A uniform rope of length $l$ lies on a table. If the coefficient of friction is $\mu$, then the maximum length

Question

A uniform rope of length $l$ lies on a table. If the coefficient of friction is $\mu$, then the maximum length $l_1$ of the part of this rope which can overhang from the edge of the table without sliding down is:

Accepted Answer

1. **System Definition:** Let the total length of the rope be $l$ and its linear mass density (mass per unit length) be $\lambda$. Let $l_1$ be the length hanging over the edge and $l_2 = l - l_1$ be the length lying on the table.
2. **Forces:**
   - **Driving Force:** The weight of the hanging portion tries to pull the rope down. $F_{driving} = m_{hanging}g = (λl_1)g$.
   - **Resisting Force:** The static friction acting on the portion of the rope on the table opposes the motion. The normal force $N$ is equal to the weight of the portion on the table: $N = m_{table}g = λ(l - l_1)g$. The maximum static friction is $f_{max} = μN = μλ(l - l_1)g$ [Source 64].
3. **Equilibrium Condition:** For the rope to be on the verge of sliding (maximum overhang), the driving force must equal the maximum static friction:
   $$ (λl_1)g = μλ(l - l_1)g $$
4. **Solving for $l_1$:**
   Cancel $λg$ from both sides:
   $$ l_1 = μ(l - l_1) $$
   $$ l_1 = μl - μl_1 $$
   $$ l_1(1 + μ) = μl $$
   $$ l_1 = rac{μl}{1 + μ} $$

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