Question

A man has fallen into a ditch of width $d$ and two of his friend are slowly pulling him out using a light rope and two fixed pulleys as shown in the figure. Show that the force (assumed equal for both the friends) exerted by each friend on the rope increases as the man moves up. Find the force when the man is at a depth $h$.

Answer 1

Let 'T' be the tension in both side rope when the friends are pulling the man ditch out.

Hence, we can say that the forces are acting on the man in the ditch by both the friends are 'T'

where mg be the gravitational force acting downwards.

Let , $\theta$ be the angle made by both the tension with the vertical.

therefore we have,

$2T\cos \theta =mg$

$T=\frac{mg}{2\cos \theta }$

$\theta$ can be increased by $\frac{\pi }{2}$ as the man pulls up and hence, T will gets increased accordingly.

Let 'h' be the height when man is in the ditch.

'L' be the length of the rope between the man in the ditch and either of the friends pulling him,

Hence, from the figure

$L=\sqrt{h^2+\left(\frac{d^2}{4}\right)}$

$\cos \theta =\frac{h}{L}=\frac{h}{\sqrt{h^2+\left(\frac{d^2}{4}\right)}}$

$T=\frac{mg}{2\cos \theta }=\frac{mg}{2\frac{h}{\sqrt{h^2+\left(\frac{d^2}{4}\right)}}}=\frac{mg}{4h}\sqrt{4h^2+d^2}$