CHAPTER
11 FLUIDS
CONCEPTUAL QUESTIONS
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1.
REASONING AND SOLUTION A pile of empty aluminum cans has a volume of 1.0 m3.
This volume includes the inner volume of each can which contains air. It also includes the
air in the spaces between the cans. Therefore, most of the volume is occupied by air, not by
aluminum. The mass of the pile of cans is not ρ AlV = (2700 kg/m3 )(1.0 m 3 ) = 2700 kg ,
because the volume of aluminum in the pile of cans is not 1.0 m3.
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2.
REASONING AND SOLUTION Initially, the net force on the plunger is zero. The
pressure in the bottle is equal to atmospheric pressure; therefore, the magnitude of the force
per unit area pushing outward on the plunger is equal to the magnitude of the force per unit
area that pushes inward on the plunger. After the first pull-and-release cycle, the force per
unit area pushing outward on the plunger is less than the force per unit area pushing inward,
because there is less pressure inside the bottle. Therefore, a larger external force is required
to pull the plunger. On the 15th pull, the force per unit area pushing outward is very close to
zero. The net force on the plunger is inward and due mostly to atmospheric pressure.
Therefore, on the 15th pull, a much greater force will be required than on previous pulls.
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3.
REASONING AND SOLUTION A person could not balance her entire weight on the
pointed end of a single nail, because it would penetrate her skin. According to Equation
11.3, the pressure exerted by the nail is P = F / A where F represents the weight of the
person, and A is the area of the tip of the nail. Since the tip of the nail has a very small
radius, its area is very small; therefore, the pressure that the nail exerts on the person is
large. The reason she can safely lie on a "bed of nails" is that the effective area of the nails
is very large if the nails are closely spaced. Thus, the weight of the person F is distributed
over all the nails so that the pressure exerted by any one nail is small.
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4.
REASONING AND SOLUTION As you climb a mountain, your ears "pop" because of the
change in atmospheric pressure.
a. As you climb up, the outside pressure becomes lower than the pressure in your inner ear.
The outward force per unit area on your eardrum is greater than the inward force per unit
area; therefore, your eardrum moves outward.
b. As you climb down, the outside pressure becomes greater than the pressure in your inner
ear. The outward force per unit area on your eardrum is less than the inward force per unit
area on your eardrum; therefore, your eardrum moves inward.
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548 FLUIDS
5.
REASONING AND SOLUTION The bottle of juice is sealed under a partial vacuum.
Therefore, when the seal is intact, the button remains depressed, because the pressure inside
the bottle is less than the atmospheric pressure outside of the bottle. The force per unit area
pushing up on the button from the inside is significantly smaller than the force per unit area
pushing down on the outside. When the seal is broken, air rushes inside the bottle. The
force pushing up on the button increases and offsets the force pushing down. The natural
springiness of the material from which the lid is made can then make the button "pop up."
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6.
REASONING AND SOLUTION Even though the tube is thin, pouring liquid into it
increases the pressure uniformly throughout the vessel by an amount ρgh , where h is the
height of the water in the tube. Apparently, before the tube is full, the outward force exerted
on the walls of the container due to the fluid pressure is greater than the container can
withstand. Therefore, the container bursts.
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7.
REASONING AND SOLUTION A closed tank is completely filled with water. A valve is
opened at the bottom of the tank and the water begins to flow out. The water at the valve is
pushed out by the force that results from the pressure exerted by the column of water above
the valve. This force must be larger than the opposing inward force caused by atmospheric
pressure at the valve. When these two forces are equal, the water will cease to exit at the
valve. Therefore, when the water stops flowing, there will still be a noticeable amount of
water in the tank.
On the other hand, if the tank were open to the atmosphere, rather than closed, the water
at the valve would experience an additional outward force due to atmospheric pressure on
the top of the water column and the tank would empty completely.
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8.
REASONING AND SOLUTION When you drink through a straw, you draw the air out of
the straw, and the external air pressure leads to the unbalanced force that pushes the liquid
up into the straw. This action requires the presence of an atmosphere. The moon has no
atmosphere, so you could not use a straw to sip a drink on the moon.
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9.
SSM REASONING AND SOLUTION The gauge pressure of a fluid enclosed in a
container is the amount by which the container pressure exceeds atmospheric pressure. The
actual pressure of the fluid, taking atmospheric pressure into account, is called the absolute
pressure. On earth, at sea level, the atmosphere exerts a pressure of 1.0 × 105 Pa on every
surface with which it is in contact.
An exam question asks, "Why does the cork fly out with a loud 'pop' when a bottle of
champagne is opened?" To answer this question, a student replies, "Because the gas
pressure in the bottle is about 0.3 × 10 5 Pa ." The cork flies out because the absolute
pressure within the bottle exceeds the atmospheric pressure on the outside. Assuming that
the student recognizes this fact, the student must be referring to a gauge pressure of
Chapter 11 Conceptual Questions
549
0.3 × 10 5 Pa . Then, the absolute pressure within the bottle would be 1.0 × 105 Pa plus
0.3 × 10 5 Pa and would indeed exceed atmospheric pressure.
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10. REASONING AND SOLUTION A scuba diver is below the surface of the water when a
storm approaches, dropping the air pressure above the water. According to Equation 11.4,
the absolute pressure at the location of the scuba diver is Pdiver = P0 + ρgh where P0 is
atmospheric pressure, ρ is the density of the water, and h is the depth of the diver below the
surface. Since the storm causes the atmospheric pressure to drop, the absolute water
pressure at the location of the diver will also drop. If the diver were wearing a sensitive
gauge designed to measure the absolute pressure, it would register the drop in pressure.
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11. REASONING AND SOLUTION According to Archimedes' principle, the water will apply
a buoyant force to the steel beam; the magnitude of the buoyant force will be equal to the
weight of the water displaced by the beam. Therefore, the buoyant force experienced by the
beam is given by ( ρ water V beam ) g where ρ water is the density of the water and V beam is the
volume of the beam. If we neglect any change in water density with depth, then, the
quantity ρ water V beam has the same value regardless of whether the beam is suspended
vertically or horizontally under the water. Therefore, both the vertically suspended and the
horizontally suspended beam experience the same buoyant force.
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12. REASONING AND SOLUTION Archimedes' principle states that any fluid applies a
buoyant force to an object that is partially or completely immersed in it; the magnitude of
the buoyant force equals the weight of the fluid that the object displaces.
A glass beaker, filled to the brim with water, is resting on a scale. A block is placed in the
water, causing some of it to spill over. The water that spills is wiped away, and the beaker is
still filled to the brim.
a. According to Table 11.1, the density of wood is 550 kg/m3,
while the density of water at 4 °C is 1.000 × 103 kg/m3 . Since the
Buoyant force
on the block
density of the wood is less than the density of water, the block will
float with part of its volume above the surface of the water.
According to Archimedes' principle, the buoyant force exerted on
Weight of
the bottom surface of the block must be equal in magnitude to the
the block
weight of water that is displaced by the block. However, since the
block floats on the surface of the water, the buoyant force must be
equal in magnitude and opposite in direction to the weight of the
block, as suggested in the free body diagram at the right. Hence,
the weight of the block must be equal to the weight of the water displaced by the block. The
weight of the contents of the beaker remains the same. Therefore, the initial and final scale
readings are the same if the block is made of wood.
550 FLUIDS
b. According to Table 11.1, the density of iron is 7860 kg/m3, while the density of water at
4 °C is 1.000 × 103 kg/m3 . Since the density of iron is greater than the density of water, the
iron block will sink to the bottom of the beaker. Let V represent the volume of water that
spills over. According to the definition of mass density (Equation 11.1) the mass of water
that spills over is equal to mwater = ρ water V . The beaker is still filled to the brim, but the
block now occupies the volume V where the water used to be. According to Equation 11.1,
the mass that now occupies the volume V is given by mblock = ρiron V . Since ρiron > ρwater,
the mass of the block is greater than the mass of the water that spilled over. The weight of
the contents of the beaker increases. Therefore, the final reading on the scale will be greater
than the initial reading.
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13. REASONING AND SOLUTION According to Archimedes' principle, any fluid applies a
buoyant force to an object that is partially or completely immersed in it; the magnitude of
the buoyant force equals the weight of the fluid that the object displaces. Therefore, the
magnitude of the buoyant force exerted on an object immersed in water is given by
FB = ρ water Vg , where ρwater is the density of water, V is the volume displaced by the
immersed object, and g is the magnitude of the acceleration due to gravity. If the
acceleration due to gravity on a distant planet is less than it is on earth, then, other factors
remaining the same, the buoyant force will be less on the planet than it is on earth.
However, the weight of the object will also be less than the weight of the object on earth.
When an object floats in water, the upward buoyant force exerted
by the water must be equal in magnitude and opposite in direction to
the weight of the object, as shown at the right. Hence, FB = mobject g .
It follows that ρ water Vg = mobject g . Notice that the acceleration due
Buoyant force
on object
to gravity, g, appears on both sides of this equation. Algebraically
canceling the g's we have ρ water V = mobject . Therefore, the object will
float so that it displaces a volume of water V, where
V = mobject / ρ water . This result is independent of g. It is the same on
Weight of object
earth as it is on the distant planet. Therefore, it would be no more
difficult to float in water on this planet than it would be on earth.
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14. REASONING AND SOLUTION An ice cube is placed in a glass, and the glass is filled to
the brim with water. When the ice cube melts, and the temperature of the water returns to its
initial value, the water level remains the same.
According to Table 11.1, the density of ice is less than that of liquid water; therefore, the
ice will float in the glass with some of the ice above the level of the water. It will displace a
volume of water Vwater that is less than the volume of the ice cube, Vice. According to
Archimedes' principle, the water will apply a buoyant force to the ice cube; the magnitude of
this force will be equal to the weight of the water that is displaced by the ice cube. Since the
ice cube floats, the buoyant force must be equal in magnitude and opposite in direction to
the weight of the ice cube. That is, mice g = mwater g ; thus, the mass of the ice cube must be
Chapter 11 Conceptual Questions
551
equal to the mass of the water that is displaced by the ice cube. When the ice cube melts,
and the water has returned to its initial temperature, the mass of the water that originated
from the ice cube occupies the volume Vwater. In other words, the water that originated
from the ice cube will occupy a volume that is exactly equal to the volume that was
displaced by the ice cube. Therefore, the water level will remain the same.
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15. REASONING AND SOLUTION As a person dives toward the bottom of a swimming
pool, the pressure increases noticeably. The buoyant force, however, remains the same.
According to Archimedes' principle, the water applies a buoyant force to the swimmer, the
magnitude of which is equal to the weight of the water displaced by the swimmer. In other
words, the buoyant force experienced by the swimmer is given by ( ρ water Vswimmer ) g , where
ρ water is the density of the water and Vswimmer is the volume of water displaced by the
swimmer. If we neglect any change in water density with depth, then the quantity
ρ water Vswimmer remains the same as the swimmer approaches the bottom of the pool.
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16. SSM The plastic cup containing the two blocks floats in the water because there is
sufficient buoyant force to balance the total weight of the cup and the blocks. According to
Archimedes’ principle, the magnitude of the buoyant force is equal to the weight of the
water that the cup displaces. Thus, the weight of the displaced water consists of the
following three parts:
Wplastic = the weight of the cup itself
Wwooden = the weight of the wooden block
Wlead = the weight of the lead block
When the wooden block is removed from the cup and placed in the water, it floats. Thus, its
weight Wwooden is still balanced by a buoyant force, which has a magnitude equal to the
weight of the water that the wooden block displaces. Thus, the weight of water displaced by
the floating block is Wwooden, just as it is when the block is in the cup. As a result, the level
of the water in the pot does not change because the same amount of water is being displaced
in both cases.
When the lead block (instead of the wooden block) is removed from the cup and placed in
the water, it sinks to the bottom. The reason it sinks is that the magnitude of the buoyant
force acting on it is less than its weight Wlead. When the lead block is in the cup, however,
its weight is balanced by a buoyant force that has a magnitude equal to Wlead. Thus, the lead
block outside the cup causes less water to be displaced than does the lead block in the cup.
The result is that the level of the water in the pot is higher when the lead block is in the cup
than when it is outside the cup. In other words, the level falls when the lead block is
removed from the cup and put into the water.
552 FLUIDS
17. REASONING AND SOLUTION When the gate is not in use, the hollow, triangularshaped gate is filled with seawater and rests on the ocean floor. As shown in the figure, the
gate is anchored to the ocean floor by a hinge mechanism, about which it can rotate. When
a dangerously high tide is expected, water is pumped out of the gate, and the structure
rotates into a position where it can protect against flooding.
When the gate is filled with sea water, its average density (including the gate material) is
greater than that of seawater, and the gate sinks to the ocean floor. When water is pumped
from the gate, the density of the "gate system" decreases. Therefore, the mass, and hence the
weight, of the "gate system" decreases. When the buoyant force exerted on the bottom of the
"gate system" by the sea water exceeds the weight of the "gate system," the gate begins to
rise. Since the left end of the gate is anchored to the hinge mechanism, the gate will rotate
about the hinge as its right end rises.
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18. REASONING AND SOLUTION In steady flow, the velocity v of a fluid particle at any
point is constant in time. On the other hand, a fluid accelerates when it moves into a region
of smaller cross-sectional area, as shown in the figure below.
X
vX
Y
vY
a. A fluid particle at X with speed vX must be accelerated to the right in order to acquire the
greater speed vY at Y. From Newton's second law, this acceleration can arise only from a net
force that acts in the direction XY. If there are no other external forces acting on the fluid,
this force must arise from the change in pressure within the fluid. The pressure at X must be
greater than the pressure at Y.
b. The definition of steady flow makes no reference as to how the velocity of a fluid particle
varies from point to point as the fluid flows. It simply states that the velocity of a fluid
particle at any particular point is constant in time. Therefore, the condition of steady flow
does not rule out the acceleration discussed in part (a).
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19. REASONING AND SOLUTION Consider a stream of water that falls from a faucet.
According to the equation of continuity, the mass flow rate ρAv must be the same at every
point along the stream (ρ = water density, A = cross-sectional area of the stream, v = speed
of stream). As the water falls, it is accelerated due to gravity; therefore, the speed of the
water increases as it falls. Since the density of water is uniform throughout the stream,
when v increases, the cross-sectional area A of the stream must decrease in order to maintain
a constant mass flow rate. Therefore, the cross-sectional area of the stream becomes smaller
as the water falls.
Chapter 11 Conceptual Questions
553
If the water is shot upward, as it is in a fountain, the velocity of the stream is upward,
while the acceleration due to gravity is directed downward. Therefore, the speed of the
stream decreases as the stream rises. Since v decreases, the cross-sectional area A must
increase in order to maintain a constant mass flow rate ρAv . Therefore, the cross-sectional
area of the stream becomes larger as the water is shot upward.
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20. REASONING AND SOLUTION Suppose you are driving your car alongside a moving
truck. You will notice that your car will be pulled toward the truck as it passes.
Consider the air at two points. One point O is on the open side of your car, that is, the
side that is not adjacent to the truck. The other point B is between the car and the truck.
Bernoulli's principle, as applied to a constant elevation (Equation 11.12), indicates that
where the air speed is smaller, the air pressure is greater. Since the car is pulled toward the
truck, the pressure at point O must be greater than at point B. Therefore, the air at point O is
not moving very fast, while the air at point B is rushing more rapidly around the curved
surfaces of the front of the vehicles and through the narrow space between them.
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21. REASONING AND SOLUTION Bernoulli's equation applies to the steady flow of a
nonviscous, incompressible fluid. Turbulent flow is an extreme type of unsteady flow. It
occurs when there are sharp obstacles or bends in the path of a fast-moving fluid.
Bernoulli's equation does not apply to turbulent flow. When water cascades down a rockstrewn spillway, the water exhibits turbulent flow. The velocity at any particular point
changes erratically from one instant to another. Therefore, Bernoulli's equation could not be
used to describe water flowing down the rock-strewn spillway.
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22. REASONING AND SOLUTION Two
sheets of paper are held by adjacent
corners, so that they hang downward.
