellungen jederzeit ändern.
Capítulo 2
ESTÁTICA DOS FLUIDOS
A ausência de movimento elimina os efeitos tangenciais e conseqüentemente a presença de
te...
Exercício 2.4
)abs(mmHg3400)abs(
cm
kgf
62,4)abs(MPa453,0)abs(
m
kgf
200.46)abs(atm47,4p
mca10atm97,0MPa098,0Pa108,9
cm
kg...
)abs(kPa55,13610055,36ppp atmMabsM =+=+=
Exercício 2.9
( )
( )
)abs(mca12,17
000.10
000.171p
h
)abs(Pa200.171200.95000.76p...
Exercício 2.12
( )
( ) ( )
m105
5,11sen
5,4
1
000.8
10
sen
D
d
p
L0Lsen
D
d
Lp
D
d
LH
4
D
H
4
d
L
Pa10001,010001,0p
0LsenH...
C44K317
100
95
200.125
050.112
373T
cm95105,01010V
050.112000.100050.12p
)abs(Pa200.125000.100200.25p
o
2
3
2
abs2
abs1
==...
( )
( )
kPa5,43Pa500.43p3480p08,0
180000.10p33,0p25,0
180p33,0p25,0
000.2p09,0p24,0p25,0
11
11
21
221
==→=
−−=
−=
−+=
Exer...
Exercício 2.20
( ) ( )
( ) ( )
( ) ( )
( )
kPa50109,39ppp)c
)abs(kPa1,60)abs(Pa100.6039908000.100p
Pa908.39
103,50
1501020...
Exercício 2.23
m4,02,06,0b
m2,0
6
h
h
2
hAh
I
hh
N920.252,1
2
2,1
000.30hhApF
m2,14,06,0
000.30
000.80
4,06,0h
6,0.4,0.h
2...
Exercício 2.26
m736,0
634.7
680.4
2,1
F
F
yxxFyF
N634.73,0
4
8,1
000.10b
4
R
F
m2,18,1
3
2
R
3
2
y
N860.43,0
2
8,1
000.10b...
6
R
Rb
2
RAh
I
hh 12
3bR
CG
11CP ===−
3
1
22
3
R
2
bR
3
R4
4
bR
3
R
2
bR
b
4
R
VF
2
bR
Rb
2
R
AhF
3
R
6
R
2
R
2
12
1
1
2
2...
Exercício 2.31
( ) ( ) N6363,06,0
4
3,0000.103,0D
4
hApF
N107,1
4
6,0
6,0000.10
4
D
hApF
2222
MMMMM
3
22
F
FFFF
=−
π
××=−
...
Exercício 2.35
2
1
h
x
h
3
x6
h
3
x
2
x
hxb
3
x
b
2
x
2
x
hxbF
3
x
xb
2
x
AhF
FF
2
1
2
2
1
2
22
1
1111
2211
=→=→=
γ
γ
×γ=×...
( )
m3,02,05,0h
m5,0
1
23,0
000.10
6250
4
D
V/G4
H
H
4
D
VGEG
22
con
2
con
=−=
=
×π
⎟
⎠
⎞
⎜
⎝
⎛
−×
=
π
−γ
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
...
Exercício 2.41
Supondo o empuxo do ar desprezível:
3
c
ccc
3
fl
fl
ap
m
N
670.26
03,0
800
V
G
VG
m03,0
000.10
300E
VVE
N30...
Exercício 2.45
( )
( )
( )
3B
B
BAbase
2
b
bc
b
base
bbase
3cAbAbc
m
N
000.25
4,02,0000.15000.13
2,06,02,0p
m
N
000.13
1
0...
Exercício 2.48
estável0m037,00467,0
5,2
103,083.2000.10
r
cm3,083.2
12
1025
12
bL
I0
G
I
r
cm67,433,05cm5yCG
cm33,05,0
3
2...
Exercício 2.50
z6
g
g5
1z
g
a
1zp
y
z Δγ=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+Δγ=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
±Δγ=Δ
Exercício 2.51
h
km
2,646,3
s
m
83,17557,3t...
