Question

A pulley system has a velocity ratio of $3$ and an efficiency of $80\%$. Calculate
(i) the mechanical advantage of the system.
(ii) the effort required to raise a load of $480\mathrm{N}$ by the system.

Answer 1

Step 1: (i) Given data,

Velocity ratio $\left(\mathrm{VR}\right)=3$

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

Step 2: Find the mechanical advantage.

The efficiency in terms of mechanical advantage and velocity ratio is,

$\text{Efficiency}\left(\eta \right)=\frac{\mathrm{Mechanical}\mathrm{advantage}\left(\mathrm{MA}\right)}{\mathrm{Velocity}\mathrm{ratio}\left(\mathrm{VR}\right)}\times 100\%$

Substitute the given values in the efficiency formula.

$\begin{array}{rcl}\text{Efficiency}\left(\eta \right)& =& \frac{\mathrm{MA}}{\mathrm{VR}}\times 100\%\\ 80\%& =& \frac{\mathrm{MA}}{3}\times 100\%\\ \mathrm{MA}& =& 2.4\end{array}$

So, the mechanical advantage is $2.4$.

Step 3: (ii) Given data,

From the part (i) the mechanical advantage $\left(\mathrm{MA}\right)=2.4$

Load $\left(\mathrm{L}\right)=480\mathrm{N}$

Step 4: Calculate the effort required.

We know that, $\mathrm{MA}=\frac{\mathrm{Load}\left(\mathrm{L}\right)}{\mathrm{Effort}\left(\mathrm{E}\right)}$

Substitute the given values in the mechanical advantage formula.

$\begin{array}{rcl}\mathrm{MA}& =& \frac{\mathrm{L}}{\mathrm{E}}\\ 2.4& =& \frac{480\mathrm{N}}{\mathrm{E}}\\ \mathrm{E}& =& \frac{480}{2.4}\\ \mathrm{E}& =& 200\mathrm{N}\end{array}$

Thus, the effort required to raise a load of $480\mathrm{N}$ by the system is$\begin{array}{l}200\mathrm{N}\end{array}$.