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APPLIED MECHANICS – GEC 214
Ajayi, O
...
MNSE, B
...
Eng (Nsukka), PhD (in View)
...
P 51
Ext
...
PRINCIPLES OF MECHANICS
I
...

It is basic to Engineering Sciences because of its use in the analysis and design of engineering
components and systems
...

I
...

Fluid Mechanics: This is the part of mechanics which deals with the behaviour of liquids and
gases under the action of balanced and unbalanced forces
...

A sub division of fluid mechanics include: Compressible fluid mechanics and incompressible
fluid mechanics
...
It studies the reaction of bodies to external forces i
...
it looks
into the resulting stress and/or deformation that arise as a result of application of external loads
(or forces)
...

Solid Mechanics: In this part of mechanics, the bodies are assumed to be rigid solids
...
These deformations are
however very small and do not cause problems to the conditions of equilibrium of such bodies
...

Sub division of solid mechanics
This include: (i) Statics and (ii) Dynamics
...
The emphasis in this course shall be on solid mechanics
...

Concept of time: This answers the question ‘when’
...

Concept of space: This answers the question ‘where
...
This
definition arises from the frame of reference that all point are defined from an imaginary origin
measured out in three direction called xyz axes
...
e
...
2kg of a body is placed on a tower
...


2
...
It could be direct or indirect (i
...
at a distance)
...
1 KINDS OF FORCES
1
...
g
...


2

2
...
It can be gravitational,
electrical or magnetic in nature
...

2
...
g
...

(ii) Direction: This is the line of action to which the force is acting (or the effect of the force can
be felt) e
...
10N North 45o South
...

(iii) Point of Action: This is the point on a body on which the force is acting
...

2
...
g
...

(2) Concentrated Force: This is a force whose effect on a body is felt on very small area
compared to the entire surface area of the body
...

2
...
Spatial or space
forces have their lines of action lying in three dimensional space
...
The following configurations are obtainable: (i) concurrent
coplanar
(ii) Non concurrent coplanar and (iii) Spatial; concurrent and non concurrent
...
5 LAWS OF SOLID MECHANICS
Owing to the contribution of Sir Isaac Newton to science, Newtonian mechanics still remains the
basis of engineering sciences
...

These laws govern the behaviours of bodies under the action of forces
...

N
...

(2) ‘Newton’s’ Second Law: States that if the resultant force acting on a particle is not zero;
the

rate of change of velocity with time is directly proportionally to the resultant force,

and takes place in the direction of the force
...

t

Thus;

where F = force, V = velocity and t = time

Fα a

Applying the principles of proportionality and dimensionality;
F = ma where m = constant of proportionality called mass
...

(3) ‘Newton’s’ Third Law: States that the force acting on a body produces a reaction which
is equal in magnitude and opposite in direction
...

Mathematically;

Fα

Mm
r2

where F = force of attraction btw bodies, M and m are masses

of the bodies and r2 is the square of their distance apart
...

(5) The parallelogram law for addition of forces: states that if two forces acting on a body
can be represented by the two sides of a parallelogram, then the resultant can be
represented by the diagonal of the parallelogram in magnitude and direction
...


3
...

Another name for resultant of a force system is ‘ EQUIVALENT FORCE’: the equivalent force
is the vector sum of the individual forces
...
1 RESULTANT OF CONCURRENT FORCES
...

Triangle Law of Forces
This law sates that ‘if two forces acting on a body can be represented by the sides of a
triangle; then the third side drawn into size can represent the resultant in magnitude and
direction
...

X
Y

If forces X and Y can be represented by the sides of a rectangle in magnitude and direction, the
third side will represent the resultant (sum of X and Y) in magnitude and direction
...
’
For example: consider the body below, acted upon by the system of coplanar forces

6

X
Y

M

X

Z

If the forces X, Y, Z and M can be used to represent the sides of a polygon, then the line which
is drawn to close up the polygon will represent the resultant (sum of X, Y, Z, and M)
...
The last side that would close the polygon is the resultant R
...
Graphical method of solution: This method makes use of the three laws explained above
...
The direction of each force is measured using a protractor
...

2
...
This method gives a more accurate solution
...
Rectangular component method: This is carried out by resolving the forces into their
rectangular components
...
e into their x and y components
...
For example, determine the resultant of the two
force system below:

7

F1

30o
10o
F2

The resultant R = ΣRx + ΣRy
And

ΣR y = ( f y ) + ( f y )
1

Where ΣRx = ( f x )1 + ( f x )2

2

Then, the magnitude of R is found by :
2
R = Rx2 + Ry

⎛R ⎞
And the direction is found by: tan −1 ⎜ y ⎟
⎝ Rx ⎠

Note: all angles are measured counterclockwise from the positive x-axis
...
0 MOMENT AND COUPLE
Moments of a force: The moment of a force about a point is defined as the product of the force
and the perpendicular distance measured from the point along the line of action to the force
...
Consider the diagram below: O is
the fixed point of reference
...


The moment of F about O is Mo = F x d

Conventional sign of moment: Clockwise moment is taken to be negative and anticlockwise
moment is taken to be positive
...

Method of determining the moment of a force
Depending on the nature of application of a force; there are three methods that can be used to
determine the moments of a force (1) by direct method: Using the definition that moment is force
multiplied by perpendicular distance
...

For this theorem to be applied adequately; the forces concerned will first have to be resolved to
their rectangular components after which each moment due to individual component is
determined and then summed in accordance with the sign convention of moment
...

WORKED EXAMPLE
Determine the moment of force F about O

F = 24N
20o
0
...

9

Project F backwards to P where a line from O meets F at 90o
F = 24N
20o
0
...
4m
Therefore, moment Mo = F x d = 24 x 1
...

Fx = Fcos20o = 24cos20o = 22
...
2N

Thus, Moment Mo = - ( Fx x 0
...


