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Physics for engineers - notes 1- 10
H 110: FUNDAMENTALS OF PHYSICS I
1.0 UNITS AND DIMENSIONS
1.1 Introduction
Measurable quantities in physics are assigned units of measurements.
Quantities are divided into 2 namely:-
Basic / fundamental quantities
Derived quantities
1.2 Basic /Fundamental quantities
They don’t depend on other quantities. These quantities are used to fully
describe other physical quantities. They include:-
Basic Quantity S. I. Unit Symbol
Length Metre m
Mass Kilogramme kg
Time second s
Amount of substance mole mol
Electric current Ampere A
Thermodynamic temp. Kelvin K
Luminous intensity candela cd
1.3 Derived quantities
They are described in terms of basic or fundamental quantities e.g.
volume, area, pressure, density etc.
Metre: It’s the distance between two points. The standard of a
metre is marked on a bar of platinum (90%) – Iridium
(10%) alloy kept at 0
o
c.
Second: It’s the duration of 9, 192, 631, 770 periods of certain
microwave radiation emitted by the ceasium atom. The
atomic clock is the most accurate and other clocks
(secondary) are set compared to it.
Kilogramme: The standard mass is the platinum. Iridium cylinder
whose mass is exactly one kilogramme
3
Note: The physical quantities, time, mass and length are
fundamental quantities we use in our study of
mechanics.
1.4 Dimension and dimension Analysis
Dimension: It is a physical property described by the words time,
length or mass. (This property is the same no matter
what units it is expressed.
Dimension: Symbol
Length: L
Time: T
Mass: M
Dimension Analysis :
It is a technique of establishing the validity of a solution to a problem,
a unit or an equation by checking for dimensional consistency.
Dimensional units must have the following properties:-
- For addition and subtraction, quantities must have the same
dimensional units.
For division and multiplication they may have different units
For equations to hold they must have the same dimensional units
on both sides.
Note: Constants and angles have their dimensional units as 1 e.g 𝜋,
Cos θ, Sin θ, Tan θ, 1, 2…., ½ , 4/ 3
, exp, ln, log, etc.
Examples
- Define and give the dimensional units for the following:-
a) Speed: It is the rate of change of displacement
Dimension of velocity =
𝐿
−
(S.I unit is metre per second)
𝑇
b) Acceleration: It is the rate of change of velocity
Dimensions of acceleration =
𝐿𝑇
−
= LT
𝑇
Density: It is the mass per unit volume
Dimensions of density
M
−
(S.I unit kg/m
3
𝐿
- The expression for kinetic energy is K.E = ½ mv
2
and potential energy is
P.E = mgh. Show that both expressions have the same dimensions, hence
can be subtracted or added from each other.
½ mv
2
= mgh
M(LT
-
2
= MLT
-
x L
ML
2
T
-
= ML
2
T
-
- The period T of a pendulum is given by the dimension equation
T = km
x
l
y
g
z
, where m is mass of the bob, l is the length of the string, g is
acceleration due to gravity and k , x, y, z, are constants. Calculate the
values of x, y, and z.
T =M
x
L
y
(LT
-
z
= M
x
L
y + z
T
_- 2z
y + z = 0, y = 1/ 2
x = 0.
2.0 VECTORS
2.1 Introduction
If a sack of flour has a mass of 10kg, that mass is not dependent on where the
flour, whether it at rest in a storeroom on land or in motion on a ship in sea.
The above statement describes only the magnitude / size (10kg) but not the
position. This shows that mass is a scalar quantity.
For a quantity like velocity it is quite different. To a passenger in Mombasa
desiring to go to Nairobi city on a bus moving at 20m/s, it obviously makes a
big difference whether the bus is moving towards Nairobi city or Malindi
town. Here both direction and size/magnitude are vitally important. Such a
quantity like velocity is a vector quantity.
2.2 Scalar and Vector quantities
Scalar quantity : It is a physical quantity that has no direction and it is
completely specified by its magnitude / size alone, e.g. mass, energy, time, etc.
Vector quantity: It is a physical quantity that is completely specified only
when both its magnitude / size and direction are given, e.g.
velocity, displacement, force, momentum, acceleration etc.
2.3 Representing vectors
A vector quantity is represented in many ways.
Pictorial representation: A vector is represented by a directed line segment
(arrow). Where length of the line is the size/magnitude while the arrow shows
direction.
