Prisoner sliding down a rope with minimum velocity

[LCSphysicist]

Homework Statement

All below.

Relevant Equations

THe tension, W = mg and F =ma

I thought in this equations
f is the man's pull\
f + dm*g = T < 600
Where dm is equal to the mass of the string that pull the up part (15-x) after descending x meters.
dm/(15-x) = m/15

And, to the man: W - f = Mx''

I can solve this, and i got ~8m/s

Is this right?

 

[haruspex]

Where dm is equal to the mass of the string that pull the up part (15-x) after descending x meters.

Doesn't all of the rope pull on the top of the rope constantly?

 

[rude man]

You kow his weight pus the weight of the rope would break the hook.
What if he just slid down the rope?
What if he took an intermediate speed, just fast enough to balance his extra weight?
(This is a good example of Einstein's famous elevator gedankenexperiment. What was he trying to show?)

 

[LCSphysicist]

You kow his weight pus the weight of the rope would break the hook.
What if he just slid down the rope?
What if he took an intermediate speed, just fast enough to balance his extra weight?
(This is a good example of Einstein's famous elevator gedankenexperiment. What was he trying to show?)

I am not sure if i got this questions.
"What if he took an intermediate speed, just fast enough to balance his extra weight?"
How can a speed balance a force?

 

[haruspex]

I am not sure if i got this questions.
"What if he took an intermediate speed, just fast enough to balance his extra weight?"
How can a speed balance a force?

Quite.
Apart from how you accounted for the rope's contribution to the tension, I think you were on the right track.

 

[rude man]

I am not sure if i got this questions.
"What if he took an intermediate speed, just fast enough to balance his extra weight?"
How can a speed balance a force?

You're right, it's not the speed that changes (reduces) his weight, it's his acceleration downwards. The problem asked for the speed with which he hits the ground. So compute his minimum downward acceleration, then compute final v from that.

 

[haruspex]

You're right, it's not the speed that changes (reduces) his weight, it's his acceleration downwards. The problem asked for the speed with which he hits the ground. So compute his minimum downward acceleration, then compute final v from that.

I think that is what @LCSphysicist intended to do, but only considered the tension immediately above the prisoner at any point, so omitted the weight of the rope above the prisoner.

 

[Delta2]

I am kind of confused with this problem regarding what is the tension of the rope and the maximum pull force without breaking the hook... Is the maximum pull force 500N?

 

[haruspex]

I am kind of confused with this problem regarding what is the tension of the rope and the maximum pull force without breaking the hook... Is the maximum pull force 500N?

It depends what you are including in "pull" force. The hook is subject to whatever the tension is in the top of the rope.

 

[Delta2]

It depends what you are including in "pull" force. The hook is subject to whatever the tension is in the top of the rope.

I mean the maximum force with which the prisoner pulls the rope, without breaking the hook.

 

[haruspex]

I mean the maximum force with which the prisoner pulls the rope, without breaking the hook.

As I posted, the pull on the hook comes from the rope, and that can be up to 600N, so yes, the max the prisoner can exert on the rope is 500N - unless the prisoner is a bit smart! (According to the question statement, he isn't.)

 

[Delta2]

As I posted, the pull on the hook comes from the rope, and that can be up to 600N, so yes, the max the prisoner can exert on the rope is 500N - unless the prisoner is a bit smart! (According to the question statement, he isn't.)

How smart can he be? What else is the smartest thing to do besides pulling the rope with 500N force?

 

[haruspex]

How smart can he be? What else is the smartest thing to do besides pulling the rope with 500N force?

Clue: he is less smart than he might have been because "the bottom end of the rope hangs..."

 

[Delta2]

Clue: he is less smart than he might have been because "the bottom end of the rope hangs..."

Ok suppose he ties the rope to a big heavy stone (hard to find one of these inside a cell) and throws down the stone. So in this configuration with the rope tied at two ends, I assume we can infer that he can double the maximum pulling force to a maximum of 1000N?

 

[haruspex]

Ok suppose he ties the rope to a big heavy stone (hard to find one of these inside a cell) and throws down the stone. So in this configuration with the rope tied at two ends, I assume we can infer that he can double the maximum pulling force to a maximum of 1000N?

The only way he could get the rope (or rather, hook) to take double the weight would be to tie both ends at the top (second hook, or whatever), but it seems clear the rope is not long enough for that.
Think about this.. how does the prisoner avoid putting the whole of his own weight on the hook? Does that suggest a refinement?

 

[Delta2]

The only way he could get the rope (or rather, hook) to take double the weight would be to tie both ends at the top (second hook, or whatever), but it seems clear the rope is not long enough for that.
Think about this.. how does the prisoner avoid putting the whole of his own weight on the hook? Does that suggest a refinement?

Ok if the rope is long enough and he has many hooks in his disposal he can multiply the maximum pulling force but suppose we have only 1 hook and limited length of rope, what is the smartest thing he can do?

 

[haruspex]

Ok if the rope is long enough and he has many hooks in his disposal he can multiply the maximum pulling force but suppose we have only 1 hook and limited length of rope, what is the smartest thing he can do?

TRy to answer my question. What does the prisoner do to avoid placing his full weight on the rope? How can we use the same trick on part of the rope's weight?

 

[Delta2]

Hmmm, the prisoner just won't hold so tight to the rope if he wants to avoid placing his full weight on the rope. I just can't understand how we can use this "on part of the rope's weight"...

 

[haruspex]

Hmmm, the prisoner just won't hold so tight to the rope if he wants to avoid placing his full weight on the rope. I just can't understand how we can use this "on part of the rope's weight"...

