PHYS122 University of Waterloo

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Physics for Physicists

Christopher ODonovan

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Copyright c 2005-2010 Christopher ODonovan

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Chapter 17

Universal Gravitation

17.1 Introduction

In this chapter the only new physics introduced is Newtons Law of

Universal Gravitation, Eq 17.1, there are many concepts and much

jargon that are used in other fields of physics, largely because in

the historical development of physics the study of the motion of the

planets occurred first.

There are two important properties of the gravitational force.

First, it is a radial force (i.e. it is directed along the line joining

the centres of the bodies) it creates no torque (i.e. r F = 0) and

so angular momentum is conserved. And, second, it is a conserva-

tive force so mechanical energy is conserved. These two properties

of gravitationally interacting systems, that both angular momentum

and energy are conserved, allow us to solve many classes of prob-

lems that would be difficult or impossible using Newtons Second

Law. Such problems are more easily tackled using Lagrangian or

Hamiltonian mechanics, which were originally developed for ex-

actly such problems. These more mathematically sophisticated forms

of mechanics are not addressed in this chapter.

69


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70 CHAPTER 17. UNIVERSAL GRAVITATION

F12m2m1

F21

Figure 17.1: The force F12 is the gravitational attraction of mass 1

on mass 2. The vector r12 is from the centre of mass 1 to that of mass

2; r12 is the unit vector in the same direction. From Newtons Third

Law, the forces on each of the two masses are equal in magnitude and

opposite in direction, F12 = F21.

17.2 Newtons Other Law

Newtons law of universal gravitation is1

F12 =Gm1m2

r212

. (17.1)

The attractive gravitational force between two point masses is pro-portional to each of the masses and inversely proportional to the

square of the distance between the objects. The force is directed

along the line joining the objects. The constant of proportionality,

G, is the universal gravitational constantand, in SI units, is equal to

6.672 1011 N m2/kg2.

For extended objects it is possible to calculate, using integral

calculus, the gravitational attraction. Certain symmetries, however,make such calculations unnecessary. Happily, one such symmetry is

that of a spherical object with uniform density, which we can treat

as a point mass with all the mass concentrated at the centre of the

sphere.

The planet Earth, due to its rotation, has a radius at the equator

slightly larger than that at the poles (it is an oblate spheroid). This

difference, however, is a fraction of a percent so for all but the mostsensitive calculations we can treat it as a sphere. Other celestial

bodies have a similar flattening that we can ignore for our purposes.

The gravitational field due to a spherical object of mass M is the

1I dont know why this isnt called Newtons fourth law.


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17.2. NEWTONS OTHER LAW 71

gravitational force per unit mass, 2

g = GM

r2r

where r is the distance from the origin at the centre of the sphere

and r is the unit vector in the radial direction.

Problem 17.2.1. Clarke Orbit

Consider the planet Earth and an artificial satellite of mass m trav-

elling in a circular orbit of radius r about it. (a) Use your knowl-

edge of the universal gravitational constant, G, the radius of Earth,

R, and the free fall acceleration at the Earths surface, g, to find an

expression for the mass of the Earth, M. Compare your answer to

the accepted value. [answer: 5.9736 1024 kg] (b) Use dimensional

analysis to derive an expression for the period of the satellite. (Your

answer will involve the universal gravitational constant, G, along

with M and r, the orbits radius.) (c) Draw the free body diagram

of the satellite. Include on your diagram the acceleration,a, andvelocity, v, vectors. Add to your diagram the radial and tangential

unit vectors, r and .3 (d) Determine the period, T, of the satellite

using appropriate equations. (Hint: For uniform circular motion the

speed is a constant equal to one circumference per period.) (e) For

what size orbit, rClarke, will the orbit be geosynchronous?4 Draw a

sketch of this orbit along with the planet Earth. Add to your sketch

the rotational axis of Earth.2Similarly, the electric field is the electric force (Coulombs Law) per unit

charge. In each case the vector field associates a vector quantity with each point

in space (except at singularities). Since the gravitational force couples with mass

the gravitational field is the force per unit mass; for the electric or Coulomb force,

the corresponding field is the force per unit charge.3These are the unit vectors for a polar coordinate system. The radial unit

vector, r, points away from the origin and the tangential unit vector, , points

in the direction of increasing , the polar angle. Interestingly, these unit vectors

move with changing and so their time derivatives are not constant, r = & = r.

4An object in a geosynchronous orbit will appear stationary to an observer on

the surface of the planet Earth. Such an orbit is commonly used for communica-

tions satellites and was first proposed for this use in 1945 by noted science fiction

author Arthur C. Clarke; consequently this orbit is also known as a Clarke orbit.


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rv

periapsis apoapsis

Figure 17.2: For an object in a circular orbit, left, the centripetal

force must be equal to the objects mass multiplied by the centripetal

acceleration, a = v2/r. For an object in an elliptical orbit, right, thingsare more complicated. When a planet is farthest from the Sun it is

said to be at aphelion; at closest approach it is at perihelion. For an

object in orbit about the Earth the corresponding terms are apogee andperigee. The general terms are apoapsis and periapsis.

17.3 Keplers Laws

Historically, Kepler deduced his three eponymous and empirical laws

in 1605, thirty-eight years before Newton was born and they werean attempt to account for the observed discrepancies in the motion

of the various planets. They came almost a century after Nicolaus

Copernicus heliocentric universe started modern astronomy by ini-

tiating the paradigm shift from the geocentric to the heliocentric uni-

verse. In Copernicus model the planets had circular orbits around

the Sun; Kepler improved this model by postulating that the orbits

were actually ellipses with the Sun at one of the foci of the ellipses.

17.3.1 Keplers First Law

All planets move in elliptical orbits with the Sun at

one focus.

The planets known to Kepler have orbits that are almost circular

so the discrepancy between their observed positions and that pre-

dicted by Copernicus circular orbits was small but measurable.

We already know from our study of uniform circular motion in

Ch. 10 that the acceleration, a, speed, v, and radius, r, are each

constant and are related by the expression v2 = ar. For elliptical


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17.3. KEPLERS LAWS 73

a = semi-major axisb = semi-minor axis

e = c/a = eccentricityc = linear eccentricity

c

a

b

y

x

foci

Figure 17.3: An ellipse, like other conic sections, is created by the

intersection of a plane with a right, circular cone (left); for an ellipse

the angle between the plane and the axis of the cone must be between/2 and half angle, . An ellipse is characterized (right) by the lengthsof its semi-major and semi-minor axes, a and b. Other quantities ofinterest for an ellipse are the linear eccentricity, c, and the eccentricity,e. These latter two quantities are not independent of the former two.

motion all three quantities vary with position and there is no sim-

ple relationship between them. However, angular momentum and

energy are conserved for either orbit so we can still perform many

calculations. In particular, at the apses, where the velocity and po-sition vectors are perpendicular, the angular momentum is easy to

calculate. (The apses are the points closest, periapsis, and farthest,

apoapsis, from the object being orbited; see Fig. 17.2.)

There are many ways to describe ellipses. The most basic is

discussed in Fig. 17.3 while another is that it is the locus of points

such that the sum of the distances from two points, called the foci,

is a constant. Using this definition we can draw an ellipse with two

pins and a length of string by placing the two pins at the foci and

creating a loop from the string with length 2a + 2c, where a is thesemi-major axis and c is the linear eccentricity (see Fig. 17.3).

Problem 17.3.1. Elliptical Questions

Consider the diagram of an ellipse shown on the right in Fig. 17.3.(a) Show that the sum of the distance of the point (b, 0) to the foci is2a + 2c. (b) By considering the triangle with apices at the two fociand (a, 0), find a relationship between a, b and c. (Hint: The sum ofthe distances of the point (0, a) to the foci may be helpful.) (c) If theeccentricity of an ellipse is e = 0.017, what is the ratio a/b?


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Problem 17.3.2. Elliptical Orbits

If a comet (which orbits the sun in a highly elliptical orbit) has an

perihelion ofrP and a period ofT (a) determine its aphelion, rA and

(b) show that for a circular orbit with rP = R you get the expected

result for rA.

17.3.2 Keplers Second Law

The radius vector drawn from the Sun to a planet

sweeps out equal areas in equal times.

Problem 17.3.3. Angular Momentum and Keplers Second Law

In timet a planet in an elliptical orbit sweeps out an area ofdA as

it moves a distance r where r is the position vector relative to the

sun which is located at one of the foci of the ellipse. (a) Draw a

diagram of the area A along with the vectors r and r. (b) On the

same diagram indicate the area r r. (c) What is the relationship

between your answers to (a) and (b)? (d) Multiply your answer to(b) by m, the mass of the planet, divide it byt and take thet 0

limit of the result. What is this quantity? (e) If there is no torque on

the planet what can you say about your answer to (d)? (f) What can

you conclude about the quantity dA found in (a)?

17.3.3 Keplers Third LawThe square of the orbital period is proportional to

the cube of the semi-major axis of the elliptical orbit.

