The questions I am having trouble with is a figure of an uneven trapazoid swimmi

Question

The questions I am having trouble with is a figure of an uneven trapazoid swimming pool. Two parralel sides one measuring 30ft and the other measures 18 feet, the length of the pool is 15 feet and then the other side has a 3ft measurement then angles out towards the 18foot side but is not given. Questions: If the pool is 10 feet deep, how much work does it take to pump all the water out? Water weights 62.4 lbs/cubic foot. Show all work. I go 35,942,400 ft lbs of work but I have a feeling it is incorrect.

Explanation / Answer

I'm assuming the pool is actually a pentagon that looks like a trapezoid with a bit of the corner broken off.

Characterize the volume as the area of the pentagon times the 10-foot depth.

Make the area easier to manage by dividing it into a 15 by 18 rectangle and a trapezoidal piece with parallel sides of length 3' and 15'. Partition this into a rectangular strip and a triangle.

Now you have a rectangle 15' by 18', a rectangle 3' by 12', and a right-isosceles triangle with sides of 12 feet.

The area of the pool is 270 ft2 + 45 ft2 + 72 ft2 = 387 ft2.

The volume is 3870 ft3

One way to characterize the necessary work is to look at removing one cubic foot of water from the pool. If you could pull out a cube of water, at what point is it "out" of the pool? If the surface of the water is at zero feet, we have to lift that cube at least one foot. Any less and the bottom of the cube is still "in the pool". Regardless of where the water goes, once a cubic foot has been lifted a foot, it's out of the pool. What about when the pool is half empty? A cube of water can be raised above the surface of water, but below the zero foot line, but until it is above the zero foot line, it's still in the pool. The work increases with depth because you have to lift the water higher to get it "out" of the pool.

For the top 387 cubic feet, you have to raise 387(62.4) pounds of water 1 foot.

You have to spend twice that raising the next 387 cubic feet, three times for the third, et cetera.

The sum of 1 to 10 is 55, so you multiply 387(62.4) by 55 to get the work necessary to raise each cubic foot to a height of 1 foot above the (initial) surface of the water.

387(62.4)(55) = 1,328,184 foot-pounds.

Remember, this number depends on my assumption of the pool's dimensions. A drawing would really help.

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