Solucionario Fundamentos de Física 9na edición Capitulo 9

9
Solids and Fluids
CLICKER QUESTIONS
Question H1.01a
Description: Exploring and interrelating weight, force, pressure, and buoyancy.
Question
A block and a beaker of water are placed side by side on a scale (case A). The block is then placed into the
beaker of water, where it fl oats (case B). How do the two scale readings compare?
1. Scale A reads more than scale B.
2. Scale A reads the same as scale B.
3. Scale A reads less than scale B.
4. Not enough information
A B
Commentary
Purpose: To explore and interrelate ideas about weight, force, pressure, and buoyancy.
Discussion: Consider the beaker, water, and block as one compound “object” or “system.” In both cases,
that system has the same mass and therefore the same weight. Since the system is not accelerating, the
net force on it must be zero, so the force the scale exerts on this system must be equal to the total weight.
Scales measure force. The forces are the same, so the scale readings are identical.
But, how does the weight of the fl oating block in case B affect the scale reading?
It is useful to look at the beaker. There are three forces on the beaker: (1) gravitation pulling down;
(2) water pushing down; and (3) scale pushing up. Since the beaker is at rest, the force of the scale pushing
up must balance the forces exerted down.
457

458 Chapter 9
Force #1 is the same for both cases; it is the weight of the beaker.
Force #2 is different for the two cases. In case A, the force pushing down is the weight of the water. In case
B, the force pushing down is larger than the weight of the water, because the water level is higher in B than
in A, so the pressure on the bottom of the beaker is larger. The amount the force is larger in B is exactly
equal to the weight of the block. That is, it is the weight of the displaced water.
Therefore, to support the beaker in B, the scale must push up with a force equal to the total weight of the
water, block, and beaker, which is exactly the force exerted by the scale in A.
Key Points:
• Some questions are easiest to answer by considering a set of objects as if they were one compound
object or system.
• A scale measures force. If the system is not accelerating and interactions with its environment are small
(e.g., via the buoyancy of air), then the scale reading is the weight of the system.
• Newton’s laws hold for a body of fl uid as well as for a solid object.
• When thinking about fl uids and forces, the concept of pressure is often useful.
For Instructors Only
This is the fi rst of three related questions. It is very easy if approached the right way, but students can get
themselves quite confused. If students give the straightforward, correct answer, we recommend challenging
them to explain how the fl oating block can infl uence the scale reading. Resolving the confusion this gener-ally
causes will strengthen their understanding of several related concepts.
For students already familiar with the concept of buoyancy and with Archimedes’ principle (perhaps from
high school), this question can be used to introduce, motivate, and provide context for the concept of
pressure.
Scales measure force, not weight. If three criteria are met, that force is the weight: (1) the system is not
accelerating; (2) buoyancy due to air is negligible; (3) the only external forces on the system are due to
gravitation and the scale.
Question H1.01b
Description: Exploring and interrelating weight, force, pressure, and buoyancy.
Question
A block and a beaker of water are placed side by side on a scale (case A). The block is then placed into the
beaker of water, where it sinks (case B). How do the two scale readings compare?
1. Scale A reads more than scale B.
2. Scale A reads the same as scale B.
3. Scale A reads less than scale B.

Solids and Fluids 459
A B
Commentary
Purpose: To explore and interrelate ideas about weight, force, pressure, and buoyancy.
Discussion: As with the previous question, the scale readings must be the same because the total mass sup-ported
by the scale, and therefore the total weight, are the same for both cases.
As with the previous question, we would like to understand how the scale readings can be the same. The
scale is only sensitive to the normal force exerted upon it by the beaker contacting it. How does it “know”
about the block?
It is useful to focus on the beaker and think about the forces on it. There are four: (1) gravitation pulling
down; (2) block pushing down; (3) water pushing down; and (4) scale pushing up. The force of the scale
must balance the other three forces exerted down.
Force #1 is the same in both cases; it is the weight of the beaker.
Force #2 is different for the two cases. In case A, it is not exerted on the beaker. (However, a force down
due to the block is exerted directly on the scale.) In case B, the block is pushing down on the beaker with a
force smaller than its weight, because part of its weight is supported by buoyancy. In case A, the full weight
of the block is pushing down on the scale.
Force #3 is also different for the two cases. In case A, a force equal to the weight of the water is pushing
down on the beaker, but in case B, the force pushing down is larger than the weight of the water, because the
water level is higher in B than it is in A, so the pressure at the bottom is higher too. The net effect is that
the force due to the water is increased by exactly the amount that the force due to the block is decreased.
The result is that the scale readings are the same!
Note that there must be a thin layer of water underneath the block, even though it is touching the bottom of
the beaker. Otherwise, there would be no buoyant force.
Key Points:
• When an object is placed in water, whether it fl oat or sinks, the water level rises, which causes the
pressure at the bottom of the water to increase.
• An object in a fl uid experiences a buoyant force, whether it fl oat or sinks.
• Scales measure force, not weight. If the system is not accelerating and interactions with its
environment are small (e.g., via the buoyancy of air), then the scale reading is the weight of the system.
• “Old” force ideas such as free-body diagrams are useful for understanding fl uids. New ideas such as
pressure and buoyancy are also useful.

460 Chapter 9
This is the second of three related questions. It is very similar to the fi rst, but understanding how the scale
reading comes to be the same (by analyzing the forces on the various bodies) introduces a new wrinkle:
the fl uid only supports part of the weight of the block, and the pressure exerted on the inside bottom of the
beaker by the water and block is not uniform.
Many students will think the scale readings are the same without fully appreciating what the fuss is all
about. They might have trouble understanding why some people are confused.
Students who think the scale readings are different might need a demonstration to be convinced of the
predicted result.
Students might think that the force exerted by the water in case B is actually smaller than in case A, perhaps
because the effective area of water in contact with the beaker is smaller. They do not realize that there must
be water beneath the block in order for there to be a buoyant force. Otherwise, we are talking about a suc-tion
cup, for which there is no water on one side, and an enormous force due to the water on the other.
Question H1.01c
Description: Honing understanding of buoyancy.
Question
Two blocks, A and B, have the same size and shape. Block A fl oats in water, but block B sinks in water.
Which block has the larger buoyant force on it?
1. Block A has the larger buoyant force on it.
2. Block B has the larger buoyant force on it.
3. Neither; they have the same buoyant force on them.
4. Impossible to determine from the given information
Commentary
Purpose: To develop your understanding of buoyancy.
Discussion: According to Archimedes’s Principle, the buoyant force on an object is equal to the weight of
the fl uid displaced by the object. Whether the block fl oats or sinks is irrelevant.
Since block B sinks, it displaces its entire volume, whereas block A displaces only part of its volume.
Since the two blocks have the same total volume, block B displaces the larger volume of water, so it also
has the larger buoyant force on it.
If block B experiences a larger buoyant force but sinks, it must have a larger mass, and therefore a larger
density. This is the only way they can have the same size and shape yet behave as they do.
Key Points:
• The buoyant force on an object is equal to the weight of the fl uid displaced by the object. It does not
depend on other factors, such as whether the object fl oats or sinks.

This is the third of three related questions. It uses a different situation, but makes a good follow-up to the
fi rst two in that it focuses attention on one specifi c difference between the fi rst two questions, helping to
resolve lingering confusion and solidify students’ understanding. It can also be used effectively as a stand-alone
question, if desired.
Many students will be overly focused on the state of an object: in this case, whether the block sinks or
fl oats. Many will think that the buoyant force must be smaller on the sinking block, and that is why it sinks.
Encourage them to consider other reasons why one might sink.
Students often assume that the masses of the two blocks are the same.
A demonstration can be useful, if only to let students see that the masses of the objects are defi nitely not the
same.
Question H1.02a
Description: Developing understanding of buoyancy.
Question
A metal block sits on top of a fl oating wooden block. If the metal block is placed on the bottom of the
beaker, what happens to the level of water in the beaker?
?
1. The level decreases.
2. The level stays the same.
3. The level increases.
Commentary
Purpose: To explore buoyancy and Archimedes’ principle.
Discussion: Archimedes’ principle states that the buoyant force on a fl oating object is equal in magnitude to
the weight of displaced fl uid. The buoyant force on the two-block “object” in the fi rst case must equal the
weight of the two blocks (so the net force on them is zero), so the amount of water displaced must have that
same weight. The displaced water must go somewhere, so the water level in the beaker rises.
In the second case, the fl oating wooden block will displace an amount of water with weight equal to the
wooden block alone. The metal block at the bottom will displace a volume of water equal to its volume,
but no more: the normal force due to the beaker’s bottom helps support it. Since the density of water is less
than the density of the metal (or it wouldn’t sink), this means the volume of water displaced will weigh
less than the metal block. So, in the second case the two blocks will displace less total water than in the fi rst
case, and the water level rises less in the second case.

462 Chapter 9
Key Points:
• Archimedes’ principle states that the buoyant force on a fl oating object is equal in magnitude to the
weight of displaced fl uid.
• A fl oating object displaces a weight of fl uid equal to its own weight.
• A submerged object displaces a volume of fl uid equal to its own volume.
• It is sometimes helpful to think of a combination of objects as a single object.
This is the fi rst of three related questions that help students develop a robust understanding of buoyancy,
weight, fl oating, sinking, and fl uid displacement. The situations lend themselves to a live demonstration,
using a “predict, observe, and reconcile” pattern.
One possible source of confusion with this question is how the weight of the metal block in the fi rst case
can displace any water, when the block is not in the water. Having students draw free-body diagrams for
each block can be helpful for resolving this.
Question H1.02b
Question
A metal block sits on top of a fl oating wooden block. If the metal block is suspended from the bottom of the
wooden block, what happens to the volume of the wooden block that is submerged in the water?
?
1. The volume decreases.
2. The volume stays the same.
3. The volume increases.
Commentary
the weight of displaced fl uid. In both cases, the water is supporting the same weight—the combined weight
of the two blocks—so the volume of water displaced must be the same.

However, in the fi rst case, all the water is displaced by the wooden block, while in the second some of the
water is displaced by the hanging metal block and the rest by the wooden block. So for the second case,
more of the wooden block will be above the water’s surface, and the submerged volume of the wooden
block has decreased.
Key Points:
This is the second of three related questions that help students develop a robust understanding of buoyancy,
Students might wonder what the difference between the two cases is: in one the small block pushes down
on the big, and in the second it pulls down. Having students draw a free body diagram for each block
should help them realize that the magnitude of that pull is less.
Question H1.02c
Question
A metal block sits on top of a fl oating wooden block. If the metal block is suspended from the bottom of the
wooden block, what happens to the level of water in the beaker?
?
1. The level decreases.
2. The level stays the same.
3. The level increases.
Commentary
the weight of displaced fl uid. In both cases, the water is supporting the same weight—the combined weight
of the two blocks—so the total volume of water displaced must be the same. Thus, the water level in the
beaker must be the same as well.

464 Chapter 9
Key Points:
This is the third of three related questions that help students develop a robust understanding of buoyancy,
This question should be rather easy for students who have grasped the ideas raised in the previous two
questions; it serves primarily to confi rm their understanding.
QUICK QUIZZES
1. (c). The mass that you have of each element is as follows:
= ρ = (19.3 ×103 kg/m3 )(1 m3 ) = 19.3 ×103 kg
m V gold gold gold
= ρ = (10.5 ×103 kg/m3 )(2 m3 ) = 21.0 ×103 kg
m V silver silver silver
= ρ = (2.70 ×103 kg/m3 )(6 m3 ) = 16.2 ×103 kg
m V aluminum aluminum aluminum
2. (a). At a fi xed depth, the pressure in a fl uid is directly proportional to the density of the fl uid.
Since ethyl alcohol is less dense than water, the pressure is smaller than P when the glass is fi lled
with alcohol.
3. (c). For a fi xed pressure, the height of the fl uid in a barometer is inversely proportional to the
density of the fl uid. Of the fl uids listed in the selection, ethyl alcohol is the least dense.
4. (b). The blood pressure measured at the calf would be larger than that measured at the arm. If we
imagine the vascular system of the body to be a vessel containing a liquid (blood), the pressure
in the liquid will increase with depth. The blood at the calf is deeper in the liquid than that at the
arm and is at a higher pressure.
Blood pressures are normally taken at the arm because that is approximately the same height as
the heart. If blood pressures at the calf were used as a standard, adjustments would need to be
made for the height of the person, and the blood pressure would be different if the person were
lying down.
5. (c). The level of fl oating of a ship is unaffected by the atmospheric pressure. The buoyant force
results from the pressure differential in the fl uid. On a high-pressure day, the pressure at all points
in the water is higher than on a low-pressure day. Because water is almost incompressible, how-ever,
the rate of change of pressure with depth is the same, resulting in no change in the buoyant
force.

