Types of Units

Every physical quantity is measured by comparing it with a fixed standard amount of the same kind. That standard is called a unit. When we say a door is 2 metres tall, a bag of rice has a mass of 25 kilograms, or a bus takes 40 minutes to reach school, the words metre, kilogram and minute are units. When scientists look closely at all the units in use, every one of them falls into one of two types.

• Fundamental unit
• Derived unit

A good way to picture this is an alphabet. A few letters stand on their own, and thousands of words are built by joining them. Fundamental units are the letters of measurement, and derived units are the words made from them. Once you know which fundamental units a unit is made of, you can read what it means, and you can also catch an equation that has been written wrongly.

Think about the metre. It is not one quantity multiplied or divided by another. It is simply the standard of length, and nothing simpler lies underneath it. The same is true of the kilogram for mass and the second for time. So the fundamental unit of length is the metre, the fundamental unit of mass is the kilogram, and the fundamental unit of time is the second.

In the SI system, the International System of Units used by scientists all over the world, there are exactly seven fundamental units, one for each of seven basic physical quantities. Every other unit in science, however complicated it looks, is made from some of these seven.

In daily life we often measure temperature in degree Celsius, but the SI fundamental unit of temperature is the kelvin. Of the seven, the metre, the kilogram and the second are the ones you will meet most often in this class, because almost every unit of motion, force and energy is built from these three alone.

Density is a good first example. Density is mass divided by volume, so its unit is kilogram per cubic metre, written \( \text{kg/m}^3 \) or \( \text{kg m}^{-3} \). Inside it are two fundamental units, the kilogram and the metre. Force is another example. Its unit is \( \text{kg m s}^{-2} \), which contains three fundamental units: the kilogram (kg), the metre (m) and the second (s).

Saying "kilogram metre per second squared" every time is clumsy, so this combined unit has been given a short name of its own, the newton (N):

\[ 1\ \text{N} = 1\ \text{kg m s}^{-2} \]

In the same way, the unit of pressure is called the pascal (Pa), and it is equal to \( \text{kg m}^{-1}\text{s}^{-2} \). Many derived units have been given such names, often after the scientist whose work they honour. For some quantities, though, there is no simpler name, and the combination of fundamental units itself is used. The unit of density is simply \( \text{kg m}^{-3} \), and the unit of velocity is simply \( \text{m/s} \).

Every derived unit comes from the formula that defines its quantity. Put the unit of each quantity into that formula, and the derived unit appears. The table below moves from simple quantities to more complex ones. Notice that each row uses units already found in the rows above it: velocity uses length and time, acceleration uses velocity, force uses acceleration, and work uses force.

To find which fundamental units make up a derived unit, analyse the definition of the physical quantity. The method is the same every time:

• Write the formula that defines the quantity.
• Replace each quantity in the formula by its unit.
• If a derived unit appears, open it up into its fundamental units.
• Multiply and divide the units, collecting the powers of each one.
• Give the result its special name, if it has one.

Example 1: The unit of area is square metre.

Step 1: Area is defined as:

\[ \text{Area} = \text{length} \times \text{breadth} \]

Step 2: Putting in the unit of length for both lengths:

\[ = \text{m} \times \text{m} \]

Step 3: Multiplying the two metres:

\[ = \text{m}^2 \]

So the unit of area is \( \text{m}^2 \). It is formed by multiplying the fundamental unit of length, the metre, by itself.

Example 2: The unit of force is newton (N).

Step 1: By the definition of force, force is mass multiplied by acceleration:

\[ F = m \times a \]

Step 2: The unit of mass is kg and the unit of acceleration is \( \text{m s}^{-2} \). Putting these in:

\[ \text{N} = \text{kg} \times \text{m s}^{-2} \]

Step 3: Writing the units together:

\[ \text{N} = \text{kg m s}^{-2} \]

Hence the kilogram, the metre and the second are the fundamental units that combine to form the unit of force, which is called the newton.

Example 3: The unit of work is joule (J).

Step 1: Work is force multiplied by distance:

\[ W = F \times d \]

Step 2: Putting in the unit of force, \( \text{kg m s}^{-2} \), and the unit of distance, m:

\[ \text{J} = \text{kg m s}^{-2} \times \text{m} \]

Step 3: Collecting the two metres together:

\[ \text{J} = \text{kg m}^2\text{s}^{-2} \]

The joule does not stand on its own. It depends on the kilogram, the metre and the second, so the joule is a derived unit.

Putting together what we have learnt, the two types of unit differ in these ways:

Many formulae and equations in science come from the conclusions of scientific studies. A formula can easily be remembered or copied wrongly. The validity and uniformity of such formulae and equations can be checked by analysing the units of the physical quantities in them.

Example 1: Check whether \( s = v \times t \) is valid. Here s is distance, v is velocity and t is time.

