CHAPTER 2 Fluid Statics

Chapter / Fluid Statics CHPTER Fluid Statics FE-tpe Exam Review Problems: Problems - to -9. (C) p = γ h = (.6 98) (8.5.54) = 96 6 Pa = 96.6 kpa Hg. (D) p = p ρ gh = 84. 9.8 4 = 44 76 Pa. (C) p = p + γ h γ h = +. 98. = 8 Pa = 8 kpa w atm x x water w.4 () p = γ H = (.6 98).6 = 5 Pa. a pa, after = 5 + = 5 =.6 98 Hafter. Hafter =.85 m.5 (B) The force acts / the distance from the hinge to the water line: 5 = 5 98 5 = 667 N=.7 kn.6 () The gate opens when the center of pressure in at the hinge:. + h I. + h b(. + h) / = + 5. p = + = + = 5 +.. (. + h) b(. + h)/ This can be solved b trial-and error, or we can simpl substitute one of the answers into the equation and check to see if it is correct. This ields h =.8 m..7 (D) Place the force hinge:.8 () W = γ V F + F at the center of the circular arc. FH passes through the H V P = F = 4.w 98 + ( π. /4) w 98 =. w = 5.6 m. V 9 9.8 = 98. 5 w. w = 6 m.9 () 5 pplug = + γ h = + 666 (. ) = 4 7 Pa 9.8 F = p = 4 7 π. =.5 N. plug plug 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part. 5

Chapter / Fluid Statics Chapter Problems: Pressure z. ΣF = ma : p z p s sinα = ρ a z z ΣFz = maz: pz p s cosα = ρ a z + ρ g Since scosα = and s sin α = z, we have p p = ρ a Let and z : then and p p = ( + g ) z p p z z ρ a z p = p = p z z z ρg V p s s p z α p = pz = p.. p = γh. a) 98 = 98 Pa or 98. kpa b) (.8 98) = 78 48 Pa or 78.5 kpa c) (.6 98) = 4 Pa or 4 kpa d) (.59 98) = 55 98 Pa or 56. kpa e) (.68 98) = 66 7 Pa or 66.7 kpa. h = p/γ. 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. = =. =. =. =.4 a) p = γh =.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) p = γh + γh = 98. + (.6 98). = 46 Pa or 4.6 kpa. b) 98.5 + 5 6.6 = 96 Pa or.96 kpa. c) 96 + 98 + (.6 98). = 6 Pa or 6. kpa..6 p = pg h =.7 9.8 ( ) = 77 Pa = 7.4 kpa. 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part. 6

Chapter / Fluid Statics.7 pg 9.8 poutside = ρog h = h = =.5 Pa RT.87 5 o pbase =.84 Pa pg 9.8 pinside = ρ = = =.67 Pa ig h h RTi.87 9 If no wind is present this p base would produce a small infiltration since the higher pressure outside would force outside air into the bottom region (through cracks)..8 p = ρgdh where h = z. From the given information S =. + h/ since S() = and S() =.. B definition ρ = S, where ρwater = kg/m. Then dp = ( + h/) gdh. Integrate: p dp = ( + h/) gdh p = 9. 8( + ) = Pa or kpa Note: we could have used an average S: Savg =.5, so that ρ avg = 5 kg/m..9 p i p j p p = + + x z k ρa i ρα j ρα k ρgk = ρ a i + a j + a k ρg k = ρa ρg = x z ( x z ) p = ρ( a + g).. atm / α R p = p [( T α z) / T ]g = [(88.65 )/88] 9.8/.65 87 = 96.49 kpa p = patm ρgh = 9.8 / = 96.44 kpa.87 88 96.44 96.49 % error = =.5% 96.49 The densit variation can be ignored over heights of m or less. g/ αr T α z p = p p = patm p T 9.8/.65 87 88.65 = 88 =.7 Pa or.7 kpa This change is ver small and can most often be ignored. atm 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part. 7

Chapter / Fluid Statics. Equation.5. gives 9 dp. = ρ, where dp = ρ gdh. d ρ Therefore, or,. ρ gdh = d ρ ρ dρ 9.8 = dh = 4.67 9 ρ. Integrating from ρ = kg/m, we get 9 9 dh =4.67 h or, or, Hence, =4.67 h =.. 9.7 = h= 4.67 ln 9.7 4.67 h =. ln 4.8 h = Now, assuming ρ = constant = kg/m, we have = h= 9.8 h= 4h= a) h = 45 m: paccurate = 455 kpa pestimate = 4547 kpa % error= 4547 455 455 =.76 b) h = 5 m: paccurate = 56 kpa pestimate = 556 kpa % error= 556 56 =. 56 c) h = 45 m: paccurate = 45954 kpa pestimate = 45469 kpa 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part. 8