6. (b). Since both lead and iron are denser than water, both objects will be fully submerged and
(since they have the same dimensions) will displace equal volumes of water. Hence, the buoyant
forces acting on the two objects will be equal.
7. (a). When there is a moving air stream in the region between the balloons, the pressure in this
region will be less than on the opposite sides of the balloons where the air is not moving. The
pressure differential will cause the balloons to move toward each other. This is demonstration of
Bernoulli’s principle in action.
ANSWERS TO MULTIPLE CHOICE QUESTIONS
1. m = ρ V = ( × )( × − ) × gold
19.3 103 kg m3 4.50 10 2 m (11.0 10−2 m)(26.0 ×10−2 m) = 24.8 kg, and
choice (a) is the correct response.
2. On average, the support force each nail exerts on the body is
F
mg
( . )( .
) =
= = 0 535
1 1 208
66 0 9 80
1 208
.
kg m s
N
2
so the average pressure exerted on the body by each nail is
1 0 535 5
F
A av
P
nail
end
N
6 2
1.00 m
= =
×
= × −
10
5 35 10
.
. Pa
and (d) is the correct choice.
3. From Pascal’s principle, F A F A 1 1 2 2 = , so if the output force is to be F2
= 1.2 × 103 N, the
required input force is
F
A
A
× 0.050 m
= 1
1 2 10
F 1
2
2 0 70
⎛
⎝ ⎜
⎞
⎠ ⎟
=
⎛
⎝ ⎜
⎞
⎠ ⎟
m
2
. 2
( . 3 N) = 86 N
making (c) the correct answer.
4. According to Archimedes’s principle, the buoyant force exerted on the bullet by the mercury is
equal to the weight of a volume of mercury that is the same as the submerged volume of the bul-let.
If the bullet is to fl oat, this buoyant force must equal the total weight of the bullet. Thus, for a
fl oating bullet,
ρ ρ mercury submerged lead bullet V g = V g and
V
V
submerged
bullet
ρ
= = × kg lead
ρ
mercury
11.3 103 m
0 831 . 3
kg m
3
13 6 10 3
.
×
=
so the correct response is (d).
5. The absolute pressure at depth h below the surface of a liquid with density ρ, and with pressure P0
at its surface, is P = P + gh 0 ρ . Thus, at a depth of 754 ft in the waters of Loch Ness,
= × 1
Pa ( )⎛⎝
P = 1.013 × 105 Pa + (1.00 × 103 kg m3 )(9.80 m s2 ) 754 ft
m
3.281 ft
⎞⎠
⎡
⎣ ⎢
⎤
⎦ ⎥
2.35 106
and (c) is the correct response.

466 Chapter 9
6. We assume that the air inside the well-sealed house has essentially zero speed and the thickness
of the roof is negligible so the air just above the roof and that just below the roof is at the same
altitude. Then, Bernoulli’s equation gives the difference in pressure just below and just above the
roof (with the pressure below being the greatest) as
2
2
P P g y y 1 2 2
1
2 1
1
2
− = ρ ( − )+ ρ ( − ) air air v v
or
ΔP = ( ) ( )⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣
1
2
1 29 95
1
2 237
.
.
kg m mi h
mi h
3 ⎢
⎤
⎦ ⎥
−
⎧⎨ ⎪
⎩⎪
⎫⎬ ⎪
⎭⎪
+ = ×
2
0 0 1.2 103 Pa
and the correct choice is (a).
7. From the equation of continuity, A A 1 1 2 2 v = v , the speed of the water in the smaller pipe is
A ⎤
1
A
v v 2
2
1
2
2
0 250
0 100
=
⎛
⎝ ⎜
⎞
⎠ ⎟
= ( )
( )
⎡
⎣ ⎢⎢
π
π
.
.
m
m ⎦ ⎥⎥
(1.00 m s) = 6.25 m s
so (d) is the correct answer.
8. All of these phenomena are the result of a difference in pressure on opposite sides of an object
due to a fl uid moving at different speeds on the two sides. Thus, the correct response to this ques-tion
is choice (e). Bernoulli’s equation can be used in the discussion of each of these phenomena.
9. The boat, even after it sinks, experiences a buoyant force, B, equal to the weight of whatever
water it is displacing. This force will support part of the weight, w, of the boat. The normal force
exerted on the boat by the bottom of the lake will be n = w − B < w will support the balance of
the boat’s weight. The correct response is (c).
10. The absolute pressure at depth h below the surface of a fl uid having density ρ is P = P + gh 0 ρ ,
where P0 is the pressure at the upper surface of that fl uid. The fl uid in each of the three vessels
has density ρ = ρwater, the top of each vessel is open to the atmosphere so that P P 0 = atmo in each
case, and the bottom is at the same depth h below the upper surface for the three vessels. Thus,
the pressure P at the bottom of each vessel is the same and (c) is the correct choice.
11. Since the pipe is horizontal, each part of it is at the same vertical level or has the same
y-coordinate. Thus, from Bernoulli’s equation (P + 1 + gy = constant)
2
ρv2 ρ , we see that the
sum of the pressure and the kinetic energy per unit volume (P + 1 )
2
ρv2 must also be constant
throughout the pipe, making (e) the correct choice.
12. Once the water droplets leave the nozzle, they are projectiles with initial speed v v 0y i =
and having speed v v f y = =0 at their maximum altitude, h. From the kinematics equation
v v y y y 2 a y
= 2 + 2 (Δ ), the maximum height reached is h = v2 2 g . Thus, if we want to quadruple the
0
i maximum height (h′ = 4h), we need to double the speed of the water leaving the nozzle (v′ = v ) i i 2 .
Using the equation of continuity, A′ ′ = A i i v v , it is seen that if v′ = v i i 2 , it is necessary to have
A′ = ′ A = A i i i (v v ) 2. This says that the area needs to be decreased by a factor of 2, and the correct
choice is (d).

ANSWERS TO EVEN-NUMBERED CONCEPTUAL QUESTIONS
2. We approximate the thickness of the atmosphere by using P = P + gh 0 ρ with P0 = 0 at the top of
the atmosphere and P = 1 atm at sea level. This gives an approximation of
h
P P
g
=
− 5
−
10 0 4
( )( ) 0 =
m 3 2 or h ~ 10 km
1
1 10
10
ρ
~
Pa
kg m m s
Because both the density of the air, ρ, and the acceleration of gravity, g, decrease with altitude,
the actual thickness of the atmosphere will be greater than our estimate.
4. Both must have the same strength. The force on the back of each dam is the average pressure of
the water times the area of the dam. If both reservoirs are equally deep, the force is the same.
6. The external pressure exerted on the chest by the water makes it diffi cult to expand the chest
cavity and take a breath while under water. Thus, a snorkel will not work in deep water.
8. A fan driven by the motor removes air and hence decreases the pressure inside the cleaner. The
greater air pressure outside the cleaner pushes air in through the nozzle toward this region of
lower pressure. This inward rush of air pushes or carries the dirt along with it.
10. The water level on the side of the glass stays the same. The fl oating ice cube displaces its own
weight of liquid water, and so does the liquid water into which it melts.
12. The higher the density of a fl uid, the higher an object will fl oat in it. Thus, an object will fl oat
lower in low density alcohol.
14. A breeze from any direction speeds up to go over the mound, and the air pressure drops at this
opening. Air then fl ows through the burrow from the lower to the upper entrance.
PROBLEM SOLUTIONS
9.1 The elastic limit is the maximum stress, F A where F is the tension in the wire, that the wire can
withstand and still return to its original length when released. Thus, if the wire is to experience
a tension equal to the weight of the performer without exceeding the elastic limit, the minimum
cross-sectional area is
A
2
= π D = F
min = elastic limit
mg
elastic limit min
4
and the minimum acceptable diameter is
D
4 mg
4(70 )(9 8 )
elastic limit min
.
= ( ) =
( × ) = × − =
π π
kg m s2
5 0 10
3
1 3 10 1 3 8
.
. .
Pa
m mm

468 Chapter 9
9.2 (a) In order to punch a hole in the steel plate, the
superhero must punch out a plug with cross-sectional
area, Acs, equal to that of his fi st and
a height t equal to the thickness of the steel
plate. The area Ashear of the face that is sheared
as the plug is removed is the cylindrical surface
with radius r and height t as shown in the sketch. Since A r cs = π 2, then r Acs = π and
c = ( )= = ( ) ×
A rt t
Acs
2 2 2 200
cm
shear 1 00 102
π π
π
π .
. m
cm
2
2
π
= 70 9 .
If the ultimate shear strength of steel (i.e., the maximum shear stress it can withstand before
shearing) is 2.50 × 108 Pa = 2.50 × 108 N m2, the minimum force required to punch out
this plug is
F = A ⋅ stress =
⎛
1
104 . ⎢⎢
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣
shear
2
2
m
cm
2 cm
70 9
⎤
⎦ ⎥⎥
2.50 108 = 1.77 ×106
×
⎛
⎝ ⎜
⎞
⎠ ⎟
N
m
N
2
(b) By Newton’s third law , the wall would exert a force of equal magnitude in the opposite
direction on the superhero, who would be thrown backward at a very high recoil speed.
9.3 Two cross-sectional areas in the plank, with one directly above the rail and one at the outer end of
the plank, separated by distance h = 2.00 m and each with area A = (2.00 cm)(15.0 cm) = 30.0 cm2,
move a distance Δx = 5.00 × 10−2 m parallel to each other. The force causing this shearing effect
in the plank is the weight of the man F = mg applied perpendicular to the length of the plank at
its outer end. Since the shear modulus S is given by
S
shear stress
shear strain
F A
x h
Fh
x A
= = = Δ (Δ )
we have
S =
( 80 . 0 )( 9 . 80 2
)( 2 .
00
)
( × − )
Pa 2 2 2 ( )( ) ⎡⎣
5 .
00 10 2 3
kg m s m
m
0 0 1 10
1 05 10 4
7
.
.
cm m cm
⎤⎦
= ×
9.4 As a liquid, the water occupied some volume Vl . As ice, the water would occupy volume 1.090Vl
if it were not compressed and forced to occupy the original volume. Consider the pressure change
required to squeeze ice back into volume Vl . Then, V V V V 0 l l = 1.09 and Δ = − 0.090 , so
Δ Δ
N
m2 09
V= − l
P B
V
V
⎛
⎝ ⎜
⎞
⎠ ⎟
= − × ⎛⎝
⎞⎠
−
0
2 00 109
0 090
1
.
.
.
1 65 108 1600
Vl
⎛
⎝ ⎜
⎞
⎠ ⎟
= . × Pa ≈ atm
9.5 Using Y = F L A L 0 (Δ ) with A = π d2 4 and F = mg, we get
Y =
( )( ) ⎡⎣
⎤⎦
( )
. 2
4 90 9 80 50
kg m s m
( × − )
π 1 . 0 10 2
m
2
3 5 108
1.6 m
Pa
( )
= . ×
r
t
Acs
Ashear

