Resolucao de-exercicios-cap 2 - franco-brunetti

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Exercícios resolvidos de mecânica dos fluidos

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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=

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