Engineering Mechanics-I (Statics)

CHAPTER - 1
1. VECTORS AND SCALARS
1.1 Introduction
Mechanics is a physical science which deals with the state of rest or motion of rigid bodies
under the action of forces. It is divided into three parts: mechanics of rigid bodies,
mechanics of deformable bodies, and mechanics of fluids. Thus it can be inferred that
Mechanics is a physical science which deals with the external effects of force on rigid
bodies. Mechanics of rigid bodies is divided into two parts: Statics and Dynamics.
Statics: deals with the equilibrium of rigid bodies under the action of forces.
Dynamics: deals with the motion of rigid bodies caused by unbalanced force acting on them.
Dynamics is further subdivided into two parts:

 Kinematics: dealing with geometry of motion of bodies without reference to the

forces Causing the motion, and
 Kinetics: deals with motion of bodies in relation to the forces causing the motion.
Basic Concepts:
The concepts and definitions of Space, Time, Mass, Force, Particle and Rigid body are
basic to the study of mechanics.
Space: - is the geometric region occupied by bodies whose position is described by linear &
angular measurement relative to a coordinate system.
Time: - is the measurement of succession of events (exists) and it is a basic quantity in
dynamics.
Mass: - is the measure of inertia of a body, which is resistance to a change of velocity or its
state of motion; this includes changes to its speed, direction or state of rest.
Force: - is the action of a body on another. A force tends to move a body in the direction of
its action and it is characterized by Magnitude, Direction of action and Point of application.
Particle: - is a body of negligible dimensions which may analyzed as a mass concentrated at
a point. A body can be treated as a particle when its dimension is irrelevant to the description
of its position or the action of the force applied to it.
Rigid body: - a body is considered as rigid when the relative movement between its parts is
negligible.
In this course, the bodies are assumed to be rigid such that whatever load applied, they
don‟t deform or change shape. But translation or rotation may exist. The loads are
assumed to cause only external movement, not internal. In reality, the bodies may deform.

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But the changes in shapes are assumed to be minimal and insignificant to affect the
condition of equilibrium (stability) or motion of the structure under load.

When we deal Statics/Mechanics of rigid bodies under equilibrium condition, we can

represent the body or system under a load by a particle or centerline. Thus, the general
response in terms of other load of the bodies can be spotted easily.

Fundamental Principles

The three laws of Newton are of importance while studying mechanics:

First Law: A particle remains at rest or continues to move in a straight line with uniform
velocity if there is no unbalanced force on it.
Second Law: The acceleration of a particle is proportional to the resultant force acting on it
and is in the direction of this force.
F  m a
Third Law: The forces of action and reaction between interacting bodies are equal in
magnitude, opposite in direction and collinear.
The first and third laws have of great importance for Statics whereas the second one is basic
for dynamics of Mechanics.
Law of gravitation: - Another important law for mechanics is the Law of gravitation by
Newton, as it usual to compute theweight of bodies. Accordingly:

F  G m1m
2
2

r
Thus the weight of a mass „m‟ is W = mg

1.2 SCALARS AND VECTORS

1.2.1 Definition and properties
After generally understanding quantities as Fundamental or Derived, we shall also treat them
as either Scalars or Vectors.
Scalar quantities: - are physical quantities that can be completely described (measured) by
their magnitude alone. These quantities do not need a direction to point out their application
(Just a value to quantify their measurability). They only need the magnitude and the unit of
measurement to fully describe them.
E.g. Time[s], Mass [Kg], Area [m2], Volume [m3], Density [Kg/m3], Distance [m], etc.

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Vector quantities: Like Scalar quantities, Vector quantities need a magnitude. But in
addition, they have a direction, and sometimes point of application for their complete
description. Vectors are represented by short arrows on top of the letters designating them.
E.g. Force [N, Kg.m/s2], Velocity [m/s], Acceleration [m/s2], Momentum [N.s, kg.m/s], etc.

1.2.2 Types of Vectors

Generally vectors fall into the following three basic classifications:
Free Vectors: are vectors whose action in space is not confined or associated with a unique
line in space; hence they are „free‟ in space.
E.g. Displacement, Velocity, Acceleration, Couples, etc.

Sliding Vectors: are vectors for which a unique line in space along the action of the quantity
must be maintained.

