Chapter / Fluid Statics CHPTER Fluid Statics FE-type Eam Review Problems: Problems - to -9. (C). (D). (C).4 ().5 () The pressure can be calculated using: p = γ h were h is the height of mercury. p= γ h= (.6 980N/m ) (8.5 in 0.054 m/in) = 96 600 Pa Hg Since the pressure varies in a vertical direction, then: Hg p= p0 ρgh= 84 000Pa.00kg/m 9.8m/s 4000m = 44 760 Pa p = p + γ h γ h = 0 + 0 000 0. 980 0. = 800 Pa w atm m m water w This is the gage pressure since we used p atm =0. Initially, the pressure in the air is pir, = γ H = (.6 980) 0.6 = 50 Pa. fter the pressure is increased we have: p = 50 + 0 000 = 50 =.6 980 H. H = 0.085 m ir, The moment of force P with respect to the hinge, must balance the moment of 5 the hydrostatic force F with respect to the hinge, that is: ( ) P = F d 5 F = γ h= 9.8 kn/m m ( m )] F = 98. kn The location of F is at ( ) I. yp = y+ =.67 + =. m d =.. =. m y.67(. ). P = 98.. P =.7 kn The gate opens when the center of pressure is at the hinge:.6 (). + h I. + h b(. + h) / y = + 5. yp = y + = + = 5 +.. y (. + h) b(. + h) / This can be solved by trial-and-error, or we can simply substitute one of the answers into the equation and check to see if it is correct. This yields h =.08 m. 4 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter / Fluid Statics.7 (D).8 () The hydrostatic force will pass through the center, and so F H will be balanced by the force in the hinge and the force P will be equal to F. P= F = 9.8 4. w+ 9.8 ( π. / 4) w= 00. w = 5.6 m. The weight is balanced by the buoyancy force which is given by F where is the displaced volume of fluid: 900 9.8 = 980 0.0 5 w. w = 6 m = γ The pressure on the plug is due to the initial pressure plus the pressure due to the acceleration of the fluid, that is: pplug = pinitial + γ gasolineδ Z where,.9 () a Δ Z =Δ g 5 p plug = 0 000 + 6660 (. ) = 4 070 Pa 9.8 F = p = 4 070 π 0.0 = 0.5 N. plug plug Pressure..4 Since p = γ h, then h = p/γ a) h = 50 000/980 = 5.5 m c) h = 50 000/(.6 980) =.874 m This requires that p = p ( γh) = ( γh) water Hg water Hg b) 980 h = (.6 980) 0.75 h = 0. m..6 Δ p = γδz Δp = 0.004. (0,000) = 77 psf or 5.7 psi..8 From the given information the specific gravity is S =.0 + z/00 since S(0) = and S(0) =.. y definition ρ = 000 S, where ρ water = 000 kg/m. Using dp = γ dz then, by integration we write: p 0 z dp = 000( + z /00) gdz = 000g z + 00 0 0 0 p = 000 9. 8( 0 + ) = 0 000 Pa or 0 kpa 00 5 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter / Fluid Statics Note: we could have used an average S: S avg =.05, so that ρ avg = 050 kg/m and so p = γ h p = 050 9.8 0 = 0 005 Pa.0 / From Eq. (.4.8): atm[( 0 ) / 0] g α p = p T α z T R = 00 [(88 0.0065 00)/88] 9.8/0.0065 87 = 96.49 kpa ssuming constant density, then 00 p= p atm ρ gh= 00 9.8 00 /000 0.87 88 = 96.44 kpa 96.44 96.49 % error = 00 = 0.05% 96.49 Since the error is small, the density variation can be ignored over heights of 00 m or less. Eq..5. gives ρ gdh = d ρ or ρ dp = ρ ut, dp = ρgdh. Therefore, d ρ dρ ρ = T g dh Integrate, using ρ 0 =.00 slug/ft, and =,000 lb/in :..4 ρ h dρ g dh ρ =. 