# College Math Teaching

## February 24, 2011

### Calculus Problem: the hanging rope problem: two models.

Filed under: calculus, class room experiment, integrals, physics, student learning — collegemathteaching @ 10:21 pm

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: $\int^{25}_{0} .5xdx = 625/4$
Work done on the bottom 25 feet of rope: $(25)(25)(.5) = 625/2$
So the total work done is $625/4 + 625 /2 =(3/4)625 ft-lb.$ which is correct.

But a student suggested the integral $\int^{50}_{25} .5xdx$ 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 $\rho xdx$

Now let’s make our problem a bit more general: we have a rope of length $L$ and we are going to lift $a$ feet of it where, of course, $a \leq L$
So the total work is $\int^{a}_{0}\rho xdx + \rho a(L-a)$ which, written in terms of integrals, becomes:
$\int^{a}_{0}\rho xdx + \int^{L}_{a} \rho adx$
We can write this as a single integral if we do a change of variable in the second integral:
Let $(a/(L-a))(x-a) = u$ which means $(a/(L-a))dx= du$ so the second integral gets changed to:
$\int^{a}_{0} \rho (L-a) du$ so the work becomes:
$\int^{a}_{0}\rho xdx + \int^{a}_{0} \rho (L-a) du$. But the variable is dummy so we can switch back to $x$ in the second integral to obtain:
$\int^{a}_{0}\rho (x + L-a) dx$.
Now to get to the student’s suggestion, do yet another change of variable:
Let $w = x + (L - a)$
Then the integral becomes $\int^{L}_{L-a} \rho wdw$

However, this integral makes no sense with the model that uses $\rho dx$ as a segment of rope that gets lifted $x$ 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 $x$ running from top to bottom with $x = 0$ representing the top of the cliff. Then think of the $dx$ denoting the amount that the rope is moved and $\rho (L-x)$ the weight of the rope that is being pulled when there is $x$ length of rope remaining to be pulled.

In other words, start with $L$ amount of rope hanging off of the cliff. Then pull up on the rope with a “jerk” that moves the rope $\Delta x$ units upward. This movement moved (approximately) the whole rope upward which means that one applied a force of $\rho L$ over $\Delta x$ distance. Now pull up again and now one moves $\rho (L - \Delta x)$ amount of rope $\Delta x$ units upward. Then repeat; this yields the integral $\int^{L}_{0} \rho (L-x) dx = \int^{L}_{0} \rho xdx$ 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 $L - a$ and the maximum is $L$ hence the integral is $\int^{L}_{L-a} \rho xdx$.