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Rujith de Silva
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Rope and weight puzzle
« on: Apr 18th, 2003, 11:55am » |
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You have a supply of ropes of different elasticities, a weight, and a hook. The weight can be hung from the hook using a single rope, multiple ropes knotted together, a web of ropes complexly knotted, etc. In general, severing one rope in the assemblage will cause the weight to descend somewhat. For example, if it's hung by a single rope, cutting it will cause the weight to fall. If the weight is hung by two equal elastic ropes joined in parallel, then cutting one will cause the other to stretch, causing the weight to descend a little. If the two ropes have no elasticity, then cutting one will have no effect on the weight, as the other rope will support it. Can you construct an assemblage such that cutting one rope in it will cause the weight to ascend, or prove that such an assemblage is impossible?
(N.B.: This is not a trick question, and all the usual simplifications apply: the ropes are thin, flexible, frictionless & light, have zero or positive elasticity; the hook is thin and frictionless; gravity is uniform, weight is positive. This implies the entire assemblage is effectively in a single line under the hook, but adding multiple hooks at the same height does not materially affect the problem.)
Not an original problem, BTW, but I can't remember where I encountered it a long time ago.
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| « Last Edit: Apr 18th, 2003, 1:32pm by Rujith de Silva » |
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cho
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Re: Rope and weight puzzle
« Reply #1 on: Apr 18th, 2003, 12:37pm » |
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Would I be permitted to use 2 hooks at the same height, run a long elastic from one over the other, stretch it very tightly between them and then use a short piece of rope to tie it to the second hook securely enough so that the elastic can't slide through the knot? So when the second rope is cut the stored up energy pulls the weight up?
Or with one hook, wrap one elastic very tightly around and around the hook before securing it with a second rope knotted around it?
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Rujith de Silva
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Re: Rope and weight puzzle
« Reply #2 on: Apr 18th, 2003, 1:28pm » |
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Responding to the ingenious suggestions by cho, I clarified that the ropes are also frictionless. Let me re-iterate: this is not a trick question, dependent on a word-play or some such "out."
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aero_guy
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Re: Rope and weight puzzle
« Reply #3 on: Apr 18th, 2003, 1:58pm » |
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I got a very good if not exactly what you were looking for solution: As the ropes are light but not massless, just hang the weight from an elastic rope and then hang a second rope from the weight. Cut the second rope and there is now less weight hanging and the mass goes up.
Oh, and this would make an excellent medium puzzle.
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| « Last Edit: Apr 18th, 2003, 1:59pm by aero_guy » |
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cho
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Re: Rope and weight puzzle
« Reply #4 on: Apr 18th, 2003, 2:08pm » |
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If the ropes are frictionless, then it's going to be real tough tying them to the weight.
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cho
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Re: Rope and weight puzzle
« Reply #5 on: Apr 18th, 2003, 2:31pm » |
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The same solution would work even if you stipulated that the ropes can only be tied to the hooks or the package a rope tied with both ends to one hook stretched to the max and looped around a short nonelastic rope with both ends tied to the other hook, and the weight rope also looped through it and over the second hook.
If we can't use tricks like this to store some tension that is not being used to carry the weight, then I can't imagine how it could be done.
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cho
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Re: Rope and weight puzzle
« Reply #6 on: Apr 19th, 2003, 3:01am » |
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Yes, I can imagine a solution using one non-elastic rope and two elastics and 3 hooks
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Boody
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Re: Rope and weight puzzle
« Reply #7 on: Apr 20th, 2003, 9:33am » |
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Without using another mass (made with ropes for example), I don't see any solution. 
I assume that we couldn't attach a rope onto the groud.
And that there 's no water:
If there 's some water:
Let us let dip a rope into some water to create a balance with the weight of the wet rope. Then we can cut this rope to disrupt the balance.
Or if we 've got a lot of time. Let's the wet rope dry without cutting it. 
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Jeremiah Smith
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Re: Rope and weight puzzle
« Reply #8 on: Apr 21st, 2003, 9:51am » |
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I've seen this before... I forget the exact answer, and probably couldn't describe it anyway. But it's possible.
Actually, the version I've seen has two weights. Whoops.
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| « Last Edit: Apr 21st, 2003, 9:54am by Jeremiah Smith » |
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aero_guy
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Re: Rope and weight puzzle
« Reply #9 on: Apr 21st, 2003, 4:30pm » |
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cho, if the ropes were frictionless not only would you not be able to tie knots, but if they are natural fiber they would instantly disintegrate. I still like my answer for simplicity.
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SWF
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Re: Rope and weight puzzle
« Reply #10 on: Apr 21st, 2003, 10:32pm » |
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Allowing more than one hook mounted on a rigid supporting structure makes this problem too easy. Also, knots don't always need friction. If you know how to make a chain of rubber bands, it involves a kind of knot (square knot?) that does not require friction.
