A uniform rope of total length l is at rest o | vedclass

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Question

$(c)$ Let mass per unit length of chain is $x$. Then, weight pulling the chain on table is

$F=f l x g$

Weight of chain on table

$=$ Normal re

Vedclass · Accepted Answer

$f=1 /(1+1 / \mu)$

Vedclass · Answer

$(c)$ Let mass per unit length of chain is $x$. Then, weight pulling the chain on table is

$F=f l x g$

Weight of chain on table

$=$ Normal reaction of table

$=l(1-f) x g$

Friction acting on chain,

$f=\mu N=\mu l(1-f) x g$

in equilibrium,

$f=F \Rightarrow \mu l(1-f) x g=f l x g$

$\Rightarrow \quad \mu(1-f)=f \Rightarrow \mu=f(1+\mu)$

$\Rightarrow \quad f=\frac{\mu}{1+\mu} \Rightarrow f=\frac{1}{\left(1+\frac{1}{\mu}\right)}$

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

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