Lesson Explainer: Quantities and Units in Mechanics Mathematics

In this explainer, we will learn how to identify fundamental and derived quantities used in mechanics, such as length, time, and velocity, and identify their units and unit conversions.

Definition: Unit of Measurement

A unit is a defined magnitude of a quantity. Any other magnitude of the same quantity may be expressed as a multiple of the unit.

For example, the kilogram is a unit of mass. Any other mass may be defined as a certain number of kilograms.

Specific units of measurement are arbitrary and may be defined by convenience. For example, radians and degrees were arbitrarily defined to be useful in certain situations. Radians are typically more useful in trigonometric problems since the base functions of sine and cosine operate naturally with radians. Quantities in degrees, minutes, and seconds, however, must first be converted into radians before inputting them into trigonometric functions, but they are a much more intuitive way for humans to think about angles.

Nevertheless, both of these units describe the same quantity, also known as a dimension. Every system of units has a unit for every physical quantity, and the most commonly used system of units is the International System of Units (SI), which defines 7 base quantities from which all other physical quantities may be derived.

Definition: The International System of Units (SI)

The International System of Units (SI) consists of 7 base units for the following base quantities:

QuantityDimension SymbolUnit
MassMKilogram (kg)
TimeTSecond (s)
LengthLMeter (m)
Electric currentIAmpere (A)
Amount of substanceNMole (mol)
Luminous intensityJCandela (cd)
Thermodynamic temperatureΘKelvin (K)

Any physical quantities not among the 7 base quantities may be expressed as a product of powers of the 7 base units. Let’s look at a simple example of how to derive the SI unit for a quantity not among the 7 base SI quantities.

Example 1: Deriving the Units of Velocity

Velocity is equal to the distance covered over a period of time. Which of the following is not a unit of velocity?

  1. cm/s
  2. m/s
  3. m/s2
  4. m/min
  5. km/h

Answer

For an object that moves a distance (or displacement as a vector quantity) in a given amount of time, the velocity is given by the equation 𝑣=𝑑𝑑, where 𝑑 is the displacement and 𝑑 is the time elapsed. The right-hand side has the dimension length/time, LT, so the left-hand side, velocity, must also have the dimension length/time.

Let’s consider the answers:

(A) cm/s is a length unit, cm, divided by a time unit, s, so this is a unit for velocity.

(B) m/s is a length unit, m, divided by a time unit, s, so this is also a unit for velocity.

(C) m/s2 is a length unit, m, divided by a time unit squared, s2, so this is not a unit for velocity.

For completeness, (D) m/min is also a length unit, m, divided by a time unit, min, so this is also a unit for velocity;

(E) km/h is a length unit, km, divided by a time unit, h, so this is also a unit for velocity.

Therefore, the answer is (C); m/s2 is not a unit for velocity.

Some quantities in fact, such as angle or percentage or slope, are called β€œdimensionless” or β€œdimension 1” because they are the product of base quantities all raised to the power of zero.

For example, the gradient of an incline is dimensionless because it is given by one length divided by another, so it has dimension LL=1.

Sometimes, there may be more than one step to deriving the SI unit for a quantity not among the 7 base quantities. Such quantities are often given a special name. The SI also consists of further 22 derived quantities with special names. For example, the derived SI unit for force is the familiar newton. Let’s look at an example of how to derive these more complex units from the 7 base SI units.

Example 2: Using Formulae to Derive Units

The force applied on an object is calculated using the formula 𝐹=π‘šπ‘Ž, where π‘Ž is the acceleration of the object (measured in metres per second squared, or m/s2) and π‘š is its mass (measured in kilograms, or kg). The force is then measured in newtons. Which of the following is the expression of newtons in terms of the base units kilograms, metres, and seconds?

  1. (β‹…/)kgms
  2. kgβ‹…s2/m
  3. kgβ‹…m/s2
  4. kgβ‹…m2/s2
  5. kgβ‹…m/s

Answer

Consider the equation for force: 𝐹=π‘šπ‘Ž.

In an equation relating physical quantities, the dimensions on both sides of the equation must be identical, so the dimension of force must be equal to the dimension of mass multiplied by the dimension of acceleration.

Mass is an SI base quantity, with SI base unit kilogram. Acceleration, however, is not an SI base quantity, but it too may be given by a standard equation π‘Ž=𝑣𝑑, where 𝑣 is the velocity and 𝑑 is the time elapsed. Again, the dimensions on both sides of the equation must be identical. On the right-hand side of this equation, we have one SI base quantity, time, but another that is still a derived quantity, velocity. However, velocity may be expressed by a standard equation 𝑣=𝑑𝑑, where 𝑑 is the displacement and 𝑑 is the time. Now, we finally have an equation where the right-hand side is expressed entirely in SI base units, length and time. Working backward from this equation, we may rewrite the equation for acceleration: π‘Ž=𝑑=𝑑𝑑.

And working backward from this equation, we can rewrite the equation for force: 𝐹=π‘šπ‘‘π‘‘.

