question_answer

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)  \[\frac{mg}{4h}\sqrt{{{d}^{2}}+4{{h}^{2}}}\]  

B)         \[\frac{mg}{4h}\times d\]

C)  \[\frac{mg}{4h}\times \sqrt{{{d}^{2}}+4{{h}^{2}}}\]

D)         \[\frac{mgh}{d}\]

Correct Answer: A

Solution :

 Let \[T\] be the force exerted by each friend on the roe. The component of each tension acting vertically upwards as clear from the figure below is Thus,     \[T\cos \theta =mg\] or,          \[T=\frac{mg}{2\cos \theta }\]                                   ? (i) Let us determine the value of \[\cos \theta \] from the figure below:                 \[\cos \theta =\frac{base}{hypotenuse}=\frac{h}{\sqrt{{{\left( \frac{d}{2} \right)}^{2}}+{{h}^{2}}}}\] Substituting, it in equation (i), we have