Chapter 9

Page 9.1

9

Solids and Fluids

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

2min

min4

D F mgA

elastic limit elastic limit

and the minimum acceptable diameter is

2

3min 8

4 70 kg 9.8 m s41.3 10 m 1.3 mm

5.0 10 Pa

mgD

elastic limit

9.2 (a) In order to punch a hole in the steel plate, the

superhero must punch out a plug with cross-

sectional area, csA , equal to that of his fist and

a height t equal to the thickness of the steel plate. The area shearA of the face that is sheared as the plug is re-

moved is the cylindrical surface with radius r and height t as shown in the sketch. Since Acs = r2, then

csr A and

2 2

2shear

1.00 10 cm2 2 2 2.00 cm 70.9 cmcsA

A r t t

If the ultimate shear strength of steel (i.e., the maximum shear stress it can withstand before shearing) is,

8 8 22.50 10 Pa 2.50 10 N m the minimum force required to punch out this plug is

2

2shear

1 m70.9 cmF A stress 8

4 2 2

N2.50 10

10 cm m

61.77 10 N

Chapter 9

Page 9.2

(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, sepa-

rated by distance 2.00 mh and each with area, 22.00 cm 15.0 cm 30.0 cmA move a distance

25.00 10 mx 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

shear stress F A FhS

shear strain x h x A

we have

2

7

2 2 2 4 2

80.0 kg 9.80 m s 2.00 m1.05 10 Pa

5.00 10 m 30.0 cm 1 m 10 cmS

9.4 As a liquid, the water occupied some volume lV . 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 lV . Then, 0 1.09 and 0.090l lV V V V , so

9 82

0

0.090N2.00 10 1.65 10 Pa 1600 atm

m 1.09

l

l

VVP B

V V

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

2

82

2

4 90 kg 9.80 m s 50 m3.5 10 Pa

1.0 10 m 1.6 mY

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

2

0 0

210 3 3

2

4

18 10 Pa 0.22 10 m 0.10 10 m22 N

4 3.1 10 m

Y d LY A LF

L L

Chapter 9

Page 9.3

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

cos30 cos30 0x xR F F F , and a y-component of sin 30º sin 30º 22 Ny yR F F F F .

Thus, the resultant force is

22 N directed down the page in the diagramR .

9.7 From0 0( ) ( ) ( ) ( )Y F A L L stress L L , the maximum compression the femur can withstand is

6

0 39

160 10 Pa 0.50 m4.4 10 m 4.4 mm

18 10 Pa

stress LL

Y

9.8 (a) When at rest, the tension in the cable equals the weight of the 800-kg object, 37.84 10 N .

Thus, from 0 ( )Y F L A L , the initial elongation of the cable is

30 3

4 2 10

7.84 10 N 25.0 m2.45 10 m= 2.5 mm

4.00 10 m 20 10 Pa

F LL

A Y

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

yF mg ma

yF m g a [1]

If 2800 kg and 3.0 m sym a , the elongation of the cable will be

20 3

4 2 10

800 kg 9.80 3.0 m s 25.0 m3.2 10 m 3.2 mm

4.00 10 m 20 10 Pa

F LL

A Y

Thus, the increase in the elongation has been

initial

3.20 mm 2.45 mm 0.75 mmincrease L L

(c) From the definition of the tensile stress, stress F A , the maximum tension the cable can withstand is

Chapter 9

Page 9.4

4 2 8 4max max

4.00 10 m 2.2 10 Pa 8.8 10 NF A stress

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

4max 3

max 2

8.8 10 N6.9 10 kg

9.8 3.0 m s

Fm

g a

9.9 From the defining equation for the shear modulus, we find the displacement, x , as

3 4 2

26 2

5

5.0 10 m 20 N 10 cm

1 m3.0 10 Pa 14 cm

2.4 10 m 0.024 mm

h F A h Fx

S S A

9.10 The shear modulus is given by

shear stress stressS

shear strain x h

Hence, the stress is

10 63

5.0 m1.5 10 Pa 7.5 10 Pa

10 10 m

xstress S

h

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.

0 00 0Cu CuAl Al

rod Al CuAl Cu Al Cu

F L LF L LFL L L

AY AY A Y Y

where 2A r with 30.20 cm 2.0 10 mr . Thus,

3

2rod 2 10 10

3

5.8 10 N 1.3 m 2.6 m1.9 10 m 1.9 cm

7.0 10 Pa 11 10 Pa2.0 10 mL

Chapter 9

Page 9.5

9.12 The acceleration of the forearm has magnitude

3

3 23

km 10 m 1 h80

h 1 km 3 600 s4.4 10 m s

5.0 10 sa

t

v

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

3 2

7

2 4 2 2

3.0 kg 4.4 10 m s5.5 10 Pa

2.4 cm 10 m 1 cm

FStress

A

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

0 3 3

3

4 3 4

M M M

V R R

The average density of the combined world is

2total

3 32 3

4 22 32

934 3

3 4

MM M M

V RRR

so

3

30

32 4 1284.74

9 3 27

M R

R M

or 04.74

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

total 3 3Au total

# karats 22 117.988 10 kg 7.322 10 kg

24 24 12

mm m

and the mass of copper is

3 4Cu total

1 17.988 10 kg 6.657 10 kg

12 12m m

Chapter 9

Page 9.6

(b) The volume of the gold present is

3

Au 7 3Au 3 3

Au

7.322 10 kg3.79 10 m

19.3 10 kg m

mV

and the volume of the copper is

4

Cu 8 3Cu 3 3

Cu

6.657 10 kg7.46 10 m

8.92 10 kg m

mV

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

3

total total 4 3av 7 3 8 3

total Au Cu

7.988 10 kg1.76 10 kg m

3.79 10 m 7.46 10 m

m m

V V V

9.15 (a) The total normal force exerted on the bottom acrobat’s shoes by the floor equals the total weight of the acro

bats in the tower. That is

2 3total 75.0 68.0 62.0 55.0 kg 9.80 m s 2.55 10 Nn m g

(b)

34

2 2 4 2total shoe

sole

2.55 10 N3.00 10 Pa

2 2 425 cm 1 m 10 cm

n nP

A A

(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 total shoe sole2A A , and hence the

pressure, will change unless all the acrobats wear the same size shoes.

