5 Questions around this concept.
Length of chain is $L$ and coefficient of static friction is $\mu$. Calculate the maximum length of the chain which can be hung from the table without sliding. Maximum length of hung chain
A uniform rope of length I lies on a table. If the coefficient of friction is $\mu$ then the maximum length $x$ of the part of this rope which can overhang from the edge of the table without sliding down is
The length of the chain is 1 m and the coefficient of static friction is $\mu$. Calculate the maximum length of chain which can be hung from the table without sliding $(\mu=0.5)$
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A uniform rope of total length $l$ is at rest on a table with fraction $f$ of its length hanging (see figure). If the coefficient of friction between the table and the chain is $\mu$ then
A uniform chain of length l is placed on the table in such a manner that its l' part is hanging over the edge of the table without sliding.


$
\mu=\frac{m_2}{m_1}=\frac{\text { mass hanging from table }}{\text { mass on table }}
$
The chain will have uniform linear density.
So the ratio of mass and ratio of length for any part of the chain will be equal.
$
\begin{aligned}
& \mu=\frac{\text { length of part hanging from table }}{\text { length of part on table }}=\frac{l^{\prime}}{l-l^{\prime}} \\
& l^{\prime}=\frac{\mu l}{(\mu+1)}
\end{aligned}
$
Where $l=$ length of chain
$l^{\prime}=$ chain hanging
$
\left(l-l^{\prime}\right)=\text { chain lying on table }
$
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