Download From our equations of motion for constant acceleration we have

CHAPTER 4
FORCES AND MOTION
Mechanics
It is now time to take a closer look at the motion of objects and what makes them
move the way they do. As part of our daily experiences we all know that to make an
object move in a particular way you must apply the appropriate force i.e. a push or
pull in the direction you wish the object to go. The harder you push or pull the faster
it moves. These simple observations can be explained by Newton’s laws of motion
which form the foundation of the study known as mechanics. They allow us to
predict the motions of the planets, the path of a projectile and many more everyday
events.
Newton’s first law of motion
Before Newton formulated his laws of motion most people held the Aristotelian view
that an object remains at rest unless a force makes it move. In other words the
‘natural’ state of an object was that of rest, not motion. This seems reasonable when
we consider some everyday examples. If the engine of a car doesn’t provide a force
the car soon comes to a stop. If you push a book across the table it too will come to a
stop. So it seems if you want to keep an object moving you need to keep applying a
force to it.
But let us perform a few more careful experiments. What if we push the book over a
highly polished table? We find that it travels further. Now try it over a skating rink
or an air track. It travels even further. We now imagine that by using even smoother
surfaces the book will travel further and further. In the limit if we have an extremely
smooth (frictionless) surface the book will continue moving without slowing down.
We now conclude that you do not need a force to keep an object moving. We see
that it is the presence of other forces (friction, air resistance etc.) that slows them
down. This leads us to a formulation of Newton’s first law of motion
NEWTON’S FIRST LAW: Consider an object on which no
net force acts. If the body is at rest, it will remain at rest. If
the body is moving with constant velocity it will continue to
do so.
Note that forces may still act on the object but the net (resultant) must be zero.
Newton’s first law is sometimes called the law of inertia and it holds only in a special
set of reference frames. A reference frame is quite simply a set of co-ordinate axes
we use to make our measurements and observations. For example we can measure
position relative to the Earth’s surface. We say the Earth is our reference frame. If
we are moving along in a train and making measurements we can use the train as our
reference frame. If the train is moving at a constant velocity we will find that
Newton’s first law of motion is true. It is an inertial reference frame. If we tried to
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do the same from a merry-go-round that was turning we would probably find that
Newton’s first law was not true. This is because objects on the merry-go-round have
a force acting on them, it keeps them moving in a circle. Strictly speaking the Earth
is not an inertial reference frame but the acceleration of objects on the Earth’s surface
is quite small.
Newton’s second law of motion
If a net force does act on an object then its effect is given by a mathematical
formulation of Newton’s second law.
NEWTON’S SECOND LAW:
F
= ma
in other words the sum of all the forces acting on the object is equal to the mass of
that object times the acceleration. Note that this is a vector equation and the
direction of the acceleration is the same as the net force. The consequences of this
law are very familiar from our everyday experiences. If we slide a book across a
table we can do it quite easily and achieve a reasonable acceleration. Try pushing on
a whole bookcase and the result is quite different. The larger the mass of the object
the smaller the acceleration.
We can now use Newton’s second law to carefully define force. Let us take the
standard kilogram as our object. We place the object on a frictionless table and pull
it along until, by trial and error, it has an acceleration of 1 m.s-2. This force is equal
to 1 kg.m.s-2 and is given a special name called the newton (abbreviated N). It is the
SI unit of force.
Force is a vector quantity and so obeys the laws of addition and subtraction for
vectors. If 2 or more forces act on a body we can find the net force by placing the tail
of one force vector on the head of the next and so on. That forces behave as vectors
is a fact that was verified by experiment. Newton’s second law is a way of
expressing this fact, it does not prove it. We can also take components of forces.
Using components we can write Newton’s second law as
F
x
= ma x ,
F
y
= ma y ,
F
z
= maz
Newton’s second law also allows us to define the mass of an object. Once we have a
constant reproducible force the acceleration that force produces for an object is
inversely proportional to the mass of that object. This means if the force produces an
acceleration of 10 m.s-2 for the standard 1 kg mass and an acceleration of 5 m.s-2 for
our test object then the mass of the test object is
F = (1 kg) (10 m.s-2) = m (5 m.s-2)
m = 2 kg
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This is also referred to as the inertial mass of an object. It is the characteristic of the
object that relates the force on the object to the resulting acceleration. Less precisely,
we can think of it as a measure of the amount of matter in the object. This is
different to the weight of an object which is a measure of the force due to gravity on
that object. Newton’s second law shows that weight and mass are related but not
equal.