9.6 From Y = F L A L 0 (Δ ), the tension needed to stretch the wire
by 0.10 mm is
F
Y A ( Δ L
) ( π
)( Δ
)
= = L
2
Y d L
L
=
0
0
( × 10
) ×
4
( ) ( × )
( × ) =
Pa π . 10 0 10 10
18 10 0 22
22
N 2
−3 2 −3
−
m m
4 3.1 10 m
.
30° 30°
The tension in the wire exerts a force of magnitude F on the tooth in each direction along
the length of the wire as shown in the above sketch. The resultant force exerted on the
tooth has an x-component of R F F F x x = Σ = − cos30° + cos30° = 0, and a y-component of
R F F F F y y = Σ = − sin 30º − sin 30º = − = −22 N.
Thus, the resultant force is

= 22 N directed down the page in the diagram .
R
9.7 From Y = (F A)(L L) = (stress)(L L) 0 0 Δ Δ, the maximum compression the femur can withstand is
ΔL
stress L
Y
=
( )( ) =
( 160 × 10 )( 0 50
)
×
0
6
Pa . m
9
18 10
Pa
= 4.4 × 10−3 m = 4.4 mm .
9.8 (a) When at rest, the tension in the cable equals the weight of the 800-kg object, 7.84 × 103 N.
Thus, from Y = F L A L 0 (Δ ), the initial elongation of the cable is
ΔL
F L
A Y
=
⋅
⋅
=
( × )( )
× −
0
3
7 . 84 10 25 .
0
4
4 .
00 10
N m
( m2 )( × )= × −
2 45 10 2 5 10
20 10
3
Pa
. m= . mm
(b) When the load is accelerating upward, Newton’s second law gives
− = , or F m g ay
F mg may
= ( + ) [1]
If m ay = 800 kg and = +3.0 m s2, the elongation of the cable will be
ΔL
0 800 kg 9.80 3.0 m s2 (25.0 m)
F L
A Y
=
⋅
⋅
=
( ) + ( ) ⎡⎣
⎤⎦
−
( × )( × ) = × = −
m 3. 2 2 mm
3 2 10 4 10
400 10 20 10
3
.
.
m Pa
Thus, the increase in the elongation has been
increase = (ΔL) − (ΔL) = − = initial 3.20 mm 2.45 mm 0.75 mm
→F
→F
continued on next page

470 Chapter 9
(c) From the defi nition of the tensile stress, stress = F A, the maximum tension the cable can
withstand is
= ⋅( ) = (4.00 × 10−4 m2 )(2.2 × 108 Pa) = 8.8 × 104 N
F A stress max max
Then, Equation [1] above gives the mass of the maximum load as
m
F
max
max
g a 2
N
9.8 3.0 m s
=
+
= ×
( + ) = × 8 8 10
6 9 10
4
. 3
. kg
9.9 From the defi ning equation for the shear modulus, we fi nd the displacement, Δx, as
Δx
h ( = FA
) = ⋅
S
h F
S A
⋅
=
( × )( )
5.0 10−3 m 20 N
3.0 106 Pa cm
×
cm
1 m
m
2
2
⎛
( )( ) 2
⎝ ⎜
⎞
⎠ ⎟
= × − =
14
10
2 4 10 0
4
. 5 .024 mm
9.10 The shear modulus is given by
S
shear stress
shear strain
stress
x h
= =(Δ )
Hence, the stress is
stress S
Δ ⎛
x
h
= ⎛⎝
⎞⎠
10 5 0 .
= ( 1 5 × 10
) ×
.
Pa
m
⎝ 10 103 m
⎞⎠
= 7.5 × 106 Pa
9.11 The tension and cross-sectional area are constant through the entire length of the rod, and the
total elongation is the sum of that of the aluminum section and that of the copper section.
ΔL ΔL ΔL
) + ( ) = 0 0 0 0 ( ) + ⎡ ( )
F L
AY
F L
AY
F
A
L
rod Al Cu
Al
Al
Cu
Cu
= + = (
⎣ ⎢⎢
⎤
⎦ ⎥⎥
Al
Al
Cu
L
Y
Cu Y
where A = π r2 with r = 0.20 cm = 2.0 × 10−3 m. Thus,
N
m
m
ΔLrod 1
7.0 10
=
( 5 8 × 10
)
( 2 0 × 10
− ) ×
1 3 3
3 2
.
.
.
2 6 = × − =
.
m
1 . 9 10 2 1 .9 cm
+ m
×
π 0 Pa 10
11 10 Pa
⎡
⎤
⎥
⎦ ⎢⎣ 9.12 The acceleration of the forearm has magnitude
a
= Δ =
t
⎛
km 3
h
⎝ ⎜
⎞
⎠ ⎟
⎛
⎝ ⎜
⎞
⎠ ⎟
v
Δ
80
10 1
5
m
1 km
h
3 600 s
.
.
0 10
4 4 10 3
3
×
= × − s
m s2
The compression force exerted on the arm is F = ma and the compressional stress on the bone
material is
Stress
F
A
= =
( )( × )
3 0 4 4 10
2 4 10
( )= 5 . 5 ×
107 −
3
4
. .
.
kg m s
cm
2
Pa 2 1 2
2 m cm
Since the stress is less than the allowed maximum, the arm should survive .

9.13 The average density of either of the two original worlds was
3
4
= M
= M
= V
R
M
R
ρ
0 4π 3 3 π 3
The average density of the combined world is
ρ
M
( )
= = π π
( ) =
′
=
⎛⎝ ⎜
⎞⎠ ⎟
V
M
R
M
R
total 2
4
3
3
4
4 2
3
32
3
2
2 3
M
9π R3
so
ρ
ρ π
π
0
32 3
9
M
= = R
3
4
3
128
27
4 74 =⎛⎝
⎞⎠
⎛
⎝ ⎜
⎞
⎠ ⎟
R
M
. or ρ = 4 74ρ0 .
9.14 (a) The mass of gold in the coin is
m
( #
karats ) = m
total
= = m Au
total
.
24
22
24
11
12
(7 988 ×10−3 kg) = 7.322 ×10−3 kg
and the mass of copper is
m m Cu total = = ( × − kg) = × − 1
12
1
12
7.988 10 3 6.657 10 4 kg
(b) The volume of the gold present is
V
= m
= ×
Au
Au
Au
kg
3 3
×
19.3 kg m
= ×
−
ρ
7 322 10
10
3 79
. 3
. 10−7 m3
and the volume of the copper is
V
= m
= ×
Cu
Cu
Cu
kg
3 3
×
8.92 kg m
= ×
−
ρ
6 657 10
10
7 46
. 4
. 10−8 m3
(c) The average density of the British sovereign coin is
ρav
m = × −
V
= =
k total
total
m
total
V V
+
Au Cu
7.988 10 3 g
= × 10− 10−
kg m 7 3 8 3
3.79 m 7.46 m
3
× + ×
1.76 104
9.15 (a) The total normal force exerted on the bottom acrobat’s shoes by the fl oor equals the total
weight of the acrobats in the tower. That is
n = m g = [( + + + ) ] total 75.0 68.0 62.0 55.0 kg (9.80 m s2 ) = 2.55 ×103 N
(b) P
= = = ×
n
A
n
A
total shoe
sole
. 3
2 55 10
2
N
2 2 425 cm 1
m
2 cm2
Pa
10
3 00 10 4
4
( ) ⎡⎣
⎤⎦
= . ×
(c) If the acrobats are rearranged so different ones are at the bottom of the tower, the total
weight supported, and hence the total normal force n, will be unchanged. However, the total
area A A total shoe sole = 2 , and hence the pressure, will change unless all the acrobats wear the
same size shoes.

472 Chapter 9
9.16 We shall assume that each chair leg supports one-fourth of the total weight so the normal force
each leg exerts on the fl oor is n = mg 4. The pressure of each leg on the fl oor is then
mg
r leg
P
n
A
leg
4 95 0 kg )(9 80 m s2
)
= = = (
. .
. 0 10
4 050 π 2 π
2 96 10
2 2
6
( × ) = × − m
. Pa
9.17 (a) If the particles in the nucleus are closely packed with negligible space between them, the
average nuclear density should be approximately that of a proton or neutron. That is
ρ
( )
( × ) ×
m ×
V
≈ proton
=
proton π nucleus
proton
m
4 r 3
3 1 67
3 ∼
. 10
4 1 10
4 10
27
15 3
17
−
−
kg
m
kg m3
π
∼
(b) The density of iron is ρFe
= 7.86 × 103 kg m3 and the densities of other solids and liquids
are on the order of 103 kg m3. Thus, the nuclear density is about 1014 times greater than
that of common solids and liquids, which suggests that atoms must be mostly empty space.
Solids and liquids, as well as gases, are mostly empty space.
9.18 Let the weight of the car be W. Then, each tire supports W 4, and the gauge pressure is
P
F
A
W
A
= =
4
Thus, W = 4 A P = 4(0.024 m2 )(2.0 × 105 Pa) = 1.9 × 104 N .
9.19 The volume of concrete in a pillar of height h and cross-sectional area A is V = Ah, and its weight
is F Ah g = ( )(5.0 × 104 N m3 ). The pressure at the base of the pillar is then
P
F
A
( Ah
)( 5 0 × 10
)= ( × ) h g = =
A
5 0 10
4
4 .
.
N m
N m
3
3
Thus, if the maximum acceptable pressure on the base is, Pmax = 1.7 × 107 Pa, the maximum
allowable height is
h
P
max
max
Pa
N m
.
.
N m3 3
=
×
= ×
×
=
5 0 10
1 7 10
4 5 0 10
7
. 4
3.4 × 102 m
9.20 Assuming the spring obeys Hooke’s law, the increase in force on the piston required to compress
the spring an additional amount Δx is ΔF = F − F = (P − P )A = k (Δx) 0 0 . The gauge pressure
at depth h beneath the surface of a fl uid is P − P = gh 0 ρ , so we have ρghA = k (Δx), or the
required depth is h = k (Δx) ρgA. If k = 1 250 N m, A = π r2 with r = 1.20 × 10−2 m, and the
fl uid is water (ρ = 1.00 × 103 kg m3 ), the depth required to compress the spring an additional
0.750 cm = 7.50 × 10−3 m is
h =
( 1 250 )( 7 50 × 10−
)
( × )
1 00 10 9 80
3
3
N m m
kg m3
.
. . m s m
m
2 ( ) × ( ) ⎡⎣
⎤⎦
=
π 1 20 10−
2 11
2 2 .
.
9.21 (a) P = P + gh = × + ( × ) 0
ρ 101.3 103 Pa 1.00 103 kg m3 (9.80 m s2 )(27.5 m) = 3.71 × 105 Pa
(b) The inward force the water will exert on the window is
F = PA = P( r ) = ( × ) ⎛ ×
⎝ ⎜
⎞
⎠
−
2
2
π 2 3 71 10
5 π
35 0 10
.
.
Pa
m
2 ⎟ = ×
3.57 104 N

