If the coefficient of static friction between a table and a uniform massive rope is μs, what fraction of the rope can hang over

Question

aleksklad

1 day ago

6

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?

Physics

2 answers:

inna [987] · Answer 1 · 2026-03-04T08:11:46+03:00

Here's the procedure explained: Assume F represents the portion of the rope that is extending over the table. In this scenario, the frictional force that holds the rope on the table can be calculated using the formula: Ff = u*(1-f)*m*g. Additionally, it is important to determine the gravitational force that attempts to pull the rope off the table, Fg, calculated through: Fg = f*m*g. You then need to set these two equations equal to each other and resolve for f: f*m*g = u*(1-f)*m*g leads to f = u*(1-f) = u - uf. Simplifying gives f + uf = u, which results in f = u/(1+u) representing the fraction of the rope. This will lead you to the final answer.

kicyunya [1K] · Answer 2 · 2026-03-04T08:15:23+03:00

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

Further clarification:

A rope with a consistent mass hangs over the edge of the table without sliding down.

Concept:

Let's analyze the problem:

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

Total length of the rope is $L$ .

The overall mass of the rope is

$M=\lambda L$ …… (1)

Assume $x$ represents the fraction of the rope's length that is hanging off the table.

The length of the rope that hangs over the table's edge is:

${l_1}=xL$ …… (2)

The mass of the portion of the rope that hangs over the table is:

${m_1}=x{l_1}$

Now insert the value of ${l_1}$ from equation (1) into this equation.

${m_1}=x\lambda L$ …… (3)

The remaining mass of rope that lies on the table is:

${m_2}=M-{m_1}$

Substitute the values of $M$ from equation (1) and $m_1$ from equation (3) into the above equation.

${m_2}=\lambda L - x\lambda L$

Simplifying the expression above gives:

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

Based on the free body diagram for the hanging mass of the rope, the total forces acting on it, both horizontal and vertical, balance out to zero.

The balance condition for the horizontal force acting on the rope is:

$\sum {F_x}=0$

The balance condition for the vertical force on the rope is:

$\sum {F_y}=0$

Under equilibrium, the horizontal net force amounts to:

$F-{m_1}g = 0$

Now simplify this equation,

$\boxed{F={m_1}g}$

Insert the value of ${m_1}$ from equation (3) into the above equation.

$\boxed{F=x\lambda Lg}$ …… (5)

According to the free body diagram, the normal force acting on the rope is counterbalanced by the weight of the section resting on the table.

In equilibrium, the vertical net force is:

$\boxed{N={m_2}g}$

Here, $N$ signifies the normal force from the table, balanced out by the weight of the rope that is on the table.

Insert the value of ${m_2}$ from equation (4) into this equation.

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

The force due to friction acting on the rope is:

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

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

Substituting the value of $N$ from equation (6) provides:

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

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

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

By setting the equations (5) and (8) equal to one another,

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

Rearranging and simplifying this expression we get.

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

Therefore, the fraction of the rope that can extend down from the edge of the table without slipping is $\boxed{\frac{{{\mu _s}}}{{\left( {1 + {\mu _s}} \right)}}}$ .

Learn more:

1. Average translational kinetic energy:

2. Acceleration of body by considering friction:

3. Conservation of energy

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.