Question

A man slides down a light rope whose breaking strength is $\eta $ times the weight of man $\left( {\eta < 1} \right)$ . The maximum acceleration of the man so that the rope just breaks is:
(A) $g\left( {1 - \eta } \right)$
(B) $g\left( {1 + \eta } \right)$
(C) $g\eta $
(D) $\dfrac{g}{\eta }$

Answer 1

Hint By the breaking strength, the question means to say that the maximum amount of tension that the rope can withstand is $\eta $ time the weight of the man. Hence, all you need to do is to find the equation of tension in the rope and substitute the value of weight and maximum tension to find the value of maximum acceleration of the man so that the rope just breaks.

Complete step by step answer
We will proceed with the solution exactly as explained in the hint section of the solution to the question. We will first try to find the equation of forces at a point in the string or rope and we will equate the downward and upwards forces to find the equation between tension in the rope, weight of the man and the force due to acceleration of the man.
Since there are only two acting forces on the man in the given question, tension and the weight of the man, the equation of forces will be:
$W - T = {F_{net}}$
Where, $W$ is the weight of the man,
$T$ is the tension in the rope and
${F_{net}}$ is the net resultant force
If we assume the mass of the man as $m{\kern 1pt} kg$ , the weight of the man becomes:
$W = mg$
As told in the question itself, breaking strength is $\eta $ times the weight of the man, we can write:
$
  {T_{\max }} = \eta W \\
   \Rightarrow {T_{\max }} = \eta mg \\
 $
The resultant force can be given as:
${F_{net}} = ma$
Substituting the values, we can write:
$mg - \eta mg = ma$
Dividing both sides by $m$ , we get:
$
  g - \eta g = a \\
   \Rightarrow a = g\left( {1 - \eta } \right) \\
 $
As we can see, the value of acceleration so that the rope just breaks is $a = g\left( {1 - \eta } \right)$ which matches with the value given in option (A)

hence, option (A) is the correct answer.

Note The main stage where many students make a mistake is while writing down the equation of forces as they consider tension or the gravitational force as the resultant force while the force due to acceleration as an internal force of the system. Always remember that the overall acceleration of a system must always relate to the net resultant force on the system.