( ) ( )
( ) ( ) Pa600.314,05,0000.10h5,0p
Pa400.614,05,0000.10h5,0p
m14,0278,05,0h
5,0
h
tg)b
5,15278,0
10
78,2
tg
O2HB
O2...
Nächste SlideShare
Wird geladen in …5
×

Resolucao de-exercicios-cap 2 - franco-brunetti

43.208 Aufrufe

Veröffentlicht am

Exercícios resolvidos de mecânica dos fluidos

Veröffentlicht in: Ingenieurwesen
  • Als Erste(r) kommentieren

Resolucao de-exercicios-cap 2 - franco-brunetti

  1. 1. 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 5 104,5 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 25 400 D D FF 4 D F 4 D F N400 1,0 2,0 200F 1,0F2,0F 2 2 1 2 2 BO2 2 2 1 BO BO BOAO ==⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ×= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ =⇒ π = π =×= ×=× Exercício 2.3 mm3681000 000.136 5000.10 h hh Hg OHOHHgHg 22 =× × = γ=γ
  2. 2. Exercício 2.4 )abs(mmHg3400)abs( cm kgf 62,4)abs(MPa453,0)abs( m kgf 200.46)abs(atm47,4p mca10atm97,0MPa098,0Pa108,9 cm kgf 1 m kgf 000.1074,0600.13hp mca2,36 000.1 200.36p h bar55,398,0 cm kgf 62,310 m kgf 200.36p MPa355,0108,9 m kgf 200.3666,2600.13hp mmHg2660 1 5,3760 p 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+ −
  3. 3. )abs(kPa55,13610055,36ppp atmMabsM =+=+= Exercício 2.9 ( ) ( ) )abs(mca12,17 000.10 000.171p h )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 kg 500 2,0 1,0 000.1 h h hh0ghp 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 45,0 xA 6 D V)c m45,03,05,0 000.8 6,04,0000.10 x5,0 x2y D m3,0 2 4,01 2 yy xyyx2 x2yx5,0D)b m4,0 000.10 5,0000.8 y y5,0)a 2 2 2 =×=××+ ×π =+ π = =−− + =−− γ +γ = = − = −′ =→′=+ +γ=++γ = × = ×γ=×γ −−
  4. 4. Exercício 2.12 ( ) ( ) ( ) m105 5,11sen 5,4 1 000.8 10 sen D d p L0Lsen D d Lp D d LH 4 D H 4 d L 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 ( ) ( ) ( ) ( ) ( ) mca7,3 000.10 000.37 p Pa000.37000.17000.20000.17pp)b absmmHg831684147p mmHg147m147,0 000.136 000.20 Pa000.20p 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 h h 4 d 2 h 4 D h phhh2p 1p10331p104 0357,00714,0 4 p31 4 0714,0 p dD 4 pF 4 D p)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 = π Δ =γ−γ+Δγ+ ×=+× − π =+ ×π − π =+ π −− −− Exercício 2.14 ( ) 1 2 11 22 222 111 arar 21 ar HgO2Har T T Vp Vp mRTVp mRTVp)c Pa050.12p0000.1361,0000.10155,0p cm5 1 10 5,0hA.hA.y)b Pa200.25000.10000.1362,0p 02,02,0p)a =⇒= = =′⇒=×−×+′ =×=Δ⇒Δ=Δ =−= =×γ−×γ+
  5. 5. C44K317 100 95 200.125 050.112 373T cm95105,01010V 050.112000.100050.12p )abs(Pa200.125000.100200.25p o 2 3 2 abs2 abs1 ==××= =×−×= =+= =+= Exercício 2.15 3 A A A atmAAabs atm OH A OH A 2222 A 212A m kg 12,1 293287 576.94 RT p )abs(Pa576.94200.95624ppp Pa200.95000.1367,0p)b mca0624,0 000.10 624p h Pa6240015,02000.8600p m0015,0 40 4 2 3,0 D d 2 h h 4 d 2 h 4 D h h2000.83,0000.103,0000.8p 0hhh2p)a 2 2 = × ==ρ =+−=+= =×= −=−= γ = −=××−−= =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =Δ→ π = π Δ 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 95 2 T T p p VV T T Vp Vp )abs(kPa1001090pkPa10Pa000.101000.10z.p)c m5,0 000.10 000.5 zz.p)b )abs(kPa95590p kPa90Pa032.90662,0000.136p Pa500016,0000.10025,0000.136p16,0025,0p)a =××==⇒= =+=′⇒==×=′γ=′ ==⇒γ= =+= ==×= =×+×=⇒×γ+×γ= Exercício 2.17 ( ) ( ) 2 3 22 2 2 1 2 3 3 2 2 2 12 2 1 1 32 21 3,0p1,05,0p5,0p 4 D pDD 4 p 4 D p 000.22,0000.10pp 000.10pp ×+−×=×→ π +− π = π =×=− =−
  6. 6. ( ) ( ) 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 kg 993.10 183,05,010 950.34 LDg G4 L 4 D g G gV G )c m183,0 5,0210 5,110005,0 L m0005,0 2 5,0501,0 2 DD Dv F LDL v F)b N5,11FFF desce196319755,0395030GsenF cimaparaN196378549817F N7854 4 5,0 000.40 4 D pF N9817 4 5,0 000.50 4 D pF)a = ××π× × = π = π ==ρ = ×π×× × = = − = − =ε πμ ε =⇒π ε μ= =−= >=×== =−= = ×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 N 200.73 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+
  7. 7. Exercício 2.20 ( ) ( ) ( ) ( ) ( ) ( ) ( ) kPa50109,39ppp)c )abs(kPa1,60)abs(Pa100.6039908000.100p Pa908.39 103,50 150102013,50000.10100 A FAApG p FApAApAApG cm3,50 4 8 4 D A;cm201 4 16 4 D A)b N15005,008,016,0 001,0 5 8,0DD v F s m.N 8,0 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××=+π ε μ= = × = μγ =ν − − − l Exercício 2.21 2 3 p p p p p p 2 p p pp2 12 m s.N 8,0 10 000.810 g m001,0 2 998,01 2 DD D vL4 pL v 4 D p LD 4 D p pistãonomédiapressãopondephp 000.10pp = × = νγ =μ = − = − =ε ε μ =→ ε μ= τπ= π ==γ+ =− − Exercício 2.22 N33933,0 4 2,1 000.10b 4 R F N160.23,02,16,0000.10AhF 22 y x =× ×π ×= π γ= =×××=γ= kPa23,25Pa230.25000.10230.15000.10pp m N 230.152000.85,769hpp Pa5,769 998,0001,0 2,02,18,04 p 21 2p2 p −=−=−−=−= =×−=γ−= = × ××× =
  8. 8. Exercício 2.23 m4,02,06,0b m2,0 6 h h 2 hAh I hh N920.252,1 2 2,1 000.30hhApF m2,14,06,0 000.30 000.80 4,06,0h 6,0.4,0.h 2 12 4h CG cp 22 p m m =−= == × ==− =××=γ== =−×=−× γ γ = γ=γ+γ N640.8 2,1 4,0 25920 h b FFbFhF pp =×==→×=× Exercício 2.24 N948.59100.115,42,1F N668.7 2 100.11100.5100.5 2,16,0F N755.285,46,0 2 100.11100.5 5,46,0 2 100.5 FFF 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 I hh 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
  9. 9. Exercício 2.26 m736,0 634.7 680.4 2,1 F F yxxFyF N634.73,0 4 8,1 000.10b 4 R F m2,18,1 3 2 R 3 2 y N860.43,0 2 8,1 000.10bR 2 R F y x CPCPCPyCPx 22 y c 2 x =×==⇒= =× ×π ×= π γ= =×== =×=••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 N 000.35 6,0 5,2000.86,05,3000.10 6,0 h6,0h 4 D 6,0h6,0 4 D 4 D hFGF 6,0 4 D G 4 D 6,0hF 4 D hF = ×−+× = γ−+′γ =γ π +′γ=× π γ+ π γ⇒=+ × π γ= π +′γ= π γ= Exercício 2.29 xCGCG γ1 γ2 R R O Fx1 F2 Fy1 21 ll = 2 bR Rb 2 R F AhF FxFF 2 1 11x 1111x 22CG1y11x γ =γ= γ= =+ ll
