SOLIDS AND FLUIDS

10
CONCEPTUAL QUESTIONS
1. SOLVE The molecules in a liquid are close to each other, but the molecules in a gas are far apart. More molecules
in a given volume means that liquids should be significantly denser than gases.
REFLECT Liquids pool at the bottom of a container. Gases expand to fill a container.
2. SOLVE The density of each of the new rods is the same as the density of the original rod.
REFLECT Density is a property of the material, not a property of the object.
3. SOLVE A vacuum pump is actually using the atmospheric pressure to hold up the water column. For the 10 m tall
water column, the difference between vacuum and the atmospheric pressure equals the weight of the water column
divided by the cross-sectional area of the water column.
REFLECT To pump water to a height of 15 m you must apply a pressure difference that is larger than atmospheric
pressure. There are several ways of doing this, for example by using a rotating turbine or by placing water in the
U-shaped pipe with a liquid denser than water in one side of the U.
4. SOLVE Ice floats on water because the buoyant force on the ice equals the gravitational force on the ice. The
buoyant force equals the weight of the displaced water and the gravitational force equals the weight of the ice
(including the part of the ice that’s above the water line). This means that the weight of the displaced water equals
the weight of the ice, so when the ice melts the melt water fits precisely into the volume the ice previously
displaced. Consequently, when the ice melts the water level in the glass stays constant and the glass does not
overflow.
REFLECT Water is unusual; in most substances the solid is denser than liquid. Had water been similar to almost
all other substances we would not have ice floating on top of frozen lakes or frozen seas. Instead, the ice would lie
on the bottom. In such a case, the solid is completely submerged in the liquid, and when the solid melts it expands
and would force a container filled to the rim to overflow.
5. SOLVE The steel ship displaces an amount of water that weighs as much as the ship. The buoyant force from the
displaced water balances the gravitational force on the ship, and the ship floats.
Another way of thinking about it is that it’s not just the density of the steel that matters, but the density of the air
inside the ship. The air inside lowers the average density of the ship far below the density of steel. If the ship were
to take in water, the density of the ship goes up. If it takes in enough water, the ship will sink.
REFLECT The same argument applies to balloons. The skin of the balloon is much denser than air, but the total
weight of the balloon can equal (or even be less than) the weight of the air displaced by the balloon. A lead
zeppelin could actually fly!
6. SOLVE When you float, the buoyant force from the water is opposing and equal in magnitude to the gravitational
force on your body. Because the buoyant force is proportional to the fluid density and to the submerged volume, a
smaller part of you will be submerged when you float on a fluid with higher density. Sea water has a higher density
than fresh water (because of the salt in the water). Consequently, less of you is submerged when you float on the
ocean than when you float on a lake. It is “easier” to float on the ocean.
REFLECT The Dead Sea has a salinity much higher than the world’s oceans and is famous for how easy it is to
float on it.

10.1
10.2 Chapter 10

7. SOLVE To a first approximation, no it doesn’t. The buoyant force equals the density of the liquid times the
submerged volume times g. When the submarine is fully submerged, none of these quantities change.
REFLECT For an extremely precise calculation, you would have to consider that water is not 100%
incompressible. Its density increases slightly the deeper you go. This would increase the buoyant force. On the
other hand, the submarine will also get compressed ever so slightly the deeper it goes. That would decrease the
buoyant force. However, both of these effects are extremely small and do not play a role in most practical
situations.
8. SOLVE Gas is very much a compressible fluid and will expand or compress with increasing or decreasing
pressure. Because the water pressure decreases when the bubble rises, the bubble’s size increases.
REFLECT If you take a deep breath and dive under water, the air in your lungs is like a gas bubble. The deeper
you swim, the more that bubble gets compressed. This means that the buoyant force on you decreases. Normally,
the buoyant force from a big breath of air would make you float to the surface if you stopped swimming. However,
there is a limit where your lungs are compressed so much that if you stop swimming you will sink farther rather
than float back up. Exactly where that limit is varies from person to person and by how big of a breath the person
takes, but it is typically around 10 m.
9. SOLVE The air inside the car is heavier than the helium balloon, so the air is “thrown back,” creating a forward
buoyant force on the balloon. The balloon moves forward.
REFLECT In an accelerating reference frame, the buoyant force is always in the direction of the acceleration.
10. SOLVE (a) At constant speed, the balloon behaves just like it would inside a stopped car — it just floats to the
ceiling. (b) When you round a curve at constant speed, your car is accelerating toward the center point of the curve
(centripetal acceleration). The buoyant force is always in the direction of the acceleration in an accelerating
reference frame, so in this case the balloon moves toward the center point of the curve.
REFLECT If you release a helium balloon inside a spacecraft in apparent weightlessness, the balloon doesn’t go
anywhere at all — there’s no buoyant force.
11. SOLVE A gas is a non-ideal fluid when its compressibility plays role, for example in a gas flow near or above the
speed of sound.
A glass of water spilled out on top of a table is a non-ideal fluid because the velocity at each point in the fluid
changes with time.
The air around the wing tips of an aircraft in flight is a non-ideal fluid because it flows in a vortex that is rotational.
Molasses poured out of a bottle is a non-ideal fluid because it’s viscous.
REFLECT There are other ways in which a fluid can be non-ideal. For example, fluids can be magnetized,
meaning their motions are affected by external and internal magnetic fields.
12. SOLVE According to Bernoulli’s principle for a flow with no elevation changes, if the flow speed increases in
one part of a flow, the pressure decreases, and vice versa. Consequently, the wind blowing across a flat roof lowers
the air pressure above the roof compared to the air pressure below (inside the building). The resulting net force is
directed upwards.
REFLECT This is how the wind can literally lift the roof off a building. (Of course, in many cases there are also
horizontal forces on edges of the roof exposed to the wind, and these forces contribute to taking a roof or part of a
roof off a building.)
13. SOLVE According to Bernoulli’s principle for a flow with no elevation changes, if the flow speed increases in
one part of a flow, the pressure decreases, and vice versa. Consequently, when you blow across the top of the
paper, you lower the air pressure above the paper compared to the air pressure below it. This results in a net force
that is directed upwards, which makes the paper rise.
REFLECT When the paper rises, the airflow is directed at a slight downward angle by Newton’s third law. This
downward airflow is what makes airplanes fly.
14. SOLVE When a mountain glacier melts it turns into water which (by way of streams and rivers) makes its way
into the ocean, raising the sea level. When floating sea ice melts the sea level does not change (compare Problem
 Solids and Fluids 10.3

5.14). This assumes that the density (and consequently the salinity) of the melt water equals the density of sea
water, which would be true for arctic sea ice.
REFLECT The greatest threat to keeping the World’s sea levels constant comes from the melting of the World’s
largest masses of ice located on land, particularly the ice of Antarctica and Greenland.

MULTIPLE-CHOICE PROBLEMS
15. SOLVE The volume of the sphere is:

V=
m
=
(10 kg ) = 3.7 × 10 −3 m 3
ρ ( 2700 kg/m3 )
From the formula for the volume of sphere we can then calculate its radius:
4π
V= 3
r3
3V
r=3 = 9.6 cm
4π
The correct answer is (d).
REFLECT We could probably have gotten the correct answer with a simple estimate. A 10-cm cube of aluminum
has a volume of 1 L and would weigh 2.7 kg. A sphere with a 10 cm radius contains approximate four 10-cm
cubes, or approximate 10 kg.
16. SOLVE Equation 10.1 relates stress and strain:
F ΔL
=Y
A L
Rewrite this equation to calculate the Young’s modulus:

Y=
F L
=
F L
=
( 4.7 kN ) ( 25 cm ) = 15 × 1010 N/m 2
A ΔL π4 d 2 ΔL 4(
1.0 cm ) ( 0.10 mm )
π 2

The correct answer is (b).

REFLECT This value for the Young’s modulus seems very reasonable for a metal rod because it is very close to
the values of both copper and steel.
17. SOLVE The stress is force divided by the cross-sectional area the force is acting on:
F
=
F
=
( 650 N ) = 8.3 × 106 N/m 2
4 (
A π4 d 2 1.0 cm )
π 2

The correct answer is (b).

REFLECT This is another value we should have been able to estimate well enough to get the correct answer.
There are 10 4 cm 2 in 1 m 2 and π is a little less than 1, so the correct answer in SI units must be a bit more than
4
6.5 × 106.
18. SOLVE The pressure difference between atmospheric pressure and the pressure at a depth of 150 m is:
ΔP = ρfluid gh = (1000 kg/m 3 )( 9.80 m/s2 )(150 m ) = 1.47 MPa

The fractional volume change of the steel spoon is given by Equation 10.2:
F ΔV
= −B
A V
ΔV
ΔP = − B
V
ΔV
=−
ΔP
=−
(1.47 MPa ) = −9.2 × 10 −6
V B ( × 1010 N/m 2 )
16

where we have used the bulk modulus for steel from Table 10.2. The negative sign means that the cube is
compressed. The correct answer is (a).
10.4 Chapter 10

REFLECT This is an example of a problem where it is difficult to know if the numerical answer makes sense or
not, because we lack everyday experience of steel getting compressed by large underwater pressures. It is always
good to double-check all the calculations in cases like these.
19. SOLVE The pressure difference between atmospheric pressure and the pressure at a depth of 1.5 km is:
ΔP = ρfluid gh = (1000 kg/m 3 )( 9.80 m/s2 )(1.5 km ) = 1.5 × 10 7 Pa

The correct answer is (c).

REFLECT You can easily figure out the correct answer by doing the multiplication in your head.
20. SOLVE The buoyant force is:
FB = ρfluid gV = ρfluid gs 3 = (1000 kg/m3 )( 9.80 m/s2 )( 3.9 cm ) = 0.58 N
3

The correct answer is (b).

REFLECT The buoyant force does not depend on the density of the submerged object. The only property of the
submerged object it depends on is the object’s volume. The buoyant force also depends on the density of the
displaced fluid.
21. SOLVE The maximum depth of the submarine is:

h=
P
=
( 4.2 MPa ) = 430 m
ρfluid g (1000 kg/m3 )( 9.80 m/s2 )
The correct answer is (d).
REFLECT We have used the formula for pressure as a function of height (or depth) in a fluid column and
rewritten it to express the depth as a function of pressure.
22. SOLVE The net upward force is the buoyant force minus the gravitational force:
Fnet = FB − Fg = ρair gV − mg = ρair gV − ρ heliumVg = Vg ( ρair − ρ helium ) = 4π
3
r 3 g ( ρair − ρ helium )
= π6 d 3 g ( ρair − ρ helium ) = ( 30 cm ) ( 9.80 m/s2 ) ( (1.28 kg/m3 ) − ( 0.179 kg/m3 ) ) = 0.15 N
π 3
6

The correct answer is (a).

REFLECT Optionally you could have calculated numerical values for the individual forces and then subtracted
them to get the same result.
23. SOLVE The volume flow rate is the flow speed times the cross-sectional area. Given the flow rate and the cross-
sectional area (or in this case the diameter of a circular cross-section) we can calculate the flow speed:

v=
Q Q
= π 2 =
( 5.2 L/min ) = 0.49 m/s
4 (
1.5 cm )
π 2
A 4d

The correct answer is (c).

