CH 4 5 NEWTON'S LAWS THEIR APPLICATION

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```					DYNAMICS,
the Theory of Forced Motion

CH 4 & 5 NEWTON'S LAWS & THEIR APPLICATION
Objectives:
At the end of this section you should be able to

•   define instantaneous acceleration
•   write down Newton's laws in algebraic form, i.e. as mathematical
expressions and in your own words
•   explain the term rate of change of momentum
•   distinguish between individual force and net force
•   define the term inertial frame (with examples)
•   comment on the difference between law and definition
•   define gravitational field g(r)
•   distinguish between g(r), gravity g and gravitational constant G
•   distinguish between weight and apparent weight
•   list and discuss contact forces
•   describe and discuss circular motion

INTRODUCTION
In the first part of these notes (copies of transparencies) you were exposed to
Kinematics, the theory of motion, which is the branch of mechanics dealing with
the motion of bodies without reference to mass or force. This section is on
Dynamics, the theory of non-inertial motion, which deals with motion with
reference to mass and force.
Mentioning 'the theory of non-inertial motion' prompts two questions, "what is
inertial motion" and "what do we mean by a theory"? We shall deal with these
questions as we go along.

4.2 / 3 Newton's first and second laws
(Sir) Isaac Newton (1642 – 1727) was born in England in the same year that
Galileo Galilei died: 1642. He attended Cambridge University which was closed in
1665 – 1666 because of the plague (a deadly disease that affected most of
England). He used this time to read the writings of Galileo (1564 - 1642) and of
Johannes Kepler (1571 – 1630). From Kepler's observations (which we shall
discuss in more detail later) he learned that the planetary orbits are best described
as elliptical and that the motion of the planets is thus not uniform, i.e. neither at
constant velocity (v ≠ const) nor at constant speed (v ≠ const). In the half of their
orbit when they are getting closer to the sun, the planets are speeding up, in the
other half they are slowing down. They are accelerating, then decelerating and then

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accelerating again, just like a stone which slows down when thrown upwards and
speeds up when it comes down again. Thus, the planets behave very much like an
object in free fall above the surface of the earth.
Newton's correct conclusion was this. Even when on a perfectly circular orbit at
constant speed, an object (like the moon orbiting the earth) is in free fall. Thus, it is
accelerating, although its speed does not increase. How could that be? This is how
Newton explained it. The figure below shows a canon, mounted on a mountain
peak, firing canon balls with increasing initial velocity v. At 'low' initial muzzle
speed the ball will land near the foot of the mountain.

v
.

With increasing velocity the ball will travel further and further until, at a certain
critical value, it will not hit the surface beneath the mountain but will return to the
mountain top on a circular orbit: the gunner would have shot himself in the back.
According to Newton, this is exactly the situation of the moon on its orbit around
the earth, as indicated by the outer (solid) circle. While travelling forward it is also
falling towards the centre of the earth: the moon is in free fall, hence is
accelerating. Thus, circular motion is not uniform motion with zero acceleration.
Although speed v = constant, acceleration a ≠ 0, hence Galileo's concept of
circular inertia (i.e. circular motion requiring no force) was wrong.
According to Newton, only uniform motion, i.e. motion with a constant velocity
along a straight line, with zero acceleration, requires no force, hence is inertial.
Newton's "discovery" of rectilinear inertial motion is so important that it was later
enunciated as a law, i.e. Newton's first law of motion, which states:
Every object remains at rest or in motion in a straight line at constant speed
unless acted upon by an unbalanced force.
It also became clear to Newton that, as accelerated motion requires the presence of
a force, so the presence of a force will lead to acceleration. Also, the greater the
force on an object, the greater its acceleration. This is expressed in Newton's
second law, which states that
the acceleration of an object is proportional to, and in the direction of, the
force acting on the object. Conversely, the force needed to
accelerate an object is proportional to the acceleration.

2
We interpret the need of a force to change the state of motion of an object as a
resistance of the object to change, i.e. as a measure of its inertia. As we shall
discuss in more detail later, this inertia or resistance to change of the state of
motion of a body is determined by the inertial mass of the body.