In order to use Newton’s second law to solve problems we have to realise that it
applies to only one object at a time. We must find all the forces acting on that body
and then we draw them in a free-body diagram. In it, the object is represented by a
dot (make sure it is clearly visible) and each force by a vector with its tail on the dot.
We only consider external forces i.e. forces exerted on the object by other objects.
Internal forces in which parts of the object exert forces on other parts are not
important here (we will see later that they cancel out).
Let’s look at some examples of Newton’s second law and forces in action.
Q.
A student pushes a loaded sled of mass m = 240 kg for a distance d of
2.3 m over the frictionless surface of a frozen lake. In doing so he
exerts a constant horizontal force F with magnitude F = 130 N. If the
sled starts from rest, what is its final velocity?
A.
Our first step to solving any force problem is to draw a free-body
diagram. This is shown below. The sled is represented by the dot
and the force by an arrow. We also set up the relevant co-ordinate
axes. In this case it is only a 1-dimensional problem so only the xaxis is drawn. We choose the force to be along the positive x-axis (we
are free to do this since we are not told otherwise). Using Newton’s
second law of motion we can find the acceleration of the sled.
m
F
x
Considering the x-component of F and the acceleration along the x-axis we
obtain from Newton’s second law (written in component form)
Fx = max
F
130 N
ax = x =
m
240 kg
= 0.542 m.s-2
From our equations of motion for constant acceleration we have
v2 = v 02 + 2a(x - x0)
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Since the sled starts from rest v0 = 0 and we identify that x - x0 = d. Our
equation simplifies to
v =
2ad
(2)(0.542 m.s-2 )(2.3 m)
=
= 1.6 m.s-1
Q.
In a 2-dimensional tug-of-war, Alex, Betty and Charles pull on an
automobile tyre as shown in the diagram below. Alex pulls with a
force FA (220 N) and Charles with a force FC (170 N). With what
force must FB does Betty pull if the tyre remains stationary?
y
Alex
Charles
FA
FC
47 o
137

x
o
Betty
FB
A.
In this example we do need to use 2-dimensions to deal with the
forces. Again we draw our free-body diagram as indicated above.
We are free to choose any orientation for the xy co-ordinate system
with respect to the actual forces. It is usually a good idea to align at
least one force with one of the axes. Now we can apply Newton’s
second law. Since the tyre is stationary the acceleration is zero so we
can write
F
= FA + FB + FC = 0
You will find that the component form of this vector equation is more useful
when solving problems. In terms of the x and y components we have
F
= FC cos FA cos 47.0 o = 0
F
= FC sin FA sin 47.0 o FB = 0
x
and
y
Note the use of minus signs to indicate a component is in the direction
opposite to the positive x or y axes. Substituting values into the first
equation we obtain
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(170 N)( cos  = (220 N) (cos 47.0 o)
(220 N) (0.682)
cos-1
= 28.0 o
170 N
Substituting into the other equation we obtain
FB = FC sin FA sin 47.0 o
= (170 N) (sin 28.0 o) + (220 N)(sin 47.0 o)
= 241 N
Some particular forces
There are many forces around us but there are some particular types that are very
important and very common. We will now take a closer look at these.
Weight
The weight of an object is the force on that object due to the local gravitational field.
If that object is near the Earth then the weight is due to the Earth’s gravitational field.
From Newton’s second law we can write the weight of an object near the Earth’s
surface as
W = mg = m (9.8 m.s-2)
Since weight is a force it has a direction. The direction is down towards the Earth’s
centre. On the moon the same object would weigh less. The magnitude of the
gravitational field on the moon is not as large as the Earth’s. The mass of the object,
however, will be the same. It is a property of the object. In our everyday speech we
tend to interchange the word weight and mass but in physics they have completely
different meanings.
Normal force
Consider a book resting flat on a table. There is a weight force acting on the book.
