FLUIDS AND ELASTICITYsrjcstaff.santarosa.edu/~Lwillia2/41/41Ch15key.pdfThe lower the velocity of fluid, the higher the pressure at a given point in the pipe (Bernoulli’s principle)

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15-1

Conceptual Questions

15.1. (a) The density will not change. Although the mass and volume both increase by 32 8 times,= their ratio is unchanged. (b) The volume V has increased by 32 8 times,= but the mass has not changed, so the new density ρ′ is

/(8 ) /8m Vρ ρ′ = = . The new density has decreased by a factor of 8.

15.2. The pressure only depends on the depth with respect to the top surface of the liquid. Since points a, b, and c are all at the same depth (i.e., with respect to point e) the pressure is the same for each, so

a b cp p p= =

15.3. The pressure only depends on the depth from the surface of the liquid. Since point d is the deepest and point e the highest, then

d f ep p p> > A point halfway between point e and point b would have a pressure about the same as the pressure at point d.

15.4. (a) The pressure at the bottom of each tank is given by ,p gdρ= so it will be the same because the depth of the water in each tank is the same. The area of the bottom of tank A is larger than the area of the bottom of tank B. From / ,p F A= we have F = pA, so the A BF F> because the bottom area of tank A is larger. (b) The pressure at any given depth is the same in both tanks because the water depth is the same. Since the area of the sides indicated in Figure Q15.4 is the same for each tank, the force on these sides is also the same, so A B.F F= This makes sense. Since there’s more water in tank A, the total force of the water on the bottom of tank A is larger.

15.5. A floating object displaces it weight in liquid, so placing the boat above point A will increase the depth of the body of water just enough to compensate for the boat’s weight. Thus, the depth of point B is increased slightly so that it is at the same pressure as point A, and A Bp p= . While a column above point A contains the ship, it also contains less water than a similar column above point B, but both columns contain the same weight (or mass if we assume a uniform gravitational field).

15.6. Archimedes’ principle tells us that the displaced liquid has the same weight as the floating object, so L L B B B L L B( / )V V V Vρ ρ ρ ρ= ⇒ =

where the subscripts L and B refer to liquid and block, respectively. The ratio L B/V V is the fraction of the block that is under the liquid, so the densest block is the one for which this fraction is the largest. Thus,

A C Bρ ρ ρ> >

FLUIDS AND ELASTICITY

15

15-2 Chapter 15


You’ve heard that only 10% of an iceberg is visible above the surface of the ocean. That means 90% of the iceberg is below the surface. Therefore the density of ice is 90% the density of seawater. You can verify this by looking in Table 15.1 and finding on the Web the density of ice: 3

ice 917 kg/mρ = .

15.7. Archimedes’ principle states that the buoyant force on an object is equal to the weight of the fluid displaced by the object. Each object displaces exactly the same amount of fluid since each is the same volume. Therefore, the buoyant force on all three objects is the same, so a b c.F F F= = Note that the buoyant force does not depend on the mass or location of the object.

15.8. For objects that are completely submerged the buoyant force is proportional to the volume of the object. If the densities of objects A, B, and C are the same, then the objects with greater mass (A and C) must also occupy a larger volume. Thus, A and C will experience a larger buoyant force than B.

A C BF F F= >

15.9. The sphere is floating in static equilibrium, so the upward buoyant force exactly equals the sphere’s weight: BF w= . But according to Archimedes’ principle, BF is the weight of the displaced liquid. That is, the weight of the missing water in B is exactly matched by the weight of the added ball. Thus, the total weights of both containers are equal.

15.10. The lower the velocity of fluid, the higher the pressure at a given point in the pipe (Bernoulli’s principle). The pressure p in the horizontal liquid-containing pipe is s ,p p gdρ= + where d is the depth of the liquid in the vertical pipes a, b, or c and sp is the gas pressure at the liquid surface. Thus, the vertical pipes with the highest liquid level d have the lowest surface pressure s ,p or the highest gas velocity. Therefore, the gas velocities are b a c.v v v> >

15.11. The pressure is reduced at the chimney due to the movement of the wind above (Bernoulli’s principle). Thus, the air will flow in the window and out the chimney. Prairie dogs ventilate their burrows this way; a small breeze above their mound lowers the pressure there and allows the air in the burrow to move between openings of different types or heights.

15.12. Equation 15.34 is / ( / ).F A Y L L= ∆ The second wire is the same material, and has length 2L L′ = and area 2 2(2 ) 4 4 .A r r Aπ π′ = = = The force required to stretch it the same length 1 mmL∆ = is

4 2 22

L L LF A Y AY AY FL L L∆ ∆ ∆ ′ = ′ = = = ′

So the force required is 4000 N.

15.13. The stress F/A on the wire is proportional to the strain ∆L/L in the elastic region. The breaking point is past the elastic limit. The elastic limit is reached at a particular strain. Since the wire is the same diameter the area A stays the same, so an equal force of 5000 N will also take the wire to its elastic limit.

Exercises and Problems

Section 15.1 Fluids

15.1. Solve: The volume of the liquid is 6

3 30.055 kg 10 mL 50 mL

1100 kg/m mm mVV

ρρ

= ⇒ = = =

Assess: The liquid’s density slightly higher than that of water 3(1000 kg/m ), so it is reasonable that it requires slightly less than 55 mL to get a mass of 55 g.

Fluids and Elasticity 15-3


15.2. Solve: The volume of the helium gas in container A is equal to the volume of the liquid in container B. That is, A BV V= . Using the definition of mass density / ,m Vρ = this relationship becomes

3 3A B He HeB He

A B He B

7000 7000 (7000)(0.18 kg/m ) 1260 kg/mm m m m ρ ρρ ρ ρ ρ

= ⇒ = ⇒ = = =

Referring to Table 15.1, we find that the liquid is glycerine.

15.3. Model: The density of water is 31000 kg/m . Visualize:

Solve: Volume of water in the swimming pool is 31

26 0 m 12 m 3 0 m (6 0 m 12 m 2 0 m) 144 mV = . × × . − . × × . =

The mass of water in the swimming pool is 3 3 5(1000 kg/m )(144 m ) 1 4 10 kgm Vρ= = = . ×

15.4. Model: The densities of gasoline and water are given in Table 15.1. Solve: (a) The total mass is

total gasoline water 0 050 kg 0 050 kg 0 100 kgm m m= + = . + . = . The total volume is

gasoline 4 3watertotal gasoline water 3 3

gasoline water

0 050 kg 0 050 kg 1 24 10 m680 kg/m 1000 kg/m

m mV V Vρ ρ

−. .= + = + = + = . ×

2 3totalavg 4 3

total

0 100 kg 8.1 10 kg/m1 24 10 m

mV

ρ −.

= = = × . ×

(b) The average density is calculated as follows: total gasoline water water water gasoline gasoline

3 3 3water water gasoline gasoline 2 3

avg 3water gasoline

(50 cm )(1000 kg/m 680 kg/m ) 8 4 10 kg/m100 cm

m m m V V

V VV V

ρ ρ

ρ ρρ

= + = +

+ += = = . ×

+

Assess: The above average densities are between those of gasoline and water, which is reasonable.

Section 15.2 Pressure

15.5. Model: The density of sea water is 31030 kg/m . Solve: The pressure below sea level can be found from Equation 15.5 as follows:

5 3 2 40

5 8 8 3

1.013 10 Pa (1030 kg/m )(9.81 m/s )(1.1 10 m)

1.013 10 Pa 1.1103 10 Pa 1.1113 10 Pa 1.1 10 atm

p p gdρ= + = × + ×

= × + × = × = ×

where we have used the conversion 51 atm 1 013 10 Pa= . × . Assess: The pressure deep in the ocean is very large.

15-4 Chapter 15


15.6. Visualize:

Solve: The pressure at the bottom of the vat is 0 1 3 atmp p gdρ= + = . . Substituting into this equation gives

5 2 5 31.013 10 Pa (9.8 m/s )(2.0 m) (1.3)(1.013 10 ) Pa 1550.5 kg/mρ ρ× + = × ⇒ =

The mass of the liquid in the vat is 2 3 2 3(0.50 m) (1550.5 kg/m ) (0.50 m) (2.0 m) 2.4 10 kgm V dρ ρπ π= = = = ×

15.7. Model: The density of water is 31000 kg/m and the density of ethyl alcohol is 3790 kg/m .

Solve: (a) The volume of water that has the same mass as 38 0 m. of ethyl alcohol is 3

3 3water alcohol alcohol alcoholwater 3

water water water

790 kg/m (8 0 m ) 6 3 m1000 kg/m

m m VV ρρ ρ ρ

= = = = . = .

