Question

With what minimum acceleration can a fireman slide down a rope whose breaking strength is 1/3 of his weight.

• A. $2g/3$
• B. $g$
• C. $g/3$
• D. $0$

Answer 1

The correct option is D $0$

Sum of the forces acting $=ma.$

These types of questions can also be easily solved by drawing a free body diagram as shown below.

If we add the inertia force (ma) to the $FBD$ then we can solve it using dynamic equilibrium $($all forces balance$).$

Since he is accelerating downward, there is an inertia force $F=ma$ upward $($opposite the direction of acceleration$).$

The only other two forces acting on the fireman are the rope tension upward and his weight $=mg$ downward$.$

equilibrium: $T-mg+ma=0$

setting rope tension $=2/3\left(mg\right)$ gives$:$

$2/3\left(mg\right)-mg+ma=0$

$a=\left(1/3\right)g$ downward

if $a=0,$ the rope tension is equal to his weight. And if he accelerates up the rope$,$ the direction of the inertia force is downward and the rope tension will be higher than his weight

$\begin{array}{c}\sum F_y=m{a}_y\\ T-mg=m\left(-a\right)\\ whichisthesameequationasabove:-T-mg+ma=0\end{array}$

Hence,

option $\left(D\right)$ is correct answer.