CHAPTER 2 Fluid Statics

Transcription

1 Chater / Fluid Statics CHAPTER Fluid Statics FE-tye Exam Review Problems: Problems - to -9. (C) h (.6 98) (8.5.54) 96 6 Pa Hg. (D) gh Pa. (C) h h Pa w atm x x water w.4 (A) H (.6 98).6 5 Pa. a a, after Hafter. Hafter.85 m.5 (B) The force acts / the distance from the hinge to the water line: ( ) P ( ) [98 ( )]. P 67 N.6 (A) The gate oens when the center of ressure in at the hinge:. h I. h b(. h) / y 5. y y 5.. Ay (. h) b(. h) / This can be solved by trial-and error, or we can simly substitute one of the answers into the equation and check to see if it is correct. This yields h =.8 m..7 (D) Place the force F F H V at the center of the circular arc. F H asses through the hinge:.8 (A) W V P F 4.w 98 (. / 4) w98. w 5.6 m. V w. w 6 m.9 (A) 5 lug h 666 (. ) 4 7 Pa 9.8 F A N. lug lug 5 Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

2 Chater / Fluid Statics Chater Problems: Pressure yz. Fy may: yz s sin a y yz yz Fz maz: zy s cos a z g Since scos y and s sin z, we have y z z z gv y s s z y y y y a Let y and z : y and g then z y z z a z y z.. = h. a) 98 = 98 Pa or 98. kpa b) (.8 98) = Pa or 78.5 kpa c) (.6 98) = 4 Pa or 4 kpa d) (.59 98) = Pa or 56. kpa e) (.68 98) = 66 7 Pa or 66.7 kpa. h = /. a) h = 5 /98 = 5.5 m b) h = 5 /(.8 98) =.9 m c) h = 5 /(.6 98) =.874 m d) h = 5 /(.59 98) = 6. m e) h = 5 /(.68 98) = 7.5 m. S = h =.. =.94. = 4.48 slug/ft..4 a) = h =.76 (.6 98) = 98 h. h =.4 m. b) (.6 98).75 = 98 h. h =. m. c) (.6 98). = 98 h. h =.6 m or.6 cm..5 a) = h + h = (.6 98). = 46 Pa or 4.6 kpa. b) = 96 Pa or.96 kpa. c) (.6 98). = 6 Pa or 6. kpa..6 = gh =.4. (,) = 77 sf or 5.7 si. 6 Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

3 Chater / Fluid Statics.7 g 9.8 outside ogh h.5 Pa RT.87 5 o base =.84 Pa g 9.8 inside.67 Pa ig h h RTi.87 9 If no wind is resent this base would roduce a small infiltration since the higher ressure outside would force outside air into the bottom region (through cracks)..8 = gdh where h = z. From the given information S =. + h/ since S() = and S() =.. By definition = S, where water = kg/m. Then d = ( + h/) gdh. Integrate: d ( h / ) gdh 9. 8( ) = Pa or kpa Note: we could have used an average S: S avg =.5, so that avg = 5 kg/m..9 i j x y z k = x y z x y z a i j k gk a i a j a k g k a g ( a g).. / R atm [( ) / ]g T z T = [(88.65 )/88] 9.8/ = kpa atm gh 9.8 / = kpa % error = =.5% The density variation can be ignored over heights of m or less. g/ R T z atm T atm 9.8/ = =.7 Pa or.7 kpa 88 This change is very small and can most often be ignored. 7 Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