They are oriented so that they are parallel
and slightly separated by a gap through
which the floor can be seen. When air is
blown strongly down through the gap (see
the drawing at the right), the sheets come
closer together.
In accord with Bernoulli's equation, the
moving air between the sheets has a lower
pressure than the stationary air on either
side of the sheets. The greater pressure
on either side of the sheets generates an
inward force on the outer surface of the
sheets, and the sheets are pushed together.
Air directed
between the sheets
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554 FLUIDS
23. REASONING AND SOLUTION The figure below shows a baseball, as viewed from the
side, moving to the right with no spin. Since the air flows with the same speed above and
below the ball, the pressure is reduced by the same amount above and below the ball. There
is no net force to cause the ball to curve in any particular direction (except for gravity which
results in the usual parabolic trajectory).
up
v
bal
right
Without spin
If the ball is given a spin that is counterclockwise when viewed from the side, as shown
below, the air close to the surface of the ball is dragged with the ball. In accord with
Bernoulli's equation, the air on the top half of the ball is "speeded up" (pressure reduced by
a greater amount), while that on the lower half of the ball is also “speeded up”, but less so
(pressure reduced by a smaller amount). Thus, the pressure on the top half of the ball is
lower than on the bottom half.
Faster air,
lower pressure
Deflection force
up
vball
right
Slower air,
high pressure
Because of the pressure difference, a deflection force is generated that is directed from the
higher pressure side of the ball to the lower pressure side of the ball. Therefore, the ball
curves upward on its way to the plate.
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24. REASONING AND SOLUTION You are traveling on a train with your window open. As
the train approaches its rather high operating speed, your ears "pop."
When the train is stationary, the air inside and outside the train is stationary. The
pressures inside and outside, therefore, are equal. When the train moves at a high speed, the
air rushes rapidly by the window. According to Bernoulli's principle as expressed in
Equation 11.12, the faster moving air outside has a lower pressure than the nearly stationary
Chapter 11 Conceptual Questions
555
air inside. Under this condition some of the inside air is pushed out the window. As a result,
the pressure within the train drops. However, the pressure of the air in your inner ear is still
at the higher level that it was when the train was stationary. Thus, your ears "pop" outward.
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25. REASONING AND SOLUTION A passenger is smoking a cigarette in the back seat of a
moving car. To remove the smoke, the driver opens a window just a bit. Relative to the car,
the air in the car is stationary, while the air outside the car is moving. According to
Bernoulli's principle, as expressed in Equation 11.12, the moving air has a lower pressure
than the stationary air. Hence, the higher-pressure air inside the car will flow through the
opening to the lower-pressure region on the outside. Thus, the smoke will be drawn from the
higher-pressure region in the car to the lower-pressure region outside.
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26. SSM REASONING AND SOLUTION As discussed in the text, the dynamic lift on
airplane wings occurs because the air travels faster over the curved upper surface of the
wing than it does over the flatter lower surface (see Figure 11.34). According to Bernoulli's
equation, the pressure above the wing is lower (higher air speed), while the pressure below
the wing is higher (lower air speed). The net upward pressure on the wing causes an upward
force that lifts the plane.
The airport in Phoenix, Arizona, has occasionally been closed to large planes because of
unusually low air density. According to Equation 11.12, the net upward pressure on the
2
2
wing of a plane is given by Pbottom − Ptop = ( 1 / 2 ) ρ air ( v top
− v bottom
) where Pbottom and Ptop
are the air pressures on the bottom and top surfaces of the wing, while vtop and vbottom are
the air speeds at the top and bottom wing surfaces. When the air density is low, the quantity
2
2
− v bottom
( v top
) must be greater than it is under normal conditions to provide the same
minimum upward force to lift the plane. The speeds, vtop and vbottom, depend on the speed
of the plane relative to the ground. If the runway is not exceptionally long, the plane may
not have sufficient distance over which to increase its speed so that the quantity
2
2
− v bottom
( v top
) is large enough for take-off.
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27. REASONING AND SOLUTION To change the oil in a car, you remove a plug beneath
the engine and let the old oil run out. Your car has been sitting in the garage on a cold day.
Before changing the oil, it is advisable to run the engine for a while.
As discussed in the text, the viscosity of liquids usually decreases as the temperature is
increased. Since the car was sitting in a cold garage, the viscosity of the oil will be higher
than it is at room temperature. Therefore, if the plug is removed, the oil will run out, but at
a speed which is small compared to its speed of flow at room temperature. If the engine is
run for a while, the oil will heat up and its viscosity will decrease. The oil will then flow
more easily, and it can be drained from the car in a much shorter time than if it were at the
ambient temperature of the garage.
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556 FLUIDS
CHAPTER 11
FLUIDS
PROBLEMS
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1.
SSM REASONING AND SOLUTION Equation 11.1 can be solved for the mass m. In
order to use Equation 11.1, however, we must know the volume of the ice. Since the lake is
circular, the volume of the ice is equal to the circular area of the lake ( A = π r 2) multiplied
by the thickness d of the ice.
m = ρV = ρ (π r 2 )d = (917 kg/m3 )π (480 m) 2 (0.010 m) = 6.6 × 106 kg
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2.
REASONING According to Equation 4.5, the pillar’s weight is W = mg . Equation 11.1
can be solved for the mass m to show that the pillar’s mass is m = ρ V . The volume V of
the cylindrical pillar is its height times its circular cross-sectional area.
SOLUTION Expressing the weight as W = mg (Equation 4.5) and substituting m = ρ V
(Equation 11.1) for the mass give
W = mg = ( ρ V ) g
(1)
The volume of the pillar is V = h π r 2 , where h is the height and r is the radius of the pillar.
Substituting this expression for the volume into Equation (1), we find that the weight is
(
)
(
)
(
)
2
W = ( ρ V ) g = ⎡ ρ h π r 2 ⎤ g = 2.2 ×103 kg/m3 ( 2.2 m ) π ( 0.50 m ) 9.80 m/s 2 = 3.7 ×104 N
⎣
⎦
Converting newtons (N) to pounds (lb) gives
(
)
⎛ 0.2248 lb ⎞
3
W = 3.7 × 104 N ⎜
⎟ = 8.3 ×10 lb
1
N
⎝
⎠
3.
REASONING AND SOLUTION 14.0 karat gold is (14.0)/(24.0) gold or 58.3%. The
weight of the gold in the necklace is then (1.27 N)(0.583) = 0.740 N. This corresponds to a
volume given by V = M/ρ = W/(ρg) . Thus,
V=
0.740 N
(19 300 kg/m )(9.80 m/s )
3
2
= 3.91 × 10−6 m3
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Chapter 11 Problems
4.
557
REASONING The density of the brass ball is given by Equation 11.1: ρ = m / V . Since
the ball is spherical, its volume is given by the expression V = ( 4 / 3) πr 3 , so that the density
may be written as
m
3m
ρ= =
V 4 πr 3
This expression can be solved for the radius r; but first, we must eliminate the mass m from
the equation since its value is not specified explicitly in the problem. We can determine an
expression for the mass m of the brass ball by analyzing the forces on the ball.
The only two forces that act on the ball are the upward tension T in the wire and the
downward weight mg of the ball. If we take up as the positive direction, and we apply
Newton's second law, we find that T − mg = 0 , or m = T / g . Therefore, the density can be
written as
3m
3T
ρ=
=
3
4π r
4 πr 3 g
This expression can now be solved for r.
SOLUTION We find that
r3 =
3T
4 πρg
Taking the cube root of both sides of this result and using a value of ρ = 8470 kg/m3 for the
density of brass (see Table 11.1), we find that
3T
3(120 N)
= 7.0 × 10 –2 m
4 πρg
4 π ( 8470 kg / m 3 )(9.80 m / s 2 )
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r =
5.
3
=
3
SSM REASONING Equation 11.1 can be used to find the volume occupied by 1.00 kg
of silver. Once the volume is known, the area of a sheet of silver of thickness d can be
found from the fact that the volume is equal to the area of the sheet times its thickness.
SOLUTION Solving Equation 11.1 for V and using a value of ρ = 10 500 kg/m3 for the
density of silver (see Table 11.1), we find that the volume of 1.00 kg of silver is
V=
m
ρ
=
1.00 kg
= 9.52 × 10−5 m3
3
10 500 kg/m
The area of the silver, is, therefore,
V 9.52 × 10 –5 m3
=
= 317 m 2
−
7
d 3.00 × 10 m
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A=
558 FLUIDS
6.
REASONING AND SOLUTION
We refer to Table 11.1 for values of
3
3
ρsoda = ρwater = 1.000 × 10 kg/m and ρAl = 2700 kg/m3 for the density of aluminum. The
mass of aluminum required to make the can is
mAl = mtotal − msoda
where the mass of the soda is
msoda = ρsodaVsoda = (1.000 × 103 kg/m3)(3.54 × 10–4 m3) = 0.354 kg
Therefore, the mass of the aluminum used to make the can is
mAl = 0.416 kg − 0.354 kg = 0.062 kg
Using the definition of density, ρ = m/V, we have
mAl
0.062 kg
= 2.3 ×10 –5 m3
3
ρ Al 2700 kg/m
______________________________________________________________________________
VAl =
7.
=
REASONING According to the definition of density ρ given in Equation 11.1, the mass m
of a substance is m = ρV, where V is the volume. We will use this equation and the fact that
the mass of the water and the gold are equal to find our answer. To convert from a volume
in cubic meters to a volume in gallons, we refer to the inside of the front cover of the text to
find that 1 gal = 3.785 × 10–3 m3.
REASONING Using Equation 11.1, we find that
ρ Water VWater = ρ GoldVGold
or
VWater =
ρ Gold VGold
ρ Water
Using the fact that 1 gal = 3.785 × 10–3 m3 and densities for gold and water from
Table 11.1, we find
VWater =
ρ GoldVGold
ρ Water
c19 300 kg / m hb 0.15 mgb 0.050 mgb 0.050 mg FG 1 gal
=
H 3.785 × 10
c1000 kg / m h
3
3
−3
m
3
IJ =
K
1.9 gal
______________________________________________________________________________
Chapter 11 Problems
8.
559
REASONING The period T of a satellite is the time for it to make one complete revolution
around the planet. The period is the circumference of the circular orbit (2π R) divided by the
speed v of the satellite, so that T = (2π R)/v (see Equation 5.1). In Section 5.5 we saw that
the centripetal force required to keep a satellite moving in a circular orbit is provided by the
gravitational force. This relationship tells us that the speed of the satellite must be
v = GM / R (Equation 5.5), where G is the universal gravitational constant and M is the
mass of the planet. By combining this expression for the speed with that for the period, and
using the definition of density, we can obtain the period of the satellite.
SOLUTION The period of the satellite is
2π R
2π R
R3
=
= 2π
T=
v
GM
GM
R
According to Equation 11.1, the mass of the planet is equal to its density ρ times its volume
V. Since the planet is spherical, V = 43 π R3 . Thus, M = ρ V = ρ 43 π R3 . Substituting this
(
)
expression for M into that for the period T gives
R3
R3
3π
= 2π
=
T = 2π
3
4
GM
Gρ
Gρ 3 π R
(
)
The density of iron is ρ = 7860 kg/m3 (see Table 11.1), so the period of the satellite is
T=
3π
=
Gρ
( 6.67 ×10
3π
−11
)(
N ⋅ m 2 /kg 2 7860 kg/m3
)
= 4240 s
______________________________________________________________________________
9.
SSM WWW REASONING The total mass of the solution is the sum of the masses of
its constituents. Therefore,
ρsVs = ρ wVw + ρgVg
(1)
where the subscripts s, w, and g refer to the solution, the water, and the ethylene glycol,
respectively. The volume of the water can be written as Vw = Vs − Vg . Making this
substitution for Vw , Equation (1) above can be rearranged to give
Vg
Vs
=
ρs − ρ w
ρg − ρ w
(2)
Equation (2) can be used to calculate the relative volume of ethylene glycol in the solution.
560 FLUIDS
SOLUTION The density of ethylene glycol is given in the problem. The density of water
is given in Table 11.1 as 1.000 × 103 kg/m3 . The specific gravity of the solution is given as
1.0730. Therefore, the density of the solution is
ρs = (specific gravity of solution) × ρ w
= (1.0730)(1.000 × 103 kg/m3 ) = 1.0730 ×103 kg/m3
Substituting the values for the densities into Equation (2), we obtain
ρs − ρ w 1.0730 ×103 kg/m3 − 1.000 × 103 kg/m3
=
=
= 0.63
Vs ρg − ρ w
1116 kg/m3 − 1.000 × 103 kg/m3
Vg
Therefore, the volume percentage of ethylene glycol is 63% .
______________________________________________________________________________
10. REASONING Pressure is the magnitude of the force applied perpendicularly to a surface
divided by the area of the surface, according to Equation 11.3. The force magnitude,
therefore, is equal to the pressure times the area.
SOLUTION According to Equation 11.3, we have
c
F = PA = 8.0 × 10 4 lb / in. 2
h b 6.1 in.gb 2.6 in.g =
1.3 × 10 6 lb
______________________________________________________________________________
11. SSM REASONING Since the inside of the box is completely evacuated, there is no air
to exert an upward force on the lid from the inside. Furthermore, since the weight of the lid
is negligible, there is only one force that acts on the lid; the downward force caused by the
air pressure on the outside of the lid. In order to pull the lid off the box, one must supply a
force that is at least equal in magnitude and opposite in direction to the force exerted on the
lid by the outside air.
SOLUTION According to Equation 11.3, pressure is defined as P = F / A ; therefore, the
magnitude of the force on the lid due to the air pressure is
F = (0.85 × 105 N/m 2 )(1.3 × 10 –2 m 2 ) = 1.1× 103 N
______________________________________________________________________________
12. REASONING The cap is in equilibrium, so the sum of all the
forces acting on it must be zero. There are three forces in the
vertical direction: the force Finside due to the gas pressure inside
the bottle, the force Foutside due to atmospheric pressure outside
the bottle, and the force Fthread that the screw thread exerts on
F
the cap. By setting the sum of these forces to zero, and using the thread
Foutside
Finside
+y
Chapter 11 Problems
561
relation F = PA, where P is the pressure and A is the area of the cap, we can determine the
magnitude of the force that the screw threads exert on the cap.
SOLUTION The drawing shows the free-body diagram of the cap and the three vertical
forces that act on it. Since the cap is in equilibrium, the net force in the vertical direction
must be zero.
ΣFy = − Fthread + Finside − Foutside = 0
(4.9b)
Solving this equation for Fthread, and using the fact that force equals pressure times area,
F = PA (Equation 11.3), we have
Fthread = Finside − Foutside = Pinside A − Poutside A
(
)(
)
= ( Pinside − Poutside ) A = 1.80 × 105 Pa − 1.01× 105 Pa 4.10 × 10−4 m 2 = 32 N
______________________________________________________________________________
13. REASONING Since the weight is distributed uniformly, each tire exerts one-half of the
weight of the rider and bike on the ground. According to the definition of pressure,
Equation 11.3, the force that each tire exerts on the ground is equal to the pressure P inside
the tire times the area A of contact between the tire and the ground. From this relation, the
area of contact can be found.
SOLUTION The area of contact that each tire makes with the ground is
(
)
1 625 N + 98 N
(
)
F 2 Wperson + Wbike
(11.3)
A= =
= 2
= 4.76 × 10−4 m 2
5
P
P
7.60 × 10 Pa
______________________________________________________________________________
1
14. REASONING According to Equation 11.3, the pressure P exerted on the ground by the
stack of blocks is equal to the force F exerted by the blocks (their combined weight) divided
by the area A of the block’s surface in contact with the ground, or P = F/A. Since the
pressure is largest when the area is smallest, the least number of blocks is used when the
surface area in contact with the ground is the smallest. This area is 0.200 m × 0.100 m.
