LECSS Physics 12 Unit 2 Notes
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Physics 12 Unit 2: Vector Dynamics
In this unit you will extend your study of forces. In particular, we will examine force as a
vector quantity; this will involve solving problems where forces must be added or
subtracted using the methods learned in the previous unit.
2.A.1: What is a Force?
In simplest terms, a force can be thought of as a “push” or “pull”. It is important to keep
in mind is that force is a vector quantity; that is, force has magnitude and definite
direction.
2.A.2: Classifying Forces
Physicists have several schemes for classifying forces. One classification scheme divides
forces into two categories: mechanical forces and field forces. Another classification
scheme divides forces according to their fundamental source.
2.A.3: Mechanical and Field Forces
Some forces require direct physical contact between objects. Such forces are called
mechanical forces. Any force that you exert on an object to push, pull or lift it is a
mechanical force.
Field forces do not require direct physical contact between objects. The best example of
this kind of force is the force of gravity. All objects near the Earth experience a force of
gravity that pulls it toward the centre of the Earth. One model that explains how gravity
works states that surrounding the Earth is a gravitational field, and it is this gravitational
field that actually exerts the force of gravity on masses placed into the field.
There are other types of force-fields. Electrically charged objects are said to be
surrounded by and electric field. Because of this, electric charges can exert forces on
electric charges without the charges being in physical contact. Similarly, magnets are
said to have a magnetic force field around them that can exert forces without physical
contact.
2.A.4: The Fundamental Forces of Nature
In Physics 11, you examined forces that come from a variety of sources, leading one to
conclude that there are many kinds of force. However, all forces may be categorized into
one of the four fundamental forces.
2.A.4.1: The Gravitational Force
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Any object that has mass exerts a gravitational force of attraction on objects around it.
This force is due to the existence of a gravitational field that is present around all
objects. As we will see later in this course, the strength of the gravitational field
decreases as the distance from an object increases.
Near the surface of the Earth, the strength of the Earth’s gravitational field is equal to
9.80 N/kg and is directed toward the Earth’s centre. Gravitational field strength is given
the symbol g. We compute the force of gravity acting on an object of mass m placed in
this gravitational field using the vector equation:
Fg = mg
When the mass m is in kg, the unit of Fg is the Newton, N. The direction of Fg is the
same as that of g, toward the centre of the Earth. Note that the force of gravity acting on
an object is sometimes referred to as the object’s weight. This is a non-scientific term
used in the textbook, and we will avoid its use.
2.A.4.2: The Electromagnetic Force
This force is sometimes just called “the electric force”, as magnetism occurs because of
the electrical nature of matter. We will look at electricity and magnetism later in this
course.
At this point we need to examine some common electric forces that you should have
studied in Physics 11. These are the Force of Friction and the Normal Force.
A. The Normal Force, FN:
When you stand on the ground, the force of gravity pushes downward on you. As a
result, you exert a downward force on the ground beneath your feet. What other forces, if
any, are acting on you?
If no other forces were acting on you, the force of gravity on you would be unbalanced,
and you would accelerate into the ground beneath you. We know this does not happen.
The downward force you exert on the ground causes the ground to be compressed
(though this may not be noticeable). This compression causes the ground to exert a
reaction force on you (Third Law of Motion). This reaction force is upward (opposite to
the downward force that you exert on the ground), and is exactly equal in magnitude to
the amount of force you are exerting on the ground.
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Normal Force, FN: the force exerted by a surface that is undergoing compression due to
the presence of an object in contact with that surface. The magnitude of this force is
exactly equal to the amount of force pushing the object and surface together. The
direction of this force is always perpendicular to the surface (hence the term “normal”),
whether the surface is horizontal, vertical or inclined.
Do not confuse the normal force, FN, with net force, Fnet.
Many assume that the normal force exerted by a horizontal surface is equal in magnitude
to the force of gravity acting on the object exerting the compressive force. This is
sometimes true, but not always!
B. The Force of Friction, Ff:
Between all real surfaces, a friction force exists. This force is due to the forces of
electrical attraction between the adjacent atoms and molecules of the surfaces.