9.22 The gauge pressure in a fl uid at a level distance h below where Pgauge = 0 is P gh gauge = ρ with h
being positive when measured in the downward direction. The difference in gauge pressures
at two levels is (P ) (P ) g ( h) gauge 2 gauge 1 − = ρ Δ or (P ) (P ) g( h) gauge 2 gauge 1 = +ρ Δ with Δh being
positive if, in going from level 1 to level 2, one is moving lower in the fl uid.
(a) In moving from the heart to the head, one is moving higher in the blood column so Δh 0
and we fi nd
P P g h gauge head gauge heart ( ) = ( ) + ρ (Δ ) = 13.3 ×103 Pa + (1 060 kg m3 )(9.80 m s2 )(−0.500 m)
or
( ) = 8.11×103 Pa = 8.11 kPa
Pgauge head
(b) In going from heart to feet, one moves deeper in the blood column, so Δh 0 and
P P g h gauge feet gauge heart ( ) = ( ) + ρ (Δ ) = 13.3 ×103 Pa + (1 060 kg m3 )(9.80 m s2 )(+1.30 m)
or
( ) = 26.8 ×103 Pa = 26.8 kPa
Pgauge feet
9.23 The density of the solution is ρ = 1.02ρ = 1.02 ×103 water
kg m3. The gauge pressure of the
fl uid at the level of the needle must equal the gauge pressure in the vein, so
P gh gauge = ρ = 1.33 ×103 Pa, and
h
= = ×
P
g
1 33 10
( × )
gauge
3 3
Pa
m ( 2 ) = 0.133
ρ 1.02 10 kg m 9 80
m
. 3
. s
9.24 (a) From the defi nition of bulk modulus, B = − ΔP (ΔV V ) 0 , the change in volume of the
1.00 m3 of seawater will be
Δ
( Δ
) ) × − = − = −( 0
V
V P
B
1 00 1 13 108 1 01
water
. m3 . Pa . 3 10
0 053 8
( × 5 )
×
= −
Pa
m 10
0.210 10 Pa
. 3
(b) The quantity of seawater that had volume V0 = 1.00 m3 at the surface has a mass of
1 030 kg. Thus, the density of this water at the ocean fl oor is
=
( − ) = × m
ρ= =
+
V
m
V V 0
1 030
0 053 8
1 09 10
Δ
kg
1.00 . m3
. 3 kg m3
(c) Considering the small fractional change in volume (about 5%) and enormous change in pressure
generated, we conclude that it is a good approximation to think of water as incompressible .
9.25 We fi rst fi nd the absolute pressure at the interface between oil and water.
= +
ρoil oil
P P gh 1 0
= × +( )
. Pa kg m3 ( . m s2 )(0.300 m) = 1.03 ×105 Pa
1 013 105 700 9 80
This is the pressure at the top of the water. To fi nd the absolute pressure at the bottom, we use
P P gh 2 1 = +ρwater water, or
P2
= 1.03 × 105 Pa + (103 kg m3 )(9.80 m s2 )(0.200 m) = 1.05 × 105 Pa

474 Chapter 9
9.26 If we assume a vacuum exists inside the tube above the wine column, the pressure at the base of
the tube (that is, at the level of the wine in the open container) is Patmo = 0 + ρgh = ρgh. Thus,
h
= = ×
P
g
Pa
. 5
1 013 10
( )( ) atmo =
ρ 984 kg m m s
3 2
9 .
80
10 5. m
Some alcohol and water will evaporate, degrading the vacuum above the column.
= , gives
9.27 Pascal’s principle, F A F A 1 1 2 2 = , or F A F A pedal Master
cylinder
brake brake
cylinder
A
A brake
F
brake cylinder
master cylinder
=
⎛
⎝ ⎜
⎞
⎠⎟ =
⎛
F ( ) = pedal
⎝ ⎜
⎞
⎠ ⎟
2
2
cm
cm
N N
6 4
1 8
44 156
.
.
This is the normal force exerted on the brake shoe. The frictional force is
f nk = μ = 0.50(156 N) = 78 N
and the torque is τ = f ⋅ r = ( )( ) = ⋅ drum 78 N 0.34 m 27 N m .
9.28 First, use Pascal’s principle, F A F A 1 1 2 2 = , to fi nd the force
piston 1 will exert on the handle when a 500-lb force
pushes downward on piston 2.
2
d
d 1
F
A
A
F
d
d
F
1
2
2
1
2 2
2
2
1
2
2
4
4
=
⎛
⎝ ⎜
⎞
⎠ ⎟
=
⎛
⎝ ⎜
⎞
⎠ ⎟
=
⎛
⎝
π
π ⎜
⎞
⎠ ⎟
= ( )
( ) ( ) =
F2
2
2
0 25
1 5
500 14
.
.
in
in
lb lb
Now, consider an axis perpendicular to the page, passing through the left end of the jack handle.
Στ = 0 yields
+(14 lb)(2.0 in) − F ⋅(12 in) = 0, or F = 2.3 lb
9.29 When held underwater, the ball will have three forces acting on it: a downward gravitational
force, mg; an upward buoyant force, B = ρ V = πρ r water water 4 3 3; and an applied force, F. If the
ball is to be in equilibrium, we have (taking upward as positive) ΣF F B mg y= + − =0, or
3 3
r = − = −
F mg B mg
r
g m
⎛
⎝ ⎜
⎞
⎠ ⎟
= −
⎛
ρ π ρ π
water water
4
3
4
3
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥
g
giving
F = − × ( ) ⎛⎝
⎞⎠
⎡
0 540 1 00 10
4
3
0 250
2
3
3
. .
.
kg kg m
3 m π
⎣ ⎢
⎤
⎦ ⎥
(9.80 m s2 ) = −74.9 N
so the required applied force is F
= 74 9 . N directed downward .
→F
→
H →V
→F
1
2.0
in
10 in
Free-Body Diagram of Handle

9.30 (a) To fl oat, the buoyant force acting on the person must equal the weight of that person, or the
weight of the water displaced by the person must equal the person’s own weight. Thus,
B = mg ⇒ gV = gV sea submerged body total ρ ρ or
V
V
submerged
total
body
sea
=
ρ
ρ
After inhaling,
V
V
submerged
total
3
= 945
kg m
= = 1 230
kg m
3
0.768 76.8%
leaving 23 2 . % above surface .
After exhaling,
V
submerged
V
total
3
= 1 020
kg m
= = 1 230
kg m
3
0.829 82.9%
leaving 17 1 . % above surface .
(b) In general, “sinkers” would be expected to be thinner with heavier bones, whereas
“fl oaters” would have lighter bones and more fat.
9.31 The boat sinks until the weight of the additional water displaced equals the weight of the truck.
Thus,
w Vg truck water
kg
m
3
( ) ( × ) m
⎡⎣
m
s2 = [ ( )]
=⎛⎝
⎞⎠
( )
ρ Δ
⎤⎦
⎛⎝
103 4.00 6.00 m 4 00 10 2 m 9 80
⎞⎠
. − .
or
wtruck = 9.41×103 N = 9.41 kN
9.32 (a)
B A(2.00 m)24.00 m2
Survivor
wr w
d
t
(b) Since the system is in equilibrium, ΣF B w w y r = − − =0 .
(c) B gV gdA w w = = ( ⋅ )
ρ ρ submerged
1 025 kg m3 (9.80 m s2 )(0.024 0 m)(4.00 m2 ) = 964 N
= ( )
− − =0,
(d) From B w wr
w B w B mg r s = − = − = 964 N − (62.0 kg)(9.80 m s2 ) = 356 N
(e) ρfoam 2
N
9.80 m s m
= =
⋅
= ( )( )
m
V
w g
t A
r
r
r 356
3 ( ) =
0.090 4.00
101
m
kg m 2

476 Chapter 9
(f )
= = ( ⋅ )
= ( )(
ρ ρ
1 025 9 80 )(0.090 0 m)(4.00 m2 ) = 3.62 × 103 N
B gV gtA max w r w
kg m3 .
m s2
max max max = = − , so
(g) The maximum weight of survivors the raft can support is w m g B wr
m
−
B w
g
r
max
= × − = 3 62 10 356
max . =
333
3N N
9.80 m s2 kg
9.33 (a) While the system fl oats, B = w = w + w total block steel, or ρ ρ w b b gV gV m g submerged steel = + .
= = 5.24 ×10−4 m3, giving
When msteel = 0.310 kg, V Vsubmerged b
ρ
ρ
w b
ρ b
b
w
b
V m
V
m
V
=
−
steel = − steel = 1.00 ×103 kg m3 −
0 310
10
408
kg m 4 3
. kg
5.24 m
3
×
= −
(b) If the total weight of the block + steel system is reduced, by having msteel 0.310 kg,
a smaller buoyant force is needed to allow the system to fl oat in equilibrium. Thus, the block
will displace a smaller volume of water and will be only partially submerged in the water.
The block is fully submerged when msteel = 0.310 kg. The mass of the steel object can
increase slightly above this value without causing it and the block to sink to the bottom.
As the mass of the steel object is gradually increased above 0.310 kg, the steel object
begins to submerge, displacing additional water, and providing a slight increase in the
buoyant force. With a density of about eight times that of water, the steel object will be able
to displace approximately 0.310 kg 8 = 0.039 kg of additional water before it becomes
fully submerged. At this point, the steel object will have a mass of about 0.349 kg and will
be unable to displace any additional water. Any further increase in the mass of the object
causes it and the block to sink to the bottom. In conclusion,
the block steel system will sink if steel + m ≥ 0.350 kg.
9.34 (a) B
Vb325 m3
mb226 kg
wHe He gVb wbmbg
= = 325 m3, and
(b) Since the balloon is fully submerged in air, V Vsubmerged b
= ρ = ( )( )( ) = air
1.29 kg m3 9.80 m s2 325 m3 4.11 × 103 N
B gVb

(c) He He He ΣF B w y = − b − w = B − mbg − ρ gVb = B − (mb + ρ V )
= × − +( )( )
g
4 11 10 226 0 179 325 3 . . N kg kg m m 3 3 ⎡⎣
⎤⎦
(9.80 m s2 ) = +1.33 ×103 N
Since ΣF ma y y = 0, aywill be positive (upward), and the balloon rises .
(d) If the balloon and load are in equilibrium, ΣF B w w w y b = ( − − )− = He load 0 and
w B w w load b He = ( − − ) = 1.33 × 103 N. Thus, the mass of the load is
m
= = kg × = 1 33 10
w
g load
load
N
2
m s
9 80
136
. 3
.
(e) If mload 136 kg, then the net force acting on the balloon + load system is upward and
the balloon and its load will accelerate upward .
(f ) The density of the surrounding air, temperature, and pressure all decrease as the balloon
rises. Because of these effects, the buoyant force will decrease until at some height the
balloon will come to equilibrium and go no higher.
ρ ρ 4π =
g r
9.35 (a) B = gV =
kg air balloon air ⎛
⎝ ⎜
⎞
⎠ ⎟
3
1 29
3
. m ms m
N
3 2 ( )( )⎛⎝
⎞⎠
( )
= × =
9 80
4
3
3 00
1 43 10
3
3
. .
.
π
1.43 kN
(b) ΣF = B − w = 1.43 × 103 N −( 15.0 kg )( 9.80 m s2
) ytotal
= +1.28 ×103 N = 1.28 kN upward
(c) The balloon expands as it rises because the external pressure (atmospheric pressure)
decreases with increasing altitude.
9.36 (a) Taking upward as positive, ΣF B mg ma y y = − = , or ma gV mg y w = ρ − .
(b) Since m = ρV, we have ρ V a ρ gV ρ V g y w = − , or
⎛
= w −
a g y
⎝ ⎜
⎞
⎠ ⎟
ρ
ρ
1
⎛
(c) ay = × −
⎝ ⎜
⎞
⎠ ⎟1 00 10 ( )
1 050
1 980
. 3
.
m kg
m kg
m s
3
3
2 = −0.467 m s2 = 0.467 m s2 downward
(d) From Δy t at y y = v + 0
2 2, with v0 0 y = , we fi nd
t
= ( ) = (− )
y
ay
−
=
2 2 800
0 467
5 85
Δ .
.
.
m
m s
s 2

478 Chapter 9
ρ ρ π 3
air balloon air gV g r ( ) = ⎛⎝
B = 600 ⋅ B = 600 600
9.37 (a) total single
balloon
4
3
⎞⎠
⎡
⎣ ⎢
⎤
⎦ ⎥
= 600 1 29 9 80
π = × = 4.0 103 N 4.0 kN
4
3
0 50 3 . . . kg m m s m 3 2 ( )( ) ( ) ⎡⎣⎢
⎤⎦⎥
(b) ΣF B m g y= − = × − ( ) total total 4.0 103 N 600 0.30 kg (9.8 m s2 ) = 2.2 ×103 N = 2.2 kN
(c) Atmospheric pressure at this high altitude is much lower than at Earth’s surface , so the
balloons expanded and eventually burst.
9.38 Note: We deliberately violate the rules of signifi cant fi gures in this problem to illustrate a point.
(a) The absolute pressure at the level of the top of the block is
ρ
P P gh top water top
kg
m
3 Pa
= +
0
= × + ⎛
1 0130 105 103
. ⎝
⎞⎠
⎛⎝
⎞⎠
( × )
= ×
9 80 5 00 10−
1 0179 10
2
5
. .
.
m
s
m
Pa
2
and that at the level of the bottom of the block is
ρ
P P gh bottom water bottom
Pa
= +
0
= × +
1 .
0130 105 103
kg
m
m
s
m 3 2
⎛⎝
⎞⎠
⎛⎝
⎞⎠
( × )
= ×
9 80 17 0 10−
1 0297
. . 2
. 105 Pa
Thus, the downward force exerted on the top by the water is
F P A top top = = (1 0179 ×105 Pa)(0 100 m)2 = 1017 9 . . . N
and the upward force the water exerts on the bottom of the block is
F P A bot bot = = (1 0297 ×105 Pa)(0 100 m)2 = 1029 7 . . . N
(b) The scale reading equals the tension, T, in the cord supporting the block. Since the block is
in equilibrium, ΣF T F F mg y= + − − = bot top 0, or
T = (10.0 kg)(9.80 m s2 )− (1029.7 − 1017.9) N = 86.2 N
(c) From Archimedes’s principle, the buoyant force on the block equals the weight of the
displaced water. Thus,
B = ( ρV )g
water block
= ( ) ( )
103 kg m3 0 100 m 2 0 120 . . m m s N 2 ( ) ⎡⎣
⎤⎦
(9.80 ) = 11.8
From part (a), F F bot top − = (1 029.7 −1 017.9) N = 11.8 N, which is the same as the buoyant
force found above.