  10. 10. 6 R Rb 2 RAh I hh 12 3bR CG 11CP ===− 3 1 22 3 R 2 bR 3 R4 4 bR 3 R 2 bR b 4 R VF 2 bR Rb 2 R AhF 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 π γ=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 I yy m06,156,05,0y m56,0 000.9 032.5p h 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 5,11 125,0 yy m125,0 12 15,1 12 b I Ay I yy)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
  11. 11. Exercício 2.31 ( ) ( ) N6363,06,0 4 3,0000.103,0D 4 hApF N107,1 4 6,0 6,0000.10 4 D hApF 2222 MMMMM 3 22 F FFFF =− π ××=− π γ== ×= ×π ××= π γ== Exercício 2.32 N230.76 2 083,1000.1205,0000.45 F083,1F5,0F2F m083,0 412 2 y12by 12/b Ay I yy N000.1205,12000.40ApFPa000.40 2 000.50000.30 p Pa000.505000.105p m3 000.10 000.30p h 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 CB bCB3M 3 3 b3 2 3 M BCAB BCAB =⇒= γ=→γ= F1 F2 1l 2l ( ) ( ) ( ) ( ) ( ) m27,6z 5,1108,225,6z5,2 5,11 5,2z 08,2 5,25,2z 5,2106,4 5,2z 08,2 5,25,2z10 m5,2 N106,4251046pAF 5,2z 08,2 5,2 55 2 53 2 1 = =+− =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − +− ××=⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − +− = ×=×××== − += l l
  12. 12. Exercício 2.35 2 1 h x h 3 x6 h 3 x 2 x hxb 3 x b 2 x 2 x hxbF 3 x xb 2 x AhF 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 134000.10 V m.kN880.2 000.1 41126000.10 M V x =⇒+=× =×=⇒= ××× = = ×××× = Exercício 2.37 O ferro estará totalmente submerso. N2183,0 4 3,0 300.10h 4 D VE 22 flfl =× ×π ×= π γ=γ= A madeira ficará imersa na posição em que o peso seja igual ao empuxo. sub 2 fl 22 mad h 4 D E N1593,0 4 3,0 500.7h 4 D GE π γ= =× ×π ×= π γ== m218,0 3,0300.10 1594 D E4 h 22 fl sub = ×π× × = πγ = Exercício 2.38 N625023,0000.25500VGG conconcil =×+=γ+= F1 F2 1l 2l
  13. 13. ( ) m3,02,05,0h m5,0 1 23,0 000.10 6250 4 D V/G4 H H 4 D VGEG 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 6,08,1 000.2b 2 CBBA VG N350.11 2 9,03,0 000.10b 2 IHCH VE 2 BA IH FGE EGF 2F 21 3 2 1 ccc OHsubOH 321 22 =−−=−= −= ×−× = − = === === =−=−= =× × ×=× × γ=γ= =× × ×=× × γ=γ= = += =+ l ll l l l lll A força deverá ser aplicada à direita do ponto B, fora da plataforma AB. Exercício 2.40 ( )( ) ( )( ) 22 dd 444 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
  14. 14. Exercício 2.41 Supondo o empuxo do ar desprezível: 3 c ccc 3 fl fl ap m N 670.26 03,0 800 V G VG m03,0 000.10 300E VVE N300500800EEGG ===γ→γ= == γ =→γ= =−=→+= Exercício 2.42 mm2,7m102,7 005,0 104,14 d V4 hh 4 d V m104,11068,21082,2V m1068,2 200.8 102,2G VVEG m1082,2 800.7 102,2G VVEG 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 N 000.324000.8p4AApGAp 2Situação m N 000.1622A4A EG1Situação ooo oo ooobase 2basebasecbasebasebasebase 3cbbc =→−+= −+= −−γ+γ= =×=→×γ=→= =γ→γ=γ→×γ=×γ =→ l lll Exercício 2.44 m6 000.61009,2 2105,4 x N1009,2 12 2 10 26 D E N105,4135,110AhF GE 2F xxE3 3 2 FxG 4 4 4 3 4 3 44 = −× ×× = ×= ×π ×= × π γ= ×=×××=γ= − × =⇒•=××+• E G F
