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Length of chain is L and coefficient of static friction is $\mu$ . Calculate maximum length of the chain which can be hung from the table without sliding.

maximum length of hung chain
Option: 1
$\frac{\mu L}{1+\mu }$

Option: 2
$\frac{(\mu+1)L}{\mu}$

Option: 3
$\frac{\mu L}{1-\mu}$

Option: 4
$\left ( \frac{1-\mu}{\mu } \right )L$

Answers (1)

best_answer

As we learned

Maximum Length of hung chain -

$\mu=\frac{m_{2}}{m_{1}}=\frac{mass\ hanging\ from\ table}{mass\ on\ table}$

$\mu=\frac{l'}{{l-l'}}$

$l'=\frac{\mu\ l}{(\mu+l)}$

- wherein

$l=$ length of chain

$l'=$ chain hanging

$(l-l')=$ chain lying

So

Let y be the maximum length of the chain can be held outside the table without sliding

weight of the length of chain on table

$W{}'= \left ( L-y \right )\frac{M}{L}g$

weight of hanging part of chain = $W= \frac{M}{L}yg$

For equilibrium

limiting friction force = weight of hanging part of the chain

$F_{L}= \mu R=W$

$\mu {W}'= W\\* or\mu \frac{M}{L}\left ( L-y \right )g= \frac{M}{L}yg\\* \mu L-\mu y= y\\* y=\frac{\mu L}{1+\mu }$

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