Solution Manual For Munson, Young and Okiishi's Fundamentals of Fluid Mechanics, 8th Edition

Problem 2.2 The deepest known spot in the oceans is the Challenger Deep in the Mariana Trench of the Pacific Ocean and is approximately 11,000 m below the surface. Assume that the salt-water density is constant at 1025kg/m 3 and determine the pressure at this depth. Solution 2.2 GIVEN: Density of the fluid is 1025 kg m3 , and the depth of the Challenger Deep is 11000 m . FIND: Pressure at the depth of 11000 m . SOLUTION:    1 N  s 2   1 Pa   1025kg   9.807m  p   gh     11000m     N   m3  s 2    1 kg  m   1  m2   1.11  108 Pa p  111 MPa gage Problem 2.3 A closed tank is partially filled with glycerin. If the air pressure in the tank is 6 lb / in.2 and the depth of glycerin is 10ft , what is the pressure in lb / ft 2 at the bottom of the tank? Solution 2.3 2 lb    lb   144in.  p   h  p0   78.6 3  10ft    6 2    ft    in.   1 ft 2  p  1,650 lb ft 2 Problem 2.4 A 3-m- diameter vertical cylindrical tank is filled with water to a depth of 11m . The rest of the tank is filled with air at atmospheric pressure. What is the absolute pressure at the bottom of the tank? Solution 2.4 Known:water filled tank, dia.  3m , depth  11m Determine:absolute pressure at tank bottom Strategy: insert information into hydrostatic pressure distribution Solution: pbottom  patmos   water hbottom kN    101 kPa   9.80 3  11 m   208.80 kPa m   pbottom  209 kPa Problem 2.5 Blood pressure is usually given as a ratio of the maximum pressure (systolic pressure) to the minimum pressure (diastolic pressure). Such pressures are commonly measured with a mercury manometer. A typical value for this ratio for a human would be 120/70, where the pressures are in mm Hg. (a) What would these pressures be in pascals? (b) If your car tire was inflated to 120 mm Hg, would it be sufficient for normal driving? Solution 2.5 p h (a) N   For 120 mm Hg : p   133  103 3   0.120 m   m   p  16.0 kPa N   p   133  103 3   0.070 m   m   p  9.31 kPa For 70 mm Hg : 2  4 lb/in.  3 N  (b) For 120 mm Hg : p  16.0 10 2  1.450 10   2.32 psi m   N/m 2   Typical tire pressure is 30  35psi , therefore. 120 mm is insufficent inflation for normal driving Problem 2.6 An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown liquid is 1.5 m and the depth of the oil specific weight  8.5 kN / m3  floating on top is 5.0 m . A pressure gage connected to the bottom of the tank reads 65 kPa . What is the specific gravity of the unknown liquid? Solution 2.6 pbottom    oil   5m     u  1.5m  where  u u  pbotton    oil   5 m  1.5 m 65  103  N m3 SG     H 2 O @ 4 C 9.81  103 N m3 u unknown liquid  N  N   8.5  103 3   5 m  2  N m m    15  103 3 1.5 m m 15  103  SG  1.53 Problem 2.7 A 30-ft- high downspout of a house is clogged at the bottom. Find the pressure at the bottom if the downspout is filled with 60 F rainwater. Solution 2.7 P   gh Inserting the density f water and the specified column height: lbm  ft    62.4 3  32.2 2   30ft  ft  s  P   ft  lbm  in 2    32.2   144 2  lb  s 2   ft   P  13.0 psig Problem 2.8 How high a column of SAE 30 oil would be required to give the same pressure as 700 mm Hg? Solution 2.8 p h For pHg  poil  Hg hHg   oil hoil or kN    133 3  m  0.700m  hoil  hHg     kN   oil  8.95   m3    Hg hoil  10.4 m Problem 2.9 Bathyscaphes are capable of submerging to great depths in the ocean. What is the pressure at a depth of 5 km, assuming that seawater has a constant specific weight of 10.1 kN / m3 ? Express your answer in pascals and psi. Solution 2.9 p   h  p0 At the surface p0  0 so that   N  N  p   10.1  103 3  5  103 m  50.5  106 2 m  m  lb   1.450  104 2   N   in.  p   50.5  106 2   N m    1 2   m    p  50.5MPa  p  7320 psi Problem 2.10 The deepest known spot in the oceans is the Challenger Deep in the Mariana Trench of the Pacific Ocean and is approximately 11,000 m below the surface. For a surface density of 1030 kg/m 3 , a constant water temperature, and an isothermal bulk modulus of elasticity of 2.3  109 N/m 2 , find the pressure at this depth. Solution 2.10 GIVEN: Ocean depth of 11,000 m, surface density of 1030 kg/m 3 , constant water temperature, and isothermal bulk modulus of elasticity EV ,T  2.3 x 109 N/m 2 . FIND: Pressure at this depth SOLUTION: dp dp Ev    d /  d  /    p  p0  p      p  p0 dp  Ev  e      ln     0    Ev Ev  0  0 p0 d where: z is positive upward, z=z 0 at the surface, and p( z  0)  p0 ,  ( z  0)  0 Substitution into the hydrostatic pressure equation yields:  p  p0    dp E      g   0 ge v  dz     p  p0   p  p  p0  E   E  e  v  dp   Ev e  v   p0     p   p  p0   z   Ev     Ev  e  1   0 g  dz   0 gz   0  p0    p  p0     gz E e  v   1 0 Ev  p  p0   0 gz     ln  1   Ev   Ev       1030 kg   9.807 m   11000 m     gz  N     m3   s2  p  p0  Ev ln 1  0   0   2.3 109 2  ln  1    Ev  m      9 N   1 kg  m    2.3  10 2   2     m 1 N s        p  1.14  108 N m2  p  114 MPa gage Problem 2.11 A submarine submerges by admitting seawater S= 1.03 into its ballast tanks. The amount of water admitted is controlled by air pressure, because seawater will cease to flow into the tank when the internal pressure (at the hull penetration) is equal to the hydrostatic pressure at the depth of the submarine. Consider a ballast tank, which can be modeled as a vertical half-cylinder ( R  8 ft , L  20ft ) for which the air pressure control valve has failed shut. The failure occurred at the beginning of a dive from 60 ft to 1000 ft . The tank was initially filled with seawater to a depth of 2 ft and the air was at a temperature of 40 F . As the weight of water in the tank is important in maintaining the boat’s attitude, determine the weight of water in the tank as a function of depth during the dive. You may assume that tank internal pressure is always in equilibrium with the ocean’s hydrostatic pressure and that the inlet pipe to the tank is at the bottom of the tank and penetrates the hull at the “depth” of the submarine. Solution 2.11 GIVEN: Ballast tank, L  20ft , R  8ft , initial condition of d  60ft , h  2ft ., Tair  40 F , Final condition d= 1,000 ft. FIND: Weight of water in ballast tank as a function of depth d during dive. The water volume in the ballast tank is determined using given information about the ballast tank,  L  2 1  2 Rh  h 2     R  h  2 Rh  h 2  R sin    R 2       2   20ft  2 1  2  8  2  2 ft  2           8ft sin 8 2 ft 2 8 2 2 ft         2  8ft      Vsw i   145 ft 3 The initial air volume is Vair i   R2L 2  V wi   2  8 ft 2  20 ft  145 ft 3  1866 ft 3 During the dive the ballast tank air pressure is assumed to be in equilibrium with the ocean hydrostatic pressure. Then Pair  Patm   sw (d  h) Using the ideal gas law, M air RTair M RT  air air  Patm   sw ( h  d ) Vair Vtank  Vsw Solving for h , hd Patm  sw  M air RTair  sw Vtank  Vsw  (2) Vsw is a function of h , given by equation (1) for h  R . For h  R , define b  2 R  h and Vsw   R2 L 2   L  2 1  2 Rb  b 2     R  b  2 Rb  b 2  R sin    2 R      The air mass M air is  lbm   M air  air iVair i   0.219 3  1866ft 3 ft    408.7 lbm  The initial weight of the water is  lb  Wsw i   swVsw i   62.4 1.03 3  145ft 3  9320lb .  ft     Pseudocode for procedural language: hmin  0 hmax  2 R h new   hmin  hmax  2 ptank  pair   sw h pocean  patm   sw d ptank  pocean  hmin  h new TRUE ptank  pocean  hmax  h new  h  h  If  max min  errtol  hmin   h  hmin  hmax  2 FALSE (3) Problem 2.12 Determine the pressure at the bottom of an open 5-m- deep tank in which a chemical process is taking place that causes the density of the liquid in the tank to vary as  h  ,  hbot 2    surf 1  sin 2  where h is the distance from the free surface and surf  1700 kg/m3 . Solution 2.12 GIVEN: H bot  5m , surf  1700  h  ;   surf 1  sin 2  , m  hbot 2  kg 3 FIND: Pressure at bottom of tank. SOLUTION: The pressure gradient is dP    dh Separating variables, substituting for  , and integrating give P h Psurf 0 2 h   dP  surf g  1  sin  hbot 2 dh This integral must be solved numerically. The Romberg method gives for h  hbot Pbot  Psurf  surf g  6.080 m  Setting Psurf  0 N  m   N  s2   Pbot  1700 3  9.807 2     6.080 m    m  s   kg  m  Pbot  101103 N m2 Problem 2.13 In a certain liquid at rest, measurements of the specific weight at various depths show the following variation: The depth h  0 corresponds to a free surface at atmospheric pressure. Determine, through dp   , the corresponding variation in pressure and show the results numerical integration of dz on a plot of pressure (in psf) versus depth (in feet). Solution 2.13 dp   dz Let z  h0  h (see figure) so that dz   dh and therefore dp   dz   dh Thus, pi hi 0 dp  0  dh or hi pi    dh 0 (1) where pi is the pressure at depth hi . Equation (1) can be integrated numerically using the trapezoidal rule, i.e., I 1 n1   yi  yi1  xi1  xi  2 i 1 where y  , x h , and n = number of data points. The tabulated results are given below, along with the corresponding plot of pressure vs. depth. h (ft) 0 10 20 30 40 50 60 70 80 90 100 ɣ (lb/ft^3) 70 76 84 91 97 102 107 110 112 114 115 Pressure, psf 0 730 1530 2405 3345 4340 5385 6470 7580 8710 9855 12000 Pressure, p(psf) 10000 8000 6000 4000 2000 0 0 20 40 Depth, h (ft) 60 80 100 Problem 2.15 Under normal conditions the temperature of the atmosphere decreases with increasing elevation. In some situations, however, a temperature inversion may exist so that the air temperature increases with elevation. A series of temperature probes on a mountain give the elevation– temperature data shown in the table below. If the barometric pressure at the base of the mountain is 12.1 psia, determine by means of numerical integration the pressure at the top of the mountain. Solution 2.15 p2 dp p g z2 dz  p1 p  ln p12   R z1 T In the table below the temperature in  R is given and the integrand Elevation, ft 5000 5500 6000 6400 7100 7400 8200 8600 9200 9900 T, ˚F 50.1 55.2 60.3 62.6 67.0 68.4 70.0 69.5 68.0 67.1 T, ˚R 509.8 514.9 520.0 522.3 526.7 528.1 529.7 529.2 527.7 526.8 1/T(˚R) 0.001962 0.001942 0.001923 0.001915 0.001899 0.001894 0.001888 0.00189 0.001895 0.001898 p dp The approximate value of the integral in  2 p1 p  ln p2 g z2 dz is 9.34 obtained using the   p1 R z1 T trapezoidal rule, i.e., I 1 n1   yi  yi1  xi1  xi  where y 2 i 1 1 , x T elevation , and n  number of data points. Thus, 9900ft  1  1 tabulated. T  R  ft 5000ft  T dz  9.34 °R so that (with g  32.2ft/s 2 and R  1716ft  lb/slug  °R ) ft   ft    32.2 2   9.34 °R  p   0.1753 s  ln p2    1 1716 ft  lb/slug  °R (1) It follows from Eq.(1) with p1  12.1psia that p2  (12.1psia)e0.1753  10.2 psia (Note: Since the temperature variation is not very large, it would be expected that the assumption of a constant temperature would give good results. If the temperature is assumed to be constant at the base temperature ( 50.1F ), p2  10.1psia , which is only slightly different from the result given above.) Problem 2.16 Often young children drink milk (   1030 kg / m 3 ) through a straw. Determine the maximum length of a vertical straw that a child can use to empty a milk container, assuming that the child can develop 75 mm Hg of suction, and use this answer to determine if you think this is a reasonable estimate of the suction that a child can develop. Solution 2.16 Known:  milk  1030 kg m3 , suction  75 mm Hg Determine: maximum length of vertical straw, is this reasonable? Strategy: compute height of equivalent milk column Solution: hmax  height of milk column lifted by suction pmax   Hg hHg  75 mm Hg pmax   milk hmilk  kg     m3   13600 m3   Hg hHg   Hg    hmax,milk    75 mm   hHg    milk  milk  1030 kg    milk  pmax hmax,milk  990.3 mm milk max. length straw  1 m Although this may seem large, adults can routinely lift water much higher through a straw. Therefore, a 1 m draw seems large, but within reason for a child. Problem 2.17 (a) Determine the change in hydrostatic pressure in a giraffe’s head as it lowers its head from eating leaves 6 m above the ground to getting a drink of water at ground level as shown in the figure below. Assume the specific gravity of blood is SG = 1. (b) Compare the pressure change calculated in part (a) to the normal 120 mm of mercury pressure in a human’s heart. Solution 2.17 (a) For hydrostatic pressure change, kN  kN  p   h   9.80 3   6m   58.8 2 m  m   p  58.8 kPa (b) To compare with pressure in human heart convert pressure in part (a) to mm Hg: 58.8 kN     Hg hHg  133 3  hHg  m m  kN 2 hHg   0.442m  (103 mm )  442mm Hg m giraffe hHg  442 mmHg human hHg  120 mmHg girrafe more than 3.5 times greater Problem 2.18 What would be the barometric pressure reading, in mm Hg, at an elevation of 4 km in the U.S. standard atmosphere? Refer to Table C.2 Properties of the U.S. Standard Atmosphere (SI Units). Solution 2.18 At an elevation of 4 km , N p  6.166  104 2 (from the table given in the Problem). Since m p h N p m 2  0.464 m  h  N  133  103 3 m 6.166  104 h  464 mm Problem 2.19 Denver, Colorado, is called the “mile-high city” because its state capitol stands on land 1 mi above sea level. Assuming that the Standard Atmosphere exists, what is the pressure and temperature of the air in Denver? The temperature follows the lapse rate ( T  T0  Bz ). Solution 2.19 GIVEN: Denver altitude = 1mile  5280 ft and standard atmosphere. T  T0  Bz FIND: Temperature and pressure in Denver. SOLUTION: The Lapse rate gives: °R   T  T0  Bz  518.67 °R   0.003566   5280 ft  ft   T  500 °R  40 °F Using Equation and Table: g   z B P  P0 1   T0   ft  lbs 2     32.2 2  32.2 s  ft lbm     53.35 ft lb  0.003566R      lbm°R  ft  °R      0.003566 ft   5280ft     14.696 psia   1   518.67°R       p  12.10 psia Note: In reasonably good agreement with table in appendix of text. Problem 2.20 Assume that a person skiing high in the mountains at an altitude of 15,000 ft takes in the same volume of air with each breath as she does while walking at sea level. Determine the ratio of the mass of oxygen inhaled for each breath at this high altitude compared to that at sea level. Solution 2.20 Let  0 denote sea level and  15 denote 15,000 ft altitude. Thus, since m  mass  V , where V  volume , m0  0 V0 m15  15 V15 , where V0  V15 m15 15 V15 15   m0 0 V0 0 If it is assumed that the air composition (e.g., % of air that is oxygen) is the same at sea level as it is at 15,000 ft , use the density values from table of Properties of the U.S. Standard Atmosphere (BG/EE Units)  0  2.377  103 slugs ft 3 and 15  1.496  103 slugs 1.496  103 3 m15 ft  0.629   m0 2.377  103 slugs ft 3 slugs ft 3 so that m15  62.9% of m0 Problem 2.21 Pikes Peak near Denver, Colorado, has an elevation of 14,110ft . (a) Determine the pressure at this elevation, based on the equation below. (b) If the air is assumed to have a constant specific weight of 0.07647 lb/ft 3 , what would the pressure be at this altitude? (c) If the air is assumed to have a constant temperature of 59 F , what would the pressure be at this elevation? For all three cases assume standard atmospheric conditions at sea level as provided in the table of Properties of U.S. Standard Atmosphere at Sea Level). Solution 2.21 (a) g   z  R p  pa  1   Ta   ft 32.174 2 g s   5.252  R  ft  lb   °R  1716   0.00357  slug  °R   ft     R   0.00357  14110ft     lb  ft    p   2116.2 2  1   518.67°R ft       (b) (c) p  pa   h  2116.2 p  pa e   p  1240 lb     0.07647 3  14110ft   ft ft   p  1040 lb 2 lb ft 2 lb ft 2 (abs) (abs) gh RTa    ft    32.174 2 14110ft   s       ft lb  518.67°R     1716  lb   slug°R   p   2116.2 2  e    5.252 ft   p  1270 lb ft 2 (abs) Problem 2.22 g   z  R Equation p  pa  1   provides the relationship between pressure and elevation in the  Ta  atmosphere for those regions in which the temperature varies linearly with elevation. Derive this equation and verify the value of the pressure given in the table of Properties of the U.S. Standard Atmosphere (SI Units) for an elevation of 5km . Solution 2.22 p2 dp g z2 dz  p p   R z T 1 Let p1 p dp 1 pa for z1  0 , p2 g z p for z2  z , and T  Ta   z . dz  p p   R 0 Ta   z a z   z  p g 1 g g  ln Ta   z   ln Ta       ln Ta   z    ln ln  1   pa R  R  Ta   0 R g   z  R p  pa  1   Ta   For z  5km : K J m , pa  101.33kPa , Ta  288.15 K , g  9.807 2 ,   0.00650 , R  287 kg  K m s m 9.807 2 s J  K K   3 0.0065   0.0065  5  10 m   287 kgK  m    m   p  101.33kPa  1       288.15K      pcomputed  5.40  104 ptabulated  5.405  10 N m2 4 N m2 Problem 2.23 As shown in the figure below for the U.S. standard atmosphere, the troposphere extends to an altitude of 11km where the pressure is 22.6 kPa  abs  . In the next layer, called the stratosphere, the temperature remains constant at 56.5 C . Determine the pressure and density in this layer at an altitude of 15km . Assume g  9.77 m / s 2 in your calculations. Compare your results with those given in Table C.2 Properties of the U.S. Standard Atmosphere (SI Units). Solution 2.23 For isothermal conditions, p2  p1e Let  g  z2  z1  RT0 z1  11km , p1  22.6 kPa , R  287 J m , g  9.77 2 , kg  K s T0  56.5 C  273.15  216.65 K . p2   22.6kPa  e   m 3 3   9.77 2 1510 m-1110 m   s       J     287  216.65K  kgK      12.1 kPa Using the ideal gas model: N 12.1  103 2 p kg m 2    0.195 3 RT  J  m  287   216.65K  kg  K   In comparison to published values: pcomputed  12.1 kPa,   0.195 kg m3 kg ptabulatted  12.11 kPa,   0.1948 3 m Problem 2.24 The record low sea-level barometric pressure ever recorded is 25.8in. of mercury. At what altitude in the standard atmosphere is the pressure equal to this value? Solution 2.24 For record low pressure,   lb   25.8in.   ft 2  lb  p   Hg hHg   847 3      12.6 2  2 ft   12 in.   144in.  in.  ft   From Table C.1 Properties of the U.S. Standard Atmosphere (BG/EE Units) @ 0ft altitude p  14.696 lb in.2 @ 5000ft altitude p  12.228 lb in.2 Assume linear variation change in pressure per foot. 14.696 Thus, pressure change per foot  lb lb lb  12.228 2 2 in. in.  4.936 104 in.2 5000ft ft lb    2 lb lb 14.696 2  a (ft)  4.936  104 in.   12.6 2 ft  in. in.    a 14.696  12.6 4.936  104 ft  a  4250 ft Problem 2.25 On a given day, a barometer at the base of the Washington Monument reads 29.97 in. of mercury. What would the barometer reading be when you carry it up to the observation deck 500 ft above the base of the monument? Solution 2.25 Let  b and  od correspond to the base and observation deck, respectively. Thus, with H = height of the monument, pb  pod   air H  7.65  102 p   Hg h , where  Hg  847 lb ft lb ft 3 3 lb  500 ft   38.5 2 ft and h = barometer reading. lb  29.97  ft    Hg hod  38.5 2 ft  12   Hg  lb 38.5 2 29.97   ft  12 in.  (29.97  0.545)in.  hod   ft   ft  12  847 lb 3 ft hod  29.43in. Problem 2.26 Aneroid barometers can be used to measure changes in altitude. If a barometer reads 30.1 in. Hg at one elevation, what has been the change in altitude in meters when the barometer reading is 28.3 in. Hg? Assume a standard atmosphere and that the equation below is applicable over the range of altitudes of interest. Solution 2.26 g   z  R p  pa 1    Ta  g   z  R z1 : p1  pa 1    Ta  R  p g z   1  1 1 Ta  pa  R T  p g  z1   a  1     pa  Ta R T p g z2   a  2     pa  Ta R R   Ta  p1  g Ta  p2  g  z2  z1          pa    pa      For Ta  288K ,   0.00650 K J m , Pa  101kPa , g  9.81 2 , R  287 m kg  K s  J  K  287 kg  K   0.00650 m  R      0.190 m g 9.81 2 s N  m   p1   Hg h1   133  103 3   30.1in.  2.540  102   102 kPa in.    m  N  m   p2   Hg h2  133  103 3   28.3in.  2.540  102   95.6 kPa in.  m    Substitution yields:  102 kPa 0.190  95.6kPa 0.190      z 2  z 1  K  101kPa  101kPa    0.00650   m 288K z 2  z 1  543 m Problem 2.27 Bourdon gages (see the figure below) are commonly used to measure pressure. When such a gage is attached to the closed water tank of figure below the gage reads 5 psi. What is the absolute air pressure in the tank? Assume standard atmospheric pressure of 14.7 psi. Solution 2.27 p   h  p0  12  p gage   ft   H 2O  pair  12  lb   lb pair   5 2  14.7 2   in.   in. 1ft   62.4 2   lb ft  in.2 144 2 ft  p  19.3 psia Problem 2.28 On the suction side of a pump, a Bourdon pressure gage reads 40 kPa vacuum. What is the corresponding absolute pressure if the local atmospheric pressure is 100 kPa (abs)? Solution 2.28 p  abs   p  gage   p  atm   40 kPa+100 kPa  p  abs   60 kPa Problem 2.29 A Bourdon pressure gage attached to the outside of a tank containing air reads 77.0 psi when the local atmospheric pressure is 760 mm Hg. What will be the gage reading if the atmospheric pressure increases to 773 mm Hg? Solution 2.29 p  abs   p  gage   p  atm  Assuming the absolute pressure of the air in the tank remains constant,  p  gage   p  atm     p  gage   p  atm   i f Where i initial state and f final state. Thus, p f  gage   pi  gage   pi  atm   p f  atm  Since, lb  ft   1ft 2    pi  atm    Hg hi   847 3   0.760m   3.281     14.7 psia m   144in.2  ft     773mm  p f  atm     14.7 psia   14.9 psia  760 mm  p  gage   77.0 psi  14.7 psi  14.9 psia  p  gage   76.8psi Problem 2.31 A U-tube manometer is used to check the pressure of natural gas entering a furnace. One side of the manometer is connected to the gas inlet line, and the water level in the other side open to atmospheric pressure rises 3 in. What is the gage pressure of the natural gas in the inlet line in in. H 2O and in lb/in 2 gage ? Solution 2.31 Patm   H 2O g h  Pgas  lbm  ft  6   lb  s 2  Pgas  0   62.4 3  32.174 2   ft    ft  s   12   32.174 ft  lbm   Pgas  31.2 lb ft 2 gage  6 in. H 2O gage Problem 2.32 A barometric pressure of 29.4 in. Hg corresponds to what value of atmospheric pressure in psia, and in pascals? Solution 2.32 lb  29.4   1ft 2   ft   (psi) p   h   847 3     ft   12   144in.2   p  14.4psia N  m   (Pa) p   h   133  103 3   29.4in.  2.540  102   in.  m    p  99.3kPa (abs ) Problem 2.33 For an atmospheric pressure of 101 kPa (abs) determine the heights of the fluid columns in barometers containing one of the following liquids: (a) mercury, (b) water, and (c) ethyl alcohol. Calculate the heights including the effect of vapor pressure and compare the results with those obtained neglecting vapor pressure. Do these results support the widespread use of mercury for barometers? Why? Solution 2.33 (Including vapor pressure) p (atm)   h  pv , where pv vapor pressure  h  p(atm)  pv  N N  1.6  101 2 2 m m  0.759m 3 N 133  10 3 m 101  103 (a) hmercury  (b) hwater  (c) halchohol  101  103  1.77  103 9.80  103 101  103  5.9  103 7.74  103 m  10.1 m  12.3 m (Without vapor pressure): p(atm)   h  h  p(atm)  N m 2  0.759m (a) hmercury  N 133  103 3 m 101  103 (b) hwater  (c) halchohol  101  103 9.80  103 101  103 7.74  103 m  10.3 m m  13.0 m hmercury  0.759 m hwater  10.1 m halchohol  12.3 m hmercury  0.759 m .vs. w/vapor pressure hwater  10.3 m halchohol  