Question

A chain of length L remains in limiting equilibrium with a length x hanging down from a rough horizontal
table. The coefficient of friction between table and surface is:

1)x/L
2)L-x/L
3)L-x/L
4)x/L-x​

Answer 1

Answer:

$\boxed{\mathfrak{(4) \ \dfrac{x}{L-x}}}$

Explanation:

Let m be the mass of complete chain.

It is given that;

Total length of the chain is L and x length of the chain is hanging down.

$\sf Mass \: per \: unit \: length = \dfrac{m}{L}$

Friction acting on chain above table will balance the weight of hanging portion. (From the figure attached)

So,

$\rm \implies \mu N = \dfrac{m}{L} xg \\ \\ \rm \implies \mu \cancel{\dfrac{m}{L}}(L - x) \cancel{g} = \cancel{\dfrac{m}{L}} x \cancel{g} \\ \\ \rm \implies \mu (L - x) = x \\ \\ \rm \implies \mu = \dfrac{x}{L - x}$

$\therefore$ Coefficient of friction between table and surface ($\mu$) = $\rm \dfrac{x}{L - x}$

Answer 2

Concept:

Friction can be defined as the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other.

Given:

A chain of length $L$ remains in equilibrium and a length $x$ is hanging down from a horizontal table.

Find:

The coefficient of friction between the table and surface.

Solution:

Let $m$ be the mass of the chain.

Now, the total length of the chain is $L$ and $x$ is the length of the chain is hanging down the table.

Mass per unit length is $\frac{m}{L}$.

So, the length of the chain on the horizontal table is $L-x$ and the friction on the table is $\frac{m}{L}(L-x)g$ as $g$ is the acceleration due to gravity.

The length of the chain hanging down is $x$ and the friction down the chain is $\frac{mxg}{L}$.

Now, the friction acting on the chain above the table balance out the force of the chain hanging.

So,

$\mu \frac{m}{L}(L-x)g=\frac{mxg}{L}$

$\mu =\frac{x}{L-x}$.

Therefore, the coefficient of friction between the table and surface is $\mu =\frac{x}{L-x}$ .

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