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Unformatted text preview: Problem 5. A plant hangs from a 3.2cm diameter suction cup
afﬁxed to a smooth horizontal surface (Fig. 1842).
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 1342 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 crosssectional
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 183.) 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 Ushsped tube open at both ends contains _
water and a quantity of oil occupying a 2.0cm
length of the tube, as shown in Fig. 1845. 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 45cm 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
massload 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 184, 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, Arithee, 2 (1 g/cm3)7r(§ x52 cm)2x (2.8 cm) = 59.5 g. FIGURE 1847 Problem 35 Solution. 39. (a) HouIr 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 184b 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 crosssectional
area is given, the average speed of ﬂow is u = Rv/A =
(18x10“ 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. 1848 a horizontal pipe of crosssectional
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 185) and
Bernoulli’s equation (Equation 186] 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 186a 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 185 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 1869. 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 crosssection of the lower
pipe, then Equation 183 gives P3 = P,l + pghg.
Combining these results, we ﬁnd P3 ~ RI 2 %p’u3+
mm = pghg, or hg : hl — 3112/29. I IUUlClIl 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. 1849? FIGURE 1849 Problem 48. Solution In order for the initial horizontal distance to equal the
height of the hole, the velocity of efﬂux (vimie in
Example 188) times the time of fall must equal 3; in
Fig. 1849. (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 188, 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 188 is called
Torricelli’s theorem.) Problem 51. The venturi ﬂowmeter shown in Fig. 1851 is used
to measure the ﬂow rate of water in a solar
collector system. The ﬁcwmeter 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 1863.) and
the continuity equation (Equation 185) 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)=§WiAi A—E‘E a 2(P1 — Pg] 01' RV = 1.11:4] : —_'—"—w‘.
pm; ~41?) [This is the same calculation as Example 189. 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.
 Spring '05
 wurthmeir

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