Q84P - University Physics with Modern Physics

Question

Question: If the coefficient of static friction between a table and a uniform, massive rope is ${\mathbf{\mu }}_{\mathbf{s}}$ , what fraction of the rope can hang over the edge of the table without the rope sliding?

Step-by-Step Solution

Answer

The fraction of the rope that can hang over the edge is $\frac{{\mu }_s}{1+{\mu }_s}$ .

Step 1: Identification of the given data

The given data is listed below as:

 

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

Step 2: Significance of the static friction

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.

Step 3: Creation of free-body diagrams

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.

Step 4: Determination of the fraction of the rope

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

 

$T=f_s$                                                                                                                       …(i)

 

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:

 

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

 

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

 

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

 

$w_2={\mu }_s{w}_1$ 

 

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

 

$$\begin{gathered} m_{2}g={\mu }_s{m}_{1}g \\ m_2={\mu }_s{m}_1 \end{gathered}$$ 

 

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 ${\mu }_s{m}_1$ in the above equation.

 

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

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