13.0 m wo/vapor pressure Mercury is a better choice because it requires less height to represent a comparable pressure difference and the effect of vapor pressure is mush smaller. Problem 2.34 The closed tank of the figure below is filled with water and is 5 ft long. The pressure gage on the tank reads 7 psi. Determine: (a) the height, h, in the open water column, (b) the gage pressure acting on the bottom tank surface AB, and (c) the absolute pressure of the air in the top of the tank if the local atmospheric pressure is 14.7 psia. Solution 2.34 p   h  p0 (a) p1   H 2O (2 ft)  pair Also p1   H 2O h so that 2 lb  lb     lb   144in.  62.4 h  62.4 2ft  7          ft 3  ft 3     in.2   ft 2  h  18.2 ft 2 2  lb   lb   144in.    1ft  (b) pAB   62.4 3   4ft    7 2        ft   in.   ft 2    144in.2   (c) pair  7psi  14.7 psia=  pair  21.7psia p  8.73 psi Problem 2.35 A mercury manometer is connected to a large reservoir of water as shown in the figure below. h Determine the ratio, w , of the distances hw and hm indicated in the figure. hm Solution 2.35 p1   w hw   w hm p1  p3   m  2 hm   w hw   w hm  2 m hm  w  hw   2 m   w  hm so that hw  2 m   w     2SGm  1 , where SGm  m  13.56 hm w w Thus, hw  2 13.56   1  26.1 hm Problem 2.36 The U-tube manometer shown in the figure below has two fluids, water and oil ( S  0.80 ). Find the height difference between the free water surface and the free oil surface with no applied pressure difference. Solution 2.36 GIVEN: Soil  0.8 (see the figure in the problem) FIND: Free surface height difference. SOLUTION: PA  0 gh0   w ghw  PA   hw  h0  0   h0 S0  w   10 cm  0.8   8 cm h  h0  hw  10 cm  8 cm h  2 cm Problem 2.37 A U-tube manometer is connected to a closed tank containing air and water as shown in the figure below. At the closed end of the manometer the air pressure is 16 psia. Determine the reading on the pressure gage for a differential reading of 4 ft on the manometer. Express your answer in psi (gage). Assume standard atmospheric pressure and neglect the weight of the air columns in the manometer. Solution 2.37 p1   gf (4 ft)   H 2O (2 ft)  pgage lb lb   in.2   lb  lb    pgage  16 2  14.7 2  144 2    90 3   4ft    62.4 3   2ft    in.   ft   ft  ft   in.   627 lb 1ft 2 ft 144in.2  2  pgage  4.67 psi Problem 2.38 The container shown in the figure below holds 60 F water and 60 F air as shown. Find the absolute pressures at locations A, B, and C. Solution 2.38 GIVEN: Figure, water and air at 60 F FIND: Absolute pressures at points A, B, C. SOLUTION:    w   62.3 lb   ft 3  lb  0.0361 3 3  3   ft   1728in  in Modelling the air as an ideal gas: lb   ft    14.7 2   P lbm in   12in     0.0000442 3 a  ft  lb  RT  in  53.35   520R   lbm °R     a  0.0000442 The hydrostatic equation gives: lb     0.0361 3   8in   in in   PA  15.0 psia lb     0.0000442 3  10in   in in   PB  15.0 psia lb     0.0361 3  14in   in in   PC  15.5 psia PA  Patm   whA  14.7 PB  PA   a hB  15.0 PC  PB   whC  15.0 lb 2 lb 2 lb 2 lb in 3 Problem 2.39 A closed cylindrical tank filled with water has a hemispherical dome and is connected to an inverted piping system as shown in the figure below. The liquid in the top part of the piping system has a specific gravity of 0.8, and the remaining parts of the system are filled with water. If the pressure gage reading at A is 60 kPa, determine (a) the pressure in pipe B, and (b) the pressure head, in millimeters of mercury, at the top of the dome (point C). Solution 2.39   (a) pA   SG   H 2O  3m    H 2O  2 m   pB N  N    pB  60 kPa+  0.8   9.81  103 3   3m    9.81  103 3   2 m   m  m    pB  103 kPa N  N  (b) pc  pA   H 2O  3 m   60 kPa   9.81 103 3   3m   30.6 103 2  m  m N 30.6  103 2 3 p m  0.230 m=0.230 m  10 mm   h  230 mm h c   m   Hg 133  103 N   3 m Problem 2.40 Two pipes are connected by a manometer as shown in the figure below. Determine the pressure difference pA  pB , between the pipes. Solution 2.40 pA   H2O  0.5m  0.6m    gf  0.6 m    H2O 1.3m  0.5m   pB pA  pB   gf  0.6m    H2O  0.5m  0.6m  1.3m  0.5m  kN  kN      2.6  9.81 3   0.6m    9.81 3  1.9m   m  m    pA  pB  3.32 kPa Problem 2.41 Find the percentage difference in the readings of the two identical U-tube manometers shown in the figure below. Manometer 90 uses 90 C water and manometer 30 uses 30 C water. Both have the same applied pressure difference. Does this percentage change with the magnitude of the applied pressure difference? Can the difference between the two readings be ignored? Solution 2.41 GIVEN: Figure, Two identical U-tube manometers. Manometer 90 uses 90C water while manometer 30 uses 30C water. Same pressure difference applied across each manometer. FIND: Percent difference in readings. Does this percent difference change with the applied pressure difference? Can the difference in the two manometer readings be ignored? SOLUTION: Apply the manometer rule: PB  PA   w gh  h90  PB  PA 90 w g h30  PB  PA 30 w g Using the 30 C water as a reference 1   1    h90  h30    90 w 30 w   100   30 w  1 100   100   1    h30   90 w    30 w   h90  h30  h h   996   1 100   90 30   3.2%   100    965   h30   h30  Note that this percent difference does not change with the applied pressure difference and the difference in the two manometer readings cannot be ignored in most cases. Problem 2.42 A U-tube manometer is connected to a closed tank as shown in the figure below. The air pressure in the tank is 0.50 psi and the liquid in the tank is oil lb (   54.0 3 ). The pressure at point A is 2.00 psi. ft Determine: (a) the depth of oil, z , and (b) the differential reading, h , on the manometer. Solution 2.42 (a) pA   oil z  pair lb   144in.2   lb   2 2  0.5 2   in.   ft 2  pA  pair  in.  z lb  oil 54.0 3 ft (a ) (b)   z  4.00 ft  pA   oil  2 ft    SG   H 2 O h  0 2 lb   lb   144in.      54.0 3   2 ft   2 2   2 ft  p   oil (2 ft)  in.   ft   h A  lb  SG   H 2 O  3.05   62.4 3  ft      h  2.08 ft Problem 2.43 For the inclined-tube manometer of the figure below, the pressure in pipe A is 0.6 psi. The fluid in both pipes A and B is water, and the gage fluid in the manometer has a specific gravity of 2.6. What is the pressure in pipe B corresponding to the differential reading shown? Solution 2.43  3   8   3  pA   H 2O  ft    gf  ft  sin 30   H 2O  ft   pB  12   12   12  8  pB  pA   gf  ft  sin 30  12  lb   in.2  lb  8      0.6 2  144 2    2.6   62.4 3   ft   0.5  in.   ft  ft   12     32.3 lb 1 ft 2 ft 144 ft 2  2  pB  0.224 psi Problem 2.44 A flowrate measuring device is installed in a horizontal pipe through which water is flowing. A U-tube manometer is connected to the pipe through pressure taps located 3 in. on either side of lb the device. The gage fluid in the manometer has a specific weight of 122 3 . Determine the ft differential reading of the manometer corresponding to a pressure drop between the taps of lb 0.5 2 . in. Solution 2.44 Let p1 and p2 be pressures at pressure taps. Apply manometer equation between p1 and p2 . p1   H2O  h1  h    gf h   H2O h1  p2 lb   in.2   0.5 144    in.2   ft 2   p1  p2 h  lb lb  gf   H 2 O 122 3  62.4 3 ft ft  h  1.45 ft Problem 2.45 The sensitivity Sen of the micromanometer shown in the figure below is defined as Sen  H . pL  pR Find the sensitivity of the micromanometer in terms of the densities  A and  B . How can the sensitivity be increased? Solution 2.45 GIVEN: Figure and sensitivity defined as: Sen  H . pL  pR DETERMINE: Sensitivity as a function of fluid densities. How can the sensitivity increase? SOLUTION: Apply manometer rule, PR   A h    B   A  H  PL PL  PR   A h    B   A  H Sen  H H  PL  PR  Ah    B   A  H  Sen  1 h    B   A  H  A The sensitivity can be increased by decreasing the denominator.  decrease density difference or h  decrease by increasing the ratio of reservoir area to tube area. H Problem 2.46 The cylindrical tank with hemispherical ends shown in the figure below contains a volatile liquid and its vapor. kg The liquid density is 800 3 , and its vapor density is m negligible. The pressure in the vapor is 120 kPa  abs  and the atmospheric pressure is 101 kPa  abs  . Determine: (a) the gage pressure reading on the pressure gage, and (b) the height, h, of the mercury, in the manometer. Solution 2.46 kg   m N  (a) Let    specific weight of liquid   800 3   9.81 2   7850 3  m  s  m pvapor (gage)  120 kPa(abs)  101kPa  abs   19 kPa pgage  pvapor     l m   19  103 N     7850 3   l m   m m   N 2 pgage  26.9 kPa (b) pvapor (gage)     lm    Hg  h   0 19  103 N  N      7850 3   l m    133  103 3   h   0  m m  m    N 2 h  0.202 m Problem 2.47 Determine the elevation difference, h , between the water levels in the two open tanks shown in the figure below. Solution 2.47 Let subscript 1 indicate the surface of the left tank, and subscrip 2 the surface of the right tank. p1   H2 Oh   SG   H2 O  0.4m    H2 O  h  0.4m    H2 O  h   p2 h  0.4m   0.9  0.4m   h  0.040 m Problem 2.48 What is the specific gravity of the liquid in the left leg of the U-tube manometer shown in the figure below? Solution 2.48 GIVEN: Figure FIND: Specific gravity S of unknown fluid SOLUTION:  h10  10cm  Let  h15  15cm  h  20cm  20 Apply manometer rule, Patm   w g  h20  h10   u gh15  Patm S u h20  h10 20cm  10cm    15cm w h15 S  0.667 Problem 2.49 For the configuration shown in the figure below what must be the value of the specific weight of the lb unknown fluid? Express your answer in 3 . ft Solution 2.49 Let  be specific weight of unknown fluid. Applying the manometer rule:   4.9  3.3    5.5  1.4     3.3  1.4   patm   H 2O  ft   patm ft     ft    H 2O  12 12 12         H O  5.5  1.4    4.9  3.3  in.  2  3.3  1.4 in. lb  4.1  1.6    62.4 3     ft   1.9   lb   82.1 3 ft Problem 2.50 The manometer shown in the figure below has an air bubble either in (a) the right horizontal line or (b) the left vertical leg. Find h1  h2 for both cases if p A  pB . Solution 2.50 GIVEN: Figure, manometer with small pockets of air. FIND: h1  h2 if (a) air pocket in horizontal line and (b) air pocket in vertical line. SOLUTION: (a) Air pocket in horizontal line. Apply the manometer rule between the left liquid surface (A) and the right liquid surface (B), PA  PB   h2   h1 h2  h1  1   PA  PB  PA  PB  14.7 psia h1  h2 (b) Air pocket in (left) vertical line. Apply the manometer rule PA  PB   h2    h1  h  PA  PB  14.7 psia h1  h2  h NOTE: For above analyses, the hydrostatic pressure of the air pocket has been neglected. Problem 2.51 The U-tube manometer shown in the figure below has legs that are 1.00 m long. When no pressure difference is applied across the manometer, each leg has 0.40 m of mercury. What is the maximum pressure difference that can be indicated by the manometer? Solution 2.51 GIVEN: Manometer in the figure in the problem. With no pressure difference applied across manometer, each mercury leg is 0.40 m high. FIND: Maximum pressure difference that can be indicated by the manometer. SOLUTION: The maximum pressure difference that can be indicated is illustrated by the sketch on the right. Applying the manometer rule, P1  P2   Hg H   w H   P1  P2  ( Hg   w ) H   Hg   w gH Using table in the Appendix, and assuming 20 º C fluid P1  P2  (13550  998)  123100  N  s2  kg  m 9.807 1.0m      m3  s2   kg  m  N m2 P1  P2  123.1 kPa Problem 2.52 Both ends of the U-tube mercury manometer of the figure below are initially open to the atmosphere and under standard atmospheric pressure. When the valve at the top of the right leg is open, the level of mercury below the valve is h1 . After the valve is closed, air pressure is applied to the left leg. Determine the relationship between the differential reading on the manometer and the applied gage pressure, pg . Show on a plot how the differential reading varies with pg for h1 = 25, 50, 75, and 100 mm over the range 0  p g  300 kPa . Assume that the temperature of the trapped air remains constant. Solution 2.52 With the valve closed and a pressure, pg , applied, pg   Hg h  pa h pg  pa  Hg (1) Where pg and pa are gage pressures. For isothermal compression of trapped air p   Constant  pi V i  pf V f where V is air volume, p is absolute pressure, i & f refer to initial and final states., respectively. patm V i   pa  patm  V f (2) For air trapped in right leg, V i  hi (Area of tube) so that Eq.(2) can be written as    hi  pa  patm   1 h  hi    2  (3)     hi   1  Substitute Eq.(3) into Eq.(1) to obtain: h   p  patm 1  h    Hg  g  hi     2   pg  patm  2 pg hi 2  Eq.(4) can be expressed in the form:  h    2hi  0  h   2 Hg   Hg  (4) The roots of this quadratic equation are  pg  patm  h   hi      2 Hg   2  pg  patm  2 pg hi  hi        Hg 2 Hg   (5) To evaluate h , the negative sign is used since h  0 for pg  0 . Tabulated values of h for various values of pg are given in the following table for different values of hi (with patm  101kPa and  Hg  133 kN/m 3 ). A plot of the data follows. Problem 2.53 The inverted U-tube manometer of the figure below contains oil ( SG  0.9 ) and water as shown. The pressure differential between pipes A and B , pA  pB , is 5 kPa . Determine the differential reading h . Solution 2.53 pA   H2 O  0.2m    oil  h    H2O  0.3m   pB h  pB  pA    H O  0.2m    H O  0.3m  2 5  103  2  oil N  N   9.80  103 3   0.1m  2  m  m   N 3 8.95  10 3 m h  0.449 m Problem 2.54 An inverted U-tube manometer containing oil ( SG  0.8 ) is located between two reservoirs as shown in the figure below. The reservoir on the left, which contains carbon tetrachloride, is closed and pressurized to 8 psi . The reservoir on the right contains water and is open to the atmosphere. With the given data, determine the depth of water, h , in the right reservoir. Solution 2.54 Let pA be the air pressure in left reserviour. Manometer equation can be written as pA   CCl4  3ft  1ft  1ft  0.7 ft    oil  0.7 ft    H2O  h  1ft  1ft   0 h pA   CCl4  0.3ft    oil  0.7ft  H O  2ft 2 2 in. lb  lb   lb      8 2  144 2    99.5 3   0.3ft    57.0 3   0.7ft  ft   ft  ft   in.     2ft = lb 62.4 3 ft h  21 ft Problem 2.55 The sensitivity Sen of the manometer shown in the figure h below can be defined as: Sen  . pL  pR Three manometer fluids with the listed specific gravities S are available: Kerosene, S  0.82 ; SAE 10 oil , S  0.87 ; and Normal octane, S  0.71 . Which fluid gives the highest sensitivity? The areas AR and AL are much larger than the cross-sectional area of the manometer tube, so H  h . Solution 2.55 GIVEN: The figure in the problem, three manometer fluids, kerosene ( S  0.82 ), SAE 10 oil ( S  0.87 ), and normal octane ( S  0.71 ). H  h . FIND: Manometer fluid that gives highest sensitivity. SOLUTION: Apply manometer rule, PR   w  H R  hR    f  hR    f  hL    w  H L  hL   PL PR   w  H R  hR  H L  hL    f  hL  hR   PL PR   w  H R  H L  hR  hL    f  h   PL PR   w  H L  H R  hL  hR    f  h   PL PR   w  H  h    f  h   PL PR  PL   w  H  h    f  h    H  h  PR  PL   w   f h Sen  h h 1   PR  PL  w   f h w g Sw  S f     Sensitivity maximized for S f closest to S w  1  SAE 10 oil . Sen  1 kg   m   1000 3   9.81 2  1  0.87   m  s  Sen  0.000784 m Pa   kg  m 1 m 3 s2 N    9810 1  0.87  kg m Pa  m 2 N  s 2 Sen  0.784 mm Pa Problem 2.56 In the figure below pipe A contains gasoline ( SG  0.7 ), pipe B contains oil ( SG  0.9 ), and the manometer fluid is mercury. Determine the new differential reading if the pressure in pipe A is decreased 25 kPa, and the pressure in pipe B remains constant. The initial differential reading is 0.30 m as shown. Solution 2.56 For the initial configuration: pA   gas  0.3m    Hg  0.3m    oil  0.4 m   pB (1) With a decrease in pA to p ‘A gage fluid levels change as shown on figure. Thus, for final configuration:    gas  0.3  a    Hg  h    oil  0.4  a   pB (2) pA Where all lengths are in m. Subtract Eq.(2) from Eq.(1) to obtain,    gas  a    Hg  0.3  h    oil  a   0 pA  pA Since 2 a  h  0.3 (see figure) then, a  (3) 0.3  h 2 and from Eq.(3),  0.3  h   0.3  h     gas  pA  pA    Hg  0.3  h    oil  0 2    2     gas  0.15    Hg  0.3   oil  0.15  pA  pA h  gas  oil  Hg   2 2 for pA  p ‘A  25 kPa 25 h kN kN  kN  kN      (0.7)  9.81 3   0.15m    133 3   0.3m    0.9   9.81 3   0.15m  2 m m  m  m     kN (0.7)  kN  (0.9)  kN  133 3   9.81 3    9.81 3  2  2  m m  m  h  0.100 m Problem 2.57 The mercury manometer of the figure below indicates a differential reading of 0.30 m when the pressure in pipe A is 30-mm Hg vacuum. Determine the pressure in pipe B. Solution 2.57 pB   oil  0.15m+0.30 m    Hg  0.3m    H2O  0.15m   pA where pA   Hg  0.030 m  Thus, pB   Hg  0.030m    oil  0.45m    Hg  0.3m    H2 O  0.15m  kN  kN  kN  kN        133 3   0.030m    8.95 3   0.45m   133 3   0.3m    9.80 3   0.15m  m  m  m  m      pB  33.4 kPa Problem 2.58 Consider the cistern manometer shown in the figure below. The scale is set up on the basis that the cistern area A1 is infinite. However, A1 is actually 50 times the internal cross-sectional area A 2 of the inclined tube. Find the percentage error (based on the scale reading) involved in using this scale. Solution 2.58 GIVEN: The figure in the problem with A1 50 A 2 . FIND: Percent error in using a scale based on A1 as infinite. SOLUTION: Apply manometer rule, using the elevation changes shown in the sketch. PH  PL    hc  hT  PH  PL   hc  hT (1) hc and hT are vertical drop & rise of fluid level from the scale’s zero. Conservation of mass requires,  hc A1  x A2 or Geometry  hT  x sin 30 Eqs. (2) and (3) into Eq. (1) PH  PL   x A2 A1 (2) (3)  A 1 x A2  x sin 30  x  2   A1  A1 2  x For an infinite cistern area, A1    hc  PH  PL  A 1  2   A1 2  x  PH  PL 2( PH  PL )    A 1  2   A1 2    A 1  x x  x   1100   2  2    1 100 Percent error = % E    100    x   x    A1 2   For 1   1 1  A2 =  % E   2     1 100 A1 50   50 2    % E  4% Problem 2.59 The cistern shown in the figure below has a diameter D that is 4 times the diameter d of the inclined tube. Find the drop in the fluid level in the cistern and the pressure difference ( p A  pB ) if the liquid in the inclined tube rises l  20 in. The angle  is 20 . The fluid’s specific gravity is 0.85 . Solution 2.59 GIVEN: The figure in the problem, D  4d , l  20in. ,   20º , S  0.85 . FIND: PA  PB SOLUTION: Conservation of mass requires that cistern level drops as the tube level rises, as show in the sketch. hT = vertical rise in tube, hc = drop in cistern fluid level. Apply manometer rule, PB  PA   (hT  hc ) Conservation of mass requires, hc Ac  lAT  AT   d 2 hc  l   l   AC   D  hT  l sin , 2  d  PA  PB    hT  hc    l sin       D    2 3   lb    1    ft  PA  PB   62.4 3   0.85  20in. sin 20        4    1728in 3  ft    PA  PB  0.248 psia Problem 2.60 The inclined differential manometer of the figure below contains carbon tetrachloride. Initially the pressure differential between pipes A and B, which contain a brine ( SG  1.1 ), is zero as illustrated in the figure. It is desired that the manometer give a differential reading of 12 in. (measured along the inclined tube) fora pressure differential of 0.1 psi . Determine the required angle of inclination,  . Solution 2.60   pB the left column falls a distance, a , and the right column When pA  pB is increased to pA rises a distance b along the inclined tube as shown in the figure. For this final configuration:    br  hi  a    CCl  a  b sin     br  hi  b sin    pB pA 4     pB   br   CCl  a  b sin    0 pA 4 (1) The differential reading, h , along the tube is h a b sin  Thus, from Eq.(1)     pB   br   CCl  h sin    0 pA 4 sin     pB    pA  br   CCl   h  4   pB  0.1psi For pA lb   in.2     0.1 2  144 2  ft   in.   sin    0.466  lb  lb   12   1.1  62.4 3   99.5 3   12 ft  ft  ft      For h  12in. ,   27.8 Problem 2.61 Determine the new differential reading along the inclined leg of the mercury manometer of the figure below, if the pressure in pipe A is decreased 10 kPa and the pressure in pipe B remains unchanged. The fluid in A has a specific gravity of 0.9 and the fluid in B is water. Solution 2.61 For the initial configuration: pA   A  0.1   Hg  0.05sin 30    H2O  0.08  pB (1) where all length are in m. When pA decreases, left column moves up a distance, a , and right column moves down a distance, a , as shown in the figure. For the final configuration:    A  0.1  a sin 30   Hg  a sin 30  0.05sin 30  a    H O  0.08  a   pB pA 2  is the new pressure in pipe A. Subtract Eq.(2) from Eq.(1) to obtain Where pA   pB   A  a sin 30   Hg a  sin 30  1   H O  a   0 pA 2 Thus, a      pA  pA  A sin 30   Hg  sin 30  1   H 2O   10 kPa For pA  pA kN m2 a  0.0540 m kN  kN  kN    0.9   9.81 3   0.5  133 3   0.5  1  9.81 3   m  m  m 10 New differential reading, h , measured along inclined tube is equal to h a 0.0540 m  0.05  a   0.05 m+0.0540 m=0.0212 m sin 30 0.5 (2) Problem 2.62 A student needs to measure the air pressure inside a compressed air tank but does not have ready access to a pressure gage. Using materials already in the lab, she builds a U-tube manometer using two clear 3-ft- long plastic tubes, flexible hoses, and a tape measure. The only readily available liquids are water from a tap and a bottle of corn syrup. She selects the corn syrup because it has a larger density ( SG  1.4 ). What is the maximum air pressure, in psia , that can be measured? Solution 2.62 Known: two 3-ft- long clear tubes, unknown length flexible hose, tape measure, corn syrup ( SG  1.4 ) Determine: Maximum compressed air pressure Strategy: reflect on possible physical considerations, apply hydrostatic pressure equation Solution: Form u-tube manometer by connecting bottom of tubes with hose; top of one tube is connected to the tank. Assume that tubes are of equal diameter. Assume that hose is not transparent  opaque hose cannot contribute to usable manometer “height” Set tubes at equal elevation. Fill tubes to bottom of tube with corn syrup  maximum difference that can be observed is 3 ft p  h  S H 2 O h  1 ft 2  lb    1.4   62.4 3   3 ft    1.82 psig=16.52 psia  144 in 2  ft     pmax  16.5 psi Problem 2.63 Determine the ratio of areas, A1 , of the two manometer A2 legs of the figure below if a change in pressure in pipe B of 0.5 psi gives a corresponding change of 1 in. in the level of the mercury in the right leg. The pressure in pipe A does not change. Solution 2.63 For the initial configuration (see the figure): pA   H2O  hi  hi    Hg  hi    oil  hi   pB (1) When pB increases the right column falls a distance, a , and the left column rises a distance, b . Since the volume of the liquid must remain constant, A a A1 b  A2 a or 1  . A2 b For the final configuration, with pressure in B equal to pB : pA   H2O  hi  hi  b    Hg  hi  a  b    oil  hi  a   pB (2) Subtract Eq.(1) from Eq.