Potential Energy Problem: Pulling a Chain up onto a Table

[Ascendant0]

Homework Statement

A chain is held on a frictionless table with one fourth of its length hanging over the edge. If the chain has length L = 0.28 m and mass m = 0.012 kg, how much work is required to pull
the hanging part back onto the table?

Relevant Equations

Potential energy dU

So, the first thing that came to mind when I was trying to figure out how to set this up is that it will be a dU problem. After trying to figure out how to set it up to no avail, I took a look at how they solved it in the solutions manual. It's making absolutely no sense to me...

They state "note that the mass of a segment is (m/L) dy". I'm completely lost on that part, as to why "L" is in the denominator? Wouldn't that setup mean that the smaller "L" is, the larger the mass, to the point where it becomes infinite if it is infinitesimally small??? I'm not seeing the sense behind how they've set it up, as from what I'm thinking, they're basically stating the shorter the length, the larger the mass. Can someone help me to view this correctly so I can understand why it is set up the way it is?

 

[Lnewqban]

It shows the dependence between the mass and the length, which can be expressed as kg/m, for example.
Since length L = 0.28 m and mass m = 0.012 kg, we can say that this chain has a linear mass of 0.012/0.28 = 0.0428 kg/m.

 

[TSny]

They state "note that the mass of a segment is (m/L) dy". I'm completely lost on that part, as to why "L" is in the denominator?

The total length of the chain is $L$ and the total mass is $m$. If $\Delta y$ is the length of a segment of the chain, the mass of this segment is a fraction of the total mass $m$. For example, suppose $\dfrac {\Delta y} L$ is $\dfrac1 {10}$ so that $\Delta y$ is one-tenth of $L$. In this case, the mass of the segment $\Delta y$ will be $\dfrac1 {10}$ of the total mass $m$. That is, $$(\text{mass of segment of length } \Delta y) = \frac {\Delta y} L \cdot m$$ This can be rewritten as $$(\text{mass of segment of length } \Delta y) = \frac m L \cdot \Delta y$$

Wouldn't that setup mean that the smaller "L" is, the larger the mass, to the point where it becomes infinite if it is infinitesimally small???

For a given type of chain, the mass of a segment of length $\Delta y$ does not depend on the total length $L$ of the chain. If $L$ were cut in half, the total mass $m$ would also be reduced by one-half. But the ratio $\dfrac m L$ would not change. So, according to the formula above, the mass of a segment $\Delta y$ is not changed when $L$ is changed.

 

[Ascendant0]

Thanks to both of you. I get it now. It's a little late for me to get back to the problem, but I'm going to revisit it tomorrow and make sure it all makes sense now. I believe it should. I appreciate the help, thank you