Problem 17.3.4. Circular Thirds

Consider an object in a circular orbit of radius r. It experiences a

centripetal force given by Newtons Law of Universal Gravitation,

Eq. 17.1. (a) Draw the free body diagram of an object of mass

m undergoing uniform circular motion about an object of mass M.

(b) Newtons Second Law along with Newtons Law of Universal

Gravitation, Eq. 17.1, and the usual expression for centripetal accel-

eration to find an expression for the square of the objects velocity,

v2. (c) For uniform circular motion the velocity is a constant equal


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17.4. ENERGY OF GRAVITATIONAL SYSTEMS 75

rU

Figure 17.4: The zero for the gravitational potential energy for a

point mass at the origin is defined as the point at infinity so that no un-

necessary and arbitrary constants are introduced, giving Eq. 17.2. As

a consequence, the gravitational potential energy is always negative,

U r1.

to one circumference per period, T. Use this to find a relationship

between the period and radius of the orbit.

17.4 Energy of Gravitational Systems

Since the work done by the gravitational force depends only upon

the initial and final configurations it is a conservative force and,

rather than calculating the work done along a trajectory, we can use,

instead, the change in the appropriate potential energy

U12 = W12 =

2

1

F dr = GMm

1

r2

1

r1

.

Since it is only a change in a potential energy that has any physical

meaning we can choose any arbitrary point as the zero of potential

energy. It is both convenient and conventional to choose the point at

infinity as the this point; the gravitational potential energy is then

U= GMm

r. (17.2)

Note that this expression is always negative as r in this equation is

always positive (were using a spherical coordinate system) and that

the total energy of a gravitationally bound system is always negative


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(i.e. if the energy is positive then the objects do not form a bound

system). See Fig. 17.4.

IfM m then we can consider the centre of the mass M as

an inertial reference frame (why?) and the mechanical energy of the

system is

E= K+ U=1

2mv

2GMm

r.

However, if the two masses are comparable in size then we cannot

assume that M m and the two objects will orbit their common

centre of mass.

IfM m then we would have to include both kinetic energies

in our calculation of the mechanical energy. If there were several

bodies the mechanical energy would have a kinetic energy for each

body as well as a potential energy for each possible pair of bodies,

E=1

2

i

miv2

i G

2

i,j,i=j

mimj

rij.

Binding Energy

The binding energy of a system is the energy re-

quired to separate the system so that all parts are in-

finitely far apart and all their velocities are zero.

Because of our choice for the zero of gravitational potential en-

ergy, the binding energy is the energy we need to add to make the

mechanical energy zero. Consequently it is the negative of the total

energy of the system.

Problem 17.4.1. Circular Binding

Consider a planet of mass m in an elliptical orbit about a star of mass

M m with a radius r. (cf. Pr. 26.5.1) (a) Draw the free body

diagram of the planet. (b) What is the acceleration, a, of the planet?

(c) What is the speed, v, of the planet in its circular orbit? (d) What is

the total mechanical energy, E= K+U, of the planet? (e) Use your

result to (c) to find the kinetic energy, K, of the planet in terms of the

radius of its orbit, r. (f) What is the kinetic energy, K, of the planet

in terms of its gravitational potential energy, U? (This is a special


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case of the virial theorem.5) (g) User your result from (f) to find a

simple expression for the energy, E. [answer: E= GMm/2r, fora more general result see Pr. 17.4.2.]

Problem 17.4.2. Elliptical Binding

This problem is on the practice exam. One of its results is that the

energy of a an object of mass m in an elliptical orbit about an objectof mass M m is E = GMm/2a, where a is the semi-majoraxis.

Example 17.4.1. Binding EnergyConsider a satellite of mass m which is in a circular orbit of radius r1about Earth (which has a mass M m). As the systems on boardthe satellite start to fail due to its advanced age it must be ejected so

as not to contribute to the growing sea of space junk about Earth.

(a) What is the initial energy, E1, of the satellite? (b) What isthe speed, v1, of the satellite? (c) A powerful rocket engine is fired

for a short time, increasing the speed of the satellite by 10%. Theinitial and final velocity vectors are parallel. What is the satellites

new energy, E2? (d) The orbit is now elliptical and the satellite is atperigee. What is the length of the semi-major axis, a? (e) What isthe maximum distance of the satellite from the centre of the Earth,

r3? (f) If, instead, the rocket increased the speed of the satellite bya factor of

2 what would be the satellites energy, E4? (g) What

would be the maximum distance of the satellite for this case?

Solution

(a) From Prob. 17.4.1 we know that the energy of an object in a

circular orbit is

E1=

GMm

2r1 .

5The virial theorem tells us that if the potential is proportional to rn then the

average kinetic energy is equal to n2

times the average potential energy. For a

circular orbit the kinetic and potential energies are constant and so this is an exact

relationship. For the case of a circular orbit this result can also be found as a result

of Newtons Second Law.


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(b) We recall that for a circular orbit the kinetic energy is

K1 = 12U1 12mv21 =GMm

2r1

so its speed is v1 =

GM/r1

(c) The new kinetic energy is K2 =1

2m(v1 + v)

2 = 12m(11

10v)2

and the potential energy is unchanged, so

E2 =121GMm

200r1 GMm

r1= 79GMm

200r1.

(d) We know from Pr. 17.4.2 that the energy of an object in an el-

liptical orbit is E= GMm/2a so

E2 =79GMm

200r1= GMm/2a a = 100

79r1.

Since 2a = ra + rp and the radius of the initial orbit is now theclosest distance, rp = r1, the maximum distance is

r3 = 2a r1 = 20079 r1 r1 = 12179 r1,

an increase of more than 50%.

(e) The new kinetic energy would be K4 =1

2m(

2v1)2 = 2K1 =

U1 and the total energy would beE4 = K4 + U1 = 0.

(f) The maximum distance would be infinite. The satellite would

have reached its escape velocity. Doubling the kinetic energy of

an object in a circular orbit added enough energy to make the

total energy zero, an amount called the binding energy.

Example 17.4.2. Escape Speed

Consider an object of mass m that is fired vertically upward withspeed v from the surface of a spherical, airless, non-rotating planetof mass M m and radius R. (a) What is the initial energyof the object, Ei? (b) What is the speed of the object, vh, when it


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is a height h above the surface of the planet? (c) What is the maxi-

mum height, hmax, of the object? (d) The escape speed is the initial

vertical velocity required for an object at the surface of a planet to

just not return to the planet. What is the escape speed, vesc, of this

planet? (e) What is the energy, Eesc, of the object if it initially hasexactly the escape speed?

Solution

(a)

E= K+ U =1

2mv

2

GMm

R(b) The energy at height h is

Eh =1

2mv

2

h

GMm

R + h.

Energy is conserved so we can equate this to the result from (a)

and rearrange to get

v2

h= v

2+ 2GM

1

R + h

1

R

.

(c) The maximum height occurs when vh = 0. Using this in our

answer to (c) and solving for h gives

h =2GMR

2GMv2

iR

.

(d) The initial velocity for the escape speed occurs when the de-

nominator of the answer to (d) is zero:

vesc =

2GM

R.

(e) Substituting the answer to (f) into that of (a) gives

Eesc =1

2m

2GM

R

GMm

R= 0.

This is exactly what we would expect: The system for this case

is just barely not bound so the total energy is zero.


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Figure 17.5: Einstein rings are formed when the rays of light from an

object are bent by an intense gravitational field like, for example, that

of a galaxy which acts as a gravitational lens. In these images from

the Hubble Space Telescope

17.5 General Relativity

There are some observations that are in slight disagreement with

Newtons Universal Gravitation. One of them is the precession of

the perihelion of the orbits of the planet Mercury, for which there is a

discrepancy of 43 arcseconds per century. A second is the deflectionof light by gravity (which can be accurately observed during a solar

eclipse) which is twice that predicted by Newtons theory.

These discrepancies are accounted for by Einsteins theory of

General Relativity. The instantaneous action at a distance of New-

tons theory is replace by mass curving space and objects travelling

on geodesic curves.6

In General Relativity Newtons Universal Gravitation is replacedby the Einstein Field Equations which in tensor form are

R 1

2gR+ g =

8G

c4T,

where R is the curvature tensor, g is the metric tensor, R is the

scalar curvature is the cosmological constant, T

is the stress-

energy tensor, c is the speed of light, and G is the same gravitational

constant as in Newtons Universal Gravitation. This is a concise

6A geodesic is the shortest path between two points. In a flat space it is a

straight line but in a curved space they are no longer straight. As a specific ex-

ample, on a sphere a great circle is a geodesic. In General Relativity an object

moving on an inertial path traces a geodesic.


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17.5. GENERAL RELATIVITY 81

Figure 17.6: In this image from the Hubble Space Telescope a parti-

cle jet is emerging from a black hole at the centre of the M87 galaxy,

fifty million light years away from our galaxy. The black hole has a

mass billions of times that of our Sun.

form of ten non-linear, partial differential equations. If the gravita-

tional field is weak then these equations reduce to Newtons Univer-

sal Gravitation.