(2) to obtain  H2O  b    Hg  a  b    oil  a   pB  pB b  pB  pB    Hg  a    oil  a   Hg   H O 2 pB  pB  0.5 psi and a  1in. lb   in.2   lb   1  lb   1    0.5 144     847 3   ft    0.8   62.4 3   ft   2  2  12 in.   ft   ft    ft   12     0.00711ft b lb lb 847 3  62.4 3 ft ft 1 ft A1 a   12 A 2 b 0,00711ft  A1  11.7 A2 Problem 2.64 Determine the change in the elevation of the mercury in the left leg of the manometer of the figure below as a result of an increase in pressure of 5 psi in pipe A while the pressure in pipe B remains constant. Solution 2.64 For the initial configuration:  18   6   12  pA   H 2O     Hg  sin 30    oil    pB  12   12   12  (1) Where all lengths are in ft.  the When pA increases to pA left column falls by the distance, a , and the right column moves up the dance, b , as shown in the figure. For the final configuration: 18 6 12    H O   a    Hg  a  sin 30  b sin 30    oil   b sin 30   pB (2) pA 2 12  12     12  Subtract Eq.(1) from Eq.(2) to obtain   pA   H O  a    Hg  a  b sin 30   oil  b sin 30  0 pA 2 The volume of liquid must be constant: A1 a  A2 b , 2 2 1  1   in.  a   in.  b  b  4 a 2  4  Thus, Eq.(3) can be written as   pA   H O  a    Hg  a  4 a sin 30   oil  4 a sin 30  0 pA 2 in.2   lb     5 2   144 2    pA    pA ft   in.   a  lb  lb  lb   H 2 O   Hg (3)   oil (2)  62.4 3   847 3  (3)  (0.9)  62.4 3  (2) ft  ft  ft   a  0.304 ft down (3) Problem 2.65 The U-shaped tube shown in the figure below initially contains water only. A second liquid with specific weight,  , less than water is placed on top of the water with no mixing occurring. Can the height, h , of the second liquid be adjusted so that the left and right levels are at the same height? Provide proof of your answer. Solution 2.65 The pressure at point (1) must be equal to the pressure at point (2) since the pressures at equal elevations in a continuous mass of fluid must be the same. p1   h p2   H 2 O h There two pressures can only be equal if    H 2O . Since    H 2O , the configuration shown in the figure is not possible. No Problem 2.66 An inverted hollow cylinder is pushed into the water as is shown in the figure below. Determine the distance,  , that the water rises in the cylinder as a function of the depth, d , of the lower edge of the cylinder. Plot the results for 0  d  H , when H is equal to 1 m. Assume the temperature of the air within the cylinder remains constant. Solution 2.66 For constant temperature compression within the cylinder, p i V i  p f V f (1) where V is the air volume, and i and f refer to the initial and final states, respectively p f    d     patm pi  patm Vi  4 D2 H V f   4 D2  H       From Eq.(1): patm  D 2 H      d     patm  D 2  H    4 4  patm  101 kPa,   9.80 kN m3 , H  1 m,   2  ( d  11.31)  d (1m)  0  For d  0 ,   0 ,  use negative sign:    d  11.31  d 2  18.61d  128 2  d  11.31  d 2  18.61 d  128 2 Tabulated data with the corresponding plot are shown below. Depth, d (m) 0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 0.1000 Water rise, l (m) 0.000 0.007 0.016 0.024 0.033 0.041 0.049 0.057 0.065 0.073 0.080 (2) Problem 2.68 The basic elements of a hydraulic press are shown in the figure below. The plunger has an area of 1 in.2 , and a force, F1 , can be applied to the plunger through a lever mechanism having a mechanical advantage of 8 to 1. If the large piston has an area of 150 in.2 , what load, F2 , can be raised by a force of 30 lb applied to the lever? Neglect the hydrostatic pressure variation. Solution 2.68 A force of 30 lb applied to the level results in a plunger force, F1 , of F1  (8)(30)  240 lb . Since F1  pA1 and F2  pA2 where p is the pressure and A1 and A2 are the areas of the plunger and piston, respectively. Since p is constant throughout the chamber, F1 F2  A1 A2 so that F2   150in.2  A2 F1   240lb    1in.2   A1   F2  36,000 lb Problem 2.69 The hydraulic cylinder shown in the figure below, with a 4-in.- diameter piston, is advertised as being capable of providing a force of F  20 tons . If the piston has a design pressure (the maximum pressure at which the cylinder should safely operate) of 2500 lb/in 2 , gage , can the cylinder safely provide the advertised force? Solution 2.69 Assuming a “ton” is a “short ton”, the advertised force is  2000 lb  Fadv  20 tons    40000 lb  ton  The maximum force that can be safely developed by the piston is lb      2 Fsafe  Pgage A  Pgage  d 2    2500 2   4 in    31400lb in  4 4    No. The cylinder cannot safely provide 20 tons of force. Problem 2.70 A Bourdon gage is often used to measure pressure. One way to calibrate this type of gage is to use the arrangement shown in the figure below (a). The container is filled with a liquid and a weight, W , placed on one side with the gage on the other side. The weight acting on the liquid through a 0.4-in. -diameter opening creates a pressure that is transmitted to the gage. This arrangement, with a series of weights, can be used to determine what a change in the dial movement,  , in the figure below (b), corresponds to in terms of a change in pressure. For a particular gage, some data are given below. Based on a plot of these data, determine the relationship between  and the pressure, p , where p is measured in psi . Solution 2.70 p W  lb  W   7.96 W  lb  Area  0.4in. 2   4 (where p is in psi ) From graph W  0.0522 So that from Eq.(1) p  psi   0.0522  7.96 p  psi   0.416  (1) Problem 2.71 A bottle jack allows an average person to lift one corner of a 4000-lb automobile completely off the ground by exerting less than 20 lb of force. Explain how a 20-lb force can be converted into hundreds or thousands of pounds of force, and why this does not violate our general perception that you can’t get something for nothing (a somewhat loose paraphrase of the first law of thermodynamics). Hint: Consider the work done by each force. Solution 2.71 Known: 20 lb applied force lifts corner of 4, 000 lb automobile Determine: Explain Strategy: Force = (pressure)(area); consider work done by pistons of different size Solution: Consider two piston/cylinders of different diameter connected by a rigid tube filled with a liquid. F1 F  p 2 A1 A2 A  d   F2  F1  2   F1  2   A1   d1  d2  4  F2  16 F1 d1 2 d2  10  F2  100 F1 d1 Therefore producing the required force multiplication is not difficult. Assuming all solid boundaries are rigid, the volume pushed out of the small cylinder must equal that entering the large cylinder.  V 1  x1 A1  x2 A2 A   x2  x1  1   A2  Therefore, the larger piston moves a smaller distance than the smaller piston. Comparing the work done on the smaller piston to the work done by the larger piston:   A    A  W2  F2 x2   F1  2    x1  1    F1x1  W1   A1     A2   Therefore, the work done on the small piston is equal to work done by large piston. Problem 2.72 Suction is often used in manufacturing processes to lift objects to be moved to a new location. A 1 4-ft by 8-ft sheet of -in. plywood weighs approximately 36 lb . If the machine’s end effector 2 has a diameter of 5 in. , determine the suction pressure required to lift the sheet, expressed in inches of H 2O suction. Solution 2.72 Known: W  36 lb ; DCUP  5 in. Determine: Suction required to lift sheet in inches H20 Strategy: Force = (pressure)(area); Solution: W  F  pA  p R 2 p W R 2  36 lb   2.5 in  2  1.833 lb in 2   H2Oh lb in 2  50.77 in H 2O h lb 1 ft 3 62.4 3  ft 1728 in 3 1.833 h  51 in H 2O Problem 2.73 A piston having a cross-sectional area of 0.07 m 2 is located in a cylinder containing water as shown in the figure below. An open U-tube manometer is connected to the cylinder as shown. For h1  60 mm and h  100 mm , what is the value of the applied force, P , acting on the piston? The weight of the piston is negligible. Solution 2.73 For equilibrium, P  p p AP where p p is the pressure acting on piston and AP is the area of the piston. Also, p p   H 2O h1   Hg h  0 kN  kN  kN   p p   Hg h   H 2O h1  133 3   0.100 m    9.80 3   0.060 m   12.7 2 m  m  m    N   p  12.7  103 2  0.07 m 2 m     p  889 N Problem 2.74 1 A 6-in.-diameter piston is located within a cylinder that is connected to a -in. -diameter 2 inclined-tube manometer as shown in the figure below. The fluid in the cylinder and the manometer is oil (specific weight = 59 lb/ft 3 ). When a weight, W , is placed on the top of the cylinder, the fluid level in the manometer tube rises from point (1) to (2). How heavy is the weight? Assume that the change in position of the piston is negligible. Solution 2.74 With piston alone let pressure on face of piston  p p . Manometer equation becomes p p   oil h1 sin 30  0 (1) With weight added pressure p p increases to pp where pp  p p  W AP ( AP area of piston) 6   pp   oil  h1  ft  sin 30  0 12   (2) Subtract Eq.(1) from Eq.(2) to obtain  6  pp  p p   oil  ft  sin 30  0  12  W  6    oil  ft  sin 30 AP  12   lb  6    59 3   ft   0.5     6   ft   12   ft  4  12  W 2 W  2.90 lb Problem 2.75 The container shown in the figure below has square cross sections. Find the vertical force on the horizontal surface, ABCD. Solution 2.75 The vertical force on surface ABCD is equal to the weight of the imaginary fluid above ABCD as show on the picture on the right, so F  WV   lbm     62.4 3    22  12  ft 2  2.5ft   ft    W  R2  r 2 h F 1470lb Problem 2.76 Find the weight W needed to hold the wall shown in the figure below upright. The wall is 10 m wide. Solution 2.76 The hydrostatic force F on the wall is found from F   ghc A  kg   m   1000 3  9.81 2   2m  4 10m 2 m  s     kg  m   kN   78500  2     s   1000 N   785 kN The force F is located one-third of the water depth from the bottom of the water. h 1  4m   1.33m 3 Summing moments about the pinned joint, FW  1.33m  785 kN  149 kN h F   H 7 m Assuming no friction between the rope and the pulley, W  FW  W  149 kN DISCUSSION Note that the atmospheric pressure acts on both sides of the wall. Therefore, the forces due to atmospheric pressure are equal and opposite, and cancel. Problem 2.77 Determine the magnitude and direction of the force that must be applied to the bottom of the gate shown in the figure below to keep the gate closed. Solution 2.77 The hydrostatic force on the gate is F   yc A kg   m   1000 3   9.81 2  1.3m  0.4 m  2 m  0.8 m  m  s    26700 N The location of the force F is y p  yc  I xc 12 yc A Using Appendix, y p  yc  bh3 h2  yc  12 yc A 12 yc 2 0.8 m    1.3  0.4  m   1.73m 12 1.3  0.4  m Summing moments about the hinge,  M hinge  FR h  F  y p  H   0 FR   F yp  H h    26700N 1.73  1.3 m  0.8m FR  14, 400 N Problem 2.78 An automobile has just dropped into a river. The car door is approximately a rectangle, measures 36 in. wide and 40 in. high, and hinges on a vertical side. The water level inside the car is up to the midheight of the door, and the air inside the car is at atmospheric pressure. Calculate the force required to open the door if the force is applied 24 in. from the hinge line. See the figure below. (The driver did not have the presence of mind to open the window to escape.) Solution 2.78 Note that the force due to atmospheric pressure acts in equal and opposite directions on two sides of the door. The hydrostatic force on the inside of the door is lb   ft 3  36in.  40in.   Fi   hc A   62.4 3   10in     260 lb 3   2   ft   1728in   The hydrostatic force on the outside of the door is Fo   hc A lb   ft 3   12in     62.4 3    20in   36  40  in 2  4ft  3   ft  ft   1728in     3540 lb Summing moments about the hinge line  M H   24in. Freq  18in. Fi  18in. Fo  18   18  Freq   Fo  Fi      3540  260  lb     24   24  Freq  2, 460lb Problem 2.79 Consider the gate shown in the figure below. The gate is massless and has a width b (perpendicular to the paper). The hydrostatic pressure on the vertical side creates a counterclockwise moment about the hinge, and the hydrostatic pressure on the horizontal side (or bottom) creates a clockwise moment about the hinge. Show that the net clockwise moment is  2 h2  u ghb      .  w  2 6  Solution 2.79 The vertical force on the horizontal side is Fv   w ghA   w gh  l  b  Constant force resultant acts at midpoint The horizontal force on the vertical side is h FH   w ghc A   w g    h  b  2 The resultant acts at 1 3 I xc h 12 bh h h 2h y p  yc       yc A 2 h bh   2 6 3 2 Summing moments about the hinge 2   l   gh b  h    M Fv x p  FH h  y p   w ghlb    w   2 3 2      l 2 h2    M   w ghb    2 6    Problem 2.80 Consider the gate shown in the figure below. The gate is massless and has a width b (perpendicular to the paper). The hydrostatic pressure on the vertical side creates a counterclockwise moment about the hinge, and the hydrostatic pressure on the horizontal side (or bottom) creates a clockwise moment about the hinge. Will the gate ever open? Solution 2.80 Sum moments about hinge    M  FV LV  FH LH l h h   w ghbl     w g   bh   2 2 3 2 2 l h    w ghb    2 6    h2 l 2  , 6 2 or h  3l If sum of moments is negative, gate will open  h2 l 2  M  0   6 2  if h  3 l , gate will open Problem 2.81 A tank contains 6 in. of oil ( S  0.82 ) above 6 in. of water ( S  1.00 ). Find the force on the bottom of the tank. See the figure below. Solution 2.81 Assume atmospheric pressure acts on outside of tank. Pressure is constant at constant elevation in a stagnant fluid. Apply hydrostatic pressure equation FNET  PBOTTOM A    oil hoil   whw  A   w  Soil hoil  S whw  A    lb     1   1    62.4 3   0.82  ft   1 ft   1 ft 2 ft    2   2   FNET  56.7 lb Problem 2.82 A structure is attached to the ocean floor as shown in the figure below. A 2-m diameter hatch is located in an inclined wall and hinged on one edge. Determine the minimum air pressure, p1 , within the container that will open the hatch. Neglect the weight of the hatch and friction in the hinge. Solution 2.82 FR   hc A where hc  10 m  1  2 m  sin 30  10.5 m 2 Thus, N     2 FR  10.1 103 3  10.5 m     2 m   3.33  105 N m   4 To locate FR , yR  10 m I xc  1m  21m  yc where yc  sin 30 yc A 4     1m  4  21 m  21.012 m yR     21m  1m 2 For equilibrium, MH  0 FR  21.012 m  20 m   p1  1m  1m  2 3.33  105 N  1.012 m   p   1  1m  1m  2 p1  107 kPa Problem 2.83 Concrete is poured into the forms as shown in the figure below to produce a set of steps. Determine the weight of the sandbag needed to keep the bottomless forms from lifting off the ground. The weight of the forms is 85 lb , and the specific weight of the concrete is 150 lb / ft 3 . Solution 2.83 From the free-body-diagram (   )  Fy  0 Ws  Wc  W f  pb A  0 (1) Ws  weight of sandbag Wc  weight of concrete W f  weight of forms pb  pressure along bottom surface due to concrete A  area of bottom surface From the data given: lb   Wc   150 3   Vol. concrete  ft   10in. 24in.  10in.16in.  10in. 8in.  lb     150 3   3ft    1500lb ft  in.2  144 2 ft W f  85 lb lb   24  lb  p A  150 3   ft   300 2 ft   12  ft   30  A   ft   3ft   7.5ft 2  12  Thus, from Eq.(1)   lb   Ws   300 2  7.5ft 2  1500lb  85lb  ft   Ws  665 lb Problem 2.84 A long, vertical wall separates seawater from fresh water. If the seawater stands at a depth of 7 m, what depth of freshwater is required to give a zero resultant force on the wall? When the resultant force is zero, will the moment due to the fluid forces be zero? Explain. Solution 2.84 For a zero resultant force FRs  FRf  s hcs As   f hcf A f Thus, for a unit length of wall kN   7 m  kN   h    10.1 3     7 m  1m    9.80 3   m   h  1m   m  2   m  2  h  7.11 m In order for moment to be zero, FRs and FRf must collinear. 1 1m  7 m 3 7 I xc  yc  12  m  4.67 m For FRs : yR  yc A 2 7   m   7 m 1m  2  1 1m  7.11m 3 I xc 7.11  yc  12  m  4.74 m Similarly for FRf : yR  yc A 2  7.11   m 7.11m 1m     2  Thus, the distance to FRs from the bottom (point 0) is 7 m  4.67 m  2.33m . For FRf this distance is 7.11m  4.74 m  2.37 m . The forces are not collinear.  No; for zero resultant force, the sum of the moments will not be zero. Problem 2.85 Forms used to make a concrete basement wall are shown in the figure below. Each 4-ft -long form is held together by four ties—two at the top and two at the bottom as indicated. Determine the tension in the upper and lower ties. Assume concrete acts as a fluid with a weight of 150 lb / ft 3 . Solution 2.85 (1)  Fx  0 , or F1  F2  FR and (2)  M 0  0 , or 1F1   2 F2   R FR , where FR  pc A   hc A Thus, FR  150 lb ft 3  5 ft 10 ft  4 ft   30000 lb Also, 1  4ft 10ft 3 I xc 12  10ft  5ft   R  10ft  yR  10ft  yc   yR  yc   10ft  hc  yc A 5ft 10ft  4ft   5 ft  1.67 ft  3.33 ft Thus, from Eq.(2):  9 ft  F1  1ft  F2   3.33ft  30000 lb   99,900 ft  lb (3) 9 F1  F2  99900 From Eq.(1), F1  F2  30,000lb , or F2  30000  F1 9 F1  30000  F1  99,900 F1  8,740 lb F2  30000lb – 8740lb F1  21, 260 lb Problem 2.86 While building a high, tapered concrete wall, builders used the wooden forms shown in the figure below. If concrete has a specific gravity of about 2.5 , find the total force on each of the three side sections (A, B, and C) of the wooden forms (neglect any restraining force of the two ends of the forms). Solution 2.86 The horizontal force FA FA   yc A lb    2.5  62.4 3   6 ft 12  8  ft 2 ft    FA  89900 lb  FB  22500 lb The force FB is horizontal FB   yc A lb    2.5  62.4 3   3ft  6  8  ft 2 ft   The horizontal force FCH lb   FCH   yc A  2.5  62.4 3   9ft  6  8  ft 2  67400 lb ft   The vertical force FCV is the weight of concrete “above” the slanted side (the dashed volume) lb   1   FCV   V  2.5  62.4 3   3  6  8  ft 3   3  6  8  ft 3   33700 lb 2  ft    The total force FC is FC  FCH 2  FCV 2   67400 2   33700 2 lb  FC  75400lb Problem 2.87 A homogeneous, 4-ft -wide, 8-ft -long rectangular gate weighing 800 lb is held in place by a horizontal flexible cable as shown in the figure below. Water acts against the gate, which is hinged at point A. Friction in the hinge is negligible. Determine the tension in the cable. Solution 2.87  6 ft  FR   hc A where hc    sin 60  2  Thus, lb   6 ft   FR   62.4 3     sin 60  6 ft  4 ft   3890 lb ft   2   To locate FR , yR  I xc  yc where yc  3ft yc A so that 1  4 ft  6ft 3 12 yR   3ft  4.0 ft  3ft  6 ft  4 ft  For equilibrium, MH  0 and T  8 ft  sin 60   W  4 ft  cos 60   FR  2 ft  T 800lb  4ft  cos 60   3890lb  2ft  8ft  sin 60   T  1350 lb Problem 2.88 A gate having the shape shown in the figure below is located in the vertical side of an open tank containing water. The gate is mounted on a horizontal shaft. (a) When the water level is at the top of the gate, determine the magnitude of the fluid force on the rectangular portion of the gate above the shaft and the magnitude of the fluid force on the semicircular portion of the gate below the shaft. (b) For this same fluid depth determine the moment of the force acting on the semicircular portion of the gate with respect to an axis that coincides with the shaft. Solution 2.88 (a) For rectangular portion,  FR r   hc A where hc  3 m  FR r   9800  N    3m  6 m  6 m  m3   FR r  1060 kN For semi-circular portion,  FR sc   hc A where hc  6 m  4  3m  4R  6m   7.27 m 3 3  FR sc   9800  N    7.27 m    3m 2  3 m  2   FR sc  1010 kN (b) For semi-circular portion yR  I xc  yc  yc A 0.1098 R 4  7.27 m=7.36 m   2  7.27 m    R 2 Thus, moment with respect to shaft, M :   M   FR  sc   7.36m  6.00m   1010  103 N 1.36m   M  1.37  106 N  m Problem 2.89 A pump supplies water under pressure to a large tank as shown in the figure below. The circularplate valve fitted in the short discharge pipe on the tank pivots about its diameter A–A and is held shut against the water pressure by a latch at B. Show that the force on the latch is independent of the supply pressure, p , and the height of the tank, h . Solution 2.89 The pressure on the gate is the same as it would be for an open tank with a depth of hc  p h  as shown in the figure.  M A  0 , or (1)  y R  yc  FR  R FB where    FR  pc A   hc  R 2   p   h   R 2  and  R4 (2) yR  yc  I xc R2 4   yc A  p   h  2 p  4  h    R     Thus, from Eqs.(1) and (2) FB   yR  yc  F  FB   R  4 R  R  p   h   R2 p  4  h    R3 , which is independent of both p and h . Problem 2.91 Find the center of pressure of an elliptical area of minor axis 2a and major axis 2b where axis 2a is vertical and axis 2b is horizontal. The center of the ellipse is a vertical distance h below the surface of the water ( h  a ). The fluid density is constant. Will the center of pressure of the ellipse change if the fluid is replaced by another constant-density fluid? Will the center of pressure of the ellipse change if the vertical axis is tilted back an angle  from the vertical about its horizontal axis? Explain. Solution 2.91 For a hydrostatic pressure distribution, using geometric information from the Appendix,   ba3    I xc  4  a 2   y p  yc  yc A h  ab  4h y p  yc  a2 a2  h 4h 4h Recognizing symmetry about minor axis, xp  0 Above expressions for x p and y p contain only geometric properties (and not fluid properties) Location of center of pressure not dependent on density. Consider the side view of the ellipse. The equation y p  yc  I xc yc A Requires that the y-coordinate lie in the plane of the surface. I xc is smaller for the smaller horizontally projected area of the tilted ellipse. Therefore, for the tilted ellipse, ( y p  yc ) is smaller for the tilted ellipse, the center of pressure is higher. Problem 2.92 The dam shown in the figure below is 200 ft long and is made of concrete with a specific gravity of 2.2 . Find the magnitude and y coordinate of the line of action of the net horizontal force. Solution 2.92 The headwater horizontal force and its line of action are lb   FHH   yc A   62.4 3   30 ft  60  200  ft 2  2.25  107 lb ft   2 2 y AP  hH   60 ft   40 ft . 3 3 The tailwater horizontal force and its line of action are lb   FTH   yc A   62.4 3  10 ft  20 ft  200 ft   2.50  106 lb ft   2 2 yTP  h T   20ft   13.3ft . 