Associated Topics || Dr. Math Home || Search Dr. Math

Monkey Climbing a Chain

```
Date: 09/22/1999 at 21:30:38
From: Huma
Subject: Physics question

A 20-lb. monkey is attached to a 50-ft. chain that weighs 0.5 lb.
per (linear) foot. The other end of the chain is attached to the
40-ft.-high ceiling of the monkey's cage. Find the amount of work the
monkey does in climbing up her chain to the ceiling.

I don't even know where to start... help!
```

```
Date: 10/14/1999 at 17:46:49
From: Doctor Fwg
Subject: Re: Physics question

Dear Huma,

Thank you for sending such an interesting question. Here is one
possible solution, but be sure to check my work to make sure I didn't
make any foolish errors in solving the monkey/chain problem.

Before attempting to solve this problem there are a few assumptions
that might be made. For example, one should probably assume that the
monkey is a "point" mass attached to the very end of the chain. This
will avoid the problem of not knowing the exact position of the
monkey's center of mass, either at the floor or at the ceiling. A
second assumption might be that there is no "loop" in the chain, once
it is lifted above the floor. This assumption allows one to let the
total length of the chain always equal the exact sum of the distances
between the monkey and the loop and the ceiling and the loop.

This problem would be relatively easy to solve if one wanted to find
the work done by an unchained monkey climbing up a 40-ft. chain. In
this case, the work (W) would be the product of the force exerted by
the monkey and the distance that the force was exerted through.
Typically, if the force (F) that must exerted is constant, the work
required is just the product of the force and the distance (X) that
the force is exerted through, or W = FX. However, a more general
method for calculating mechanical work, even when the force exerted
changes with distance or time, involves the following differential
expression:

1) dw = Fdx = Mgdx.

In this case, integrating the expression above to get the total work
performed by the monkey would involve using Mg in place of F, where M
is the mass of the monkey (in slugs, this value of M = 20/32), and g
is the acceleration due to gravity. F (or Mg) is the weight of the
monkey and is also the minimum muscular force that the monkey must
exert in order to overcome her weight and thus lift herself. M and g
are both constants and the integration limits for dw are between 0 and
W, and for dx between 0 and 40 ft. The integration step itself is
trivial and the final result is:

2) W = Mg(40 - 0) = 40Mg = 40(20/32)(32) = 40(20) = 800 ft-lb

where the mass of the monkey is in "slugs" (i.e., 20/32) and
g = 32 ft/sec^2. This same result could have been obtained just by
using W = FX = MgX, but illustrating how this equation is obtained,
using integration, is the purpose of the preceding paragraphs.

Now, in the present case, the mass that the monkey must lift increases
as she climbs the chain because the monkey supports more and more of
the chain's mass as she climbs. Furthermore, as soon as the monkey
climbs above the height of 10 ft., the increase in chain mass
supported by the monkey changes rather abruptly because the length of
the chain is no longer the same as the monkey's height above the
floor. So now one must account for the mass of the monkey as well as
the increasing mass of the chain that must also be lifted by the
monkey. One result of all of this is that the minimum work that must
be performed by the monkey, in climbing as well as lifting part of her
attached chain, must be greater than 800 ft-lb.

For illustrative purposes, first consider the work that the monkey
would have to do if the chain were only piled up on the floor and not
attached to the ceiling at all. In this situation the integration
limits for dw are also between 0 and W, and for dx between 0 and
40 ft, and the differential work expression would be:

3) dw = Fdx = (M + (0.5x/32))gdx,

where the term (M + (0.5x/32)) gives the total mass that the monkey
must lift (in slugs) as a function of "x," the height climbed. M is
the monkey's mass and the (0.5x/32) term gives the additional mass of
the chain supported by the monkey at height x.

Performing the integration produces:

4) W = Mg(x - 0) + (0.5/32)(x^2 - 0)(g/2)
= Mg(40 - 0) + (0.5/32)(40^2 - 0)(g/2)
= 40(20/32)(32) + (0.5/32)(1600)(32/2), so:

5) W = 800 ft-lb + 400 ft-lb = 1,200 ft-lb.

Now, this amount of "mechanical" work is more than the actual amount
of work that would be required to lift a chain that is partially
supported by a ceiling attachment, as in this case. So, the true
answer to this problem must lie between the 800 and 1,200 ft-lb values
calculated above.

It so happens that Eq. 3 will also give the work that must be supplied
by the monkey to climb the first 10 ft. of the chain, because this
section of chain is, in fact, lifted from the floor and none of that
portion of the chain's weight is supported by the ceiling. In this
case the integration limits on dx are between 0 and 10 ft. and the
integration step is straightforward. The work for this portion of the
climb should come out to 225 ft-lb.

After the 10 ft. height is exceeded, the total length of the chain is
no longer the same as the height climbed by the monkey because the
chain begins to pull upward and away from the floor after the 10 ft.
limit is reached. So now it would be helpful to have another
expression for the chain mass as a function of the height climbed at
all heights between 10 and 40 ft. and to then integrate (i.e. add up)
the work during that portion of the climb. See the following table:

Table 1.
Relation Between Monkey's Height and Length of Supported Chain

Columns Below Are:  (1) Monkey's Height, (2) Length of Chain Supported
by Monkey, (3) Length of Chain Supported by Ceiling, (4) Total Length
of Chain - all in feet.

(1)  (2)  (3)  (4)
---  ---  ---  ---
0    0   40   50
10   10   40   50
20   15   35   50
30   20   30   50
40   25   25   50

From this table it can be seen that the length of chain supported by
the monkey, only at or after the 10 ft. height limit is reached, is
equal to (x/2 + 5), where x is either greater than or equal to 10 ft.
So the work that the monkey needs to do, between 10 and 40 ft, may be
found from:

6) dw = Fdx = {M + (0.5/32)(x/2 + 5)}gdx,

where the integration limits here, on dx, are between 10 and 40 ft.

If the results using Eq. 3 (with integration limits on dx between 0
and 10 ft.) and Eq. 6 (with integration limits on dx between 10 and 40
ft.) are added together, the minimum work that must be exerted by the
monkey to climb the chain will be found. Note: the final sum here must
be between 800 and 1,200 ft-lb. It is believed that the numerical
result of integrating Eq. 6 between the limits of 10 and 40 ft. will
produce a result of 862.5 ft-lb. Adding this result to the 225 ft-lb
result above will produce a final answer of 1087.5 ft-lb for the total
amount of "mechanical" work that must be performed by the monkey.

I hope that seeing this solution helps you to understand this problem.

- Doctor Fwg, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Calculus
College Physics
High School Calculus
High School Physics/Chemistry

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search