E.g. Force acting on rigid bodies.

NB: From the above we can see that a force can be applied any where along its line of action
on arigid body with out altering its external effect on the body. This principle is known as
Principle of Transmissibility.
Fixed Vectors: are vectors for which a unique and well-defined point of application is
specified to have the same external effect.
E.g. Force acting on non-rigid (deformable) bodies.

1.2.3 Representation of Vectors

A) Graphical representation
Graphically, a vector is represented by a directed line segment headed by an arrow. The
length of the line segment is equal to the magnitude of the vector to some predetermined
scale and the arrow indicates the direction of the vector.
Length of the line equals, to some scale, the
magnitude of the vector and the arrow indicates the
direction of the vector

NB: The direction of the vector may be measured by an angle θ from some known reference
direction.

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B) Algebraic (arithmetic) representation

Algebraically a vector is represented by the components of the vector along the three
dimensions. For example:

A  Ax i  Ay j  Az k

Where Ax , Ay and Az are the components of the vector A along the x, y and z axes

respectively.

The vectors i, j and k are the unit vectors along the three respective axes.

Ax  A cos  x  Al , l  cos  x

Ay  A cos y  Am , m  cos y

Az  A cos  z  An , n  cos  z

Where l , m and n are the directional cosines of the vector A . Thus:

A2  Ax 2  Ay 2  Az 2  l 2  m2  n2  1

Properties of vectors
Equality of vectors: Two free vectors are said to be equal if and only if they have the same
magnitude and direction.

The Negative of a vector: is a vector which has equal magnitude to a given vector but
opposite in direction.

Null vector: is a vector of zero magnitude. A null vector has an arbitrary direction.

Unit vector: is any vector whose magnitude is unity.

A unit vector along the direction of a certain vector, say vector A (denoted by u A ) can then
be found by dividing vector A by its magnitude.

A
uA 
A

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Generally, any two or more vectors can be aligned in different manner. But they may be:
 Collinear-Having the same line of action.
 Coplanar- Lying in the same plane.
 Concurrent- Passing through a common point

1.3 Operations with Vectors

Scalar quantities are operated in the same way as numbers are operated. But vectors are not
and have the following rules:

1.3.1 Vector Addition or Composition of Vectors

Composition of vectors is the process of adding two or more vectors to get a single vector, a
resultant, which has the same external effect as the combined effect of individual vectors on
the rigid body they act.
There are different techniques of adding vectors
A) Graphical Method
I. The parallelogram law
The law states, “if A and B are two free vectors drawn on scale, the resultant (the equivalent
vector) of the vectors can be found by drawing a parallelogram having sides of these vectors,
and the resultant will be the diagonal starting from the tails of both vectors and ending at the
heads of both vectors.”

Once the parallelogram is drawn to scale, the magnitude of the resultant can be found by
measuring the diagonal and converting it to magnitude by the appropriate scale. The direction
of the resultant with respect to one of the vectors can be found by measuring the angle the
diagonal makes with that vector.

Note: As we can see in the above figure.

A B  R  B  A ⇒ Vector addition is commutative

The other diagonal of the parallelogram gives the difference of the vectors, and depending
from which vertex it starts, it represents either A – B or B – A

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Since the two diagonal vectors in the above figure are not equal, of course one is the negative
vector of the other, vector subtraction is not commutative.

i.e A B  B  A

NB. Vector subtraction is addition of the negative of one vector to the other.

II. The Triangle rule

The Triangle rule is a corollary to the parallelogram axiom and it is fit to be applied to more
than two vectors at once. It states “If the two vectors, which are drawn on scale, are placed tip
(head) to tail, their resultant will be the third side of the triangle which has tail at the tail of
the first vector and head at the head of the last.”

R  A B

Thus the Triangle rule can be extended to more than two vectors as, “If a system of vectors
are joined head to tail, their resultant will be the vector that completes the polygon so
formed, and it starts from the tail of the first vector and ends at the head of the last vector.”

R  A B C

NB. From the Triangle rule it can easily be seen that if a system of vectors when joined
head to tail form a closed polygon, their resultant will be a null vector.
III. Analytic method.
The analytic methods are the direct applications of the above postulates and theorems in
which the resultant is found mathematically instead of measuring it from the drawings as in
the graphical method.