3 3 Fnet  FHH  FTH  2.25 107 lb  2.50 106 lb Fnet  2.00 107 lb acting to right. The line if action is located by taking moments about the base of the dam. 2.25  107 lb   20 ft    2.50  106 lb   6.7 ft   ynet   2.00 107 lb  ynet  21.7 ft above base Problem 2.93 The dam shown in the figure below is 200 ft long and is made of concrete with a specific gravity of 2.2 . Find the magnitude and x coordinate of the line of action of the vertical force on the dam resulting from the water. Solution 2.93 The only vertical force due to the water is on the headwater side of the dam. This vertical force equals the weight of the water above the surface and the force acts through the centroid of the water volume. Therefore lb   1   FHV   V   62.4 3     40 ft  60 ft  200 ft  ft   2   FHV  1.50  107 lb The x -location of this force is the centroid of the 40ft  60ft triangle which gives xp  2  40 ft  or x p  26.7 ft 3 . Problem 2.94 The figure below is a representation of the Keswick gravity dam in California. Find the magnitudes and locations of the hydrostatic forces acting on the headwater vertical wall of the dam and on the tailwater inclined wall of the dam. Note that the slope given is the ratio of the run to the rise. Consider a unit length of the dam ( b  1ft ). Solution 2.94 Consider a unit length of the dam. The headwater force is  bhH h  FHH   yc A    H   bhH   2  2  2 lb  2   62.4 3  1ft 100 ft  ft   2 FHH  312,000 lb The location y p is 2 2 y p  hH  100 ft  3 3 y p  66.7 ft The tailwater force is  hT  FT   yc A     A  2  where A   30ft 1ft  1.02  0.7 2  36.6 ft 2 1.0 so  lb   2  62.4 3   30 ft  36.6ft ft  FT   2  FT  34,300 lb or . The location y p is 2 1.02  0.72 2 1.02  0.72 yp  hT   30ft  3 1.0 3 1.0 or yp  24.4 ft Problem 2.95 The Keswick dam shown in the figure below is made of concrete and has a specific weight of 150 lb/ft 3 . The hydrostatic forces and the weight of the dam produce a total vertical force of the dam on the foundation. Find the magnitude and location of this total vertical force. Consider a unit length of the dam (b  1 ft). Solution 2.95 The hydrostatic vertical force is due to the tailwater. Its magnitude for a dam unit length is 1 FTV   V      b  h T  0.7 h T   0.35 bhT 2 2 lb   2  0.35  62.4 3  1ft  30ft  ft   FTV  19700 lb The dam weight consists of W1  W2 . Now lb   W1   C V   150 3  1ft  21.5 ft  595.0  491.0  ft  335400 lb ft   lb   1  52.1  ft   290800 lb . W2   c V2  150 3  1ft     52.1ft   ft  2  0.7   The total force F is F  FTV  W1  W2  19700  335400  290800  lb F  645,900 lb . Problem 2.96 The Keswick dam shown in the figure below is made of concrete and has a specific weight of 150 lb/ft 3 . The coefficient of friction  between the base of the dam and the foundation is 0.65 . Is the dam likely to slide downstream? Consider a unit length of the dam ( b  1 ft ). Solution 2.96 The total vertical force acting downward is F  W1  W2  FTV . Using the result of Problem 2.51, F  645900lb . The horizontal force resisting movement of the dam is  F  0.65  645900 lb   419800 lb The net force causing the dam to move downstream is  FHH  FTH  . Using the result of Problem 2.50,   1.0 1.0 FTH  FT   28100 lb    34300 lb   2 2  2 2 1.0  0.7  1.0  0.7  Then FHH  FTH   312000  28100  lb  283900 lb . Since FHH  FTH   F , the dam will not slide downstream. Problem 2.97 The figure below is a representation of the Altus gravity dam in Oklahoma. Find the magnitudes and locations of the horizontal and vertical hydrostatic force components acting on the headwater wall of the dam and on the tailwater wall of the dam. Note that the slope given is the ratio of the run to the rise. Consider a unit length of the dam (b  1ft). Solution 2.97 First consider the headwater hydrostatic force components. h  FHH   yc A    H   bhH   2    bhH2 2 lb   2  62.4 3  1ft  87 ft  ft   2 FHH  236, 200 lb . w1  0.11555.0  1475.0  ft 2 2 y p  hH   87 ft  3 3 y p  58 ft =8ft. . FHV   V ‘   V ” lb    1ft  7 ft  87 ft  ft 3    3490 lb  V ‘   62.4 lb  1ft  8ft  87  7  ft   2 ft 3    19970 lb  V ”   62.4 FHV  3490 lb  19970 lb FHV  23460 lb . The location x p of FHV is found from x p   V ‘ x’p   V ” x”p  V ‘   V ” where x ‘p and x ”p are the horizontal locations of  V ‘ and  V ” respectively. The numerical values give xp   3490 lb    8ft   19970 lb    8ft  1 2   3490  19970  lb x p  5.14 ft 2 3  . The tailwater hydrostatic force components h  FTH   yc A    T   bhT   2    bhT2 2 lb  2   62.4 3  1ft  4ft  ft   2  FTH  499 lb   0.6  4 ft   2.4 ft. 2 2 y p  hT   4ft  or 3 3 y p  2.67 ft . lb   62.4 3  1ft  2.4ft  4ft    h   bhT  ft  FTV   V   b  T    2 2 2   2 2 x p     2.4ft  or 3 3 x p  1.6 ft .  FTV  300 lb Problem 2.98 The Altus dam in the figure below is made of concrete with a density of 150 lbm/ft 3 . The coefficient of friction  between the base of the dam and the foundation is 0.65. Is the dam likely to slide downstream? Consider a unit length of the dam  b = 1 ft  . Solution 2.98 The total vertical force acting downward is F  FHV  FTV  W1  W2  W3 . Using the results of Problem 2.97 FH  23, 460 lb FTV  300 lb . w1  0.11555.0  1475.0  ft=8ft The dam weights are w 3  0.6 1553.0  1475.0  ft  46.8ft lb  1   W1   c V 1   150 3    1ft  8ft 1555.0  1475.0  ft=48,000lb ft   2   lb   W2   c V 2   150 3  1ft 10ft 1564.0  1475.0  ft  133,500lb ft   lb  1   W3   c V 3  150 3    1ft  46.8ft 1553.0  1475.0  ft  273,800lb . ft   2   F   23460  300  48000  133500  273800  lb  479,000lb Horizontal force resisting sliding movement of the dam is:  F  0.65  479000 lb   311, 400 lb . Net force acting to slide dam downstream is  FHH  FTH  . Using the results of Problem 2.53  FHH  FTH    236200  499  lb=235700 lb .  FHH  FTH    F , the dam will not slide downstream. Problem 2.99 Find the magnitude and location of the net horizontal force on the gate shown in the figure below. The gate width is 5.0 m. Solution 2.99 The hydrostatic horizontal force FH’ is FH’   whc A   w ghc A kg   m   1000 3  9.81 2   2.0 m  2.0  5.0  m 2 m  s    196000 N=196 kN The location h ‘ of FH’ is h ‘  3.0 m  y p  I   3.0 m   yc  xc  yc A   1   5.0 m  2.0 m 3    3.0 m   2.0 m  12   2.0 m  2.0  5.0  m 2      0.833m The hydrostatic horizontal force FH” is FH”   w hc A   w ghc A kg   m   1000 3  9.81 2   3.0 m  2.0  5.0  m 2 m  s    294000 N  294 kN The location of h of FH is  I  h ”  4.0 m  y p  4.0 m   yc  xc  yc A   1   5.0 m  2.0 m 3    4.0 m   3.0 m  12   3.0 m  2.0  5.0  m 2      0.889 m. The magnitude of the net horizontal force FH FH  FH”  FH’  294kN -196kN FH  98 kN The location of the net horizontal force F above the base is denoted by h and is found by noting the moment of the resultant is equal to the moment of the components or Fh  FH” h”  FH’ h’ , FH” h”  FH’ h’ F 294 kN   0.889 m   196 kN  0.833m   98 kN h h  1.00 m . DISCUSSION Note that the resultant of the two opposing forces is not located between the two forces; this is only true for two forces that are acting in the same direction. Again note that the atmospheric pressure force was not considered as it acts uniformly on both sides of the gate and cancels out. Also note that the horizontal hydrostatic forces do not depend on the 45o angle. Problem 2.100 Find the magnitude and location of the net vertical force on the gate shown in the figure below. The gate width is 5.0 m. Solution 2.100 The hydrostatic vertical force FV’ is the weight of the water above the gate to the level A-A. Then FV’  W3  W2   w g  V 3  V 2  2 kg   m  1   N s  1000 3   9.81 2    2.0  2.0  5.0  m3   2.0 1.0  5.0  m3  m  s 2   kg  m FV’  98100 N+98100 N  196200 N The hydrostatic vertical force FV” is the weight of the water above the gate to the level B-B. FV”  W3  W2  W1  w g  V 3  V 2  V 1   N  s2  kg   m  FV”  98100 N+98100 N+ 1000 3   9.81 2   2.0  1.0  5.0  m3   kg  m   m  s     98100 N+98100 N+98100 N =294300 N The magnitude of the vertical force FV FV  FV”  FV’  294300  196200 N FV  98100 N acting downward. The location of F is found by first finding the locations of FV’ and FV” . First ’  W3 3  W2 2 . W3  W2 Recognizing that  3    and  2  gives 3 2  2.0   2.0  98100 N    98100 N    3m   2m  ’  98100 N  98100 N  ‘  0.833 m Also  ”  W3 3  W2 2  W11 W3  W2  W1 Recognizing that 1   gives 2  2.0   2.0   2.0  98100 N    98100 N    98100 N   3m  2m  2m     ”   98100 N  98100 N  98100 N  ”  0.889 m . The location of the resultant force from the left side of the gate is denoted by V and is found from FV” ”  FV’ ’ V  FV  294300 N  0.889 m   196200 N  0.833m  294300 N  196200 N V  1.00 m from left side of gate. DISCUSSION Noting that the resultant vertical force, FV  98100 N is the weight of a volume of water measuring   2.0 m , width 5.0m , and height  1.0 m , is there a quick way to find FV ? Problem 2.101 Find the total vertical force on the cylinder shown in the figure below. Solution 2.101 The net force F on the cylinder is due to the water and is F  F1  F2  p1 A1  p2 A2 . Since the atmospheric pressure does not contribute to the net force, p1 and p2 will be considered gage pressures. kg   m  1000 3  9.81 2  18  5  cm N m  s  p1   w gh    1275 2 cm   kg  m   m 100   2 m  N  s   kg   m  1000 3  9.81 2   3 cm N m  s  p2   w gh    294 2 cm   kg  m   m 100  2  m   N  s  Then 2 N  m   N  2 2 m  2  2 F   1275 2   3cm      294 2  6  3 cm    m 4 m 4  100 cm    100 cm  or F  1.52 N .   2 Problem 2.102 A 3-m -wide, 8-m -high rectangular gate is located at the end of a rectangular passage that is connected to a large open tank filled with water as shown in the figure below. The gate is hinged at its bottom and held closed by a horizontal force, FH , located at the center of the gate. The maximum value for FH is 3500 kN . (a) Determine the maximum water depth, h, above the center of the gate that can exist without the gate opening. (b) Is the answer the same if the gate is hinged at the top? Explain your answer. Solution 2.102 For gate hinged at bottom MH  0 so that  4 m  FH  FH (see figure) (1) and kN   FR   hc A   9.80 3   h  3 m  8 m    9.80  24h  kN m   1  3m 8 m 3 I xc 5.33 yR   yc  12 h h yc A h  3m  8 m  h 5.33  5.33  Thus,   m   h  4    h  4  h  h  and from Eq.(1)  4 m  3500 kN    4   so that h  16.2 m 5.33    9.80  24  h  kN h  For gate hinged at top MH  0 so that  4 m  FH  1 FH (see figure) (1) where 5.33  5.33  1  y R   h  4     4    h  4  4 h  h  Thus, from Eq.(1)  4 m  3500 kN    5.33   4   9.80  24  h  kN  h  and h  13.5 m Maximum depth for gate hinged at top is less than maximum depth for gate hinged at bottom. Problem 2.103 A gate having the cross section shown in the figure below is 4 ft wide and is hinged at C. The gate weighs 18, 000 lb , and its mass center is 1.67 ft to the right of the plane BC. Determine the vertical reaction at A on the gate when the water level is 3 ft above the base. All contact surfaces are smooth. Solution 2.103 F1   hc A where hc  1.5ft lb   Thus, F1   62.4 3  1.5 ft  3ft  4 ft   1120 lb ft   The force F1 acts at a distance of 1ft from the base of the gate. F2  p2 A2 where p2   H2O  3ft  Thus, lb   F2   62.4 3   3ft  5 ft  4 ft   3740 lb ft   and acts at the center of the bottom gate surface. For equilibrium,  Mc  0 and F1 11ft   F2  2.5ft   FA  5ft   W 1.67 ft  so that FA  18, 000lb 1.67 ft   1120lb 11ft    3740lb  2.5ft   1680lb 5ft Problem 2.104 The massless, 4-ft -wide gate shown in the figure below pivots about the frictionless hinge O. It is held in place by the 2000 lb counterweight, W . Determine the water depth, h . Solution 2.104 FR   hc A where hc  h 2 Thus, h h2 FR   H2O  h  b    H 2O  4 ft  2 2 To locate FR ,   1  4 ft  h3 h 2 I xc 12  yc    h yR  h yc A  4 ft  h  2 3 2 For equilibrium,  M 0  0 FR d  W  3ft  where d  h  yR  so that h  3  2000 lb  3ft  2 Thus,  2    H O   h2   4 ft  h3   3 2000 lb  3ft  lb   1    62.4 3   2   4 ft  ft     h  5.24 ft h 3 Problem 2.105 A 200  lb homogeneous gate 10 ft wide and 5 ft long is hinged at point A and held in place by a 12-ft -long brace as shown in the figure below. As the bottom of the brace is moved to the right, the water level remains at the top of the gate. The line of action of the force that the brace exerts on the gate is along the brace. (a) Plot the magnitude of the force exerted on the gate by the brace as a function of the angle of the gate,  , for 0   90 . (b) Repeat the calculations for the case in which the weight of the gate is negligible. Comment on the result as   0 . Solution 2.105 (a) For the free-body-diagram of the gate (see figure),  FA  0 so that    FR    W  cos     FB cos    sin     FB sin    cos   3 2  (1) Also,  sin   L sin  (assuming hinge and end of brace at same elevation) or sin    sin  L and   sin   FR   hc A      w   2  where w is the gate width. Thus, Eq.(1) can be written as  3  W cos   FB   cos  sin   sin  cos     sin   w  6 2     so that   2 w    2w  W W   sin   cos    tan   6  2 6  2  (2) FB   cos  sin   sin  cos  cos  tan   sin  For   62.4 lb ft 3 ,   5 ft , w  10 ft , and W  200 lb , lb  2   62.4 3   5ft  10 ft  200 lb ft   tan   6 2  2600 tan   100 FB  cos  tan   sin  cos  tan   sin  Since sin   sin   (3)  sin  and   5 ft , L  12 ft L 5 sin  12 and for a given  ,  can be determined. Thus, Eq.(3) can be used to determine FB for a given  . (b) For W  0 , Eq.(3) reduces to FB  2600 tan  cos  tan   sin  (4) and Eq.(4) can be used to determine FB for a given  . Tabulated data of FB vs.  for both W  200 lb and W  0 lb are given below. θ, deg 90.0 85.0 80.0 75.0 70.0 65.0 60.0 55.0 50.0 45.0 40.0 35.0 30.0 25.0 20.0 15.0 10.0 5.0 2.0 F(B), lb (W=200 lb) 2843 2745 2651 2563 2480 2403 2332 2269 2213 2165 2125 2094 2075 2069 2083 2130 2250 2646 3858 F(B), lb (W=0 lb) 2843 2736 2633 2536 2445 2360 2282 2210 2144 2085 2032 1985 1945 1911 1884 1863 1847 1838 1836 As   0 the value of FB can be determined from Eq.(4), FB  2600 tan  cos  tan   sin  Since sin   5 sin  12 it follows that 2 5 cos   1  sin   1    sin 2   12  2 and therefore FB  2600 tan  2 5 5 1    sin 2  tan   sin  12  12  2600  2 5 5 1    sin 2   cos  12  12  Thus, as   0 FB  2600  1840 lb 5 1 12 Physically this result means that for   0 , the value of FB is indeterminate, but for any “very small” value of  , FB will approach 1840 lb . Problem 2.106 An open tank has a vertical partition and on one side contains gasoline with a density   700 kg / m3 at a depth of 4 m , as shown in the figure below. A rectangular gate that is 4 m high and 2 m wide and hinged at one end is located in the partition. Water is slowly added to the empty side of the tank. At what depth, h , will the gate start to open? Solution 2.106 FRg   g hcg Ag ; where g refers to gasoline. kg   m  FRg   700 3  9.81 2   2 m  4 m  2 m  m  s    110 103 N=110 kN FRw   w hcw Aw ; where w refers to water. N  h   FRw   9.80  103 3     2 m  h  ; where h is depth of water.  m  2    FRw  9.80 103 h 2 For equilibrium,  M H  0  FRw w  FRg  g w  h 4 ; g  m 3 3 9.80 103  h2   h3   110 103 N   43 m  h  3.55 m which is the limiting value for h. Problem 2.107 A horizontal 2-m-diameter conduit is half filled with a liquid ( SG  1.6 ) and is capped at both ends with plane vertical surfaces. The air pressure in the conduit above the liquid surface is 200 kPa . Determine the resultant force of the fluid acting on one of the end caps, and locate this force relative to the bottom of the conduit. Solution 2.107 Fair  pA, where p is air pressure Thus, N    2  Fair   200 103 2     2 m   200 103 N  m  4  Fliquid   hc A2 where hc  4R 3 (see the figure below) Thus, N   4 1m    1      2 3 Fliquid  1.6   9.81103 3         2 m   10.5 10 N  m   3   2   4  For Fliquid , yR  I xc  yc yc A2 where I xc  0.1098R4 and yc  hc  4R 3 (see the figure below) Thus, yR  0.1098 1m  4  4 1m    1    2       2 m   3   2  4   4 1m   0.5891m 3 Since Fresultant  Fair  Fliquid   200  10.5   103 N=639 kN, we can sum moments about O to locate resultant to obtain Fresultant  d   Fair 1m   Fliquid 1m  0.5891m  So that 200 103 N  1m   10.5 103 N   0.4109 m   d 639 103 N  0.990 m above bottom of conduit. Problem 2.108 A 4-ft by 3-ft massless rectangular gate is used to close the end of the water tank shown in the figure below. A 200-lb weight attached to the arm of the gate at a distance ℓ from the frictionless hinge is just sufficient to keep the gate closed when the water depth is 2 ft , that is, when the water fills the semicircular lower portion of the tank. If the water were deeper, the gate would open. Determine the distance . Solution 2.108 FR   hc A where hc  4R 3 (see the figure below) Thus,  4R    R FR   H 2O    3   2 2   2 lb   4  2 ft      2 ft       333lb   62.4 3     2 ft   3     To locate FR , yR   I xc  yc yc A 0.1098 R 4 4R  2  4 R    R  3     3   2  (see the figure below) 4 0.1098  2 ft  4  2 ft      1.178ft 2 3  4  2 ft     2 ft    3   2 For equilibrium, MH  0 So that W   FR 1ft+y R  And   333lb 1ft  1.178ft   3.63ft 200lb Problem 2.109 A thin 4-ft-wide, right-angle gate with negligible mass is free to pivot about a frictionless hinge at point O, as shown in the figure below. The horizontal portion of the gate covers a 1-ftdiameter drain pipe that contains air at atmospheric pressure. Determine the minimum water depth, h , at which the gate will pivot to allow water to flow into the pipe. Solution 2.109 For equilibrium,  M0  0 FR1  1  FR2   2 (1) FR1   hc1 A1 lb  h     62.4 3     4 ft  h   125 h 2 ft   2   For the force on the horizontal portion of the gate (which is balanced by pressure on both sides except for the area of the pipe) lb  2  2     FR2   h   1ft    62.4 3   h    1ft  ft  4 4   49.0 h Thus, from Eq. (1) with 1  125 h2   h3    49.0 h 3ft  h  1.88ft h and  2  3ft 3 Problem 2.110 The closed vessel of the figure below contains water with an air pressure of 10 psi at the water surface. One side of the vessel contains a spout that is closed by a 6-in.-diameter circular gate that is hinged along one side as illustrated. The horizontal axis of the hinge is located 10 ft below the water surface. Determine the minimum torque that must be applied at the hinge to hold the gate shut. Neglect the weight of the gate and friction at the hinge. Solution 2.110 Let F1 force due to air pressure, and F2 force due to hydrostatic pressure distribution of water. Thus, 2 lb   in.2      6   F1  pair A   10 2  144 2     ft   283lb ft   4   12   in.   and F2   hc A 1  3  6   where hc  10ft     ft   10.15ft 2  5  12   So that 2 lb      6  F2   62.4 3  10.15ft    ft   124 lb ft   4  12   Also, yR2  I xc  yc yc A where yc  10 ft 1  6    ft   16.92 ft 3 2  12  5 So that 4    3    ft   4  12  yR2   16.92 ft  16.92 ft 2    6  16.92 ft    ft   4  12  For equilibrium,  M0  0 And    10 ft   3  C  F1  ft   F2  yR2  3   12    5      10 ft   3   102 ft  lb C   283lb   ft   124 lb  16.92 ft  3   12    5   Problem 2.111 (a) Determine the horizontal hydrostatic force on the 2309-m-long Three Gorges Dam when the average depth of the water against it is 175 m . (b) If all of the 6.4 billion people on Earth were to push horizontally against the Three Gorges Dam, could they generate enough force to hold it in place? Support your answer with appropriate calculations. Solution 2.111 (a) N  175 m   FR   hc A   9.80 103 3   175 m  2309 m  m  2    3.46 1011 N (b) Required average force per person  3.46 1011 N 6.4 109 N  lb   54.1 12.2  person  person  Yes. It is likely that enough force could be generated since required average force per person is relatively small. Problem 2.113 A 2-ft-diameter hemispherical plexiglass “bubble” is to be used as a special window on the side of an above-ground swimming pool. The window is to be bolted onto the vertical wall of the pool and faces outward, covering a 2-ft-diameter opening in the wall. The center of the opening is 4 ft below the surface. Determine the horizontal and vertical components of the force of the water on the hemisphere. Solution 2.113  Fx  0 , or FH  FR  pc A Thus, FH   hc A  62.4 lb ft 3   4 ft   2 ft 2  784 lb  to right  4 and  Fy  0 , or FV  W   V   4  R3 , 3 2 where R  1ft Thus, lb  4 1ft      131lb  down on bubble   6 ft 3   3 FV  62.4 Problem 2.114 Consider the curved surface shown in the figure below (a) and (b). The two curved surfaces are identical. How are the vertical forces on the two surfaces alike? How are they different? Solution 2.114 In both cases the magnitude of the vertical force is the weight of shaded section shown on the right. In addition, the location of the vertical force is the same (the centroid of the shaded section.) Therefore: Alike: magnitude and location of vertical forces same. However, the two vertical forces are different in that the force in (a) is acting upward and the force in (b) is acting downward. Therefore: Different: direction of vertical forces opposite. Problem 2.115 The figure below shows a cross section of a submerged tunnel used by automobiles to travel under a river. Find the magnitude and location of the resultant hydrostatic force on the circular roof of the tunnel. The tunnel is 4 mi long. Solution 2.115 Due to symmetry, there is no net horizontal force on the roof. The vertical force is equal to the weight of fluid above the tunnel. This vertical force acts through the centroid of the fluid volume. Then for a tunnel length ,    F   V     2 Rh  R 2  2   lb  ft      2   62.4 3   4 mi   5280   2  20 ft  70 ft    20 ft   mi 2 ft      F  2.86 109 lb . This force acts downward through the point “O”. Problem 2.116 The container shown in the figure below has circular cross sections. Find the vertical force on the inclined surface. Also find the net vertical force on the bottom, EF. Is the vertical force equal to the weight of the water in the container? Solution 2.116 The vertical force on the inclined surface is equal to the weight of the water “above” it. This “water volume” is V  V cyl  V hole  V frustrum   1 V   ro2   ri2    h    h ro2  ri2  ro ri . 3 V    2 ft   3ft    1ft   3  1 ft 2 2 1 2 2   1ft   2 ft   1ft    2 ft 1ft    24.1ft 3   3 The vertical force FVi is ri  1″, ro  2″, h  1″,   3″.   lb   FVi   V   62.4 3  24.1ft 3  FVi  1500lb ft   The pressure is uniform over the bottom EF so lb   2 FVb  pA   hA   62.4 3   7 ft    2 ft  ft   or FVb  5490 lb This force FVb is not equal *CRC Standard Math Tables to the weight of water in the container. . Problem 2.117 The 18-ft-long lightweight gate of the figure below is a quarter circle and is hinged at H . Determine the horizontal force, P , required to hold the gate in place. Neglect friction at the hinge and the weight of the gate. Solution 2.117 For equilibrium (from free-body-diagram of fluid mass),  Fx  0 So that Similarly, FH  F1   hc1 A1 lb  6 ft     62.4 3     6 ft 18ft   20200 lb ft   2    Fy  0 So that lb     2  FV  W   H 2O   volume of fluid    62.4 3    6 ft   18 ft   31800 lb 4 ft     Also, x1  4  6 ft  8  ft  3 (see the figure below) y1  and 6 ft  2 ft 3 For equilibrium (from free-body-diagram of gate)  M0  0 So that P  6 ft   FH  y1   FV  x1  or P  20200 lb  2 ft    31800 lb   6 ft  ft      20200 lb 8 Problem 2.118 The air pressure in the top of the 2-liter pop bottle and the figure below is 40 psi , and the pop depth is 10 in. The bottom of the bottle has an irregular shape with a diameter of 4.3 in. (a) If the bottle cap has a diameter of 1 in. what is the magnitude of the axial force required to hold the cap in place? (b) Determine the force needed to secure the bottom 2 in. of the bottle to its cylindrical sides. For this calculation assume the effect of the weight of the pop is negligible. (c) By how much does the weight of the pop increase the pressure 2 in. above the bottom? Assume the pop has the same specific weight as that of water. Solution 2.118 (a) lb      2 Fcap  pair  Area cap   40 2    1in.  31.4 lb  in.   4  (b)  Fvertical  0 Fsides  F1   pressure @ 2 in. above bottom    Area  lb     2   40 2     4.3in.  in.   4   581lb (c) p  pair   h     lb  lb  8  1   40 2   62.4 3   ft   in.  ft   12   in.2   144 2  ft   lb lb  40 2  0.289 2 in. in. Thus, the increase in pressure due to weight  0.289 psi (which is less that 1% of air pressure). Problem 2.119 In drilling for oil in the Gulf of Mexico, some divers have to work at a depth of 1300 ft. (a) Assume that seawater has a constant density of 64 lb/ft 3 and compute the pressure at this depth. The divers breathe a mixture of helium and oxygen stored in cylinders, as shown in the figure below, at a pressure of 3000 psia . (b) Calculate the force, which trends to blow the end cap off, that the weld must resist while the diver is using the cylinder at 1300 ft . (c) After emptying a tank, a diver releases it. Will the tank rise or fall, and what is its initial acceleration? Solution 2.119 (a) The hydrostatic pressure is  ft 2   lb  p   sw h   64 3  1300 ft   or  144in.2   ft    (b) p  578 psig The net horizontal force on the end caps is FN  Fin  Fout  pin Ain  pout Aout and   wall stress=  FN FN  Awall Aout  Ain 2 pin Ain  pout Aout pin Din2  pout Dout  2 Aout  Ain Dout  Din2 lb  lb 2 2   3000 2   6in.  14.7  578  2  8in. in.  in.  2 2 8in.   6in. and   2500 psi. (c) The net vertical force on an empty tank and Newton’s second law give   Fvert  FBuoy  W  ma or a FBuoy  W  m FBuoy m g where m is the mass of the tank. Now       FBuoy   sw V   sw    Dout 2    Dout 3  6  4   where   30 in.  6 in.  24 in. Also       m  steel    Dout 2  Din 2    Dout 3  Din3  . 