Pumping Work and Lifting Problems
When Every Drop Travels a Different Distance
Imagine emptying a swimming pool by pumping water over the edge. Water at the bottom must travel farther than water near the top. A cable hanging from a building gets lighter as you pull it up. In these problems, different parts of the object travel different distances: that’s what makes them interesting.
The key insight: slice the object into thin horizontal layers, figure out how much work each layer requires, and integrate.
Prerequisite Map
Before You Start
Quick self-check. These problems test the prerequisites. If you struggle, review before proceeding.
1. Work Integral Setup: A 30-ft chain weighing 2 lb/ft hangs from a ceiling. If a piece at depth $x$ (measured from top) must be lifted distance $x$, what integral gives the total work?
Check Your Answer
$$W = \int_0^{30} 2x\,dx$$
Each small piece has weight $(2\,dx)$ lb and travels distance $x$, so $dW = 2x\,dx$.
✅ Got it? You understand the work-distance setup!
❌ Stuck? Review Work with Integrals: the chain problem there is essential background.
2. Similar Triangles: A cone has radius 4 m at the top and height 10 m (vertex at bottom). At depth $x$ from the top, what is the radius $r(x)$?
Check Your Answer
Using similar triangles: $\displaystyle\frac{r}{10-x} = \frac{4}{10}$, so $r(x) = \frac{4(10-x)}{10} = \frac{2(10-x)}{5}$
At depth 0 (top): $r = 4$ m ✓ At depth 10 (bottom): $r = 0$ ✓
✅ Got it? Similar triangles are key for conical tanks!
❌ Stuck? This geometry is essential. Draw the cone and label the triangles. The ratio of radius to “height from vertex” is constant.
3. Cross-Sectional Area: For a cone with $r(x) = \frac{2(10-x)}{5}$, what is the cross-sectional area $A(x)$?
Check Your Answer
$$A(x) = \pi r^2 = \pi \left(\frac{2(10-x)}{5}\right)^2 = \frac{4\pi(10-x)^2}{25}$$
✅ Got it? You’re ready for tank problems!
❌ Made an error? Remember: area of a circle is $\pi r^2$, and you must square the entire expression for $r$.
Quick Reference
| Property | Value |
|---|---|
| Section | Chapter 5, Section 4 |
| Difficulty | Advanced |
| Time | ~25 minutes |
Key Concepts
The Pumping Work Strategy
For pumping fluid from a tank or lifting distributed weight:
- Set up coordinates: usually $x$ pointing down from top of tank
- Slice: divide into thin horizontal layers of thickness $dx$
- For each slice, find:
- Volume or length
- Weight (mass × gravity, or weight per unit length)
- Distance it must travel
- Write work for one slice: $dW = (\text{weight}) \times (\text{distance})$
- Integrate over all slices
General Formula for Pumping
$$\boxed{W = \int_a^b \rho g \cdot A(x) \cdot d(x) \, dx}$$
where:
- $\rho$ = fluid density (kg/m³ or lb/ft³)
- $g$ = gravitational acceleration (9.8 m/s² in SI, or use $\rho g$ directly as weight density)
- $A(x)$ = cross-sectional area at depth $x$
- $d(x)$ = distance this slice must be lifted
- $dx$ = thickness of slice
Common Tank Shapes
| Shape | Cross-Section $A(x)$ | Key Geometry |
|---|---|---|
| Rectangular | $A = \ell \cdot w$ (constant) | Length × width |
| Cylindrical | $A = \pi r^2$ (constant) | Circular cross-section |
| Conical | $A = \pi [r(x)]^2$ | Radius varies with depth |
| Spherical | $A = \pi [r(x)]^2$ | Use $r^2 = R^2 - x^2$ |
| Trapezoidal | $A = w(x) \cdot \ell$ | Width varies linearly |
Setting Up Coordinates
Option 1: Origin at top (x pointing down)
- Water at depth $x$ travels distance $x$ to reach top
- Common for “pump to top of tank” problems
x=0 ───────── ← top (outlet)
│~~~~~│
x │~~~~~│ ← slice at depth x travels distance x
│~~~~~│
x=h ───────── ← bottomOption 2: Origin at bottom (y pointing up)
- Water at height $y$ travels distance $(h - y)$ to reach top at height $h$
- Common when tank shape is easier to describe from bottom
Density Reference
| Substance | SI (kg/m³) | US (lb/ft³) |
|---|---|---|
| Water | 1000 | 62.5 |
| Seawater | 1025 | 64 |
| Oil (typical) | 800-900 | 50-56 |
Weight density: In SI, multiply by $g = 9.8$ m/s² to get N/m³. In US, pounds are already weight, so 62.5 lb/ft³ can be used directly.
Cable/Chain Lifting
For lifting a cable or chain of weight $w$ lb/ft (or density $\rho$ kg/m):
$$W = \int_0^L w \cdot x \, dx \quad \text{or} \quad W = \int_0^L \rho g \cdot x \, dx$$
where $x$ = distance from top of cable (so section at $x$ must travel distance $x$).
