Quantities, SI units and dimensions in mechanics

A physical quantity is a measurable property written as a numerical value and a unit. In mechanics, the unit is part of the answer:

$$v=12\ \mathrm{m\,s^{-1}}$$

means a velocity of twelve metres per second. Writing only $v=12$ loses information. It also makes it harder to detect an incorrect calculation.

This lesson develops three essential habits:

1. use consistent SI units before substituting into a formula;
2. decide whether a quantity has direction;
3. check that both sides of an equation have the same dimensions.

Prerequisites

You should be able to:

• multiply and divide with powers of ten;
• use index laws, including negative indices;
• rearrange formulae;
• round to a stated number of significant figures;
• recognise simple vectors and magnitudes.

Review standard form and accuracy or units and compound measures if conversions involving powers of ten are insecure.

A quantity is not the same as its unit

Mass is a quantity. The kilogram is its SI unit. The symbol $m$ often represents a particular mass, while $\mathrm{kg}$ labels the scale on which it is measured.

Keep these roles separate. In

$$m=3.5\ \mathrm{kg},$$

$m$ is the quantity symbol, $3.5$ is the numerical value and $\mathrm{kg}$ is the unit symbol.

Unit symbols have fixed forms:

• they do not take plurals, so write $8\ \mathrm{kg}$, not $8\ \mathrm{kgs}$;
• unit names are lower case in prose, but symbols named after people are capitalised, such as newton and $\mathrm N$;
• a space separates the value from the unit, as in $20\ \mathrm m$;
• $\mathrm{m\,s^{-1}}$ means metres per second and $\mathrm{m\,s^{-2}}$ means metres per second per second.

Fundamental and derived quantities

At A level, most mechanics units are built from three SI base units:

Fundamental quantity	Common symbol	SI base unit	Unit symbol
mass	$m$	kilogram	$\mathrm{kg}$
length	$x$, $s$, $r$, $h$	metre	$\mathrm m$
time	$t$	second	$\mathrm s$

A derived quantity is defined using other quantities. Its unit follows from that definition.

Quantity	Defining relationship	SI unit
area	length $\times$ length	$\mathrm{m^2}$
volume	length $\times$ length $\times$ length	$\mathrm{m^3}$
speed or velocity	displacement $\div$ time	$\mathrm{m\,s^{-1}}$
acceleration	change in velocity $\div$ time	$\mathrm{m\,s^{-2}}$
force	mass $\times$ acceleration	$\mathrm N$
moment	force $\times$ perpendicular distance	$\mathrm{N\,m}$

Newton’s second law, $F=ma$, gives

$$1\ \mathrm N =1\ \mathrm{kg}\times1\ \mathrm{m\,s^{-2}} =1\ \mathrm{kg\,m\,s^{-2}}.$$

The newton is therefore a convenient name for a compound unit. It is not a new fundamental unit.

Mass and weight are different quantities

Mass measures the amount of matter and is measured in kilograms. Weight is the gravitational force on that mass and is measured in newtons:

$$W=mg.$$

A mass of $5\ \mathrm{kg}$ has weight $5g=49\ \mathrm N$ when $g=9.8\ \mathrm{m\,s^{-2}}$.

Worked example 1: derive a unit from a formula

The resultant force on a $750\ \mathrm{kg}$ car is $1800\ \mathrm N$. Find its acceleration.

From $F=ma$,

$$a=\frac Fm =\frac{1800\ \mathrm N}{750\ \mathrm{kg}}.$$

Replace the newton by base units:

$$a =\frac{1800\ \mathrm{kg\,m\,s^{-2}}}{750\ \mathrm{kg}} =2.4\ \mathrm{m\,s^{-2}}.$$

The kilograms cancel. This confirms that the remaining unit is appropriate for acceleration.

Self-check 1

A force of $36\ \mathrm N$ acts on a particle of mass $4.5\ \mathrm{kg}$. Find the acceleration, showing how the units cancel.

Answer

$$a=\frac Fm =\frac{36\ \mathrm{kg\,m\,s^{-2}}}{4.5\ \mathrm{kg}} =\boxed{8\ \mathrm{m\,s^{-2}}}.$$

Converting simple units

Common metric prefixes are powers of ten.

Prefix	Symbol	Multiplier
kilo	$\mathrm k$	$10^3$
centi	$\mathrm c$	$10^{-2}$
milli	$\mathrm m$	$10^{-3}$

Thus

$$1\ \mathrm{km}=1000\ \mathrm m, \qquad 1\ \mathrm{cm}=0.01\ \mathrm m, \qquad 1\ \mathrm{mm}=0.001\ \mathrm m.$$

A reliable conversion multiplies by a fraction equal to $1$, arranged so the unwanted unit cancels.