6  4       Substituting into the equation for a gives 1 4 1  6  a 1 1  2 2 steel   Dout  Din  Dout 3  Din 3  6 4   sw   Dout 2  Dout 3      g. The numerical values give   ft  lbm   lb   1 2 1 3  3  64 3   4  24  8  6 8  in.  32.2   lb  sec 2   32.2 ft  a   ft   lbm   1 1 3  sec 2 2 2 3  3 489 24 8  6  8  6 in.      6 ft 3   4   or a  23.1 ft sec2 tank will fall since a  0.    Problem 2.120 Hoover Dam is the highest arch-gravity type of dam in the United States. A cross section of the dam is shown in the figure below (a). The walls of the canyon in which the dam is located are sloped, and just upstream of the dam the vertical plane shown in the figure below (b) approximately represents the cross section of the water acting on the dam. Use this vertical cross section to estimate the resultant horizontal force of the water on the dam, and show where this force acts. Solution 2.120 Break area into 3 parts as shown. For area 1: lb  1   1 FR1   hc A1   62.4 3     715ft     295ft  715ft  ft   3  2   1.57 109 lb For area 3 : FR3  FR1  1.57 109 lb For area 2 : lb  1   FR2   hc A2   62.4 3     715ft  290 ft  715ft  ft   2    4.63 109 lb Thus, FR  FR1  FR2  FR3  1.57 109 lb  4.63 109 lb  1.57 109 lb  7.77 109 lb Since the moment of the resultant force about the base of the dam must be equal to the moments due to FR1 , FR2 , and FR3 , it follows that 2 1 2 FR  d  FR1    715ft   FR2    715ft   FR3    715ft  3 2 3 and d 1.57 109 lb   23   715ft    4.63 109 lb   12   715ft   1.57 109 lb   32   715ft  7.77  109 lb  406 ft Thus, the resultant horizontal force on the dam is 7.77 109 lb acting 406 ft up from the base of the dam along the axis of symmetry of the area. Problem 2.121 A plug in the bottom of a pressurized tank is conical in shape, as shown in the figure below. The air pressure is 40 kPa , and the liquid in the tank has a specific weight of 27 kN / m3 . Determine the magnitude, direction, and line of action of the force exerted on the curved surface of the cone within the tank due to the 40  kPa pressure and the liquid. Solution 2.121 d tan 30  2 1  d  2 tan 30  1.155m volume of cone =  d  2   1 32 For equilibrium,  Fvertical  0 So that Fc  pair A  aw where Fc is the force the cone exerts of the fluid. Also,   pair A   40 kPa    d 2 4   2   40 kPa    1.155 m   41.9 kN 4   And 2    d   3m 1m  W    d 2  3m     1m     d 2   32 12   4  4   kN  22    27 3   1.155 m   m   75.4 kN   m  3   Thus, Fc  41.9 kN  75.4 kN  117 kN And the force on the cone has a magnitude of 117 kN and is directed vertically downward along the cone axis. Problem 2.122 The homogeneous gate shown in the figure below consists of one quarter of a circular cylinder and is used to maintain a water depth of 4 m . That is, when the water depth exceeds 4 m , the gate opens slightly and lets the water flow under it. Determine the weight of the gate per meter of length. Solution 2.122 Consider the free body diagram of the gate and a portion of the water as shown.  M o  0 , or (1)  2W  1W1  FH  3  FV  4  0 , where (2) FH   hc A  9.8 103 N m3  3.5 m 1m 1m   34.3kN since for the vertical side, hc  4 m  0.5 m  3.5 m Also, (3) FV   hc A  9.8 103 N m3  4 m 1m 1m   39.2 kN Also, (4) N   3 2  W1   1m     1m   1m   9.8  103 3 1   m3  2.10 kN 4  m  4 (5) Now,  4  0.5 m and (6) 1 1m 1m 3 I xc 12  3  0.5 m+  yR  yc   0.5 m   0.5 m   0.524 m 3.5 m 1m 1m  yc A (7) and  2  1m  4 1m  4R  1  0.576 m 3 3 To determine 1 , consider a unit square that consist of a quarter circle and the remainder as show in the figure. The centroids of areas (1) and (2) are as indicated. Thus, 4    0.5   A2   0.5  1  A1 3   So that with A2   4 12   4 and A1  1   4 this gives 4      0.5     0.5  1  1   3  4   4 or (8) 1  0.223m Hence, by combining Eqs.(1) through (8):  0.576 m W   0.223m  2.10 kN    34.3kN  0.524 m    39.2 kN  0.5m   0 or W  64.4 kN Problem 2.123 The concrete (specific weight = 150 lb / ft 3 ) seawall of the figure below has a curved surface and restrains seawater at a depth of 24 ft . The trace of the surface is a parabola as illustrated. Determine the moment of the fluid force (per unit length) with respect to an axis through the toe (point A). Solution 2.123 The components of the fluid force acting on the wall are F1 and W as shown on the figure where lb   24 ft   F1   hc A   64.0 3     24 ft  1ft  ft   2    18400 lb and y1  24 ft  8ft 3 Also, W  V To determine V find area BCD. Thus,  24  y  dx  0  24  0.2 x 2  dx 0 A x0 x0 x 0  0.2 x3    24 x   3   0 And with x0  120 , A  175ft 2 so that V  A 1ft  175ft 3 Thus,   lb   W   64.0 3  175ft 3  11200 lb ft   To locate centroid of A:  24  y  x dx  0  24 x  0.2 x3  dx  12 x02  0 x0 xc A   x dA   0 x0 x0 and xc  12  120   2 0.2  120  4 175 4  4.11ft Thus, M A  F1 y1  W 15  xc   18400 lb  8ft   11200 lb 15ft  4.11ft   25200 ft  lb 0.2 x04 4 Problem 2.124 A step-in viewing window having the shape of a half-cylinder is built into the side of a large aquarium. See the figure below. Find the magnitude, direction, and location of the net horizontal forces on the viewing window. Solution 2.124 Due to symmetry, the net force parallel to the wall is zero or Fz  0 The net horizontal force perpendicular to the wall is  lb  Fx   hc A   64 3   25  5  ft 10 ft  10 ft   ft  Fx  1.92  105 lb The vertical location of Fx is 1 3 bh I xc 10 ft 2 or h2 y p  yc   yc  12  yc   30 ft  12 yc 12  30 ft  yc A yc bh The net horizontal force also acts through the coordinate z  0 and acts in an outward direction. y p  30.3ft Problem 2.125 Find the magnitude, direction, and location of the net vertical force acting on the viewing window in Problem 2.124. Solution 2.125 The net vertical force must equal the weight of fluid inside the viewing window. Then Fy  25100 lb, 2    lb    Fy   V   h  R 2    64 3  10 ft     5ft  or acting upword. 2   ft  2 This net vertical force acts through the centroid of the window volume. Using Appendix B gives x 4 R 4  5ft   3 3 or x  2.12 ft Problem 2.126 A l0-m -long log is stuck against a dam, as shown in the figure below. Find the magnitudes and locations of both the horizontal force and the vertical force of the water on the log in terms of the diameter D . The center of the log is at the same elevation as the top of the dam. Solution 2.126 Consider the water forces on the log as shown on the right. The horizontal forces FH is on the top portion only and is  D  D  FH        4  2  where  is the log length. Assuming 10 C water, Table A.5 gives kg   m  FH  1000 3   9.81 2   0.25 m  0.5 m 10 m   12300 N  FH m  s   The location of FH is 2  D  2  0.5  yp      m   0.167 m  y p 3 2  3 2  The vertical force FV is the weight of water “above» the bottom of the log minus the weight of water above the top half of the log. This is  D2  D   D 2  D 2    D3  3  FV      D    1      8 2 4 16 4 4         kg   m 2  1000 3  9.81 2  10 m 1.0 m  3   m  s    1  4 4   FV  82300 N The location x of FV is found by first locating the centroid of area A1 by x1  A1 2 x1 2  A2 x2 . A1 Using Table B x1  2 2  D   D    D   D 2D            2   4   16   2 3  2 2  D D    16 2  1   1 2   16  16  2  3     D  1       4 16  0.112 D and is the location of FV 1 . The location of FV 2 is D . The location of FV is 2 D FV 2    FV 1  0.112 D  2 x FV  D2 D2   D   D2  D2          0.112 D  2   2   4 16   B   D2 D2  D2  D2      2  4 16  B 1   1      1  0.112  1   4 4   4  4  D   1 1      1   1   2 4  4 4  x  0.525 D FV 2  weight of water “above” bottom portion of log. FV 1  weight of water above top left portion of log. Problem 2.127 Find the net horizontal force on the 4.0-m -long log shown in the figure below. Solution 2.127 The force FL on the left side of the log is the horizontal force on the horizontally projected area of the log. This horizontally projected area measures D  1.0 m by 4.0 m and gives FL   ghc A  N  s2  kg   m   1000 3  9.81 2   0.5 m 1.0 m  4.0m    kg  m  m  s      19600 N  19.6 kN The force FR on the right side of the log is the horizontal force on the horizontally projected area of the lower half of the log. This horizontally projected area measures and gives  N  s2  kg   m  FR   ghc A  1000 3   9.81 2   0.25 m  0.5m  4.0 m      m  s   kg  m   4910 N  4.91 kN The net horizontal force is F  FL  FR  19.6 kN  4.91kN F  14.7 kN, acting to right. D  0.5m by 4.0 m 2 Problem 2.128 An open tank containing water has a bulge in its vertical side that is semicircular in shape as shown in the figure below. Determine the horizontal and vertical components of the force that the water exerts on the bulge. Base your analysis on a 1-ft length of the bulge. Solution 2.128 FH horizontal force of wall on fluid FV vertical force of wall on fluid W   H 2OVvol 2 lb     3ft      1ft    62.4 3   2 ft      882 lb lb   F1   hc A   62.4 3   6ft  3ft  6ft  1ft   3370lb ft   For equilibrium, FV  W  882 lb  and FH  F1  3370 lb  The force the water exerts on the bulge is equal to, but opposite in direction to FV and FH above. Thus,  FH wall  3370 lb   FV wall  882 lb  Problem 2.129 A closed tank is filled with water and has a 4-ft-diameter hemispherical dome as shown in the figure below. A U-tube manometer is connected to the tank. Determine the vertical force of the water on the dome if the differential manometer reading is 7 ft and the air pressure at the upper end of the manometer is 12.6 psi. Solution 2.129 For equilibrium,  Fvertical  0 So that FD  pA  W Eq. (1) Where FD force exerted by dome on the fluid and p is water pressure at the dome base. From the manometer, p A   gf  7 ft    H 2O  4ft   p lb   in.2  lb  lb     p  12.6 2  144 2    3.0   62.4 3   7 ft    62.4 3   4 ft    in.   ft  ft  ft      2880 lb ft 2 Thus, from Eq.(1) with volume of sphere   6  diameter 3 lb     lb   2 1  3 FD   2880 2     4ft     4ft    62.4 3   35100lb 26 ft   4  ft    The vertical force that the water exerts on the dome is 35100lb . Problem 2.130 A 3-m-diameter open cylindrical tank contains water and has a hemispherical bottom as shown in the figure below. Determine the magnitude, line of action, and direction of the force of the water on the curved bottom. Solution 2.130 Force  weight of water supported by hemispherical bottom   H 2O  volume of cylinder    volume of hemisphere    9.80  kN    3m 2  8 m    3m 3  3 4 12 m    485 kN The force is directed vertically downward, and due to symmetry it acts on the hemisphere along the vertical axis of the cylinder. Problem 2.131 Three gates of negligible weight are used to hold back water in a channel of width b as shown in the figure below. The force of the gate against the block for gate (b) is R. Determine (in terms of R) the force against the blocks for the other two gates. Solution 2.131 For case (b) 2 h b h FR   hc A      h  b   2 2 and yR  2 h 3 Thus, MH  0 So that 2  h 2b 2   2   h b  hR   h  FR   h    R  3 3   3   2  R  h 2b 3 (1) For case (a) on free-body-diagram shown FR  yR  and  h 2b 2 2 h 3 (from above) and 4h (see the figure below) 3 2   h 2       h2b 2 W    V o       b    4  16     Thus,  M H  0 So that  h 4h  2  W    FR  h   FB h  2 6  3   h2b  h 4h   h2b  2      h   FB h 16  2 6  2 3  FB   h2b(0.390)  3R(0.390)  FB  1.17 R For case (c), for the free-body-diagram shown, the force FR1 on the curved section passes through the hinge and therefore does not contribute to the moment around H. On bottom part of gate  3h   h  3 FR2   hc A       b    h 2b  4  2  8 and 3 1 h b    I 12  2   3h yR2  xc  yc  yc A  3h  h  4    b   4  2  28  h 36 Thus, MH  0 So that  28  FR2  h   FB h  36  or 3   28  7 FB    h 2b      h 2b 8   36  24 From Eq.(1)  h 2b  3R , thus FB  7 R  0.875 R 8 Problem 2.133 An iceberg (specific gravity 0.917 ) floats in the ocean (specific gravity 1.025 ). What percent of the volume of the iceberg is under water? Solution 2.133 For equilibrium, W  weight of iceberg  FB  buoyant force or V ice ice  V sub ocean , where V sub  volume of ice submerged . Thus,  SGice V sub 0.917  ice    0.895  89.5% V ice  ocean SGocean 1.025 Problem 2.134 A floating 40-in.- thick piece of ice sinks 1 in. with a 500-lb polar bear in the center of the ice. What is the area of the ice in the plane of the water level? For seawater, S  1.03 . Solution 2.134 Without the polar bear on the ice, the submerged depth d of the ice is found by equating the weight of the ice and the buoyant force. Denoting the pure water specific weight by  and the ice area by A gives FB  Wice or Wice   SAd . The ice sinks an additional depth d  with the bear in the center of the ice. Equating the new buoyant force to the weight of the ice plus bear gives FB  Wice  Wbear ,  SA  d  d     SAd  Wbear , or A Wbear 500 lb  or A  93.4 ft 2 lb   Sd   1   62.4 3  1.03  12 ft    ft   Problem 2.135 A spherical balloon filled with helium at 40°F and 20 psia has a 25  ft diameter. What load can it support in atmospheric air at 40°F and 14.696 psia ? Neglect the balloon’s weight. Solution 2.135 For static equilibrium, the buoyant force must equal the load. Neglecting the weight of the balloon and assuming air and helium to be ideal gases, the load is L  FB    air   He  V    air   He  g V  p   p    g  4            R3    R air  R  He   T  3 Using Table A.4, the numerical values give  ft  4  3 lb lb    20 144  2   32.2 2   3  12.5ft   14.696 144  2  sec   ft  ft   L   ft lb ft lb 32.2 ft  lbm        53.35  500 o R 386   o   o    lbm  R   lbm  R    lb  sec2   or L  527 lb  Problem 2.136 A river barge, whose cross section is approximately rectangular, carries a load of grain. The barge is 28 ft wide and 90 ft long. When unloaded, its draft (depth of submergence) is 5 ft and with the load of grain the draft is 7 ft . Determine: (a) the unloaded weight of the barge, and (b) the weight of the grain. Solution 2.136 (a) For equilibrium,  Fvertical  0 So that Wb  FB   H 2O  (submerged volume) lb )  5ft  28ft  90 ft  ft 3  786000 lb  (62.4 (b)  Fvertical  0 WB  Wg  FB   H 2O  (submerged volume) lb )  7 ft  28ft  90 ft   786, 000 lb ft 3  315000 lb Wg  (62.4 Problem 2.137 A barge is 40 ft wide by 120 ft long. The weight of the barge and its cargo is denoted by W .   When in salt-free riverwater, it floats 0.25 ft deeper than when in seawater   64 lb/ft 3 . Find the weight W . Solution 2.137 In both cases, the weight W must equal the weight of the displaced water or W   SFW A  d  0.25ft    SW Ad Soling for d gives  SW Ad   SFW A  d  0.25ft  or  0.25ft   62.4 3  lb  0.25ft   SFW  ft   9.75ft .  d lb  SW   SFW  64.0  62.4  3 ft Then lb   W   SW Ad   64.0 3   40 120  ft 2  9.75ft  ft    short ton  W  3.00 106 lb  ,  2000 lb  or W  1500 short tons. Problem 2.138 When the Tucurui Dam was constructed in northern Brazil, the lake that was created covered a large forest of valuable hardwood trees. It was found that even after 15 years underwater the trees were perfectly preserved and underwater logging was started. During the logging process a tree is selected, trimmed, and anchored with ropes to prevent it from shooting to the surface like a missile when cut. Assume that a typical large tree can be approximated as a truncated cone with a base diameter of 8 ft , a top diameter of 2 ft , and a height of 100 ft . Determine the resultant vertical force that the ropes must resist when the completely submerged tree is cut. The specific gravity of the wood is approximately 0.6 . Solution 2.138 W weight , FB buoyant force , T tension in ropes For equilibrium,  Fvertical  0 T  FB  W For a truncated cone, Volume  h r12  r1r2  r22   3 where: r1  base radius r2  top radius h  height Thus, V tree   100ft   4ft 2  4ft  1ft  1ft 2   2200ft 3         3 For buoyant force, FB   H 2O  V tree  (62.4 lb ft 3   ) 2200 ft 3  137000 lb For weight,   lb   W   tree  V tree   0.6   62.4 3  2200 ft 3  82400 lb ft   Therfore, T  137000 lb  82400 lb  54600 lb Problem 2.140 An inverted test tube partially filled with air floats in a plastic water-filled soft drink bottle as shown in the figure below. The amount of air in the tube has been adjusted so that it just floats. The bottle cap is securely fastened. A slight squeezing of the plastic bottle will cause the test tube to sink to the bottom of the bottle. Explain this phenomenon. Solution 2.140 Where the test tube is floating the weight of the tube, W , is balanced by the buoyant force, FB , as shown in the figure. The buoyant force is due to the displaced volume of water as shown. This displaced volume is due to the air pressure, p , trapped in the tube where p  po   H 2O h. When the bottle is squeezed, the air pressure in the bottle, po , is increased slightly and this in turn increases p , the pressure compressing the air in the test tube. Thus, the displaced volume is decreased with a subsequent decrease in FB . Since W is constant, a decrease in FB will cause the test tube to sink. Problem 2.141 A child’s balloon is a sphere 1 ft. in diameter. The balloon is filled with helium (  = 0.014 lbm/ft 3 ). The balloon material weighs 0.008 lbf/ft 2 of surface area. If the child releases the balloon, how high will it rise in the Standard Atmosphere. (Neglect expansion of the balloon as it rises.) Solution 2.141 A force balance in the vertical direction for the balloon gives    Fz  0  air g V   He g V  wA for the balloon at rest at its highest elevation. Then  air   He g V  wA gV  2  w D wA   He    He  gV   D3  g  6    lb  ft  lbm   6  0.008 2  32.2  6w lbm ft  lb  sec2    He   0.014 3   ft  gD  ft  32.2 2  1.0 ft  sec   lbm  0.062 3 . ft Interpolating Table A.2 for the Standard Atmosphere,  0.06590  0.062  z  elevation=5000ft  5000ft     0.06590  0.05648  z  7,070ft Problem 2.142 A 1-ft-diameter, 2-ft-long cylinder floats in an open tank containing a liquid having a specific weight γ. A U-tube manometer is connected to the tank as shown in the figure below. When the pressure in pipe A is 0.1 psi below atmospheric pressure, the various fluid levels are as shown. Determine the weight of the cylinder. Note that the top of the cylinder is flush with the fluid surface. Solution 2.142 From a free-body-diagram of the cylinder  Fvertical  0 So that 2   W  FB     1ft   2 ft  4  (1)  2 A manometer equation gives,     3.5 ft    SG   H O  2.5 ft    H O 1ft   p A   2   3.5ft   1.5   62.4   80.6 2 lb  lb  lb   in.2    2.5ft 62.4 1ft 0.1 144               ft 3  ft 3  in.2   ft 2    lb ft 3 lb    Therfore, W   ft 3   80.6 3   127 lb ft   2  Problem 2.143 A not-too-honest citizen is thinking of making bogus gold bars by first making a hollow iridium  S  22.5 ingot and plating it with a thin layer of gold  S  19.3 of negligible weight and volume. The bogus bar is to have a mass of 100 lbm . What must be the volumes of the bogus bar and of the air space inside the iridium so that an inspector would conclude it was real gold after weighing it in air and water to determine its density? Could lead  S  11.35  or platinum  S  21.45 be used instead of iridium? Would either be a good idea? Solution 2.143 S x  22.5  iridium  ; SG  19.3  gold  ; V BB  V x  V AS ; mBB  mx  100lbm Neglect the weight of air in the air space and the buoyant force of air on the bar. The volume of a pure gold bar would be V GB  WGB . G The bogus bar must have the same volume and weight as the pure gold bar so it will weigh like a solid gold bar in water. The volume condition gives V GB  V BB  V AS  Vx . Since WGB  Wx , V AS  V x  V GB  WGB G  Wx G , V AS  V x  xV x , G S    V AS  V x  x  1  V x  x  1 .  G   SG  The numerical value of the iridium volume is Vx Wx x  WGB x  100 lb lb    22.5  62.4 3  ft    0.0712 ft 3 .  22.5   1 or The air space volume is V AS  0.0712 ft 3   19.3  V AS  0.0118ft 3 . The bogus bar volume is V BB  V AS  V x   0.0118  0.0712  ft 3 or V BB  0.0830 ft 3 . Lead will not work because it is less dense than gold. Platinum could work because it is more dense than gold. However, platinum is more expensive per unit weight than gold, so it would be a foolish choice. Problem 2.144 A solid cylindrical pine  S = 0.50  spar buoy has a cylindrical lead  S = 11.3 weight attached, as shown in the figure below. Determine the equilibrium position of the spar buoy in seawater (i.e., find d ). Is this spar buoy stable or unstable? For seawater, S = 1.03 . Solution 2.144 The equilibrium position is found by equating the buoyant force and the body weight. FB  W  swdA      A   p  p A d       p p  sw  S    S p  p S sw 11.3  0.5ft   0.50 16ft  1.03  d  13.3ft Since d  13.8 ft (the total length of the spar buoy), the spar buoy floats. We now have to check the stability of the buoy. I  4   radius 4  1ft 4  0.7854 ft 4 , 4  c  distance from bottom of buoy to center of gravity of buoy         A        p   p A p     cW   cpWcp  2  2    c  W  W p   A    p A p   p     S       S p  p      2  11.3  0.5  0.25   0.5 16  0.5  8   2    ft S    S p  p 11.3  0.5   0.5 16   c =5.09 ft d d 13.3ft   6.65ft , n   c   5.09 ft  6.65ft  1.56 ft , 2 2 2 m V s  Ad   1ft  13.3ft   41.8ft 3 , 2 I 0.7854ft 4 n   1.56ft   1.58ft Vs 41.8ft 3 m0  buoy is stable Problem 2.145 When a hydrometer (see the figure below) having a stem diameter of 0.30 in. is placed in water, the stem protrudes 3.15 in. above the water surface. If the water is replaced with a liquid having a specific gravity of 1.10 , how much of the stem would protrude above the liquid surface? The hydrometer weighs 0.042 lb . Solution 2.145 When the hydrometer is floating its weight, W , is balanced by the buoyant force, FB .,  Fvertical  FB  W  0  H O  1  W   SG   H O  2 2 2  2 1 SG For water, V1  W H O 2  0.042 lb  6.73 104 ft 3 lb 62.4 3 ft For other liquid, V2  6.73  104 ft 3  6.12 104 ft 3 1.10 Therefore, V1  V2   6.73  6.12   104 ft 3  0.61 104 ft 3 The change in submergence depth occurs with only the stem protruding from the surface. in.3  2   4 3      0.30in. 0.61 10 ft 1728        4 ft 3       1.49in. With the new liquid the stem would protrude 3.15in.