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Let's say we want to pull a uniform ( density $\rho$ ) rope of length $a$ up so that it hangs in equilibrium on a nail. This means $a/2$ is hanging on the left and $a/2$ on the right. Let the direction down be positive. In order to pull a segment $dx$ of the rope up, a work $dW$ is done against gravity, as in $$dW = -mgdx=-\rho xgdx$$

Then the total work done in pulling the rope up an an mount of $a/2$ is $$W = \int_0^{a/2}-\rho xgdx=-\rho g\frac{1}{4}a^2=-(\rho a)g\frac{1}{4}a=-mg\frac{1}{4}a$$

Is this correct? The model of work has to somehow account for the fact, that gravity is not constant. This is because whenever the rope is pulled back up, there is less mass on the one side. The nail is supporting the part which has been pulled up and so does not contribute to gravity.

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As is often the case, a work/energy approach is the simplest method here.

Can we assume that the nail is smooth, so no energy is lost due to friction between the rope and the nail ? If that is so, all we need to know to solve the problem is the height of the nail off the ground.

By how far has the centre of mass of the rope been raised once it is hanging from the nail (this is where we have to know the height of the nail) ? So how much has the potential energy of the rope been increased by ? Since the kinetic energy of the rope is zero at the beginning and zero at the end, the increase in its potential energy is equal to the work done on the rope.

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  • $\begingroup$ Yes, the nail is assumed to be smooth. Initially, the rope is entirely on one side of the nail. The center of mass is at $a/2$. Then in total, the CM is moved up a distance of $a/2$. But why consider the motion of the CM? $\endgroup$ Commented Jun 2, 2021 at 12:55
  • $\begingroup$ @variations We want to find the increase in the potential energy of the rope. The simplest way to do this is to consider where the rope's CoM is at the start; where its CoM is at the end; and so how far its CoM has been raised. Note that the CoM is not a fixed point on the rope - the position of the CoM depends on the configuration of the rope as a whole. $\endgroup$
    – gandalf61
    Commented Jun 2, 2021 at 14:07

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