Mo = -22
...
7 + 8
...

( c) By Principle of Transmissibility: Take the force to Q, a convenient point along the line of
action of b F
Moment Mo = F x d = Fy x 4 = 33Nm

y
O

Fy
Q

0
...
Consider the
diagram below
...

F1

x

a
d

a

y

F2

Moment of a couple: This is defined as the product of one of the force and the perpendicular
distance between the forces
...
e
...

Equivalent couples: Two couples acting on the plane or parallel planes are said to be equivalent
if they have the same moment acting in the same direction
...
1 Replacing a force with a force – couple system
...
2

Consider the diagram above
...
Then, two equal and
opposite forces ( both equal to F) will be placed at Y
...
See diagram below (fig
...


11

F

x

a
d

a

y

F

F

The two Fs at Y counsels each other out and thus, maintains the mechanical effect of F at x
...
The moment of the couple is F x d
...

WORKED EXAMPLE
Consider the couple system below
...
determine the magnitude of the
forces
...
5kN
Fx = 5cos30o = 4
...
5KN
4
...
3KN

2
...
3 x 4) + (-2
...
2 = - 12
...

Moment is 12
...


13

(2)

x

A

d

F
y

F

4m

c

4m

To cal
...


[ xy ] = [ xc ] + [ yc ]
2

2

= 42 + 42 = 5
...
8kNm
F
...
8
F = 12
...
7 = 2
...
0 RESULTANT OF NON- CONCURRENT COPLANAR FORCE SYSTEM

As was defined earlier for non-concurrent coplanar forces; there’s not the same point of action
for non- concurrent coplanar forces
...

Estimating the resultant and its line of action

Resultant: To find the resultant, its magnitude and direction, can be calculated using the
rectangular component method introduced earlier
...


Line of Action of the Resultant: To find the line of action which can also be called the location of
the resultant; the principle of moment is employed
...
e
...
x

where ∑ M = sum of moments and x =

distance of resultant From point of reference
...

F1

F3
F4
∅3

∅4

F5
F2
To Estimate The Resultant
Resolve the forces to their rectangular components
...


Rx = F1x + F2x + F3x + F4x + F5x
Rx = 0 + 0 + F3Cos ∅3 + F4Cos∅4 + 0
And:
Ry = F1y + F2y + F3y + F4y + F5y
Ry = -F1 + F2 + F3Sin∅3 + F4Sin∅4 - F5
Hence, knowing the values of F1,F2,F3,F4 and F5 can lead to the final determination of Rx and
Ry and thus, R in magnitude and direction
...

-

Find the individual moment of the forces about a reference point

-

Sum up all the values of moment found above

15

-

Equate this sum to the moment |R|
...

Thus;

From diagram above; taking moment about point A
...
x = |∑MA|

Use the positive value of ∑MA

Hence, x = |∑MA|
|Ry|

And |Rx|
...
y = |∑MA|
|Rx|
WORKED EXAMPLES

1
...

20N

7N
3m

80N

1m

100N
2m

2m
B

A
Solution

The question is all about finding the resultant and location of resultant
(a) To find the resultant, resolve the forces to its rectangular components
Rx = 0 i
...
no force in the horizontal direction
Ry = -7 - 20 - 80 + 100 = -7N
R = p(02 +72) = 7N

16

R = 7N in the downward vertical
(b) To find the location of Resultant:
Take moment about A
∑MA = 0 + (-20) x 3 + (-30 x 4) + 100 x 6
= 0 - 60 - 320 + 600 = 220 Nm

...
y = 220

No location since Rx is zero
...
x = 220
x = 220 = 31
...
3m on the left side of A
2
...

Fcos30

12N
70m

60o

30o

17N

O
80m

Solution

Taking moment about point O
...
6F = 1360 +1320 = 2680

...
6 = 44
...
Determine the resultant of the forces on the truss and its location along AB
...
2N
Ry = -8 - 35 - 10 -7Cos370 = -58
...
R = p(4
...
6)2 = 58
...
8 N

α = 2740

For the location:
Take moment about A
∑MA = (-8 x 3) + (-35 x 5) + ((-7Cos 370) x 5) + (-10 x 7)
= -24 - 175 – 28
...
∑MA = 297 Nm clockwise
Ry
...
1m
58
...
1m to the right of A
6
...
It occurs when the load applied is not concentrated at a point, but is spread over an
area, along a line or throughout the entire solid body e
...
Weight of a beam; loads caused by
wind pressure and liquid pressure etc
...
Triangular Load: see figure below:
F = ½ Pb
P

b

This is define as a distributed load whose intensity (or effect) varies uniformly from zero to a
maximum intensity P
...
g
...
Uniformly distributed Load: see figure below
P

b
F = Pb
The resultant of a distributed load can be determined by replacing each distributed load by its
equivalent concentrated load, placed at its centroid
...

19

WORKED EXAMPLE

Determine the equivalent resultant force of the distributed loads of the figures below
...

5KN/m
2KN/m

8m

2m

3m

2
...

8KN

1m

6KN

1m

2m

4m

CENTROID OF AREAS OF COMMON SHAPES

1
...
TRIANGLE

2

/ 3h

h
c
1

/ 3h

1

2

/ 3b

/ 3b
A = ½ bh

3
...
SEMICIRCLE

c
r
4r

/3

o

A = 1/2 π r2

Assignment: Write and draw other centroids of areas of common shapes into your note
...

(i)

Divide the loading diagram into the component load distributions
...


(iii)

Evaluate the centroids of the associated area of its loading diagram
...


(v)

Determine the location or line of action of resultant from the total moment about a
fixed point
...
The loading diagram can be divided into a triangle and a rectangle
...
3m

3
...

∴ R = 23kN downwards
...

