Chapter 13
Fluids
25 ••
One sphere is made of gold and has a radius rAu and another sphere is
made of copper and has a radius rCu. If the spheres have equal mass, what is the
ratio of the radii, rAu/ rCu?
Picture the Problem We can use the definition of density to find the ratio of the
radii of the two spheres. See Table 13-1 for the densities of gold and copper.
Use the definition of density to
express the mass of the gold sphere:
3
mAu = ρ AuVAu = 43 πρ Au rAu
The mass of the copper sphere is
given by:
3
mCu = ρ CuVCu = 43 πρ Cu rCu
Dividing the first of these equations
by the second and simplifying yields:
3
ρ
m Au 43 πρ Au rAu
= 4
= Au
3
mCu 3 πρ Cu rCu ρ Cu
Solve for rAu/rCu to obtain:
rAu
m ρ
= 3 Au Cu
rCu
mCu ρ Au
Because the spheres have the same
mass:
rAu
ρ
= 3 Cu
rCu
ρ Au
Substitute numerical values and
evaluate rAu/rCu:
rAu 3 8.93 kg/m 3
=
= 0.773
rCu
19.3 kg/m 3
⎛ rAu ⎞
⎟⎟
⎜⎜
⎝ rCu ⎠
3
29 •
When at cruising altitude, a typical airplane cabin will have an air
pressure equivalent to an altitude of about 2400 m. During the flight, ears often
equilibrate, so that the air pressure inside the inner ear equalizes with the air
pressure outside the plane. The Eustachian tubes allow for this equalization, but
can become clogged. If an Eustachian tube is clogged, pressure equalization may
not occur on descent and the air pressure inside an inner ear may remain equal to
the pressure at 2400 m. In that case, by the time the plane lands and the cabin is
repressurized to sea-level air pressure, what is the net force on one ear drum, due
to this pressure difference, assuming the ear drum has an area of 0.50 cm2?
Picture the Problem Assuming the density of the air to be constant, we can use
the definition of pressure and the expression for the variation of pressure with
depth in a fluid to find the net force on one’s ear drums.
The net force acting on one ear drum
F = ΔPA
1259
1260 Chapter 13
is given by:
where A is the area of an ear drum.
Relate the pressure difference to the
pressure at 2400 m and the pressure
at sea level:
ΔP = Psea level − P2400 m = ρgΔh
Substituting for ΔP in the expression
for F yields:
F = ρgΔhA
Substitute numerical values and evaluate F:
(
)(
)
(
)
F = 1.293 kg/m 3 9.81 m/s 2 (2400 m ) 0.50 cm 2 = 1.5 N
37 •• Many people have imagined that if they were to float the top of a
flexible snorkel tube out of the water, they would be able to breathe through it
while walking underwater (Figure 13-33). However, they generally do not take
into account just how much water pressure opposes the expansion of the chest and
the inflation of the lungs. Suppose you can just breathe while lying on the floor
with a 400-N (90-lb) weight on your chest. How far below the surface of the
water could your chest be for you still to be able to breathe, assuming your chest
has a frontal area of 0.090 m2?
Picture the Problem The depth h below the surface at which you would be able
to breath is related to the pressure at that depth and the density of water ρw
through P = ρ w gh .
Express the pressure due to a column
of water of height h:
P = ρ w gh ⇒ h =
Express the pressure at depth h in
terms of the weight on your chest:
P=
Substituting for P yields:
h=
P
ρw g
won your chest
Aof your chest
won your chest
Aof your chest ρ w g
Substitute numerical values and evaluate h:
h=
400 N
= 45 cm
⎛⎜ 0.090 m 2 ⎞⎟ ⎛⎜1.00 × 10 3 kg/m 3 ⎞⎟ ⎛⎜ 9.81 m/s 2 ⎞⎟
⎠
⎠⎝
⎠⎝
⎝
Fluids 1261
39 ••
A 5.00-kg lead sinker is accidentally dropped overboard by fishermen
in a boat directly above the deepest portion of the Marianas trench, near the
Philippines. By what percentage does the volume of the sinker change by the time
it settles to the trench bottom, which is 10.9 km below the surface?
