2.2: Working with units
There are several rules when working with units that are not always obvious, and sometimes not correctly applied, even in standard textbooks. Errors with units have led to serious confusions and accidents like the collapse of a Mars orbiter and an aeroplane running out of fuel in mid-air. It is therefore important for every engineer to be able to correctly work with units and we repeat the most important rules here.
In mechanics, we often deal with scalar variables that represent physical quantities that have a measurement unit, like forces, velocities, or positions. Scalar variables consist of the product of two elements: a number and a measurement unit.
Force can be expressed as the product of a numerical value and the measurement unit newton:
\[F=3 \cdot \mathrm{N}=3 \mathrm{~N}. \nonumber\]
Following the ISO norm [5] and the "Red Book" [2], for quantities that have a measurement unit, the numerical value of a scalar variable \(Q\) that represents such quantities is denoted by \(\{Q\}\), and the measurement unit is denoted by \([Q]\), such that
\[Q=\{Q\} \cdot[Q] \tag{2.6} \label{2.6}\]
A mass \(m=3.5 \mathrm{~kg}\) has the numerical value \(\{m\}=3.5\) and the unit \([m]=\mathrm{kg}\). The same quantity value can be expressed as \(3500 \mathrm{~g}\). Then, the numerical value is \(\{m\}=3500\), and \([m]=\mathrm{g}\).
Note that square brackets only have meaning when placed around the physical quantity. For axes labels one commonly uses round brackets \([4,8]\) around the unit like for a time axis: \(t\) (s). Another option [1] is to use division: \(t / \mathrm{s}\). Generally, the letters of units are printed in roman (upright), those of scalar variables in italic.
For time \(t\),
- good use is: \(t=4 \mathrm{~s}, t=4(\mathrm{~s}), t / \mathrm{s}=4\), or \([t]=\mathrm{s}\),
- not good use is: \(\cancel{t \Rightarrow 4}\), or \(\cancel{t}\).
Besides paying attention to the units, also you need to make sure that if you have a vector on the left side of the \(=\) sign, you also need to have a vector on the right side.
From Equation 2.6 it can be seen that every scalar equation in mechanics can be split into two equations, one for the numerical values, and one for the units. For example, an equation like \(F=m a\), with \(F=1 \mathrm{~N}, m=1 \mathrm{~kg}\) and \(a=1 \mathrm{~m} / \mathrm{s}^{2}\) can be split into:
\[\begin{align} F & =m \cdot a \tag{2.7} \label{2.7}\\[4pt] \{F\}[F] & =\{m\}[m] \cdot\{a\}[a] \tag{2.8} \label{2.8}\\[4pt] \{F\} & =\{m\} \cdot\{a\} \rightarrow 1=1 \cdot 1 \tag{2.9} \label{2.9}\\[4pt] {[F] } & =[m] \cdot[a] \rightarrow \mathrm{N}=\mathrm{kg} \cdot \mathrm{m} / \mathrm{s}^{2} \tag{2.10} \label{2.10}\end{align}\]
When making a calculation it is important that in the end result both the numerical values and the units are correct since obviously, in equations, the units on both sides of the equal sign need to be identical, otherwise the equality can never hold. Normally, the units automatically work out well if you work with SI units. In some cases, one has to deal with equations like \(a=c_{1} t^{2}\). In that case the unit of the constant \(c_{1}\) is not given, but can be obtained from the equation \([a]=\left[c_{1}\right]\left[t^{2}\right]\), which can be written as \(\mathrm{m} / \mathrm{s}^{2}=\left[c_{1}\right] \cdot \mathrm{s}^{2}\), from which it follows that the unit of \(c_{1}\) is: \(\left[c_{1}\right]=\mathrm{m} / \mathrm{s}^{4}\).
A general advice is to work as long as possible with variables, and only to substitute numbers and units in the final solution. Units almost always help as a part of plausibility checks for calculations. Moreover, writing the numerical value and unit in each step of the derivation is much more work, and errors in numerical values are harder to find than in equations.
In this textbook we mostly work with the International System of Units, also called SI units, the SI system, or metric units, which are the world’s most widely used system of measurement. If you have to work with non-SI units, as
a habit first convert it to the SI unit as a first step to prevent problems later on. Often this conversion is just a matter of multiplying the non-SI unit value by a constant factor, e.g. \(1.0 \mathrm{inch}=1.0 \mathrm{inch} \times 0.0254 \mathrm{~m} / \mathrm{inch}\), however in some cases the conversion is more complex, like when converting temperature in Fahrenheit to degrees Kelvin. Preferably also work in the base SI unit, remove any factors related to prefixes (like converting \(\mathrm{mm}\) to \(\mathrm{m}\) by dividing by a factor 1000) before starting calculations, using scientific notation (e.g. \(3.02 \times 10^{8}\) ) for very large or small values.