4 Chater / Fluid Statics d. Eq..5. gives, 44. But, d = gdh. Therefore, d gdh d or Integrate, using =. slug/ft : Now, h 7 d d. dh. = 7. 7 h or h h dh h g g 7 gdh dh ln( 4.4 h) h 4.4 Assume = const: gh.. h 64.4h a) For h = 5 ft: accurate = 96,7 sf and estimate = 96,6 sf. 96, 6 96, 7 % error. % 96, 7 b) For h = 5 ft: accurate =, sf and estimate =, sf.,, % error.7 %, c) For h = 5, ft: accurate = 976,6 sf and estimate = 966, sf. 966, 976, 6 % error.85 % 976, 6. Use the result of Examle.: = e gz/rt. a) = e 9.8 /87 7 = 8.9 kpa. b) = e 9.8 /87 88 =.8 kpa. c) = e 9.8 /87 58 = 6.9 kpa Use Eq..4.8: = (.65 z / 88) a) z =. = 69.9 kpa. b) z = 6. = 47. kpa. c) z = 9. =.6 kpa. d) z =. =.5 kpa. 8 Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

5 Chater / Fluid Statics /.5 Use the result of Examle.: = e gz RT. gz..z ln. ln. RT z =,7 ft. Manometers.6 = h = (.6 98).5 = 5 Pa or.5 kpa..7 a) = h. 45 = (.6 98) h. h =.7 m b) = (.6 98) h. Use = 45, then h =.7 m The % error is. %..8 Referring to Fig..6a, the ressure in the ie is = gh. If = 4 Pa, then 4 = gh = 9.8h or h 4 a) = 68 kg/m. gasoline b) = 899 kg/m. benzene c) = 999 kg/m. water d) = 589 kg/m. carbon tetrachloride Referring to Fig..6a, the ressure is = wgh = a V a) V = V =.9 m/s b) c) d). Then V.94. / V =,4. V = 5 ft/sec V = 595. V = 9.9 m/s / V =,87. V = 48 ft/sec.8 wgh. a 9 Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

6 Chater / Fluid Statics. See Fig..6b: = h + H. = = sf or 4.5 si.. gh gh gh 4gh4 = = 64 Pa or.64 kpa. (Use gh) = 9.8(.) H H =.74 m or 7.4 cm. (Use gh) 4 4 o w = 99.8(.) (.) = Pa or. kpa = 98(.) (.4) + 98(.) = 5 6 Pa or 5.6 kpa.5 w = o. w o = 94 Pa or.94 kpa..6 w = o. With w = 5, o = 4 Pa or 4. kpa..7 a) + 98 = = 678 Pa or 6.8 kpa. b) = = 8 85 Pa or 8.84 kpa. c) = /. = 9.5 sf or.65 si. d) = /. = 44 sf or.6 si =. = 7 89 Pa or 7.89 kpa. Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

7 Chater / Fluid Statics = H. H =.68 m H new = =.956 m. H =.7 = ( ) = = Pa or. kpa. H H H = = 587 Pa or 5.87 kpa. Note: In our solutions we usually retain significant digits in the answers (if a number starts with then 4 digits are retained). In most roblems a material roerty is used, i.e., S =.59. This is only significant digits! only are usually retained in the answer!.4 The equation for the manometer is.7..9sin 4 A water B oil HG Solve for B :.7.9sin 4. B A water HG oil.7.6.9sin A water water water.7.6.9sin A water kpa.7.6.9sin kn/m. kpa.4 The distance the mercury dros on the left equals the distance along the tube that the mercury rises on the right. This is shown in the sketch. Oil (S =.87) B Water h cm A 9 cm 7 cm h 4 o From the revious roblem we have Mercury B A water HG oil For the new condition.7.9sin 4. ().7.sin 4. sin 4 h h () B A water HG oil where h in this case is calculated from the new manometer reading as h h / sin 4 9 h.78 cm Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