Symbol representation :
Vector A can be represented as in:-
A - Arrow on top
A - wavy line below
A - Bold face
Vectors can also be represented in terms of i, j and k.
i.e OA = [
] ; OA = 3 i + 4 j
OA = [
] ; OA = 3 i + 4 j + 6 k
Position vector : It is a vector drawn from the origin of some coordinate system to a
point in space to indicate position of object with respect to
origin, i.e
OA = [ ] or OA = [ 4 ]
Displacement vector : It is a directed line segment (arrow) whose length indicates the
magnitude of the displacement and whose direction is the
direction of displacement.
2.4 Operation on vectors
2.4.1 Vector addition
In addition it means two vectors are added to get another vector, i.e
A + B = C
There are two ways of doing this:-
Triangle method: If A and B are drawn to scale with tail of B at the
tip of A , then C is a vector from the tail of A to the
tip of B.
Tip – to – tip Method (polygon): It is an extension of the triangle method to
two or more than two vectors.
2.4.2 : Vector subtraction
The negative of a vector of equal magnitude but different direction.
A = - B
Vector subtraction is vector addition of opposite vectors.
A - B = A + ( - B )
Example:
- Given that A = 5 i + 3 j and B = 2 i - 4 j
Find: a) A + B b) A - B
a) A + B = 7i - j b) A - B = 3i - 7j
Example
- Find A. B and B. A given that
A = 2 i + 3 j + 4 k and B = -j – 2 j + k
A.B = (2x – 1) + (3 x –2) + (4 x 1) =
B.A = (
1 x 2) + (
2 x 3) + (1 x 4) =
- Given that, | A | = √14, | B| = √ 16 and the angle between A and B is 30
o
, find A. B
A. B = | A || B | Cos θ = √ 14 x √ 16 Cos 30
0
Cross Product
Given that two vectors A and B the cross product of A and B is defined as
A x B = | A||B| Sin θ
Consider two vectors
A = a 1
i + a 2
j + a 3
k and B = b 1
i + b 2
j + b 3
k
Represent in matrix form A x B = |
1
2
3
A x B = i [(a 2
b 3
) – (b 2
a 3
)] + j [ (b 1
a 3
) – (a 1
b 3
)] + k [(a 1
b 2
) – (b 1
a 2
)]
Example.
Given that A = 2 i + 3 j - k and B = - i + j + 2 k
Find A x B
A x B = | 2 3 −
A x B = i [(3 x 2) – (1 x -1)] + j [ (-1 x -1) – (2 x 2)] + k [(2 x 1) – (3 x - 1)]
= 7 i – 3 j + 6 k
2.4.4 : Multiplication with scalars
Consider vectors A, B and scalar S then
S ( A + B ) = S A + S B
2.4.5 : Magnitude and Direction of a vector
Given that A = a 1
i + a 2
j + a 3
k , then
|A| = √[(a 1
)2 + (a 2
)2 + (a 3
)2]
Example:
- Given that A = 2 i + 3 j + 4 k, find | A|
|A| = √[(2)
2
2
2
] = 5.39 units
- If A + B + C = 0 and A = 2 i + 3 j + 4 k , B = 5 j + 6 j + 7 k. What is C , | C| and angle
between C and x axis.
C = - A – B = -7 i - 9 j - 11 k
= √[(-7)
2
2
2
] = 15.84 units
θ= Tan (-9/-7) =52.
o
2.4.6 : Angle between vectors
We find angles between vectors by using the dot product. This is because
dot product gives the result of a scalar.
A.B = | A||B| Cos θ, θ = Cos
𝑨.𝑩
|𝑨||𝑩|
2.4.7. Angle between vector and axes
Consider vector A as shown:-
2
(
5
Three dimension ( x,y,z) co-ordinate and spherical co-ordinate.
The rectangular co-ordinate ( x, y, z ) and spherical co-ordinates ( r, θ, 𝜙 ) are related by:
x
=r Sin θ Cos 𝜙 , y = r Sin θ Sin 𝜙 , z = r Cos θ, r = √𝑥
2
2
2
, Tan θ =
𝑥
2
2
𝑍
and Tan 𝜙 =y/x
Examples.
- Find the magnitude and direction of the following vectors.
a) A = 5 i + 3 j b) B = 10 i – 7 j c) C = -2 i - 3 j + 4 k
Solution
a) | A| = r = √
2
2
= 5.
θ = Cos
( ) = 30.
0
b) | B| = r = √
2
2
= 12.
θ = Cos
(
10
) = 35.
0
c) | C| = r = √−
2
2
2
= 5.
θ = Tan
√−
2
−
2
(
√
) = 42.
