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Measurement & Units

Source lecture(s): SC133 Lec 1

Intuition

Physics is a quantitative science: every statement must survive a comparison with a number read off an instrument. Before we can say how the world moves, we agree on what we measure it against — a system of units — and how well we know it — significant figures and uncertainty. A result without units is meaningless; a result without an error bar is a guess wearing a lab coat.

Why it matters

Getting units right catches most first-year mistakes for free (a velocity had better come out in m/s), and dimensional analysis lets you reconstruct half-forgotten formulas and sanity-check every answer in this course — from projectile range to the pendulum period.

The SI base units

Quantity Unit Symbol
length metre m
mass kilogram kg
time second s
electric current ampere A
temperature kelvin K
amount mole mol
luminous intensity candela cd

Every mechanical quantity is a product of powers of m, kg, s. Force, for example, has dimensions \([\text{M L T}^{-2}]\) and the derived unit newton, \(1\,\text{N} = 1\,\text{kg·m/s}^2\).

Dimensional analysis

Write each quantity's dimensions in \([\text{M}^a \text{L}^b \text{T}^c]\) and demand both sides of any equation match. A valid physical equation is dimensionally homogeneous.

\[[\text{speed}] = \frac{\text{L}}{\text{T}}, \qquad [\text{acceleration}] = \frac{\text{L}}{\text{T}^2}, \qquad [\text{energy}] = \frac{\text{M L}^2}{\text{T}^2}\]

This single tool reappears, in full force, in PC316 fluid mechanics where G. I. Taylor estimated a nuclear yield from a photograph with nothing else.

Worked example: is \(v = \sqrt{2gh}\) dimensionally sound?

\[[\sqrt{gh}] = \sqrt{\frac{\text{L}}{\text{T}^2}\cdot \text{L}} = \sqrt{\frac{\text{L}^2}{\text{T}^2}} = \frac{\text{L}}{\text{T}} = [v]\ \checkmark\]

Both sides are a speed, so the formula could be right — dimensional analysis can't supply the factor of 2, but it can reject anything with the wrong units.

Significant figures & orders of magnitude

  • Significant figures track precision: \(9.8\,\text{m/s}^2\) (2 s.f.) vs \(9.81\,\text{m/s}^2\) (3 s.f.). A result is only as precise as its least precise input.
  • Fermi estimates (order-of-magnitude reasoning) get you within a factor of ten with almost no data — a physicist's superpower.

Common mistakes

  • Dropping units mid-calculation, then guessing them at the end. Carry units through every line; they self-check the algebra.
  • Reporting 8 digits from 2-digit data. Significant figures are honesty about what you actually know.
  • Confusing mass and weight. Mass (kg) is intrinsic; weight (\(mg\), in newtons) is a force that depends on where you are.

Knowledge graph position

Prerequisites: none — this is the entry point of the course. Leads to: kinematics, vectors, and every quantitative page that follows.

Quiz

Q1 (conceptual). Why can dimensional analysis rule a formula out but never fully confirm it?

Answer

It checks units, not dimensionless factors. \(v=\sqrt{2gh}\) and \(v=\sqrt{gh}\) are both dimensionally valid; only a derivation or experiment fixes the factor of 2.

Q2 (computational). The period of a pendulum might depend on length \(L\), mass \(m\), and gravity \(g\). Use dimensions to find the only possible combination.

Answer

\(T \propto L^a m^b g^c\). Matching \([\text{T}]\): mass can't appear (\(b=0\)), and \(\sqrt{L/g}\) is the only length/gravity combination with units of time. So \(T \propto \sqrt{L/g}\) — mass drops out, as observed. See pendulum.

Q3 (multiple choice). \(1\,\text{N}\) in base units is: (a) \(\text{kg·m/s}\) (b) \(\text{kg·m/s}^2\) (c) \(\text{kg·m}^2/\text{s}^2\)

Answer

(b) — force = mass × acceleration.