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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> | <math> 1.664(4-h^2)^\frac{3}{2}(tan^{-1}(\frac{\sqrt{4-h^2}}{h}) - h\sqrt{4-h^2}) = 4t </math> | ||
| − | Ugh. | + | Ugh. See discussion. |
Revision as of 06:46, 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 height:
$ 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} $
$ z = \frac{41.6(4-h^2)^\frac{3}{2}}{100}=.416(4-h^2)^\frac{3}{2} $
The depth varies as a function of time; to determine this relationship, we must use the fact that
$ \frac{dv}{dt} = 4 $
$ V = 4t+C=4t $
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:
$ A_{water} = 4tan^{-1}(\frac{\sqrt{4-h^2}}{h}) - h\sqrt{4-h^2} $
The volume is then
$ V = zA_{water} = 1.664(4-h^2)^\frac{3}{2}(tan^{-1}(\frac{\sqrt{4-h^2}}{h}) - h\sqrt{4-h^2}) $
$ 1.664(4-h^2)^\frac{3}{2}(tan^{-1}(\frac{\sqrt{4-h^2}}{h}) - h\sqrt{4-h^2}) = 4t $
Ugh. See discussion.