8 Chater / Fluid Statics Subtracting Eq.() from Eq.() yields h.sin 4 h sin 4 B B water HG oil Substituting the values of h and B gives B B sin sin kpa.4 Before ressure is alied the air column on the right is 48" high. After ressure is alied, it is (4 H/) ft high. For an isothermal rocess V V using absolute ressures. Thus, A = (4 H / )A or = 8467 / (4 H/ ) From a ressure balance on the manometer (ressures in sf): = H +, 4 H / or H 5.59 H =. H =.7 or. ft..44 a).45 a) = 98(.6.) + 5 6(..6) + 4H + 5 6(.7 H). H =.76 m or.76 cm b).644 = 6.4( /) ( /) + 849H (.5/ H). H =.6 ft or.48 in. H D / d ( ) D / d (./.5) H ( 4 5 6)(./.5) 6 H =.4 m or.4 mm 6 b) (4 /.) H ( )(4 /.) =.5 ft or.8 in..46 ( oil = 4. kpa from No..) = 98(. + z) + 68(. z) + 86(. z). z =.45 m or 4.5 cm.47 a) air = = 56 Pa ( + z) 6 9.8(. + z) =. z =.5. h =. + z =.5 m or 5 cm Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

9 Chater / Fluid Statics b) air = = 68 Pa (.8 + z) 6 9.8(. + z) =. z =.75 m h =.+ z =.4 or.4 cm c) air = = 8.4 sf (6 + z).6 6.4(4/ + z) =. z =.558 ft. h = 4/ + (.558) =.445 ft or 4. in. d) air = = 485 sf ( + z).6 6.4(8/ + z) =. z =.67 ft. h = 8/ + (.67) =.75 ft or 8.65 in. Forces on Plane Areas.48 F ha = 98. /4 = 694 N P N P a) F = c A = 98 4 = 6 N or.6 kn b) F c A 98 ( 4) N or 98. kn c) F = c A = 98 4 = 9 N or.9 kn d) F = c A = 98 4/.866 = 9 5 N or 9.5 kn.5 For saturated ground, the force on the bottom tending to lift the vault is F = c A = 98.5 ( ) = 9 4 N The weight of the vault is aroximately W g V walls 4 9.8[(.5.) + (.) + (.8..)] = 8 4 N. The vault will tend to rise out of the ground..5 F = c A = 666 = 67 4 N or 67.4 kn Find in Table B.5 in the Aendix..5 a) F = c A = 98 (.88/) (.88 /) = 5 N or 5 kn where the height of the triangle is ( ) / =.88 m. b) F = c A = 98 (.88 /) = 77 N or 77. kn c) F = c A = 98 (.866/) (.88 /) = 54 5 N or 54.5 kn Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

10 Chater / Fluid Statics.5 a) F ha ,9 lb. y 68 /6 7. = 7.46 ft. y = 7.46 =.54 ft /5.46 = /x. x =.5. (.5,.54) ft. b) F = = 44,9 lb. The centroid is the center of ressure. 8 y =.667 ft.. x =. ft (.,.667) ft. 5. x c) F = 6.4 ( ) 4 = 4, lb. y 6 8 / = 9.86 ft. y = =.57 ft /5.4 = /x. x =.4 ft. (.4,.57) ft x.54 a) F ha 98 6 = 79 7 N or 79.7 kn. 4 y y I / Ay 6 /4(4 6) = 6.67 m. (x, y) = (,.67) m y (x, y) x b) F ha 98 6 = 69 8 N or 69.8 kn. y 4 / 8 6 = 6.67 m. x + y = 4 6 x xf da (6 ) (4 )(6 ). x y xdy y y dy x6 (4 4 6 ). y y y dy x =.8488 m (x, y) = (.8488,.67) m dy y da (x, y) x c) F = 98 (4 + 4/) 6 = 9 N or.9 kn. y 5 4 / = 5.5 m. y =.5 4/.5 =.5 x. x =.975. (x, y) = (.975,.5) m y x d) F 98 ( 4 4sin 6.9 ) 6 = N y /6(6 5.6) = m. y =.4 m cos 5. =.8,.5.8 =.7,.4/.57 =. 7 / x. x =.6. x = =.4. (x, y) = (.4,.4) m o 4 Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

