Why a Hanging Cable Requires an Integral
A rigid object lifted vertically moves as one piece: every part rises the same distance. A hanging cable behaves differently. The part initially closest to the edge travels only a short distance before reaching the platform, while parts farther down must travel farther. To calculate the work, we must represent this changing lift distance by position.
We will model a cable hanging vertically over an edge and being pulled onto a platform. The cable is inextensible, gravity is constant, and the lift is slow enough that changes in kinetic energy can be neglected. We also ignore friction and the work needed to bend the cable over the edge. Under these assumptions, the work supplied by the lifting force equals the increase in the cable’s gravitational potential energy.
Position and Lift Distance
Let the cable have length \(L\). Set \(x=0\) at the edge of the platform, and measure \(x\) vertically downward along the cable. Thus a small segment initially at position \(x\) is \(x\) metres below the platform. When the entire cable is lifted onto the platform, that segment rises a distance \(x\).
This is the key geometric fact: the segment’s lifting distance is not one fixed number for the whole cable. It is the position variable \(x\). We will first consider a cable whose weight is spread uniformly along its length.
Deriving the Work Integral for a Uniform Cable
Consider a short segment of length \(\Delta x\) whose initial position is approximately \(x\). Its weight is approximately \(w\Delta x\), and in a complete lift its displacement is approximately \(x\). Since work against gravity for a vertical rise is weight multiplied by rise, the work for this segment is approximately \(wx\Delta x\).
Partition the cable into shorter and shorter segments and add their work contributions. The sums approach the definite integral of the work contributions along the cable. The integral accounts for the fact that different positions have different lift distances.
Because \(w\) is constant for a uniform cable, it can be taken outside the integral. Evaluating gives a compact formula.
The formula can also be checked using the cable’s total weight and its centre. Its total weight is \(wL\), and its midpoint begins \(L/2\) below the edge. Thus total weight multiplied by midpoint rise is \(wL(L/2)=wL^2/2\), agreeing with the integral. The integral derivation is more flexible: it still works when different parts have different weights per unit length.
Worked Example: Lifting a Uniform Cable
Worked Example: Lifting a Uniform Cable
A 12-metre cable hangs vertically over a platform. Its linear weight density is \(15\ \text{N/m}\). Find the work required to lift the whole cable onto the platform.
Let \(x\) be the initial depth below the platform. A segment at depth \(x\) rises \(x\) metres, so the work integral is:
Evaluate the integral, retaining the units:
For a check, the cable’s total weight is \((15\ \text{N/m})(12\ \text{m})=180\ \text{N}\). Its midpoint rises \(12/2=6\ \text{m}\), giving \(180\ \text{N}\cdot6\ \text{m}=1080\ \text{J}\), as before.
Cables Made of Sections with Different Densities
A cable need not be uniform throughout. One useful model is a cable made from sections, each with its own constant linear weight density. The position coordinate remains the same: a segment initially at depth \(x\) still rises \(x\) metres when the whole cable is lifted. What changes is the weight per unit length in the work integrand.
Suppose the linear weight density at position \(x\) is \(w(x)\), where \(w(x)\) is constant on each of a finite number of intervals. A short segment at \(x\) has weight approximately \(w(x)\Delta x\), so its work contribution is approximately \(w(x)x\Delta x\). Integrating over the sections gives the total work. Equivalently, integrate separately on each section and add the results.
For a piecewise-constant density, the integral is evaluated piece by piece. The bounds for a section must match that section’s position along the cable. In particular, a lower section is lifted farther than an upper section, so it cannot be assigned the upper section’s lifting distance.
Worked Example: A Cable with Two Uniform Sections
Worked Example: A Cable with Two Uniform Sections
A 10-metre cable hangs over a platform. From the edge to a depth of 4 metres, its linear weight density is \(8\ \text{N/m}\). From a depth of 4 metres to a depth of 10 metres, its density is \(12\ \text{N/m}\). Find the work required to lift the entire cable onto the platform.
Use \(x=0\) at the edge. Each part at depth \(x\) rises \(x\) metres. The first section occupies \(0\le x\le4\); the second occupies \(4\le x\le10\). Therefore:
Evaluate the first section’s contribution:
Evaluate the second section’s contribution:
Add the section contributions:
Check by treating each section separately. The first section weighs \((8\ \text{N/m})(4\ \text{m})=32\ \text{N}\), and its midpoint is 2 metres below the edge, so its work is \(32\cdot2=64\ \text{J}\). The second section weighs \((12\ \text{N/m})(6\ \text{m})=72\ \text{N}\), and its midpoint is 7 metres below the edge, so its work is \(72\cdot7=504\ \text{J}\). Their sum is \(568\ \text{J}\), matching the integral.
When the Cable Is Lifted Only Partway
The same position-based method also describes a partial lift. Suppose the free end is pulled upward a distance \(h\), where \(0<h<L\). A segment initially at depth \(x<h\) reaches the platform after rising \(x\) metres. A segment initially at depth \(x\ge h\) has not yet reached the platform and rises \(h\) metres. Thus the rise distance is the smaller of \(x\) and \(h\).
For a uniform cable with linear weight density \(w\), split the integral where the expression for the rise changes. The upper part has \(0\le x\le h\); the lower part has \(h\le x\le L\).
This formula assumes the cable is pulled over the edge and laid on the platform, and that the portion still hanging moves upward by \(h\). The split in the integral is essential: the two portions have different displacements.
Worked Example: Partially Lifting a Uniform Cable
Worked Example: Partially Lifting a Uniform Cable
An 8-metre cable has uniform linear weight density \(10\ \text{N/m}\). The free end is pulled upward 3 metres. Find the work done against gravity.
For \(0\le x\le3\), the segment rises \(x\) metres. For \(3\le x\le8\), it rises 3 metres. Therefore:
The part that reaches the platform contributes:
The part that remains hanging contributes:
The total work is:
As a check, the first 3 metres weigh \(30\ \text{N}\) and their midpoint rises \(1.5\ \text{m}\), giving \(45\ \text{J}\). The remaining 5 metres weigh \(50\ \text{N}\) and rise 3 metres, giving \(150\ \text{J}\). The total is again \(195\ \text{J}\).
A Common Mistake: Giving Every Part the Same Lift Distance
A frequent error is to multiply the cable’s total weight by its full length, as if every part of the cable rose by that full length. For the 12-metre cable in the first example, that incorrect calculation would be:
This is wrong because the upper end starts at the edge and travels almost no distance, while the bottom travels 12 metres. The cable does not move as a rigid object. For a uniform cable, its average rise is \(L/2\), not \(L\), which is why the correct calculation is \(180\ \text{N}\cdot6\ \text{m}=1080\ \text{J}\). More generally, use the displacement appropriate to each position and integrate the segment contributions.
Check Your Understanding
Use the position of each segment to decide its lifting distance before setting up an integral.
- A uniform 6-metre cable has linear weight density \(5\ \text{N/m}\). Set up the integral for lifting the entire cable, and evaluate it.
- For a full lift, why is the lifting distance of a segment at depth \(x\) equal to \(x\), rather than the full cable length?
- A cable has density \(7\ \text{N/m}\) on \(0\le x\le2\) and \(11\ \text{N/m}\) on \(2\le x\le5\). Write the work integral for lifting it completely.
- A uniform cable of length \(L\) is pulled upward by \(h\), where \(0<h<L\). State the displacement of a segment at depth \(x\) in each of the ranges \(0\le x<h\) and \(h\le x\le L\).
- Explain why multiplying the total cable weight by \(L\) does not give the work for a full lift.
