(New page: "Water pours into the tank here at the rate of 4 cubic ft/min. The tank's cross-sections are 4-ft diameter semicircles. One end of the tank is movable, but moving it to increase the volu...)
 
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<math>z = \frac{F_{fluid}}{k}</math>
 
<math>z = \frac{F_{fluid}}{k}</math>
  
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We can calculate the force of the water, and therefore z, as a function of height:
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We can calculate the force of the water, and therefore z, as a function of the distance d between the top of the container and the surface of the water:
  
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<math>F_{fluid} = \delta \int_{-2}^h (width)(depth) dy = 62.4 \int_{-2}^h -2y\sqrt{4-y^2}dy=41.6(4-h^2)^\frac{3}{2}</math>
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<math>F_{fluid} = \delta \int_{-2}^d (width)(depth) dy = 62.4 \int_{-2}^d -2y\sqrt{4-y^2}dy=41.6(4-d^2)^\frac{3}{2}</math>
  
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<math>z = \frac{41.6(4-h^2)^\frac{3}{2}}{100}=.416(4-h^2)^\frac{3}{2}</math>
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<math>z(d) = \frac{41.6(4-d^2)^\frac{3}{2}}{100}=.416(4-d^2)^\frac{3}{2}</math>
  
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The depth varies as a function of time; to determine this relationship, we must use the fact that
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Now, solving for <math>z</math> when <math>d=0</math>
  
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<math>\frac{dv}{dt} = 4</math>
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<math>z(0) = 3.328</math>
  
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<math>V = 4t+C=4t</math>
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Therefore, the tank will overflow, because when the tank is full (d = 0), the spring will have only compressed 3.328 ft.
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We can assume that C = 0, as the volume at time 0 should also be 0.  The area of the water against the plunger, in terms of the height, can be found by subtracting the triangle formed by two radii and a chord from a slice of a circle:
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<math> A_{water} = 4tan^{-1}(\frac{\sqrt{4-h^2}}{h}) - h\sqrt{4-h^2}</math>
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The volume is then
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<math> V = zA_{water} = 1.664(4-h^2)^\frac{3}{2}(tan^{-1}(\frac{\sqrt{4-h^2}}{h}) - h\sqrt{4-h^2}) </math>
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<math> 1.664(4-h^2)^\frac{3}{2}(tan^{-1}(\frac{\sqrt{4-h^2}}{h}) - h\sqrt{4-h^2}) = 4t </math>
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Ugh.
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Latest revision as of 07:56, 28 September 2008

"Water pours into the tank here at the rate of 4 cubic ft/min. The tank's cross-sections are 4-ft diameter semicircles. One end of the tank is movable, but moving it to increase the volume compresses a spring. The spring constant is $ k = 100 $ lb/ft. If the end of the tank moves 5 ft against the spring, the water will drain out of a safety hole in the bottom at the rate of 5 cubic ft/min. Will the movable end reach the end before the tank overflows?"

We know that, according to Hooke's Law,

$ F_{spring} = kz $

where $ z = $ distance compression of the spring. We also know that

$ F_{fluid} = F_{spring} $

Therefore

$ z = \frac{F_{fluid}}{k} $

We can calculate the force of the water, and therefore z, as a function of the distance d between the top of the container and the surface of the water:

$ F_{fluid} = \delta \int_{-2}^d (width)(depth) dy = 62.4 \int_{-2}^d -2y\sqrt{4-y^2}dy=41.6(4-d^2)^\frac{3}{2} $

$ z(d) = \frac{41.6(4-d^2)^\frac{3}{2}}{100}=.416(4-d^2)^\frac{3}{2} $

Now, solving for $ z $ when $ d=0 $

$ z(0) = 3.328 $

Therefore, the tank will overflow, because when the tank is full (d = 0), the spring will have only compressed 3.328 ft.

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