11 Chater / Fluid Statics.55 F ha 6.4 ( 6 ) = 4,8 lb. y I y ya 6 / =.758 ft. 6 (6.758) 4,8 = P. P = 7,47 lb. y F P.56 F ha 98 6 = N, or 77 kn. y I y Ay /. = m ( 7.778) 77 = 5 P. P = 5 kn..57 F ha 98 =.54 6 N, or 54 kn. y I y Ay / = 5.9 m. 5 ( ) 54 = 5 P. P = kn. I H bh.58 y y H H H Ay /. y is measured from the surface. bh H / 6 From the bottom, H y H H H. Note: This result is indeendent of the angle, so it is true for a vertical area or a sloed area l sin 4. ( ) sin 4. ( ). F l l F l P l l P a) 98 = ( + )P. P = 98 N b) 98 4 = (4 + )P. P = 5 N c) 98 5 = (5 + )P. P = 87 6 N h..4 =.4 m. A = =.8 m Use forces: F h A (..4) = 757 N c.4 F hc A 98 (.4.4) = 67 N y ( 4. ). I / 6 y y =.5657 m Ay.4 (.4 / ) (.4 / ) M hinge : 757.4/ 67 ( ).4P =. P = 46 N. 5 Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

12 Chater / Fluid Statics.6 To oen, the resultant force must be just above the hinge, i.e., y must be just less than h. Let y = h, the condition when the gate is about to oen: y ( h H) /, A ( h H), I [ ( h H)]( h H) / 6 4 h H ( h H) / 6 h y H h H h H ( h H) ( h H) / 6 h H a) h. h = H =.9 m b) h = H =. m c) h = H =.5 m.6 The gate is about to oen when the center of ressure is at the hinge. a) b) c) b.8 / y. H (.8/ H). (.9 H).8b H =. b / y. H (./ H). H =.6667 m. ( H)b b. / y. H (./ H). (. H).b H =.9 m..6 H F bh bh F H b b H H bh bh. H a) H =.464 m b) H =.7 m c) H =.9' d) H = 5.96' F H/ l/ F.64 A free-body-diagram of the gate and block is sketched. Sum forces on the block: T F W T F y where F B is the buoyancy force which is given by B F sto T F B R ( H) Take moments about the hinge: F H y F B T.5 F ( y ) where F H is the hydrostatic force acting on the gate. It is, using h.5 m and A 6 m, H R x Ry W FH ha 9.8 kn/m.5 m 6 m 88.9 kn 6 Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

13 Chater / Fluid Statics From the given information, y I y.5 / m ya T 5. kn.5 F W T R H B kn kn H m.55 m 9.8 kn/m m.65 The dam will tole if the moment about O of F and F exceeds the restoring moment of W and F. Assume m dee a) W (.4 98)( / ) =.9 6 N O 7 66 F d w = 9.67 m. (dw is from O to W.) 6 F = = N. d =.697 m. 45 F = N. d = 5 m. (d is from O to F.) 45 F 98 = N. d = 8.8 m Wdw F d N m 6 F d F d 96. N m will not tole. F W F b) W = (.4 98) ( ) = N dw = 9 78 = 9.67 m. 6 F. 54 N. d. 7 m. F = 98 6 = N. d = m. 6 F 98 = N. d Wdw F d N m 6 F d F d N m it will tole. = 8.57 m. c) Since it will tole for H = 6, it certainly will tole if H = 75 m. 7 Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