Worked example 2: convert a length and a time

Convert $3.75\ \mathrm{km}$ to metres:

$$3.75\ \mathrm{km} \times\frac{1000\ \mathrm m}{1\ \mathrm{km}} =3750\ \mathrm m.$$

Convert $2.4\ \mathrm{min}$ to seconds:

$$2.4\ \mathrm{min} \times\frac{60\ \mathrm s}{1\ \mathrm{min}} =144\ \mathrm s.$$

The unit to remove appears once above and once below the fraction bar.

Converting compound units

For speed, both the distance unit and the time unit must be converted. Since

$$1\ \mathrm{km}=1000\ \mathrm m \quad\text{and}\quad 1\ \mathrm h=3600\ \mathrm s,$$

we have

$$1\ \mathrm{km\,h^{-1}} =\frac{1000\ \mathrm m}{3600\ \mathrm s} =\frac{5}{18}\ \mathrm{m\,s^{-1}}.$$

Therefore

$$\boxed{\mathrm{km\,h^{-1}}\ \text{to}\ \mathrm{m\,s^{-1}}: \text{ multiply by }\frac5{18}},$$

and in reverse,

$$\boxed{\mathrm{m\,s^{-1}}\ \text{to}\ \mathrm{km\,h^{-1}}: \text{ multiply by }\frac{18}{5}}.$$

These factors should be understood, not merely memorised.

Worked example 3: kilometres per hour to metres per second

Convert $90\ \mathrm{km\,h^{-1}}$ to $\mathrm{m\,s^{-1}}$.

$$90\times\frac5{18}=25.$$

Hence

$$\boxed{90\ \mathrm{km\,h^{-1}}=25\ \mathrm{m\,s^{-1}}}.$$

This is sensible: the numerical value becomes smaller because one metre per second is faster than one kilometre per hour.

Worked example 4: metres per second to kilometres per hour

A sprinter’s speed is $10.4\ \mathrm{m\,s^{-1}}$. In kilometres per hour,

$$10.4\times\frac{18}{5}=37.44.$$

So the speed is

$$\boxed{37.4\ \mathrm{km\,h^{-1}}}$$

to $3$ significant figures.

Self-check 2

Convert:

1. $72\ \mathrm{km\,h^{-1}}$ to $\mathrm{m\,s^{-1}}$;
2. $15\ \mathrm{m\,s^{-1}}$ to $\mathrm{km\,h^{-1}}$.

Answer

$$72\times\frac5{18}=\boxed{20\ \mathrm{m\,s^{-1}}},$$ $$15\times\frac{18}{5}=\boxed{54\ \mathrm{km\,h^{-1}}}.$$

Powers in units must also be converted

If a length is squared or cubed, its conversion factor is squared or cubed too.

$$1\ \mathrm{cm^2} =(10^{-2}\ \mathrm m)^2 =10^{-4}\ \mathrm{m^2},$$

not $10^{-2}\ \mathrm{m^2}$. Similarly,

$$1\ \mathrm{cm^3} =(10^{-2}\ \mathrm m)^3 =10^{-6}\ \mathrm{m^3}.$$

Worked example 5: area conversion

Convert $320\ \mathrm{cm^2}$ to square metres.

$$320\ \mathrm{cm^2} =320(10^{-4})\ \mathrm{m^2} =\boxed{0.032\ \mathrm{m^2}}.$$

The most common area conversion error

Since $1\ \mathrm m=100\ \mathrm{cm}$,

$$1\ \mathrm{m^2}=(100\ \mathrm{cm})^2=10\,000\ \mathrm{cm^2}.$$

The conversion factor belongs to every length direction, so it must be squared.

Convert before substituting

Mechanics formulae work with any coherent system of units, but the inputs must belong to the same system. At A level, convert to SI units unless the question clearly supports another consistent choice.

Worked example 6: a mixed unit calculation

A cyclist travels at a constant speed of $24\ \mathrm{km\,h^{-1}}$ for $45\ \mathrm s$. Find the distance travelled in metres.

The formula $s=vt$ cannot yet be used directly because hours and seconds are mixed. First convert the speed:

$$v=24\times\frac5{18} =\frac{20}{3}\ \mathrm{m\,s^{-1}}.$$

Then

$$s=vt =\frac{20}{3}\times45 =\boxed{300\ \mathrm m}.$$

An alternative is to convert $45\ \mathrm s$ to $45/3600\ \mathrm h$ and obtain $0.3\ \mathrm{km}$. Both approaches are consistent and give the same distance.

Self-check 3

A car travels at $108\ \mathrm{km\,h^{-1}}$ for $25\ \mathrm s$. How far does it travel?

Answer

First convert the speed:

$$108\times\frac5{18}=30\ \mathrm{m\,s^{-1}}.$$

Therefore

$$s=vt=30(25)=\boxed{750\ \mathrm m}.$$

Scalars and vectors

A scalar has magnitude only. A vector has magnitude and direction.

Scalars	Vectors
mass	displacement
time	velocity
distance	acceleration
speed	force

Units alone do not distinguish a scalar from a vector. Distance and displacement are both measured in metres. Speed and velocity are both measured in $\mathrm{m\,s^{-1}}$. Their meanings differ.

