Let a machine overcome a load $L$ by the application of an effort $E$ . In time $t$ , let the displacement of effort be $d_{E}$ and the displacement of load be $d_{L}$ . $\; \text{ Work input } \;\;=\; \text{ Effort } \; \times \; \text{ displacement of effort } \;$ $\;= E \times d_{E}$Work output $=$ Load $\times$ displacement of load $= L \times d_{L}$Efficiency $=\; \frac{\; \text{ Work output }\;}{\; \text{ Work input }\;}$ $=\; \frac{L \times d_{L}}{E \times d_{E}} =\; \frac{L}{E} \times \frac{d_{L}}{d_{E}}$ $=\; \frac{L}{E} \times \frac{1}{\frac{d_{E}}{d_{L}}}$But $\; \frac{\; \text{ Load (L) }\;}{\; \text{ Effort (E) }\;} =$ M.A. and $\; \frac{d_{E}}{d_{L}} =$ V.R. $\therefore \;\; \text{ Efficiency } \; \eta \;=\; \frac{\; \text{ M.A. }\;}{\; \text{ V.R. }\;}$$\; \text{ or } \;\;\; \text{ M.A. } \;\;=\; \text{ V.R. } \; \times \eta$Thus, the mechanical advantage of a machine is equal to the product of its efficiency and Velocity Ratio.