5.6: Dimensional Analysis

5.6 Dimensional Analysis — Verifying Equations

Dimensional analysis is one of the simplest ways engineers can check an equation for mistakes. The rule is straightforward: both sides of a physically meaningful equation must have the same dimensions.

This does not prove that an equation is correct. Instead, it tells us whether the equation could be correct. If the dimensions on the two sides do not match, the equation is definitely wrong.

Before You Start — Units vs. Dimensions

Recall that a dimension describes the type of physical quantity being measured, while a unit describes how we measure it.

For example, velocity has dimensions of length divided by time:

\[ [v] = \frac{L}{T} \nonumber \]

Meters per second (m/s), kilometers per hour (km/h), and feet per second (ft/s) are all different units that can describe that same dimension.

In this section, the brackets \([\,]\) mean “the dimensions or units of.” For example, \([F]\) means “the dimensions or units of force.” The brackets do not mean that we are calculating a numerical value.

The Fundamental Dimensions Used Here

Most of the examples in this chapter can be reduced to three fundamental dimensions:

• \(M\) = mass
• \(L\) = length
• \(T\) = time

For example, acceleration has dimensions \(LT^{-2}\), and force has dimensions \(MLT^{-2}\).

✓ Worked Example 5.5 — Verifying \(\sigma = F/A\)

What the equation means:

• \(\sigma\) (Greek letter sigma) represents stress.
• \(F\) represents force.
• \(A\) represents area.

Stress describes how much force is distributed over an area. Its SI unit is the pascal (Pa).

Left side:

\[ [\sigma] = \text{Pa} = \frac{\text{N}}{\text{m}^2} = \frac{\text{kg}\cdot\text{m/s}^2}{\text{m}^2} = \frac{\text{kg}}{\text{m}\cdot\text{s}^2} \nonumber \]

Remember that one newton is:

\[ 1\text{ N} = 1\frac{\text{kg}\cdot\text{m}}{\text{s}^2} \nonumber \]

Right side:

\[ \left[\frac{F}{A}\right] = \frac{\text{N}}{\text{m}^2} = \frac{\text{kg}}{\text{m}\cdot\text{s}^2} \nonumber \]

We can also express both sides using fundamental dimensions:

\[ [\sigma] = ML^{-1}T^{-2} \qquad\text{and}\qquad \left[\frac{F}{A}\right] = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2} \nonumber \]

Conclusion: Both sides have identical dimensions. The equation is dimensionally consistent. ✓

✓ Worked Example 5.6 — Verifying \(P = V^2/R\)

What the equation means: This equation comes from electrical engineering.

• \(P\) represents electrical power, measured in watts (W).
• \(V\) represents voltage, measured in volts (V).
• \(R\) represents electrical resistance, measured in ohms (Ω).
• \(I\) represents electric current, measured in amperes (A).

You will study these quantities in detail in the circuits chapter. For now, we only need two unit relationships:

\[ 1\Omega = \frac{1\text{ V}}{1\text{ A}} \qquad\text{and}\qquad 1\text{ W}=1\text{ V}\cdot1\text{ A} \nonumber \]

Check the right side:

\[ \left[\frac{V^2}{R}\right] = \frac{\text{V}^2}{\text{V/A}} = \text{V}^2 \times \frac{\text{A}}{\text{V}} = \text{V}\cdot\text{A} = \text{W} \nonumber \]

Conclusion: The left side is power in watts, and the right side also reduces to watts. The equation is dimensionally consistent. ✓

⚠ Watch Out — Consistent Does Not Mean Correct

Dimensional consistency is a necessary condition for a physical equation, but it is not sufficient to prove the equation is correct.

For example, both of these equations are dimensionally consistent:

\[ d=\frac{1}{2}at^2 \qquad\text{and}\qquad d=37at^2 \nonumber \]

Both equations produce units of length, but dimensional analysis cannot tell us whether the numerical coefficient is correct. It is an error-checking tool, not a replacement for the physical model.

Practice — Spot the Invalid Equation

For each equation, determine whether it is dimensionally valid. If it is not, identify what is wrong. You do not need to know whether each equation is a correct physical law — only whether its units are consistent.

a) \(F = mv\), where \(F\) is force, \(m\) is mass, and \(v\) is velocity

b) \(P = VI\), where \(P\) is electrical power, \(V\) is voltage, and \(I\) is current

c) \(V = I + R\), where \(V\) is voltage, \(I\) is current, and \(R\) is resistance

d) \(T = mg\theta\), where \(T\) is a force, \(m\) is mass, \(g\) is acceleration, and \(\theta\) is an angle measured in radians

Answers - click to expand

1. \(F = mv\) is dimensionally invalid.

   Force has units of newtons:

   \[ [F] = \text{N} = \text{kg}\cdot\frac{\text{m}}{\text{s}^2} \nonumber \]

   The right side is mass multiplied by velocity:

   \[ [mv] = \text{kg}\cdot\frac{\text{m}}{\text{s}} \nonumber \]

   The dimensions do not match. The right side has the dimensions of momentum, not force. Newton's second law uses acceleration: \(F=ma\).

2. \(P = VI\) is dimensionally valid.
   \[ [VI] = \text{V}\cdot\text{A} = \text{W} \nonumber \]

   The right side reduces to watts, which is the unit of electrical power. Both sides are consistent.

3. \(V = I + R\) is dimensionally invalid.

   The left side is voltage, measured in volts:

   \[ [V]=\text{V} \nonumber \]

   The right side attempts to add current and resistance:

   \[ [I+R] = \text{A}+\Omega \nonumber \]

   Quantities with different dimensions cannot be added directly. Ohm's Law multiplies current by resistance: \(V=IR\).

4. \(T = mg\theta\) is dimensionally valid.

   Mass multiplied by acceleration has units of force:

   \[ [mg] = \text{kg}\cdot\frac{\text{m}}{\text{s}^2} = \text{N} \nonumber \]

   An angle measured in radians is dimensionless, so multiplying by \(\theta\) does not change the dimensions:

   \[ [mg\theta] = \text{N}\cdot1 = \text{N} \nonumber \]

   Both sides therefore have the dimensions of force. However, this only tells us that the equation is dimensionally possible. It does not prove that \(T=mg\theta\) is the correct physical model.

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