# Falling chain with friction

I really need some help with a physics problem, and I guess my doubt is more conceptual, than question based. But still, let me pose it to you:

A uniform chain of mass $M$ and length $L$ lies on a rough table, with coefficient of friction $k$. When 1/3rd part of the chain is hanging off the edge, it starts slipping. Find the work done by the frictional force on the chain, at the instant the last link falls off.

Now, to find the work, I know we must use integration. The work done would be

$$\int k \, p \, T \, g\, dp \tag{1}$$

Where $k$ is the coefficient of friction, $p$ is the varying length, $T$ is the mass per unit length ($= \frac{M}{l}$), and $g$ is the acceleration due to gravity. After applying appropriate limits, we get the answer. The integrand in $(1)$ simply should mean force $\times$ displacement. Now does my question come in. In the integrand, the Force quantity is defined by the variable frictional quantity, which is equal to $(k \, p \, T \, g)$, so where did the quantity of displacement of the chain go? Shouldn't there be an extra $p$ in the intgrand, such that it becomes $(k \, p^2 \, T \, g)$? Or is the notation $dp$ itself used to signify the displacement?

I know the question is a bit complicated, but please help me out. It will be highly appreciated.

• Yes $dp$ is your missing "symbol" with unit length. You use $dp$ instead of $p$, because you integrate of $p$. – fibonatic Jul 22 '13 at 17:59
• @user26216 When the pending edit goes through, take a look at how the integral was formatted (by clicking "edit") - it is quite simple. We use Latex notation for mathematical symbols here, implemented via Mathjax. For an overview see this guide. – user10851 Jul 22 '13 at 18:08
• A quick Google will find you many pedagogical articles on this problem. See for example wiki.brown.edu/confluence/download/attachments/2752884/… – John Rennie Jul 22 '13 at 18:42
• Wouldn't a uniform chain be like a string or rope, without the stranding? – John Alexiou Jul 22 '13 at 20:44
• I understand now. I was forgetting the fact that we integrate force, which is a function of the displacement, to find the work done. I'll definitely learn the Latex notation. Thanks! – Lexicon Jul 23 '13 at 5:52

The quantity of displacement that you are looking for is $dp$ in the integral.
At any instant the friction on the chain is $kpTg$, instantaneous work done by friction when the chain moves $dp$ length would be $-kpTg \times dp$ (force x displacement). Negative sign because work is done against direction of motion or because friction is opposing motion.
To find out the total work done against friction, integrate $-kpTgdp$ with $p$ varying from $2l/3$ to zero, here $p$ is the length of the chain on the table.
On integrating, the expression that you get should be $$-\frac{k(p^2)Tg}2$$ Put the limits from $2l/3$ to zero.
You left off the limits of your integral: $\int_0^{p_0}dp\ p\ A$ where $p_0$ is the longest length of chain on the table and $A$ is all your other factors lumped together.
If you know how to integrate, I think you'll find an extra factor of $p$....