Introduction to Dimensions in Mechanics
Welcome to the study of Dimensions! In CCEA AS 2 (Section B: Mechanics 2), dimensional analysis is one of the most powerful and reliable tools you will learn. Whether you are checking if a complicated mechanical formula makes sense, finding the units of a physical constant, or predicting how different physical quantities link together, dimensional analysis gives you a foolproof method to verify your work.
Don't worry if this seems abstract at first. At its heart, dimensional analysis simply comes down to tracking the basic physical "ingredients" that make up any mechanical quantity. Think of it like baking: you cannot end up with a cake if you only mix flour and sugar without measuring your quantities properly. Similarly, in mechanics, an equation for velocity must always balance out to give a velocity on both sides.
Quick Review Box:
• Dimension: The fundamental physical nature of a quantity (independent of specific measurement units).
• Unit: A specific standard used to measure a dimension (e.g. metres, seconds, kilograms).
• In Mechanics 2, we build everything from just three base dimensions: Mass, Length, and Time.
1. Fundamental Base Dimensions
In classical mechanics, every quantity can be expressed in terms of three fundamental dimensions:
• Mass: Represented by the dimension symbol \(\text{M}\) (or \([M]\))
• Length: Represented by the dimension symbol \(\text{L}\) (or \([L]\))
• Time: Represented by the dimension symbol \(\text{T}\) (or \([T]\))
We write square brackets around a variable, such as \([v]\), to mean "the dimensions of \(v\)".
Dimensions vs. SI Units: A Crucial Distinction
A very common error in exam papers is mixing up dimensions with SI units:
• Dimensions use capital letters: \(\text{M}\), \(\text{L}\), \(\text{T}\).
• SI Base Units use specific units: \(\text{kg}\), \(\text{m}\), \(\text{s}\).
For example, the dimension of speed is \(\text{L}\text{T}^{-1}\), whereas its SI unit is \(\text{m}\,\text{s}^{-1}\). Never write \(\text{kg}\) when asked for the dimension of mass!
2. Derived Mechanical Quantities
By using standard definitions and physical laws from mechanics, we can express all other quantities in terms of \(\text{M}\), \(\text{L}\), and \(\text{T}\). Let's build them step-by-step:
Geometry and Kinematics
• Displacement, Length, Radius, Distance (\(s, x, r, l\)): \([s] = \text{L}\)
• Area (\(A\)): \(\text{Length} \times \text{Length} \implies [A] = \text{L} \times \text{L} = \text{L}^2\)
• Volume (\(V\)): \(\text{Length}^3 \implies [V] = \text{L}^3\)
• Velocity / Speed (\(v, u\)): \(\left[\frac{\text{d}s}{\text{d}t}\right] = \frac{\text{L}}{\text{T}} = \text{L}\text{T}^{-1}\)
• Acceleration (\(a, g\)): \(\left[\frac{\text{d}v}{\text{d}t}\right] = \frac{\text{L}\text{T}^{-1}}{\text{T}} = \text{L}\text{T}^{-2}\)
• Frequency (\(f\) or \(n\)): \(\left[\frac{1}{\text{period}}\right] = \frac{1}{\text{T}} = \text{T}^{-1}\)
Mass Densities: Take Special Care!
• Volumetric Density (\(\rho\)): \(\left[\frac{\text{Mass}}{\text{Volume}}\right] = \frac{\text{M}}{\text{L}^3} = \text{M}\text{L}^{-3}\)
• Linear Density / Mass per unit length (\(\rho\) or \(\mu\)): \(\left[\frac{\text{Mass}}{\text{Length}}\right] = \frac{\text{M}}{\text{L}} = \text{M}\text{L}^{-1}\)
Examiner Tip: Always read the question carefully to see whether \(\rho\) is defined as density (\(\text{M}\text{L}^{-3}\)) or mass per unit length (\(\text{M}\text{L}^{-1}\)).
Forces, Energy, and Momentum
• Force, Weight, Tension, Resistance (\(F, W, T, R\)): Using Newton's Second Law (\(F = ma\)):
\([F] = [m][a] = \text{M} \times \text{L}\text{T}^{-2} = \text{M}\text{L}\text{T}^{-2}\)
• Work, Energy (Kinetic, Potential, Total) (\(W, E\)): Using \(\text{Work} = \text{Force} \times \text{distance}\):
\([E] = [F][s] = (\text{M}\text{L}\text{T}^{-2})(\text{L}) = \text{M}\text{L}^2\text{T}^{-2}\)
• Power (\(P\)): Using \(\text{Power} = \frac{\text{Work}}{\text{time}}\):
\([P] = \left[\frac{W}{t}\right] = \frac{\text{M}\text{L}^2\text{T}^{-2}}{\text{T}} = \text{M}\text{L}^2\text{T}^{-3}\)
• Linear Momentum (\(p\)) and Impulse (\(I\)): Using \(p = mv\) or \(I = Ft\):
\([p] = [m][v] = \text{M}(\text{L}\text{T}^{-1}) = \text{M}\text{L}\text{T}^{-1}\)
\([I] = [F][t] = (\text{M}\text{L}\text{T}^{-2})(\text{T}) = \text{M}\text{L}\text{T}^{-1}\)
• Pressure and Stress (\(p, \sigma\)): Using \(\text{Pressure} = \frac{\text{Force}}{\text{Area}}\):
\([p] = \left[\frac{F}{A}\right] = \frac{\text{M}\text{L}\text{T}^{-2}}{\text{L}^2} = \text{M}\text{L}^{-1}\text{T}^{-2}\)
• Surface Tension / Tension per unit length (\(S\)): Using \(S = \frac{\text{Force}}{\text{Length}}\):
\([S] = \left[\frac{F}{L}\right] = \frac{\text{M}\text{L}\text{T}^{-2}}{\text{L}} = \text{M}\text{T}^{-2}\) (or \(\text{M}\text{L}^0\text{T}^{-2}\))
Dimensionless Quantities
Some quantities have no physical dimensions. We say their dimension is \(1\) or \(\text{M}^0\text{L}^0\text{T}^0\).
