access to transport is a feature of today’s society. Most people access some
form of transport for travel to and from school or work and for leisure outings
at weekends or on holidays. When describing journeys that they may have taken in
buses or trains, they usually do so in terms of time or their starting point and
their destination. When describing trips they may have taken in planes or cars,
they normally use the time it takes, distance covered or the speed of the
vehicle as their reference points. While distance, time and speed are fundamental to
the understanding of kinematics and dynamics, very few people consider a trip in
terms of energy, force or the momentum associated with the vehicle, even at low or
The faster a vehicle is travelling, the further it will go
before it is able to stop. Major damage can be done to other vehicles and to the
human body in collisions, even at low speeds. This is because during a collision some or all
of the vehicle’s kinetic energy is dissipated through the vehicle and the
object with which it collides. Further, the materials from which vehicles are
constructed do not deform or bend as easily as the human body. Technological
advances and systematic study of vehicle crashes have increased understanding of
the interactions involved, the potential resultant damage and possible ways of
reducing the effects of collisions. There are many safety devices now installed
in or on vehicles, including seat belts and air bags. Modern road design takes
into account ways in which vehicles can be forced to reduce their speed.
The branch of Physics that deals with phenomena such as
those mentioned above is called mechanics.
Note: Some internet browsers (eg Firefox) do not
accurately display text symbols such as Greek letters used to represent
quantities in Physics. For example, capital delta is displayed as D in
Firefox. So, if you see below something like Ds/Dt for speed or velocity,
it is meant to appear as delta s/delta t. This is just something to be
aware of in case you do come across such issues. The square root sign is
another one not displayed properly by some browsers. Any symbols used in
equations produced by equation editors will of course display properly.
The branch of Physics that is concerned with the motion and
equilibrium of bodies in a particular frame of reference is called “mechanics”.
Mechanics can be divided into three branches: (i) Statics –
which deals with bodies at rest relative to some given frame of reference, with
the forces between them and with the equilibrium of the system; (ii) Kinematics
- the description of the motion of bodies without reference to mass or force;
and (iii) Dynamics – which deals with forces that change or produce the
motions of bodies.
Some common terms used in the study of mechanics (and
indeed many other branches of Physics) are: scalars, vectors and SI Units.
Scalars – A scalar is a physical quantity defined in terms of magnitude
(size) only - eg temperature,
mass, volume, density, distance.
Vectors – A vector is a physical quantity defined in terms of both
magnitude and direction - eg force, velocity, acceleration, electric field
Diagramatically we can represent a vector by a straight line with an
arrow on one end. The length of the
line represents the magnitude of the vector quantity and the direction in which
the arrow is pointing represents the direction of the vector quantity.
We will say much more about vectors later in this topic.
Systeme International (SI) Units – The internationally agreed
system of units. There are seven
fundamental units. The three that
we will use in this topic are the metre (length), the kilogram (mass) and the
second (time). Various prefixes
are used to help express the size of quantities eg a nanometre (1 nm) = 10-9
of a metre, a gigametre (1 Gm) = 109 metres. See Appendix C,
p.243 of Excel Preliminary Physics by Warren for a list of some common SI units
and prefixes. Most texts will contain such information.
Since describing the features of something is usually a lot
simpler than explaining how or why it works, we will start with a look at
The following terms are commonly used to describe motion.
Displacement – is the distance of a body from a given point in a
given direction. It is a vector
quantity. The SI unit of
displacement is the metre (m).
Speed – The speed of a body is the rate at which it is covering
distance. It is a scalar quantity.
The SI units are m/s, which can also be written as ms-1.
= average speed, d = total distance travelled and t = total time taken to travel
Velocity – The velocity of a body is its speed in a given
direction. In other words, velocity
is the rate of change of displacement with time.
It is a vector quantity with the same SI units as speed.
where vav =
average velocity, Dr = change in
displacement and Dt = change in
time taken to achieve that change in displacement.
Another way to express average velocity is as the average of the initial and
= average velocity, u = initial velocity of the body and v = final velocity of
the body. Note that this
equation applies ONLY when the velocity of the body is increasing or decreasing
at a constant rate.