There are 7 base SI units, and the only base SI units relevant in this textbook are \(\mathrm{m}, \mathrm{s}\) and \(\mathrm{kg}\). There are 22 derived SI units of which we use 6 in this textbook: the \(\mathrm{rad}=\mathrm{m} / \mathrm{m}, \mathrm{Hz}=1 / \mathrm{s}, \mathrm{N}=\mathrm{kg} \mathrm{m} / \mathrm{s}^{2}, \mathrm{~Pa}=\mathrm{N} / \mathrm{m}^{2}, \mathrm{~J}=\mathrm{Nm}\), and \(\mathrm{W}=\mathrm{J} / \mathrm{s}\). The unit to use for a certain quantity or parameter can often be determined by evaluating known dynamics equations. As an example, since \(x\) has unit \(\mathrm{m}\) and \(t\) has unit \(\mathrm{s}\), we can use the equation for acceleration \(a=\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}\) to determine that acceleration \(a\) has the unit \(\mathrm{m} / \mathrm{s}^{2}\) (note how the square sign \({ }^{2}\) is placed behind the \(t\) but before the \(x\) in the second derivative notation). Similarly, one obtains from \(F=m \times a\) that \(1 \mathrm{~N}=1 \mathrm{~kg} \times 1 \mathrm{~m} / \mathrm{s}^{2}=1 \mathrm{~kg} \mathrm{~m} / \mathrm{s}^{2}\). This method can often help to find the right unit by using a known equation, such that units don’t have to be memorised. Keep in mind that the equations in this textbook only work correctly if one uses SI units or units derived from SI units (like \(\mathrm{N}\) or W).
In some cases the scalar variable represents a unitless quantity. Examples of unitless quantities are integer numbers (for example the number of wheels under a car) or ratios of quantities with the same unit (for example the ratio of two distances). Most mathematical functions, like trigonometric or logarithmic functions, only accept unitless quantities. For example there is no mathematical way to calculate \(\tan (1 \mathrm{~m})\). So, always make sure only unitless numbers or ratios are used. E.g. \(\log \left(x_{2} / x_{1}\right)\) is fine, but \(\log \left(x_{2}\right)\) cannot be calculated if \(\left[x_{2}\right]=\mathrm{m}\). It is important to note that additions and subtractions can only be performed between scalar quantities of the same unit, while multiplications, divisions and powers can be taken on quantities of different units as long as the operation is also applied to the respective units.
A special case are angles, even though they usually have degrees \({ }^{\circ}\) or rad behind them, these angle designations are actually unitless numbers, which measure e.g. the ratio of the length of a circular arc divided by the length of the perimeter of a full circle. The degree or radian symbol only indicates how the trigonometric function should be applied. When not otherwise specified in this textbook we work with radians. Since angles are unitless, most mathematical functions can accept angles as an argument.
When dealing with vectors, one also needs to specify the direction of the vector using one or more unit vectors, so for specifying a vector one uses
the product of a number, a measurement unit and at least one unit vector. Subtraction and addition is only possible for vectors with the same unit, scalar products are possible as long as the units are also multiplied, so for instance \(\overrightarrow{\boldsymbol{v}} \times t=\Delta \overrightarrow{\boldsymbol{x}}\) is a correct equation, because \(\mathrm{m} / \mathrm{s} \times \mathrm{s}=\mathrm{m}\) and because quantities on both sides of the \(=\) sign are vectors.
Expressions and equations may contain mixtures of quantities given with their numerical value and quantities given as variables. In all cases one needs to ensure that the units on both sides of the equal sign \(=\) are equal, and that one cannot equate a scalar to a vector or matrix. If a vector quantity evaluates to zero one we use the zero vector \(\overrightarrow{\mathbf{0}}\), which is a vector with magnitude zero and no direction.
For a particle of mass \(m=3 \mathrm{~kg}\) having acceleration vector \(\overrightarrow{\boldsymbol{a}}=1 \mathrm{~m} / \mathrm{s}^{2} \hat{\boldsymbol{i}}\), the resultant force vector \(\overrightarrow{\boldsymbol{F}}\) on the particle can be calculated as:
Correct is: \(\overrightarrow{\boldsymbol{F}}=m \overrightarrow{\boldsymbol{a}}=3 \mathrm{~kg} \times 1 \mathrm{~m} / \mathrm{s}^{2} \hat{\boldsymbol{\imath}}=3 \mathrm{~kg} \mathrm{~m} / \mathrm{s}^{2} \hat{\boldsymbol{\imath}}=3 \mathrm{~N} \hat{\boldsymbol{\imath}}\).
