Solucionario Fundamentos de Física 9na edición Capitulo 9

9Solids 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

56157_09_ch09_p457-498.indd 45756157_09_ch09_p457-498.indd 457 1/4/08 9:53:16 PM1/4/08 9:53:16 PM

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. 4. Not enough information


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. 4. Not enough information

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. 4. Not enough information

Commentary


Discussion: Archimedes’ principle states that the buoyant force on a fl oating object is equal in magnitude to 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, weight, fl oating, sinking, and fl uid displacement. The situations lend themselves to a live demonstration, using a “predict, observe, and reconcile” pattern.

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. 4. Not enough information

Commentary


Discussion: Archimedes’ principle states that the buoyant force on a fl oating object is equal in magnitude to 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, weight, fl oating, sinking, and fl uid displacement. The situations lend themselves to a live demonstration, using a “predict, observe, and reconcile” pattern.

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:

m Vgold gold gold3 3kg/m m= = ×( )( ) =ρ 19 3 10 1 193. .33 103× kg

m Vsilver silver silver3 3kg/m m= = ×( )ρ 10 5 10 23. (( ) = ×21 0 103. kg

m Valuminum aluminum aluminum3kg/m= = ×ρ 2 70 103.(( )( ) = ×6 16 2 103m kg3 .

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 V= = ×( ) ×( ) ×−ρgold3kg m m19 3 10 4 50 10 11 0 13 2. . . 00 26 0 10 24 82 2− −( ) ×( ) =m m kg. . , and

choice (a) is the correct response.

2. On average, the support force each nail exerts on the body is

Fmg

1 1 208

66 0 9 80

1 2080 535= =

( )( )=

. ..

kg m sN

2

so the average pressure exerted on the body by each nail is

PF

Aavnailend

6 2

N

1.00 m= =

×= ×−

1 50 535

105 35 10

.. Pa

and (d) is the correct choice.

3. From Pascal’s principle, F A F A1 1 2 2= , so if the output force is to be F231 2 10= ×. N, the

required input force is

FA

AF1

1

22 0 70

1 2 10=⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

×0.050 m

m

2

2.. 33 86N 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 ρmercury submerged lead bulletV g V g= and V

Vsubmerged

bullet

lead

mercury

kg= = ×ρρ

11 3 103. mm

kg m

3

313 6 100 8313..

×=

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,

P = × + ×( )( )1 013 10 1 00 10 9 80 7545 3. . .Pa kg m m s3 2 ftm

3.281 ftPa( )⎛⎝

⎞⎠

⎡⎣⎢

⎤⎦⎥= ×1

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

P P g y y1 2 22

12

2 1

1

2− = −( ) + −( )ρ ρair airv v

or

∆P = ( ) ( )⎛⎝⎜

⎞⎠⎟

⎡

⎣

1

21 29 95

1

2 237.

.kg m mi h

mi h3

⎢⎢⎤

⎦⎥ −

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪+ = ×

2

30 0 1 2 10. Pa

and the correct choice is (a).

7. From the equation of continuity, A A1 1 2 2v v= , the speed of the water in the smaller pipe is

v v21

21

2

2

0 250

0 100=⎛⎝⎜

⎞⎠⎟

=( )( )

⎡

⎣⎢⎢

⎤A

A

ππ

.

.

m

m ⎦⎦⎥⎥( ) =1 00 6 25. . m s 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 ρ= water, the top of each vessel is open to the atmosphere so that P P0 = 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 gy constant+ + =1

22ρ ρv , we see that the

sum of the pressure and the kinetic energy per unit volume ( )P + 12

2ρv 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 v0 y i= and having speed v vf y= = 0 at their maximum altitude, h. From the kinematics equation v vy y ya y2

02 2= + ( )∆ , the maximum height reached is h gi= v2 2 . Thus, if we want to quadruple the

maximum height ( )′ =h h4 , we need to double the speed of the water leaving the nozzle ( )′ =v vi i2 . Using the equation of continuity, ′ ′ =A Ai iv v , it is seen that if ′ =v vi i2 , it is necessary to have ′ = ′ =A A Ai 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

hP P

g=

− −( )( ) =

05

1410 0

1 1010

ρ~

Pa

kg m m sm

3 2 or h ~ 10 km

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

AD F

elastic limit

mg

elastic limitminmin= = =π 2

4

and the minimum acceptable diameter is

Dmg

elastic limitmin

.=

( ) =( )( )4 4 70 9 8

π πkg m s2

55 0 101 3 10 1 3

83

.. .

×( ) = × =−

Pam 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 rcs = π

2, then r A cs= π and

A r t tAcs

shear cmc= ( ) = = ( ) ×

2 2 2 2 001 00 102

π ππ

π .. mm

cm2

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 10 2 50 108 8. .× = ×Pa N m2, the minimum force required to punch out this plug is

F A stress= ⋅ =⎛

⎝⎜⎞

⎠⎟⎡

⎣shear

22

2cmm

cm70 9

1

104.⎢⎢⎢

⎤

⎦⎥⎥

×⎛⎝⎜

⎞⎠⎟= ×2 50 10 1 77 108 6. .

N

mN

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 15 0 30 0. . .cm cm cm2, move a distance ∆x = × −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

Sshear stress

shear strain

F A

x h

Fh

x A= = = ( )∆ ∆

we have

S =( )( )( )×( )−

80 0 9 80 2 00

5 00 10 32

. . .

.

kg m s m

m

2

00 0 1 101 05 10

4

7

..

cm m cmPa

2 2 2( )( )⎡⎣ ⎤⎦= ×

9.4 As a liquid, the water occupied some volume Vl . As ice, the water would occupy volume 1 090. Vl 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 Vl l0 1 09 0 090= = −. .and ∆ , so

∆ ∆P B

V

V

Vl= −⎛⎝⎜

⎞⎠⎟= − ×⎛

⎝⎞⎠

−

0

92 00 100 090

1.

.

.

N

m2 0091 65 10 16008

Vl

⎛⎝⎜

⎞⎠⎟= × ≈. Pa atm

9.5 Using Y F L A L= 0 ( )∆ with A d= π 2 4 and F mg= , we get

Y =( )( )⎡⎣ ⎤⎦( )

×( )−

4 90 9 80 50

1 0 10 2

kg m s m

m

2.

.π22

83 5 101.6 m

Pa( )

= ×.

r

t

Acs

Ashear

r

t

Acs

Ashear

r

t

Acs

Ashear

r

t

Acs

Ashear



9.6 From Y F L A L= 0 ( )∆ , the tension needed to stretch the wire by 0.10 mm is

FY A L

L

Y d L

L=

( )=

( )( )

=×( ) ×

∆ ∆

0

2

0

10

4

18 10 0 22

π

π Pa . 110 0 10 1022

3 2 3− −

−

( ) ×( )×( ) =

m m

4 3.1 10 m N

2

.

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 Fx x= = − ° + ° =Σ cos cos30 30 0, and a y-component of R F F F Fy y= = − − = − = −Σ sin sin30 30 22º º N.

Thus, the resultant force is

R��= 22 N directed down the page in the diagrram .

9.7 From Y F A L L stress L L= =( )( ) ( )( )0 0∆ ∆ , the maximum compression the femur can withstand is

∆Lstress L

Y=( )( )

=×( )( )

×0

6

9

160 10 0 50

18 10

Pa m.

Pam mm= × =−4 4 10 4 43. . .

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 ( )∆ , the initial elongation of the cable is

∆LF L

A Y=

⋅⋅

=×( )( )

× −0

3

4

7 84 10 25 0

4 00 10

. .

.

N m

m2(( ) ×( ) = × −

20 102 45 10 2 5

103

Pam = mm. .

(b) When the load is accelerating upward, Newton’s second law gives

F mg may− = , or F m g ay= +( ) [1]

If m ay= = +800 3 0kg and m s2. , the elongation of the cable will be

∆LF L

A Y=

⋅⋅

=( ) +( )⎡⎣ ⎤⎦0800 9 80 3 0 25 0kg m s m2. . .(( )

×( ) ×( ) = × =−−

4 00 10 20 103 2 10

4 103

..

m Pam 3.

