Trouble in Calculating Force required to pull a chain over an inclined Plane

Question

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.

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.
$$W=int dw=int Fdx$$

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.

Ruben Campos Delgado · Accepted Answer

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  gsintheta =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.

Trouble in Calculating Force required to pull a chain over an inclined Plane

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