anonymous
  • anonymous
2nd Course Calculus Force Problem: A cubical metal tank has a parabolic gate bolted at the bottom of one lateral face. The gate is 2 feet across the top and 1 foot in height. The liquid stored in the tank has a density of 50 lb/ft^3. Assign a coordinate system to the parabola and find the exact fluid force on the gate when the liquid is 2 feet deep in the tank. THEN - The bolts on the gate are designed to withstand a fluid force of 160lbs without rupturing. What is the maximum height to which the container can be filled without exceeding its design limit?
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
5 ft.
anonymous
  • anonymous
For the first part, I set up the integral as \[50\int\limits_{-1}^{1} (2-y)\sqrt{y} dy\] Then I changed the integral to 100 * the integral from 0 to 1... Am I on the right track?
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anonymous
  • anonymous
xD It's alright. I attached a screenshot of the formula I used from my textbook, if that helps...
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dumbcow
  • dumbcow
haha thx i had to look up fluid force....anyway your integral is correct :)
dumbcow
  • dumbcow
did you need help with integrating? answer should be 280/3
anonymous
  • anonymous
Alriiigghht! I just wanted to make sure I wasn't going off in the wrong direction - for some reason I can plug in to the formula, but it just doesn't click in my head (understanding-wise) as it should. I got confused as to what I was doing. I got now it though, both answers - the second should be three... I think. Lol Thanks!
dumbcow
  • dumbcow
yep , height of 3 ft has force of 160
dumbcow
  • dumbcow
side note: since integral is in terms of "y" you would never have limits of -1 to 1 the height "y" varies from 0 to 1 y = x^2 ---> x = sqrt(y) so L(y) is 2sqrt(y) that is why it becomes 100 int (2-y)sqrt(y) dy
anonymous
  • anonymous
Thanks for pointing that out!! (And for helping, but had you not mentioned that I probably would've gone and messed something up on my test tomorrow. Note to self: Pay more attention! xD)

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