9.39 Constant velocity means that the submersible is in equilibrium under the gravitational force, the
upward buoyant force, and the upward resistance force:
ΣF ma y y = =0
−(1.20 ×104 kg + ) + +1100 N = 0 sea water m g ρ gV
where m is the mass of the added sea water and V is the sphere’s volume.
Thus,
π 100
3 3 1 . .
m = × ⎛⎝
⎞⎠
⎡ ( )
⎛⎝
⎞⎠
1 03 10 +
⎣ ⎢
⎤
⎦ ⎥
4
3
1 50
kg
m
m 3
1 20 104 N
9.80 m s
kg 2− . ×
or
m = 2.67 × 103 kg
9.40 At equilibrium, ΣF B F mg y= − − = spring 0, so the spring force is
F B mg V mg spring water block = − = ( )− ⎡⎣
⎤⎦
ρ
where
V
m
block
5 00
= = = ×10− m
ρ
wood
kg 3 3
3
650 kg m
7 69
.
.
Thus, Fspring
3 3 3 kg m 10 m kg = ( ) × ( )− ⎡⎣
103 7.69 − 5.00 ⎤⎦(9.80 m s2 ) = 26.4 N.
The elongation of the spring is then
Δx
F
k
= = = = spring N
160 N m
m cm
26 4
0 165 16 5
.
. .
9.41 (a) The buoyant force is the difference between the weight in air and the apparent weight when
immersed in the alcohol, or B = 300 N − 200 N = 100 N. But, from Archimedes’s principle,
this is also the weight of the displaced alcohol, so B = (ρ V )g alcohol . Since the sample is fully
submerged, the volume of the displaced alcohol is the same as the volume of the sample.
This volume is
V
B
N
100
= =( =
ρg
700 kg m 3 )( 9 80
m s
2
) alcohol
1 46
.
. ×10−2 m3
(b) The mass of the sample is
m
= weight in air
= 300
= g
30 6
N
9.80 m s
kg 2 .
and its density is
ρ= =
. 3
×
= × −
m
V
30 6
1 46 10
2 10 10 2
.
.
kg
m
kg m 3
3

480 Chapter 9
9.42 The difference between the weight in air and the apparent weight when immersed is the buoyant
force exerted on the object by the fl uid.
(a) The mass of the object is
m
= weight in air
= 300
= g
30 6
N
9.80 m s
kg 2 .
The buoyant force when immersed in water is the weight of a volume of water equal to the
volume of the object, or Bw = (ρwV )g. Thus, the volume of the object is
V
= = −
B
w
g
w
( )( ) =
ρ
300 265
9 80
3 5
N N
103 kg m3 . m s2
. 7 × 10−3 m3
and its density is
ρobject 3
. kg
3
= = kg m
m
×
= × −
m
V
30 6
3 57 10
8 57 10 3
.
. 3
(b) The buoyant force when immersed in oil is equal to the weight of a volume
V = 3.57 × 10−3 m3 of oil. Hence, B V g oil oil = (ρ ) , or the density of the oil is
ρoil
= = −
oil
N N
m m
( × − 3
)
B
Vg
300 275
3 ( ) = 714
3.57 10 3 9.80 s
kg m 2
9.43 The volume of the iron block is
V
m = =
.
kg
2 . 54 10 4 m3
2 00
iron = × −
×
iron
ρ 7.86 10 kg m
3 3
and the buoyant force exerted on the iron by the oil is
B = (ρ V )g = ( )( × − ) oil
916 kg m3 2.54 10 4 m3 (9.80 m s2 ) = 2.28 N
Applying ΣFy = 0 to the iron block gives the support force exerted by the upper scale (and hence
the reading on that scale) as
F m g B upper iron = − = 19.6 N − 2.28 N = 17.3 N
From Newton’s third law, the iron exerts force B downward on the oil (and hence the beaker).
Applying ΣFy = 0 to the system consisting of the beaker and the oil gives
F B m m g lower oil beaker − − ( + ) = 0
The support force exerted by the lower scale (and the lower scale reading) is then
F B m m g lower oil beaker N = + + ( ) = + + ( ) 2 28 2 00 1 00 . . . kg m s N 2 ⎡⎣
⎤⎦
(9.80 ) = 31.7
9.44 (a) The cross-sectional area of the hose is A = π r2 = π d2 = π ( )2 4 2.74 cm 4, and the volume
fl ow rate (volume per unit time) is Av = 25 0 . L 1.50 min. Thus,
25 0 25 0 4
L 1.50 min L
A 1.50 min π 74
v= =
⎛
⎝ ⎜
⎞
⎠ ⎟
⋅
2
. .
.
⎛ 3
1 10
3
⎝ ⎜
2 1
( )
⎡
⎣ ⎢⎢
⎤
⎦ ⎥⎥
⎛
⎝ ⎜
⎞
⎠ ⎟
cm
min
60 s
cm
2 L
⎞
⎠ ⎟
= ( )⎛⎝ ⎜
⎞⎠ ⎟
1
m
47 1 =
. cm s 0.471
m s 2
10 cm

(b)
A
A
π d
= ⎛
d
d
d
2
1
2
2
2
1
2
1
2
4
4 1
3
=
⎛
⎝ ⎜
⎞
⎠ ⎟
⎛
⎝ ⎜
⎞
⎠ ⎟
=
⎛
⎝ ⎜
⎞
⎠ ⎟
π ⎝
⎞⎠
2 1
=
9
or A
A
2
1
9
=
Then from the equation of continuity, A A 2 2 1 1 v = v , we fi nd
= ( ) = A
A
= 1
9 0 471 4 24
1 v v 2
2
⎛
⎝ ⎜
⎞
⎠ ⎟
. m s . m s
9.45 (a) The volume fl ow rate is Av, and the mass fl ow rate is
ρ Av = (1.0 g cm3 )(2.0 cm2 )(40 cm s) = 80 g s
(b) From the equation of continuity, the speed in the capillaries is
=
aorta
aorta v v capiliaries
capillaries
⎛
⎝ ⎜
⎞
⎠ ⎟
A
A
=
2 0 ( )
3 0 10
×
⎛
⎝ ⎜
⎞
40 3
⎠ ⎟
.
.
cm
cm
cm s
2
2
or vcapiliaries = 2.7 × 10−2 cm s = 0.27 mm s .
9.46 (a) From the equation of continuity, the fl ow speed in the second pipe is
1
⎛
v v 2
2
1
1
2
= 2
⎝ ⎜
⎞
⎠ ⎟
=
⎛
⎝ ⎜
⎞
⎠ ⎟
A
A
0.0 cm
.50 cm
.75
2
2 ( m s) = 11.0 m s
(b) Using Bernoulli’s equation and choosing y = 0 along the centerline of the pipes gives
1 65 10 = + − ( ) = × + × ρ v v . . Pa kg m m s m s 3 ( ) ( ) − ( ) ⎡⎣
1 2
2 5 3
2
P P 2 1 1
2
1 20 10
1
2
⎤⎦
2 75 11 0 2 2 . .
or P2
= 2.64 × 104 Pa .
9.47 From Bernoulli’s equation, choosing y = 0 at the level of the syringe and needle,
+ ρ v 2
= + ρ v2, so the fl ow speed in the needle is
P P 2
12
2
1
12
1
2 1 2 2
= +
v v 2 1
(P − P )
ρ
In this situation,
F
A 1 2 1 1
P P P P P
1
2 00
. N
2 50 10
− = − =( ) = =
atmo gauge × −
. 5
8 00 104
m
Pa 2= . ×
Thus, assuming v1 ≈ 0,
v2
( × 4
)
×
2 8 00 10
1 00 10
= + 12 6
3 0
=
.
.
.
Pa
kg m
m s 3
9.48 We apply Bernoulli’s equation, ignoring the very small change in vertical position, to obtain
2 2 − = − ( ) = ( ) − ⎡⎣
P P 1 2
12
2
2
2 12
1
1
2
2 3
1
2 1
⎤⎦
ρ v v ρ v v = ρ v , or
ΔP = ( )( × − ) = × − 3
2
1.29 kg m3 15 10 2 m s 2 4.4 10 2 Pa

482 Chapter 9
9.49 (a) Assuming the airplane is in level fl ight, the net lift (the difference in the upward and
downward forces exerted on the wings by the air fl owing over them) must equal the weight
of the plane, or (P P )A mg lower
surface
wings − = . This yields
upper
surface
mg
A lower
surface
P P
upper
surface wings
−
⎛
⎝ ⎜
⎞
⎠ ⎟
= =
( 8 66 × 10 )( 9 80
)= ×
90 0
9 43 10
4
3 . .
.
.
kg m s
m
Pa
2
2
(b) Neglecting the small difference in altitude between the upper and lower surfaces of the
wings and applying Bernoulli’s equation yields
P P lower air lower upper air upper + = + 1
2
1
2
ρ v2 ρ v2
so
( − ) 2 = ( 2
= + m s
v v upper lower
lower upper
air
225
P P
( × ) = 2
ρ ) +
2 9 43 103
1 29
255
.
.
Pa
kg m
m s 3
(c) The density of air decreases with increasing height , resulting in a smaller pressure
difference. Beyond the maximum operational altitude, the pressure difference can no longer
support the aircraft.
9.50 For level fl ight, the net lift (difference between the upward and downward forces exerted on
the wing surfaces by air fl owing over them) must equal the weight of the aircraft, or
(P P ) A Mg lower
surface
− = . This gives the air pressure at the upper surface as
upper
surface
= −
Mg
upper
surface
A P P
lower
surface
9.51 (a) Since the pistol is fi red horizontally, the emerging water stream has initial velocity
components of (v v , v ) 0 0 0 x y = = nozzle . Then, Δy t at y y = v + 0
2 2, with a g y = − , gives the time
of fl ight as
t
= ( ) = (− )
y
ay
−
=
2 2 150
9 80
0 553
Δ .
.
.
m
m s
s 2
(b) With ax x = 0 = 0 and nozzle v v , the horizontal range of the emergent stream is Δx = v t nozzle
where t is the time of fl ight from above. Thus, the speed of the water emerging from the
nozzle is
vnozzle
= Δx
= m
= m s 0.553 s
t
8 00
14 5
.
.
(c) From the equation of continuity, A A 1 1 2 2 v = v , the speed of the water in the larger cylinder is
v v v 1 2 1 2 2 1 = (A A ) = (A A ) nozzle, or
2
2
v v v 1
2
1
2
1
2
=
⎛
⎝ ⎜
⎞
⎠ ⎟
=
⎛
⎝ ⎜
⎞
⎠ ⎟
π
π
r
r
r
r nozzle nozzle =⎛⎝ ⎜
⎞⎠ ⎟
( ) = 1 00
14 5 0 145
2 .
. .
mm
10.0 mm
m s m s
(d) The pressure at the nozzle is atmospheric pressure, or P2
= 1.013 × 105 Pa .

