Question

Derive a relationship between $\mathrm{MA},\mathrm{VR}\mathrm{and}\mathrm{\eta }$ of a machine.

Answer 1

Mechanical Advantage: It is the ratio of the load$\left(\mathrm{L}\right)$ to the effort$\left(\mathrm{E}\right)$.

$\mathrm{MA}=\frac{\mathrm{L}}{\mathrm{E}}$……..(1)

Velocity Ratio$\left(\mathrm{VR}\right)$: It is defined as the ratio of displacement of the effort$\left(d_{\mathrm{E}}\right)$ to the displacement of the load $\left(d_L\right)$.

$VR=\frac{d_E}{d_L}$………(2)

Efficiency $\mathrm{\eta }$$=\frac{\mathrm{Workdone}\mathrm{output}\left({\mathrm{W}}_{\mathrm{output}}\right)}{\mathrm{Workdone}\mathrm{input}\left({\mathrm{W}}_{\mathrm{input}}\right)}$

Consider a practical machine that overcomes a load $\left(\mathrm{L}\right)$ by the application of an effort $\left(\mathrm{E}\right)$ in time $\mathrm{t}$. Suppose the displacement of effort be ${\mathrm{d}}_{\mathrm{e}}$ and the displacement of load be ${\mathrm{d}}_{\mathrm{l}}$

Now we can write

$$\begin{gathered} {\mathrm{W}}_{\mathrm{input}}=\mathrm{E}\mathrm{x}{\mathrm{d}}_{\mathrm{e}} \\ {\mathrm{W}}_{\mathrm{output}}=\mathrm{L}\mathrm{x}{\mathrm{d}}_{\mathrm{l}} \\ \mathrm{\eta }=\frac{{\mathrm{W}}_{\mathrm{output}}}{{\mathrm{W}}_{\mathrm{input}}}=\frac{\mathrm{L}\mathrm{x}{\mathrm{d}}_{\mathrm{l}}}{\mathrm{E}\mathrm{x}{\mathrm{d}}_{\mathrm{e}}}=\frac{\mathrm{L}}{\mathrm{E}}\mathrm{x}\frac{{\mathrm{d}}_{\mathrm{l}}}{{\mathrm{d}}_{\mathrm{e}}}=\frac{\mathrm{L}}{\mathrm{E}}\mathrm{x}\frac{1}{{\displaystyle \frac{{\mathrm{d}}_{\mathrm{e}}}{{\mathrm{d}}_{\mathrm{l}}}}} \end{gathered}$$

From equations (1) and (2),

$$\begin{gathered} \mathrm{\eta }=\frac{\mathrm{MA}}{\mathrm{VR}} \\ \mathrm{MA}=\mathrm{\eta }\mathrm{x}\mathrm{VR} \end{gathered}$$

Thus, the relationship between $\mathrm{MA},\mathrm{VR}\mathrm{and}\mathrm{\eta }$ of machines is $\mathrm{MA}=\eta \times VR$.