W05 - Examples

Fluid force on a triangular plate

Fluid force on a triangular plate

Find the total force on the submerged vertical plate with the following shape: Equilateral triangle, sides $2\mathrm{m}$, top vertex at the surface, liquid is oil with density $\rho =900kg/{\mathrm{m}}^3$.

Solution

(1) Write the width function:

Establish coordinate system: $y=0$ at water line (also the vertex), and $y$ increases going down.

Method 1: Geometry of similar triangles

Top triangle with base at $w(y)$ is similar to total triangle with base $\sqrt{3}$.

Therefore, corresponding parts have the same ratios.

Therefore:

$$\begin{array}{c}\frac{w(y)}{y}=\frac{2}{\sqrt{3}}\\ \\ \gg \gg w(y)=\frac{2}{\sqrt{3}}y\end{array}$$

Method 2: Quick linear interpolation function

$$\begin{array}{c}w(y)=0+\frac{2-0}{\sqrt{3}}\cdot y\\ \\ \gg \gg w(y)=\frac{2}{\sqrt{3}}y\end{array}$$

Generalized “quick linear interpolation function”

Generalization:

$$w(y)=e_1+\frac{e_2-e_1}{\text{Height}}(\pm y-a)$$

where:

• $e_1$ is edge 1 and $e_2$ is edge 2 (increasing order of $y$)
• $\pm y$ is $+y$ when $e_1$ comes earlier (smaller $y$), and $-y$ if it comes later
• $a$ is created to force the quantity $(\pm y-a)$ to equal $0$ for the given $y$ value at $e_1$

---

(2) Compute integral using width function:

Bounds:

• $y=0$ (shallow)
• $y=\sqrt{3}$ (deep)

Integral formula:

$$\begin{array}{c}F=\rho g{\int }_0^{\sqrt{3}}yw(y)dy\\ \\ \gg \gg 900\cdot 9.8{\int }_0^{\sqrt{3}}y\left(\frac{2y}{\sqrt{3}}\right)dy\gg \gg 10184.5{\int }_0^{\sqrt{3}}{y}^{2}dy\\ \\ \gg \gg 10184.5\frac{y^3}{3}{|}_0^{\sqrt{3}}\gg \gg 17640\end{array}$$ Link to original

Weight of water on a dam

Weight of water on angled dam

Find the total hydrostatic force on an angled dam with the following geometric description: Tilted trapezoid. Base $=2,000\mathrm{m}$, Top $=3,000\mathrm{m}$, and vertical height $185\mathrm{m}$. The base is tilted at an angle of $\theta =55^{\circ }$.

Solution

(1) Write the width function:

Establish coordinate system: $y=0$ at water line (also the top edge), and increases going down.

“Quick linear interpolation function”:

$$\begin{array}{c}w(y)=3000+\frac{2000-3000}{185}y\\ \\ \gg \gg 3000-\frac{1000}{185}y\end{array}$$

---

(2) Incorporate angle of incline in strip thickness:

$$dz=\csc 55^{\circ }dy$$

So the area of a strip is:

$$\begin{array}{c}dA=w(y)dz\\ \\ \gg \gg \left(3000-\frac{1000}{185}y\right)\csc 55^{\circ }dy\end{array}$$

---

(3) Compute total force using integral formula.

Plug data into formula:

$$\begin{array}{c}F=\rho g{\int }_a^{b}h(y)w(y)dz\\ \\ \gg \gg \rho g\csc 55^{\circ }{\int }_0^{185}y\left(3,000-\frac{1,000}{185}y\right)dy\\ \\ \gg \gg 4.777\times 10^{11}\end{array}$$Link to original

Pumping water from spherical tank

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.

---

(2) Calculate weight of single slice:

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

---

(3) Work to lift out single slice:

Distance to raise a slice:

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

Then:

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

---

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

$$\begin{array}{c}\int dW\gg \gg {\int }_{-5}^{+5}\left(9800\frac{\mathrm{kg}}{{\mathrm{m}}^2{\mathrm{s}}^2}\right)\pi (5^2-y^2)(5-y)dy\\ \\ \gg \gg \approx 2.6\times 10^7\mathrm{J}\end{array}\text{\hspace{1em}(Note A)}$$

---

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

---

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?

Link to original

Water pumped from a frustum

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 $\rho =62.5lb/f{\mathrm{t}}^3$.

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 $\pi r^2$:

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

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

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

---

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

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

Work to raise layer at $y$:

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

---

(3) Integrate over all layers.

Integrate from top to bottom of frustum:

$$\begin{array}{c}{\int }_0^8\rho \pi y{\left(6-\frac{3}{8}y\right)}^{2}dy\gg \gg 528\pi \rho \\ \gg \gg 528\pi \cdot 62.5\gg \gg \approx 1.04\times 10^5\mathrm{ft}\text{-}\mathrm{lb}\end{array}$$ Link to original

Raising a building

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 $1500kg/{\mathrm{m}}^3$.

Solution

(1) Weight of each layer:

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

---

(2) Work to lift layer into place:

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

---

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

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

Raising a chain

Raising a chain

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

Solution

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

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

---

(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:

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

---

(3) Integrate over the chain for total work:

$$\begin{array}{c}\int dW\gg \gg {\int }_0^{80}(80-y)\cdot 0.5dy\\ \\ \gg \gg 1600\mathrm{ft}\text{-}\mathrm{lb}\end{array}$$ Link to original

Raising a leaky bucket

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 $2ft/sec$ and is leaking water weight at a constant rate of $0.2lb/sec$. 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:

$$\begin{array}{c}\frac{\text{rate of leak}}{\text{rate of lift}}=\text{leaked weight per foot}\\ \\ \gg \gg \frac{0.2lb/sec}{2ft/sec}\gg \gg 0.1lb/ft\end{array}$$

Total water weight at height $y$:

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

---

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

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

---

(3) Total work by integrating $dW$:

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

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).

Link to original