Solving the Einstein Field Equations is quite difficult and is bestleft for a more advanced course.

17.5.1 Black Holes

A black hole is an object that is so massive that the escape speed

is greater than the speed of light. The radius from such an object

for which the escape speed is exactly the speed of light is called

the Schwarzschild Radius. The spherical surface with this radius is

called the event horizon as events beyond this surface are not ob-

servable.

The presence of a black hole can be inferred by its interaction

with objects nearby. Most dramatically, the effect called gravita-

tional lensing in which a multiple images of, for example, a galaxy

behind a large compact object are formed in what are known as Ein-

stein rings (see Fig. 17.5). Astronomers are confident that there is a

black hole at the centre of our galaxy with a mass millions of times

the mass of our Sun.

Problem 17.5.1. Dark Events


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The event horizon of a black hole is the geometric surface where the

escape velocity is equal to the speed of light, c. (a) If a black holehas a mass N times that of the Sun, M, what will be the radiusof the event horizon, reh? (b) If an astronaut of height h is orbiting

this black hole with an orbit of radius r and her feet toward thisblack hole what will be the difference,g, between the gravitationalfields felt by her head and her feet? (Assume that her centre of mass

is at her geometric centre.) (c) If this is more than a few times the

gravitational field on the surface of the Earth, gE, it will be quiteuncomfortable. At approximately what distance from the black hole,

rp, will g = gE ifN = 10 and h = 2 m? (M = 2 1030 kg and

the binomial expansion, (1 x)n = 1 + nx + 12n(n + 1)x2 +

1

6n(n + 1)(n + 2)x3 + ..., |x| < 1, may be useful) (d) What is the

ratio rp/reh?

17.6 Further Problems

Problem 17.6.1. Astronomical Unit

The Earths orbit is roughly circular (e = 0.017) and has a period ofone year. The centre to centre average Earth-Sun distance is called

an Astronomical Unit (AU). (a) Determine the length of an AU in

SI units given that the mass of the Sun is M = 21030 kg. (b) De-

termine the radius of Mars orbit, in AU, given that it is roughly

circular and that its period is about two years.

Problem 17.6.2. Halleys Comet

Halleys comet has a perihelion of 0.6 AU and a period of 76 years.

(a) What is its aphelion, rA? (b) What is the eccentricity of thisorbit? (c) What is the speed of the comet at perihelion? (d) What

is the speed of the comet at aphelion relative to that at perihelion,

rA/rP?

Problem 17.6.3. Pulsar

A binary pulsar consists of two neutron stars orbiting their mutual

centre of mass. If each star has a mass ofM and a period T deter-mine the separation, d, between the two stars (a) using dimensionalanalysis and (b) analytically.


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17.6. FURTHER PROBLEMS 83

RE rB

vf

vi

Figure 17.7: Lunar Baseball

Problem 17.6.4. Geosynchronous Satellite

A geosynchronous satellite is a satellite whose orbital period is the

same as that of the earth, T = 24 hours. If the mass of the earth is

MEarth = 5.98 1024 kg determine the speed at which the satellite

travels. (Answer in m/s.) Check the dimensionality of your result.

Problem 17.6.5. Lunar Baseball

A baseball pitcher located on the surface of the Earth (radius RE)

pitches a baseball horizontally to a batter in circular orbit aroundthe Earth, at a radial distance rB from the centre of the Earth. See

diagram; ignore air resistance and the rotation of the Earth. Thrown

with an initial speed vi the baseball enters an elliptical orbit, and is

moving with a speed vf when it reaches the batter. Notice that the

batters orbital speed in the clockwise direction is greater than vf,

so from his point of view the baseball is coming straight at him, as

in the usual game of baseball! (a) Determine the required pitch

speed, vi, and express it as a fraction of vesc, the escape speed of

any object from the surface of the Earth. (b) Determine vf using

conservation of angular momentum, and express your answer as a

fraction ofvi. (c) Considering the period of the elliptical orbit, how

long, t, will it take for the ball to reach the batter? (d) What is the

binding energy, Eb, of the ball while on its elliptical trajectory from

pitcher to batter? (e) Notice that if the batter simply bunts the ball

when it crosses home plate, i.e. bats the ball so that it comes to

rest relative to him, he has effectively hit a home run: the ball is now

in a circular orbit about the Earth at a radial distance rB! Determine

E, the resulting change in the balls total energy.


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Problem 17.6.6. Light Headed

Consider an object of mass m on the surface of a spherical, airless,non-rotating planet of mass M m and radius R. (a) Whatis the free-fall acceleration due to gravity, g(R)? (b) If the ob-

ject is raised to a height h above the surface of the planet whatwould be the change in the free fall acceleration due to gravity,

g = g(R + h) g(R)? (c) Perform a binomial expansion onyour answer to (b) assuming that R h, and so find an expressionfor the fractional change in the free-fall acceleration due to gravity,

g/g, as a function of height above the surface, h. (d) What is g(r),the free-fall acceleration due to gravity at a distance r from the cen-tre of the planet? Assume r > R. (e) What is the natural logarithmof your result from (d), ln (g(r))? Write your answer as y = mx + bwhere y = ln g and x = ln r. (f) What is d

drln (g(r))? (g) For small

values ofr we can write dr r and dg g. Use this andyour result from (e) to find an expression for the fractional change

in the free-fall acceleration due to gravity, g/g, due to a change inr. Compare your answer to that of part (c).

Problem 17.6.7. Gravitational Tunnelling

It is straightforward to show, using integral calculus, that the gravi-

tational force on an object of mass m at a depth x inside a uniform,spherical object of density and radius R is equal the gravitationalforce on the same object caused by a uniform sphere of the same

density and centre with a radius equal to the distance to the centre,

R x (i.e. only the part of the sphere below the object has a netgravitational effect upon the object). (a) What is the mass of a

sphere, M(x), of density and radius Rx? (Hint: Check that youget the expected result for x = 0.) (b) What is the magnitude of thenet force, F(x), the object as a function of depth? Choose downas the positive direction. (Hint: Check that you get the expected

results for x = R and x = 0.) (c) What is the acceleration, x, ofthe object? (Hint: Check that you force is positive in the down

direction.) (d) If there was a tunnel through the centre of a uniform,

non-rotating, airless planet of mass M and radius R and an object ofmass m was dropped from the end of the tunnel, what is the objectsposition as a function of time, x(t)? Take t = 0 at the moment it isdropped (i.e. x(0) = x(0) = 0). (Hint: The substitution y = R x


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may be helpful in finding x(t). Alternatively, plotting what you ex-pect x(t) to look like may be helpful dont forget that down is thepositive direction.) (e) In order to deliver mail to the alien beings on

the opposite end of this tunnel the mail is dropped at one end. How

long, t, does it take to arrive at the opposite end?


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Chapter 19

Elastic Properties of Solids

19.1 Introduction

Stress is the force per unit area applied to an object and is an intrinsic

quantity. There are three different types of stress (see Fig. 19.1),

tensile, shear and bulk. When a stress is applied to an object theresulting fractional deformation is called a strain; strain is also an

intrinsic property.

For small stresses we expect, in the spirit of Hookes Law, the

strain to be proportional to the stress. The constant of proportionality

is called the elastic modulus.

elastic modulus =stress

strain(19.1)

and, as the ratio of two intrinsic quantities, is also an intrinsic quan-

tity. As such, it is a property of the material the object is made from

rather than of the specific object. A short table of elastic moduli is

included on p. 99.

Stress can be applied to an object in different ways: tensile stress,

shear stress or bulk stress. If a material is not isotropic then the

elastic moduli will be tensors rather than a scalars the stress vector

and the resulting strain vector will not necessarily be parallel. Also,

the stresses can be applied in some combination at the same time

resulting in a more complicated deformation. For this chapter we

will assume that the materials are isotropic and that the stresses are

applied one at a time.

93


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94 CHAPTER 19. ELASTIC PROPERTIES OF SOLIDS

TensileStress

ShearStress Stress

Bulk

Figure 19.1: There are three different ways to apply stress to an ob-ject. For tensile stress the applied forces are equal, opposite and co-

linear. For shear stress the applied forces are equal, opposite and not

co-linear. For bulk stress the applied forces are everywhere perpendic-

ular to the faces of the object.

19.2 Tensile Stress

If the forces are parallel and co-linear then the elastic modulus is

know as Youngs Modulus, Y, the stress is the force per unit area,in which the area, A. is perpendicular to the applied forces, and theresulting strain is is the fractional change in length, /, parallel tothe forces

Y =

F/A

/ . (19.2)

We would expect that when a material is stretched in one direc-

tion it would contract in the other two dimensions. The ratio of the

longitudinal to transverse strain is known as Poisons Ratio. A per-

fectly incompressible material would have no change in volume and

a Poissons Ratio of 0.5; in practice when a material is deformed the

molecular bonds change in length and the volume of the materialdoes change by a generally tiny amount.1 For this chapter we will

assume that the strains are small and any change in cross-sectional

areas can be neglected.