REFLECT Always be careful to do the unit conversions correctly.
24. SOLVE The volume flow rate is the volume of the bucket divided by the time it takes to fill it. Given the flow rate
and the cross-sectional area (or in this case the diameter of a circular cross-section) we can calculate the flow
speed:

V (12 L )
Q
v= = πt 2 =
( 90 s ) = 0.75 m/s
4 (
1.5 cm )
π 2
A 4d

The correct answer is (c).

REFLECT Compare to Problem 10.23.

PROBLEMS
25. ORGANIZE AND PLAN From the density given in Table 10.1 we can calculate the mass of 1 L of liquid water.
With the mass known we can calculate the volume of gas (vapor).
 Solids and Fluids 10.5

Known: Vl = 1 L; ρl = 1000 kg/m 3 ; ρ g = 0.804 kg/m 3 .

SOLVE The mass of water is:
m = Vl ρl = (1 L )(1000 kg/m 3 ) = 1 kg

The volume of this mass of water in the gas state is:

Vg =
m
=
(1 kg ) = 1 m3
ρg ( 0.804 kg/m 3 )
REFLECT The volume expands by approximately 3 orders of magnitude.
26. ORGANIZE AND PLAN To calculate the mass of a certain volume if liquid, we need to know its density. The
density of gasoline is given in Table 10.1.
Known: Vl = 56 L; ρl = 680 kg/m 3 .
SOLVE The mass of gasoline is:
m = Vl ρl = ( 56 L )( 680 kg/m 3 ) = 38 kg

REFLECT Most liquids have densities similar to water, approximate 1 kg/L. A small difference in density is
important, however, because it determines which liquid will float on top of the other if two liquids are placed in the
same container (assuming the liquids don’t mix).
27. ORGANIZE AND PLAN Two objects with the same size and shape have the same volume. If the volume is the
same, the mass ratio equals the density ratio. We can get the densities from Table 10.1.
Known: ρ Pb = 11,300 kg/m 3 ; ρU = 19,100 kg/m 3 .
SOLVE The mass of the uranium bullet divided by the mass of the lead bullet is:
mU ρU (19,100 kg/m 3 )
= = =1.6903
mPb ρ Pb (11,300 kg/m 3 )

The uranium bullet is approximately 1.7 times heavier than a lead bullet of the same size and shape.
REFLECT There are elements that are even heavier than uranium. Why are those elements not used as armor-
penetrating bullets?
28. ORGANIZE AND PLAN With the density of gasoline from Table 10.1 we can calculate the mass of 1 L of gasoline.
Then, with the density of water we can calculate the volume of water with this mass.
Known: Vgasoline = 1 L; ρ gasoline = 680 kg/m 3 ; ρ water = 1000 kg/m 3 .
SOLVE The mass of 1.0 L of gasoline is:
m = Vgasoline ρ gasoline = (1 L )( 680 kg/m 3 ) = 0.7 kg

The volume of 0.7 kg of water is:

Vwater =
m
=
( 0.7 kg ) 0.7 L
ρ water (1000 kg/m3 )
REFLECT Gasoline floats on top of water. This can make it difficult to extinguish a gasoline fire.
29. ORGANIZE AND PLAN The person’s goal is to change his volume so that it equals 65 L (that is the volume of 65
kg of water). If we first calculate the volume of the person with 2.4 L in his lungs, the amount of air he has to let
out is the difference between this volume and 65 L.
Known: m = 65 kg; ρ 2.4 L = 990 kg/m 3 ; ρ water = 1000 kg/m 3 .
SOLVE The person’s goal is to change his volume so it equals:

V=
m
=
( 65 kg ) = 65 L
ρ water (1000 kg/m 3 )
The person’s initial volume equals:

V2.4 L =
m
=
( 65 kg ) = 66 L
ρ 2.4 L ( 990 kg/m3 )
10.6 Chapter 10

The amount of air to let out is:

ΔV = V2.4 L − V = ( 66 L ) − ( 65 L ) = 0.66 L

REFLECT The starting volume of air in the person’s lungs did not enter into the calculation, but we should of
course check that he has enough air to let out to achieve the density change. He can let out 0.66 L of air because
this number is smaller than what he started with (2.4 L).
30. ORGANIZE AND PLAN The stretch can be calculated from Equation 10.1 if we know the Young’s modulus of
copper. We can find the Young’s modulus in Table 10.2.
Known: L = 0.35 m; d = 7.0 mm; F = 1.2 kN; Y = 11 × 1010 N/m 2 .
SOLVE Equation 10.1 relates stress and strain:
F ΔL
=Y
A L
Rewrite this equation to calculate the stretch:
F F (1.2 kN )
ΔL = L= L= ( 0.35 m ) = 99 μm
4 (
7.0 mm ) (11 × 1010 N/m 2 )
π π 2
AY 4
d 2Y

REFLECT The quantity F / ( AY ) is dimensionless and equals the strain.

31. ORGANIZE AND PLAN The force compressing the block is the gravitational force on the mass of the man. From
the force we can calculate the compression using Equation 10.1 if we know the Young’s modulus of concrete. We
can find the Young’s modulus in Table 10.2.
Known: L = 28 cm; m = 95 kg; Y = 3 × 1010 N/m 2 .
SOLVE Equation 10.1 relates stress and strain:
F ΔL
=Y
A L
Rewrite this equation to calculate the compression, keeping in mind that the cross-sectional area A of a cube
equals L2 :

ΔL =
F mg
L= 2 L=
mg
=
( 95 kg )( 9.80 m/s2 ) = 1.1 × 10−7 m
AY LY LY ( 28 cm )( 3 × 1010 N/m 2 )

REFLECT This is an example of a problem where it is difficult know if the numerical answer makes sense or not,
because we can’t tell by our own senses how much a block of concrete compresses as we climb up on it. It is
always good to double-check all the calculations in cases like these.
32. ORGANIZE AND PLAN The force causing the stretch is the gravitational force on the 10.0-kg mass. The stretch
can be calculated from Equation 10.1 if we know the Young’s modulus of copper. We can find the Young’s
modulus in Table 10.2.
Known: L = 1.05 m; d = 1.50 mm; m = 10.0 kg; Y = 11 × 1010 N/m 2 .
SOLVE Calculate the stretch from Equation 10.1:
F ΔL
=Y
A L
F mg (10.0 kg )( 9.80 m/s2 )
ΔL = L= π 2 L= (1.05 m ) = 5.29 × 10 −4 m
4 (
1.50 mm ) (11 × 1010 N/m 2 )
π 2
AY 4
dY

REFLECT If you like you can split the calculation into intermediate steps and calculate values for the force F and
the area A.
33. ORGANIZE AND PLAN The required force can be calculated from Equation 10.1 if we know the Young’s modulus
of steel. We can find the Young’s modulus in Table 10.2. The mass to hang on the rod is the required force divided by g.
Known: L = 1.5 m; d = 1.2 mm; ΔL = 0.50 mm; Y = 20 × 1010 N/m 2 .
 Solids and Fluids 10.7

SOLVE Calculate the required force from Equation 10.1:

F ΔL
=Y
A L
ΔL π 2 ΔL ( 0.50 mm ) = 75 N
(1.2 mm ) ( 20 × 1010 N/m 2 )
π 2
F = AY = 4dY =
L L 4
(1.5 m )
The mass to stretch the steel rod 0.50 mm is:

m=
F
=
( 75 N ) = 7.7 kg
g ( 9.80 m/s2 )

REFLECT Equation 10.1 tells us that if we had placed the mass on top of the rod instead, we would have
compressed the rod by the same amount, 0.50 mm. While this is mathematically correct, it would be difficult to do
this in practice because the rod is very thin and would likely bend rather than compress.
34. ORGANIZE AND PLAN The compression can be calculated from Equation 10.1, keeping in mind that the woman
has two femurs, each holding the gravitational force from only half of the upper body mass.
Known: L = 29 cm; d = 3.8 cm; m = 12 × 48 kg; Y = 5.0 × 109 N/m 2 .
SOLVE Calculate the compression from Equation 10.1:
F ΔL
=Y
A L
F mg ( 12 × 48 kg ) ( 9.80 m/s2 ) 29 cm = 1.2 × 10−5 m
ΔL = L= π 2 L= ( )
4 (
3.8 cm ) ( 5.0 × 10 9 N/m 2 )
π 2
AY 4
dY

REFLECT Instead of dividing the mass by two, you could have kept the full mass and said that it is supported by
two femurs, i.e., doubled the area instead. Obviously, this gets you the same result.
35. ORGANIZE AND PLAN The stretch can be calculated from Equation 10.1 if we know the Young’s modulus of
steel. We can find the Young’s modulus in Table 10.2.
Known: L = 73 cm; d = 0.15 mm; F = 15 N; Y = 20 × 1010 N/m 2 .
SOLVE Calculate the stretch from Equation 10.1:
F ΔL
=Y
A L
F F (15 N )
ΔL = L= L= ( 73 cm ) = 0.31 cm
4 (
0.15 mm ) ( 20 × 1010 N/m 2 )
π 2 π 2
AY 4
d Y

REFLECT We have assumed that the material in the guitar string is not stressed beyond its elastic limit. The
stretch is about 0.42% of the original length of the guitar string. This should be within the elastic limit.
36. ORGANIZE AND PLAN The compression can be calculated from Equation 10.1 if we know the Young’s modulus
of concrete. We can find the Young’s modulus in Table 10.2.
Known: L = 8.9 m; d = 45 cm; m = 12,500 kg; Y = 3 × 1010 N/m 2 .
SOLVE Calculate the compression from Equation 10.1:
F ΔL
=Y
A L
F mg (12,500 kg )( 9.80 m/s2 ) 45 cm = 1 × 10−5 m
ΔL = L= π 2 L= ( )
4 (
45 cm ) ( 3 × 1010 N/m 2 )
π 2
AY 4
dY

REFLECT A structural engineer may also have to take into account the mass of the concrete itself (about 3400 kg).
37. ORGANIZE AND PLAN The gravitational force on the mass of a column of liquid causes a pressure difference
between the top of the liquid and the bottom, as described by Equation 10.4. In a barometer, the top of the fluid is
at zero pressure ( P0 = 0) because the liquid is contained in an evacuated tube. Other than this we only need to
know the density of the liquid to calculate the height of the liquid column in a barometer.
Known: P = 1 atm; P0 = 0; ρ = 1000 kg/m 3 .
10.8 Chapter 10

SOLVE Calculate the height of the water column in a barometer from Equation 10.1:
P = P0 + ρ gh

h=
P − P0
=
(1 atm ) − ( 0 ) = 10 m
ρg (1000 kg/m3 )( 9.80 m/s2 )
REFLECT A water barometer is not practical if you follow the standard design of a mercury barometer. However,
by building a barometer in a U-shape where the two columns have different cross-sectional area, you can build a
water barometer similar to the hydraulic lift in Figure 10.7.
38. ORGANIZE AND PLAN The pressure difference between sea level ( P0 = 1 atm ) and the ocean trench is given by
Equation 10.4.
Known: h = 10.9 km; P0 = 1 atm; ρ = 1000 kg/m 3 .
SOLVE (a) The pressure at a depth of 10.9 km is:
P = P0 + ρ gh = (1 atm ) + (1000 kg/m 3 )( 9.80 m/s2 )(10.9 km ) = 107 MPa