4.4 / 5 Newton's law of gravitation & the third law
Newton's next contribution to an understanding of the motion of the planets around
the sun (and of the moon around the earth) on near circular orbits was his law of
gravitation. He had argued that, in its orbit around the earth, the moon was
continuously accelerating towards the earth, just like an apple falling off a tree and
accelerating to the ground. He also surmised that the force responsible for the
circular (hence accelerated) motion of the moon around the earth and for the
uniform acceleration of the apple was the same. It is the attraction of the earth, as
indicated above. But what was the origin of this force?
Although the 'nature' of this force was obscure, there was a name for it:
gravity/gravitation (in Latin, which was the language used amongst natural
philosophers, gravitas means 'heaviness' or 'weight'). Massive objects are heavy;
they have a tendency to fall down, i.e. to move closer towards the centre of the
earth. How massive or heavy a body appears to be is determined by its
gravitational mass (The distinction between mass and weight is sometimes still
misunderstood. We often express body-weight in kg - few people refer to their
body-mass or would ask, "what is your mass?", although this would, in fact, be
correct!).
So, it is the mass of a stone that determines the attraction it experiences towards
earth. In the same manner the mass of the moon determines its attraction towards
earth, and the mass of the earth its attraction towards the sun. In other words: the
attractive force an object experiences due to gravitation is proportional to its
gravitational mass.

Newton also realised that, amongst physical objects and bodies, attraction
is mutual (if you pull your friend's hand, you can feel your friend's hand pulling
you) i.e. the forces two bodies exert on each other have equal magnitude, though
opposite direction. This is Newton’s third law,

if object A exerts a force on object B, then object B will exert an ‘equal but
opposite’ force on object A.

Therefore, if the moon (mass m) experiences a pull from earth proportional to m,
then (from Newton’s third law) the earth must experience a pull from the moon
proportional to m. Likewise if the earth (mass M) experiences a pull from the
moon proportional to M then the moon must experience a pull from the earth
proportional to M. The logical conclusion is that the gravitational pull, Fg,
“between” moon and earth must be proportional to the product of the individual
masses, Fg ∝ mM.

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It took Newton a bit longer to figure out that Fg decreases with distance squared,
i.e. that Fg ∝ 1/r2. This we will discuss later.

Newton's laws as formulae
Newtons second law (N2) is easily cast into mathematical form, F = m a, where F
is the net force (a vector) acting on a body, a is its acceleration (a vector) and m is
its inertial mass (a scalar). Using the definition of momentum, p = m v (to be
discussed later), and the definition of a = ∆v/∆t, we may write this as F = m ∆v/∆t
= ∆p/∆t. Now, N2 does not require a to be constant. It holds for any type of force,
hence any type of acceleration (provided v does not approach the speed of light, a
complication dealt with in relativity, which does not concern us here). Thus, if we
want to express N2 in terms of p we'd better use the definition of instantaneous
acceleration, i.e.
∆v dv
a = lim        =    ,
∆ t → 0 ∆t   dt

which we call the rate of change of velocity. Thus, we may express F = ma as F =
m dv/dt = d mv/dt = dp/dt, and say that force equals the rate of change of
momentum.

We may express the fact that F is the net (or resultant) force acting on a body, by
writing it as the vector-sum of the individual forces (Fk) acting on the body, F =
ΣkFk, where Σk is the symbol for (and is read as) "sum over all k of…."
For the special case that F = 0 (hence F = 0), N2 reduces to ma = 0, hence
a = 0 (given that m ≠ 0) which implies v = const. Conversely, if v = const, then a = 0,
hence ma = F = 0, i.e. there cannot be a (net) force acting on the body. This is
Newton's first law, N1. Thus N2 implies N1, while the opposite is not true. The
statement, if F = 0 then v = const, hence a = 0 does not imply N2. The relationship
between F, m and a could easily be more complicated than F = ma (in the theory of
special relativity, which applies at high speeds, it is more complicated).

Thus we express N1 separately in mathematical form, though not in the form of an
equation. For "v = const (e.g. = 0) unless there is a force F ≠ 0 acting" or, "if v = const
(e.g. = 0) then F = 0, and vice versa", we write

v = const (e.g. = 0) ⇔ F = 0,

where ⇔ means "implies in both directions".

Newton's third law (N3) may be expressed in words as: "If object 1 exerts a force F21
on object 2, then object 2 exerts an equal but opposite force F12 on object 1, i.e. F21 =
- F12.

Finally, Newton's law of gravitation (NG) combines the two observations, F12 = F21
=: Fg (remember, "=:" means denoted or defined as), Fg ∝ m1 m2 and Fg ∝ 1/r2. Here

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m1 and m2 are for instance the masses of earth and moon and r is the distance from
centre to centre, while F12 = F21 =: Fg is the magnitude of the gravitational force
"between" them (force between two bodies always refers to the force either body
exerts on the other one!). Thus, we may write Fg ∝ m1 m2/r2, hence Fg = G m1 m2/r2,
where the proportionality constant G is called the gravitational constant or constant of
general gravitation.

