Physics 430: Lecture 1
Mass, Force, Newton’s Laws
Dale E. Gary
NJIT Physics Department
1.1 Classical Mechanics
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First began with Galileo (1584-1642), whose experiments with falling
bodies (and bodies rolling on an incline) led to Newton’s 1st Law.
Newton (1642-1727) then developed his 3 laws of motion, together
with his universal law of gravitation.
This is where your previous experience in mechanics doubtless ends,
but the science of mechanics does not end there, as you’ll see in this
course.
Two additional, highly mathematical frameworks were developed by
the French mathematician Lagrange (1736-1813) and the Irish
mathematician Hamilton (1805-1865).
Together, these three alternative frameworks by Newton, Lagrange,
and Hamilton make up what is generally called Classical Mechanics.
They are distinct from the other great forms of non-classical
mechanics, Relativistic Mechanics and Quantum Mechanics, but both
of these borrow heavily from Classical Mechanics.
August 31, 2010
This Course
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The first part of this course will be a review in terms of the basic
ideas of mechanics, which you should have already thoroughly
learned in your introductory physics courses.
However, we will be looking again at the ideas with more advanced
mathematical tools, and it is really these tools that we are trying to
hone during the first 5 chapters.
The 6th chapter introduces some ideas from the Calculus of Variations
that are the basis for the Lagrange formalism, which you will learn in
chapter 7.
The Lagrange approach then becomes another of our mathematical
tools for looking at more complex mechanical systems involving
rotations of rigid bodies, oscillating systems, and so on, through
chapter 11.
Unfortunately, we will not have time to study Hamilton’s approach,
although I will try to get to it briefly in the final lecture.
August 31, 2010
1.2 Space and Time
We live in a three-dimensional world, and
for the purpose of this course we can
consider space and time to be a fixed
framework against which we can make
measurements of moving bodies.
 Each point P in space can be labeled with a
distance and direction from some arbitrarily
chosen origin O. Expressed in terms of unit
vectors xˆ , yˆ , zˆ

Ways of writing vector notation
F  ma


F  ma
F  ma
z axis
r  x xˆ  y yˆ  z zˆ
 It is equivalent to write the vector as an
ordered triplet of values
r  ( x, y , z )

We can also write components of vectors
using subscripts v  (v x , v y , v z ) a  (a x , a y , a z )
r
y axis
z
x
y
x axis
August 31, 2010
Other Vector Notations
You will be used to unit vector notation i, j, k, but we will follow the text
and use the xˆ , yˆ , zˆ notation.
 At times, it is more convenient to use notation that makes it easier to
use summation notation, so we introduce the equivalents:
r1  x, r2  y, r3  z
e1  xˆ , e 2  yˆ , e3  zˆ
which allows us to write
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3
r  r1e1  r2e 2  r3e3   ri ei
i 1

In the above example, this form has no real advantage, but in other
cases we will meet, this form is much simpler to use. The point is that
we may choose any convenient notation, and you should become
tolerant of different, but consistent forms of notation.
August 31, 2010
Vector Operations
Sum of vectors r  (r1 , r2 , r3 ); s  ( s1 , s2 , s3 )  r  s  (r1  s1 , r2  s2 , r3  s3 )
 Vector times scalar
cr  (cr1 , cr2 , cr3 )
 Scalar product, or dot product

r  s  rs cos 
s
3
 r1s1  r2 s2  r3 s3   rn sn

n 1
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r
Vector product, or cross product
p  r  s;
p x  ry s z  rz s y
p y  rz s x  rx s z
p z  rx s y  ry s x
r  s  rs sin 
 xˆ

r  s  det  rx
sx

yˆ
ry
sy
zˆ  xˆ

rz   rx
s z  s x
yˆ
ry
sy
zˆ
rz
sz
August 31, 2010
Fundamental Nature of the
End-of-Chapter Problems

In the main chapter, just after equation (1.7), there is a note that says
“see Problem 1.7 at the end of the chapter for a proof that
3
”
rs cos   r1s1  r2 s2  r3 s3   rn sn
n 1
Problem 1.7 says:
 Prove that the two definitions of the scalar product are equal. One
way to do this is to choose your x axis along the directionyof r.
[Strictly speaking you should first make sure that the definition (1.7)
s
is independent of the choice of axes. If you like to worry about such
niceties, see Problem 1.16.]

