A heavy uniform chain lies on a horizontal table top. If the coefficient of friction between the chain and the table surface is 0.25, then the maximum fraction of the length of the chain that can hang over one edge of the table is

The correct answer is 20%.

Let us analyze the problem step-by-step to find the maximum fraction of the chain's length that can hang over the edge of the table without slipping.

1. Define the Variables:
Let $L$ be the total length of the uniform chain.
Let $M$ be the total mass of the chain.
The linear mass density (mass per unit length) of the chain is:
$\lambda =\frac{M}{L}$
Let $f$ be the fraction of the total length of the chain hanging over the edge. Therefore, the hanging length is $fL$, and the length remaining on the horizontal table top is $(1-f)L$.

2. Identify the Forces Acting on Each Part of the Chain:
For the hanging part of the chain:
The mass of the hanging part is $m_{\text{hang}}=fM$.
The downward gravitational force (weight) pulling the chain down is:
$W_{\text{hang}}=fMg$
For the part of the chain on the table:
The mass of this part is $m_{\text{table}}=(1-f)M$.
The normal force $N$ exerted by the table on this part of the chain balances its weight:
$N=(1-f)Mg$

3. Set Up the Condition for Equilibrium:
The static friction force opposing the downward slide of the hanging part is given by:
$f_s=\mu N$
where $\mu$ is the coefficient of static friction between the chain and the table surface. Substituting the expression for the normal force, we get:
$f_s=\mu (1-f)Mg$
For the chain to remain in static equilibrium, the pulling force of the hanging weight must not exceed the maximum static friction force:
$W_{\text{hang}}\le f_s$
$fMg\le \mu (1-f)Mg$

4. Solve for the Maximum Fraction $f$:
We can divide both sides of the inequality by $Mg$:
$f\le \mu (1-f)$
$f\le \mu -\mu f$
Rearranging the inequality to solve for $f$:
$f+\mu f\le \mu$
$f(1+\mu )\le \mu$
$f\le \frac{\mu }{1+\mu }$

5. Substitute the Given Value of Coefficient of Friction:
Given that the coefficient of friction $\mu =0.25$, we substitute this value into the inequality:
$f_{\text{max}}=\frac{0.25}{1+0.25}$
$f_{\text{max}}=\frac{0.25}{1.25}=\frac{1}{5}=0.20$
Expressing the fraction as a percentage:
$0.20\times 100\%=20\%$

Thus, the maximum fraction of the length of the chain that can hang over the edge of the table is 20%.