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# VECTORS AND EQUILIBRIUM

Learning Objectives

After you have completed this chapter, you will be able to:
1. Understand and use rectangular coordinate system.
2. Understand the idea of unit vector, null vector and position vector.
3. Represent a vector as tow perpendicular components (rectangular components).
4. Understand the rule of vector addition and extend it to add vectors using rectangular
components.
5. Understand multiplication of vectors and solve problems.
6. Define the moment of force or torque.
7. Appreciate the use of the torque due to force.
8. Show an understanding that when there is no resultant force and no resultant torque, a system
is in equilibrium.
9. Appreciate the applications of the principle of moments.
10. Apply the knowledge gained to solve problems on statics.

P
hysical quantities that have both numerical and directional properties are represented by vectors. This
chapter is concerned with the vector algebra and its applications in problem of equilibrium of forces
and equilibrium of torques.

## 2.1 BASIC CONCEPTS OF VECTORS

(i) Vectors
As we have studied in school Physics, there are some physical quantities which require both
magnitude and direction for their complete description, such as velocity, acceleration and force.
They are called vectors. In books, vectors are usually denoted by bold face characters such as A,
d, r and v. If we wish to refer only to the magnitude of a vector d we use light face type such as d.
A vector is represented graphically by a direction line segment with an arrow head. The length of
the line segment, according to the chosen scale, corresponds to the magnitude of the vector.

## (ii) Rectangular coordinate system

Two lines drawn at right angles to each other as shown in Fig. 2.1 (a) are known as coordinate
axes and their point of intersection is known as origin. This system of coordinate axes is called as
Cartesian or rectangular coordinate system.
Y Y

X X X X
O O

Y Y
Fig. 2.1(a) Fig. 2.1(b)
One of the lines is named as x-axis and the other the y-axis. Usually the x-axis is taken as the
horizontal axis, with the positive direction to the right, and the y-axis as the vertical axis with the
positive direction upward.
Z Z

A P(a,b,c)
Y O
O Y
X 1 X
Fig. 2.2(a) Fig. 2.2(b)
The direction of a vector in a plane is denoted by the angle which the representative the of the
vector makes with positive x-axis in the anti-clock wise direction, as shown in Fig. 2.1(b). The
point P shown in Fig 2.1(b) has coordinates (a,b). This notation means that if we start at the origin,
we can reach P by moving a units along the positive x-axis and then b units along the positive y-
axis.
The direction of a vector in space requires another axis which is at right angle to both x and y-
axes, as shown in Fig. 2.2 (a). The third axis is called z-axis.
The direction of a vector in space is specified by the three angles which the representative line of
the vector makes with x, y and z-axes respectively as shown in Fig 2.2(b). The point P of a vector
A is thus denoted by three coordinates (a,b,c).
Given two vectors A and B as shown in Fig 2.3(a), their sum is obtained by drawing their
representative lines in such a way that tail of vector B coincides with the head of the vector A.
Now if we join the tail of A to the head of B, as shown in the Fig 2.3(b), it will represent the vector
sum (A+B) in magnitude and direction. This is known as head to tail rule of vector addition. This
rule can be extended to find the sum of any number of vectors. Similarly the sum B+A is illustrated
by dotted lines in Fig 2.3(c). It is seen that
A+B=B+A . (2.1)
So the vector addition is said to be commutative. It means that when vectors are added, the result
is the same for any order of addition.
B B -B

A A A A A
B B
Fig. 2.3(a) Fig. 2.3(b) Fig. 2.3(c) Fig. 2.3(d)
(iv) Resultant Vector
The resultant of a number of similar vectors force vectors for example, is that single vector which
would have the same effect as all the original vectors taken together.
(v) Vector Subtraction
The subtraction of a vector is equivalent to the addition of the same vector with its direction
reversed. Thus, to subtract vector B from vector A, reverses the direction of B and add it to A, as
shown in Fig 2.3(d).
A-B=A+(-B)
(vi) Multiplication of a Vector by a Scalar
The product of a vector A and a number n>0 is defined to be a new vector nA having the same
direction as A but a magnitude n times the magnitude of A as illustrated in Fig. 2.4. If the vector is
multiplied by a negative number, then its direction is reversed.
In the event that n represents a physical quantity with dimension, the product nA will correspond
to a new physical quantity and the dimension of the resulting vector will be the same as the
product of the dimensions of the two quantities which were multiplied together. For example when
velocity is multiplied by scalar mass m, the product is a new vector quantity called momentum
having the same dimensions as those of mass and velocity.

A 2A -2A

Fig. 2.4
(vii) Unit Vector
A unit vector in a given direction is a vector with magnitude one in that direction. It is used to
represent the direction of a vector.
, which we read as A hat, thus
A unit vector in the direction of A is written as A
A=A A
=A
A . (2.2)
A
The direction along x, y and z axes are generally represented by unit vectors i , j and k
respectively (Fig. 2.5 a). The use of unit vectors is not restricted to Cartesian coordinate system
only. Unit vectors may be defined for any direction. Two of the more frequently used unit vectors
are the vector r which represents the direction of a vector r (Fig. 2.5 b) and the vector n which
represents the direction of a normal drawn on a specified surface as shown in Fig. 2.5(c).