0. ft/s = h= 7. 0 ρ, 000 ( lb/in ) 44 in ft This gives ρ = 7 0.5 7. 0 h Now, h h g g 7 p = ρ gdh = dh ln(0.5 7. 0 h) 7 7 0.5 7. 0 h = 7. 0 0 0 If we assume ρ = const: p = ρ gh =.0. h = 64.4h b) For h = 5000 ft: p accurate =,00 psf and p estimate =,000 psf., 000, 00 % error = 00 = 0.7 %, 00 Use Eq..4.8: p = 0(88 0.0065 / 88) 7 h 9.8 z 0.0065 87 6 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter / Fluid Statics a) for z = 000 p = 69.9 kpa. c) for z = 9000 p = 0.6 kpa. Manometers.6.8.0..4.6.8 Referring to Fig..7b, the pressure in the pipe is: p = γh = (.6 980) 0.5 = 50 Pa or.5 kpa. Referring to Fig..7a, the pressure in the pipe is p = ρgh. If p = 400 Pa, then 400 = ρgh = ρ 9.8h or 400 ρ = 9.8h 400 a) ρ = = 680 kg/m The fluid is gasoline 9.8 0.6 400 c) ρ = = 999 kg/m The fluid is water 9.8 0.45 See Fig..7b: The pressure in the pipe is given by p = γ h + γ H p = 0.86 6.4 5 95. +.6 6.4 p p = γ 0. + γ H γ 0. water oil oil Hg water = 649.5 psf or 4.5 psi 40 000 6 000 = 90 9.8 0.+ 600 9.8 H 000 9.8 0. Solving for H we get: H = 0.74 m or 7.4 cm ( ) p = γ 0.6 γ (0.0) γ (0.04) γ (0.0) water Hg water Hg water Using γ water p water = 5.6 kpa = 9.8 kn/m and γ =.6 9.8 kn/m Hg p water 9.8 0. 0.68 9.8 0. + 0.86 9.8 0. = p oil With p water = 5 kpa, p oil = 4 kpa pgage = pair + γ water 4 where, pair = patm γ Hg H p = γ H + γ 4 gage Hg water p gage =.6 9.8 0.6 + 980 4 = 7.89 kpa Note: we subtracted atmospheric pressure since we need the gage pressure. 7 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter / Fluid Statics.40 p + 980 0.05 +.59 980 0.07 0.8 980 0. =.6 980 0.05 p = 587 Pa or 5.87 kpa. The distance the mercury drops on the left equals the distance along the tube that the mercury rises on the right. This is shown in the sketch. Oil (S = 0.87) Water Δh 0 cm 9 cm 7 cm Δh 40 o Mercury.4.44 From the previous problem we have: ( p) p γwater γhg γoil = + 0.07 0.09sin 40 0. =. kpa () For the new condition: ( ) = + γ ( +Δ ) γ γ ( Δ ) p p h h () water 0.07 HG 0.sin 40 oil 0. sin 40 where Δh in this case is calculated from the new manometer reading as: Δ h+δ h/ sin 40 = 9 cm Δ h= 0.78 cm Subtracting Eq.() from Eq.() yields: ( p ) ( p ) = γ ( Δh) γ γ ( Δh ) water HG oil 0.0sin 40 sin 40 Substituting the value of Δh gives: ( ) ( ) ( ) p =.+ 0.0078.6 0.0sin 40 0.87 0.0078sin 40 9.8 = 0.5 kpa a) Using Eq. (.4.6): ( ) ( ) p = γ z z + γh+ γ γ H where h= z5 z=7 6 = cm 4000 = 9800(0.6 0.) + 5 600(0.0) + ( 400 5 600)H H = 0.076 m or.76 cm 8 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter / Fluid Statics.46 From No..0: p oil = 4.0 kpa From No..6: p oil = p water 9.8 (0.