Taking moment about A
∑MA = - 20 x 5 + (- 3 x 7
...
9 = -121
...
9kNm clockwise
...
x = 121
...
9/23 = 5
...
R = 23kN downwards at 5
...

2
...

Hence;

25

21KN

4KN/m

1
...
5m

3
...
25m

Equivalent load of rectangle = 3kN/m x 7 = 21kN
“

“

“ Spandrel = 1/3 x 1 x 4 = 1
...


Centroids: For rectangular = ½ b = ½ x 7 = 3
...
75m
Magnitude of resultant
Ry = -21 – 1
...
3kN
R = 22
...

Location of the line of action
...
5) + ( - 1
...
25) = - 82
...
405kNm clockwise
...
x = 82
...
405/22
...
7m
...
3kN downwards at 3
...
The distributed load is of 2-types: a rectangle and a triangle
...
30m

2
...

Centroids : for triangles = 2/3x b = 2/3 x 4 = 2
...
7) + (-12 x 5 ) + (-8x7) = -148
...
4KNm clockwise
Thus:
/Ry/
...
4
x = 148
...
6m
...
0 APPLICATION OF VECTORS TO RESOLUTION OF VECTORS

27

Vector quantities are quantities which have both magnitude and direction
...

The i and j can also be said to be unit vectors which are in the direction of the positive
sense of x and y axes respectively
...
A force of 500N is exerted on a particle as shown in the diagram (a) below
...

28

2
...
(i) Determine the
horizontal and vertical components of the Tension
...

y

30o

T

3m

x

4m

500N

b

a

Solutions

1
...
Representing the Tension in its triangular form:
Computing ∅
Tan ∅ = ¾ = 0
...
75 = 36
...
870) and Ty = -TSin∅
Tx = 64N

and Ty = -48N

(i) The Horizontal Component = Txi = 64Ni

29

The Vertical Component = Tyj = -48Nj
(ii) Direction of Tension is:
α = tan-1 Ty = tan-1 -48
Tx =

64

α = 3230 and T = (64i – 48j) N
8
...

Since the resultant of a set of forces acting on a body is the sum total of the individual effect of
each force making up the system of forces; the resultant R can then be analysed as:
n

R = ∑ Fi where Fi is the individual force making up the system of forces
...

Therefore, component wise, the resultant R will be:
R = ∑ Rx i + ∑ Ry j --------------------------------------------------------------- (1)

For example: Assume the forces acting on a body are T, P and Q
The Resultant R = T + P + Q
Resolving each force into its rectangular components gives:
Rxi + Ryj = (Txi + Tyj) + (Pxi + Pyj) + (Qxi + Qyj)
= (Tx + Px + Qx)i + (Ty + Py + Qy)j
It follows by vector rule that:
Rx = Tx + Px + Qx and Ry = Ty + Py + Qy -------------------------------- (2)
Equations (2) and (1) therefore agrees
...

RULE FOR ANALYSIS

30

To use the vector form of analysis to solve for the resultant of a system of forces; the following
rule must be adhered to for easy estimation
...
Resolve the individual force in the system to its rectangular components
2
...
Using the values derived from above step; write out the resultant R as R = Rxi + Ryj
2
4
...
Compute the direction of R as tan α =

Ry
Rx

where α is the required direction in its proper

sense
...


2
...


3
...
If the Resultant of the system of forces shown is zero
...


5
...


T= 50N

Q

Q

S= 150N
o

50

30o

50o
10o
P
6
...
Determine the values of P and Q

32

T= 80N

Solutions

1
...
30N
F1y = (40Sin 450)N = 28
...
5N
F2y = -(35Cos 300)N = -30
...
5N
F3y = -(62Cos 500)N = - 39
...
Rx = F1x + F2x + F3x = (28
...
5 + (- 47
...
7N
Ry = F1y + F2y + F3y = (28
...
3) + (-39
...
9N
Writing the vector
R = (1
...
9j) N

i
...
7)2 + (-41
...
9N
Direction
Tan α = Ry = -41
...
647
Ry

-1
...
647) = 87
...
80 = 267
...
T1x = 0N
T1y = 50N
T2x = -(80Sin500)N = -61
...
4N

...
3N)) = -61
...
4N)) = -1
...
3i – 1
...
|R| = p(-61
...
4)2 = 61
...
4 ⎤
o
⎥ = 1
...
3 ⎦

α = tan −1 ⎢


...
30
3
...
3N
Rx = T1x + T2x = 10 + (-40) = -30N
Ry = T1y + T2y = 0 + (-69
...
3N
R = (-30i – 69
...
3)
2

α = tan −1

2

= 75
...
6o
−30

Direction = 1800 + α = 246
...
T1x = -30Sin 100 = -5
...
5 N
T2x = 10Sin 500 = 7
...
4 N
T3x = 10Cos 150 = 9
...
6 N
Rx = T1x + T2x + T3x = (-5
...
7 + 9
...
2 N
Ry = T1y + T2y + T3y = (29
...
4 + (-2
...
3 N

...
2i + 33
...
2 ) + ( 33
...
5 N

33
...
9o
12
...
90 (in the 1st Quadrant)

34

5
...
3 N
Ty = 50Sin 300 = 25 N
Qx = 0 and Qy = Q N
Px = -PSin 100 N and Py = -PCos 100 N
Rx = Tx + Qx + Px = 43
...
4 – PSin100 = 0
PSin 100 = 43
...
3
= 249
...
4Cos 100 = 0
Q = 249
...
6 N
Thus: P = 249
...
6 N
6
...