A. Trigonometric rules:
The resultant of two vectors can be found analytically from the parallelogram rule by

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applying the cosine and the sine rules.

Consider the following parallelogram. And let θ be the angle between the two vectors

Consider triangle ABC

From cosine law,

 R  A  B  2 A B cos  
2 2 2

A  B  2 A B cos  
2 2
 R 

This is the magnitude of the RESULTANAT of the two vectors

Similarly, the inclination, β, of the resultant vector from A can be found by using sine law
which is the angle the resultant makes with vector A.

sin    sin  
 
B R

 B 
   sin 1 sin   
 R 
 

Decomposition of vector: is the process of getting the components of a given vector along
some other different axis. Practically decomposition is the reverse of composition.
Consider the following vector A. And let our aim be to find the components of the vector
along the n and t axes.

Fromtriangle ABC at (b),  =180 -    

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An A sin 
=  An = A
sin  sin  sin 
sin 
Similarly, At = A
sin 
The above are general expressions to get the components of a vector along any axis. In most
cases though, components are sought along perpendicular axes, i.e.

 
 =180 -    90  sin   1

An = A sin   A cos 

At = A sin   A cos 

B. Component method of vector addition

This is the most efficient method of vector addition, especially when the number of vectors to
be added is large. In this method first the components of each vector along a convenient axis
will be calculated. The sum of the components of each vector along each axis will be equal to
the components of their resultant along the respective axes. Once the components of the
resultant are found, the resultant can be found by parallelogram rule as discussed above.
Example 1: Two vectors of the indicated magnitudes are given as shown on the
diagram.
i. Perform the composition of the vectors.
ii. Determine an angle θ formed between the resultant R and vertical y-axis.
iii. Express the resultant R in component method and calculate its unit vector.
iv. Find the vector difference, D from A-B

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Solution:
i) First complete the two sides of parallelogram and create the resultant vector R
along the diagonal starting from the tails and ending at the heads of two vectors as
shown in the figure below.

Using cosine rule:

2 2 2
R  A  B  2 A B cos( ) ,   70 from the geometry of parallelogram

 402  202  2  40  20  cos(70)

 2000  1600  0.342 

R  38.12 Units

The magnitude of the resultant can be also measured directly from the scale drawing of the
vectors.

ii) The angle  formed between the resultant R and vertical y-axis is determined by
using sin rule as follows:
sin(  50) sin(70)

A R
sin(  50) sin(70)

40 38.12
40  sin(70)
sin(  50) 
38.12
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 40  0.94 
  50  sin 1  
 38.12 
  50  sin 1  0.986 
  80.41  50
  30.41

iii) In order to write the resultant vector R in a component form, the vector is resolved
in to its horizontal and vertical components as shown in the diagram below.

R  Rx i  Ry j

   R cos(59.59)  i   R sin(59.59)  j

  38.12  cos(59.59)  i   38.12  sin(59.59)  j

R  19.29i  32.86 j
Unit vector of the resultant vector R is given by:
R
UR 
R

19.29i  32.86 j

 19.29    32.86 
2 2

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19.29i  32.86 j

38.12
U R  0.51i  0.86 j

iv) To carry out the vector difference D, create a vector in the second diagonal
direction of the parallelogram as follows:

2 2 2
D  A  B  2 A B cos(110)

D  402  202  2  40  20  cos(110)

 2000  1600  0.342 

D  50.47 Units

The angle  that the vector difference D forms with horizontal can be determined using sin
rule as follows.
sin(   40) sin(110)

A D

A
 sin(   40)  sin(110)
D

 40 
   40  sin 1 sin(110)
 50.47 

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   48.13  40  8.13

The vector difference D can be resolved to component form using the angle  it forms with
horizontal.

D  Dxi  Dy j

  D cos    i  D sin    j

 50.47cos 8.13 i  50.47sin 8.13 j

D  49.96i  7.14 j

NB: This vector difference can also be determined by first converting the individual vectors
A and B to their component form and then subtracted algebraically.

