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The given problem states that

A rope is fastened at both ends to a ceiling, one end is allowed to fall freely. Find the maximum force that the fastening at the other end must be capable of handling.

enter image description here

Now I was easily able to calculate the velocity of the center of mass of the falling portion of the rope when it has become straight and in line with the fastened part. Now according to me is the point with the maximum force, but I didn't understand how to calculate that force as the only thing I could think of is impulse of that force, but I didn't understand how to apply it as no interval was given.

Can someone help me with this problem?

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  • $\begingroup$ Someone put a reference about this problem. The cited paper says that my thinking in the final paragraph is off. “An important concept...” In other words, the assumption of zero tension on the slack side of the “U” is not an innocuous assumption! In reality it will be much higher tension and fall faster than free fall. $\endgroup$
    – Al Brown
    Aug 20 at 19:08
  • $\begingroup$ You might like this Micheal youtu.be/bcsb1xAv7XA $\endgroup$
    – Al Brown
    Aug 23 at 20:41
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Edit:

While it’s interesting and points out a simple common trick with basic derivatives... apparently my solution makes an unwarranted assumption: that the tension on the free-falling side of the “U” is zero. Such chains actually go down faster than free fall.

Two relevant papers are listed below*. One says:

Calkin and March went on to measure the falling motion of an actual chain.. [it] does indeed fall faster than free fall...[T]ension T at the fixed support of the chain is particularly interesting... the experimental tension.. maximum value of about 25Mg, where M is the total mass of the chain. This maximum tension is of course far in excess of the maximum value of only 2Mg expected when the falling end is falling freely.

This video may be the latest and greatest in physicists trying unofficially to understand chains: https://youtu.be/bcsb1xAv7XA

Even so, the solution with the assumption is interesting:

End of edit


Assume rope length $2L$. Let $s:= \frac{m}{2L}$ be mass per unit length. Let $y$ be how much of the rope has been added to the still, hanging portion on the left. At $y=0$, none of the rope has fallen. Hence $y=L$ at the end of the fall. Let the distance that the end of the rope on the right has fallen be $x$. $$x=2y \text{ , } 0 \le x \le 2L$$

Some portion of rope is falling at speed $v(t)=\frac{dx}{dt}$, and $v(0)=0$.

We will take a small time $dt$ during which a small bit of rope length $dy$ with a mass of $dm$ is added to the left-hand hanging portion, i.e. is decelerated instantaneously (under impulse) from the falling speed $v(t)$ to $0$. The impulse makes our mass $dm$ stop, or change in speed $\Delta v=v$ during a time $dt$. $$F = ma = (dm) \left(\frac{\Delta v}{dt}\right) = (dm) \left(\frac{v}{dt}\right)$$

$$dm = sdy = \tfrac{1}{2}sdx$$

$$F = (\tfrac{1}{2}sdx) \left(\frac{v}{dt}\right) = \tfrac{1}{2}sv \frac{dx}{dt} = \tfrac{1}{2} sv^2$$

We have to add the weight $w(t)$ of the hanging rope which goes from $mg/2$ to $mg$. $$F= \tfrac{1}{2} s v^2 + ({L} + \tfrac{x}{2})sg $$

An important concept is that, at any time during the fall (such that $0<x \le 2L$), the portion of rope that is still falling has been falling unencumbered, because the tension only develops between the tiny portion $dy$ that’s being stopped and the motionless length that’s hanging from the ceiling. So at no point can that tension affect whatever portion of the rope is still falling. Hence: $$ v(t) = gt \text{ , } x= \tfrac{1}{2} gt^2 \le 2L$$

$$F= \tfrac{1}{2} s v^2 + (\tfrac{2L+x}{2})sg $$

$$F_{\text{max}}= 2mg \text{ , at } x=x_{\text{max}}=2L $$

*See C. W. Wong, K. Yasui, Falling chains, American Journal of Physics 74 (2006) 490, and the earlier M. G. Calkin, R. H. March, The dynamics of a falling chain I, American Journal of Physics 57 (1989)

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    $\begingroup$ Wow this is great thanks a lot. I think the mistake I was doing was not adding the mass to the other end. $\endgroup$ Aug 20 at 12:59
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    $\begingroup$ Youre intuition was right though. And thanks $\endgroup$
    – Al Brown
    Aug 20 at 13:02
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    $\begingroup$ Although I didn't understand how did u write dm=1/2sdx? $\endgroup$ Aug 20 at 13:07
  • $\begingroup$ The x=2y is from the free side falling twice as far as whatever is added to the hanging rope. So the mass per unit length s, over the amount of length added to the hanging part, dy. Equals s dy. Or 1/2 s dx $\endgroup$
    – Al Brown
    Aug 20 at 13:10
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    $\begingroup$ @mikestone you might fond this interesting too youtu.be/bcsb1xAv7XA $\endgroup$
    – Al Brown
    Aug 23 at 20:40

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