DIMENSIONAL ANALYSIS

DERIVED QUANTITIES AND UNITS

Applications of Dimensional Analysis

Dimensional analysis is a fundamental aspect of measurement and

is applied in real-life physics. We make use of dimensional

analysis for three prominent reasons:

●To check the consistency of a dimensional equation

●To derive the relation between physical quantities in physical
phenomena

●To change units from one system to another

●

DIMENSIONAL ANALYSIS
Dimensional analysis is simply means the relationship between the three
major fundamental quantities and other quantities in dimensions
(1)

Mass is represented in parenthesis by letter [M]

(2)

Length is represented in parenthesis by letter [L]

(3)

Time is represented in parenthesis by letter [T]

Multiple Choice

You know 1000 mg = 1 g.  Convert 2.5 grams into milligrams using dimension analysis.

• 25mg

• 250 mg

• 2500 mg

• 25 000 mg

SIMPLE CALCULATIONS OF SOME DIMENSIONS

These are the three major symbols used in dimension. So let’s see how it is used in dimensioning;

FORCE: This is the product of mass and acceleration
F = ma
Where: m = mass = [M]
a = acceleration = [LT-²] ..…………… (i)
So the dimension for Force = [M LT-²]

MOMENTUM: This is the product of mass and velocity
P = MV
Where: m = mass = [M]
V = Velocity = [LT-1] ………………………. (ii)
So the dimension for Momentum is = [MLT-1]

PRESSURE: This is force per unit area
Where: F = force = [M LT-²] as in (i) above
A = area = [L²] ….……………. (iii)
So the dimension for Pressure is = F / A
= [M LT-²] * [L-²] = [M L-1T-²]

VOLUME: is the quantity of three-dimensional space enclosed by some closed boundary.
V = [L³]………………………. (iv)

DENSITY: This is mass per unit volume
Where: M = mass = [M]
V = volume = [L³]………………. (v)
The dimension for Density is = [ML-³]

VELOCITY: This is distance per unit time
Where: d = Distance = [L]
T = Time = [T] …………………………. (vi)
The dimension for velocity is = [LT-1]

ACCELERATION: This is velocity per unit time
Where: V = velocity = [LT-1] as in (vi) above
T = time = [T] …………………………… (vii)
The dimension for acceleration is = V/T = [LT-1] * [T-1]
= [LT-²]

IMPULSE: This is the product of force and time
Where: F = force = [M LT-²] as in (i) above
T = time = [T] ……..………… (viii)
The dimension for impulse is = F * T
= [M LT-²] * [T]
= [MLT-1]

DIMENSIONS

Multiple Select

Which of the following expressions are the dimensionally correct? Given that s, u, v, a and t correspond to the kinematic quantities of displacement, initial velocity, final velocity, acceleration and time elapsed.

• s = u t + 2a t

• a = (2u - v) / t

• t = u /s ²

• u ² = 2 a s

• v ² = 3 a t

Multiple Choice

What will the resulting units be?

• inches

• feet

• meters/ft

• inches/ft

Multiple Choice

What is deci

• One-hundredth
• One-fifth
• Ten
• One-tenth

1. Using Dimension shows if the formula v = u + at is correct.
2. Show using dimensional analysis that Work done is equal to

energy.

3. Which of the following is equivalent to kgm/s

(A) N/s (B) Nms (C ) Ns (D) J/s,

4. If the dimension of force is MLT-2 and area is L^2, then
pressure which is the force over an area will have a dimension

ASSIGNMENT