(b) Compared to atmospheric pressure:

P (107 MPa ) (1.07 × 108 Pa )
= = = 1.06 × 103
P0 (1 atm ) (1.01 × 105 Pa)
i.e., the pressure in the deepest ocean trench is more than 1,000 times the atmospheric pressure!
REFLECT Despite the immense pressure at 10.9-km depth, the Mariana trench has been visited by humans in
submersibles.
39. ORGANIZE AND PLAN The pressure difference between sea level ( P0 = 1 atm ) and the ocean trench is given by
Equation 10.4. The fractional volume change due to compression forces can be calculated from Equation 10.2,
where the force per unit area is the pressure. We also need to know the bulk modulus of steel, and that is listed in
Table 10.2.
Known: h = 5.75 km; P0 = 1 atm; ρ = 1000 kg/m 3 ; B = 16 × 1010 N/m 2 .
SOLVE (a) The pressure at a depth of 5.75 km is:
P = P0 + ρ gh = (1 atm ) + (1000 kg/m 3 )( 9.80 m/s2 )( 5.75 km ) = 56.5 MPa

(b) The fractional volume change of the steel spoon is given by Equation 10.2:
F ΔV
= −B
A V
ΔV
P = −B
V
ΔV P
=− =−
( 5.65 × 107 N/m 2 ) = −3.53 × 10 −4 m
V B (16 × 1010 N/m 2 )
REFLECT The spoon shrinks due to the compression forces.
40. ORGANIZE AND PLAN The pressure difference in a column of fluid with constant density equals the height of the
column times the density times g.
Known: h = 100 m; ρ = 1.28 kg/m 3 .
SOLVE For an altitude increase of 100 m, the air pressure decreases by:
ΔP = ρ gh = (1.28 kg/m 3 )( 9.80 m/s2 )(100 m ) = 1.25 kPa

REFLECT This is equivalent of 1.24% of an atmosphere. Is the assumption of constant air density a good one?
For a small elevation change like 100 m, it is fairly good, but it does not hold for larger elevation changes.
Passenger aircraft typically fly at altitudes of 10,000 m. If you assumed a constant air density, how much would the
air pressure decrease for an altitude increase of 10,000 m?
41. ORGANIZE AND PLAN The pressure difference between sea level ( P0 = 1 atm ) and an ocean depth is given by
Equation 10.4. We can rewrite this equation to solve for a depth given a pressure.
 Solids and Fluids 10.9

Known: P = 3 atm; P0 = 1 atm; ρ = 1000 kg/m 3 .

SOLVE Rewrite Equation 10.4 to solve for the depth:
P = P0 + ρ gh

h=
P − P0
=
( 3 atm ) − (1 atm ) = 2 × 101 m
ρg (1000 kg/m3 )( 9.80 m/s2 )
REFLECT When diving, the pressure increases by about 1 atm for every 10 m depth.
42. ORGANIZE AND PLAN A gauge pressure is the pressure difference from atmospheric pressure and equals the
height of a column of liquid times the density of the liquid times g.
Known: ΔP = 4 atm; ρ = 1000 kg/m 3 .
SOLVE The height of the water tower should be:
ΔP = ρ gh

h=
ΔP
=
( 4 atm ) = 4 × 101 m
ρ g (1000 kg/m 3 )( 9.80 m/s2 )
REFLECT Water pressure increases by about 1 atm for every 10 m depth.
43. ORGANIZE AND PLAN The fractional volume change due to compression forces is given by Equation 10.2. Since
pressure is force per unit area we can use this equation (and the bulk modulus of steel from Table 10.2) to calculate
the required pressure to compress the steel ball. The pressure at a certain depth is given by Equation 10.4, which
we will rewrite to solve for the depth.
Known: ΔV / V = 0.015%; P0 = 1 atm; ρ = 1000 kg/m 3 ; B = 16 × 1010 N/m 2 .
SOLVE The required pressure to compress the steel ball is:
F ΔV
P= = −B = − (16 × 1010 N/m 2 )( −0.015% ) = 2.4 × 10 7 N/m 2
A V
The ocean depth with this pressure is:
P = P0 + ρ gh
P − P0 ( 2.4 × 10 7 N/m 2 ) − (1 atm )
h= = = 2.4 km
ρg (1000 kg/m 3 )( 9.80 m/s2 )
REFLECT The problem text didn’t say whether the original volume of the steel ball was measured in vacuum or
in atmospheric pressure. Does that change the numerical answer?
44. ORGANIZE AND PLAN The water exerts a force on the bottom of the pool equal to its pressure times the area of
the bottom of the pool. The pressure can be calculated from Equation 10.4.
Known: h = 3.0 m; A = ( 25 m ) × (15 m ) = 3.8 × 10 2 m 2 ; P0 = 1 atm; ρ = 1000 kg/m 3 .
SOLVE The pressure at the bottom of the pool is:
P = P0 + ρ gh = (1 atm ) + (1000 kg/m 3 )( 9.80 m/s2 )( 3.0 m ) = 1.3 × 105 Pa

The force is:

F = PA = (1.3 × 105 Pa )( 3.8 × 102 m 2 ) = 4.9 × 107 N

REFLECT At this relatively shallow depth, most of the pressure (and most of the force) still comes from the
atmospheric pressure.
45. ORGANIZE AND PLAN The diastolic pressure is a gauge pressure, i.e., relative to the pressure of the surrounding
atmosphere. A gauge pressure equals the height of a column of liquid times the density of the liquid times g. Since
we know the pressure we can calculate the height.
Known: ΔP = 70 mm Hg; ρ = 1060 kg/m 3 .
SOLVE The diastolic pressure should equal the pressure difference ΔP over a column with height h :
ΔP = ρ gh
10.10 Chapter 10

Solve for h:
⎛ 101325 Pa ⎞
( 70 mm Hg ) ⎜ 760 mm Hg ⎟
h=
ΔP
=
( 70 mm Hg ) = ⎝ ⎠ = 90 cm
ρ g (1060 kg/m 3 )( 9.80 m/s2 ) (1060 kg/m3 )( 9.80 m/s2 )
REFLECT The ratio between 90 cm and 70 mm equals the ratio between the density of mercury and the density of
blood.
46. ORGANIZE AND PLAN We want to know what force is required to hold the car at 2.0 m elevation above the piston
where we apply the force. When the car was level with our piston, the required force was 590 N. Now the force
must hold both the mass of the car and the added mass of 2.0 m of hydraulic fluid. This added mass is the volume
of the fluid times its density, which we know. Following Example 10.6, the added force must equal the
gravitational force on the added mass times the ratio of the piston areas.
Known: A2 = 20 A1; mcar = 1200 kg; h = 2.0 m; A1 = 0.10 m 2 ; ρ = 850 kg/m 3 .
SOLVE The added mass on the right-hand side is:
mfluid = ρV = ρ hA2 = 20 ρ hA1 = 20 ( 850 kg/m 3 )( 2.0 m )( 0.10 m 2 ) = 3400 kg

This means that the force on the left-hand side must now hold up a total mass of:
mtotal = mcar + mfluid = (1200 kg ) + ( 3400 kg ) = 4600 kg

The required force on the right-hand side is given by the pressure-balance equation of Example 10.6:
A1 A 1
F1 =
A2
F2 = 1 mtotal g =
A2 20
( 4600 kg )( 9.80 m/s2 ) = 2.3 kN
REFLECT This type of hydraulic lift is very useful in that it reduces the force required to lift a heavy object, such
as a car. However, when the height of the lift becomes large, the mass of the hydraulic fluid itself increases and at
some point you would be better off lifting the object directly, foregoing the lift. In this example, with a moderate
lift of 2.0 m, the force is still reduced compared to lifting the car directly (but it’s only reduced to 20%, not 5% as
in Example 10.6).
47. ORGANIZE AND PLAN The pressure on each piston is the air pressure plus the applied force on that piston divided
by the piston area. The pressures on the two pistons are equal when the system is in equilibrium. Because the
pistons are at the same height, the air pressure is the same on both pistons.
Known: A1 = 0.50 m 2 ; A2 = 5.60 m 2 ; F1 = 2.0 kN.
SOLVE The system is in equilibrium when:
F1 F2
=
A1 A2

This means that the larger piston can support a force:

A2 ( 5.60 m 2 ) 2.0 kN = 22 kN
F2 = F1 = ( )
A1 ( 0.50 m 2 )
i.e., it can support a mass:

m2 =
F2
=
( 22 kN ) = 2.3 × 103 kg
g ( 9.80 m/s2 )

REFLECT The first equation in our solution can be rewritten:

F1 A1
=
F2 A2
 Solids and Fluids 10.11

48. ORGANIZE AND PLAN The pressure difference from the top to the bottom of a column of fluid equals the density of the
fluid times the height of the column times g. When one column is placed on top of another, the pressures add.
Known: hglycerin = 30 cm; ρglycerin = 1260 kg/m 3 ; hwater = 35 cm; ρ water = 1000 kg/m 3 .
SOLVE The pressure difference between the bottom of the glycerin and the top of the glycerin is:
ΔPglycerin = ρ glycerin ghglycerin = (1260 kg/m 3 )( 9.80 m/s2 )( 30 cm ) = 3.7 kPa

The pressure difference between the bottom of the water and the top of the water is:
ΔPwater = ρ water ghwater = (1000 kg/m 3 )( 9.80 m/s2 )( 35 cm ) = 3.4 kPa

The pressure difference between the bottom of the cylinder and the top of the cylinder is:
ΔP = ΔPglycerin + ΔPwater = ( 3.7 kPa ) + ( 3.4 kPa ) = 7.1 kPa
REFLECT Water is lighter than glycerin and sits on top of the glycerin. The total pressure at the top of the
cylinder is the atmospheric pressure P0 . The total pressure at the bottom of the water, which is also the top of the
glycerin, is P0 + ΔPwater . The total pressure at the bottom of the glycerin, which is also the bottom of the cylinder, is
P0 + ΔPwater + ΔPglycerin .

49. ORGANIZE AND PLAN The buoyant force is given by Archimedes’s principle, Equation 10.5. We can calculate
the buoyant force since we know that the density of water is 1000 kg/m 3 .
Known: V = 185 m 3 ; ρ fluid = 1000 kg/m 3 .
SOLVE Insert the known values into Equation 10.5 to calculate the buoyant force:
FB = ρ fluid gV = (1000 kg/m 3 )( 9.80 m/s2 )(185 m 3 ) = 1.81 × 106 N

REFLECT This is a large force. It’s difficult to submerge a submarine!