(b) The pressure at the bottom of the cubic tank is 0 water :p p gdρ= +

5 3 2 1 3 51 013 10 Pa (1000 kg/m )(9 8 m/s )(6 3) 1 2 10 Pa/p = . × + . . = . ×

where we have used the relation 1/3water( )d V= .

15.8. Model: 3 3oil waterThe density of oil is 900 kg/m and the density of water is 1000 kg/m .ρ ρ= =

Visualize:

Solve: The pressure at the bottom of the oil layer is 1 0 oil 1,p p gdρ= + and the pressure at the bottom of the water layer is

2 1 water 2 0 oil 1 water 25 3 2 3 2 5

2 (1 013 10 Pa) (900 kg/m )(9 8 m/s )(0 5 m) (1000 kg/m )(9 8 m/s )(1 2 m) 1 2 10 Pa

p p gd p gd gd

p

ρ ρ ρ= + = + +

= . × + . . + . . = . ×

Assess: A pressure of 51 2 10 Pa 1 2 atm. × = . is reasonable.



15.9. Model: The density of seawater is 2seawater 1030 kg/m .ρ =

Visualize:

Solve: The pressure outside the submarine’s window is out 0 seawater ,p p gdρ= + where d is the maximum safe depth for the window to withstand a force F. This force is out in/ ,F A p p= − where A is the area of the window. With

in 0,p p= we simplify the pressure equation to

out 0 seawaterseawater

6

2 2 21 0 10 N 3 2 km

(0 10 m) (1030 kg/m )(9 8 m/s )

F Fp p gd dA A g

d

ρρ

π

− = = ⇒ =

. ×= = .

. .

Assess: A force of 61 0 10 N. × corresponds to a pressure of 6

21 0 10 N 314 atm

(0 10 m)FA

ρπ

. ×= = =

.

A depth of 3 km is therefore reasonable.

15.10. Visualize:

We assume that the seal is at a radius of 5 cm. Outside the seal, atmospheric pressure presses on both sides of the cover and the forces cancel. Thus, only the 10-cm-diameter opening inside the seal is relevant, not the 20 cm diameter of the cover. Solve: Within the 10 cm diameter area where the pressures differ,

to left atmos to right gasF p A F p A= =

where 2 3 27 85 10 mA rπ −= = . × is the area of the opening. The difference between the forces is 3 2

to left to right atmos gas( ) (101,300 Pa 20,000 Pa)(7 85 10 m ) 0 64 kNF F p p A −− = − = − . × = .

Normally, the rubber seal exerts a 0.64 kN force to the right to balance the air pressure force. To pull the cover off, an external force must pull to the right with a force 0 64 kN> . .

15-6 Chapter 15


Section 15.3 Measuring and Using Pressure

15.11. Model: The density of water is 31000 kg/m .ρ = Visualize: Please refer to Figure 15.16. Solve: From the figure and the equation for hydrostatic pressure, we have

0 atmosp gh pρ+ =

Using 0 0 atm,p = and 5atmos 1 013 10 Pa,p = . × we get

3 2 50 Pa (1000 kg/m )(9 81 m/s ) 1 013 10 Pa 10 3 mh h+ . = . × ⇒ = .

Assess: This large value of h is due to water having a much smaller density than mercury.

15.12. Model: Assume that the oil is incompressible. Its density is 3900 kg/m . Visualize: Please refer to Figure 15.18. Because the liquid is incompressible, the volume displaced in the left cylinder of the hydraulic lift is equal to the volume displaced in the right cylinder. Solve: Equating the two volumes,

2 22 2 2

1 1 2 2 1 1 2 2 1 21

0 040 m( ) ( ) (0.20 m) 3.2 m0 010 m

rA d A d r d r d d dr

π π . = ⇒ = ⇒ = = = .

15.13. Model: Assume that the vacuum cleaner can create zero pressure. Visualize:

Solve: The gravitational force on the dog is balanced by the force resulting from the pressure difference between the atmosphere and the vacuum hose( 0)p = in the hose. The force applied by the hose is

atmos hose atmos2

4 25

( )

(10 kg)(9 8 m/s ) 9 7 10 m1 013 10 Pa

F p p A p A mg

A −

= − = =

.= = . ×

. ×

Since 2( /2) ,A dπ= the diameter of the hose is 2 / 0 035 m 3 5 cmd A π= = . = . .

Section 15.4 Buoyancy

15.14. Model: The buoyant force on the sphere is given by Archimedes’ principle. Visualize:



Solve: The sphere is in static equilibrium because it is neutrally buoyant. That is,

B G l l s0 N 0 NyF F F V g m gρ∑ = − = ⇒ − =

The sphere displaces a volume of liquid equal to its own volume, l s ,V V= so

2 3s sl 3 34 4

s s3 3

0 0893 kg 7 9 10 kg/m(0 030 m)

m mV r

ρπ π

.= = = = . ×

.

A density of 3790 kg/m in Table 15.1 identifies the liquid as ethyl alcohol. Assess: If the density of the fluid and an object are equal, we have neutral buoyancy.

15.15. Model: The buoyant force on the cylinder is given by Archimedes’ principle. Visualize:

cylV is the volume of the cylinder and wV is the volume of the water displaced by the cylinder. Note that the volume displaced is only from the part of the cylinder that is immersed in water. Solve: The cylinder is in static equilibrium, so B GF F= . The buoyant force is the weight w wV gρ of the displaced water. Thus

wB w w G cyl cyl w w cyl cyl cyl w

cyl

3 2 3cyl

(0 040 m)(1000 kg/m ) 6 7 10 kg/m(0 060 m)

VF V g F mg V g V VV

AA

ρ ρ ρ ρ ρ ρ

ρ

= = = = ⇒ = ⇒ =

.= = . ×

.

where A is the cross-sectional area of the cylinder. Assess: cyl wρ ρ< for a cylinder floating in water is an expected result.


Solve: The sphere is in static equilibrium. The free-body diagram on the sphere shows that

B G B G G G G

3 3sphere sphere sphere sphere

1 40 N3 3

4 3 3 (1000 kg/m ) 750 kg/m3 4 4

y

w w

F F T F F T F F F F

V g V gρ ρ ρ ρ

∑ = − − = ⇒ = + = + =

= ⇒ = = =

15-8 Chapter 15


15.17. Model: The buoyant force on the rock is given by Archimedes’ principle. Visualize:

Solve: Because the rock is in static equilibrium, Newton’s first law gives

net B G rock

rock waterrock rock water rock rock water rock rock water rock

rock rock

( ) 0 N

1 1 1 12 2 2 2

F T F F

m gT V g V g V g m gρρ ρ ρ ρ ρ ρρ ρ

= + − =

= − = − = − = −

Using 3rock 4800 kg/mρ = and rock 5 0 kg,m = . we get 44 NT = .

15.18. Model: The buoyant force on the aluminum block is given by Archimedes’ principle. The density of aluminum and ethyl alcohol are 3 3

Al ethyl alcohol2700 kg/m and 790 kg/m .ρ ρ= = Visualize:

The buoyant force BF and the tension due to the string act vertically up, and the gravitational force on the aluminum block acts vertically down. The block is submerged, so the volume of displaced fluid equals Al,V the volume of the block. Solve: The aluminum block is in static equilibrium, so

B G f Al Al Al Al Al f

6 3 2 3 3

0 N 0 N ( )

(100 10 m )(9 81 m/s )(2700 kg/m 790 kg/m ) 1 9 N

yF F T F V g T V g T V g

T

ρ ρ ρ ρ−

∑ = + − = ⇒ + − = ⇒ = −

= × . − = .

where we have used the conversion 3 2 3 4 3 100 cm 100 (10 m) 10 m− −= × = . Assess: The gravitational force on the aluminum block is Al Al 2 65 NV gρ = . . A similar order of magnitude for T is reasonable.