This, however, cannot be the only force since the net force must be zero, the book is
stationary. There is a force exerted by the table on the book. This force is normal
(perpendicular) to the surface of the table and is called the normal force (usually
indicated by N). This is shown in the diagram below.
y
Normal force N
N
x
W
Weight W
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We must remember that the normal force is always perpendicular to the surface and
that the normal force is not always equal to the weight force.
Friction
If we try to slide the book over the table (or any surface) we find that the motion is
resisted by a force. This force is called friction (usually indicated by f). It is parallel
to the surface and opposite in direction to the intended motion. Friction is due to
attraction (bonding) between the surface and the object. Sometimes we simply a
problem by assuming there is no (very little) friction. Such a surface is called
frictionless.
Tension
When an object is being pulled along by a rope the rope is under tension. There is a
force exerted by the rope on the object. This force is called tension (usually
indicated by T). The direction of the tension is along the rope, away from the object
and toward the centre of the rope. This is indicated in the figure below.
T
T
T
T
The rope is usually assumed to be massless. This means its mass is much less than
the mass of the object. Any pulleys are also assumed to be massless and frictionless.
This means its mass is also much smaller than the object and that the friction on the
pulley axle is very small.
Fundamental forces
All the forces we have discussed so far and all forces so far known to physicists are
manifestations of the four fundamental forces of nature. They are
 gravitational
 electromagnetic
 weak nuclear
 strong nuclear
Most everyday forces (such as friction, tension, normal force) are due to
electromagnetic forces.
Newton’s third law of motion
The most familiar of Newton’s laws of motion is probably his third and final law of
motion. We are all aware that when we push on an object it pushes back. For
instance to get out of a chair you may push down on that chair (arm rests or
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somewhere) and it pushes back on you. Newton’s third law talks about the pair of
forces acting between 2 bodies.
NEWTON’S THIRD LAW: If a body A exerts a force FBA
on a body B then that body exerts and equal and opposite
force FAB on body A.
Since force is a vector and has direction opposite means in the opposite direction. So
the two forces have the same magnitude but are in the opposite direction.
Mathematically we can write Newton’s third law as
FAB = FBA
Here the notation FAB means the force exerted on body A by body B. Sometimes you
will hear people talk about action and reaction forces and they state Newton’s third
law as “the reaction force is equal and opposite to the action force”. There is nothing
wrong with this but it is not always clear which force is the action and which is the
reaction.
You may now think that if the two forces are equal and opposite they should cancel
out and the net force will always be zero. If this is so then nothing will accelerate.
The point to keep in mind here that the 2 forces are acting on different objects. In
this case the forces do not cancel out.
Let us take a closer look at a few examples of action-reaction pairs. Imagine a
satellite orbiting around the Earth. The only force acting on the satellite is FSE the
force exerted on the satellite by the gravitational attraction of the Earth. By
Newton’s third law the satellite should exert an equal and opposite force on the
Earth. This is the gravitational pull of the satellite on the Earth. Gravity is the
mutual attraction of 2 objects. The force is equal in magnitude to that on the satellite
but because the mass of the Earth is so large Newton’s second law tells us that the
acceleration will be very small.
FSE
FES
Now consider a book resting on a table. We can draw the forces acting on the book.
There is the gravitational pull of the Earth on the book FBE(weight). Since the book
is not accelerating this force is cancelled by the normal force of the table on the book
FBT. However, these 2 forces do not form an action-reaction pair. They are equal
and opposite but do not act on the same pair of objects. The ‘reaction’ to FBE is the
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gravitational pull on the Earth due to the satellite FEB. The ‘reaction’ of the normal
force due to the table is FTB, the force of the book on the table.
FBT (normal from table)
FBE
FBT
FEB
FTB
FBE (weight of book)
Friction
We are all familiar with the effects of friction. They are unavoidable in our everyday
lives. However, the effects are not always to our disadvantage. Without friction we
would not be able to walk, or write and objects would slide around. Friction provides
the force that makes all these possible. There are different types of friction but in our
case we will consider the friction between 2 dry solid surfaces (usually an object and
some surface).
Let us begin our investigation of friction with a few simple experiments. First we
slide a book over the surface of a table. We observe that the book gradually slows
down and eventually comes to rest. This is the first property of friction, it opposes
motion. If we want to keep the book moving at a constant velocity we have to push it
with a force that is equal and opposite to the frictional force.
