Hanging over a pulley there is a rope, with a weight at one end.

Additional material

[ Editor: from a 1994 thread, after the entry above was posted in reply
to Mark Kantrowitz's <mkant+@cs.cmu.edu> version of the puzzle. ]

Apparently I am not the only one who finds this answer less than
completely satisfactory.

For one thing, why does "equal mass" guarantee that the tension will be
"just enough" so that the weight rises at exactly "the same rate"?  (At
the end of this article, I'll post a variant puzzle to explore this
further.)

For another thing, when is the tension in the rope increased?  Suppose
the monkey weighs 10 Newtons and is climbing the rope at a constant
rate, say, 10 cm/s, the tension in the rope is then 10 Newtons, the
same as when the monkey is not climbing.  Depending on how you model
the monkey's act of beginning to climb, the tension might increase only
during an initial acceleration period, or might even be zero at all
times except for the instant at which the monkey starts out, during
which it's infinite.  Of course anyone who understands freshman physics
will figure out some such solution, but the solution is worded as if
rate of rise at any instant were a function of tension at that instant.

And another thing, the puzzle makes an unstated assumption about the
initial conditions, namely that both the weight and the monkey are
initially at rest in the reference frame.  Suppose that initially the
weight at the end of the rope is going down, while the monkey hangs
from the other end (which is going up), all consistent with the problem
statement.  Then the weight will end up gaining the same upward
velocity as the monkey, but whether this results in a net up or down
movement depends on unstated quantities.

I'd like to propose first of all that the weight and monkey are
explicitly at rest initially (though you can certainly generalize the
problem by solving in terms of u, the initial upward velocity of the
monkey, and v, its rate of climbing the rope).  Then I have two
proposed methods of solution:

(1) By conservation of angular momentum.  Note that the total linear
momentum of weight + monkey is not necessarily conserved, since the
pivot of the pulley (which is attached to the ceiling) can deliver
varying forces and impulses to the system.  But their total *angular*
momentum about the pivot *is* conserved, since the gravitational forces
and moment arms are the same, and the pivot itself (being frictionless)
doesn't contribute.

Now the initial angular momentum is zero, because nothing is moving.
In general the total angular momentum is mv - mV, where v is the
monkey's upward velocity and V is the weight's upward velocity.  So
v = V.  Since the rope between the weight and the monkey is shortening
as the monkey climbs, at least one of them must be rising, so they are
both rising at the same rate.

(2) By converting to an equivalent problem of linear momentum.  The
tension in the rope is a superposition of the weight of the monkey (or
equivalently of the weight), which exactly balances the gravitation
force on each of the two bodies, plus any forces or impulses the monkey
generates in order to start climbing.  So "unwind" the whole apparatus
so that the gravitational component is removed: that is, put the
weight and the monkey on a flat, frictionless surface with the rope
extending barely taut between them (i.e., with zero tension but not
sagging).  The pulley becomes irrelevant and can be removed.  Now the
linear momentum of weight + monkey is zero, and their center of gravity
never moves; this requires the weight to move toward the center of
gravity at exactly the same rate as the monkey as the rope between them
shortens.  Translating this back to the original apparatus, both
objects move toward the midpoint of the rope (which might be at or on
either side of the pulley) at the same rate.

Of course these solutions describe only what happens before the time
when either the monkey or the weight reaches the pulley.

Now for the ObPuzzle that I promised:

A monkey is initially hanging onto a rope that is wound about a
combination of weights and pulleys as shown:

------------------------------------ ceiling
         |        |
         |        |
         |       _|_
         |      / | \  pulley
         |     |  +  |
         |     |\___/|
         |     |     |
         |     |     |
         |     |     |
         |     |     |
         |     |     |\_O
         | ___ |     |  | monkey
  pulley |/   \|     |/\|
         |  +  |     |   \_/
          \_|_/
            |
            |
            |
          ,-+-.
         /     \
        | weight|
        |       |
        +-------+

One end of the rope and one of the pulleys are fixed to the ceiling,
which does not move.  The weight is suspended from the pivot of the
other pulley.  Only the weight and the monkey have mass, and the
pulleys are frictionless.  Initially everything is at rest and in
equilibrium.

If the monkey starts climbing the rope, does the weight go up or down?
At what speed relative to the monkey?

-- David A. Karr <karr@cs.cornell.edu>

If the rope were inextensible and massless, the monkey and weight start
at the same height, both start unmoving, and the pulley had no friction
nor any angular inertia, then the weight would go up and down along
with the monkey.  Any acceleration of the monkey is induced by changing
the tension on the rope, which will equally accelerate the weight.
Thus, both the monkey and the weight have the same accelerations.  They
start at the same height with the velocity, so they remain at the same
height as each other at any given time.

The same would occur if one changes the assumptions so that they start
at different heights but that the gravitational acceleration is
constant with height.  In this case, the difference in the heights
remains the same.

Anyway, what happens in the "real" situation?  First off, I assume that
the pulley is frictionally locked to the rope and won't slip.

The monkey pulls on the rope which sends tension along the rope and
accelerates the monkey upward.  If this is done gently enough, then the
friction of the pulley will allow the monkey to climb with the weight
and rope unmoving (in height).