15.19. Model: The buoyant force on the steel cylinder is given by Archimedes’ principle. Visualize:

The length of the cylinder above the surface of mercury is d. Solve: The cylinder is in static equilibrium with B GF F= . Thus

B Hg Hg G cyl cyl Hg Hg cyl cyl Hg cyl

3cyl

3Hg

(0 20 m ) (0 20 m)

7900 kg/m0 20 m (0 20 m) (0 20 m) 1 0 084 m 8 4 cm13,600 kg/m

F V g F mg V g V V A d A

d

ρ ρ ρ ρ ρ ρ

ρρ

= = = = ⇒ = ⇒ . − = .

= . − . = . − = . = .

That is, the length of the cylinder above the surface of the mercury is 8.4 cm.

15.20. Model: The buoyant force is determined by Archimedes’ principle. Ignore any compression the air in the beach ball may undergo as a result of submersion. Solve: The mass of the beach ball is negligible, so the force needed to push it below the water is equal to the buoyant force.

3 3 3 24 4(1000 kg/m ) (0 30 m) (9 8 m/s ) 1 1 kN3 3B wF p R gπ π = = . . = .

Assess: It would take a 113 kg (250 lb) person to push the ball below the water. Two people together could do it. This seems about right.


Solve: For the Styrofoam sphere and the mass not to sink, the sphere must be completely submerged and the buoyant force BF must be equal to the sum of the gravitational force on the Styrofoam sphere and the attached mass. Neglecting the volume of the hanging mass, the volume of displaced water equals the volume of the sphere, so

( )3 3 24B water water 3

3 3 24G Styrofoam Styrofoam Styrofoam 3

(1000 kg/m ) (0 25 m) (9 8 m/s ) 641 4 N

( ) (150 kg/m ) (0 25 m) (9 8 m/s ) 96 2 N

F V g

F V g

ρ

ρ π

π= = . . = .

= = . . = .

Because G Styrofoam B( ) ,F mg F+ =

B G Styrofoam2

( ) 641 4 N 96 2 N 55 6 kg9 8 m/s

F Fm

g− . − .

= = = ..

To two significant figures, the mass is 56 kg.

15-10 Chapter 15


Section 15.5 Fluid Dynamics

15.22. Model: Treat the water as an ideal fluid. The hose is a flow tube, so the equation of continuity applies. Solve: The volume flow rate is

3 33 3600 L 600 10 m 1 25 10 m /s

8 0 min 8 0 60 sQ

−−×

= = = . × . . ×

Using the definition 2( /2) ,Q vA v dπ= = we get

3 34 4(1 25 10 m /s) 2 0 cm(4 0 m/s)

Qdvπ π

−. ×= = = .

.

Assess: This is a reasonable diameter for a typical garden hose.

15.23. Model: Treat the water as an ideal fluid. The pipe itself is a flow tube, so the equation of continuity applies. Visualize:

Note that 1 2 3 1 2 3, , and and , , and A A A v v v are the cross-sectional areas and the speeds in the first, second, and third segments of the pipe, respectively. Solve: (a) The equation of continuity is

2 2 2 2 2 21 1 2 2 3 3 1 1 2 2 3 3 1 1 2 2 3 3

2 2 22 3

2 2

2 3

(0.0050 m) (4.0 m/s) (0.010 m) (0.0025 m)

0.0050 m 0.0050 m(4.0 m/s) 1.0 m/s (4.0 m/s) 16 m/s0.010 m 0.0025 m

A v A v A v r v r v r v r v r v r v

v v

v v

π π π= = ⇒ = = ⇒ = =

= =

= = = =

(b) The volume flow rate through the pipe is 2 4 3

1 1 (0 0050 m) (4 0 m/s) 3 1 10 m /sQ A v π −= = . . = . ×

15.24. Model: Treat the water as an ideal fluid so that the flow in the tube follows the continuity equation. Visualize:

Solve: The equation of continuity is 0 0 1 1,v A v A= where 20A L= and ( )21

1 2 .A Lπ= The above equation simplifies to 2

20 1 1 0 0

4 1 272Lv L v v v vπ

π = ⇒ = = .



15.25. Model: Treat the oil as an ideal fluid obeying Bernoulli’s equation. Consider the path connecting point 1 in the lower pipe with point 2 in the upper pipe a streamline. Visualize: Please refer to Figure EX15.25. Solve: Bernoulli’s equation is

2 2 2 21 1 12 2 2 1 1 1 2 1 1 2 1 22 2 2 ( ) ( )p v gy p v gy p p v v g y yρ ρ ρ ρ ρ ρ+ + = + + ⇒ = + − + −

Using 51 200 kPa 2 0 10 Pa,p = = . × 3900 kg/m ,ρ = 2 1 10 0 m,y y− = . 1 2 0 m/s,v = . and 2 3 0 m/s,v = . we get

52 1 096 10 Pa 110 kPap = . × = .

Section 15.6 Elasticity

15.26. Model: Turning the tuning screws on a guitar string creates tensile stress in the string. Solve: The tensile stress in the string is given by T/A, where T is the tension in the string and A is the cross-sectional area of the string. From the definition of Young’s modulus,

//

T A T LY LL L A Y

= ⇒ ∆ = ∆

Using 22000 N, 0 80 m, (0 00050 m) ,T L A π= = . = . and 10 220 10 N/mY = × (from Table 15.3), we obtain 0 010 m 1.0 cm.L∆ = . =

Assess: 1.0 cm is a large stretch for a length of 80 cm, but 2000 N is a large tension.

15.27. Model: The dangling mountain climber creates tensile stress in the rope. Solve: Young’s modulus for the rope is

/ stress/ strain

F AYL L

= =∆

The tensile stress is 2

62

(70 kg)(9 8 m/s ) 8 734 10 Pa(0 0050 m)π

.= . ×

.

and the strain is 0 080 m/50 m 0 00160. = . . Dividing the two quantities yields 9 25 5 10 N/mY = . × .

15.28. Model: The hanging mass creates tensile stress in the wire. Solve: The force (F) pulling on the wire, which is simply the gravitational force (mg) on the hanging mass, produces tensile stress given by F/A, where A is the cross-sectional area of the wire. The fractional change in the wire’s length is ∆L/L = 0.010. From the definition of Young’s modulus, we have

2 3 2 10 2

2/ ( ) ( / ) (0 50 10 m) (7 10 N/m )(0 010) 60 kg/ 9 8 m/s

mg A r Y L LY mL L g

π π −∆ . × × .= ⇒ = = =

∆ .

15.29. Model: The load supported by a concrete column creates compressive stress in the concrete column. Solve: The gravitational force on the load produces tensile stress given by F/A, where A is the cross-sectional area of the concrete column and F equals the gravitational force on the load. From the definition of Young’s modulus,

2

2 10 2/ (200,000 kg)(9 8 m/s ) 3 0 m 1 mm/ (0 25 m) 3 10 N/m

F A F LY LL L A Y π

. . = ⇒ ∆ = = = ∆ . ×

Assess: A compression of 1.0 mm of the concrete column by a load of approximately 200 tons is reasonable.

15.30. Model: Water is almost incompressible and it applies a volume stress. Solve: (a) The pressure at a depth of 5000 m in the ocean is

5 3 2 7 70 sea water (5000 m) 1.013 10 Pa (1030 kg/m )(9.8 m/s )(5000 m) 5.057 10 Pa 5.1 10 Pap p gρ= + = × + = × = ×

15-12 Chapter 15


(b) Using the bulk modulus of water, 7

105 057 10 Pa 0 025 0 2 10 Pa

V pV B

∆ . ×= = = .

. ×2 2 2

(c) The volume of a mass of water decreases from V to 0.975V. Thus the water’s density increases from ρ to /0 975ρ . . The new density is

33

5000 m1030 kg/m 1056 kg/m

0 975ρ = =

.

15.31. Solve: The pressure p at depth d in a fluid is 0p p gdρ= + . Using 31 29 kg/m. for the density of air, 3 2 3

bottom top air bottom top (1 29 kg/m )(9 8 m/s )(16 m) 202 Pa 1.99 10 atmp p gd p pρ −= + ⇒ − = . . = = ×

Assuming bottom 1 atm,p =

3bottom top

bottom

1 99 10 atm 0 201 atm

p pp

−− . ×= = . ,

15.32. Model: We assume that there is a perfect vacuum inside the cylinders with 0 Pap = . We also assume that the atmospheric pressure in the room is 1 atm. Visualize: Please refer to Figure P15.32. Solve: (a) The flat end of each cylinder has an area 2 2 2(0 30 m) 0 283 m .A rπ π= = . = . The force on each end is thus

5 2 4atm 0 (1 013 10 Pa)(0 283 m ) 2 86 10 NF p A= = . × . = . ×

To two significant figures, the force on each end is 42 9 10 N. × . (b) The net vertical force on the lower cylinder when it is on the verge of being pulled apart is

4atm G players G players atm

4

2

( ) 0 N ( ) 2 86 10 N

2 86 10 Nnumber of players 29 2(100 kg)(9 8 m/s )

yF F F F F∑ = − = ⇒ = = . ×

. ×= = .