In our second experiment we gently push on the book. Eventhough we exert a force
on the book it does not move. As we push even harder the book still does not move.
The frictional force seems to adjust so as to be equal but opposite to the force we
exert. What if we push really hard? The book does begin to move. At first it
accelerates but as we reduce our push it moves with a constant velocity. So there is a
maximum force that friction can exert but once the object moves then the frictional
force is less.
N
N
fs
F
W
Book resting on table
a
N
W
W
Frictional force equal
but opposite to push F
fk
F
Frictional force equal to
maximum value
v
N
fs
F
N
fk
F
W
W
Book accelerates initially as
F is greater than fk
F is reduced until it is
equal to fk
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By performing such experiments carefully physicists have been able to formulate 2
expressions to account for the above behaviour. In the case where the object is not
moving we call such friction the static frictional force and denote it by fs. The
maximum value (magnitude) of this force is given by
fs,max = sN
where s is the static coefficient of friction (it has no units, it is dimensionless) and
N is the normal force acting on the object. If the external force that is pushing is less
than this maximum frictional force then the frictional force is equal to the external
force i.e.
fs  sN
Once the object is moving then the frictional force decreases. It is called the kinetic
frictional force and is denoted by fk . The magnitude of this force is given by
fk = kN
where k is the coefficient of kinetic friction.
Note that neither of the expressions for the frictional force say anything about
direction. They are scalar expressions. The directions of fs and fk are always parallel
to the surface and tend to oppose any other forces parallel to the surface or oppose
any motion. These expressions are empirical (based on experimental data) and not
fundamental equations like Newton’s second and third laws. Under some situations
our expressions for frictional forces may not be accurate, at high speed for instance.
They serve as approximations to what actually happens.
The coefficients of friction depend on the nature of the 2 surfaces in contact. We
usually state these as the coefficient of static friction between a rubber tyre and
asphalt. Tables of such useful values exist. Friction is basically due to the attractive
forces between the 2 surfaces.
Q
A 100 kg chest standing on a floor has a coefficient of static friction
of 0.6. A boy approaches the chest and leans horizontally against it.
(a) What is the frictional force exerted by the floor on the chest before
the boy comes into contact with it?
(b) What is the maximum force the boy can exert before the chest
begins to slip?
(c) What is the frictional force exerted by the floor when the exerts a
force of 100 N on the chest?
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A
Drawing a free-body diagram we have
N
W
(a) Initially the only forces acting on the chest are the weight and
normal force. If there was friction too it would be the only force in the
horizontal direction. This would result in an acceleration in the
horizontal direction. However, the chest is at rest so there is no
friction.
(b) Now we add the push of the boy and the frictional force in the free
body diagram
Taking components in the horizontal direction we obtain
F
x
N
= fs  F = 0
fs
F
and in the vertical direction we have
W
F
y
= N  W = 0
From the first equation we obtain
fs  F
so F is a maximum when fs is a maximum i.e. F = fs,max
From the last equation we have N = W.
Our expression for the static frictional force is
fs,max = sN
So we obtain
F = fs,max = sW
substituting we obtain
F = 0.6 x (100 kg) (9.8 m.s-2)
= 588 N
(c) If the boy exerts a force of 100 N this is less than the maximum
frictional force (588 N) so the block remains at rest and the frictional
force is equal and opposite to the force the boy exerts.
fs = 100 N
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Q
A block is at rest on an inclined plane as shown below. The
coefficient of static friction between the block and incline is s. What
is the maximum possible angle of inclination max of the surface for
which the block will not slip?
N
fs


W
A
In this problem we choose our co-ordinate axes to be parallel and
perpendicular to the inclined surface. This will make the mathematics
easier, the physics is always going to be the same.
Taking components parallel to the surface and applying Newton’s
second law
fs  W sin  = 0
(taking up the incline as positive)
taking components perpendicular to the incline we obtain
N  W cos  = 0
(taking upwards as positive)
The block begins to slip when fs = fs,max . Our expression for fs,max is
fs,max = sN
Substituting for fs and N we obtain
fs,max = W sin s W cos 
rearranging we obtain
sin 
s =
= tan 
cos 
So max = tan-1 s
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