 This example shows that the problems are part of the text—even
r
problems that are not assigned. Many of the problems guide you to
further learning, and you are encouraged to try some of them.
 With r along x, we have r  (r ,0,0); s  ( s , s ,0)  ( s cos  , s sin  ,0)
x
y
r s  r1s1  r2 s2  r3 s3  rs cos   0  0  rs cos 
 Uses the fact that we are free to choose our axes without changing the
result. But what about Problem 1.16?
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August 31, 2010
x
Problem 1.16
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3
(a) Defining the scalar product r  s by Equation (1.7), r  s   rn sn
n 1
show that Pythagoras’s theorem implies that the magnitude of any vector
r is r  r  r .
r  r12  r22  r32 (Pythagora s' s Thm)
3
3
n 1
n 1
r  r   rn rn  rn2  r12  r22  r32
(b) It is clear that the length of a vector does not depend on our choice of
coordinate axes. Thus, the result of part (a) guarantees that the scalar
product r  r, as defined by (1.7), is the same for any choice of orthogonal
axes. Use this to prove that r  s as defined by (1.7), is the same for any
choice of orthogonal axes. [Hint: Consider the length of the vector r  s ].
(r  s)  (r  s)  r  s  r 2  s 2  2(r  s)
2


r  s  r  s  r 2  s2 / 2
Everything on the right is a length (squared), which is the same for any
choice of axis, so the same is true of r s .
2
August 31, 2010
Lesson About the Problems
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I cannot stress enough that doing problems is going to allow you to learn
10 times more than just reading the book. In the last problem, for
example, we learned how to distribute dot products, how to express the
length of a vector in terms of its dot product with itself, and generally how
to deal with proofs in terms of simple, easy-to-do steps.
The problems are written in a way to help you discover the answers. You
should pay attention to references in the text to the problems, and have a
look at the problems for further understanding.
The problems I assign may assume knowledge you can only gain from
following the thread from the text into the unassigned problems.
August 31, 2010
Differentiation of Vectors
This course makes heavy use of Calculus (and differential equations, and
other forms of advanced mathematics). In general, we will refresh your
memory about the techniques you will need as they come up, but we will
do so from a Physics perspective—only paying lip-service to the underlying
mathematical proofs.
 What we need now is a simple form of something called Vector Calculus.
As long as you remember that vectors are just triplets of numbers, and
vector equations can be thought of as three separate equations, you will be
fine.
 For now, consider only the derivative of the position vector r(t), which you
should know gives the velocity v(t) = dr(t)/dt. Likewise, the derivative of
the velocity (the second derivative of the position) gives the acceleration:
a(t) = dv(t)/dt = d2r(t)/dt2. Formally:
dx
x
 lim
whe re x  x(t  t)  x(t)
for scalars:
dt t 0 t

for vectors:
dr
r
 lim
dt t 0 t
whe re r  r (t  t )  r (t )
August 31, 2010
Differentiation of Vectors
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By the usual rules of differentiation, the derivative of a sum of vectors is
d
dr ds
(r  s) 

dt
dt dt
 and the derivative of a scalar times a vector follows the usual product rule
d
dr df
( fr )  f
 r
dt
dt dt
dr dx
dy
dz
ˆ  y yˆ  z zˆ so
v  vx xˆ  v y yˆ  vz zˆ
 Also note: r  x x
 xˆ  yˆ  zˆ
dt dt
dt
dt
implies that the unit vectors are constant (i.e.

dxˆ dyˆ dzˆ


 0 ).
dt dt dt
However, we will find in other coordinate systems the unit vectors are NOT
constant!
How can this be?
August 31, 2010
1.3 Mass and Force
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What is the difference between mass and weight?
Mass has to do with inertial force (ma). Weight has to do with gravitational
force (mg). In the first case, the mass is “resistance to changes in motion”
while in the second case it is a rather mysterious “attractive property” of
matter. In fact, these two different properties of mass are identical, which
is what Galileo’s experiments showed (dropping masses off the leaning
tower of Pisa). But it was not understood until Einstein’s theory of General
Relativity.
m1
Inertial balance:
Allows measurement of inertial mass
without mixing in gravitational force.
m2
Try this in the Phun Physics Engine
August 31, 2010
Point Mass (Particle)
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For now, we want to focus on the concept of a point mass, or particle. This
is an approximation, which is worthwhile to look at carefully. It basically
refers to a body that can move through space but has NO internal degrees
of freedom (rotation, flexure, vibrations).
Later we will talk about bodies as collections of particles, or a continuous
distribution of mass, and in considering such bodies the laws of motion are
considerably more complicated.
Despite this being an approximation, the approximation is still useful in
many cases, such as for elementary particles (protons, neutrons, electrons),
or even planets and stars (sometimes).
August 31, 2010
1.4 Newton’s Three Laws
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Law of Inertia
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Force Law
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In the absence of forces, a particle moves with constant velocity v.
(An object in motion tends to remain in motion, an object at rest tends
to remain at rest.)
For any particle of mass m, the net force F on the particle is always
equal to the mass m times the particle’s acceleration: F = ma.
Conservation of Momentum Law
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If particle 1 exerts a force F21 on particle 2, then particle 2 always
exerts a reaction force F12 on particle 1 given by F12 =  F21.
(For every action there is an equal and opposite reaction.)
August 31, 2010
Aside: Dot Notation
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Dot Notation:
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We will be using this dot notation extensively. It means
differentiation with respect to time, t (only!).
You may have seen “prime” notation, but if the differentiation is not
with respect to time, it is NOT equivalent to y-dot.