These include:
• Pure numbers and fractions (such as \(2\), \(\frac{1}{2}\), \(\pi\))
• Angles (\(\theta\)), because angle in radians is \(\frac{\text{arc length}}{\text{radius}} = \frac{\text{L}}{\text{L}} = 1\)
• Trigonometric functions (\(\sin\theta, \cos\theta, \tan\theta\)) and exponential functions
Section Takeaway: Memorising the base derivations for Force (\(\text{M}\text{L}\text{T}^{-2}\)) and Energy (\(\text{M}\text{L}^2\text{T}^{-2}\)) allows you to work out almost any other dimension in seconds.
3. Principle of Dimensional Homogeneity (Consistency)
The Principle of Dimensional Homogeneity states that for any physical equation to be valid, every single additive term on both sides of the equation must have the exact same dimensions.
Think of it this way: You can add \(3\text{ metres} + 5\text{ metres} = 8\text{ metres}\), but you cannot add \(3\text{ metres} + 5\text{ seconds}\). That would be meaningless!
Important Rules for Dimensional Homogeneity:
1. If \(A = B + C - D\), then \([A] = [B] = [C] = [D]\). You do not add dimensions together (i.e. \([B+C]\) is simply \([B]\), not \(2[B]\)).
2. Pure mathematical constants and numerical coefficients (like \(\frac{1}{2}\) or \(2\pi\)) have dimension \(1\) and are ignored when testing consistency.
3. If an equation is dimensionally consistent, it might be physically correct (it shows the structure is valid, though numerical constants cannot be verified by dimensions alone). If it is not dimensionally consistent, it is definitely incorrect.
Step-by-Step Example: Checking Consistency
Problem: Show that the kinematic equation \(v^2 = u^2 + 2as\) is dimensionally consistent, where \(u, v\) are velocities, \(a\) is acceleration, and \(s\) is displacement.
Step 1: Find the dimensions of the Left-Hand Side (LHS):
\([v^2] = [v]^2 = (\text{L}\text{T}^{-1})^2 = \text{L}^2\text{T}^{-2}\)
Step 2: Find the dimensions of each term on the Right-Hand Side (RHS):
• First term: \([u^2] = [u]^2 = (\text{L}\text{T}^{-1})^2 = \text{L}^2\text{T}^{-2}\)
• Second term: \([2as] = [2] \times [a] \times [s] = (1) \times (\text{L}\text{T}^{-2}) \times (\text{L}) = \text{L}^2\text{T}^{-2}\)
Step 3: Compare:
Dimensions of LHS = \(\text{L}^2\text{T}^{-2}\)
Dimensions of every term on RHS = \(\text{L}^2\text{T}^{-2}\)
Since all terms have identical dimensions (\(\text{L}^2\text{T}^{-2}\)), the equation is dimensionally consistent.
4. Determining Dimensions of Unknown Constants
Examiners often ask you to find the dimensions or SI units of a physical constant appearing in an equation.
Step-by-Step Example: The Universal Gravitational Constant (\(G\))
Problem: Newton's law of universal gravitation is given by \(F = \frac{G m_1 m_2}{r^2}\), where \(F\) is force, \(m_1, m_2\) are masses, and \(r\) is distance. Determine the dimensions of \(G\).
Step 1: Rearrange the equation to make \(G\) the subject:
\(G = \frac{F r^2}{m_1 m_2}\)
Step 2: Take dimensions of both sides:
\([G] = \frac{[F] [r]^2}{[m_1] [m_2]}\)
Step 3: Substitute the known dimensions:
\([G] = \frac{(\text{M}\text{L}\text{T}^{-2})(\text{L}^2)}{(\text{M})(\text{M})} = \frac{\text{M}\text{L}^3\text{T}^{-2}}{\text{M}^2}\)
Step 4: Simplify powers using basic laws of indices:
\([G] = \text{M}^{1-2}\text{L}^3\text{T}^{-2} = \text{M}^{-1}\text{L}^3\text{T}^{-2}\)
5. Deriving Relationships (The Method of Dimensions)
If we know (or suspect) that a physical quantity \(Q\) depends on variables \(A\), \(B\), and \(C\), we can use dimensional analysis to find the exact power to which each variable must be raised!