2
n
k r
O
i j Y r

X
Fig. 2.5(a) Fig. 2.5(b) Fig. 2.5(c)
(viii) Null Vector
It is a vector of zero magnitude and arbitrary direction. For example the sum of a vector and its
negative vector is a null vector.
A+ (-A) =0 .. (2.3)
(ix) Equal Vectors
Two vectors A and B are said to be equal if they have the same magnitude and direction,
regardless of the position of their initial points.
This means that parallel vectors of the same magnitude are equal to each other.
(x) Rectangular Components of a Vector
A component of a vector is the effective value in a given direction. A vector may be considered as
the resultant of its component vectors along the specified directions. It is usually convenient to
resolve a vector into components along mutually perpendicular directions called rectangular
components.
Y

N P

A
Ay j

X
O Ax i M
Fig. 2.6
Let there is a vector A represented by OP making angle with the x-axis. Draw projection OM of
vector OP on x-axis and projection ON of vector OP on y-axis as shown in Fig. 2.6. Projection OM
being along x-direction is represented by Ax i and projection ON=MP along y-direction is
represented by= Ay j . By head and tail rule
A = Ax i + Ay j . (2.4)
Thus Ax i and Ay j are the components of vector A. Since these are at right angle to each other,
hence, are called rectangular components of A. Considering the right angled triangle OPM, the
magnitude of Ax i or x-components of A is
Ax = A cos .. (2.5)

And that of Ay j or y-component
Ay = A sin .. (2.6)
(xi) Determination of a Vector from its Rectangular Components
If the rectangular components of a vector, as shown in Fig. 2.6, are given, we can find out the
magnitude of the vector using Pythagorean Theorem.
In the right angled OPM.
OP2 = OM2 + MP2
2 2
Or A2 = Ax + Ay .. (2.7)
And direction is given by
MP Ay
tan
OM Ax
Ay
tan 1
Or Ax .. (2.8)
(xii) Position Vector
The position vector r is a vector that describes the location of a particle with respect to the origin.
It is represented by a straight line drawn in such a way that its tail coincides with the origin and the
head with point P (a,b) as shown in Fig. 2.7(a). The projections of position vector r on the x and y
axes are the coordinates a and b and they are the rectangular components of the vector r. Hence
r=a i +b j .. (2.9)
In three dimensional space, the position vector of a point P (a,b,c) is shown in Fig. 2.7(b) and is
represented by
r=a i +b j + c k .. (2.9)

3
Y

P(a,b) Z
P(a,b,c)
r r
b
O Y
a
O X X

## Fig. 2.7(a) Fig. 2.7(b)

2.2 VECTOR ADDITION BY RECTANGULAR COMPONENTS
Let A and B be two vectors which are represented by two directed lines OM and ON
respectively. The vector B is added to A by the head to tail rule of vector addition (Fig 2.9). Thus
the resultant vector R = A + B is given, in direction and magnitude by the vector OP.
Y

N B
By
B Ry
M Bx
A S
Ay Ay
Ax By
O Q R X
Rx
Fig. 2.9
In the Fig. 2.9 Ax, Bx and Rx are the x components of the vectors A, B and R and their
magnitudes are given by the lines OQ, MS and OR respectively, But
OR = OQ + QR
Or OR = OQ + MS
Or Rx = Ax + Bx . (2.11)
Or Rx = Rx i = (Ax + Bx) i
This means that the sum of the magnitudes of x-components of two vectors which are to be added
is equal to the x-component of the resultant. Similarly the sum of the magnitudes of y-components
of two vectors is equal to the magnitude of y-component of the resultant, that is
Ry = Ay + By . (2.12)
Or
Ry = Ry j = (Ay + By) j
Since Rx i and Ry j are the rectangular components of the resultant vector R, hence
R = Rx i + Ry j
Or R = (Ax + Bx) i + (Ay + By) j
The magnitude of the resultant vector R is thus given as
R = ( Ax + Bx )2 +( Ay + By )2 . (2.13)
And the direction of the resultant vector is determined from
R ( A + By )
And tan 1 y tan 1 y . (2.14)
Rx ( Ax + Bx )
Similarly for any number of coplanar vectors A, B, C we can write
R = ( Ax + Bx + C x +...)2 + ( Ay + By + C y +...)2 . (2.15)
( Ay + By + C y +...)
And tan 1 . (2.16)
( Ax + Bx + C x +...)
The vector addition by rectangular components consists of the following steps.
i. Find x and y components of all given vectors.
ii. Find x-component Rx of the resultant vector by adding the x-components of all the vectors.
iii. Find y-component Ry of the resultant vector by adding the y-components of all the vectors.
iv. Find the magnitude of resultant vector R using
R = R x 2 + Ry 2