I have a solution assuming the 'ropes' are rubber bands tied with this kind of knot, and having one hook on the weight and one to hang the whole thing from. I will assume the weight does not have infinite density, so it has some volume, say it is a cube. The starting configuration looks like:
A
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B|______________D
|......C......|
|.............|
|.............|
|.............|
|.............|
|.............|
|.............|
F|_____________|E
A is the hook supporting the whole thing. C is the hook on the weight.
There are three rubberbands:- From A to B, under tension force equal to the weight of the cube.
- From B to C, stretched under tension more than the weight. Knotted to AB at B.
- From C around corners D, E, and F. Under same load as BC. Knotted at B.
Of course it won't hang as shown above (that is only for ASCII art convenience).The cube will hang at 45 degrees with B and E directly below A. If s is the side length of the cube and L is the length of AB when supporting the weight, the center of mass of the cube is initially distance L+s*sqrt(2)/2 below A.
When the rubber band from C to D to E to F to B is cut, the cube hangs with side BCD parallel to the ground and C directly below A. The center of mass is distance L+s/2+x below A, where x is the length of BC when it carries a load equal to the weight. If BC was sufficently stretched at first such that x is less than s*(sqrt(2)-1)/2, then the center of mass will have ascended.
If touching the weight in more than one place is considered cheating, I have found another solution which touches the weight once and the supporting hook once.
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Rujith de Silva
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Re: Rope and weight puzzle
« Reply #11 on: Apr 22nd, 2003, 7:45am » |
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There's been lots of "ingenious" constructions proposed, so let me try
to define this more precisely. The problem is entirely
one-dimensional, being situated in a vertical line under point H
(hook). The assemblage consists of a set of ropes, and a set of
attachments. Each rope has a constant of elasticity k >= 0, whereby
every length on it stretches by a factor kf, where f is the stretching
force applied to that length. Each attachment is of three kinds: (1)
A point on a rope is permanently attached to a point on another rope
(2) A point on a rope is permanently attached to the hook (3) A point
on a rope is permanently attached to the weight. There is at least
one continuous pathway from the hook to the weight via the ropes and
attachments. The weight stretches the ropes, according to the usual
laws of physics; the ropes themselves have no weight. At equilibrium,
the weight is situated at rest some distance below the hook.
Modifying the assemblage by severing one rope, and then permitting the
new assemblage to again reach equilibrium, results in the weight being
situated at rest at a smaller distance below the hook.
By the way, the above excludes slip-knots (where a point on a rope is
permanently attached to another rope, but not to a particular point on
that rope), even though they probaby don't make a difference one way
or another.
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| « Last Edit: Apr 22nd, 2003, 7:51am by Rujith de Silva » |
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James Fingas
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Re: Rope and weight puzzle
« Reply #12 on: Apr 22nd, 2003, 8:01am » |
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SWF,
I don't think your solution works the way you say it does. The location of B wouldn't change, but I think the vertical distance from B to C would increase, and the vertical distance from C to the center of the cube would definitely increase.
To see why the vertical distance from B to C would have to increase, think about the properties of the rubber band from B to C, k>0 and l>0. l is the unstretched length, and k is the spring constant. If l were zero, then the rubber band would be perfectly linear. That is to say, the vertical extension would be proportional to the vertical force, and the horizontal extension would be proportional to the horizontal force. For l>0, the vertical extension for a given force decreases with increasing horizontal extension.
Before the other rope is cut, the rope BC hangs at an angle alpha to the vertical. If l=0, then the vertical force can be a maximum of half the weight of the cube, so the vertical extension can be a maximum of x/2 (x is the vertical extension of BC under the full weight). After the rope is cut, the vertical extension will be exactly x. Having l>0 is even worse, because, as I said before, the vertical extension for a given force decreases with increasing horizontal extension. The vertical extension in the first case will therefore be even less than x/2 if l>0.
Here is my idea:
______
| 8
|A 8
8\ 8
8 \8B
8 |
8 |
[IIII] (weight F)
The sections indicated with 8s are weightless, frictionless chains (ropes that aren't stretchy). The vertical sections of rope have a relatively small k value, and the diagonal section (which in real life is also vertical) has a relatively large k value.
The chains don't stretch at all, so B will not change. However, putting the diagonal rope in transfers more of the force from the chains to the stretchy ropes, increasing the value of A. The weight hangs at A+B. Cutting the diagonal rope restores half the force of the weight to each leg, bringing the weight up by decreasing A. I have done the math, but I think it would bore you, so I didn't include it here. Try it out if you want, with k=1, l=0 for all three ropes, B=2, and F=2 (you can also use l>0 if you want, but it makes the math harder).