222 mm

Thus, the increase in the elongation has been

increase L L= ( ) − ( ) = − =∆ ∆

initial mm mm3 20 2 45 0. . ..75 mm

→F

→F

30° 30°→F

→F

30° 30°→F

→F

30° 30°→F

→F

30° 30°

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

F A stressmax max2m Pa= ⋅( ) = ×( ) ×−4 00 10 2 2 104 8. .(( ) = ×8 8 104. N

Then, Equation [1] above gives the mass of the maximum load as

mF

g amaxmax

2

N

9.8 3.0 m s=

+= ×

+( ) = ×8 8 106 9 10

43.

. kg

9.9 From the defi ning equation for the shear modulus, we fi nd the displacement, ∆x, as

∆xh F A

S

h F

S A=

( ) = ⋅⋅

=×( )( )×

−5 0 10 203. m N

3.0 106 PPa cm

cm

1 m

m

2

2

2( )( )⎛⎝⎜

⎞⎠⎟

= × =−

14

10

2 4 10 0

4

5. .0024 mm

9.10 The shear modulus is given by

Sshear stress

shear strain

stress

x h= = ( )∆

Hence, the stress is

stress Sx

h= ⎛

⎝⎞⎠ = ×( ) ×

⎛∆1 5 10

5 010..

Pam

10 10 m3⎝⎝⎞⎠ = ×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.

∆ ∆ ∆L L LF L

AY

F L

AY

F

A

Lrod Al Cu

Al

Al

Cu

Cu

= + =( )

+( )

=0 0 00 0( )+( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

Al

Al

Cu

CuY

L

Y

where A r= π 2 with r = = × −0 20 2 0 10 3. .cm m. Thus,

∆Lrod 1

N

m

m

7.0 10=

×( )×( ) ×−

5 8 10

2 0 10

1 33

3 2

.

.

.

π00 10 Pa

m

11 10 Pa m +

×⎡⎣⎢

⎤⎦⎥= × =−2 6

1 9 10 1 92.. . ccm

9.12 The acceleration of the forearm has magnitude

a

t= ∆ =

⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟v

∆

8010 1

5

3kmh

m1 km

h3 600 s

...

0 104 4 103

3

×= ×− s

m s2

The compression force exerted on the arm is F ma= and the compressional stress on the bone material is

StressF

A= =

( ) ×( )−

3 0 4 4 10

2 4 10

3

4

. .

.

kg m s

cm

2

2 mm cmPa

2 215 5 107

( ) = ×.

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

ρπ π0 3 34 3

3

4= = =M

V

M

R

M

R

The average density of the combined world is

ρπ π

=′=

⎛⎝⎜

⎞⎠⎟

=( )( ) =M

V

M

R

M

Rtotal 2

4

3

3

4

4 2

3

323

2

2 3

MM

R9 3π

so

ρρ π

π0

3

332

9

4

3

128

274 74= ⎛⎝

⎞⎠⎛⎝⎜

⎞⎠⎟= =M

R

R

M. or ρ ρ= 4 74 0.

9.14 (a) The mass of gold in the coin is

mm

mAutotal

total

karats=( )

= =#

.24

22

24

11

127 988 ××( ) = ×− −10 7 322 103 3kg kg.

and the mass of copper is

m mCu total kg= = ×( ) = ×− −1

12

1

127 988 10 6 657 103 4. . kg

(b) The volume of the gold present is

Vm

AuAu

Au3 3

kg

19.3 kg m= = ×

×= ×

−

ρ7 322 10

103 79

3.. 110 7− m3

and the volume of the copper is

Vm

CuCu

Cu3 3

kg

8.92 kg m= = ×

×= ×

−

ρ6 657 10

107 46

4.. 110 8− m3

(c) The average density of the British sovereign coin is

ρavtotal

total

total

Au Cu

k= =+

= × −m

V

m

V V

7 988 10 3. gg

3.79 m 7.46 mkg m7 3 8 3

3

× + ×= ×− −10 10

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 kg m s75 0 68 0 62 0 55 0 9 80. . . . . 22 N( ) = ×2 55 103.

(b) Pn

A

n

A= = = ×

total shoesole

2

N

2 425 cm m2

2 55 10

1

3.22 2cm

Pa10

3 00 104

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 Atotal 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

Pn

A

mg

rlegleg

2kg m s= = =

( )( )4 95 0 9 80

4 0 502π π

. .

. 00 102 96 10

2 26

×( )= ×

− mPa.

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

ρπnucleus

proton

proton

proton≈ =×m

V

m

r4 3

3 1 673 ∼

. 110

4 1 104 10

27

15 317

−

−

( )×( )

×kg

mkg m3

π∼

(b) The density of iron is ρFe3kg m= ×7 86 103. 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

PF

A

W

A= =

4

Thus, W A P= = ( ) ×( ) = ×4 4 0 024 2 0 10 1 9 105 4. . .m Pa N2 .

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 Ahg = ( ) ×( )5 0 104. N m3 . The pressure at the base of the pillar is then

PF

A

Ah

Ahg= =

( ) ×( )= ×( )5 0 10

5 0 104

4.

.N m

N m3

3

Thus, if the maximum acceptable pressure on the base is, Pmax Pa= ×1 7 107. , the maximum allowable height is

hP

maxmax

3 3N m

Pa

N m=

×= ×

×=

5 0 10

1 7 10

5 0 104

7

4.

.

.33 4 10. × 2 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 ∆x is ∆ ∆F F F P P A k x= − = −( ) = ( )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 x= ( )∆ , or the required depth is h k x gA= ( )∆ ρ . If k = 1 250 N m, A r r= = × −π 2 21 20 10with m. , and the fl uid is water ρ = ×( )1 00 103. kg m3 , the depth required to compress the spring an additional 0 750 7 50 10 3. .cm m= × − is

h =( ) ×( )

×( )−1 250 7 50 10

1 00 10 9 80

3

3

N m m

kg m3

.

. . m s mm

2( ) ×( )⎡⎣

⎤⎦=

−π 1 20 102 11

2 2.

.

9.21 (a) P P gh= + = × + ×( )03 3101 3 10 1 00 10 9 80ρ . . .Pa kg m m3 ss m Pa2( )( ) = ×27 5 3 71 105. .

(b) The inward force the water will exert on the window is

F PA P r= = ( ) = ×( ) ×⎛⎝⎜

⎞⎠

−

π π2 52

3 71 1035 0 10

..

Pam

2 ⎟⎟ = ×2

43 57 10. N



9.22 The gauge pressure in a fl uid at a level distance h below where Pgauge = 0 is P ghgauge = ρ with h being positive when measured in the downward direction. The difference in gauge pressures

at two levels is ( ) ( ) ( )P P g hgauge gauge2 1− = ρ ∆ or ( ) ( ) ( )P P g hgauge gauge2 1= + ρ ∆ with ∆h 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 ∆h < 0 and we fi nd

P P g hgauge head gauge heart( ) = ( ) + ( ) = ×ρ ∆ 13 3 103. PPa kg m m s m3 2+ ( )( ) −( )1 060 9 80 0 500. .

or

Pgauge headPa kPa( ) = × =8 11 10 8 113. .

(b) In going from heart to feet, one moves deeper in the blood column, so ∆h > 0 and

P P g hgauge feet gauge heart( ) = ( ) + ( ) = ×ρ ∆ 13 3 103. PPa kg m m s m3 2+ ( )( ) +( )1 060 9 80 1 30. .

or

Pgauge feetPa kPa( ) = × =26 8 10 26 83. .

