This will be a diversion from the Lebesgue measure/integration posts.

During today’s question and answer period, a student asked the following question: if one has a 50 foot long rope hanging off of a long cliff and one pulls up 25 feet of rope to the top, how much work does one do? Assume the rope weighs .5 pounds per linear foot.

My “instinct” told me to break this into two parts:

Work done on the first 25 feet:

Work done on the bottom 25 feet of rope:

So the total work done is which is correct.

But a student suggested the integral which also works. The question is: why?

To discover why, let us remind ourselves of the model that we are using: to compute the work done by pulling up a rope, one divides the rope into small segments, each of equal weight. Each segment gets lifted a different amount so the integrand becomes

Now let’s make our problem a bit more general: we have a rope of length and we are going to lift feet of it where, of course,

So the total work is which, written in terms of integrals, becomes:

We can write this as a single integral if we do a change of variable in the second integral:

Let which means so the second integral gets changed to:

so the work becomes:

. But the variable is dummy so we can switch back to in the second integral to obtain:

.

Now to get to the student’s suggestion, do yet another change of variable:

Let

Then the integral becomes

However, this integral makes no sense with the model that uses as a segment of rope that gets lifted units.

But there is another model that actually conforms more to what we actually feel if we pull up a rope ourselves.

First, set our variable running from top to bottom with representing the top of the cliff. Then think of the denoting the amount that the rope is moved and the weight of the rope that is being pulled when there is length of rope remaining to be pulled.

In other words, start with amount of rope hanging off of the cliff. Then pull up on the rope with a “jerk” that moves the rope units upward. This movement moved (approximately) the whole rope upward which means that one applied a force of over distance. Now pull up again and now one moves amount of rope units upward. Then repeat; this yields the integral with the equality following from a simple change of variable.

This conforms to how the rope gets easier to pull as we pull the rope up; there is less weight to pull when the amount of hanging rope is less. So here we minimum amount of rope left to be pulled will be and the maximum is hence the integral is .

[…] Today’s posting is light due to my post about this calculus problem: pulling a rope up a cliff. It turns out that this problem can be handled two different ways: in one way, one moves small bits […]

Pingback by 24 February 2011 pm « blueollie — February 25, 2011 @ 2:58 am

E muito dificil ,sou da 6º sere não do 3 º ano do ensino medio !!

Comment by isabelle — March 20, 2012 @ 11:17 pm