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

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$F (1 - l / L)$

$F (1 + l / L)$

$F / (1 - l / L)$

$F / (1 + l / L)$

answer is A.

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Detailed Solution

Let M be the mass of the rope, Then force F is applied at one end, thon acceleration in the rope a = (F / m). Mass of the rope above length I.

$\begin{array}{l} M^{'} = \frac{M}{L} (L - l) and F^{'} = M^{'} a \\ ∴ F^{'} = \frac{M}{L} (L - l) \times \frac{F}{M} = F (\frac{L - l}{L}) = F (1 - \frac{l}{L}) \end{array}$

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