14 Chater / Fluid Statics.66 The dam will tole if there is a net clockwise moment about O. a) W W W. W ( 6 4 ) = 8,64 lb. W ( 4 4 / ) = 77,8 lb. W ( 4. / ) 6.4 = 7,87 ft. F 6.4 ( 4 ) = 49,9 4/ ft. F ( ) = ft W F Assume m dee W F O F F F F = 8,7 5 ft = 8,8 ft M O : (49,9)(4/) + (8,7)(5) + (8,8)() 8,64)() won t ti. b) W = = 56,6 lb. W = (4 6/) =, lb. F =, lb. W (6.86/) 6.4 = 4,79 lb. F = lb F 6.4 = 8,7 lb. F / = 46,8 lb. M O : (,)() + (8,7)(5) + (46,8)() will ti. c) Since it will tole for H = 6 ft., it will also tole for H = 8 ft. Forces on Curved Surfaces.67 M hinge =..5P d w W d F =. d w W P 4 P = 6 7 N Note: This calculation is simler than that of Examle.7. Actually, We could have moved the horizontal force F H and a vertical force F V (equal to W) simultaneously to the center of the circle and then.5p = F H.=F. This was outlined at the end of Examle.7. F d.68 Since all infinitesimal ressure forces ass thru the center, we can lace the resultant forces at the center. Since the vertical comonents ass thru the bottom oint, they roduce no moment about that oint. Hence, consider only horizontal forces: ( F ) 98 (4 ) 784 8N H water ( F ) N H oil M: P P = 66. kn. 8 Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

15 Chater / Fluid Statics.69 Place the resultant force F F H V at the center of the circular arc. F H asses thru the hinge showing that P. F V a) P F V 98( 6 4 4) 594 N or 594. kn. b) P = F V = 6.4 ( ) =, lb..7 a) A free-body-diagram of the volume of water in the vicinity of the surface is shown. Force balances in the horizontal and vertical directions give: F H F F W F V where FH and FV are the horizontal and vertical comonents of the force acting on the water by the surface AB. Hence, FH F 9.8 kn/m kn The line of action of F H is the same as that of F. Its distance from the surface is To find F V we find W and F : y I y m ya 98 W V 9.8 kn/m 4.7 kn 4 F 9.8 kn/m kn FV F W kn To find the line of action of F V, we take moments at oint A: where F x F d W d V V R d m, and d.55 m: 4 4 F d W d xv.8 m F 66 V Finally, the forces F H and F V that act on the surface AB are equal and oosite to those calculated above. So, on the surface, F H acts to the right and F V acts downward. b) If the water exists on the oosite side of the surface AB, the ressure distribution would be identical to that of Part (a). Consequently, the forces due to that ressure distribution would have the same magnitudes. The vertical force F V = 66 N would act uward and the horizontal force F H = 76. N would act to the left. F B. W F V F F H x V.A 9 Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

16 Chater / Fluid Statics.7 Place the resultant FH FV at the circular arc center. F H asses thru the hinge so that P F V. Use the water that could be contained above the gate; it roduces the same ressure distribution and hence the same F V. We have P F V = 98 ( ) = 98 7 N or 98.7 kn..7 Place the resultant FH FV at the center. F V asses thru the hinge (98 ) =.8 P. P = 7 7 N or 7.7 kn..7 The incremental ressure forces on the circular quarter arc ass through the hinge so that no moment is roduced by such forces. Moments about the hinge gives: P =.9 W =.9 4. P = N..74 The resultant FH FV of the unknown liquid acts thru the center of the circular arc. F V asses thru the hinge. Thus, we use only ( F H ) oil. Assume m wide: a) M R R R R S R R : R x R x 458 N/m b) M R R R R S R R :.. R x R x 9. lb/ft.75 The force of the water is only vertical (F V )w, acting thru the center. The force of the oil can also be ositioned at the center: a) P ( F H ) o (. 8 98).. 6 = 8476 N. F W ( F ) ( F ) y V o V w = S (.8 98) b) gv W. = 996 lb S F W ( F ) ( F ) y V o V w = S S Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