1Interestingly, there are materials with a negative Poissons Ratio materials

that expand in the transverse direction when they are stretched.


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19.2. TENSILE STRESS 95

stress

strain

fracture

F

F

neck

(a) (b)

Figure 19.2: As the stress on an object is increased the strain is

initially proportional to the stress (blue portion of curve on left) but

shortly afterwards the material becomes permanently deformed (green

region). For a brittle material the stress reaches a maximum as the

material fractures. For a ductile material the the stress reaches a maxi-

mum and necking occurs (red region on left, cartoon on right) before

the material fractures.

19.2.1 Tensile Strength

As the stress applied to an object is increased the corresponding

strain will initially be linear (blue region of curve in Fig. 19.2.a) but

then, as the stress is further increased, becomes non-linear (green re-

gion of curve). Shortly afterward entering this non-linear region thematerial becomes permanently deformed (i.e. it will not return to its

original shape if the stress is removed); the stress that causes perma-

nent deformation is called the yield strength of the material. What

happens next depends upon the type of material the object is made

from. If it is brittle then its ultimate strength, the stress at which the

material fractures, occurs shortly afterwards. If, instead, the material

is ductile then the stress will reach a maximum, called the ultimatestrength, and then decrease as the material begins to neck (i.e. the

cross-sectional area at one region of the material starts to decrease,

see Fig. 19.2.b) before ultimately fracturing.

Many materials have very differ stress-strain curves for compres-

sion. Materials such as concrete or stone are much stronger in com-


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h

x

F

F

A

Figure 19.3: The geometry for a shear force is more complicated than

that of either a tensile of bulk force. The displacement,x, is parallelto the applied forces, F, while the length h is the shortest distancebetween the parallel forces. The area A is perpendicular to the lengthh.

pression than in tension. And, of course, you cannot push a string.2

Problem 19.2.1. Youngs ModulusConsider a cable made of a material with Youngs modulus Y andlength . (a) If it is to stretch no more than 0.1% when a mass

m1 is suspended from its end what must be its radius, r1? (b) Ifit is to stretch by 5% when a mass m2 is suspended from its endwhat must be its radius, r2? (c) For the case in (a), what is thecorresponding spring constant, k1, and what would be the angular

frequency, 1, of any resulting oscillation? (d) For the case in (b),what is the corresponding spring constant, k2, and what would bethe angular frequency, 2, of any resulting oscillation?

19.3 Shear Stress

If the applied forces are parallel but not co-linear then the materialwill undergo a shear stress and the shear modulus will be the relevant

elastic modulus,

G =F/A

x/h. (19.3)

2This is the first law of String Theory.


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19.4. BULK STRESS 97

Here, h is the distance between the applied shear forces and x isthe displacement of the material in the direction of the forces. Note

that this strain, x/h, is intrinsic.3

19.3.1 Shear Strength

Materials can also fail due to excess shear stress. In general, ductile

materials have lower shear strengths than brittle materials some-

thing known to anyone that has tried to cut cookies with scissors.4

Problem 19.3.1. Hole Punch

Consider a sheet of metal of thickness d. The maximum shear stressof the metal is (i.e. if a larger shear stress is applied the metal will

shear). (a) Determine the force, F1, required to punch a hole ofradius r through the metal sheet. (b) What force, F3, is required topunch a hole with twice the radius, 2r, through the sheet of metal?(c) What force, F2, is required to punch the original size hole, radiusr, through a sheet of metal twice as thick? (d) Are your answers

consistent with the maximum shear stress of the metal, , being anintrinsic property?

19.4 Bulk Stress

If the force is applied uniformly to the entire surface of an object,

then the bulk modulus will be the relevant elastic modulus (although,really, this is just the three dimensional version of tensile stress).

This occurs when a solid object is submerged in a fluid or when a

fluid is compressed by, for example, a piston.5 The resulting stress is

the change in pressure, P, caused by being submerged in the fluid.

3It may be helpful to think of holding a book horizontally between your palms

and pushing one hand to the left and the other to the right. The resulting shear onthe book will slide the pages past one-another, each page by a small amount.4Proper cookies that is, not the horrible uncooked pieces of dough that often

pass for cookies now-a-days. Uncooked cookies are both oxymoronic and dis-

gusting.5Interestingly, for an ideal gas at constant temperature the bulk modulus is

equal to the pressure a property easily derived from Boyles Law, PV =constant, and the definition of the bulk modulus, Eq. 19.4.


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And the resulting strain is the fractional change in the volume of the

object, V/V.

B = PV /V

(19.4)

The bulk modulus is the reciprocal of the compressibility of the ob-

ject. The minus sign is present to keep the bulk modulus positive

for positive P, V will (usually) be negative.

Problem 19.4.1. Cold Pop

When water freezes it expands by about 10%. What is the pres-

sure increase, P, when a pop can freezes. Assume that the cans

dimensions remain unchanged. The bulk modulus of ice is 2 GPa.Comment on the assumptions made in this question.

19.4.1 Bulk Strength

As the bulk pressure on a gas is increased it will (usually) con-

dense into a liquid. If the pressure is increased further the liquid

will (usually) increase further in density and become a solid. Of-

ten this process can be continued, with the solid taking on more and

more compact crystalline structures. These types of transitions are

called phase transitions and occur suddenly as the pressure reaches

a critical value. A phase diagram usually shows the different phases

of a material as a function of temperature and pressure.

As some types of stars run out of fuel to feed the fusion reactionthat powers them they explode as supernovae. After this spectacu-

lar explosion the remnant is often a neutron star. The density of a

neutron star is similar to that of an atomic nucleus, about 1014 timesdenser than the precursor star. This phase transition is more dramatic

than most other types.

19.5 Sound Waves

When a sound wave travels through a fluid the wave is a longitudinal

wave and it is the bulk modulus that provides the restoring force that

all oscillations must have. The square of the wave speed is equal to

the bulk modulus divided by the density of the fluid. When sound


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19.5. SOUND WAVES 99

Y G Bethanol - - 1

water - - 2.2

Mercury - - 27

glass 70 26 40aluminium 69 25 76

diamond 1220 480 440

Table 19.1: Young (Y), shear (G) and bulk (B) moduli for selectedmaterials. All quantities in units of giga-Pascals, GPa. The values

for glass are approximate because there are several different kinds ofglass.

travels through a solid material the wave can be either a longitudinal

or transverse wave.6 The longitudinal wave will obey the same for-

mula as for fluids but the transverse wave will have the shear modu-lus in the formula for its speed. Since the two moduli have different

values the two types of sound waves in solids will travel at different

speeds. In Table 19.1 some elastic moduli are listed and as we can

see, the bulk and shear moduli are often quite different.

The different arrival time of these two waves are used to help

locate the epicentres of earthquakes the different arrival times ofthe two waves at a seismic monitors will give the distance to the

epicentre (there are also types of surface waves that can provide ad-

ditional information). This type of data was used to deduce the inner

structure of the Earth, including that it has a molten core. Small un-

derground explosions (aka seismic testing) can provide information

about the presence of oil or gas deposits since transverse wave do

not travel through liquids.

6The longitudinal and transverse are known as P (for primary) and S (for shear)

waves, respectively. This nomenclature has nothing to do with the orbitals of the

same names in chemistry.


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Problem 19.6.1. Two Wires

Two wires of circular cross section and made from the same ma-

terial are subjected to the same tensile force. The second wire isthree times as long and its diameter is twice as large as the first

wire. (a) Determine the ratio of their changes in length,2/1.(b) What would the ratio be if the two wires had square cross sec-

tions? (Interpret the diameter of the square as its diagonal.)

Problem 19.6.2. Atwoods Machine

The wire connecting the two masses in an Atwoods machine has aradius r, a length and a Youngs modulus Y. (a) If the twomasses are m1 and m2 and the pulley is light, by how much is thewire stretched while the masses are in motion? (b) If the pulley had

a radius R and mass m3, by how much is the wire stretched whilethe masses are in motion?

Problem 19.6.3. Young Hooke

A wire of length is made out of a material with a Youngs modulusY and cross sectional area A is stretched by an amount x by anapplied force F at one end (the other end is attached to a wall).(a) What is the force, F, in terms of the other quantities? (b) What isthe work done, W, in stretching the wire? (c) Hookes law is givenby F = kx. Derive an expression for Hookes constant, k, in termsof the other constants.