Chapter / Fluid Statics % error= 45469 45954 45954. Use the result of Example.: p = e gz/rt. a) p = e 9.8 /87 7 = 8.9 kpa. b) p = e 9.8 /87 88 =.8 kpa. c) p = e 9.8 /87 58 = 6.9 kpa. 9.8.4 Use Eq..4.8: p = (.65 z / 88).65 87. =. a) z =. p = 69.9 kpa. b) z = 6. p = 47. kpa. c) z = 9. p =.6 kpa. d) z =. p =.5 kpa..5 Use the result of Example.: Manometers p gz ln =. p RT p = e p gz / RT 7 9.8z ln. =.5 87 5.6 p = γh = (.6 98).5 = 5 Pa or.5 kpa..7 a) p = γh. 45 = (.6 98) h. h =.7 m T = 5 k R = 87 J/kg k b) p +.78.5 = (.6 98) h. Use p = 45, then h =.7 m The % error is. %. z = 7,9 m..8 Referring to Fig..6a, the pressure in the pipe is p = ρgh. If p = 4 Pa, then 4 = ρgh = ρ 9.8h or 4 ρ =. 9.8h 4 a) ρ = = 68 kg/m. gasoline 9.8.6 4 b) ρ = = 899 kg/m. benzene 9.8.7 4 c) ρ = = 999 kg/m. water 9.8.45 4 d) ρ = = 589 kg/m. carbon tetrachloride 9.8.54 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part. 9

Chapter / Fluid Statics.9 Referring to Fig..6a, the pressure is p = ρwgh =. av 9.8.6 a) V = = 957. V =.9 m/s. b) c) d) ρ Then V = 9.8.75 V = = 96.4. V = 4.6 m/s. 9.8. V = = 595. V = 9.9 m/s. 9.8.5 V = = 994. V = 44.6 m/s.. See Fig..6b: p = γh + γh. p =.86 9.8.5 +.6 9.8.4 = 695 Pa =.96 kpa. p p ρ gh ρ gh ρ gh ρ gh. = + + + + 4 4 ρwgh. ρ = + 97 9.8. + 9.8. + 58 9.8.5 + 59 9.8.8 = 64 Pa or.64 kpa. p p ( p p ) ( p p ) ( p p ) = + + (Use p = ρ g h) 4 4 4 6 = 9.8(.) + 6 9.8 H + 9 9.8.. H =.74 m or 7.4 cm. p p ( p p ) ( p p ) ( p p ) = + + (Use p = ρ g h) 4 4 po pw = 9 9.8(.) + 6 9.8(.) + 9.8.5 = Pa or. kpa.4 p p = ( p p ) + ( p p ) + ( p p ) + ( p p ) 5 4 4 5 p = 98(.) + 6 9.8(.4) + 98(.) + 6 9.8.6 = 5 6 Pa or 5.6 kpa.5 pw + 98.5.6 98..68 98. +.86 98.5 = po. pw po = 94 Pa or.94 kpa..6 pw 98..68 98. +.86 98. = po. With pw = 5, po = 4 Pa or 4. kpa. a 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part.

Chapter / Fluid Statics.7 a) p + 98 =.6 98.. p = 678 Pa or 6.8 kpa. b) p + 98.8 =.6 98.. p = 8 85 Pa or 8.84 kpa. c) p + 98.8 =.6 98.. p = 46 Pa = 4.6 kpa. d) p + 98.6 =.6 98.. p = 797 Pa =.8 kpa..8 p 98 4 +.6 98.6 =. p = 7 89 Pa or 7.89 kpa..9 8 + 98.5 =.59 98 H. H =.68 m Hnew =.68 +.7 =.956 m. H =.7 =.65. p + 98 (.5 +.65) =.59 98.956. p = Pa or. kpa. H H H.4 p + 98.5 +.59 98.7.8 98. =.6 98.5. p = 587 Pa or 5.87 kpa. Note: In our solutions we usuall retain significant digits in the answers (if a number starts with then 4 digits are retained). In most problems a material propert is used, i.e., S =.59. This is onl significant digits! onl are usuall retained in the answer!.4 The equation for the manometer is p + γ.7 = p + γ.+ γ.9sin 4 water B oil HG Solve for pb: p = p + γ.7 γ.9sin 4 γ. B water HG oil = p + γ.7.6γ.9 sin 4.87γ. water water water (.7.6.9sin 4.87.) γ ( ) = p + water = kpa +.7.6.9 sin 4.87. 9.8 kn/m =. kpa.4 The distance the mercur drops on the left equals the distance along the tube that the mercur rises on the right. This is shown in the sketch. Oil (S =.87) B Water h cm 9 cm 7 cm h 4 o Mercur 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part.