In Physics 11, you should have learned that friction between objects that are in motion
relative to one another produces heat energy (and reduces kinetic energy). While this is
true, friction also exists between adjacent surfaces that are not moving! We will examine
this case later.
In all cases, it seems logical that the force of friction between two surfaces will increase
if the amount of force pushing the forces together increases. The magnitude of this force
is equal to the magnitude of the normal force, FN. Mathematically:
Ff FN
This proportionality becomes an equality if we insert a constant of proportionality. This
constant is called the coefficient of friction,  Thus:
Ff = FN
This not a vector equation, as the two forces do not have the same direction!
The Two Kinds of Friction:
1. Kinetic Friction: As we know, friction exists between two touching surfaces moving
relative to one another. This kind of friction is called kinetic friction, Ffk. The
coefficient of friction that relates friction to the normal force is called the coefficient of
kinetic friction, k. Thus, Ffk = kFN.
2. Static Friction: As mentioned above, friction exists between two touching surfaces
that are not moving relative to one another.
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If you place a brick on a horizontal wooden plank that has one end elevated, the brick
may not slide down the incline. The force that holds it in place is called static friction,
Ffs. We cannot write an equation that relates Ffs to FN as we did for kinetic friction,
because the amount of static friction is not a constant value. As the end of the plank in
our example is raised higher and higher, the force of static friction gets larger and larger.
Eventually, the force of static friction reaches a maximum value, and can no longer hold
the brick in place. Because the force of static friction may be less than its maximum
value, the equation relating Ffs to FN is usually written as:
Ffs  sFN
If you have ever tried to move a large stationary object, you may have noticed that once
you get it moving, less force is required to keep it moving than to get it in motion. This is
because the coefficient of static friction between two surfaces is usually always greater
than the coefficient of kinetic friction!
The Direction of the Force of Friction:
For objects in motion, the direction of the force of kinetic friction is always directly
opposite the direction of motion.
The direction of the force of static friction is said to be in the direction of impending
motion. For example, if you exert a force to try to push a stationary object to the left on a
rough surface, but it does not move, it is because the force of static friction is directed to
the right – exactly opposite the direction that the object would move if no friction acted
on it.
2.A.4.3: The Strong Nuclear Force
In most atomic nuclei, there is a strong electrical force of repulsion between the protons.
Balancing this force is the strong nuclear force. This force has a very short range, so
when the protons in a nucleus are set into vibration, the distance between them make
become so large that they are no longer held together. A nuclear explosion is the result.
2.A.4.4: The Weak Nuclear Force: This force is involved in nuclear transmutations
(sometimes called nuclear decay).
2.B: The Effect of Forces: Newton’s Laws
The effect that forces have on objects is determined by whether or not the forces acting
on an object are balanced or unbalanced. The magnitude and direction of the net force
acting on the object determine whether the forces acting on an object are balanced or
unbalanced.
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2.B.1: The Meaning of Net Force, Fnet
The net force on an object is equal to the vector sum of all the forces acting on the
object. If a number of forces act on an object, Fnet = F1 + F2 + F3 + .... We can represent
this vector sum as follows:
Fnet = F
Where the symbol  literally means “sum of”.
2.B.2: Motion with Balanced Forces: Newton’s First Law of Motion
When Fnet = 0, then we have a very special case. When all the forces on an object are
balanced, the situation is the same as is there were no forces acting on the object at all.
Isaac Newton described this situation this way:
Newton’s First Law of Motion: Every body continues in its state of rest or its state of
motion with a uniform speed in a straight line unless it is compelled to change that state
by an unbalanced force impressed upon it.
In Newton’s description, it is seen that every object has a tendency to maintain its state of
rest or of uniform motion. Newton called this natural tendency Inertia. Because of this,
Newton’s First Law is often called The Law of Inertia.
We see the effect of inertia all the time. If we are traveling with a high speed in a car that
comes to a sudden stop, we feel as though we are being thrown forward. In fact, no force
is throwing you forward. Rather, your inertia compels you to remain in your previous
state of motion in a straight line.
