Prescribed Readings
1. College Physics, Raymond A. Serway and Chris Vuille, International Edition
published by BROOKS/COLE CENGAGE Learning.
2. F.J.Bueche, Principles of Physics, 6th Edition. 1995, McGraw-Hill. ISBN 007-113854-4. and 5th Edition. 1988. ISBN 0-07-100150-6.
3. A.Beiser, Physics, 5th Edition. Addison Wesley Publishing Company. 1992.
Unit 1 :
Scalars and vectors
Physical quantities involved in the fields of science and technology are classified either
as scalars or vectors.
Scalars:
A scalar is any physical quantity which has magnitude and units but has no direction.
Thus to specify a scalar quantity completely only a number and a unit are required.
Examples are mass, distance, speed, temperature, energy, time and electrical charge.
These quantities require only a number and associated units to specify them.
Vectors:
A vector is a physical quantity which has both magnitude and direction. Hence to specify
a vector quantity completely, its direction besides a number and the unit is to be quoted.
For example the displacement of a body is its change of position in a particular direction.
Thus it is a vector quantity. On the other hand, distance is a scalar. Displacement,
velocity, acceleration, force and momentum are some examples of vector quantities.
Representation of vectors
Essentially, there are two ways of representing vectors:
(a) Using symbols. In most textbooks and modules, vector quantities are represented by
letters with an arrow at their top for example
⃗a
. In books, it is usually represented
by bold face type letters such as a. The magnitude of a vector is called its modulus
and it is indicated as mod a or just a. It gives the size of a vector.
(b) Graphical. In this method, a straight line is drawn with an arrow-head at one end.
The arrow-head indicates the direction of the vector and the length of the line is
chosen to represent the magnitude of the vector. This is done by establishing a scale
as is done for a graph or map drawing. For example, if we have to draw the vector
diagram of a 20 newton (20N) force acting towards east and another force of
30newton (30N) force directed south, we may choose a scale of 10N to represent
1cm. Thus the first force is a line of length 2cm, drawn west-east with the arrow head
at the eastern end, as
shown in figure 1.15.1(a).
Similarly, the other force is
represented by a 3 cm line drawn north-south with the arrow-head at the southern end
as in figure 1.15.1(b).
20N : 2
cm
(a)
)
30N :
(b)
Figure 1.15.1
The arrow-head is referred to as 'head' and the other end as 'tail'.
Describing vectors
Vectors come in many forms. It is therefore necessary to describe those that
occur often in physics.
Unit vector - A vector whose magnitude or modulus is unity is defined as a unit
vector.
Null vector - A vector whose modulus is zero is called a null vector.
Equal vectors:
Two vectors are said to be equal when they have the same
magnitude, units and direction.
Negative vectors. If one vector has the same magnitude and units as another vector
a say, but is directed in the opposite direction to that of a it is denoted -a, the
negative of a, figure
2
a
a
a
(ɑ))
-a
(b)
Equal vectors
negative vectors
Figure 1.16.1
Resultant vector
When two or more vectors are added, the resulting vector is called the resultant of
the added vectors. For example if vectors A and B are added to give vector C, the
latter is the resultant of A and B.
Combination of scalars and vectors
Addition and subtraction of scalars are fairly simple operations. Simple arithmetic
is all that one needs to accomplish this . For example if a 2 kg mass is added to
another 2 kg mass, the answer is a 4kg mass. On the other hand the combined
effect of two vectors is not found by simple arithmetic but by a process called
vector addition.
This process is essentially a geometrical operation as the
direction of the vectors is to be taken into account.
Addition of vectors
There are two methods that you need to know in order to correctly add two or more
vectors; the graphical and resolution methods.
(a) graphical method
This process is more easily illustrated by taking an example involving two
displacements. Starting from a point represented by 'O', an airplane flies due east
for 150 km up to point A and then changes direction and goes due north for 200
km to reach B. What is its actual resultant displacement?
