Question

A uniform chain of mass m and length L is held vertically in such a way that its lower end just touches the horizontal floor. The chain is released from rest in this position. Any portion that strikes the floor comes to rest. Assuming the chain does not form a heap on the floor. Calculate the force exerted by it on the floor when the whole chain is on the floor.
A. \[\dfrac{{3m}}{ L \\ \\ }\]
B. $\dfrac{{3mgx}}{L}$
C. $\dfrac{{gx}}{L}$
D. $\dfrac{{3gx}}{L}$

Answer 1

Hint: Given that the chain does not form a heap which means that the subsequent parts of chain fall individually on the ground. The whole chain lies on the floor. So here the force exerted by the chain will be equal to the sum of weight of the lying part of chain and the force exerted by the part which just came to rest by hitting the ground(due to momentum transfer).

Complete step by step answer:
Let us consider a small segment of chain at height x of length dx as shown in the figure:

The mass per unit length of the chain is $\dfrac{m}{L}$
So, mass of the small segment dx $dm = \dfrac{{m.dx}}{L}$
The velocity with which the segment falls on the ground is $v = \sqrt {2gx} $(By Newton's 3rd law of motion)
So, the momentum transferred to the ground $dp = dm.v = \dfrac{{m.dx.\sqrt {2gx} }}{L}$
And $v = \dfrac{{dx}}{{dt}} = \sqrt {2gx} \Rightarrow dt = \dfrac{{dx}}{{\sqrt {2gx} }}$
Therefore, the force exerted to the ground ${F_1} = \dfrac{{dp}}{{dt}} = \dfrac{{m.dx.\sqrt {2gx.} \sqrt {2gx.} }}{{L.dx}} \Rightarrow {F_1} = \dfrac{{2mgx}}{L}$
And the force exerted by the weight of the already lying part on the ground ${F_2} = (\dfrac{m}{L})x.g$
Hence, the total force exerted is $F = {F_1} + {F_2} = \dfrac{{2mgx}}{L} + \dfrac{{mgx}}{L} \Rightarrow F = \dfrac{{3mgx}}{L}$

So, the correct answer is “Option B”.

Note:
Here if we carefully examine the options given in the question only the correct option B has the dimensions of force rest all options (A-mass per unit length, C-acceleration, D-acceleration) represent different physical quantities. Hence, we can answer this type of question without solving the whole question.