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? express your answer in terms of μₛ.

Explanation:

Step1: Define Variables

Let the total length of the rope be \( L \) and the mass per unit length be \( \lambda \). Let the length of the rope hanging over the edge be \( x \), so the length on the table is \( L - x \). The mass of the hanging part is \( m_h=\lambda x \) and the mass on the table is \( m_t = \lambda(L - x) \).

Step2: Analyze Forces

The normal force on the rope on the table is \( N = m_tg=\lambda(L - x)g \). The maximum static friction force is \( f_s=\mu_s N=\mu_s\lambda(L - x)g \). The weight of the hanging part is \( F = m_hg=\lambda xg \).

Step3: Equilibrium Condition

For the rope not to slide, the friction force must balance the weight of the hanging part: \( f_s = F \). So, \( \mu_s\lambda(L - x)g=\lambda xg \).

Step4: Solve for \( \frac{x}{L} \)

Cancel out \( \lambda g \) from both sides: \( \mu_s(L - x)=x \). Expand: \( \mu_s L-\mu_s x = x \). Bring terms with \( x \) to one side: \( \mu_s L=x + \mu_s x=x(1 + \mu_s) \). Then, \( \frac{x}{L}=\frac{\mu_s}{1 + \mu_s} \).

Answer:

\(\frac{\mu_s}{1 + \mu_s}\)