9.23 The density of the solution is ρ ρ= = ×1 02 1 02 103. .water3kg m . The gauge pressure of the

fl uid at the level of the needle must equal the gauge pressure in the vein, so P ghgauge Pa= = ×ρ 1 33 103. , and

hP

g= = ×

×( )gauge

3 3

Pa

1.02 10 kg m mρ1 33 10

9 80

3.

. ssm

2( ) = 0 133.

9.24 (a) From the defi nition of bulk modulus, B P V V= − ( )∆ ∆ 0 , the change in volume of the 1 00. m3 of seawater will be

∆∆

VV P

B= −

( )= −

( ) × −0

81 00 1 13 10 1 01

water

3m Pa. . . 33 100 053 8

5×( )×

= −Pa

0.210 10 Pam10

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

ρ = =+

=−( ) = ×m

V

m

V V0

1 030

0 053 81 09 10

∆kg

1.00 m3.. 33 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 waterr as incompressible .

9.25 We fi rst fi nd the absolute pressure at the interface between oil and water.

P P gh1 0

51 013 10 700 9 80

= +

= × + ( )

ρoil oil

3Pa kg m. . m s m Pa2( )( ) = ×0 300 1 03 105. .

This is the pressure at the top of the water. To fi nd the absolute pressure at the bottom, we use P P gh2 1= + ρwater water, or

P251 03 10 1 9 80 0 200= × + ( )( )(. . .Pa 0 kg m m s m3 3 2 )) = ×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 P gh ghatmo = + =0 ρ ρ . Thus,

hP

g= = ×

( )( ) =atmo

3 2

Pa

984 kg m m sρ1 013 10

9 80

5.

.110 5. m

Some alcohol and water will evaporate, degrading the vacuum above the column.

9.27 Pascal’s principle, F A F A1 1 2 2= , or F A F Apedal Mastercylinder

brake brakecylinder

= , gives

FA

Abrakebrake cylinder

master cylinder

=⎛

⎝⎜⎞

⎠⎟⎟=⎛⎝⎜

⎞⎠⎟( ) =Fpedal

2

2

cm

cmN N

6 4

1 844 156

.

.

This is the normal force exerted on the brake shoe. The frictional force is

f nk= = ( ) =µ 0 50 156 78. N N

and the torque is τ = ⋅ = ( )( ) = ⋅f rdrum N m N m78 0 34 27. .

9.28 First, use Pascal’s principle, F A F A1 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.

FA

AF

d

dF

d

d11

22

12

22 2

12

22

4

4=⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

=⎛⎝

ππ ⎜⎜

⎞⎠⎟

= ( )( ) ( ) =

F2

2

2

0 25

1 5500 14

.

.

in

inlb lb

Now, consider an axis perpendicular to the page, passing through the left end of the jack handle. Στ = 0 yields

+( )( ) − ⋅ ( ) =14 2 12 0lb .0 in inF , 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 V r= =ρ πρwater water4 33 ; and an applied force, F. If the

ball is to be in equilibrium, we have (taking upward as positive) ΣF F B mgy = + − = 0, or

F mg B mgr

g mr= − = −

⎛⎝⎜

⎞⎠⎟

= −⎛

ρ π ρ πwater water

4

3

4

3

3 3

⎝⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥g

giving

F = − ×( ) ⎛⎝

⎞⎠

⎡0 540 1 00 10

4

3

0 250

23

3

. ..

kg kg mm3 π

⎣⎣⎢

⎤

⎦⎥( ) = −9 80 74 9. .m s N2

so the required applied force is F��= 74 9. N directed downward .

→F

→H →

V→F1

2.0

in10 in

Free-Body Diagram of Handle

→F

→H →

V→F1

2.0

in10 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 totalρ ρ or V

Vsubmerged

total

body

sea

=ρρ

After inhaling,

V

Vsubmerged

total

3

3

kg m

kg m= = =945

1 2300 768 76. .. %8

leaving 23 2. % above surface .

After exhaling,

V

Vsubmerged

total

3

3

kg m

kg m= = =1 020

1 2300 829 8. 22 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 V gtruck water

3

kg

mm

= ( )[ ]

= ⎛⎝⎞⎠ ( )

ρ ∆

10 4 00 63 . .000 4 00 10 9 802m mm

s2( ) ×( )⎡⎣ ⎤⎦⎛⎝

⎞⎠

−. .

or

wtruck N kN= × =9 41 10 9 413. .

9.32 (a) A� (2.00 m)2�4.00 m2B

Survivor

wr w

d

t

(b) Since the system is in equilibrium, ΣF B w wy r= − − = 0 .

(c) B gV g d Aw w= = ⋅( )

= ( )ρ ρsubmerged

3kg m m s1 025 9 80. 22 2m m N( )( )( ) =0 024 0 4 00 964. .

(d) From B w wr− − = 0,

w B w B m gr s= − = − = − ( )( ) =964 62 0 9 80 356N kg m s2. . NN

(e) ρfoam 2

N

9.80 m s m= =

⋅= ( )( )

m

V

w g

t Ar

r

r 356

0 090 4. .000101

mkg m

23

( ) =continued on next page


476 Chapter 9

(f )

kg m m s3 2

B gV g t Aw r wmax

.

= = ⋅( )

= ( )(ρ ρ

1 025 9 80 ))( )( ) = ×0 090 0 4 00 3 62 103. . .m m N2

(g) The maximum weight of survivors the raft can support is w m g B wrmax max max= = − , so

mB w

gr

maxmax .=

−= × − =3 62 10 356

3333 N N

9.80 m s2 kkg

9.33 (a) While the system fl oats, B w w w= = +total block steel, or ρ ρw b bgV gV m gsubmerged steel= + .

When msteel kg= 0 310. , V Vbsubmerged3m= = × −5 24 10 4. , giving

ρ ρ ρbw b

bw

b

V m

V

m

V=

−= − = × −steel steel kg m1 00 103 3.

00 310

10408

. kg

5.24 mkg m4 3

3

×=−

(b) If the total weight of the block + steel system is reduced, by having msteel kg< 0 310. , 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 kg= 0 310. . 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 0 039. .kg 8 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 00 350. kg.

9.34 (a) B

Vb�325 m3

mb�226 kg

wHe� �He gVbwb�mbg

(b) Since the balloon is fully submerged in air, V Vbsubmerged3m= = 325 , and

B gVb= = ( )( )( ) =ρair3 2 3 kg m m s m1 29 9 80 325 4 1. . . 11 103× N




(c) He He HeΣF B w w B m g gV B m Vy b b b b= − − = − − = − +ρ ρ(( )= × − + ( )( )

g

4 11 10 226 0 179 3253. .N kg kg m m3 3⎡⎡⎣ ⎤⎦( ) = + ×9 80 1 33 103. .m s N2

Since ΣF may y= > 0, aywill be positive (upward), and the balloon rises .

(d) If the balloon and load are in equilibrium, ΣF B w w wy b= − −( ) − =He load 0 and w B w wbload He N= − −( ) = ×1 33 103. . Thus, the mass of the load is

mw

gloadload

2

N

m skg= = × =1 33 10

9 80136

3.

.

(e) If mload kg< 136 , then the net force acting on the balloon + load system is upward and

the balloon and its load will accelerate upwward .

(f ) The density of the surrounding air, temperatture, 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.

9.35 (a) kgair balloon airB gV gr= =

⎛⎝⎜

⎞⎠⎟=ρ ρ π4

31 29

3

. mm m s m

N

3 2( )( )⎛⎝⎞⎠ ( )

= × =

9 804

33 00

1 43 10

3

3

. .

.

π

11 43. kN

(b) ΣF B wy = − = × − ( )( )total2N kg m s1 43 10 15 0 9 803. . .

== + × =1 28 10 1 283. .N 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 may y= − = , or ma gV mgy w= −ρ .