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B
200 km
A
O
150 km
Figure 1.18.1
Solution
Displacement is defined as the straight line drawn from the starting point A of a
motion to the end point B of the same motion.
displacement is the vector
In figure 1.18.1, the first
⃗
OA . To accurately specify this, a scale of 1 cm:
50km is chosen. You can choose any other scale as long as it is suitable for the
vectors at hand. Using this scale, a horizontal line OA = 3 cm in length is drawn to
represent the150 km displacement. A second line AB = 4cm is drawn in the
northward direction to represent the 200km displacement, as shown in figure 1.18.2.
By joining O to B, the vector triangle OAB is completed. It is clear from the figure
that the final displacement is given by OB, the hypotenuse of the triangle. The
length of OB = 5 cm, which is equivalent to 250 km on the chosen scale. The
direction of the resultant vector is the angle ∠ AOB
which is nearly 53
o
in this
case. This can be measured using an ordinary protractor. Thus the direction of the
resultant displacement OB is 53
o
north of east.
B
5 cm
4 cm
A
O
3 cm
Figure 1.18.2
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The process illustrates the addition of two displacements to get resultant
displacement and is applied for addition of any two vectors such as force, velocity etc.
The vector addition is represented by the equation,
OA + AB = OB . . . . . . . . . . . . . . .. (1.18.1)
Note that
OA  + AB is not the same as OB.
To sum up, the following steps are followed in the addition of two vectors:
(i)
A suitable scale is chosen for the vectors to be added.
(ii)
The vectors involved a and b are drawn according to this scale.
(iii)
Starting from any arbitrary origin, the vector a is drawn, see figure 1.18.3
.
b
b
R
a
(b
)
(a)
a
Figure 1.18.3
(iv) The vector b is drawn from the head of a i.e. the tail of b starts from the
head of a. Care must be taken to maintain the angle between the two vectors.
(v) The resultant vector is the vector drawn from the tail of a to the head of
b. In figure1.17.3 (b), resultant R = a + b
For addition of three or more vectors, the same procedure applies; the vectors are
drawn one after another.
Subtraction of vectors
It is sometimes necessary to subtract one vector from another. To subtract vector a
from vector b, we first form the negative of a, denoted -a, a vector that has the same
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length as a but pointing in the opposite direction. We then add - a to b in the usual
manner.
-a
b
a
=
b
+
-a
a-b
b
Figure 1.19.1, vector subtraction
Example 1.19.1 Two vectors of 5.0 units and 7.0 units make on angle of 60o with each
other. What is the vector sum?
Solution:
The problem is solved graphically using the triangle method. Choosing a
convenient scale draw vector a, 5.0 units long. At the end (head) of a, draw vector b,
7.0 units long so that there is a 60 o angle between the directions of a and b. The
resultant vector R is the line joining A to C. On measurement, R is found to be nearly
10.4 units and makes on angle 24o 30’ with the vector a.
Resolution of vectors
Just as we can add two or more vectors to give a single resultant vector, we can break a
vector into two or more vectors. The latter process is called resolving the vector. The
new vectors are called the components of the original vector. It is possible to have any
number of components for a given vector.
However, the most useful choice of
components is usually that in which both components act in mutually perpendicular
directions or rectangular components. Of the two, one is the horizontal component or
the x-component and the other is the vertical component or y-component.
6
Illustration: Resolve vector, a , which makes an angle  with the horizontal into
two components.
B
C
R
a
b
ay
A
O
B
A
ax
a
Figure 1.20.1 (b)
Figure 1.20.1(a)
Procedure
From the tail of vector a, draw a horizontal line.
From the head of a, drop a
perpendicular onto this horizontal line (call this point B ) to get AB. It is evident that if
OA represents a vector ax and AB represents a vector ay, then graphical addition shows
a = ax + ay . . . . . . . . . . . . . . . . . . . . . . .(1.20.1)
In other words, ax and ay are the two rectangular components into which vector a has
been resolved.
In the triangle OAB,
sin = |AB|/|OB| and cos = |OA|/|OB| . . . .(1.20.2)
The magnitude of
ax = |OA| = |OB| cos = a cos
ax = a  cos . . . . . . . . . . . . . . . . . . . .