4
v. Find the direction of resultant vector R by using
R
tan 1 y
Rx
Where is the angle, which the resultant vector makes with positive x-axis. The signs of Rx and Ry
determine the quadrant in which resultant vector lies. For that purpose proceed as given below.
Ry
Irrespective of the sign of Rx and Ry, determine the value of tan
1
from the calculator or
Rx
by consulting trigonometric tables. Knowing the value of , angle is determined.
a) If both Rx and Ry components are positive, then the resultant lies in the first quadrant and
its direction is = .
b) If Rx is negative and Ry component is positive, then the resultant lies in the second
quadrant and its direction is =180- .
c) If both Rx and Ry components are negative, then the resultant lies in the third quadrant and
its direction is =180+ .
d) If Rx is positive and Ry is negative, then the resultant lies in the fourth quadrant and its
direction is =360- .
Table 2.1
II Y I
tan - tan +
Rx - Rx +
Ry + Ry +

X X
tan + tan -
Rx - Rx +
Ry - Ry -
III Y IV
Y Y
= =180-

O X X O X
st

X O X X O X

=180+ =360-

Y Y
2.3 PRODUCT OF TWO VECTORS
There are two types of vector multiplication. The product of these two types is known as scalar
product and vector product. As the name implies, scalar product of two vector quantities is a
scalar quantity, while vector product of two vector quantities is a vector quantity.
Scalar or Dot Product
The scalar product of two vectors A and B is written as A.B and is defined as
A.B = AB cos . (2.17)
Where A and B are the magnitudes of vectors A and B and is the angle between them.

B B

Bcos
A A
Fig. 2.10(a) Fig. 2.10(b)
For physical interpretation of dot product of two vectors A and B, these are first brought to a
common origin (Fig. 2.10 a), then
A.B = A (projection of B on A)
A.B = A (magnitude of component of B in the direction of A)
= A (B cos) = AB cos
Similarly B.A = B (A cos) = BA cos
5
We come across this type of product when we consider the work done by a force F whose point of
application moves distant d in a direction making an angle with the line of action of F, as shown
in Fig. 2.11.
F

F cos
d
Work done = (effective component of force in the direction of motion) distance moved
= (F cos) d = Fd cos
Using vector notation
F.d = Fd cos = work done
Characteristics of Scalar Product
1. Since A.B = AB cos and B.A = BA cos = AB cos hence, A.B = B.A, the order of
multiplication is irrelevant. In other words, scalar product commutative.
2. The scalar product of tow mutually perpendicular vectors is zero. A.B = AB cos90 =0
In case of unit vectors i , j and k , since they are mutually perpendicular, therefore,
i . j = j . k = k . i =0.. (2.18)
3. The scalar product of two parallel vectors is equal to the product of their magnitudes. Thus for
parallel vector ( =0)
A.B = AB cos0=AB
In case of unit vectors
i . i = j . j = k . k = 1 .. (2.19)
And for antiparallel vectors (=180)
A.B = AB cos180= -AB
4. The self product of a vector A is equal to square of its magnitude.
A.A = AA cos0 = A2
5. Scalar product of two vector A and B in terms of their rectangular components
A.B = ( Ax i + Ay j + Azk).(
B i + B j + B k)
x y z

## Or A.B = Ax Bx + Ay By + AzBz (2.20)

Equation 2.17 can be used to find the angle between two vectors: since,
A.B = AB cos = Ax Bx + Ay By + Az Bz
Ax Bx + Ay By + AzBz
Therefore cos= .. (2.21)
AB
Vector or Cross Product
The vector product of two vectors A and B, is a vector which is defined as
A B AB sin n . (2.22)
Where n is a unit vector perpendicular to the plane containing A and B as shown in Fig. 2.12 (a).

n AB AB B

A

B B
B
A A
A BA
Fig. 2.12(a) Fig. 2.12(b) Fig. 2.12(c) Fig. 2.12(d)
Its direction can be determined by right hand rule. For that purpose place together the tails of
vectors A and B to define the plane of vectors A and B. The direction of the product vector is
perpendicular to this plane. Rotate vector A into B through the smaller of the two possible angles
with the curl of the fingers of the right hand, keeping the thumb erect. The direction of the product
vector will be along the erect thumb, as shown in the Fig. 2.12 (b). Because of this direction rule,
B A is a vector opposite in sing to A B . Hence
A B = B A . (2.23)
Characteristics of Cross Product
1. Since A B is not same asB A , the cross product is non commutative.
2. the cross product of two perpendicular vectors has maximum magnitude A B =AB sin90
n = AB n
In case of unit vectors, since they form a right handed system and are mutually
perpendicular.

6
i j = k , j k = i , k i = j
3. the cross product of two parallel vector is null vector, because for such vectors =0 or
180, Hence
A B =AB sin0 n = AB sin180 n =0
As a consequence A A = 0
Also i i = j j = k k = 0 .. (2.19)
4. cross product of two vectors A and B in terms of their rectangular component is :
A B = ( Ax i + Ay j + Azk) (B i + B j + B k)
x y z

