hw7 - Problem 5. A plant hangs from a 3.2-cm diameter...

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Unformatted text preview: Problem 5. A plant hangs from a 3.2-cm diameter suction cup affixed to a smooth horizontal surface (Fig. 18-42). What is the maximum weight that can be suspended (a) at sea level and (b) in Denver, where atmospheric pressure is about 0.80 atm? -9 FIGURE 13-42 Problem 5. Solution (a) The force exerted on the suction cup by the atmosphere is F = PA = Pam(7rcf2/4) = (1.013x 11)5 Pa)rr(0.016 1n)2 = 81.5 N (perfect vacuum inside cup assumed). This is equal to the maximum weight. (b) At Denver, P = 0.88m.” so the maximum weight is 80% of that in part (a), or 65.2 N (a slight variation in g with altitude is neglected). Problem 11. A paper clip is made from wire 1.5 mm in diameter. You unbend a paper clip and push the end against the well. What force must you exert to give a pressure of 120 atm? Solution An average pressure of 120 atm over the cross-sectional area of the wire, findz, results in a force ofF = PA = (120x 101.3 kPa)§«(1.5x10"3 m}2 = 21.5 N. Problem 13. When a couple with a total mass of 120 kg lies on a water bed, the pressure in the bed increases by 4700 Pa. What surface area of the two bodies is in contact with the bed? Solution The pressure increase times the average horizontal contact surface area. equals the weight of the couple. or A3... = rug/AP = (120x9.8 N)/(4700 Pa) = 0.250 m2. Problem 19. Scuba equipment provides the diver with air at the same pressure as the surrounding water. But at pressura greater than about 1 We, the nitrogen in air becomes dangerously narcotic. At what depth does nitrogen naroosis become a hazard? Solution In fresh water (,0 2 103 leg/m3), the pressure is 1 MPa at a depth ofh = (P -- Po)/pg :7 [1 MP5- o.103 MPe)/(9.8x103 N/ma) = 91.7 m, where P0 is atmospheric pressure at the surface. (See Equa- tion 18-3.) The depth is a little less in salt water since its density is slightly greater.‘ Problem 21. A vertical tube open at the top contains 5.0 cm of oil (density 0.82 g/cm3) floating on 5.0 cm of water. Find the gauge pressure atthe bottom of the tube. Solution The pressure at the top of the tube is atmospheric pressure. Pa. The absolute pressure at the interface of the oil and water is R- : R1 + ponghoii, and at the bottom is P = 'l' Pwaterghwater = Po ‘l' poilghoil'l' pwamghwm, (see Equation 18—3). Therefore, the gauge pressure at the bottom is P — Pa = (poiihoifi Pweterhwaterlg = (0.32 + 1.000003 kg/m3)(0.05 m)x (9.8 Iii/$2) : 892 Pa (gauge). Problem 21 Solution. Problem 25. A U-shsped tube open at both ends contains _ water and a quantity of oil occupying a 2.0-cm length of the tube, as shown in Fig. 18-45. If the oil‘s density is 0.32 times that of water, what is the height difference it? Solution From Equation 1&3, the pressure at points at the same level in the water is the some, P1 = P2. Now, P1 = PM... + page“? cm - h) and P2 = Pam. + Ponle 0m), 50 h = (2 cm)(1* Pail/Pine) = (2 ml)" [1 — 0.82) = 3.6 mm (h is positive as shown). Problem 29. A garage lift has a 45-cm diameter piston supporting the load. Compressed air with a maximum pressure of 500 lcPa is applied to a small piston at the other end of the hydraulic system. \Nhat is the maximum mass the lift can support? Solution If we neglect the variation of pressure with height in the hydraulic system (which is usually small compared to the applied pressure), the fluid pressure is the same throughout, or Pappl = F/A (for either the small or large cylinders). Thus, Fm“ = (500 kPaHi‘rx {0.45 In)? = 79.5 RN, which corresponds to a mass-load of Frau/g = 8.11 tonnes (metric tons). 