SOLUTION The pressure exerted by N blocks stacked on top of one another is
P=
F N Wone block
=
A
A
(11.3)
where Wone block is the weight of one block. The least number of whole blocks required to
produce a pressure of two atmospheres (2.02 × 105 Pa) is
N=
PA
2.02 × 105 Pa ) ( 0.200 m × 0.100 m )
(
=
= 24
Wone block
169 N
______________________________________________________________________________
562 FLUIDS
15. REASONING Pressure is the magnitude of the force applied perpendicularly to a surface
divided by the area of the surface, according to Equation 11.3. To find the force that the
suitcase applies to the floor, we first find the upward normal force FN that the floor applies
to the suitcase. By Newton’s action-reaction law, the force that the suitcase applies to the
floor has the same magnitude as FN, but points downward. In determining FN, we will use
Newton’s second law to account for the fact that the suitcase is accelerating.
SOLUTION Assuming that upward is the positive direction, we apply Newton’s second
law as follows:
ΣF = FN − mg = ma or FN = mg + ma
1 1 1
Net force
Normal
force
Weight
According to Equation 11.3, the pressure, then, is
P=
FN
=
b
m g+a
g = b16 kg gc 9.80 m / s + 1.5 m / s h =
b 0.50 mgb 0.15 mg
2
2
2 .4 × 10 3 Pa
A
A
______________________________________________________________________________
16. REASONING Pressure is defined in Equation 11.3
as the magnitude of the force acting perpendicular to
a surface divided by the area over which the force
acts. The force acting perpendicular to the slope is
due to the component of the skier’s weight W that is
directed perpendicular to the slope. In the drawing
at the right, this component is labeled Wperpendicular.
Note that the fact that the skier is moving is of no
importance.
SOLUTION
1
W
2 perpendicular
Wperpendicular
W
35°
35°
We assume that each ski bears the same amount of force, namely
(see the drawing). According to Equation 11.3, the pressure that each ski
applies to the snow is
P=
1
W
2 perpendicular
A
where A is the area of each ski in contact with the snow. From the drawing, we see that
Wperpendicular = W cos 35° , so that the pressure exerted by each ski on the snow is
P=
1
W
2 perpendicular
A
(
)
)
2
W cos 35° ( 58 kg ) 9.80 m/s cos 35°
=
=
= 1.8 × 103 Pa
2
2A
2 0.13 m
(
where we have used the fact that W = mg (Equation 4.5).
Chapter 11 Problems
563
17. SSM REASONING AND SOLUTION Both the cylinder and the hemisphere have
circular cross sections. According to Equation 11.3, the pressure exerted on the ground by
the hemisphere is
W
W
P = h = h2
Ah π rh
where Wh and rh are the weight and radius of the hemisphere. Similarly, the pressure
exerted on the ground by the cylinder is
P=
Wc
=
Ac
Wc
π rc2
where Wc and rc are the weight and radius of the cylinder. Since each object exerts the
same pressure on the ground, we can equate the right-hand sides of the expressions above to
obtain
Wh
Wc
=
π rh2 π rc2
Solving for rh 2 , we obtain
Wh
rh2 = rc2
(1)
Wc
The weight of the hemisphere is
Wh = ρgV h = ρg
1
2
c
4
3
π rh3
h=
2
3
ρgπ rh3
where ρ and V h are the density and volume of the hemisphere, respectively. The weight of
the cylinder is
Wc = ρgVc = ρgπ rc2 h
where ρ and Vc are the density and volume of the cylinder, respectively, and h is the height
of the cylinder. Substituting these expressions for the weights into Equation (1) gives
r =r
2
h
2
c
Wh
Wc
=r
2
c
ρg π rh3
ρg π rc2 h
2
3
Solving for rh gives
rh = 23 h = 23 ( 0.500 m ) = 0.750 m
______________________________________________________________________________
564 FLUIDS
18. REASONING AND SOLUTION The figures at the right
show all the forces that act on each mass. T is the tension
in the rope, FB is the upward force exerted on mass 1 due to
the pressure beneath the piston, and F0 is the downward
force exerted on mass 1 by the atmosphere.
mass 1
mass 2
T
T
F
B
Applying Newton's second law to mass 1, and taking "up"
as the positive direction, we have
FB + T – m1g – F0 = m1a
(1)
F0
m g
2
m g
1
Similarly for mass 2 we have
T – m2g = – m2a
(2)
Subtracting Equation (2) from Equation (1) gives
FB + m2g – m1g – F0 = (m1 + m2) a
(3)
Solving Equation (3) for FB gives
FB = m1(g + a) + m2(a – g) + F0
(4)
We must now find the acceleration a and the force F0. From kinematics, we recall that
y = v0 y t + 12 a y t 2 . Solving for ay and setting voy = 0, we have
ay =
2 y 2(1.25 m)
=
= 0.230 m/s 2
2
2
t
(3.30 s)
From the definition of pressure (P = F/A), the force F0 is
F0 = P0A = P0(π r2) = π (1.013 × 105 N/m2)(2.50 × 10–2 m)2 = 1.99 × 102 N
Substituting these values into Equation (4) gives:
FB = (0.500 kg)(9.80 m/s2 + 0.230 m/s2)
+ (9.50 kg)(0.230 m/s2 – 9.80 m/s2) + 1.99 × 102 N = 113 N
Thus, the pressure beneath the piston is
FB
F
113 N
= B2 =
= 5.75 × 104 Pa
–2
2
A πr
π (2.50 ×10 m)
______________________________________________________________________________
PB =
Chapter 11 Problems
565
19. SSM REASONING Since the faucet is closed, the water in the pipe may be treated as a
static fluid. The gauge pressure P2 at the faucet on the first floor is related to the gauge
pressure P1 at the faucet on the second floor by Equation 11.4, P2 = P1 + ρ gh .
SOLUTION
a. Solving Equation 11.4 for P1 , we find the gauge pressure at the second-floor faucet is
P1 = P2 − ρ gh = 1.90 × 10 5 Pa – (1.00 × 10 3 kg / m 3 )( 9.80 m / s 2 )( 6.50 m ) = 1.26 × 10 5 Pa
b. If the second faucet were placed at a height h above the first-floor faucet so that the
gauge pressure P1 at the second faucet were zero, then no water would flow from the
second faucet, even if it were open. Solving Equation 11.4 for h when P1 equals zero, we
obtain
P2 − P1
1.90 × 10 5 Pa − 0
= 19.4 m
ρg
(1.00 × 10 3 kg / m 3 )(9.80 m / s 2 )
______________________________________________________________________________
h=
=
20. REASONING The pressure P2 at a lower point in a static fluid is related to the pressure P1
at a higher point by Equation 11.4, P2 = P1 + ρ gh, where ρ is the density of the fluid, g is
the magnitude of the acceleration due to gravity, and h is the difference in heights between
the two points. This relation can be used directly to find the pressure in the artery in the
brain.
SOLUTION Solving Equation 11.4 for pressure P1 in the brain (the higher point), gives
(
)(
)
P1 = P2 − ρ gh = 1.6 × 104 Pa − 1060 kg/m3 9.80 m/s 2 ( 0.45 m ) = 1.1× 104 Pa
______________________________________________________________________________
21. REASONING The magnitude of the force that would be exerted on the window is given
by Equation 11.3, F = PA , where the pressure can be found from Equation 11.4:
P2 = P1 + ρgh . Since P1 represents the pressure at the surface of the water, it is equal to
atmospheric pressure, Patm . Therefore, the magnitude of the force is given by
F = ( Patm + ρgh ) A
where, if we assume that the window is circular with radius r, its area A is given by A = π r2.
SOLUTION
a. Thus, the magnitude of the force is
566 FLUIDS
F = 1.013 × 10 5 Pa + (1025 kg / m 3 )( 9.80 m / s 2 )( 11 000 m) π (0.10 m) 2 = 3.5 × 10 6 N
b. The weight of a jetliner whose mass is 1.2 × 10 5 kg is
W = mg = (1.2 × 10 5 kg)(9.80 m / s 2 ) = 1.2 × 10 6 N
Therefore, the force exerted on the window at a depth of 11 000 m is about three times
greater than the weight of a jetliner!
______________________________________________________________________________
22. REASONING As the depth h increases, the pressure increases according to Equation 11.4
(P2 = P1 + ρgh). In this equation, P1 is the pressure at the shallow end, P2 is the pressure at
the deep end, and ρ is the density of water (1.00 × 103 kg/m3, see Table 11.1). We seek a
value for the pressure at the deep end minus the pressure at the shallow end.
SOLUTION Using Equation 11.4, we find
PDeep = PShallow + ρgh
or
PDeep − PShallow = ρgh
The drawing at the right shows that a value for h can
be obtained from the 15-m length of the pool by using
the tangent of the 11° angle:
tan11° =
h
15 m
or
15 m
11°
h
h = (15 m ) tan11°
PDeep − PShallow = ρ g (15 m ) tan11°
(
)(
)
= 1.00 × 103 kg/m3 9.80 m/s 2 (15 m ) tan11° = 2.9 × 104 Pa
______________________________________________________________________________
23. SSM REASONING AND SOLUTION Using Equation 11.4, we have
Pheart − Pbrain = ρ gh = (1.000 × 103 kg/m3 )(9.80 m/s 2 )(12 m) = 1.2 × 105 Pa
______________________________________________________________________________
24. REASONING Since the diver uses a snorkel, the pressure in her lungs is atmospheric
pressure. If she is swimming at a depth h below the surface, the pressure outside her lungs is
atmospheric pressure plus that due to the water. The water pressure P2 at the depth h is
related to the pressure P1 at the surface by Equation 11.4, P2 = P1 + ρ gh, where ρ is the
density of the fluid, g is the magnitude of the acceleration due to gravity, and h is the depth.
This relation can be used directly to find the depth.
Chapter 11 Problems
567
SOLUTION We are given that the maximum difference in pressure between the outside
1 1.01× 105 Pa .
and inside of the lungs is one-twentieth of an atmosphere, or P2 − P1 = 20
(
)
Solving Equation 11.4 for the depth h gives
(
)
1 1.01× 105 Pa
P2 − P1
20
h=
=
= 0.50 m
ρg
1025 kg/m3 9.80 m/s 2
(
)(
)
______________________________________________________________________________
25. REASONING
The drawing at the
Evacuated so that
right shows the situation. As discussed
pressure is zero,
in Conceptual Example 6, the job of
P1 = 0
the pump is to draw air out of the pipe
that dips down into the water. The
atmospheric pressure in the well then
pushes the water upward into the pipe.
In the drawing, the best the pump can
do is to remove all of the air, in which
h
case, the pressure P1 at the top of the
water in the pipe is zero. The pressure
P2 at the bottom of the pipe at point A
B
is the same as that at the point B,
P2 =1.013×105 Pa
A
namely, it is equal to atmospheric
5
pressure (1.013 × 10 Pa ), because the
two points are at the same elevation,
and point B is open to the atmosphere. Equation 11.4, P2 = P1 + ρgh can be applied to
obtain the maximum depth h of the well.
}
SOLUTION Setting P1 = 0 Pa, and solving Equation 11.4 for h, we have
P1
1.013 × 10 5 Pa
= 10.3 m
ρg (1.000 × 10 3 kg / m 3 ) (9.80 m / s 2 )
______________________________________________________________________________
h=
=
26. REASONING Pressure is defined as the magnitude of the force acting perpendicular to a
surface divided by the area of the surface. Thus, the magnitude of the total force acting on
the vertical dam surface is the pressure times the area A of the surface. But exactly what
pressure should we use? According to Equation 11.4, the pressure at any depth h under the
water is P = Patm + ρ gh , where Patm is the pressure of the atmosphere acting on the water at
the top of the reservoir. Clearly, P has different values at different depths. As a result, we
need to use an average pressure. In Equation 11.4, it is only the term ρgh that depends on
depth, and the dependence is linear. In other words, the value of ρgh is proportional to h.
568 FLUIDS
Therefore, the average value of this term is ρ g
water in the full reservoir and
1
2
( H ) , where H is the total depth of the
1
2
H is the average depth. The average pressure acting on the
vertical surface of the dam in contact with the water is, then,
P = Patm + ρ g
( H)
1
2
(1)
SOLUTION According to the definition of pressure given in Equation 11.3, the magnitude
Ftotal of the total force acting on the vertical surface of the dam is Ftotal = PA , where P is
given by Equation (1). Using Equation (1) to substitute for P gives
Ftotal = PA = ⎡ Patm + ρ g
⎢⎣
( H )⎤⎥⎦ A
1
2
The area is A = (120 m)(12 m), and Patm = 1.01× 105 Pa . Table 11.1 gives the density of
water as ρ = 1.00 × 103 kg/m3 . With these values, we find that
Ftotal = ⎡ Patm + ρ g
⎣⎢
( H )⎤⎦⎥ A
1
2
(
)(
= ⎡1.01× 105 Pa + 1.00 ×103 kg/m3 9.80 m/s 2
⎣
) 12 (12 m )⎤⎦ (120 m )(12 m ) = 2.3 ×108 N
27. SSM REASONING Let the length of the tube be denoted by L, and let the length of the
liquid be denoted by A . When the tube is whirled in a circle at a constant angular speed
about an axis through one end, the liquid collects at the other end and experiences a
centripetal force given by (see Equation 8.11, and use F = ma ) F = mrω 2 = mLω 2 .
Since there is no air in the tube, this is the only radial force experienced by the liquid, and it
results in a pressure of
F mLω 2
P= =
A
A
where A is the cross-sectional area of the tube. The mass of the liquid can be expressed in
terms of its density ρ and volume V: m = ρV = ρAA . The pressure may then be written as
P=
ρ AALω 2
A
= ρ ALω 2
(1)
Chapter 11 Problems
569
If the tube were completely filled with liquid and allowed to hang vertically, the pressure at
the bottom of the tube (that is, where h = L ) would be given by
P = ρgL
(2)
SOLUTION According to the statement of the problem, the quantities calculated by
Equations (1) and (2) are equal, so that ρALω 2 = ρgL . Solving for ω gives
g
9.80 m / s 2
=
= 31.3 rad / s
A
0.0100 m
______________________________________________________________________________
ω=
28. REASONING AND SOLUTION The mercury, being more dense, will flow from the right
container into the left container until the pressure is equalized. Then the pressure at the
bottom of the left container will be P = ρwghw + ρmghmL and the pressure at the bottom of
the right container will be P = ρmghmR. Equating gives
ρwghw + ρmg(hmL − hmR) = 0
(1)
Both liquids are incompressible and immiscible so
hw = 1.00 m and hmL + hmR = 1.00 m
Using these in (1) and solving for hmL gives, hmL = (1/2)(1.00 − ρw/ρm) = 0.46 m.
So the fluid level in the left container is 1.00 m + 0.46 m = 1.46 m from the bottom.
______________________________________________________________________________
29. REASONING AND SOLUTION The pressure at the bottom of the container is
P = Patm + ρwghw + ρmghm
We want P = 2Patm, and we know h = hw + hm = 1.00 m. Using the above and rearranging
gives
hm =
Patm − ρ w gh
( ρm −
ρw ) g
=
(
)(
)
1.01 × 105 Pa − 1.00 × 103 kg/m3 9.80 m/s 2 (1.00 m )
(
)(
13.6 × 103 kg/m3 − 1.00 × 103 kg/m3 9.80 m/s 2
)
= 0.74 m
______________________________________________________________________________
30. REASONING AND SOLUTION The force exerted on the top surface is
F2 = P2A2 = Patmπ R22
570 FLUIDS
The force exerted on the bottom surface is F1 = P1A1 = (Patm + ρgh)π R12. Equating and
rearranging yields
R22 = R12(1 + ρgh/Patm)
or
R22 = 1.485 R12
(1)
Consider a right triangle formed by drawing a vertical line from a point on the
circumference of the bottom circle to the plane of the top circle so that two sides are equal
to R2 – R1 and h. Then tan 30.0° = (R2 – R1)/h.
R1 = R2 −h tan 30.0° = R2 – 2.887 m
a. Now,
(2)
Substituting (2) into (1) and rearranging yields
0.485 R22 – 8.57 R2 + 12.38 = 0
which has two roots, namely, R2 = 16.1 m and 1.59 m. The value R2 = 1.59 m leads to a
negative value for R1. Clearly, a radius cannot be negative, so we can eliminate the root
R2 = 1.59 m, and we conclude that R2 = 16.1 m .
b. Now that R2 is known, Equation (2) gives R1 = 13.2 m .
