Question

A homogenous chain lies in limiting
equilibrium on a horizontal table of
coefficient of friction 0.3 with part of it
hanging over the edge of the table. The
fractional length of the chain hanging
down the edge of the table is​

Answer 1

Answer:

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Explanation:

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Answer 2

Given:

A homogenous chain lies in limiting equilibrium on a horizontal table of coefficient of friction 0.3 with part of it hanging over the edge of the table.

To find:

Fraction off the chain hanging down the edge of the table.

Calculation:

Since the chain is in translational equilibrium we can say that the weight of the hanging part is balanced by the friction experienced by the part present on the horizontal table.

Let mass of chain be m and length of hanging part be x:

$\therefore \: weight = friction$

$= > ( \dfrac{m}{l} \times x)g = \mu(normal \: reaction)$

$= > ( \dfrac{m}{l} \times x)g = \mu \bigg \{\dfrac{m}{l} \times (l - x) g\bigg \}$

$= > xg = \mu \bigg \{ (l - x) g\bigg \}$

$= > xg = \mu lg - \mu xg$

$= > xg (1 + \mu)= \mu lg$

$= > x(1 + \mu)= \mu l$

$= > \dfrac{x}{l} = \dfrac{ \mu}{1 + \mu}$

So, final answer is:

$\boxed{ \bf{fraction \: of \: hanging \: part = \dfrac{x}{l} = \dfrac{ \mu}{1 + \mu} }}$