31. On land, the most massive concrete block you can carry is 25 kg. How massive a block could you carry underwater, if the density of concrete is 2300 Jig/ma? Solution The 25x 9.8 N force you exert underwater is equal to the apparent weight of the most massive block of concrete when submerged, Wan], = W — Fb :. WU —pw/pc), where pC/pw is the ratio of the densities of concrete and water (also known as the specific gravity of concrete). (This relation is derived in Example 18-4, since the buoyant force on an object submerged in fluid is Pa, = Wpfiuid/p.) Thus, W : Warp“ — Pan/Perl: 01' m = “’79 = (25 kng (1 — 1/2.3)-1 = 44.2 kg. Problem 35. A partially full beer bottle with interior diameter 52 mm is floating upright in water, as shown in Fig. 18—47. A drinker takes a swig and replaces the bottle in the water, where it now floats 28 mm higher than before. Honr much beer did the drinker drink? Solution Archimedes’ principle implies that the weight of the beer swallowed equals the difference in the weight of water displaced by the bottle, before and after. Therefore Anthea, = pfizo AV, where “g” was canceled from both sides. The difference in the volume of water displaced equals the ewes—sectional area of the bottle- times 28 mm. If we ignore the thicknem of the walls of the bottle, Ari-thee, 2 (1 g/cm3)7r(-§ x52 cm)2x (2.8 cm) = 59.5 g. FIGURE 18-47 Problem 35 Solution. 39. (a) Hou-Ir much helium (density 0.18 kg/ma) is needed to lift a balloon carrying two people in a basket, if the total mass of people, basket, and balloon (but not gas) is 280 kg? (b) Repeat for a hot air balloon, whose air density is 10% less than that of the surrounding atmosphere. Solution The buoyant force must exceed the weight of the load (mass M, including the balloon) plus the gas (mass m), F5 2 [M+m)g. But Fg, :— pajrgV, and if we neglect the volume of the balloon’s skin etc. compared to that of the gas it contains, V = m/pgas, therefore m : Pgasv = Pgaleb/Pairgl 2 (Peas/PairllM + m) 01‘ m 2 2"‘.4,c3,;,,_.-,/(,oa,,r — p335). (a) When the gas is helium, pat/p”... = 1.2/0.18, and m 2 (280 kg)(6.67 —1)-1 = 49.4 kg. (b) For hot air, pg,s = 0.99m. and m 3 (280 kg)(0.9/0.i) = 2520 kg. (Note: these masses correspond to gas volumes of 275 m3 for helium and 2330 m3 for hot air.) Problem 43. A typical mass flow rate for the Mississippi River is 1.8x107 kg/s. Find (a) the volume flow rate and (b) the flow speed in a region where the river is 2.0 km wide and an average of 6.1 m deep. Solution (3) The mass flow rate and the volume flow rate are related by Equations 18-4b and 5, namely Rm : va : pHv. Therefore, RV = (1.8x107 kg/s)% (111}3 kg/ma) = 1.8x 10‘1 mafs for the Mississippi. (b) At a point in the river where the cross-sectional area is given, the average speed of flow is u = Rv/A = (1-8x10“ m3/5)/{2x103 x 5.1 m2) = 1.48 m/s (= 5.31 km/h = 3.30 mph). The actual flow rate of any river varies with the season, local weather and vegetation conditions, and human water consumption. Froolem 47. In Fig. 18-48 a horizontal pipe of cross-sectional area A is joined to a lower pipe of crosssectional area git. The entire pipe is full of liquid with density p, and the left end is at atmospheric pressure Pa. A small open tube extends upward from the lower pipe. Find the height hg of liquid in the small tube (a) when the right end of the lower pipe is closed, so the liquid is in hydrostatic equilibrium, and (b) when the liquid flows with speed i; in the upper pipe. Solution The continuity equation [Equation 18-5) and Bernoulli’s equation (Equation 18-6] can be applied to