17 Chater / Fluid Statics.76 The ressure in the dome is a) = = 4 87 Pa or 4.87 kpa. The force is F = A rojected = ( ) 4.87 = 4.4 kn. b) From a free-body diagram of the dome filled with oil: F weld + W = A Using the ressure from art (a): F weld = 4 87 (.8 98) 4 = 4 N or.4 kn A W F weld.77 A free-body diagram of the gate and water is shown. y H F d w W H P. a) H = m. F = 98 4 = 9 4 N. / y W 98 xdy 98 dy 98 / = 6 6 N. / x xdy x dx d w x xdy 4 / 4 =.75 m. / 4x dx P = 7 98 N or 7.98 kn. F h/ x da=xdy b) H = 8 ft. F = = 7987 lb d w W 6 4 xdy x dx / = 66 lb. 4 x dx 6/4 8 x =.75 ft. P lb 8/ 8 4x dx Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

18 Chater / Fluid Statics Buoyancy.78 W = weight of dislaced water. a) + 5 = 98 (6 d d /). d + d 8.5 =. d =.7 m. b) 7 =. 98 (6 d d /). d + d 7.8 =. d =.6 m F B =. F B = 75 = 98 V. V = m =. or 7645 cm = 8 N/m..8 6 = 5 d 6.4. d =.846' or 4.6" = ( + 8h ) 98 h =.465 m. distance from to =.465 =.55 m.8 T + F B = W. (See Fig.. c.) T = = 884 N or 8.84 kn..8 The forces acting on the balloon are its weight W, the buoyant force F B, and the weight of the air in the balloon F a. Sum forces: 4 4 F B = W + F a or R g R a g T a = 5.4 K or 77.4C T a.84 The forces acting on the blim are the ayload F, the weight of the blim W, the buoyant force F B, and the weight of the helium F h : F B = F + W + F h = F +. F F P = and N eole = =. 6 8 Of course equiment and other niceties such as gyms, ools, restaurants, etc., would add significant weight. Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

19 Chater / Fluid Statics.85 Neglect the bouyant force of air. A force balance yields Density: F B = W + F = 5 + = 6 = 98 V. V =.6 m gv W = 5. = 8.5 kg/m Secific wt: = g = = 867 N/m Secific gravity: 8.5 S S =.85 water.86 From a force balance F B = W + A. F B a) The buoyant force is found as follows (h > 6'): cos h 5 R, Area = R (h 5 R) R sin R F B = 6.4[R R + (h 5 R) R sin]. W A F B = 5 + ha. The h that makes the above F B s equal is found by trial-anderror: h = 6.5: 859? 577 h = 6.8: 866? 858 R h 5 h = 7.: 87? 96 h = 6.8 ft. b) Assume h > 6. ft and use the above equations with R =. ft: h = 6.4: 857? 85 h = 6.4 ft. c) Assume h < 6.667ft. With R =.667 ft, F B = 6.4[R (R h + 5) R sin] F B = 5 + ha. Trial-and-error for h: cos R h R 5 h = 6: 849? 74 h = 6.: 85? 765 R h 5.87 a) W F B h = 6.4: 857? 7 h = 6.5 ft.. h / V V m. h = 7.6 m 4 4 m Hg.6 h.5 / 4 =.769 kg Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

20 Chater / Fluid Statics.5.5 b) ( ) 9.8 = S x c) ( ) 9.8 = 98.5 S x. S x = S x = (. m Hg ) m Hg = a) ( ) 9.8 = 98 4 b) m Hg =.886 kg..5 S x. S x =.89. Stability 4 4 d (/).89 a) Io =.67 ft W ( 5 / ) / V =.46. deth = r HO (5 /) =.8 ft GM.67 /.46 (.5.4) =.457'. It will not float with ends horizontal. b) I o =.67 ft 4, V =.66 ft, deth =.6667 ft GM.67 /.66 (5 4) / =.8 ft. It will not float as given c) V =.99, deth = 6.4", GM = =.47 ft. It will float With ends horizontal Io d / 64. The dislaced volume is V d h 5 x / xd since h = d. The deth the cylinder will sink is deth = V xd / d / 4. xd A The distance CG is CG h 5. x d /. Then 4 d / 64 d 5 GM. xd / d x This gives (divide by d and multily by x ): x x >. Consequently, x > 869 N/m or x < 45 N/m 4 Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