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 floor is 4n mg . The pressure of each leg on the floor is then

2

6leg 22

2leg

95.0 kg 9.80 m s42.96 10 Pa

4 0.500 10 m

n mgP

A r

9.17 (a) If the particles in the nucleus are closely packed with negligible space between them, the average nuclear

Chapter 9

Page 9.7

density should be approximately that of a proton or neutron. That is

27proton proton 17 3

nucleus 3315proton

3 1.67 10 kg 4 10 kg m

4 3 4 1 10 m

m m

V r

(b) The density of iron is 3 3Fe 7.86 10 kg m kg/m3 and the densities of other solids and liquids are on

the order of 3 310 kg m . Thus, the nuclear density is about 1410 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 4W , and the gauge pressure is

4

F WP

A A

Thus, 2 5 44 4 0.024 m 2.0 10 Pa 1.9 10 NW A P .

9.19 The volume of concrete in a pillar of height h and cross-sectional area A is V Ah , and its weight is

4 35.0 10 N mgF Ah . The pressure at the base of the pillar is then

4 3

4 35.0 10 N m

5.0 10 N mg AhF

P hA A

Thus, if the maximum acceptable pressure on the base is, 7max 1.7 10 PaP , the maximum allowable height is

7

max 2max 4 3 4 3

1.7 10 Pa3.4 10 m

5.0 10 N m 5.0 10 N m

Ph

9.20 Assuming the spring obeys Hooke’s law, the increase in force on the piston required to compress the spring an ad-

ditional amount x is 0 0F F F P P A k x . The gauge pressure at depth h beneath the surface

of a fluid is 0P P gh , so we have, ghA k x or the required depth is h k x gA , If

1 250 N mk , 2 2 with 1.20 10 mA r r , with 3 31.00 10 kg m , and the fluid is water (

= 1.00 103 kg/m3), the depth required to compress the spring an additional is 30.750 cm 7.50 10 m is

Chapter 9

Page 9.8

3

23 3 2 2

1 250 N m 7.50 10 m2.11 m

1.00 10 kg m 9.80 m s 1.20 10 m

h

9.21 (a) 3 3 3 2 50 101.3 10 Pa 1.00 10 kg m 9.80 m s 27.5 m 3.71 10 PaP P gh

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

2

22 5 4

35.0 10 m3.71 10 Pa 3.57 10 N

2F PA P r

9.22 The gauge pressure in a fluid at a level distance h below where gauge 0P is

gaugeP gh with h being positive

when measured in the downward direction. The difference in gauge pressures at two levels is

gauge 2 gauge 1( ) ( ) ( )P P g h or (Pgauge)2 = (Pgauge)1 + g (h) with h being positive if, in going from level 1

to level 2, one is moving lower in the fluid.

(a) In moving from the heart to the head, one is moving higher in the blood column so 0h and we find

3 3 2gauge gauge

head heart13.3 10 Pa 1 060 kg m 9.80 m s 0.500 mP P g h

or

3gauge

head8.11 10 Pa 8.11 kPaP

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

3 3 2gauge gauge

feet heart13.3 10 Pa 1 060 kg m 9.80 m s 1.30 mP P g h

or

3gauge

feet26.8 10 Pa 26.8 kPaP

9.23 The density of the solution is 3 3water1.02 1.02 10 kg m . The gauge pressure of the fluid at the level

of the needle must equal the gauge pressure in the vein, so 3gauge 1.33 10 PaP gh , and

Chapter 9

Page 9.9

3gauge

3 3 2

1.33 10 Pa0.133 m

1.02 10 kg m 9.80 m s

Ph

g

9.24 (a) From the definition of bulk modulus, 0B P V V , the change in volume of the 31.00 m of seawater

will be

3 8 5

0 310

water

1.00 m 1.13 10 Pa 1.013 10 Pa0.053 8 m

0.210 10 Pa

V PV

B

(b) The quantity of seawater that had volume 30 1.00 mV at the surface has a mass of 1 030 kg. Thus, the

density of this water at the ocean floor is

3 33

0

1 030 kg1.09 10 kg m

1.00 0.053 8 m

m m

V V V

(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 first find the absolute pressure at the interface between oil and water.

1 0 oil oil

5 3 2 51.013 10 Pa 700 kg m 9.80 m s 0.300 m 1.03 10 Pa

P P gh

This is the pressure at the top of the water. To find the absolute pressure at the bottom, we use, P2 = P1 + waterghwa-

ter, or

5 3 3 2 52 1.03 10 Pa 10 kg m 9.80 m s 0.200 m 1.05 10 PaP

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 + gh = gh. Thus,

5atmo

3 2

1.013 10 Pa10.5 m

984 kg m 9.80 m s

Ph

g

Chapter 9

Page 9.10

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

9.27 Pascal’s principle, 1 1 2 2F A F A , or pedal Master brake brakecylinder cylinder

F A F A , gives

2

brake cylinder

brake pedal 2master cylinder

6.4 cm44 N 156 N

1.8 cm

AF F

A

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

0.50 156 N 78 Nkf n

and the torque is drum 78 N 0.34 m 27 N mf r .

