FluidsWhat Is a Fluid?

{College Physics 7th ed. pages 288-289/8th ed. pages 290-291)The study offluids, inasmuch as is required for AP Physics B, outlines two

primary branches of this subject at the elementary level: fluids that arestationary (hydrostatics) and fluids that are in motion [hydrodynamics). Inboth topics, the relationship between forces causing fluid pressure, howfluid pressure is proportional to fluid velocity, and how fluid pressurevaries with depth in a liquid are all ofprimary importance. In addition, thefeatures ofthe fluid dictate its behavior under certain conditions. Thus, interms of understanding the basics of fluid behavior, it is best to considerthe fluid as an ideal fluid, which satisfies the following constraints:

1. The fluid is incompressible. Such a fluid has a constant densitythroughout.2. The fluid is nonviscous. Such a fluid possesses no internal friction

between layers of the fluid that could impede its motion, whichallows it to undergo streamlined motion.3. The fluid moves with no turbulence. Such a fluid has no irregular

motion such that elements ofthe fluid do not rotate, but simply moveforward (i.e., translate).4. The fluid motion is steady. Such a fluid has velocity, density, and

pressure at each point in the fluid that do not change with time.

Hydrostatics: What Relationship Exists AmongPressure. Density, and Depth Within a Fluid?

(College Physics 7th ed. pages 274-282/8th ed. pages 276-284)What is heavier, a pound of feathers or a pound of lead? It is importantto ask which of these materials takes up more space rather than whichis heavier. Of course, lead is the more compact material; thus, an equal

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210 Chapter 9

amount (or mass, m, in kilograms) oflead takes up less space (or volume, V,in cubic meters) than an equal amount of feathers. The combinationof mass m and volume V leads to the a concept of fluids, that of density, p:

The density of a particular material or fluid (measured in units ofkilograms percubic meter, orkg/m3) depends directlyon howmuch matter(or mass) is present in a particular volume occupied by the substance.Gold, for example, with a density of 19,300 kg/m3, has nearly nine timesthe amount of mass per cubic meter than does concrete, with a densityof 2200 kg/m3, meaning that an equal mass of gold takes up much lessvolume than does an equal mass ofconcrete. (For comparison, the densityofwater at standard temperature and pressure is 1000 kg/m3.) In fluids, itis also important to realize that the mass ofa substance may immediatelybe calculated using the equation above when only the density and volumeare given. Mass, m, is also written as m = pV. For example, the mass of2 m3 of concrete is m = (2200 kg/m3)(2 m3) = 4400 kg.

Although an ideal fluid's mass is uniform for a particular givenvolume, the force exerted on the bottom of a container holding a fluid

depends on both the height h of the liquid above the container bottomand the area A over which the force is distributed. The combination oftheforce F of the fluid exerted over a certain area A (in square meters) givesthe fluid's pressure P at that location. Thus, the definition of pressure is

In the above equation, the unit for pressure is newtons per squaremeter (N/m2), which is called the pascal (Pa). As with the densityequation, the pressure equation can be solved for force F to giveF = PA. For common comparison, one pascal (1 N/m2) is equivalent to

1.4 x 10"4 pounds per square inch (lb/in2), the English unit for pressure.Therefore, the pressure at sea level due to the weight of Earth's

atmosphere, which is about 15 lb/in.2, is 1.01 x 105 Pa, or 1 atmosphere(1 atm). Similarly, a car tire with 30 lb/in.2 has air that is pressurized to2.02 x 105 Pa, and air on Venus's surface is pressurized to nearly 1.01 x

107 Pa, nearly 100 times that of air pressure at Earth's surface. Noticealso that pressure and area are inversely related, implying that as thearea of a surface under the same force decreases, the total pressurewill increase and vice versa. Hence, it is easier to walk on snow whenwearing snowshoes (with a very wide surface contact area) than whenwearing normal street shoes (with a small surface contact area).

Consider a "block" of fluid of mass m whose top and bottom havearea A. We can derive the relationship between pressure and depth ina fluid, where F, and F2 are forces on the mass due to the surroundingfluid, as shown.

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Fluids 211

Applying IF = 0, we have F2- F^- mg = 0. Using the pressure anddensity equations given above, we can rewrite this expression as P^z -PtA, - pVg = 0. Because V = Ah (where h = the vertical height of the

liquid block) and all areas A are equal, we then have P2-P^- p(h)g = 0,or P2 = P1 +pgh.

If the block is at the very top of the liquid, then P, becomes thedownward pressure due to the air above the liquid, which is 1.01 x 10s Pa,written as Po. Therefore, P2 is called the absolute or total pressure on the

block. The term pgh is called the gauge pressure on the block, whereh is the depth of the block in the fluid and p is the fluid's density. Asexpected, our fictitious block of liquid can be replaced by a real blockof any material beneath the surface of the liquid, which thereforeexperiences both an absolute and a gauge pressure. To summarize:

P = P0 + pgh (absolute or total pressure) Pg = pgh (gauge pressure)

If you ever swam far below the surface of the water in a swimmingpool, you may remember feeling gauge pressure exerted on your bodyfrom the weight of the overlying water. Other examples of such submerged objects in water are submarines and all sea life that lives beneaththe top of a water layer. Notice that gauge pressure is independent ofthe shape of the container holding the fluid.