.

That is, 30 players are needed to pull the two cylinders apart.

15.33. Model: Assume that the oil is incompressible and its density is 3900 kg/m . Visualize: Please refer to Figure P15.33. Solve: (a) The pressure at depth d in a fluid is 0p p gdρ= + . Here, pressure 0p at the top of the fluid is due both to the atmosphere and to the gravitational force on the piston. That is, 0 atm G p( ) / .p p F A= + At point A,

G PA atm

25 3 2

2

2A A

( ) (1 00 m 0 30 m)

(10 kg)(9 8 m/s )1 013 10 Pa (900 kg/m )(9 8 m/s )(0 70 m) 185,460 Pa(0 020 m)

(185,460 Pa) (0 10 m) 5 8 kN

Fp p gA

F p A

ρ

π

π

= + + . − .

.= . × + + . . =

.

= = . = .

(b) In the same way, G P

B atm B( ) (1 30 m) 190,752 Pa 6 0 kNFp p g F

Aρ= + + . = ⇒ = .

Assess: BF is larger than A,F because Bp is larger than A.p



15.34. Model: Assume that blood is incompressible. Solve: When lying down, the pressure at both the heart and the brain is at 118 mm Hg (gauge pressure) or 118 mm Hg+ 760 mm Hg = 878 mm Hg absolute pressure. Upon standing up, the pressure in the heart remains at 878 mm Hg, but the pressure in the brain is reduced according to the hydrostatic pressure equation:

H Bp p ghρ= +

where B and H refer to brain and heart, respectively. Inserting h = 0.40 m and solving for Bp gives

3 2B H

760 mm Hg878 mm Hg (1060 kg/m )(9.8 m/s )(0.40 m) 847 mm Hg101.3 kPa

p p ghρ

= − = − =

In gauge pressure, the blood pressure in the brain is B 847 mmp = Hg − 760 mm Hg = 87 mm Hg. Assess: This blood pressure is below the 90 mm Hg that can cause fainting, which explains why people whose blood vessels do not constrict to compensate for this effect feel faint when standing from a prone position.

15.35. Model: The tire flattens until the pressure force against the ground balances the upward normal force of the ground on the tire. Solve: The area of the tire in contact with the road is 2(0 15 m)(0 13 m) 0 0195 mA = . . = . . The normal force on each tire is

2G (1500 kg)(9 8 m/s ) 3675 N4 4

Fn .= = =

Thus, the pressure inside each tire is

inside 23675 N 14 7 psi188,500 Pa 1 86 atm 27 psi

1 atm0 0195 mnpA

.= = = = . × =

.

15.36. Visualize:

Solve: (a) Because the patient’s blood pressure is 140/100, the minimum fluid pressure needs to be 100 mm of Hg above atmospheric pressure. Since 760 mm of Hg is equivalent to 1 atm and 1 atm is equivalent to 51 013 10 Pa,. ×

the minimum pressure is 4100 mm 1 333 10 Pa= . × . The excess pressure in the fluid is due to force F pushing on the internal 6.0-mm-diameter piston that presses against the liquid. Thus, the minimum force the nurse needs to apply to the syringe is

4 2 fluid pressure area of plunger (1 333 10 Pa) (0 0030 m) 0 38 NF π = × = . × . = .

(b) The flow rate is ,Q vA= where v is the flow speed of the medicine and A is the cross-sectional area of the needle. Thus,

6 3

3 22.0 10 m /2.0 s 20 m/s

(0.125 10 m)QvA π

−

−×

= = =×

Assess: Note that the pressure in the fluid is due to F that is not dependent on the size of the plunger pad. Also note that the syringe is not drawn to scale.

15-14 Chapter 15


15.37. Solve: The fact that atmospheric pressure at sea level is 2101 3 kPa 101,300 N/m. = means that the weight of the atmosphere over each square meter of surface is 101,300 N. Thus the mass of air over each square meter is m =

2 2(101,300 N)/ (101,300 N)/(9 80 m/s ) 10,340 kg per m .g = . = Multiplying by the earth’s surface area will give the

total mass. Using 6e 6 27 10 mR = . × for the earth’s radius, the total mass of the atmosphere is

2 6 2 2 18air earth e(4 ) 4 (6 37 10 m) (10,340 kg/m ) 5 27 10 kgM A m R mπ π= = = . × = . ×

15.38. Visualize: Let d be the atmosphere’s thickness, p the atmospheric pressure on the earth’s surface, and 0( 0 atm)p = the pressure beyond the earth’s atmosphere.

Solve: The pressure at a depth d in a fluid is 0p p gdρ= + . This equation becomes 5

air 3 2air

1 atm 1 013 10 Pa1 atm 0 atm 8 0 km(1 3 kg/m )(9 8 m/s )

gd dg

ρρ

. ×= + ⇒ = = = .

. .

15.39. Model: Oil is incompressible and has a density 3900 kg/m . Visualize: Please refer to Figure P15.39. Solve: (a) The pressure at point A, which is 0.50 m below the open oil surface, is

3 2A 0 oil (1 00 m 0 50 m) 101,300 Pa (900 kg/m )(9 8 m/s )(0 50 m) 106 kPap p gρ= + . − . = + . . =

(b) The pressure difference between A and B is 3 2

B A 0 B 0 A B A( ) ( ) ( ) (900 kg/m )(9 8 m/s )(0 50 m) 4 4 kPap p p gd p gd g d dρ ρ ρ− = + − + = − = . . = . Pressure depends only on depth, and C is the same depth as B. Thus C A 4 4 kPap p− = . also, even though C isn’t directly under A.

15.40. Model: Assume that oil is incompressible and its density is 3900 kg/mρ = . Visualize: Please refer to Figure P15.40. Solve: (a) The hydraulic lift is in equilibrium and the pistons on the left and the right are at the same level. Therefore, Equation 15.10 simplifies to

left piston right piston G elephantG student2 2

left piston right piston student elephant

G studentstudent elephant

G elephant

( )( )( ) ( )

( ) (70 kg)( ) (1.0 m) 0.2415 m( ) (1200 kg)

F F FFA A r r

F gr rF g

π π= ⇒ =

= = =

The diameter of the piston the student is standing on is therefore 2 0.2415 m 0.48 m.× = (b) With the second student added, the pistons are no longer at the same height. Therefore, Equation 15.10 gives

right piston left piston

right piston left piston

F Fgh

A Aρ= −

With left piston s (70 kg)F m g g= + and h = 0.35 m, we can solve for the second students mass s:m

elephant left pistonleft piston

right piston

23 2

2

70 kg

(1200 kg)(0.2415 m) 70 kg (900 kg/m )(0.35 m) (0.2415 m) 58 kg(1.0 m)

sm A

m hAA

ρ

π

= − +

= − + =



15.41. Model: Assume that the oil is incompressible and its density is 3900 kg/m . Visualize:

Solve: The pressures 1p and 2p are equal. Thus,

1 2 1 20 0

1 2 1 2

F F F Fp p gh ghA A A A

ρ ρ+ = + + ⇒ = +

With 2 21 1 2 2 1 1 2 2, 4 , , and ,F m g F m g A r A rπ π− = = = we have

1/21/21 2 2 1

22 2 21 2 1

4 4m g m g m g m ggh r ghr r r

ρ ρπ π

− = + ⇒ = − π π

Using 31 2 155 kg, 110 kg, 0 08 m, 900 kg/m ,m m r ρ= = = . = and 1 0 m,h = . the calculation yields 2 0 276 mr = . .

The diameter is 55 cm. Assess: Both pistons are too small to hold the people as shown, but the ideas are correct.