## A B = ( Ax Bz - AzBy )i + (AzBx - AxBz )j + ( AxBy AyBx )k

.. (2.25)
The result obtained can be expressed fro memory in determinant form as below:
i j k
A B = Ax Ay Az
Bx By Bz
5. The magnitude of A B is equal to area of the parallelogram formed with A and B as two
Examples of Vector Product
i. When a force F is applied on a rigid body at a point whose position vector is r from
any point of the axis about which the body rotates, then the turning effect of the
force, called the torque is given by the vector product of r and F.
=rF
ii. The force on a particle of charged q and velocity v in a magnetic field of strength B
is given by vector product.
F = q (v B)
2.4 TORQUE
We have already studied in school Physics that a turning effect is produced when a nut is
tightened with spanner (Fig. 2.13). The turning effect increased when you push harder on the
spanner. It also depends on the length of the spanner: the longer the handle of the spanner, the
greater the tuning effect of an applied force. The turning effect of a force is called its movement or
torque and is defined as the product of force F and the perpendicular distance from its line of
action to the pivot which is the point around which the body (spanner) rotates. This distance OP is
called moment arm l. Thus the magnitude of torque represented by is
= lF . (2.26)
When the line of action of the applied force passes through the pivot point, the value of
moment arm l=0, so in this case torque is zero.

The nut is easy to turn with a spanner It is easier still if the spanner has a long handle
Fig. 2.13
We now consider the torque due to a force F acting on a rigid body. Let the force F can be
resolved into two rectangular components, F sin perpendicular to r and F cos along the
direction of r (Fig. 2.14 a). The torque due to F cos about pivot O is zero as its line of action
passes through point O. Therefore, the torque due to F is equal to the torque due to F sin only
about O. It is given by
= (F sin)r = rF sin . (2.27)
Alternatively the moment arm l is equal to the magnitude of the component of r perpendicular to
the line of action of F as illustrated in Fig. 2.14 (b). Thus
= (r sin)F = rF sin . (2.28)
Where is the angle between r and F

F
F

Fsin
P rsin
r
O r P
O

7
From Eq. 2.27 and Eq. 2.28 it can be seen that the torque can be defined by the vector product of
position vector r and the F
= (rF sin) n .. (2.29)
Where (rF sin) is the magnitude of the torque. The direction of the torque represented by n is
perpendicular to the plane containing r and F given by right hand rule for the vector product of two
vectors.
The SI unit for torque is Newton meter (N m).
Torque is the counter part of force for rotational motion. Just as force determines the linear
acceleration produced in a body, the torque acting on a body determines its angular acceleration.
If the body is at rest or rotating with uniform angular velocity, the angular acceleration will be zero.
In this case the torque acting in the body will be zero.
2.5 EQUILIBRIUM OF FORCES
We have studied in school Physics that if a body, under the action of a number of forces, is at rest
or moving with uniform velocity, it is said to be in equilibrium.
First Condition of Equilibrium
A body at rest or moving with uniform velocity has zero acceleration. Form Newtons Law of
motion the vector sum of all force acting on it must be zero.
This is known as first condition of equilibrium. Using the mathematical symbol F for the sum of all
forces we can write
F = 0 .. (2.30)
In case of coplanar forces, this condition is expressed usually in terms of x and y components of
the forces. We have studied that x-component of the resultant force F equals the sum of x-
directed or x-components of all the forces acting on the body. Hence
Fx = 0 .. (2.31)
Similarly for the y-directed forces, the resultant of y-directed forces should be zero. Hence
Fy = 0 .. (2.32)
It may be noted that if the rightward forces are taken as positive then leftward forces are taken as
negative. Similarly if upward forces are taken as positive then downward forces are taken as
negative.
2.6 EQUILIBRIUM OF TORQUES
Second Condition of Equilibrium
Let tow equal and opposite forces are acting on a rigid body as shown in Fig. 2.17.

Fig. 2.17
Although the first condition of equilibrium is satisfied, yet it may rotate having clockwise turning
effect. As discussed earlier, for angular acceleration to be zero, the net torque acting on the body
should be zero. Thus for a body in equilibrium, the body should be zero. This is known as second
condition of equilibrium. Mathematically it is written as
= 0 .. (2.33)
By convention, the counter clockwise torqueses are taken as positive and clockwise torques as
negative. An axis is chosen for calculating the torques. The position of the axis is quite arbitrary.
Axis can be chosen anywhere which is convenient in applying the torque equation. A most helpful
point of rotation is the one through which lines of action of several forces pass.
We are now in a position to state the complete requirements for a body to be in equilibrium, which
are
(1) F = 0 i.e Fx = 0 and Fy = 0
(2) = 0
When 1st condition is satisfied, there is no linear acceleration and body will be in translational
equilibrium. When 2nd condition is satisfied, there is no angular acceleration and body will be in
rotational equilibrium.
For a body to be in complete equilibrium, both conditions should be satisfied, i.e., both linear
acceleration and angular acceleration should be zero.
If a body is at rest, it is said to be in static equilibrium but if the body is moving with uniform
velocity, it is said to be in dynamic equilibrium.

8
We will restrict the applications of above mentioned conditions of equilibrium to situations in which
all the forces lie in a common plane. Such forces are said to be coplanar. We will also assume that
these forces lie in the xy-plane.
If there is more than one object in equilibrium in a given problem, one object is selected at a time
to apply the conditions of equilibrium.
The content of the provided text is a detailed explanation of vectors and equilibrium, including definitions, properties, and operations related to vectors, as well as the conditions for equilibrium of forces and torques. The text does not contain any specific mathematical questions or problems that require solving.