9.28 First, use Pascal’s principle, F1/A1 = F2/A2, to find the force

piston 1 will exert on the handle when a 500-lb force

pushes downward on piston 2.

2 21 1 1

1 2 2 22 22 2 2

2

2

4

4

0.25 in500 lb 14 lb

1.5 in

A d dF F F F

A d d

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

14 lb 2.0 in 12 in 0F , 2.3 lbF

9.29 When held underwater, the ball will have three forces acting on it: a downward gravitational force, mg; an upward

buoyant force, B = waterV = 4waterr3/3; and an applied force, F. If the ball is to be in equilibrium, we have (taking

upward as positive) 0yF F B mg , or

3 3

water water

4 4

3 3

r rF mg B mg g m g

giving

Chapter 9

Page 9.11

3

3 3 24 0.250 m

0.540 kg 1.00 10 kg m 9.80 m s 74.9 N3 2

F

so the required applied force is . 74.9 N directed downwardF

9.30 (a) To float, 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,

sea submerged body total B mg gV gV or submerged body

total sea

V

V

After inhaling,

3

submerged

3total

945 kg m0.768 76.8%

1 230 kg m

V

V

leaving .23.2% above surface.

After exhaling,

3

submerged

3total

1 020 kg m0.829 82.9%

1 230 kg m

V

V

leaving .17.1% above surface.

(b) In general, “sinkers” would be expected to be thinner with heavier bones, whereas “floaters” 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,

truck water

3 23 2

kg m10 4.00 m 6.00 m 4.00 10 m 9.80

m s

w V g

or

3truck 9.41 10 N 9.41 kNw

Chapter 9

Page 9.12

9.32 (a)

(b) Since the system is in equilibrium, 0 .y rF B w w

(c)

submerged

3 2 21 025 kg m 9.80 m s 0.024 0 m 4.00 m 964 N

w wB gV g d A

(d) From 0rB w w ,

2964 N 62.0 kg 9.80 m s 356 Nr sw B w B m g

(e) 3

foam 2 2

356 N101 kg m

9.80 m s 0.090 m 4.00 m

r r

r

m w g

V t A

(f)

max

3 2 2 31 025 kg m 9.80 m s 0.090 0 m 4.00 m 3.62 10 N

w r wB gV g t A

(g) The maximum weight of survivors the raft can support is max max max rw m g B w so

3

maxmax 2

3.62 10 N 356 N333 kg

9.80 m s

rB wm

g

9.33 (a) While the system floats, total block steelB w w w .or

w g submerged bV g steelbV m g .

When steel 0.310 kgm , 4 3submerged 5.24 10 mbV V giving

steel steel 3 3 3

4 3

0.310 kg1.00 10 kg m 408 kg m

5.24 10 m

w bb w

b b

V m m

V V

(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 float 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 in-

Chapter 9

Page 9.13

creased 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 mstee 0.350 kg.

9.34 (a)

(b) Since the balloon is fully submerged in air, Vsubmerged = Vb = 325 m3, and

3 2 3 3air 1.29 kg m 9.80 m s 325 m 4.11 10 NbB gV

(c)

He He He

3 3 3 2 34.11 10 N 226 kg 0.179 kg m 325 m 9.80 m s 1.33 10 N

y b b b bF B w w B m g gV B m V g

Since Fy = may 0, ay will be positive (upward), and the balloon rises.

(d) If the balloon and load are in equilibrium, He load 0y bF B w w w and

3load He 1.33 10 Nbw B w w . Thus, the mass of the load is

3

loadload 2

1.33 10 N136 kg

9.80 m s

wm

g

(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

Chapter 9

Page 9.14

these effects, the buoyant force will decrease until at some height the balloon will come to equilibrium and go

no higher.

9.35 (a) 3

33 2air balloon air

3

4 41.29 kg m 9.80 m s 3.00 m

3 3

1.43 10 N 1.43 kN

rB gV g

(b) 3 2total

3

1.43 10 N 15.0 kg 9.80 m s

1.28 10 N 1.28 kN upward

yF B w

(c) The balloon expands as it rises because the external pressure (atmospheric pressure) decreases with increas-

ing altitude.

9.36 (a) Taking upward as positive, y yF B mg ma , or

y wma gV mg .

(b) Since m = V, we have V y wa gV V g , or

1wya g

(c) 3 3

2 2 23

1.00 10 m kg1 9.80 m s 0.467 m s 0.467 m s downward

1 050 m kgya

(d) From 20 2y yy t a t v with 0y = 0, we find

2

2 2 8.00 m5.85 s

0.467 m sy

yt

a

Chapter 9

Page 9.15

3total single air balloon air

balloon

33 2 3

4 (a) 600 600 600

3

4600 1.29 kg m 9.80 m s 0.50 m 4.0 10 N 4.0 kN

3

B B gV g r

9.37

(b) 3 2 3total total 4.0 10 N 600 0.30 kg 9.8 m s 2.2 10 N 2.2 kNyF B m g

(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 significant figures in this problem to illustrate a point.