15.42. Model: Water and mercury are incompressible and immiscible liquids. Visualize:

The water in the left arm floats on top of the mercury and presses the mercury down from its initial level. Because points 1 and 2 are level with each other and the fluid is in static equilibrium, the pressure at these two points must be equal. If the pressures were not equal, the pressure difference would cause the fluid to flow, violating the assumption of static equilibrium. Solve: The pressure at point 1 is due to water of depth w 10 cm:d =

1 atmos w wp p gdρ= + Because mercury is incompressible, the mercury in the left arm goes down a distance h while the mercury in the right arm goes up a distance h. Thus, the pressure at point 2 is due to mercury of depth Hg 2 :d h=

2 atmos Hg Hg atmos Hg2p p gd p ghρ ρ= + = +

Equating 1p and 2p gives

3w

atmos w w atmos Hg w 3Hg

1 1 1000 kg/m2 (10 cm) 3 7 mm2 2 13,600 kg/m

p gd p gh h dρρ ρρ

+ = + ⇒ = = = .

The mercury in the right arm rises 3.7 mm above its initial level.

15-16 Chapter 15


15.43. Model: Glycerin and ethyl alcohol are incompressible and do not mix. Visualize:

Solve: The alcohol in the left arm floats on top of the denser glycerin and presses the glycerin down distance h from its initial level. This causes the glycerin to rise distance h in the right arm. Points 1 and 2 are level with each other and the fluids are in static equilibrium, so the pressures at these two points must be equal:

1 2 0 eth eth 0 gly gly eth gly

3eth

3gly

(20 cm) (2 )

1 1 790 kg/m(20 cm) (20 cm) 6 27 cm2 2 1260 kg/m

p p p gd p gd g g h

h

ρ ρ ρ ρ

ρρ

= ⇒ + = + ⇒ =

= = = .

You can see from the figure that the difference between the top surfaces of the fluids is 20 cm 2 20 cm 2(6 27 cm) 7 46 cm 7 5 cmy h∆ = − = − . = . ≈ .

15.44. Model: Assume the liquid is incompressible. Visualize: Solve: (a) Can 2 has moved down with respect to Can 1 since the water level in Can 2 has risen. Since the total volume of water stays constant, the water level in Can 1 has fallen by the same amount. The water level is equalized in the two cans at the middle of the height change, so the change in height of the water is half the relative change in height of the cans. Can 2 has moved relative to Can 1 (6 5 cm 5 0 cm) 2 3 0 cm. − . × = . down. (b) The water level in Can 1 has fallen by the same amount. The new level is 5.0 − 1.5 cm = 3.5 cm. Assess: The two cans are an inexpensive method of measuring relative changes in height.

15.45. Model: The water is in hydrostatic equilibrium. Visualize:

Solve: (a) The force of the liquid on the bottom is the pressure p gDρ= times the area A = WL. This gives

F pA gDWLρ= =



(b) The pressure on the front panel of the aquarium can be found by integrating the force on a thin strip dy (see figure above). Using F = pA, we find

212

0

D

dF gyLdy

F gyLdy gD L

ρ

ρ ρ

=

= =Ñ

(c) For a 100-cm-long, 35-cm-wide, 40-cm-deep aquarium filled with water, the force from the liquid on the front window is

3 2 212 (1000 kg/m )(9 8 m/s )(0 40 m) (1 0 m) 784 N 0 78 kNF = . . . = ≈ .

The force on the bottom is 3 2(1000 kg/m )(9 8 m/s )(0 40 m)(0 35 m)(1 0 m) 1372 N 1 4 kNF = . . . . = ≈ .

15.46. Visualize:

The figure shows a small column of air of thickness dz, of cross-sectional area 21 m ,A = and of density ( ).zρ The column is at a height z above the surface of the earth. Solve: (a) The atmospheric pressure at sea level is 51 013 10 Pa. × . That is, the weight of the air column with a 21 m

cross section is 51 013 10 N. × . Consider the weight of a 21 m slice of thickness dz at a height z. This slice has volume 2(1 m ) ,dV Adz dz= = so its weight is 0/2 2

0( ) (1 m ) (1 m ) .z zdw dV g g dz e g dzρ ρ ρ −= = = The total weight of

the 21 m column is found by adding all the dw. Integrating from 0z = to ,z = ∞

0

0

/20

0

/20 0 0

20 0

(1 m )

(1 m )

(1 m )

z z

z z

w g e dz

g z e

g z

ρ

ρ

ρ

∞−

∞−

=

= −

=

∫

Because 20 0101,300 N (1 m ) ,w g zρ= =

30 3 2 2

101,300 N 8 1 10 m(1 28 kg/m )(9 8 m/s )(1 0 m )

z = = . ×. . .

(b) Using the density at sea level from Table 15.1, 3 33 (8 08 10 m) 3 1600 m/(8 08 10 m) 3(1 28 kg m ) (1 28 kg m ) 1 05 kg mz// e / e /ρ − . × − . ×= . = . = .

This is 82% of 0.ρ

15-18 Chapter 15


15.47. Model: Ignore the change in water density over lengths comparable to the size of the fish. Visualize:

Solve: The buoyant force is the force of gravity on the volume of water displaced by the fish times, or

B w F b( )F V V gρ= +

where the subscripts w, F, and b indicate water, fish, and bladder, respectively. The force due to gravity on the fish is

G F air bFF V g V gρ ρ= +

For neutral buoyancy, the two forces must have equal magnitude. Equating them and solving for the bV gives

w F b F air b3 3

w Fb F F F3 3

air w

( )

1000 kg/m 1080 kg/m 0 08011 19 kg/m 1000 kg/m

FV V g V g V g

V V V V

ρ ρ ρ

ρ ρρ ρ

+ = +

− −= = = .

− . −

so the fish needs to increase its volume by 8.01%.

15.48. Model: The buoyant force on the ceramic statue is given by Archimedes’ principle. Visualize:

Solve: The gravitational force on the statue is the 28.4 N registered on the scale in air. In water, the gravitational force on the statue is balanced by the sum of the buoyant force BF and the spring’s force on the statue; that is,

statueG statue B spring on statue w statue statue

w statue3

3 3statuestatue

11.4 N( ) 28.4 N 17.0 N

( ) (28.4 N)(1000 kg/m ) 2.49 10 kg/m(11.4 N) (11.4 N)

w

mF F F V g Vg

m g

ρρ ρ

ρρ

= + ⇒ = + ⇒ = =

= = = ×




Solve: (a) Initially, as it floats, the cylinder is in static equilibrium, with the buoyant force balancing the gravitational force on the cylinder. The volume of displaced liquid is Ah, so

B liq G( )F Ah g Fρ= = Force F pushes the cylinder down distance x, so the submerged length is h x+ and the volume of displaced liquid is

( ).A h x+ The cylinder is again in equilibrium, but now the buoyant force balances both the gravitational force and force F. Thus

B liq G[ ( )]F A h x g F Fρ= + = +

Since liq G( ) ,Ah g Fρ = we’re left with

liqF Agxρ=

(b) The amount of work dW done by force F to push the cylinder from x to x dx+ is liq( ) .dW Fdx Agx dxρ= = To

push the cylinder from i 0 mx = to f 10 cm 0 10 mx = = . requires work f f

i i

2 21liq liq i f2

3 2 2 212

( )

(1000 kg/m ) (0 020 m) (9 8 m/s )(0 10 m) 0 62 J

x x

x xW Fdx Ag xdx Ag x xρ ρ

π

= = = −

= . . . = .

Ñ Ñ


Let 1d be the length of the cylinder in the less-dense liquid with density 1,ρ and 2d be the length of the cylinder in the more-dense liquid with density 2p . Solve: The cylinder is in static equilibrium, so

B1 B2 G 1 1 2 2 1 2

11 2 1 2

2 2

0 N ( ) ( ) ( )

( )

yF F F F Ad g Ad g A d d g

d d d d

ρ ρ ρ

ρ ρρ ρ

∑ = + − = ⇒ + = +

+ = +

Since 1 2 1 2,l d d d l d= + ⇒ = − we can simplify the above equation to obtain

12

2 1d lρ ρ

ρ ρ −

= −

15-20 Chapter 15


That is, the fraction of the cylinder in the more dense liquid is 1 2 1( ) ( ).f /ρ ρ ρ ρ= − − Assess: As expected 0f = when ρ is equal to 1,ρ and 1f = when 2.ρ ρ=


Solve: The tube is in static equilibrium, so

B G tube G Pb liquid

2 3liquid 2

( ) ( ) 0 N (0 25 m) (0 030 kg) (0 250 kg)

(0 280 kg) (0 280 kg) 8 9 10 kg/m(0 25 m) (0 020 m) (0 25 m)

yF F F F A g g g

A

ρ

ρπ

∑ = − − = ⇒ . = . + .