(b) Since m V= ρ , we have ρ ρ ρV a g V V gy w= − , or

a gyw= −

⎛⎝⎜

⎞⎠⎟

ρρ

1

(c) ay =× −

⎛⎝⎜

⎞⎠⎟( )1 00 10

1 0501 9 80

3..

m kg

m kgm s

3

32 == − =0 467 0 467. .m s m s downward2 2

(d) From ∆y t a ty y= +v02 2, with v0 0y = , we fi nd

ty

ay

=( ) = −( )

−=

2 2 8 00

0 4675 85

∆ .

..

m

m ss2


478 Chapter 9

9.37 (a) total singleballoon

B B= ⋅ =600 6000 6004

33ρ ρ π

air balloon airgV g r( ) = ⎛⎝

⎞⎠

⎡⎣⎢

⎤⎦⎥

= 6600 1 29 9 804

30 50 3. . .kg m m s m3 2( )( ) ( )⎡

⎣⎢⎤⎦⎥

π == × =4 0 10 4 03. .N kN

(b) ΣF B m gy = − = × − ( )total total N kg4 0 10 600 0 30 93. . .88 2 2 10 2 23m s N kN2( ) = × =. .

(c) Atmospheric pressure at this high altitude iis 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

P P ghtop water top

3Pakg

m

= +

= × + ⎛

0

5 31 0130 10 10

ρ

. ⎝⎝⎞⎠⎛⎝

⎞⎠ ×( )

= ×

−9 80 5 00 10

1 0179 10

2

5

. .

.

m

sm

Pa

2

and that at the level of the bottom of the block is

P P ghbottom water bottom

Pa

= +

= × +

0

5 31 0130 10 10

ρ

.kg

m

m

sm3 2

⎛⎝

⎞⎠⎛⎝

⎞⎠ ×( )

= ×

−9 80 17 0 10

1 0297

2. .

. 1105 Pa

Thus, the downward force exerted on the top by the water is

F P Atop top Pa m= = ×( )( ) =1 0179 10 0 100 1017 95 2. . . N

and the upward force the water exerts on the bottom of the block is

F P Abot bot Pa m= = ×( )( ) =1 0297 10 0 100 1029 75 2. . . 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 mgy = + − − =bot top 0, or

T = ( )( ) − −( ) =10 0 9 80 1029 7 1017 9 86. . . . .kg m s N2 22 N

(c) From Archimedes’s principle, the buoyant force on the block equals the weight of the displaced water. Thus,

B V g= ( )

= ( ) ( )

ρwater block

3kg m m10 0 100 0 1203 2. . m m s N2( )⎡⎣ ⎤⎦( ) =9 80 11 8. .

From part (a), F Fbot top N 11.8 N− = −( ) =1 029 7 1 017 9. . , 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 may y= = 0

− × +( ) + + =1 20 10 1100 04. kg Nsea waterm g gVρ

where m is the mass of the added sea water and V is the sphere’s volume.

Thus,

m = ×⎛⎝

⎞⎠⎛⎝

⎞⎠ ( )⎡

⎣⎢⎤⎦⎥+1 03 10

4

31 50

13 3. .kg

mm3

π 11001 20 104N

9.80 m skg2 − ×.

or

m = ×2 67 103. kg

9.40 At equilibrium, ΣF B F mgy = − − =spring 0, so the spring force is

F B mg V m gspring water block= − = ( ) −⎡⎣ ⎤⎦ρ

where

Vm

blockwood

33 3kg

650 kg m10 m= = = × −

ρ5 00

7 69.

.

Thus, Fspring3 3 3 kg m 10 m kg= ( ) ×( ) −⎡⎣ ⎤−10 7 69 5 003 . . ⎦⎦( ) =9 80 26 4. . m s N.2

The elongation of the spring is then

∆xF

k= = = =spring N

160 N mm cm

26 40 165 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 = − =300 200 100N N N. But, from Archimedes’s principle, this is also the weight of the displaced alcohol, so B V g= ( )ρalcohol . Since the sample is fully submerged, the volume of the displaced alcohol is the same as the volume of the sample. This volume is

VB

g= = ( )( ) =ρalcohol

3 2

N

kg m m s

100

700 9 801 46

.. ×× −10 2 m3

(b) The mass of the sample is

mweight in air

g= = =300

30 6N

9.80 m skg2 .

and its density is

ρ = =×

= ×−m

V

30 6

1 46 102 10 102

3.

..

kg

mkg m3

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

mweight in air

g= = =300

30 6 N

9.80 m s kg2 .

The buoyant force when immersed in water is the weight of a volume of water equal to the volume of the object, or B V gw w= ( )ρ . Thus, the volume of the object is

VB

gw

w

= = −( )( ) =ρ

300 265

9 803 5

N N

10 kg m m s3 3 2.. 77 10 3× − m3

and its density is

ρobject 3

kg

m kg m= =

×= ×−

m

V

30 6

3 57 108 57 103

3.

.. 33

(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 goil oil= ( )ρ , or the density of the oil is

ρoiloil

3

N N

m m= = −

×( )−

B

Vg

300 275

3 57 10 9 803. . ss kg m

23

( ) = 714

9.43 The volume of the iron block is

Vm= =

×= × −iron

iron3 3

kg

7.86 10 kg mρ2 00

2 54 10 4.. m3

and the buoyant force exerted on the iron by the oil is

B V g= ( ) = ( ) ×( )−ρoil3 3 216 kg m m m s9 2 54 10 9 804. .(( ) = 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 Bupper iron N N N= − = − =19 6 2 28 17 3. . .

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 glower oil beaker− − +( ) = 0

The support force exerted by the lower scale (and the lower scale reading) is then

F B m m glower oil beaker N= + +( ) = + +( )2 28 2 00 1 00. . . kg m s N2⎡⎣ ⎤⎦( ) =9 80 31 7. .

9.44 (a) The cross-sectional area of the hose is A r d= = = ( )π π π2 2 24 2 74 4. cm , and the volume fl ow rate (volume per unit time) is Av = 25 0. L 1.50 min. Thus,

v = =

⎛⎝⎜

⎞⎠⎟ ⋅

25 0 25 0 4

2

. .

.

L 1.50 min L

1.50 minA π 774

1 10

12

3

( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎛⎝⎜

⎞⎠⎟ cm

min

60 s

cm

L2

3⎛⎛⎝⎜

⎞⎠⎟

= ( )⎛⎝⎜⎞⎠⎟ =47 1

10 471. . cm s

m

10 cm m s2




(b) A

A

d

d

d

d2

1

22

12

2

1

2

4

4 1

3=⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟=⎛⎝⎜

⎞⎠⎟= ⎛π

π ⎝⎝⎞⎠ =

2 1

9 or A

A2

1

9=

Then from the equation of continuity, A A2 2 1 1v v= , we fi nd

v v21

21 9 0 471 4 24=

⎛⎝⎜

⎞⎠⎟

= ( ) =A

A. . 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 2 0 40 80. . g cm cm cm s g s3 2

(b) From the equation of continuity, the speed in the capillaries is

v vcapiliariesaorta

capillariesaorta=

⎛

⎝⎜⎞

⎠⎟A

A==

×⎛⎝⎜

⎞⎠⎟( )2 0

3 0 10403

.

.

cm

cm cm s

2

2

or vcapiliaries cm s mm s= × =−2 7 10 0 272. . .

9.46 (a) From the equation of continuity, the fl ow speed in the second pipe is

v v21

21

1

22=

⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

A

A

0.0 cm

.50 cm.75

2

2 mm s m s( ) = 11 0.

(b) Using Bernoulli’s equation and choosing y = 0 along the centerline of the pipes gives

P P2 1 12

22 5 31

21 20 10

1

21 65 10= + −( ) = × + ×ρ v v . . Pa kgg m m s m s3( ) ( ) − ( )⎡

⎣⎤⎦2 75 11 0

2 2. .

or P242 64 10= ×. Pa .

9.47 From Bernoulli’s equation, choosing y = 0 at the level of the syringe and needle, P P2

12 2

21

12 1

2+ = +ρ ρv v , so the fl ow speed in the needle is

v v2 12 1 22

= +−( )P P

ρ

In this situation,

P P P P PF

A1 2 1 11

2 00

2 50 10− = − = ( ) = =

× −atmo gauge

N.