21 Chater / Fluid Statics.9 W S d water V water 4 water S d. W S d water V S d. h = Sd. d / S GM ( d / Sd / ) d. Sd S If GM = the cube is neutral and 6S 6S + =. 6 S 6 4 =.7887,.. The cube is unstable if. < S < Note: Try S =.8 and S =. to see if GM. This indicates stability. water water G C h.9 As shown, y /(6+6) = 6.5 cm above the bottom edge SA 4 G = 6.5 cm SA S A = S A. S A = a) y = 4. x For G: y = = x =.64. G must be directly under C..6 tan. = C G b) y =. x For G: y y =.4, x = = =.4. x =.8.68 tan. = The centroid C is.5 m below the water surface. CG =.5 m. Using Eq..4.47: GM 8 / The barge is stable. 5 Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

22 Chater / Fluid Statics y Using Eq..4.47: Linearly Accelerating Containers =.8 m. CG =. m / GM Stable H.96 a) tan. H = 8.55 m. max = 98 ( ) = 99 6 Pa b) max = (g + a z ) h = (9.8 + ) = 59 6 Pa c) max =.94 6 ( ).94 (. + 6) ( 6) = 47 sf or 7.5 si d) max =.94 (. + 6) ( 6) = 7 sf or 7.45 si.97 The air volume is the same before and after. z A h.5 8 = hb/. tan b h. h. h =.856. Use dotted line. B.5w w =.74 m. a) A = ( 7.66) = 5 74 Pa or 5.74 kpa b) B = ( 7.66) = 76 6 Pa or 76.6 kpa c) C =. Air fills the sace to the dotted line. b h w C x.98 Use Eq..5.: Assume an air-water surface as shown in the above figure. 8a a) 6 = a x ( 8) x = h a 6 = 8 a x x or a x 4.45 =.74 a x. a x 9.8 a x. a x = a x =.64, 7.46 m/s b) 6 = a x ( 8) (9.8 + ). 5 8a x a 6 = 8 a x x or a x. =.574 a x. 9.8 a x 5. a x +.44 = a x =.5, 4.8 m/s 6 Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

23 Chater / Fluid Statics c) 6 = a x ( 8) ( ) (.5 + 8a x ) a 6 = 8 a x x or a x.875 =.6 a x. 4.8 a x 7.6 a x = a x =., 6.8 m/s.99 a) a x =.866 = 7. m/s, a z = m/s. Use Eq..5. with the ee hole as osition. The x-axis is horizontal assing thru A. We have A = 7. (.) (9.8 + ) (.866) = 58 9 Pa b) A = 8.66 (.848) ( ) (.799) = Pa c) The ee hole is located at (.696, 5.598). Use Eq..5.: A = (.696).94 (. + ) ( 5.598) = 48 sf d) The ee hole is located at (4.98, 7.464). Use Eq..5.: A = ( 4.98).94 (. + 5) ( 7.464) = 9 sf. a) The ressure on the end AB (z is zero at B) is, using Eq..5., (z) = ( 7.66) 9.8(z) = z. 5 FAB ( z) 4dz = 64 N or 64 kn b) The ressure on the bottom BC is (x) = (x 7.66) = 76 6 x FBC ( 76 6 x) 4dx =.6 6 N or 6 kn c) On the to (x) = (x 5.74) where osition is on the to surface: Fto ( 5 74 x) 4dx = N or 55 kn. a) The ressure at A is 58.9 kpa. At B it is B = 7. (.7.) (9.8) (.866) = 8495 Pa. Since the ressure varies linearly over AB, we can use an average ressure times the area: F AB. 5 = N or. kn b) D =. C = 7. (.5.) 9.8( ) = 49 8 Pa. F CD = 74 7 N or 74.7 kn c) A = 58 9 Pa. C = 49 8 Pa. F AC. 5 = 8.8 kn. z x 7 Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