Chapter / Fluid Statics From the previous problem we have ( B ) γ water γhg γoil For the new condition p = p +.7.9sin 4. () ( ) = + γ (.7 + ) γ.sin 4 γ (. sin 4) p p h h () B water HG oil where h in this case is calculated from the new manometer reading as h + h / sin 4 = 9 h =.78 cm Subtracting Eq.() from Eq.() ields ( p ) ( p ) = γ ( h) γ.sin 4 γ ( h sin 4) B B water HG oil Substituting the values of h and ( pb ) gives ( p B ) ( ) ( ) ( ) p B =.+.78.6.sin 4.87.78sin 4 9.8 =.5 kpa.4 Before pressure is applied the air column on the right is cm high. fter pressure is applied, it is (. H/) m high. For an isothermal process p V = p V using absolute pressures. Thus,.56.. = p(. H / ) or p = (. H/) From a pressure balance on the manometer (pressures in kpa): 98.56 7 +. =.6 +,. H/ or H 4.7H +.7 = H =. m or.7 m..44 p p = ( p p ) + ( p p ) + ( p p ) + ( p p ) 5 4 4 5 4 = 98(.6.) + 5 6(..6) + 4H + 5 6(.7 H). H =.76 m or 7.6 mm.45 H D / d = p γ + γ + ( γ γ ) D / d (./.5) = = 8.487 H 98 + 5 6 + ( 4 5 6)(./.5) 6 H = 8.487 4 =.4 m or.4 mm 6.46 p p ( p p ) ( p p ) ( p p ) = + + (poil = 4. kpa from No..) 4 4 5 5 4 = 98(. + z) + 68(. z) + 86(. z). z =.45 m or 45. mm 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part.

Chapter / Fluid Statics.47 a) pair = 65 + 65 = 56 Pa. 56 + 98( + z) 6 9.8(. + z) =. z =.5. h =. + z =.5 m or 5 cm b) pair = 8 8 + 88 = 68 Pa. 68 + 98(.8 + z) 6 9.8(. + z) =. z =.75 m h =.+ z =.4 or.4 cm c) pair = 46 + 4.6 = 884 Pa. 884 + 98(.8 + z) 6 9.8(. + z) =. z =.6 m = 6 mm. d) pair = 8 + 8 = 88 Pa 88 + 98(.6 + z) 6 9.8(. + z) z = 8. m = 8. mm. Forces on Plane reas.48 F = γ h = 98 π. /4 = 694 N..49 5 5 5 P = 98. = 67 N P a) F = pc = 98 4 = 6 N or.6 kn b) F = pc = 98 ( 4) + 98 + 98 = 98 N or 98. kn c) F = pc = 98 4 = 9 N or.9 kn d) F = pc = 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 = pc = 98.5 ( ) = 9 4 N The weight of the vault is approximatel W = ρgv walls = 4 9.8[(.5.) + (.) + (.8..)] = 8 4 N. The vault will tend to rise out of the ground..5 F = pc = 666 π = 67 4 N or 67.4 kn Find γ in Table B.5 in the ppendix..5 a) F = pc = 98 (.88/) (.88 /) = 5 N or 5 kn where the height of the triangle is ( ) / =.88 m. b) F = pc = 98 (.88 /) = 77 N or 77. kn c) F = pc = 98 (.88.866/) (.88 /) = 54 5 N or 54.5 kn 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part.

Chapter / Fluid Statics (.4.8).5 a) F = γ h = 98 8. = 74,995 N..8 (.4) /6 p = 8. + = 8.4 m. = 9 8.4 =.76 m. 8..6.4.9 Now, = =.6 m..64 x x Thus, the force acts at (.6m,.76 m). b) F = 98 9 (.4.8) = 9,7 N. The centroid is the center of pressure..4.9 =.8 m. = =.6 m..6 x x Thus, the location of force is (.6m,.8m). c) F = 98(9.8.77) (.4.8) = 79,997 N. p.8 (.4) /6 =.9 + =.97 m..9.6.4.9 Now, = =.6 m..6 x x Thus, the location is (.6m,.77 m)..54 a) F = γ h = 98 6 π = 79 7 N or 79.7 kn. p 4 =.74.97 =.77 m I π = + = 6 + = 6.67 m. (x, ) p = (,.67) m 4(4 π 6) (x, ) x b) F = γh = 98 6 π = 69 8 N or 69.8 kn. 4 π / 8 p = 6 + = 6.67 m. x + = 4 π 6 x γ γ xpf = pd = (6 ) = (4 )(6 ). x xd d γ xpγ 6 π = (4 4 6 + ) = γ. d xp =.8488 m (x, ) p = (.8488,.67) m d d (x, ) x c) F = 98 (4 + 4/) 6 = 9 N or.9 kn. 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part. 4