______________________________________________________________________________
31. SSM REASONING According to Equation 11.4, the initial pressure at the bottom of the
c h
pool is P0 = Patm
0
c h
+ ρgh , while the final pressure is Pf = Patm
f
+ ρgh . Therefore, the
change in pressure at the bottom of the pool is
∆P = Pf − P0 =
cP h
atm
f
+ ρgh −
cP h
atm
0
c h −cP h
+ ρgh = Patm
f
atm
0
According to Equation 11.3, F = PA , the change in the force at the bottom of the pool is
∆F = ( ∆P ) A =
cP h −cP h
atm
atm
f
0
A
SOLUTION Direct substitution of the data given in the problem into the expression above
yields
133 Pa
∆F = 765 mm Hg − 755 mm Hg (12 m)(24 m)
= 3.8 × 10 5 N
1.0 mm Hg
b
g
FG
H
IJ
K
Note that the conversion factor 133 Pa = 1.0 mm Hg is used to convert mm Hg to Pa.
______________________________________________________________________________
Chapter 11 Problems
571
32. REASONING We designate F1 and A1, respectively, as the magnitude of the force applied
to the input piston and the circular cross-sectional area of the piston. The analogous
quantities for the output plunger are F2 and A2. Since the difference in height between the
input piston and output plunger is being neglected, the relation between the force
magnitudes and areas is F2 / F1 = A2 / A1 (Equation 11.5). Using the fact that the area of
each circular cross-section is π r 2 , we can utilize the data given for the ratio of the radii of
the input piston and output plunger and determine the force F2 applied to the car.
SOLUTION According to Equation 11.5, we have
F2 A2
=
F1 A1
Substituting A1 = π r12 and A2 = π r22 , we find that
F2 π r22
=
F1 π r12
2
or
⎛r ⎞
2
F2 = F1 ⎜⎜ 2 ⎟⎟ = ( 45 N )( 8.3) = 3100 N
⎝ r1 ⎠
33. REASONING We label the input piston as “2” and the output plunger as “1.” When the
bottom surfaces of the input piston and output plunger are at the same level, Equation 11.5,
F2 = F1 ( A2 / A1 ) , applies. However, this equation is not applicable when the bottom surface
of the output plunger is h = 1.50 m above the input piston. In this case we must use Equation
11.4, P2 = P1 + ρ gh , to account for the difference in heights. In either case, we will see that
the input force is less than the combined weight of the output plunger and car.
SOLUTION
a. Using A = π r2 for the circular areas of the piston and plunger, the input force required to
support the 24 500-N weight is
(
⎡
−3
⎛ A2 ⎞
⎢ π 7.70 × 10 m
F2 = F1 ⎜ ⎟ = ( 24 500 N ) ⎢
2
⎜A ⎟
⎝ 1⎠
⎢⎣ π ( 0.125 m )
)
2⎤
⎥
⎥ = 93.0 N
⎥⎦
(11.5)
b. The pressure P2 at the input piston is related to the pressure P1 at the bottom of the
output plunger by Equation 11.4, P2 = P1 + ρgh, where h is the difference in heights.
( )
( )
Setting P2 = F2 / A2 = F2 / π r22 , P1 = F1 / π r12 , and solving for F2, we have
572 FLUIDS
⎛ π r2 ⎞
F2 = F1 ⎜ 22 ⎟ + ρ gh π r22
⎜πr ⎟
⎝ 1 ⎠
( )
(11.4)
(
⎡
−3
⎢ π 7.70 × 10 m
= ( 24 500 N ) ⎢
2
⎢⎣ π ( 0.125 m )
(
)
2⎤
⎥
⎥
⎥⎦
)(
)
(
+ 8.30 × 102 kg/m3 9.80 m/s 2 (1.30 m ) π 7.70 ×10−3 m
)
2
= 94.9 N
______________________________________________________________________________
34. REASONING Equation 11.5 gives the force F2 of the output plunger in terms of the force
F1 applied to the input piston as F2 = F1(A2/A1), where A2 and A1 are the corresponding
areas. In this problem the chair begins to rise when the output force just barely exceeds the
weight, so F2 = 2100 N. We are given the input force as 55 N. We seek the ratio of the
radii, so we will express the area of each circular cross section as π r2 when we apply
Equation 11.5.
SOLUTION According to Equation 11.5, we have
A2
A1
=
F2
or
F1
π r22 F2
=
π r12 F1
Solving for the ratio of the radii yields
r2
r1
=
F2
F1
=
2100 N
= 6.2
55 N
______________________________________________________________________________
35. REASONING Since the piston and the plunger are at the same height, Equation 11.5,
F2 = F1 ( A2 / A1 ) , applies, and we can find an expression for the force exerted on the
spring. Then Equation 10.1, Fxapplied = kx, can be used to determine the amount of
compression of the spring.
SOLUTION Substituting the right hand side of Equation 11.5 into Equation 10.1, we find
that
A2
F1
= kx
A1
F I
GH JK
Chapter 11 Problems
573
From the drawing in the text, we see that the force on the right piston must be equal in
magnitude to the weight of the rock, or F1 = mg . Therefore,
mg
F A I = kx
GH A JK
2
1
Solving for x, we obtain
F I = (40.0 kg)(9.80 m / s ) FG 15 cm IJ =
GH JK
1600 N / m
H 65 cm K
mg A2
x=
k A1
2
2
2
5.7 × 10 –2 m
______________________________________________________________________________
36. REASONING AND SOLUTION From Pascal's principle, the pressure in the brake fluid at
the master cylinder is equal to the pressure in the brake fluid at the plungers: PC = PP, or
FC
F
= P
AC AP
⎛r
A
π r2
FP = FC P = FC P2 = FC ⎜ P
⎜r
AC
π rC
⎝ C
or
⎞
⎟
⎟
⎠
2
The torque on the pedal is equal to the torque that is applied to the master cylinder so that
F A = FCA C
or
FC = F
A
AC
Combining the expression for FC with the expression for FP above, we have
2
2
⎛ rp ⎞
⎛ 0.150 m ⎞ ⎛ 1.90 × 10 –2 m ⎞
⎜⎜ ⎟⎟ = (9.00 N) ⎜
⎟ = 108 N
⎟⎜
⎝ 0.0500 m ⎠ ⎝⎜ 9.50 × 10 –3 m ⎠⎟
⎝ rC ⎠
______________________________________________________________________________
A
FP = F
AC
37. SSM REASONING The pressure P ′ exerted on the bed of the truck by the plunger is
P ′ = P − Patm . According to Equation 11.3, F = P ′A , so the force exerted on the bed of
c
h
the truck can be expressed as F = P − Patm A . If we assume that the plunger remains
perpendicular to the floor of the load bed, the torque that the plunger creates about the axis
shown in the figure in the text is
c
h
c
h
τ = FA = P − Patm AA = P − Patm ( π r 2 ) A
SOLUTION Direct substitution of the numerical data into the expression above gives
c
h
τ = 3.54 × 10 6 Pa − 1.01 × 10 5 Pa π ( 0.150 m ) 2 ( 3.50 m ) = 8.50 × 10 5 N ⋅ m
______________________________________________________________________________
574 FLUIDS
38. REASONING Since the duck is in equilibrium, its downward-acting weight is balanced by
the upward-acting buoyant force. According to Archimedes’ principle, the magnitude of the
buoyant force is equal to the weight of the water displaced by the duck. Setting the weight
of the duck equal to the magnitude of the buoyant force will allow us to find the average
density of the duck.
SOLUTION
Since the weight Wduck of the duck is balanced by the magnitude FB of the
buoyant force, we have that Wduck = FB. The duck’s weight is Wduck = mg = (ρduckVduck)g,
where ρduck is the average density of the duck and Vduck is its volume. The magnitude of the
buoyant force, on the other hand, equals the weight of the water displaced by the duck, or
FB = mwaterg, where mwater is the mass of the displaced water. But mwater = ρ water 14 Vduck ,
(
)
since one-quarter of the duck’s volume is beneath the water. Thus,
(
)
ρduckVduck g = ρ water 14 Vduck g
Weight of duck
Magnitude of
buoyant force
Solving this equation for the average density of the duck (and taking the density of water
from Table 11.1) gives
ρduck =
1
4
ρ water =
1
4
(1.00 ×103 kg/m3 ) = 250 kg/m3
______________________________________________________________________________
39. SSM REASONING According to Archimedes principle, the buoyant force that acts on
the block is equal to the weight of the water that is displaced by the block. The block
displaces an amount of water V, where V is the volume of the block. Therefore, the weight
of the water displaced by the block is W = mg = ( ρ waterV ) g .
SOLUTION The buoyant force that acts on the block is, therefore,
F = ρ waterV g = (1.00 × 103 kg/m3 )(0.10 m × 0.20 m × 0.30 m)(9.80 m/s 2 ) = 59 N
______________________________________________________________________________
40. REASONING The ice with the bear on it is floating, so that the upward-acting buoyant
force balances the downward-acting weight Wice of the ice and weight Wbear of the bear.
The magnitude FB of the buoyant force is the weight WH
according to Archimedes’ principle.
2
O
of the displaced water,
Thus, we have FB = WH
2
O
= Wice + Wbear , the
expression with which we will obtain Wbear. We can express each of the weights WH
2
O
and
Chapter 11 Problems
575
Wice as mass times the magnitude of the acceleration due to gravity (Equation 4.5) and then
relate the mass to the density and the displaced volume by using Equation 11.1.
SOLUTION Since the ice with the bear on it is floating, the upward-acting buoyant force
FB balances the downward-acting weight Wice of the ice and the weight Wbear of the bear.
The buoyant force has a magnitude that equals the weight WH
2
O
of the displaced water, as
stated by Archimedes’ principle. Thus, we have
FB = WH
2
O
= Wice + Wbear
Wbear = WH
or
2
O
− Wice
In Equation (1), we can use Equation 4.5 to express the weights WH
2
O
(1)
and Wice as mass m
times the magnitude g of the acceleration due to gravity. Then, the each mass can be
expressed as m = ρ V (Equation 11.1). With these substitutions, Equation (1) becomes
Wbear = mH O g − mice g = ( ρ H OVH O ) g − ( ρiceVice ) g
2
2
(2)
2
When the heaviest possible bear is on the ice, the ice is just below the water surface and
displaces a volume of water that is VH O = Vice . Substituting this result into Equation (2),
2
we find that
Wbear = ( ρ H OVice ) g − ( ρiceVice ) g = ( ρ H
2
(
)(
2
O
− ρice )Vice g
)(
)
= 1025 kg/m3 − 917 kg/m3 5.2 m3 9.80 m/s 2 = 5500 N
41. REASONING The paperweight weighs less in water than in air, because of the buoyant
force FB of the water. The buoyant force points upward, while the weight points
downward, leading to an effective weight in water of WIn water = W – FB. There is also a
buoyant force when the paperweight is weighed in air, but it is negligibly small. Thus, from
the given weights, we can obtain the buoyant force, which is the weight of the displaced
water, according to Archimedes’ principle. From the weight of the displaced water and the
density of water, we can obtain the volume of the water, which is also the volume of the
completely immersed paperweight.
SOLUTION We have
WIn water = W − FB
or
FB = W − WIn water
According to Archimedes’ principle, the buoyant force is the weight of the displaced water,
which is mg, where m is the mass of the displaced water. Using Equation 11.1, we can write
the mass as the density times the volume or m = ρV. Thus, for the buoyant force, we have
FB = W − WIn water = ρVg
576 FLUIDS
Solving for the volume and using ρ = 1.00 × 103 kg/m3 for the density of water (see
Table 11.1), we find
V=
W − WIn water
ρg
=
c
6.9 N − 4 .3 N
hc
1.00 × 10 3 kg / m 3
9 .80 m / s 2
h
= 2 .7 × 10 −4 m 3
______________________________________________________________________________
42. REASONING According to Equation 11.1, the density of the life jacket is its mass divided
by its volume. The volume is given. To obtain the mass, we note that the person wearing
the life jacket is floating, so that the upward-acting buoyant force balances the downwardacting weight of the person and life jacket. The magnitude of the buoyant force is the
weight of the displaced water, according to Archimedes’ principle. We can express each of
the weights as mg (Equation 4.5) and then relate the mass of the displaced water to the
density of water and the displaced volume by using Equation 11.1.
SOLUTION According to Equation 11.1, the density of the life jacket is
mJ
ρJ =
(1)
VJ
Since the person wearing the life jacket is floating, the upward-acting buoyant force FB
balances the downward-acting weight WP of the person and the weight WJ of the life jacket.
The buoyant force has a magnitude that equals the weight WH
2
O
of the displaced water, as
stated by Archimedes’ principle. Thus, we have
FB = WH
2
O
= WP + WJ
(2)
In Equation (2), we can use Equation 4.5 to express each weight as mass m times the
magnitude g of the acceleration due to gravity. Then, the mass of the water can be
expressed as mH O = ρ H OVH O (Equation 11.1). With these substitutions, Equation (2)
2
2
2
becomes
mH O g = mP g + mJ g
( ρ H OVH O ) g = mP g + mJ g
or
2
2
2
Solving this result for mJ shows that
mJ = ρ H OVH
2
2
O
− mP
Substituting this result into Equation (1) and noting that the volume of the displaced water is
VH O = 3.1× 10−2 m3 + 6.2 ×10−2 m3 gives
2
Chapter 11 Problems
ρJ =
ρ H OVH
2
2
O
− mP
VJ
577
1.00 × 103 kg/m3 )( 3.1×10−2 m3 + 6.2 × 10−2 m3 ) − 81 kg
(
=
= 390 kg/m3
3.1×10−2 m3
43. SSM WWW REASONING AND SOLUTION Under water, the weight of the person
with empty lungs is Wempty = W − ρ water gVempty , where W is the weight of the person in air
and Vempty is the volume of the empty lungs. Similarly, when the person's lungs are
partially full under water, the weight of the person is Wfull = W − ρ water gVfull . Subtracting
the second equation from the first equation and rearranging gives
Vfull − Vempty =
Wempty − Wfull
ρ water g
=
40.0 N − 20.0 N
(1.00 ×10
3
kg/m
3
)(9.80 m/s )
2
= 2.04 × 10−3 m3
______________________________________________________________________________
44. REASONING When an object is completely submerged within a fluid, its apparent weight
in the fluid is equal to its true weight mg minus the upward-acting buoyant force. According
to Archimedes’ principle, the magnitude of the buoyant force is equal to the weight of the
fluid displaced by the object. The weight of the displaced fluid depends on the volume of
the object. We will apply this principle twice, once for the object submerged in each fluid,
to find the volume of the object.
SOLUTION The apparent weights of the object in ethyl alcohol and in water are:
Ethyl
alcohol
15.2
N
= N
mg − ρalcohol gV
Water
13.7
N
mg − ρ water gV
= N
Weight in
alcohol
Weight
in water
True
weight
True
weight
(1
Magnitude of
buoyant force
(2
Magnitude of
buoyant force
These equations contain two unknowns, the volume V of the object and its mass m. By
subtracting Equation (2) from Equation (1), we can eliminate the mass algebraically. The
result is
15.2 N − 13.7 N = gV ( ρ water − ρalcohol )
Solving this equation for the volume, and using the densities from Table 11.1, we have
V=
15.2 N − 13.7 N
1.5 N
=
= 7.9 ×10−4 m3
2
3
3
3
g ( ρ water − ρalcohol )
9.80 m/s 1.00 ×10 kg/m − 806 kg/m
(
)(
)
578 FLUIDS
______________________________________________________________________________
45. REASONING AND SOLUTION The figure at the right shows the two
forces that initially act on the box. Since the box is not accelerated, the
two forces must have zero resultant: FB0 − Wbox = 0 . Therefore,
FB0 = Wbox
(1)
F
B0
W
box
From Archimedes' principle, the buoyant force on the box is equal to
the weight of the water that is displaced by the box:
FB0 = Wdisp
(2)
Combining (1) and (2) we have Wbox = Wdisp, or mbox g = ρwatergVdisp. Therefore,
mbox = ρwaterVdisp
Since the box floats with one-third of its height beneath the water, Vdisp = (1/3)Vbox, or
Vdisp = (1/3)L3. Therefore,
mbox =
ρ water L3
(3)
3
The figure at the right shows the three forces that act on the box after
water is poured into the box. The box begins to sink when
Wbox + Wwater > FB
F
B
(4)
The box just begins to sink when the equality is satisfied. From
Archimedes' principle, the buoyant force on the system is equal to the
weight of the water that is displaced by the system: FB = Wdisplaced.