50. ORGANIZE AND PLAN The buoyant force is given by Archimedes’s principle, Equation 10.5. We can calculate
the buoyant force since we know from Table 10.1 that the density of air is 1.28 kg/m 3 .
Known: d = 17.5 cm; ρ fluid = 1.28 kg/m 3 .
SOLVE First we calculate the volume of the balloon:

4π r 3 π d 3 π (17.5 cm )
3

V= = = = 2.81 × 103 cm 3 = 2.81 L

3 6 6
Then we use Equation 10.5 to calculate the buoyant force:
FB = ρ fluid gV = (1.28 kg/m 3 )( 9.80 m/s2 )( 2.81 L ) = 0.0352 N

REFLECT This is a small force. The balloon will not be able to lift very much.
51. ORGANIZE AND PLAN To calculate the buoyant force using Equation 10.5 we need to know the parachutist’s
volume, which is his mass divided by his density. We also need to know the density of air: 1.28 kg/m 3 .
Known: m = 70 kg; ρ = 1050 kg/m 3 ; ρ fluid = 1.28 kg/m 3 .
SOLVE First we calculate the volume of the parachutist:

V=
m
=
( 70 kg ) = 0.067 m 3
ρ (1050 kg/m 3 )
Then we use Equation 10.5 to calculate the buoyant force:
FB = ρ fluid gV = (1.28 kg/m 3 )( 9.80 m/s2 )( 0.067 m 3 ) = 0.84 N

The weight of the parachutist is:

W = mg = ( 70 kg )( 9.80 m/s2 ) = 6.9 × 10 2 N
The weight of the parachutist is obviously much larger than the buoyant force from the air. To compare the two we
can calculate their ratio:
W ( 6.9 × 10 2 N )
= = 8.2 × 10 2
FB ( 0.84 N )
REFLECT The ratio we calculated equals ρ / ρfluid .
10.12 Chapter 10

52. ORGANIZE AND PLAN The acceleration is given by Newton’s second law equals the net force on the object
divided by its mass. The net force is the sum of the forces acting on the object. These forces are the gravitational
force and the buoyant force.
Known: m = 2.50 g; ρ = 7140 kg/m 3 ; ρfluid = 1000 kg/m 3 .
SOLVE With our usual convention of upwards being the positive direction, the gravitational force on the penny is:
Fg = −mg = − ( 2.50 g )( 9.80 m/s2 ) = −24.5 mN

The buoyant force can be calculated from Equation 10.5:

FB = ρ fluid gV = ρfluid g
m
= (1000 kg/m 3 )( 9.80 m/s2 )
( 2.50 g ) = 3.43 mN
ρ ( 7140 kg/m3 )
The acceleration is the net force divided by the mass:
Fnet Fg + FB ( −24.5 mN ) + ( 3.43 mN )
a= = = = −8.43 m/s2
m m ( 2.50 g )
REFLECT Combining all the expressions without calculating intermediate values for the forces, we find that the
acceleration is:
⎛ρ ⎞
a = g ⎜ fluid − 1⎟
⎝ ρ ⎠
That means we didn’t need to know the mass of the penny.
53. ORGANIZE AND PLAN Since the iceberg is not accelerating, the net force acting on the iceberg must be zero. The
net force is the sum of the gravitational force and the buoyant force. That means the buoyant force is equal in
magnitude to the gravitational force but in the opposite direction. Once we know the buoyant force we can use
Archimedes’s principle to calculate the volume of water displaced by the iceberg. Dividing this volume with the
total volume of the iceberg we get the fraction of the iceberg’s volume that is below the water line.
Known: m = 6500 kg; ρ ice = 931 kg/m 3 ; ρ fluid = 1030 kg/m 3 .
SOLVE (a) The buoyant force is:
FB = − Fg = − ( − mg ) = mg = ( 6500 kg )( 9.80 m/s2 ) = 63.7 kN

(b) The volume of displaced water is calculated from Equation 10.5:

FB = ρ fluid gV

V=
FB
=
( 63.7 kN ) = 6.311 m 3
ρ fluid g (1030 kg/m3 )( 9.80 m/s2 )
(c) The fraction of the iceberg’s volume that is below that water line equals the volume of displaced water divided
by the total volume of the iceberg:
V ρ V ( 931 kg/m 3 )( 6.311 m 3 )
= ice = = 90.4%
m m ( 6500 kg )
ρ ice
REFLECT The fraction equals ρ ice / ρfluid .

54. ORGANIZE AND PLAN In Problem 10.53 we learned that the fraction of ice that is below that water line is
ρ ice / ρfluid , which means that the fraction that is above the water’s surface equals:
ρ ice
1−
ρfluid
Known: ρ ice = 931 kg/m 3 ; ρfluid = 1000 kg/m 3 .
 Solids and Fluids 10.13

SOLVE The answer is:

1−
ρ ice
=1−
( 931 kg/m3 ) = 6.90%
ρfluid (1000 kg/m3 )
REFLECT Compare to the result of Problem 10.53. Because of sea water’s higher density, ice floats higher in sea
water than in fresh water.
55. ORGANIZE AND PLAN In Problem 10.53 we learned that the fraction of a floating object that is below that water
line is ρsolid / ρ fluid . From this we can calculate the density of the wood ball.
Known: ρfluid = 1000 kg/m 3 ; ρsolid / ρfluid = 12 .
SOLVE The answer is:

ρsolid =
ρfluid
=
(1000 kg/m3 ) = 500.0 kg/m3
2 2
REFLECT In this problem (and in Problems 10.53 and 10.54) we have neglected the buoyant force from the air.
This is reasonable because the buoyant force from air is 3 orders of magnitude smaller than the other forces.
However, it would be prudent to round off our calculated density of the wood ball to 500 kg/m 3 .
56. ORGANIZE AND PLAN The scale measures the force Fnet acting on the scale, giving the measurement as an
apparent mass ma = − Fnet /g. (We have introduced a minus sign here because a scale measures the net downward
force, and we usually think of the downward direction as the negative y-direction.) The net force in this case is the
gravitational force acting on the person reduced by the buoyant force acting on the person. The buoyant force can
be calculated knowing that the person’s volume is his mass divided by his density.
Known: m = 89.2 kg; ρ = 1025 kg/m 3 ; ρfluid = 1000 kg/m 3 .
SOLVE The gravitational force is:
Fg = − mg = − ( 89.2 kg )( 9.80 m/s2 ) = −874 N

The buoyant force is calculated from Equation 10.5:

FB = ρ fluid gV = ρfluid g
m
= (1000 kg/m 3 )( 9.80 m/s2 )
(89.2 kg ) = 853 N
ρ (1025 kg/m3 )
The scale then reads:

ma =
(
− Fnet − Fg + FB
= =
) (
− ( −874 N ) + ( 853 N )
= 2.18 kg
)
g g ( 9.80 m/s2 )
REFLECT Without calculating the forces in intermediate steps, you can express the final answer only in terms of
m, ρ , and ρ fluid , without using g. Does this mean that the scale would read the same result if the person were
submerged in a swimming pool on the Moon?
57. ORGANIZE AND PLAN The scale measures the net downward force acting on the scale. The net force in this case
is the gravitational force acting on the person reduced by the buoyant force acting on the person. Since we know
what the scale reads and the person’s mass, we can calculate the buoyant force. From the buoyant force we can
calculate the person’s volume and then his density.
Known: m = 69.5 kg; Fnet = −22.0 N; ρfluid = 1000 kg/m 3 .
SOLVE The net force on the scale is Fnet = Fg + FB where the gravitational force is:

Fg = −mg = − ( 69.5 kg )( 9.80 m/s2 ) = −681 N

and the buoyant force is:

FB = ρ fluid gV
We can solve for the person’s volume:
Fnet = Fg + ρ fluid gV

V=
Fnet − Fg
=
( −22.0 N ) − ( −681 N ) = 6.73 × 10−2 m3
ρ fluid g (1000 kg/m3 )( 9.80 m/s2 )
10.14 Chapter 10

Finally, from the person’s volume and mass we can calculate the person’s density:

ρ=
m
=
( 69.5 kg ) = 1.03 × 103 kg/m3
V ( 6.73 × 10 −2 m 3 )

REFLECT The net force must be in the same direction as the gravitational force; otherwise the person wouldn’t
be able to sit on the scale but would float to the surface!
58. ORGANIZE AND PLAN Archimedes’s principle tells us that the buoyant force on the mattress is equal to the
weight of the displaced fluid. The mattress can support a total weight (including its own weight) that equals the
buoyant force. To calculate the volume of the mattress we will assume that it is shaped like a rectangular block.
Known: mmattress = 0.39 kg; V = (1.90 m )( 0.75 m )( 0.11 m ) = 0.16 m 3 ; ρ fluid = 1000 kg/m 3 .
SOLVE The buoyant force is calculated from Equation 10.5:
FB = ρ fluid gV = (1000 kg/m 3 )( 9.80 m/s2 )( 0.16 m 3 ) = 1.5 kN

The total mass that can be supported is:

m=
FB
=
(1.5 kN ) = 1.6 × 102 kg
g ( 9.80 m/s2 )

which includes the air mattress’s own mass and the mass of the air inside the mattress. Both of these masses are too
small to impact our final answer; the air mattress weighs only 0.39 kg, which is less than our smallest significant
digit for m above, and the air inside the mattress weighs even less!
REFLECT Because a human body has approximately the same density as water, you can only sink an air mattress
when you sit on it if the volume of the air mattress is smaller than your volume.
59. ORGANIZE AND PLAN The scale measures the force Fnet acting on the scale, giving the measurement as an
apparent mass ma = − Fnet /g. The net force in this case is the gravitational force reduced by the buoyant force.
Known: d = 5.0 cm; ρ = 2700 kg/m 3 ; ρair = 1.28 kg/m 3 ; ρ water = 1000 kg/m 3 .
SOLVE (a) The gravitational force is:
Fg = −mg = − ρVg = − ρ d 3 g = − ( 2700 kg/m 3 )( 5.0 cm ) ( 9.80 m/s2 ) = −3.3 N
3

The buoyant force in air is calculated from Equation 10.5:

FB = ρair gV = ρair gd 3 = (1.28 kg/m 3 )( 9.80 m/s2 )( 5.0 cm ) = 1.5 mN
3

In air, the scale reads:

ma =
(
− Fnet − Fg + FB
= =
) (
− ( −3.3 N ) + (1.5 mN ) )
= 0.34 kg
g g ( 9.80 m/s2 )
(b) The buoyant force in water is:
FB = ρ water gV = ρ water gd 3 = (1000 kg/m 3 )( 9.80 m/s2 )( 5.0 cm ) = 1.2 N
3

In water, the scale reads:

ma =
(
− Fnet − Fg + FB
= =
) (
− ( −3.3 N ) + (1.2 N ) )
= 0.21 kg
g g ( 9.80 m/s2 )
REFLECT As you can see, the buoyant force from air is very small, typically 3 orders of magnitude smaller than
the gravitational force on a solid. Only for unusual materials with very low densities does the buoyant force from
air matter (see Problem 10.62 for an example).
60. ORGANIZE AND PLAN To lift the person, the buoyant force on the balloon must be equal to or larger than the
combined gravitational forces on the person, the skin of the balloon, and the helium in the balloon.
Known: mperson = 60 kg; mskin = 2 kg; ρ helium = 0.179 kg/m 3 ; ρair = 1.28 kg/m 3 .
SOLVE The total mass of person plus balloon is:
m = mperson + mskin + mhelium = mperson + mskin + ρ heliumV
 Solids and Fluids 10.15

where V is the volume of the spherical balloon. The buoyant force is:
FB = ρair gV
To lift the person, the net upward force on the person must be positive:
Fnet = Fg + FB = − mg + ρair gV ≥ 0