(a) The absolute pressure at the level of the top of the block is

top 0 water top

5 3 23 2

5

kg m1.0130 10 Pa 10 9.80 5.00 10 m

m s

1.0179 10 Pa

P P gh

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

bottom 0 water bottom

5 3 23 2

5

kg m1.0130 10 Pa 10 9.80 17.0 10 m

m s

1.0297 10 Pa

P P gh

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

25

top top 1.0179 10 Pa 0.100 m 1017.9 NF P A

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

25

bot bot 1.0297 10 Pa 0.100 m 1029.7 NF P A

(b) The scale reading equals the tension, T, in the cord supporting the block. Since the block is in equilibrium,

bot top 0yF T F F mg , or

Chapter 9

Page 9.16

210.0 kg 9.80 m s 1029.7 1017.9 N 86.2 NT

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

water block

23 3 210 kg m 0.100 m 0.120 m 9.80 m s 11.8 N

B V g

From part (a), bot top 1 029.7 1 017.9 N 11.8 NF F , 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:

0y yF ma

4sea water1.20 10 kg 1 100 N 0m g gV

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

Thus,

33 4

3 2

kg 4 1 100 N1.03 10 1.50 m 1.20 10 kg

m 3 9.80 m sm

or

32.67 10 kgm

9.40 At equilibrium, spring 0yF B F mg , so the spring force is

spring water blockF B mg V m g

where

3 3

block 3wood

5.00 kg7.69 10 m

650 kg m

mV

Chapter 9

Page 9.17

Thus, 3 3 3 3 2spring 10 kg m 7.69 10 m 5.00 kg 9.80 m s 26.4 N.F

The elongation of the spring is then

spring 26.4 N

0.165 m 16.5 cm160 N m

Fx

k

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 = (alcoholV)g. Since the sample is fully submerged, the volume of the displaced alco-

hol is the same as the volume of the sample. This volume is

2 3

3 2alcohol

100 N1.46 10 m

700 kg m 9.80 m s

BV

g

(b) The mass of the sample is

2

300 N30.6 kg

9.80 m s

weight in airm

g

and its density is

3 32 3

30.6 kg2.10 10 kg m

1.46 10 m

m

V

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 fluid.

(a) The mass of the object is

2

300 N30.6 kg

9.80 m s

weight in airm

g

The buoyant force when immersed in water is the weight of a volume of water equal to the volume of the

object, or Bw = (wV)g. Thus, the volume of the object is

Chapter 9

Page 9.18

3 3

3 3 2

300 N 265 N3.57 10 m

10 kg m 9.80 m s

w

w

BV

g

and its density is

3 3object 3 3

30.6 kg8.57 10 kg m

3.57 10 m

m

V

(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

Boil= (oilV)g, or the density of the oil is

oil 3

oil 3 3 2

300 N 275 N714 kg m

3.57 10 m 9.80 m s

B

Vg

9.43 The volume of the iron block is

iron 4 3

3 3iron

2 .00 kg2 .54 10 m

7.86 10 kg m

mV

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

3 4 3 2oil 916 kg m 2.54 10 m 9.80 m s 2 .28 NB V g

Applying Fy = 0 to the iron block gives the support force exerted by the upper scale (and hence the reading on that

scale) as

upper iron 19.6 N 2.28 N 17.3 NF m g B

From Newton’s third law, the iron exerts force B downward on the oil (and hence the beaker). Applying Fy = 0 to

the system consisting of the beaker and the oil gives

lower oil beaker 0F B m m g

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

2lower oil beaker 2.28 N 2.00 1.00 kg 9.80 m s 31.7 NF B m m g

Chapter 9

Page 9.19

9.44 (a) The cross-sectional area of the hose is 22 2 4 2.74 cm 4A r d , and the volume flow rate

(volume per unit time) is A =25.0 L/1.50 min. Thus,

25.0 L 1.50 min 25.0 L

A v

1.50 min 2 2

4 1 min

2.74 cm

3 310 cm

60 s 1 L

2

1 m47.1 cm s 0.471 m s

10 cm

(b)

2 222 2 2

21 1 1

4 1 1

4 3 9

A d d

A d d

or 12

9

AA

Then from the equation of continuity, 2 2 1 1A Av v , we find

12 1

2

9 0.471 m s 4.24 m sA

A

v v

9.45 (a) The volume flow rate is A, and the mass flow rate is

3 21.0 g cm 2.0 cm 40 cm s 80 g sA v

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

2

aortacapiliaries aorta 3 2

capillaries

2 .0 cm40 cm s

3.0 10 cm

A

A

v v

or 2capiliaries 2.7 10 cm s 0.27 mm s . v

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

2

12 1 2

2

10.0 cm2.75 m s 11.0 m s

2.50 cm

A

A

v v

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

2 22 2 5 3 3

2 1 1 2

1 11.20 10 Pa 1.65 10 kg m 2.75 m s 11.0 m s

2 2P P

v v

or 42 2.64 10 PaP .

Chapter 9

Page 9.20

9.47 From Bernoulli’s equation, choosing at y = 0 the level of the syringe and needle, so the flow speed in the needle,

1 12 22 2 1 12 2

P P v v is

1 22

2 1

2 P P

v v

In this situation,

41 2 1 atmo 1 gauge 5 2

1

2 .00 N8.00 10 Pa

2 .50 10 m

FP P P P P

A

Thus, assuming 0,

4

2 3 3

2 8.00 10 Pa0 12.6 m s

1.00 10 kg m

v

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

2 31 12 2 2 2

1 2 2 1 1 1 12 2 22P P

v v v v v , or

2

3 2 23

1.29 kg m 15 10 m s 4.4 10 Pa2

P

9.49 (a) Assuming the airplane is in level flight, the net lift (the difference in the upward and downward forces ex-

erted on the wings by the air flowing over them) must equal the weight of the plane, or

lower upper wingssurface surface

( )P P A mg . This yields

4 2

3lower upper 2surface surface wings

8.66 10 kg 9.80 m s9.43 10 Pa

90.0 m

mgP P

A

(b) Neglecting the small difference in altitude between the upper and lower surfaces of the wings and applying

Bernoulli’s equation yields

2 2lower air lower upper air upper

1 1

2 2P P v v

Chapter 9

Page 9.21

so

3lower upper 22

upper lower 3air

2 2 9.43 10 Pa225 m s 255 m s

1.29 kg m

P P

v v

(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 flight, the net lift (difference between the upward and downward forces exerted on the wing surfaces by

air flowing over them) must equal the weight of the aircraft, or lower uppersurface surface