**Analysis:**
no_question

the text is primarily expository, providing definitions, explanations, and examples of vector operations and equilibrium conditions. it does not pose any specific mathematical questions or provide answers to such questions.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
Sei sulla pagina 1di 9

# VECTORS AND EQUILIBRIUM

Learning Objectives

After you have completed this chapter, you will be able to:
1. Understand and use rectangular coordinate system.
2. Understand the idea of unit vector, null vector and position vector.
3. Represent a vector as tow perpendicular components (rectangular components).
4. Understand the rule of vector addition and extend it to add vectors using rectangular
components.
5. Understand multiplication of vectors and solve problems.
6. Define the moment of force or torque.
7. Appreciate the use of the torque due to force.
8. Show an understanding that when there is no resultant force and no resultant torque, a system
is in equilibrium.
9. Appreciate the applications of the principle of moments.
10. Apply the knowledge gained to solve problems on statics.

P
hysical quantities that have both numerical and directional properties are represented by vectors. This
chapter is concerned with the vector algebra and its applications in problem of equilibrium of forces
and equilibrium of torques.

## 2.1 BASIC CONCEPTS OF VECTORS

(i) Vectors
As we have studied in school Physics, there are some physical quantities which require both
magnitude and direction for their complete description, such as velocity, acceleration and force.
They are called vectors. In books, vectors are usually denoted by bold face characters such as A,
d, r and v. If we wish to refer only to the magnitude of a vector d we use light face type such as d.
A vector is represented graphically by a direction line segment with an arrow head. The length of
the line segment, according to the chosen scale, corresponds to the magnitude of the vector.

## (ii) Rectangular coordinate system

Two lines drawn at right angles to each other as shown in Fig. 2.1 (a) are known as coordinate
axes and their point of intersection is known as origin. This system of coordinate axes is called as
Cartesian or rectangular coordinate system.
Y Y

X X X X
O O

Y Y
Fig. 2.1(a) Fig. 2.1(b)
One of the lines is named as x-axis and the other the y-axis. Usually the x-axis is taken as the
horizontal axis, with the positive direction to the right, and the y-axis as the vertical axis with the
positive direction upward.
Z Z

A P(a,b,c)
Y O
O Y
X 1 X
Fig. 2.2(a) Fig. 2.2(b)
The direction of a vector in a plane is denoted by the angle which the representative the of the
vector makes with positive x-axis in the anti-clock wise direction, as shown in Fig. 2.1(b). The
point P shown in Fig 2.1(b) has coordinates (a,b). This notation means that if we start at the origin,
we can reach P by moving a units along the positive x-axis and then b units along the positive y-
axis.
The direction of a vector in space requires another axis which is at right angle to both x and y-
axes, as shown in Fig. 2.2 (a). The third axis is called z-axis.
The direction of a vector in space is specified by the three angles which the representative line of
the vector makes with x, y and z-axes respectively as shown in Fig 2.2(b). The point P of a vector
A is thus denoted by three coordinates (a,b,c).
Given two vectors A and B as shown in Fig 2.3(a), their sum is obtained by drawing their
representative lines in such a way that tail of vector B coincides with the head of the vector A.
Now if we join the tail of A to the head of B, as shown in the Fig 2.3(b), it will represent the vector
sum (A+B) in magnitude and direction. This is known as head to tail rule of vector addition. This
rule can be extended to find the sum of any number of vectors. Similarly the sum B+A is illustrated
by dotted lines in Fig 2.3(c). It is seen that
A+B=B+A . (2.1)
So the vector addition is said to be commutative. It means that when vectors are added, the result
is the same for any order of addition.
B B -B

A A A A A
B B
Fig. 2.3(a) Fig. 2.3(b) Fig. 2.3(c) Fig. 2.3(d)
(iv) Resultant Vector
The resultant of a number of similar vectors force vectors for example, is that single vector which
would have the same effect as all the original vectors taken together.
(v) Vector Subtraction
The subtraction of a vector is equivalent to the addition of the same vector with its direction
reversed. Thus, to subtract vector B from vector A, reverses the direction of B and add it to A, as
shown in Fig 2.3(d).
A-B=A+(-B)
(vi) Multiplication of a Vector by a Scalar
The product of a vector A and a number n>0 is defined to be a new vector nA having the same
direction as A but a magnitude n times the magnitude of A as illustrated in Fig. 2.4. If the vector is
multiplied by a negative number, then its direction is reversed.
In the event that n represents a physical quantity with dimension, the product nA will correspond
to a new physical quantity and the dimension of the resulting vector will be the same as the
product of the dimensions of the two quantities which were multiplied together. For example when
velocity is multiplied by scalar mass m, the product is a new vector quantity called momentum
having the same dimensions as those of mass and velocity.

A 2A -2A

Fig. 2.4
(vii) Unit Vector
A unit vector in a given direction is a vector with magnitude one in that direction. It is used to
represent the direction of a vector.
, which we read as A hat, thus
A unit vector in the direction of A is written as A
A=A A
=A
A . (2.2)
A
The direction along x, y and z axes are generally represented by unit vectors i , j and k
respectively (Fig. 2.5 a). The use of unit vectors is not restricted to Cartesian coordinate system
only. Unit vectors may be defined for any direction. Two of the more frequently used unit vectors
are the vector r which represents the direction of a vector r (Fig. 2.5 b) and the vector n which
represents the direction of a normal drawn on a specified surface as shown in Fig. 2.5(c).