W
box
W
water
The equality in (4) can be written as
mboxg + mwaterg = mdisplacedg
(5)
When the box begins to sink, the volume of the water displaced is equal to the volume of the
box; Equation (5) then becomes
mbox + ρwaterVwater = ρwaterVbox.
The volume of water in the box at this instant is Vwater = L2h, where h is the depth of the
water in the box. Thus, the equation above becomes
Chapter 11 Problems
579
mbox + ρwaterL2h = ρwaterL3
Using Equation (3) for the mass of the box, we obtain
ρ water L3
3
+ ρ water L2 h = ρ water L3
Solving for h gives
2
3
2
3
h = L = (0.30 m) = 0.20 m
______________________________________________________________________________
46. REASONING AND SOLUTION The buoyant force exerted by the water must at least
equal the weight of the logs plus the weight of the people,
FB = WL + WP
ρwgV = ρLgV + WP
Now the volume of logs needed is
V=
MP
ρW − ρL
=
4 ( 80.0 kg )
1.00 × 10 kg/m − 725 kg/m
3
3
3
= 1.16 m3
The volume of one log is
VL = π(8.00 × 10–2 m)2(3.00 m) = 6.03 × 10–2 m3
The number of logs needed is
N = V/VL = (1.16)/(6.03 × 10–2) = 19.2
Therefore, at least 20 logs are needed .
______________________________________________________________________________
47. SSM REASONING The height of the cylinder that is in the oil is given by
hoil = V oil / ( π r 2 ) , where Voil is the volume of oil displaced by the cylinder and r is the
radius of the cylinder. We must, therefore, find the volume of oil displaced by the cylinder.
After the oil is poured in, the buoyant force that acts on the cylinder is equal to the sum of
the weight of the water displaced by the cylinder and the weight of the oil displaced by the
cylinder. Therefore, the magnitude of the buoyant force is given by
580 FLUIDS
F = ρ water gV water + ρ oil gVoil . Since the cylinder floats in the fluid, the net force that acts on
the cylinder must be zero. Therefore, the buoyant force that supports the cylinder must be
equal to the weight of the cylinder, or
ρ water gV water + ρ oil gVoil = mg
where m is the mass of the cylinder. Substituting values into the expression above leads to
V water + ( 0.725)Voil = 7 .00 × 10 –3 m 3
(1)
From the figure in the text, Vcylinder = Vwater + Voil . Substituting values into the expression
for Vcylinder gives
V water + Voil = 8.48 × 10 –3 m 3
(2)
Subtracting Equation (1) from Equation (2) yields Voil = 5.38 × 10 –3 m 3 .
SOLUTION The height of the cylinder that is in the oil is, therefore,
Voil
5.38 × 10 –3 m 3
= 7.6 × 10 –2 m
πr2
π ( 0.150 m) 2
______________________________________________________________________________
hoil =
=
48. REASONING AND SOLUTION The figure at the right shows the
forces that act on the balloon as it holds the passengers and the ballast
stationary above the earth. W0 is the combined weight of the balloon,
the load of passengers, and the ballast. The quantity FB is the
buoyant force provided by the air outside the balloon and is given by
F
B
W
FB = ρairgVballoon
(1)
0
Since the balloon is stationary, it follows that FB − W0 = 0, or
FB = W0
(2)
Then the ballast is dropped overboard, the balloon accelerates upward through a distance y
2y
in a time t with acceleration ay, where (from kinematics) ay = 2 .
t
Chapter 11 Problems
The figure at the right shows the forces that act on the balloon while it
accelerates upward.
581
F
B
Applying Newton's second law, we have FB − W = may, where W is the
weight of the balloon and the load of passengers. Replacing m by (W/g)
we have
FB − W =
W
W
a
g y
Thus,
FB = W +
⎛
W ⎛ 2y ⎞
2y ⎞
⎜ 2 ⎟ = W ⎜1 + 2 ⎟
g⎝t ⎠
⎝ gt ⎠
Solving for W gives
W=
⎛ gt 2 ⎞
= FB ⎜⎜ 2
⎟
⎛
gt + 2 y ⎟⎠
2y ⎞
⎝
⎜1 + 2 ⎟
⎝ gt ⎠
FB
The amount of ballast that must be thrown overboard is therefore [using Equations (1) and
(2)]
⎛ gt 2 ⎞ ⎛
gt 2 ⎞
∆W = W0 − W = FB − FB ⎜⎜ 2
⎟⎟ = ⎜⎜1 − 2
⎟⎟ ρairgVballoon
⎝ gt + 2 y ⎠ ⎝ gt + 2 y ⎠
∆W = rair g
⎛
⎞
3
( 43 ρ rballoon
) ⎜⎜1 − gt 2gt+ 2 y ⎟⎟
2
⎝
⎠
⎡
⎤
(9.80 m/s 2 )(15.0 s)2
∆W = (1.29 kg/m3 )(9.80 m/s 2 ) ⎡⎣ 43 π (6.25 m)3 ⎤⎦ ⎢1 −
⎥
2
2
⎣ (9.80 m/s )(15.0 s) + 2(105 m) ⎦
= 1120 N
______________________________________________________________________________
49. REASONING AND SOLUTION The upward buoyant force must equal the weight of the
shell if it is floating, ρwgV = W. The submerged volume of the shell is V = (4/3)π R23,
where R2 is its outer radius. Now R23 = (3/4)m/(ρwπ ) gives
R2 =
3
3m
4π ρ w
=
3
(
3 (1.00 kg )
4π 1.00 × 10 kg/m
3
3
)
= 6.20 × 10−2 m
The weight of the shell is W = ρgg(V2 − V1) so R13 = R23 − (3/4)m/(π ρg), and
582 FLUIDS
R1 =
3
3m ⎛ 1
1 ⎞
⎜
⎟ =
−
4π ⎜ ρ w ρg ⎟
⎝
⎠
3
⎞
3 (1.00 kg ) ⎛
1
1
−
⎜
⎟
3
3
4π
2.60 × 103 kg/m3 ⎠
⎝ 1.00 × 10 kg/m
= 5.28 × 10−2 m
______________________________________________________________________________
50. REASONING AND SOLUTION The volume flow rate is given by
Q = Av = π r2v = π (0.305 m)2(1.22 m/s) = 0.356 m3/s
The number of gallons that flows in one day (8.64 × 104 s) is
⎛
⎞
gal
8.64 × 104 s ) =
( 0.356 m3/s ) ⎜ 3.791.0
−3 3 ⎟ (
× 10 m
8.12 × 106 gal
⎝
⎠
______________________________________________________________________________
51. SSM REASONING AND SOLUTION The mass flow rate Qmass is the amount of fluid
mass that flows per unit time. Therefore,
m ρV (1030 kg/m3 )(9.5 × 10 –4 m3 ) ⎛ 1.0 h ⎞
−5
Qmass = =
=
⎜
⎟ = 4.5 × 10 kg/s
t
t
6.0 h
⎝ 3600 s ⎠
______________________________________________________________________________
52. REASONING
a. According to Equation 11.10, the volume flow rate Q is equal to the product of the crosssectional area A of the artery and the speed v of the blood, Q = Av. Since Q and A are
known, we can determine v.
b. Since the volume flow rate Q2 through the constriction is the same as the volume flow
rate Q1 in the normal part of the artery, Q2 = Q1. We can use this relation to find the blood
speed in the constricted region.
SOLUTION
a. Since the artery is assumed to have a circular cross-section, its cross-sectional area is
A1 = π r12 , where r1 is the radius. Thus, the speed of the blood is
v1 =
Q1 Q1
3.6 ×10−6 m3 / s
= 2 =
= 4.2 × 10−2 m/s
2
A1 π r1 π 5.2 × 10−3 m
(
)
(11.10)
b. The volume flow rate is the same in the normal and constricted parts of the artery, so
Q2 = Q1. Since Q2 = A2v2 , the blood speed is v2 = Q2/A2 = Q1/A2. We are given that the
Chapter 11 Problems
583
radius of the constricted part of the artery is one-third that of the normal artery, so r2 = 13 r1.
Thus, the speed of the blood at the constriction is
Q1
Q
Q1
3.6 × 10−6 m3 / s
= 12 =
=
= 0.38 m/s
2
A2 π r2 π 1 r 2
−3
1
⎡
⎤
π 3 5.2 ×10 m
3 1
⎣
⎦
______________________________________________________________________________
v2 =
( )
(
)
53. REASONING The length L of the side of the square can be obtained, if we can find a value
for the cross-sectional area A of the ducts. The area is related to the volume flow rate Q and
the air speed v by Equation 11.10 (Q = Av). The volume flow rate can be obtained from the
volume V of the room and the replacement time t as Q = V/t.
SOLUTION For a square cross section with sides of length L, we have A = L2. And we
know that the volume flow rate is Q = V/t. Therefore, using Equation 11.10 gives
Q = Av
V
= L2 v
t
or
Solving for L shows that
(a) Air speed = 3.0 m / s
L=
V
=
tv
(b) Air speed = 5.0 m / s
L=
V
=
tv
b
120 m 3
= 0.18 m
1200 s 3.0 m / s
b
gb
g
120 m 3
= 0.14 m
1200 s 5.0 m / s
gb
g
______________________________________________________________________________
54. REASONING The number N of capillaries can be obtained by dividing the total crosssectional area Acap of all the capillaries by the cross-sectional area acap of a single capillary.
2
We know the radius rcap of a single capillary, so acap can be calculated as acap = π rcap
. To
find Acap, we will use the equation of continuity.
SOLUTION The number N of capillaries is
N=
Acap
acap
=
Acap
2
π rcap
(1)
2
.
where the cross-sectional area acap of a single capillary has been replaced by acap = π rcap
To obtain the total cross-sectional area Acap of all the capillaries, we use the equation of
continuity for an incompressible fluid (see Equation 11.9). For present purposes, this
equation is
Aaorta vaorta
Aaorta vaorta = Acap vcap or Acap =
(2)
vcap
584 FLUIDS
2
, and with this substitution,
The cross-sectional area of the aorta is Aaorta = π raorta
Equation (2) becomes
Aaorta vaorta
Acap =
vcap
=
2
π raorta
vaorta
vcap
Substituting this result into Equation (1), we find that
N=
Acap
2
π rcap
⎛ π r2 ⎞
⎜ aorta ⎟ vaorta
2
⎜ vcap ⎟
raorta
vaorta
⎠
=⎝
=
2
2
π rcap
rcap
vcap
2
1.1 cm ) ( 40 cm/s )
(
=
=
2
−4
( 6 ×10 cm ) ( 0.07 cm/s )
2 × 109
55. SSM REASONING AND SOLUTION
a. The volume flow rate is given by Equation 11.10. Assuming that the line has a circular
cross section, A = π r 2 , we have
Q = Av = (π r 2 )v = π (0.0065 m) 2 (1.2 m/s) = 1.6 × 10 –4 m3 / s
b. The volume flow rate calculated in part (a) above is the flow rate for all twelve holes.
Therefore, the volume flow rate through one of the twelve holes is
Qhole =
Q
2
= (π rhole
)vhole
12
Solving for vhole we have
vhole =
Q
=
1.6 × 10 –4 m3 /s
−4
= 2.0 × 101 m/s
12π
12π (4.6 × 10 m)
______________________________________________________________________________
2
rhole
2
56. REASONING AND SOLUTION Using Bernoulli's equation
∆P = P1 − P2 = (1/2)ρv22 − (1/2)ρv12 = (1/2)(1.29 kg/m3)[(8.5 m/s)2 − (1.1 m/s)2]
∆P = 46 Pa
Chapter 11 Problems
585
The air flows from high pressure to low pressure (from lower to higher velocity), so it
enters at B and exits at A .
______________________________________________________________________________
57. SSM REASONING AND SOLUTION
a. Using Equation 11.12, the form of Bernoulli's equation with y 1 = y 2 , we have
(
)
P1 − P2 = ρ v22 − v12 =
1
2
1.29 kg/m3 ⎡
2
(15 m/s) 2 − ( 0 m/s ) ⎤⎥ = 150 Pa
⎢
⎣
⎦
2
b. The pressure inside the roof is greater than the pressure on the outside. Therefore, there
is a net outward force on the roof. If the wind speed is sufficiently high, some roofs are
"blown outward."
______________________________________________________________________________
58. REASONING We assume that region 1 contains the constriction and region 2 is the normal
region. The difference in blood pressures between the two points in the horizontal artery is
given by Bernoulli’s equation (Equation 11.12) as P2 − P1 = 12 ρ v12 − 12 ρ v22 , where v1 and
v2 are the speeds at the two points. Since the volume flow rate is the same at the two points,
the speed at 1 is related to the speed at 2 by Equation 11.9, the equation of continuity:
A1v1 = A2v2, where A1 and A2 are the cross-sectional areas of the artery. By combining these
two relations, we will be able to determine the pressure difference.
SOLUTION Solving the equation of continuity for the blood speed in region 1 gives
v1 = v2A2/A1. Substituting this result into Bernoulli’s equation yields
P2 − P1 =
Since A1 =
1
4
1 ρ v2
1
2
−
1 ρ v2
2
2
⎛v A ⎞
= 12 ρ ⎜ 2 2 ⎟
⎜ A ⎟
⎝ 1 ⎠
2
−
1
2
ρ v22
A2 , the pressure difference is
2
P2 − P1 =
1
2
=
1
2
⎛v A ⎞
ρ ⎜ 12 2 ⎟ −
⎜ A ⎟
⎝4 2⎠
1
2
ρ v22 =
1
2
ρ v22 (16 − 1)
(1060 kg/m3 ) ( 0.11 m/s )2 (15) = 96 Pa
We have taken the density ρ of blood from Table 11.1.
______________________________________________________________________________
586 FLUIDS
59. SSM REASONING AND SOLUTION Let the speed of the air below and above the
wing be given by v 1 and v 2 , respectively. According to Equation 11.12, the form of
Bernoulli's equation with y 1 = y 2 , we have
c
h
P1 − P2 = 21 ρ v 22 − v 12 =
1.29 kg / m 3
( 251 m / s) 2 − ( 225 m / s) 2 = 7.98 × 10 3 Pa
2
From Equation 11.3, the lifting force is, therefore,
c
h
F = P1 − P2 A = ( 7 .98 × 10 3 Pa)(24.0 m 2 ) = 1.92 × 10 5 N
______________________________________________________________________________
60. REASONING The pressure P, the fluid speed v, and the elevation y at any two points in an
ideal
fluid
of
density
ρ
are
related
by
Bernoulli’s
equation:
P1 + ρ v12 + ρ gy1 = P2 + ρ v22 + ρ gy2 (Equation 11.11), where 1 and 2 denote, respectively,
1
2
1
2
the first and second floors. With the given data and a density of ρ = 1.00 × 103 kg/m3 for
water (see Table 11.1), we can solve Bernoulli’s equation for the desired pressure P2.