Chapter / Fluid Statics 4 /6 p = 5. + = 5.5 m. =.5 5. 6 4/.5 =.5. x =.975. (x, )p = (.975,.5) m x x d) F = 98 ( 4 + 4sin 6.9 ) 6 = N 5.4 /6 p = 5.6 + = 5.657 m. =.4 m 6 5.6 cos 5. =.8,.5.8 =.7,.4/.57 =. / x. x =.6. 7 x =.8 +.6 =.4. (x, )p = (.4,.4) m. 4 5. o.55 F = γ h = 98 (.8 +.5) (.8 ) = 766 N. p I.8 / = + =. + =.5 m...8 (4.8.5) 766 = P. P = 74,5 N..56 F = γh = 98 6 =.777 6 N, or 77 kn. I p = + = 7.5 + 7.5 4 5 / = 7.778 m. ( 7.778) 77 = 5 P. P = 5 kn..57 a) = h =98 8 =.57 N=57 kn = + =9.5+ 4 5 9.5 =9.7 m 9.7 57=5 = 76 kn b) = h =98 =.57 N=96 kn = + =.5+ 4 5.5 =.68 m 4.68 96=5 = 9 kn c) = h =98 =.57 N=54 kn p F P 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part. 5

Chapter / Fluid Statics = + =.5+ 4 5.5 =.65 m 6.65 54=5 = 6 kn.58.59.6 I H bh / H H p = + = + = + = H. p is measured from the surface. bh H/ 6 From the bottom, H p = H H = H. Note: This result is independent of the angle α, so it is true for a vertical area or a sloped area. sin 4 l. ( ) sin 4. γ ( ). F = γ 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. =..4 +.4.4 =.8 m Use forces: F = γh = 98.5657 (..4) = 757 N c.4 F = γhc = 98 (.4.4) = 67 N = p ( 4. ). I.4.4.4 / 6 p = + = + =.5657 m.4 (.4/) (.4/) ΣM hinge = : 757 4. + 67 ( 4.. 5657) 4. P =. P = 46 N..6 To open, the resultant force must be just above the hinge, i.e., p must be just less than h. Let p = h, the condition when the gate is about to open: = h + H = h + H I = h + H h + H ( )/, ( ), [( )]( ) /6 h + H ( h + H ) 4 /6 h + H h + H h + H p = + = + = ( h + H ) ( h + H ) / 6 h + H a) h =. h = H =.9 m b) h = H =. m c) h = H =.5 m 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part. 6

Chapter / Fluid Statics.6 The gate is about to open when the center of pressure is at the hinge. b.8 / a) p =. + H = (.8/ + H ) +. (.9 + H ).8b H =. b / b) p =. + H = (./ + H ) +. ( + H )b H =.6667 m. b. / c) p =. + H = (./ + H ) +. (. + H ).b H =.9 m..6 H F bh bh = γ = γ F H b b H H l γbh = γblh. H = l a) H = =.464 m b) H =.7 m c) H =. m d) H =.56 m F H/ l/ F.64 free-bod-diagram of the gate and block is sketched. Sum forces on the block: T Σ F = W = T + F where FB is the buoanc force which is given b B F stop T F B = γ π R ( H ) Take moments about the hinge: F H p F B T.5 = F ( ) where FH is the hdrostatic force acting on the gate. It is, using h =.5 m and = = 6 m, H p R x R W FH ( )( ) = γ h = 9.8 kn/m.5 m 6 m = 88.9 kn From the given information, p ( ) I / = + =.5 + = m.5 6 ( ) 88.9 T = = 5. kn.5 ( ) 7 5. 44.77 kn. γπ 44.77 F = W T = = R H = B 44.77 kn H = m =.55 m ( 9.8 kn/m ) π ( m) ssume m deep 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part. 7

Chapter / Fluid Statics.65 The dam will topple if the moment about O of F and F exceeds the restoring moment of W and F. F W F a) W = (.4 98)(6 5 + 4 5/) =.9 6 N O 7 + 6 6 F d w = = 9.67 m. (dw is from O to W.) + 6 F = 98 5.9 =.544 6. 9 N. d = =.697 m. 45 F = 98 45 = 9.9 6 N. d = 5 m. (d is from O to F.) 45 + F = 98 = 8.9 6. 94 5 + 55. N. d = = 8.8 m.. 94 + 55. 6 Wd w + F d = 48. 8 N m will not topple. 6 F d + F d = 96. N m b) W = (.4 98) (6 65 + 65 ) = 7.55 6 N. 9 7 + 78 6 dw = = 9.67 m. 9 + 78 6 F. 54 N. d. 7 m. F = 98 6 = 7.66 6 N. d = m. 6 + F = 98 =. 6. 94 5 + 7. 58 N. d =. 94 + 7. 58 6 Wd w + F d = 54. 9 N m it will topple. 6 F d + F d = 544. 5 N m = 8.57 m. c) Since it will topple for H = 6, it certainl will topple if H = 75 m..66 The dam will topple if there is a net clockwise moment about O. a) W W W W = +. = (.8.9 ) 98.4 = 546.7 kn..9 W = 7. 98.4 = 9.4 kn. W = ( 6.7/) 98 = 4. kn @ 6.7 m. F = 98 6 ( ) = 76. kn @ 4 m. F = 98.5 ( ) = 44.4 kn @ m. F F = F p p = 64.9 kn @ 4.5 m = 97. kn @ 6 m W F ssume m deep W F O F Net moment about : 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part. 8

