A chain of mass 0.80 kg and length 1.5 m rests on a rough surface table so that one of its ends hangs over the edge the chain starts sliding off the table all by itself provided the overhanging part equals n=1+3 of the chain length . What will be the total forces pperofrmed by the friction forces chain ? Given ans=-2.66j

Dear Student,

Given, mass of chain= 0.8 kg or M (say)

Mass/ unit length= M/L

Mass of hanging part= $\frac{M}{L}\times \frac{L}{3}=\frac{M}{3}$
Lets take a small length= x

Weight of this fragment is = Mgx/l.....(i)

We know friction force= $\mu N where \mu is co-efficient of friction and N is the normal reaction force to the weight$

plugging (i) we get frictional force f= $\mu \frac{Mgx}{L}$...(ii)
Given that when the length of the hanging part is l/3 the chain starts slipping, or in other words maximum frictional force offered is=
 
$\frac{\mu 2Mg}{3L}=\frac{Mg}{3} \Rightarrow \mu =\frac{1}{2}$...(iii)



$$\begin{gathered} dW=f.dS \\ Where dS=-dx a small element of the chain, - sign because displacement is in direction opposite to that of applied force \\ W=-\frac{Mg}{2L}{\int }_0^{\frac{2L}{3}}xdx\Rightarrow -\frac{Mg}{2L}\times \frac{4\times L^2}{9}\Rightarrow -\frac{0.8\times 10}{2}\times \frac{4\times 1.5}{2\times 9}=-1.33 J \\ Comment 1: In a laymans language we can put f=\frac{Mg}{3} as max friction force and displacement as dS=-\frac{2L}{3} which will yield -2.66 J but that is wrong as frictional \\ is not cons\tan t over the length of the chain, as the normal reaction force keeps changing as more and more of the chain falls off. \\ In a pure conservative field there would not be any work loss, but \sin ce there is friction there is some work loss which is released in the system in form of heat and sound. \\ Comment 2: You have asked for force but the answer is in J . Take care of your units. \end{gathered}$$

REGARDs