Incorrect is: \(\overrightarrow{\boldsymbol{F}}=\overrightarrow{3} \cdot \boldsymbol{a} \mathbb{N}\), because the vector \(\overrightarrow{\boldsymbol{a}}\) has unit \(\mathrm{m} / \mathrm{s}^{2}\) such that the units on both side of the equal sign are different.
A particle performs a harmonic oscillation, such that its location coordinate \(x\) is a function of amplitude \(x_{0}\), angular frequency \(\Omega\), and time \(t\) :
\[x=x_{0} \sin (\Omega t) \tag{2.11} \label{2.11}\]
Fill in suitable units for \(x, x_{0}, \Omega\), and \(t\) and check the equation for consistency.
Which of these expressions make correct use of units, for mass \(m\), time \(t\), force \(F\), angular velocity \(\omega\), and angle \(\theta\) ? Correct the mistakes.
A. \(\quad m=20\)
B. \(\quad m=3 \mathrm{~N}\)
C. \(\quad F=3 \cdot m[\mathrm{~N}]\)
D. \(\quad \mathrm{F}=20 N\)
E. \([m]=\mathrm{kg}\)
F. \(\quad m /[m]=20\)
G. \(\quad t=3 \mathrm{~s}\)
H. \(\quad \omega=2 \mathrm{rad} / \mathrm{s}\)
I. \(\quad \theta=\omega \cdot t=2 \cdot t[\mathrm{~s}]\)
J. \(\quad\{m\}=20\)
K. \(\quad \omega=t^{2}\)
L. \(\quad \omega=1 \mathrm{rad} / \mathrm{s} \cdot t^{2}\)
M. \(\quad F(t)=\cos (1 \mathrm{rad} / \mathrm{s} \cdot t) \mathrm{N}\)
N. \(\quad F(t)=3 e^{-t} \mathrm{~N}\)
O. \(\quad F(t)=\ln \left(t^{2} / \mathrm{s}^{2}\right) \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}^{2}\)
\[\begin{align} & \text { Exemplary solution } \\[4pt] & \qquad \begin{array}{l} m \neq 20 \\[4pt] m \neq 3 \mathrm{~N} \\[4pt] F \neq 3 \cdot m[\mathrm{~N}] \\[4pt] \\[4pt] \\[4pt] \mathrm{F} \neq 20 \mathrm{~N} \\[4pt] \\[4pt] {[m]=\mathrm{kg}} \\[4pt] m /[\mathrm{m}]=20 \\[4pt] t=3 \mathrm{~s} \\[4pt] \omega=2 \mathrm{rad} / \mathrm{s} \\[4pt] \theta=\omega \cdot t \neq 2 \cdot t[\mathrm{~s}] \\[4pt] \{m\}=20 \\[4pt] \omega \neq t^{2} \\[4pt] \omega \neq 1 \mathrm{rad} / \mathrm{s} \cdot t^{2} \\[4pt] F(t)=\cos (1 \mathrm{rad} / \mathrm{s} \cdot t) \mathrm{N} \\[4pt] F(t) \neq 3 e^{-t} \mathrm{~N} \\[4pt] F(t)=\ln \left(t^{2} / \mathrm{s}^{2}\right) \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}^{2} \end{array} \end{align}\]
unit omitted; \(m=20 \mathrm{~kg}\)
wrong unit for mass quantity; \(m=3 \mathrm{~kg}\) unit omitted and unit inserted with wrongly used square brackets; \(F=\) \(3 \mathrm{~m} / \mathrm{s}^{2} \cdot m\)
roman variable, italic unit symbol; \(F=\) \(20 \mathrm{~N}\)
correct!
correct!
correct!
correct!
wrong unit and \([\mathrm{s}]\) is wrong; \(\theta=2 \mathrm{rad} / \mathrm{s} \cdot t\) correct!
unit mismatch; e.g. \(\omega=1 \mathrm{rad} / \mathrm{s}^{3} \cdot t^{2}\)
unit mismatch; e.g. \(\omega=1 \mathrm{rad} / \mathrm{s}^{3} \cdot t^{2}\)
correct !
exponent should not have units; e.g. \(F(t)=e^{-t / \mathrm{s}} \mathrm{N}\)
correct !