Why Variable Density Changes the Work Model
A hanging cable does not necessarily have the same mass in every metre. It might be made from materials of different thicknesses, or have a coating that becomes heavier toward one end. When the cable is lifted, each small piece has its own weight, and pieces at different heights travel different distances. The work must account for both variations.
The useful starting point is to label each piece by its position along the cable. Its density tells us how much mass it contains per unit length; its position tells us how far it must be lifted. Multiplying these quantities locally and adding over the whole cable gives the work.
Linear Mass Density and Coordinates
Assume a cable of length \(L\) hangs vertically from a fixed level surface, such as the edge of a platform. We lift it steadily until it is all at the level of that surface. Let \(x\) measure distance down the cable from the surface, so \(x=0\) is the top and \(x=L\) is the bottom. The density \(\lambda(x)\) refers to the material that began at position \(x\).
A segment that began at \(x\) must rise a distance \(x\) to reach the surface. This is the essential geometric fact in the model: the segment’s displacement is not the same for every position. We assume \(\lambda(x)\) is nonnegative and continuous, or at least piecewise continuous, so the accumulated work can be represented by an integral. We also assume constant gravitational acceleration \(g\), a taut inextensible cable, and no energy lost to friction or other effects.
Why the Integral Gives the Work
Consider a small segment near position \(x\), with length \(dx\). Its mass is approximately \(\lambda(x)\,dx\), so its weight is approximately \(g\lambda(x)\,dx\). It rises through distance \(x\). Since work by a constant force in the direction of motion is force times distance, the work on this segment is approximately \(gx\lambda(x)\,dx\).
To justify adding these contributions, partition the interval from \(0\) to \(L\) into small pieces. For a piece of length \(\Delta x_i\) represented by a point \(x_i^*\), its mass is approximately \(\lambda(x_i^*)\Delta x_i\), its weight is approximately \(g\lambda(x_i^*)\Delta x_i\), and its rise is approximately \(x_i^*\). Thus the total work is approximated by the sum
As the largest piece length tends to zero, these sums converge to the definite integral of their continuous (or piecewise continuous) work-per-length function. Therefore
This derivation explains why density alone is not enough: it supplies the mass of each segment, while the factor \(x\) supplies that segment’s lifting distance.
Worked Examples
Worked Example: Density Increases Toward the Bottom
A \(4\)-metre cable hangs from a platform. Its density, measured from the top, is \(\lambda(x)=(2\ \text{kg/m})+(0.5\ \text{kg/m}^2)x\). Find the work to lift it to the platform, using \(g=9.8\ \text{m/s}^2\).
At position \(x\), the segment rises \(x\) metres. Substitute the density and length into the model:
Inside the integral, \(x\lambda(x)\) has numerical expression \(2x+0.5x^2\), with the units supplied by the density and length. Integrate and evaluate:
The integral has units of kilogram-metres, so multiplying by \(g\) gives joules:
Check using the total mass and its average lifting distance. The cable’s mass is \(\int_0^4(2+0.5x)\,dx=8+4=12\) kg. Its mass-weighted distance from the top is \(\frac{1}{12}\int_0^4x(2+0.5x)\,dx=\frac{80}{36}=\frac{20}{9}\) m. Hence the work is \(12(9.8)(20/9)=784/3\) J, agreeing with the integral.
Worked Example: Density Given from the Bottom
A \(3\)-metre cable has density \(\lambda(s)=(1\ \text{kg/m})+(1\ \text{kg/m}^2)s\), where \(s\) is measured upward from the bottom. Find the work to lift the cable to the top, using \(g=9.8\ \text{m/s}^2\).
A segment \(s\) metres from the bottom is \(3-s\) metres below the top, so it must rise \(3-s\) metres. Use \(s\) as the integration variable and include this displacement:
Expand the product, find an antiderivative, and evaluate at both limits:
The integral is \(9\ \text{kg}\cdot\text{m}\), giving
For a check, the total mass is \(\int_0^3(1+s)\,ds=3+4.5=7.5\) kg. The mass-weighted lifting distance is \(9/7.5=1.2\) m. Thus \(7.5(9.8)(1.2)=88.2\) J, as before. The factor \(3-s\), rather than \(s\), is correct because \(s\) is measured from the bottom while the cable is lifted to the top.
Worked Example: A Quadratic Density Profile
A \(2\)-metre cable has density \(\lambda(x)=(1\ \text{kg/m})+(1\ \text{kg/m}^3)x^2\), measured from its top. Find the work to lift it to the top, using \(g=9.8\ \text{m/s}^2\).
Each segment rises \(x\) metres, so
To check, the mass is \(\int_0^2(1+x^2)\,dx=2+8/3=14/3\) kg, and the mass-weighted distance from the top is \(6/(14/3)=9/7\) m. Therefore \(Mg(9/7)=(14/3)(9.8)(9/7)=58.8\) J, matching the computed work.
Choosing the Coordinate Carefully
The coordinate is a bookkeeping choice, but it determines the expression for each segment’s displacement. When measuring down from the top, a segment at \(x\) rises \(x\). When measuring up from the bottom, a segment at \(s\) rises \(L-s\). The density function must use the same coordinate as the integral. If a density is supplied using a different coordinate, either rewrite the density in the chosen coordinate or use the supplied coordinate and express the displacement accordingly.
For instance, if \(s\) is measured from the bottom and the density is \(\lambda(s)\), the corresponding formula is
It is not necessary to change the density to a new variable if this formula directly matches the information given. The key is to pair each piece’s mass with the distance that same piece travels.
A Common Mistake: Giving Every Segment the Full Lift
A tempting but incorrect calculation is to find the cable’s total weight and multiply it by its full length. That treats every segment as though it rose \(L\) metres. In fact, the segment at the top rises almost zero distance, while only the bottom segment rises close to \(L\).
For the first example, the cable’s total mass is \(12\) kg. The incorrect calculation would be
This is too large because it assigns a \(4\)-metre rise to every part of the cable. The correction is to multiply each segment’s weight by its own rise \(x\), then integrate: \(9.8\int_0^4x(2+0.5x)\,dx=784/3\) J. The segment-by-segment model retains the variation in both mass and displacement.
Interpreting the Work Integral
The integral can also be read through the cable’s center of mass. If the total mass is \(M=\int_0^L\lambda(x)\,dx\), then its center of mass is at distance \(\bar{x}=\frac{1}{M}\int_0^Lx\lambda(x)\,dx\) below the top. The work formula becomes \(W=Mg\bar{x}\): the total weight times the rise of the center of mass. This is a useful check, but the direct integral is often the clearest way to model a given density.
Variable-density models are useful whenever the amount of material per unit length changes along an object. The same modeling discipline applies: identify a small piece, determine its mass, determine the distance it travels, and integrate the resulting work contributions. For this cable setup, the entire problem is captured by the product \(x\lambda(x)\), with the coordinate chosen so that the segment’s rise is represented correctly.
Check Your Understanding
For each question, identify the coordinate and the segment displacement before evaluating or interpreting an integral.
- A cable hangs \(5\) metres from a platform, with density \(\lambda(x)\) measured down from the platform. What distance does a segment at position \(x\) rise?
- If density is given as \(\lambda(s)\), where \(s\) is measured up from the bottom of a cable of length \(L\), what work integral models lifting it to the top?
- Why does a short segment of length \(dx\) near \(x\) have mass approximately \(\lambda(x)\,dx\)?
- What physical assumption allows the work of all the cable’s small segments to be added using one gravitational acceleration \(g\)?
- Why does multiplying the cable’s total weight by \(L\) generally overestimate the work?