This gives us an equation we can solve for the balloon’s volume:

− mg + ρair gV ≥ 0
− m + ρairV ≥ 0
− mperson − mskin − ρ heliumV + ρairV ≥ 0
( ρair − ρhelium )V ≥ mperson + mskin
m
V ≥ person
+ mskin
=
( 60 kg ) + ( 2.0 kg ) = 56 m 3
ρair − ρ helium (1.28 kg/m 3 ) − ( 0.179 kg/m 3 )
Since the balloon was specified to be spherical, we can also express our answer as a radius for the ballon:
4π r 3
V=
3
3V 3 3 ( 56 m 3 )
r=3 = = 2.4 m.
4π 4π
REFLECT The density of hydrogen is only half the density of helium. Does that mean a hydrogen balloon
produces twice the lift of an identical-sized helium balloon?
61. ORGANIZE AND PLAN We are only asked to estimate the percentage body fat, not calculate it precisely.
A reasonable estimate would be to place the woman’s density on a linear interpolation between the density of
100% fat and the density of 0%.
Known: ρ fat = 900 kg/m 3 ; ρ nonfat = 1100 kg/m 3 ; ρ = 1060 kg/m 3 .
SOLVE The estimate of the woman’s body fat is:
ρ − ρ nonfat
× 100% =
(1060 kg/m3 ) − (1100 kg/m3 ) × 100% = 20%
ρ fat − ρ nonfat ( 900 kg/m 3 ) − (1100 kg/m3 )
REFLECT A precise calculation is possible. Begin by dividing the woman’s mass with her volume:
m mfat + mnonfat m + mnonfat
ρ= = = fat
V Vfat + Vnonfat mfat mnonfat
+
ρfat ρ nonfat
Divide this equation with ρ nonfat on both sides and define a variable x = mfat / mnonfat to rewrite the equation:
ρ x +1
=
ρ nonfat ρ nonfat
x +1
ρfat
This equation can be solved for x. The solution is:
ρ 1−
(1060 kg/m3 )
1−
x=
ρ nonfat
=
(1100 kg/m3 ) = 0.205
ρ
−1 (1060 kg/m3 ) − 1
ρfat ( 900 kg/m3 )
We can now calculate the precise value for the woman’s percentage body fat:
mfat mfat 1 1 1
= = = = = 0.170 = 17%
m mfat + mnonfat 1 + mnonfat 1 + 1 1 + 1
mfat x 0.205
Our simple estimate of 20% was reasonably close to the precise value of 17%.
10.16 Chapter 10

62. ORGANIZE AND PLAN The apparent weight equals the gravitational weight minus the buoyant force. If we
subtract the apparent weight from the actual weight and then divide by the actual weight we get the fraction we are
looking for.
Known: ρ = 16 kg/m 3 ; ρfluid = 1.28 kg/m 3 .
SOLVE The apparent weight is wa = mg − FB . Subtract this from the actual weight and divide to get a fraction
x of the actual weight:
w − wa ( mg ) − ( mg − FB ) FB
x= = =
w mg mg
We use Equation 10.5 for the buoyant force FB to get our final expression for the fraction:
F ρ gV ρ fluidV ρfluid
x = B = fluid = =
mg mg m ρ
where we have also used ρ = m / V for the Styrofoam piece. Insert known values to calculate the fraction:
ρfluid (1.28 kg/m 3 )
x= = = 8.0%
ρ (16 kg/m3 )
REFLECT The expression for the fraction makes a lot of sense. For example, if we asked the same question for a
piece of plastic with density 1000 kg/m in water, we would expect that x = 100% because the plastic is apparently
3

weightless in the water.

63. ORGANIZE AND PLAN The acceleration is given by Newton’s second law and is the net upward force divided by
3
the submarine’s mass. The net upward force equals the displaced weight of 1.5 m of sea water. We can figure out
the submarine’s mass because we know that prior to “blowing the tank” the buoyant force and the gravitational
force were in balance.
Known: V = 135 m 3 ; Vfluid = 1.5 m 3 ; ρ fluid = 1030 kg/m 3 .
SOLVE We can calculate the mass of the submarine from knowing that initially the apparent weight was zero:
0 = wa = mg − FB = mg − ρfluid gV
m = ρ fluidV = (1030 kg/m 3 )(135 m 3 ) = 1.39 × 103 kg
The net upward force on the submarine after “blowing the tank” is:
Fnet ρfluid gVfluid (1030 kg/m 3 )( 9.80 m/s2 )(1.5 m 3 )
a= = = = 0.11 m/s2
m m (1.39 × 103 kg )
REFLECT If the submarine had been in a fresh water lake instead of sea water, would the acceleration have been
any different?
64. ORGANIZE AND PLAN The total volume of the fish must displace an amount of sea water such that the apparent
weight (gravitational weight minus buoyant force) of the fish is zero. From the average density of the fish’s tissues
we can calculate the volume of the fish not including the volume of the bladder. Subtracting this volume from the
total volume gives us the required volume of gas in the swim bladder.
Known: ρ = 1050 kg/m 3 ; m = 9.5 kg; ρfluid = 1028 kg/m 3 .
SOLVE We can calculate total volume of the fish because we know that its apparent weight is zero:
0 = wa = mg − FB = mg − ρ fluid gV

V=
m
=
( 9.5 kg ) = 9.2 L
ρfluid (1028 kg/m3 )
The volume of the fish not including the gas in the swim bladder is:

Vtissue =
m
=
( 9.5 kg ) = 9.0 L
ρ (1050 kg/m3 )
The required volume of gas in the swim bladder is the difference between these two volumes:
Vgas = V − Vtissue = ( 9.2 L ) − ( 9.0 L ) = 0.19 L
REFLECT Was it okay to neglect the mass of the bladder gas?
 Solids and Fluids 10.17

65. ORGANIZE AND PLAN The volume flow rate is the flow speed times the cross-sectional area.
Known: d = 2.75 cm; v = 0.450 m/s.
SOLVE The volume flow rate is:

( 2.75 cm ) ( 0.450 m/s ) = 2.67 × 10−4 m3 /s

2
Q = Av = π4 d 2 v = π
4

REFLECT Depending on the application, it may be more useful to express the volume flow rate in units of liters
per second. In this case Q = 0.267 L/s.
66. ORGANIZE AND PLAN The volume flow rate is the flow speed times the cross-sectional area.
Known: d = 4.00 cm; Q = 1.20 × 10 −4 m 3 /s.
SOLVE The flow speed is:

v=
Q Q
= π 2 =
(1.20 × 10−4 m3 /s ) = 0.0955 m/s
4 (
4.00 cm )
π 2
A 4d

REFLECT You can check that the formula is correct by making sure the units come out to meters per second (or
distance per time).
67. ORGANIZE AND PLAN The volume flow rate is the same through the hose as it is through the nozzle, and it equals
the flow speed times the cross-sectional area.
Known: da = 2.25 cm; va = 0.320 m/s; db = 0.30 cm.
SOLVE (a) The volume flow rate is:

( 2.25 cm ) ( 0.320 m/s ) = 1.27 × 10−4 m3 /s

2
Q = Aa va = π4 da 2 va = π
4

(b) The flow speed through the nozzle is:

vb =
Q Q
= π 2 =
(1.27 × 10 −4 m3 /s ) = 18.0 m/s
4 (
0.30 cm )
π 2
Ab 4 db

REFLECT Because the flow rate is constant, the flow speed in the nozzle can also be expressed as:
Aa d2
vb = va = va a 2
Ab db

68. ORGANIZE AND PLAN The volume flow rate remains constant. If we first calculate the volume flow rate for the
wider part of the artery, we can then use that to calculate the flow speed in the narrower part.
Known: d1 = 1.45 mm; v1 = 2.65 cm/s; d2 = 1.36 mm.
SOLVE The volume flow rate is:

(1.45 mm ) ( 2.65 cm/s ) = 4.38 × 10 −8 m 3 /s

2
Q = A1v1 = π4 d12 v1 = π
4

The flow speed through the narrower part of the artery is:

v2 =
Q Q
= π 2 =
( 4.38 × 10−8 m3 /s ) = 3.01 cm/s
4 (
1.36 mm )
π 2
A2 4 d2

REFLECT As expected, the flow speed increases when the cross-sectional area decreases.
69. ORGANIZE AND PLAN Because the total cross-sectional area of the two small pipes is smaller than the cross-
sectional area of the large pipe, the flow will fill both of the small pipes, meaning half the volume flow rate goes
through each of the small pipes after the large pipe branches.
Known: Q1 = 1.20 × 10 −4 m 3 /s; d1 = 2.0 cm; d2 = 1.0 cm.
SOLVE The volume flow rate in each of the smaller pipes is:
Q1 (1.20 × 10 −4 m 3 /s )
Q2 = = = 6.00 × 10 −5 m 3 /s
2 2
REFLECT If a single small pipe had branched into two large pipes, the exact geometry of the branch would
become important, because all the water could continue in just one of the big pipes, or some fraction could go in
one pipe with the rest in the other pipe.
10.18 Chapter 10

70. ORGANIZE AND PLAN The volume of the bathtub divided by the time it takes to fill it is the volume flow rate,
which also equals the flow speed times the cross-sectional area of the pipe.
Known: d = 1.2 cm; V = 250 L; t = 6 min.
SOLVE The volume flow rate is:
V ( 250 L )
Q= = = 7 × 10 −4 m 3 /s
t ( 6 min )

The speed of water in the pipe is:

v=
Q Q
= π 2 =
( 7 × 10−4 m3 /s ) = 6 m/s
4 (
1.2 cm )
π 2
A 4d

REFLECT When the faucet is not open, hot water in the pipes gets cold fast, but as long as water in your water
heater remains hot, you should have hot water through the faucet in just a few seconds.
71. ORGANIZE AND PLAN Bernoulli’s equation relates fluid pressure to flow speed and elevation. We know all the
needed quantities to calculate gauge pressure at point B.
Known: v A = 1.55 m/s; pA = 180 kPa; yB − y A = 7.50 m; vB = 1.75 m/s; ρ = 900 kg/m 3 .
SOLVE Bernoulli’s equation from this problem is:
( P0 + PA ) + 12 ρ vA2 + ρ gyA = ( P0 + PB ) + 12 ρ vB 2 + ρ gyB
Here, P0 is the surrounding ambient pressure and can be dropped since it occurs on both sides of the equation. If
we rewrite the equation we can calculate the gauge pressure at B:
PB = PA − ρ ( (v
1
2 B
2
− 12 v A 2 ) + g ( yB − y A ) )
= (180 kPa ) − ( 900 kg/m 3 ) ( ((1.75 m/s) − (1.55 m/s) ) + ( 9.80 m/s )( 7.50 m ))
1
2
2 2
2