24 Chater / Fluid Statics. Use Eq..5. with osition at the oen end: a) A = since z = z. B = = 89 Pa. C = 89 Pa. z A b) A = (.9 ) = 9 Pa. B = (.9) 9.8(.6) = 4 Pa C = 9.8 (.6) = 5886 Pa. c) A = (.9) = 8 Pa. B =.9 9.8(.6) = 6 Pa. 5 d) A =. B =.94 (.-6) = sf. C = sf. e) A = = 64 sf. B = = 4 sf. C = = sf. 7 5 f) A =.94. = 8 sf. 7 5 B =.94( ) = 4 sf. C = = 5 sf. Rotating Containers C B C = 89 Pa x. Use Eq..6.4 with osition at the oen end: 5 = 5.6 rad/s. 6 a) ( 5.6 /) (.6.5) = Pa. A z A B = 6 99 Pa. C = 98.6 = 5886 Pa. C B r b) A = 495 Pa. B = 8859 Pa. C = 98.4 = 94 Pa. 8 Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

25 Chater / Fluid Statics c) A = 59.7 sf. B = 89.7 sf. C = 6.4 5/ = sf. d) A = 9.5 sf. B = 7.5 sf. C = 6.4 5/ = 78 sf..4 Use Eq..6.4 with osition at the oen end. a) A (.9 ) = 4 5 Pa. z A B = = 4 6 Pa. C = 98.6 = 5886 Pa. b) A (.6 ) = 8 Pa. r C B B = = 4 8 Pa. C = 98.4 = 94 Pa. c) A = 947 sf. B = / = 87 sf. d) A = 4 sf. B = / = 6 sf. C = 6.4 5/ = sf. C = 6.4 5/ = 78 sf..5 Use Eq..6.4 with osition at the oen end and osition at the origin. Given: =. a) = (.45 ) 98 (.6). = 7.6 rad/s. b) = (. ) 98 (.4). = 9.4 rad/s. c) = d) = = 7.4 rad/s. 5. = 9.57 rad/s r z 9 Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

26 Chater / Fluid Statics.6 The air volume before and after is equal. r h. 6.. r h =.44. a) Using Eq..6.5: r 5 / = 9.8 h h =.48 m h z r A = (.7) = 849 Pa. A r b) r 7 / = 9.8 h. h =.6 m. A = = 78 Pa. c) For =, art of the bottom is bared. Using Eq..6.5: r. 6. r h r h. g h, r g g g. 44 h h. 44 h h h. Also, h h =.8..6h.64 =.79. h =.859 m, r =.8 m. A = (.6.8 )/ = 7 4 Pa. or A h h z r r d) Following art (c): h h h.64 =.96. h =.5 m. A = (.6.65 )/ = 57 9 Pa r =.65 m.7 The answers to Problem.5 are increased by 5 Pa. a) 5 Pa b) 5 78 Pa c) 4 4 Pa d) 8 9 Pa 4 Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

27 Chater / Fluid Statics.8 ( r) r g[ (.8 h)]. ( r) 5 r 98(.8 h) if h <.8. ( r) ( r r ) if h > ( 5 65 ) = 667 N. a) F rdr r r dr (We used h =.48 m).6 b) F rdr (4 5r 96 r) dr = 7 N. (We used h =.6 m) da = rdr dr c) d).6 (5 (.8 ) = 95 N. (We used r =.8 m).8 F rdr r r dr.6 ( (.65 ) = 6 4 N. (We used r =.65 m).65 F rdr r r dr 4 Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter

28 Chater / Fluid Statics 4 Cengage Learning. All Rights reserved. May not be scanned, coied or dulicated, or osted to a ublicly accessible website, in whole or in art. Full file at htt://testbank6.eu/solution-manual-mechanics-of-fluids-4th-edition-otter