Step 1: The left-hand side is a distance, whose unit is:

\[ s \rightarrow \text{m} \]

Step 2: Multiplying the units of velocity and time on the right-hand side:

\[ v \times t \rightarrow \text{m s}^{-1} \times \text{s} = \text{m} \]

Step 3: Comparing the two sides:

\[ \text{m} = \text{m} \]

Both sides have the unit metre, so this equation is valid.

Example 2: Suppose someone claims that \( s = \frac{v}{t} \). We check it the same way.

Step 1: The left-hand side is a distance, so its unit is m.

Step 2: Dividing the unit of velocity by the unit of time on the right-hand side:

\[ \frac{v}{t} \rightarrow \frac{\text{m s}^{-1}}{\text{s}} = \text{m s}^{-2} \]

Step 3: Comparing the two sides:

\[ \text{m} \neq \text{m s}^{-2} \]

The unit on the left-hand side is not equal to the unit on the right-hand side, so \( s = \frac{v}{t} \) is invalid, or wrong. In fact \( \text{m s}^{-2} \) is the unit of acceleration. The wrong equation claims that a distance equals an acceleration, which can never be true.

Unit analysis also tells us which quantities may be added or subtracted. Physical quantities can be added or subtracted only if they have the same composition of fundamental units. Add 2 metres of rope to 3 metres of rope and you have 5 metres of rope, but adding 2 litres of milk to 3 kilograms of rice does not give 5 of anything.

In the three cases below, u and v are velocities, a is acceleration, s is distance and t is time:

• \( u + v \) is possible, because both are velocities with the same unit, \( \text{m s}^{-1} \).
• \( s - at^2 \) is also possible. The unit of s is m, and when simplified the unit of \( at^2 \) is \( \text{m s}^{-2} \times \text{s}^2 = \text{m} \) as well.
• \( s + at \) is not possible. The unit of s is m, but the unit of at is \( \text{m s}^{-2} \times \text{s} = \text{m s}^{-1} \). They cannot be added because their fundamental units are different.

Let us test the validity of two equations by unit analysis. Here u is the initial velocity and v the final velocity, both with the unit \( \text{m s}^{-1} \). a is acceleration, with the unit \( \text{m s}^{-2} \). s is distance in m and t is time in s.

Example 1: \( v^2 = u^2 + 2as \)

Step 1: Finding the unit of the left-hand side:

\[ v^2 \rightarrow (\text{m s}^{-1})^2 = \text{m}^2\text{s}^{-2} \]

Step 2: Finding the unit of the first term on the right:

\[ u^2 \rightarrow (\text{m s}^{-1})^2 = \text{m}^2\text{s}^{-2} \]

Step 3: Finding the unit of the second term. The number 2 has no unit, so only the units of a and s are multiplied:

\[ 2as \rightarrow \text{m s}^{-2} \times \text{m} = \text{m}^2\text{s}^{-2} \]

Step 4: Putting all the units into the equation:

\[ \text{m}^2\text{s}^{-2} = \text{m}^2\text{s}^{-2} + \text{m}^2\text{s}^{-2} \]

The units of the physical quantities are uniform on either side, so this is a valid equation.

Example 2: \( s = ut + \frac{1}{2}a^2t \)

Step 1: The left-hand side is a distance, so its unit is m.

Step 2: Finding the unit of the first term on the right:

\[ ut \rightarrow \text{m s}^{-1} \times \text{s} = \text{m} \]

Step 3: The number \( \frac{1}{2} \) in the second term has no unit, so writing only the units of \( a^2 \) and t:

\[ a^2t \rightarrow (\text{m s}^{-2})^2 \times \text{s} \]

Step 4: Simplifying the powers:

\[ = \text{m}^2\text{s}^{-4} \times \text{s} = \text{m}^2\text{s}^{-3} \]

Step 5: Putting all the units into the equation:

\[ \text{m} = \text{m} + \text{m}^2\text{s}^{-3} \]

There is no similarity in the base units of the physical quantities on either side. The last term is not in metres, so it cannot even be added to a distance. So this equation is not valid.

Analysis of units is needed for three main jobs:

• To prove the accuracy of an equation, as in the examples above.
• To find out the inter-relation of physical quantities. For example, a joule is a newton multiplied by a metre, so \( 1\ \text{J} = 1\ \text{N m} \).
• To convert units from one measurement system to another.

Here is an example of the third job. The CGS system measures mass in grams and length in centimetres. Since \( 1\ \text{kg} = 1000\ \text{g} \) and \( 1\ \text{m} = 100\ \text{cm} \), one newton becomes \( 1000 \times 100 = 100000\ \text{g cm s}^{-2} \). The CGS unit of force, \( \text{g cm s}^{-2} \), is called the dyne, so 1 newton equals 100000 dynes. This conversion was possible only because the newton could be opened up into its fundamental units.