1
$\begingroup$

Question:- A uniform chain of length $l$ and mass $m$ is hanging vertically at edge of an inclined plane AB of length $l$ which is making an angle of $30$ degrees with horizontal. Find work done in pulling Chain slowly to AB.

enter image description here

Using concept of centre of mass and gravitational potential energy I was able to find work done as $\frac {3mgl}{4}$

I also tried to find total work done by calculating differential work done and then integrating it over desired limits.

$$W=\int dw=\int Fdx$$

enter image description here

According to me $$F=\frac{mgx}{l}+\frac {mg(l-x)}{2l}$$

But after integrating it over limits $-l$ to $0$. I got wrong answer.

How can I calculate correct value of $F$. Also can somebody tell me what is the significance of pulling chain slowly.

$\endgroup$
4
  • $\begingroup$ @Why are you reading my name That means Chain is pulled with constant velocity. I think that this is required because we have to write change in kinetic energy as well with change in potential energy in work equation. $\endgroup$
    – user1000
    Feb 9, 2021 at 16:40
  • $\begingroup$ Also To write change in kinetic energy we should be given acceleration with which chain is pulled. In that case $F_{net}$ won't be $0$. $\endgroup$
    – user1000
    Feb 9, 2021 at 16:47
  • 2
    $\begingroup$ @ user1000, the significance of 'pulling the chain slowly' is that the system is moving quasistatically. That's a fancy way of saying that is covering very small distances in long periods of time. So there is little to no velocity of chain. So if you can apply Work-Energy Theorem, you have to take change in kinetic energy as zero. This is primarily because kinetic energy of chain approaches zero since velocity approaches zero. $\endgroup$ Feb 9, 2021 at 17:38
  • $\begingroup$ @Why are you reading my name You were correct. $F$ was calculated wrong because I have taken mistakenly sine of 30 degrees as $\frac {√3}{2}$ $\endgroup$
    – user1000
    Feb 9, 2021 at 18:17

1 Answer 1

2
$\begingroup$

An infinitesimal mass element can be written as

\begin{equation} dm=\rho dx \end{equation}

where $\rho$ is the density, assumed to be constant. This means that a finite mass is obtained by multiplying the density with a finite length.

The total work is given by a sum of two contributions. First, we have to consider when the chain is vertical, and second, when it lies along the incline. Moreover, if a piece with length $x$ is on the incline, then obviously the other piece has length $l-x$. I also assume that there is no friction.

The works are then \begin{equation} W_1=\int_0^{l}dx\,\rho(l-x)g=\frac{\rho gl^2}{2} \end{equation}

\begin{equation} W_2=\int_0^{l}dx\, \rho x g\sin\theta =\frac{\rho gl^2}{4} \end{equation}

The total work done is then

\begin{equation} W=W_1+W_2=\frac{3}{4}\rho l^2=\frac{3}{4}\left(\frac{m}{l}\right)l^2=\frac{3}{4}mlg \end{equation}

We move the chain slowly so that we can neglect possible deformations and internal stresses that could arise during the journey. In that case, the density would be a function of $x$, $\rho(x)$ so the result would obviously change.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.