correct answer monkey business 3?

[Ghost Post]

Wow, does the monkey really always go down? I don't think so.
Depending on the situation, it might just not ascend as fast as the bananas.

In the ideal situation, the monkey and the bananas go up at the same speed. The monkey has a speed relative to the rope, which is twice the speed of the rope.

In the not so ideal situation (the rope starts falling because of its own weight), the monkey can still ascend, as long as his relative climbing speed is greater than the falling speed of the rope.
This means that there are three situations:

1. As described by the Minotaur. The rope has so much mass, that it falls faster than the monkey can climb.

2. The monkey stays put! His relative climbing speed is exactly the speed of the rope (=the speed of the bananas).

3. The monkey goes up. He is still faster than the rope, but loses on the bananas.

These three solutions have all in common, that the bananas will reach the top first. Then the monkey can put all his climbing effort into his getting to those bananas. Ah, how great is the enjoyment after hard work.

Erik the Dutch

PS.
And to discuss even further. If the rope is really so heavy that the monkey cannot climb fast enough (relative to the rope I mean), it will mean that the rope will have quite some inertia. And that means that the little animal will start going up anyway, before the rope has gathered enough speed to tear him down again.

[Wonko the Sane]

His relative climbing speed can't get that fast because he climbs by pulling the rope down. The rope will fall as he pulls it down, so he'll always descend. It believe this was explained fairly clearing in an early post on the main thread.

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It's not the size of the spork, it's whether or not it's made of #7 recyclable plastic.

[Ghost Post]

Wonko, I have looked through the threads and could not find a clear answer.

In the ideal situation, the monkey and the bananas go up at the same speed.

My problem is simple:
Consider an almost ideal situation (very light rope).
Why should the monkey suddenly go down, instead of going up a littel bit slower than the bananas?

_________
Does Minotaur like bananas or does he prefer shooting monkeys?

[Ewan]

easy, The rope has wieght and the pully is still friction less.

monkey pulls on rope to accelerate himself up it. since the pully is frictionless this is the same as pulling the bannans up.

the monkey gose up, the bannans go up, a little (quantum) bit of rope moves to the otherside of the pully.

the little bit of rope now means that the moneky side is heavier, and hence will start to fall at 9.8 m/s^2 (remeber pully is frictionless)

the monkey realising he is about to fall to his death (lets say there are spikes under him) pulls harder and harder however the only upward force that he generates is that caused by accelerating the rope downwards (like the gasses from the back of a rocket)

so a VERY strong (or light) monkey could conciveably use this force to launch himself into orbit

Q.2 what happens if the money climbs DOWN the rope? hehe

(PS lets also say that the pully has spikes on)

[Ewan]

oops! I forgot that the bannans will crash into the pully! just saving the monkey from thge spikes PHEW! (also I should say that the mass of the rope has no bearing on how fast it falls, (as long as the pully is frictionless) it is not possible for the monkey to climb faster than the rope falls as stated in the top post. he can only go up using the rocket motor principle)

[Ewan]

hmm actualy I think im wrong about the 9.8 m/s^2 the force on the monkey side of the rope is Mk (mass of monkey)+Mr(mass of rope)/2 +dMr(mass of rope pulled through) so the total force on the rope is 2dMr so the rope falls at 2dMr/(Mr+2Mk)

of course this increases as more rope falls throught the pully

now for the rocket effect the monkey produces a force by expelling the tensionless rope behind him at a greater acclertaion than that at which the rope is falling but since this rope must first be pulled through the pully and hence speed the systems descent I dont think its possible to produce a force with it.

SO. in conclusion, the monkey can go up but the bannanas will reach the top first?

[Ewan]

OK ive worked it through and get

Asys= (2dMt/(Mr+2Mk)) Amk

where
Asys = the acceleration of the system
dM = the unit mass of the rope
Mr = total mass of rope
Mk = mass of monkey(and also bannans)
Amk = acceleration of monkey relative to rope
t = time

now 2dM/(Mr+2Mk) is less than 1 so when the monkey starts pulling it will go up, but as t increases the systems acceleration will overtake the monkey accleration and hence the money will go down.

So weather the monkey can reach the top depends on the density of the rope, the length of the rope and the mass of the monkey. But NOT on how hard the monkey pulls!

[Artalan]

It's not a maths problem it's a logic problem.
Clearly as the monkey starts to climb he will unbalance the rope and it will run through the pulley until the bananas reach the top and then the monkey will be able to climb the rest of the way. What I take issue with is the statement in the solution that the monkey does not do any significant work whether the rope is weightless or not. It seems obvious that he lifts an increasing weight of rope, because it's fastened to him, until at the top when he supports the weight of half the rope less a length approximating to one quarter the circumference of the pulley sheave. This could be a significant burden for a monkey light enough to balance a bunch of bananas, although since the weight of the bunch is not stated it is evidently considered unimportant.
I mentioned the size of the pulley sheave because this must be of sufficient diameter to place the bananas out of reach at the beginning. Any sensible monkey would just reach across the gap rather than climb - unless of course a transparent barrier is interposed. If this is the case and it reaches right up to the pulley the poor old monkey will still have a job to reach the fruit always supposing he had the sense to climb in the first place, which rather begs the question of the problem.
Whilst writing this reply it occurs to me that if the rope were indeed weightless the monkey should be able to climb to the top, getting further away from the bananas, without them moving. Whereupon he would have to haul them up and if he clung to the pulley whilst so doing, that would entail considerable work and he would need the well-earned meal to replenish his energies!

