Introduction
Raising a bucket from a well requires work against gravity. The calculation becomes more interesting when the bucket loses water as it rises: its weight then decreases during the lift. If the supporting rope has mass, the rope’s contribution changes too, because less of it hangs in the well as the bucket approaches the top.
The central modelling step is to express the force that must be overcome at each position. Once that force is written as a function of height, the work is its integral over the bucket’s upward displacement. The integral accounts for the entire lift, including changes in the bucket’s load and in the length of hanging rope.
Choose a Position Variable
Let \(H\) be the depth of the bucket below the well opening at the start of the lift. Measure height \(y\), in metres, upward from the bucket’s starting position. Thus \(y=0\) at the start and \(y=H\) when the bucket reaches the opening.
Let \(m_b\) be the bucket’s mass, in kilograms, and let \(m_w(y)\) be the mass of water still in the bucket when it has risen a distance \(y\). The water mass function must satisfy \(m_w(y)\ge 0\) over the lift. It may be constant if there is no leak, or decrease with \(y\) if water escapes as the bucket rises.
Suppose the rope is uniform, has mass density \(\mu\) kilograms per metre, and hangs vertically between the well opening and the bucket. At height \(y\), the hanging length is \(H-y\), so its mass is \(\mu(H-y)\). This formula applies to the portion of rope still hanging in the well, not to the rope’s original full length at every position.
For work in SI units, use \(g=9.8\ \text{m/s}^2\), unless a problem specifies another value. If the rope’s mass is negligible, set \(\mu=0\). If the bucket does not leak, \(m_w(y)\) is constant.
Why the Work Is an Integral
Over a small upward displacement \(\Delta y\), the force is approximately the force at that position, and the work is approximately \(F(y)\Delta y\). Adding these small contributions over the lift gives sums of the form \(\sum F(y_i)\Delta y_i\). As the displacements become arbitrarily small, these sums approach the definite integral of the force.
This model describes the work against gravity during the lift. With leaking water, the mass used at each position is the mass still in the bucket at that position. Water that has already escaped is no longer part of the lifted load. The model assumes that the draining process does not introduce significant extra effects such as drag or kinetic energy that must also be included.
Worked Example: A Bucket Losing Water at a Constant Rate per Metre
Worked Example: A Bucket Losing Water at a Constant Rate per Metre
A bucket of mass \(12\ \text{kg}\) starts \(10\ \text{m}\) below the well opening. It initially contains \(30\ \text{kg}\) of water. As it rises, it loses water at a rate of \(2\ \text{kg}\) per metre. Ignore the rope’s mass. Find the work required to raise the bucket to the opening.
Use \(y\) for the distance risen. The remaining water mass is $$ m_w(y)=30-2y,\qquad 0\le y\le 10. $$ At the top, \(m_w(10)=30-20=10\ \text{kg}\), so the water mass stays nonnegative throughout the lift. With \(\mu=0\), the force is $$ F(y)=9.8\bigl(12+30-2y\bigr)=9.8(42-2y)\ \text{N}. $$ Therefore, $$ \begin{aligned} W &=\int_0^{10}9.8(42-2y)\,dy\\ &=9.8\bigl[42y-y^2\bigr]_0^{10}\\ &=9.8\bigl((420-100)-0\bigr)\\ &=9.8(320)\\ &=3136\ \text{J}. \end{aligned} $$ To check the model, the force starts at \(9.8(42)=411.6\ \text{N}\) and ends at \(9.8(22)=215.6\ \text{N}\), as expected when water is lost. Differentiating the antiderivative \(42y-y^2\) gives \(42-2y\), the factor in the force integral. The work is \(3136\ \text{J}\).
Worked Example: Water Loss That Depends on Height
Worked Example: Water Loss That Depends on Height
A \(10\ \text{kg}\) bucket starts \(8\ \text{m}\) below the opening. Its water mass at height \(y\) is \(m_w(y)=24(1-y/8)^2\ \text{kg}\), and the rope’s mass is negligible. Find the work for the lift.
The water mass is \(24\ \text{kg}\) at \(y=0\) and \(0\ \text{kg}\) at \(y=8\). It is nonnegative throughout the interval because it is a positive constant times a square. The required force and work are $$ F(y)=9.8\left(10+24\left(1-\frac{y}{8}\right)^2\right) $$ and $$ W=9.8\int_0^8\left(10+24\left(1-\frac{y}{8}\right)^2\right)\,dy. $$ Expand the square to evaluate the integral: $$ \left(1-\frac{y}{8}\right)^2 =1-\frac{y}{4}+\frac{y^2}{64}. $$ An antiderivative of this expression is \(y-y^2/8+y^3/192\). Hence $$ \begin{aligned} W &=9.8\left(80+ 24\left[y-\frac{y^2}{8}+\frac{y^3}{192}\right]_0^8\right)\\ &=9.8\left(80+24\left(8-8+\frac{512}{192}\right)\right)\\ &=9.8\left(80+24\cdot\frac{8}{3}\right)\\ &=9.8(80+64)\\ &=9.8(144)\\ &=1411.2\ \text{J}. \end{aligned} $$ For a check, differentiating \(y-y^2/8+y^3/192\) gives \(1-y/4+y^2/64\), which is the expanded square. The bucket contributes \(10\cdot8=80\ \text{kg}\cdot\text{m}\) inside the integral, and the water contributes \(64\ \text{kg}\cdot\text{m}\), for a total of \(144\ \text{kg}\cdot\text{m}\). Multiplication by \(g=9.8\ \text{m/s}^2\) gives the stated work.
Including the Weight of the Rope
For a uniform rope, the amount of rope hanging in the well decreases from \(H\) metres to zero. Its weight is therefore not a constant force during the lift. At position \(y\), its contribution to the force is \(g\mu(H-y)\), which decreases linearly as the bucket rises.