There is no equation to compute the amount of inertia possessed by an object, as it is
actually a non-scientific quantity. Newton claimed that the greater an object’s mass, the
greater its inertia.
Newton’s First Law as stated above is actually somewhat inaccurate. To state that
something is in a “state of rest” has no true meaning. As we learned in unit 1, all
velocities are measured relative to some frame of reference that is assumed to be at rest.
There is no “true” state of rest.
In summary, Newton’s First Law can be stated mathematically as:
When Fnet = 0, v = a constant.
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2.B.3: Fnet  0 , Newton’s Second Law of Motion
From the above, it is easy to conclude that if the net force on an object is not zero, its
velocity is not constant. This leads us assume that in this case, the object will accelerate.
Newton’s Second Law of Motion: When a quantity of unbalanced (or “net”) force acts
upon an object, it will accelerate in the direction of the unbalanced force at a rate that is
directly proportional to the quantity of unbalanced force and inversely proportional to the
object’s mass. Mathematically,
F
a  net
m
When modern metric units for mass (kg), acceleration (m/s2), and force (N) are used, the
above relationship becomes an equality:
a
Fnet
or Fnet  ma
m
This is a vector equation that implicitly tells us that acceleration is in the direction of the
net force.
The reason that the relationship between a, Fnet, and m becomes an equality when metric
units are used is the direct result of how the metric unit of force is defined. 1 N of
unbalanced force is defined to be the quantity of force required to accelerate a 1 kg mass
at a rate of 1 m/s2. In fact the units of m/s2 and N/kg are exactly identical. This explains
why the gravitational field strength near the Earth’s surface is numerically equal to the
acceleration due to gravity near the Earth’s surface.
At this point you should study Unit 2 Sample Problems 1-5. Ensure you understand
the solutions thoroughly.
2.B.4: Newton’s Third Law: The Law of Action and Reaction
When an object has a force exerted on it, that force must be applied by another object.
This tells us that the exertion of forces involves an interaction between two or more
objects. Newton’s Third Law describes this interaction.
Newton’s Third Law of Motion: During the interaction of two objects, A and B,
whenever object A exerts a force on object B, object B exerts an equal, but oppositely
directed force on object A. Mathematically, this is stated:
FAB = -FBA
In this equation, FAB is the force that A exerts on B, FBA is the force that B exerts on A,
and the negative sign tells us that the two vectors have opposite directions.
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You can think of many examples of the Third Law. When you hit the head of a nail with
a hammer, the hammer and the nail both undergo a change in their states of motion. This
does not mean that both objects need to be in motion for the third law to apply. When
you sit in a chair, gravity pushes down on you. As a result, you exert a downward force
on the chair, which, in turn, exerts an upward force on you.
Perhaps the most difficult dynamics problems to solve in Physics 11 involve the Third
Law. Typically, these problems involve Atwood machines or a system of two or more
objects accelerating together on a surface due to the application of an unbalanced force.
At this point you should examine Sample Problems 6-9. Ensure you understand the
solutions thoroughly.
Assignment 1:
Readings: Giancoli Second Edition, read Sections 3-1 to 3-8.
Questions and Problems: Giancoli Second Edition pages 66-68, problems 1, 2, 3, 8, 9, 15
and 16.
Hand-in Assignment: Complete and submit Dynamics Assignment 1.
2.C: Solving Complex Vector Force Problems
To be successful at solving dynamics problems, there are some very important things to
keep in mind.
1. ALWAYS draw a “free-body diagram” showing all the forces involved.
2. Be aware of problems where all the forces are balanced. The clue to these is that the
velocity of a moving object is constant.
3. In certain situations, it will be necessary to resolve all the forces into orthogonal
components. The directions of the components will be determined by the situation as
described below.
2.C.1: Motion on a Horizontal Surface
When an object is moving along a horizontal surface, it is usually necessary to resolve all
forces into horizontal (x) and vertical
(y) components. This is necessary
because there can be no unbalanced
force acting vertically. This is
evidenced by the fact that an object on
a horizontal surface may only
accelerate in the x-direction.