Sample Problem 1The Super Kamiokande neutrino experiment in Japan is a massive

water-filled, cylindrical underground mine that holds about 11,000sphere-like glass phototubes that detect light emitted when neutrinospass through the water. The water has a density of 1000 kg/m3, andthe bottom-most phototube is 45 m below the surface of the water.

(a) Determine the gauge pressure on one of the bottom-mostphototubes.(b) Determine the absolute pressure on one of the bottom-most

phototubes.(c) Determine the ratio ofthe absolute pressure to sea-level pressure

on one of the bottom-most phototubes.(d) If the front of one of the bottom-most phototubes has an area of

0.2 m2, determine the force exerted on the phototube front.

Solution to Problem 1(a) Because the fluid density p and the liquid depth h are both

given, the gauge pressure can immediately be calculated fromPg = P9^ = (1000 kg/m3)(10 m/s2)(45 m) = 4.5 x 10s Pa = gaugepressure Pg.(b) The absolute or total pressure on one of the bottom-most

phototubes is simply the gauge pressure plus the addition of theoverlying air pressure, 1.01 x 105 Pa. Therefore, P = Po + pgh =

1.01 x 10s Pa + 4.5 x 105 Pa, or total or absolute pressureP = 5.51 x 105 Pa.(c) The ratio of the answer in part (b) to the pressure at sea level

will state how much larger P is relative to 1 atm. Thus, (5.51 x10s Pa)/(1.01 x 105) Pa = 5. Therefore, the absolute pressureon a bottom-most phototube is five times the sea-level airpressure on Earth.

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212 Chapter 9

(d) According to the pressure equation, the force can be written asF= PA. Therefore, the total force on the phototube face is F= PA =(5.51 x 105 Pa)(0.2 m2) - 1 x 105 N.

How Do Fluids Act to Providea Support Force?

(College Physics 7th ed. pages 282-288/8th ed. pages 284-290}What happens to a fluid when an object is placed in it? When an objectfloats on or near the surface of a liquid, such as in a water bath, two

important results occur. First, the object is clearly supported by anupward force exerted by the liquid, called the buoyant force, FB. Second,the fluid is pushed aside, or displaced, and the weight of this displacedfluid is equal to the buoyant force (which is known as Archimedes's

principle).How does the buoyant force compare with the "dry" weight of theobject? If the object is hung by a spring scale in air (as shown here in theleft-hand diagram) and then in water (right-hand diagram), the reading

on the spring scale clearly decreases by an amount equal to the buoyantforce.

T I

r~\

In air

Tl\

In air

cm^_

A

--

, *

,-

In water

In water

V

For example, if the scale "in air" in the diagram reads 22 N andthen reads 14 N "in water," IF = 0 applied to that situation (see the

free-body diagrams) shows that the buoyant force FB must be 8 N:FB- = 0, or 14 N FB-22 N = 0. therefore, F..=IF = 0 = 7 +22 N - 14 N = 8 N.

If, on the other hand, the object is completely submerged (and floats)and if IF = 0, we can write F - w = 0, or F = mg = (pV)objectg.

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Fluids 213

/

I. . .. . 1

F\

'if

By Archimedes's principle, the weight of the displaced water>s equivalent to the weight of the submerged objectW'wmcn is equal to the buoyant force. Thus, for a fully immersedfloating object, the buoyant force FB is written

FB = pgV (buoyant force)

where p is the density of the fluid and V is the volume of the displacedliquid (which, again, for a fully submerged object, is equal to the volumeof the object). In the case in which the object is partially submerged,however, the weight of the displaced liquid is, of course, less than if theobject were fully submerged; therefore, FB is lower. Thus, the quantity

V in the equation above is only the volume of the submerged material,not the volume of the entire object. Study Sample Problem 2 to see thisfurther.

When using FB = pVg, be sure to differentiate between the volumeof the entire object and the volume of that part of the object that issubmerged. When the object is fully submerged, V is the total objectvolume. When it is partially submerged, however, V representsonly the volume of the part of the object that is submerged.

Sample Problem 2Plank

Water

A plank of pine wood (density of 550 kg/m3) of dimensions 4.0 m x4.0 m x 0.3 m is placed in a water bath whose density is 1000 kg/m3

as shown.(a) Sketch the forces acting on the floating plank.(b) Verify by calculation that this plank must float.(c) Determine the buoyant force on this plank while it is floating.(d) Determine the percentage of the plank that is not submerged.