Acceleration – The acceleration of a body is the rate of change of
the velocity of the body with time. It is a vector quantity, with units of
(metres/second)/second, written as ms-2.
where aav = average acceleration, Dv
= change in velocity of the body and Dt
= change in time over which the change in velocity took place.
Alternatively, we may write:
where aav = average acceleration,
v = final velocity of the body, u =
initial velocity of the body, and t = time over which the change in velocity
Note that a body accelerates when:
It speeds up;
It slows down;
It changes direction.
Often the most effective way to describe the motion of a
body is to graph it. Note that in this section the variable “s”
will be used to represent displacement instead of “r”. You
will find in Physics that these two variables are both in common use to denote
These may be used to gain information about the
displacement of an object at various times or about the velocity of the object
at various times.
Clearly, the gradient (slope) of a displacement-time graph
gives the velocity.
Note that a positive gradient implies a positive velocity
and a negative gradient implies a negative velocity.
For a curved displacement-time graph, the gradient of the tangent to the
curve at a particular point equals the gradient of the curve at that point,
which in turn equals the velocity of the object at that particular time.
Such a velocity, that is, the velocity at a particular instant in
time, is called the instantaneous velocity.
An example of an instrument that measures instantaneous
velocity is the speedometer in a car. In
older cars the speedometer was linked mechanically to the
transmission. These days, however, a device located in the transmission produces
a series of electrical pulses whose frequency varies in proportion to the
vehicle's speed. The electrical
pulses are sent to a calibrated device that translates the pulses into the speed
of the car. This information is sent to a device that displays the vehicle's
speed to the driver in the form of a deflected speedometer needle or a digital
Note that a straight line displacement-time
graph implies that velocity is constant. A
curved line displacement-time graph implies that velocity is changing with time
(ie the object is accelerating).
These may be used to gain information about the
displacement, velocity and acceleration of an object at various times.
The gradient is clearly the acceleration of the object.
Gradient = Dv/Dt = acceleration
Note that a positive gradient implies a positive
acceleration and a negative gradient implies a negative acceleration.
Also, the area under the graph,
in the case above, has units of: seconds x metres per
second = metres. Thus, the area
under a velocity-time graph is equal to the displacement travelled by the object
in the time Dt.
Note that a horizontal straight line velocity-time graph
implies that acceleration is zero – ie velocity remains constant.
A non-horizontal, straight line velocity-time graph implies
that acceleration is constant and non-zero.
A curved line velocity-time graph implies that acceleration
These may be used to gain information about the velocity
and acceleration of an object at various times.
The area under an acceleration-time graph gives the
velocity of an object during the time interval Dt. Check the
units of the area: (ms-2 x s = ms-1).
A horizontal straight line acceleration-time graph implies
that velocity is varying at a constant rate (ie velocity is increasing or
decreasing by the same amount each second).
That is, acceleration is constant.
Often it is necessary to compare the velocity of one object
to that of another. For instance,
two racing car drivers, A and B, may be travelling north at 150 km/h and 160
km/h respectively. We could say
that the velocity of car B relative to car A is 10 km/h north. In other words, driver A would see driver B pull away from
her with a velocity of 10 km/h north.
Likewise, two jet aircraft, C and D, flying directly at
each other in opposite directions (hopefully as part of an aerobatics display)
may have velocities of 900 km/h north and 1000 km/h south respectively.
We could say that the velocity of D relative to C is 1900 km/h south.
In other words, jet C will observe jet D flying towards it at a speed of
Clearly, when the objects are travelling in the same
direction, the velocity of one relative to the other is the difference between
their speeds, taking due care to state the appropriate direction.
When the objects are travelling in opposite directions the velocity of
one relative to the other is the sum of their speeds, again taking due care to
state the correct direction. There
is a vector equation which can be used to calculate the relative
velocities of objects, even when the objects travel at various angles to one
another but this equation is outside the scope of the present syllabus (for some
"Flight of Fantasy"
What is Force?
In simple terms a force can be defined as a push or a pull.
We experience examples of forces every day.
If we push a stationary lawn mower (with enough force) it begins to move
– that is, it accelerates and its velocity increases.
