Question

A man has fallen into a ditch of width d and two of his friends are slowly pulling him out using a light rope and two fixed pulleys as shown in figure (5−E4). Show that the force (assumed equal for both the friends) exerted by each friend on the road increases as the man moves up. Find the force when the man is at a depth h.
Figure

Answer 1

The force when the man is at a depth h is $=\frac{m g}{4 h} \sqrt{d^{2}+4 h^{2}}$.

Explanation:

(a) Let the ropes make an angle θ with vertical at any depth.

From the diagram of the free body,

F cosθ + F cosθ - mg = 0

2F cosθ = mg

$\mathrm{F}=\frac{\mathrm{mg}}{2 \cos \theta}$

As the man slightly moves towards up, the angle θ increases, i.e cosθ  decreases.. Thus there is an rise in F.

(b) When the man reaches depth h,

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

$F=\frac{m g}{2 \cos \theta}$

$=\frac{m g}{2\left[\frac{h}{\sqrt{\left(\frac{d}{2}\right)^{2}+h^{2}}}\right]}$

$=\frac{m g}{4 h} \sqrt{d^{2}+4 h^{2}}$