Therefore,
Px = PSin 500 N
Py = PCos 500 N
Qx = -QSin 500 N
Qy = QCos 500 N
Tx = 80N, Ty = 0N
Sx = 0 N, Sy = 150 N

...
B: Tan∅ = Sin∅

-150 – Qcos 500

Cos∅

1
...
192)(-150-QCos500) = -80 + QSin 500
-178
...
8Q = -80 + QSin 500
Collect like terms:
-0
...
8
-Q(0
...
8
Q=

98
...
8 N
−1
...
8)Sin 500 = -80

36

PSin 500 + 61
...
PSin 500 = -80 -61
...
8sin 50o −127
...
2 N
sin 50o
0
...
2N in opposite direction to diagram and Q = 61
...


9
...
e
...
Hence, summarily, it could be said that “for a
system in equilibrium; sum of forces in one direction must equal sum of forces in opposite
direction
...
1 VECTOR REPRESENTATION OF EQUILIBRIUM CONDITION

Coplanar forces (forces in x-y plane) are represented in magnitude and direction by showing its x
and y components in the proper sense of i and j unit vectors
...
2 MOMENT CONDITION FOR EQUILIBRUIM

Just as was stated for the force conditions of equilibrium, there is also a similar condition
adapted to the moment of forces keeping the system in equilibrium
...
This means that, the moment condition for equilibrium can also be
stated as that “the sum of moments in one direction is equal to the sum of moments in the other
direction”
...
e
...

Mathematically;
For a system in equilibrium;
Total moment about a fixed point = 0
∑M = 0
Hence,
Sum of clockwise moments = Sum of anticlockwise moments
9
...

Static Equilibrium: this is the state of a body at rest
...
e
...

Dynamic Equilibrium: this is the state of a body moving with uniform velocity; in which case

the acceleration remains zero i
...
the rate of change of velocity with time is zero
...

Whatever is done after to alter the equilibrium condition of a body will also alter simultaneously
the velocity state of the body
...
Determine the magnitude and direction of the force F1 which will keep the box of 50kg,
acting on an inclined plane of 20º to the horizontal in equilibrium
...

Completing and using the triangle of force resulting from the system:

sin 20o =

F
Mg

F = MgSin20º
= 50 x 10 x Sin20º = 171
...
0 N in direction Ø = 20º to the horizontal
2
...


39

T2

15m

T1

6m
γ

α
7m

9m

100N
Solution
Using the triangles ACB and DCE to find the angles:
Consider triangle ACB
Tan α = 6
7
α = Tan-1 (6/7) = 40
...
0
Hence,
Σfx = -T2Cos40
...
6º
T2 cos 40
...
5 T2 ------------------------- (1)
Also, ΣFy = T2Sin40
...
60 + T1Sin590 = 100 --------------- (2)

40

Put (1) into (2)
T2Sin40
...
5T2) Sin590 = 100
T2 =

100
0

= 51
...
6 + 1
...
6 N
PRACTICE QUESTION

Two ropes attached to the top of two poles (as shown in the diagram) are used to hold a boat on
the sea in position
...
6Kg
...
(Take g = 10m/s2)

T

T

4m

3m

3m

41

4m

10
...

In solving a problem involving equilibrium, it is important that only all the forces
(external) acting in the system be considered
...

A properly drawn free body diagram gives a detailed account of all the external forces
acting on the body in question
...

The free body is only a schematic diagram of the real body isolated from its supports and
represented by lines
...
1 COMPONENTS OF A FREE BODY DIAGRAM

1
...
They must be shown by their
magnitude and direction and point of action
...
Reactions: These represent the constraints of supports or connected bodies
...

3
...
It is always
vertically downward through the centre of gravity of the body in question
...
Dimensions and Angles: The significant line and angle distances must be shown clearly
by a free body diagrams
...
2 FREE BODY DIAGRAM OF THREE FORCE SYSTEM IN EQUILIBRIUM

42

These forces are shown to point round a triangle in a direction head to tail
...
Hence, system is in equilibrium
...
0 INTRODUCTION

The word dynamics is also known as dynamo which means motion
...
It is divided into two broad
headings: Kinematics and Kinetics
...
0 KINEMATICS

This is study of a body in motion without reference to the force producing it
...
It considers the displacement, distance, speed,
velocity and acceleration of a given body in relation to time
...
It considers the energy responsible for a given motion
...

Particle kinematics, rigid body kinematics, Kinetics of particles and rigid body kinetics
respectively
...
1 PARTICLE KINEMATICS

The concept of particles does not mean that reference is directed to small matter but it stretch to
mean the consideration of a whole body as a single entity, thus, the behavior of a body as a
whole single unit will be considered
...

The principle of particle kinematics is the performing of the first and second order differential
equation of a function which maps the position of the particle at any given instance with respect
to an inertial frame of reference, to obtain the velocity and acceleration of the body at such
instance
...
These three are
discussed later
...
0 THE GEOMETRY OF MOTION IN A STRAIGHT LINE

By the word geometry of motion, reference is made to the position, velocity and acceleration of
motion with respect to the time
...
Thus to define the position, velocity and acceleration of such a motion, a reference point
which at times may be called ‘the fixed point’ is chosen and the motion is described in relation to
that fixed point
...

Displacement of a body: This is the distance a body moves from the reference point in a
particular direction
...

Velocity: This is the time rate of change of displacement
...

Acceleration: This is the time rate of change of velocity
...
It is a vector quantity
...
It is mathematically expressed as X= Xf – Xi

where Xf = final position, Xi

=initial position
Instantaneous displacement: This is defined as the magnitude of the change in position of a
particle in a specified direction at any instant of time
...
If x is the instantaneous displacement dependent on the time, t, it can be
mathematically expressed as:
X (t) = Xf (t) – Xi (t)

m

Average Velocity: This is defined as the change of average displacement with time
...

It is expressed mathematically as:
V = dx / dt

m/s

45

Average acceleration: This is defined as the change in average velocity with time
...

