Question

A fireman wants to slide down a rope. The breaking load for the rope is ${\left( {\dfrac{3}{4}} \right)^{{\text{th}}}}$ of the weight of the man. With what minimum acceleration should the fireman slide down? (Acceleration due to gravity is g)
${\text{A}}{\text{.}}$ Zero
$
  {\text{B}}{\text{. }}\dfrac{{\text{g}}}{4} \\
  {\text{C}}{\text{. }}\dfrac{{{\text{3g}}}}{4} \\
  {\text{D}}{\text{. }}\dfrac{{\text{g}}}{2} \\
 $

Answer 1

Hint: Here, we will proceed by finding the frictional force acting on the firearm. Then, we will draw the free body diagram of the firearm. Finally, we will apply Newton's second law of motion.

Step By Step Answer:

Formulas Used: W = mg, Force of reaction = -(Force of action) and Net force acting in the direction of the acceleration = (Mass of the body)(Acceleration of the body).

Let the mass of the fireman be m

As we know that weight of any object of mass m is given by W = mg where g is the acceleration due to gravity

Weight of the fireman W = mg

Given, Breaking load for the rope = $\dfrac{3}{4} \times $(Weight of the fireman) = $\dfrac{3}{4} \times $W = $\dfrac{{3{\text{mg}}}}{4}$

Since, the breaking load for any rope refers to the maximum tension which will be there in that rope

i.e., Maximum tension in the rope ${{\text{T}}_{{\text{max}}}} = \dfrac{{3{\text{mg}}}}{4}$

In order for the fireman to slide down a rope, the force of friction acting on the fireman due to the rope, acts in the upward direction.

According to Newton’s third law of motion, to every action there occurs an equal and opposite reaction.

i.e., Force of reaction = -(Force of action)
The negative sign in the above equation represents that the directions of action and reaction forces act in the opposite direction.

Due to the above law, an equal amount of the force of friction acts in the downward direction in the rope.

According to equilibrium in the rope,

Force of friction = Maximum tension acting in the rope

$ \Rightarrow $ Force of friction = ${{\text{T}}_{{\text{max}}}} = \dfrac{{3{\text{mg}}}}{4}$

Let the minimum acceleration with which the fireman which slide down be a

The free body diagram of the man is shown in the figure
According to the Newton’s second law of motion,

Net force acting in the direction of the acceleration = (Mass of the body)(Acceleration of the body)

 $ \Rightarrow $ (Weight of the fireman) – (Force of friction) = (Mass of the fireman)(Acceleration of the fireman)
$ \Rightarrow $ W - $\dfrac{{3{\text{mg}}}}{4}$ = ma
$ \Rightarrow $ mg - $\dfrac{{3{\text{mg}}}}{4}$ = ma
$ \Rightarrow $ $\dfrac{{4{\text{mg}} - 3{\text{mg}}}}{4}$ = ma
$ \Rightarrow $ ma = $\dfrac{{{\text{mg}}}}{4}$
$ \Rightarrow $ a = $\dfrac{{\text{g}}}{4}$

The minimum acceleration with which the fireman slides down is $\dfrac{{\text{g}}}{4}$.

Hence, option B is correct.

Note: In this particular problem, when the fireman is sliding downwards the force of friction acting on the fireman will be acting in the upward direction because the force of friction always acts in a direction opposite to that of the motion. Since, force of friction is a type of force which opposes motion of a body.