Solve for the length of a rope when you know the work required to reel it in.

[Pull and Twist]

Hey guys, I'm currently working a take home problem that's a little different then what I have been practicing in my Calc. book. I thought I set it up correctly, but yet I am not getting the correct answer.

Can someone show me where I am going wrong?

This is what I have so far.

View attachment 4017

 

[MarkFL]

I think what I would do is first compute the work increment:

\(\displaystyle dW=2.5(\ell-y)\,dy\) where $0<\ell$

Now, add up the elements and equate that to the given work total:

\(\displaystyle 2.5\int_{0}^{\frac{\ell}{3}} \ell-y\,dy=225\)

Divide through by 2.5:

\(\displaystyle \int_{0}^{\frac{\ell}{3}} \ell-y\,dy=90\)

Let:

\(\displaystyle u=\ell-y\,\therefore\,du=-dy\)

\(\displaystyle \int_{\frac{2\ell}{3}}^{\ell} u\,du=90\)

\(\displaystyle \ell^2=18^2\)

And so:

\(\displaystyle \ell=18\)

 

[Pull and Twist]

I think what I would do is first compute the work increment:

\(\displaystyle dW=2.5(\ell-y)\,dy\) where $0<\ell$

Now, add up the elements and equate that to the given work total:

\(\displaystyle 2.5\int_{0}^{\frac{\ell}{3}} \ell-y\,dy=225\)

Divide through by 2.5:

\(\displaystyle \int_{0}^{\frac{\ell}{3}} \ell-y\,dy=90\)

Let:

\(\displaystyle u=\ell-y\,\therefore\,du=-dy\)

\(\displaystyle \int_{\frac{2\ell}{3}}^{\ell} u\,du=90\)

\(\displaystyle \ell^2=18^2\)

And so:

\(\displaystyle \ell=18\)

How did you come up with \(\displaystyle \ell-y\)?

I'm confused as my math book has never set up a problem any other way then the way I did... so I'm not sure how you deduced your integrand.

 

[MarkFL]

How did you come up with \(\displaystyle \ell-y\)?

I'm confused as my math book has never set up a problem any other way then the way I did... so I'm not sure how you deduced your integrand.

If $y$ is the distance through which the rope has been hauled, that is, the bottom of the rope is $y$ units above the ground, then the rope hauler is at that moment working against the weight of $\ell-y$ length of rope.

Does that make sense?

 

[Dethrone]

You have made a small error in your workings. Notice that the rest of the rope only moves up a distance of $L/3$, along with the rest of the rope. If you set up your second integral as $\int_{L/3}^{L} 2.5\frac{L}{3}\,dx$, you will get your desired answer. Indeed, Mark's method is a bit more clever.

 

[Pull and Twist]

You have made a small error in your workings. Notice that the rest of the rope only moves up a distance of $L/3$, along with the rest of the rope. If you set up your second integral as $\int_{L/3}^{L} 2.5\frac{L}{3}\,dx$, you will get your desired answer. Indeed, Mark's method is a bit more clever.

Thank you... that makes sense in respect to how I have been working the problems. Although I do agree that Mark's method is clever and requires less steps.