Chapter / Fluid Statics Σ M O = 76. 4 + 64.9 4.5 + 97. 6 546.7.9 9.4 4. 4. 6.7 44.4 = 85. kn-m < and hence the dam won t tip. b) W = (.8 8.9 ) 98.4 = 8 kn. 8.9 W = 7. 98.4 = 6 kn. 6.86 W = 8 98 = 65.7 kn. F = 98 9 8 = 589 kn. F = 98.5 = 44.4 kn. 9 Fp = 98 9 = 65 kn; F 98 5 66. kn. p = = Net moment about : Σ M O = 589 6 + 65 4.5 + 66. 6 8.9 6 4. 65.7 6.7 = 9 kn-m > Hence, the dam will tip. c) Since it will topple for H = 8 m., it will also topple for H = 4 m. Forces on Curved Surfaces.67 ΣM hinge =..5P dw W d F =. d w W P 4 P = + π 98 8 98. 5 π 4 4 = 6 7 N Note: This calculation is simpler than that of Example.7. ctuall, We could have moved the horizontal force F H and a vertical force F V (equal to W) simultaneousl to the center of the circle and then.5p = F H.=F. This was outlined at the end of Example.7. F d.68 Since all infinitesimal pressure forces pass thru the center, we can place the resultant forces at the center. Since the vertical components pass thru the bottom point, the produce no moment about that point. Hence, consider onl horizontal forces: ( F ) = 98 (4 ) = 784 8N H water ( F ) =.86 98 = 68 7N H oil ΣM: P = 784. 8 68. 7. P = 66. kn..69 a) Place the resultant force FH + FV at the center of the circular arc. F H passes through the hinge showing that P =. F V 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part. 9

Chapter / Fluid Statics P = = 98( 6 4 + π 4) = 594 N or 594. kn. F V b) Repeating the process from part a: = =98 9 4+ 4 = 8 kn.7 a) free-bod-diagram of the volume of water in the vicinit 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 components of the force acting on the water b the surface B. Hence, FH ( )( )( ) = F = 9.8 kn/m 8 + 4 = 76 kn F B. W F V F F H x V. The line of action of FH is the same as that of F. Its distance from the surface is p ( ) I 4 = + = 9 + = 9.7 m 9 8 To find FV we find W and F: π W = γ V = ( 9.8 kn/m ) ( ) 4 =.7 kn 4 F ( ) = 9.8 kn/m 8 4 = 68 kn FV = F + W =.7 + 68 = 66 kn To find the line of action of FV, we take moments at point : F x = F d + W d V V where R d = m, and d = = =.55 m: 4 4 ( π ) ( π ) F d + W d 68 +.7.55 xv = = =.8 m F 66 V Finall, the forces FH and FV that act on the surface B are equal and opposite to those calculated above. So, on the surface, FH acts to the right and FV acts downward. 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part.

Chapter / Fluid Statics b) If the water exists on the opposite side of the surface B, the pressure distribution would be identical to that of Part (a). Consequentl, the forces due to that pressure distribution would have the same magnitudes. The vertical force FV = 66 N would act upward and the horizontal force FH = 76. N would act to the left..7 Place the resultant FH + FV at the circular arc center. F H passes thru the hinge so that P =. Use the water that could be contained above the gate; it produces the same F V pressure distribution and hence the same F V. We have P = F V = 98 (6 4 + 9π) = 98 7 N or 98.7 kn..7 Place the resultant FH + FV at the center. F V passes thru the hinge (98 ) =.8 P. P = 7 7 N or 7.7 kn..7 The incremental pressure forces on the circular quarter arc pass through the hinge so that no moment is produced b 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 passes thru the hinge. Thus, we use onl ( F H ) oil. ssume m wide: ΣM R R R R S R R : R x R. 98 4 π + 98 = 4 γ γ π x = 458 N/m.75 The force of the water is onl vertical (FV)w, acting thru the center. The force of the oil can also be positioned at the center: a) P = ( F H ) o = (. 8 98).. 6 = 8476 N. Σ F = = W + ( F ) ( F ) V o V w = S 98π.6.6π 6 +.6 6 (.8 98) 98π.8 6 4 b) ρ g V = W. 98.8.6 6 S =.955. Σ F = = W + ( F ) ( F ) V o V w = S 98π π + 4 (.8 98) 98π 98.8. S =.955. 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part.

Chapter / Fluid Statics.76 The pressure in the dome is a) p = 6 98.8 98 = 4 87 Pa or 4.87 kpa. The force is F = pprojected = (π ) 4.87 = 4.4 kn. b) From a free-bod diagram of the dome filled with oil: Fweld + W = p Using the pressure from part (a): Fweld = 4 87 π (.8 98) 4 π = 4 N or.4 kn p W F weld.77 free-bod diagram of the gate and water is shown. H F + d w W = H P. a) H = m. F = 98 4 = 9 4 N. / W = 98 xd = 98 d = / 98 = 6 6 N. / x xd x dx d = w x = xd = 4 = / 4 =.75 m. / 4x dx P = + 9. 75 4 6 6 = 7 98 N or 7.98 kn. F h/ x d=xd b) H =.4 m. F = 98. (.4.) = 9 N.4.4 W = 98.xd = 98. d = 84 (.4) = 6 N. / 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part.