It is expressed mathematically as:
a = dv/ dt m/s2
In terms of instantaneous displacement, it is expressed as:
a = d2x / dt2 m/s2
Thus, the instantaneous velocity and acceleration of a particle motion can be obtained by
differentiating the instantaneous displacement
...
i
...

x =∫v dt = ∫∫a dt
Class Examples

1
...
Determine the position, the velocity and
acceleration when t = 5s
...
the motion of a particle is defined by the relation
x = 10 t3/3 − 7t2 / 4
3
...
5t2), where x and t are expressed in m and seconds
...

4
...
5s
...
Starting from rest and at zero displacement a particle motion is given by

a = 10√ (v2 +

15) m/s2
...


46

6
...

When t = 1s, the particles position is 10m from the reference point
...

7
...

Find the position the particle reach in 10s
...
A car has an initial speed of 25m/s and a constant deceleration of 5m/s2
...
the speed of a particle at an instant when t= 1s is 12m/s
...
Determine (a)The acceleration of the particle at this time and the total
distance it has moved
...
A train leaves train station at 10:05am for a city 100km away
...
What is the acceleration of the train, and at what time will the
train reach its destination
...
0 GENERAL CURVILINEAR MOTION

Curvilinear motion occurs when a particular object (in this case a particle) moves along a curved
path
...
These
include: Rectangular, Cylindrical and Polar coordinates respectively
...
1 RECTANGULAR COORDINATES

In this, the coordinates are represented by x, y, and z
...

By the equation (1) above, velocity and acceleration of the particle can be composed to be:
V=

dx dy
dz
i+
j+ k
dt
dt
dt

V= Vx i+Vy j+Vz k
dv y
dv
dv d 2 r dvx
a=
= 2 =
i+
j+ z k
dt dt
dt
dt
dt

a=a x i + a y j + az k

Where v and a are velocity and acceleration respectively
...
A particle undergoes a curvilinear motion defined by x = 9t m along the horizontal with
the equation of the trajectory being y = X2/16
...

(b) The magnitude of the velocity for this time (c) The magnitude of the acceleration and
(d) The directions of the velocity and acceleration respectively
...


r = xi + yj

When t=2s, x =9(2) =18m and y =

182
= 20
...
25)2] = 27
...

dt

Vy =

dy 2 xvx 2(18)(9)
=
=
= 20
...
25 j

49

V=

(9

2

)

+ 20
...
16m / s

(c) a = ax i + a y j
d 2x
a x = 2 = 0m / s 2
dt

ax =

d 2 y 2 xax 2vx vx 2(18)(0) 2(9)(9)
=
+
=
+
= 10
...
13 j
a=

(d)

(0

2

)

+ 10
...
13m / s 2

Direction of velocity:
θ = tan-1 (vy/vx) = tan-1 (20
...
040
Direction of acceleration:
θ = tan-1 (ay/ax) = tan-1 (10
...
A fly moves a curvilinear motion tracing the path described by y = 10 t3 m and x = 5 t2 m
...
Determine the distance moved by the fly in 10 s, and the magnitudes
of the velocity and acceleration in this time
...
67 = 3km

(Solve for the acceleration on your own)
Practice Question

1 At the instant shown below, the particle A travels with a velocity of Vx = 10m/s
and acceleration of a =2 m/s
...
At
what speed does it strike A
...
3 m/s, vb = 81
...
A particle travels from A to B in 1s
...
5s to go from A to C determine its average
velocity when it goes from B to C
...
A particle moves with a velocity v (t) = 0
...
Determine the magnitude and
coordinate direction of the acceleration
...


14
...


...

dR
...



...


But ε r = θ ε θ and k =0
Therefore;

...



...



...


...


...


...

= r ε r + r ε r + r θ εθ + r θ εθ + r θ εθ + z k + z k
dt

...


But ε θ = − θ ε r

...



...



...


a = r ε r + r θ εθ + r θ εθ + r θ εθ − r θ ε r + z k

⎛
...


...

⎞
a = ⎜ r − r θ ⎟ ε r + ⎜ 2 r θ + r θ ⎟ ε θ + z k --------------- (4)
⎜
⎟
⎝
⎠
⎜
⎟
⎝
⎠

53

a = arε r + aθ ε θ + az k
Therefore, equation (1), (2) and (4) are the equations of motion in a cylindrical coordinate
...
However, there are special cases when the
particle is performing motion in r-θ plane
...
This
is a modification to the cylindrical coordinate given above
...
e plane Z) does not exist and that


...


Thus, in a polar coordinate, the equation above becomes:

R = rε r

...


V = r ε r + r θ εθ

...

⎞
a = ⎜ r − r θ ⎟ε r + ⎜ r θ + 2rθ ⎟ε θ
⎜
⎟
⎝
⎠
⎝
⎠

The 2- component system is called the radial and transverse component
...

This is derived by modifying the cylindrical component system and it is given by assuming that

...


the Z- plane does not exist and that r = z = 0
...


V = r θ εθ

54

V = rωε θ
And


...

Where ε r , ε θ and K are units vectors in the direction specified
14
...
This in retrospect, is
similar to the phenomenon of motion in a circle, in which, there can be the normal component
(acting towards the centre) and the tangential component (acting along the tangent to the circle)
...


n

o
Un

Ut

The velocity is simply given by:

R=r
V=vut =
a=

dr
ut
dt

---------------------- 1

dv
= vut + vut
dt

55

t

But ut = θ un
a = vut + vθ un

------------------------ 2

θ Can be expressed as

Hence, a = vut + v
Note: θ = ω =

v2

ρ

v

ρ

un ------------------- 3

v

ρ

Where ρ = radius of curvature of circle of motion
Equation (1) can be written as a = at ut + anun
Class Examples

1
...
if the speed of the vehicle was
increased at a constant rate of 7m/s2, determine the time for it to reach the maximum
acceleration of 10m/s2
...
What is the speed at this time?
Solution

This is the situation of motion in a circular path
...