0
ϕ = Tan
−
−
=56.
0
- The rectangular components of the vectors which lie in x – y plane have their magnitudes and
directions given below. Find the x and y components of the vectors.
a) r = 10 and θ = 30
0
b) r = 7 and θ = 60
0
)
Solution
a) x = r Cos θ = 10 Cos 30
0
= 8.66, y = r Sin θ = 10 Sin 30
0
= 5
b) x = r Cos θ = 7 Cos 60
0
= 3.5, y = r Sin θ = 7 Sin 60
0
= 6.
- a) Find the magnitude and direction of the resultant vector A = 5 i + 3 j and B = 2 i – 4 j
Solution
R = A + B = 7 i – j
|R| = √(
2
2
) = 7.
Tan θ =
𝑦
𝑥
−
7
θ = - 8.
0
2.4.8 Resolution of Vectors
A component of a vector is the effective part of a vector in that direction.
Consider a Force F pulling in the direction as shown.
X component of F is F Cos θ
Y component of F is F Sin θ
Example
Consider two forces F 1
and F 2
pulling as shown below. Find the X and Y components
of the forces given that
|F
1
| = 2.88 and | F
2
| = 3.
3.0: FORCE
Introduction
Force is defined as pull or push in on a body, it is a vector quantity it is measured in
Newtons. A Newton is the force that gives a mass of 1kg an acceleration of 1m/s
2
Task:
(1) Give five effects of force.
(2) Name and explain at least 10 different types of force.
3.1 : Resolution of forces
Forces is a vector quantity which can also be expressed in x and y on
rectangular coordinates
Consider a force F pulling a load along a surface at an angle θ
The horizontal component of the force is F Cos θ while the vertical component is
F Sin θ. Also consider a load sliding along an inclined plane at a constant acceleration
Example
Find the tension in each cord if the weight of suspended object is 490N
Solution
∑ F
y
T
3
∑ F
x
T
2
Cos400 – T
1
Cos 600 = 0
T
2
= 0.653T
1
∑ Fy = 0, T 2
Sin400 + T 1
Sin 600 – 490N = 0
(0.653T
1
) Sin 400 +T 1
Sin 600 = 490
T
1
= 381N
T
2
= 0.653 (381) = 249N
Sliding (Kinetic friction) :- The frictional force exerted by one surface on another when
one surface slides over another surface.
F
sliding/kinetic
=F
k
= μ k
R- μ k
- coefficient of kinetic/sliding friction
F
static
=F
s
≤ μ s
R – μ s
coefficient of static friction
Task:
(1) State the Laws of friction
Examples
- A block of wood of mass 20kg requires a horizontal force of 50N to pull it with a
uniform velocity along a horizontal surface. Calculate the coefficient of friction
between the block and the surface.
R = mg = 20 x 10 =200N,
F
r
= μR
μ = 50/ 200
- A mass of 5kg is placed on a plane inclined at an angle of 30
0
to be horizontal.
Calculate the force required to pull the mass up the plane at uniform velocity if
μ = 0.
R = mgCos
0
= 50Cos
0
= 43.3N
F
r
= μR = μmgCosθ = 0.5 x 50Cos300 = 21.65N
There are two forces opposing motion, F
r
and mgSinθ
F
net
= F – (F
r
We have uniform motions, net force is zero
F = F
r
+ mgSinθ
F = 21.65 + 50Sin
0
= 46.65N
3.3 : Newton Laws
First Law (law of inertia): - every body continues to be in state of rest or to move
with uniform velocity unless a resultant force acts on it.
Implication of the 1
st
law – causes inertia
Inertia:- is the property of an object that resists change of motion
Second law:– The rate of change of momentum is directly proportional to the change
causing it (resultant force) and takes place in the direction of force.
Momentum :- – is defined as the product of mass of a body and its velocity.
p =m v.................................................. (1)
Consider a force F , acting on a body of mass, m for a time t , causing a change in
velocity from u to v , then:-
∆ p = m v – m u..................................... (2)
From 2
nd
law
∆𝑷
∆𝑡
𝑚𝒗−𝑚𝒖
∆𝑡
∆𝑷
F ∝
∆𝑡
F = 𝐾
∆
𝑷
∆
𝑡
however, K = 1
F = 𝐾
∆
𝑷
∆
𝑡
∆𝑡
𝑚(𝒗−𝒖)
∆𝑡
But a =
(𝒗−𝒖)
∆𝑡
Therefore F =m a........................................... (3)