dv
d 2r
a
 v  2  r
dt
dt
y 

y-dot means dy/dt only.
dy
 y
dx
August 31, 2010
Equivalence of First Two Laws
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The Law of Inertia and the Force Law can be stated in equivalent ways.
Obviously, if F  ma , then in the absence of forces F  ma  0
dv
 0  v  v0
dt
Thus, the velocity is constant (objects in motion tend to remain in motion)
and could be zero (objects at rest tend to remain at rest).
The second law can be rewritten in terms of momentum:
p  mv
In Classical Mechanics, the mass of a particle is constant, hence
p  mv  ma
.
 So we can write F  p
 In words, forces cause a change in momentum, and conversely any change
in momentum implies that a force is acting on the particle.
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August 31, 2010
The Equation of Motion
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Newton’s Second Law is the basis for much of Classical Mechanics, and the
equation F  ma has another name—the equation of motion.
The typical use of the equation of motion is to write
ma   Forces
where the right hand side lists all of the forces acting on the particle.
 In this text, an even more usual way to write it is:
mr   Forces
which is perhaps an easier way to understand why it is called the equation
of motion. This relates the position of the particle vs. time to the forces
acting on it, and obviously if we know the position at all times we have an
equation of motion for the particle.
August 31, 2010
Differential Equations
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Most of you should have had a course in differential equations by now, or
should be taking the course concurrently.
A differential equation is an equation involving derivatives, in this case
derivatives of the particle position r(t).
Consider the one-dimensional equation for the position x(t) of a particle
under a constant force:
F
x(t )  0
m
This equation involves the second derivative (with respect to time) of the
position, so to get the position we simply integrate twice:
F
x (t )   x(t )dt  v0  0 t
m
F
x(t )   x (t )dt  x0  v0t  0 t 2
2m
This was so easy we did not actually need to know the theory of differential
equations, but we will meet with lots of more complicated equations where
the DiffEQ theory is needed, and will be introduced as needed.
August 31, 2010
Example: Problem 1.24

Statement of Problem:

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In case you haven’t studied any differential equations before, I shall be
introducing the necessary ideas as needed. Here is a simple example to get you
started. Find the general solution of the first-order equation df/dt = f for an
unknown function f(t). [There are several ways to do this. One is to rewrite the
equation as df / f = dt and then integrate both sides.] How many arbitrary
constants does the general solution contain? [Your answer should illustrate the
important general theorem that the solution of any nth-order differential
equation contains n arbitrary constants.]
Solution:

First note that we cavalierly treat df/dt as if it were a quotient with numerator
and denominator, when it is really an operator d/dt operating on a function f. We
get away with this by considering that
df
f
dt
 lim
t  0
t
so we do our manipulation before taking the limit
f
f
 f  f  ft 
 t
t
f
f
df
 lim t 
 dt
t 0 f
t 0
f
lim
August 31, 2010
Example: Problem 1.24, cont’d

Solution, cont’d:



For the “well-behaved” functions we typically deal with in Physics, it is okay to do
this. Mathematicians would scream, because the limit may not exist, or will
behave badly, for some discontinuous functions, but in Physics we typically do
not have to worry about that.
After taking the limit, the f and t become “differentials” df and dt.
Once the equation is in the form
df
 dt
f
it is easy to integrate both sides to get:


1
df   dt with solution ln f  c1  t  c2
f
Here “ln” is the natural logarithm, whose inverse is the exponential function e.
The two arbitrary constants can be combined into a single constant c = c2c1.
Taking the “inverse log” of both sides:
t c
t c
t
f (t )  e
 e e  ae
where I have introduced a new constant a = ec. So we are left with 1 constant
for a 1st-order equation, agreeing with n constants for an nth-order equation.
August 31, 2010
Inertial Reference Frames
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We saw in the example of problems 1.7 and 1.16 that we are free to choose
our axes (our coordinate system) any way we wish. This is called a
reference frame.
What may not be obvious is that we can even choose a reference frame that
is moving (changing with time). In fact, it is pretty much impossible to do
anything else. If we choose a reference frame fixed to a lab bench, for
example, our reference frame is rotating with the Earth, orbiting the Sun,
and moving around the galaxy.
We define an inertial frame as any non-accelerating frame (one moving at
constant velocity—both magnitude and direction). In such frames, Newton’s
first law holds—objects not subject to forces move in straight lines.
A non-inertial frame is one that is accelerating. In such a frame, objects not
subject to forces appear to accelerate. We describe such effects as
“fictitious” forces, which we’ll meet in Chapter 9—Centrifugal Force and
Coriolis Force.
You’ll note that our lab bench frame is not inertial! It is accelerating as
Earth moves. However, if our experiment does not take too long, and the
distance over which we make our measurements is not too great, we can
approximate the frame as inertial.
August 31, 2010
1.5 The Third Law and
Conservation of Momentum
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
Newton’s first two laws refer to forces acting on a single particle. The Third
Law, by contrast, explicitly refers to two particles interacting—the particle
being accelerated, and the particle doing the forcing.
Introduce notation F21 (F-on-by) to represent the force on particle 2 by
particle 1. Then
Newton’s Third Law
If particle 1 exerts a force F21 on particle 2, then particle 2 always
exerts a reaction force F12 on particle 1 given by F12 =  F21.
F12
F21
2
F2ext
1
F1ext
Fext1
Fext2
p 1  F1  F12 ; p 2  F2  F21;
P  F1  F2  F12  F21  0
ext
ext
p 1  F1  F1  F12 ; p 2  F2  F2  F21;
ext
ext
ext
ext
P  F1  F2  F1  F12  F2  F21  F1  F2  F ext
ext
p 1  F1  F1  F12 ;
p ext
ext
p 2  F2  F2  F21;
 Fext  F ext 1  F ext 2 ; P  0
August 31, 2010
Multi-Particle Systems


It should be fairly obvious how to extend this to systems of N particles,
where N can be any number, including truly huge numbers like 1023.
Let a or b designate one of the particles. Both a and b can take any value
1, 2, …, N. The net force on particle a is then
Fa 
Fab  Fa

b a
ext

 p a
where the sum runs over all particles except a itself (a particle does not
exert a force on itself).
a :
 The total force on the system of particles is just the sum of all of the p
P   p a
a
P    Fab   Faext   Faext  F ext
a b a

Each term Fab can be paired with Fba:
a
a
Fab    Fab  Fba   0


a b a
a b a


August 31, 2010
Conservation of Momentum
This final result: P  F ext says that the internal forces on a system of
particles do not matter—they all cancel each other, so the change in total
momentum of a system of particles is only due to external forces.
 In particular, if there are no external forces, the momentum cannot change,
i.e. it is constant. This is the principle of conservation of momentum:

Principle of Conservation of Momentum
If the net external force Fext on an N-particle system is zero,
the system’s total momentum P is constant.
This is one of the most important results in classical physics, and is true
also in relativity and quantum mechanics.
 It may be helpful to write out the sums from the previous slide explicitly for
3 and 4 particles (Problems 1.28 and 1.29), to convince yourself of the pairwise canceling of the internal forces.
 We showed that Newton’s 3rd Law implies Conservation of Momentum. You
can also show (Problem 1.31) that Conservation of Momentum implies
Newton’s 3rd Law.

August 31, 2010
Example: Problem 1.30

Statement of Problem:


Conservation laws, such as conservation of momentum, often give a surprising
amount of information about the possible outcome of an experiment. Here is
perhaps the simplest example: Two objects of masses m1 and m2 are subject to
no external forces. Object 1 is traveling with velocity v when it collides with the
stationary object 2. The two objects stick together and move off with common
velocity v. Use conservation of momentum to find v in terms of v, m1 and m2.
Solution:

Conservation of momentum says the momentum before the collision must be the
same as the momentum after the collision:
m1v1  m2 v 2  m1  m2 v

Since v1 = v, and v2 = 0 (second object is stationary), we simply solve for v to
find:
v 
m1
v
m1  m2
August 31, 2010
Trying Things Out in Phun

Things to try:



Show why Aristotle thought that the natural state of things is “rest,” and why he
thought “earth” bodies sought the ground while “air” bodies sought the heavens.
Show how Galileo was able to overcome the idea of Aristotle that the natural
state was “rest,” which led to Newton’s 1st Law.
Show that by removing external forces one can observe conservation of
momentum through collisions between two bodies.
August 31, 2010