+Δz) 0.68 9.8 (0. Δz) + 0.86 9.8 (0. Δz) Δz = 0.045 m or 4.5 cm. Forces on Plane reas.48.50 The hydrostatic force is calculated using: F = γ h where, h = 0 m, and = πr = π ( 0.5 m) hence, F π ( ) = 980 0 0.5 = 694 N. For saturated ground, the force on the bottom tending to lift the vault is: F = p c = 9800.5 ( ) = 9 400 N The weight of the vault is approimately: W = ρ g walls ( ) ( ) ( ) W = 400 9.8.5 0. + 0. + 0 0.8. 0. = 8 400 N. The vault will tend to rise out of the ground..5 b) Since the triangle is horizontal the force is due to the uniform pressure at a depth of 0 m. That is, F = p, where p= γ h= 9.8 0 = 98. kn/m The area of the triangle is = bh =.88 / =.88 m.54 F = 98..88 = 77.4 kn. a) F = γh= 9.8 6 π = 79.7 kn 4 I π /4 yp = y+ = 6 + = 6.67 m (, y) p = (0, 0.67) m y 4π 6 c) F = 9.8 (4 + 4/) 6 =.9 kn y p = 5 + 4 /. 6 = 5.50 m y =.5 5. 6 4/.5 =.5. = 0.975 (, y) p = (0.975,.5) m y.56 F = γh = 980 6 0 =.777 0 6 N, or 77 kn 9 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter / Fluid Statics y = I p y + y = + 4 5 / 75. = 7.778 m 75. 0 Σ = 0 (0 7.778) 77 = 5 P P = 5 kn M Hinge The vertical height of water is h =. 0.4 =.4 m The area of the gate can be split into two areas: = + or =..4 + 0.4.4 =.80 m Use forces: F = γ h = 980 0.5657 (..4) = 754 N.60.4 F = γ h = 980 (0.4.4) = 674 N The location of F is at y p = (.4) = 0. 754 m, and F is at y p I.4 0.4.4 / 6 = y+ = + = 0.5657 m y 0.4 (.4 / ) (.4 / ).4 ΣM hinge = 0 : 754 + 674 (.4 0.5657).4P = 0 P = 46 N The gate is about to open when the center of pressure is at the hinge..6 b) b / yp =. + H = (.0/ + H) + ( + H)b H = 0.6667 m free-body-diagram of the gate and block is sketched. T Sum forces on the block: Σ F = 0 W = T + F y F stop 0 T.64 where F is the buoyancy force which is given by F = γ πr ( H) F H y p F W Take moments about the hinge: R T.5 = F ( y ) H p Ry where F H is the hydrostatic force acting on the gate. It is, using 0 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter / Fluid Statics h = = = FH.5 m and 6 m, ( )( ) = γ h= 9.8 kn/m.5 m 6 m = 88.9 kn From the given information, y p ( ) I / = y+ =.5 + = m y.5 6 ( ) 88.9 T = = 5. kn.5 F = W T = 70 5. = 44.77 kn. γπ R H = 44.77 44.77 kn H = m =.55 m ( 9.8 kn/m ) π ( m) ( ).66 The dam will topple if there is a net clockwise moment about O. The weight of the dam consists of the weight of the rectangular area + a triangular area, that is: W = W+ W. The force F acting on the bottom of the dam can be divided into two forces: Fp due to the uniform pressure distribution and Fp due to the linear pressure distribution. W =.4 6.4 6 6 = 56, 609 lb assume m deep b) W =.4 6.4 6 4 / =, 9 lb W = 6.4 (60.86/) = 4, 794 lb F = 6.4 0 60 =,0 lb F = 6.4 5 0 =,0 lb F p = 6.4 0 0 = 8,70 lb W F W F O F F p = 6.4 50 0 / = 46,800 lb Σ M O : (,0)(0) + (8,70)(5) + (46,800)(0) (56,609)(),0(0/) (,9)(4) 4,794(.8) = 740,78 > 0. will tip. 