We thus may express the four laws N1 – NG, in mathematical form, as

(1)                     v = const (e.g. = 0) ⇔ F = 0,                    (N1)
dp
(2)                     F = ma =       , where F = ∑ Fk ,                (N2)
dt              k
(3)                     F21 = - F12,                                     (N3)
mm
(4)                     Fg = − G 1 2 2 r12
ˆ                               (NG)
r
ˆ
(where r12 is a unit vector pointing in the direction
from m2 to m1, as discussed later)

Be sure you interpret the left side of the implication in N1 correctly. "v = const
(e.g. = 0)" refers to "a body moving at constant velocity or remaining at rest", i.e.
remaining at v = 0.
Consider the following situation. A stone is thrown up vertically, i.e. it rises, comes to
rest at its highest point, and falls down again. Express its acceleration as a = ay ˆ ,
j
where ˆ is a unit vector in the upwards (y-) direction. Asked whether ay is > 0, < 0 or
j
= 0, (i) on the way up, (ii) at the highest point, and (iii) on the way down, you should
answer with confidence that ay < 0 in all three cases. In particular, a ≠ 0 at the highest
point, where v = 0 momentarily (i.e. for a very short moment). You can check this,
when approaching this problem from a gravitational point of view, as discussed later.

N1 and inertial frames (see 4.2)
If you drop a bottle (or whatever) in a train, it will fall straight down to the floor,
whether the train is at rest or is moving, provided it is moving at constant velocity.
You may have noticed that it is almost impossible to tell whether a lift is moving,
especially whether it is moving up or down. Only from the way it starts moving or
comes to a stop, i.e. from the way it accelerates or decelerates, is it obvious that
you are moving and in which direction. We say, a lift, a car, a train, a ship etc.,
moving at constant velocity (not only speed!) constitute an inertial frame.
Consider an object (e.g. a particle) moving with constant velocity v in the absence
of any force, such that N1 applies. Then, in a frame moving with the particle at
velocity v, the particle will be at rest, again in the absence of any force. Thus N1 is

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still valid. Obviously, this frame is an inertial frame, and we may define: a frame,
in which N1 applies, is an inertial frame.
Conversely, if you sit in an accelerating car you feel yourself pushed into your seat
by a force, which is absent when the car is stationary. Thus, an accelerating car (or
train, ship or aircraft) is not an inertial frame.
The important implication is this. The laws of physics as usually formulated (e.g.
N1 – NG) only apply to inertial frames.

N2 – a law or a definition?
If a mass m accelerates with acceleration a, then according to N1 there is a force
acting on it, while N2 tells us that this force is given by ma. Could we say that ma
'defines' the force (as is suggested in some text books), such that N2 is a definition
rather than a law? The answer is an emphatic NO.

The claim that doubling the force on an object of constant mass doubles its
acceleration is an empirical observation, which can be tested by experiment.
Let mass M be attached to a vertical string (see Fig), which is guided over a
(frictionless) pulley.
Let the other end of the string be
attached to mass m, which is resting on                         m          tensiometer
a frictionless table. If m is free to
move, the tension T in the string, which
is measured with a tensiometer (i.e. a
calibrated spring) will accelerate m
M
horizontally with acceleration a = T/m.
Now, if M is increased until T is doubled (as measured by the tensiometer while m
and M are moving), it is observed that a is doubled too, i.e. is again given by T/m.
This did not have to be so but it is, except for objects moving close to the speed of
light c (like an electron in a particle accelerator).

N3 - a misinterpretation
You may have heard or read the farmer and the donkey story. Farmer "Freddy" is
sitting on his cart and asks his donkey to set the cart into motion. The donkey
argues that this is impossible, saying: The more I pull on the cart (with force F
exerted on the rope attached to the cart) the more the cart will pull back on me (via
the tension T in the rope attached to the donkey) with the same force. Thus I cannot
set the cart into motion.
Question/Exercise

How would you have countered the donkey's argument, such that it would
have no option but to set the cart into motion? Be as explicit as possible (a
sketch might help).

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NG - Newton's law of gravitation (cf. CH 12)
This is how we wrote NG in mathematical (vector) notation:

m1m2
Fg = − G          ˆ
r12 .
r2

Before discussing the notation and some of the
features, I would like to rewrite it slightly, by                         ˆ
r
inserting the mass m of the moon for m1, the                        m
mass M of the earth for m2. Also, let me simplify
ˆ                                            r      Fg
the unit vector r12 (which points
from m2 = M to m1= m) to r . ˆ                         M
Thus, we have
mM
Fg = − G 2 r . ˆ
r

Now, have a look at the sketch. The position of the moon is specified by the
position vector r, pointing from the centre of the earth to the centre of the moon.
The force on the moon, exerted by the earth, is denoted as Fg. Its direction is
ˆ
opposite to the unit vector r , which points in the direction of r. This explains, why

Let me expand on this. A vector A, pointing in the x-direction, may be written as A
= A ˆ , where A is the component of A in the direction of the unit vector ˆ .
i                                                                      i
ˆ
Similarly, we may write a vector F, pointing in the radial direction, as F = F r ,
where F is the component of F in the radial direction.