The General Method:
1. Set up the relationship: Write \(Q = k A^x B^y C^z\), where \(k\) is a dimensionless constant (\([k] = 1\)), and \(x, y, z\) are unknown indices.
2. Write down dimensional equation: \([Q] = [A]^x [B]^y [C]^z\)
3. Substitute the dimensions for all variables.
4. Collect powers of \(\text{M}\), \(\text{L}\), and \(\text{T}\) on both sides.
5. Equate powers (indices) of \(\text{M}\), \(\text{L}\), and \(\text{T}\) to create simultaneous linear equations.
6. Solve the equations for \(x, y, z\) and write the final formula.
Step-by-Step Worked Example: Period of a Simple Pendulum
Problem: The period of oscillation \(T\) of a simple pendulum is assumed to depend on the mass of the bob \(m\), the length of the string \(l\), and the acceleration due to gravity \(g\). Use dimensional analysis to find an expression for \(T\).
Step 1: Set up the model equation
Let \(T = k m^x l^y g^z\), where \(k\) is a dimensionless constant.
Step 2: Express in dimensions
\([T] = [m]^x [l]^y [g]^z\)
Substitute standard dimensions:
\(\text{M}^0\text{L}^0\text{T}^1 = (\text{M})^x (\text{L})^y (\text{L}\text{T}^{-2})^z\)
Step 3: Combine powers on the RHS
\(\text{M}^0\text{L}^0\text{T}^1 = \text{M}^x \text{L}^{y+z} \text{T}^{-2z}\)
Step 4: Equate powers of \(\text{M}\), \(\text{L}\), and \(\text{T}\)
• For \(\text{M}\): \(x = 0\)
• For \(\text{T}\): \(-2z = 1 \implies z = -\frac{1}{2}\)
• For \(\text{L}\): \(y + z = 0 \implies y + \left(-\frac{1}{2}\right) = 0 \implies y = \frac{1}{2}\)
Step 5: Substitute indices back into the original formula
\(T = k m^0 l^{1/2} g^{-1/2}\)
\(T = k \sqrt{\frac{l}{g}}\)
Notice how dimensional analysis revealed that the period does not depend on the mass \(m\) at all (\(x = 0\))!
6. Common Pitfalls & How to Avoid Them
• Pitfall 1: Confusing Dimensions with SI Units.
Mistake: Writing \(\text{kg}\text{m}\text{s}^{-2}\) when asked for dimensions of Force.
Correction: Always write \(\text{M}\text{L}\text{T}^{-2}\).
• Pitfall 2: Forgetting the Definition of Linear Density.
Mistake: Using \(\text{M}\text{L}^{-3}\) whenever the word "density" appears.
Correction: Check if it states "mass per unit length" or "linear density" \(\implies \text{M}\text{L}^{-1}\).
• Pitfall 3: Negative Sign Errors in Indices.
Mistake: When equating powers of \(\text{T}\) for acceleration \((\text{L}\text{T}^{-2})^z\), writing \(2z\) instead of \(-2z\).
Correction: Take an extra line of working to expand brackets carefully.
• Pitfall 4: Treating Dimensionless Constants as 0 instead of 1.
Mistake: Setting \([k] = 0\).
Correction: A dimensionless constant has value \(1\) in dimensional multiplication (\([k] = 1 = \text{M}^0\text{L}^0\text{T}^0\)). Multiplying by \(0\) would incorrectly wipe out the term!
• Pitfall 5: Adding Dimensions Across Addition Signs.
Mistake: Writing \([u + at] = \text{L}\text{T}^{-1} + \text{L}\text{T}^{-1} = 2\text{L}\text{T}^{-1}\).
Correction: Dimensions do not add arithmetically. The dimension of the sum is simply \(\text{L}\text{T}^{-1}\).
Summary & Revision Checklist
• Base Dimensions: Mass (\(\text{M}\)), Length (\(\text{L}\)), Time (\(\text{T}\)).
• Key Quantities: Velocity (\(\text{L}\text{T}^{-1}\)), Acceleration (\(\text{L}\text{T}^{-2}\)), Force (\(\text{M}\text{L}\text{T}^{-2}\)), Energy/Work (\(\text{M}\text{L}^2\text{T}^{-2}\)), Power (\(\text{M}\text{L}^2\text{T}^{-3}\)), Pressure (\(\text{M}\text{L}^{-1}\text{T}^{-2}\)).
• Homogeneity: Every separate additive term on the LHS and RHS must have identical dimensions.
• Indices Method: Set \(Q = k A^x B^y C^z\), equate powers of \(\text{M}\), \(\text{L}\), \(\text{T}\), and solve the linear system for \(x, y, z\).