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 Monkey Bidness Goto page Previous  1, 2, 3
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Page 3
Guest

 Posted: Tue Apr 11, 2000 4:13 pm    Post subject: 81 HAHAHAH and i even got thanked for deleting my post that really took us to page 3.
HyToFry
Drama queen

 Posted: Tue Apr 11, 2000 4:15 pm    Post subject: 82 I thought about that to scirroco, however the riddle says, "He ADJUSTED his aim and fired", if what you were thinking was true, it would have said "How did the hunter trick the monkey so that he could shoot him?" Or something of that nature. Oh, although i am Page 3 in this case, I have NEVER gone by any alias, (except page 3 this time)
Ghost Post
Icarian Member

 Posted: Tue Apr 11, 2000 11:07 pm    Post subject: 83 Why will two equal weights on either side of a pulley be in equilibrium regardless of height ? I thought it would be fairly simple physics. Consider the forces on the monkey : he has his own weight pulling him down, since he is 1 kg in weight and gravity is 9.8 m/s^2 then he has 9.8 Newtons of force pulling him downwards. He is also holding onto the rope which is proving pull upwards, in this case 9.8 Newtons of force provided by the bananas on the other side of the pulley. The pulley just changes the direction of the force. Likewise the bananas are feeling the same amount of force, 9.8 Newtons down from gravity and 9.8 Newtons up from the monkey. Height on the rope is completely irrelevant. Unlike a seesaw or balancing scales the position of the weights on a pulley doesn't matter. There is no leverage to be gained from being higher or lower on the rope. Gravity isn't a factor over a distance of only 5 metres. You don't feel lighter after climbing a set of stairs do you ? The Earth is so huge that the effect of just five metres has a negligable effect of gravity. Whatever tiny change occurs will be handled by the friction in the pulley.
worm
Guest

 Posted: Wed Apr 12, 2000 3:34 am    Post subject: 84 heehee, now I get it. whippet and aarondalf are both so adamant about their positions because they are both right. when aarondalf says they DON'T come to equilibrium, he/she's saying they don't go from different heights to the same height (right?). this thinking equates equilibrium with equal heights. I brought up this idea of equilibrium b/c of a feeling in my gut along the lines of the scales that whippet mentioned...it was probably just gas. when whippet says they DO come to equilibrium, he/she's saying they stay at their respective heights (right?). that is, they are in equilibrium even though they're at different heights. both of these arguments say the same thing. they just have two notions of "coming to equilibrium". i think monkey III should now be shot with a dart and put to bed (figuratively speaking, of course).
Aarondalf
the original GL stud

 Posted: Wed Apr 12, 2000 8:57 am    Post subject: 85 Yes, we are talking of two different things. I meant equilibrium imn the sence of equal heights, not forces. So this was just a misinterpretation on my part. I have spoke to my physics teacher on the matter of the monkey and nannas and he says the monkeys will stay put, but the nannas go up. Because he moves up at x meters per sec, and the rope goes down at x meters per sec and therefore he cant move no matter how great the force. The bannanas on the other hand are attached to the rope which is moving at x meters a sec so they go up. So all the energy that the monkey uses will go into increasing the kinetic energy of the bannanas. Imagine if he were on the ground and trying to "hoist" the bannanas upwards, and they started at say 5 meters above ground. The maximum weight he could lift without being flung upwards him self is 1kg (or whatever he weighs). So when he pulls, all the energy goes into the rope and therefore bannanas. This is as good as i can do to explain it, other that to use all those scooby cartoon. PS, I am aware that those cartoons are not real and cant be used as examples. PPS, If someone wants to challenge this, please do so without saying "by symmetry this happens....". Please explain it for the dim wits like myself.
worm
Guest

Posted: Wed Apr 12, 2000 12:58 pm    Post subject: 86

aarondalf, you said

 Quote: Because he moves up at x meters per sec, and the rope goes down at x meters per sec and therefore he cant move no matter how great the force.

humor me here. i understand that the monkey is moving up (with respect to the rope) at the same speed that it's going down. within that system there is no net upward velocity for the monkey.

what if that entire system (monkey and rope) is being pulled up by the bananas on the other side?