FIGURE 18—48 Problem :17. an incompressible fluid whether it is at rest or flowing steadily. (a) In hydrostatic equilibrium, the flow speed is zero everywhere. Since the pressure is Pa at the left end of the upper horizontal pipe and at the top of the liquid in the small vertical tube, Equation 18-6a for these points gives Pa + O + pghl = Pa + 0 + pghg, or h1 = hg‘ where we measured the heights y from the lower horizoutal tube. (h) In steady flow, Equa— tion 18-5 gives the flow speed in the lower pipe as n“ = u(A/%A) = 21:, where u is the speed in the upper pipe and the cross~sectional areas are given. Then Equa- tion 18-69. gives P,1 + éprfi + pghl = P3 + %p(2u)2 + 0, where P; is the pressure anywhere in the lower pipe. (Since the lower pipe is horizontal. y = D, and uniform in crossesction. u : constant, hence P3 = constant.) Now, even when liquid flows in the pipes, the liquid in the small vertical tube is stagnant. If we assume the pressure is constant over the cross-section of the lower pipe, then Equation 18-3 gives P3 = P,l + pghg. Combining these results, we find P3 ~ RI 2 -%p’u3+ mm = pghg, or hg : hl — 3112/29. I IUUlCl-Il 48. A can of height h is full of water. At what height 3,: should a small hole he cut so the water initially goes as far horizontally as it does vertically, as shown in Fig. 18-49? FIGURE 18-49 Problem 48. Solution In order for the initial horizontal distance to equal the height of the hole, the velocity of efflux (vimie in Example 18-8) times the time of fall must equal 3; in Fig. 18-49. (The Water is in free fall with initial velocity whole : um.) Then 1:; = é—gt2 = uholet, or y = QUIEDIB/g. Using the approximate value of shore = 1/2901 - y} from Example 18-8, where h — y has been substituted for the vertical distance between the height of the fluid and the hole, we find that y = Wall?! — so = 4o — y), or y = 4h/5- {The expression for shop, in Example 18-8 is called Torricelli’s theorem.) Problem 51. The venturi flowmeter shown in Fig. 18-51 is used to measure the flow rate of water in a solar collector system. The fic-wmeter is inserted in a pipe with diameter 1.9 cm; at the venturi of the flowmeter the diameter is reduced to 0.64 cm. The manometer tube contains 01] with density 0.82 times that of water. If the difierence in oil levels on the two sides of the manometer tube is 1.4 cm1 what is the volume flow rate? Solution If we apply Bernoulli’s equation (Equation 18-63.) and the continuity equation (Equation 18-5) to points 1 and 2 in the flowmeter, we can calculate the volume rate of flow: 1 1 P] + -'= P2 + and mill = V2142 imply l 2 2 1 2 1 1 Pl—P225Pivn‘vi)=§Wi-Ai A—E‘E a 2(P1 — Pg] 01' RV = 1.11:4] : -—_'—"—w‘. pm; ~41?) [This is the same calculation as Example 18-9. Note that pressure variation with height in the. flowmeter is assumed negligible.) The pressure difference is related to the difference in height and the density of oil in the manometer (where the fluid is assumed stagnant): P1 = P3 + 991;: and P2 = P3 +pyy2 + Poilgh imply P1 — P2 = [p — poi|)gh, since 3;; — y2 = it. If we use A = §de2 for each part of the flowmeter, we finally _obtain RV = fin 29h(1— Pail/pi/(dgd " did) = 1'20 cm3/s, when the given numerical values are substituted (we used h,d1,d2 in cm and g z 980 CHI/52). ...
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This note was uploaded on 12/03/2009 for the course PHYS 2c taught by Professor Wurthmeir during the Spring '05 term at UCSD.

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hw7 - Problem 5. A plant hangs from a 3.2-cm diameter...

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