Solution to Problem 2(a)

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214 Chapter 9

(b) To determine by calculation if the plank floats (other than simplycomparing the plank's density with that ofwater), it is necessary to

show that the maximum buoyant force (FBmax) exceeds the weight(mg) ofthe plank. IfFBmax > w, the plank floats. So, FBniax = pliquidgVmax= (1000 kg/m3)(10 m/s2)[(4 m)(4 m)(0.3 m)] = 4.8 x 104 N = FBmax. The

plank's weight is w = mg = (pV) vg = (550 kg/m3)[(4 m)(4 m)(0.3m)](10 m/s2) = 2.6 x 104 N = w. Therefore, the plank must floatbecause FBmax>w.(c) The buoyant force on the plank, based on parts (a) and (b) (and

because the plank is at rest), is simply the weight ofthe displacedwater, which is balanced by the plank's weight. Thus, FB = 2.6 x

104N.(d) To determine the percentage ofthe plank that is above water, it is

necessary to calculate the volume ofthe plank that is submergedso that the ratio (Vtolal - Vsubmenjed)/Vtotal may be determined. The

total volume of the plank that is submerged may be foundfrom the buoyant force: FB = PliqMgVsubmerged. Solving for volumegives

fb 2.6xlO4N 3^submerged pg (l000kg/m2)(l0m/s2)

Therefore,

(4 m)(4 m)(0.3 m) - 2.6 m3 _ 4.8 m3 - 2.6 m3 _ .Vw* (4m)(4m)(0.3m) 4.8 mwhich means that 46% of the plank is above water.

What Happens to the Pressureof a Liquid in a Closed Pipe?

{College Physics 7th ed. pages 279-281/8th ed. pages 281-283)It has been shown that the pressure within a fluid depends on thedepth of the fluid. For a confined fluid, a pressure change must betransmitted evenly throughout the entire fluid, which is true even ifthe fluid is enclosed in a tube or pipe that has a varying diameter.

An example of such a device is a hydraulic lift found at an automechanic shop. In this case, a fluid remains enclosed within a pipeof varying diameters, which therefore affects the force on the fluid.This property, first noticed in the early 17th century by French

scientist Blaise Pascal, has come to be known as Pascal's principle.It specifically states that any pressure change applied to a fluid in

an enclosed pipe is transmitted undiminished through the fluid andto the walls of the container. By employing Pascal's principle, it is

possible to input a fairly small force and amplify it, providing a verylarge output force, as is the case with a hydraulic lift that can raiseand lower automobiles.

Because pressure is constant throughout the fluid, we can write P =FJA^ = F.JA2 at two different locations within the fluid. Thus, Pascal's

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Fluids 215

principle written mathematically becomes F^A2 = F^\v where A is thecross-sectional area of the pipe enclosing the fluid. For a circular crosssection, A = nr2. So,

F%A2 = F^ (Pascal's principle)

Hydrodynamics: Is the Mass ofFluid Constant During Fluid Flow?

{College Physics 7th ed. pages 289-291/8th ed. pages 290-293)How does water behave while passing through the nozzle of a gardenhose (where it travels from a wide-diameter tube to a small-diameteropening in the nozzle)? In a section of pipe whose cross-sectional areaAj is nr\, the fluid moves a distance d = v, At in time At. As we have

previously seen, the mass m of the fluid may be expressed as pV; thus,the product of area A, and the distance dwill be substituted for volume

V to give A^ At. Therefore, the mass m of fluid passing through thissection of the pipe is pV = p(A^v^ At). In a separate location of the pipe,however, where the cross-sectional area A2is different, the same mass ofwater must pass through this location because the pipe is closed and the

fluid is assumed to be noncompressible to maintain mass conservation.Therefore, in the same time interval At, mass m, must equal mass m2,

which is pfiA^v^ At) = p2{A2v2 At). For the case of an incompressible fluidthat is unchanged throughout the pipe (and if the pipe has no leaks), thedensity p cancels from each side to give

A.v. = A,v (equation of continuity)

Because the mass flow rate (Am/At = pAV) is constant, the productAv is constant, signifying that an inverse relation exists between area A

and speed v. When a pipe is narrowly constricted such that A is small,the corresponding effect is to increase v and vice versa. Notice that the

mass flow rate has units of kilograms per second, whereas Av has unitsof cubic meters per second, helping you remember that the former

is indeed a rate of mass movement and the latter is a rate of volumemovement.

How Are Fluid Speed and Pressure Related?(College Physics 7th ed. pages 291-297/8th ed. pages 293-299)

In addition to describing the mass flow rate of a fluid through a pipeof varying diameter and the fluid's associated change of speed, therealso exists a relationship between fluid pressure and speed in such apipe, particularly when the pipe changes elevation, causing a change influid potential energy. The basis of this relationship, called Bernoulli's

equation, lies in the application of energy conservation as applied toan ideal fluid. As a space constraint, its derivation is not shown here

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216 Chapter 9

(refer to the textbook pages referenced above), but its use is employedin the Sample Problem 3 as well as in Free-Response Question 2.

