Re: Wolf of the Red Moon's question regarding the computation of work
Hello Wolf of the Red Moon,
I prefer to work problems like this in general terms, and then plug our given data into the resulting formula.
First, let's let:
$r$ = the radius of the tank
$\ell$ = the length of the tank
$q$ = the distance above the top of the tank the fluid must be pumped
Now, let's imagine slicing the contents of the tank horizontally into rectangular sheets. The length of each sheet is constant, given by the length of the tank $\ell$. The width $w$ of each sheet will be a function of its vertical position within the tank.
So, let's orient a vertical axis, with its origin at the bottom of the tank, as in the diagram:
View attachment 1011
We wish to find $w$ as a function of $y$. We should observe that we may state:
$$(y-r)^2+\left(\frac{w}{2} \right)^2=r^2$$
Hence:
$$w(y)=2\sqrt{r^2-(y-r)^2}$$
and so the volume of an arbitrary sheet is:
$$dV=\ell\cdot w(y)\,dy=2\ell\sqrt{r^2-(y-r)^2}\,dy$$
Next, we want to determine the weight $\omega$ of the arbitrary sheet. So let's define $\rho$ to be the weight density of the fluid, where:
$$\rho=\frac{\omega}{dV}\,\therefore\,\omega=\rho\,dV$$
Hence:
$$\omega=2\rho\ell\sqrt{r^2-(y-r)^2}\,dy$$
Next, we want to determine the distance $d$ the arbitrary sheet must be lifted. This is:
$$d=q+(2r-y)$$
Thus, using the fact that work is the product of the applied force and the distance through which this force is applied, we find the work done to lift the arbitrary sheet is:
$$dW=\omega d=(q+(2r-y))\left(2\rho\ell\sqrt{r^2-(y-r)^2}\,dy \right)$$
In order to utilize the given useful hint, we may arrange this as:
$$dW=\omega d=((q+r)-(y-r))\left(2\rho\ell\sqrt{r^2-(y-r)^2}\,dy \right)$$
Now, summing by integration from the bottom of the tank $y=0$ to the top $y=2r$, we may state:
$$W=2(q+r)\rho\ell\int_0^{2r}\sqrt{r^2-(y-r)^2}\,dy-2\rho\ell\int_0^{2r}(y-r)\sqrt{r^2-(y-r)^2}\,dy$$
Now, by geometry, we should observe that:
$$2\int_0^{2r}\sqrt{r^2-(y-r)^2}\,dy=\pi r^2$$
Notice this is simply the area of a circle of radius $r$.
And we may also observe that the integral:
$$\int_0^{2r}(y-r)\sqrt{r^2-(y-r)^2}\,dy$$
may be rewritten using the substitution:
$$u=y-r\,\therefore\,du=dy$$
and we have:
$$\int_{-r}^{r} u\sqrt{r^2-u^2}\,du$$
Since the integrand is an odd function, and the interval of integration symmetric about the origin, we may state by the odd function rule:
$$\int_{-r}^{r} u\sqrt{r^2-u^2}\,du=0$$
Putting this all together, we have:
$$W=\pi(q+r)r^2\rho\ell$$
Now, plugging in the given data:
$$q=5\text{ ft}$$
$$r=\frac{3}{2}\text{ ft}$$
$$\rho=42\,\frac{\text{lb}}{\text{ft}^3}$$
$$\ell=4\text{ ft}$$
we find:
$$W=\pi\left(\left(5+\frac{3}{2} \right)\text{ ft} \right)\left(\frac{3}{2}\text{ ft} \right)^2\left(42\,\frac{\text{lb}}{\text{ft}^3} \right)\left(4\text{ ft} \right)=2457\pi\text{ ft}\cdot\text{lb}$$