2
n
k r
O
i j Y r

X
Fig. 2.5(a) Fig. 2.5(b) Fig. 2.5(c)
(viii) Null Vector
It is a vector of zero magnitude and arbitrary direction. For example the sum of a vector and its
negative vector is a null vector.
A+ (-A) =0 .. (2.3)
(ix) Equal Vectors
Two vectors A and B are said to be equal if they have the same magnitude and direction,
regardless of the position of their initial points.
This means that parallel vectors of the same magnitude are equal to each other.
(x) Rectangular Components of a Vector
A component of a vector is the effective value in a given direction. A vector may be considered as
the resultant of its component vectors along the specified directions. It is usually convenient to
resolve a vector into components along mutually perpendicular directions called rectangular
components.
Y

N P

A
Ay j

X
O Ax i M
Fig. 2.6
Let there is a vector A represented by OP making angle with the x-axis. Draw projection OM of
vector OP on x-axis and projection ON of vector OP on y-axis as shown in Fig. 2.6. Projection OM
being along x-direction is represented by Ax i and projection ON=MP along y-direction is
represented by= Ay j . By head and tail rule
A = Ax i + Ay j . (2.4)
Thus Ax i and Ay j are the components of vector A. Since these are at right angle to each other,
hence, are called rectangular components of A. Considering the right angled triangle OPM, the
magnitude of Ax i or x-components of A is
Ax = A cos .. (2.5)

And that of Ay j or y-component
Ay = A sin .. (2.6)
(xi) Determination of a Vector from its Rectangular Components
If the rectangular components of a vector, as shown in Fig. 2.6, are given, we can find out the
magnitude of the vector using Pythagorean Theorem.
In the right angled OPM.
OP2 = OM2 + MP2
2 2
Or A2 = Ax + Ay .. (2.7)
And direction is given by
MP Ay
tan
OM Ax
Ay
tan 1
Or Ax .. (2.8)
(xii) Position Vector
The position vector r is a vector that describes the location of a particle with respect to the origin.
It is represented by a straight line drawn in such a way that its tail coincides with the origin and the
head with point P (a,b) as shown in Fig. 2.7(a). The projections of position vector r on the x and y
axes are the coordinates a and b and they are the rectangular components of the vector r. Hence
r=a i +b j .. (2.9)
In three dimensional space, the position vector of a point P (a,b,c) is shown in Fig. 2.7(b) and is
represented by
r=a i +b j + c k .. (2.9)

3
Y

P(a,b) Z
P(a,b,c)
r r
b
O Y
a
O X X

## Fig. 2.7(a) Fig. 2.7(b)

2.2 VECTOR ADDITION BY RECTANGULAR COMPONENTS
Let A and B be two vectors which are represented by two directed lines OM and ON
respectively. The vector B is added to A by the head to tail rule of vector addition (Fig 2.9). Thus
the resultant vector R = A + B is given, in direction and magnitude by the vector OP.
Y

N B
By
B Ry
M Bx
A S
Ay Ay
Ax By
O Q R X
Rx
Fig. 2.9
In the Fig. 2.9 Ax, Bx and Rx are the x components of the vectors A, B and R and their
magnitudes are given by the lines OQ, MS and OR respectively, But
OR = OQ + QR
Or OR = OQ + MS
Or Rx = Ax + Bx . (2.11)
Or Rx = Rx i = (Ax + Bx) i
This means that the sum of the magnitudes of x-components of two vectors which are to be added
is equal to the x-component of the resultant. Similarly the sum of the magnitudes of y-components
of two vectors is equal to the magnitude of y-component of the resultant, that is
Ry = Ay + By . (2.12)
Or
Ry = Ry j = (Ay + By) j
Since Rx i and Ry j are the rectangular components of the resultant vector R, hence
R = Rx i + Ry j
Or R = (Ax + Bx) i + (Ay + By) j
The magnitude of the resultant vector R is thus given as
R = ( Ax + Bx )2 +( Ay + By )2 . (2.13)
And the direction of the resultant vector is determined from
R ( A + By )
And tan 1 y tan 1 y . (2.14)
Rx ( Ax + Bx )
Similarly for any number of coplanar vectors A, B, C we can write
R = ( Ax + Bx + C x +...)2 + ( Ay + By + C y +...)2 . (2.15)
( Ay + By + C y +...)
And tan 1 . (2.16)
( Ax + Bx + C x +...)
The vector addition by rectangular components consists of the following steps.
i. Find x and y components of all given vectors.
ii. Find x-component Rx of the resultant vector by adding the x-components of all the vectors.
iii. Find y-component Ry of the resultant vector by adding the y-components of all the vectors.
iv. Find the magnitude of resultant vector R using
R = R x 2 + Ry 2