(e) With the two cylinders horizontal, y y 1 2 = and gravity terms from Bernoulli’s equation
+ ρ v 2
2 = + ρ v2 2 so the needed pressure in the larger
can be neglected , leaving P1 w 1 P w
2 2
cylinder is
= + ρ ( 2
− ) = × + 1 00 ×
10 P P w
1 2 2
2 5
v v .
1
3
2
1 013 10
.
Pa
kg m
m s m s
3
2
14 5 0 145 2 2 . . ( ) − ( ) ⎡⎣
⎤⎦
or
P1
= 2.06 × 105 Pa
(f ) To create an overpressure of ΔP = 2.06 × 105 Pa − 1.013 × 105 Pa = 1.05 × 105 Pa in the
larger cylinder, the force that must be exerted on the piston is
= (Δ ) = (Δ )(π 2 ) = (1.05 × 105 Pa)π 1.00 × 10−2 2 ( m) = 33.0 N
F PA P r 1 1 1
9.52 (a) From Bernoulli’s equation,
2
2
+ ρ v 1
+ ρ = + ρ v
+ ρ P w gy P gy
w
w
1 w
1 2
2
2 2 2
or
2
2 1 − = 2 ⎡ − − ( − )
v v 2
2 1 2
1
⎣ ⎢
⎤
⎦ ⎥
P P
g y y
w ρ
and using the given data values, we obtain
2
Point
⎡ − ( 2 )( )
− = × − ×
v v 2
2
1
5 5
1 . 75 10 1 .
20 10
3 2
×
1 .
00 10
Pa Pa
kg m
m s m 3
⎣ ⎢
⎤
⎦ ⎥
9.80 2.50
and
v 2
v 2
− 2 = 61.0 m2 s2 [1]
1
From the equation of continuity,
1
2
v v v 2
2
1
1
2 1
2
1
2
=
⎛
⎝ ⎜
⎞
⎠ ⎟
=
⎛
⎝ ⎜
⎞
⎠ ⎟=
⎛
⎝ ⎜
⎞
⎠ ⎟
A
A
r
r
r
r
π
π
2
1 v
2
v v 2
1
3 00
1 50
=⎛⎝ ⎜
⎞⎠ ⎟.
.
cm
cm
or v v 2 1 = 4 [2]
( − )v2 = . m2 s2, or
Substituting Equation [2] into [1] gives 16 1 61 0 1
v1
= 61 .
0
= 2 02 15
.
m s
m s
2 2
(b) Equation [2] above now yields v2 = 4(2.02 m s) = 8.08 m s .
(c) The volume fl ow rate through the pipe is: flow rate = A = A 1 1 2 2 v v.
Looking at the lower point:
flow rate = (π r 2
) = π ( × − 2 m) 2
( m s) 1
1
v 3.00 10 2.02 = 5.71×10−3 m3 s
Point
Δx2
Δx1
y1
y2
→v
2
→v
1
P1A1
P2A2
1
2
Figure 9.29

484 Chapter 9
9.53 First, consider the path from the viewpoint of projectile motion to fi nd the speed at which the
water emerges from the tank. From Δy = v yt + ayt 0
12
2 with v0 0 y = , we fi nd the time of fl ight as
t
= ( ) = (− )
y
ay
−
=
2 2 100
9 80
0 452
Δ .
.
.
m
m s
s 2
From the horizontal motion, the speed of the water coming out of the hole is
v v 2 0
Δ .
0 600
x
t
= = = = 1 .
33 x
m
0.452 s
m s
We now use Bernoulli’s equation, with point 1 at the top of the tank and point 2 at the level of the
hole. With P P P 1 2 = = atmo and v1 ≈ 0, this gives
ρ gy ρ 2
ρ gy 1 2
2
1
2
= v +
or
= − = = (
h y y
2 2
2
g
)
2
× 1 2
( )= 1 33
2 9 80
9 00 10
v .
.
.
m s
m s2
−2 m = 9.00 cm
9.54 (a) Apply Bernoulli’s equation with point 1 at the open top of the tank and point 2 at the
opening of the hole. Then, P P P 1 2 = = atmo and we assume v1 ≈ 0. This gives
12
ρ v 2
+ ρ gy = ρ gy , or
2
2 1 v2 1 2 = 2 g(y − y ) = 2(9.80 m s2 )(16.0 m) = 17.7 m s
(b) The area of the hole is found from
A
flow rate
× − = = m min
1 2
2
2 50 10 3
17 7
m s
m2 ⎛⎝ ⎜
min
6
3
v
.
. 0 s
⎞⎠ ⎟= × − 2 35 10 . 6
The diameter is then
d
A
2
2
( × −
6
)= × =
4 4 2 35 10 3
−
.
m
= = 1 . 73 10 1 .
73
π π
m m
2
m

9.55 First, determine the fl ow speed inside the larger portions from
v1
= = ×
1
−
4
. 3
2 2
1 80 10
2 50 10
( × −
)
flow rate
A
m s
m
π . 4
= 0.367 m s
The absolute pressure inside the large section on the left is P P gh 1 0 1 = +ρ , where h1 is the height of
the water in the leftmost standpipe. The absolute pressure in the constriction is P P gh 2 0 2 = +ρ , so
P P gh h g 1 2 1 2 − = ρ ( − ) = ρ (5.00 cm)
The fl ow speed inside the constriction is found from Bernoulli’s equation with y y 1 2 = . This gives
2
2
v v v 2
1
2
1 2 1
1 2
2
= + ( − ) = + 2 ( − ) ρ
P P g h h , or
v2
= (0.367 m s)2 + 2(9.80 m s)(5.00 ×10−2 m) = 1.06 m s
The cross-sectional area of the constriction is then
A
= flow rate
= × = 1 71 ×
10 2
2
1 80 10 4
1 06
− m 3
s
m s
v
.
.
. −4 m2,
and the diameter is
d
A
2
2
( × −
4
)= × =
4 4 1 71 10 2
−
.
m
= = 1 . 47 10 1 .
47
π π
m c
2
m
9.56 (a) For minimum pressure, we assume the fl ow is very slow. Then, Bernoulli’s equation gives
P gy P gy + + ⎛⎝ ⎜
⎞⎠ ⎟
+ + ⎛⎝ ⎜
⎞⎠ ⎟
1
2
1
2
ρ v2 ρ =
ρ v2 ρ
river rim
P gy y river min rim river ( ) + 0 = 1 atm + 0 + ρ ( − )
or
kg
m ( ) = × + ⎛⎝ ⎜
Priver min 3 Pa
⎞⎠ ⎟
1.013 105 103 9.80
m
s
m m 2
⎛⎝ ⎜
⎞⎠ ⎟
(2096 − 564 )
Priver min ( ) = (1.013 ×105 +1.50 ×107 ) Pa = 1.51×107 Pa = 15.1 MPa
(b) The volume fl ow rate is flow rate = Av =(π d2 4)v. Thus, the velocity in the pipe is
v = ( ) = ( )
d 3
4 4 4500 1
( )
flow rate
π d
2 π 2
86 400 s
m d
0.150 m
m s
⎛
⎝ ⎜
⎞
⎠ ⎟
= 2.95

486 Chapter 9
(c) We imagine the pressure being applied to stationary water at river level, so Bernoulli’s
equation becomes
P g y y river rim river rim atm + = + − ( ) ⎡⎣
⎤⎦
0 1 +
1
2
ρ ρ v2
or
P P P river river min rim river min = ( ) + =( ) + 1
2
1
2
ρ v2 10 2 95
4
3
2
kg
m
m
s
river min
3
⎛⎝ ⎜
⎞⎠ ⎟
⎛⎝ ⎜
⎞⎠ ⎟
= ( ) +
.
P .35 kPa
The additional pressure required to achieve the desired fl ow rate is
ΔP = 4.35 kPa
9.57 (a) For upward fl ight of a water-drop projectile from geyser vent to fountain-top,
= 2 + 2 (Δ ), with vy = 0 when Δy = Δymax, gives
v0 0 2 2 9 8 40 0 28 0 y y= − a (Δy) = − (− )( ) = max . m s2 . m . m s
v v y y y 2 a y
0
(b) Because of the low density of air and the small change in altitude, atmospheric pressure at
the fountain top will be considered equal to that at the geyser vent. Bernoulli’s equation,
with vtop = 0, then gives
1
2
ρ v2 0 ρ vent top vent = + g(y − y )
or
= 2 g(y − y ) = 2(9.80 m s2 )(40.0 m) = 28.0 m s
vvent top vent
(c) Between the chamber and the geyser vent, Bernoulli’s equation with vchamber ≈ 0 yields
1
2
(P + 0 + gy) = P + + gy
ρ ρ 2 ρ chamber atm vent vent v
or
⎡ 1
P − P = ρ + g(y − y )
atm vent vent chamber ⎣ ⎢
⎤
⎦ ⎥
=
2
10
v2
3
⎡
28 0
2 . ( m) MPa 2
9 80 175
kg
m
m s
m
s
3 2
⎛⎝ ⎜
⎞⎠ ⎟
( ) +⎛⎝ ⎜
⎞⎠ ⎟
.
⎣ ⎢⎢
⎤
⎦ ⎥⎥
= 2.11
or
P P P gauge atmo
= − = (2.11×106 Pa)(1 atm 1.013 ×105 Pa) = 20.8 atmospheres

9.58 (a) Since the tube is horizontal, y y 1 2 = and the gravity
terms in Bernoulli’s equation cancel, leaving
1 2
2
+ ρ v 2
= + ρ v
P P 1 1
1
2
2 2
or
2
v v 2
2 1 2
1
3 2 2 1 20 10
10
− =
( − ) =
( × )
×
P P
ρ
. Pa
7.00 2 kg m3
and
2
− 2 = 3.43 m2 s2 [1]
v v 2
1
From the continuity equation, A A 1 1 2 2 v = v , we fi nd
A
1
= A
v v v 2
2
1
1
2
2
1
2 40
1 20
=
⎛
⎝ ⎜
⎞
⎠ ⎟
=
⎛
⎝ ⎜
⎞
⎠ ⎟
r
r
.
.
cm
cm
⎛⎝ ⎜
⎞⎠ ⎟
2
1 v
or
v v 2 1 = 4 [2]
v2 = . m2 s2 and v1 = 0.478 m s.
Substituting Equation [2] into [1] yields 15 3 43 1
Then, Equation [2] gives v2 = 4(0.478 m s) = 1.91 m s .
(b) The volume fl ow rate is
v = v = (π 2
)v = π (1.20 ×10− m) (1.91 m s) = 8.64 ×10−4 m3 s
A A r 1 1 2 2 2
2
2 2
= + where Fy is the
9.59 From ΣF T mg F y y = − − =0, the balance reading is found to be T mg Fy
vertical component of the surface tension force. Since this is a two-sided surface, the surface
tension force is F = γ (2 L) and its vertical component is F L y = γ (2 )cosφ where φ is the contact
angle. Thus, T = mg + 2γ Lcosφ .
T = 0.40 N when φ = 0° ⇒ mg + 2γ L= 0.40 N [1]
T = 0.39 N when φ = 180° ⇒ mg − 2γ L= 0.39 N [2]
Subtracting Equation [2] from [1] gives
γ = − = −
0 . 40 0 . 39 0 . 40 0 .
39
( × −
3 .
0 10 2
N N
4
N N
L 4 m)= 8.3 × 10−2 N m
9.60 Because there are two edges (the inside and outside of the ring), we have
γ
F
L
= =
= =
π
1.61 1
total
F
circumference
F
r
2
4
( )
−
−
× 0 N
4 1.75 10 m
N m
2
2
−
( × )= ×
π
7.32 10 2
P2
P1
A2
A1
→v
1 1
2 →v
2
Figure 9.30(a)