. .= = = . ×

. . .

Assess: This is a reasonable value for a liquid.

15.52. Model: Archimedes’ Principle determines the buoyant force. Visualize:

Solve: The plastic hemisphere will hold the most weight when its rim is at the surface of the water. The buoyant force balances the gravitational force on the bowl and rock.

B G rock G bowl( ) ( ) 0 NyF F F F∑ = − − =

Thus

bowl

3 3 2 2w bowl

1 4(1000 kg/m ) (0 040 m) (9 8 m/s ) (0 021 kg)(9 8 m/s ) 0 16 kg2 3

mg p V g m g mπ = − = . . − . . ⇒ = .

Assess: Putting a rock as big as 160 g in an 8-cm-diameter bowl before it sinks is reasonable.



15.53. Model: The buoyant force is determined by Archimedes’ principle. The spring is ideal. Visualize:

Solve: The spring is stretched by the same amount that the cylinder is submerged. The buoyant force and spring force balance the gravitational force on the cylinder.

2

3 2 2

0 N

(1 0 kg)(9 8 m/s )(1000 kg/m ) (0 025 m) (9 8 m/s ) 35 N/m

0 181 m 18 cm

y B S

w

w

F F F mg

p Ayg ky mg

mgyp Ag k π

∑ = + − =

+ =

. .= =

+ . . += . =

Assess: This is difficult to assess because we don’t know the height h of the cylinder and can’t calculate it without the density of the metal material.

15.54. Model: The buoyant force is determined by Archimedes’ principle. The cross-sectional area of the boat is denoted by A. Visualize:

Solve: For the boat to have neutral buoyancy, the buoyant force must have the same magnitude as the force due to gravity:

B ( )gF F dAg M nm gρ= ⇒ = + where ρ is the density of the liquid, M is the boat’s mass, and n is an integer. Rearranging this expression gives

nm dA Mρ= − If we plot the mass nm added to the boat as a function of the depth d, the y-intercept will be the negative of the boat’s mass M and the slope s will be related to the liquid’s density as / .s Aρ=

15-22 Chapter 15


From the fit to the data we find M = 29 g and s = 2.667 kg/m. This gives a density of 3 3 3

4 22 667 kg/m 1067 kg/m 1 1 10 kg/m

25 10 msA

ρ −.

= = = = . ××

Assess: The mass of the boat is some 6 times greater than the mass of the blocks, which seems reasonable. The density of the liquid is slightly greater than the density of water, which is also a reasonable result.

15.55. Model: The buoyant force on the can is given by Archimedes’ principle. Visualize:

The length of the can above the water level is d, the length of the can is L, and the cross-sectional area of the can is A. Solve: The can is in static equilibrium, so

1B G can G water water water2( ) ( ) 0 N ( ) (0 020 kg)yF F F F A L d g g m gρ∑ = − − = − = . +

The mass of the water in the can is 6 3

3canwater water

waterwater

355 10 m(1000 kg m ) 0 1775 kg2 2

0 1975 kg( ) 0 020 kg 0 1775 kg 0 1975 kg 0 0654 m

Vm /

A L d d LA

ρ

ρρ

−× = = = .

.− = . + . = . ⇒ − = = .2

Because 2 6 3can (0 031 m) 355 10 m ,V Lπ −= . = × L = 0.1176 m. Using this value of L, we get 0 0522 m 5 2 cmd = . ≈ . .

Assess: / 5 22 cm 11 76 cm 0 444,d L /= . . = . thus 44.4% of the length of the can is above the water surface. This is reasonable.



15.56. Model: The buoyant force on the boat is given by Archimedes’ principle. Visualize:

The minimum height of the boat that will enable the boat to float in perfectly calm water is h. Solve: The boat barely floats if the water comes completely to the top of the sides. In this case, the volume of displaced water is the volume of the boat. Archimedes’ principle in equation form is B w boatF V gρ= . For the boat to float B G boat( ) .F F= Let us first calculate the gravitational force on the boat:

G boat G bottom G side 1 G side 23 3

G bottom steel bottom3 3

G side 1 steel side 1

G side 2 steel si

( ) ( ) 2( ) 2( ) , where

( ) (7900 kg/m )(5.0 10 0.020 m ) 7900 N

( ) (7900 kg/m )(5.0 0.0050 m ) 197.5 N

( )

ρ

ρ

ρ

= + +

= = × × =

= = × × =

=

F F F F

F V g g g

F V g h g gh

F V 3 3de 2

G boat

(7900 kg/m )(10 0.0050 m ) 395( ) [7900 2(197.5 ) 2(395 )] N (7900+1185 ) N

= × × =

= + + =

g h g ghF g gh gh h g

Going back to the Archimedes’ equation and remembering that h is in meters, we obtain

w boat (7900 1185 ) N (1000)[10 5 0 ( 0 020)] 7900 118514 cm

V g h g h hh

ρ = + ⇒ × . × + . = +=

15.57. Model: The two pipes are identical. Visualize:

Solve: The water speed is the same in both pipes. The flow rate is 6

6 3 3

2

3 0 10 L/min 2( )1(3 0 10 )(10 ) m /s60 3 5 m/s

2 2 (1 5 m)

Q vA

QvA π

−

= . × =

. × = = = .

.

15-24 Chapter 15


15.58. Model: Treat the liquid as an ideal liquid obeying Bernoulli’s equation (Equation 15.27). Visualize:

Solve: (a) Bernoulli’s equation gives 2 21 1

0 0 0 1 1 12 2p v gy p v gyρ ρ ρ ρ+ + = + +

Because the pipe expands equally in the positive and negative y directions, the contributions from the gravitational-potential-energy term cancel by symmetry, so Bernoulli’s equation reduces to

2 21 10 0 1 12 2p v p vρ ρ+ = +

Expressing 1p in terms of 0p gives 2 21

1 0 0 12 ( )p p v vρ= + −

Finally, we can express the velocity 1v in terms of 0v using the continuity equation (Equation 15.17): 20

0 0 1 1 1 0 21

dv A v A v vd

= ⇒ =

Inserting this into the expression for 0p gives 4

2 011 0 02 4

11 dp p v

dρ

= + −

(b) Inserting the given values gives 4

3 211 2

16 8 cm50 kPa (1000 kg/m )(10 0 m/s) 1 50 kPa 25 1 kPa 75 kPa20 0 cm

p . = + . − = + . = .

Assess: We find that the pressure increases where the velocity has decreased, which is in accordance with Bernoulli’s equation.

15.59. Solve: Treat the sap as incompressible and assume sap is almost entirely made of water. Solve: To replace the water lost to the atmosphere, the vessels must supply 110 g/h of sap. The sap provided by each vessel is 2( /2) ,v dρ π so for N = 2000 sap vessels, the flow rate in each vessel is

2

3 6 24(0 110 kg/h)(1 h 3600 s)0 110 kg/h 187 nm/s

2 (1040 kg/m )(2000)(100 10 m)d /N v vρ π

π −. = . ⇒ = = ×

15.60. Model: Treat the water as an ideal fluid obeying Bernoulli’s equation. A streamline begins in the bigger size pipe and ends at the exit of the narrower pipe. Visualize: Please see Figure P15.60. Let point 1 be beneath the standing column and point 2 be where the water exits the pipe. Solve: (a) The pressure of the water as it exits into the air is 2 atmosp p= . (b) Bernoulli’s equation, Equation 15.26, relates the pressure, water speed, and heights at points 1 and 2:

2 2 2 21 1 11 1 1 2 2 2 1 2 2 1 2 12 2 2 ( ) ( )p v gy p v gy p p v v g y yρ ρ ρ ρ ρ ρ+ + = + + ⇒ − = − + −

From the continuity equation, 4 2 4 2 4 3

1 1 2 2 1 1(4 m/s)(5 10 m ) (10 10 m ) 20 10 m /s 2 0 m/sv A v A v v− − −= = × ⇒ × = × ⇒ = . Substituting into Bernoulli’s equation,

3 2 2 311 2 1 atmos 2 (1000 kg/m )[(4 0 m/s) (2 0 m/s) ] (1000 kg/m )(9 8 m/s)(4 0 m)

6000 Pa 39,200 Pa 45 kPa

− = − = . − . + . .