+1.49in.  4.64in. above surface Problem 2.146 A 2-ft-thick block constructed of wood  SG  0.6  is submerged in oil  SG  0.8 and has a 2-ft-thick aluminum (specific weight = 168lb / ft 3 ) plate attached to the bottom as indicated in the figure below. Determine completely the force required to hold the block in the position shown. Locate the force with respect to point A. Solution 2.146 Equilibrium:  Fvertical  F  Ww  FBw  Wa  FBa  0   Ww   SGw   H 2O Vw lb  1     0.6   62.4 3    10 ft  4 ft  2 ft   1500 lb ft   2   lb   Wa   168 3   0.5ft  10 ft  2 ft   1680 lb ft     FBw   SGoil   H 2 O Vw lb  1     0.8   62.4 3    10ft  4ft  2ft   2000lb ft   2   lb   FBa   SGoil   H 2O Va   0.8   62.4 3   0.5ft  10 ft  2 ft   499 lb ft     Therefore, F  1500 lb  2000 lb+1680 lb  499 lb  F  681 lb upward Equilibrium:  10   M a  0  F   3 ft  Ww  FBw    5ft  Wa  FBa   10    681lb    ft  1500lb  2000lb    5ft 1680lb  499lb   6.22 ft  3  F acts 6.22 ft to the right of point A Problem 2.147 How much extra water does a 147  lb concrete canoe displace compared to an ultralightweight 38  lb Kevlar canoe of the same size carrying the same load? Solution 2.147 For equilibrium,  Fvertical  0 and W  FB   H 2O V and V is displaced volume. For concrete canoe, lb   147 lb   62.4 3  Vc ft   Vc  2.36ft 3 For Kevlar canoe, lb   38lb   62.4 3  Vk ft   Vk  0.609ft 3 Extra water displacement  2.36 ft 3  0.609 ft 3  1.75ft 3 Problem 2 .148 A submarine is modeled as a cylinder with a length of 300 ft , a diameter of 50 ft , and a conning tower as shown in the figure below. The submarine can dive a distance of 50 ft from the floating position in about 30 sec . Diving is accomplished by taking water into the ballast tank so the submarine will sink. When the submarine reaches the desired depth, some of the water in the ballast tank is discharged leaving the submarine in “neutral buoyancy” (i.e., it will neither rise nor sink). For the conditions illustrated, find (a) the weight of the submarine and (b) the volume (or mass) of the water that must be in the ballast tank when the submarine is in neutral buoyancy. For seawater, S = 1.03 . Solution 2.148 (a) Denoting the cylinder radius by R , the submarine weight is equal to the buoyant force so W  FB   V submerged      R 2 1.03 when the submarine is in the partially submerged position. The numerical values give  lb  2 W   64 3    25ft   300 ft 1.03 or  ft  W  3.88  107 lb (b) For neutral buoyancy at the lower depth, the submarine weight W plus the ballast weight WB must equal the buoyant force so   W  WB  FB    R 2 1.10  or   WB    R 2 1.10   W . The ballast volume V B    V B   R 2  1.10   V B  41700 ft 3 W  WB  so    25ft   300 ft 1.10   2 3.88  107 lb  lb   64 3   ft  Problem 2.150 When an automobile brakes, the fuel gage indicates a fuller tank than when the automobile is traveling at a constant speed on a level road. Is the sensor for the fuel gage located near the front or rear of the fuel tank? Assume a constant deceleration. Solution 2.150 accelerating automobile so sensor located in front of fuel tank. decelerating  braking  automobile Problem 2.151 An open container of oil rests on the flatbed of a truck that is traveling along a horizontal road at 55 mi / hr . As the truck slows uniformly to a complete stop in 5 s , what will be the slope of the oil surface during the period of constant deceleration? Solution 2.151 slope  ay  ay dz  g  az dy final velocity – initial velocity time interval m     0   55 mph   0.4470 s  mph      4.92 m  5s s2 Thus, m  4.92 2   dz s   0.502  m dy 9.81 2  0 s Problem 2.152 A 5-gal, cylindrical open container with a bottom area of 120 in.2 is filled with glycerin and rests on the floor of an elevator. (a) Determine the fluid pressure at the bottom of the container when the elevator has an upward acceleration of 3 ft / s2 . (b) What resultant force does the container exert on the floor of the elevator during this acceleration? The weight of the container is negligible. (Note: 1 gal  231 in.3 ) Solution 2.152 h A  volume  231in.3  h 120in.2   5gal    gal      h  9.63in. p     g  az  z (a) Thus, 0 pb 0 dp     g  az h dz and pb    g  a z  h slugs   ft ft   9.63   ft    2.44 3   32.2 2  3 2   ft   s s   12   lb  68.9 2 ft (b) From free-body-diagram of container, F f  pb A  1ft 2  lb     68.9 2  120in.2   144 in.2  ft        57.4 lb Thus, force of container on floor is 57.4 lb downward . Problem 2.153 A plastic glass has a square cross section measuring 2½ in. on a side and is filled to within ½ in. of the top with water. The glass is placed in a level spot in a car with two opposite sides parallel to the direction of travel. How fast can the driver of the car accelerate along a level road without spilling any of the water? Solution 2.153 Slope of water surface a   car g or acar   g  slope  ft   1.0in.      32.2 2     sec   2.5in.   or acar  12.9 ft sec2 Problem 2.154 The cylinder in the figure below accelerates to the left at the rate of 9.80 m/s 2 . Find the tension in the string connecting at rod of circular cross section to the cylinder. The volume between the rod and the cylinder is completely filled with water at 10°C . Solution 2.154 First find the pressure difference in the water over a length   8.0 cm . Since gravity is perpendicular to the rod, Eq.(2.41) gives dp    a x d x For the x-direction. Integrating gives p2  p1    ax  x2  x1  . For 10°C water, Table A.5 gives  m  kg   m N  p2  p1   1000 3  9.80 2   8.0 cm     784 2  m  s  m  100 cm  We next apply Newton’s second law to the rod    Fx  ma x , T   p1  p2  A  ma x , T   p2  p1  A  ma x . Using the specified information,   D2   m   w Srod A   w S rod   4    3  1m  kg     2  1000 3   2.0    1.0cm   8.0cm     0.0126 kg m   4  100 cm   D2  2 2 m  5 2  1.0 cm   A   7.854 10 m . 4 4 100 cm   Therefore,   N  m   T   784 2  7.854  105 m 2   0.0126 kg   9.80 2  m  s    T  0.062 N Problem 2.155 A closed cylindrical tank that is 8 ft in diameter and 24 ft long is completely filled with gasoline. The tank, with its long axis horizontal, is pulled by a truck along a horizontal surface. Determine the pressure difference between the ends (along the long axis of the tank) when the truck undergoes an acceleration of 5ft / s 2 . Solution 2.155 p  ay y Thus, 24 p2  p dp    a y 0 dy 1 Where p  p1 at y  0 and p  p2 at y  24 ft , and p2  p1    a y  24 ft  slugs  ft     1.32 3  5 2   24 ft  ft  s   lb  158 2 ft or p1  p2  158 lb ft 2 Problem 2.156 The cart shown in the figure below measures 10.0 cm long and 6.0 cm high and has rectangular cross sections. It is half-filled with water and accelerates down a 20 incline plane at a  1.0 m/s 2 . Find the height h . Solution 2.156 Unfortunately, there are 2 x-directions in the problem statement. Noting that the gravisty vector is in the negative z-direction, change the label on the axis normal to the z-direction to be “n”. Resolving the acceleration along the plane into n,z components: a z  a sin  , an  a cos  ,   20o For rigid-body motion of the fluid in the n,z coordiantes:: dp    an dn    g  a z  dz dp    a cos  dn    g  a sin   dz  0  along free surface p  patm Using trignometirc relationships this equation can be converted into x,y coordinates. dn  dx cos   dy sin  dz  dy cos   dx sin    a cos  dn    g  a sin   dz  0   a cos   dx cos   dy sin      g  a sin    dy cos   dx sin    0     a cos 2  dx    a cos  sin   dy     g  a sin   cos   dy     g  a sin    sin    dx  0    a cos 2     g  a sin   sin   dx     a cos  sin     g  a sin   cos   dy  0          a cos 2   sin 2    g sin   dx     g cos   dy  0      a   g sin   dx    g cos   dy  0  a  g sin   dx   g cos   dy  0 Integration yields:  a  g sin   x   g cos   y  C  a sin   y    xC  g cos  cos   The constant of integration can be determined by noting that the container is ½-full:  water   y dx 0    sin   a      x  C dx  0  g cos  cos      2 a    tan    C   g cos   2 C  water  a   tan     g cos  2  m     s2    1 2  10.0 cm  10.0 cm  6.0 cm  / 2   s  tan 20o  C 2 10.0 cm    9.81 m  cos 20o   1.723 cm Solving for the the requested length:   a y    tan   x  C  g cos     1 o h tan 20   10 cm   1.723 cm  4.277 cm   9.81 cos 20o    h  4.28 cm Problem 2.157 The U-tube manometer in the figure below is used to measure the acceleration of the cart on which it sits. Develop an expression for the acceleration of the cart in terms of the liquid height h , the liquid density  , the local acceleration of gravity g , and the length  . Solution 2.157 Writing Newton’s second law in the horizontal direction (x-direction) for the bottom leg of the manometer gives  Fx  ma x , p A  pr A   Aa , or a p  pr  Applying the manometer rule to the two legs of the manometer gives p  patm   gh and pr  patm   gh Subtracting gives p  pr   g  h  hr    gh so a  gh  or h a  g   Problem 2.158 A tank has a height of 5.0 cm and a square cross section measuring 5.0 cm on a side. The tank is one third full of water and is rotated in a horizontal plane with the bottom of the tank 100 cm from the center of rotation and two opposite sides parallel to the ground. What is the maximum rotational speed that the tank of water can be rotated with no water coming out of the tank? Solution 2.158 dp    gdz   2 r dr Since dp  0 along the free surface, the free surface is identified by the equation 0    gdz   2 r dr or 0   gdz   2 r dr Integrating gives z r  r1 0   g  b dz   2  r dr , 2 b  2 2 2  0  g  z    r  r1 , 2 2    or b 2 2 2 z  r  r1 . 2 2g   Recognizing that the volume of water in the rotating tank must equal b2h gives 6  r1  h r1  h  b  2 b2h    zb dr  b  r 2  r12  dr ,  r1 r1 6  2 2 g   r h  br  2  r 3 2   1 b2h  b     r1 r   6  2 2 g  3   r1 ,   bh  2   r  h 3 r 3  b2h  1  b    1  r12 h   ,  6 3 3  2 2g     3  2bh  2   r1  h  r13     r12 h  ,  3 2g  3 3   or  4bhg   r  h 3 r 3   1  r12 h  3 1   3 3   . The numerical values give  cm   4  5cm  5cm   981 2  s    100 cm 3  95cm 3  2    95cm   5cm   3   3 3   rad   rev   60s     3.68    or s   2 rad   min     35.1rpm DISCUSSION Note the that when r  r1  h ,   b 2 b 2 z  2r1h  h2 .  r1  h 2  r12    2 2g 2 2g   The numerical values give 2 rad   3.68   5 s   2 z  cm   2  95cm  5cm    5cm     cm  2  2  981 2  s    4.23cm or the assumption indicated in the above figures that the water does not reach the uppermost side of the tank is correct. Problem 2.159 An open 1-m-diameter tank contains water at a depth of 0.7 m when at rest. As the tank is rotated about its vertical axis the center of the fluid surface is depressed. At what angular velocity will the bottom of the tank first be exposed? No water is spilled from the tank. Solution 2.159 Equation for surfaces of constant pressure: z  2r 2 2g  constant For free surface with h  0 at r  0 , h  2r 2 2g The volume of fluid in rotating tank is given by R V f   2 rh dr  0 2 2 R 3  2 R 4  r dr 2 g 0 4g Since the initial volume, Vi   R2hi , must equal the final volume, V f  Vi So that  2 R 4 4g   R 2 hi or m  4  9.81 2   0.7 m  4 ghi rad s     10.5  2 2 s R  0.5 m  Problem 2.160 The U-tube in the figure below rotates at 2.0 rev/sec. Find the absolute pressures at points C and B if the atmospheric pressure is 14.696 psia . Recall that 70 °F water evaporates at an absolute pressure of 0.363 psia . Determine the absolute pressures at points C and B if the U-tube rotates at 2.0 rev/sec . Solution 2.160 Applying the manometer rule to one of the legs and using the data in Table A.6, pB  patm + gh lb    62.3 3  1in. ft  , pB  14.696 psia    in.3   1728 3  ft   pB  14.732 psia Section 2.6.2 gives: p   r 2 . r Integrating from r  0 to r  R gives:  pB pC R dp   2  r dr or o pB  pC   2 R 2 2 . Therefore, 2 lbm  rev  2  62.3 3   2.0 2 2   2.5ft    R sec ft     pc  pB   14.732 psia 2 2 2  144in.   rev   32.2ft  lbm  2  ft 2   2 rad   lb  sec2    pc  8.10 psia Check for phase change: pc  0.363psia  no evaporation  above answer is correct. DISCUSSION Note that if pc were calculated to be less than 0.363psia , some of the water would vaporize and pc would be 0.363 psia . Problem 2.161 A child riding in a car holds a string attached to a floating, helium-filled balloon. As the car decelerates to a stop, the balloon tilts backwards. As the car makes a right-hand turn, the balloon tilts to the right. On the other hand, the child tends to be forced forward as the car decelerates and to the left as the car makes a right-hand turn. Explain these observed effects on the balloon and child. Solution 2.161 A floating balloon attached to a string will align itself so that the string it normal to lines of constant pressure. Thus, if the car is not accelerating, the lines of p  constant are horizontal (gravity acts vertically down), and the balloon floats “straight up” (i.e.   0 ). If forced to the side (   0 ), the balloon will return to the vertical (   0 equilibrium position in which the two forces T and FB -W line up. Consider what happens when the car decelerates with an amount a y  0 . As show by the equation, slope  ay dz  , g  az dy the lines of constant pressure are not horizontal, ay ay dz   0 dy g  az g since a z  0 and ay  0 . Again, the balloon’s equilibrium position is with the string normal to p  const. lines. That is, the balloon tilts back as the car stops. V2 (the centrifugal acceleration), the lines of p  const. are as shown, R and the balloon tilts to the outside of the curve. When the car turns, a y  Problem 2.162 A closed, 0.4-m-diameter cylindrical tank is completely filled with oil  SG  0.9  and rotates about its vertical longitudinal axis with an angular velocity of 40 rad / s . Determine the difference in pressure just under the vessel cover between a point on the circumference and a point on the axis. Solution 2.162 Pressure in a rotating fluid varies in accordance with the equation, p  2 r 2 2   z  constant Since z A  z B , pB  p A   2 rB2  rA2   2 2 kg  rad   (0.9)  103 3  40  s    m    0.2 m 2  0  2  28.8 kPa Problem 2.163 The largest liquid mirror telescope uses a 6-ft-diameter tank of mercury rotating at 7 rpm to produce its parabolic-shaped mirror as shown in the figure below. Determine the difference in elevation of the mercury, h ,between the edge and the center of the mirror. Solution 2.163 For free surface of rotating liquid, z  2r 2 2g  constant Let z  0 at r  0 and therefore constant  0 . Thus, h  z for r  3 ft and with   rad  0.733 s    7 rpm   2 rad   1min    rev   60s  It follows that 2 rad  2   0.733   3ft  s  h    0.0751ft ft   2  32.2 2  s   Problem 2.101 Find the total vertical force on the cylinder shown in the figure below. Solution 2.101 The net force F on the cylinder is due to the water and is F  F1  F2  p1 A1  p2 A2 . Since the atmospheric pressure does not contribute to the net force, p1 and p2 will be considered gage pressures. kg   m   1000 3  9.81 2  18  5  cm N m  s   1275 2 p1   w gh   cm   kg  m   m  100   2 m  N  s   kg   m   1000 3  9.81 2   3 cm N m  s   294 2 p2   w gh   cm   kg  m   m 100   m  N  s 2   Then 2 N  m   N  2 2 m  2  2 F   1275 2   3cm      294 2  6  3 cm    m 4 m 4  100 cm    100 cm  or F  1.52 N .   2 Problem 2.102 A 3-m -wide, 8-m -high rectangular gate is located at the end of a rectangular passage that is connected to a large open tank filled with water as shown in the figure below. The gate is hinged at its bottom and held closed by a horizontal force, FH , located at the center of the gate. The maximum value for FH is 3500 kN . (a) Determine the maximum water depth, h, above the center of the gate that can exist without the gate opening. (b) Is the answer the same if the gate is hinged at the top? Explain your answer. Solution 2.102 For gate hinged at bottom MH  0 so that  4 m  FH   FH (see figure) (1) and kN   FR   hc A   9.80 3   h  3m  8 m    9.80  24h  kN  m  1  3m 8 m 3 I xc 5.33 yR   yc  12 h h yc A h  3m  8 m  h 5.33  5.33   h  4  Thus,   m   h  4   h  h  and from Eq.(1)  4 m  3500 kN    4   so that h  16.2 m 5.33    9.80  24  h  kN h  For gate hinged at top MH  0 so that  4 m  FH  1 FH (see figure) (1) where 5.33  5.33  1  yR   h  4     4    h  4  4 h  h  Thus, from Eq.(1)  4 m  3500 kN    5.33   4   9.80  24  h  kN  h  and h  13.5 m Maximum depth for gate hinged at top is less than maximum depth for gate hinged at bottom. Problem 2.103 A gate having the cross section shown in the figure below is 4 ft wide and is hinged at C. The gate weighs 18, 000 lb , and its mass center is 1.67 ft to the right of the plane BC. Determine the vertical reaction at A on the gate when the water level is 3 ft above the base. All contact surfaces are smooth. Solution 2.103 F1   hc A where hc  1.5ft lb   Thus, F1   62.4 3  1.5ft  3ft  4 ft   1120 lb ft   The force F1 acts at a distance of 1ft from the base of the gate. F2  p2 A2 where p2   H2O  3ft  Thus, lb   F2   62.4 3   3ft  5ft  4 ft   3740 lb ft   and acts at the center of the bottom gate surface. For equilibrium,  Mc  0 and F1 11ft   F2  2.5ft   FA  5ft   W 1.67 ft  so that FA  18, 000lb 1.67 ft   1120lb 11ft    3740lb  2.5ft   1680lb 5ft Problem 2.104 The massless, 4-ft -wide gate shown in the figure below pivots about the frictionless hinge O. It is held in place by the 2000 lb counterweight, W . Determine the water depth, h . Solution 2.104 FR   hc A where hc  h 2 Thus, FR   H 2O h h2 h b      H2O  4 ft  2 2 To locate FR ,   1  4 ft  h3 h 2 I xc yR   yc  12   h h yc A 4 ft  h   2 3 2 For equilibrium,  M 0  0 FR d  W  3ft  where d  h  yR  so that Thus, h  3  2000 lb  3ft   h3    h2    4 ft   2   H O  2  3 2000 lb  3ft  lb   1    62.4 3   2   4 ft  ft     h  5.24 ft h 3 Problem 2.105 A 200  lb homogeneous gate 10 ft wide and 5 ft long is hinged at point A and held in place by a 12-ft -long brace as shown in the figure below. As the bottom of the brace is moved to the right, the water level remains at the top of the gate. The line of action of the force that the brace exerts on the gate is along the brace. (a) Plot the magnitude of the force exerted on the gate by the brace as a function of the angle of the gate,  , for 0   90 . (b) Repeat the calculations for the case in which the weight of the gate is negligible. Comment on the result as   0 . Solution 2.105 (a) For the free-body-diagram of the gate (see figure),  FA  0 so that    FR    W  cos     FB cos    sin     FB sin    cos   3 2  (1) Also,  sin   L sin  (assuming hinge and end of brace at same elevation) or sin    sin  L and   sin   FR   hc A      w   2  where w is the gate width. Thus, Eq.(1) can be written as  3  W cos   FB   cos  sin   sin  cos     sin   w  2  6   so that   2 w    2 w  W W   sin   cos    tan   6  2 6  2  FB   (2) cos  sin   sin  cos  cos  tan   sin  For   62.4 lb ft 3 ,   5 ft , w  10 ft , and W  200 lb , lb   2  62.4 3   5ft  10 ft  200 lb ft   tan   6 2  2600 tan   100 FB  cos  tan   sin  cos  tan   sin  Since sin   sin   (3)  sin  and   5 ft , L  12 ft L 5 sin  12 and for a given  ,  can be determined. Thus, Eq.(3) can be used to determine FB for a given  . (b) For W  0 , Eq.(3) reduces to FB  2600 tan  cos  tan   sin  (4) and Eq.(4) can be used to determine FB for a given  . Tabulated data of FB vs.  for both W  200 lb and W  0 lb are given below. θ, deg 90.0 85.0 80.0 75.0 70.0 65.0 60.0 55.0 50.0 45.0 40.0 35.0 30.0 25.0 20.0 15.0 10.0 5.0 2.0 F(B), lb (W=200 lb) 2843 2745 2651 2563 2480 2403 2332 2269 2213 2165 2125 2094 2075 2069 2083 2130 2250 2646 3858 F(B), lb (W=0 lb) 2843 2736 2633 2536 2445 2360 2282 2210 2144 2085 2032 1985 1945 1911 1884 1863 1847 1838 1836 As   0 the value of FB can be determined from Eq.(4), FB  2600 tan  cos  tan   sin  Since sin   5 sin  12 it follows that 2  5 cos   1  sin   1    sin 2   12  2 and therefore FB  2600 tan  2 5  5 1    sin 2  tan   sin  12  12  2600  2 5  5 1    sin 2   cos  12  12  Thus, as   0 FB  2600  1840 lb 5 1 12 Physically this result means that for   0 , the value of FB is indeterminate, but for any “very small” value of  , FB will approach 1840 lb . Problem 2.106 An open tank has a vertical partition and on one side contains gasoline with a density   700 kg / m3 at a depth of 4 m , as shown in the figure below. A rectangular gate that is 4 m high and 2 m wide and hinged at one end is located in the partition. Water is slowly added to the empty side of the tank. At what depth, h , will the gate start to open? Solution 2.106 FRg   g hcg Ag ; where g refers to gasoline. kg   m  FRg   700 3   9.81 2   2 m  4 m  2 m  m  s    110  103 N=110 kN FRw   whcw Aw ; where w refers to water. N  h   FRw   9.80 103 3    2 m  h  ; where h is depth of water. m  2     FRw  9.80  103 h 2 For equilibrium,  M H  0  FRw w  FRg  g w  h 4 ; g  m 3 3 9.80 103  h2   h3   110 103 N   34 m  h  3.55 m which is the limiting value for h. Problem 2.107 A horizontal 2-m-diameter conduit is half filled with a liquid ( SG  1.6 ) and is capped at both ends with plane vertical surfaces. The air pressure in the conduit above the liquid surface is 200 kPa . Determine the resultant force of the fluid acting on one of the end caps, and locate this force relative to the bottom of the conduit. Solution 2.107 Fair  pA, where p is air pressure Thus, N     2 Fair   200 103 2     2 m   200 103 N  m  4  Fliquid   hc A2 where hc  4R 3 (see the figure below) Thus, N   4 1m    1      2 3 Fliquid  1.6   9.81 103 3         2 m   10.5  10 N  m   3   2   4  For Fliquid , yR  I xc  yc yc A2 where I xc  0.1098R4 and yc  hc  4R 3 (see the figure below) Thus, yR  0.1098 1m  4  4 1m    1    2       2 m   3   2  4   4 1m   0.5891m 3 Since Fresultant  Fair  Fliquid   200  10.5  103 N=639 kN, we can sum moments about O to locate resultant to obtain Fresultant  d   Fair 1m   Fliquid 1m  0.5891m  So that 200  103 N  1m   10.5  103 N   0.4109 m   d 639  103 N  0.990 m above bottom of conduit. Problem 2.108 A 4-ft by 3-ft massless rectangular gate is used to close the end of the water tank shown in the figure below. A 200-lb weight attached to the arm of the gate at a distance ℓ from the frictionless hinge is just sufficient to keep the gate closed when the water depth is 2 ft , that is, when the water fills the semicircular lower portion of the tank. If the water were deeper, the gate would open. Determine the distance . Solution 2.108 FR   hc A where hc  4R 3 (see the figure below) Thus,  4R    R FR   H 2O    3   2 2   2 lb   4  2 ft      2 ft       333lb   62.4 3     2 ft   3     To locate FR , yR   I xc  yc yc A 0.1098 R 4 4R  2  4 R    R  3     3   2  (see the figure below) 4 0.1098  2 ft  4  2 ft      1.178ft 2 3  4  2 ft     2 ft    3   2 For equilibrium, MH  0 So that W   FR 1ft+y R  And   333lb 1ft  1.178ft   3.63ft 200lb Problem 2.109 A thin 4-ft-wide, right-angle gate with negligible mass is free to pivot about a frictionless hinge at point O, as shown in the figure below. The horizontal portion of the gate covers a 1-ftdiameter drain pipe that contains air at atmospheric pressure. Determine the minimum water depth, h , at which the gate will pivot to allow water to flow into the pipe. Solution 2.109 For equilibrium,  M0  0 FR1  1  FR2   2 (1) FR1   hc1 A1 lb  h     62.4 3     4 ft  h   125 h 2 ft   2   For the force on the horizontal portion of the gate (which is balanced by pressure on both sides except for the area of the pipe) lb  2  2     FR2   h   1ft    62.4 3   h    1ft  ft  4 4   49.0 h Thus, from Eq. (1) with 1  125 h2   h3    49.0 h 3ft  h  1.88ft h and  2  3ft 3 Problem 2.110 The closed vessel of the figure below contains water with an air pressure of 10 psi at the water surface. One side of the vessel contains a spout that is closed by a 6-in.