Chapter / Fluid Statics d w.4 x x d d = x = = =.4 m.4 x d d.4 P = 9 +.4 6 = 484 N.4 Buoanc.78 W = weight of displaced water. a) + 5 = 98 (6 d + d /). d + d 8.5 =. d =.7 m. 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part. b) 7 =. 98 (6 d + d /). d + d 7.8 =. d =.6 m..79 5 + FB =. FB = 75 = 98 V. V = 7.645 m γ 7.645 =. or 7645 cm γ = 8 N/m..8 5 6 = 7.5 9 d 98. d =.6 m =.6 cm..8 9.8 + 6 = ( + 8h ) 98 h =.465 m. distance from top =.465 =.55 m.8 T + FB = W. (See Fig.. c.) T = 4.59 98 = 884 N or 8.84 kn..8 The forces acting on the balloon are its weight W, the buoant force FB, and the weight of the air in the balloon Fa. Sum forces: 4 4 FB = W + Fa or πr ρg = + πr ρ a g 4 9.8 4 9.8 π 5 = + π 5. Ta = 5.4 K or 77.4 C.87 9.87T a.84 The forces acting on the blimp are the paload Fp, the weight of the blimp W, the buoant force FB, and the weight of the helium Fh: FB = Fp + W + Fh 9.8 5π 5 = Fp +. Fp + 5 π 5 9. 8.87 88. 77 88 8 FP = 9.86 8 9. 86 and Npeople = =. 6 8

Chapter / Fluid Statics Of course equipment and other niceties such as gms, pools, restaurants, etc., would add significant weight..85 Neglect the bouant force of air. force balance ields Densit: FB = W + F = 5 + = 6 = 98 V. V =.6 m ρ g V = W. ρ 9.8.6 = 5. ρ = 8.5 kg/m Specific wt: γ = ρg = 8.5 9.8 = 867 N/m Specific gravit: ρ 8.5 S = = S =.85 ρ water.86 From a force balance FB = W + p..87 a) W F B a) The buoant force is found as follows (h > 4.8 m): h 4.5 R cos θ =, rea = θr (h 4.5 R) R sinθ R FB = 98[πR θr + (h 4.5 R) R sinθ]. FB = 667 + γh. The h that makes the above FB s equal is found b trial-anderror: h = 4.95: 86? 7 h = 5.4: 895? 88 h = 5.: 87? 8685 h = 5.4 m. b) ssume h > 4.9 m and use the above equations with R =.4m: h = 4.9: 849? 84 h = 4.9 m. c) ssume h < 5 m. With R =.5 m, FB = 98[θR (R h + 4.5) R sinθ] FB = 667 + γh. Trial-and-error for h: cos θ = R h + 4.5 R h = 4.8: 8? 66 h = 4.86: 89? 7855 h = 4.9: 875? 957 h = 4.875 m. =. [ hπ ]. +. 6. 5 / 4 9. 8 = 98 V. = + π. 5 π. 5 5 V. 5. 6 =. 769 m. h = 7.6 m 4 4 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part. 4

Chapter / Fluid Statics =.6 hπ.5 / 4 =.769 kg m Hg π.5 π.5 b) (. +.769) 9.8 = 98.5 +. S x. 4 4 π.5 c) (. +.769) 9.8 = 98.5 Sx. Sx =.45. 4 Sx =.959..88 π.5 π.5 (. + m Hg )9.8 = 98.5 +.. mhg =.886. 4 4 π.5 a) (. +.886) 9.8 = 98 4 b) mhg =.886 kg..5 Sx. Sx =.89. Stabilit.89 a) I o 4 4 π d π (.5) 4 4 = = =.9 m. 64 64 W.8 98 π (.5). V = = =. m γ 98 H O. depth = =.44 m π (.5) 4.9 GM = (.5.) =. m. It will not float with ends horizontal. b) Io =.9 4 m 4 = =.8 98.5. =.785 m 98 depth=.785.5 =.6 m =.9.785..8 =.45> In this case the clinder will float with ends horizontal. c) = =.8 98.5.5 98 =.98 m depth=.98.5 =. m 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part. 5

