Question

With what minimum acceleration can a fireman slide down a rope while breaking strength of the rope is \( \frac{2}{3} \) of the weight?

• A. \( \frac{2}{3}g \)
• B. \( g \)
• C. \( \frac{1}{3}g \)
• D. zero

Hint:

When dealing with tension and breaking strength problems, always set up Newton's second law and solve inequalities to find limits on acceleration or force.

Answer 1

The Correct Option is C

Let the mass of the fireman be \( m \), and let the tension in the rope be \( T \). As the fireman slides down with acceleration \( a \), we apply Newton’s second law: \[ mg - T = ma \quad \Rightarrow \quad T = m(g - a) \] Given that the maximum breaking strength of the rope is \( \frac{2}{3} \) of the fireman's weight: \[ T \leq \frac{2}{3}mg \] Substituting for \( T \): \[ m(g - a) \leq \frac{2}{3}mg \Rightarrow g - a \leq \frac{2}{3}g \Rightarrow a \geq \frac{1}{3}g \] So, the minimum acceleration the fireman must slide down with to prevent the rope from breaking is \( \frac{1}{3}g \).