We now have force expressed entirely in SI base quantities. Substituting in the dimensions for each of these quantities, we have 𝐹==β‹…/,MLTMLT and by substituting in the SI units for these quantities, we obtain the SI unit for force, the newton, in terms of the base SI units: 𝐹=β‹…/,kgms which is answer (C).

We can follow this process to find the derived SI unit for any physical quantity. Let’s look at another, more complicated example.

Example 3: Using Formulae to Derive Units of Kinetic Energy

The kinetic energy, which is measured in joules, is given by the rule 𝑇=12π‘šπ‘£οŠ¨. Which of the following units is equal to the joule?

  1. kg/m/s2
  2. kgβ‹…m/s
  3. kgβ‹…m2/s2
  4. kg/m2/s2
  5. kgβ‹…m/s2

Answer

Consider the equation for kinetic energy: 𝑇=12π‘šπ‘£.

In an equation relating physical quantities, the dimensions on both sides of the equation must be identical, so the dimension of energy must be equal to the dimension of 12 multiplied by the dimension of mass, mutliplied by the dimension of velocity squared.

12 is dimensionless, and π‘š is already a base SI quantity, so we need only find the dimension of π‘£οŠ¨.

Velocity is given by the change in displacement or as an SI base quantity, length, over time: 𝑣=𝑙𝑑.

Therefore, velocity squared is given by 𝑣=𝑙𝑑.

Therefore, 𝑇=12π‘šπ‘£οŠ¨ is given in base SI quantities by 𝑇=12π‘šπ‘™π‘‘.

Using the SI units for mass, kilogram, length, metre, and time, second, the right-hand side, and therefore the unit of 𝑇, the joule, has dimension: 1=1=1β‹…/.Jkgmskgms

Therefore, the answer is (C).

Units of quantities not included in either the 7 base quantities or the 22 derived quantities may be expressed as a product of either the base units, the derived units, or both. For example, the unit for torque, 𝜏, may be expressed as follows: 𝜏==/.Nmkgms

Sometimes, of course, it is more convenient to use units with different magnitudes. For example, the SI unit for speed is metres per second, but when considering the speed of a moving vehicle, kilometres per hour is more convenient.

However, it is often necessary to work with quantities given in different units. In such cases, we usually convert all quantities to SI units first. Let’s look at an example of how to convert units of different magnitudes into SI units.

Example 4: Converting Dynes into Newtons

What is 2.87Γ—10 dynes in newtons?

Answer

To start with, recall that 1 dyne (dyn) is defined by 1=1β‹…/.dyngcms

Since the newton is a derived SI unit, where 1=1β‹…/Nkgms, we can now proceed by converting all units on the right-hand side of the equation into SI units: 1=10,1=10.gkgcmm

So, we have 1=ο€Ή10⋅10/.dynkgms

Collecting the powers of 10, 1=10β‹…/=10.dynkgmsN

Multiplying both sides by the given magnitude, 2.87Γ—10=2.87Γ—10Γ—10=287.dynNN

Sometimes, converting units to SI units may not be as straightforward as multiplying by a power of 10. In the final example, let’s look at how to convert from units with nonstandard magnitudes to SI units, and vice versa.

Example 5: Converting between Speed Units

Which of the following is false?

  1. 72 km/h = 20 m/s
  2. 3 km/min = 50 cm/s
  3. 15 m/s = 54 km/h
  4. 42 m/min = 70 cm/s

Answer

To check each equation, we can convert the units on the left-hand side into the units on the right-hand side.

Starting with (A), we need to convert kilometres per hour into metres per second. It can help to write the units as fractions: 72/=72=72ο€Ή103600=720003600=20/.kmhkmhmsmsms

Therefore, (A) is true.

Next, for (B), we need to convert kilometres per minute into centimetres per second: 3/=3=3ο€Ή1060=300060=3000ο€Ή1060=5000/.kmminkmminmsmscmscms

Therefore, (B) is false. We will proceed with the others to confirm they are true.

For (C), we need to convert metres per second into kilometres per hour: 15/=15=15ο€Ή10=3600Γ—15Γ—10=54/.msmskmhkmhkmh

Therefore, (C) is true.

Finally, for (D), we need to convert metres per minute into centimetres per second: 42/=42=42ο€Ή1060=420060=70/.mminmmincmscmscms

Therefore, (D) is true.

Let’s finish by recapping some of the key points from this explainer.

Key Points

  • The International System of Units (SI) consists of 7 base units for physical quantities. Every other physical quantity has a unit that may be expressed in terms of a product of powers of these 7 base units.
  • The unit for any physical quantity in terms of the SI units may be found from an equation for that quantity by converting all quantities on the right-hand side of the equation into base quantities and replacing them with their SI units to give a product of powers.
  • Units of nonstandard magnitudes may be converted into SI units by multiplying by the appropriate factor. For example, hours may be converted into seconds by multiplying by 3β€Žβ€‰β€Ž600.

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