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1.4 Vector Multiplication: Dot and Cross products

1.4.1 Multiplication of vectors by scalars
Let n be a non-zero scalar and A be a vector, then multiplying A by n gives as a vector whose

magnitude is n A and whose direction is in the direction of A if n is positive or is in

opposite direction to A if n is negative.

Multiplication of vectors by scalars obeys the following rules:-

i. Scalars are distributive over vectors.

 
n A  B  n A  nB

ii. Vectors are distributive over scalars.

 n  m  A  n A  mA
iii. Multiplication of vectors by scalars is associative.

 nm  A  n  m A   m  n A 
1.4.2 Multiplication of vector by a vector
In mechanics there are a few physical quantities that can be represented by a product of
vectors. Eg. Work, Moment, etc
There are two types of products of vector multiplication

1.4.2 Dot Product: Scalar Product

The scalar product of two vectors A and B which are θ degrees inclined from each other

denoted by A.B (A dot B) will result in a scalar of magnitude A . B cos   .

i.e A.B  A . B cos  

If the two vectors are represented analytically as

A  axi  a y j  az k and B  bxi  by j  bz k , then

A.B  axbx  a yby  az bz

1.4.3 Cross Product: Vector Product

The vector product of two vectors A and B that are θ degrees apart denoted by AxB (A
cross B) is avector of magnitude ⃗ ⃗⃗ and direction perpendicular to the
plane formed by the vectors A and B. The sense of the resulting vector can be determined

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by the right-hand rule.

A  B  A B sin  , perpendicular to the plane formed by A and B

If the two vectors are represented analytically as,

A  axi  a y j  az k and B  bxi  by j  bz k , then

the cross product A  B will be the determinant of the three by three matrix as,

A  B  B  A

NB. Vector product is not commutative; in fact, ⃗ ⃗⃗ ⃗⃗ ⃗

Moment of a Vector
The moment of a vector V about any point O is given by:

M   r V

Where: r is a position vector from point O to any point on the line of action of the vector.

M    ryVz  rzVy  i   rzVx  rxVz  j   rxVy  ryVx  k

Position vector r is defined as a fixed vector that locates a point in space relative to another
point in space.

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Example-2:

Vector A with magnitude of 10 originates at point (1, 0, 1) and passes through (1, 1, -1).
Vector B with magnitude of 12 pass through the points (2, 2, 0) and (4, 3, -1).
a. Determine the moment of each vector about point O(5, -3, 4)
b. Calculate the projection of each vector along an axis passing through point (2, -1, 2)
and (-1, 2, -2).
Solution:
a. The component form of vector A passing through the points (1, 0, 1) and (1, 1, -1), is
calculated by multiplying magnitude of vector A with its unit vector u A through the
line of action (two points).

A  A uA

 
 x  x  i   y2  y1  j   z2  z1  k
 A  2 1 
 x x 2 y y 2 z z 2 
  2 1  2 1  2 1 

 10 
1  1 i  1  0  j   1  1 k 
 
 0   1   2 
2 2 2
 
 j  2k 
 10  
 5 
 1 2 
 10  j k
 5 5 

 10 0.447 j  0.894k 

A  4.47 j  8.94k

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The moment of vector A is the cross product of position vector r OA from point O(5, -3, 4) to
any point on the line of action vector A , consider point (1, 0, 1)

M A  r OA  A
  x2  x1  i   y2  y1  j   z2  z1  k   A

 1  5  i   0   3  j  1  4  k    4.47 j  8.94k 

  4i  3 j  3k    4.47 j  8.94k 

i j k
M A  r OA  A  4 3 3
0 4.47 8.94

 3  8.94    3 4.47  i  3  0    4  8 j   4  4.47  3  0  k

  26.82  13.41 i   35.76 j   17.88 k

M A  13.41i  35.76 j  17.88k

Similarly, the component form of vector B passing through the points (2, 2, 0) and (4, 3, -1),
is calculated by multiplying magnitude of vector B with its unit vector u B through the line of
action (two points).