ˆ
Thus, we may write the force on the moon as Fg = F(r) r , where F(r) is the radial
component of F, which is a function of r and is negative, because F is pointing in
ˆ
the opposite direction of r .
In fact, F(r) is given by F (r ) = − G 2 . Thus, F(r) is not the magnitude of the
mM
r
vector F, which cannot be negative, but the radial component of F, which is
negative, as F is a force of attraction.

If we interpret m to be the mass of a planet, M the mass of the sun and r the sun –
planet distance, then F(r) represents the force between planet and sun. The
gravitational constant G is the same, not only for all planets and the sun, the moon
and the earth, but for any two masses in the universe (i.e. also on earth). It is
therefore referred to as 'the constant of universal gravitation'. Its value was
measured by Sir Henry Cavendish in 1798 and is given by

G = 6.670 × 10-11 N m2 kg-2.

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The gravitational field
Kepler believed in a vis motorix, a driving force pushing the planets along their
orbits, which he thought to be of magnetic origin. Galileo believed that the circular
motion of the planets was inertial, hence did not require any force to keep them
going. Newton showed that Kepler was right in assuming that the sun did exert a
force on the planets, but that this force is not of a magnetic origin. Newton had no
difficulty with the concept of a force at a distance. We feel nowadays, that Galileo
was right in seeing a problem. Here is why.
We ask: how can the earth exert a force on the moon, how can the earth experience
a force exerted by the sun? Neither the sun nor the earth has arms with which they
could pull at each other. So, how is this pull transmitted?
You may ask, if we can 'see' at a distance why should we not 'feel' at a distance? Well,
we can see at a distance because light is reflected from distant objects and enters our
eyes, where it produces an image of the objects on the retina of each eye. And we have
a perception of distance, because our two eyes produce slightly different images which
are superimposed by our brain to a three-dimensional image. Thus, the process of
seeing a distant object takes place here, in my eyes and my brain; it is a local process,
far away from the object. So, how can a planet experience the force of the sun locally?
Our answer is this. The space surrounding the sun is filled with a field, the
gravitational field of the sun, which extends to the planets and beyond, getting
weaker with distance. Similarly, the earth is surrounded by a gravitational field, and
this is how the moon experiences a force locally, as exerted by the earth.
The gravitational force is proportional to the gravitational field strength and to the
gravitational mass of the object experiencing the force. Using NG, we express this
as follows.
Consider the force experienced by m in the field produced by M,

    M
F (r ) = − G        = m − G 2  =: mg (r ) ,
mM
r 2
    r 
where g(r) is the gravitational field. Obviously, g(r) = F(r)/m, hence we define in
general:
gravitational field:        g = Fg/m,                   (D.1)
i.e. the gravitational field is gravitational force per mass. According to this
definition, the units of g are N/kg. The magnitude g(r) of this field is often called
the gravitational field strength.

As F is a vector so g must be a vector. Using NG in vector-form, we thus may write
the field produced by M (where M is the mass of the sun, the earth or of any other
spherically symmetric mass distribution)
M
g = −G      ˆ
r = g(r).
r2

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The symbol g(r) indicates that g is a function of the position vector r. We cannot
write g = g(r), because as a vector, g does not only depend on the distance r from
ˆ
M, i.e. the length of its position vector r, but also on its direction (as given by r ).

For the gravitational field near the surface of the earth we have
M
g = −G 2 r , ˆ
R

where R is the radius of the earth (R ~ 6400 km). According to this expression, its
magnitude g=: g, called gravity, given by
M
g= G 2 ,
R
depends a bit on position, but is given, to good accuracy, by g = 9.81 N/kg ≈
10 N/kg. Thus, the weight of any object near the surface of the earth (up to some
hundreds of km) is given by Fg =: W = m g. For instance, the weight of an object
with mass m = 100 g is W ≈ 0.1 kg × 10 N/kg = 1 N.

Gravity and acceleration due to gravity
Galileo observed (quite correctly) that all objects fall with the same acceleration,
provided air resistance can be neglected. Newton’s laws, NG and N2, easily explain
why.

Drop a stone, a bottle, whatever, mass m. Its weight is W = m g, where m is the
gravitational mass. The gravitational force F = W accelerates m according to

F = m a, where m is the inertial mass.