Chapter / Fluid Statics =.9.98.5. =.54 It will not float with ends horizontal. 4.9 With ends horizontal Io = π d /64. The displaced volume is V = γ π /4 98 = 8.4 5γ x d h xd since h = d. The depth the clinder will sink is V depth = 5 5 = 8.4 γxd / πd /4 =. γxd h The distance CG is 5 CG =. γxd /. Then 4 πd /64 d 5 GM = +. γ /. 5 xd > 8.4 γ d x This gives (divide b d and multipl b γx): 6.5.5 γx + 5. 5 γ x >. Consequentl, γx > 869 N/m or γx < 45 N/m.9 W Sγ d water V = = = γ water 4 γ water S d. W S d water V = = γ = 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 <.7887. Note: Tr S =.8 and S =. to see if GM >. This indicates stabilit. γ water γ water G C h 6 9 + 6 4.9 s shown, = = 6.5 cm above the bottom edge. 6 + 6 4γ 9.5 + 6γ 8.5 + 6S γ 4 G = = 6.5 cm..5γ 8 + γ 8 + S γ 6 + 4 S = 74 + 64 S. S =.. 6 4 + 8 + 8 7 6 + 8 4 + 8 4.9 a) = = 4. x = 6 + 8 + 8 6 + 8 + 8. 6 4 +.5 8 +.5 8 7 For G: = = 4.68.. 6 +.5 8 +.5 8 =.5. 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part. 6

Chapter / Fluid Statics. 6 +.5 8 4 +.5 8 4 x =. 6 +.5 8 +.5 8 =.64. G must be directl under C..6 tan θ =. θ =...68 C.6 G.68 b) 5 5 +.5.5 +.5 8.75 = 5 +.5 +.5 = 5; 5.5 +.5 5 +.5 5 x = 5 +.5 +.5 =.5. 5 5 +.5.5.5 +.5.5 8.75 For G: = = 5.85. 5 +.5.5 +.5.5. 5.5 +.5.5 5 +.5.5 5 x = =.95. 5 +.5.5 +.5.5.75 =.85, x =.75. tan θ =. θ =.6..85.94 The centroid C is.5 m below the water surface. CG =.5 m. l 8 / Using Eq..4.47: GM =.5 =.777.5 =.77 >. l 8 The barge is stable. 8.485.44 + 6. 97.95 = =.8 m. CG = 8.. 5 =. m. 8.485 + 6. 97 l 8.485 / Using Eq..4.47: GM =. =.46. =.6. Stable. 4.97l Linearl ccelerating Containers H.96 a) tanα = =. H = 8.55 m. pmax = 98 (8.55 + ) = 99 6 Pa = 99.6 kpa 9. 8 4 b) pmax = ρ(g + az) h = (9.8 + ) = 59 6 Pa = 59.6 kpa c) pmax = 8 (.6) + (9.8 + 8) (.8) = 486 Pa = 5 kpa d) pmax = (9.8 + 8).8 = 5 6 Pa = 5. kpa 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part. 7

Chapter / Fluid Statics.97 The air volume is the same before and after. z h.5 8 = hb/. tanα = =. 9. 8 b 9 8 4 = h. h. h =.856. Use dotted line. B.5w +.5.45 = 4. w =.74 m. a) p = ( 7.66) 9.8.5 = 5 74 Pa or 5.74 kpa b) pb = ( 7.66) = 76 6 Pa or 76.6 kpa c) pc =. ir fills the space to the dotted line. b h α w C x.98 Use Eq..5.: ssume an air-water surface as shown in the above figure. 8a a) 6 = ax ( 8) 9.8. 5 x 9. 8 4 = h 9. 8 8a 6 = 8 ax + 4.5 9.8 x or ax 4.45 =.74 a x. a x 9.8 a x. ax + 9.67 = ax =.64, 7.46 m/s b) 6 = ax ( 8) (9.8 + ). 5 + 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part. a x 8 8a x. 9. 8 8 6 = 8 ax + 49.5 9.8 or ax. =.574 a x. 9.8 a x 5. ax +.44 = ax =.5, 4.8 m/s c) 6 = ax ( 8) (9.8 + 5) (.5 + a x 8a x ). 4. 8 8 6 = 8 ax + 7. 4.8 or ax.875 =.6 a x. 4.8 a x 7.6 ax + 8.66 = ax =., 6.8 m/s.99 a) ax =.866 = 7. m/s, az = m/s. Use Eq..5. with the peep hole as position. The x-axis is horizontal passing thru. We have p = 7. (.) (9.8 + ) (.866) = 58 9 Pa b) p = 8.66 (.848) (9.8 + 5) (.799) = 57 46 Pa. a) The pressure on the end B (z is zero at B) is, using Eq..5., p(z) = ( 7.66) 9.8(z) = 76 6 98 z. 5 FB = ( 76 6 98z) 4dz = 64 N or 64 kn b) The pressure on the bottom BC is

