W05 Regular

Due date: Sunday 2/15, 11:59pm

Hydrostatic pressure

01

02

Fluid force on a parabolic plate

A parabolic plate is submerged vertically in water as in the figure:

The shape of the plate is bounded below by $y=x^2$ and above by the line $y=1$. (Note that $y$ increases going up in this coordinate system.)

Compute the total fluid force on this plate.

(Hint: your integrand should contain $(1-y)$ as a factor.)

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Solution

Solutions - 0100-02

(1) Integral formula:

$$F={\int }_a^b\rho gh(y)w(y)dy$$

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(2) Integrand components:

$$w(y)=2\sqrt{y},h(y)=1-y$$

So we have:

$$F=\rho g{\int }_0^{1}2\sqrt{y}(1-y)dy$$

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(3) Integrate:

$$\begin{array}{c}\gg \gg \rho g{\int }_0^{1}2y^{1/2}-2y^{3/2}dy\\ \\ \gg \gg {\rho g(\frac{4}{3}{y}^{3/2}-\frac{4}{5}{y}^{5/2})|}_0^1\\ \\ \gg \gg \rho g\frac{8}{15}\gg \gg 5227\mathrm{k}\mathrm{g}\mathrm{m}/{\mathrm{s}}^2\end{array}$$

(Assuming $\rho =1000$ and $g=9.8$.)

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02 $★$

03

Fluid force on triangular plates

For diagrams 1, 2, 3 (L to R) below, set up an integral to compute the hydrostatic force on the plate.

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Solution

Solutions - 0100-03

(1) Integral formula:

$$F={\int }_a^b\rho gh(y)w(y)dy$$

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Option 1:

(2) Using $y=0$ at water line, increasing downwards: (a) Left:

$$F={\int }_0^4\rho gy(3+\frac{0-3}{4}y)dy$$

(b) Center:

$$F={\int }_2^5\rho gy(0+\frac{2-0}{3}(y-2))dy$$

(c) Right:

$$F={\int }_0^4\rho gy(4+\frac{0-4}{5}(y+1))dy$$

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Option 2:

(2) Using $y=0$ at top of shape, increasing downwards: (a) Left:

$$F={\int }_0^4\rho gy(3+\frac{0-3}{4}y)dy$$

(b) Center:

$$F={\int }_0^3\rho g(y+2)(0+\frac{2-0}{3}y)dy$$

(c) Right:

$$F={\int }_1^5\rho g(y-1)(4+\frac{0-4}{5}y)dy$$Link to original

03 $★$

05

Fluid force on circular plates

For diagrams 1, 2, 3 (L to R) below, set up an integral to compute the hydrostatic force on the plate.

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Solution

Solutions - 0100-05

(1) Integral formula:

$$F={\int }_a^b\rho gh(y)w(y)dy$$

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Option 1:

(2) Using $y=0$ at water line, increasing downwards: (a) Left:

$$F={\int }_1^5\rho gy\left(2\sqrt{2^2-{(y-3)}^2}\right)dy$$

(b) Center:

$$F={\int }_0^3\rho gy\left(2\sqrt{3^2-y^2}\right)dy$$

(c) Right:

$$F={\int }_0^2\rho gy\left(2\sqrt{4^2-{(2-y)}^2}\right)dy$$

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Option 2:

(2) Using $y=0$ at center of shape, increasing downwards: (a) Left:

$$F={\int }_{-2}^2\rho g(y+3)2\sqrt{2^2-y^2}dy$$

(b) Center:

$$F={\int }_0^3\rho gy\left(2\sqrt{3^2-y^2}\right)dy$$

(c) Right:

$$F={\int }_{-2}^0\rho g(y+2)2\sqrt{4^2-y^2}dy$$Link to original

Work performed

04 $★$

03

Work to raise a leaky bucket

A bucket of water is raised by a chain to the top of a $50$-foot building. The water is leaking out, and the chain is getting lighter.

The bucket weighs $4\mathrm{lbs}$, the initial water weighs $30\mathrm{lbs}$, and the chain weighs $0.25\mathrm{l}\mathrm{b}/\mathrm{f}\mathrm{t}$, and the water is leaking at a rate of $0.3\mathrm{l}\mathrm{b}\mathrm{s}/\mathrm{s}\mathrm{e}\mathrm{c}$ as the bucket is lifted at a constant rate of $3\mathrm{f}\mathrm{t}/\mathrm{s}\mathrm{e}\mathrm{c}$.