Take acceleration = 10m/s2
aT =

(a

2

t

+ an 2

But aT = 7 m

s2

)

(1)

, an =

v2

ρ

Calculating for V
v = u + a tt
v = 0 + 7t
an =

7t 2 49t 2
=
= 1
...
63t ) ) = 10
2

2

2

56

t= 2
...
09 = 14
...
The motion of a particle is described by the relations:
2 = 1- 5t2, r = 2t – 5t2
...


Calculate the (a) magnitude and direction of the velocity (b) magnitude and direction of the
acceleration after it has moved for about 30o
(This is a motion in which the particle is instantaneously rotating and translating
simultaneously)
...
524rad
6

θ = 1 – 5t2 = 0
...
095s
Since, the z-component is not represented, a polar coordinate results
V = rε r + rθεθ ------------------------------------- 1
r = 2 − 10t = 2 − 10 ( 0
...
048 m
2

s

r = 2t − 5t 2 = 2 ( 0
...
095 ) = 0
...
095)2 = 0
...
095) = 0
...
048ε r + 0
...
05 ) εθ = 1
...
0089ε θ m/s
V=

(1
...
00892 = 1
...
0089 ⎞
o
Direction θ: = tan-1 ⎜
⎟ = 0
...
048 ⎠
⎝

...

⎞
a = ⎜ r − r θ ⎟ ε r + ⎜ r θ + 2rθ ⎟ ε θ
⎝
⎠
⎝
⎠

57

(Substitute in values and carry out the solution on your own)
15
...
The kinematics of rigid body deals with relations existing between time,
positions, velocities and acceleration of various particles making up a rigid body
...

1
...
This can either be rectilinear translation (if the paths are straight lines) or rectilinear
curvilinear (if the paths are curved lines)
...
Rotation about a fixed axis: The motion of the particles moves in circles about the same
fixed axis
...

3
...
g
...

4
...
(Do not confuse this for rotation)
...
General motion: The motion of a rigid body which cannot be categorized into any of the
above is called a general motion
...
This can
either be rectilinear translation (if the paths are straight lines) or rectilinear curvilinear (if the
paths are curved lines
...
Let the position vector from a reference
frame of A and B be rA and rB, with rA/B the vector joining A and B
...
(ii)
But rA/B is always constant, since there is no relative change in position of the particles or
points making up a rigid body
...

(Beer et al
...

Rotation about A Fixed Axis
Consider a rigid body about a fixed (the Z- axis), having a point p on its body
...


59

Z

N
θ

P
r

Y

X
The position of p on the rigid body is defined by the agle θ the position vector makes with the
Z plane
...
The distance
s, moved by p when the body rotates through θ is given as
S = (NP) θ = (r sin) θ
...

Written in vector form, gives:
V = ds/dt = ω x r = dr/dt………………
...
(ii)
a = α x r + ω x V ……………………………
...
(iv)

Resolving (iv) to normal and tangential components gives at = αk x r = rα and an = -ω2r = rω2
Equation of rotation of a rigid body
ω = dθ/dt…………
...
(vi)
α = ω (dω/dθ)…………………
...

Therefore,
θ = θo + ωt…………
...

…………………
...

Class examples

1
...

(a) Find the angular velocity and angular acceleration when t = 0 and t = 2
...
(b) Find
the angular distance and acceleration when the ω = 0
...

Solution

θ = 4t3 - 12t2 + 15
Angular velocity is given as:
ω = dθ/dt = 12t2 – 24t
(i) When t = 0; ω = 12(0)2 – 24(0) = 0 rad/s
(ii) When t = 2
...
5)2 – 24(2
...

Angular acceleration is given as:
α = dω/dt = 24t – 24
(i) When t = 0, α = 24(0) – 24 = - 24 rad/s2
(ii) When t = 2
...
5) - 24 = 36 rad/s2
(b) when ω = 0
ω = dθ/dt = 12t2 – 24t = 0
Factorizing gives
12t (t – 2) = 0
t = 0 or t = 2s
...

When t = 2s, θ = 4(2)3 – 12(2)2 + 15 = -1 rad
Angular acceleration is given as;
α = 24t – 24
When t = 0, α = 24(0) – 24 = -24 rad/s2
When t = 2s α = 24(2) – 24 = 24 rad/2
x
2
...
The load A and B are moved by the double pulley as shown
...
The radii of the outer and inner
circles of the pulley are given as 50cm and 30cm respectively
...
5m/s2
...

(c) The acceleration of A at t = 1s
...
Thus, VA = 3m/s and aA = 0
...
5 = 6rad/s
...
5/0
...
5 rad
...
5/2π = 1
...
3(ω)……………
...
3(7) = 2
...
3(6
...
95m upwards
The acceleration of A
The acceleration of a = the tangential acceleration of inner circle
aA = rα = 0
...
3m/s2
General Plane Motion
A general motion can always be considered as the sum of a translation and a rotation
...

Consider the rod AB in the diagram below:

B

l
ω
VB
A
VA

63

If B is the fixed point
...
The length of the rod is l
...
(1)
VA = VB + ωk x rA/B……
...

From equation (1) and (2) it is clear that the mathematical relationship between VA/B and rA/B is
given in vector form as;
VA/B = ωk x rA/B………… (3)

Note: the angular velocity ω of a rigid body in plane motion is independent of the reference
point
...

However, the sense of the relative velocity changes i
...
VA/B = VB/A
...
Thus, the change is in the direction of r
...
1
...
2
...


Consider fig
...
If point A is fixed and B moves with velocity VB = 20i
...
However, if the point can move with
velocity VA = 10j
...


For the diagram of fig
...


Determine the angular velocity of the connecting rod BC and the velocity of point C, if the point
A is fixed
...

Solution

1
...
33 ω j + 2
...
5 ω
ω = - 20/2
...
36 m/s
Direction, θ = tan-1(10/20) = 26
...