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter / Fluid Statics Forces on Curved Surfaces.68.70 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, they produce no moment about that point. Hence, consider only horizontal forces: ( γ h) ( γ h) ( F ) = = 9.8 (4 0) = 784.8 kn H water water ( F ) = = 0.86 9.8 0 = 68.7 kn H oil oil ΣM: P = 7848. 687.. P = 66. kn 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: FH = F F = W + F where FH and F are the horizontal and vertical components of the force acting on the water by the surface. Hence, FH = F = 9.8 kn/m 8 + 4 = 706. kn ( )( )( ) The line of action of F H is the same as that of F. Its distance from the surface is y p ( ) I 4 = y+ = 9 + = 9.07 m y 9 8 F. W F F F H. To find F we find W and F : π W = γ = ( 9.8 kn/m ) ( ) 4 =.7 kn 4 F ( ) = 9.8 kn/m 8 4 = 68 kn F = F + W =.7 + 68 = 66 kn To find the line of action of F, we take moments at point : F = F d + W d where R d = m, and d = = =.55 m: 4 4 ( π) ( π) 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter / Fluid Statics F d+ W d 68 +.7.55 = = =.08 m F 66 Finally, the forces F H and F that act on the surface are equal and opposite to those calculated above. So, on the surface, F H acts to the right and F acts downward..7 Place the resultant FH + F at the center. F passes through the hinge. The moment of F H must equal the moment of P with respect to the hinge: (9.8 0) =.8 P. P = 70. kn. The resultant F + F H of the unknown liquid acts thru the center of the circular arc. F passes through the hinge. Thus, we use only F ). ssume m wide: ( )( ) F = γ h= γ R R = γ R H ( H.74 F = γ h= γ R R = γ R The horizontal force due to the water is ( )( ) The weight of the gate is W Sγw γw( πr ) w w w w = = 0. 4 Summing moments about the hinge: ( ) ( 4 π ) F R + W R = F R w H R R R R R a) π 4 980 + 0. 980 = γ R 4 π γ = 4580 N/m.76 The pressure in the dome is: a) p = 60 000 980 0.8 980 = 4 870 Pa or 4.87 kpa The force is F = p projected = (π ) 4.87 = 40.4 kn b) From a free-body diagram of the dome filled with oil: W F weld + W = p p Using the pressure from part (a): F weld F weld = 4 870 π (0.8 980) 4 π = 400 or.4 kn 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter / Fluid Statics uoyancy.78 Under static conditions the weight of the barge + load = weight of displaced water. (a) 0 000 + 50 000 = 980 (6 d + d /). d + d 8.5 = 0 d =.7 m.80.8 The weight of the cars will be balanced by the weight of displaced water: 000 60 = 6.4(5 00 Δd ) Δd = 0.846 ft or 4.6 in T + F = W (See Fig.. c.) T = 40 000.59 980 = 8804 N or 8.804 kn t the limit of lifting: F = W+p where p is the pressure acting on the plug. (b) ssume h> 5 + Rand use the above equation with W F.86 R =. ft and h = 6.4 ft: ( ) F = γ = γ 0 πr = 6.4 0.977 = 858 lb w w segment ( ) W + p= 500 lb + 6.4 6.4 π 4 = 857 lb Hence, the plug will lift for h >6.4 ft. θ p R h 5 (a) When the hydrometer is completely submerged in water:.88 W = γ w π 0.05 π 0.005 (0.0 + m Hg)9.8 = 980 0.5 + 0. 