( )P P A Mg . This gives the air

pressure at the upper surface as

upper lowersurface surface

MgP P

A

9.51 (a) Since the pistol is fired horizontally, the emerging water stream has initial velocity components of (0x =

nozzle, 0y = 0) Then, y = 0yt + ayt2/2, with ay = g, gives the time of flight as

2

2 2 1.50 m0.553 s

9.80 m sy

yt

a

(b) With ax = 0, and 0x = nozzle, the horizontal range of the emergent stream is x = nozzlet where t is the time of

flight from above. Thus, the speed of the water emerging from the nozzle is

nozzle

8.00 m14.5 m s

0.553 s

x

t

v

(c) From the equation of continuity,A11 = A22, the speed of the water in the larger cylinder is 1 = (A2/A1) 2 =

(A2/A1)nozzle, or

2 22

2 21 nozzle nozzle2

1 1

1.00 mm14.5 m s 0.145 m s

10.0 mm

r r

r r

v v v

(d) The pressure at the nozzle is atmospheric pressure, or P2 = 1.013 105 m/s .

(e) With the two cylinders horizontal, y1 = y2 and gravity terms from Bernoulli’s equation can be neglected,

Chapter 9

Page 9.22

leaving 2 21 1 2 22 2w wP P v v so the needed pressure in the larger cylinder is

3 3

2 22 2 51 2 2 1

1.00 10 kg m1.013 10 Pa 14.5 m s 0.145 m s

2 2

wP P

v v

or

51 2.06 10 PaP

(f) To create an overpressure of P = 2.06 105 Pa = 1.05 105 Pa in the larger cylinder, the force that must be

exerted on the piston is

2

2 5 21 1 1 1.05 10 Pa 1.00 10 m 33.0 NF P A P r

9.52 (a) From Bernoulli’s equation,

2 21 2

1 1 2 22 2

w ww wP gy P gy

v v

or

Figure 9.29

1 22 22 1 2 12

w

P Pg y y

v v

and using the given data values, we obtain

5 5

2 2 22 1 3 3

1.75 10 Pa 1.20 10 Pa2 9.80 m s 2.50 m

1.00 10 kg m

v v

and [1]

2 2 2 22 1 61.0 m s v v

From the equation of continuity,

22

1 1 12 1 1 12

2 2 2

A r r

A r r

v v v v

Chapter 9

Page 9.23

2

2 1

3.00 cm

1.50 cm

v v or 2 = 41 [2]

Substituting Equation [2] into [1] gives 2 2 2116 1 61.0 m s v , or

2 2

1

61.0 m s2.02 m s

15 v

(b) Equation [2] above now yields 2 = 4(2.02 m/s) = 8.08 m/s.

(c) The volume flow rate through the pipe is: flow rate = A11 = A22.

Looking at the lower point:

2

2 2 3 31 1 3.00 10 m 2.02 m s 5.71 10 m sflow rate r v

9.53 First, consider the path from the viewpoint of projectile motion to find the speed at which the

water emerges from the tank. From 1 20 2y yy t a t v with 0y = 0, we find the time of flight as

2

2 2 1.00 m0.452 s

9.80 m sy

yt

a

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

2 0

0.600 m1.33 m s

0.452 sx

x

t

v v

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 P1 =

P2 = Patmo and, 1 0. This gives

21 2 2

1

2gy gy v

or

222 2

1 2 2

1.33 m s9.00 10 m 9.00 cm

2 2 9.80 m sh y y

g

v

Chapter 9

Page 9.24

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, P1 = P2 = Patmo and we assume 1 0. This gives 1 22 2 12

gy gy v , or

22 1 22 2 9.80 m s 16.0 m 17.7 m sg y y v

(b) The area of the hole is found from

3 3

6 22

2

2 .50 10 m min 1 min2 .35 10 m

17.7 m s 60 s

flow rateA

v

The diameter is then

6 2

2 32

4 2.35 10 m41.73 10 m 1.73 mm

Ad

9.55 First, determine the flow speed inside the larger portions from

4 3

1 221

1.80 10 m s0.367 m s

2.50 10 m 4

flow rate

A

v

The absolute pressure inside the large section on the left is P1 = P0 + gh1, where h1 is the height of the water in the

leftmost standpipe. The absolute pressure in the constriction is P2 = P1 + gh2, so

1 2 1 2 5.00 cmP P g h h g

The flow speed inside the constriction is found from Bernoulli’s equation with y1 = y2. This gives

2 2 22 1 1 2 1 1 2

22P P g h h

v v v

2 22 0.367 m s 2 9.80 m s 5.00 10 m 1.06 m s v

The cross-sectional area of the constriction is then

4 3

4 22

2

1.80 10 m s1.71 10 m

1.06 m s

flow rateA

v,

Chapter 9

Page 9.25

and the diameter is

4 2

2 22

4 1.71 10 m41.47 10 m 1.47 cm

Ad

9.56 (a) For minimum pressure, we assume the flow is very slow. Then, Bernoulli’s equation gives

2 2

river rim

1 1=

2 2P gy P gy

v v

river rim rivermin0 1 atm 0P g y y

or

5 3river 3 2min

kg m1.013 10 Pa 10 9.80 2096 m 564 m

m sP

5 7 7river min

1.013 10 1.50 10 Pa =1 .51 10 Pa 15.1 MPaP

(b) The volume flow rate is flow rate =A=( d2 /4). Thus, the velocity in the pipe is

3

22

4 4500 m d4 1 d2.95 m s

86 400 s0.150 m

flow rate

d

v

(c) We imagine the pressure being applied to stationary water at river level, so Bernoulli’s equation becomes