4
v. Find the direction of resultant vector R by using
R
tan 1 y
Rx
Where is the angle, which the resultant vector makes with positive x-axis. The signs of Rx and Ry
determine the quadrant in which resultant vector lies. For that purpose proceed as given below.
Ry
Irrespective of the sign of Rx and Ry, determine the value of tan
1
from the calculator or
Rx
by consulting trigonometric tables. Knowing the value of , angle is determined.
a) If both Rx and Ry components are positive, then the resultant lies in the first quadrant and
its direction is = .
b) If Rx is negative and Ry component is positive, then the resultant lies in the second
quadrant and its direction is =180- .
c) If both Rx and Ry components are negative, then the resultant lies in the third quadrant and
its direction is =180+ .
d) If Rx is positive and Ry is negative, then the resultant lies in the fourth quadrant and its
direction is =360- .
Table 2.1
II Y I
tan - tan +
Rx - Rx +
Ry + Ry +

X X
tan + tan -
Rx - Rx +
Ry - Ry -
III Y IV
Y Y
= =180-

O X X O X
st

X O X X O X

=180+ =360-

Y Y
2.3 PRODUCT OF TWO VECTORS
There are two types of vector multiplication. The product of these two types is known as scalar
product and vector product. As the name implies, scalar product of two vector quantities is a
scalar quantity, while vector product of two vector quantities is a vector quantity.
Scalar or Dot Product
The scalar product of two vectors A and B is written as A.B and is defined as
A.B = AB cos . (2.17)
Where A and B are the magnitudes of vectors A and B and is the angle between them.

B B

Bcos
A A
Fig. 2.10(a) Fig. 2.10(b)
For physical interpretation of dot product of two vectors A and B, these are first brought to a
common origin (Fig. 2.10 a), then
A.B = A (projection of B on A)
A.B = A (magnitude of component of B in the direction of A)
= A (B cos) = AB cos
Similarly B.A = B (A cos) = BA cos
5
We come across this type of product when we consider the work done by a force F whose point of
application moves distant d in a direction making an angle with the line of action of F, as shown
in Fig. 2.11.
F

F cos
d
Work done = (effective component of force in the direction of motion) distance moved
= (F cos) d = Fd cos
Using vector notation
F.d = Fd cos = work done
Characteristics of Scalar Product
1. Since A.B = AB cos and B.A = BA cos = AB cos hence, A.B = B.A, the order of
multiplication is irrelevant. In other words, scalar product commutative.
2. The scalar product of tow mutually perpendicular vectors is zero. A.B = AB cos90 =0
In case of unit vectors i , j and k , since they are mutually perpendicular, therefore,
i . j = j . k = k . i =0.. (2.18)
3. The scalar product of two parallel vectors is equal to the product of their magnitudes. Thus for
parallel vector ( =0)
A.B = AB cos0=AB
In case of unit vectors
i . i = j . j = k . k = 1 .. (2.19)
And for antiparallel vectors (=180)
A.B = AB cos180= -AB
4. The self product of a vector A is equal to square of its magnitude.
A.A = AA cos0 = A2
5. Scalar product of two vector A and B in terms of their rectangular components
A.B = ( Ax i + Ay j + Azk).(
B i + B j + B k)
x y z

## Or A.B = Ax Bx + Ay By + AzBz (2.20)

Equation 2.17 can be used to find the angle between two vectors: since,
A.B = AB cos = Ax Bx + Ay By + Az Bz
Ax Bx + Ay By + AzBz
Therefore cos= .. (2.21)
AB
Vector or Cross Product
The vector product of two vectors A and B, is a vector which is defined as
A B AB sin n . (2.22)
Where n is a unit vector perpendicular to the plane containing A and B as shown in Fig. 2.12 (a).

n AB AB B

A

B B
B
A A
A BA
Fig. 2.12(a) Fig. 2.12(b) Fig. 2.12(c) Fig. 2.12(d)
Its direction can be determined by right hand rule. For that purpose place together the tails of
vectors A and B to define the plane of vectors A and B. The direction of the product vector is
perpendicular to this plane. Rotate vector A into B through the smaller of the two possible angles
with the curl of the fingers of the right hand, keeping the thumb erect. The direction of the product
vector will be along the erect thumb, as shown in the Fig. 2.12 (b). Because of this direction rule,
B A is a vector opposite in sing to A B . Hence
A B = B A . (2.23)
Characteristics of Cross Product
1. Since A B is not same asB A , the cross product is non commutative.
2. the cross product of two perpendicular vectors has maximum magnitude A B =AB sin90
n = AB n
In case of unit vectors, since they form a right handed system and are mutually
perpendicular.