. 5548 00 10

m Pa2 = ×.

Thus, assuming v1 0≈ ,

v2

4

302 8 00 10

1 00 1012 6= +

×( )×

=.

..

Pa

kg m m s3

9.48 We apply Bernoulli’s equation, ignoring the very small change in vertical position, to obtain

P P1 212 2

212 1

2 1

2

12 3

2 122− = −( ) = ( ) −⎡

⎣⎤⎦ =ρ ρ ρv v v v v , or

∆P = ( ) ×( ) = ×− −3

21 29 15 10 4 4 102 2 2. .kg m m s Pa3


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 mglowersurface

uppersurface

wings− = . This yields

P Pmg

Alowersurface

uppersurface wings

−⎛⎝⎜

⎞⎠⎟= =

88 66 10 9 80

90 09 43 10

43

. .

..

×( )( )= ×

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 Plower air lower upper air upper+ = +1

2

1

22 2ρ ρv v

so

v vupper lower

lower upper

air

m s= +−( )

= (22

225P P

ρ)) +

×( )=2

32 9 43 10

1 29255

.

.

Pa

kg m m s3

(c) The density of air decreases with increasingg 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 Mglowersurface

uppersurface

− = . This gives the air pressure at the upper surface as

P PMg

Auppersurface

lowersurface

= −

9.51 (a) Since the pistol is fi red horizontally, the emerging water stream has initial velocity components of ( , )v v v0 0 0x y= =nozzle . Then, ∆y t a ty y= +v0

2 2, with a gy = − , gives the time of fl ight as

ty

ay

=( ) = −( )

−=

2 2 1 50

9 800 553

∆ .

..

m

m ss2

(b) With ax x= =0 0 and nozzlev v , the horizontal range of the emergent stream is ∆x t= vnozzle where t is the time of fl ight from above. Thus, the speed of the water emerging from the nozzle is

vnozzle

m

0.553 s m s= = =∆x

t

8 0014 5

..

(c) From the equation of continuity, A A1 1 2 2v v= , the speed of the water in the larger cylinder is v v v1 2 1 2 2 1= =( ) ( )A A A A nozzle, or

v v v122

12

2

1

2

=⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

ππ

r

r

r

rnozzle nozzle ==⎛⎝⎜

⎞⎠⎟ ( ) =1 00

14 5 0 1452

.. .

mm

10.0 mm m s m s

(d) The pressure at the nozzle is atmospheric pressure, or P251 013 10= ×. Pa .




(e) With the two cylinders horizontal, y y1 2= and gravity terms from Bernoulli’s equation

can be neglected , leaving P Pw w1 12

2 222 2+ = +ρ ρv v so the needed pressure in the larger

cylinder is

P P w1 2 2

212 5

3

21 013 10

1 00 10= + −( ) = × + ×ρv v .

. Pa

kg mm m s m s

3

214 5 0 145

2 2. .( ) − ( )⎡

⎣⎤⎦

or

P152 06 10= ×. Pa

(f ) To create an overpressure of ∆P = × − × = ×2 06 10 1 013 10 1 05 105 5 5. . .Pa Pa Pa in the larger cylinder, the force that must be exerted on the piston is

F P A P r1 1 12 51 05 10 1 00 10= ( ) = ( )( ) = ×( ) × −∆ ∆ π π. .Pa 22 2

33 0m N( ) = .

9.52 (a) From Bernoulli’s equation,

P gy P gyww

ww1

12

1 222

22 2+ + = + +ρ ρ ρ ρv v

or

v v22

12 1 2

2 12− = − − −( )⎡

⎣⎢

⎤

⎦⎥

P Pg y y

wρ

and using the given data values, we obtain

v v22

12

5 5

321 75 10 1 20 10

1 00 10− = × − ×

×. .

.

Pa Pa

kg mm m s m3

2− ( )( )⎡

⎣⎢

⎤

⎦⎥9 80 2 50. .

and

v v22

12 61 0− = . m s2 2 [1]

From the equation of continuity,

v v v21

21

12

22 1

1

2

=⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

A

A

r

r

r

r

ππ

22

1v

v v2

2

1

3 00

1 50= ⎛⎝⎜

⎞⎠⎟

.

.

cm

cm or v v2 14= [2]

Substituting Equation [2] into [1] gives 16 1 61 012−( ) =v . m s2 2, or

v1

61 0

152 02= =..

m s m s

2 2

(b) Equation [2] above now yields v2 4 2 02 8 08= ( ) =. . m s m s .

(c) The volume fl ow rate through the pipe is: flow rate = =A A1 1 2 2v v .

Looking at the lower point:

flow rate r m m s= ( ) = ×( ) ( )−π π12

12 2

3 00 10 2 02v . . == × −5 71 10 3. m s3

Point

∆x2

∆x1

Point

y1

y2

→v2

→v1

P1A1

P2 A2

1

2

Figure 9.29

Point

∆x2

∆x1

Point

y1

y2

→v2

→v1

P1A1

P2 A2

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 ∆y t a ty y= +v0

12

2 with v0 0y = , we fi nd the time of fl ight as

ty

ay

=( ) = −( )

−=

2 2 1 00

9 800 452

∆ .

..

m

m ss2

From the horizontal motion, the speed of the water coming out of the hole is

v v2 0

0 6001 33= = = =x

x

t

∆ ..

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 P1 2= = atmo and v1 0≈ , this gives

ρ ρ ρgy gy1 22

2

1

2= +v

or

h y y

g= − = =

( )( ) = ×1 2

22 2

2

1 33

2 9 809 00 10

v .

..

m s

m s2−− =2 9 00 m 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 P1 2= = atmo and we assume v1 0≈ . This gives

12 2

22 1ρ ρ ρv + =gy gy , or

v2 1 22 2 9 80 16 0 17 7= −( ) = ( )( ) =g y y . . . m s m m s2

(b) The area of the hole is found from

Aflow rate

22

32 50 10

17 7

1= = × − m min

m s

min

6

3

v.

. 00 s m2⎛

⎝⎜⎞⎠⎟ = × −2 35 10 6.

The diameter is then

dA

22

634 4 2 35 10

1 73 10 1 73= =×( )

= × =−

−

π π.

. .m

m m2

mm



9.55 First, determine the fl ow speed inside the larger portions from

v11

4

2 2

1 80 10

2 50 10= = ×

×( )−

−

flow rate

A

m s

m

3.

.π 440 367= . m s

The absolute pressure inside the large section on the left is P P gh1 0 1= + ρ , where h1 is the height of the water in the leftmost standpipe. The absolute pressure in the constriction is P P gh2 0 2= + ρ , so

P P g h h g1 2 1 2 5 00− = −( ) = ( )ρ ρ . cm

The fl ow speed inside the constriction is found from Bernoulli’s equation with y y1 2= . This gives

v v v22

12

1 2 12

1 2

22= + −( ) = + −( )

ρP P g h h , or

v2

2 20 367 2 9 80 5 00 10 1 06= ( ) + ( ) ×( ) =−. . . . m s m s m m s

The cross-sectional area of the constriction is then

Aflow rate

22

41 80 10

1 061 71 10= = × = ×

− m s

m s

3

v.

.. −−4 m2,

and the diameter is

dA

22

424 4 1 71 10

1 47 10 1 47= =×( )

= × =−

−

π π.