SOLUTION
Solving Bernoulli’s equation for P2 and taking the elevation at the first floor to be y1 = 0 m ,
we have
(
)
P2 = P1 + ρ v12 − v22 + ρ g ( y1 − y2 )
1
2
= 3.4 × 105 Pa +
1
2
(1.00 ×103 kg/m3 ) ⎡⎣⎢( 2.1 m/s )2 − (3.7 m/s )2 ⎤⎦⎥
(
)(
)
+ 1.00 × 103 kg/m3 9.80 m/s 2 ( 0 m − 4.0 m ) = 3.0 × 105 Pa
61. REASONING
a. The drawing shows two 1
points, labeled 1 and 2, in
the fluid. Point 1 is at the
top of the water, and point 2
is where it flows out of the
dam
at
the
bottom. y
1
Bernoulli’s
equation,
Equation 11.11, can be used
to determine the speed v2 of
the water exiting the dam.
P1 = 1 atmosphere
v1 = 0 m/s
2 P2 = 1 atmosphere
v2
y2
Chapter 11 Problems
587
b. The number of cubic meters per second of water that leaves the dam is the volume flow
rate Q. According to Equation 11.10, the volume flow rate is the product of the crosssectional area A2 of the crack and the speed v2 of the water; Q = A2v2.
SOLUTION
a. According to Bernoulli’s equation, as given in Equation 11.11, we have
P1 + 12 ρ v12 + ρ gy1 = P2 + 12 ρ v22 + ρ gy2
Setting P1 = P2, v1 = 0 m/s, and solving for v2, we obtain
(
)
v2 = 2 g ( y1 − y2 ) = 2 9.80 m/s 2 (15.0 m ) = 17.1 m/s
b. The volume flow rate of the water leaving the dam is
(
)
Q = A2 v2 = 1.30 × 10−3 m 2 (17.1 m/s ) = 2.22 × 10−2 m3 /s
(11.10
)
______________________________________________________________________________
62. REASONING The number of gallons per minute of water that the fountain uses is the
volume flow rate Q of the water. According to Equation 11.10, the volume flow rate is
Q = Av , where A is the cross-sectional area of the pipe and v is the speed at which the water
leaves the pipe. The area is given. To find a value for the speed, we will use Bernoulli’s
equation. This equation states that the pressure P, the fluid speed v, and the elevation y at
any two points (1 and 2) in an ideal fluid of density ρ are related according to
P1 + ρ v12 + ρ gy1 = P2 + ρ v22 + ρ gy2 (Equation 11.11). Point 1 is where the water leaves
1
2
1
2
the pipe, so we are seeking v1. We must now select point 2. Note that the problem specifies
the maximum height to which the water rises. The maximum height occurs when the water
comes to a momentary halt at the top of its path in the air. Therefore, we choose this point
as point 2, so that we can take advantage of the fact that v2 = 0 m/s.
SOLUTION
We know that P1 = P2 = Patmospheric and v2 = 0 m/s.
Substituting this
information into Bernoulli’s equation, we obtain
Patmospheric + ρ v12 + ρ gy1 = Patmospheric + ρ ( 0 m/s ) + ρ gy2
1
2
1
2
2
Solving for v1 gives v1 = 2 g ( y2 − y1 ) , which we now substitute for the speed v in the
expression Q = Av for the volume flow rate:
588 FLUIDS
Q = Av1 = A 2 g ( y2 − y1 )
(
= 5.00 ×10−4 m 2
) 2 (9.80 m/s2 ) (5.00 m − 0 m ) = 4.95 ×10−3 m3 / s
To convert this result to gal / min, we use the facts that 1 gal = 3.79 × 10−3 m3 and
1 min = 60 s:
⎛
m3 ⎞ ⎛
1 gal
⎟⎜
Q = ⎜ 4.95 ×10−3
⎜
⎜
⎟
s ⎝ 3.79 × 10−3 m3
⎝
⎠
⎞ ⎛ 60 s ⎞
= 78.4 gal / min
⎟⎜
⎟ ⎝ 1 min ⎟⎠
⎠
63. SSM REASONING Since the pressure difference is known, Bernoulli's equation can be
used to find the speed v 2 of the gas in the pipe. Bernoulli's equation also contains the
unknown speed v 1 of the gas in the Venturi meter; therefore, we must first express v 1 in
terms of v 2 . This can be done by using Equation 11.9, the equation of continuity.
SOLUTION
a. From the equation of continuity (Equation 11.9) it follows that v 1 = A2 / A1 v 2 .
c
h
Therefore,
v1 =
0.0700 m 2
v = ( 1.40 ) v 2
0.0500 m 2 2
Substituting this expression into Bernoulli's equation (Equation 11.12), we have
P1 + 21 ρ ( 1.40 v 2 ) 2 = P2 + 21 ρ v 22
Solving for v 2 , we obtain
v2 =
2 ( P2 − P1 )
ρ ( 1.40 ) − 1
2
=
2( 120 Pa)
(1.30 kg / m 3 ) (1.40) 2 − 1
= 14 m / s
b. According to Equation 11.10, the volume flow rate is
Q = A2 v 2 = (0.0700 m 2 )( 14 m / s) = 0.98 m 3 / s
______________________________________________________________________________
64. REASONING AND SOLUTION We have
P1 + (1/2)ρv12 + ρgy1 = P2 + (1/2)ρv22 + ρgy2
Chapter 11 Problems
589
Taking v1 = 0 m/s, y2 = 0 m, P1 = 1.01 × 105 Pa, and noting that v2 is a maximum when
P2 = 0 Pa, we can solve for v2. We thus obtain,
v2 =
2P1
ρ
+ 2gy1 =
(
) + 2 9.80 m/s2 (12 m )
(
)
1.00 × 103 kg/m3
2 1.01 × 105 Pa
= 21 m/s
______________________________________________________________________________
65. REASONING The top and bottom surfaces of the roof are at the same height, so we can
use Bernoulli’s equation in the form of Equation 11.12, P1 + 12 ρ v12 = P2 + 12 ρ v22 , to
determine the wind speed. We take point 1 to be inside the roof and point 2 to be outside the
roof. Since the air inside the roof is not moving, v1 = 0 m/s. The net outward force ΣF acting
on the roof is the difference in pressure P1 − P2 times the area A of the roof, so
ΣF = (P1 − P2)A.
SOLUTION Setting v1 = 0 m/s in Bernoulli’s equation and solving it for the speed v2 of the
wind, we obtain
v2 =
2 ( P1 − P2 )
ρ
Since the pressure difference is equal to the net outward force divided by the area of the
roof, (P1 − P2) = ΣF /A, the speed of the wind is
2 ( ΣF )
2(22 000 N)
= 33 m/s
ρA
(1.29 kg/m3 )(5.0 m × 6.3 m)
______________________________________________________________________________
v2 =
=
66. REASONING AND SOLUTION As seen in the figure, the lower pipe is at the level of zero
potential energy.
v2
A2
h
A1
v1
590 FLUIDS
If the flow rate is uniform in both pipes, we have (1/2)ρv12 = (1/2)ρv22 + ρgh (since
P1 = P2) and A1v1 = A2v2. We can solve for v1, i.e., v1 = v2(A2/A1), and plug into the
previous expression to find v2, so that
v2 =
2gh
2
⎛ A2 ⎞
⎜ A ⎟ −1
⎝ 1⎠
=
(
)
2 9.80 m/s 2 (10.0 m )
2
⎡ π ( 0.0400 m )2 ⎤
⎢
⎥ −1
⎢ π ( 0.0200 m )2 ⎥
⎣
⎦
= 3.61 m/s
The volume flow rate is then given by
Q = A2v2 = π r22v2 = π (0.0400 m)2(3.61 m/s) = 1.81× 10 –2 m3 /s
______________________________________________________________________________
67. REASONING In level flight the lift force must balance the plane’s weight W, so its
magnitude is also W. The lift force arises because the pressure PB beneath the wings is
greater than the pressure PT on top of the wings. The lift force, then, is the pressure
difference times the effective wing surface area A, so that W = (PB – PT)A. The area is
given, and we can determine the pressure difference by using Bernoulli’s equation.
SOLUTION According to Bernoulli’s equation, we have
PB + 12 ρ vB2 + ρ gyB = PT + 12 ρ vT2 + ρ gyT
Since the flight is level, the height is constant and yB = yT, where we assume that the wing
thickness may be ignored. Then, Bernoulli’s equation simplifies and may be rearranged as
follows:
PB + 12 ρ vB2 = PT + 12 ρ vT2 or PB − PT = 12 ρ vT2 − 12 ρ vB2
Recognizing that W = (PB – PT)A, we can substitute for the pressure difference from
Bernoulli’s equation to show that
(
)
W = ( PB − PT ) A = 12 ρ vT2 − vB2 A
=
1
2
(1.29 kg/m3 ) ⎡⎣⎢( 62.0 m/s )2 − (54.0 m/s )2 ⎤⎦⎥ (16 m2 ) =
9600 N
We have used a value of 1.29 kg/m3 from Table 11.1 for the density of air. This is an
approximation, since the density of air decreases with increasing altitude above sea level.
______________________________________________________________________________
Chapter 11 Problems
591
68. REASONING AND SOLUTION
a. Since the volume flow rate, Q = Av, is the same at each point, and since v is greater at the
lower point, the upper hole must have the larger area .
b. Call the upper hole number 1 and the lower hole number 2 (the surface of the water is
position 0). Take the zero level of potential energy at the bottom hole and then write
Bernoulli's equation as
P1 + (1/2)ρv12 + ρgh = P2 + (1/2)ρv22 = P0 + ρg(2h)
in which P1 = P2 = P0 and from which we obtain
v1 = 2 gh
and v2 = 4 gh
or v2 / v1 = 2
Using the fact that Q1 = A1v1 = Q2 = A2v2 , we have v2/v1 = A1/A2. But since the ratio of the
areas is A1/A2 = r12/r22, we can write that
r1
=
r2
A1
v
= 2 =
A2
v1
4
2 = 1.19
______________________________________________________________________________
69. SSM REASONING AND SOLUTION Bernoulli's equation (Equation 11.12) is
P1 + 21 ρv 12 = P2 + 21 ρv 22
If we let the right hand side refer to the air above the plate, and the left hand side refer to the
air below the plate, then v 1 = 0 , since the air below the plate is stationary. We wish to find
v 2 for the situation illustrated in part b of the figure shown in the text. Solving the equation
above for v 2 ( with v 2 = v 2 b and v 1 = 0 ) gives
v2 b =
2 ( P1 − P2 )
ρ
In Equation (1), P1 is atmospheric pressure and P2 must be determined. We must first
consider the situation in part a of the text figure.
(1)
592 FLUIDS
The figure at the right shows the forces that act on the rectangular plate
in part a of the text drawing. F1 is the force exerted on the plate from
the air below the plate, and F2 is the force exerted on the plate from the
air above the plate. Applying Newton's second law, we have (taking
"up" to be the positive direction),
F
1
mg
F1 − F2 − mg = 0
F
2
F1 − F2 = mg
Thus, the difference in pressures exerted by the air on the plate in part a of the drawing is
P1 − P2 =
F1 − F2
A
=
mg
A
(2)
where A is the area of the plate. From Bernoulli's equation ( Equation 11.12) we have, with
v2 = v2a and v1a = 0 ,
P1 − P2 = 21 ρ v 22 a
(3)
where v 2 a is the speed of the air along the top of the plate in part a of the text drawing.
Combining Equations (2) and (3) we have
mg 1 2
= ρv
A 2 2a
(4)
The figure below, on the left, shows the forces that act on the plate in part b of the text
drawing. The notation is the same as that used when the plate was horizontal (part a of the
text figure). The figure at the right below shows the same forces resolved into components
along the plate and perpendicular to the plate.
F
1
F
1
mg cosθ
F
F
2
2
θ
mg
mg sinθ
Applying Newton's second law we have F1 − F2 − mg sin θ = 0 , or
F1 − F2 = mg sin θ
Thus, the difference in pressures exerted by the air on the plate in part b of the text figure is
Chapter 11 Problems
P1 − P2 =
F1 − F2
A
=
593
mg sin θ
A
Using Equation (4) above,
P1 − P2 = 21 ρv 22 a sin θ
Thus, Equation (1) becomes
v2 b =
2
c
1
2
ρv 22 a sin θ
h
ρ
Therefore,
v 2 b = v 22 a sin θ = (11.0 m / s) 2 sin 30.0 ° =
7.78 m / s
______________________________________________________________________________
70. REASONING The volume flow rate Q is governed by Poiseuille’s law: Q =
π R 4 ( P2 − P1 )
8η L
(Equation 11.14). We will obtain the factor Rdilated/Rnormal by applying this law to the
dilated and to the normal blood vessel.
SOLUTION Solving Poiseuille’s law for the radius R gives R = 4
8η LQ
. When the
π ( P2 − P1 )
vessel dilates, the viscosity η, the length L of the vessel, and the pressures P1 and P2 at the
ends of the vessel do not change. Thus, applying this result to the dilated and normal vessel,
we find that
8η LQdilated
4
π ( P2 − P1 )
Rdilated
Q
=
= 4 dilated = 4 2 = 1.19
Rnormal
Qnormal
8η LQnormal
4
where we have used
Qdilated
Qnormal
π ( P2 − P1 )
= 2 , since the effect of the dilation is to double the volume
flow rate.
71. SSM WWW REASONING AND SOLUTION
a. If the water behaves as an ideal fluid, and since the pipe is horizontal and has the same
radius throughout, the speed and pressure of the water are the same at all points in the pipe.
594 FLUIDS
Since the right end of the pipe is open to the atmosphere, the pressure at the right end is
atmospheric pressure; therefore, the pressure at the left end is also atmospheric pressure, or
1.01 × 10 5 Pa .
b. If the water is treated as a viscous fluid, the volume flow rate Q is described by
Poiseuille's law (Equation 11.14):
Q=
π R 4 ( P2 − P1 )
8η L
Let P1 represent the pressure at the right end of the pipe, and let P2 represent the pressure
at the left end of the pipe. Solving for P2 (with P1 equal to atmospheric pressure), we
obtain
P2 =
8η LQ
+ P1
π R4
Therefore,
8(1.00 × 10 −3 Pa ⋅ s)(1.3 m)(9.0 × 10 −3 m 3 / s )
+ 1.013 × 10 5 Pa = 1.19 × 10 5 Pa
π ( 6.4 × 10 −3 m) 4
______________________________________________________________________________
P2 =
72. REASONING In this problem, we are treating air as a viscous fluid. According to
Poiseuille's law, a fluid with viscosity η flowing through a pipe of radius R and length L
has a volume flow rate Q given by Equation 11.14: Q = πR 4 ( P2 – P1 ) / ( 8ηL ) . This
expression can be solved for the quantity P2 – P1 , the difference in pressure between the
ends of the air duct. First, however, we must determine the volume flow rate Q of the air.
SOLUTION
Since the fan forces air through the duct such that 280 m 3 of air is
replenished every ten minutes, the volume flow rate is
Q=
FG 280 m IJ FG 1.0 min IJ = 0.467 m
H 10.0 min K H 60 s K
3
3
/s
The difference in pressure between the ends of the air duct is, according to Poiseuille's law,
8η LQ 8(1.8 × 10 –5 Pa ⋅ s)(5.5 m)(0.467 m 3 / s)
=
= 4.4 Pa
π R4
π (7.2 × 10 –2 m) 4
______________________________________________________________________________
P2 – P1 =
Chapter 11 Problems
595
73. REASONING The volume flow rate Q of a viscous fluid flowing through a pipe of radius R
π R 4 ( P2 − P1 )
is given by Equation 11.14 as Q =
, where P2 − P1 is the pressure difference
8η L
between the ends of the pipe, L is the length of the pipe, and η is the viscosity of the fluid.
Since all the variables are known except L, we can use this relation to find it.
SOLUTION Solving Equation 11.14 for the pipe length, we have
(
)(
)(
)
4
π 5.1× 10−3 m 1.8 ×103 Pa
π R 4 ( P2 − P1 )
=
= 1.7 m
L=
8η Q
8 1.0 × 10−3 Pa ⋅ s 2.8 × 10−4 m3 /s
(
)
______________________________________________________________________________
74. REASONING This is an application of Poiseuille’s law, as given in Equation 11.14.
According to this law, a pressure difference P2 – P1 is needed to make the viscous blood
flow through the needle.