In one dimensional motion, a sign can encode direction after a positive direction has been chosen. If east is positive, then

$$v=-6\ \mathrm{m\,s^{-1}}$$

means a velocity of $6\ \mathrm{m\,s^{-1}}$ west. The corresponding speed is

$$|v|=6\ \mathrm{m\,s^{-1}}.$$

A scalar cannot be made into a vector merely by attaching a minus sign. For example, a negative distance has no physical meaning in elementary mechanics, while a negative displacement can be meaningful.

Learn these distinctions in depth in position, displacement, velocity and acceleration.

Dimensions

Dimensions describe the type of a quantity independently of the unit chosen. Use

$$[M]\quad\text{for mass}, \qquad [L]\quad\text{for length}, \qquad [T]\quad\text{for time}.$$

Square brackets here mean “the dimensions of”. They do not mean numerical units.

For example,

$$[v]=[LT^{-1}],$$

because velocity is length divided by time. Similarly,

$$[a]=[LT^{-2}],$$

and from $F=ma$,

$$[F]=[M][LT^{-2}]=[MLT^{-2}].$$

Quantities may be added or equated only if they have the same dimensions. This is the principle of dimensional homogeneity.

Worked example 7: check a constant acceleration equation

Consider

$$s=ut+\frac12at^2.$$

The left side has dimension $[L]$. On the right,

$$[ut]=[LT^{-1}][T]=[L],$$

and

$$[at^2]=[LT^{-2}][T^2]=[L].$$

Every term has dimension $[L]$, so the equation is dimensionally consistent.

Now consider the incorrect equation

$$s=u+\frac12at^2.$$

Here $[s]=[L]$ but $[u]=[LT^{-1}]$. A displacement cannot equal a velocity plus a displacement, so the equation must be wrong.

Worked example 8: determine possible powers

Suppose a time $t$ is proposed to depend on distance $s$ and constant acceleration $a$ through

$$t=ks^p a^q,$$

where $k$ is dimensionless. Find $p$ and $q$.

Taking dimensions gives

$$[T]=[L]^p[LT^{-2}]^q =[L^{p+q}T^{-2q}].$$

Match powers of $L$ and $T$:

$$p+q=0, \qquad -2q=1.$$

Thus

$$q=-\frac12, \qquad p=\frac12.$$

Therefore the proposed dependence has the form

$$t=k\sqrt{\frac{s}{a}}.$$

Dimensional reasoning determines the powers, but it cannot determine the dimensionless constant $k$.

What dimensional analysis can and cannot prove

Consistent dimensions are necessary for a physical equation, but they do not prove the equation is correct. Both $s=ut+at^2$ and $s=ut+\tfrac12at^2$ are dimensionally consistent. Dimensions cannot detect the missing factor $1/2$.

Self-check 4

Which of these formulae could be dimensionally valid? Here $v$ and $u$ are velocities, $a$ is acceleration, $s$ is displacement and $t$ is time.

1. $v=u+at$
2. $v^2=u^2+2as$
3. $s=ut+at$

Answer

1. Valid dimensionally, because

$$[at]=[LT^{-2}][T]=[LT^{-1}]=[v]=[u].$$

2. Valid dimensionally, because

$$[as]=[LT^{-2}][L]=[L^2T^{-2}]=[v^2]=[u^2].$$

3. Invalid dimensionally. Although $[ut]=[L]$, we have $[at]=[LT^{-1}]$, which is velocity rather than displacement.

Common misconceptions

• “Kilograms measure weight.” Kilograms measure mass. Weight is a force measured in newtons.
• “Acceleration is measured in metres per second.” That is a velocity unit. Acceleration is measured in $\mathrm{m\,s^{-2}}$.
• “To convert $\mathrm{km\,h^{-1}}$ to $\mathrm{m\,s^{-1}}$, multiply by $1000$ and by $3600$.” Hours occur in the denominator, so converting the denominator divides by $3600$. The net factor is $1000/3600=5/18$.
• “A negative vector has negative magnitude.” Magnitude is non-negative. A negative component indicates direction relative to the chosen axis.
• “Matching dimensions proves a formula.” It only shows that the formula has passed one necessary check.
• “Units may be added at the final line.” Carrying units through the calculation exposes conversion and formula errors earlier.

Exam checklist

Before finalising a mechanics answer, ask:

1. Have I converted all data into a consistent set of units?
2. Does the requested quantity need a direction as well as a magnitude?
3. Is the final unit appropriate for the quantity?
4. Are all terms in each equation dimensionally compatible?
5. Is the numerical scale reasonable after conversion?
6. Have I rounded only at the end and followed the requested accuracy?

Next steps

Continue with modelling assumptions in mechanics to see how physical situations are simplified. Then study position, displacement, velocity and acceleration before using the constant acceleration equations. For forces and the unit newton, move on to forces and free body diagrams and Newton’s laws.