Although I draw it in more than one dimension, you would of course contstruct it in a single dimension. I just didn't think this diagram would help you:
Hook
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|A
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|B
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Weight
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Rujith de Silva
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Re: Rope and weight puzzle
« Reply #13 on: Apr 22nd, 2003, 2:12pm » |
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Very good! Here's an alternative construction/explanation, very
similar to James', but maybe a little simpler to understand. It uses
two identical stretchy ropes (elastics) and three non-stretchy ropes.
First, connect a stretchy rope HA, a non-stretchy rope AB and the
other stretchy rope BW:
H
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A
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B
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W
W will stretch HA and BW by some amount. To keep things simple,
assume that HA, AB and BW are each of length 1 AFTER being stretched.
Now attach non-stretchy ropes HB and AW, of length 2.00001. Doing
this will not affect the assemblage, and the new ropes HB and AW will
not bear any weight - they will be just slightly slack.
H
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A |
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| B
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W
Now sever AB. The result will look like:
H
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A |
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| B
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W
Note that HB and AW are still the same length, but HA and BW have
contracted somewhat, because now each side is bearing only HALF the
weight. Also note that the distance AB has INCREASED. This is the
key to the puzzle!
I highly recommend constructing this using some rubber-bands and
string. I used a paper-clip as the "severable link," and a hammer as
the weight. It was amazing to watch the hammer descend when I
expected it to ascend, and vice versa! That yields a follow-up
question: why does our intuition suggest so strongly that
constructions of ropes should NOT behave like this?
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SWF
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Re: Rope and weight puzzle
« Reply #14 on: Apr 22nd, 2003, 7:35pm » |
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James Fingas, I cannot follow your reasoning. Particularly, when you say:
Quote:| the vertical force can be a maximum of half the weight of the cube, |
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I see no reason for the vertical force to be limited. Are you forgetting the force that corners of the cube apply to the ropes, and that ropes around the cubes can be tightened to squeeze on the cube as much as you want without changing the net vertical force?
Before the conditions of the problem were clarified, the other method I was thinking of was pretty simple. Hang the weight from a single rope. By very tightly wrapping a second rope around the first rope (spiraling around many times along the length), it will squeeze from the sides and the orginal rope with be forced to lengthen, thus lowering the weight. If the second rope is cut, the weight will rise.
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B1gB4dB3n
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Re: Rope and weight puzzle
« Reply #15 on: Apr 23rd, 2003, 4:19pm » |
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Would it not also be possible (using two hooks) to do it with one elastic and one static rope? My solution is two-dimensional, though, and therefore might not be part of this problem..
e=elastic rope s=static rope h=hook w=weight
he e e e e e eh
s e
s e
s e
W
if the elastic rope is stretched to begin with (stretched with a force grater than the weight of the weight) and then the weight is attached, stretching it further as shown, when the static rope is cut, the weight will be pulled up to the hook on the right.
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James Fingas
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Re: Rope and weight puzzle
« Reply #16 on: Apr 25th, 2003, 1:46pm » |
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Quote:| ropes around the cubes can be tightened to squeeze on the cube as much as you want without changing the net vertical force |
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I'm not sure what you're saying. The vertical force in rope BC is limited precisely because you can't change the net vertical force. Sure, you can apply as much horizontal force as you want, but the sum of the vertical forces in ropes BC and BF has to equal the force of gravity on the weight. If we assume that B and E lie directly under A, then the BF rope always carries more than half of the vertical weight, leaving less than half for rope BC.
If l=0, the horizontal force doesn't matter at all (increasing the horizontal forces raises the weight the same amount before and after cutting), and if l>0, then increasing horizontal force raises the weight more before cutting than it does after cutting.
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SWF
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Re: Rope and weight puzzle
« Reply #17 on: Apr 26th, 2003, 3:28pm » |
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James Fingas, suppose the block weighs one pound and ropes BC and CDEF are tied so that the tension in each is 1000 pounds. A, B, and E are in a straight line, so the vertical component of force due either rope BC or CDEF is 1000/sqrt(2) pounds. The whole thing is still in equilibrium.
Vertical forces acting on intersection point B of the ropes:- Upward force of 1 pound from rope AB
- Downward force of 2*1000/sqrt(2) due to the combined effect of ropes BC and CDEF
- Upward force of 2*1000/sqrt(2)-1 due to corner of the block pressing against the ropes.
Vertical forces acting on the block:- Downward 2*1000/sqrt(2)-1 on corner B
- Upward 2*1000/sqrt(2) on corner E
- Downward 1 pound due to gravity.
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James Fingas
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Re: Rope and weight puzzle
« Reply #18 on: Apr 28th, 2003, 2:14pm » |
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SWF,
I see what you mean now. The ropes are tight against the block. I guess I wasn't thinking clearly about what happens when you tighten the ropes a lot. Yeah, and the block rises too! Sorry 
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