Back when I was in Junior High (1973 & thereabouts) on a day that was either the last day before summer vacation or the last day before spring break, our math teacher unleashed upon us a godawful word problem to bruise our brains.
I’d like to have a copy of the thing so as to inflict it on folks whenever the subject of truly intimidating word problems comes up, or to compare to things like the Einstein puzzle (see thread); unfortunately, I don’t have a copy!
It involved a monkey…in a room with a rope running up and over a pipe and down the other side…the rope had a length, the monkey had both a weight and an age, plus a mother who had both of these as well, and the room had a height, and there was a distance from the floor to the pipe, and so on. The terms were written out in phrases like “The Age of the monkey minus the number of years of the monkey’s mother that exceed the length of the portion of the rope that is greater than the difference between the weight of the monkey and the height of the room minus the shorter end of the rope is less than the distance between the top of the monkey’s head and the end of the longer end of the rope divided by the sum of the differences in the two monkey’s ages from which the weight of the monkey’s mother plus the distance from the pipe to the ceiling has been subtracted” and went on for several similar sentences before asking something like the age of the mother at the time of the monkey’s birth.
I remember trying to set up a variable string and work on both sides of an equation to solve for something and ending up with terms that appeared to represent things like cubic monkeys per rope-year, but I never did solve the thing and neither did anyone else in my class.
Anyone that might have run across this brain-buster (or better yet actually has a copy) would be doing me a favor by posting it here.
Haven’t taken the time to write out all the variables and relationships, but I have to ask one question.
Is this a trick? If the banana weighs a certain amount per inch, but the weight on the other side is exactly equal to the monkey AND the rope is the same length on both sides (so there isn’t more rope weight on one side then the other) doesn’t that mean the banana has to be 0 (in length and weight)?
Hot damn, that’s it, all right!
(OK, so my memory of something I hadn’t seen in person since 1973 was a little off)
Yep, doesn’t look any easier than it used to…
::recalculating cubic monkeys per rope-year in light of new parameters::
Yup. Marilyn Vos Savant did this one in a column a while back. This was, of course, before I found Cecil.
That’s true only if you’re talking about frictionless ropes and fences. Of course, we’re talking about a 12-year old monkey that weighs only 18 oz., and can carry a banana more than half his weight (or more : see below), anything might be possible.
This question revolves around the rather lengthy (but only mildly convoluted) form of the sentence about ages. If you understand algebra, it’s only talking ratios so it reduces to one variable on one side and one on the other, with a bunch of numerical coefficients in between. This one’s even easier than the ‘Einstein’ one, though neither have any tricks about them.
If you do redistribute this, please please don’t leave in the ambiguity in the statement : “One-half the weight of the monkey plus the weight of the banana …”. It’s not entirely clear whether this is (1/2 Wm) + Wb or if it’s 1/2(Wm + Wb). If it’s the latter, the banana weighs more than the monkey. Better to say “The weight of the banana plus one-half the weight of the monkey …”
I did in fact notice the ‘equal length’ specification and that the weight on each side is not equal; so I made a simple estimate of the frictional force necessary. The force the rope/fence needs to balance is the gravitational force on the banana. Since we solved for the weight of the banana in ozm (called Wb) we know the extra force on the rope is Wb ozf (we could use the banana’s mass (0.002232 slugs) and work backwards, but why?).
Assume (in this crazy monkey-world, you can assume a lot, can’t you?) that all of this force needs to be balanced by friction, and that the rope is distributed flatly over the fence (so that the normal force, which produces friction, is straight down). The normal force is the weight of everything in the problem except the monkey-mother, i.e. (monkey + weight + rope + banana).
The frictional force is F= u*N, where u is the coefficient of static friction that we’re solving for and N is the normal force.
So we need u > F/N to keep the rope from sliding. If we take the banana’s length as 4.73 attoparsecs (the solution on the site linked to), the Wb (ozf) / Wb + 2*Wm + Wr (ozf) equals 0.10, which is plenty low for most interfaces. Even if you take the larger possible length for the banana of 2.277 millirods, you still only need u = 0.19; this is probably good enough as long as your rope isn’t slick nylon and the fence isn’t polished metal.
Here I thought this was going to be about Araucaria araucana.
panama jack
my other car is a monkey