Ghost Post
Icarian Member

 Posted: Wed Apr 12, 2000 3:22 pm    Post subject: 87 Haha Noggin is back in the fray. Just ignore all previous, wrong, arguments. Aarondalf: If the monkey and banans where in space, attached to a straight piece of rope for simplicity and the monkey started pulling on the rope, whould they not 'both' move towards each other? So what's the difference on earth?
HyToFry
Drama queen

 Posted: Wed Apr 12, 2000 6:55 pm    Post subject: 88 Good point noggin, however this doesn't prove Aarondolf wrong, you are correct that they will move towards each other at the speed that the monkey is pulling on the rope, but if the bannanas are ascending towards the pulley at the same speed that the monkey is pulling on the rope, they would still be moveing towards each other. Aarondolf is wrong, but in order for him to believe the space theory, he would have to agree in the first place.. see my point? worm has the best explanation, if the monkey was to "lift" the nannas up at speed x, would the monkey fall at 2x and the nannas rise at 2x, NO, the monkey would fall at speed x and the nannas would rise at speed x, until the monkey stopped lifting, the same would be true, if the monkey were to reach over, and push down on the nannas, monkey rises at the same speed that the nannas decend. I like to look at it like this, if the monkey was to somehow let go of the rope (and assumeing hes not tied), the monkeys and nannas would fall at the same speed. Same is true if there is a force applied by one of them, half of the force goes to the monkey, and half to the nannas, causeing them both to rise towards the pully. if the monkey and banannas, were at the pulley, and the monkey started to climb down the rope, (assumeing that their is a length of rope under the monkey) you (Aarondalf) are stateing that the monkey will remain up by the pulley, and the nannas will descend.. if this doesn't make sense to you, than you now know your wrong. Hope this makes scense
Ghost Post
Icarian Member

 Posted: Wed Apr 12, 2000 10:09 pm    Post subject: 89 That's correct Worm, that's exactly what I meant. It's probably because I'm an engineer by training and when looking at the forces on a static object we rely on the fact that the object is in equilibrium ie. it isn't moving or deforming in any way.
Aarondalf
the original GL stud

 Posted: Thu Apr 13, 2000 11:35 am    Post subject: 90 Sorry, i worked through it again from the beggining again and i realise they will go up at the same rate. I realised that gravity is constant (practically) on both sides and therefore if you were to use the a treadmill and have somebody walk along it, the same result would occur. Just assume you have a treadmill 10m long, on one end a 100kg man, and a 100kg block on the other. It is really easy to work the question if you look at it this way. If he pushes with force 100n, then he moves at 1ms^-2, and the tread mill goes at .5ms^-2 because there is 200kgs on it. And the mans 1ms-0.5ms =0.5 ms. Which is the same as the block which is moving at the same speed as the treadmill. So for the record I was wrong, you were right. But this explanation might help the mentally challenged like myself.
Aarondalf
the original GL stud

 Posted: Thu Apr 13, 2000 11:35 am    Post subject: 91 Sorry, i worked through it again from the beggining again and i realise they will go up at the same rate. I realised that gravity is constant (practically) on both sides and therefore if you were to use the a treadmill and have somebody walk along it, the same result would occur. Just assume you have a treadmill 10m long, on one end a 100kg man, and a 100kg block on the other. It is really easy to work the question if you look at it this way. If he pushes with force 100n, then he moves at 1ms^-2, and the tread mill goes at .5ms^-2 because there is 200kgs on it. And the mans 1ms-0.5ms =0.5 ms. Which is the same as the block which is moving at the same speed as the treadmill. So for the record I was wrong, you were right. But this explanation might help the mentally challenged like myself.
Ghost Post
Icarian Member

 Posted: Thu Apr 13, 2000 5:42 pm    Post subject: 92 I assume you guys haven't thought about the mass of the rope?
HyToFry
Drama queen

 Posted: Thu Apr 13, 2000 5:49 pm    Post subject: 93 There's one in every crowd isn't there
worm
Guest

 Posted: Thu Apr 13, 2000 5:58 pm    Post subject: 94 yup, somebody has to be that guy (or girl, as the case may be)
Aarondalf
the original GL stud