Common Pitfalls
| Mistake | Why It Happens | How to Avoid |
|---|---|---|
| Wrong distance formula | Confusing “depth” with “distance to travel” | Draw the tank. Label the outlet. Distance = how far the slice moves to reach the outlet. |
| Forgetting $g$ in SI units | Using 1000 kg/m³ as weight | Density (kg/m³) is mass. Multiply by $g = 9.8$ to get weight density. In US units (lb/ft³), this is already done. |
| Similar triangles error | Getting radius formula backwards | At the TOP of an inverted cone, radius is maximum. Check: your formula should give $r = R$ at top, $r = 0$ at vertex. |
| Wrong integration limits | Integrating over the whole tank when it’s not full | Read carefully: “filled to depth 6 m” means water surface is 6 m from the bottom. Convert to your coordinate system. |
| Outlet not at tank top | Assuming water pumps to the rim | If outlet is 2 m ABOVE the tank, add 2 to every distance. Slice at depth $x$ travels $(x + 2)$, not just $x$. |
| Forgetting to square $r(x)$ | Writing $A = \pi r$ instead of $A = \pi r^2$ | Always double-check: area of circle = $\pi r^2$. |
Historical Note: The Steam Engine and Work
The concept of “work” in physics was developed largely to analyze steam engines. In 1824, Sadi Carnot studied how much useful work could be extracted from a given amount of heat, launching thermodynamics. James Watt (of watt fame) measured his engines’ output by how many horses they could replace, giving us “horsepower” (1 hp = 550 ft-lb/s = 746 W). Pumping work calculations were literally how engineers compared engine efficiency in the Industrial Revolution.
Still confused about coordinate setup? The key choice is where to put the origin. Placing it at the outlet (where water exits) often makes the distance formula simplest. Review Work with Integrals for the foundation.
Practice Problems
A rectangular tank is 3 m long, 2 m wide, and 1.5 m deep. It is filled with water (density 1000 kg/m³). How much work is required to pump all the water to the top of the tank?
A 40-foot chain weighing 4 lb/ft hangs vertically from the top of a building. How much work is done in pulling the entire chain to the top?
A tank in the shape of an inverted cone (point at bottom) has height 8 m and top radius 3 m. It is filled with water to a depth of 6 m. Find the work required to pump all the water out over the top rim. Use $\rho = 1000$ kg/m³ and $g = 9.8$ m/s².
An 80-ft rope weighing 1.5 lb/ft hangs from the top of a well, with a 35-lb bucket of water attached at the bottom.
- How much work is required to lift the bucket 25 ft? (The rope winds onto a spool as you lift.)
- How much work is required to lift the bucket all the way to the top?
- What percentage of the total work (part b) is done in the first 25 ft (part a)?
A hemispherical tank of radius 4 m is full of water. The water must be pumped to a point 2 m above the top of the tank. Set up and evaluate the integral for the work required.
Use $\rho = 1000$ kg/m³ and $g = 10$ m/s² for simpler arithmetic.
Conceptual Questions (CCI-Style)
Tank A is a cylinder of height $h$ and radius $r$. Tank B is an inverted cone of height $h$ and top radius $r$. Both are filled with water. Which requires more work to empty by pumping to the top?
- Tank A (cylinder)
- Tank B (cone)
- They require the same work
- Cannot determine without knowing $r$ and $h$
Common Misconceptions
the total work to pump fluid from a tank equals the total weight of the fluid times the height of the tank.
This is the rate-as-fixed-number error. The formula $W = (\text{total weight}) \times h$ assumes every particle of fluid travels the full height $h$, but in reality only the fluid at the bottom travels that far. Fluid near the top barely moves. For a rectangular tank of height 1.5 m filled with water, $W = F \cdot d$ using total weight $\rho g A h$ times $h = 1.5$ m gives double the correct answer; the correct work $\int_0^{1.5} \rho g A x\,dx = \frac{1}{2}\rho g A h^2$ shows that the factor of $\frac{1}{2}$ arises because the fluid is distributed over the full depth with each layer traveling a different distance.
Mastery Checklist
Mental Model
The Bucket Brigade:
Imagine emptying a tank using a human chain passing buckets. Each bucket at the bottom must travel the full height. Buckets near the top barely need to move. The total work is the sum of all individual bucket-lifts.
For a cone that’s wide at the top, most buckets start near the top and travel short distances, giving less total work. For a cylinder, buckets are evenly distributed, so more buckets travel the full distance.
The integral $\int \rho g \cdot A(x) \cdot x \, dx$ is just adding up all those bucket-lifts: weight of slice × distance it travels.
Fun Fact: Pumping the Thames
London’s Thames Water pumps about 2.6 billion liters of water daily to supply the city. If we assume an average pumping height of 50 meters, that’s roughly: $$W = (2.6 \times 10^9 \text{ kg})(9.8)(50) \approx 1.3 \times 10^{12} \text{ J/day}$$
That’s about 15 megawatts of continuous power, just for pumping! Municipal water systems spend 2-3% of a nation’s total electricity on pumping water uphill to reservoirs and homes.
Connections
Looking back:
- Work with Integrals: the general framework
- Spring Work: simpler variable force problems
- Volumes by slicing: same “slice and integrate” approach
Looking ahead:
- Hydrostatic pressure uses similar slicing techniques
- Center of mass involves weighted averages over slices
Real-world connections:
- Municipal water systems calculate pumping costs
- Oil extraction requires lifting fluid from deep wells
- Elevator motor sizing depends on cable-lifting work
| Previous | Up | Next |
|---|---|---|
| Spring Work | Ch5 Sec4 Skills | Section Overview |
Last updated: 2026-01-23