What is the total work required to raise the bucket of water?

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Solution

Solutions - 0120-03

(1) Integral formula:

$$\begin{array}{c}W=\int dW\gg \gg {\int }_a^{b}F(y)dy\end{array}$$

Let $y=0$ at the ground and increase going up.

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(2) Compute force:

The force on the rope (at the top) when the bucket is at height $y$ is:

$$F=F_{\text{bucket}}+F_{\text{water}}+F_{\text{chain}}$$

We know $F_{\text{bucket}}=4\mathrm{lb}$.

Water is leaking at ${\displaystyle \frac{0.3}{3}}\mathrm{l}\mathrm{b}/\mathrm{f}\mathrm{t}$. Therefore:

$$F_{\text{water}}=30-\frac{0.3}{3}y$$

The weight of chain remaining is:

$$F_{\text{chain}}=0.25\cdot (50-y)$$

Put together:

$$\begin{array}{c}F=4+30-\frac{0.3}{3}y+0.25\cdot (50-y)\\ \\ \gg \gg 46.5-0.35y\end{array}$$

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(3) Integrate:

$$W={\int }_0^{50}(46.5-0.35y)dy\gg \gg 1887.5\mathrm{f}\mathrm{t}\cdot \mathrm{l}\mathrm{b}$$Link to original

05

04

Work to pump water from cylindrical tank

A cylindrical tank is full of water and the water is pumped out the top. (See figure.) The length of the tank is $7\mathrm{m}$ and the radius is $5\mathrm{m}$.

(a) Set up an integral for the total work performed assuming the tank is initially completely full.

(b) Set up an integral for the total work performed assuming the tank is initially full to $3\mathrm{m}$ and the water is pumped out of a spigot extending $1\mathrm{m}$ above the top of the tank.

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Solution

Solutions - 0120-04

(1) Integral formula:

$$\begin{array}{c}W=\int dW\gg \gg \int h(y)\rho gdV\\ \\ \gg \gg \int \rho gh(y)A(y)dy\end{array}$$

Set $y=0$ at the center of the tank.

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(a) (2) Geometry:

So:

$$w(y)=2\sqrt{5^2-y^2}$$

Tank length is $7\mathrm{m}$ so a horizontal slice is a rectangle with area $w(y)\cdot 7$.

Depth is:

$$h(y)=y+5$$

Therefore:

$$W={\int }_{-5}^5\rho g(y+5)7\cdot 2\sqrt{25-y^2}dy$$

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(b) (2) Change bounds and height:

$$W={\int }_2^5\rho g(y+6)7\cdot 2\sqrt{25-y^2}dy$$Link to original

06 $★$

05

Work to build a pyramid

The Great Pyramid of Giza is $140\mathrm{m}$ tall and has a square base with $230\mathrm{m}$ on each side. It is built of stone with mass density $2000\mathrm{k}\mathrm{g}/{\mathrm{m}}^3$.

Set up an integral that expresses the work (against gravity) required to build the pyramid.

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Solution

Solutions - 0120-05

(1) Integral formula:

$$\begin{array}{c}W=\int dW\gg \gg \int h(y)\rho gdV\\ \\ \gg \gg \int \rho gh(y)A(y)dy\end{array}$$

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(2) Integrand components:

Option 1: Set $y=0$ at the base, going up.

Take a cross-sectional slice with a vertical plane. This intersects the surface of the pyramid in a triangle whose width $w(y)$ is the side length of the square (the horizontal cross section) at height $y$.

$$w(y)=230+\frac{0-230}{146}y$$

Note that $A(y)=w{(y)}^2$. So we have:

$$W={\int }_0^{146}\rho gy{(230-\frac{230}{146}y)}^{2}dy$$

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Option 2:

Set $y=0$ at the vertex, going down.

$$w(y)=0+\frac{230-0}{146}y$$

And:

$$W={\int }_0^{146}\rho g(146-y){\left(\frac{230}{146}y\right)}^{2}dy$$Link to original