-diameter circular gate that is hinged along one side as illustrated. The horizontal axis of the hinge is located 10 ft below the water surface. Determine the minimum torque that must be applied at the hinge to hold the gate shut. Neglect the weight of the gate and friction at the hinge. Solution 2.110 Let F1 force due to air pressure, and F2 force due to hydrostatic pressure distribution of water. Thus, 2 lb   in.2      6   F1  pair A   10 2  144 2     ft   283lb ft   4   12   in.   and F2   hc A 1  3  6   where hc  10ft     ft   10.15ft 2  5  12   So that 2 lb      6  F2   62.4 3  10.15ft     ft   124 lb  4   12  ft   Also, yR2  I xc  yc yc A where yc  10 ft 1  6    ft   16.92 ft 3 2  12  5 So that 4    3     ft   4   12   16.92 ft  16.92 ft yR2  2    6  16.92 ft    ft   4  12  For equilibrium,  M0  0 And    10 ft   3  C  F1  ft   F2  yR2  3   12    5      10 ft   3  C   283lb   ft   124 lb  16.92 ft   102 ft  lb 3   12    5   Problem 2.111 (a) Determine the horizontal hydrostatic force on the 2309-m-long Three Gorges Dam when the average depth of the water against it is 175 m . (b) If all of the 6.4 billion people on Earth were to push horizontally against the Three Gorges Dam, could they generate enough force to hold it in place? Support your answer with appropriate calculations. Solution 2.111 (a) N   175 m   FR   hc A   9.80 103 3    175 m  2309 m   m  2   3.46 1011 N (b) Required average force per person  3.46 1011 N 6.4 109 N  lb   54.1 12.2  person  person  Yes. It is likely that enough force could be generated since required average force per person is relatively small. Problem 2.113 A 2-ft-diameter hemispherical plexiglass “bubble” is to be used as a special window on the side of an above-ground swimming pool. The window is to be bolted onto the vertical wall of the pool and faces outward, covering a 2-ft-diameter opening in the wall. The center of the opening is 4 ft below the surface. Determine the horizontal and vertical components of the force of the water on the hemisphere. Solution 2.113  Fx  0 , or FH  FR  pc A Thus, FH   hc A  62.4 lb ft 3   4 ft   2 ft 2  784 lb  to right  4 and 4  R3  Fy  0 , or FV  W   V   3 2 , where R  1ft Thus, lb  4 1ft      131lb  down on bubble   6 ft 3   3 FV  62.4 Problem 2.114 Consider the curved surface shown in the figure below (a) and (b). The two curved surfaces are identical. How are the vertical forces on the two surfaces alike? How are they different? Solution 2.114 In both cases the magnitude of the vertical force is the weight of shaded section shown on the right. In addition, the location of the vertical force is the same (the centroid of the shaded section.) Therefore: Alike: magnitude and location of vertical forces same. However, the two vertical forces are different in that the force in (a) is acting upward and the force in (b) is acting downward. Therefore: Different: direction of vertical forces opposite. Problem 2.115 The figure below shows a cross section of a submerged tunnel used by automobiles to travel under a river. Find the magnitude and location of the resultant hydrostatic force on the circular roof of the tunnel. The tunnel is 4 mi long. Solution 2.115 Due to symmetry, there is no net horizontal force on the roof. The vertical force is equal to the weight of fluid above the tunnel. This vertical force acts through the centroid of the fluid volume. Then for a tunnel length ,    F   V     2 Rh  R 2  2    lb  ft    2    62.4 3   4 mi   5280   2  20 ft  70 ft    20 ft   mi 2    ft   F  2.86 109 lb . This force acts downward through the point “O”. Problem 2.116 The container shown in the figure below has circular cross sections. Find the vertical force on the inclined surface. Also find the net vertical force on the bottom, EF. Is the vertical force equal to the weight of the water in the container? Solution 2.116 The vertical force on the inclined surface is equal to the weight of the water “above” it. This “water volume” is V  V cyl  V hole  V frustrum   1 V   ro2   ri2    h    h ro2  ri2  ro ri . 3 V    2 ft   3ft    1ft   3  1 ft 2 2 1 2 2   1ft   2 ft   1ft    2 ft 1ft    24.1ft 3   3 The vertical force FVi is ri  1″, ro  2″, h  1″,   3″.   lb   FVi   V   62.4 3  24.1ft 3  FVi  1500lb ft   The pressure is uniform over the bottom EF so lb   2 FVb  pA   hA   62.4 3   7 ft    2 ft  ft   or FVb  5490 lb This force FVb is not equal *CRC Standard to the weight of water Math Tables in the container. . Problem 2.117 The 18-ft-long lightweight gate of the figure below is a quarter circle and is hinged at H . Determine the horizontal force, P , required to hold the gate in place. Neglect friction at the hinge and the weight of the gate. Solution 2.117 For equilibrium (from free-body-diagram of fluid mass),  Fx  0 So that Similarly, FH  F1   hc1 A1 lb  6 ft     62.4 3     6 ft 18ft   20200 lb ft   2    Fy  0 So that lb     2  FV  W   H 2O   volume of fluid    62.4 3    6 ft  18ft   31800 lb 4 ft     Also, x1  4  6 ft  8  ft  3 (see the figure below) y1  and 6 ft  2 ft 3 For equilibrium (from free-body-diagram of gate)  M0  0 So that P  6 ft   FH  y1   FV  x1  or P  20200 lb  2 ft    31800 lb   6 ft  ft      20200 lb 8 Problem 2.118 The air pressure in the top of the 2-liter pop bottle and the figure below is 40 psi , and the pop depth is 10 in. The bottom of the bottle has an irregular shape with a diameter of 4.3 in. (a) If the bottle cap has a diameter of 1 in. what is the magnitude of the axial force required to hold the cap in place? (b) Determine the force needed to secure the bottom 2 in. of the bottle to its cylindrical sides. For this calculation assume the effect of the weight of the pop is negligible. (c) By how much does the weight of the pop increase the pressure 2 in. above the bottom? Assume the pop has the same specific weight as that of water. Solution 2.118 (a) lb     2 Fcap  pair  Area cap   40 2    1in.  31.4 lb  in.   4  (b)  Fvertical  0 Fsides  F1   pressure @ 2 in. above bottom    Area  lb     2   40 2     4.3in.  in.   4   581lb (c) p  pair   h     lb  lb  8  1   40 2   62.4 3   ft   in.  ft   12   in.2   144 2  ft   lb lb  40 2  0.289 2 in. in. Thus, the increase in pressure due to weight  0.289 psi (which is less that 1% of air pressure). Problem 2.119 In drilling for oil in the Gulf of Mexico, some divers have to work at a depth of 1300 ft. (a) Assume that seawater has a constant density of 64 lb/ft 3 and compute the pressure at this depth. The divers breathe a mixture of helium and oxygen stored in cylinders, as shown in the figure below, at a pressure of 3000 psia . (b) Calculate the force, which trends to blow the end cap off, that the weld must resist while the diver is using the cylinder at 1300 ft . (c) After emptying a tank, a diver releases it. Will the tank rise or fall, and what is its initial acceleration? Solution 2.119 (a) The hydrostatic pressure is  ft 2   lb  p   sw h   64 3  1300 ft   or  144in.2   ft    (b) p  578 psig The net horizontal force on the end caps is FN  Fin  Fout  pin Ain  pout Aout and   wall stress=  FN FN  Awall Aout  Ain 2 pin Ain  pout Aout pin Din2  pout Dout  2 Aout  Ain Dout  Din2 lb  lb 2 2   3000 2   6in.  14.7  578  2  8in. in.  in.  2 2  8in.   6in. and   2500 psi. (c) The net vertical force on an empty tank and Newton’s second law give   Fvert  FBuoy  W  ma or a FBuoy  W  m FBuoy m g where m is the mass of the tank. Now       FBuoy   sw V   sw    Dout 2    Dout 3  6  4   where   30in.  6in.  24in. Also       m  steel    Dout 2  Din 2    Dout 3  Din 3  . 6  4       Substituting into the equation for a gives 1 4 1  6  a 1 1  2 2 steel   Dout  Din  Dout 3  Din 3  6 4   sw   Dout 2  Dout 3      g. The numerical values give   lb   1 ft  lbm   2 1 3  3  64 3   4  24  8  6 8  in.  32.2   lb  sec 2   32.2 ft  a   ft   lbm   1 1 3  sec 2 2 2 3  3    489 24 8 6 8 6 in.      6 ft 3   4   or a  23.1 ft sec2 tank will fall since a  0.    Problem 2.120 Hoover Dam is the highest arch-gravity type of dam in the United States. A cross section of the dam is shown in the figure below (a). The walls of the canyon in which the dam is located are sloped, and just upstream of the dam the vertical plane shown in the figure below (b) approximately represents the cross section of the water acting on the dam. Use this vertical cross section to estimate the resultant horizontal force of the water on the dam, and show where this force acts. Solution 2.120 Break area into 3 parts as shown. For area 1 : lb  1   1 FR1   hc A1   62.4 3     715ft     295ft  715ft  ft   3  2   1.57  109 lb For area 3 : FR3  FR1  1.57  109 lb For area 2 : lb   1   FR2   hc A2   62.4 3     715ft  290 ft  715ft  ft   2    4.63  109 lb Thus, FR  FR1  FR2  FR3  1.57  109 lb  4.63  109 lb  1.57 109 lb  7.77  109 lb Since the moment of the resultant force about the base of the dam must be equal to the moments due to FR1 , FR2 , and FR3 , it follows that 2 1 2 FR  d  FR1    715ft   FR2    715ft   FR3    715ft  3 2 3 and d 1.57 109 lb   23   715ft    4.63 109 lb   12   715ft   1.57 109 lb   32   715ft  7.77 109 lb  406 ft Thus, the resultant horizontal force on the dam is 7.77 109 lb acting 406 ft up from the base of the dam along the axis of symmetry of the area. Problem 2.121 A plug in the bottom of a pressurized tank is conical in shape, as shown in the figure below. The air pressure is 40 kPa , and the liquid in the tank has a specific weight of 27 kN / m3 . Determine the magnitude, direction, and line of action of the force exerted on the curved surface of the cone within the tank due to the 40  kPa pressure and the liquid. Solution 2.121 d tan 30  2 1  d  2 tan 30  1.155 m volume of cone =  d  2   1 32 For equilibrium,  Fvertical  0 So that Fc  pair A  aw where Fc is the force the cone exerts of the fluid. Also,   pair A   40 kPa    d 2 4 2     40 kPa    1.155 m   41.9 kN 4   And 2    d   3m 1m   W    d 2  3m     1m     d 2  32 12   4  4   kN  22    27 3   1.155 m   m   75.4 kN   3  m   Thus, Fc  41.9 kN  75.4 kN  117 kN And the force on the cone has a magnitude of 117 kN and is directed vertically downward along the cone axis. Problem 2.122 The homogeneous gate shown in the figure below consists of one quarter of a circular cylinder and is used to maintain a water depth of 4 m . That is, when the water depth exceeds 4 m , the gate opens slightly and lets the water flow under it. Determine the weight of the gate per meter of length. Solution 2.122 Consider the free body diagram of the gate and a portion of the water as shown.  M o  0 , or (1)  2W  1W1  FH  3  FV  4  0 , where (2) FH   hc A  9.8 103 N m3  3.5 m 1m 1m   34.3kN since for the vertical side, hc  4 m  0.5 m  3.5 m Also, (3) FV   hc A  9.8  103 N m3  4 m 1m 1m   39.2 kN Also, (4) N   3 2  W1   1m     1m   1m   9.8  103 3 1   m3  2.10 kN 4  m  4 (5) Now,  4  0.5 m and (6) 1 1m 1m 3 I xc  3  0.5 m+  yR  yc   0.5 m   0.5 m  12  0.524 m yc A 3.5 m 1m 1m  (7) and  2  1m  4 1m  4R  1  0.576 m 3 3 To determine 1 , consider a unit square that consist of a quarter circle and the remainder as show in the figure. The centroids of areas (1) and (2) are as indicated. Thus, 4    0.5   A2   0.5  1  A1 3   So that with A2   4 12   4 and A1  1   4 this gives 4      0.5     0.5  1  1   3  4   4 or (8) 1  0.223 m Hence, by combining Eqs.(1) through (8):  0.576 m W   0.223m  2.10 kN    34.3kN  0.524 m    39.2 kN  0.5 m   0 or W  64.4 kN Problem 2.123 The concrete (specific weight = 150 lb / ft 3 ) seawall of the figure below has a curved surface and restrains seawater at a depth of 24 ft . The trace of the surface is a parabola as illustrated. Determine the moment of the fluid force (per unit length) with respect to an axis through the toe (point A). Solution 2.123 The components of the fluid force acting on the wall are F1 and W as shown on the figure where lb  24 ft   F1   hc A   64.0 3     24 ft 1ft  ft   2    18400 lb and y1  24 ft  8ft 3 Also, W  V To determine V find area BCD. Thus,  24  y  dx  0  24  0.2 x 2  dx 0 A x0 x0 x 0  0.2 x3    24 x   3   0 And with x0  120 , A  175ft 2 so that V  A  1ft  175ft 3 Thus,   lb   W   64.0 3  175ft 3  11200 lb ft   To locate centroid of A:  24  y  x dx  0  24 x  0.2 x3  dx  12 x02  0 x0 xc A   x dA   0 x0 x0 and xc  12 Thus,  120   2 0.2 175  120  4 4  4.11ft 0.2 x04 4 M A  F1 y1  W 15  xc   18400 lb  8ft   11200 lb 15ft  4.11ft   25200 ft  lb Problem 2.124 A step-in viewing window having the shape of a half-cylinder is built into the side of a large aquarium. See the figure below. Find the magnitude, direction, and location of the net horizontal forces on the viewing window. Solution 2.124 Due to symmetry, the net force parallel to the wall is zero or Fz  0 The net horizontal force perpendicular to the wall is  lb  Fx   hc A   64 3   25  5  ft 10 ft 10 ft   ft  Fx  1.92 105 lb The vertical location of Fx is 1 3 bh I 10 ft 2 or h2 y p  yc  xc  yc  12  yc   30 ft  yc A yc bh 12 yc 12  30 ft  The net horizontal force also acts through the coordinate z  0 and acts in an outward direction. y p  30.3ft Problem 2.125 A step-in viewing window having the shape of a half-cylinder is built into the side of a large aquarium. See the figure below. Find the magnitude, direction, and location of the net horizontal forces on the viewing window. Find the magnitude, direction, and location of the net vertical force acting on the viewing window. Solution 2.125 The net vertical force must equal the weight of fluid inside the viewing window. Then Fy  25100 lb,    lb    2 Fy   V   h  R 2    64 3  10 ft     5ft  or acting upword. 2   ft  2 This net vertical force acts through the centroid of the window volume. Using Appendix B gives x 4 R 4  5ft   3 3 or x  2.12 ft Problem 2.126 A l0-m -long log is stuck against a dam, as shown in the figure below. Find the magnitudes and locations of both the horizontal force and the vertical force of the water on the log in terms of the diameter D . The center of the log is at the same elevation as the top of the dam. Solution 2.126 Consider the water forces on the log as shown on the right. The horizontal forces FH is on the top portion only and is  D  D  FH        4  2  where  is the log length. Assuming 10 C water, Table A.5 gives kg   m  FH   1000 3  9.81 2   0.25 m  0.5 m 10 m   12300 N  FH  m  s  The location of FH is 2  D  2  0.5  yp      m   0.167 m  y p 3 2  3 2  The vertical force FV is the weight of water “above» the bottom of the log minus the weight of water above the top half of the log. This is   D2  D   D 2  D 2    D3  3  FV      D    1      8 2 4 16 4 4         kg   m 2   1000 3  9.81 2  10 m 1.0 m  3   m  s    1  4  4  FV  82300 N The location x of FV is found by first locating the centroid of area A1 by x1  A1 2 x1 2  A2 x2 . A1 Using Table B x1  2 2  D   D    D   D 2D            2   4   16   2 3  2 2  D D    16 2  1   1 2   16  16  2  3     D  1       4 16  0.112 D and is the location of FV 1 . The location of FV 2 is D . The location of FV is 2 D FV 2    FV 1  0.112 D  2 x FV   D2 D2   D   D2  D2          0.112 D  2   2   4 16   B   D2 D2  D2  D2      B 2  4 16  1   1      1  0.112  1   4 4   4  4  D   1 1      1  1   2 4  4 4  x  0.525 D FV 2  weight of water “above” bottom portion of log. FV 1  weight of water above top left portion of log. Problem 2.127 Find the net horizontal force on the 4.0-m -long log shown in the figure below. Solution 2.127 The force FL on the left side of the log is the horizontal force on the horizontally projected area of the log. This horizontally projected area measures D  1.0 m by 4.0 m and gives FL   ghc A  N  s2  kg   m    1000 3  9.81 2   0.5 m 1.0 m  4.0 m    kg  m   m  s     19600 N  19.6 kN The force FR on the right side of the log is the horizontal force on the horizontally projected area of the lower half of the log. This horizontally projected area measures and gives  N  s2  kg   m  FR   ghc A  1000 3  9.81 2   0.25 m  0.5m  4.0 m    kg  m   m  s     4910 N  4.91 kN The net horizontal force is F  FL  FR  19.6 kN  4.91kN F  14.7 kN, acting to right. D  0.5m by 4.0 m 2 Problem 2.128 An open tank containing water has a bulge in its vertical side that is semicircular in shape as shown in the figure below. Determine the horizontal and vertical components of the force that the water exerts on the bulge. Base your analysis on a 1-ft length of the bulge. Solution 2.128 FH horizontal force of wall on fluid FV vertical force of wall on fluid W   H 2OVvol 2 lb     3ft      1ft    62.4 3   2 ft      882 lb lb   F1   hc A   62.4 3   6 ft  3ft  6 ft 1ft  ft    3370 lb For equilibrium, FV  W  882 lb  and FH  F1  3370 lb  The force the water exerts on the bulge is equal to, but opposite in direction to FV and FH above. Thus,  FH wall  3370 lb   FV wall  882 lb  Problem 2.129 A closed tank is filled with water and has a 4-ft-diameter hemispherical dome as shown in the figure below. A U-tube manometer is connected to the tank. Determine the vertical force of the water on the dome if the differential manometer reading is 7 ft and the air pressure at the upper end of the manometer is 12.6 psi. Solution 2.129 For equilibrium,  Fvertical  0 So that FD  pA  W Eq. (1) Where FD is the force the dome exerts on the fluid and p is the water pressure at the base of the dome. From the manometer, pA   gf  7ft    H2O  4ft   p So that lb   in.2  lb  lb     p  12.6 2  144 2    3.0   62.4 3   7 ft    62.4 3   4 ft  in.   ft  ft  ft      2880 lb ft 2 Thus, from Eq.(1) with volume of sphere   6  diameter 3 lb     lb   2 1  3 FD   2880 2     4 ft     4 ft    62.4 3  26  ft   4  ft    35100 lb The force that the vertical force that the water exerts on the dome is 35100 lb . Problem 2.130 A 3-m-diameter open cylindrical tank contains water and has a hemispherical bottom as shown in the figure below. Determine the magnitude, line of action, and direction of the force of the water on the curved bottom. Solution 2.130 Force  weight of water supported by hemispherical bottom   H 2O  volume of cylinder    volume of hemisphere    9.80  kN   2 3 3m   8 m    3m   3 4 12 m    485 kN The force is directed vertically downward, and due to symmetry it acts on the hemisphere along the vertical axis of the cylinder. Problem 2.131 Three gates of negligible weight are used to hold back water in a channel of width b as shown in the figure below. The force of the gate against the block for gate (b) is R. Determine (in terms of R) the force against the blocks for the other two gates. Solution 2.131 For case (b) 2 h b h FR   hc A      h  b   2 2 and yR  2 h 3 Thus, MH  0 So that 2  hR   h  FR 3  2  2   h b  hR   h     3   2  R  h 2b 3 (1) For case (a) on free-body-diagram shown 4h (see the figure below) 3 2 FR  yR   h 2b 2 (from above) and 2 h 3 and W  V o   h 2      2     b   4       h 2b 16 Thus, MH  0 So that  h 4h  2  W    FR  h   FB h  2 6  3  and  h2b  h 4h   h2b  2      h   FB h 16  2 6  2 3  It follows that FB   h2b(0.390) From Eq.(1)  h 2b  3 R , thus FB  1.17 R For case (c), for the free-body-diagram shown, the force FR1 on the curved section passes through the hinge and therefore does not contribute to the moment around H. On bottom part of gate  3h  h  3 FR2   hc A      b    h 2b  4  2  8 and 3 1 h  b    3h I xc 12 2   yc  yR2  yc A  3h  h  4    b   4  2  28  h 36 Thus, MH  0 So that  28  FR2  h   FB h  36  or 3  28  7 FB    h 2b     h 2b 8  36  24 From Eq.(1)  h 2b  3R , thus FB  7 R  0.875 R 8 Problem 2.133 An iceberg (specific gravity 0.917 ) floats in the ocean (specific gravity 1.025 ). What percent of the volume of the iceberg is under water? Solution 2.133 For equilibrium, W  weight of iceberg  FB  buoyant force or V ice ice  V sub ocean , where V sub  volume of ice submerged . Thus,  SGice V sub 0.917  ice    0.895  89.5% V ice  ocean SGocean 1.025 Problem 2.134 A floating 40-in.- thick piece of ice sinks 1 in. with a 500-lb polar bear in the center of the ice. What is the area of the ice in the plane of the water level? For seawater, S  1.03 . Solution 2.134 Without the polar bear on the ice, the submerged depth d of the ice is found by equating the weight of the ice and the buoyant force. Denoting the pure water specific weight by  and the ice area by A gives FB  Wice or Wice   SAd . The ice sinks an additional depth d  with the bear in the center of the ice. Equating the new buoyant force to the weight of the ice plus bear gives FB  Wice  Wbear ,  SA  d  d     SAd  Wbear , or A Wbear 500 lb  or A  93.4 ft 2 lb   Sd   1   62.4 3  1.03  12 ft  ft     Problem 2.135 A spherical balloon filled with helium at 40°F and 20 psia has a 25  ft diameter. What load can it support in atmospheric air at 40°F and 14.696 psia ? Neglect the balloon’s weight. Solution 2.135 For static equilibrium, the buoyant force must equal the load. Neglecting the weight of the balloon and assuming air and helium to be ideal gases, the load is L  FB    air   He  V    air   He  g V  p   p    g  4            R3    R air  R  He   T  3 Using Table A.4, the numerical values give  ft  4  3 lb lb    20 144  2   32.2 2   3  12.5ft   14.696 144  2  sec   ft  ft   L   53.35 ft  lb   386 ft  lb   500 o R  32.2 ft  lbm        lbm  o R   lbm  o R    lb  sec2     or L  527 lb  Problem 2.136 A river barge, whose cross section is approximately rectangular, carries a load of grain. The barge is 28 ft wide and 90 ft long. When unloaded, its draft (depth of submergence) is 5 ft and with the load of grain the draft is 7 ft . Determine: (a) the unloaded weight of the barge, and (b) the weight of the grain. Solution 2.136 (a) For equilibrium,  Fvertical  0 So that Wb  FB   H 2O  (submerged volume) lb )  5ft  28ft  90 ft  ft 3  786000 lb  (62.4 (b)  Fvertical  0 WB  Wg  FB   H 2O  (submerged volume) lb )  7 ft  28ft  90 ft   786, 000 lb ft 3  315000 lb Wg  (62.4 Problem 2.137 A barge is 40 ft wide by 120 ft long. The weight of the barge and its cargo is denoted by W .   