. . . . . (1.20.3)
.(1.20.4)
Similarly, the magnitude of ay = |B| = |OB| sin = |a |sin
ay = a sin
. . . . . . . . . .. . . . . . . . . . . (1.20.5)
Thus a single vector a is replaced by its component two vectors ax and ay. The method
of vector resolution is widely used in addition and / or subtraction of vectors.
Now that we have ended up with a right-angle triangle, we can easily calculate the
length of the resultant vector(the hypotenuse) using Pythagoras theorem. Thus
7
|OB|2 = |OA|2 + |AB|2 . . . . . . . . . . . . . . . . . . . . .(1.20.6)
Or
|a|2 = |ax|2 + |ay|2 . . . . . . . . . . . . . . . . . . . . . . .. . . .(1.20.7)
The angle θ
is easily found using trigonometric relations as follows
|a |
tan θ= y
|a x|
. . . . . . . . . .. . . . . . . . . . . . . . . . . . .
.(1.20.8)
In navigation, the graphical method is more popular. So far we have only dealt with a
single vector which we have resolved into components.. Now let us take a case
involving two vectors A and B which must be added using the method of vector
resolution, figure 1.19.3.
+
A
-
+x
B
-
Figure 1.20.3
8
Procedure
The first thing you must do is to obtain expressions for the x
and y components of the
two vectors. This is done in the same way as that for the single vector in figure 1.19.1.
Thus
Ax = Acos θ1
Bx = - Bcos θ2
..........
. . (1.20.9)
Ay = Asin θ1
By = - Bsin θ2
........ ....
(1.20.10)
The negative signs arise in vector B components because they are directed in the negative
x
and
y
axes. Now one of the important equations or facts to remember is that
when vectors are added, corresponding vectors are added or subtracted just like scalars.
This is so because corresponding vectors are either parallel or anti-parallel. What this
means is the following: Since the
x
component of vector A, i.e. Ax is parallel to Bx,
they can be added together as follows:
Ax + Bx. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1.20.11)
The
y
components can be added in the same way i.e.
Ay + By . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .(1.20.12)
Now addition of two or more vectors result into another vector called the resultant vector.
Let us call this vector C. Therefore, we can write the vector sum as follows:
A + B = C . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1.20.13)
This sum can be written in terms of vector components as follows:
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(Ax + Ay )+ (Bx + By) = Cx + Cy
. . . . . . . . . . . .. . . . . . . .
(1.20.14)
From the equations above, it can be concluded that
Ax + Bx = Cx . . . . . . . . . . . . . . . . . . . . . . . . . . .(1.20.15)
and
Ay + By = Cy
. . . . . . . . . . . . . . . . . . . . . . . . . . (1.20.16)
The magnitude of the resultant is found using Pythagoras theorem i.e.
|C|2 = |Cx|2 +|Cy|2
. . . . . . . . . . . . . . . . . . . . . . . (1.20.17)
And its direction is found from
tan
θ=
|C y|
|C x|
................ .........
. (1.20.18)
Example 1.20.1 below illustrates the use of the method of vector resolution in the
addition of two vectors.
Study example 1.20.1 for a feel of how this method can be
applied in real situations.
Example 1.20.1
Two vectors are shown in figure 1.20.3. Use the method of vector resolution to compute
their resultant.