The rope’s total work can be separated from the bucket’s and water’s work by splitting the integral: $$ W =g\int_0^H\bigl(m_b+m_w(y)\bigr)\,dy +g\mu\int_0^H(H-y)\,dy. $$ The second integral is $$ \int_0^H(H-y)\,dy =\left[Hy-\frac{y^2}{2}\right]_0^H =\frac{H^2}{2}. $$ Thus, the work against the weight of a uniform hanging rope is \(g\mu H^2/2\). This term reflects that each piece of rope rises by a different distance: pieces near the bucket rise farther than pieces near the opening.
Worked Example: A Leaking Bucket and a Uniform Rope
Worked Example: A Leaking Bucket and a Uniform Rope
A \(8\ \text{kg}\) bucket starts \(12\ \text{m}\) below the opening. It contains \(24\ \text{kg}\) of water initially, and the remaining water mass is \(m_w(y)=24-1.5y\ \text{kg}\). The uniform rope has mass density \(\mu=1.5\ \text{kg/m}\). Find the work required to raise the bucket to the opening.
The water mass decreases from \(24\ \text{kg}\) to \(24-1.5(12)=6\ \text{kg}\), so it remains nonnegative. The hanging rope length is \(12-y\), and its mass is \(1.5(12-y)\ \text{kg}\). The total force is $$ \begin{aligned} F(y) &=9.8\bigl(8+(24-1.5y)+1.5(12-y)\bigr)\\ &=9.8(8+24+18-3y)\\ &=9.8(50-3y)\ \text{N}. \end{aligned} $$ Integrating gives $$ \begin{aligned} W &=9.8\int_0^{12}(50-3y)\,dy\\ &=9.8\left[50y-\frac{3y^2}{2}\right]_0^{12}\\ &=9.8\left(600-\frac{3(144)}{2}\right)\\ &=9.8(600-216)\\ &=9.8(384)\\ &=3763.2\ \text{J}. \end{aligned} $$ The antiderivative differentiates to \(50-3y\), as required. Also, the initial force is \(9.8(50)=490\ \text{N}\). At the opening, no rope is hanging and \(6\ \text{kg}\) of water remains, so the final force is \(9.8(8+6)=137.2\ \text{N}\), matching \(9.8(50-3(12))\). As a further check, the bucket contributes \(8(12)=96\ \text{kg}\cdot\text{m}\); the water contributes \(\int_0^{12}(24-1.5y)\,dy=288-108=180\ \text{kg}\cdot\text{m}\); and the rope contributes \(1.5(12^2/2)=108\ \text{kg}\cdot\text{m}\). Their sum is \(96+180+108=384\ \text{kg}\cdot\text{m}\), which gives the same work after multiplication by \(9.8\ \text{m/s}^2\).
A Common Mistake: Treating the Rope’s Weight as Constant
A tempting but incorrect model is to take the rope’s initial mass, \(\mu H\), and count its full weight over the entire lift. That would give a rope work of \(g\mu H\cdot H=g\mu H^2\). It overestimates the work because the hanging rope gets shorter as the bucket rises.
For example, for a \(10\ \text{m}\) lift with \(\mu=2\ \text{kg/m}\), the incorrect constant-force calculation gives $$ 9.8(2\cdot10)(10)=1960\ \text{J}. $$ The correct calculation uses the changing hanging length: $$ 9.8\int_0^{10}2(10-y)\,dy =9.8\cdot2\left[10y-\frac{y^2}{2}\right]_0^{10} =9.8(100) =980\ \text{J}. $$ The correct value is half the incorrect one because the hanging length decreases linearly from \(10\) metres to zero. A related error is to keep the water mass at its initial value even when the problem says water leaks. Always use the mass present at the particular height being considered.
Setting Up the Model Reliably
Before evaluating an integral, check that the variables describe the physical situation. In this model, \(y\) measures upward displacement from the starting point, so the bounds are \(0\) and \(H\). The hanging rope length is \(H-y\), not \(y\). The remaining water mass must describe the bucket at height \(y\), and should not become negative on the interval.
Set \(y=0\) at the bucket’s initial position and \(y=H\) at the opening.
Write \(m_w(y)\) for water remaining and, for a uniform rope, \(\mu(H-y)\) for the hanging rope.
Multiply the total mass being supported at height \(y\) by \(g\).
Calculate \(\int_0^H F(y)\,dy\), keeping units consistent.
Confirm that the endpoint forces match the physical loads and that the antiderivative differentiates to the integrand.
Why This Model Is Useful
The integral keeps track of changing loads without pretending that one force acts throughout the whole lift. It can accommodate different leak patterns by changing \(m_w(y)\), and it can include or omit the rope by choosing an appropriate value of \(\mu\). The essential principle is to describe the load at each position first, then accumulate the work over the displacement.
Check Your Understanding
For each question, consider how the force depends on the bucket’s height before attempting any calculation.
- A bucket starts \(6\ \text{m}\) below the opening with \(15\ \text{kg}\) of water and loses \(1\ \text{kg}\) per metre of lift. What function describes the water mass at height \(y\), and what are its values at the start and top?
- A uniform rope has mass density \(\mu\) and a hanging length \(H\) initially. What is its hanging mass when the bucket has risen \(y\) metres?
- Why does using the rope’s initial weight as a constant force over the entire lift overestimate the rope’s work?
- If a bucket has mass \(m_b\), retains water mass \(m_w(y)\), and is attached to a uniform rope of density \(\mu\), write the work integral for a lift through height \(H\).
- For the integral in the previous question, what are the units of the force integrand and of the final work?