Bernoulli's equation, written P + \ pv2 + pgy = constant, containsterms that are essentially those of energy conservation. First, notice thatfluid pressure P is present, as are | pv2 and pgy, which are analogous to

kinetic energy (per unit volume) and gravitational potential energy (perunit volume), respectively. Therefore, notice that as the pressure P of thefluid increases, the associated fluid speed v decreases and vice versa.

A + P9Yi + p pv\ = P2 + pgyz + -pv\ (Bernoulli's equation)

An example is the Venturi tube, which is a horizontal tube of varyingdiameter used to measure the speed of fluid flow (similar tubes, calledpitot tubes, exist on aircraft to measure airflow). There are countless

examples of this principle in our ordinary lives, one of which is evidentin Sample Problem 3.

Sample Problem 3A person suffering from shortness of breath visits a doctor, who

discovers that blood flow in an artery (shown here) is severelyrestricted, noting that the flow at position 2 in the artery is threetimes faster than at position 1. (For the purposes of this problem,

assume human blood is an ideal fluid.)

(a) Based on the information given, comment qualitatively onthe diameter of this artery at position 2 relative to that at

position 1.(b) Describe the differences in the blood pressure at position

2 relative to position 1. What can occur as a result of thisdifference?(c) Determine the ratio of r, to r2.(d) If r, = 1.0 cm, calculate the value of r2.(e) If the average density of human blood is 1060 kg/m3 and the

blood's speed at position 1 is 0.1 m/s, determine the pressuregradient AP between positions 1 and 2 (assume the heights of 1and 2 are equal).

Solution to Problem 3(a) It is stated that blood flow is faster at position 2 than at

position 1, which signifies, by the volume flow-rate condition,that the cross-sectional area (and therefore the diameter)

at position 2 must be much narrower than at position 1.Therefore, the blood vessel has a constriction at position 2 that

causes the increase in blood speed.(b) By Bernoulli's equation, as the speed of the fluid increases,

an inverse relation exists with the pressure of the fluid. Thus,because the blood speed has increased at position 2, the bloodpressure must be lower at position 2 than at position 1. Thislower pressure may cause a collapse in the artery because thereis a greater pressure outside the vessel than inside at position 2.

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Fluids 217

(c) The ratio can be determined using the volume flow-rate conditionA, v, = A2v2, where each cross-sectional areaA is nr^note also that

it is stated that v2 = 3v,). We then have r*vt = r22v2, and solving forthe ratio of radii gives r*/r2z = v/v, or rt/r2 = ^v2 / v, = yJ3vt I v, =

1.7. Therefore, the ratio of rt to r2 is 1.7, or r, = 1.7r2.(d) Based on the calculation in part (c), r, = 1.7r2, or r2 = 1.0 cm/1.7.

Therefore, r2= 0.59 cm.(e) The change of pressure AP between positions 1 and 2 can be

found by applying Bernoulli's equation to the two locations andcalculating the final pressure change Pt - P2. Because the heightof the artery is constant, the terms pgy^ and pgy2 will drop out,

and we can solve for P, - P2as follows: Pt + \pv? = P2 + \pv22.Rearranging and collecting similar terms on each side gives P, -P2 = \pv22 -\pv?. Because it is given that v2 = 3v,, we cansubstitute to further reduce the equation to Px - P2 = \ p(3v,)2 -

\ pv2 = \ p(9v,2 - v,2) = i p(8v,2) = 4pv,2. Substituting for thedensity and speed will finally give Px - P2 = 4(1060 kg/m2)(0.1 m/s)2 = 42 Pa. Therefore, the pressure change AP withinthe artery between points 1 and 2 is 42 Pa.

Fluids: Student Objectives for the AP Exam You should understand the nature and meaning of mass density

(p = m/V) and how to substitute pV to solve for mass m. You should understand the nature and meaning of pressure P, where

P = F/A, and how to substitute PA to solve for force F. You should understand and be able to calculate gauge pressure

(P = pgh) due to a liquid of depth (or height) h. You should understand and be able to calculate absolute pressure P,

atmospheric pressure {P^ plus gauge pressure (pgh), written as P = Po+ pgh. You should understand and be able to perform calculations using

Pascal's principle {Ffi2 = F^A,). You should understand the nature of floating objects and how they

are affected by a buoyant force FB. You should understand, with respect to floating objects, the application

of Archimedes's principle and that (1) the volume of liquid that isdisplaced by a submerged object equals the volume of the materialthat is submerged and (2) the weight of the displaced material is equalto the (upward) buoyant force exerted by the liquid on the immersedobject (i.e., Archimedes's principle). You should understand and be able to calculate the maximum buoyant

force (FBmax=p gVobject). You should understand the nature and characteristics of an ideal

fluid. You should be able to perform calculations with and should understand

the meaning of the equation of continuity (A,v, = A2v2), which pertainsto the speed of an ideal fluid as it passes through constrictions ofvarying cross-sectional area. You should understand and be able to perform calculations using

Bernoulli's equation, which states that at any two points (1 and 2) in aflowing ideal fluid, P + pgy + i pv2 = constant.

