Lifting a Cable or Chain

Why Lifting a Chain Requires an Integral

A box lifted a fixed height has one lifting distance: every part of it rises by the same amount. A hanging chain hauled onto a ledge is different. A slice near the ledge rises only a short distance, while a slice farther down rises farther. To find the total work, we add the contributions from many small slices, accounting for both each slice’s weight and its own lifting distance.

We will model a uniform chain, so its weight per unit length is constant. The chain is lifted slowly, and we will calculate the work done against gravity. We neglect friction at the ledge and any change in kinetic energy. These assumptions let us focus on the work required to raise the chain.

Hauling the Whole Chain onto a Ledge

Suppose a chain of length \(L\) hangs vertically below a ledge. Its upper end is at ledge height, and the chain is hauled up until it lies on the ledge. Measure \(x\) metres downward from the ledge. A small slice at depth \(x\) has weight approximately \(w\,dx\), and it must rise a distance \(x\) to reach the ledge.

The work for that slice is its weight multiplied by its lifting distance. Summing the slices gives the integral below. The variable \(x\) runs from \(0\), at the ledge, to \(L\), at the chain’s lower end.

This result depends on the chain being uniform and on the destination being the ledge height. It also shows why multiplying the chain’s total weight by its full length is not correct: each slice rises a different distance.

Hauling Only Part of the Chain

Now suppose only \(a\) metres of a hanging chain of length \(L\) are hauled onto the ledge, where \(0\leq a\leq L\). The different slices do not all rise by the same amount. A slice initially at depth \(x<a\) reaches the ledge, so it rises \(x\) metres. A slice initially at depth \(x\geq a\) remains hanging and rises \(a\) metres as the chain moves upward.

We split the integral at \(x=a\), because the lifting distance changes there. The first integral accounts for slices that reach the ledge; the second accounts for slices that remain hanging.

Evaluating each part gives:

For \(a=0\), the expression gives zero work, as expected when no chain is moved. For \(a=L\), it reduces to \(wL^2/2\), the result for hauling the entire chain onto the ledge.

Why the Slice Integral Works

The integral comes from adding the work for small pieces of the chain. Divide the hanging length into short intervals. A slice at depth \(x_i\) has approximate weight \(w\Delta x\), and if it rises \(d(x_i)\) metres, its work is approximately \(w\,d(x_i)\Delta x\). Adding those estimates and taking the limit as the slices become arbitrarily short gives the integral.

For partial hauling, there is a second way to verify the formula. After \(s\) metres have been hauled onto the ledge, \(L-s\) metres remain hanging. The hanging part weighs \(w(L-s)\) newtons. In a small additional pull \(ds\), that hanging weight gives the force against which the chain is lifted. Therefore the work is the integral of this changing force over the haul distance.

This matches the slice-by-slice result. The two descriptions organize the same work differently: one follows individual chain slices and their lifting distances; the other tracks the changing weight of the hanging portion as the chain is hauled in.

A Common Mistake: Using One Lifting Distance for Every Slice

A tempting but incorrect calculation for a chain of length \(L\) is to multiply its total weight \(wL\) by \(L\), as though every part of it rose the full length. That would give \(wL^2\), twice the correct work for hauling the whole chain onto the ledge. The slice at the ledge rises almost zero distance, while the slice at the bottom rises almost \(L\) metres; the correct average lifting distance is \(L/2\), not \(L\).

For example, for the \(18\)-metre chain with \(w=4\) newtons per metre, the incorrect calculation gives \((72\text{ N})(18\text{ m})=1296\text{ J}\). The correct value is \(648\text{ J}\), because the slice lifting distances average \(9\) metres. A reliable setup identifies the lifting distance of a typical slice before integrating.

Choosing the Model Carefully

Before evaluating an integral, establish what the chain is doing and where each slice ends. The formulas here assume a uniform chain hanging vertically from the ledge, hauled onto a horizontal surface at ledge height, with no frictional losses. If the destination height or initial arrangement differs, the slice’s lifting distance may differ too, and the integral must represent that geometry.

Keep units consistent. A weight density in newtons per metre multiplied by a slice length in metres gives a force in newtons. Multiplying that force by a lifting distance in metres gives joules. If the data instead give mass density, convert it to weight density using \(w=\mu g\) before using the work formulas.