4 4 m Hg = 0.0886 kg When the hydrometer without the stem is submerged in a fluid: W = γ S =.089 ( 0.05) π (0.0+ 0.089)9.8 = S 980 0.5 4 4 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter / Fluid Statics Stability.90 With ends horizontal ( γ π ) 4 Io = π d 64 The displaced volume is = W γ water = d h/ 4 / 980 = 8.0 0 γ d since h = d. 5 The depth the cylinder will sink is depth = 5 8.0 0 γ d = = 0.0 0 π d /4 5 γ d The distance CG is CG h 5 = 0. 0 γ d /. Then GM = I O 4 πd 64 CG = 5 8.0 0 γ d / d 5 + 0. 0 γ d / > 0 This gives (divide by d and multiply by γ ): 6.8 0.5 γ + 5. 0 5 γ > 0. Consequently, γ > 868 N/m or γ < 46 N/m.9.94 6 9 + 6 4 s shown, y = = 6.5 cm above the bottom edge. 6 + 6 4γ 9.5 + 6γ 8.5 + 6Sγ 4 G = = 6.5 cm. 0.5γ 8 + γ 8 + S γ 6 0 + 04 S = 74 + 64 S. S =. The centroid C is.5 m below the water surface. CG =.5 m Using Eq..4.47: 8 / GM =.5 =.777.5 = 0.77 0 8 > The barge is stable. Linearly ccelerating Containers.96 (a) p ma = 000 0 (0 4) 000 (9.8) (0 ) = 99 60 Pa (c) p ma =.94 60 (0 ).94 (. + 60) (0 6) = 470 psf or 7.5 psi.98 Use Eq..5.: b) 60 000 = 000 a ( 8) 000 (9.8 + 0).5 + 8a 9.8 5 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter / Fluid Statics 8a 60 = 8 a + 49.5 9.8 or a. =.574 9.8 a 5. a +.44 = 0 a = 0.5, 4.8 m/s a.00 a) The pressure on the end (z is zero at ) is, using Eq..5., p(z) = 000 0 ( 7.66) 000 9.8(z) = 76 60 980 z 5. F = ( 76 60 980z) 4dz = 640 000 N or 640 kn 0 b) The pressure on the bottom C is p() = 000 0 ( 7.66) = 76 60 0 000. 7. 66 FC = ( 76 60 0 000) 4d =.6 0 6 N or 6 kn 0 Use Eq..5. with position at the open end: b) p = 000 0 (0.9 0) = 9000 Pa. z p = 000 0 (0.9) 000 9.8( 0.6) = 4 Pa.0 p C = 000 9.8 ( 0.6) = 5886 Pa. e) p =.94 60 7. 5 = 64 psf. C p =.94 60 7. 5.94. 5 = 4 psf. p C =.94. 5 = 0 psf. Rotating Containers Use Eq..6.4 with position at the open end: z a) p = 000 0 (0 0.9 ) = 40 500 Pa.04 p = 40 500 + 980 0.6 = 4 600 Pa p C = 980 0.6 = 5886 Pa r C ω c) p =.94 0 0 7. 5 44 = 947 psf 6 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter / Fluid Statics p = 947 + 6.4 5 = 87 psf p C = 6.4 5 = 0 psf The air volume before and after is equal. πrh 0 = π. 6.. rh 0 = 0.44. h z r 0 (a) Using Eq..6.5: r 0 5 / = 9.8 h h = 0.48 m p = 000 5 0.6 980 ( 0.7) r.06 = 849 Pa (c) For ω = 0, part of the bottom is bared. π Using Eq..6.5: ω 0 g π π 0.6 0. = r0 h r h r ω r = h, = h g g g 0. 44 = h h ω ω h h 0.44 0 = 9.8 or h lso, h h = 0.8..6h 0.64 = 0.79. h = 0.859 m, r = 0.08 m p = 000 0 (0.6 0.08 ) = 7,400 Pa h z r 0 r pr () = [0 (0.8 )] ρω r ρg h d = π r d r pr ( ) = 500ω r + 980(0.8 h) if h < 0.8 d r.08 pr () = ω ( r r ) if h > 0.8 500 a) 0.6 π π ( 500 650 ) = 6670 N 0 F = p rdr = r + r dr (We used h = 0.48 m) 7 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter / Fluid Statics c) 0.6 π π (50 000( 0.08 ) = 950 N 0.08 F = p rdr = r r dr (We used r = 0.08 m) 8 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.