Question

Taking force, length and time to be the fundamental quantities, find the dimensions of
(a) density,
(b) pressure,
(c) momentum and
(d) energy.

Answer 1

Let F be the dimension of force.

(a) Density$=\frac{m}{v}=\frac{\left(\mathrm{force}/\mathrm{acceleration}\right)}{\mathrm{volume}}$
$$\begin{gathered} \left[\mathrm{Acceleration}\right]=\left[{\mathrm{LT}}^{-2}\right]\mathrm{and}\left[\mathrm{volume}\right]=\left[{\mathrm{L}}^3\right] \\ \therefore \left[\mathrm{Density}\right]=\frac{\left[\mathrm{F}/{\mathrm{LT}}^{-2}\right]}{\left[{\mathrm{L}}^3\right]}=\frac{\mathrm{F}}{{\mathrm{L}}^4{\mathrm{T}}^{-2}}=\left[{\mathrm{FL}}^{-4}{\mathrm{T}}^2\right] \end{gathered}$$

(b) Pressure$=\frac{\mathrm{force}}{\mathrm{area}}$
$$\begin{gathered} \left[\mathrm{Area}\right]=\left[{\mathrm{L}}^2\right] \\ \therefore \left[\mathrm{Pressure}\right]=\frac{\left[\mathrm{F}\right]}{\left[{\mathrm{L}}^2\right]}=\left[{\mathrm{FL}}^{-2}\right] \end{gathered}$$

(c) Momentum = mv = (force/acceleration) × velocity
$\mathrm{Acceleration}=\left[{\mathrm{LT}}^{-2}\right]\mathrm{and}\mathrm{velocity}=\left[{\mathrm{LT}}^{-1}\right]$
$\therefore \left[\mathrm{Momentum}\right]=\frac{\left[\mathrm{F}\right]}{\left[{\mathrm{LT}}^{-2}\right]}\times \left[{\mathrm{LT}}^{-1}\right]=\left[\mathrm{FT}\right]$

(d) Energy$=\frac{1}{2}m{v}^2=\frac{\mathrm{force}}{\mathrm{acceleration}}\times {\left(\mathrm{velocity}\right)}^2$
$\therefore \left[\mathrm{Energy}\right]=\frac{\left[\mathrm{F}\right]}{\left[{\mathrm{LT}}^{-2}\right]}\times {\left[{\mathrm{LT}}^{-1}\right]}^2=\left[\mathrm{FL}\right]$