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 the figure. What is the force (assumed equal for both the friends) exerted by each friends on the slope ?

• A. $\frac{mg}{4h}\sqrt{d^2+4h^2}$
• B. $\frac{mgd}{4h}$
• C. $\frac{mg}{h}\sqrt{d^2+4h^2}$
• D. $\frac{mgh}{d}$

Answer 1

The correct option is A $\frac{mg}{4h}\sqrt{d^2+4h^2}$

Consider the given diagram,

The man is pulled out slowly (i.e. with constant velocity ) which means there is no vertical acceleration. Hence, the net force on the man along vertical direction will be zero.

$2\ast T\ast cos\theta =m\ast g$
$cos\theta =2h\ast \sqrt{(d^2+4h^2)}$

As the rope is light, tension will be same throughout the rope . So the force exerted by the friends is given by
$T=(m\ast g\ast \sqrt{d^2+4h^2})/4h$