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218 Chapter 9

Multiple-Choice Questions1. Water flows with a speed of 5.0 m/s from a 2.0-cm-diameter pipe

into a 6.0-cm-diameter pipe. In the 6.0-cm-diameter pipe, what isthe approximate speed of the water?

(A) 0.1 m/s(B) 0.6 m/s(C) 2.0 m/s(D) 3.5 m/s(E) 5.0 m/s

2. Water is pumped into one end of a long pipe at the rate of 10.0 m3/s.It emerges at the other end at a rate of 6.0 m3/s. What is the mostlikely reason for the decrease in the flow rate of this water?

(A) The water is being pumped uphill.(B) The water is being pumped downhill.(C) The diameter of the pipe is not the same at the two ends.(D) There is a tremendous friction force on the inside walls of the

pipe.(E) There is a leak in the pipe.

3. In a classroom demonstration, a 73.5-kg physics professor lies ona "bed of nails" that consists of a large number of evenly spaced,

relatively sharp nails mounted in a board so that the points extendvertically upward from the board. While the professor is lying down,

approximately 1,900 nails make contact with his body. If the area ofcontact at the point of each nail is 1.26 x 10"6 m2, what is the averagepressure at each contact point?(A) 1.59 x 104 Pa(B) 5.71 x 108 Pa(C) l.llxlO12Pa

(D) 1.11 xlO6Pa(E) 3.01 x 105 Pa

4. A frog is at rest at the bottom of a lake at a depth y below the surface.If the top surface of the frog has area A, which of the followingexpressions correctly describes the total downward force F exertedon the frog?(A)(P0 +(B) P0A(C) pgyA

(DHP^(E) Po + pgy

5. What are the units of volume flow rate?(A) seconds(B) kilograms per second(C) square meters per second(D) cubic meters per second

(E) cubic meters

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Fluids 219

6.

7.

8.

Liquid water (of density 1000 kg/m3) flows with a speed of 8.0 m/sin a horizontal pipe of diameter 0.2 m. As the pipe narrows to adiameter of 0.04 m, what is the approximate mass flow rate of thewater in this constriction (in units of mass flow rate)?

(A) 0.01(B) 0.25(C) 0.8(D) 240(E) 900

Rope

A 2-kg wooden block displaces 10-kg of water when it is forciblyheld fully immersed (the block is less dense than the water). Theblock is then tied down such that part of it is submerged as shown,and it displaces only 5 kg of water. What is the approximate tensionin the string?(A) 10 N(B) 20 N(C) 30 N

(D) 70 N(E) 100 N

For Questions 8 and 9, refer to the diagram, which depicts a horizontal piping system, viewed from directly overhead, that delivers a

constant flow of water through pipes of varying relative diameterslabeled 1 through 5.

At which of the labeled points is the water in the pipe moving withthe lowest speed?

(A)l(B) 2(C)3(D)4

(E) 5

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220 Chapter 9

9. At which of the labeled points is the water in the pipe under thelowest pressure?(A)l(B)2

(C)3(D)4(E)5

10. A hydraulic press has one piston of radius /?, and another piston ofradius Rr If a 400.0-N force is applied to the piston of radius R, andthe resulting force exerted on the other piston is 1600.0 N, which ofthe following is a correct mathematical statement concerning eachpiston radius?(A) R2 = R,(B) R, = 2R2(QR1=^R2(D) R2 = 2Rt

11. Using the value of atmospheric pressure at sea level, 1.0 x 10s Pa,what is the approximate mass of Earth's atmosphere that is above aflat building that has a rooftop area of 5.0 m2?(A) 2.0x10^ kg(B) 4.0 x lO"2 kg(C) 9.0 x 102 kg(D) 5.0 x 104 kg

(E) 5.0 x 105 kg

12. A person is standing near the edge of a railroad track when a highspeed train passes. By Bernoulli's equation, what happens to theperson?(A) The person is pushed away from the train.(B) The person increases in mass as the train approaches and then

decreases in mass as the train recedes.(C) The person is pushed upward into the air.(D) The person is unaffected by the train.(E) The person is pushed toward the train.

13. A fluid is undergoing "incompressible" flow. Which of the followingbest applies to this statement?(A) The pressure at a given point cannot change with time.(B) The velocity at a given point cannot change with time.(C) The density cannot change with time or location.

(D) The pressure must be the same everywhere in the fluid.(E) The velocity must be the same everywhere in the fluid.

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Fluids 221

14. A wooden block of density 450 kg/m3 floats on the surface of a poolof stationary liquid water. Which of the following is a correct free-body diagram for this situation, where F=buoyant force, N= normal

force, and w = weight?(A) ^N/

/

\

VI

(B) /

[1\

r F i

(Q/

i

\i '

(D) /

(E)

r1s

15. An object with a volume of 2 x 10"2 m3 floats in a tank of water with70% of its volume exposed above the water. If the density of water

is 1000 kg/m3, what is the approximate weight of this object?(A)3N(B)6N(C) 12 N(D) 30 N(E)60N

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222 Chapter 9

Free-Response Problems

i.

