I need someone to explain the intuition and logic behind this particular problem

Question

A rectangular tank, with its top at ground level, is used to catch runoff water. Assume that the water weights 62.5lb/ft3. how much work does it take to empty the tank by pumping the water back to ground level once the tank is full. The length of the rectangular tank is 14 ft and the width is 16 ft. the total height of the tank is 30 ft

I think i just need an intuitive explanation on this. I know I'm supposed to find a force function and take the integral of that function to find the work. Down below is the solution that i got from my textbook

ΔF= 62.5 * 16 * 14 Δy
ΔF = 14000y

I then take the definite integral of the function above from 0 to 30 and i get the work, which is 6,300,000

why do i need to find the area of the tank (length * width) and multiply it by the weight of the water(which in this case is the force of the system)? what does the result of this represent?

My questions are more about the intuition and logic behind the steps taken in finding the work. I'm not interested in the how's, but more in the why's.

I just need an explanation on how to find the function of force here.

any help will be appreciated

Answer 1

We know that (loosely) $$W = \int F \cdot d$$ where $F$ is the force and $d$ is the distance.

The force is the weight of the water, and the weight is given by the volume times the weight per volume. So we want the volume times the weight per volume (i.e. the force), and the distance that we are moving each strip of water within the tank, which is given by $(h-y)$, where $h$ is the height of the tank and $y$ is the height of a strip of water within the tank.

So we have $$W = \int (\rho \, \mathrm dv)(h-y) $$ where $\rho$ is the weight per volume. Substitute:

$$W = \int(62.5 \, \mathrm dv)(30-y)$$

Clearly $\mathrm dv = l \times w \times \mathrm dy$, and so we have

$$W = \int_0^{30} (62.5 \cdot l \cdot w\cdot \mathrm dy)(30-y)$$ $$W = \int_0^{30} (62.5 \cdot 14 \cdot 16\cdot \mathrm dy)(30-y)$$ $$W = \int_0^{30} (420000 - 14000 y) \, \mathrm dy = 6300000$$

Answer 2

A horizontal slab of water, which is $y$ feet below ground level and $\Delta y$ feet thick, has volume $16\cdot 14 \Delta y$ cubic feet, and hence weighs $62.5\cdot 16\cdot 14 \Delta y$ pounds. Pounds is force. This many pounds must be moved $y$ feet. So the work involved in pumping out this thin slab is

$$\mbox{force } \times \mbox{ distance} = (62.5\cdot 16\cdot 14 \Delta y)\times (y).$$

The total work is to add up all these pieces of work from each possible level $y$.

$$\int_0^{30} 62.5\cdot 16\cdot 14 y \; dy.$$