Picture the Problem This problem is an application of the definition of the bulk
modulus. The change in volume of the sinker is related to the pressure change and
ΔP
.
the bulk modulus by B = −
ΔV V
From the definition of bulk modulus:
Express the pressure at a depth h in
the ocean:
The difference in pressure between
the pressure at depth h and the
surface of the water is:
Substitute for ΔP in equation (1) to
obtain:
ΔV
ΔP
=−
V
B
(1)
P (h) = Pat + ρgh
ΔP = P (h) − Pat = ρgh
ΔV
ρgh
=−
V
B
Substitute numerical values and evaluate the fractional change in volume of the
sinker:
(
)(
)
ΔV
1025 kg/m 3 9.81 m/s 2 (10.9 km )
=−
= − 1.4%
7.7 GPa
V
Buoyancy
41 •
A 500-g piece of copper (specific gravity 8.96) is suspended from a
spring scale and is submerged in water (Figure 13-34). What force does the spring
scale read?
Picture the Problem The scale’s reading will be the difference between the
weight of the piece of copper in air and the buoyant force acting on it.
Express the apparent weight w′ of
the piece of copper:
w' = w − B
1262 Chapter 13
Using the definition of density and
Archimedes’ principle, substitute for
w and B to obtain:
w' = ρ CuVg − ρ wVg
= (ρ Cu − ρ w )Vg
Express w in terms of ρCu and V
and solve for Vg:
w = ρ CuVg ⇒ Vg =
Substitute for Vg in the expression
for w′ to obtain:
w' = (ρ Cu − ρ w )
Substitute numerical values and
evaluate w′:
1 ⎞
⎛
2
w' = ⎜1 −
⎟ (0.500 kg ) 9.81m/s
⎝ 8.96 ⎠
w
ρ Cu
w
ρ Cu
⎛
ρ ⎞
= ⎜⎜1 − w ⎟⎟ w
⎝ ρ Cu ⎠
(
)
= 4.36 N
47 ••
A large piece of cork weighs 0.285 N in air. When held submerged
underwater by a spring scale as shown in Figure 13-35, the spring scale reads
0.855 N. Find the density of the cork.
Picture the Problem The forces acting on the cork are B, the upward force due to
the displacement of water, mg, the weight of the piece of cork, and Fs, the force
exerted by the spring. The piece of cork is in equilibrium under the influence of
these forces.
Apply
∑F
y
= 0 to the piece of cork:
B − w − Fs = 0
(1)
or
B − ρ corkVg − Fs = 0
(2)
Express the buoyant force as a
function of the density of water:
B = wdisplaced = ρ wVg ⇒ Vg =
Substitute for Vg in equation (2) to
obtain:
B − ρ cork
Solve equation (1) for B:
B = w + Fs
Substitute in equation (3) to obtain:
w + Fs − ρ cork
fluid
or
B
ρw
− Fs = 0
w + Fs
ρw
− Fs = 0
B
ρw
(3)
Fluids 1263
w − ρ cork
Solving for ρcork yields:
w + Fs
ρ cork = ρ w
ρw
=0
w
w + Fs
Substitute numerical values and evaluate ρcork:
ρ cork = (1.00 × 103 kg/m 3 )
0.285 N
= 250 kg/m 3
0.285 N + 0.855 N
48 ••
A helium balloon lifts a basket and cargo of total weight 2000 N under
standard conditions, at which the density of air is 1.29 kg/m3 and the density of
helium is 0.178 kg/m3. What is the minimum volume of the balloon?
Picture the Problem Under minimum-volume conditions, the balloon will be in
equilibrium. Let B represent the buoyant force acting on the balloon, wtot represent
its total weight, and V its volume. The total weight is the sum of the weights of its
basket, cargo, and helium in its balloon.