488 Chapter 9
9.61 From h = 2γ cosφ ρ gr, the surface tension is
γ ρ
h gr
2
φ
=
=
cos
( × − )( )
. m kg m3 ( . m s2 )( × )
2 1 10 2 1 080 9 80
°
= ×
−
− 5 0 10
5 6 10
4
2 .
.
m
2 cos0
N m
9.62 The height the blood can rise is given by
h
= = ( ) °
γ φ
ρ
2 cos 2 0 . 058 cos
0
gr
( )
m ( 2 )( × ) = 2 0 10−
1 050 9 .
80
N m
5 6 . 6
kg m2 m s m
.
9.63 From the defi nition of the coeffi cient of viscosity, η = F L Av, the required force is
F
A
L
= =
× ⋅ ( ) ( )( ) ⎡⎣
− η v 1 79 10 0 800 1 20 3 . . . N s m m m 2 ⎤⎦
( )
×
0 50
= −
8 6
N 3
.
.
m s
0.10 10 m
9.64 From the defi nition of the coeffi cient of viscosity, η = F L Av, the required force is
F
( × − ⋅ )⎡( )( ) η v 1 500 10 0 010 0 040 3 N s m m m 2 . . ⎣ ⎤⎦
A
L
= =
( )
×
0 30
= −
5
0 12
N 3
.
.
.
m s
1 10 m
9.65 Poiseuille’s law gives flow rate
π
η
P P R
L
=
( − ) 1 2
4
8
and P P 2 = atm in this case. Thus, the desired gauge pressure is
− = ( ) =
P P
8 L flow rate
8 0 12 50
R 1 4
( ⋅ )
atm
2 η N s m m
π
. ( )( .
× )
( × )
−
−
8 6 10
0 50 10
5
2 4
.
m s
m
3
π
or
− = 2 1×106 = 2 1 atm . Pa . MPa
P P 1
9.66 From Poiseuille’s law, the fl ow rate in the artery is
flow rate
Δ P π
R
( ) ( × − ) η
L
4
4 Pa π 3 m
= ( ) =
8
400 2.6 10
( × ⋅ )( × )= × − −
3 2 10 3 2
m s 2
8 2 7 10 8 4 10
5
. .
.
−
N s m m
3
Thus, the fl ow speed is
v= = ×
−
3
3 2 10
( × ) =
−
flow rate
A
m s
3
2.6 10 m
1 5
5
2
.
.
π
m s

9.67 If a particle is still in suspension after 1 hour, its terminal velocity must be less than
vt ( ) =⎛⎝ ⎜
⎞⎠ ⎟
⎛
⎝ ⎜
⎞
⎠ ⎟
max
cm
h
h
3 600 s
⎜
⎛⎝ m
m s 100
5 0
1 1
.
cm
⎞⎠ ⎟
= 1.4 ×10−5 .
Thus, from vt f = 2 r2 g(ρ − ρ ) 9η, we fi nd the maximum radius of the particle:
r
η
ρ ρ
9
2
g
v
t
f
max
max
N s m2
= ( )
( − )
=
( × − ⋅ )
( × 5 )
9 1 00 10 3 1
. .4 10
( ) − ( ) ⎡⎣
2 9 80 1 800 1 000
− m s
⎤⎦
. m s2 kg m3
= 2.8 ×10−6 m = 2.8 μm
9.68 From Poiseuille’s law, the excess pressure required to produce a given volume fl ow rate of fl uid
with viscosity η through a tube of radius R and length L is
Δ
8 ( Δ Δ
)
= P
L V t
R
4
η
π
If the mass fl ow rate is (Δm Δt ) = 1.0 × 10−3 kg s, the volume fl ow rate of the water is
Δ
Δ
V Δ Δ
t
m t = = ×
×
= ×
−
−
ρ
1 0
1 0 10
3
1 0 10 3
.
.
.
10 kg s
kg m
3
6 m3 s
and the required excess pressure is
ΔP =
8(1.0 × 10−3 Pa ⋅ s)(3.0 × 10−2 m)(1.0 × 10−6 m3 s)
( × ) = ×
π 0 15 10−
1 5 10
3 4
5
.
.
m
Pa
9.69 With the IV bag elevated 1.0 m above the needle, the pressure difference across the needle is
ΔP = ρgh = (1.0 × 103 kg m3 )(9.8 m s2 )(1.0 m) = 9.8 × 103 Pa
and the desired fl ow rate is
Δ
Δ
V
t
= ( )
cm m 6 cm
min 60 s 1 min
3 3 3
500 1 10
30
( ) =
2.8 × 10−7 m3 s
Poiseuille’s law then gives the required diameter of the needle as
= = ( )
D R
( )( × )( × 3 )
η
1 4 L V t
8 1 0 10 3 π
( P
)
⎡
⎣ ⎢
⎤
⎦ ⎥
=
× − ⋅
2 2
8
2
Δ Δ
Δ
2 5 10− 2 8 10−
9 8 10
2 7
. Pas . m .
m s
( × 3
Pa
)
π .
⎡
⎣ ⎢⎢
1 4
⎤
⎦ ⎥⎥
or
D = 4.1 × 10−4 m = 0.41 mm

490 Chapter 9
9.70 We write Bernoulli’s equation as
P gy P gy out out out in in in + + = + + 1
2
1
2
ρ v2 ρ ρ v2 ρ
or
P = P − P = ρ ⎡ 1
( v2 − v2
)+ gy ( − y ) gauge in out out in out in 2
⎣ ⎢
⎤
⎦ ⎥
Approximating the speed of the fl uid inside the tank as vin ≈ 0, we fi nd
Pgauge
= (1 00 ×10 kg m3 ) ( m s) + m s2
3 30 0 2 9 80 . ⎡ . ( . )( )
1
2
⎣ ⎢
⎤
⎦ ⎥
0.500 m
or
Pgauge = 4.55 ×105 Pa = 455 kPa
9.71 The Reynolds number is
RN
ρ
v d =( 1 050 kg m3 )( 0 . 55 m s )( 2 . 0 × 10
− 2 m
) = η
. × ⋅
2 7
= 10−3 N s m2 4.3 × 103
In this region (RN 3 000), the fl ow is turbulent .
9.72 From the defi nition of the Reynolds number, the maximum fl ow speed for streamlined
(or laminar) fl ow in this pipe is
vmax
η
( × − N ⋅ s m2
)( ) max
=
⋅( ) =
RN
d
ρ
. 3
1 0 10 2 000
1 000
( )( )= 0 .080 m s = 80 .
cm s kg m 3 2 . 5 × 10−
2
m
9.73 The observed diffusion rate is 8.0 ×10−14 kg 15 s = 5.3 ×10−15 kg s. Then, from Fick’s law, the
difference in concentration levels is found to be
− =( )
C C
diffusion rate L
DA 2 1
( )
5 3 10 15 0
=
( × − )
. kg s .10
( × )( × )= × − −
1 8 10 10 4
5 0 10 6 0 10
m
m2 s m2
−
. .
. 3 kg m3
9.74 Fick’s law gives the diffusion coeffi cient as D = diffusion rate A⋅(ΔC L), where ΔC L is the
concentration gradient.
Thus,
−
5 .
7 ×
10
15
D = ( × )⋅( ×
− −
4 2
2 . 0 10 3 .
0 10
kg s
m2 kg m4 )= 9.5 × 10−10 m2 s
9.75 Stokes’s law gives the viscosity of the air as
η
= = ×
π π
−
13
.
N s m2 ( )= 1.4 ×10−5 ⋅
( × ) ×
− −
F
6 r
3 0 10
v 6 4
6 25 . 10 45 .
10
N
m m s
→v
out
0.500 m

9.76 Using vt =2 r2 g(ρ − ρ f ) 9η, the density of the droplet is found to be ρ = ρ + η f t 9 v 2 r2 g.
Thus, if r = d 2 = 0.500 × 10−3 m and vt= 1.10 × 10−2 m s when falling through 20 °C water
(η = 1.00 × 10−3 N⋅ s m2 ), the density of the oil is
ρ= +
( × − ⋅ )( × −
1 000
9 1 00 10 3 1 10 10 2 kg
m
N s m m s
3
. 2 . )
( × ) ( ) = × 2 5 00 10− 9 80
1 02 10
4 2
3
. .
.
m ms
kg m
2
3
9.77 (a) Both iron and aluminum are denser than water, so both blocks will be fully submerged.
Since the two blocks have the same volume, they displace equal amounts of water and
the buoyant forces acting on the two blocks are equal.
(b) Since the block is held in equilibrium, the force diagram at the
right shows that
ΣF T mg B y= 0 ⇒ = −

is the same for the two blocks, so the spring scale
The buoyant force B

is largest for the iron block , which has a higher density, and
reading T
hence weight, than the aluminum block.
(c) The buoyant force in each case is
B = (ρ V )g = ( × )( ) water
1.0 103 kg m3 0.20 m3 (9.8 m s2 ) = 2.0 × 103 N
For the iron block:
= (ρ ) − = (7.86 × 103 kg m3 )(0.20 m3 )(9.8 m s2 )− B
T Vg B iron iron
or
Tiron = 1.5 × 104 N − 2.0 × 103 N = 13 × 103 N
For the aluminum block:
= (ρ ) − = (2.70 × 103 kg m3 )(0.20 m3 )(9.8 m s2 )− B
T Vg B aluminum aluminum
or
Taluminum = 5.2 × 103 N − 2.0 × 103 N = 3.3 × 103 N
9.78 In going from the ocean surface to a depth of 2.40 km, the increase in pressure is
ΔP = P − P = gh = ( × )( ) × 0
T
ρ 1.025 103 kg m2 9.80 m s2 (2.40 103 m) = 2.41 × 107 Pa
The fractional change in volume of the steel ball is given by the defi ning equation for bulk
modulus, ΔP = −B(ΔV V ), as
ΔV Δ
V
= − = − ×
P
B
×
= −
steel
Pa
Pa
2 41 10
16 0 10
1 5
7
10
.
.
. 1×10−4
Fg mg
B

492 Chapter 9
9.79 (a) From Archimedes’s principle, the granite
continent will sink down into the peridotite
layer until the weight of the displaced
peridotite equals the weight of the continent.
Thus, at equilibrium,
ρ ρ g p At g Ad g ( ) ⎡⎣
⎤⎦
= ( ) ⎡⎣
⎤⎦
or ρ ρ g p t = d .
Area A
d
Peridotite
Continent
(granite)
(b) If the continent sinks 5.0 km below the surface of the peridotite, then d = 5 0. km, and the
result of part (a) gives the fi rst approximation of the thickness of the continent as
⎛
t = d
⎝ ⎜
⎞
⎠ ⎟
= ×
.
. ⎟ ( ) = 5.0 km 5.9 km
×
⎛
⎝ ⎜
⎞
⎠
ρ
ρ
p
g
3
3
kg m
kg m
3 3 10
2 8 10
3
3
9.80 (a) Starting with P = P + gh 0 ρ , we choose the reference level at the level of the heart, so
P P0 H = . The pressure at the feet, a depth hH below the reference level in the pool of blood
in the body is P P gh F H H = +ρ . The pressure difference between feet and heart is then
P P gh F H H − =ρ .
(b) Using the result of part (a),
P P F H − = (1.06 × 103 kg m3 )(9.80 m s2 )(1.20 m) = 1.25 × 104 Pa
= π 2 4 and that of a single capillary is A d c = π 2
9.81 The cross-sectional area of the aorta is A d 1 1
2 4.
If the circulatory system has N such capillaries, the total cross-sectional area carrying blood from
the aorta is
A NA
2
4
N d
2 c
2
= = π
From the equation of continuity,
=
1
1 A A 2
2
⎛
⎝ ⎜
⎞
⎠ ⎟
v
v
, or
⎛
⎞
Nπ d 2
π d 2
v
v
1
2
2
1
=
⎝ ⎜
⎠ ⎟
4 4
,
which gives
N
d
d
=
⎛
⎝ ⎜
⎞
⎠ ⎟
⎛
⎝ ⎜
⎞
⎠ ⎟
=
⎛
.
⎝ ×
−
v
v
1
2
1
2
2
2
1 0
1 .
0 10
m s
⎜ m s
⎞
⎠ ⎟
×
×
⎛
⎝ ⎜
⎞
⎠ ⎟
= ×
−
−
0 50 10
10 10
2 5 10
2
6
2
. 7
.
m
m
9.82 (a) We imagine that a superhero is capable of producing a perfect vacuum above the water
in the straw. Then P = P + gh 0 ρ , with the reference level at the water surface inside the
straw and P being atmospheric pressure on the water in the cup outside the straw, gives the
maximum height of the water in the straw as
h
= P
− atm
= = . × 0 1 013 105 N m
ρ g
ρ
P
g max
water
atm
water
2
10 3 . 3
kg m3 9.80 m s2
m
1 00 10
.
( × )( ) =
(b) The moon has no atmosphere so Patm = 0, which yields hmax = 0 .
t
g
p