2river rim river rim

10 1 atm

2P g y y v

or

2

2 3river river rim rivermin min

river min

1 1 kg m10 2.95

2 2 m s

4.35 kPa

P P P

P

v

Chapter 9

Page 9.26

The additional pressure required to achieve the desired flow rate is

P = 4.35 kPa

9.57 (a) For upward flight of a water-drop projectile from geyser vent to fountain-top, 2 20 2y y ya y v v , with y

= 0, when y = ymax, gives

20 max

0 2 2 9.8 m s 40.0 m 28.0 m sy ya y v

(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 top = 0, then gives

2vent top vent

10

2g y y v

or

2vent top vent2 2 9.80 m s 40.0 m 28.0 m sg y y v

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

2atm vent ventchamber

10

2P gy P gy v

or

2atm vent vent chamber

2

33 2

1

2

28.0 m skg m10 9.80 175 m 2.11 MPa

m 2 s

P P g y y

v

or

6 5gauge atmo 2.11 10 Pa 1 atm 1.013 10 Pa 20.8 atmospheresP P P

9.58 (a) Since the tube is horizontal, y1 = y2 and the gravity

Chapter 9

Page 9.27

terms in Bernoulli’s equation cancel, leaving

2 21 1 2 2

1 1

2 2P P v v

or Figure 9.30(a)

3

1 22 22 1 2 3

2 1.20 10 Pa2

7.00 10 kg m

P P

v v

and

2 2 2 22 1 3.43 m s v v [1]

From the continuity equation, A11 = A22, we find

2 2

1 12 1 1 1

2 2

2.40 cm

1.20 cm

A r

A r

v v v v

or

2 14v v [2]

Substituting Equation [2] into [1] yields 2 2 2115 3.43 m sv and 1 = 0.478 m/s.

Then, Equation [2] gives 2 = 4(0.478 m/s) = 1.91 m/s.

(b) The volume flow rate is

2

2 2 4 31 1 2 2 2 2 1.20 10 m 1.91 m s 8.64 10 m sA A r v v v

9.59 From Fy = T mg Fy = 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 = (2L) and its vertical

component is F = (2L)cos where is the contact angle. Thus, T = mg + 2Lcos .

0.40 NT when 0 2 0.40 Nmg L [1]

0.39 NT when 180 2 0.39 Nmg L [2]

Subtracting Equation [2] from [1] gives

Chapter 9

Page 9.28

2

2

0.40 N 0.39 N 0.40 N 0.39 N8.3 10 N m

4 4 3.0 10 mL

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

total

22

2

= = 2 ( )

1.61 10 N= = 7.32 10 N m

4 4 1.75 10 m

F F

L circumference

F

r

9.61 From h = 2cos/gr, the surface tension is

2 3 2 4

2

2 cos

2.1 10 m 1 080 kg m 9.80 m s 5.0 10 m5.6 10 N m

2 cos0

h gr

9.62 The height the blood can rise is given by

2 2 6

2 0.058 N m cos 02 cos5.6 m

1 050 kg m 9.80 m s 2.0 10 mh

gr

9.63 From the definition of the coefficient of viscosity, = FL/A, the required force is

3 2

3

1.79 10 N s m 0.800 m 1.20 m 0.50 m s8.6 N

0.10 10 m

AF

L

v

9.64 From the definition of the coefficient of viscosity, = FL/A, the required force is

3 2

3

1 500 10 N s m 0.010 m 0.040 m 0.30 m s0.12 N

1.5 10 m

AF

L

v

Chapter 9

Page 9.29

9.65 Poiseuille’s law gives 4

1 2

8

P P Rflow rate

L

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

2 5 3

1 atm 442

8 0.12 N s m 50 m 8.6 10 m s8

0.50 10 m

L flow rateP P

R

or

61 atm 2.1 10 Pa 2.1 MPaP P

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

434

5 3

3 2 2

400 Pa 2 .6 10 m 3.2 10 m s

8 8 2 .7 10 N s m 8.4 10 m

P Rflow rate

L

Thus, the flow speed is

5 3

23

3.2 10 m s1.5 m s

2.6 10 m

flow rate

A

v

9.67 If a particle is still in suspension after 1 hour, its terminal velocity must be less than

5max

cm 1 h 1 m5.0 1.4 10 m s

h 3 600 s 100 cmt

v .

Thus, from 1 = 2r2 g(f)/9, we find the maximum radius of the particle:

maxmax

3 2 5

6

2 3

9

2

9 1.00 10 N s m 1.4 10 m s2.8 10 m 2.8 m

2 9.80 m s 1 800 1 000 kg m

t

f

rg

v

Chapter 9

Page 9.30

9.68 From Poiseuille’s law, the excess pressure required to produce a given volume flow rate of fluid with viscosity

through a tube of radius R and length L is

4

8 L V tP

R

If the mass flow rate is (m/t) = 1.0 103 kg/s, the volume flow rate of the water is

3

6 33 3

1.0 10 kg s1.0 10 m s

1.0 10 kg m

V m t

t

and the required excess pressure is

3 2 6 3

54

3

8 1.0 10 Pa s 3.0 10 m 1.0 10 m s1.5 10 Pa

0.15 10 mP

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

3 3 2 31.0 10 kg m 9.8 m s 1.0 m 9.8 10 PaP gh

and the desired flow rate is

3 3 6 3

7 3500 cm 1 m 10 cm

2.8 10 m s30 min 60 s 1 min

V

t

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

1 41 4 3 2 7 3

3

8 1.0 10 Pa s 2.5 10 m 2.8 10 m s82 2 2

9.8 10 Pa

L V tD R

P

or

44.1 10 m 0.41 mmD

9.70 We write Bernoulli’s equation as

Chapter 9

Page 9.31

22out out out in in in

1 1

2 2P gy P gy v v

or

22gauge in out out in out in

1

2P P P g y y

v v

Approximating the speed of the fluid inside the tank as in 0, we find

23 3 2

gauge

11.00 10 kg m 30.0 m s 9.80 m s 0.500 m

2P

or

5gauge 4.55 10 Pa 455 kPaP

9.71 The Reynolds number is

3 2

3 2

1 050 kg m 0.55 m s 2 .0 10 m

2 .7 10 N s m

dRN

v4.3 × 103

In this region (RN > 3 000), the flow is turbulent.