. .m

m c2

mm

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

22 2ρ ρ ρ ρv v

river rim

=

P g y yriver min rim river atm( ) + = + + −( )0 1 0 ρ

or

Priver min 3 Pa kg

m( ) = × + ⎛⎝⎜

⎞⎠⎟1 013 10 10 9 805 3. .

m

s m m2

⎛⎝⎜

⎞⎠⎟ −( )2096 564

Priver min Pa =( ) = × + ×( ) ×1 013 10 1 50 10 1 515 7. . . 1107 Pa 15.1 MPa=

(b) The volume fl ow rate is flow rate A d == v v( )π 2 4 . Thus, the velocity in the pipe is

v =( ) = ( )

( )4 4 4500 1

2 2

flow rate

d

m d

0.150 m

d3

π π 886 400 s 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 yriver rim river rim atm+ = + −( )⎡⎣ ⎤⎦ +0 11

22ρ ρ v

or

P P Priver river min rim river min= ( ) + = ( ) +1

2

1

212ρv 00 2 95

4

32

kg

m

m

s

river min

3⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟

= ( ) +

.

.P 335 kPa

The additional pressure required to achieve the desired fl ow rate is

∆P = 4 35. kPa

9.57 (a) For upward fl ight of a water-drop projectile from geyser vent to fountain-top, v vy y ya y2

02 2= + ( )∆ , with vy = 0 when ∆ ∆y y= max, gives

v0 0 2 2 9 8 40 0 28 0y ya y= − ( ) = − −( )( ) =∆max

. . . m s m 2 mm s

(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

202ρ ρvvent top vent= + −( )g y y

or

vvent top vent2 m s m= −( ) = ( )( ) =2 2 9 80 40 0 28g y y . . ..0 m s

(c) Between the chamber and the geyser vent, Bernoulli’s equation with vchamber ≈ 0 yields

P gy P gy+ +( ) = + +01

22ρ ρ ρ

chamber atm vent ventv

or

P P g y y− = + −( )⎡

⎣⎢⎤⎦⎥

=

atm vent vent chamberρ 1

2

10

2v

33

228 0

29 80 175

kg

m

m s m

s 3 2

⎛⎝⎜

⎞⎠⎟( )

+ ⎛⎝⎜⎞⎠⎟

.. mm MPa( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 2 11.

or

P P Pgauge atmo5 Pa atm 1.013 = − = ×( ) ×2 11 10 1 106. PPa atmospheres( ) = 20 8.



9.58 (a) Since the tube is horizontal, y y1 2= and the gravity terms in Bernoulli’s equation cancel, leaving

P P1 12

2 221

2

1

2+ = +ρ ρv v

or

v v22

12 1 2

32 2 1 20 10

10− =

−( )=

×( )×

P P

ρ. Pa

7.00 kg2 mm3

and

v v22

12 3 43− = . m s2 2 [1]

From the continuity equation, A A1 1 2 2v v= , we fi nd

v v v21

21

1

2

2

1

2 40

1 20=⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

=A

A

r

r

.

.

cm

cmm⎛⎝⎜

⎞⎠⎟

2

1v

or

v v2 14= [2]

Substituting Equation [2] into [1] yields 15 3 4312v = . m s2 2 and v1 0 478= . m s.

Then, Equation [2] gives v2 4 0 478 1 91= ( ) =. . m s m s .

(b) The volume fl ow rate is

A A r1 1 2 2 22

22 2

1 20 10 1 91v v v= = ( ) = ×( ) ( )−π π . . m m s == × −8 64 10 4. m s3

9.59 From ΣF T mg Fy y= − − = 0, the balance reading is found to be T mg Fy= + where Fy is the vertical component of the surface tension force. Since this is a two-sided surface, the surface tension force is F L= ( )γ 2 and its vertical component is F Ly = ( )γ φ2 cos where φ is the contact angle. Thus, T mg L= + 2γ φcos .

T = 0 40. N when φ = ° ⇒0 mg L+ =2 0 40γ . N [1]

T = 0 39. N when φ = ° ⇒180 mg L− =2 0 39γ . 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

4 mL )) = × −8 3 10 2. N m

9.60 Because there are two edges (the inside and outside of the ring), we have

γ

π

= =

= =1.61 1

total

F

L

F

circumference

F

r

2

4

( )

× 00 N

4 1.75 10 m N m

2

2

−

−−

×( ) = ×π

7 32 10 2.

P2P1

A2

A1

→v112 →v2

Figure 9.30(a)

P2P1

A2

A1

→v112 →v2

Figure 9.30(a)


488 Chapter 9

9.61 From h gr= 2γ φ ρcos , the surface tension is

γ ρ

φ=

=×( )( )−

h gr

2

2 1 10 1 080 9 802

cos

. .m kg m m s3 2(( ) ×( )°

= ×−

−5 0 105 6 10

42

..

m

2 cos0N m

9.62 The height the blood can rise is given by

hgr

= =( ) °

( )2 2 0 058 0

1 050 9 80

γ φρcos . cos

.

N m

kg m2 m s mm

2( ) ×( ) =−2 0 105 6

6..

9.63 From the defi nition of the coeffi cient of viscosity, η = F L A v, the required force is

FA

L= =

× ⋅( ) ( )( )⎡⎣−η v 1 79 10 0 800 1 203. . . N s m m m2 ⎤⎤⎦( )

×=−

0 508 6

..

m s

0.10 10 m N3

9.64 From the defi nition of the coeffi cient of viscosity, η = F L A v, the required force is

FA

L= =

× ⋅( ) ( )( )⎡−η v 1 500 10 0 010 0 0403 N s m m m2 . .⎣⎣ ⎤⎦( )×

=−

0 30

50 12

.

..

m s

1 10 m N3

9.65 Poiseuille’s law gives flow rateP P R

L=

−( )1 24

8

πη

and P P2 = atm in this case. Thus, the desired gauge pressure is

P PL flow rate

R1 4

8 8 0 12 50− =

( ) = ⋅( )atm

2 N s m mη

π. (( ) ×( )

×( )−

−

8 6 10

0 50 10

5

2 4

.

.

m s

m

3

π

or

P P162 1 10 2 1− = × =atm Pa MPa. .

9.66 From Poiseuille’s law, the fl ow rate in the artery is

flow rateP R

L

Pa m= ( )

=( ) ×( )−∆ π

ηπ4 3 4

8

400 2 6 10.

88 2 7 10 8 4 103 2 10

3 25

. ..

× ⋅( ) ×( ) = ×− −−

N s m mm s

23

Thus, the fl ow speed is

v = = ×

×( )=

−

−

flow rate

A

m s

2.6 10 m

3

3

3 2 101 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

100 5 0

1 1.

ccm m s⎛

⎝⎜⎞⎠⎟ = × −1 4 10 5. .

Thus, from vt fr g= −2 92 ( )ρ ρ η, we fi nd the maximum radius of the particle:

rg

t

f

maxmax

2 N s m

=( )−( )

=× ⋅( )−

9

2

9 1 00 10 13

ηρ ρv

. .44 10

2 9 80 1 800 1 000

5×( )( ) −( )⎡⎣ ⎤⎦

− m s

m s kg m2 3.== × =−2 8 10 2 86. . m mµ

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

∆∆ ∆

PL V t

R=

( )84

ηπ

If the mass fl ow rate is ∆ ∆m t( ) = × −1 0. 10 kg s3 , the volume fl ow rate of the water is

∆∆

∆ ∆V

t

m t= = ××

= ×−

−

ρ1 0

1 0 101 0 103

.

..

10 kg s

kg m

3

366 m s3

and the required excess pressure is

∆P =× ⋅( ) ×( ) ×− − −8 1 0 10 3 0 10 1 0 103 2 6. . . Pa s m m s3(( )

×( )= ×

−π 0 15 101 5 10

3 45

..

m Pa

9.69 With the IV bag elevated 1.0 m above the needle, the pressure difference across the needle is

∆P gh= = ×( )( )( ) = ×ρ 1 0 10 9 8 1 0 9 8 13. . . . kg m m s m3 2 003 Pa

and the desired fl ow rate is

∆∆V

t=

( )( ) =

500 1 10

30

6 cm m cm

min 60 s 1 min

3 3 3

22 8 10 7. × − m s3

Poiseuille’s law then gives the required diameter of the needle as

D RL V t

P= =

( )( )

⎡

⎣⎢

⎤

⎦⎥ =

× ⋅−

2 28

28 1 0 10

1 4 3ηπ∆ ∆∆

. Pa ss m m s

Pa

3( ) ×( ) ×( )×( )

− −2 5 10 2 8 10

9 8 10

2 7

3

. .