Taking gravitational mass and inertial mass being identical we thus have
W = mg = F = mag, where ag is the acceleration due to gravity. Thus, we have ag =
g, i.e. the acceleration due to gravity is equal to gravity. As ag is in the same
direction as g, we also have that ag = g.

In general we use the same symbol for acceleration due to gravity and for gravity,
namely g. However, conceptually ag and g are very different things. Gravity g
(measured in N kg-1) is the source of a force, W = mg, while ag (measured in m s-2)
is the response to this force, ag = W/m = mg/m = g. Obviously, as ag = g, we must
have that [ag] = [g], i.e. that m s-2 = N kg-1, which is easily established.

To appreciate the difference between acceleration due to gravity and gravity
consider a brick placed on a table. Obviously, the acceleration a of this brick is
zero, a = 0, while gravity g at the position of the brick is certainly not zero, such
that W = mg ≠ 0. Obviously it would be quite misleading to refer to g, which is
acting on m, resulting in its weight, as 'acceleration due to gravity'.

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Weight and apparent weight (see 4.4 and 12.2)
Mark Shuttleworth, the first Afronaut (i.e. African Astronaut or Cosmonaut) went
on a mission to the International Spacestation. He got quite a bit of exposure on
TV and you may have seen him floating in the space station and drifting from wall
to wall. You may have seen similar shots of American astronauts in the space
shuttle or of Russian cosmonauts in (the Russian space station) MIR. Apparently,
in these situations, the astronauts and cosmonauts are weightless. Does that mean
that they really have no weight, i.e. that W = mg = 0? How could that be? Although
the space stations are some hundred kilometres above the surface of the earth, g ≠ 0
there, and these space travellers are certainly not massless. So, what is the story?
If you want to check on your weight you step onto a bathroom scale. Strictly
speaking you want to determine your mass, but the scale responds to your weight.
How? There are some springs mounted in the housing of the scale which are
compressed by your weight, and the compression is indicated on the dial of the
scale (calibrated in kg). Let us analyse this a bit further.
When a spring is compressed (or expanded) through a distance x by an applied
force Fa, it pushes (or pulls) back with a spring force Fs = - Fa. We say that the two
forces are in equilibrium and N1 applies. According to Hooke’s law, Fs ∝ x, i.e. Fs
= k x, where k is the spring constant. Thus, x is a measure of Fs and hence of Fa.
P = -mg
Now, when you step on your bathroom scale
Fa = W = mg, i.e. your weight (magnitude mg). The
response of the compressed springs of the scale results
in a push P = - W. As P is an upwards force, we have
P = mg. Thus, the distance by which the scale springs
W = mg           are compressed is a measure of your weight and hence
Now assume you step onto a scale in a lift that is moving up or down with constant
velocity. As you are moving with the lift at constant velocity there is no net force
acting on you (according to N1). There are two individual forces acting on you,
your weight W = m g (downwards) and the push P of the scale (upwards).
According to N1 we must have P + W = 0, hence P = - W, and P = mg, as before.
Assume now that the lift is accelerating upwards with acceleration a. As you are
accelerating with the lift, there must now be a net upwards force F acting on you,
such that F = ma. The only two forces acting on you are P (up) and W (down).
Thus, we must have that P + W = ma, or P – mg = ma. Hence P = mg + ma, and
finally
P = m(g + a) =: Wa.

Thus, the scale reading does not reflect your true weight W = mg anymore or, as it
is calibrated in kg, your true mass. It is now a measure of your apparent weight

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In summary: We call mg the (true) weight W, i.e. W = mg, which is ≠ 0, unless g =
0, which is rarely the case (there are regions in space where g = 0. Can you say
where?). The apparent weight is given by Wa = P, i.e. it is the reading that a scale
would indicate if it were calibrated in newton. If calibrated in kg (as usually) a
scale reads the apparent mass, ma = Wa/g.

Forces in general (see 4.1)
Most textbooks on basic physics list four forces of nature, i.e. the (1) gravitational, (2)
electromagnetic, (3) strong nuclear, and (4) weak nuclear force. Strictly speaking this
is not true anymore. In recent years high energy (or particle) physicists have put
forward a theory that combines the electromagnetic force with the weak force into an
electroweak force.

Something similar happened two centuries ago. For centuries electric and magnetic
forces were considered to be separate forces and independent from each other. At
some stage, however, they were recognised to be connected and were combined into
one electromagnetic force. For some time now it has been a dream in modern physics
to find a grand unified theory (GUT) that will combine all known forces (or
interactions, as they are sometimes called). But for now this appears to be a dream.

We shall meet these interactions known today in some detail as we go along. Here
we are mainly concerned with gravitational forces and the gravitational field, and
(indirectly) with electromagnetic forces. However, before looking further into the
subtleties of gravity and the gravitational interaction, I would like to discuss some
attributes of forces and introduce some typical types of mechanical forces.