  15. 15. Exercício 2.45 ( ) ( ) ( ) 3B B BAbase 2 b bc b base bbase 3cAbAbc m N 000.25 4,02,0000.15000.13 2,06,02,0p m N 000.13 1 000.1016,0000.5 A FA6,0 A FG p FGAp 2Situação m N 000.15000.5332,0A6,0AEG 1Situação =γ ×γ+×= −×γ+×γ= = +×× = +××γ = + = += =×=γ=γ→×γ=×γ→= Exercício 2.46 ( ) ( ) N171.10 6 12 1085,7132,110 6 D gG 1085,7 293400.41 200.95 TR p m kg 132,1 293287 200.95 TR p Pa200.957,0000.1367,0p 3 3 3 2Har 3 2H 2H 2H 3 ar ar ar Hgatm = ×π ××−×= π ρ−ρ= ×= × ==ρ = × ==ρ =×=×γ= − − Exercício 2.47 79,0x 21,0x 62 16466 x:Raízes 01x6x6 0 2 x 2 1 x12 1 xFazendo0 22 1 12 0 2 b 2 b b 2 b 2 b 0 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
  16. 16. Exercício 2.48 estável0m037,00467,0 5,2 103,083.2000.10 r cm3,083.2 12 1025 12 bL I0 G I r cm67,433,05cm5yCG cm33,05,0 3 2 yCC cm5,0 10 5,2 L V h hL 2 bh 2V m105,2 000.10 5,2G V 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 H 0 R H 2.x R H 2 x 1 :RportudodividindoexFazendo 0H2H2R0 2 H 2 H H4 R 0HH 2 1 HR4 R HH 2 1 2 h 2 H 4 R IHRG 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 CC0,5cm
  17. 17. Exercício 2.50 z6 g g5 1z g a 1zp y z Δγ=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +Δγ=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ±Δγ=Δ Exercício 2.51 h km 2,646,3 s m 83,17557,3tav)b s m 57,320tg8,9a20tgga g a x z )a x 2 o x o x x =×=×== =×=→=→= Δ Δ Exercício 2.52 oo o x 4130tg 30cos8,9 45,2 tg cosg a tg =θ⇒+ × =α+ α =θ Exercício 2.53 ( ) 2x 3 x 3 Hg s m 72,1 5,1 257,0 10 x z ga m257,0 000.136 10140175 z g a x z )b m29,1 000.136 10175p h)a =×= Δ Δ = = ×− =Δ→= Δ Δ = × = γ = Exercício 2.54 )abs(kPa106 10 6,010000.1 100ghpp )abs(kPa7,125 10 6,010000.1 7,119ghpp )abs(kPa7,119100106,0 2 5,10 000.1p s rd 5,10 60 100 2n2pr 2 p 3atmC 3AB 32 2 A atm 2 2 A = ×× +=ρ+= = ×× +=ρ+= =+×⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ××= =×π×=π=ω→+Δ ω ρ= − Exercício 2.55 2x x s m 78,2 10 6,3 100 t v a g a tg)a ===→=α 140 175 Pa zΔ
  18. 18. ( ) ( ) ( ) ( ) Pa600.314,05,0000.10h5,0p Pa400.614,05,0000.10h5,0p m14,0278,05,0h 5,0 h tg)b 5,15278,0 10 78,2 tg 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 m 8,530tg10a g a 30tg m73,1 30tg 1 30tg h L L h 30tg m11011hm11 10 10110p h m10 10 10100p h =×=⇒= == Δ =⇒ Δ = =−=Δ⇒ × = γ = = × = γ = Exercício 2.57 s5 4 6,3 72 a v t t v a s m 4 5,0 2,0 10a g a tg x x 2x x ===→= =×= =α Exercício 2.58 ( ) kN6,13N600.131010006,31000GmaFmaGF s m 6,31 000.10 200.27600.13 g1 z pp a g a 1zpp Pa600.131,0000.1361,0p Pa200.272,0000.1362,0p 2 12 y y 12 Hg2 Hg1 −=−=×−−×=−=⇒=+ −=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + Δγ − =⇒⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +Δγ=− =×=×γ= =×=×γ=

×