Tablelop

A student hangs an unknown material in the shape of a cube 2.1 cmon a side from a sensitive spring scale that is at rest at sea level onEarth as shown. The scale reads 0.245 N. The student is given thefollowing list of densities of common solids and liquids.

Solids (in kg/m3)AluminumBrass

ConcreteDiamondGoldLead

SilverWood (pine)

2,7008,470

2,2003,520

19,30011,30010,500

550

Liquids (in kg/m3)Blood (at 37C)Hydraulic oil

MercuryWater (at 4C)

1,060800

13,6001,000

(a) According to the table, determine the material the student'scube is most likely made of.(b) Determine the percent deviation in the student's measurement

of the quantity found in part (a).

Tablelop

Unknown liquid

(C)

The cube is now fully immersed into a container with anunknown ideal liquid as shown here, and the scale readingis now 0.173 N.Sketch the free-body diagram of the cube as it remainssuspended in the liquid.

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Fluids 223

(d) What liquid does the student determine is in the container?Support your conclusion with appropriate calculations.

Ground

\ Pipe

2. A fountain emitting a single stream of water (density 1000 kg/m3)at a playground is fed from a vertical pipe that is below ground butwhose opening is at ground level as shown. At the ground-levelopening, the pipe's diameter is 0.06 m and the water exits the pipewith a velocity of 9.0 m/s upward.(a) Determine the maximum height attained by the water after it

exits the pipe.(b) Determine the volume flow rate of the water as it exits the pipe.(c) Determine the mass flow rate of the water as it exits the pipe.(d) The fountain's lower end ofthe underground pipe has a diameter

of 0.12 m and is 6.0 m below the ground. Determine the absolutepressure in the underground pipe at a depth of 6.0 m.(e) The owner of the fountain wishes to launch the water so that

it reaches a height of 10.0 m above the ground with the samevolume flow rate. She decides to do so by attaching a new

nozzle to the pipe at the ground-level opening. Determine whatthe diameter the new nozzle must be to achieve this height.

AnswersMultiple-Choice Questions

1. B The mass flow rate is pAv = constant throughout (i.e., equation ofcontinuity), so A,vv = A2v2 can be used to calculate the new speed inthe 6.0-cm-diameter pipe (bear in mind that diameter, not radius, isgiven, but because in the ratio the units of area cancel, the units ofcentimeter do not need to be changed to meters). We can solve forv2, substitute A = nr2 for each pipe, and solve for v2:

r2

Substituting 1 cm for r, and 3 cm for r2 gives

_(lcm)25.0m/sV* ~ (3cm)2

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224 Chapter 9

which simplifies to v2 = [1(5.0 m/s)]/9 =f, or v2 0.6 m/s, whichis choice (B) (College Physics 7th ed. pages 288-291/8th ed. pages

290-293).

2. E Again, mass flow rate is pAv = constant throughout (i.e., equationof continuity). Therefore, Atv, = A2v2. Because it is stated that the

initial flow rate is 10.0 m3/s, this quantity must remain constantthroughout the pipe (assuming it is closed). Therefore, because it is

stated that the outflow is 6.0 m3/s, the only possible explanation forthe discrepancy is that liquid left the pipe via a leak, or choice (E)

[College Physics 7th ed. pages 288-291/8th ed. pages 290-293).

3. E This question is a basic application of the definition of pressure,P = F/A, where F is the force exerted on one nail (not the entire 1,900

of them). Therefore, the entire weight of the professor [mg) must bedistributed over all the nails. Once that value is found, the averagepressure per nail can be calculated:

F ing/1900 73.5 kg(10.0m/s2)/1900 735/1900P~A'"l.26xl0-6m2~ 1.26x10^ m2 1.26x10^

(7.4xl02)/(2xl02)-i '-^ ^ = 3.0xl05 Pa1.26 xlO"6or P = 3.0 xlO5 Pa, which is the approximate value of choice (E){College Physics 7th ed. pages 274-276/8th ed. pages 276-278).

4. A The total force exerted on the frog is derived from the totalpressure exerted on the frog due to both the overlying column ofwater above the frog as well as the overlying column of air abovethat water, according to PmA = F. Here P is the absolute pressure

given by Ptot = Po + pgh. Therefore, substituting gives PwlA = F =(Po + pgh)A. The depth of the water, however, is y, which replacesh to give F = (P0 + pgy)A, which is choice (A) (College Physics 7th

ed. pages 274-276, 277-282/8th ed. pages 276-278, 279-284).

5. D The equation of continuity gives mass flow rate as Av, or volumeper second. Therefore, the units are square meters multiplied bymeters per second, or cubic meters per second, as shown by choice

(D) (College Physics 7th ed. pages 288-291/8th ed. pages 290-293).

6. D The equation providing mass flow rate is m/At, or simply pAV.Substituting values gives (1000 kg/m3)(3.14)(l x 10"1 m)2(8 m/s) -(3000X8X1 x 10"2) ~ 240 kg/s, which is choice (D) (College Physics

7th ed. pages 288-291/8th ed. pages 290-293).