If we push on the brakes of a moving bicycle, it slows down – that is
it decelerates (or undergoes negative acceleration).
If we apply sufficient force to an aluminum can, by squeezing it with
our hand, we can change the shape of the can.
So a force can cause a change in the state of motion of
an object or a change in the shape of an object.
In fact, all accelerations (and decelerations) are caused by forces.
Does every force cause acceleration?
Again, from our everyday experience, we know the answer to
this question is “no”. If a person pushes on the brick wall of a house, the
house does not accelerate. Sometimes
when we want to push or pull an object from one place to another we find that no
matter how hard we push or pull, we just cannot move (accelerate) the object.
In previous Science courses, a qualitative
relationship was established between force, mass and acceleration.
In the Preliminary Physics Course we need to establish a quantitative
relationship between these three quantities.
What is the relationship between force and
We could perform an experiment to determine the
relationship between the size of a force applied to an object at rest on a
laboratory bench and the change in velocity experienced by the object over a set
period of time (ie the acceleration). Such
an experiment would produce results as shown below.
The graph above shows that:
The change in velocity does not happen instantaneously. A certain
amount of force is required before the object begins to accelerate. This makes sense, since the force of friction between the
bench and the object must be overcome before the object can move.
So, we can say that a net external force is required in order to
change the velocity of an object.
The acceleration produced is directly proportional to the force
applied. If we repeated the
experiment on a frictionless surface (eg using a dry-ice puck on a very smooth,
polished table top) the straight-line graph would even pass through the origin.
What is the relationship between acceleration and
We could measure the accelerations produced when the same
sized force is applied to different objects.
Such an experiment would produce results like those below.
The graph above suggests that there is an inverse
relationship between acceleration and mass.
A plot of acceleration versus the reciprocal of mass, using the same
data, would produce a graph similar to that below.
This graph clearly shows that:
The acceleration produced by a given force is inversely
proportional to the mass of the object.
LAWS OF MOTION
Newton’s First and Second
Newton’s Third Law:
of Mass and Weight:
OF VECTOR DIAGRAMS:
Many of the quantities with which we deal in Physics
are vectors. Sometimes
we need to add a number of vectors together.
For instance, we may be trying to calculate the total or resultant force
acting on a car when several forces act on the car simultaneously – the wind,
friction, gravity and the force supplied by the engine.
Sometimes we need to subtract two vectors. For instance, we may be trying to calculate the change in
velocity of a car as it goes around a bend in the road.
The change in velocity of the car equals the final velocity of the car
minus initial velocity of the car.
When the need arises to add or subtract vector
quantities, this proves to be easy only when the vector quantities act along the
same straight line. If the vectors
act at an angle to each other we really need to draw a vector diagram to assist
in solving the problem.
Vector analysis is an extremely
important aspect of Physics and there are several different methods available to
add, subtract and even multiply vectors. Unfortunately,
the current Syllabus requires that you have only a very basic understanding of
vector analysis. So, we will
examine a single, simple but very useful method of adding and subtracting
The method we will use is called the “Vector
Polygon” method. To find the
sum of a number of vectors draw each vector in the sum, one at a time, in the
appropriate direction, placing the tail of the second vector so that it just
touches the head of the first. Continue
in this fashion until all of the vectors in the sum have been included in the
diagram. Note that it does not
matter which vector you start with.
The vector that closes the vector polygon in the same sense
as the component vectors is called the equilibrant.
It is the vector which when drawn into the diagram gets you back to where
you started. The vector that
closes the vector polygon in the opposite sense to the component vectors is
called the resultant.
The resultant is the answer to the sum of all the vectors.
Its size can be calculated mathematically or measured using a ruler if
the vector polygon has been drawn to scale.
The direction of the resultant can be calculated mathematically or can be
measured using a protractor if the vector polygon has been drawn to scale.
Either way, the direction of the resultant must be stated in an
Sometimes in Physics our vector additions only involve two
vectors at a time. In this case,
the polygon formed is a triangle, making the mathematical calculation of the
magnitude and direction of the resultant quite straight forward.
So, keep your wits about you and bring your
knowledge of triangle geometry and trigonometry into the Physics classroom.
