Mq1–Mq5 - Dimensions, units and consistency

Mq1–Mq5 - Dimensions, units and consistency

Describe mechanical quantities using mass, length and time, then use that description to check units and equations. Mq1–Mq3 develop dimensions and their relation to units, including dimensionless quantities; Mq4 changes unit systems, and Mq5 tests whether a proposed relationship is dimensionally consistent.

Build dimensions from mass, length and time

How can pressure and energy per unit volume have exactly the same dimensions? One pushes on a surface; the other measures stored energy. Their definitions look different, but reducing each to mass, length and time reveals the connection.

You already know

Speed is distance divided by time, acceleration is change of velocity divided by time, and force is mass times acceleration. Work done by a constant force along a displacement is force times displacement; power is energy transferred per unit time.

A unit is a chosen measuring scale. A dimension describes the kind of quantity being measured. Metres and centimetres are different units of length, but both have dimension LL. A mass measured in kilograms or grams still has dimension MM.

Dimensions

The dimensions of a quantity QQ, written [Q][Q], express it using powers of the base dimensions mass MM, length LL and time TT. For example, [v]=LT−1[v]=LT^{-1} for speed. The SI unit m s−1\mathrm{m\,s^{-1}} is a unit, not a dimensional expression.

Products, quotients and powers follow the ordinary laws of indices. A numerical multiplier such as 22 does not change dimensions.

Combining dimensions

For physical quantities AA, BB and a numerical power nn,

[AB]=[A][B],[AB]=[A][B],[An]=[A]n[AB]=[A][B],\qquad \left[\frac{A}{B}\right]=\frac{[A]}{[B]},\qquad [A^n]=[A]^n

Thus [force]=M(LT−2)=MLT−2[\text{force}]=M(LT^{-2})=MLT^{-2}.

Not in the formula booklet: learn it.

Use definitions to rebuild a result if you forget it. “Density” without further qualification means mass per unit volume; mass per unit area is a different quantity.

QuantityDefining relationDimensionsSI unit
Speeddistance divided by timeLT−1LT^{-1}m s−1\mathrm{m\,s^{-1}}
Accelerationvelocity change divided by timeLT−2LT^{-2}m s−2\mathrm{m\,s^{-2}}
Momentummass times velocityMLT−1MLT^{-1}kg m s−1\mathrm{kg\,m\,s^{-1}}
Forcemass times accelerationMLT−2MLT^{-2}N=kg m s−2\mathrm{N}=\mathrm{kg\,m\,s^{-2}}
Impulseforce times timeMLT−1MLT^{-1}N s\mathrm{N\,s}
Energy or workforce times distanceML2T−2ML^2T^{-2}J=kg m2 s−2\mathrm{J}=\mathrm{kg\,m^2\,s^{-2}}
Powerenergy divided by timeML2T−3ML^2T^{-3}W=kg m2 s−3\mathrm{W}=\mathrm{kg\,m^2\,s^{-3}}
Densitymass divided by volumeML−3ML^{-3}kg m−3\mathrm{kg\,m^{-3}}
Mass per unit areamass divided by areaML−2ML^{-2}kg m−2\mathrm{kg\,m^{-2}}
Mass per unit lengthmass divided by lengthML−1ML^{-1}kg m−1\mathrm{kg\,m^{-1}}

These dimensions and unit relations are not listed in the formula booklet: learn how to derive them.

Worked example 1: pressure from its definition

Question. Pressure is force per unit area. Find its dimensions and express the pascal in SI base units.

Plan. Divide the dimensions of force by those of area, then replace the base dimensions by their SI units.

[p]=[F][A]=MLT−2L2force per area=ML−1T−2subtract length powers1 Pa=1 N m−2=1 kg m−1 s−2expand the newton\begin{aligned} [p]&=\frac{[F]}{[A]}=\frac{MLT^{-2}}{L^2} &&\text{force per area} \\ &=ML^{-1}T^{-2} &&\text{subtract length powers} \\ 1\,\mathrm{Pa}&=1\,\mathrm{N\,m^{-2}}=1\,\mathrm{kg\,m^{-1}\,s^{-2}} &&\text{expand the newton} \end{aligned}

Why this step? Dividing by an area removes two powers of length, rather than one.

Answer. [p]=ML−1T−2[p]=ML^{-1}T^{-2}; the SI base unit is kg m−1 s−2\mathrm{kg\,m^{-1}\,s^{-2}}.

Check. (kg m s−2)/m2=kg m−1 s−2(\mathrm{kg\,m\,s^{-2}})/\mathrm{m^2}=\mathrm{kg\,m^{-1}\,s^{-2}}, agreeing with the dimensional calculation.

Change the definition to energy per unit volume. The extra length in energy is cancelled by an extra length in the denominator.

Worked example 2: energy density

Question. Energy density is energy per unit volume. Find its dimensions and compare them with pressure.

Plan. Derive energy from force times distance, then divide by volume.

[E]=(MLT−2)L=ML2T−2force times distance[energy density]=ML2T−2L3energy per volume=ML−1T−2cancel length powers\begin{aligned} [E]&=(MLT^{-2})L=ML^2T^{-2} &&\text{force times distance} \\ [\text{energy density}]&=\frac{ML^2T^{-2}}{L^3} &&\text{energy per volume} \\ &=ML^{-1}T^{-2} &&\text{cancel length powers} \end{aligned}

Why this step? The powers of length differ in both the numerator and denominator, so comparing names alone would hide the equal dimensions.