7. C It is stated that the block displaces 5 kg of water, which suggeststhat the weight of the displaced water is w = mg = 5 kg(10 m/s2), or

50 N, which, by Archimedes's principle, is also equal to the buoyantforce FB. Next, sketch the free-body diagram and apply the conditions

of equilibrium: ZF = 0 gives FB - T- mg = 0. So, FB-mg = T. whichbecomes 50 N - (2 kg)(10 m/s2) = T, or T = 30 N, which is choice (C)

(College Physics 7th ed. pages 282-288/8th ed. pages 284-290).

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Fluids 225

8. A The mass flow rate is pAv = constant throughout (i.e., equation ofcontinuity). Therefore, A,vt = A2v2 shows that the speed v is inverselyproportional to the cross-sectional area A of the pipe. So, when A isthe largest, the speed is the lowest. The largest pipe diameter shownis 1; therefore, it has the water with the lowest speed {College Physics7th ed. pages 288-291/8th ed. pages 290-293).

9. B By Bernoulli's equation, an inverse relationship exists betweenthe square of the speed of the fluid flow and the surroundingpressure according to P + pgy + pv2 = constant. Because P ~ 1/v2,the pressure P will decrease when the speed ofthe fluid increases. As

seen from the solution to Question 8, the speed of the fluid will be thelargest when the area of the pipe is the smallest (i.e., the narrowest).

Because the narrowest pipe is pipe 2, it will possess the fastestmoving water with the lowest water pressure. Notice also that thispipe is viewed from overhead; thus, all locations are the same heightabove the ground (College Physics 7th ed. pages 288-291, 291-294/8th ed. pages 290-293, 293-296).

10. D The pressure is uniform within the press, so, by Pascal's principle,F^A2 = F/iy which contains each piston's value of R under area A.The area is nr2, which becomes F,(;rfl22) = Fz{nR}). Solving for the ratio

of the radii gives R\IR\ = FJFV which becomesR2/R, = ^1600/400.Substituting the values of the forces gives R2/R, =^1600/400 orR/R, = 2. Therefore, R2 = 2Ry which is choice (D) (College Physics

7th ed. pages 277-281/8th ed. pages 279-283).

11. D Using the relation P = FIA, we can calculate the force F on therooftop and then, realizing that this force is equal to the weight ofthe overlying air, equate this result to mg by the relation w = mg. So,P = F/A becomes PA = F= mg, where m is the mass of the air above

the rooftop, which simplifies to PA/g = m. Substituting gives [(1.0 x105 Pa)(5.0 m2)]/10.0 m/s2, so m = 5.0 x 104 kg, which is choice (D).(College Physics 7th ed. pages 274-276/8th ed. pages 276-278).

12. E By Bernoulli's equation, an inverse relationship exists betweenthe speed of the fluid flow and the surrounding pressure according

to P + pgy + \ pv2 = constant. Because P l/v2, the pressure P willdecrease when the air moving with the speeding train passes byat a higher-than-normal speed. Therefore, this high-speed air willcreate a region of low pressure on the train side of the person. Thus,the higher air pressure on the platform side of the person creates apressure gradient, causing the standing person to be moved slightlytoward the moving train (College Physics 7th ed. pages 291-294/8th ed. pages 293-296).

13. C An ideal fluid is nonviscous, possesses a steady motion (i.e.,velocity, density, and pressure within the fluid do not change with

time), does not experience turbulence, and remains incompressible.An incompressible fluid is one in which the fluid density remains

constant. Therefore, choice (C) is correct (College Physics 7th ed.pages 288-290/8th ed. pages 290-291).

2010 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

226 Chapter 9

14. D For an object floating in a liquid, the upward support force oftheliquid is the buoyant force (FB), which is equal to the gravitationalforce (w=mg) experienced by the object, but is opposite in direction.Therefore, choice (D) is correct {College Physics 7th ed. pages 282-

288/8th ed. pages 284-290).

15. E To solve this question, it is necessary to use the relation forbuoyant force, which relates the submerged volume to the buoyant

force, which is equivalent to the weight of the object. We are toldthat 70% of the object is exposed, and we need to determine how

much of the object is submerged according to FB= p liquid gVsubmergedTherefore, 70%(2 x 10"2 m3) = 1.4 x 10"2 m3, which signifies that

0.6 x 10"2 m3 of the object is submerged. The buoyant force (and thusthe weight of the object) is FB= (1000 kg/m3)(10 m/s2}(0.6 x 10"2 m3) =

1 x 104(0.6 x 10"2) = 60 N, which is choice (E) {College Physics 7th ed.pages 282-288/8th ed. pages 284-290).

Free-Response Problems1. (a) With the information provided, the student needs to calculate the

density of the cube and compare it with those given, finding thevolume by the dimensions given and the mass from the weight

shown (w = mg) on the scale. According to the density equationp = m/V, we have p = {w/g)/V= w/gV= (0.245 N)/[10/s2)(0.021 m)3Lwhich gives a density ofp = 2650 kg/m3. Therefore, the cube is most

likely made of aluminum {College Physics 7th ed. pages 274-276/8th ed. pages 276-278).