1: A fighter pilot flies her F-14D Tomcat jet with a true
airspeed of 400 km/h
crosswind from the East blows at 300 km/h relative to the ground.
Calculate the jet’s resultant velocity relative to the ground.
For aircraft, the true airspeed (TAS) is the actual speed of the aircraft
through the air (the speed of the aircraft relative to the air). The wind speed
is usually measured relative to the ground. Groundspeed is the speed of the
aircraft relative to the ground. The groundspeed of the aircraft is the
vector sum of the true airspeed and the wind speed.
Obviously, a vector diagram would be very useful in solving
Example Problem 1. See the diagram
By Pythagoras’ Theorem, the magnitude of
the resultant velocity of the jet is:
and the direction can be found using basic trigonometry
So, the velocity of the jet relative to the ground is
500 km/h N36.9oW.
Note that if the angle between the two vectors being added
together is other than 90o, Cosine Rule and Sine Rule
can be used to solve the problem mathematically.
Note also the use of the compass in the diagram to establish
2: In the previous problem, in which direction should the pilot
head and with what airspeed in order to actually fly north at 400 km/h relative to
Again a vector diagram is useful.
Our intuition tells us that the pilot must fly into the wind.
So, when we draw a diagram that shows all of the information that we know
to be true, we obtain the diagram shown below.
Clearly, if the pilot flies N36.9oE with an
airspeed of 500 km/h, the wind will bring her back to a heading of due North at a
speed of 400 km/h relative to the ground. Remember
also, there is usually more than one way to give the direction. The direction the pilot should fly in this example could just
as correctly be given as E53.1oN or as a True Bearing of 36.9o.
3: Four children pull on a small tree stump firmly stuck in
the ground. Looking down on the
tree stump from above, the forces applied by the children are as shown below.
Determine the resultant force applied to the tree stump.
To solve this problem mathematically we would need to add
two of the vectors together, then add our answer to the third vector and finally
add our answer to that addition to the fourth vector.
It is actually far quicker and easier to solve this problem graphically.
To do this we construct a vector polygon, using the rules stated above
and simply measure the size and direction of the resultant.
See the diagram below.
The resultant force, R, is found by measurement to be 3.1 N at an angle of
39o clockwise from the direction of the 4.5 N force.
Note that when using a graphical approach, the scale
must be clearly stated on the diagram.
Always choose a sensible numerical scale.
Also, choose a scale that will produce a large diagram.
The larger the diagram, the more accurate the answer.
For the example problem above, the scale used was 1 N = 1.5 cm.
This is certainly the smallest scale I would use for this particular
problem. Anything less is too
inaccurate. A scale of 1 N = 2 cm
would be preferable. The smaller
scale was used here to fit the diagram neatly onto this page.
also, that depending on which printer is used to print these notes, there may be
a small discrepancy between the stated scale and the actual scale on the page.
If vector A is as shown below:
then vector –A is a vector of the same magnitude as A
but opposite direction.
In order to find the difference between two vectors, add
the negative of the second vector to the first.
4: A Maserati (car) is moving due East at 20 ms-1.
A short time later it is moving due North at 20 ms-1.
Calculate the change in velocity of the Maserati.
To find the change in size of any quantity, you subtract
the initial size of the quantity from the final size. Obviously, with vector quantities you must do a vector
subtraction not just an arithmetic one, since vectors possess both size and
Change in velocity = final
velocity – initial velocity
This should really be written as:
Change in velocity = final
velocity + (– initial velocity)
since that is how we draw the vector diagram.
We literally add the negative of the initial velocity to the final
velocity. Study the vector diagram
below to ensure you understand the process of vector subtraction.
Using Pythagoras’ Theorem and basic trigonometry as shown
in Example 1 above, we find that the change in velocity of the Maserati is
28.3 ms-1 at an angle of 45o West of North.
Note that even though the car has the same initial and
final speeds, because the direction of the car has changed, so too has its
we are only interested in part of a vector rather than all of it.
For instance, if we push a car that has run out of petrol, we apply a
force to the car. However, if we
are not careful some of the force we apply pushes down vertically on the car and
the rest of it pushes horizontally on the car.