25N
45o
70o
15N
10
Solution
The first activity here is to find the x and y
Thus the x component of the 25N force is
F25x =25cos45 = +17.7N
components of the given vectors.
and F25y = 25sin45 = +17.7N
Note that both components are positive because they are directed in the positive x
and
y directions. Compare these with the components of the 15N force as calculated
below
F15x = -15cos70o = - 0.342×15 = -5.13N and F15y = - 0.9397×15 = -14.10N
Graphically, the four components can now be represented as follows, figure 1.20.4:
+
F25y = +17.7
-5.13N
F25x = +17.7N
-14.10N
Figure 1.20.4
We are now in position to find the sums of the corresponding components as discussed
+
above. We let the resultant vector to FR. Thus
F Rx =F 25 x +F 15 x =17 . 7 N +(−5. 13 N )=12. 57 N
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and
F Ry =F 25 y +F 15 y =17 .7 N + (-14.10N) = 3.6N
The magnitude of F R is found using Pythagoras theorem as follows:
F R=√ F Rx +F Ry =√(12 .57 )2 +(3. 6 )2
From which expression we get
F R= 13.1 N approximately. The final step is to find
+x
the direction that this vector makes with the
draw a graphical representation of the x
figure 1.20.5.
and
y
axis.
At this point, it is helpful to
components we have just calculated,
Figure 1.20.6 shows the graphical representation of the resultant vector,
FR and its direction.
+
3.6N
-
12.57N
+
-
Figure 1.20.5
FR = +13.1N
3.6N
ϕ
=
-
12.57 N
12
+
-
Figure 1.20.6
φ=
3.6
=0 .288
12. 5
o
Tan
or φ=16.1 .
The answer can now be stated as follows: resultant is of magnitude 13.1N oriented at an
angle of 16.1o to the + x
axis.
Student activity
Assess your understanding of vector addition by
(i)
drawing the vector diagram of the following four vector forces:
F1 = 10N oriented (anticlockwise) at an angle of 45o to the + x
axis
F2 = 5N oriented (anticlockwise) at an angle of 135o to the + x
F3 = 6N oriented at an angle of 60o below the −x axis
F4 = 6N oriented at an angle of 30o below the + x
(ii)
axis
axis
finding the resultant of the forces cited above.
Answer: (i) book-work
(ii) |FR| = 6.22N oriented at an angle of 22.48o above the
+x
axis.
Vector subtraction
Vector subtraction follows the same rules. The fact to know is that subtraction is simply
the addition of a negative vector.
For example, subtraction of vector B from vector A is
the same as adding – B to A i.e.
A – B = A + (-B) . . . . . . . . . . . . . . . . .(1.21.1)
Take note. A – B is not the same as B – A. In other words subtraction is not
commutative.
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Another form of presenting vectors and their components when adding is to use a table.
Let us present the vectors and their components used in example 1.20.1 see table 1.21.1.
Note that the same table would still be valid for the subtraction of F 15 from F25 but
the signs of the components of F15 will be opposite to that in table 1.21.1, see table
1.21.2. In this case, the resultant |FR| = 39.15N which is very different from the one we
obtained before.
Table 1.21.1
Vector (N)
x- component (N)
25
15
17.7
17.7
-5.13
- 14.10
12.57
FR
y – component (N)
3.6
Table 1.21.2
Vector (N)
x- component (N)
17.7
25
15
5.13
FR
22.83
y – component (N)
17.7
14.10
31.8
Self-assessment questions and exercises
1.0 Suppose that the displacement of an object is related to time according to
the expression x = ct3 , what are the dimensions of the constant c?
14
v
2.0 Show that the equation ( v f)2 = ( i )2 + 2ax is dimensionally correct,
where x is the distance ,
respectively, a
3.0
v f and v i are the final and initial velocities
is the acceleration and t is the time .
Vector A is 3 units in length and points along the positive x-axis: Vector
B is 4 units in length and points in the negative y axis . Use the graphical
method to find the magnitude and direction of the vectors (a) A+B and
(b) A-B.
4.0 A burglar attempts to move a safe by pulling on it with a rope at an
angle
of 40o degrees upward from the horizontal. If the horizontal component of
the force exerted by the rope is 125N, what force is the burglar exerting?
5.0
A man pulls a sledge up a hill by exerting a force of 150N at an angle of
40o with the hill. What are the components of this force perpendicular to
the hill and along it slope?
6.0
A force A is added to a force that has x and y components of 3N
and –5N respectively. The resultant of the two forces is in the
negative x direction and has a magnitude of 4N. Find the x and y
components of A.
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