Apply
∑F
y
= 0 to the balloon:
B − wtot = 0
1264 Chapter 13
Express the total weight of the
balloon:
wtot = 2000 N + ρ HeVg
Express the buoyant force due to the
displaced air:
B = wdisplaced = ρ airVg
Substitute for B and wtot to obtain:
ρ airVg − 2000 N − ρ HeVg = 0
Solving for V yields:
V=
fluid
2000 N
(ρ air − ρ He )g
Substitute numerical values and evaluate V:
V =
2000 N
= 183 m 3
3
2
1.29 kg/m − 0.178 kg/m 9.81 m/s
(
)(
3
)
49 ••
[SSM] An object has ″neutral buoyancy″ when its density equals that
of the liquid in which it is submerged, which means that it neither floats nor sinks.
If the average density of an 85-kg diver is 0.96 kg/L, what mass of lead should, as
dive master, suggest be added to give him neutral buoyancy?
Picture the Problem Let V = volume of diver, ρD the density of the diver, VPb the
volume of added lead, and mPb the mass of lead. The diver is in equilibrium under
the influence of his weight, the weight of the lead, and the buoyant force of the
water.
Apply
∑F
y
= 0 to the diver:
Substitute to obtain:
B − wD − wPb = 0
ρ wVD + Pb g − ρ DVD g − mPb g = 0
or
ρ wVD + ρ wVPb − ρ DVD − mPb = 0
Rewrite this expression in terms
of masses and densities:
ρw
Solving for mPb yields:
mPb =
mD
ρD
+ ρw
mPb
ρ Pb
− ρD
mD
ρD
ρ Pb (ρ w − ρ D ) mD
ρ D (ρ Pb − ρ w )
− mPb = 0
Fluids 1265
Substitute numerical values and evaluate mPb:
(11.3 ×10 kg/m )(1.00 ×10 kg/m − 0.96 ×10 kg/m )(85 kg ) =
=
(0.96 ×10 kg/m )(11.3 ×10 kg/m − 1.00 ×10 kg/m )
3
m Pb
3
3
3
3
3
3
3
3
3
3
3
3.9 kg
55 •
[SSM] Water is flowing at 3.00 m/s in a horizontal pipe under a
pressure of 200 kPa. The pipe narrows to half its original diameter. (a) What is
the speed of flow in the narrow section? (b) What is the pressure in the narrow
section? (c) How do the volume flow rates in the two sections compare?
Picture the Problem Let A1 represent the cross-sectional area of the largerdiameter pipe, A2 the cross-sectional area of the smaller-diameter pipe, v1 the
speed of the water in the larger-diameter pipe, and v2 the velocity of the water in
the smaller-diameter pipe. We can use the continuity equation to find v2 and
Bernoulli’s equation for constant elevation to find the pressure in the smallerdiameter pipe.
(a) Using the continuity equation,
relate the velocities of the water to
the diameters of the pipe:
A1v1 = A2 v2
or
πd12
πd 22
d12
v1 =
v2 ⇒ v2 = 2 v1
d2
4
4
Substitute numerical values and
evaluate v2:
⎛ d ⎞
v 2 = ⎜⎜ 1 1 ⎟⎟ (3.00 m/s ) = 12.0 m/s
⎝ 2 d1 ⎠
(b) Using Bernoulli’s equation for
constant elevation, relate the
pressures in the two segments of the
pipe to the velocities of the water in
these segments:
P1 + 12 ρ w v12 = P2 + 12 ρ w v22
Solving for P2 yields:
P2 = P1 + 12 ρ w v12 − 12 ρ w v22
2
(
= P1 + 12 ρ w v12 − v22
)
Substitute numerical values and evaluate P2:
(
)[
]
P2 = 200 kPa + 12 1.00 × 103 kg/m 3 (3.00 m/s ) − (12.0 m/s ) = 133 kPa
2
2
1266 Chapter 13
π d12
(c) Using the continuity equation,
evaluate IV1:
I V1 = A1v1 =
Using the continuity equation,
express IV2:
I V2 = A2v2 =
Substitute numerical values and
evaluate IV2:
Thus, as we expected would be the
case:
v1 =
4
π d 22
4
π d12
4
(3.00 m/s)
v2
2
I V2
⎛d ⎞
π⎜ 1 ⎟
2
= ⎝ ⎠ (12.0 m/s )
4
π d12
(3.00 m/s)
=
4
I V1 = I V2
57 ••
[SSM] Blood flows at at 30 cm/s in an aorta of radius 9.0 mm.