Instead of letting the whole rope out in advance, he could, in principle, pay it out as he goes down. Thus, the part not yet supporting him is accelerating down with the prisoner, so not contributing quite so much to the tension.
However, this is all tongue-in-cheek. It wouldn't actually help.

 

[Delta2]

Instead of letting the whole rope out in advance, he could, in principle, pay it out as he goes down. Thus, the part not yet supporting him is accelerating down with the prisoner, so not contributing quite so much to the tension.
However, this is all tongue-in-cheek. It wouldn't actually help.

If I understand this process correctly the best he could do is to increase the maximum pulling force by 100N. In this process the maximum pulling force would depend on the position of the prisoner above the ground, its an interesting variation of the problem.

 

[Delta2]

I get a non-linear ODE if I try to solve the problem using the process @haruspex describes at post #19. Specifically I get $$200=(70+(15-x)\frac{10}{15})\ddot{x}$$

 

[jbriggs444]

However, this is all tongue-in-cheek. It wouldn't actually help.

Right. If the rope is payed out by the prisoner as he descends, one must account for the impulsive upward acceleration of the infinitesimal rope section making the transition from falling to stationary. To first order, the result is a wash -- the rope starts and ends at rest. There is no delta p to harvest.

However, paying the rope out turns out to be sub-optimal. The prisoner's acceleration profile is shifted to a lower acceleration to start and a larger acceleration later. That means that his time aloft has increased. Gravity has more time to work. The delta p from that source is increased and impact velocity must increase accordingly.

A sufficiently enterprising prisoner can realize that he is subject matter of a physics homework problem and that his physical prowess is unlimited. He unhooks the rope, rolls it into a ball and leaps from the window, unsupported. Just before impact he hurls the rope downward with sufficient velocity that he is brought to a stop.

 

[Delta2]

A sufficiently enterprising prisoner can realize that he is subject matter of a physics homework problem and that his physical prowess is unlimited. He unhooks the rope, rolls it into a ball and leaps from the window, unsupported. Just before impact he hurls the rope downward with sufficient velocity that he is brought to a stop.

This cartoon-like scenario is amusing but theoretically possible. He just would have to shoot the ball of rope at very high speed (my rough calculations are for subsonic speeds of 150-200m/s) which makes it practically infeasible, unless of course he caries with him that good old ACME ball of rope cannon ...

 

[LCSphysicist]

I think that is what @LCSphysicist intended to do, but only considered the tension immediately above the prisoner at any point, so omitted the weight of the rope above the prisoner.

You are right, now i got nine ~9,2m/s something like this, don't remember now, and since the answer is 9m/s, maybe is right

 

[Delta2]

I get 9.2582m/s all this stems from that the maximum pulling force of the prisoner is 500N. Also I took g=10m/s^2 in my calculations.

 

[rude man]

Instead of letting the whole rope out in advance, he could, in principle, pay it out as he goes down. Thus, the part not yet supporting him is accelerating down with the prisoner, so not contributing quite so much to the tension.
However, this is all tongue-in-cheek. It wouldn't actually help.

?
How could he pay out the rope and descend at the same time?

 

[hmmm27]

?
How could he pay out the rope and descend at the same time?

Stick it in his pocket ? And, it would help... but, the question clearly(ish) states the rope is already in place.

 

[jbriggs444]

How could he pay out the rope and descend at the same time?

The idea is that the prisoner wants to be extra clever. @haruspex laid out the strategy in response #19 above.

To go into painful detail, one starts with the rope dangling from the hook as described in the problem statement. The prisoner pulls up the rope. Leaving the top end attached at the hook, he stashes the remainder of the rope in one of his pockets. As he climbs down the rope, more is continuously pulled from his pocket.

The expected benefit is that the portion of the rope in the man's pockets is accelerating downward along with the man. Not all of its weight needs to be supported by the hook. Unfortunately, that supposed benefit disappears on more careful analysis. As @haruspex hinted.

 

[Lnewqban]

You are right, now i got nine ~9,2m/s something like this, don't remember now, and since the answer is 9m/s, maybe is right

Could you show your work?
For some reason, my calculations differ.
Perhaps I am seeing the problem incorrectly:

The hook is holding the dead weight of the rope (10 Kg x g).
The prisoner stands on the window (his initial velocity equals zero) and jumps out.
While he falls, he holds the rope not tightly enough to slowdown much or stop (which would break the hook) but not loose enough to hit the ground too fast.

By doing so, he "selects" a maximum falling constant acceleration (resulting from the balance between the weight and the hands-rope-friction forces), which value is less than g but greater than the maximum permissible by the strength of the hook.

 

[LCSphysicist]

Could you show your work?
For some reason, my calculations differ.
Perhaps I am seeing the problem incorrectly:

The hook is holding the dead weight of the rope (10 Kg x g).
The prisoner stands on the window (his initial velocity equals zero) and jumps out.
While he falls, he holds the rope not tightly enough to slowdown much or stop (which would break the hook) but not loose enough to hit the ground too fast.

By doing so, he "selects" a maximum falling constant acceleration (resulting from the balance between the weight and the hands-rope-friction forces), which value is less than g but greater than the maximum permissible by the strength of the hook.

"By doing so, he "selects" a maximum falling constant acceleration"
I think he sellect a minimum acceleration, so he will come to the ground with the minimum velocity possible, and, to do this, he needs to pull the rope with the maximum force that the rope can support without breaking.
So the force need to be maximum, to the acceleration be minimum. {this force will retard his free falling}