91Ó°ÊÓ

The given data is listed below as:

The coefficient of the static friction is, ${\mathrm{μ}}_s$

The static friction is described as the force that occurs due to friction which is applied on a particular object. Mainly static friction keeps an object in rest.

The free body diagram of the problem has been drawn below in order to better understand the problem.

Here, the mass $m_2$of the rope is hanging while the mass $m_1$is on the table.

The free body diagram of the vertical rope is drawn below:

From the above figure, it can be observed that the force is mainly acting in the y direction of the rope.

The equation of the forces acting in the y direction is expressed as:

$T=w_2$

Here, T is the tension of the rope and $w_2$is the weight of the hanging mass.

The free body diagram of the horizontal rope is drawn below:

From the above figure, it can be observed that the force is mainly acting in both the directions of the rope.

The equation of the summation of the forces acting in the x direction is expressed as:

$T=f_s$ …(¾±)

Here, $f_s$ is the frictional force of the rope.

The equation of the summation of the forces acting in the y direction is expressed as:

$n=w_1$

Here, n is the normal force and $w_1$is the weight of the mass that is above the table.

The equation of the frictional force is expressed as:

$$\begin{gathered} f_s={\mathrm{μ}}_{s}n \\ ={\mathrm{μ}}_s{w}_1 \end{gathered}$$

Here, $f_s$is the static frictional force and ${\mathrm{μ}}_s$ is the coefficient of static friction.

Substitute ${\mathrm{μ}}_s{w}_1$for $f_s$and $w_2$for T in the equation (i).

$w_2={\mathrm{μ}}_s{w}_1$

Substitute $m_{2}g$for $w_2$and $m_{1}g$for $w_1$in the above equation.

$$\begin{gathered} m_{2}g={\mathrm{μ}}_s{m}_{1}g \\ {m}_2={\mathrm{μ}}_s{m}_1 \end{gathered}$$

The equation of the fraction of the rope can be expressed as:

$f=\frac{m_2}{m_1+m_2}$

Here, f is the fraction of the rope, $m_2$ is the mass that is hanging from the table and $m_1$is the mass that is on the table.

Substitute ${\mathrm{μ}}_s{m}_1$in the above equation.

$$\begin{gathered} f=\frac{{\mathrm{μ}}_s{m}_1}{m_1+{\mathrm{μ}}_s{m}_1} \\ =\frac{{\mathrm{μ}}_s{m}_1}{m_1\left(1+{\mathrm{μ}}_s\right)} \\ =\frac{{\mathrm{μ}}_s}{1+{\mathrm{μ}}_s} \end{gathered}$$

Thus, the fraction of the rope that can hang over the edge is$\frac{{\mathrm{μ}}_s}{1+{\mathrm{μ}}_s}$ .