65

Acceleration of a general plane motion
v A = vB + v A / B
v A = vB + ω × rA / B
v A = vB + ω × rA / B + ω × rA / B

v A = vB + α × rA / B + ω × (ω × rA / B )
a A = a B + ar

From equation above:

a B = vB

ar = α × rA / B + ω × (ω × rA / B )
ar = at + an
ar = at 2 + an 2

v A = vB + v A / B

5

vA = vB + ω × rA / B

6

Differentiating velocity to get acceleration gives:
vA = vB + ω × rA / B + ω × rA / B

7

Thus,
v A = vB + α × rA / B + ω × (ω × rA / B )

8

This equation 8 is the equation for the acceleration of a general plane motion
...
Thus, it can be summarized as:
a A = aB + ar

9

Where the linear component is given as:
a B = vB

10

And the rotational component is given as:
ar = α × rA / B + ω × (ω × rA / B )

11

The rotational component can be further broken down to tangential and radial components
respectively as:
ar = at + an

12

The magnitude is given as:
ar = at 2 + an 2

13

Class example: Solve for acceleration in the last examples

66

16
...
These equations take
their references from the same point and the same instant
...

This leads to the phenomenon of relative displacement, relative velocity, and relative
acceleration respectively
...
If the position coordinates of particles A and B are xA and xB respectively, moving in
the direction as shown in the figure, and measured from the same reference point
...

Hence equation 1 above is the relative displacement of B to A
Relative velocity
The relative velocity is obtained from the relative displacement and is defined as the rate of
change of relative displacement
...

Relative acceleration
This is the rate of change of relative velocity
...

Class Examples

1) The geometry of the particle motions are given by the relation xA=5t3- 2t2 + 5 m and VB=7t 8t2 + 9t3 m
...
Find
a) The distance of B relative to A
...

c) The velocity of B relative to A
d) The magnitude of the acceleration of B relative to A when the time elapsed, t is 10s
...
Find the
a) Instantaneous velocity of B
...

c) The magnitude and direction of VB
d) The magnitude and direction of aB/A
...
Find
a) Relative velocity of B to A
...
Given that the interval is 20s
...
0 DEPENDENT MOTION OF PARTICLES

This arises when the position of a particle depends upon the position of another particle, thereby,
meaning that the location, movement, speed, acceleration, etc of the particle will depend mostly
on the geometry of the other particle(s) motion
...
Consider the figure below:

xA

xB
A

B

The geometry of A is linked to B by the relation:
xA + 3xB = l

(1)

Where xA = displacement of A, xB = displacement of B and l = total length of the rope
...


69

Worked examples

1) Consider the diagram below:

D
y
8m
o

A

E
F

x

C

B

Particles A and C are connected by a cable via three pulleys D, E and F through particle B
...
At t = ts,
particle A moves from rest from y to o with a constant acceleration and velocity 4m/s
...

b) The velocity and acceleration of C when particle A is at o
...


Therefore,

70

VA= UA + aAt= 0 + t
t = VA/1 = 4s
...


Therefore, from equation (1)
xA+ 2xB + xC = l
The l remains unchanged despite the position of A, B, C; hence,
Δxa+2Δxb+Δxc= 0
8 + 2(8) + xc =0
xc = - 24m
...

b) The velocity
Va +2Vb +Vc = 0
4 + 2(2) + Vc = 0
4 + 4 + Vc = 0
Vc = -8m/s
Vc = 8m/s upwards
The acceleration
aA +2aB + ac = 0
1+ 2(0) + ac = 0
ac = -1ms2
ac = 1m/s2 upwards
2) Consider the diagram below: the displacement of A and B are given as xA= 2t2 + 5t and xB= 4t
+ 5 respectively
...

b) The velocity and acceleration of C
...


A

C
B

3) Block C starts from rest and moves downward with a constant acceleration
...
5m/s
...
5m/s
...


A

C
B

(Solve number 2 and 3 on your own)
...
0 Instantaneous Centre of Rotation

The instantaneous centre of rotation is the point about which you can assume a body is rotating
at a given instant, as you determine the velocities of the point of the body at that instant (Beer et
al
...
The instantaneous centre of rotation method of finding the velocities of points or
particles in a rigid body is an alternative way of solving problems involving the velocities of the
various points of a rigid body undergoing general plane motion
...
We have
dealt with vector analysis method while going through previous topics
...

Location of the instantaneous centre of rotation
Case 1: when the velocity of a point, say A, and the angular velocity ω, of the body are known
...
Produce a diagrammatic representation of the body
b
...

c
...

Class example

Determine the position of the instantaneous centre for the body with point A, having absolute
and angular velocities as 2 m/s and 0
...


ω

A

VA

y

73

Solution

ω
VA

A

Therefore, r = VA/ ω = 2/ 0
...

Hence, the instantaneous centre is located 4 m from point A
...

a
...
From points A and B, draw perpendicular lines to VA and VB respectively
...
However, if the velocity of one of
the two points is know, the angular velocity of the body ω, can be determined by using:
ω = V/ r, when it has been drawn to scale in a graphical representation
...
2 below
...
if the body rotates clock wisely with angular velocity 2
...
2a

74

A

4m

3m
B
C

30o

ω
VA
Fig
...


B

C

VB
VA

A

From VA = ω (AC)
a
...
5 = 2 m
...

b
...
46 m
Therefore, VB = 2
...
46) = 8
...


C

2
...
28m and BC = 3
...
17m
AC
vB = ( BC ) ωBA = 4
...
3 below
...
Produce a diagrammatic representation of the body
b
...
Draw a second line through the tips of points A and B to meet the first line
...
C is the position of the instantaneous centre of rotation for the body at the instant
shown
...
Calculate VB

19
...
1 Introduction

Kinetics has been previously defined as the study of the forces producing a particular motion
...