Answer. Energy density and pressure both have dimensions ML−1T−2ML^{-1}T^{-2}.

Check. J m−3=(N m)m−3=N m−2=Pa\mathrm{J\,m^{-3}}=(\mathrm{N\,m})\mathrm{m^{-3}}=\mathrm{N\,m^{-2}}=\mathrm{Pa}. Equal dimensions do not make the two physical quantities identical.

A volume contributes three powers of length, even when the quantity being measured is unfamiliar.

Watch out

Wrong: [energy density]=ML2T−2/L=MLT−2[\text{energy density}]=ML^2T^{-2}/L=MLT^{-2}. This divides by length, not volume. Correct: [energy density]=ML2T−2/L3=ML−1T−2[\text{energy density}]=ML^2T^{-2}/L^3=ML^{-1}T^{-2}.

This is a useful surprise: a pascal can also be written as a joule per cubic metre. Dimensional analysis recognises common structure without erasing physical meaning.

Exam tip

If dimensions are requested, finish with capital MM, LL, TT, such as [p]=ML−1T−2[p]=ML^{-1}T^{-2}. If units are requested, give units such as kg m−1 s−2\mathrm{kg\,m^{-1}\,s^{-2}}. A correct unit expression does not answer a request for dimensions.

Try another pair of definitions whose length powers cancel in different ways.

Dimensionless does not mean meaningless

Two quantities of the same dimension can form a ratio whose dimensions cancel. Strain, defined as extension divided by original length, is one example: L/L=1L/L=1. It records a physical change, even though it has no remaining powers of mass, length or time.

Dimensionless quantity

A quantity is dimensionless if its dimensions are M0L0T0=1M^0L^0T^0=1. For example, a ratio of two lengths is dimensionless. A length measured in metres is not dimensionless merely because its numerical value happens to be 11.

An angle in radians is arc length divided by radius, so [θ]=L/L=1[\theta]=L/L=1. The named unit radian remains useful: it tells you that the dimensionless number measures an angle. Counting cycles is also dimensionless, but counting cycles per unit time introduces a time factor.

Angle, frequency and angular speed

With angle θ\theta in radians, frequency ff in cycles per unit time and angular speed ω\omega,

[θ]=1,[f]=T−1,[ω]=T−1[\theta]=1,\qquad [f]=T^{-1},\qquad [\omega]=T^{-1}

The SI units are rad\mathrm{rad}, Hz=s−1\mathrm{Hz}=\mathrm{s^{-1}} and rad s−1\mathrm{rad\,s^{-1}}, respectively. Frequency and period are different: a period has dimension TT.

Not in the formula booklet: learn it.

Dimensionless coefficients can still carry important information. For example, a friction coefficient compares two forces, so its dimensions cancel. You cannot infer that every letter called a “constant” is dimensionless: gravitational acceleration is constant in many models but has dimensions LT−2LT^{-2}.

Key idea

Look at the defining ratio or equation, rather than the numerical value or the word “constant”. A dimensionless number can describe an angle, a proportion or a property of a material.

The following choices separate the time in a frequency from the count of cycles.

Let a given relationship define an unfamiliar quantity

You do not need to recognise a coefficient's name to find its dimensions. Rearrange the given relationship, replace each known quantity by its dimensions, and simplify. The same symbol can have different dimensions in different models.

Worked example 3: resistance proportional to speed

Question. A resistance force has magnitude R=kvR=kv, where vv is speed. Find the dimensions and SI base units of kk.

Plan. Rearrange to k=R/vk=R/v, then divide the dimensions of force by those of speed.

[k]=[R][v]=MLT−2LT−1rearrange the model=MT−1cancel length and subtract time powers\begin{aligned} [k]&=\frac{[R]}{[v]}=\frac{MLT^{-2}}{LT^{-1}} &&\text{rearrange the model} \\ &=MT^{-1} &&\text{cancel length and subtract time powers} \end{aligned}

Why this step? Calling kk a constant says its value stays fixed in the model, not that it has dimension 11.

Answer. [k]=MT−1[k]=MT^{-1}; SI base unit kg s−1\mathrm{kg\,s^{-1}}.

Check. (MT−1)(LT−1)=MLT−2(MT^{-1})(LT^{-1})=MLT^{-2}, the dimensions of RR.

Change only the power of speed from 11 to 22. The coefficient must now supply a different set of dimensions.

Worked example 4: resistance proportional to speed squared

Question. In a different model, R=kv2R=kv^2. Find the dimensions and SI base units of this model's kk.

Plan. Divide the force dimensions by the square of the speed dimensions.

[k]=[R][v]2=MLT−2(LT−1)2square both powers=MLT−2L2T−2=ML−1simplify\begin{aligned} [k]&=\frac{[R]}{[v]^2}=\frac{MLT^{-2}}{(LT^{-1})^2} &&\text{square both powers} \\ &=\frac{MLT^{-2}}{L^2T^{-2}}=ML^{-1} &&\text{simplify} \end{aligned}

Why this step? Squaring speed doubles both its length power and its time power.

Answer. [k]=ML−1[k]=ML^{-1}; SI base unit kg m−1\mathrm{kg\,m^{-1}}.

Check. (ML−1)(L2T−2)=MLT−2(ML^{-1})(L^2T^{-2})=MLT^{-2}, again a force.

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