Units
The unit of force is [F] = N (newton), where "[ ]" stands for unit of. It is defined
with the help of N2 as follows. Obviously, as F = m a, we have [F] = [m][a]. With
[m] = kg and [a] = ms-2 , where kg, m, and s are basic SI units, we have [F] = N =
m kg s-2. Thus, the newton is said to be a derived unit, and its definition may be
expressed in words as follows:

A force of 1 newton will accelerate a mass of 1 kg by 1 m/s2.

Its symbol (N) is a capital letter as it is named after a person, Isaac Newton. Other
units with capital letters as symbols are the current unit ampere (A), the absolute
temperature unit kelvin (K), which both are basic SI units, the derived energy unit
joule (J), the power unit watt (W), the electric potential unit volt (V), the magnetic
field unit tesla (T), etc.

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The vector - character of F
A vector, multiplied with a scalar, e.g. a number, results in another vector, which
points in the same or the opposite direction. If vector B = 3.5 A, then B is 3 ½ times
longer than A and points in the same direction. Thus, according to N2, as acceleration
is a vector, force must be a vector too. In a plane, F may be specified by its two
components Fx and Fy (in three dimensions we need Fz as an additional component). Or
it may be specified by its magnitude (or length) F = (Fx2+ Fy2)1/2 and the angle θ it
makes with the horizontal, where tanθ = Fy /Fx , hence θ = arc tan (Fy /Fx) .
There is an important implication, which I must stress. If forces are to be added,
they must be added as vectors, i.e. not algebraically, unless they all lie along the
same line, i.e. point into the same or the opposite direction.
For example, if F = P + W, unless P and W are parallel, F ≠ P + W. If they are
parallel (i.e. all pointing in the same positive direction), then F = P + W, if
antiparallel (i.e. parallel but in opposite directions), then F = P – W. Here, P and W
are interpreted as the magnitudes of P and W. However, if W > P, such that F < 0,
then F has to be interpreted as a component of F, where F is pointing in the
opposite direction of P, i.e. the negative direction. Check these statements with the
help of some sketches.
In any case we have that Fx = Px + Wx and Fy = Py + Wy. As these are components,
each one of them may be either positive or negative.
Let me illustrate the vector addition of forces.
Consider an object, e.g. a lamp (mass m, weight W = mg)
suspended from the ceiling on two cables as shown. Tl is the                             R
tension in the left cable, Tr the tension in the right cable. As                Tl           Tr
point P is at rest, the forces must be in equilibrium, i.e. must                     P
add up to zero (due to N1), Tl + Tr + W = 0. Thus, the                                   W
resultant R = Tl + Tr must be equal in length but opposite to
W, such that R + W = 0 (see Fig). Now, force equilibrium
implies that all x components must add up to zero and so must
all y-components. As Wx = 0 we must have Rx = 0, while Wy =                                   mg
- mg implies Ry = R = mg, as shown.

Contact forces
Tension is a contact force. The tension in the left cable and the tension in the right
cable exert forces Tl and Tr on the vertical cable, because they are attached to the
vertical cable at P. The vertical cable exerts a force R on the lamp because it is in
contact with the lamp (R is the tension in the vertical cable).

The push exerted on the soles of your shoes by the top of the bathroom scale is
another example of a contact force. As this push is perpendicular (or normal) to the
surface of the scale we refer to it as a normal force (symbol N). Let us discuss this
concept, which plays an important role in the topic of friction, in more detail.

12
N

Consider a block placed on a table as shown. It
experiences the gravitational pull of the earth (as
transmitted by the gravitational field), i.e. it has a weight              W
W. As the block does not move, there must be a force N                         P

on the block which compensates its weight, N = -W.

This force, exerted by the table on the block, is normal to the surface of the table, it
is a normal force. It is the response to the push P of the block on the table: the
harder the block pushes on the table, the harder the table will push back, N = -P
(Newton's third law, N3). Both, N and P, are due to the contact between the block
and the table: they are contact forces.

Now, although the three forces, W, P and N, are equal in magnitude (W = P = N),
they are of different origin and they act on different objects. Thus, you must not
confuse them. For instance, W is a force exerted by the earth (via the gravitational
field). P is a force exerted by the block on the table, and N is a force exerted in
return by the table on the block.

Thus, N and W act on the same object (the block), are equal in magnitude, but
have opposite direction. Thus they add up to zero, i.e. they compensate. On the
other hand, N and P do not compensate, although they have equal magnitude and
opposite direction. They act on different objects, hence it does not make sense to
add them up. Thus, here is a rule: Only add up forces that act on the same body.