[ewan]

Artalan: hmmmmmmmmmmm no i think its a maths or rather physics, problem, but lets do this a bit at a time.


1, is the monkey attached to the rope? I assumed he was mearly holding onto it. even if this is true he still dose work climbing up the rope in both the heavy and light cases

2, if the pully is wide then it makes the problem more difficult as the monkey will have to climb over the top of the pully to get to the bannans, lets ignore this and say the pully is infinitly small and the monkey is stupid

3, why do you say that if the rope is wieghtless the monkey can climb while the bannanas remain stationary?

[Andy]

It's obviously[!] a logic, physics, and math problem. Since we aren't given enough actual numbers, we can't finish the math - but we can set up equations (if we're ambitious enough) or at least describe the components and their resulting possibilities qualitatively.

The forces initially are gravity on the monkey, gravity on the bananas, gravity on the rope; all resisted appropriately so nothing moves.

Once the monkey begins 'climbing' the rope, the monkey is exerting an additional downward force on his end of the rope and a corresponding upward force on himself. This combination causes the monkey to rise (relative to the rope), at a rate determined by the weight and exertion of the monkey. The initial application of force will cause the monkey to accelerate against gravity (move upward). Since the force is also applied to the rope and via the rope to the bananas, the rope will begin to move through the pulley and the bananas will begin to rise. Because of the greater mass of the rope-and-bananas as compared to the monkey, the rope and bananas will be accelerated less than the monkey, and the monkey will begin to rise relative to the ground. If the pulley is not frictionless, the friction at the pulley will also slow (and perhaps stop) the acceleration of the rope-and-bananas but will have no direct effect on the monkey.

Assuming that the force applied to the rope-and-bananas overcomes friction in the pulley, the rope will begin to move downward (on the monkey's side). This will increase the mass of rope on the monkey's side, which will increase the acceleration of the rope-and-bananas without any direct effect on the monkey. As the rope-and-bananas continue to experience gravitational acceleration in addition to the acceleration caused by the monkey's exertions, the monkey's acceleration relative to the ground will decrease.

Here are some of the obvious possibilities:
1) The pulley's friction is enough to prevent any movement of the rope (the rope is effectively clamped). The monkey will reach the pulley, climb through or around it, and continue down the other side of the rope until he reaches the bananas.
2) The rope is massive compared to the monkey and bananas, and the monkey climbs quickly. The monkey will reach the pulley ahead of the bananas. The monkey may wait for the bananas to reach the pulley also. Question: is there any combination of values that would allow the monkey to grab the pulley housing and release the rope, and the weight of the bananas then be sufficient to overcome the inertia and now-unbalanced weight of the rope so that the bananas would stop rising and begin to fall?
3) The rope is extremely light, so that its inertia and gravitational acceleration once it becomes unbalanced are both negligible. The monkey and the bananas will reach the pulley at nearly the same time. Either may be first, depending on specifics.
4) The rope is moderately heavy, and the monkey climbs slowly. The monkey's ascent will stop as the rope becomes more and more unbalanced; the monkey will begin to descend at a faster and faster rate until the bananas become wedged in the pulley. The monkey will then continue to climb until he reaches the pulley (and the bananas). Depending on specifics, the monkey may descend below his original altitude.

[Ghost Post]

Andy, you are a man of my heart.

HyToFry, isn't it time to adjust the "official" answer?

Cheerio!

[Joe Schmoe]

This scenario assumes the rope has mass. If the rope is evenly draped over the pully so that it's weight is equal on both sides, and the monkey (lets assume he is hanging on to the rope)is equal to the weight of the bananas. Does it matter where on the rope the monkey is, bottom or middle or near the top? Is not the total weight on one side of the pulley still equal to the other. If Mr monkey maintains an extremely steady grip, he should be able to climb all the way up to the top of the rope without the bananas moving at all. Should he momentarily release his grip as he is reaching for a higher grasp, the bananas will weigh more and pull a slight bit of rope through the pulley causing an imbalance favoring the bananas which, once it occurs will cause the banana side of the pulley to weigh more than monkeys side which will cause a chain reaction and the monkey will accelerate towards the pulley even if he stops climbing.
Conversely, if the monkey holds firmly to the rope and drops his body, his momentum should pull a bit of rope through the pully to his side which will cause the bananas to rocket toward the pulley. At that point the monkey would have to climb up the rope to retreive a banana, assuming he even wants one.

[HyToFry]

Erik, what did you mean by "Isn't it time to adjust the official answer"? Why did you direct it to me?

Just curious.

[Ghost Post]

Oops, wrong guess from my side.

HyToFry, you call yourself "moderator" so I guessed you were the - incognito - minotaur.

Cheers

Erik