Question

A pulley system has a velocity ratio $5$. Draw a neat labeled diagram of the pulley system to lift a load by applying the effort in a convenient direction. Mark the tension in your diagram.
(i) If the efficiency of the system is $80\%$, find its mechanical advantage.
(ii) If a load of $10\mathrm{kgf}$ is pulled up by a distance of $2\mathrm{m}$ in $10\mathrm{s}$, calculate the power developed by the effort.

Answer 1

Step 1: (i) Given data.

Velocity ratio $\left(\mathrm{VR}\right)=5$

Efficiency $\left(\eta \right)=80\%=0.8$

Step 2: Draw the required diagram

1. The labeled diagram is shown below.
2. Tension is shown by the symbol $T$, effort is denoted by $\left(E\right)$, and load is denoted by $\left(L\right)$.

Step 3: Find the mechanical advantage $\left(\mathrm{MA}\right)$.

$\begin{array}{rcl}\mathrm{MA}& =& \left(\mathrm{VR}\right)\times \left(\eta \right)\\ & =& 5\times 0.8\\ & =& 4\end{array}$

Thus, the mechanical advantage is $4$.

Step 4: (ii) Given data.

Load $\left(\mathrm{L}\right)=10\mathrm{kgf}$

Distance $\left(\mathrm{d}\right)=2\mathrm{m}$

Time $\left(t\right)=10\mathrm{s}$

Acceleration due to gravity $\left(\mathrm{g}\right)=10\mathrm{m}/{\mathrm{s}}^2$.

Step 5: Calculate the power developed by the effort.

Convert load to newton,

$\begin{array}{rcl}\mathrm{L}& =& 10\times 10\mathrm{N}\\ & =& 100\mathrm{N}\end{array}$

We know that, $\left(\mathrm{power}\mathrm{output}=\frac{\mathrm{Load}\times \mathrm{distance}}{\mathrm{time}}\right)$.

Substitute the given values in the above formula.

$\begin{array}{rcl}\mathrm{power}\mathrm{output}& =& \frac{100\times 2}{10}\\ & =& \frac{200}{10}\\ & =& 20\mathrm{W}\end{array}$

Step 6: Calculate the power input (or power developed by the effort).

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

$\begin{array}{rcl}\mathrm{Power}\mathrm{input}& =& \frac{20}{0.8}\\ & =& 25\mathrm{W}\end{array}$

Thus, the power developed by the effort is $25\mathrm{W}$.