.π

⎡⎡

⎣⎢⎢

⎤

⎦⎥⎥

1 4

or

D = × =−4 1 10 0 414. .m mm


490 Chapter 9

9.70 We write Bernoulli’s equation as

P gy P gyout out out in in in+ + = + +1

2

1

22 2ρ ρ ρ ρv v

or

P P P g y ygauge in out out in out in= − = −( ) + −( )ρ 1

22 2v v⎡⎡

⎣⎢⎤⎦⎥

Approximating the speed of the fl uid inside the tank as vin ≈ 0, we fi nd

Pgauge3 2 kg m m s m s= ×( ) ( ) +1 00 10

1

230 0 9 803 2

. . .(( )( )⎡⎣⎢

⎤⎦⎥

0 500. m

or

Pgauge Pa kPa= × =4 55 10 4555.

9.71 The Reynolds number is

RNd= =

( )( ) ×( )−ρηv 1 050 0 55 2 0 10

2 7

2 kg m m s m3 . .

. ×× ⋅=−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

vmaxmax

2 N s m=

⋅( )=

× ⋅( )( )−ηρRN

d

1 0 10 2 000

1 000

3.

kg m m.080 m s cm s

3( ) ×( ) = =−2 5 100 8 0

2..

9.73 The observed diffusion rate is 8 0 10 5 3 1014 15. .× = ×− − kg 15 s kg s. Then, from Fick’s law, the difference in concentration levels is found to be

C Cdiffusion rate L

DA2 1

155 3 10 0

− =( )

=×( )−

kg s. .110

5 0 10 6 0 101 8 10

10 4

m

m s m2 2

( )×( ) ×( ) = ×− −

−

. .. 33 kg m3

9.74 Fick’s law gives the diffusion coeffi cient as D diffusion rate A C L= ⋅( ) ∆ , where ∆C L is the concentration gradient.

Thus,

D = ××( ) ⋅ ×(

−

− −

5 7 10

2 0 10 3 0 10

15

4 2

.

. .

kg s

m kg m2 4 )) = × −9 5 10 10. m s2

9.75 Stokes’s law gives the viscosity of the air as

ηπ π

= = ××( ) ×

−

− −

F

r6

3 0 10

6 2 5 10 4 5 10

13

6 4v.

. .

N

m m ss N s m2

( ) = × ⋅−1 4 10 5.

→vout

0.500 m

→vout

0.500 m



9.76 Using vt fr g= −( )2 92 ρ ρ η, the density of the droplet is found to be ρ ρ η= +f t r g9 2 2v .

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 0009 1 00 10 1 10 103 2

kg

m

N s m m s3

2. . ))×( ) ( )

= ×−2 5 00 10 9 80

1 02 104 2

3

. ..

m m skg 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 By = ⇒ = −0

The buoyant force B��

is the same for the two blocks, so the spring scale

reading T��

is largest for the iron block , which has a higher density, and

hence weight, than the aluminum block.

(c) The buoyant force in each case is

B V g= ( ) = ×( )( )ρwater3 3 2kg m m m s1 0 10 0 20 9 83. . .(( ) = ×2 0 103. N

For the iron block:

T V g Biron iron

3 3kg m m= ( ) − = ×( )( )ρ 7 86 10 0 20 93. . ..8 m s2( ) − B

or

Tiron N N N= × − × = ×1 5 10 2 0 10 13 104 3 3. .

For the aluminum block:

T V g Baluminum aluminum3kg m= ( ) − = ×( )ρ 2 70 10 03. .220 9 8m m s3 2( )( ) −. B

or

Taluminum N N N= × − × = ×5 2 10 2 0 10 3 3 103 3 3. . .

9.78 In going from the ocean surface to a depth of 2 40. km, the increase in pressure is

∆P P P gh= − = = ×( )( ) ×031 025 10 9 80 2 40ρ . . .kg m m s2 2 110 2 41 103 7m Pa( ) = ×.

The fractional change in volume of the steel ball is given by the defi ning equation for bulk modulus, ∆ ∆P B V V= − ( ), as

∆ ∆V

V

P

B= − = − ×

×= −

steel

Pa

Pa

2 41 10

16 0 101 5

7

10

.

.. 11 10 4× −

Fg � mg

TB

Fg � mg

TB


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 ρg pAt g Ad g( )⎡⎣ ⎤⎦ = ( )⎡⎣ ⎤⎦

or ρ ρg pt d= .

(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=⎛

⎝⎜⎞

⎠⎟= ×

×⎛⎝⎜

⎞⎠

ρρ

p

g

3

3

kg m

kg m

3 3 10

2 8 10

3

3

.

. ⎟⎟ ( ) =5 0 5 9. .km km

9.80 (a) Starting with P P gh= +0 ρ , we choose the reference level at the level of the heart, so P PH0 = . The pressure at the feet, a depth hH below the reference level in the pool of blood in the body is P P ghF H H= + ρ . The pressure difference between feet and heart is then

P P ghF H H− = ρ .

(b) Using the result of part (a),

P PF H− = ×( )( )( ) =1 06 10 9 80 1 20 1 23. . . .kg m m s m3 2 55 104× Pa

9.81 The cross-sectional area of the aorta is A d1 12 4= π and that of a single capillary is A dc = π 2

2 4. If the circulatory system has N such capillaries, the total cross-sectional area carrying blood from the aorta is

A NAN d

c222

4= = π

From the equation of continuity,

A A21

21=

⎛⎝⎜

⎞⎠⎟

vv

, or N d dπ π2

21

2

12

4 4=⎛⎝⎜

⎞⎠⎟

vv

,

which gives

Nd

d=⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟=

×⎛⎝ −

vv

1

2

1

2

2

2

1 0

1 0 10

.

.

m s

m s⎜⎜⎞⎠⎟

××

⎛⎝⎜

⎞⎠⎟= ×

−

−0 50 10

10 102 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

hP

g

P

gmax

.= − = = ×atm

water

atm

water

N m0 1 013 105

ρ ρ

22

3 2 kg m 9.80 m s m

1 00 1010 3

3..

×( )( ) =

(b) The moon has no atmosphere so Patm = 0, which yields hmax = 0 .

Area A

Peridotite

Continent(granite)

d

t�g

�p

Area A

Peridotite

Continent(granite)

d

t�g

�p



9.83 (a) P g= = ( )160 160mm of H O mm2 H O2ρ

= ⎛⎝⎞⎠⎛⎝

⎞⎠ ( ) =10 9 80 0 1603 kg

m

m

sm3 2. . 1 57. kPa

P = ×( ) ×

⎛⎝

⎞⎠ =1 57 10

13. Paatm

1.013 10 Pa5 1 55 10 2. × − atm

The pressure is P gh gh= =ρ ρH O H O Hg Hg2 2, so

h hHgH O

HgH O

3

32

2

kg m

kg m=⎛

⎝⎜⎞

⎠⎟=

×ρρ

10

13 6 10

3

3.

⎛⎛⎝⎜

⎞⎠⎟( ) =160 11 8mm 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 ρf gV gVdisplaced rod= 0 . But, assuming a cylindrical rod, V r Lrod = π

2 . The volume of fl uid

displaced is the same as the volume of the rod that is submerged, or V r L hdisplaced = −( )π 2 .

Thus, ρ π ρ πf g r L h g r L20

2−( )⎡⎣ ⎤⎦ = ⎡⎣ ⎤⎦, which reduces to

ρ ρf

L

L h=

−⎛⎝

⎞⎠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

P g h L

P

A w

B

+ + −( )= +

0

1

2

ρ θsin

ρρw Bv2 0+

Recognize that P P PA B= = atm since both points are open to the atmosphere. Thus, we obtain

vB g h L= −( ) = ( ) − ( )2 2 9 80 10 0 2 00sin . . . siθ m s m m2 nn . .30 0 13 3º m s⎡⎣ ⎤⎦ =

Now the problem reduces to one of projectile motion with

v v0 30 0 6 64y B= =sin . .º m s

At the top of the arc, vy y y= =0, and max.