Problem 19.6.4. New Springs

Consider a rectangular prism with length h and a square cross-sectionof side a which is made from a material with a Youngs modulus Y.(a) As a tensile force, F, is applied to the material, what is its in-crease in length, h? (b) If the volume of the prism is to remainconstant, what is the corresponding decrease in its cross-sectional

area, A? (c) If the two sides of length a decrease by the sameamount, what is the change, a, in the side? (d) What is the ratio, ,of the strain on the length, h/h, to that of the side, a/a? This ratiois called Poissons ratio. (e)

Problem 19.6.5. Newer Springs

Consider a rectangular prism with height h and a square cross-section


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of side a which is made from a light material with a Youngs mod-ulus Y. (a) If one end of the prism is attached to the ceilingand a mass, m, is attached to the other end the mass will undergooscillations. What is the corresponding spring constant, k1? (b) If a

second, identical, prism was attached in parallel to the first and themass m hung from both as in (a), what will be the correspondingspring constant, k2? (c) What is the relationship between k1 and k2?(d) If, instead, the second prism is attached end-to-end with the first

and the mass m is hung from the bottom as in (a), what will be thecorresponding spring constant, k3? (e) What is the relationship be-tween k1 and k3? Can you rearrange the result into a tidy equation?

Problem 19.6.6. New Scissors

Consider a rectangular prism with height h and a square cross-sectionof side a which is made from a light material with a shear modulusG. (a) If one end of the prism is attached to a wall and mass m, isattached to other end, what is the equilibrium lateral displacement,

x, of the end. If the end with the mass undergoes oscillations, what

is the corresponding spring constant, k1? (b) If a second, identical,prism was attached in parallel to the first and the mass m attachedto both as in (a), what will be the corresponding spring constant, k2?(c) What is the relationship between k1 and k2? Can you rearrangethe result into a tidy equation? (d) If, instead, the second prism is

attached end-to-end with the first and the mass m is attached to theend as in (a), what will be the corresponding spring constant, k3?

(e) What is the relationship between k1 and k3?

Problem 19.6.7. Skateboarder

A skateboarder of mass m moves through a turn of radius r at speedv. The wheels of the skateboard have radius a and width w, and aremade of a material with shear modulus S. The bottom half of thewheels below the axle will shear as indicated in Fig. 19.4 (a) Draw

a free body diagram of a wheel, showing the horizontal forces only.

What is the magnitude of the static friction force on each wheel?

(This force may vary from wheel to wheel; consider an average over

the four wheels.) (b) Estimate the shear angle, , by using a simpli-

fied model in which the wheels are considered to be square (of side

length 2a) instead of round. (c) What is the limit of your expression


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Figure 19.4: Front view of a skateboard wheel as the boarder executes

a tight turn. The lateral forces from the axel and the road create a shear

in the lower part of the wheel.

for as the width, w, approaches zero? Give a very brief explana-tion of why this is a reasonable answer. (d) What is the limit of your

expression for as the shear modulus, S, approaches infinity? Givea very brief explanation of why this is a reasonable answer.

Problem 19.6.8. Bulky with Gas

Show that the bulk modulus for a gas obeying Boyles Law, PV =constant, is equal to the pressure.

Problem 19.6.9. Broken Spring

Consider an ideal spring of length and spring constant k. It iscut into two parts, one with length f and the other with length (1 f). (a) What is the spring constant, k1, of the spring of length f?(b) What is the spring constant, k2, of the spring of length (1

f)?

(c) Show that you get the expected results for your answers to (a)

and (b) in the f 0 limit. (d) Show that you get the expectedresults for your answers to (a) and (b) in the f 1 limit. (e) Forsprings connected in series the effective spring constant is given by

the expression

i k1i

1. Show that you get the expected result

when the two parts are connected in series for any value off.

19.7 Further Interesting Problems

Problem 19.7.1. Young and Bulky

Consider a solid object made from a material with Youngs modulus

Y, bulk modulus B, and Poissons ratio, , with dimensions x by y


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19.7. FURTHER INTERESTING PROBLEMS 103

by z. (a) What is the volume, V, of the object? (b) If the object isstretched along its x dimension by an amountx, what is the changein volume, V, of the object? Use Poissons ratio to eliminate thechanges in the y and zdimensions. (Poissons ratio is the ratio of the

longitudinal to transverse strain.) (c) What is the fractional changein the objects volume, V/V? (d) Expand your result from (c) andneglect higher powers ofx/x since this is a small quantity. (e) Nowconsider the same object undergoing hydrostatic pressure, P. Use

your expression for the fractional change in volume from (d) to find

a relationship between the bulk modulus, B, Youngs modulus, Y,and Poissons ratio, . [answer: Y = 3B(1

2)]


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Chapter 20

Fluid Mechanics

20.1 Introduction

In this chapter we will examine static fluids as well as fluids mov-ing at a constant flow rate. For static fluids we will examine how

pressure changes with depth as well as the effects of buoyant forces.

We next study motion through a fluid and so will need to understand

how the viscosity of a fluid affects such motion. We will find that

there are two types of motion, laminar and turbulent, and the distinc-

tion between the two is characterized by Reynolds Number. Next,

we will make an approximation that the fluids are ideal fluids and

study their motion.1

Because we will be treating fluids as continuous systems it is

convenient to deal with intensive properties such as pressure or den-

sity. As we shall see, when we consider the energy of a fluid in

motion, it is convenient to use the energy density, the energy per

unit volume, and this results in the Bernoulli Equation, Eq. 20.6.

1You may wonder why Fluid Dynamics is in the Statics section of this book.

We will only be examining fluids flowing at constant rates static motion if you

will.

105


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106 CHAPTER 20. FLUID MECHANICS

20.2 Fluid Statics

It is difficult for a fluid to sustain tensile or shear stresses it is, how-

ever, able to apply a compressive force perpendicular to a surface.

Pressure, P, is a measure of this compressive force2

P =F

A,

and is the force per unit area and is a scalar quantity.3 It has SI

units pascal (1 Pa = 1 N/m2) and is always positive (i.e. there is nosuch thing as negative pressure a pressure of zero is known as a

vacuum). Often, however, it is the pressure difference, P, that is

important and, of course, this quantity can be negative.

If the force is not constant over a surface then we have, instead,

P =dF

dA. (20.1)

Other units of pressure are quite common. One atmosphere isthe air pressure at sea level and is defined as

Patm = 1 atm = 101 325 Pa

although the pressure varies on a day-to-day basis with the weather.

Because this is close to 105 Pa another common unit is the bar(from

baros, the Greek for weight) which is just this quantity.Yet more common units of pressure are pound per square inch

(psi), mm of mercury (mmHg), and torr4 (1/760 of an atmosphere or1 mmHg).

Patm = 14.6959 psi = 29.72 mmHg.

2I will use the symbol P for pressure so that it is not confused with power, for

which I use the symbol P. This is particularly important when we talk about thepower transmitted by a pressure wave in Ch. 23.

3We can understand this equation as a vector equation if we write it as F = P Aand interpret the vector area as having magnitude A and a direction perpendicularto the surface. Writing an area as a vector in this manner is actually quite common.

It is particularly convenient when dealing with infinitesimal areas, d A.4The torr is named after the inventor of the barometer, Evangelista Torricelli.

A barometer is a device used for measuring pressure and is discussed in Ex. 20.2.3.


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20.2. FLUID STATICS 107

F1

PatmA PatmAwAxg

F1(b)(a)

Figure 20.1: The free body diagram for the top face of the cube in

air is shown on the left. On the right is the same face when the cube

is under water. For the latter case the weight of both the air and water

above the top face are acting on that face.

Example 20.2.1. Under Pressure

Anyone who has dived more than a meter or two under water hasdirectly experienced the rapid change in pressure with depth. Simi-

larly, anyone who has visited an high altitude region like the top of a

mountain has directly experienced the less rapid decrease in pressure

with altitude. (For this problem use w = 1000 kg/m3 and air = 1

kg/m3 as the densities of water and air, respectively, the atmospheric

pressure at sea level, Patm = 1 bar, and the free fall acceleration at

the Earths surface, g, as 10 m/s2.)Consider a cube, each side of which has an area, A, of one

square metre, which is sitting on the horizontal ground at sea level.

(a) Draw the free body diagram of the top surface of the cube. In-

clude the weight of the atmosphere directly above the cube. Label

the balancing, internal force F1. (b) Using Newtons Second Lawcalculate the mass, m1, of air directly above the cube. Use yourknowledge that the atmospheric pressure at sea level is about one

bar. (c) Considering that a typical human has a cross sectional area

of about 1/10 m2, what is the mass resting upon each of us? Howdo you manage to carry this around? (d) Now consider the same

cube at a depth of x of water. Again draw the free body diagramfor the top surface of the cube. Label the balancing, internal force


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F2. (e) What is the pressure, P(x), on the top surface of the cube?(f) Sketch P(x); include x < 0 on your plot (i.e. for above the wateras well as below). (g) If the air had a constant density, for what value

ofx would vacuum occur?

Solution

(a) See Fig. 20.1.a.

(b) For the cube we have Ma = 0 = F1 m1g where M is the massof the top face of the cube and m1 is the mass of air directly

above the cube. Since the pressure at the top of the cube is theatmospheric pressure multiplied by the area of the top of the

cube we have

m1g = PatmA m1 = PatmAg

=(105)(1)

(10)= 104 kg.

