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4-2.Friction
hard
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
$f=\mu$
B
$f=1 /(1+\mu)$
C
$f=1 /(1+1 / \mu)$
D
$f=1 /(\mu+1 / \mu)$
(KVPY-2017)
Solution
$(c)$ 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,
$f=F \Rightarrow \mu l(1-f) x g=f l x g$
$\Rightarrow \quad \mu(1-f)=f \Rightarrow \mu=f(1+\mu)$
$\Rightarrow \quad f=\frac{\mu}{1+\mu} \Rightarrow f=\frac{1}{\left(1+\frac{1}{\mu}\right)}$
Standard 11
Physics
