Question

The mechanical advantage of a machine is $4$ and its efficiency is $60\%$. It is used to lift a load of $300\mathrm{kgf}$ to a height of $15\mathrm{m}$. Calculate
(i) The effort required.
(ii) The work done on the machine.
(Take, $\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2$)

Answer 1

Step 1: (i) Given data.

Mechanical advantage$\left(\mathrm{MA}\right)=4$

Load $\left(\mathrm{L}\right)=300\mathrm{kgf}$

Step 2: Find the effort required.

We know that mechanical advantage is calculated by

$$\begin{gathered} \mathrm{MA}=\frac{Load\left(\mathrm{L}\right)}{\mathrm{Effort}\left(\mathrm{E}\right)} \\ or \\ Effort\left(E\right)=\frac{Load\left(L\right)}{MA} \end{gathered}$$
Substitute the given values in the above formula.
$\begin{array}{rcl}\mathrm{E}& =& \frac{\mathrm{L}}{\mathrm{MA}}\\ & =& \frac{300}{4}\\ & =& 75\mathrm{kgf}\end{array}$

Hence, the effort required is $75\mathrm{kgf}$.

Step 3: (ii) Given data

Load$\left(\mathrm{L}\right)=300\mathrm{kgf}$

Height or displacement of the load $\left(\mathrm{d}\right)=15\mathrm{m}$

Efficiency $\left(\eta \right)=60\%$

Step 4: Convert load from kgf to N.

$\begin{array}{rcl}\mathrm{L}& =& 300\mathrm{kgf}\\ & =& 300\times 10\\ & =& 3000\mathrm{N}\end{array}$

Step 5: Find the work output.

We know that, $\mathrm{Work}\left(W\right)=\mathrm{Load}\left(\mathrm{L}\right)\times \left(\mathrm{d}\right)$

Substitute the given values in the above formula.

$\begin{array}{rcl}W& =& 3000\mathrm{N}\times 15\mathrm{m}\\ & =& 45,000\mathrm{J}\end{array}$

Step 6: Find the work input.

We know that, $\left(\mathrm{Efficiency}=\frac{\mathrm{Work}\mathrm{output}}{\mathrm{Work}\mathrm{input}}\right)$ or $\left(\mathrm{Work}\mathrm{input}=\frac{\mathrm{Work}\mathrm{output}}{\mathrm{Efficiency}}\right)$.

Substitute the given values in the above formula.

$\begin{array}{rcl}\mathrm{Work}\mathrm{input}& =& \frac{45000}{0.6}\\ & =& 75000\mathrm{J}\end{array}$

Thus, the input work done is$\begin{array}{l}75000\mathrm{J}\end{array}$.