Examples

Pumping water from spherical tank

Calculate the work done pumping water out of a spherical tank of radius $R=5\mathrm{m}$.

Solution

(1) Slice the tank of water into horizontal layers:

Coordinate $y$ is $y=0$ at the center of the sphere, increasing upwards.

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(2) Calculate weight of single slice:

$$\begin{array}{cc}\text{area of slice}& =A(y)=\pi r^2\gg \gg \pi (5^2-y^2)\\ & \\ \text{volume of slice}& =dV=A(y)dy\gg \gg \pi (5^2-y^2)dy\\ & \\ \text{weight of slice}& =dF=g\rho dV=\rho g\pi (5^2-y^2)dy\end{array}$$

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(3) Work to lift out single slice:

Distance to raise a slice:

$$h(y)=5-y$$

Then:

$$\text{work to lift out slice}=dW=h(y)dF=\rho g\pi h(y)(5^2-y^2)dy$$

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(4) Total work by integrating $dW$ over all slices:

$$\begin{array}{ccc}& \int dW\gg \gg {\int }_{-5}^{+5}\left(9800\frac{\mathrm{kg}}{{\mathrm{m}}^2{\mathrm{s}}^2}\right)\pi (5^2-y^2)(5-y)dy& \text{(Note A)}\\ & & \\ & \gg \gg \approx 2.6\times {10}^7\mathrm{J}& \end{array}$$

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Note A: The integration runs over all slices, which start at $y=-5$ (bottom of tank), and end at $y=+5$ (top of tank).

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Extra question: what if the spigot sits $2\mathrm{m}$ above the tank?

Extra question: what if the tank starts at just $3\mathrm{m}$ of water depth?

Water pumped from a frustum

Find the work required to pump water out of the frustum in the figure. Assume the weight of water is $\rho =62.5\mathrm{l}\mathrm{b}/\mathrm{f}{\mathrm{t}}^3$.

Solution

(1) Find weight of a horizontal slice.

Coordinate $y=0$ at top, increasing downwards.

Use $r(y)$ for radius of cross-section circle.

Linear decrease in $r$ from $r(0)=6$ to $r(8)=3$:

$$r(y)=6-\frac{3}{8}y$$

Area is $\pi r^2$:

$$\text{Area}(y)=\pi {(6-\frac{3}{8}y)}^2$$

$\text{Weight}=\text{density}\times \text{area}\times \text{thickness}$:

$$\text{weight of layer}=\rho \pi {(6-\frac{3}{8}y)}^{2}dy$$

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(2) Find work to pump out a horizontal layer.

Layer at $y$ is raised a distance of $y$.

Work to raise layer at $y$:

$$\rho \pi y{(6-\frac{3}{8}y)}^{2}dy$$

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(3) Integrate over all layers.

Integrate from top to bottom of frustum:

$$\begin{array}{c}{\int }_0^8\rho \pi y{(6-\frac{3}{8}y)}^{2}dy\gg \gg 528\pi \rho \\ \gg \gg 528\pi \cdot 62.5\gg \gg \approx 1.04\times {10}^5\mathrm{f}\mathrm{t}\text{-}\mathrm{l}\mathrm{b}\end{array}$$

Raising a building

Find the work done to raise a cement columnar building of height $5\mathrm{m}$ and square base $2\mathrm{m}$ per side. Cement has a density of $1500\mathrm{k}\mathrm{g}/{\mathrm{m}}^3$.

Solution

(1) Weight of each layer:

$$\begin{array}{cc}dV& =A(y)dy\gg \gg 4dy\\ & \\ dM& =\rho dV\gg \gg 1500\cdot 4dy\\ & \\ dF& =gdM\gg \gg 9.8\cdot 6000dy\end{array}$$

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(2) Work to lift layer into place:

$$dW=\text{weight}\times \text{distance raised}\gg \gg y\cdot 58800dy$$

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(3) Find total work as integral over the layers:

$$\begin{array}{c}W=\int dW\gg \gg {\int }_0^{5}58800ydy\\ \\ \gg \gg 735\mathrm{kJ}\end{array}$$

Raising a chain

An $80\mathrm{ft}$ chain is suspended from the top of a building. Suppose the chain has weight density $0.5\mathrm{l}\mathrm{b}/\mathrm{f}\mathrm{t}$. What is the total work required to reel in the chain?

Solution

(1) Compute weight of a ‘link’ (vertical slice of the chain):

$$\begin{array}{cc}dF& =\text{density}\times \text{length}\\ & \\ & =0.5dy\end{array}$$

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(2) Work $dW$ to raise link to top:

Each link (slice) is raised from height $y$ to height 80:

$$h(y)=(80-y)\mathrm{ft}$$

Then:

$$dW=(80-y)\cdot 0.5dy$$

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(3) Integrate over the chain for total work:

$$\begin{array}{c}\int dW\gg \gg {\int }_0^{80}(80-y)\cdot 0.5dy\\ \\ \gg \gg 1600\mathrm{f}\mathrm{t}\text{-}\mathrm{l}\mathrm{b}\end{array}$$

Raising a leaky bucket

Suppose a bucket is hoisted by a cable up an $80\mathrm{ft}$ tower. The bucket is lifted at a constant rate of $2\mathrm{f}\mathrm{t}/\mathrm{s}\mathrm{e}\mathrm{c}$ and is leaking water weight at a constant rate of $0.2\mathrm{l}\mathrm{b}/\mathrm{s}\mathrm{e}\mathrm{c}$. The initial weight of water is $50\mathrm{lb}$. What is the total work performed against gravity in lifting the water? (Ignore the bucket itself and the cable.)

Solution

(1) Compute total force from water $F(y)$:

Choose coordinate $y=0$ at base, $y=80$ at top.

Rate of water weight loss per unit height:

$$\begin{array}{c}\frac{\text{rate of leak}}{\text{rate of lift}}=\text{leaked weight per foot}\\ \\ \gg \gg \frac{0.2\mathrm{l}\mathrm{b}/\mathrm{s}\mathrm{e}\mathrm{c}}{2\mathrm{f}\mathrm{t}/\mathrm{s}\mathrm{e}\mathrm{c}}\gg \gg 0.1\mathrm{l}\mathrm{b}/\mathrm{f}\mathrm{t}\end{array}$$

Total water weight at height $y$:

$$F(y)=(50-0.1y)\mathrm{lb}$$

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(2) Work to raise bucket by $dy$:

$$dW=F(y)dy\gg \gg (50-0.1y)dy$$

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(3) Total work by integrating $dW$:

$$\begin{array}{c}W={\int }_0^{80}dW\gg \gg {\int }_0^{80}(50-0.1y)dy\\ \\ \gg \gg 50y-0.05y^2{|}_0^{80}\gg \gg 3680\mathrm{f}\mathrm{t}\text{-}\mathrm{l}\mathrm{b}\end{array}$$

Change of method and integral formula!

For this example, we use the formula $\int F(y)dy$ rather than the formula $\int h(y)dF$ used in the earlier examples.

• This integral sums over the work $dW$ to lift macroscopic material through each microscopic $dy$ as if in sequence, and $dy$ thus represents distance lifted.
• Earlier examples summed over the work $dW$ to lift microscopic material through the macroscope $h(y)$ (all the way up).