9.83 (a) P = 160 mm of H O = g(160 mm) 2 H2O ρ
=⎛⎝
103 kg 9 80 (0 160 ) =
⎞⎠
⎛⎝
⎞⎠
m
m
s
m 3 2 . . 1.57 kPa
3 1 . Pa
⎛⎝
atm
P = ( × ) ×
⎞⎠
1.013 105 Pa 1.55 × 10−2 atm
1 57 10 =
The pressure is P = ρ gh = ρ gh , so
H2O H2O Hg Hg
H O
h h Hg
Hg
H O
3
3
2
2
kg m
kg m
=
⎛
⎝ ⎜
⎞
⎠ ⎟
=
3
×
ρ
ρ
10
. 3
13 6 10
⎛
⎝ ⎜
⎞
⎠ ⎟
(160 mm) = 11.8 mm of Hg
(b) The fl uid level in the tap should rise.
(c) Blockage of fl ow of the cerebrospinal fl uid.
9.84 When the rod fl oats, the weight of the displaced fl uid equals the weight of the rod, or
ρ ρ f gV gV displaced rod = 0 . But, assuming a cylindrical rod, V rL rod = π 2 . The volume of fl uid
displaced is the same as the volume of the rod that is submerged, or V r L h displaced= π 2 ( − ).
Thus, ρ π ρ π fg r2 L h g r L
( − ) 2 0
⎡⎣
⎤⎦
= ⎡⎣
⎤⎦
, which reduces to
L
L h
ρ =
ρ f
−
⎛⎝
⎞⎠
0
9.85 Consider the diagram and apply Bernoulli’s
equation to points A and B, taking y = 0 at the
level of point B, and recognizing that vA ≈ 0.
This gives
+ + ( − )
ρ sinθ
P gh L
= P
+
A w
B
0
1
2
ρw B v2 + 0
Recognize that P P P A B = = atm since both points
are open to the atmosphere. Thus, we obtain
A
h
vB g h L = − ( ) = ( ) − ( ) 2 2 9 80 10 0 2 00 sin . . . si θ m s m m 2 n . . 30 0 13 3 º m s ⎡⎣
⎤⎦
=
Now the problem reduces to one of projectile motion with
v v 0 30 0 6 64 y B = sin . º = . m s
At the top of the arc, vy = 0, and y = ymax.
Then, v v y y y 2 a y
= 2 + 2 (Δ ) gives 0 = ( 6 . 64 m s) 2 + 2 (− 9 . 80 m s2 )( − 0 )
0
max y , or
ymax = 2.25 m above the level of point B .
L B
Valve

494 Chapter 9
9.86 When the balloon comes into equilibrium, the weight of the displaced air equals the weight of the
s
fi lled mballoon plus the weight of string that is above ground level. If mand L are the total mass
s and length of the string, the mass of string that is above ground level is ( h L ) . Thus,
= + + ⎛
h
L
⎛⎝ ⎜
⎞⎠ ⎟
ρ ρ air balloon balloon helium balloon gV m g gV
m g s
which reduces to
h
V m
m
L
s
=
⎡( − ) −
⎣ ⎢
⎤
⎦ ⎥
ρ ρ air helium balloon balloon
This yields
h =
− ( ) ( ) ⎡⎣
⎤⎦
1.29 kg m3 0.179 kg m3 4 0.40 m 3 3 − 0. π 25
0 050
2 0 1 9
kg
kg
m m
.
( . ) = .
9.87 When the balloon fl oats, the weight of the displaced air equals the combined weights of the fi lled
balloon and its load. Thus,
ρ ρ air balloon balloon helium balloon l gV m g gV m = + +o
oad g,
or
V
= m +
m
kg balloon
balloon load
−
air helium
= +
ρ ρ
600 4 000
1 29 0 179
4 14 103 kg
kg m
( − ) = ×
m 3
3
. .
.
9.88
L
L
h
H
A B
→v
A B
C D
(b) (c)
Shield
(a) Consider the pressure at points A and B in part (b) of the fi gure by applying P P gh f = + 0 ρ .
Looking at the left tube gives atm water P P gL h A= + ρ ( − ), and looking at the tube on the right,
atm oil P P gL B= +ρ .
Pascal’s principle says that P P B A = . Therefore, P gL P gL h atm oil atm water + ρ = + ρ ( − ), giving
⎛
ρ
ρ
h = − L
⎝ ⎜
⎞
⎠ ⎟
⎛
⎝
= −
oil
1 1
water
3
3
750 kg m
⎜ 1 000 kg m
⎞
⎠ ⎟
(5.00 cm) = 1.25 cm

(b) Consider part (c) of the diagram showing the situation when the air fl ow over the left
tube equalizes the fl uid levels in the two tubes. First, apply Bernoulli’s equation to points
A and B. This gives
P gy P gy A A A B B B + + = + + 1
2
1
2
ρ 2 ρ ρ 2 ρ air air air air v v
Since y y A B A B = , v = v, and v = 0, this reduces to
P P B A − =1
2
ρ 2 air v [1]
Now use P P gh f = + 0 ρ to fi nd the pressure at points C and D, both at the level of the
oil–water interface in the right tube. From the left tube, water P P gL C A = +ρ , and from the
right tube, oil P P gL D B = +ρ .
Pascal’s principle says that P P D C = , and equating these two gives
oil water P gL P gL B A + ρ = + ρ , or water oil P P gL B A − = (ρ − ρ ) [2]
Combining Equations [1] and [2] yields
v =
( − )
=
ρ ρ
water oil
ρ
air
( − )(
2
2 1 000 750 9 80
m s2
gL
. )( 5 00 × 10−
) =
1 29
13 8
. 2
.
.
m
m s
9.89 While the ball is submerged, the buoyant force acting on it is B Vg w = (ρ ) . The upward accelera-tion
of the ball while under water is
a
Σ ρ π4
y w = = − = ⎛⎝
F
m
⎡ −
B mg
m m
⎞⎠
r g y
⎣ ⎢
⎤
⎦ ⎥ =
3
1
1 000
3
kg m
1.0 kg
0 10 1 9 80 π 3
m m
3 ( )⎛⎝
⎞⎠
( ) −
⎡
⎣ ⎢⎢
⎤
⎦ ⎥⎥
4
3
. ( . s2 ) = 31 m s2
Thus, when the ball reaches the surface, the square of its speed is
= 2 + 2 (Δ ) = 0 + 2(31 m s2 )(2.0 m) = 125 m2 s2
v v y y y 2 a y
0
When the ball leaves the water, it becomes a projectile with initial upward speed of
v0 125 y = m s and acceleration of a g y = − = −9.80 m s2 . Then, v v y y y 2 a y
= 2 + 2 (Δ ) gives the
0
maximum height above the surface as
ymax
= 0 −
125
m 2 s
2
(− ) = 6 .
4 m 2 9.80 m s
2

496 Chapter 9
9.90 Since the block is fl oating, the total buoyant force
must equal the weight of the block. Thus,
A x g A x g 4 00 . − ( ) ⎡⎣
ρ ρ oil water cm
⎤⎦
+ [ ⋅ ]
cm wood = ( ) ⎡⎣
ρ A 4.00 g
⎤⎦
where A is the surface area of the top or bottom of the
rectangular block.
Solving for the distance x gives
ρ ρ ( ) = −
ρ ρ
x = −
−
⎛
⎝ ⎜
⎞
⎠ ⎟
wood oil
water oil
4 00 cm
960
.
930
1 000 930
4.00 cm x
4 00 1 71
−
⎛
⎝ ⎜
⎞
⎠ ⎟
( . cm) = . cm
9.91 A water droplet emerging from one of the holes
becomes a projectile with v v v 0 0 0 y x = and = . The
time for this droplet to fall distance h to the fl oor is
found from Δy t at y y = v + 0
12
2 to be
t
= 2
h
g
The horizontal range is
R t
h
g
= v = v
2 .
If the two streams hit the fl oor at the same spot,
it is necessary that R R 1 2 = , or
2 h 2
1
2 g
v v 1
2
h
g
=
With h h 1 2 = 5.00 cm and = 12.0 cm, this reduces to
2
v v v 1 2
1
2
12 0
5 00
= = h
h
.
.
cm
cm
2
, or v v 1 2 = 2.40 [1]
Apply Bernoulli’s equation to points 1 (the lower hole) and 3 (the surface of the water).
The pressure is atmospheric pressure at both points and, if the tank is large in comparison to
the size of the holes, v3 ≈ 0. Thus, we obtain
P gh P gh atm atm + + = + + 1
2
ρ v 2
ρ 0 ρ , or v1
1 3 1
2
3 1 = 2 g(h − h ). [2]
Similarly, applying Bernoulli’s equation to point 2 (the upper hole) and point 3 gives
P gh P gh atm atm + + = + + 1
2
ρ v 2
ρ 0 ρ , or v2
2 3 2
2
3 2 = 2 g(h − h ). [3]
Square Equation [1] and substitute from Equations [2] and [3] to obtain
2 2 40 2 3 1 3 2 g h h g h h − ( ) = − ( ) ⎡⎣
⎤⎦
.
Solving for h3 yields
h
= − = ( ) −
cm cm
.
h h
= 3
2 1 2 40
1 40
2 40 12 0 5 00
1 40
.
. . .
.
17 0. cm,
so the surface of the water in the tank is 17.0 cm above floor level .
4.00 cm
Oil
Water
x
R1R2
1
3
h2
h1
h3

9.92 When the section of walkway moves downward distance ΔL, the cable is stretched distance ΔL
and the column is compressed distance ΔL. The tension force required to stretch the cable and
the compression force required to compress the column this distance is
F
= Δ
Y A L
steel cable
L cable
cable
and F
= Δ
Y A L
Al column
L column
column
Combined, these forces support the weight of the walkway section:
F F Fg
cable + column = = 8 500 N or
Δ + Δ = 8 500 N
Y A L
L
Y A L
L
steel cable
cable
Al column
column
giving
ΔL
Y A
L
8 500 N
Y A
L
=
+
steel cable
cable
Al column
column
The cross-sectional area of the cable is
A
2
π D
2 π ( × − 2 ) 4
cable
m
= =
1 27 10
4
.
and the area of aluminum in the cross section of the column is
A
π D 2 π D 2 π D 2 − D
2
cable
outer inner outer inner = − =
4 4
( ) =
( ) − ( ) ⎡⎣
⎤⎦
4
π 0 . 162 4 m 2 0 . 161 4
m
2 4
Thus, the downward displacement of the walkway will be
ΔL =
( × ) ( × − )
8 500
10 2 2
π .
( . ) +
20 10 1 27 10
4 575
N
Pa m
m
× ( ) ( ) − ( ) ⎡⎣7 0 1010 0 162 4 2 0 161 4 2 . . . Pa m m π ⎤⎦
4 (3.25 m)
or
ΔL = 8.6 × 10−4 m = 0.86 mm

Solucionario Fundamentos de Física 9na edición Capitulo 9

Solucionario Fundamentos de Física 9na edición Capitulo 9

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Solucionario Fundamentos de Física 9na edición Capitulo 9