9.72 From the definition of the Reynolds number, the maximum flow speed for streamlined (or laminar) flow in this

pipe is

3 2max

max 3 2

1.0 10 N s m 2 0000.080 m s 8.0 cm s

1 000 kg m 2.5 10 m

RN

d

v

9.73 The observed diffusion rate is 8.0 1014 kg/15 s = 5.3 1015 kg/s. Then, from Fick’s law, the difference in con-

centration levels is found to be

2 1

15

3 3

10 2 4 2

5.3 10 kg s 0.10 m1.8 10 kg m

5.0 10 m s 6.0 10 m

diffusion rate LC C

DA

Chapter 9

Page 9.32

9.74 Fick’s law gives the diffusion coefficient as D = diffusion rate/A.(C/L), where C/L is the concentration gradient.

Thus,

1510 2

4 2 2 4

5.7 10 kg s9.5 10 m s

2.0 10 m 3.0 10 kg mD

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

135 2

6 4

3.0 10 N1.4 10 N s m

6 6 2 .5 10 m 4.5 10 m s

F

r

v

9.76 Using 1 =2r2 g(f)/9, the density of the droplet is found to be = f + 9t/2r2 g. Thus, if r = d/2 = 0.500

103 m t = 1.10 102 m/s and when falling through 20 °C water ( = 1.00 103 N.s/m2) the density of the oil is

3 2 2

3 323

4 2

9 1.00 10 N s m 1.10 10 m skg1 000 1.02 10 kg m

m 2 5.00 10 m 9.80 m s

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

0 yF T mg B

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

3 3 3 2 3water 1.0 10 kg m 0.20 m 9.8 m s 2.0 10 NB V g

For the iron block:

3 3 3 2iron iron 7.86 10 kg m 0.20 m 9.8 m sT V g B B

or

Chapter 9

Page 9.33

4 3 3iron 1.5 10 N 2.0 10 N 13 10 NT

For the aluminum block:

3 3 3 2aluminum aluminum 2.70 10 kg m 0.20 m 9.8 m sT V g B B

or

3 3 3aluminum 5.2 10 N 2.0 10 N 3.3 10 NT

9.78 In going from the ocean surface to a depth of 2.40 km, the increase in pressure is

3 2 2 3 70 1.025 10 kg m 9.80 m s 2.40 10 m 2.41 10 PaP P P gh

The fractional change in volume of the steel ball is given by the defining equation for bulk modulus,

P =B(V/V), as

7

410

steel

2.41 10 Pa1.51 10

16.0 10 Pa

V P

V B

9.79 (a) From Archimedes’s principle, the granite continent will

sink down into the peridotite layer until the weight

f the displaced peridotite equals the weight of the continent.

Thus, at equilibrium,

g pAt g Ad g

or g pt d .

(b) If the continent sinks 5.0 km below the surface of the peridotite, then d = 5.0, and the result of part (a) gives

the first approximation of the thickness of the continent as

3 3

p

3 3g

3.3 10 kg m5.0 km 5.9 km

2.8 10 kg mt d

9.80 (a) Starting with P = P0 + gh, we choose the reference level at the level of the heart, so P0 = PH. The pressure at

the feet, a depth hH below the reference level in the pool of blood in the body is PF = PH + ghH. The pressure

Chapter 9

Page 9.34

difference between feet and heart is then

PF = PH + ghH .

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

3 3 2 41.06 10 kg m 9.80 m s 1.20 m 1.25 10 PaF HP P

9.81 The cross-sectional area of the aorta is 21 1 4A d and that of a single capillary is 2

2 4cA d If the circula-

tory system has N such capillaries, the total cross-sectional area carrying blood from the aorta is

22

24

c

N dA NA

From the equation of continuity,

1

2 12

A A

v

v, or

2 22 1 1

24 4

N d d

v

v,

which gives

2 22

1 1 72 6

2 2

1.0 m s 0.50 10 m2.5 10

1.0 10 m s 10 10 m

dN

d

v

v

9.82 (a) We imagine that a superhero is capable of producing a perfect vacuum above the water in the straw. Then

P = P0 + gh, 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

5 2atm atm

max 3 3 2water water

0 1.013 10 N m10.3 m

1.00 10 kg m 9.80 m s

P Ph

g g

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

9.83 (a) 22 H O160 mm of H O 160 mmP g

33 2

kg m10 9.80 0.160 m

m s

1.57 kPa

Chapter 9

Page 9.35

35

1 atm1.57 10 Pa

1.013 10 PaP

21.55 10 atm

The pressure is 2 2H O H O Hg HgP gh gh

2

2

3 3H O

Hg H O 3 3Hg

10 kg m160 mm 11.8 mm of Hg

13.6 10 kg mh h

(b) The fluid level in the tap should rise.

(c) Blockage of flow of the cerebrospinal fluid.

