2.A Dimensions and Units
There are 7 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.
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.
- Momentum is the product of mass and velocity.
- Jerk is the rate at which acceleration changes with respect to time.
- 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.
- \(s = vt + \frac{1}{2}at^2\) where \(s\) is displacement, \(v\) is velocity and \(a\) is acceleration.
- \(P = E/A\) 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”.
Convert a density of 3 g/mL into pounds/in\(^3\).
A car is going 35 miles per hour. How many feet per second is it traveling?
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
Dimensionally Feasible
- \([s]=L\) 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.
- \([P] = ML^{-1}T^{-2}\) 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”.
- The density is
\[ \left( 3 \, \frac{g}{mL} \right) \left( \frac{1}{1000} \, \frac{kg}{g} \right)\left( \frac{1}{0.45} \, \frac{pound}{kg} \right)\left( \frac{16.39}{1} \,\frac{mL}{in^3} \right) = 0.11 \frac{pound}{in^3} \]
- The car is traveling at
\[ \left( 35 \, \frac{mi}{hr} \right) \left( \frac{5280}{1} \, \frac{ft}{mi} \right)\left( \frac{1}{3600} \, \frac{hr}{sec} \right) = 51.33 \frac{ft}{sec} \]
- The energy is
\[ \left( 250 \frac{W}{m^2} \right) \left( 10 \, m^2 \right) \left( 7 \, days \right) \left( \frac{1}{1} \, \frac{J/sec}{W} \right) \left( \frac{24}{1} \, \frac{hr}{day} \right) \left( \frac{3600}{1} \frac{sec}{hr} \right) = 1.512 \times 10^9 \, J \]