Solution to Question 4
Problem Statement Analysis:
We are given that a rope hanging vertically breaks under its own weight if the length exceeds $l_0 = 2 \text{ m}$. This defines the maximum breaking tension the material can withstand.
$$ T_{\text{break}} = \lambda l_0 g $$We need to find the maximum total length $L$ of a piece of this rope that can slide off a frictionless table without breaking.
Step 1: Equation of Motion
Let $y$ be the length of the rope hanging over the edge. The driving force is the weight of the hanging part, $\lambda y g$. The total mass is $\lambda L$.
$$ a = \frac{F_{\text{net}}}{M_{\text{total}}} = \frac{\lambda y g}{\lambda L} = \frac{y}{L}g $$Step 2: Tension at the Edge
The tension is maximum at the corner (edge of the table) because it must pull the horizontal portion of the rope. Consider the horizontal section of length $(L-y)$:
$$ T = m_{\text{horiz}} \cdot a = [\lambda (L-y)] \cdot \left( \frac{y}{L} g \right) $$ $$ T(y) = \frac{\lambda g}{L} (Ly – y^2) $$Step 3: Maximizing Tension
To find the maximum tension experienced during the slide, we differentiate $T$ with respect to $y$:
$$ \frac{dT}{dy} = \frac{\lambda g}{L} (L – 2y) = 0 \implies y = \frac{L}{2} $$Substituting $y = L/2$ back into the tension equation:
$$ T_{\text{max}} = \frac{\lambda g}{L} \left( L \frac{L}{2} – \frac{L^2}{4} \right) = \frac{\lambda g}{L} \left( \frac{L^2}{4} \right) = \frac{\lambda g L}{4} $$Step 4: Comparison with Breaking Limit
The rope must not break, so $T_{\text{max}} \le T_{\text{break}}$.
$$ \frac{\lambda g L}{4} \le \lambda l_0 g $$ $$ \frac{L}{4} \le l_0 $$ $$ L \le 4l_0 $$Given $l_0 = 2 \text{ m}$:
$$ L \le 4(2) = 8 \text{ m} $$Answer: The maximum length the piece can have is $8 \text{ m}$.