= 114 kPa
REFLECT In this case, the difference in elevation made a much larger impact on the pressure difference than
what the speed difference made. During design it is often important to identify what aspects of a problem matter
and which can be ignored.
72. ORGANIZE AND PLAN Bernoulli’s equation relates fluid pressure to flow speed and elevation. We know all the
needed quantities to calculate pressure in the upstairs pipe. We will use subscript 1 for quantities at the pressure
regulator, subscript 2 for the upstairs pipe.
Known: v1 = 0.850 m/s; p1 = 450 kPa; y2 − y1 = 3.70 m; p2 = 414 kPa; ρ = 1000 kg/m 3 .
SOLVE Rewrite Bernoulli’s equation to solve for the flow speed in the pipe:
P1 + 12 ρ v12 + ρ gy1 = P2 + 12 ρ v2 2 + ρ gy2
( P1 − P2 ) + 12 ρ v12 − ρ g ( y2 − y1 ) = 12 ρ v22
2
v2 = ⎡( P1 − P2 ) + 12 ρ v12 − ρ g ( y2 − y1 ) ⎤⎦
ρ⎣
Insert our known values to calculate the flow speed in the pipe:

v2 =
2
(1000 kg/m 3 ) ⎢⎣
(
⎡ ( 450 kPa ) − ( 414 kPa ) + ) (1000 kg/m )( 0.850 m/s ) − (1000 kg/m )( 9.80 m/s )( 3.70 m )⎤⎥⎦
1
2
3
2
3 2

= 0.450 m/s
REFLECT To which floor (or to what height) can this pressure regulator supply water?
 Solids and Fluids 10.19

73. ORGANIZE AND PLAN We will use Bernoulli’s equation to relate the fluid pressure, the flow speed, and the
elevation difference between the top of the container (subscript 1) and the location of the small hole (subscript 2).
The pressures on both sides of the equation are the same (equal to the atmospheric pressure) and cancel. If the
diameter of the hole is much smaller than the diameter of the container, we know from the continuity equation that
the flow speed over the cross-sectional area of the container is extremely small, so we will neglect it (set it to
approximately zero).
Known: y1 − y2 = 0.75 m; v1 = 0; ρ = 1000 kg/m 3 .
SOLVE Bernoulli’s equation with the pressures cancelling each other and v1 = 0 is:
ρ gy1 = 12 ρ v2 2 + ρ gy2
Rewrite this equation to calculate the flow speed through the hole:
gy1 = 12 v2 2 + gy2
v2 = 2 g ( y1 − y2 ) = 2 ( 9.80 m/s2 )( 0.75 m ) = 3.8 m/s

REFLECT The density of the fluid didn’t matter. We would get the same result for any type of fluid as long as we
can neglect the viscosity (see Section 10.6).
74. ORGANIZE AND PLAN We will use Bernoulli’s equation to relate the fluid pressure, the flow speed, and the
elevation difference between the top of the container (subscript 1) and the location of the small hole (subscript 2).
The pressures on both sides of the equation are the same (equal to the atmospheric pressure) and cancel. If the
diameter of the hole is much smaller than the diameter of the container, we know from the continuity equation that
the flow speed over the cross-sectional area of the container is extremely small, so we will neglect it (set it to
approximately zero).
Known: y1 − y2 = h; v1 = 0.
SOLVE Bernoulli’s equation with the pressures cancelling each other and v1 = 0 is:
ρ gy1 = 12 ρ v2 2 + ρ gy2
Rewrite this equation to calculate the flow speed through the hole:
gy1 = 12 v2 2 + gy2
v2 = 2 g ( y1 − y2 ) = 2 gh

REFLECT Evangelista Torricelli was an Italian physicist who invented the barometer. He lived in the
17th century. The unit for pressure called the torr (which equals a millimeter of mercury) is named after him.
75. ORGANIZE AND PLAN From Bernoulli’s equation we see that the difference in wind speed between outdoors and
indoors (where the wind speed is zero) creates a fluid pressure on the window. If we multiply this pressure with the
area we get the force. We will use subscript 1 for indoors, subscript 2 for outdoors.
Known: v2 = 90 km/h; A = 4.5 m 2 ; ρ = 1.28 kg/m 3 .
SOLVE The elevation is the same on either side of the window, and the indoors wind speed v1 = 0. With the
terms that equal zero removed, Bernoulli’s equation is:
P1 = P2 + 12 ρ v2 2

We see that the faster the wind blows, the lower the outdoors pressure P2 must be for the right-hand side of the
equation to equal P1. Consequently, the direction of the force on the window must be outwards. The magnitude of
this force is the difference between the indoors and outdoors pressures, multiplied by the surface area of the
window:
Fnet = ( P1 − P2 ) A = 21 ρ v2 2 A = (1.28 kg/m 3 )( 90 km/h ) ( 4.5 m2 ) = 1.8 kN
1 2
2

REFLECT Do you think a window could break just from the wind blowing past it?
10.20 Chapter 10

76. ORGANIZE AND PLAN The lift force from the wings must oppose the gravitational force on the aircraft. Since we
know the area of the wings, we can divide the force by the area to get the required difference between the pressure
under the wings and the pressure over the wings. From Bernoulli’s equation we can then calculate the airflow
speed over the wings. We will use subscript 1 for quantities under the wings, subscript 2 over the wings.
Known: m = 230,000 kg; v1 = 75 m/s; A = 427 m 2 ; ρ = 1.28 kg/m 3 .
SOLVE The required lift force from the wings is:
F = mg = ( 230,000 kg )( 9.80 m/s2 ) = 2.25 MN

Dividing by the area of the wings we get the required pressure difference:
F ( 2.25 MN )
P1 − P2 = = = 5.28 kPa.
A ( 427 m 2 )
The elevation difference between either side of the wing is very small and can be neglected. In this case
Bernoulli’s equation is:
P1 + 21 ρ v12 = P2 + 21 ρ v2 2

Rewrite this equation and solve for the airflow speed over the wings:
2 2
( P1 − P2 ) + v12 ( 5.28 kPa ) + ( 75 m/s ) = 1.2 × 102 m/s
2
v2 = =
ρ (1.28 kg/m )
3

REFLECT From the equations above you can see that the faster the aircraft moves (the larger v1 is), the smaller
the required difference between v2 and v1 becomes. This fact is reflected in the design of an aircraft when you
look at the shape of a wing’s cross-section. Slow aircraft tend to have “fat” wings where the upper side of the wing
is curved and thus longer than the lower side of the wing. Fast aircraft tend to have wings where the upper and
lower sides are almost equal in length. When a fast aircraft has to fly slowly, for example when taking off and
landing, it often requires the use of devices such as “flaps” that will modify the wing shape to look more like a
wing shape of a slow aircraft.
77. ORGANIZE AND PLAN We can use Poiseulle’s law to calculate what change in radius or diameter reduces the
volume flow rate by 10%, i.e., reduces the volume flow rate to 90% of the original value. We will use subscript 0
for the initial quantities and subscript 1 for quantities after the blood flow has been reduced.
Known: Q1 / Q0 = 0.90.
SOLVE Poiseulle’s law before the blow flow has been reduced is:
π R0 4 ΔP
Q0 =
8η L
The pressure difference ΔP, the viscosity η , and the length L of the artery remains constant as we change the
radius of the artery. Poiseulle’s law after the blow flow has been reduced is:
π R14 ΔP
Q1 =
8η L
If we divide the second equation with the first we get:
Q1 R14
=
Q0 R0 4
Which we can solve for the ratio between the artery radius after and before the flow change:
1/ 4
R1 ⎛ Q1 ⎞
=⎜ ⎟ = 0.901 / 4 = 0.97
R0 ⎜⎝ Q0 ⎟⎠

Consequently, the radius (or the diameter) of an arterial wall has to decrease by 3% to reduce the blood flow rate
by 10%.
REFLECT The technique demonstrated here, dividing an equation of “after-values” with the same equation of
“before-values” is often useful to determine how fractional changes in one quantity varies with fractional changes
in another quantity.
 Solids and Fluids 10.21

78. ORGANIZE AND PLAN We can use Poiseulle’s law to calculate what change in systolic pressure is required to
keep the blood flow rate constant. We will use subscript 0 for the initial quantities and subscript 1 for quantities
with plaque build-up.
Known: R1 / R0 = 0.950.
SOLVE Poiseulle’s law before plaque build-up is:
π R0 4 ΔP0
Q=
8η L
The blood flow rate Q, the viscosity η , and the length L of the artery remains constant as we change the radius
of the artery. Poiseulle’s law with plaque build-up is:
π R14 ΔP1
Q=
8η L
If we divide the second equation with the first we get:
R14 ΔP1
1=
R0 4 ΔP0
Which we can solve for the ratio between systolic pressure difference after and before the plaque build-up:
−4
ΔP1 ⎛ R1 ⎞
= ⎜ ⎟ = 0.950 −4 = 1.23
ΔP0 ⎜⎝ R0 ⎟⎠
This means the systolic pressure has to go up by about 23% to compensate for the decreased artery radius. If
ΔP0 = 120 mm Hg the new required systolic pressure is:
ΔP1 = 1.23ΔP0 = 1.23 (120 mm Hg ) = 147 mm Hg
REFLECT This is a dangerous pressure increase for just a 5% change in the artery diameter.
79. ORGANIZE AND PLAN We can use Poiseulle’s law to calculate the pressure difference, but we need to know the
viscosity of water at 50°C. Table 10.3 lists the viscosities for water at 20°C and 100°C. We can interpolate
between these two values to get a good estimate for the viscosity of water at 50°C.
Known:
T50 = 50°C; L = 2.50 km; d = 10 cm; Q = 12 L/min; η20 = 1.0 × 10−3 Pa ⋅ s; T20 = 20°C; η20 = 2.8 × 10−4 Pa ⋅ s; T100 = 100°C.
SOLVE First we estimate the viscosity of water at 50° C by interpolation:
T −T
η50 = (η100 − η20 ) 50 20 + η20 =
T100 − T20

) ((100°C)) −(( 20°C)) + (1.0 × 10

50°C − 20°C
(
= ( 2.8 × 10 −4 Pa ⋅ s ) − (1.0 × 10 −3 Pa ⋅ s ) −3
Pa ⋅ s ) = 7.3 × 10 −4 Pa ⋅ s

Rewrite Poiseulle’s law to calculate the pressure difference between the ends of the pipe:
π R 4 ΔP
Q=
8η50 L
8η50 LQ 128η50 LQ 128 ( 7.3 × 10 −4 Pa ⋅ s )( 2.50 km )(12 L/min )
ΔP = = = = 1.5 × 10 2 Pa
π R4 π d4 π (10 cm )
4

REFLECT This is a small pressure difference that a well-engineered pumping station should have no problem
with.
80. ORGANIZE AND PLAN We can use Poiseulle’s law to calculate the pressure difference.
Known: T = 50°C; η = 0.016 Pa ⋅ s; Q = 0.50 m 3 /s; L = 20 km; d = 0.76 m.
SOLVE Rewrite Poiseulle’s law to calculate the pressure difference between the ends of the pipeline:
π R 4 ΔP
Q=
8η L
8η LQ 128η LQ 128 ( 0.016 Pa ⋅ s )( 20 km )( 0.50 m 3 /s )
ΔP = = = = 20 kPa
π R4 π d4 π ( 0.76 m )
4
10.22 Chapter 10

REFLECT Oil is much more viscous than water, so heating the oil to lower the viscosity can make a big
difference.
81. ORGANIZE AND PLAN The surface area in contact with the bed is the normal force divided by the pressure. The
normal force is equal in magnitude to the gravitational force.
Known: m = 130 kg; P = 4.7 kPa.
SOLVE The normal force equals:
N = mg = (130 kg )( 9.80 m/s2 ) = 1.27 kN

The normal force is distributed over the surface area in contact with the bed, creating the given pressure. The
surface area is:
N (1.27 kN )
A= = = 0.27 m 2
P ( 4.7 kPa )

REFLECT We have assumed that the bodies don’t sink into an only partially filled water bed, so that the surface
area normal is approximately vertical everywhere.
82. ORGANIZE AND PLAN The total surface area in contact with the road is the normal force divided by the gauge
pressure. The normal force is equal in magnitude to the gravitational force. The surface area of each tire is the total
surface area divided by four, because SUVs have four tires.
Known: m = 130 kg; P = 4.7 kPa.
SOLVE The normal force equals:
N = mg = ( 3800 kg )( 9.80 m/s2 ) = 37.2 kN

The normal force is distributed over the surface area in contact with the road, and must be supported by the gauge
pressure in the tires. The surface area of each tire in contact with the road is:

A=
N
=
( 37.2 kN ) = 3.88 × 10 −2 m 2
4 P 4 ( 240 kPa )

REFLECT The palm of a hand is about 10 −2 m so this answer seems reasonable.