(b) The percent deviation is determined by substituting values:

o. .._ measured-accepted .__.% difference = xl00%accepted= 2650 kg/m3-2700kg/m3 x

2700 kg/m3which is approximately 2%.

(c) A sketch is as shown.

{College Physics 7th ed. pages 282-288/8th ed. pages 284-290)

(d) To determine the type of liquid, it is necessary to calculate itsdensity (from the buoyant force equation) and compare it with

2010 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Fluids 227

those given in the chart. Also, because the cube is in equilibrium,we need to apply IF=0 to find FB, the buoyant force, in terms ofthe

forces given. So, ZF=0gives FB=mg- T. Therefore, byFB=pgV,wehavemg-T=pgV. As for the volume V, it is stated that the cube is fully

immersed; therefore, the volume ofdisplaced liquid is equivalenttothe volume ofthe cube (length x width x height). Solving fordensityp now gives (mg - T)/gV= p. Substituting values gives the solutionfortheliquiddensity:[(0.245N)-(0.173N)]/[(l0.0m/s2)(0.02lm)]3=p,which gives 777 kg/m3 as the liquid density. According to thetable given, this liquid is most likely hydraulic oil {College Physics7th ed. pages 282-288/8th ed. pages 284-290).

2. (a) This problem is a kinematics question that simply deals withthe maximum height cf reached by an object projected straight

upward that is in free fall. We can use the kinematics equationv2 = vf + 2ad for constant acceleration, keeping in mind that atthis maximum height, the water velocity is momentarily zero.

Solving for d gives d = vf2 - vf/2a = [0 - (9.0 m/s)2]/[2(10.0 m/s2),which gives d = 4m. We could also solve this problem by energyconservation, where Kbouom = U [College Physics 7th ed. pages41-45/8th ed. pages 42-46).

(b)The volume flow rate, in cubic meters per second, is given byA,v, = A,v, comparing two locations in a pipe. Here we simply

need the product of the pipe area A and the speed of the waterv2 at the pipe opening, being careful to realize that pipe diameterD (not radius) was given in the problem. At the pipe opening, we

have A2v2 = ;rr22(v2) = tt(D/2)2(v2) = (3.14)[(0.06 m)/2]2(9.0 m/s) = 2.5 x10~2 m3/s = volume flow rate. For ease of calculation in part (d),we have chosen subscript 2 to represent the location at which

the water leaves the pipe at ground level (College Physics 7th ed.pages 288-291/8th ed. pages 290-293).

(c) The mass flow rate is mass per unit time flow, or (pV)/At. It issimply the product of the volume flow rate and the fluid density,

here 1000 kg/m3 (water), or pAxvt. Therefore, pA,v,= (1000 kg/m3)(2.5 x 102 m3/s) = 25 kg/s = mass flow rate (College Physics 7thed. pages 288-291/8th ed. pages 290-293).

(d) Because the fluid is in motion, it is necessary to use Bernoulli'sequation to determine the pressure at the 6.0-m depth. The locationbelow ground at 6.0 m is labeled position 1 (where y, = 0 m), andthe point at which the water exits the pipe above the ground islabeled position 2 (where y2 = 6.0 m). Here, it is necessary to solve

for P, at the 6.0-m depth (where P2 is the atmospheric pressurejust at ground level, where the fountain water leaves the pipe). So,

P, + P9Y, + 2 Pv\2 = P2 + P9Y2 + i Pv22- (Notice that v, = volume flow rate/pipe area). Leaving units off for clarity and solving, we then have

1 1P, = P2 +pgy2 + jpv22 -pgy, --p (2.5 xlO"2)[ *( IO 2010 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in pan.

228 Chapter 9

Substituting values gives

[2 5xlO~2"l'(D/2)24) + (4.05xl04)-(l.llxl03)

or P, = 2.00 x 10s Pa.

{College Physics 7th ed. pages 291-294/8th ed. pages 293-296)

(e) Just as in part (a), it is necessary to find the velocity of thewater as it leaves the pipe at ground level using kinematics (orenergy conservation) relationships. Then, with the same volume

flow rate, we solve for the new pipe radius r and finally solvefor the pipe diameter (by doubling r). So, vr2 = vf + 2ad. Solving

for v, gives yjv2f-2ad = v or V0-2(-10 m/s2)(10 m) = v, = 14 m/s.Because 2.5 x 102 kg/s = volume flow rate = Av, we can solve forr and substitute values (omitting units for clarity). So,

p^ pxjOl = 0,024mV jcv V (3.14X14)Thus, the diameter of the new nozzle is 0.048 m, or 4.8 cm

{College Physics 7th ed. pages 41-45,288-291/8th ed. pages 42-46,290-293).

O 2010 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or poslcd to a publicly accessible website, in whole or in part.