Obviously, we are trying to maximise the component of the force that is
applied horizontally. The angle at which we apply force to the car will determine
how much of our force is applied horizontally.
vector may be broken into two component vectors at right angles to each other.
These components are called the rectangular components of the vector.
The rectangular components of a vector add up to the original vector.
the example we used above. We may
push on the car with a force
at an angle of q
to the horizontal, as shown below.
The force F may be broken
into its vertical and horizontal components as shown below.
The magnitude of each component can then be calculated
using simple trigonometry. The size
of the vertical component of F is Fsinq.
The size of the horizontal component of F,
the one that must overcome the force of friction if we are to move the car
forward, is Fcosq.
A block of mass 20 kg is being pulled up an inclined plane by a rope inclined at
30o to the plane’s surface as shown in the diagram below.
The plane is inclined at 45o to the
horizontal. The friction force, F,
opposing the block’s motion is 10 N. Determine
the tension, T, in the rope if the net acceleration, a,
of the block up the plane is 4 ms-2.
You will observe that we have resolved two vectors in the diagram into
rectangular components – the tension, T, in the rope and
the weight, W, of the block.
The rectangular components we have chosen are those acting parallel to
and perpendicular to the inclined plane. These
components are the most useful ones in a situation like this.
Now the total force acting down the plane is the
sum of the friction force, F, and the component of the
weight force of the block acting down the plane (W
So, from the diagram we have:
= F + mg sinq
(since W = mg)
= 10 + 20 x 9.8 x sin 45o
= 148.59 N
Total force acting up the plane must be:
ma + FD
FU = (20 x 4) + 148.59
FU = 228.59 N
Note that the logic we have used to obtain an expression
for FU is
as follows. The force up the plane MUST
be sufficient to overcome the 148.59 N force down the plane and to provide the
required force of 80 N to give the block the correct acceleration up the plane.
Now we are in a position to calculate the tension in
the rope. The total force up
the plane FU
is actually the component of the tension force parallel to the plane.
This is the part of the tension force that is applied parallel to the plane.
Therefore, we have:
cos 30o =
FU / T
T = FU
T = 228.59 / cos 30o
T = 263.95 N
So, the tension in the rope works out to be 264 N.
You are now ready to try
Analysis Worksheet No.1.
You may also like to check out this
which helps you to understand vector addition & subtraction. When you
get to the site, click on the "Applet Menu" button at the
top left of the page. The menu items will appear. Select "Some
Math", "Vectors", "Vector Addition". Read the
instructions & run the applet.
Tensions In Strings (Extension
Consider a mass, m,
supported by a thin, inextensible string of negligible mass.
The two forces acting on the mass are T,
the tension in the string acting upwards and mg,
the weight force acting downwards on the mass.
The vector equation describing the net
force acting on the mass is best studied in three separate cases.
The mass is supported by the string and there is no acceleration.
The mass could be at rest or could be moving up or down at constant
velocity. In each case the net force
equation is the same. The tension
upwards is exactly balancing the weight force downwards.
Case 2: The
mass is accelerating downwards with net acceleration, a.
The tension upwards in the string is not sufficient to fully balance the
weight force downwards.
Clearly, with the mass accelerating
downwards, the tension, T = m ( g – a ).
The mass is accelerating upwards with net acceleration, a.
Here, the tension upwards in the string is doing two jobs.
It is fully balancing the weight force downwards and supplying the
required force upwards to accelerate the mass with the net acceleration of a.
Clearly, with the mass accelerating
upwards, the tension, T = m ( g + a ).
Many problems in Physics can be
solved by applying the knowledge summarized above.
Consider the following two examples.
1: Two masses X of 10kg and Y of
24kg are connected by a light inextensible string.
X and Y hang on opposite sides of a
frictionless pulley as shown above. Determine:
Assume that the
acceleration due to gravity is 9.8 ms-2.
net acceleration of the system of masses
magnitude of the tension, T, in the string.
There are two slightly different approaches possible.
determine the net force on the system, the total mass of the system and then
obtain the net acceleration from F
= ma. Once
the net acceleration of the system is known the tension in the string can be
found by realising that the tension upwards on the left side of the pulley must
balance the weight force down on the mass & supply sufficient force to
accelerate the mass upwards with the net acceleration of the system.