SOLUTION The pressure at the input of the needle is P2 = PAtmosphere + ρgh, as given by
Equation 11.4, whereas the pressure at the output of the needle is P1 = PAtmosphere.
Therefore, we have
P2 − P1 = PAtmosphere + ρgh − PAtmosphere = ρgh
Using Poiseuille’s law, we find
Q=
c
π R 4 P2 − P1
8ηL
h = π R ρgh
4
8 ηL
Solving for h gives
c
hc
hc
hb
g
h
4 .5 × 10 −8 m 3 / s 8 4.0 × 10 −3 Pa ⋅ s 0.030 m
Q 8 ηL
h=
=
= 0.34 m
π R 4 ρg π 2 .5 × 10 −4 m 4 1060 kg / m 3 9 .80 m / s 2
c
hc
______________________________________________________________________________
75. SSM REASONING AND SOLUTION
a. Using Stoke's law, the viscous force is
F = 6πη Rv = 6π (1.00 × 10 –3 Pa ⋅ s)( 5.0 × 10 –4 m)( 3.0 m / s) = 2.8 × 10 –5 N
596 FLUIDS
b. When the sphere reaches its terminal speed, the net force on the sphere is zero, so that
the magnitude of F must be equal to the magnitude of mg, or F = mg . Therefore,
6πη Rv T = mg , where v T is the terminal speed of the sphere. Solving for v T , we have
(1.0 × 10 –5 kg) (9.80 m / s 2 )
= 1.0 × 10 1 m / s
6πη R 6π ( 1.00 × 10 –3 Pa ⋅ s) (5.0 × 10 –4 m)
______________________________________________________________________________
vT =
mg
=
76. REASONING AND SOLUTION Since water is a viscous fluid, its behavior is described by
Poiseuille's law (Equation 11.14):
πR 4 ( P2 − P1 )
Q=
8 ηL
Solving for R gives:
R
L 8ηLQ OP
=M
Nπ ( P − P ) Q
2
1/ 4
1
In this expression P2 is the pressure just below the left vertical tube, P1 is the pressure just
below the right vertical tube, and L is the horizontal distance between the centers of the
vertical tubes. The difference in pressures is given by
P2 – P1 = ρg(y2 – y1) = ρg∆y
Therefore,
1/4
⎡ 8η LQ ⎤
R=⎢
⎥
⎣ π (P2 − P1 ) ⎦
1/4
⎡ 8(1.00 × 10 –3 Pa ⋅ s)(0.70 m)(0.014 m3 /s) ⎤
–2
=⎢
⎥ = 1.5 × 10 m
3
3
2
⎣ π (1.000 × 10 kg/m )(9.80 m/s )(0.045 m) ⎦
______________________________________________________________________________
77. REASONING AND SOLUTION The pump must generate an upward force to counteract
the weight of the column of water above it. Therefore, F = mg = (ρhA)g. The required
pressure is then
P = F/A = ρgh = (1.00 × 103 kg/m3)(9.80 m/s2)(71 m) = 7.0 × 105 Pa
______________________________________________________________________________
78. REASONING We apply Bernoulli’s equation as follows:
Chapter 11 Problems
597
PS + 12 ρ vS2 + ρ gyS = PO + 12 ρ vO2 + ρ gyO
At surface of
vaccine in reservoir
At opening
SOLUTION The vaccine’s surface in the reservoir is stationary during the inoculation, so
that vS = 0 m/s. The vertical height between the vaccine’s surface in reservoir and the
opening can be ignored, so yS = yO. With these simplifications Bernoulli’s equation
becomes
PS = PO + 12 ρ vO2
Solving for the speed at the opening gives
vO =
2 ( PS − PO )
=
(
2 4.1× 106 Pa
)=
86 m/s
1100 kg/m3
______________________________________________________________________________
ρ
79. SSM REASONING The buoyant force exerted on the balloon by the air must be equal in
magnitude to the weight of the balloon and its contents (load and hydrogen). The magnitude
of the buoyant force is given by ρairV g . Therefore,
ρairV g = Wload + ρ hydrogenV g
where, since the balloon is spherical, V = (4 / 3)π r 3 . Making this substitution for V and
solving for r, we obtain
1/ 3
⎡
⎤
3Wload
r=⎢
⎥
⎢⎣ 4π g(ρair − ρ hydrogen ) ⎥⎦
SOLUTION Direct substitution of the data given in the problem yields
1/ 3
⎡
⎤
3(5750 N)
r=⎢
= 4.89 m
2
3
3 ⎥
−
π
4
(9.80
m/s
)
(1.29
kg/m
0.0899
kg/m
)
⎣
⎦
______________________________________________________________________________
80. REASONING AND SOLUTION The gauge pressure of the solution at the location of the
vein is
P = ρgh = (1030 kg/m3)(9.80 m/s2)(0.610 m) = 6.16 × 103 Pa
Now
1.013 × 105 Pa = 760 mm Hg
Then
so
1 Pa = 7.50 × 10–3 mm Hg
598 FLUIDS
⎛ 7.50 × 10−3 mm Hg ⎞
P = 6.16 × 103 Pa ⎜⎜
⎟⎟ = 46.2 mm Hg
1
Pa
⎝
⎠
______________________________________________________________________________
(
)
81. REASONING The weight W of the water bed is equal to the mass m of water times the
acceleration g due to gravity; W = mg (Equation 4.5). The mass, on the other hand, is equal
to the density ρ of the water times its volume V, or m = ρ V (Equation 11.1).
SOLUTION Substituting m = ρ V into the relation W = mg gives
W = mg = ( ρ V ) g
(
)
(
)
= 1.00 × 103 kg/m3 (1.83 m × 2.13 m × 0.229 m ) 9.80 m/s 2 = 8750 N
We have taken the density of water from Table 11.1. Since the weight of the water bed is
greater than the additional weight that the floor can tolerate, the bed should not be
purchased.
______________________________________________________________________________
82. REASONING AND SOLUTION
a. We will treat the neutron star as spherical in shape, so that its volume is given by the
familiar formula, V = 43 π r 3 . Then, according to Equation 11.1, the density of the neutron
star described in the problem statement is
m
m
3m
3(2.7 × 1028 kg)
ρ= =
=
=
= 3.7 × 1018 kg/m3
3
3
3
3
4
V 3πr
4π r
4π (1.2 × 10 m)
b. If a dime of volume 2.0 × 10 –7 m3 were made of this material, it would weigh
W = mg = ρVg = (3.7 × 1018 kg/m3 )(2.0 × 10 –7 m3 )(9.80 m/s 2 )= 7.3 × 1012 N
This weight corresponds to
⎛ 1 lb ⎞
12
7.3 × 1012 N ⎜
⎟ = 1.6 × 10 lb
⎝ 4.448 N ⎠
______________________________________________________________________________
83. SSM REASONING AND SOLUTION Using Equation 11.10 for the volume flow rate,
we have Q = Av , where A is the cross-sectional area of the pipe and v is the speed of the
fluid. Since the pipe has a circular cross section, A = π r 2 . Therefore, solving Equation
11.10 for v, we have
Chapter 11 Problems
599
Q
Q
1.50 m3 /s
=
=
= 1.91 m/s
A π r 2 π (0.500 m) 2
______________________________________________________________________________
v=
The Reynold's number, Re, can be written as
84. REASONING AND SOLUTION
Re = 2v ρ R / η . To find the average speed v ,
(
)
( 2000 ) 4.0 × 10−3 Pa ⋅ s
(Re)η
=
= 0.5 m/s
v=
2 ρ R 2 1060 kg/m3 8.0 × 10−3 m
(
)(
)
______________________________________________________________________________
85. SSM REASONING AND SOLUTION
a. The pressure at the level of house A is given by Equation 11.4 as P = Patm + ρ gh . Now
the height h consists of the 15.0 m and the diameter d of the tank. We first calculate the
radius of the tank, from which we can infer d. Since the tank is spherical, its full mass is
given by M = ρV = ρ [(4 / 3)π r 3 ] . Therefore,
3M
r =
4π ρ
3
1/ 3
or
⎛ 3M ⎞
r =⎜
⎟
⎝ 4π ρ ⎠
1/3
⎡ 3(5.25 × 105 kg) ⎤
= ⎢
3
3 ⎥
⎣ 4π (1.000 × 10 kg/m ) ⎦
= 5.00 m
Therefore, the diameter of the tank is 10.0 m, and the height h is given by
h = 10.0 m + 15.0 m = 25.0 m
According to Equation 11.4, the gauge pressure in house A is, therefore,
P − Patm = ρ gh = (1.000 × 103 kg/m3 )(9.80 m/s 2 )(25.0 m) = 2.45 × 105 Pa
b. The pressure at house B is P = Patm + ρ gh , where
h = 15.0 m + 10.0 m − 7.30 m = 17.7 m
According to Equation 11.4, the gauge pressure in house B is
P − Patm = ρ gh = (1.000 × 103 kg/m3 )(9.80 m/s 2 )(17.7 m) = 1.73 × 105 Pa
______________________________________________________________________________
86. REASONING The density ρ of an object is equal to its mass m divided by its volume V, or
ρ = m/V (Equation 11.1). The volume of a sphere is V = 43 π r 3 , where r is the radius.
According to the discussion of Archimedes’ principle in Section 11.6, any object that is
600 FLUIDS
solid throughout will float in a liquid if the density of the object is less than or equal to the
density of the liquid. If not, the object will sink.
SOLUTION
a. The average density of the sun is
ρ=
m
m
1.99 × 1030 kg
=
=
V 43 π r 3 4 π 6.96 ×108 m
3
(
)
3
= 1.41× 103 kg/m3
b. Since the average density of the solid object (1.41 × 103 kg/m3) is greater than that of
water (1.00 × 103 kg/m3, see Table 11.1), the object will sink .
c. The average density of Saturn is
m
m
5.7 × 1026 kg
ρ= =
=
V 43 π r 3 4 π 6.0 × 107 m
3
(
)
3
= 0.63 × 103 kg/m3
The average density of this solid object (0.63 × 103 kg/m3) is less than that of water (1.00 ×
103 kg/m3), so the object will float .
______________________________________________________________________________
87. REASONING AND SOLUTION Let rh represent the inside radius of the hose, and rp the
radius of the plug, as suggested by the figure below.
hose
plug
rh
rp
Then, from Equation 11.9, A1v1 = A2v2 , we have π rh2 v1 = (π rh2 − π rp2 )v2 , or
v1
v2
2
rh2 − rp2 )
⎛ rp ⎞
(
=
= 1− ⎜ ⎟
⎜r ⎟
⎝ h⎠
rh2
or
rp
rh
= 1−
v1
v2
According to the problem statement, v2 = 3v1, or
rp
rh
= 1−
v1
2
=
= 0.816
3v1
3
______________________________________________________________________________
Chapter 11 Problems
601
88. REASONING AND SOLUTION
a. The force exerted on the piston (surface area A = π r2 = 1.96 × 10−3 m2) is
F = PA = (1.013 × 105 N/m2)(1.96 × 10−3 m2) = 199 N
The spring is therefore compressed by an amount x, where
x = F/k = (199 N)/(3600 N/m) = 0.055 m
b. The work done by the spring is
W = (1/2) kx2 = (1/2)(3600 N/m)(0.055 m)2 = 5.4 J
______________________________________________________________________________
89. SSM REASONING AND SOLUTION The weight of the coin in air is equal to the sum
of the weights of the silver and copper that make up the coin:
e
j e
j
Wair = mcoin g = msilver + mcopper g = ρ silver Vsilver + ρ copper Vcopper g
(1)
The weight of the coin in water is equal to its weight in air minus the buoyant force exerted
on the coin by the water. Therefore,
Wwater = Wair − ρ water (Vsilver + Vcopper ) g
(2)
Solving Equation (2) for the sum of the volumes gives
Vsilver + Vcopper =
Wair − Wwater
ρ water g
=
mcoin g − Wwater
ρ water g
or
Vsilver + Vcopper =
(1.150 × 10 –2 kg) (9.80 m / s 2 ) − 0.1011 N
= 1.184 × 10 –6 m 3
3
2
3
(1.000 × 10 kg / m ) (9.80 m / s )
From Equation (1) we have
Vsilver =
or
mcoin − ρ copper Vcopper
ρ silver
=
1.150 × 10 –2 kg − (8890 kg / m 3 )Vcopper
10 500 kg / m 3
602 FLUIDS
Vsilver = 1.095 × 10 –6 m 3 − ( 0.8467 )Vcopper
(3)
Substitution of Equation (3) into Equation (2) gives
Wwater = Wair − ρ water 1.095 × 10 –6 m 3 − ( 0.8467 )V copper + Vcopper g
Solving for Vcopper gives Vcopper = 5.806 × 10 –7 m 3 . Substituting into Equation (3) gives
Vsilver = 1.095 × 10 –6 m 3 − ( 0.8467 )( 5.806 × 10 –7 m 3 ) = 6.034 × 10 –7 m 3
From Equation 11.1, the mass of the silver is
msilver = ρsilverVsilver = (10 500 kg/m3 )(6.034 × 10−7 m3 ) = 6.3 × 10−3 kg
______________________________________________________________________________
90. REASONING In each case the pressure at point A in Figure 11.12 is atmospheric pressure
and the pressure in the tube above the top of the liquid column is P. In Equation 11.4 (P2 =
P1 + ρgh), this means that P2 = Patmosphere and P1 = P. With this identification of the
pressures and a value of 13 600 kg/m3 for the density of mercury (see Table 11.1),
Equation 11.4 provides a solution to the problem.
SOLUTION Rearranging Equation 11.4, we have P2 – P1 = ρgh. Applying this expression
to each setup gives
P2 − P1 Mercury = PAtmosphere − P = ρgh Mercury
c
h
cP − P h
2
1
Unknown
= PUnknown
b g
− P = b ρgh g
Unknown
Since the left side of each of these equations is the same, we have
ρ Mercury ghMercury = ρ Unknown ghUnknown
ρ Unknown = ρ Mercury
Fh
GH h
Mercury
Unknown
I = c13 600 kg / m hF h
GH 16h
JK
3
Mercury
Mercury
I=
JK
850 kg / m 3
______________________________________________________________________________
91. SSM WWW REASONING According to Equation 11.4, the pressure Pmercury at a point
7.10 cm below the ethyl alcohol-mercury interface is
Chapter 11 Problems
Pmercury = Pinterface + ρ mercury ghmercury
603
(1)
where Pinterface is the pressure at the alcohol-mercury interface, and hmercury = 0.0710 m .
The pressure at the interface is
Pinterface = Patm + ρ ethyl ghethyl
(2)
Equation (2) can be used to find the pressure at the interface. This value can then be used in
Equation (1) to determine the pressure 7.10 cm below the interface.
SOLUTION Direct substitution of the numerical data into Equation (2) yields
Pinterface = 1.01 × 10 5 Pa + (806 kg / m 3 )( 9.80 m / s 2 )( 1.10 m) = 1.10 × 10 5 Pa
Therefore, the pressure 7.10 cm below the ethyl alcohol-mercury interface is
Pmercury = 1.10 × 10 5 Pa + (13 600 kg / m 3 )( 9.80 m / s 2 )( 0.0710 m) = 1.19 × 10 5 Pa
______________________________________________________________________________
92. REASONING AND SOLUTION If the concrete were solid, it would have a mass of
M = ρV = (2200 kg/m3)(0.025 m3) = 55 kg
The mass of concrete removed to make the hole is then 55 kg − 33 kg = 22 kg. This
corresponds to a volume V = (22 kg)/(2200 kg/m3) = 0.010 m3. Since the hole is spherical
V = (4/3)π r3 so
(
3
)
3 0.010 m
3V
3
=
= 0.13 m
r=
4π
4π
______________________________________________________________________________
3
93. REASONING AND SOLUTION
a. Taking the nozzle as position 1 and the top of the tank as position 2 we have, using
Bernoulli's equation, P1 + (1/2)ρv12 = P2 + ρgh, we can solve for v1 so that
v12 = 2[(P2 − P1)/ρ + gh]
v1 = 2[(5.00 × 105 Pa)/(1.00 × 103 kg/m3 ) + (9.80 m/s 2 )(4.00 m)] = 32.8 m/s
b. To find the height of the water use (1/2)ρv12 = ρgh so that
604 FLUIDS
h = (1/2)v12/g = (1/2)(32.8 m/s)2/(9.80 m/s2) = 54.9 m
______________________________________________________________________________
94. REASONING AND SOLUTION Only the weight of the block compresses the spring.
Applying Hooke's law gives W = kx. The spring is stretched by the buoyant force acting on
the block minus the weight of the block. Hooke's law again gives FB – W = 2kx.
