2.A Dimensions and Units

There are 7 fundamental dimensions.

Fundamental Dimensions
Dimension Symbol Sample Units
Length \(L\) meter (m), foot (ft), mile (mi)
Time \(T\) second (s), day, hour (hr)
Mass \(M\) gram (g), kilogram (kg)
Amount \(N\) mole (mol)
Temperature \(\theta\) Kelvin (K), Celsius (\(^\circ \mbox{C}\))
Electric Current \(I\) Ampere (amp)
Luminosity \(J\) Candela (cd)

Everything else is a derived dimension. Here are some examples.

Derived Dimensions
Dimension Symbol Intuition Sample Units
Area L\(^2\) length \(\times\) width square foot (ft\(^2\)), square meter (m\(^2\)), acre
Volume \(L^3\) length \(\times\) width \(\times\) height cubic meter (m\(^3\)), liter(L), gallon (gal)
Velocity \(\frac{L}{T}\) distance/time meter/second, miles/hour
Acceleration \(\frac{L}{T^2}\) velocity/time meters/second\(^2\)
Force \(\frac{ML}{T^2}\) mass \(\times\) acceleration Newton (N), pound (lb)
Energy (Work) \(\frac{ML^2}{T^2}\) force \(\times\) displacement Joule (J), calorie, kilowatt hour (kwh), foot-pound
Power \(\frac{ML^2}{T^3}\) energy/time Watt (W), horsepower
Density \(\frac{M}{L^3}\) mass/volume kilogram/meter\(^3\)
Pressure \(\frac{M}{LT^2}\) force/area newton/meter\(^2\), pascal (Pa), psi, atm
Electric Charge \(I T\) current \(\times\) time coulomb

Activities

Find the Dimension

Find the dimension of the following quantities.

  1. Momentum is the product of mass and velocity.
  2. Jerk is the rate at which acceleration changes with respect to time.
  3. Voltage is the work needed per unit charge to move a test charge between two points. (You need to look up both “work” and “charge” in our list of derived dimensions.)

Dimensionally Feasible

A formula is dimensionally feasible when the units for the two sides match. Decide whether each of the following formulas is dimensionally feasible or infeasible.

  1. \(s = vt + \frac{1}{2}at^2\) where \(s\) is displacement, \(v\) is velocity and \(a\) is acceleration.
  2. \(P = E/A\) where \(P\) is pressure, \(E\) is energy and \(A\) is area.

Geometric Formulas for a Cylinder

A student is trying to remember the formulas for the surface area \(A\) and the volume \(V\) of a cylinder of radius \(r\) and height \(h\). They come across the following two formulas: \[ \pi r^2 h\] and \[ 2\pi r (r+h).\] What are the dimensions for each of these formulas? Why does this tell you which one is \(A\) and which one is \(V\)?

Unit Conversions

Solve the following problems by multiplying by the appropriate unit conversions. This unit conversion chart will be helpful, or you can use Google to find a particular conversion, for example just search for “km to miles”.

  1. Convert a density of 3 g/mL into pounds/in\(^3\).

  2. A car is going 35 miles per hour. How many feet per second is it traveling?

  3. Sunlight deposits energy at a rate of 250 W/m\(^2\) on a 10 m\(^2\) garden. How much energy (in Joules) is deposited in a week? Note that 1 Watt is 1 Joule per second)

Solutions

Find the Dimension

  1. \(M L T^{-1}\)
  2. \(L T^{-3}\)
  3. \(ML^2T^{-2} / IT = ML^2 T^{-3}I^{-1}\)

Dimensionally Feasible

  1. \([s]=L\) and \(\left[vt + \frac{1}{2}at^2\right] = [vt] + [at^2] = L + LT^{-2}T^2 = L + L = L\). So this formula is dimensionally feasible.
  2. \([P] = ML^{-1}T^{-2}\) and \([E/A] = ML^2T^{-2}L^{-2} = MT^{-2}\). This formula is not dimensionally feasible.

Geometric Formulas for a Cylinder

We know that \([V]=L^3\) and \([A]=L^2\).

We have \([\pi r^2h] = L^3\) and \([2\pi r (r+h)] = L (L+L) = L^2\). So dimensional analysis tells us that \(V=\pi r^2h\) and \(A=2\pi r (r+h)\).

Unit Conversions

Solve the following problems by multiplying by the appropriate unit conversions. This unit conversion chart will be helpful, or you can use Google to find a particular conversion, for example just search for “km to miles”.

  1. The density is

\[ \left( 3 \, \frac{g}{mL} \right) \left( \frac{1}{1000} \, \frac{kg}{g} \right)\left( \frac{1}{0.45} \, \frac{pound}{kg} \right)\left( \frac{16.39}{1} \,\frac{mL}{in^3} \right) = 0.11 \frac{pound}{in^3} \]

  1. The car is traveling at

\[ \left( 35 \, \frac{mi}{hr} \right) \left( \frac{5280}{1} \, \frac{ft}{mi} \right)\left( \frac{1}{3600} \, \frac{hr}{sec} \right) = 51.33 \frac{ft}{sec} \]

  1. The energy is

\[ \left( 250 \frac{W}{m^2} \right) \left( 10 \, m^2 \right) \left( 7 \, days \right) \left( \frac{1}{1} \, \frac{J/sec}{W} \right) \left( \frac{24}{1} \, \frac{hr}{day} \right) \left( \frac{3600}{1} \frac{sec}{hr} \right) = 1.512 \times 10^9 \, J \]