(a) Calculate the volume flow rate in liters per minute. (b) Although the crosssectional area of a capillary is much smaller than that of the aorta, there are many
capillaries, so their total cross-sectional area is much larger. If all the blood from
the aorta flows into the capillaries and the speed of flow through the capillaries is
1.0 mm/s, calculate the total cross-sectional area of the capillaries. Assume
laminar nonviscous, steady-state flow.
Picture the Problem We can use the definition of the volume flow rate to find
the volume flow rate of blood in an aorta and to find the total cross-sectional area
of the capillaries.
(a) Use the definition of the volume
flow rate to find the volume flow
rate through an aorta:
I V = Av
Substitute numerical values and
evaluate IV:
I V = π 9.0 × 10 −3 m 3
(
= 7.634 × 10 −5
) (0.30 m/s)
2
m 3 60 s
1L
×
× −3 3
s min 10 m
= 4.58 L/min = 4.6 L/min
(b) Use the definition of the volume
flow rate to express the volume flow
rate through the capillaries:
I V = Acap vcap ⇒ Acap =
IV
vcap
Fluids
Substitute numerical values and
evaluate Acap:
Acap =
1267
7.63 × 10−5 m 3 /s
= 7.6 × 10−2 m 2
0.0010 m/s
59 ••
The $8-billion, 800-mile long Alaskan Pipeline has a maximum
volume flow rate of 240,000 m3 of oil per day. Along most of the pipeline the
radius is 60.0 cm. Find the pressure P′ at a point where the pipe has a 30.0-cm
radius. Take the pressure in the 60.0-cm-radius sections to be P = 180 kPa and the
density of oil to be 800 kg/m3. Assume laminar nonviscous steady-state flow.
Picture the Problem Let the subscript 60 denote the 60.0-cm-radius pipe and the
subscript 30 denote the 30.0-cm-radius pipe. We can use Bernoulli’s equation for
constant elevation to express P′ in terms of v60 and v30, the definition of volume
flow rate to find v60, and the continuity equation to find v30.
P
P'
r60 = 60.0 cm
r30 = 30.0 cm
v30
v 60
Using Bernoulli’s equation for
constant elevation, relate the
pressures in the two pipes to the
velocities of the oil:
2
2
P + 12 ρv60
= P' + 12 ρv30
Solving for P′ yields:
2
2
P' = P + 12 ρ v 60
− v30
Use the definition of volume flow
rate to express v60:
v 60 =
Using the continuity equation, relate
the speed of the oil in the halfstandard pipe to its speed in the
standard pipe:
A60v60 = A30v30 ⇒ v30 =
Substituting for v60 and A30 yields:
v30 =
(
Substitute for v60 and v30 and simplify to obtain:
)
(1)
IV
I
= V2
A60 π r60
A60
v 60
A30
A60 I V
I
I
= V = V2
A30 A60 A30 π r30
1268 Chapter 13
⎛ ⎛ I ⎞2 ⎛ I ⎞2 ⎞
ρI 2 ⎛ 1 1 ⎞
P' = P + ρ ⎜ ⎜⎜ V2 ⎟⎟ − ⎜⎜ V2 ⎟⎟ ⎟ = P + V2 ⎜⎜ 4 − 4 ⎟⎟
⎜ ⎝ π r60 ⎠ ⎝ π r30 ⎠ ⎟
2π ⎝ r60 r30 ⎠
⎝
⎠
1
2
Substitute numerical values in equation (1) and evaluate P′:
3
⎛
1d
1h ⎞
5 m
⎟⎟
⎜
×
×
800 kg/m ⎜ 2.40 × 10
d
24
h
3600
s
⎠
⎝
P' = 180 kPa +
2π 2
⎛
⎞
1
1
⎟
× ⎜⎜
−
4
4 ⎟
⎝ (0.600 m ) (0.300 m ) ⎠
(
= 144 kPa
3
)
2