Question

Sneha has to lift a load of mass $50\text{kg}$ to a height $1\text{m}$. She can exert a maximum effort of $250\text{N}$, so she uses an inclined plane to lift the load up. What should be the minimum length of the plank used by her? $(\text{take}g=10\text{N/kg})$

Answer 1

Given, maximum effort $\left(E\right)=250\text{N}$
Load $\left(L\right)=mg=50\times 10=500\text{N}$
Height $\left(\text{AC}\right)=1\text{m}$

Let angle made by inclined plane with the horizontal be $\theta$.

From the figure above,

$E=L\sin \theta$
$\Rightarrow 250=500\sin \theta$

$\sin \theta =\frac{250}{500}=\frac{1}{2}$ ...$\left(\text{i}\right)$

Also, from triangle $\text{ABC}$,
$\sin \theta =\frac{\text{height}}{\text{length of the plane}}$
$=\frac{\text{AC}}{\text{AB}}$ ...$\left(\text{ii}\right)$

From equation $\left(\text{i}\right)$ and $\left(\text{ii}\right)$, we get

$\frac{\text{AC}}{\text{AB}}$ = $\frac{1}{2}$

$\Rightarrow \frac{1}{\text{AB}}$ = $\frac{1}{2}$

$\Rightarrow \text{AB}=2\text{m}$