83. ORGANIZE AND PLAN The mass is the volume of the barrel multiplied with the density of oil. From the formula
for the volume of a circular cylinder we can calculate the diameter. The pressure difference is the density times the
height times g.
Known: V = 42 gallons; h = 32 inches; ρ = 900 kg/m 3 .
SOLVE (a) The mass of the oil in the barrel is:
m = V ρ = ( 42 gallons )( 900 kg/m 3 ) = ( 42 gallons )( 3.8 × 10 −3 m 3 /gallon )( 900 kg/m 3 ) = 1.4 × 10 2 kg

(b) The inside diameter of the barrel is:

V = π4 d 2 h

4V 4 ( 42 gallons ) 4 ( 42 gallons )( 3.8 L/gallon )

d= = = = 0.50 m
πh π ( 32 inches ) π ( 32 inches )( 2.54 cm/inch )

(c) The pressure difference between the top and the bottom of the barrel is:
ΔP = ρ gh = ( 900 kg/m 3 )( 9.80 m/s2 )( 32 inches ) = ( 900 kg/m 3 )( 9.80 m/s2 )( 32 inches )( 2.54 cm/inch ) = 7.2 kPa.

REFLECT All these calculations are straightforward, but it is important to do the unit conversions correctly.
Particularly it is important to be aware that there are several different “gallons” used around the world!
84. ORGANIZE AND PLAN From the masses we can calculate the volume of each type of fluid. With the volume
known we can calculate the height of each fluid column. The gauge pressure at the bottom of the oil is the density
of oil times the height of the oil times g. The gauge pressure at the bottom of the water equals the pressure at the
bottom of the oil plus the density of water times the height of the water times g.
Known: d = 1.0 cm; moil = 25 g; mwater = 25 g; ρ oil = 900 kg/m 3 ; ρ water = 1000 kg/m 3 .
 Solids and Fluids 10.23

SOLVE (a) The volume of oil is:

Voil =
moil
=
( 25 g ) = 28 mL
ρoil ( 900 kg/m 3 )
The height of the oil column is:
Voil = π4 d 2 hoil
4Voil 4 ( 28 mL )
hoil = = = 0.35 m
πd π (1.0 cm )
2 2

The volume of water is:

Vwater =
mwater
=
( 25 g ) = 25 mL
ρ water (1000 kg/m 3 )
The height of the water column is:
Vwater = π4 d 2 hwater
4Vwater 4 ( 25 mL )
hwater = = = 0.32 m
π d2 π (1.0 cm )
2

(b) The gauge pressure at the bottom of the oil is:

ΔPoil = ρoil ghoil = ( 900 kg/m 3 )( 9.80 m/s2 )( 0.35 m ) = 3.1 kPa

The gauge pressure at the bottom of the water is:

ΔPwater = ΔPoil + ρ water ghwater = ( 3.1 kPa ) + (1000 kg/m 3 )( 9.80 m/s2 )( 0.32 m ) = 6.2 kPa

REFLECT Because the mass of the oil and the mass of the water were equal,
ρ oil ghoil = ρ water ghwater
i.e., each fluid contributes an equal amount to the gauge pressure at the bottom of the tube.
85. ORGANIZE AND PLAN The required force is the difference in pressure multiplied by the surface area.
Known: A = ( 90 cm )( 50 cm ) = 0.45 m 2 ; P1 = 0.75 atm; P2 = 0.25 atm.
SOLVE The required force is:
( )
F = ( P1 − P2 ) A = ( 0.75 atm ) − ( 0.25 atm ) ( 0.45 m 2 ) = 23 kN

REFLECT Not even the world’s strongest human could pull this escape window inward. (At least not under
normal circumstances; faced with a life-threatening situation, some humans have been documented to exhibit
extraordinary strength. However, even in such a situation it is difficult to imagine anyone being able to make this
pull.)
86. ORGANIZE AND PLAN The pressure in the fluid increases with elevation and equals the elevation multiplied by
the fluid density and g.
Known: ΔP = 140 mm Hg; ρ = 1000 kg/m 3 .
SOLVE The bottle should be placed at a height above the arm equal to:

h=
ΔP
=
(140 mm Hg ) = 1.9 m
ρ g (1000 kg/m 3 )( 9.80 m/s2 )
REFLECT In most cases the patient would be on a bed or another surface approximately 1 m above the floor. This
means the IV bottle would have to be positioned much higher than most nurses can reach. To get around this
problem, the bottle is placed on something called an IV pole, which is extendable in height. This has a design
consequence for hospitals — the ceilings must be high enough to accommodate the IV poles!
87. ORGANIZE AND PLAN The force must equal the blood pressure multiplied by the cross-sectional area of the
plunger. The speed of the plunger is the volume flow rate divided by the cross-sectional area of the plunger. The
speed of the emerging fluid is the volume flow rate divided by the cross-sectional area of the needle.
10.24 Chapter 10

Known: dplunger = 1.2 cm; P = 130 mm Hg; Q = 1.5 mL/s; dneedle = 220 μm.
SOLVE (a) The health provider must push on the plunger with a force:
F = PA = P π4 dplunger 2 = (130 mm Hg ) π4 (1.2 cm ) = 2.0 N
2

(b) The plunger is moving at a speed:

v=
Q
=
Q
=
(1.5 mL/s ) = 1.3 cm/s
4 (
1.2 cm )
π 2 π 2
A 4
dplunger

(c) The fluid emerges from the needle at a speed:

v=
Q
=
Q
=
(1.5 mL/s ) = 39 m/s
4 (
220 μm )
π 2 π 2
A d
4 needle

REFLECT These are fairly straightforward calculations, but it is important to pay attention to all the various unit
conversions and make sure the answers seem reasonable.
88. ORGANIZE AND PLAN Volume equals mass divided by density, so volume flow rate equals mass flow rate
divided by density. The flow speed is the volume flow rate divided by the cross-sectional area of the river. Power
is energy per unit time and kinetic energy is one-half times mass times speed squared. That means power can be
expressed as one-half times the mass flow rate times speed squared.
Known: A = ( 4.5 m )(1.3 km ) = 5.9 × 103 m 2 ; M = 1.5 × 10 7 kg/s; Pelectricity / P = 5%; ρ = 1000 kg/m 3 .
SOLVE (a) The volume flow rate is:

Q=
M
=
(1.5 × 107 kg/s ) = 1.5 × 10 4 m3 /s
ρ (1000 kg/m3 )
(b) The flow speed of the river is:
Q (1.5 × 10 4 m 3 /s )
v= = = 2.6 m/s
A ( 5.9 × 103 m 2 )

(c) The total power of the moving water is:

(1.5 × 107 kg/s )( 2.6 m/s )

2
P = 12 Mv 2 = 1
2
= 49 MW

If 5% of the kinetic energy can be harnessed as electricity, the electric power would be:
Pelectric = 0.05P = 0.05 ( 49 MW ) = 2.7 MW

REFLECT If you want to get more power out of the river, it would be better to harness the power where the water
has more kinetic energy, i.e., where the flow speed is higher.
89. ORGANIZE AND PLAN To lift the payload, the buoyant force on the balloon must be equal to or larger than the
combined gravitational forces on the payload and the helium in the balloon.
Known: mpayload = 340 kg; ρ helium = 0.179 kg/m 3 ; ρair = 1.28 kg/m 3 .
SOLVE (a) The total mass of payload plus balloon is:
m = mpayload + mhelium = mpayload + ρ heliumV

where V is the volume of the balloon. The buoyant force is:

FB = ρair gV
To lift the payload, the net upward force on the payload must be positive:
Fnet = Fg + FB = −mg + ρair gV ≥ 0
 Solids and Fluids 10.25

This gives us an equation we can solve for the balloon’s volume:

− mg + ρair gV ≥ 0
− m + ρairV ≥ 0
− mpayload − ρ heliumV + ρairV ≥ 0
( ρair − ρhelium )V ≥ mpayload
V≥
mpayload
=
( 340 kg ) = 309 m 3
ρair − ρ helium (1.28 kg/m3 ) − ( 0.179 kg/m3 )
If we multiply this volume by the density of helium we get the amount of helium required:
mhelium = ρ heliumV = ( 0.179 kg/m 3 )( 309 m 3 ) = 55.3 kg

(b) If the balloon is spherical, the volume of the balloon is related to its diameter by:
π d3
V=
6
from which we can calculate balloon’s diameter:

6V 6 ( 309 m 3 )
d=3 =3 = 8.39 m
π π
REFLECT If you want to double the payload, you would have to double the amount of helium. In other words,
there is a fixed ratio between the mass of the payload and the required amount of helium:
mpayload
=
( 340 kg ) = 6.15
mhelium ( 55.3 kg )
90. ORGANIZE AND PLAN The volume flow rate of the blood must remain constant, because there is no change in the
oxygen demand from the body. If we first calculate the volume flow rate for a typical aorta, we can then use that to
calculate the flow speed in an aorta with reduced diameter.
Known: d1 = 1.8 cm; v1 = 35 cm/s; d2 = 50% × d1 = 0.90 cm.
SOLVE The volume flow rate is:

(1.8 cm ) ( 35 cm/s ) = 8.9 × 10 −5 m3 /s

2
Q = A1v1 = π4 d12 v1 = π
4

The flow speed through an aorta with reduced diameter is:

v2 =
Q Q
= π 2 =
(8.9 × 10−5 m3 /s ) = 1.4 m/s
4 (
0.90 cm )
π 2
A2 4 d2

REFLECT The answer makes sense. It is easy to see from the formula that halving the diameter, i.e., reducing the
cross-sectional area to one-fourth, results in a quadrupled flow speed.
91. ORGANIZE AND PLAN We can calculate the new gauge pressure from Bernoulli’s equation, which relates
pressure to flow speeds (and also to elevation changes, but there’s no elevation change in this problem). We need
to know the density of blood, which is listed in Table 10.1.
Known: P1 = 120 mm Hg; v1 = 35 cm/s; v2 = 1.4 m/s; ρ = 1060 kg/m 3 .
SOLVE Bernoulli’s equation (without the terms that apply to elevation changes) is:
P1 + 21 ρ v12 = P2 + 21 ρ v2 2