(c) Since a human has one tenth the cross sectional area she sup-

ports one tenth this mass, or 1000 kg of air. It is not the mag-

nitude of the forces acting upon a body but the net force that

is important. In this case, the internal pressure of a human is

slightly larger than atmospheric pressure (this is why they leak

when punctured) and because the pressure acts upon all surfaces

of the human the net force is almost zero. (As we shall see, the

net force is the buoyant force which will be examined in Ex.20.2.4.)

(d) See Fig. 20.1.b.

(e) If the mass of water directly above the cube is m2 then the forceon the top of the cube is

F2 = (m1 + m2)g =PatmA

g+ Ax

g = PatmA + wAxg

where the volume of water above the cube is V = Ax = 1 m3.In general, the pressure at a depth x is then

P(x) = Patm + gx (20.2)


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air

water

P

x

Figure 20.2: A sketch of the pressure, P, as a function of depth, x;x = 0 is the surface of the water. The rate of change of pressurewith depth (i.e. the slope) changes dramatically at the surface of the

water (x > 0) because its density is 1000 times greater than that of air(x < 0).

and for this particular case we get

P(x) = 104(10 + x) Pa

where x is in metres.

(f) See Fig. 20.2.

(g) If the air had a uniform density and an height x then we wouldhave

Patm = agx x = Patmag

=105

(1)(10)= 10 km.

The edge of space (i.e. vacuum) is generally taken as 100 km

and the density does indeed change as a function of height.

If Patm gx then we can often neglect the second term inEq. 20.2. An example of such a situation is changes in atmospheric

pressure near the Earths surface. The density of air is approximately

1.2 kg/m3 so a 10 m change of height would change the pressure by

P

Patm=

gx

Patm=

1.2 9.8 10101325

= 0.116%


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m1

m2

A1 A2

Figure 20.3: A hydraulic lift consists of two pistons in two connected

cylinders. The cylinders are filled with an incompressible fluid of den-

sity and the two pistons have different cross sectional areas, A1 andA2.

which is negligible considering that the daily variation in air pressureis a few percent anyway.

20.2.1 Pascals Law

A change in the pressure applied to a fluid is trans-

mitted to every point of the fluid and its container.

As a consequence, we have P = gh, the difference in pres-sure at various places in a fluid depends upon the vertical distance

between the points, h, and not the shape of the container holdingthe fluid. Here is the density of the fluid and g is the free fallacceleration due to gravity.

Example 20.2.2. Hydraulic Lift

A hydraulic lift consists of two cylinders of cross section areas A1and A2 connected by a pipe (see Fig. 20.3). In each cylinder is a lightpiston and the system is filled with an incompressible fluid (such as

oil). (a) If a mass m1 is placed on the first piston what mass, m2,must be placed on the second piston to maintain equilibrium? (As-

sume the pistons have the same thickness and are at equal heights.)


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(b) The pistons are now at heights h1 and h2 measured from the floor.If the density of the fluid is what mass, m2, must be placed on thesecond piston to maintain equilibrium? (c) What is the ratio m2/m2?(d) Is it meaningful to include the change in atmospheric pressure at

the two pistons in your solution to part (b)? (Take = 2000 kg/m3,m2 = 1000 kg, A2 = 10

2 m2, make reasonable assumptions forthe remaining quantities and calculate the percent change in your

answer.)

Solution

(a) The pressure at the points in the fluid just under the pistons are,by Pascals Law, equal. The free body diagram for each of the

pistons then tells us that 0 = P1A1 m1g and 0 = P2A2 m2g. Solving each equation for the pressure, Pi and equatingthe results gives

m1g

A1=

m2g

A2 m2 =

A2

A1m1

(b) The difference in the two pressures is P = gh so

g(h2 h1) = P2 P1 = m2g

A2 m1g

A1

m2 = A2(h2 h1) +A2A1 m

1.

(c)m2m2

= 1 +A1(h2 h1)

m1

Example 20.2.3. Barometer

A barometer is a device to measure pressure. One common ex-

ample (see Fig. 20.4) is a glass tube with one end sealed which is

then filled with a liquid and inverted, creating a vacuum at the top

(where the pressure is zero). (a) If the distance between the

top of the fluid and the surface of the fluid exposed to the air is h asshown in Fig. 20.4, what is the atmospheric pressure, Patm? (b) If the


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hPbottom = Patm

Ptop = 0

Figure 20.4: A liquid barometer is a tube sealed at one end and filled

with a liquid. The tube is then inverted, creating a vacuum at the sealed

end. An equation similar to Eq. 20.2 can then be used to determine,

for example, the atmospheric pressure.

fluid was mercury, Hg, which has a density of 13.5951 g/cm3, then

what height, h, will be observed if standard atmospheric pressure ispresent?

Solution

(a) Using Pascals Law we get

P = Ptop Pbottom = gh

Patm = gh

where we have taken Ptop = 0 and Pbottom = Patm.

(b) Rearranging our answer to (a) we get

h =Patm

g=

(101, 325)

(13.5951 103)(9.81) = 760 mm.

This is exactly what we expect since one of the other units forpressure is mmHg and in these units one standard atmosphere is

760 mmHg. The origin of this unit is, of course, mercury filled

barometers.

Often pressure is measured with a pressure gauge which mea-

sures the gauge pressure. This quantity is the difference between


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F2

F1

F5

F3

mg

Patm

b

a + b

Patm + gb

x

(a) (b)

P

Patm + g(a + b)

Figure 20.5: On the left is a plot ofP

(x) for a cube of side a andmass m which is suspended by a string and submerged in a fluid ofdensity , as shown on the right. Note that the diagram on the right

is an incomplete free body diagram since three of the four horizontal

forces are not shown.

two pressures and is what we commonly use to measure, for exam-

ple, the pressure in our bicycle tires. For a pressure gauge used formeasuring the pressure in tires the reading it gives is the difference

in the pressures inside and outside the tire; this is why a flat tire has

zero pressure.

Example 20.2.4. Cubically Buoyant

A uniform, cubic block of side a and mass m is suspended from alight, inextensible string attached to the centre of one face and is

submerged in a fluid of density . The top surface of the cube is at

a depth b. (a) Sketch the pressure on one of the vertical facesof the cube as a function of depth, x. (b) Calculate the net force,F1, on one of the vertical faces of the cube. You may need to useEq. 20.1 or you can be clever and find a shortcut. (c) Calculate the


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magnitude of the net force on the vertical faces of the cube. (d) Cal-

culate the net force, F2, on the top face of the cube, caused by thefluid. (e) Calculate the net force, F3, on the bottom face of the cube.(f) Calculate the net force, F4, on the cube, caused by the fluid. This

force is called the buoyant force. (g) What is the tension, F5, in thestring? (h) Under what conditions will the tension in the string be

zero?

Solution

(a) The pressure as a function of depth is P(x) = Patm + gx; see

Fig. 20.5.a.

(b) The force on an infinitesimal strip of width a, height dx and atdepth x is

dF = P(x)adx

so the force on one of the vertical faces has magnitude

F1 =

a+b

b

(Patm + gx) adx = aPatmx +

12gx2

a+bb

= Patma2 + 1

2ga3 + ga2b.

Alternatively, we can recognize that the force on the side is the

shaded area on the graph from (a) multiplied by the width of the

cube, a.(c) There are four vertical faces. Opposite faces will have forces

that are equal and opposite so the net force on the vertical faces

is zero.

(d) The top face of the cube experiences a pressure Patm + gb sothe net force on this face is

F2 = Patma

2 + gba2

.

(e) Similarly, the net force on the bottom face is

F3 =

Patma

2 + g(a + b)a2

.


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20.3. FLUID DYNAMICS 115

(f) The forces on the sides of the cube cancel so the net force on the

cube caused by the fluid is

F4 = F2 + F3 = ga3.

This is equal to the weight of the displaced fluid.

This is known as Archimedes principle: The buoyant force is

equal to the weight of displaced fluid,

FB = gV, (20.3)

in which V is the volume of the submerged object (or fractionthereof) and is the density of the fluid.

(g) Applying Newtons Second Law we get (remembering that the

net horizontal force is zero)

0 = F2 + F3 + F5 + mg F5 = (mg ga3)

where we have used g = g. We note that we get the expectedresult in the 0 limit.(h) The tension in the string is zero ifm = a3 which occurs if the

density of the cube, m/a3 is equal to the density of the fluid, .This is called neutral buoyancy.

20.3 Fluid Dynamics

Fluid flow is traditionally divided in to two regimes: turbulent and

laminar. In laminar flow the particles of the fluid follow smooth

paths and in turbulent flow they dont follow such simple paths

the flow contains whirlpools and eddies. The Reynolds Number (see

20.3.2) characterizes this distinction.5 In this chapter we will ex-

amine laminar flow only the NavierStokes equations, continuummechanics and computational fluid dynamics are used in study of

turbulent flow. It is very interesting.