5.3 Friction
Frictional forces are contact forces. They are forces between two surfaces (in
contact) which are sticking to each other (static friction) or are sliding against each
other (kinetic friction). These sticky forces are due to molecular interactions
between the two surfaces.
N
Static friction                                                        P
m v=
Push against a crate on the floor. Unless you push hard
enough the crate will not move. Why not? You are                        Fs        W = mg
exerting a (pushing) force P on the crate. Thus,
according to N1 and N2 it should move.

Answer: there is a (static) frictional force on the crate compensating your push, Fs =
- P. The harder you push, the stronger the frictional force opposes your push. Note
that this is not an example of N3 (can you explain why not?). Only when you push
hard enough, i.e. when P > Fs,max will the crate start moving. This maximum
frictional force, Fs,max, is proportional to the weight of the crate, Fs,max ∝ N, where N
= mg is the normal force exerted by the floor on the crate (N3!). The proportionality
constant µs =: Fs,max /N is a pure number, called the coefficient of static friction. Its
magnitude depends on the roughness of the two surfaces in contact, but is
independent of the area.

13
Thus, when pushing against the crate until it just starts moving, we have
(considering magnitudes of forces only)

P = Fs ≤ Fs,max = µs N = µs mg, or shorter: Fs ≤ µs N.

Note that you must not write this as a vector equation, because Fs and N are not in
the same (or opposite) direction but are normal to each other.

Kinetic friction                                                                 N
P
v
Once you have set the crate in motion you will have to keep                       m
pushing, even if it is to move at constant v. If you stop
pushing it will come to rest very soon. The reason: there is a         Fk
W = mg
(kinetic) frictional force Fk on the crate in the opposite
direction of v.

Again, this frictional force is proportional to the normal force, Fk ∝ N = mg. The
proportionality constant is the coefficient of kinetic friction, µk =: Fk/N, hence
Fk =µk N = µk mg.

Again, you must not write this as a vector equation!

Implications: If you push with a force the P = Fk =µk N = µk mg, then you will be
moving the crate at constant velocity. If P > Fk, the crate will accelerate, if P < Fk
it will slow down.
Comparing static friction with kinetic friction
Fs ≤ µs N says that Fs will compensate any push on the crate up to a maximum value
Fs,max =µs N (Note: This means that Fs is not constant; it increases with P until it
reaches its maximum value). If P > µs N = µs mg, the crate will start moving.

Fk = µk N = µk mg says that, if the crate is moving with velocity v, then there will be
a frictional force Fk on it, always acting in the opposite direction to v. This force
will slow the crate down, unless a push P ≥ Fk is applied to the crate.

Note that, in most cases, µs > µk i.e. once the object has started moving, the
frictional force decreases.

5.4 Dynamics of circular motion
Let us briefly revise circular motion and moving frames (see KINEMATICS). An
object, moving on a circular path of radius r, is changing its direction all the time.
Thus, its acceleration a is not constant, even if its speed v is constant. If v =
constant, then a is always normal to v and is directed towards the centre of the
circular path. The magnitude of a is given by
v2
a=       .
r

14
As a is pointing in a radial direction towards the centre of the circle, we call a the
ˆ
Introducing a unit vector r at P, in the direction of r, or expressing v in terms of
the angular velocity ω = 2π /T, where T is the period, we may write a as
v2
a=−        r = − ω 2 r.
ˆ
r
According to F = ma, the object of mass m must experience a force F in the
direction of a, i.e. towards the centre. Such a force is called a centripetal force
(meaning towards the centre). As a = v 2/r = ω 2r, we have

v2                   v2
F=m    = mrω , or F = − m r = − mω 2 r .
2
ˆ
r                    r
What is the origin of this force? There is a misconception that mv2/r is the
centripetal force, that it defines F. This is a wrong interpretation. The right hand side
of the above equation is the response to F, not its cause. Thus, the origin of F,
which depends on the case considered, has to be examined.
For the planets, orbiting the sun, F is obviously the gravitational attraction of the
sun. Similarly, for the moon, F is the gravitational force exerted on the moon by
the earth at the centre of the orbit.

The acceleration at the equator
Let me pose the following question. If an object (e.g. a book placed on a table) at the
equator has a radial acceleration of a = 3 cm/s2 due to the rotation of the earth, and if
the acceleration due to gravity at the equator without rotation were 983 cm/s2, with
what acceleration would the object approach the ground when dropped? With 986
cm/s2, 983 cm/s2 or 980 cm/s2?
In order to answer this question about a book being dropped from a table in the City
library in Entebbe, i.e. at the equator and falling towards the floor, we have the
following. Assume that the acceleration of the book due to gravity alone (i.e.
without rotation of the earth) is ag = 983 cm/s2. Now, the floor of the library has an
acceleration afg = 3.35 ≈ 3 cm/s towards the center of the earth, as we worked out
earlier. Thus, the relative acceleration af of the book w.r.t. the floor is given by af =
ag- afg. Considering magnitudes, we have af = ag – afg = (983 – 3) cm/s2 = 980 cm/s2.
Thus, the book will hit the ground a bit later than it would have done, if the earth
were not rotating.