Eliminating kx gives FB = 3W. Now FB = ρwgV, so that the volume of the block is
V = 3M/ρw = 3(8.00 kg)/(1.00 × 103 kg/m3) = 2.40 × 10–2 m3
The volume of wood in the block is
Vw = M/ρb = (8.00 kg)/(840 kg/m3) = 9.52 × 10–3 m3
The volume of the block that is hollow is V – Vw = 1.45 × 10−2 m3. The percentage of the
block that is hollow is then
100(1.45 × 10–2)/(2.40 × 10–2) = 60.3 % .
______________________________________________________________________________
95. REASONING AND SOLUTION The force exerted by a pressure is F = PA and is
perpendicular to the surface. The force on the outside of each half of the roof is
⎛ 133 N/m 2 ⎞
4
F = (10.0 mm Hg) ⎜⎜
⎟⎟ (14.5 m × 4.21 m ) = 8.11 × 10 N
⎝ 1 mm Hg ⎠
Adding the forces vectorially gives for the vertical force
Fv = 2F cos 30.0° = 1.41 × 105 N
and for the horizontal force Fh = 0 N. The net force is then 1.41× 105 N .
The direction of the force is downward , since the horizontal components of the forces on
the halves cancel.
______________________________________________________________________________
96. REASONING AND SOLUTION
a. At the surface of the water (position 1) and at the exit of the hose (position 2) the
pressures are equal (P1 = P2) to the atmospheric pressure. If the hose exit defines y = 0 m,
we have from Bernoulli's equation (1/2)ρv12 + ρgy = (1/2)ρv22. If we take the speed at the
surface of the water to be zero (i.e., v1 = 0 m/s) we find that
Chapter 11 Problems
v2 =
605
2gy
b. The siphon will stop working when v2 = 0 m/s, or y = 0 m , i.e., when the end of the
hose is at the water level in the tank.
c. At point A we have
PA + ρg(h + y) + (1/2)ρvA2 = P0 + (1/2)ρv22
But vA = v2 so that
PA = P0 – ρ g ( y + h)
______________________________________________________________________________
97. CONCEPT QUESTIONS
a. The total mass is the sum of the mass of the gold and the mass of the quartz:
mT = mG + mQ.
b. The total volume is the sum of the volume of the gold and volume of the quartz:
VT = VG + VQ.
c. For any pure substance, the volume is related to the mass and the density according to the
definition of density given in Equation 11.1 (ρ = m/V). Thus, the volume is V = m/ρ.
SOLUTION The total mass of the rock is
mT = mG + mQ
(1)
VT = VG + VQ
(2)
The total volume of the rock is
Using Equation 11.1 to express the individual volumes in terms of the individual masses and
densities, we have from Equation (2) that
mQ
m
(3)
VT = G +
ρG
ρQ
Solving Equation (1) for mQ and substituting into Equation (3) gives
VT =
mG
ρG
+
Solving this result for mG, we find
mT − mG
ρQ
⎛ 1
1 ⎞ mT
⎟+
= mG ⎜
−
⎜ ρG ρQ ⎟ ρQ
⎝
⎠
606 FLUIDS
VT −
mG =
mT
ρQ
⎛ 1
1 ⎞
−
⎜
⎟
⎜ ρG ρQ ⎟
⎝
⎠
=
12.0 kg
( 4.00 ×10−3 m3 ) − 2660
kg/m3
1
1
−
19 300 kg/m3 2660 kg/m3
= 1.6 kg
The densities for gold and quartz have been taken from Table 11.1.
______________________________________________________________________________
98. CONCEPT QUESTIONS
a. The pressure outside the tube at the surface of the basting sauce is the atmospheric
pressure of 1.013 × 105 Pa. The pressure in the bulb must be less than this value. If it were
greater than atmospheric pressure, there would be a greater force on the top of the liquid in
the tube than at the bottom, and the liquid could not have risen in the tube.
b. The atmospheric pressure outside the tube pushes the sauce up the tube, to the extent that
the smaller pressure in the bulb allows it. The smaller the pressure in the bulb, the higher
the sauce will rise. Conversely, the greater the pressure in the bulb, the less the sauce will
rise. Thus, in the second trial the smaller value for h means that the pressure in the bulb is
greater than in the first trial shown in the drawing.
SOLUTION According to Equation 11.4, we have
PAtmospheric = PBulb + ρ gh
a. Solving for the pressure in the bulb for the first trial shown in the drawing, we find
PBulb = PAtmospheric − ρ gh
(
)(
)
= 1.013 ×105 Pa − 1200 kg/m3 9.80 m/s 2 ( 0.15 m ) = 9.95 × 104 Pa
b. Solving for the pressure in the bulb for the second trial, we find
PBulb = PAtmospheric − ρ gh
(
)(
)
= 1.013 ×105 Pa − 1200 kg/m3 9.80 m/s 2 ( 0.10 m ) = 1.001× 105 Pa
______________________________________________________________________________
99. CONCEPT QUESTIONS
a. When the tube is floating, it is in equilibrium and there can be no net force acting on it.
Therefore, the upward-directed buoyant force must have a magnitude that equals the
magnitude of the tube’s weight, which acts downward. This is true in either battery acid or
antifreeze.
Chapter 11 Problems
607
b. According to Archimedes’ principle, the buoyant force has a magnitude that equals the
weight of the displaced fluid. Since the antifreeze has the smaller density (smaller mass per
unit volume), a greater volume of it is required to yield the displaced mass of fluid that has
the necessary weight.
c. The farther up from the bottom of the tube the mark is, the greater is the volume of
displaced fluid to which it corresponds. Since a greater volume of antifreeze must be
displaced to float the tube, the antifreeze mark should be placed higher up the tube.
SOLUTION Since the magnitude of the buoyant force FB equals the weight W of the tube,
we have FB = W. According to Archimedes’ principle, the magnitude of the buoyant force
equals the weight of the displaced fluid, which is the mass m of the displaced fluid times the
acceleration due to gravity, or mg. But according to the definition of the density ρ, the mass
is the density of the fluid times the displaced volume V, or m = ρV. The result is that the
weight of the displaced fluid is ρVg. Therefore, FB = W becomes
ρVg = W
The volume V of the displaced fluid equals the cross-sectional area A of the tube times the
height h beneath the fluid surface, or V = Ah. With this substitution, our previous result
becomes
ρAhg = W
a. Solving for hacid, we find that
hacid =
W
ρacid Ag
=
5.88 × 10−2 N
(1280 kg/m )( 7.85 ×10
3
−5
m
2
)(9.80 m/s )
2
= 5.97 × 10−2 m
b. Solving for hantifreeze, we find that
hantifreeze =
=
W
ρantifreeze Ag
5.88 × 10−2 N
(1073 kg/m )( 7.85 ×10
3
−5
m
2
)(9.80 m/s )
2
= 7.12 × 10−2 m
______________________________________________________________________________
100. Concept Questions
a. Since the effects of air resistance and viscosity are being ignored, the water can be
treated as a freely-falling object, as Chapter 2 discusses. It accelerates with the acceleration
608 FLUIDS
due to gravity. Therefore, it has a greater speed at the lower point than it did upon leaving
the faucet.
b. The volume flow rate in cubic meters per second is the same as it was when the water
left the faucet. This is because no water is added to or taken out of the stream after the
water leaves the faucet.
c. The cross-sectional area of the water stream is less than it was when the water left the
faucet. With the volume flow rate unchanging, the equation of continuity applies in the
form of Equation 11.9. The volume flow rate is the cross-sectional area of the stream times
the speed of the water. When the speed increases, as it does when the water falls, the crosssectional area decreases.
SOLUTION Using the equation of continuity as stated in Equation 11.9, we have
A1v1
N
Below faucet
= A2v2
N
At faucet
or
A1 =
A2v2
v1
To find the cross-sectional area A1, we must find the speed v1. To do this, we use Equation
2.9 from the equations of kinematics:
v12 = v22 + 2ay
or
v1 = v22 + 2ay
In using this result, we choose downward as the positive direction. Substituting into the
equation of continuity gives
A1 =
A2v2
v22
+ 2ay
=
(1.8 ×10−4 m2 ) ( 0.85 m/s )
=
2
2
0.85
m/s
2
9.80
m/s
0.10
m
+
)
(
)
(
)(
9.3 ×10−5 m 2
______________________________________________________________________________
101. Concept Questions
a. The fact that the pipe is horizontal means that the elevation of the water in the pipe and
the water exiting the nozzle are the same.
b. The speed at which the water flows in the pipe is smaller than the speed of the water
emerging from the nozzle. We can resort to the equation of continuity, which relates the
cross-sectional areas A and the speeds v at two different points in the stream according to
Equation 11.9 (A1v1 = A2v2). With its larger radius, the pipe has a greater cross-sectional
area than that of the nozzle. According to Equation 11.9, the speed is smaller where the
area is greater.
Chapter 11 Problems
609
c. Knowing that the flow speed in the pipe is smaller than the flow speed in the nozzle, we
can conclude that the absolute pressure of the water in the pipe is greater than the
atmospheric pressure at the nozzle opening. This follows from Bernoulli’s equation, which
indicates for a horizontal pipe that the pressure is greater when the flow speed is smaller.
SOLUTION According to Bernoulli’s equation as given in Equation 11.11, we have
P1 + 12 ρ v12 + ρ gy1 = P2 + 12 ρ v22 + ρ gy2
Pipe
Nozzle
The pipe and nozzle are horizontal, so that y1 = y2 and Bernoulli’s equation simplifies to
P1 + 12 ρ v12 = P2 + 12 ρ v22
We have values for the pressure P2 at the nozzle opening and the speed v1 in the pipe.
However, to solve this expression for the pressure P1 in the pipe, we also need a value for
the speed v2 at the nozzle opening. We obtain this value using the equation of continuity, as
given in Equation 11.9:
A1v1 = A2v2
or
π r12v1 = π r22v2
or
v2 =
r12v1
r22
Here, we have also used that fact that the area of a circle is π r2. Substituting this result for
v2 into Bernoulli’s equation, we find that
⎛ r12 v1 ⎞
2
1
1
P1 + 2 ρ v1 = P2 + 2 ρ ⎜ 2 ⎟
⎜ r ⎟
⎝ 2 ⎠
2
⎛ r4
⎞
P1 = P2 + 12 ρ ⎜ 14 − 1⎟ v12
⎜r
⎟
⎝ 2
⎠
(
= 1.01 × 105 Pa + 12 (1.00 × 103 kg/m3 )
(
⎡
−2
⎢ 1.9 × 10 m
⎢
−3
⎢⎣ 4.8 × 10 m
) − 1⎤⎥ ( 0.62 m/s )2 =
⎥
4
) ⎥⎦
4
1.48 × 105 Pa
The density of water is taken from Table 11.1 in the text.
______________________________________________________________________________
102. CONCEPT QUESTIONS
a. The amount of water per second leaking into the hold is the volume flow rate Q.
According to Equation 11.10, the volume flow rate is the product of the area A of the hole
and the speed v1 of the water entering the hold, Q = Av1.
610 FLUIDS
b. The speed of the water at the surface of the lake is approximately 0 m/s. Since the amount
of water in the lake is large, the water level at the surface drops very, very slowly as water
enters the hold of the ship. Thus, to a very good approximation, the speed of the water at the
surface is zero.
c. As water moves downward from the surface to the hole, it is accelerated by the
gravitational force, just like a ball that is dropped from rest at the top of a building.
SOLUTION
If we can find a value for the speed v1 of the water entering the hold, the volume flow rate Q
can be obtained from Equation 11.10 (Q = Av1), since the area A of the hole is known. The
place where the water enters the hold is just like point 1 in Figure 11.36a, whereas the lake
surface, 2.0 m above, is just like point 2 in the figure. Bernoulli’s equation, Equation 11.11
relates the pressure P, water speed v, and elevation y of these two points:
P1 + 12 ρ v12 + ρ gy1 = P2 + 12 ρ v22 + ρ gy2
Setting P1 = P2 (since the hold is open to the atmosphere), v2 = 0 m/s, and solving for v1, we
obtain v1 = 2 g ( y2 − y1 ) . The volume flow rate of the water entering the hold is
(
Q = Av1 = A 2 g ( y2 − y1 ) = 8.0 × 10−3 m 2
) 2 (9.80 m/s2 ) ( 2.0 m ) = 5.0 ×10−2 m3/s
______________________________________________________________________________
103. CONCEPT QUESTIONS
a. The weight Whot air of the hot air is equal to its mass mhot air times the magnitude g of the
acceleration due to gravity, or Whot air= mhot airg. According to Equation 11.1, the mass is
equal to the product of the density ρhot air and the volume V that the mass occupies (i.e., the
volume of the balloon), so mhot air= ρhot airV. Combining these two relations, the weight of
the hot air can be expressed as Whot air = ρhot airV g.
b. According to Archimedes’ principle, the magnitude of the buoyant force is equal to the
weight of the fluid displaced by the balloon. In this case the fluid is the surrounding cool air.
Thus, the buoyant force depends on the density of the cool air outside the balloon.
c. The free-body diagram shows the two forces acting on the
balloon, its weight W and the buoyant force FB. Newton’s
FB
second law, Equation 4.2b, states that the net force ΣFy in the
y direction is equal to the mass m of the balloon times its
acceleration ay in that direction:
W
Free-body diagram
for the balloon
+y
Chapter 11 Problems
611
FB − W = ma y
ΣFy
SOLUTION Solving Newton’s second law for the acceleration of the balloon gives
F −W
. Since we are neglecting the weight of the balloon fabric and the basket, the
ay = B
m
only weight is that of the hot air inside the balloon. Thus, m = mhot air = ρhot airV and
W = mhot airg = (ρhot airV)g . The magnitude FB of the buoyant force is equal to the weight of
the cool air that the balloon displaces, so FB = mcool airg = (ρcool airV)g. Substituting these
expressions for m, W, and FB into Newton’s second law gives
ay =
FB − W
m
=
ρcool airVg − ρ hot airVg ( ρcool air − ρ hot air ) g
=
ρ hot airV
ρ hot air
1.29 kg/m3 − 0.93 kg/m3 )( 9.80 m/s 2 )
(
=
= 3.8 m/s 2
0.93 kg/m3
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612 FLUIDS
104. CONCEPT QUESTIONS
a. In Poiseuille’s law, the upstream pressure P2 is the same for both hoses, because it is the
pressure at the outlet. The downstream pressure P1 is also the same for both hoses, since
each opens to the atmosphere at the exit end. Therefore, the term P2 – P1 is the same for
both hoses.
b. The volume flow rate Q is related to the radius R of a hose and the speed v of the water
in the hose according to Equation 11.10: Q = Av = π R2v. Here, we have used the fact that
the circular cross-sectional area of a hose is π R2.
SOLUTION Using Poiseuille’s law as given in Equation 11.14, we have
π R 4 ( P2 − P1 )
Q=
8η L
Then, from Equation 11.10 and A = π R2 for the area of a circle, we have
Q = Av = π R 2v
Substituting this expression for Q into Poiseuille’s law gives
π R 4 ( P2 − P1 )
πR v=
8η L
2
or
v=
R 2 ( P2 − P1 )
8η L
To find the ratio vB/vA of the speeds, we apply this result to each hose, recognizing that the
pressure difference P2 – P1, the length L, and the viscosity η are the same for both hoses:
RB2 ( P2 − P1 )
vB
vA
=
RB2
8η L
2
=
= (1.50 ) = 2.25
2
2
RA ( P2 − P1 ) RA
8η L
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