6
i j = k , j k = i , k i = j
3. the cross product of two parallel vector is null vector, because for such vectors =0 or
180, Hence
A B =AB sin0 n = AB sin180 n =0
As a consequence A A = 0
Also i i = j j = k k = 0 .. (2.19)
4. cross product of two vectors A and B in terms of their rectangular component is :
A B = ( Ax i + Ay j + Azk) (B i + B j + B k)
x y z

## A B = ( Ax Bz - AzBy )i + (AzBx - AxBz )j + ( AxBy AyBx )k

.. (2.25)
The result obtained can be expressed fro memory in determinant form as below:
i j k
A B = Ax Ay Az
Bx By Bz
5. The magnitude of A B is equal to area of the parallelogram formed with A and B as two
Examples of Vector Product
i. When a force F is applied on a rigid body at a point whose position vector is r from
any point of the axis about which the body rotates, then the turning effect of the
force, called the torque is given by the vector product of r and F.
=rF
ii. The force on a particle of charged q and velocity v in a magnetic field of strength B
is given by vector product.
F = q (v B)
2.4 TORQUE
We have already studied in school Physics that a turning effect is produced when a nut is
tightened with spanner (Fig. 2.13). The turning effect increased when you push harder on the
spanner. It also depends on the length of the spanner: the longer the handle of the spanner, the
greater the tuning effect of an applied force. The turning effect of a force is called its movement or
torque and is defined as the product of force F and the perpendicular distance from its line of
action to the pivot which is the point around which the body (spanner) rotates. This distance OP is
called moment arm l. Thus the magnitude of torque represented by is
= lF . (2.26)
When the line of action of the applied force passes through the pivot point, the value of
moment arm l=0, so in this case torque is zero.

The nut is easy to turn with a spanner It is easier still if the spanner has a long handle
Fig. 2.13
We now consider the torque due to a force F acting on a rigid body. Let the force F can be
resolved into two rectangular components, F sin perpendicular to r and F cos along the
direction of r (Fig. 2.14 a). The torque due to F cos about pivot O is zero as its line of action
passes through point O. Therefore, the torque due to F is equal to the torque due to F sin only
about O. It is given by
= (F sin)r = rF sin . (2.27)
Alternatively the moment arm l is equal to the magnitude of the component of r perpendicular to
the line of action of F as illustrated in Fig. 2.14 (b). Thus
= (r sin)F = rF sin . (2.28)
Where is the angle between r and F

F
F

Fsin
P rsin
r
O r P
O

7
From Eq. 2.27 and Eq. 2.28 it can be seen that the torque can be defined by the vector product of
position vector r and the F
= (rF sin) n .. (2.29)
Where (rF sin) is the magnitude of the torque. The direction of the torque represented by n is
perpendicular to the plane containing r and F given by right hand rule for the vector product of two
vectors.
The SI unit for torque is Newton meter (N m).
Torque is the counter part of force for rotational motion. Just as force determines the linear
acceleration produced in a body, the torque acting on a body determines its angular acceleration.
If the body is at rest or rotating with uniform angular velocity, the angular acceleration will be zero.
In this case the torque acting in the body will be zero.
2.5 EQUILIBRIUM OF FORCES
We have studied in school Physics that if a body, under the action of a number of forces, is at rest
or moving with uniform velocity, it is said to be in equilibrium.
First Condition of Equilibrium
A body at rest or moving with uniform velocity has zero acceleration. Form Newtons Law of
motion the vector sum of all force acting on it must be zero.
This is known as first condition of equilibrium. Using the mathematical symbol F for the sum of all
forces we can write
F = 0 .. (2.30)
In case of coplanar forces, this condition is expressed usually in terms of x and y components of
the forces. We have studied that x-component of the resultant force F equals the sum of x-
directed or x-components of all the forces acting on the body. Hence
Fx = 0 .. (2.31)
Similarly for the y-directed forces, the resultant of y-directed forces should be zero. Hence
Fy = 0 .. (2.32)
It may be noted that if the rightward forces are taken as positive then leftward forces are taken as
negative. Similarly if upward forces are taken as positive then downward forces are taken as
negative.
2.6 EQUILIBRIUM OF TORQUES
Second Condition of Equilibrium
Let tow equal and opposite forces are acting on a rigid body as shown in Fig. 2.17.

Fig. 2.17
Although the first condition of equilibrium is satisfied, yet it may rotate having clockwise turning
effect. As discussed earlier, for angular acceleration to be zero, the net torque acting on the body
should be zero. Thus for a body in equilibrium, the body should be zero. This is known as second
condition of equilibrium. Mathematically it is written as
= 0 .. (2.33)
By convention, the counter clockwise torqueses are taken as positive and clockwise torques as
negative. An axis is chosen for calculating the torques. The position of the axis is quite arbitrary.
Axis can be chosen anywhere which is convenient in applying the torque equation. A most helpful
point of rotation is the one through which lines of action of several forces pass.
We are now in a position to state the complete requirements for a body to be in equilibrium, which
are
(1) F = 0 i.e Fx = 0 and Fy = 0
(2) = 0
When 1st condition is satisfied, there is no linear acceleration and body will be in translational
equilibrium. When 2nd condition is satisfied, there is no angular acceleration and body will be in
rotational equilibrium.
For a body to be in complete equilibrium, both conditions should be satisfied, i.e., both linear
acceleration and angular acceleration should be zero.
If a body is at rest, it is said to be in static equilibrium but if the body is moving with uniform
velocity, it is said to be in dynamic equilibrium.

8
We will restrict the applications of above mentioned conditions of equilibrium to situations in which
all the forces lie in a common plane. Such forces are said to be coplanar. We will also assume that
these forces lie in the xy-plane.
If there is more than one object in equilibrium in a given problem, one object is selected at a time
to apply the conditions of equilibrium.
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