We can rewrite this to calculate the new gauge pressure P2 in the aorta with plaque build-up:

P2 = P1 + 21 ρ ( v12 − v2 2 ) = (120 mm Hg ) + 1
2 (1060 kg/m3 ) ( ( 35 cm/s ) − (1.4 m/s )
2 2
) = 113 mm Hg
REFLECT This result may seem surprising at first: shouldn’t a person with so much plaque build-up have an
increased blood pressure? Does the answer depend on if we are looking at the body’s largest blood vessel, the
aorta, or at smaller blood vessels?
10.26 Chapter 10

92. ORGANIZE AND PLAN The volume flow rate of the blood going through the aorta must also go through the
capillaries. By dividing volume flow rate of the aorta by the individual volume flow rate of a single capillary, we
get the number of capillaries in the body. For this calculation we will use subscript 1 for the aorta and subscript 2
for a capillary. The volume flow rate of the aorta equals the amount of blood in the body divided by the time it
takes to fully circulate.
Known: d1 = 1.8 cm; d2 = 10 μm; v1 = 1.0 m/s; v2 = 1.0 cm/s; V = 5.5 L.
SOLVE (a) The volume flow rate in the aorta is:

(1.8 cm ) (1.0 m/s ) = 2.5 × 10 −4 m3 /s

2
Q1 = A1v1 = π4 d12 v1 = π
4

The volume flow rate in a single capillary is:

(10 μm ) (1.0 cm/s ) = 7.9 × 10 −13 m 3 /s

2
Q2 = A2 v2 = π4 d2 2 v2 = π
4

This means the number of capillaries in the body equals:

Q1 ( 2.5 × 10 −4 m 3 /s )
= = 3.2 × 108
Q2 ( 7.9 × 10 −13 m 3 /s )

(b) The time it takes for the blood to circulate completely is:

t=
V
=
( 5.5 L ) = 21 s
Q1 ( 2.5 × 10 −4 m 3 /s )

REFLECT How long does it take before a blood cell circulates through the same capillary a second time?
93. ORGANIZE AND PLAN The apparent weight of the crown under water is the gravitational force reduced by the
buoyant force from the displaced water. If the crown is 90% gold and 10% silver by weight, it is as if we are
weighing two different objects simultaneously, one made of gold and one silver.
Known: W = 25.0 N; ρ Au =19,300 kg/m 3 ; ρ Ag =10,500 kg/m 3 ; ρ H2O = 1000 kg/m 3 .
SOLVE (a) If the crown is pure gold, its mass is:

m=
W
=
( 25.0 N ) = 2.55 kg
g ( 9.80 m/s2 )

Its volume is:

V=
m
=
( 2.55 kg ) = 0.132 L
ρ Au (19,300 kg/m 3 )
The buoyant force of this crown under water is:
FB = ρ fluid gV = (1000 kg/m 3 )( 9.80 m/s2 )( 0.132 L ) = 1.30 N

The scale would read:

Wa = W − FB = ( 25.0 N ) − (1.30 N ) = 23.7 N

(b) If the crown is 90% gold and 10% silver by weight, it is as if we are weighing two different objects. The first
object is pure gold weighing WAu = 0.9 × W = 0.9 × ( 25.0 N ) = 22.5 N and the second object is pure silver weighing
WAg = 0.1 × W = 0.1 × ( 25.0 N ) = 2.50 N. The scale will read the combined apparent weight of these two objects, so
all we need to do to get the final answer is repeat part (a) for these two objects and then add the apparent weights
together.
The mass of the gold is:

m Au =
WAu
=
( 22.5 N ) = 2.30 kg
g ( 9.80 m/s2 )
 Solids and Fluids 10.27

Its volume is:

VAu =
m Au
=
( 2.30 kg ) = 0.119 L
ρ Au (19,300 kg/m3 )
The buoyant force on the gold is:
FB, Au = ρ fluid gVAu = (1000 kg/m 3 )( 9.80 m/s2 )( 0.119 L ) = 1.17 N

The apparent weight of the gold is:

Wa , Au = WAu − FB, Au = ( 22.5 N ) − (1.17 N ) = 21.3 N

The mass of the silver is:

m Ag =
WAg
=
( 2.50 N ) = 0.255 kg
g ( 9.80 m/s2 )
Its volume is:

VAg =
m Ag
=
( 0.255 kg ) = 0.0243 L
ρ Ag (10,500 kg/m 3 )
The buoyant force on the silver is:
FB, Ag = ρ fluid gVAg = (1000 kg/m 3 )( 9.80 m/s2 )( 0.0243 L ) = 0.238 N

The apparent weight of the silver is:

Wa , Ag = WAg − FB, Ag = ( 2.50 N ) − ( 0.238 N ) = 2.26 N

The scale would read the combined apparent weight of the gold and the silver:
Wa = Wa , Au + Wa , Ag = ( 21.3 N ) + ( 2.26 N ) = 23.6 N

REFLECT Archimedes’ suggested method would require a very precise scale to be put into practice.
94. ORGANIZE AND PLAN Initially the pencil is floating in an equilibrium state with the downward gravitational
force and the upward buoyant force balancing each other. If the pencil is pushed down a small distance from its
equilibrium position, there is an increase in the upward buoyant force. Similarly, if the pencil is lifted up a small
distance, there is a decrease in the upward buoyant force. The gravitational force never changes. The change in the
buoyant force is linear with the displacement, so the situation is analogous to that of a spring. This means that the
pencil undergoes simple harmonic motion just like a spring when released. To calculate the period of the motion
all we need to do is calculate the equivalent spring constant.
Known: L.
SOLVE If the pencil is pushed down a small distance Δy from its equilibrium position, the increase in the
buoyant force is the increase in the additionally displaced fluid:
ΔFB = ρ fluid gΔV = ρ fluid gAΔy
where ρ fluid is the density of the displaced fluid and A is the cross-sectional area of the pencil. The spring constant
is the force divided by the displacement:
ΔFB ρfluid gAΔy
k= = = ρ fluid gA
Δy Δy
The period of a simple harmonic motion is given by Equation 7.3:
m
T = 2π
k
To get the answer that the problem text asked for we must express our spring constant in terms of the mass m and
the submerged length L of the pencil. This can be done by comparing the forces on the pencil in the equilibrium
position. The buoyant force is:
FB = ρ fluid gV = ρfluid gAL
10.28 Chapter 10

or in terms of the spring constant:

FB = ρ fluid gAL = kL
The gravitational force is:
Fg = − mg

with the minus sign reflecting the fact that the gravitational force is directed downward. In equilibrium the sum of
these forces is zero. This gives us the expression for the spring constant that we need:
FB + Fg = 0
kL − mg = 0
mg
k=
L
Finally, the period of the motion is:
m m L
T = 2π = 2π = 2π
k mg g
L
REFLECT The expression seems reasonable. We know from Chapter 7 that heavier objects oscillate more slowly.
In this case, the longer the submerged length of the pencil the more “mass” in the oscillation.
95. ORGANIZE AND PLAN The volume flow rate is the cross-sectional area times the flow speed, so we want to find
the flow speed. Bernoulli’s equation relates the flow speed to differences in elevation and in pressure for two
points in the flow. We will choose these two points to be where the thin tube connects to the pipe, as shown in the
figure below. The elevation difference between these two points is known. The pressure difference can be
calculated be comparing the gauge pressures of the columns of fluid above each point.
Known: d A = 1.9 cm; d B = 0.64 cm; ρoil / ρ water = 0.82; h2 = 1.4 cm.
SOLVE Bernoulli’s equation for points A and B is:
PA + 12 ρ water v A2 + ρ water gy A = PB + 12 ρ water vB 2 + ρ water gyB

where the difference in elevation between the two points is

h4 = y A − yB = 1
2 ( d A − dB ) = 12 ( (1.9 cm ) − ( 0.64 cm ) ) = 0.63 cm
as shown in the figure below. The gauge pressure at the top of the thin tube can be used to relate the pressures at
points A and B. The gauge pressure at the top of the thin tube equals:
PA + ρ water g ( h3 + h2 ) + ρ oil gh1 = PB + ρ water g ( h4 + h3 ) + ρ oil g(h2 + h1 )

This expression can be simplified by removing identical terms from either side of the equal sign:
PA + ρ water gh2 = PB + ρ water gh4 + ρoil gh2
Rewrite this expression to obtain the difference between the pressures at the points A and B:
PA − PB = ρ water gh4 − ( ρ water − ρoil ) gh2

Substitute this expression in Bernoulli’s equation above:

PA + 12 ρ water v A 2 + ρ water gy A = PB + 12 ρ water vB 2 + ρ water gyB
PA − PB + 12 ρ water v A 2 + ρ water gh4 = 12 ρ water vB 2
2 ρ water gh4 − ( ρ water − ρ oil ) gh2 + 12 ρ water v A 2 = 12 ρ water vB 2

The only unknowns in this expression are the flow speeds at points A and B. These are related because the volume
flow rate is the same at both points:
Q = π4 d A 2 v A = π4 dB 2 vB
dB 2
vA = vB
dA2
 Solids and Fluids 10.29

Substitute for v A :
2
⎛ d 2⎞
2 ρ water gh4 − ( ρ water − ρoil ) gh2 + 12 ρ water ⎜ vB B 2 ⎟ = 12 ρ water vB 2
⎝ dA ⎠
We can solve this expression for the flow speed at point B:
⎛ d 4⎞
2 ρ water gh4 − ( ρ water − ρoil ) gh2 = 12 ρ water vB 2 ⎜ 1 − B 4 ⎟
⎝ dA ⎠
−1
⎛ ⎛ ρ ⎞ ⎞⎛ dB 4 ⎞
vB = 2 g ⎜ 2h4 − ⎜⎜ 1 − oil ⎟⎟ h2 ⎟⎟ ⎜ 1 − 4 ⎟
⎜ ρ
⎝ ⎝ water ⎠ ⎠⎝ dA ⎠

Insert our known values to calculate the flow speed:

−1
⎛ ( 0.64 cm )4 ⎞
vB = 2 ( 9.80 m/s 2
)( 2 ( 0.63 cm ) − (1 − 0.82 )(1.4 cm ) ⎜ 1 − ) ⎟ = 0.45 m/s
⎜ (1.9 cm ) ⎟
4
⎝ ⎠
Finally, from the flow speed we can calculate the volume flow rate:

( 0.64 cm ) ( 0.45 m/s ) = 1.4 × 10 −5 m/s

2
Q = π4 d B 2 vB = π
4

REFLECT The higher flow speed at the narrow part of the pipe is sucking the oil in the thin tube toward point B.
When this suction has moved the oil 0.7 cm (so that the difference in between the oil levels on the two sides of the
tube is 1.4 cm) the net gravitational force on the two fluid columns is in equilibrium with the net suction force.

h1

h2
h3
PA h4

dA PB dB