 Posted: Thu Apr 13, 2000 9:46 pm    Post subject: 95 Mass of rope and friction of pulley are variables that are not given, so we just assume they are 0 to make the puzzle actually solveable.
worm
Guest

 Posted: Thu Apr 13, 2000 9:54 pm    Post subject: 96 that's not what htf and i were moaning over. this is: we have considered the mass of the rope and discussed it at length. check out page 2 of this thread, rubberpaw.
Ghost Post
Icarian Member

 Posted: Fri Apr 14, 2000 12:26 am    Post subject: 97 I read your discussions on page 2(thanks for pointing them out, even though I did read them) I don't have a conclusive answer, but I think if the monkey and the bannanas can be at equilibrium at any height(ie the rope and pully cancel out) then they will rise together. If the downward force of the extra rope on the monkey's side is more than the friction of the pulley, then the the upward velocity of the monkey caused by the reaction of him pushing down on the rope(ie pull) would either cause him to go up at a slower rate or stay put. Anyway, I think it is certainly an interesting physics problem Ahh. It is these times that I wish I had all of the units laid out. Of course, that would make it a word problem, and anyone with a decent brain would be able to calculate it all.
worm
Guest

Posted: Fri Apr 14, 2000 10:38 pm    Post subject: 98

rubberpaw i agree w/ what you said:

 Quote: If the downward force of the extra rope on the monkey's side is more than the friction of the pulley, then the the upward velocity of the monkey caused by the reaction of him pushing down on the rope(ie pull) would either cause him to go up at a slower rate or stay put.

i believe that these conditions would slow monkey's upward velocity with a corresponding increase in banana's velocity. so they really wouldn't rise together. this may be implied in the earlier discussion of a rope, but it's worth stating emphatically. of course when the monkey stops "climbing", it'll go down.

anyone else ready for a new problem?
Aarondalf
the original GL stud

 Posted: Sat Apr 15, 2000 10:26 am    Post subject: 99 Yeah, its been years since they put up a new puzzle. A monkeyless puzzle, or at least less monkey.
Wonko the Sane
Daedalian Member

 Posted: Sun Apr 23, 2000 2:06 am    Post subject: 100 Okay...so did we end up agreeing at a solution? Any some respond so we can have 100 replies to this thread ;-) ~Wonko
everyone
Guest

 Posted: Sun Apr 23, 2000 5:01 pm    Post subject: 101 yeah i agree with the solution
ori
Guest

 Posted: Wed Apr 26, 2000 10:12 am    Post subject: 102 the puzzle did not mention the hight of the room. If it is higher then 5, both bananas and monkey will be at the same hight, and at the same place - under the pulley, the shotest distance from the floor. the monkey will not climb at all for the bananas will be with him! If it is lower than 5, then both babanas and monkey are on the floor. then the monkey won't climb either for he can just walk over to the bananas
HyToFry
Drama queen

 Posted: Wed Apr 26, 2000 4:55 pm    Post subject: 103 Minotaur, we REALLY REALLY NEED a new puzzle man, come on
Ghost Post
Icarian Member

 Posted: Fri Apr 28, 2000 6:52 am    Post subject: 104 I think we have to assume that the friction of the pulley and the mass of the rope are negligible, otherwise it would all depend on what those values happened to be. Now suppose that the money weighs 0.5 kg. In this case, when the monkey clims, he goes up the rope. Right? Now, let's suppose that the monkey weighs 2 kg. Here when he climbs, he only pulls the crate up. Right? (along with the obvious imballance causing the monkey to rise/fall in relation to the crate) But wait a minute! If at one point he goes up, and at another he goes down, when will he stay still? When they have the same mass. Therefore, the monkey will stay stationary in relation to the ground, and the crate will move up in relation to the ground. -Wow, 2:52 am no not really; actually 12:52 that had me a little worried for a second [This message has been edited by Mathieu (edited 04-28-2000).]
Ghost Post
Icarian Member

 Posted: Fri Apr 28, 2000 11:05 am    Post subject: 105 Mathieu: Try a slight twist on your argument. If the monkey weighs 0.5 kg, he moves up in relation to the bananas. If the monkey weighs 2 kg, he moves down in relation to the bananas. If at one weight he oves up in relation to the bananas, and at another weight he moves down in relation to the bananas, there must be some weight in between where he remains stationary in relation to the bananas. What weight is that?
Wonko the Sane
Daedalian Member