9.84 When the rod floats, the weight of the displaced fluid equals the weight of the rod, or displaced 0 rodf gV gV .

But, assuming a cylindrical rod, 2rodV r L . The volume of fluid displaced is the same as the volume of the rod

that is submerged, or 2displacedV r L h . Thus, 2 2

0f g r L h g r L , which reduces to

0f

L

L h

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 A 0. This gives

2

0 sin

1 0

2

A w

B w B

P g h L

P

v

Recognize that PA = PB = Patm since both points

are open to the atmosphere. Thus, we obtain

22 sin 2 9.80 m s 10.0 m 2 .00 m sin 30.0º 13.3 m sB g h L v

Now the problem reduces to one of projectile motion with

0 sin 30.0º 6.64 m sy B v v

Chapter 9

Page 9.36

At the top of the arc,y = 0, and y = ymax.

Then, 2 20 2y y ya y v v gives

2 2max0 6.64 m s 2 9.80 m s 0y , or

max 2.25 m above the level of point By .

9.86 When the balloon comes into equilibrium, the weight of the displaced air equals the weight of the filled 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,

air balloon balloon helium balloon s

hgV m g gV m g

L

which reduces to

air helium balloon balloon

s

V mh L

m

This yields

33 31.29 kg m 0.179 kg m 4 0.40 m 3 0.25 kg2.0 m 1.9 m

0.050 kgh

9.87 When the balloon floats, the weight of the displaced air equals the combined weights of the filled balloon and its

load. Thus,

air balloon balloon helium balloon loadgV m g gV m g ,

or

balloon load 3 3balloon 3

air helium

600 kg 4 000 kg4.14 10 m

1.29 0.179 kg m

m mV

9.88

Chapter 9

Page 9.37

(a) Consider the pressure at points A and B in part (b) of the figure by applying P = P0 + fgh. Looking at the left

tube gives atm water AP P g L h , and looking at the tube on the right, atm oil BP P gL .

Pascal’s principle says that PB = PA. Therefore, atm oil atm waterP gL P g L h , giving

3

oil

3water

750 kg m1 1 5.00 cm 1.25 cm

1 000 kg mh L

(b) Consider part (c) of the diagram showing the situation when the air flow over the left tube equalizes the fluid

levels in the two tubes. First, apply Bernoulli’s equation to points A and B. This gives

2 2air air air air

1 1

2 2A A A B B BP gy P gy v v

Since , , and 0A B A By y v v v , this reduces to [1]

2air

1

2B AP P v

Now use P = P0 + fgh to find 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 C AP P gL , and from the right

tube, oil D BP P gL .

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

oil water B AP gL P gL or water oil B AP P gL [2]

Combining Equations [1] and [2] yields

water oil

air

2 2

2

2 1 000 750 9.80 m s 5.00 10 m13.8 m s

1.29

gL

v

9.89 While the ball is submerged, the buoyant force acting on it is B = (wV)g. The upward acceleration of the ball while

under water is

Chapter 9

Page 9.38

3

33 2 2

41

3

1 000 kg m 40.10 m 1 9.80 m s 31 m s

1.0 kg 3

y wy

F B mga r g

m m m

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

2 2 2 2 20 2 0 2 31 m s 2.0 m 125 m sy y ya y v v

When the ball leaves the water, it becomes a projectile with initial upward speed of 0 125 m sy v and accel-

eration of 29.80 m sya g . Then, 2 20 2y y ya y v v gives the maximum height above the surface as

2 2

max 2

0 125 m s6.4 m

2 9.80 m sy

9.90 Since the block is floating, the total buoyant force

must equal the weight of the block. Thus,

oil water

wood

4.00 cm

4.00 cm

A x g A x g

A g

where A is the surface area of the top or bottom

of the rectangular block.

Solving for the distance x gives

wood oil

water oil

960 9304.00 cm 4.00 cm 1.71 cm

1 000 930x

9.91 A water droplet emerging from one of the holes

becomes a projectile with 0y = 0 and 0x = 0. The

time for this droplet to fall distance h to the floor is

found from 1 20 2y yy t a t v to be

Chapter 9

Page 9.39

2 h

tg

The horizontal range is

2 h

R tg

v v .

If the two streams hit the floor at the same spot,

it is necessary that R1 = R2, or

1 2

1 2

2 2h h

g gv v

With h1 = 5.00 cm and h2 = 12.0 cm, this reduces to

2

1 2 21

12.0 cm

5.00 cm

h

h v v v or 1 2 2 .40v v [1]

Apply Bernoulli’s equation to points 1 (the lower hole) and 3 (the surface of the water). The pressure is atmos-

pheric pressure at both points and, if the tank is large in comparison to the size of the holes,3 0. Thus, we obtain

2atm 1 1 atm 3

10

2P gh P gh v or 2

1 3 12 g h h v . [2]

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

2atm 2 2 atm 3

10

2P gh P gh v or 2

2 3 22 g h h v . [3]

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

3 1 3 22 2 .40 2g h h g h h

Solving for h3 yields

2 1

3

2 .40 12 .0 cm 5.00 cm2 .4017.0 cm

1.40 1.40

h hh

,

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

Chapter 9

Page 9.40

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 com-

press the column this distance is

steel cable

cablecable

Y A LF

L

and

Al columncolumn

column

Y A LF

L

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

cable column 8 500 NgF F F or steel cable Al column

cable column

8 500 NY A L Y A L

L L

giving

steel cable Al column

cable column

8 500 NL

Y A Y A

L L

The cross-sectional area of the cable is

2

22

cable

1.27 10 m

4 4

DA

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

2 2

2222outer innerouter inner

cable

0.162 4 m 0.161 4 m

4 4 4 4

D DD DA

Thus, the downward displacement of the walkway will be

2 22 1010 2

8 500 N

7.0 10 Pa 0.162 4 m 0.161 4 m20 10 Pa 1.27 10 m

4 5.75 m 4 3.25 m

L

or

48.6 10 m 0.86 mmL