Then, v vy y ya y202 2= + ( )∆ gives 0 6 64 2 9 80 0

2= ( ) + −( ) −( ). .m s m s2maxy , or

ymax m above the level of point B= 2 25. .

�Valve

A

h

L B

�Valve

A

h

L B


494 Chapter 9

9.86 When the balloon comes into equilibrium, the weight of the displaced air equals the weight of the fi lled balloon plus the weight of string that is above ground level. If ms and L are the total mass

and length of the string, the mass of string that is above ground level is ( )h L ms. Thus,

ρ ρair balloon balloon helium balloongV m g gVh

L= + + ⎛⎛⎝⎜

⎞⎠⎟ m gs

which reduces to

hV m

mL

s

=−( ) −⎡

⎣⎢

⎤

⎦⎥

ρ ρair helium balloon balloon

This yields

h =−( ) ( )⎡⎣ ⎤⎦ −1 29 0 179 4 0 40 3 0. . . .kg m kg m m3 3 3π 225

0 0502 0 1 9

kg

kgm 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 ρair balloon balloon helium balloon lgV m g gV m= + + ooad g,

or

Vm m

balloonballoon load

air helium

kg= +−

= +ρ ρ

600 44 000

1 29 0 1794 14 103 kg

kg m m3

3

. ..

−( ) = ×

9.88

L

L

h

H

A B

A B

C D

→v

(b) (c)

Shield

(a) Consider the pressure at points A and B in part (b) of the fi gure by applying P P ghf= +0 ρ .Looking at the left tube gives atm waterP P g L hA = + −( )ρ , and looking at the tube on the right,

atm oilP P gLB = + ρ .

Pascal’s principle says that P PB A= . Therefore, P gL P g L hatm oil atm water+ = + −( )ρ ρ , giving

h L= −⎛⎝⎜

⎞⎠⎟

= −⎛⎝

1 1ρρ

oil

water

3

3

750 kg m

1 000 kg m⎜⎜⎞⎠⎟( ) =5.00 cm cm1 25.




(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 gyA A A B B B+ + = + +1

2

1

22 2ρ ρ ρ ρair air air airv v

Since y yA B A B= = =, , and v v v 0, this reduces to

P PB A− = 1

22ρair v [1]

Now use P P ghf= +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, waterP P gLC A= + ρ , and from the right tube, oilP P gLD B= + ρ .

Pascal’s principle says that P PD C= , and equating these two gives

oil waterP gL P gLB A+ = +ρ ρ , or water oilP P gLB A− = −( )ρ ρ [2]

Combining Equations [1] and [2] yields

v =−( )

=−( )(

2

2 1 000 750 9 80

ρ ρρ

water oil

air

2 m s

gL

. )) ×( )=

−5 00 10

1 2913 8

2.

..

m m s

9.89 While the ball is submerged, the buoyant force acting on it is B V gw= ( )ρ . The upward accelera-tion of the ball while under water is

aF

m

B mg

m mr gy

y w= = − = ⎛⎝

⎞⎠ −

⎡⎣⎢

⎤⎦⎥

=

Σ ρ π4

31

1 000

3

kg mm

1.0 kgm m

3( ) ⎛⎝

⎞⎠ ( ) −

⎡

⎣⎢⎢

⎤

⎦⎥⎥

4

30 10 1 9 803π

. . ss m s2 2( ) = 31

Thus, when the ball reaches the surface, the square of its speed is

v vy y ya y202 2 0 2 31 2 0 125= + ( ) = + ( )( ) =∆ m s m m s2 2. 22

When the ball leaves the water, it becomes a projectile with initial upward speed of v0 125y = m s and acceleration of a gy = − = −9 80. m s2. Then, v vy y ya y2

02 2= + ( )∆ gives the

maximum height above the surface as

ymax

2 2

2

m s

2 9.80 m sm= −

−( ) =0 125

6 4.


496 Chapter 9

9.90 Since the block is fl oating, the total buoyant force must equal the weight of the block. Thus,

ρ ρoil water cm

A x g A x g4 00. −( )⎡⎣ ⎤⎦ + ⋅[ ] cmwood= ( )⎡⎣ ⎤⎦ρ A g4 00.

where A is the surface area of the top or bottom of the rectangular block.

Solving for the distance x gives

x = −−

⎛⎝⎜

⎞⎠⎟( ) = −ρ ρ

ρ ρwood oil

water oil

cm4 00960

.9930

1 000 9304 00 1 71

−⎛⎝⎜

⎞⎠⎟( ) =. . cm cm

9.91 A water droplet emerging from one of the holes becomes a projectile with v v v0 00y x= = and . The time for this droplet to fall distance h to the fl oor is

found from ∆y t a ty y= +v012

2 to be

th

g= 2

The horizontal range is

R th

g= =v v

2 .

If the two streams hit the fl oor at the same spot, it is necessary that R R1 2= , or

v v11

222 2h

g

h

g=

With h h1 25 00 12 0= =. .cm and cm, this reduces to

v v v1 22

12

12 0

5 00= =h

h

.

.

cm

cm, or v v1 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 ghatm atm+ + = + +1

201

21 3ρ ρ ρv , or v1

23 12= −( )g h h . [2]

Similarly, applying Bernoulli’s equation to point 2 (the upper hole) and point 3 gives

P gh P ghatm atm+ + = + +1

202

22 3ρ ρ ρv , or v2

23 22= −( )g h h . [3]

Square Equation [1] and substitute from Equations [2] and [3] to obtain

2 2 40 23 1 3 2g h h g h h−( ) = −( )⎡⎣ ⎤⎦.

Solving for h3 yields

hh h

32 12 40

1 40

2 40 12 0 5 00

1 40= − =

( ) − =.

.

. . .

.

cm cm117 0. cm,

so the surface of the water in the tank is 17.0 cm above floor level .

4.00 cm � x

4.00 cm

Oil

Water

x

4.00 cm � x

4.00 cm

Oil

Water

x

R1 � R2

1

2

3

h2

h1

h3

R1 � R2

1

2

3

h2

h1

h3



9.92 When the section of walkway moves downward distance ∆L, the cable is stretched distance ∆L and the column is compressed distance ∆L. The tension force required to stretch the cable and the compression force required to compress the column this distance is

FY A L

Lcablesteel cable

cable

= ∆ and F

Y A L

LcolumnAl column

column

= ∆

Combined, these forces support the weight of the walkway section:

F F Fgcable column N+ = = 8 500 or Y A L

L

Y A L

Lsteel cable

cable

Al column

column

∆ ∆+ = 8 5000 N

giving

∆LY A

L

Y A

L

=+

8 500 N

steel cable

cable

Al column

colummn

The cross-sectional area of the cable is

AD

cable

m= =

×( )−π π2 2 2

4

1 27 10

4

.

and the area of aluminum in the cross section of the column is

AD D D D

cableouter inner outer inner= − =

−π π π2 2 2 2

4 4

(( )=

( ) − ( )⎡⎣

⎤⎦

4

0 162 4 0 161 4

4

2 2π . . m m

Thus, the downward displacement of the walkway will be

∆L =×( ) ×( )−

8 500

20 10 1 27 10

4 5 75

10 2 2

N

Pa m

m

π .

.(( ) +×( ) ( ) − ( )⎡

⎣7 0 10 0 162 4 0 161 410 2 2. . . Pa m mπ ⎤⎤

⎦( )4 3 25. m

or

∆L = × =−8 6 10 0 864. . m mm


Solucionario Fundamentos de Física 9na edición Capitulo 9

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Transcript of Solucionario Fundamentos de Física 9na edición Capitulo 9