Question

If the coefficient of static friction between a table and a uniform massive rope is μs, what fraction of the rope can hang over the edge of the table without the rope sliding?

Answer 1

Let me give you the procedure like this:
Lets say that F is the fraction of the rope hanging over the table
If its like that then we have to take into account that the friction force keeping on table is given by the following formula:
Ff = u*(1-f)*m*g 
and we need to know aso that gravity force pulling off the table Fg is given by this other formula:
Fg = f*m*g 
What you need to do is Equate the two and solve for f: 

f*m*g = u*(1-f)*m*g 
=> f = u*(1-f) = u - uf 
=> f + uf = u 
=> f = u/(1+u) = fraction of rope
With that you can find the answer

Answer 2

The fraction of rope can hang over the edge of the table without the rope sliding is $\boxed{\frac{{{\mu _s}}}{{\left( {1 + {\mu _s}} \right)}}}$.

Further explanation:

The rope, of uniform mass, is hanging over the edge of the table. The rope is hanging without sliding.

Concept:

Let us consider,

The mass per unit length of the rope  is $\lambda$.

The total length of the rope is $L$.

The total mass of rope is

$M=\lambda L$                                                                 …… (1)

Consider $x$ is the fraction of length of the rope hanging over the edge of the table.

The length of the rope hanging over the edge of the table is:

${l_1}=xL$                                                                             …… (2)

The mass of rope hang over the edge of the table is:

${m_1}=x{l_1}$    

Substitute the value of ${l_1}$ from equation (1) in the above equation.

${m_1}=x\lambda L$                                                                        …… (3)

The remaining mass of rope resting on the top of the table is:

${m_2}=M-{m_1}$

   

Substitute the value of  $M$ from equation (1) and $m_1$ from equation (3) in the above equation.

${m_2}=\lambda L - x\lambda L$    

Simplify above expression.

${m_2}=\left( {1 - x} \right)\lambda L$                                              …… (4)

From free body diagram of hanging mass of rope, the net force on the rope i.e., horizontal and vertical components of the net force acting on the rope is zero.

The equilibrium condition for horizontal component of force on rope is:

$\sum {F_x}=0$

The equilibrium condition for vertical component of the force on the rope is:

$\sum {F_y}=0$

At equilibrium, the horizontal component of net force is:

$F-{m_1}g = 0$    

Simplify the above equation,

$\boxed{F={m_1}g}$    

Substitute the value of ${m_1}$  from equation (3) in above equation.

$\boxed{F=x\lambda Lg}$                                                                                          …… (5)

From the free body diagram, the normal force acting on the rope is balanced by the weight on the part of the rope lying on the top of the table.

At equilibrium, the vertical component of net force is:

$\boxed{N={m_2}g}$    

Here, $N$ is the normal reaction on the table is balanced by weight of the rope which is resting on the table.

Substitute the value of ${m_2}$ from equation (4) in above equation.

$N=\left( {1 - x} \right)\lambda Lg$                                                                …… (6)

The friction force acting on the is:

${F_r}={\mu _s}N$

   

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

Substitute the value of $N$ from equation (6) in above equation.

${F_r}={\mu _s}\left( {1 - x} \right)\lambda Lg$                                             …… (7)

   

Substitute the value of ${F_r}$ from equation (7) in above equation.

$\boxed{F={\mu _s}\left( {1 - x} \right)\lambda Lg}$                                                                                                    …… (8)

Equating equation (5) and equation (8),

$x\lambda Lg={\mu _s}\left( {1 - x} \right)\lambda Lg$

Rearrange and simplify the above expression.

$\fbox{\begin\\x=\dfrac{{{\mu _s}}}{{\left( {1 + {\mu _s}} \right)}}\end{minispace}}$

Thus, the fraction of rope can hang over the edge of the table without the rope sliding is $\boxed{\frac{{{\mu _s}}}{{\left( {1 + {\mu _s}} \right)}}}$.    

Learn more:

1. Average translational kinetic energy: brainly.com/question/9078768

2. Acceleration of body by considering friction: brainly.com/question/7031524

3. Conservation of energy brainly.com/question/3943029

Answer detail:

Grade: High school

Subject: Physics

Chapter: Friction

Keywords:

Rope on the table, fraction of rope, length, rope without slide, coefficient of friction, static friction, uniformly massive rope, weight, equilibrium condition, mu/1+mu, mu/(1+mu), mass per unit length.