 Posted: Fri Apr 28, 2000 4:18 pm    Post subject: 106 I think I finally figured out why this puzzle seems intuitively wrong Let's look at what happens when the monkey climbs. He pulls down on the rope and his body moves up. However, the monkey weights 1kg. So does the crate. Pulling down with a force of 9.8N will render the monkey's body and the crate essentially weightless. That's basically the situation when the monkey and the crate are just sitting there. Except that the monkey's hands are still holding the entire weight of the crate. Therefore the monkey's hands will move down, pulling the crate up, and leaving the monkey's body in the same place. Then it grabs higher on the rope with it's legs or whatever and that brings the monkey back to where it started relative to the room, but lower relative to the bananas. Given then, the bananas will hit the pulley eventually and the monkey could climb down the rope to the ground, right? Wrong, as soon as the monkey starts climbing down, the bananas and the monkey weigh the same relative to each other, but this time the monkey is not working against gravity to keep the bananas up, so they'll move down until they center out again. This sound right? It sounds kind of weird, but it seems to be okay. Maybe I'm wrong.
CrystyB
Misunderstood Guy

Posted: Sat Apr 29, 2000 4:48 pm    Post subject: 107

 Quote: the monkey's hands are still holding the entire weight of the crate. Therefore the monkey's hands will move down, pulling the crate up

Actually, as i saw on some other place, the monkey's hands are holding 2kg-force, both his and the banannas' weight.

and actually i can't understand why did you say "THEREFORE something happens"...
Aarondalf
the original GL stud

 Posted: Sat Apr 29, 2000 8:32 pm    Post subject: 108 Wonko, I am afraid you are wrong, as the monkey and the bannanas WILL NOT CENTRE if they are at different heights. It took me a while to figure out just how wrong I was, but I finnaly came up with a satisfying solution to the question. Here goes: Assuming gravity is constant in regards to the small distance which they move, then we can toss it out the window, because it will cause the same force on either side. Now, when the monkey pulls with a force, lets say 1N. This causes the rope to accelerate at 0.5m/s^2, because he is pulling 2kgs of mass. Now the rope is tied to the bannanas, so whatever speed it goes at, so do the bannanas. Now to pull himself up, he only has to lift 1kg, so he accelerates at 1m/s^2 RELATIVE TO THE ROPE, but we know the rope moves at 0.5m/s^2, so his acceleration relative to the ground is 1-0.5= 0.5m/s^2 which is the same as the bannanas, therefore he moves at THE SAME SPEED AS THE BANNANAS. You can check this for different forces if you wish, but it is always the same. Whatever force the monkey puts in will give him equal acceleration to an object of equal mass on the other side.
Ghost Post
Icarian Member

 Posted: Mon May 01, 2000 12:29 pm    Post subject: 109 Hmm. I think we're missing something here. I read a while back that in space, the both of them would move equally toward each other if there was no pulley. That is an example of when there is no force pushing them apart(or pulling them apart). Since (#1)we have gravity exerting in 1 direction pulling both down and (#2) we have a pulley redirecting the line between them, then it is as if they were holding a straight line in between them with two equal forces pulling them away from each other. The pulley confuses us into thinking we're dealing with a two dimensional type thing, but we are only dealing with one-dimensional forces. The weight difference might be a little and the height difference might be enough to have the bannana reach there first, but it will be only just barely. Now that I have thought of it, the weight of the rope only really matters when the monkey climbs slowly. Either way, I don't think that the monkey can possibly reach there first. At best, they will reach the top simultaneously, but the bannana might reach there first. However, the question doesn't ask "who will reach there first," it asks "what will happen" I think I can say this for sure: !!The monkey will reach the top!! ;-)
Wonko the Sane
Daedalian Member

 Posted: Thu May 04, 2000 3:11 pm    Post subject: 110 I think there was a bit of minsinterpretation. I say that the monkey will stay the same relative to the floor. The bananas will rise and he'll just sort of float. I don't know if that's right, but it seems intuitively correct.
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