Question

Hey guys, are integral calculus questions allowed here?

Answer 1

Yep. fire away.

Answer 2

Here's the textbook question. I'm working on part (d):

"A swimming pool is 20 ft wide and 40 ft long and its bottom is an inclined plane, the shallow end having a depth of 3 ft and the deep end, 9ft. If the pool is full of water, find the hydrostatic force on (a) the shallow end, (b) the deep end, (c) one of the sides, and (d) the bottom of the pool."

I figured (d) to be exactly 300,000 lb.

Water: 62.5 lb/cu. ft.

Textbook solution is slightly different. Any takers?

Answer 3

Wow. Physics. Whats the formula for hydrostatic force? Sorry cant remember my physics class.

Answer 4

62.5 lb/cu. ft. That means one cu. ft. of water weighs 62.5 lb. Does that help?

Answer 5

At each depth, the hydrostatic pressure is just
\[ P = \rho g h \]
Now is that what you want to calculate; i.e., hydrostatic pressure vs. hydrostatic force?

Answer 6

Or rather, this gives you the pressure. Now integrate the pressure over the area.

Answer 7

Here's how I started:\[62.5\int\limits_{0}^{40}20(9-3x/20)dx\]

I'd be interested in your solutions. I can't find a flaw in mine.

Answer 8

@the shallow side \[F = \int\limits_{A}^{} p dA = \int\limits_{0}^{20}\int\limits_{0}^{6} \rho gh (dh) (ds)\]\[=\int\limits_{0}^{20}18\rho g (ds) = 18(2)(65)(32.16) = 752,544 N\]

Answer 9

oops the 2 is 20. typo

Answer 10

20 what?

Answer 11

Newtons involves meters, not feet; the question is in terms of feet. Are you sure your units are straight?

Answer 12

oh right sorry about that. rework:
\[F = \int\limits_{A}^{} P (dA) = \int\limits_{0}^{20}\int\limits_{0}^{3} \rho g h (dh) (ds) = \int\limits_{0}^{20} 4.5\rho g = 20(4.5)(6.5)(32.16) = 188,136 lb\]

Answer 13

Awesome! Thanks, rlsadiz. But could you briefly explain your method? For example, where is the 62.5 lb./cu.ft.?

Answer 14

I'm not recognizing the numbers in your equations...

Answer 15

aw dammit sorry again the 6.5 is 62.5 that the density. ^_^

Answer 16

I have to look for my glasses XD

Answer 17

And the 32.16?

Answer 18

g in ft /s^2

Answer 19

Ah! ;-) Understand, I have those, too.

Answer 20

You don't need gravity, though - that's already included in the 62.5 lb./cu.ft, right?

Answer 21

thats the density, the rho

Answer 22

rho is the *mass density* 62.5lb/cu.ft. is the *weight density* (I know, I had that problem when I began working on this problem, too. I'm too used to working with SI units!)

Answer 23

wow haha Imperial units sucks.

Answer 24

Would you mind posting a fresh version of your equation rlsadiz? It would help me double-check my own work.

Answer 25

\[F = \int\limits_{A}^{}P (dA) = \int\limits_{0}^{20}\int\limits_{0}^{3} \rho g h = \int\limits_{0}^{20} 4.5 \rho g = 20(4.5)(62.5) = 5,625 lb\]

Answer 26

What's the 4.5? Could you explain your method? I'm not understanding your integral.