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The diagram to the left shows a possible situation. An object is
being pulled along a horizontal surface by an applied force F
which is angled upward at some angle  to the horizontal.
Before the forces can be analyzed, F must be resolved into its x
and y components, as shown in the diagram below.
In this situation, the magnitudes of the components of F are
given by:
Fx  F cos
Fy  F sin 
When F has been resolved, the vector diagram is
changed as shown in the diagram to the right.
Since there can be no net vertical force, Fg must be
balanced by Fy and FN. The net force on the object (if
any) will be the resultant of Ff and Fx. Notice in this
example, the magnitude of FN is not equal to the
magnitude of Fg.
Assignment 2:
Problems: Giancoli pages 66-68, Problems 21-24, 31, 45.
2.C.2: Forces on a Simple Pendulum
When a pendulum is first released from
rest, it accelerates in the direction
perpendicular to the string. It is usual to
resolve all the forces acting on the
pendulum bob into components that are
perpendicular to the string (x) and parallel
to the string (y). In this case, all
components in the y-direction must
balance.
In the case of a simple pendulum (one in
which the only forces acting on the bob
are the force of gravity, Fg, and the force
of tension, FT), the only force that needs
to be resolved is Fg. The magnitudes of
its two components are given by:
Fgx  Fg sin 
Fgy  Fg cos
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Notice in this case that the magnitude of the xcomponent is not along the adjacent side of the known
angle as it has been previously, and thus its magnitude
is determined using the sine ratio and not the cosine
ratio.
Since the forces in the y-direction must balance, FT
balances Fgy. Fgx is unbalanced and will accelerate the
bob at right angles to the string.
In cases other than that of a simple pendulum, it may
be necessary to consider other forces acting on the bob,
including the force of friction.
2.C.3: Motion on an Incline
Being able to solve force problems
involving objects on an inclined plane is a
very important learning outcome in Physics
12. Since an object on an incline can only
accelerate in the direction parallel to the
incline, any unbalanced force must act in
this direction. For this reason, all forces
acting on an object on an incline are
resolved into components which are
parallel to the incline (x) and perpendicular
to the incline (y). Like the cases above, all
the y-components of the forces must
balance. Any unbalanced force will be in
the x-direction.
The diagram above shows a possible case. The object is acted on by the force of gravity,
Fg, the normal force, FN, exerted by the surface, a friction force Ff, and an applied
horizontal force, F. Note the direction of the friction force. Clearly, its direction is the
result of the fact that the object is moving up the incline. If the object were sliding down
the incline, Ff, would act up the incline.
Analysis of the forces in this case
requires that we resolve Fg and F into
its x and y components. The
magnitudes of the components are
given by:
Fx  F cos
Fgx  Fg sin 
Fy  F sin 
Fgy  Fg cos
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Where  is the angle that the incline makes with the horizontal. Again, notice the
differences in the trigonometric ratios used to determine the magnitudes of the
components.
Since any unbalanced force must be
directed parallel to the incline, FN must
balance Fy and Fgy. Any unbalanced force
will be given by the resultant of the vector
sum of Fx, Ff, and Fgx.
During the review of Newton’s Third Law,
you were presented with problems
involving Atwood’s Machines and
modified Atwood’s machines. These are
the problems that have two masses
connected by a light string over a pulley.
In this course, it is common to complicate Atwood’s Machine problems even further by
having one or both of the two objects accelerating up or down an incline. These types of
problems are, perhaps, some of the most challenging in this unit.
At this point, you should study Sample Problems 10-17. Ensure you understand the
solutions thoroughly.
Assignment 3:
Readings: Giancoli Second Edition, Section 3-9.
Questions and Problems: Giancoli Second Edition pages 66-68, Problems 28, 37, 40 and
49.
Hand-in Assignment: Complete Dynamics Assignment 2 and the Unit 2 Review
Assignment and submit your solutions for marking.
It is now time to write the Unit 2 Test.