With what minimum acceleration can a fireman slide down a rope while the breaking strength of the rope is $\fr

Question

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

Accepted Answer

1. **Analyze Forces:** The forces acting on the fireman descending the rope are:
   - Weight ($W = mg$) acting vertically downwards.
   - Tension ($T$) acting vertically upwards (provided by the rope).
2. **Equation of Motion:** According to Newton's Second Law, the net force downward produces the acceleration ($a$):
   $$ mg - T = ma $$
   $$ T = m(g - a) $$
3. **Constraint:** The rope breaks if the tension exceeds its breaking strength. The maximum tension the rope can withstand is given as $\frac{2}{3}$ of the weight.
   $$ T_{max} = \frac{2}{3}mg $$
4. **Minimum Acceleration:** To slide down safely, the tension must be less than or equal to $T_{max}$. A lower acceleration requires a higher tension (holding tighter). Therefore, the *minimum* acceleration corresponds to the *maximum* possible tension (just on the verge of breaking).
   $$ m(g - a_{min}) = T_{max} $$
   $$ m(g - a_{min}) = \frac{2}{3}mg $$
5. **Calculation:** Cancel mass $m$ and solve for $a_{min}$:
   $$ g - a_{min} = \frac{2}{3}g $$
   $$ a_{min} = g - \frac{2}{3}g $$
   $$ a_{min} = \frac{1}{3}g $$

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