# Work done to pull half of the rope up

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.

• 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? Commented Jun 2, 2021 at 12:55