When in salt-free riverwater, it floats 0.25 ft deeper than when in seawater   64 lb/ft 3 . Find the weight W . Solution 2.137 In both cases, the weight W must equal the weight of the displaced water or W   SFW A  d  0.25ft    SW Ad Soling for d gives  SW Ad   SFW A  d  0.25ft  or  0.25ft   SFW  d  SW   SFW  0.25ft   62.4 3  lb ft   9.75ft .  lb  64.0  62.4  3 ft Then lb   W   SW Ad   64.0 3   40 120  ft 2  9.75ft  ft    short ton  W  3.00  106 lb  ,  2000 lb  or W  1500 short tons. Problem 2.138 When the Tucurui Dam was constructed in northern Brazil, the lake that was created covered a large forest of valuable hardwood trees. It was found that even after 15 years underwater the trees were perfectly preserved and underwater logging was started. During the logging process a tree is selected, trimmed, and anchored with ropes to prevent it from shooting to the surface like a missile when cut. Assume that a typical large tree can be approximated as a truncated cone with a base diameter of 8ft , a top diameter of 2 ft , and a height of 100 ft . Determine the resultant vertical force that the ropes must resist when the completely submerged tree is cut. The specific gravity of the wood is approximately 0.6 . Solution 2.138 W weight FB buoyant force T tension in ropes For equilibrium,  Fvertical  0 So that T  FB  W (1) For a truncated cone, Volume  h 3  r12  r1r2  r22  where: r1  base radius r2  top radius h  height Thus, V tree   100 ft   4 ft 2  4 ft 1ft  1ft 2          3  2200 ft 3 For buoyant force, FB   H 2O  V tree  (62.4 lb ft 3   ) 2200 ft 3  137000 lb For weight,   lb   W   tree  V tree   0.6   62.4 3  2200 ft 3  82400 lb ft   From Eq.(1) T  137000 lb  82400 lb  54600 lb Problem 2.140 An inverted test tube partially filled with air floats in a plastic water-filled soft drink bottle as shown in the figure below. The amount of air in the tube has been adjusted so that it just floats. The bottle cap is securely fastened. A slight squeezing of the plastic bottle will cause the test tube to sink to the bottom of the bottle. Explain this phenomenon. Solution 2.140 Where the test tube is floating the weight of the tube, W , is balanced by the buoyant force, FB , as shown in the figure. The buoyant force is due to the displaced volume of water as shown. This displaced volume is due to the air pressure, p , trapped in the tube where p  po   H 2O h. When the bottle is squeezed, the air pressure in the bottle, po , is increased slightly and this in turn increases p , the pressure compressing the air in the test tube. Thus, the displaced volume is decreased with a subsequent decrease in FB . Since W is constant, a decrease in FB will cause the test tube to sink. Problem 2.141 A child’s balloon is a sphere 1 ft. in diameter. The balloon is filled with helium (  = 0.014 lbm/ft 3 ). The balloon material weighs 0.008 lbf/ft 2 of surface area. If the child releases the balloon, how high will it rise in the Standard Atmosphere. (Neglect expansion of the balloon as it rises.) Solution 2.141 A force balance in the vertical direction for the balloon gives    Fz  0   air g V   He g V  wA for the balloon at rest at its highest elevation. Then  air   He g V  wA gV  2  w D wA   He    He  gV   D3  g  6    lb   ft  lbm   6  0.008 2   32.2  6w lbm ft   lb  sec2    He   0.014 3   ft  gD  ft  32.2 2  1.0 ft  sec   lbm  0.062 3 . ft Interpolating Table A.2 for the Standard Atmosphere,  0.06590  0.062  z  elevation=5000 ft  5000 ft    0.06590  0.05648  z  7070 ft Problem 2.142 A 1-ft-diameter, 2-ft-long cylinder floats in an open tank containing a liquid having a specific weight γ. A U-tube manometer is connected to the tank as shown in the figure below. When the pressure in pipe A is 0.1 psi below atmospheric pressure, the various fluid levels are as shown. Determine the weight of the cylinder. Note that the top of the cylinder is flush with the fluid surface. Solution 2.142 From a free-body-diagram of the cylinder  Fvertical  0 So that 2   W  FB     1ft   2 ft  4  (1)  2 A manometer equation gives,     3.5 ft    SG   H O  2.5 ft    H O 1ft   p A 2 2 So that     3.5ft   1.5   62.4 and   80.6 lb ft 3 lb  lb  lb   in.2       2.5ft 62.4 1ft 0.1 144           ft 3  ft 3  in.2   ft 2    Thus, from Eq.(1) lb    W   ft 3  80.6 3   127 lb  2  ft  Problem 2.143 A not-too-honest citizen is thinking of making bogus gold bars by first making a hollow iridium  S  22.5 ingot and plating it with a thin layer of gold  S  19.3 of negligible weight and volume. The bogus bar is to have a mass of 100 lbm . What must be the volumes of the bogus bar and of the air space inside the iridium so that an inspector would conclude it was real gold after weighing it in air and water to determine its density? Could lead  S  11.35  or platinum  S  21.45 be used instead of iridium? Would either be a good idea? Solution 2.143 S x  22.5  iridium  SG  19.3  gold  V BB  V x  V AS mBB  mx  100 lbm Neglect the weight of air in the air space and the buoyant force of air on the bar. The volume of a pure gold bar would be V GB  WGB G . The bogus bar must have the same volume and weight as the pure gold bar so it will weigh like a solid gold bar in water. The volume condition gives V GB  V BB  V AS  Vx . Since WGB  Wx , V AS  V x  V GB  WGB G  Wx G , V AS  V x  x V x , G or   S  V AS  V x  x  1  V x  x  1 .  G   SG  The numerical value of the iridium volume is Vx Wx x  WGB x  100 lb lb    22.5  62.4 3  ft    0.0712 ft 3 . The air space volume is  22.5   1 or V AS  0.0712 ft 3   19.3  V AS  0.0118ft 3 . The bogus bar volume is V BB  V AS  V x   0.0118  0.0712  ft 3 or Also, lead will not work since it is less dense that gold And platinum will work since it is more dense than gold but would only be used by a not-too-bright citizen as platinum is more expensive than gold. V BB  0.0830 ft 3 . Problem 2.144 A solid cylindrical pine  S = 0.50  spar buoy has a cylindrical lead  S = 11.3 weight attached, as shown in the figure below. Determine the equilibrium position of the spar buoy in seawater (i.e., find d ). Is this spar buoy stable or unstable? For seawater, S = 1.03 . Solution 2.144 The equilibrium position is found by equating the buoyant force and the body weight (see the sketch below). FB  W  sw dA      A   p  p A or d        p  p S    S p  p  S sw  sw 11.3  0.5ft   0.50 16 ft   13.3ft  d 1.03 Since d  13.8ft (the total length of the spar buoy), the spar buoy floats. We now have to check the stability of the buoy. I  4   radius 4  1ft 4  0.7854 ft 4 , 4  c  distance from bottom of buoy to center of gravity of buoy c   cW   cpWcp W  W p p        p A p      A        2   2     A    p A p   p     S       S p  p      2   2   S    S p  p c  11.3  0.5  0.25   0.5 16  0.5  8  ft 11.3  0.5   0.5 16   c  5.09 ft , d 13.3ft   6.65ft , 2 2 n  c  d  5.09 ft  6.65ft  1.56 ft , 2 V s  Ad   1ft  13.3ft   41.8ft 3 , 2 m I 0.7854 ft 4 n    1.56 ft   1.58ft . Vs 41.8ft 3 Since m  0 , the buoy is stable Problem 2.145 When a hydrometer (see the figure below) having a stem diameter of 0.30in. is placed in water, the stem protrudes 3.15in. above the water surface. If the water is replaced with a liquid having a specific gravity of 1.10 , how much of the stem would protrude above the liquid surface? The hydrometer weighs 0.042 lb . Solution 2.145 When the hydrometer is floating its weight, W , is balanced by the buoyant force, FB . For equilibrium,  Fvertical  0 Thus, for water FB  W  H O  V1  W (1) 2 Where V1 is the submerged volume. With the new liquid  SG    H O  V2  W (2) 2 Combining Eqs.(1) and (2) with W constant  H O  V1   SG   H O  V2 2 And 2 V2  V1 SG (3) From Eq.(1) V1  W H O 2  0.042 lb  6.73 104 ft 3 lb 62.4 3 ft So that from Eq.(3) V2  6.73  104 ft 3  6.12 104 ft 3 1.10 Thus, V1  V2   6.73  6.12  104 ft 3  0.61104 ft 3 To obtain this difference the change in length,  , is in.3    2 4 3      0.30in. 0.61 10 ft 1728      4 ft 3       1.49in. With the new liquid the stem would protrude 3.15in.+1.49in.  4.64in. above the surface. Problem 2.146 A 2-ft-thick block constructed of wood  SG  0.6  is submerged in oil  SG  0.8  and has a 2- ft-thick aluminum (specific weight = 168lb / ft 3 ) plate attached to the bottom as indicated in the figure below. Determine completely the force required to hold the block in the position shown. Locate the force with respect to point A. Solution 2.146 For equilibrium,  Fvertical  0 So that F  Ww  FBw  Wa  FBa where:   Ww   SGw   H 2O Vw lb  1     0.6   62.4 3    10 ft  4 ft  2 ft   1500 lb ft   2   lb   Wa  168 3   0.5ft 10 ft  2 ft   1680 lb ft   lb  1   FBw   SGoil   H 2O Vw   0.8   62.4 3    10 ft  4 ft  2 ft   2000 lb ft   2     lb   FBa   SGoil   H 2O Va   0.8   62.4 3   0.5ft 10 ft  2 ft   499 lb ft     Thus, F  1500 lb  2000 lb+1680 lb  499 lb  681 lb upward Also,  Ma  0 So that  10  F   ft  Ww  FBw    5ft  Wa  FBa   3  or  10    681lb    ft  1500 lb  2000 lb    5ft 1680 lb  499 lb   3  and   6.22 ft to right of point A Problem 2.147 How much extra water does a 147  lb concrete canoe displace compared to an ultralightweight 38  lb Kevlar canoe of the same size carrying the same load? Solution 2.147 For equilibrium,  Fvertical  0 and W  FB   H 2O V and V is displaced volume. For concrete canoe, lb   147 lb   62.4 3  Vc ft   Vc  2.36ft 3 For Kevlar canoe, lb   38lb   62.4 3  Vk ft   Vk  0.609ft 3 Extra water displacement  2.36 ft 3  0.609 ft 3  1.75ft 3 Problem 2 .148 A submarine is modeled as a cylinder with a length of 300 ft , a diameter of 50 ft , and a conning tower as shown in the figure below. The submarine can dive a distance of 50 ft from the floating position in about 30 sec . Diving is accomplished by taking water into the ballast tank so the submarine will sink. When the submarine reaches the desired depth, some of the water in the ballast tank is discharged leaving the submarine in “neutral buoyancy” (i.e., it will neither rise nor sink). For the conditions illustrated, find (a) the weight of the submarine and (b) the volume (or mass) of the water that must be in the ballast tank when the submarine is in neutral buoyancy. For seawater, S = 1.03 . Solution 2.148 (a) Denoting the cylinder radius by R , the submarine weight is equal to the buoyant force so W  FB   V submerged      R 2 1.03 when the submarine is in the partially submerged position. The numerical values give  lb  2 W   64 3    25ft   300 ft 1.03 or  ft  W  3.88 107 lb (b) For neutral buoyancy at the lower depth, the submarine weight W plus the ballast weight WB must equal the buoyant force so   W  WB  FB    R 2 1.10  or   WB    R 2 1.10   W . The ballast volume V B  WB  so   V B   R 2 1.10   V B  41700 ft 3 W     25ft   300 ft 1.10   2 3.88  107 lb  lb   64 3   ft  Problem 2.150 When an automobile brakes, the fuel gage indicates a fuller tank than when the automobile is traveling at a constant speed on a level road. Is the sensor for the fuel gage located near the front or rear of the fuel tank? Assume a constant deceleration. Solution 2.150 accelerating automobile so sensor located in front of fuel tank. decelerating  braking  automobile Problem 2.151 An open container of oil rests on the flatbed of a truck that is traveling along a horizontal road at 55 mi / hr . As the truck slows uniformly to a complete stop in 5 s , what will be the slope of the oil surface during the period of constant deceleration? Solution 2.151 slope  ay  ay dz  dy g  az final velocity – initial velocity time interval m     0   55 mph   0.4470 s  mph      4.92 m  5s s2 Thus, m  4.92 2   dz s   0.502  m dy 9.81 2  0 s Problem 2.152 A 5-gal, cylindrical open container with a bottom area of 120 in.2 is filled with glycerin and rests on the floor of an elevator. (a) Determine the fluid pressure at the bottom of the container when the elevator has an upward acceleration of 3 ft / s 2 . (b) What resultant force does the container exert on the floor of the elevator during this acceleration? The weight of the container is negligible. (Note: 1 gal  231 in.3 ) Solution 2.152 h A  volume  231in.3  h 120in.2   5gal    gal      h  9.63in. (a) p     g  az  z Thus, pb 0 0 dp     g  az h dz and pb    g  a z  h slugs  ft ft   9.63     2.44 3  32.2 2  3 2   ft  ft  s s   12   lb  68.9 2 ft (b) From free-body-diagram of container, F f  pb A 2  lb   2  1ft   68.9 2  120 in.   144 in.2  ft        57.4 lb Thus, force of container on floor is 57.4 lb downward . Problem 2.153 A plastic glass has a square cross section measuring 2½ in. on a side and is filled to within ½ in. of the top with water. The glass is placed in a level spot in a car with two opposite sides parallel to the direction of travel. How fast can the driver of the car accelerate along a level road without spilling any of the water? Solution 2.153 Slope of water surface a   car g or acar   g  slope  ft   1.0in.      32.2 2     sec   2.5in.   or acar  12.9 ft sec 2 Problem 2.154 The cylinder in the figure below accelerates to the left at the rate of 9.80 m/s 2 . Find the tension in the string connecting at rod of circular cross section to the cylinder. The volume between the rod and the cylinder is completely filled with water at 10 °C . Solution 2.154 FIND Tension in string. SOLUTION First find the pressure difference in the water over a length   8.0 cm . Since gravity is perpendicular to the rod, Eq.(2.41) gives dp    ax d x For the x-direction. Integrating gives p2  p1    a x  x2  x1  . For 10°C water, Table A.5 gives  m  kg   m N  p2  p1   1000 3   9.80 2   8.0 cm     784 2 m  s  m   100 cm  We next apply Newton’s second law to the rod    Fx  ma x , T   p1  p2  A  ma x , or T   p2  p1  A  ma x . m Assuming the string is not elastic, a x,rod  9.80 2 . s Now   D2   m   w Srod A   w S rod   4    3  m  kg  2     1000 3   2.0    1.0 cm   8.0 cm     0.0126 kg  4 m   100 cm  and A  D2  2 2 m  5 2  1.0 cm     7.854  10 m . 4 4 100 cm   Then   N  m   T   784 2  7.854 105 m 2   0.0126 kg   9.80 2   m  s   or T  0.062 N Problem 2.155 A closed cylindrical tank that is 8ft in diameter and 24 ft long is completely filled with gasoline. The tank, with its long axis horizontal, is pulled by a truck along a horizontal surface. Determine the pressure difference between the ends (along the long axis of the tank) when the truck undergoes an acceleration of 5 ft / s 2 . Solution 2.155 p   ay y Thus, p2 24  p dp    a y 0 dy 1 Where p  p1 at y  0 and p  p2 at y  24 ft , and p2  p1    a y  24 ft  slugs  ft     1.32 3  5 2   24 ft  ft  s   lb  158 2 ft or p1  p2  158 lb ft 2 Problem 2.156 The cart shown in the figure below measures 10.0 cm long and 6.0 cm high and has rectangular cross sections. It is half-filled with water and accelerates down a 20 incline plane at a  1.0 m/s 2 . Find the height h . Solution 2.156 Unfortunately, there are 2 x-directions in the problem statement. Noting that the gravisty vector is in the negative z-direction, change the label on the axis normal to the z-direction to be “n”. Resolving the acceleration along the plane into n,z components: a z  a sin  , an  a cos  ,   20o For rigid-body motion of the fluid in the n,z coordiantes:: dp    an dn    g  az  dz dp    a cos  dn    g  a sin   dz  0  along free surface p  patm Using trignometirc relationships this equation can be converted into x,y coordinates. dn  dx cos   dy sin  dz  dy cos   dx sin    a cos  dn    g  a sin   dz  0   a cos   dx cos   dy sin      g  a sin    dy cos   dx sin    0     a cos 2  dx    a cos  sin   dy     g  a sin   cos   dy     g  a sin    sin    dx  0    a cos 2     g  a sin   sin   dx     a cos  sin     g  a sin   cos   dy  0          a cos 2   sin 2    g sin   dx     g cos   dy  0      a   g sin   dx    g cos   dy  0  a  g sin   dx   g cos   dy  0 Integration yields:  a  g sin   x   g cos   y  C  sin   a y    xC  g cos  cos   The constant of integration can be determined by noting that the container is ½-full:  water   y dx 0    a sin   x  C dx       0  g cos  cos      2 a    tan    C   g cos   2 C  water  a   tan     g cos  2  m     s2    1 2  10.0 cm  10.0 cm  6.0 cm  / 2   s C  tan 20o  2 10.0 cm    9.81 m  cos 20o   1.723 cm Solving for the the requested length:   a y    tan   x  C  g cos     1 o h  tan 20  10 cm   1.723 cm  4.277 cm   9.81 cos 20o    h  4.28 cm Problem 2.157 The U-tube manometer in the figure below is used to measure the acceleration of the cart on which it sits. Develop an expression for the acceleration of the cart in terms of the liquid height h , the liquid density  , the local acceleration of gravity g , and the length  . Solution 2.157 Writing Newton’s second law in the horizontal direction (x-direction) for the bottom leg of the manometer gives  Fx  ma x , p A  pr A   Aa , or a p  pr  Applying the manometer rule to the two legs of the manometer gives p  patm   gh and pr  patm   gh Subtracting gives p  pr   g  h  hr    gh so a  gh  or h a  g   Problem 2.158 A tank has a height of 5.0 cm and a square cross section measuring 5.0 cm on a side. The tank is one third full of water and is rotated in a horizontal plane with the bottom of the tank 100 cm from the center of rotation and two opposite sides parallel to the ground. What is the maximum rotational speed that the tank of water can be rotated with no water coming out of the tank? Solution 2.158 dp    gdz   2 r dr Since dp  0 along the free surface, the free surface is identified by the equation 0    gdz   2 r dr or 0   gdz   2 r dr Integrating gives z r 0   g  b dz   2  r dr ,  r1 2 b  2 2 2  0  g  z    r  r1 , 2 2    or b 2 2 2 z  r  r1 . 2 2g   Recognizing that the volume of water in the rotating tank must equal b2h gives 6  r1  h r1  h  b  2 b2h zb dr  b  r 2  r12  dr ,     r1 r1 6  2 2 g   r h  br  2  r 3 1 b2h 2   b     r1 r   6  2 2 g  3   r1 ,   bh  2   r  h 3 r 3  b2h  1  b    1  r12 h   ,  6 3 3  2 2g     3  2bh  2   r1  h  r13     r12 h  ,  3 2g  3 3   or  4bhg   r  h 3 r 3   1  r12 h  3 1   3 3   . The numerical values give  cm   4  5cm  5cm   981 2  s    100 cm 3  95cm 3  2    95cm   5cm   3   3 3   rad   rev   60s     3.68    or s   2 rad   min     35.1rpm DISCUSSION Note the that when r  r1  h ,   b 2 b 2 z   r1  h 2  r12    2r1h  h2 . 2 2g 2 2g   The numerical values give 2 rad   3.68   5 s   2 z  cm   2  95cm  5cm    5cm    cm   2  2  981 2  s    4.23cm or the assumption indicated in the above figures that the water does not reach the uppermost side of the tank is correct. Problem 2.159 An open 1-m-diameter tank contains water at a depth of 0.7 m when at rest. As the tank is rotated about its vertical axis the center of the fluid surface is depressed. At what angular velocity will the bottom of the tank first be exposed? No water is spilled from the tank. Solution 2.159 Equation for surfaces of constant pressure: z  2r 2 2g  constant For free surface with h  0 at r  0 , h  2r 2 2g The volume of fluid in rotating tank is given by R V f   2 rh dr  0 2 2 R 3  2 R 4 r dr  2 g 0 4g Since the initial volume, Vi   R 2 hi , must equal the final volume, V f  Vi So that  2 R 4 4g or   R2 hi  4 ghi R2  m  4  9.81 2   0.7 m  rad s    10.5 2 s  0.5 m  Problem 2.160 The U-tube in the figure below rotates at 2.0 rev/sec. Find the absolute pressures at points C and B if the atmospheric pressure is 14.696 psia . Recall that 70 °F water evaporates at an absolute pressure of 0.363 psia . Determine the absolute pressures at points C and B if the U-tube rotates at 2.0 rev/sec . Solution 2.160 Apply the manometer rule to one of the legs to get pB  patm + gh Using Table A.6, lb    62.3 3  1in. ft  pB  14.696 psia   ,  in.3   1728 3  ft   pB  14.732 psia Section 2.6.2 gives p   r 2 . r Integrating from r  0 to r  R gives pB p C R dp   2  r dr or Then o pB  pC   2 R 2 . 2 2 lbm  rev   2 62.3 3   2.0 2.5ft     2 2  R sec  ft    pc  pB   14.732 psia  2 2 2  144in.   rev   32.2 ft  lbm  2  ft 2   2 rad   lb  sec2    or pc  8.10 psia Since pc  0.33psia . DISCUSSION Note that if pc were calculated to be less than 0.363psia , some of the water would vaporize and pc would be 0.363psia . Problem 2.161 A child riding in a car holds a string attached to a floating, helium-filled balloon. As the car decelerates to a stop, the balloon tilts backwards. As the car makes a right-hand turn, the balloon tilts to the right. On the other hand, the child tends to be forced forward as the car decelerates and to the left as the car makes a right-hand turn. Explain these observed effects on the balloon and child. Solution 2.161 A floating balloon attached to a string will align itself so that the string it normal to lines of constant pressure. Thus, if the car is not accelerating, the lines of p  constant pressure are horizontal (gravity acts vertically down), and the balloon floats “straight up” (i.e.   0 ). If forced to the side (   0 ), the balloon will return to the vertical (   0 ) equilibrium position in which the two forces T and FB -W line up. Consider what happens when the car decelerates with an amount a y  0 . As show by the equation, slope  ay dz  , g  az dy the lines of constant pressure are not horizontal, but have a slope of ay ay dz   0 dy g  az g since a z  0 and a y  0 . Again, the balloon’s equilibrium position is with the string normal to p  const. lines. That is, the balloon tilts back as the car stops. V2 (the centrifugal acceleration), the lines of p  const. are as shown, R and the balloon tilts to the outside of the curve. When the car turns, a y  Problem 2.162 A closed, 0.4-m-diameter cylindrical tank is completely filled with oil  SG  0.9  and rotates about its vertical longitudinal axis with an angular velocity of 40 rad / s . Determine the difference in pressure just under the vessel cover between a point on the circumference and a point on the axis. Solution 2.162 Pressure in a rotating fluid varies in accordance with the equation, p  2 r 2 2   z  constant Since z A  z B , pB  p A   2 rB2  rA2   2 2 kg  rad   (0.9)  103 3  40  s    m    0.2 m 2  0  2  28.8 kPa Problem 2.163 The largest liquid mirror telescope uses a 6-ft-diameter tank of mercury rotating at 7 rpm to produce its parabolic-shaped mirror as shown in the figure below. Determine the difference in elevation of the mercury, h ,between the edge and the center of the mirror. Solution 2.163 For free surface of rotating liquid, z  2r 2 2g  constant Let z  0 at r  0 and therefore constant  0 . Thus, h  z for r  3ft and with   rad  0.733 s    7 rpm   2 rad   1min    rev   60s  It follows that 2 rad  2   0.733   3ft  s  h    0.0751ft ft   2  32.2 2  s  