5A flowing fluid has two main forces acting upon it, viscous and inertial. If the

viscous forces are dominant then the flow will be laminar and if the inertial forces

are dominant then the flow will be turbulent.


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bottom plate

top plate

fluid

Figure 20.6: A fluid is trapped between two parallel plates a distance

d apart. The top plate is moving with speed vtop and the bottom plateis stationary. The fluid directly adjacent to each plate moves with the

speed of that plate. If the speed of the fluid changes linearly with

the position between the plates (a reasonable first approximation) then

v(x) = vtopx/d, where x is measured from the bottom to the top plate.Shown schematically are the velocity vectors at different positions.

20.3.1 Viscosity

Viscosity is a measure of the internal resistance to flow of a fluid

this is what causes a drag force as an object moves through a fluid.6

A thin liquid such as water has a low viscosity and a thick fluid

such as oil has a high viscosity. We can subject a fluid to both shearand tensile stress.

A simple model of the effects of viscosity is a fluid trapped be-

tween two parallel plates, one of which is moving with a speed vtoprelative to the other (see Fig. 20.6). The fluid in direct contact with

each plate moves with the speed of that plate and the speed of the

fluid in between varies linearly from zero at the bottom to vtop at the

top. Because the fluid is viscous the top plate needs a constant forceto keep it moving at a constant speed relative to the bottom plate.

6Interestingly, for a superfluid the viscosity is zero. There are many interesting

phenomena resulting from superfluidity, amongst which is that the vortices (i.e.

the whirlpools and eddies) are quantized; the quantum of rotation is called the

roton.


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It is found experimentally that, for a given fluid, this force, F, isproportional to the relative speed of the plates, vtop, and to the area,A, of the plate and inversely proportional to the distance betweenthe plates, d. The constant of proportionality is called the viscosity,

, of the fluid,

F = vtopA

d. (20.4)

Equivalently, we could define the viscosity as the ratio of the shear

stress, F/A, to the velocity gradient, v/d. This definition would bemore useful in a a more complicated geometry (see Ex. 20.3.1). The

SI unit for viscosity is the Pascal-second, Pas.

Example 20.3.1. Viscometer

It is possible to measure viscosity with a viscometer, a device which

consists of two coaxial cylinders with the fluid in question trapped

between them. The outer cylinder is spun at a constant angular ve-

locity, , and the torque, , required to prevent the inner cylinder

from rotating is measured. (a) If the height of fluid between the

cylinders is h, the inner radii of the outer cylinder is r2, and the outerradius of the inner cylinder is r1, what is the shear force, F/A, thatacts on the inner cylinder? (Hint: The torque balances this force and

keeps the inner cylinder stationary.) (b) What is the velocity gradi-

ent, v/r, assuming that it is a constant? (c) In terms of the givenquantities, what is the viscosity, , of the fluid?

Solution

(a) The wetted area of the outer surface of the inner cylinder is A =2r1h and the torque is the force, F, multiplied by the momentarm, r1 (the two are perpendicular), so F = /r1. Therefore theshear force is

F

A=

2r21h.

(b) The speed of the inner surface of the outer cylinder is v = r2and the distance between the cylinders is r2 r1 so the (assumedconstant) velocity gradient is

v

r=

r2r2 r1 .


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(c) The viscosity is the ratio of the shear stress to the velocity gra-

dient

=

2r21h

r2 r1r2

=(r2 r1)2r21r2h

.

A similar problem related to the viscosity of a fluid is that of a

small sphere falling through a viscous fluid. If the motion is slow

enough the Reynolds number will be small and the flow of the fluid

around the sphere will be laminar. While the geometry of this sys-

tem is now a more interesting three-dimensional problem, we would

expect the drag force on the sphere to be proportional to its speed rel-

ative to the fluid and to the viscosity (since the viscosity is the ratioof the shear stress to the velocity gradient). George Stokes was the

first to analyze this problem and the result is now known as Stokes

Law

Fdrag = 6rv

where r is the radius of the sphere and v is its speed relative to thefluid.

This result is only valid for small speeds when the flow around

the sphere is laminar (i.e. a small Reynolds number). If the speed,

v, is too large the flow will be turbulent (i.e. a large Reynolds num-ber) and an approximate expression for the drag force, initially pro-

posed by Isaac Newton, is

Fdrag

= 12

CAv2

where A is the cross-sectional area of the object perpendicular to thevelocity vector, is the density of the fluid, and C is a dimensionlessconstant known as the drag coefficient. For a sphere, C 0.47, andfor a modern car it can be as low as 0.2 but 0.3 is more typical.7

Problem 20.3.1. Terminal Drag

Consider a metal sphere of radius r, drag coefficient C, and densitys falling in a fluid with density and viscosity . (a) Draw the

free body diagram of the sphere. (b) What is the acceleration, v1, of

7For a car the rolling resistance caused by the tires is more important at low

speeds. This resistance behaves more like a conventional friction (i.e. proportional

to the normal force).


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the sphere? Assume that the flow of the fluid around the sphere is

laminar. (c) Show that your result from (b) has the solution

v1(t) = a1(1

eb1t)

and find the constants a1 and b1. (d) As the sphere accelerates thedrag force increases until the net force is zero. What is the speed of

the sphere, vt1, when this occurs? [answer: 2(s )gr2/9] Thissystem can be used as another type of viscometer (i.e. to measure

the viscosity of a fluid). (e) If, instead, the flow was turbulent what

would be the acceleration, v2? (f) Show that your result from (e) has

the solutionv2(t) = a2 tanh(b2t)

and find the constants a2 and b2. (g) As the sphere accelerates thedrag force increases until the net force is zero. What is the speed

of the sphere, vt2, when this occurs? [answer:

8(s )rg/3C](h) Plot v1(t) and v2(t) together on the same graph. Take all of the

constants as one. Which curve approaches the asymptote faster andwhy? [answer: see Fig. 20.7]

Listing 20.3.1. Terminal Velocity

A short Octave listing to plot a couple of functions. As usual, we

spend twice as much effort making it pretty than doing the actual

work.

%===========================

% Plot the speed of a falling object which is subject to a

%Stokes, v1(t), and Newton, v2(t), drag force given by

% v1 = a1(1 eb1t)% v2 = a2 tanh(b2t)

t = 0 : 0 . 1 : 8 ;

v1 = 1 exp ( t ) ;v2 = tan h ( t ) ;

h = p l o t ( t , v1 , ; S t o k e s ; , t , v2 , +; N ew to n ; , m a r k e

%===========================

%Make it pretty


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0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5

v

t

Figure 20.7: The solution for Pr. 20.3.1.h, plots of v1(t) and v2(t).This plot compares the velocity of a falling sphere in a fluid which pro-

vides a Stokes (red) and Newton (green) drag force. Both curves ap-

proach the terminal velocity (herev = 1

) asymptotically but since the

Newton drag increases more rapidly (it is proportional to the square

of the velocity) it is smaller initially (so the velocity increases more

quickly) and grows more quickly (so it slows down more rapidly) as it

approaches the terminal velocity. The Octave code to create this plot

is in List. 20.3.1


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a x i s ( [ 0 8 0 1 . 1 8 ] ) ;

g r i d ( on ) ;%set( h, linewidth, 3 ) ;

%set( h, color, green ) ;

% use LATEX for the labels

x l a b e l ( $t $ ) ;y l a b e l ( $v$ ) ;

%===========================

%save the plot in xfig format with LATEX fonts (wont work in MatLab)

p r i n t ( d r a g p l o t . f i g ,

t e x t s p e c i a l ) ;

20.3.2 Reynolds Number

Reynolds number, Re, is a dimensionless constant that characterizes

the flow of a fluid, whether it is laminar or turbulent. The type of

flow depends upon the relative importance of the inertial and viscous

properties of the fluid. The viscous forces tend to damp the inertial

motion and, if this damping is sufficient, the flow will be laminar.

Otherwise it will be turbulent.

Reynolds number is given by the expression

Re =vd

where and are the density and viscosity of the fluid, v is thespeed of the fluid, d is the characteristic size of the system (for afluid flowing in a pipe d would be the diameter of the pipe, for anobject moving through a fluid d would be the size of the object, etc).

If Reynolds number is greater than about 2000 the flow will be

turbulent. The about is because the onset of turbulence depends

upon factors such as the smoothness of the walls of the pipe or the

steadiness of the flow rate.

From the definition of Reynolds number it is clear that, for any

fluid, if the flow speed, v, is slow enough then the flow will be lam-inar. We can see an example of this when we open a water faucet:

when the flow speed is small the water comes out of the tap in a


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fluid

pipe wall

pipe wall

fluid

pipe wall

pipe wall

inviscidviscous

Figure 20.8: As a viscous fluid flows along a pipe (left) the shear

forces create a velocity gradient that increases uniformly from zero at

the pipe walls to a maximum at the centre of the pipe. For an inviscid

fluid (right) there is a narrow boundary layer at the pipe walls where

the velocity increases rapidly, but then it remains relatively constant

for the rest of the pipe volume.

laminar stream but when the flow speed is increased there is a point

when the flow becomes turbulent.