Otherwise, the rope and pulley and weight start moving as well (the
weight rises).  Due to friction (of the pulley axle) the angular
momentum of the pulley, and the inertia of the rope, the weight will
accelerate upwards by less than the monkey.  The rope now adds more and
more weight to the monkey side, which will tend to make the weight
accelerate more and more; there are a whole lot of conflicting
causes/effects here and the exact outcome depends heavily on the
specifics.

-- Isaac Kuo <isaackuo@lhasa.berkeley.edu>

The first thing to realize here is by the addition of the lower pulley
and second connecting place to the ceiling, the tension in both of the
leftmost ropes is now m/2 (half the weight of the mass)  Therefore, we
must now assume that since we are starting from equilibrium and it is
not going to change without an impulse from the monkey, that the weight
of the monkey is m/2.  This gives us equal tensions in the two
rightmost ropes.  Now as the same problem before, if the monkey starts
climbing (using self induced acceleration) the mass on the other side
will move up as before, but this time at half the rate of the monkey.
(Vmonkey = 2 Vmass).  This is easily explained by showing that for each
foot of rope that the monkey pulls through himself to get to the top, a
foot will move on the middle rope.  But since this is on a suspended
pulley now, the effect will only move the mass half a foot upwards
(since by pulling a foot, we subtract a half foot from each side of the
suspended pulley.

So the conclusions are
a) the mass moves upwards
b) it moves upwards at half the speed of the monkey.

-- Philip Sherman <psherma1@cc.swarthmore.edu>

Assuming the "ideal" situation...

The weight is exactly twice as massy as the monkey.  For any given
tension, exactly twice as much force is applied to the weight as the
force applied to the monkey.  The monkey and weight go up at the same
speed at any given time.

-- Isaac Kuo

> This is easily explained by showing that for each foot of rope that
> the monkey pulls through himself to get to the top, a foot will move
> on the middle rope.

That's false.  Consider what happens if for some reason the pulleys are
all stuck.  Then the middle rope never moves regardless of how much or
little the monkey pulls himself up the rope.  There's no reason why 1
foot of rope pulled by the monkey would imply that 1 foot will move on
the middle rope, and indeed it's false.

-- Isaac Kuo

Ok. Isaac, I realize that my my quick assumptions that the pulley works
(hehe) may not be correct.  But are you saying I am wrong about Vmass=
1/2 Vmonkey?

That part is totally correct and I will explain why.  This system is
initially at rest or equilibrium.  So if we consider this a free body
with a equilibrium between the monkey and the mass, we know that if the
monkey exerts a force to pull himself upwards 1 foot, then if we
considered the original problem of just the mass hanging (no 2nd
pulley) it would also move up a foot.  but is is simple arithmetic that
if you pull one foot of a pulley arrangement on the left, the mass will
only move 1/2 foot.

Look

   |    | (a)  -
   |    |      |
   |    |      |
   |    |      |
   /----\ (b)  -
   |    |
   \____/

Imagine we pulled from the right side, enough rope to move b to a.  We
have pulled the length x from the ceiling, of rope.  But we still have
the same amount x left on the left side.  We never pulled all this
through.  In any case, the pulley hangs from the remaining x length
(from the left side) and dividing by two (because the hanging pulley
has two ropes holding it), we get a height of x/2.  How do you think a
100 pound man can lift a 500 lb engine with a pulley engine hoist?  The
force reduction of the weight is directly proportional to the amount of
times extra length of chain he has to pull compared to the height
needed to raise it.  In other words, with a certain pulley system, a
man can pull a rope 10 to raise the weight 10 ft with no mechanical
advantage.  But a man can also pull 50 feet to raise the weight ten
feet, but by only needing a force of m/5 rather than the weight of the
entire mass.  Same thing here with the monkey.  The monkey will reach
the top while the mass has only moved halfway up.  Read your physics
book.

-- Philip Sherman

> But are you saying I am wrong about Vmass= 1/2 Vmonkey?

No, that's right.

> ...  but is is simple arithmetic that if you pull one foot of a
> pulley arrangement on the left, the mass will only move 1/2 foot.

That is completely wrong!  It is _only_ correct if the person pulling
on the rope is not moving.  For instance, a person standing on the
ground pulling one foot of rope will pull up the mass 1/2 foot.  Think
about it, please.  My point about the stuck pulleys is that if the
person pulling on the rope moves, then a completely different
conclusion results...

When the monkey pulls 1 foot of rope, how much will the _monkey itself_
move?  You don't get some obvious answer!

Looking at distances of rope is the wrong way to approach the problem.

One must look at what forces are involved.  Because we assume that the
tension of the rope between the monkey and the ceiling anchor is
constant (at any given moment), it is trivial to see that twice the
force is applied at any time to the mass as to the monkey.  Since the
mass is twice as massy as the monkey, this results in the same
acceleration on the mass as the monkey at any given time.  This will
result in the mass and the monkey moving in concert.  _This_ is the
solution you'd find in a physics book.

-- Isaac Kuo <isaackuo@math.berkeley.edu>

[ Editor: Kuo's "No, that's right" answers "are you saying I am wrong";
his conclusion in the same posting is that the weight and the monkey
move at the same speed, not Vmass = 1/2 Vmonkey. ]