Laws Of Motion Question 321

Question: A horizontal uniform rope of length L, resting on a frictionless horizontal surface, is pulled at one end by force F. What is the tension in the rope in a distance l from the end where the force is applied?

Options:

A) (a)$ F( 1-\frac{l}{L} ) $

B) $ 2F( 1-\frac{l}{2L} ) $

C) $ \frac{F}{L} $

D) $ \frac{F}{l}( 1-\frac{l}{L} ) $

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Answer:

Correct Answer: A

Solution:

[a] Let a be the acceleration of the rope and M be its total mass.

Then $ T=\frac{M}{L}( L-l )a $ ..(i) $ \text{and }F-T=\frac{M}{L}\times \ell a $ -(ii) Dividing (i) and (ii),

$ \frac{F-T}{T}=\frac{\ell }{L-\ell }\Rightarrow F(L-\ell )-T(L-\ell )=T\ell $

$ \Rightarrow F(L-\ell )=T(L-\ell +\ell )=T\times L $

$ \Rightarrow T=F( 1+\frac{\ell }{L} ) $



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