However, the fundamental laws governing the subject of kinetics are those of the scientist called
Sir Isaac Newton
...
They are three
laws
...
These laws also form the basis for dynamics in conditions when concentration
would have to be focused on motion under changing velocities
...


77

Newton’s laws of motion
This is divided into three:
Newton’s first law of motion: This state that a body will continue in its state of rest or if moving
continues in its uniform motion unless acted upon by external forces
...
This law is sometimes referred to as the law of inertia
...

Mathematically, this is expressed as
F = ma

N

Where F = force, m = mass, a = acceleration
This law is the foundation of particle and rigid body kinematics and kinetics
It relates kinetics to kinematics
...

Newton’s second law
F = ma

This states that, as the force changes the acceleration changes in a proportional relationship
...
Thus, suggesting that a uniform motion is a motion of zero force
...
If it is originally moving under the condition of no force, it will
continue to move with constant velocity
...
2 WORK

Work is said to be done when a force moves in particular direction
...

U=F×r
Consider a particle p which is originally at position r
...
then, the work done by the force in moving the

particle from r to r1 is given as
...
It
is therefore written as:
dU = F
...


F

θ

dr

r/

r

fig
...
I unit of work is the Joule
...
Then work is simply expressed as;
dU = F dr ------------------------------------------- 2
79

From the relations (1) and (2) above we can deduce that;
dU/ dr = F

(differentiating work with respect to distance gives the force)

Also, U = ∫Fdr

(Integrating force with respect to distance gives work)

Relating the phenomenon of work to different aspects
1
...


x

Fig
...

If this spring force is used to do a particular work from a point x1, to another point x2, the work
done by the force is given as
dU = – Fdx
...
dx = − ∫ kx
...
Work done by the force of gravity:
The effect of gravity on the mass of a particle is the weight of the particle
...


x2

U = − ∫ W
...

3 work of a constant force in a linear motion
dU = F
...
dx
This is obtained when the force moves in the direction of the change in displacement dx
...
work done by gravitational force
The gravitation law (Newton’s law of universal gravitation), is given as;

F=

GMm
r2

Where F = force of gravitational attraction between masses M and m at a

distance of r apart
...

When this force does work, its work is expressed as:

dU = − F
...
0 Kinetic Energy of a Particle

Energy is defined as the ability to do work
...
The kinetic
energy which is a form of mechanical energy is defined as the energy used up by a body (particle
in this case) in moving from one place to another
...

20
...

Consider a particle of mass m, acted upon by a force F and moving in the direction as shown in
figure below, from position 1 to 2
...
= mv
dt
ds dt
ds

Giving:
Fds = mvdv

Integrating from position 1 to 2 gives:
s2

v2

∫ Fds = m ∫ vdv =

s1

v1

1 2
mv
2

v2
v1

=

1 2 1 2
mv2 − mv1
2
2

……………… (1)

The left hand side of the equation (1) above is the work done by the force F in moving from
position s1 to s2
...

Class examples

10m

1
...
Determine (a) the velocity of the box and (b)the work done
by the force
2
...
If the
orbital speed is 25,000km/h, what is the kinetic energy of the satellite?
3
...
(a)
Find the K
...
(b) The height h from which it was dropped
...
The force acting on a particular body is changing with displacement as F = 5s 2 + 2s + 5
...
Calculate (a) the work done in moving the body from a
distance of 10m to 15m
...
0 Impulse and Momentum

Impulse of a force can be defined as the product of force and the time
...
I unit of impulse is Ns
...
It is mathematically expressed as:
M = mv

The S
...
The resulting motion is called an
impulsive motion
...
An example of impulsive force is the force of impact
...
1 Principle of Impulse and Momentum

The principle governing impulse and moment relates the impulse of a force to the momentum of
a particle caused by the force
...
However, considering the equation (1), and assuming that there is no impulsive force,
then the momentum is said to be conserved i
...


t2

∫ Fdt = mv

2

− mv1 = 0

t1

84

Therefore,
mv2 = mv1
Hence, momentum is said to be conserved when the initial momentum is just equal to the final
momentum
Class examples

1
...
When it hits
an object at an instant of t = 0
...
determine the impulsive force exerted on the ball
...

1

2

30o

The box (1) on the inclined plane shown above moves downward with a speed of v1 = 10m/s
...
Both travel together with a velocity 5
...
(a) what is the impulse of the force
...


3
...
2m/
B

mB = 5kg

Consider the diagram above
...
(a) Find the velocity of A after the collision
...

85

Solution

1
...
1s, v2 = -20m/s (by making the initial direction, the positive direction)
Ft = mv2 – mv1
Ft = m(-20 – 30) = - m(50) = - 50m
...
1

N

If the mass had been given, the value of F would be known
...

2
...
5m/s
(a) Ft = miv1 – (m1 + m2) v2
I = 5 × cos 30o − ( 5 + 5 ) × 5
...
70 Ns

(b) 5t = 11
...
70
2
...

(c) K
...
5 J
2

K
...
5 = 27
...
5 – 27
...
(a)
...


86

m1v1i + m2 v2i = m1v1 f − m2 v2 f
2 ( 3) + 5 ( 0 ) = 2 ( v1 f ) + 5 ( 0
...
5m / s
2

(b)
...
52 + 5 × ( 0
...
25 + 0
...
35J

Energy lost = 9 – 6
...
65 J

Reference

Beer, F
...
R
...
, E
...
Eisenberg and W
...
Clausen, 2004
...
, McGraw-Hill, New York
...
Statics and Strength of Materials, 2nd edn
...

[TA 351
...
C
...
Engineering Mechanics: Statics & Dynamics, 10th edn
...
P
...
Mark’s Mechanics, Problem solving companion, McGrawHill, New York
...
S65]

87

88



---