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B  B uB

 
 x  x  i   y2  y1  j   z2  z1  k
 B  2 1 
 x x 2 y y 2 z z 2 
  2 1  2 1  2 1 
 
 12   4  2  i   3  2  j   1  0  k 
 
 2   1   1
2 2 2
 
 2i  j  k 
 12  
 6 

 2 1 1 
 12  i j k
 6 6 6 
 12 0.82i  0.41 j  0.41k 

B  9.84i  4.92 j  4.92 k

The moment of vector B is the cross product of position vector r OB from point O(5, -3, 4) to
any point on the line of action vector B , let‟s take point (2, 2, 0)

M B  r OB  B
  x2  x1  i   y2  y1  j   z2  z1  k   B

  2  5  i   2   3  j   0  4  k   9.84i  4.92 j  4.92k 

  3i  5 j  4k   9.84i  4.92 j  4.92k 

i j k
M B  r OB  B  3 5 4
9.84 4.92 4.92

 5  4.92    4  4.92 i  4  9.84    3 4.92  j   3 4.92  5  9.84  k

  24.6  19.68 i   39.36  14.76 j   14.76  49.2 k

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M B  4.92i  54.12 j  63.96k

In other way, we can take another point on the vector B which is (4, 3, -1). Therefore,

position vector r OB will be from point O(5, -3, 4) to (4, 3, -1) and the moment is then
calculated as:
M B  r OB  B
  x2  x1  i   y2  y1  j   z2  z1  k   B

  4  5  i   3   3  j   1  4  k   9.84i  4.92 j  4.92k 

  i  6 j  5k   9.84i  4.92 j  4.92k 

i j k
M B  r OB  B  1 6 5
9.84 4.92 4.92

 6  4.92    5 4.92 i  5  9.84    1 4.92  j   1 4.92   6  9.84  k

  29.52  24.6 i   49.2  4.92 j   4.92  59.04 k

M B  4.92i  54.12 j  63.96k

(This result using the second point is same with the above one)
The total moment about point O will be the summation of moments due to vector A and B:
MT  M A  M B
  13.41i  35.76 j  17.88k    4.92i  54.12 j  63.96k 

M T  18.33i  89.88 j  81.84 k

b. The projection of a vector on to the axis is the dot product of that vector and the unit

vector u L along the given axis or line.

Accordingly, the projection of vector A on to the line passing through the points (2, -
1, 2) and (-1, 2, -2) will be:

18 AMU Lecture Note (A. G) Faculty of Civil Engineering

Engineering Mechanics-I (Statics)

PAL  A.u L

 
 x  x  i   y2  y1  j   z2  z1  k
 A.  2 1 
 x x 2 y y 2 z z 2 
  2 1  2 1  2 1 
 1  2 i  2  1 j  2  2 k 
       
  4.47 j  8.94k  . 
 
 3   3   4 
2 2 2
 
 3i  3 j  4k 
  4.47 j  8.94k  .  
 44 
 3 3 4 
  4.47 j  8.94k  .  i j k
 44 44 44 

  4.47 j  8.94k  . 0.452i  0.452 j  0.603k 

  0  0.452    4.47  0.452    8.94  0.603

 0  2.02  5.39
PAL  7.41Units (The result of dot product is scalar magnitude)

The scalar magnitude can be expressed in a vector form by multiplying it again with the unit
vector u L .

P AL  PAL .u L

 7.41 0.452i  0.452 j  0.603k 

P AL  3.349i  3.349 j  4.468k

19 AMU Lecture Note (A. G) Faculty of Civil Engineering

Engineering Mechanics-I (Statics)

Again the projection of vector B on to the line passing through points (2, -1, 2) and (-1, 2, -2)
will be:

PBL  B.u L

 
 x  x  i   y2  y1  j   z2  z1  k
 B.  2 1 
 x x 2 y y 2 z z 2 
  2 1  2 1  2 1 
 1  2 i  2  1 j  2  2 k 
       
  9.84i  4.92 j  4.92k  . 
 
 3   3   4 
2 2 2
 
  9.84i  4.92 j  4.92k  . 0.452i  0.452 j  0.603k 

  9.84  0.452    4.92  0.452    4.92  0.603

 4.447  2.223  2.967
PBL  0.743Units (The result of dot product is scalar magnitude)

The scalar magnitude can also be expressed in a vector form by multiplying it again with the

unit vector u L .

P BL  PBL .u L

 0.743  0.452i  0.452 j  0.603k 

P BL  0.336i  0.336 j  0.448k

20 AMU Lecture Note (A. G) Faculty of Civil Engineering