Problems
Revise the remarks on solving problems. In problems involving forces point 3 is
particularly relevant, i.e. " If vectors are involved draw them in the correct
direction and roughly with the correct length". In particular, the following

15
1. Identify and draw (as vectors) all forces acting on an object. Omit forces
which are not acting on the body you are investigating.

2. If two (or more) bodies are interconnected, e.g. via strings or rods, draw all
forces acting on either object, but analyse them separately, i.e. as if the
other object(s) were not there.
This approach, known as "free body diagram" technique, is particularly useful and
you should practice it. It is best explained with the help of some examples.

Worked Problems
1. Interconnected body motion: Sliding block problem
N       a
Block M is on a frictionless table. It is
M               T                                  connected, with a massless string via a
frictionless pulley to mass m. Given m
T
and M, what is the acceleration of either
W                                  a               block and the tension in the string?
w=mg
Solution:
Given: m, M and (implicitly) g.
Wanted: a and T.
Expectation: As W is normal to any motion of M, and is compensated by N anyhow
(both W and N act on M), it does not contribute to the acceleration a. The only
driving force which does contribute is w = mg. The (inertial) mass of the two-block
system is M + m, thus we expect mg = (M + m) a, hence a = mg/(M + m).
Free body diagram technique: Consider m and M separately. The forces on m are
mg (supporting a) and T (resisting it). Take the direction of a as positive. Thus mg
– T = ma (1). The only accelerating force on M is T. Again, take the direction of a
as positive. Hence, T = Ma (2). Adding the two equations such that T and –T
cancel we have mg = (m + M) a, hence
a = mg/(M + m), as expected. Multiplying a with M (i.e. substituting a into
expression 2) we find T = Mmg/(M+m).

2. The 'Atwood machine'
o
This "machine" is actually a device to determine g.
Two blocks of mass m and M > m, are connected
g
through a massless string via a mass- and friction-less
T                                      pulley. What is the acceleration of either block and
a                                               what is the tension?
T
Solution:
mg                       a
Given: m, M and (implicitly) g
Mg                                     Wanted: a and T.

16
Considering M we have Mg – T = Ma (1), while
considering m we have T – mg = ma (2)
Add (1) and (2):         (M – m)g = (M + m) a
M −m
Solving for a we obtain:        a=         g . (Obviously a = 0 for m = M)
M +m
[Note: You could rearrange the above equation to make g the subject. Then, by
measuring a in an experiment, you could determine g.]
Solving (1) for T we get T = M(g - a), while solving (2) we get T = m(g + a).
Mm
Substituting a into either expression we find T = 2        g . (Check this!)
M +m

3. Single object: Block sliding down an inclined plane
N
A block (mass m = 800 g) slides down an inclined plane
(length s = 90 cm), which makes an angle θ = 30o with
the horizontal. Friction may be neglected. If the block         s            Fs
starts from rest at the top of the plane what is its velocity
(or speed) at the bottom? Take g = 10 N/kg.                              θ        θ
mg
Solution:
Given: m = 0.8 kg, s = 0.9 m, θ = 30o, µ = 0, u = 0, g = 10 N/kg (it is usually
sensible to express all data given in SI units).
Wanted: v.
In the absence of friction, the only two forces on the block are W = mg and the
normal force N (magnitude N = mg cosθ, as N compensates the normal component
of W). Thus, Fs = mg sinθ is the resultant of W and N, i.e. Fs = W + N. The
acceleration of the block is a = Fs /m = g sinθ (i.e. independent of m). Using the
fourth equation for uniformly accelerated motion, i.e. v2 = u2 + 2 a s = 2 a s = 2 g
sinθ s (since u = 0) we get
v = (2 g sinθ s)1/2. Inserting sin30o = ½ and the other values as given above, we
finally obtain v = (g s)1/2 = (10 × 0.9)1/2 m/s = 3 m/s. Note that m dropped out of
the calculation, i.e. v is independent of m. Hence, specifying its value was
redundant information.

Do not spend too much time on a particular
problem. If you get stuck it might be a good
idea to move on and come back later. If you are
fairly confident that you know how to solve a
problem refrain from going into detail and
doing the final calculations (you should
undertake the occasional spot-check, though)

17

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