= + =

p p p p



But 1 2 ,p p ghρ− = where h is the height of the standing water column. Thus 3

3 245 10 Pa 4 6 m

(1000 kg/m )(9 8 m/s )h ×

= = ..

15.61. Model: Treat the water as an ideal fluid obeying Bernoulli’s equation. A streamline begins at the faucet and continues down the stream. Visualize:

The pressure at point 1 is 1p and the pressure at point 2 is 2p . Both 1p and 2p are atmospheric pressure. The velocity and the area at point 1 are 1v and 1A and they are 2v and 2A at point 2. Let d be the distance of point 2 below point 1. Solve: The flow rate is

6 3 4 34 3

1 1 22 0 1000 10 m 2 0 10 m /s2 0 10 m /s 1 0 m/s

10 s (0 0080 m)Q v A v

π

− −−. × × . ×

= = = . × ⇒ = = ..

Bernoulli’s equation at points 1 and 2 is 2 2 2 21 1 1

1 1 1 2 2 2 2 12 2 2 ( )p v gy p v gy gd v vρ ρ ρ ρ ρ ρ+ + = + + ⇒ = −

From the continuity equation, 3 2 3 2

1 1 2 2 2 2(1 0 m s) (8 0 10 m) (5 0 10 m) 2 56 m sv A v A / v v /π π− −= ⇒ . . × = . × ⇒ = . Going back to Bernoulli’s equation, we have

2 212 [(2 56 m s) (1 0 m s) ] 0 283 m 28 cmgd / / d= . − . ⇒ = . ≈

15.62. Model: Treat the air as an ideal fluid obeying Bernoulli’s equation. Solve: (a) The pressure above the roof is lower due to the higher velocity of the air. (b) Bernoulli’s equation, with inside outside,y y≈ is

22 2 31

inside outside air air21 1 130 1000 m(1 28 kg m ) 835 Pa2 2 3600 s

p p v p v /ρ ρ × = + ⇒ ∆ = = . =

The pressure difference is 0.83 kPa (c) The force on the roof is 4( ) (835 Pa)(6 0 m 15 0 m) 7 5 10 N.p A∆ = . × . = . × The roof will blow up, because pressure inside the house is greater than pressure on the top of the roof.

15.63. Model: The ideal fluid obeys Bernoulli’s equation. Visualize: Please refer to Figure P15.63. There is a streamline connecting point 1 in the wider pipe on the left with point 2 in the narrower pipe on the right. The air speeds at points 1 and 2 are 1v and 2v and the cross-sectional area of the pipes at these points are 1A and 2A . Points 1 and 2 are at the same height, so 1 2y y= .

Solve: The volume flow rate is 6 31 1 2 2 1200 10 m /s.Q A v A v −= = = × Thus

15-26 Chapter 15


6 3

2 21200 10 m /s 95 49 m/s

(0 0020 m)v

π

−×= = .

.

6 3

1 21200 10 m /s 3 82 m/s

(0 010 m)v

π

−×= = .

.

Now we can use Bernoulli’s equation to connect points 1 and 2: 2 21 1

1 1 1 2 2 22 22 2 3 2 21 1

1 2 2 1 2 12 2( ) ( ) (1 28 kg m )[(95 49 m s) (3 82 m s) ] 0 Pa 5 83 kPa

p v gy p v gy

p p v v g y y / / /

ρ ρ ρ ρ

ρ ρ

+ + = + +

− = − + − = . . − . + = .

Because the pressure above the mercury surface in the right tube is 2p and in the left tube is 1,p the difference in the pressures 1p and 2p is Hgghρ . That is,

3

1 2 Hg 3 25 83 10 Pa5 83 kPa 4 4 cm

(13,600 kg/m )(9 8 m/s )p p gh hρ . ×

− = . = ⇒ = = ..

15.64. Model: The ideal fluid (i.e., air) obeys Bernoulli’s equation. Visualize: Please refer to Figure P15.64. There is a streamline connecting points 1 and 2. The air speeds at points 1 and 2 are 1v and 2,v and the cross-sectional areas of the pipes at these points are 1A and 2A . Points 1 and 2 are at the same height, so 1 2y y= . Solve: (a) The height of the mercury is 10 cm. So, the pressure at point 2 is larger than at point 1 by

3 2Hg 2 1(0 10 m) (13,600 kg/m )(9 8 m/s )(0 10 m) 13,328 Pa 13,328 Pag p pρ . = . . = ⇒ = +

Using Bernoulli’s equation, 2 2 2 21 1 1

1 air 1 air 1 2 air 2 air 2 2 1 air 1 22 2 2

2 2 2 22 11 2 3

air

( )

2( ) 2(13,328 Pa) 20,825 m /s(1 28 kg/m )

p v gy p v gy p p v v

p pv v

ρ ρ ρ ρ ρ

ρ

+ + = + + ⇒ − = −

−− = = =

.

From the continuity equation, we can obtain another equation connecting 1v and 2:v 2

21 1 2 2 1 2 2 22

1

(0 0050 m) 25(0 0010 m)

AA v A v v v v vA

ππ

.= ⇒ = = =

.

Substituting 1 225v v= in the Bernoulli equation, we get 2 2 2 2

2 2 2(25 ) 20,825 m /s 5 78 m sv v v /− = ⇒ = .

Thus 2 5 8 m/sv = . and 21 225 1 4 10 m/sv v= = . × .

(b) The volume flow rate 2 3 32 2 (0 0050 m) (5 78 m/s) 4 5 10 m /s.A v π −= . . = . ×

15.65. Model: Treat the water as an ideal fluid that obeys Bernoulli’s equation. There is a streamline from the top of the water to the hole. Visualize: The top of the water 1(at )y h= and the hole 2(at )y y= are at atmospheric pressure. The speed of the water at the top is zero because the tank is kept filled.



Solve: (a) Bernoulli’s equation connecting the two points gives 21

20 2 ( )gh v gy v g h yρ ρ ρ+ = + ⇒ = −

(b) For a particle shot horizontally from a height y with speed v, the range can be found using kinematic equations. For the y-motion, using 0 0 s,t = we have

2 21 11 0 0 1 0 1 0 1 1 12 2( ) ( ) 0 m (0 m/s) ( ) 2y yy y v t t a t t y t g t t y/g= + − + − ⇒ = + + − ⇒ =

For the x-motion, 21

1 0 0 1 0 1 0 12( ) ( ) 0 m 0 m 2x xx x v t t a t t x vt x v y/g= + − + − ⇒ = + + ⇒ =

(c) Combining the results of (a) and (b), we obtain 2 ( ) 2 4 ( )x g h y y/g y h y= − = −

To find the maximum range relative to the vertical height, 10 [4( ) 4 ] 0

24 ( )dx hh y y ydy y h y

= ⇒ − − = ⇒ =−

12With , the maximum range isy h=

max 42 2h hx h h = − =

15.66. Model: Treat the water as an ideal fluid that obeys Bernoulli’s equation. There is a streamline from the top of the water to the hole. Visualize: The top of the water and the hole are at atmospheric pressure. The speed of the water at the top is not zero because the tank is emptying.

Solve: (a) Bernoulli’s equation connecting the two points gives 2 21 1

atm top atm hole2 2 ( )p gh v p v g h dρ ρ ρ ρ+ + = + + −

The continuity equation relates the flow rates at the surface and at the hole: 2

top top hole hole top holerv A v A v vR

= ⇒ =

Inserting this into Bernoulli’s equation and solving for holev gives

hole 42

1 ( )gdvr/R

=−

(b) Inserting the given values, we find that the water level will initial at the following rate: 2 2 3 2

top hole 4 3 42 2 0 10 m 2(9 8 m/s )(1 0 m) 1 1 mm/min

1 0 m1 ( ) 1 [(2 0 10 m) /(1 0 m)]r r gdv vR R r/R

− . × . . = = = = . .− − . × . 2

Assess: For the given dimensions, the result seems reasonable.

15-28 Chapter 15


15.67. Model: The aquarium creates tensile stress. Solve: Weight of the aquarium is

3 3 2 4G water (1000 kg/m )(10 m )(9 8 m/s ) 9 8 10 NF mg Vgρ= = = . = . ×

where we have used the conversion 3 31 L 10 m= .2 The weight supported by each wood post is 414 (9 8 10 N). × =

42 45 10 N. × . The cross-sectional area of each post is 2 3 2(0 040 m) 1 6 10 m .A −= . = . × Young’s modulus for the wood is

10 2

43

3 2 10 2

/1 10 N m/

(2 45 10 N)(0 80 m) 1 23 10 m 1 mm(1 6 10 m )(1 10 N/m )

F A FLY /L L A L

FLLAY

−−

= × = =∆ ∆

. × .∆ = = = . × =

. × ×

Assess: A compression of 1 mm due to a weight of 42 45 10 N. × is reasonable.

