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6.7 Physical Applications (part 2)

Exam 1 covers up to this lesson (inclusive)

work examples / applications

example An aquarium length 2m, width 1m, height 1m, is full of water. Find the work done in pumping all the water over the top.

Just like with the chain, we will find the work to move one "slice" of water, then integrate to accumulate all slices.

mass of slice: density \(\cdot\) volume
density of water: \(\rho\) (rho)
volume: \((2)(1) dy = 2 dy\)
force or weight: mass \(\cdot\) gravity
weight of slice: \(2 \rho dy \cdot g\)
work to move it to top:

Figure: Diagram of a rectangular aquarium with dimensions 2 by 1 by 1, showing a thin horizontal slice at height y with thickness dy.

thin slice of water thickness \(dy\) at \(y\) m from bottom

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work to move it: \((\text{weight})(\text{distance to go})\)

\(= (2 \rho g dy)(1 - y)\)

\(= 2 \rho g (1 - y) dy\)

we can take infinitely many of these starting at \(y = 0\) (bottom) to \(y = 1\) (surface of water)

Accumulate all

\[ \int_{0}^{1} 2 \rho g (1 - y) dy = 2 \rho g \int_{0}^{1} (1 - y) dy = \dots = \rho g \text{ (J)} \]

\(\rho\) for water: \(1000 \text{ kg/m}^3\)

\(g = 9.81 \text{ m/s}^2\)

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Example: Pumping Water from a Conical Tank

The tank is in the shape of a cone vertex up. Height is 8m and radius at base is 1m. Tank is full of water. Find how much work to pump all water out over top?

To solve this, we first analyze a horizontal slice of the water:

Draw slice at height \( y \)
Thickness \( dy \), needs to travel up \( 8 - y \) m

\[ \text{mass} = \text{volume} \cdot \text{density} \]\[ = \pi (\text{radius})^2 \cdot dy \cdot \rho \]

\[ \text{weight} = \text{mass} \cdot \text{gravity} \]

Notice radius changes with \( y \).

Need to find a relationship between radius and height \( y \).

Figure: Diagram of an inverted cone with height 8m and base radius 1m, showing a horizontal slice at height y.

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Finding the Radius Relationship

Side View (Linear Equation Method)

Notice the outline is a triangle. The side is a line through \( (0, 8) \) and \( (1, 0) \).

\[ y = 8 - 8r \]

Solve for \( r \):

\[ y - 8 = -8r \]\[ 8r = 8 - y \]

\[ r = 1 - \frac{1}{8}y \]

Figure: Coordinate graph showing a line from (0,8) to (1,0) representing the side of the cone.

Similar Triangles Method

Outline of cone compared to a smaller triangle representing a different radius.

Ratios of sides are equal:

\[ \frac{1}{8} = \frac{r}{8 - y} \]\[ r = \frac{1}{8}(8 - y) \]\[ r = 1 - \frac{1}{8}y \]

Figure: Diagram comparing a large right triangle (height 8, base 1) to a smaller similar triangle (height 8-y, base r).

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mass of one slice: \( \pi (\text{radius})^2 \cdot dy \cdot \rho \)

\( = \pi (1 - \frac{1}{8}y)^2 \cdot dy \cdot \rho \)

weight: \( \pi \rho (1 - \frac{1}{8}y)^2 dy \cdot g = \pi \rho g (1 - \frac{1}{8}y)^2 dy \)

work: \( \pi \rho g (1 - \frac{1}{8}y)^2 dy \cdot (8 - y) = \pi \rho g (1 - \frac{1}{8}y)^2 (8 - y) dy \)

accumulate from bottom \( (y=0) \) to water surface \( (y=8) \)

\[ \int_{0}^{8} \pi \rho g (1 - \frac{1}{8}y)^2 (8 - y) dy = \dots = \boxed{16 \pi \rho g} \text{ (J)} \]

\( = 156800 \pi \)

what if tank is half full based on height

same shape, different water surface level (upper limit of integration)

\[ \int_{0}^{4} \pi \rho g (1 - \frac{1}{8}y)^2 (8 - y) dy = \dots = 147,000 \pi \]

Figure: Sketch of a truncated cone (frustum) with a shaded bottom section of height 4.

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example hydrostatic pressure or force

A small dam is in the shape of a trapezoid

Figure: Trapezoid with top width 10, bottom width 20, and height 15 meters.

all in meters

water level is even with the top

what is the force on this gate?

hydrostatic force = (hydrostatic pressure) \( \cdot \) (area)

hydrostatic pressure = (density of water) (gravity) (depth below water surface)

Figure: Coordinate graph showing a trapezoid on an x-y plane with a horizontal strip at height y and thickness dy.

one strip of gate
\( 15 - y \) below surface
need area of strip
height: \( dy \)