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# With what minimum acceleration can a fireman slide down a rope whose breaking strength is two third of his weight?A) \$\dfrac{g}{2}\$B) \$\dfrac{2g}{3}\$C) \$\dfrac{3g}{2}\$D) \$\dfrac{g}{3}\$

Last updated date: 20th Jun 2024
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Hint: Breaking strength of a rope is nothing but a tension in a rope. Now use Newton’s second law of motion to form a relation and find the minimum acceleration.

Formula Used:
\$F\$ = \$ma\$

We have been given a rope whose breaking strength is two third of the foreman who is sliding down on it , therefore we can say that the rope has a tension of two third of a weight of a foreman in upwards direction.
Now let m be the mass of a foreman
Thus we have W = mg (downward direction), T = \$\dfrac{2}{3}mg\$ (upwards direction)
Where W is weight of a foreman and T is a tension
Now applying Newton’s Second Law of Motion on a foreman sliding downwards
F = ma
F is here is T – W (net force)
Substituting this in the above formula we get
\$
\dfrac{2}{3}mg - mg = ma \\
\\
\$
Cancelling out m and solving LHS we get
\$a = - \dfrac{g}{3}\$
Therefore the minimum acceleration is \$\dfrac{g}{3}\$ in the downwards direction.

Since only magnitude is given in the options we can say option D is the correct answer.

Note: It is important to understand the concept of plus and minus signs in force related problems. These signs denote nothing but the direction of application of force. It is also important to know about Newton's Laws to solve these types of problems.