20.3.3 Ideal Fluid

In our study of fluid dynamics we will make the following approxi-

mations:1. the fluid is inviscid

2. the flow is steady and laminar

3. the fluid is incompressible

4. the flow is irrotational

A fluid that obeys these approximations is called an ideal fluid. Clearly

this is an approximation, but for low flow rates (i.e. small Reynolds

numbers), low compressibility fluids it is a reasonable approxima-

tion.

On the right in Fig. 20.8 is a schematic representation of the

flow of an ideal fluid in a pipe. The arrows represent the velocity

as a function of position in the pipe. For an inviscid, laminar flow

the velocity is uniform except at a narrow boundary layer at the pipe


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wall. On the left is a schematic representation of a viscous flow. The

fluid velocity is largest at the centre of the pipe and decreases to zero

at the pipe walls. The ideal fluid approximation will not be valid if

the flow is fast enough or the pipe is narrow enough.

20.3.4 Continuity

A consequence of the incompressible approximation is a continuity

equation. Consider the rate of volume flow through a pipe, R (theflow rate has dimensions of volume per unit time). The volume of

fluid passing point 1 of a pipe with cross sectional area is A1 andvelocity v1 in timet is A1v1t so the volume flow rate is given byR = A1v1. The flow rate at some other point in the pipe of crosssectional area A2 with velocity v2 must be the same as mass can notbe created or destroyed along the pipe, so

R = A1v1 = A2v2. (20.5)

This is the continuity equation.

20.3.5 Bernoullis Equation

Consider a particular volume of fluid with density moving to the

right through a pipe which changes in both diameter and height asdepicted schematically in Fig. 20.9, the ends of which have cross-

sectional areas A1 and A2. The pressure at the lower end, P1 appliesa force, F1 = P1a1, to the volume, causing that end to move a dis-tance x1. The work done on the left end of the volume of fluid isthen W1 = F1 x1 = +P1A1x1. Similarly, at the other end thework done on the right end of the fluid is W2 =

P2A2x2 which

is negative since the force opposes the motion. The net change in

the gravitational potential energy of this volume of fluid is a gain of

m2gh2 = A2x2gh2 and a loss ofm1gh1 = A1x1gh1. Fi-nally, the net change in the kinetic energy of this volume of fluid con-

sists of a gain of 12m2v

22 =

12A2x2v

22 and a loss of

12m1v

21 =

12

A1x1v21 . Since the work done on this volume of fluid is equal to


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h1 h2

A2

A1

F1

F2

Figure 20.9: A volume of fluid moves to the right through a pipe

which changes diameter and height. At one end of the volume the

height is h1 and the cross-section is A1. There is a force, F1, actingupon this end (caused by the pressure of the fluid to the left) causing

the motion and this end of the fluid is moving with speed v1. At theother end of the volume we have h2, A2, F2 and v2; note that F2 isacting in the opposite direction as F1 the pressure of the fluid to theright pushes against the motion.

the change in its mechanical energy,

W = K+U, we get

P1A1x1 P2A2x2 = 12A2x2v22 12A1x1v21+A2x2gh2 A1x1gh1.

We then note that, by conservation of mass and the incompressibility

assumption, A1x1 = A2x2. We then rearrange and divide by this

quantity to give

P1 +12v21 + gh1 = P2 +

12v22 + gh2.

Since we chose arbitrary points for the ends of the volume of fluid

this expression must be true for any such choice, leading to Bernoullis

equation1

2v2 + gh + P = constant (20.6)

is a constant throughout any volume of fluid. What Bernoullis equa-

tion tells us is that the energy density (i.e. the energy per unit vol-

ume) is a constant throughout a fluid.

Example 20.3.2. Venturi Effect

Consider a fluid of density moving with a flow rate R through an


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Figure 20.10: A fluid moves through an horizontal pipe of cross-

sectional area A which contains a constriction where the cross-sectional area is 14 A.

horizontal pipe with a cross-sectional area A. There is a constriction

in the pipe and the diameter is reduced by one half (see Fig. 20.10).(a) What is the velocity, v, of the fluid in the unconstricted sectionof the pipe? (b) What is the velocity, v, of the fluid in the constrictedsection of the pipe? (c) If the pressure in the unconstricted section

of the pipe is P, what is the pressure, P, in the constricted section?

Solution

(a) The flow rate is equal to the cross-sectional area multiplied by

the velocity so

v =R

A.

(b) The flow rate is unchanged so

v = R14

A= 4R

A.

(c) Bernoullis equation, 20.6, tells us

P + 12v2 = P +

12v2

P = P + 12

R2

A2 16R

2

A2

= P 152

R2

A2.

This is an example of the Venturi effect: As a fluid moves more

quickly (for example, when a pipe is constricted) the pressure

decreases. Many people find this counterintuitive.


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Problem 20.4.1. Dam

A water dam is of height h and width w. Atmospheric pressure at

the top of the water is Patm and the density of water is . (a) Whatis the pressure on the inner dam wall as a function of height, P(x)?(Measure x from the bottom of the dam.) (b) What is the net force,Fi, on the inner dam wall? (Hint: Unless you are clever you willneed to use the equation for a non-constant force, Eq. 20.1, and cal-

culate a dam integral, F =

damdF.) (c) What is the net force, Fo,

on the outer dam wall? (Assume that the density of the air is con-

stant and one thousand times less than that of water.) (d) What is thenet force, F, on the dam wall? (e) How would your answer to (d)change if you considered the air pressure constant as a function of

height? Calculate the fractional change in the net force, F/F.

Problem 20.4.2. Suspended Block

Consider a light beaker containing a volume Vf of fluid of density

f which sits on a scale. A block of mass m and volume V is sus-pended in the fluid by a light string. Assume the density of the block

is greater than the density of the fluid and that the block is fully sub-

merged. (a) Draw the free body diagram for the block. (b) Draw

the free body diagram for the beaker and the fluid together. (c) What

is the tension, FT, in the string? (d) What is the reading, W, on thescale? (answer in Newtons) (e) What is the reading on the scale,

Wcut, when the string is cut? (f) In an experiment to measure thedensity of an object it mass is first found to be m but, due to its ir-regular shape, its volume is unknown. It is then suspended in a fluid

as described above. If the change in the scales reading after the

string is cut is W, what is the density, , of the object?

Problem 20.4.3. Floating Sphere

An empty hollow sphere of inner radius r and outer radius R whichis made from a material with density floats so that exactly one half

is submerged in a fluid of density f. (a) If /f = 3 what is theratio r/R? (Empty means that there is nothing inside the sphere,not even air.) (b) If the sphere, instead of being empty, was filled

with a different fluid of density a what is the ratio r/R? (c) If the


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F

Figure 20.11: Hypodermic Needle

fluid in the sphere is air at atmospheric pressure (a = 1.3 kg/m3),

the fluid in which it floats is water (f = 103 kg/m3), the sphere is

made from aluminum ( = 2.6 103 kg/m3) and the outer radius ofthe sphere is r = 2 m what is the thickness of the hull, t = R r?Problem 20.4.4. Cylindrical Tank

A cylindrical tank of radius r1 is filled to a depth d1 with a fluid ofdensity f. The nozzle of a hose leading from the bottom of the tank,

of inner radius r2, rises a distance d2. (a) What is the speed v2 ofthe fluid exiting the hose? (b) In the r2 r1 limit what is v2? (c)Under what circumstances will this limit occur?

Problem 20.4.5. Holey Tanks!

A rectangular tank has a square cross section of width w and is filledto a height h with a fluid of density and is placed on a horizontaltable. A small hole of radius r is made a distance h from the bot-tom. (a) With what velocity, v, will the fluid jet from the hole?(b) At what distance, x ,from the base of the tank will the fluid strikethe table? (c) At what hole height, h, will the water travel the max-imum horizontal distance before striking the table? (d) What is this

maximum horizontal distance, hmax?

Problem 20.4.6. Pipes

A fluid of density is moving at a rate R (i.e. volume per unit time)through a pipe whose diameter changes from 2d at the bottom to dat the top as it ascends a distance h. (a) What is the changein the pressure between the bottom and the top of the pipe, P =

Ptop Pbottom? (b) What is the change in pressure in the h 0 limit?Is this what you expected?

Problem 20.4.7. Syringe

A hypodermic syringe contains a fluid with density . The (horizon-

tal) barrel of the syringe has a cross-sectional area A, and the needle


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h AB

A2A1

Figure 20.12: Venturi Tube

has a cross-sectional area a. See Fig. 20.11. (a) If a force Facts on the plunger, determine the volume flow rate, R, of the fluidejected from the end of the needle. (Think carefully about whether

a change in atmospheric pressure should affect you

PHYS122 University of Waterloo

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