Chapter / Fluid Statics p(x) = (x 7.66) = 76 6 x. 7. 66 FBC = ( 76 6 x) 4dx =.6 6 N or 6 kn c) On the top p(x) = (x 5.74) where position is on the top surface: 5. 74 Ftop = ( 5 74 x) 4dx = 5.5 5 N or 55 kn. a) The pressure at is 58.9 kpa. t B it is pb = 7. (.7.) (9.8) (.866) = 8495 Pa. Since the pressure varies linearl over B, we can use an average pressure times the area: 58 9 + 8495 F B =. 5 = N or. kn z x b) pd =. pc = 7. (.5.) 9.8(.866.866) = 49 8 Pa. F CD = 49 8. 5 = 74 7 N or 74.7 kn. 58. 9 + 49. 8 c) p = 58 9 Pa. pc = 49 8 Pa. F C =. 5 = 8.8 kn.. Use Eq..5. with position at the open end: a) p = since z = z. pb = 9.8.6 = 89 Pa=.9 kpa pc = 89 Pa =.9 kpa z b) p = ax = m/s (.9 ) = 9 Pa = -9 kpa pb = (.9) 9.8(.6) = 4 Pa = -. kpa pc = 9.8 (.6) = 5886 Pa = 5.89 kpa c) p = (.9) = 8 Pa = -8 kpa pb =.9 9.8(.6) = 6 Pa = -6. kpa pc =.9 kpa d) p =. pb = (9.8 8).65 = 59 Pa = -5. kpa. pc = 59 Pa = -5. kpa e) p = 8 (-.5.65) = -6.87 kpa pb = 8 (-.5.65) - 9.8 (-.65) = -.74 kpa pc = - 9.8 (-.65) = 6. kpa f) p = (-9) (-.5.65) = 8.44 kpa pb = (-9) (.5.65) - 8.8 (-.65) =. kpa pc = - 8.8 (-.65) =.76 kpa C B x 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part. 9

Chapter / Fluid Statics Rotating Containers. Use Eq..6.4 with position at the open end: 5 π ω = = 5.6 rad/s. 6 a) p = ( 5.6 /) (.6.5) =. kpa z ω p B = 5.6.9 + 98.6 = 7. kpa pc = 98.6 = 5.89 kpa C B r b) p = 5.6.6 = 4.9 kpa p B = 5.6.6 + 98.4 = 8.86 kpa pc = 98.4 =.9 kpa c) =. d) =..45 =.78 kpa = 5.6.45 +98.= 5.7 kpa =98.=.94 kpa.75 =.9 kpa = 5.6.75 +98.5= 4.8 kpa =98.5=.45 kpa.4 Use Eq..6.4 with position at the open end. a) p = (.9 ) = 4.5 kpa. z pb = 4 5 + 98.6 = 4.6 kpa. pc = 98.6 = 5.89 kpa. ω b) p = (.6 ) = 8. kpa. r C B pb = 8 + 98.4 = 4. kpa. pc = 98.4 =.9 kpa. c) =.45 =. kpa = Pa+98.= 7.8 kpa =98.=.94 kpa 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part. 4

Chapter / Fluid Statics d) =.75 = 7. kpa = Pa+98.5= 4.58 kpa =98.5=.45 kpa.5 Use Eq..6.4 with position at the open end and position at the origin. Given: p =. a) = ω (.45 ) 98 (.6). ω = 7.6 rad/s. b) = ω (. ) 98 (.4). ω = 9.4 rad/s. c) = ω (.5 ) 98 (.). ω =.8 rad/s. d) = ω (.875 ) 98 (.5). ω =.8 rad/s. r z ω.6 The air volume before and after is equal. πr h = π. 6.. r h =.44. a) Using Eq..6.5: r 5 / = 9.8 h h =.48 m b) p = 5.6 98 (.7) = 849 Pa. r 7 / = 9.8 h. h =.6 m. p = 7.6 + 98. = 78 Pa. c) For ω =, part of the bottom is bared. Using Eq..6.5: r π. 6. = πr h πr h. ω g = h, ω r g g g. 44 = h h ω ω. 44 h h =. 9. 8 = h. lso, h h =.8..6h.64 =.79. h =.859 m, r =.8 m. p = (.6.8 )/ = 7 4 Pa. or h h h z z r r r r 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part. 4

Chapter / Fluid Statics. 44 d) Following part (c): h h = 9. 8..6h.64 =.96. h =.5 m. p = (.6.65 )/ = 57 9 Pa r =.65 m.7 The answers to Problem.6 are increased b 5 Pa. a) 5 Pa b) 5 78 Pa c) 4 4 Pa d) 8 9 Pa.8 pr ( ) = ρω r ρg[ (.8 h)]. pr ( ) = 5ω r + 98(.8 h) if h <.8. p( r) = ω ( r r ) if h >.8. 5.6 = 667 N. a) F = pπrdr = π ( 5r + 65 rdr ) (We used h =.48 m).6 b) F = pπrdr = π (4 5r + 96 rdr ) = 7 N. (We used h =.6 m) dr d = πrdr c) d).6 π π (5 (.8 ) = 95 N. (We used r =.8 m).8 F = p rdr = r rdr.6 π π ( (.65 ) = 6 4 N. (We used r =.65 m).65 F = p rdr = r rdr 7 Cengage Learning. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part. 4