15.68. Model: The weight of the person creates tensile stress in the disk cartilage. Solve: The cross-sectional area of each disk of cartilage is 2.rπ Inserting this into Equation 15.34 gives

2

6 2 2(33 kg)(9.8 m/s )(0.0050 m) 1.3 mm(1.0 10 N/m ) (0.020 m)

F LYA L

FL mgLLYA YA π

∆=

∆ = = = =×

Assess: The cartilage experiences a significant compressive axial strain (~25%).

15.69. Model: Pressure applies a volume stress to water in the cylinder. Solve: The volume strain of water due to the pressure applied is

63

10

3 3 3 3

2 10 Pa 1 100 2 10 Pa

(1 10 )(1 30 m ) 1 30 10 m 1 L

V pV B

V V V

−

− −

∆ ×= − = − = − ×

. ×

∆ = ′ − = − × . = − . × ≈ −

As the safety plug on the top of the cylinder bursts, the water comes back to atmospheric pressure. To the precision of the data (one significant figure), one liter of water comes out.

15.70. Model: Air is an ideal gas and obeys Boyle’s law. Visualize: Please refer to Figure CP15.70. The quantity h is the length of the air column when the mercury fills the cylinder to the top. A is the cross-sectional area of the cylinder. Solve: For the column of air, Boyle’s law is 0 0 1 1,p V p V= where 0p and 0V are the pressure and volume before any mercury is poured, and 1p and 1V are the pressure and volume when mercury fills the cylinder above the air. Using

1 0 Hg (1 0 m ),p p g hρ= + . − Boyle’s law becomes

0 0 0 Hg 1 0 0 Hg

50

0 Hg 3 2Hg

[ (1 0 m )] (1 0 m) [ (1 0 m )]

1 013 10 Pa(1 0 m ) (1 0 m ) 0 76 m 76 cm(13,600 kg/m )(9 8 m/s )

p V p g h V p A p g h Ah

pp h g h h hg

ρ ρ

ρρ

= + . − ⇒ . = + . −

. ×. − = . − ⇒ = = = . =

.



15.71. Model: The buoyant force on the cone is given by Archimedes’ principle. Visualize:

Solve: It may seem like we need the formula for the volume of a cone. You can use that formula if you know it, but it isn’t essential. The volume is clearly the area of the base multiplied by the height multiplied by some constant. That is, the cone shown above has V aAl= where a is some constant. But the radius of the base is tan ,r l a= where

a is the angle of the apex of the cone, and 2,A rπ= making A proportional to 2l . Thus the volume of a cone of

height l is 3,V cl= where c is a constant. Because the cone is floating in static equilibrium, we must have B GF F= .

The cone’s density is 0,ρ so the gravitational force on it is 3G 0 0 .F Vg cl gρ ρ= = The buoyant force is the

gravitational force on the displaced fluid. The volume of displaced fluid is the full volume of the cone minus the volume of the cone of height h above the water, or 3 3

disp .V cl ch= − Thus 3 3B f disp f ( ) ,F V g c l h gρ ρ= = − and the

equilibrium condition is 1/33

3 3 3 0B G f 0 f 03

f( ) 1 1h hF F c l h g cl g

llρρ ρ ρ ρρ

= ⇒ − = ⇒ − = ⇒ = −

15.72. Model: The grinding wheel is a uniform disk. We will use the model of kinetic friction and hydrostatics. Visualize: Please refer to Figure CP15.72. Solve: This is a three-part problem. First find the desired angular acceleration, then use that to find the force applied by each brake pad, then finally the needed oil pressure. The angular acceleration required to stop the wheel is found using rotational kinematics.

f i 20 rad/s 900 2 (1 60) rad/s 18 85 rad/s5 0 s

/t t

ω ωω πa−∆ − × ×

= = = = − .∆ ∆ .

Each brake pad applies a frictional force k kf nµ= to the wheel. The normal force is equal to the force applied by the piston by Newton’s third law. Rotational dynamics can be used to find the magnitude of the force. kf is applied 12 cm from the rotation axis on both sides of the disk.

net

2k

2 2 2

k

12 (0.12 m)2

(15 kg)(0 13 m) ( 18 85 rad/s ) 16 6 N4 (0 12 m) 4(0 60)(0 12 m)

I

f MR

MRn

t a

a

aµ

=

=

. − .= = = .

. . .

2

2 2

The oil pressure required to generate this much force at each brake pad is

216 6 N 53 kPa

(0 010 m)FpA π

.= = =

.

relative to atmospheric pressure. The absolute pressure is 53 kPa + 101.3 kPa = 154.3 kPa = 1.5 × 510 Pa to two significant figures. Assess: The required oil pressure is about half an atmosphere, which is quite reasonable.

15-30 Chapter 15



A is a cross-sectional area of the cylinder. Solve: (a) The gravitational force on the cylinder is G 0( )F Al gρ= and the weight of the displaced fluid is

B f ( ) .F Ah gρ= In static equilibrium, B GF F= and we can write

f 0 0 f( )Ahg Alg h / lρ ρ ρ ρ= ⇒ =

(b) Now the volume of the displaced liquid is ( ).A h y− Applying Newton’s second law in the y-direction gives

G B 0 f ( )yF F F Alg A h y gρ ρ∑ = − + = − + −

Using f 0Ahg Algρ ρ= from part (a), we find

net f( )yF Agyρ= −

(c) The result in part (b) is ,F ky= − where f .k Agρ= This is Hooke’s law. (d) Since the cylinder’s equation of motion is determined by Hooke’s law, the angular frequency for the resulting simple harmonic motion is / ,k mω = and the period is

0

f f

2 2 2 2 2m m Al hT Tk Ag Ag g

π ρπ π π πω ρ ρ

= = = = = =

where we have used the expression for h from part (a). (e) The oscillation period for the 100-m-tall iceberg 3

ice( 917 kg/m )p = in sea water is

30

3 2f

(917 kg/m )(100 m)2 2 2 18 9 s(1030 kg/m )(9 80 m/s )

h lTg g

ρπ π πρ

= = = = ..

15.74. Model: A streamline connects every point on the surface of the liquid to a point in the drain. The drain diameter is much smaller than the tank diameter ( ).r R Visualize:



Solve: The pressures at the surface and drain (points 1 and 2) are equal to one atmosphere. When the liquid is a depth y, Bernoulli’s principle connecting points 1 and 2 is

2 21 1 2 2

2 22 1

1 1 (0)2 2

2

p v gy p v g

v v gy

ρ ρ ρ ρ+ + = + +

= +

The flow rate through the drain is the same as through a horizontal layer in the tank: 2

1 1 2 2 1 21

AQ v A v A v vA

= = ⇒ =

Thus the velocity of the liquid through the drain is 2

2 22 2 2 4

1

221 ( / )

A gyv v gy vA r R

= + ⇒ =

−

Since ,r R 2 2 ,v gy≈ and the flow rate through the drain is 2 2Q r gyπ≈

The flow rate gives the volume of liquid flowing out per unit time. The inverse gives the time needed for a unit volume of liquid to flow out. For the volume of liquid to decrease by dV requires a time

2 2

2 2( )

2 2dV R dy R dydtQ yr gy r g

ππ− −

= = =

Note that 0dV < implies the volume of liquid is decreasing. Integrating both sides from the initial condition ( 0, )t y d= = to the final condition ( , 0)t y = yields

1 12 2

02 2 20

2 2 22 2

2 2 dd

R R R dt y dy ygr g r g r

−= − = − =∫

Assess: The time for the tank to drain depends on the ratio of the cross-sectional areas of the tank to drain, which makes sense, as well as on the strength of the acceleration due to gravity. Note that, if g were larger (say we were on Jupiter), the time to drain would be shorter, whereas if the tank were taller (larger d), the time would be longer.

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FLUIDS AND ELASTICITYsrjcstaff.santarosa.edu/~Lwillia2/41/41Ch15key.pdfThe lower the velocity of fluid, the higher the pressure at a given point in the pipe (Bernoulli’s principle)