(a) Force of gravity on 10kg mass = 10 x
9.8 = 98N down on left side of pulley.
Force of gravity on 24kg mass = 24 x 9.8 =
235.2N down on right side of pulley.
Thus, the net force, F, applied to the
system of two masses by gravity:
= 235.2 – 98 =
137.2 N down on right side of pulley.
mass of system upon which this net force acts = 10 + 24 = 34kg.
from F = ma, the net acceleration of the system of two masses is:
a = F / m = 137.2 / 34
= 4.035 ms-2.
Note that as
you get used to using this method, it really only takes a couple of lines of
working at the most.
Once the acceleration is known the tension can be calculated from either mass.
Let’s use the 10kg mass first. Clearly,
the tension in the string on this side of the pulley must be sufficient to
balance the acceleration due to gravity down on the 10kg mass and to accelerate
the 10kg mass upwards at 4.035 ms-2.
Tension, T = (10 x 9.8) + (10 x 4.035) = 138.35 N upwards.
if we decided to use the 24kg mass instead - the tension in the string on the
right side of the pulley must be sufficient to balance the acceleration due to
gravity down on the 24kg mass less the 4.035 ms-2 that the mass is being permitted to accelerate downwards already.
Tension, T = (24 x 9.8) – (24 x 4.035) = 138.36 N upwards.
is the same (to one decimal place) as the answer we obtained using the other
mass. This must be the case.
It does not matter which mass you use to calculate the tension, you must
get the same answer in both cases because there is only one string and can
therefore be only one tension.
tension in the string is therefore 138.4 N
to one decimal place. Any
discrepancy in the above answers after the first decimal place is simply due to
rounding off the (137.2 / 34) calculation for the net acceleration in the first
stress that the whole solution (parts a & b) I have demonstrated here would
normally take no more than four to five lines of calculation.
It is the explanation I have made during the solution that has greatly
increased the space used.
The Mathematical Approach
First we write down the two vector equations of motion for the masses.
mass X: T
– mxg = mxa -
mass Y: myg
– T = mya - (2)
we solve these equations simultaneously. So,
adding equation 1 and 2 together we have:
(myg – mxg) / (mx + my
) = (24 x 9.8 – 10 x 9.8) / (10 + 24)
= 4.035 ms-2
Then from either
equation 1 or 2, we can calculate the value of T.
T = (10 x 9.8) + (10 x 4.035) = 138.35 N
second solution is probably the more mathematically pleasing to the eye.
For my liking though, the previous solution is the more physically
intuitive method. Both solutions are
equally acceptable and in the end it’s really only the setting out that
differs. Suit yourself as to which
one you use. You will find the more
mathematical approach is safer as the problems become more complex.
2: Three masses of 2kg, 4kg and 6kg
are connected by three light inextensible strings, X, Y and Z as shown below.
The masses are supported from the roof of a lift of mass 1000kg.
The lift is accelerating downwards with a net acceleration of 3 ms-2.
The tension in string X.
The tension in string Y.
The tension in string Z.
The tension in the supporting cable.
Note that since the lift is accelerating
downwards at 3 ms-2, we can write for any string
supporting a mass inside the lift or indeed for the supporting cable itself
m (g – a), see Case 2 in the notes above.
(a) Tension in X = 12 x (9.8 – 3) =
81.6 N upwards
(b) Tension in Y = 10 x (9.8 – 3) = 68 N upwards
(c) Tension in Z = 6 x (9.8 – 3) =
40.8 N upwards
(d) Tension in the supporting cable =
1012 x (9.8 – 3) = 6881.6 N
where ac is
called the centripetal ("centre-seeking") acceleration, v =
speed of the object and r = radius of the circular path.
As the name implies, centripetal acceleration is directed towards the
centre of the circle.
car of mass 900 kg moves at a constant speed of 25 m/s around a circular curve
of radius 50 m. Calculate the centripetal force acting on the car.
(11250 N, towards the centre of the circle)
OF MOMENTUM DURING COLLISIONS