0
$\begingroup$

A tank in the shape of an inverted right circular cone has height 12 meters and radius 11 meters. It is filled with 6 meters of hot chocolate. Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. Note: the density of hot chocolate is $\delta = 1500 kg/m^3$

I have tried several different things but it's obvious I am overlooking something important.

This is my latest logic (method):

Volume of each slice: $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (\frac{11y}{12})^2 dy = \frac{121\pi y^2}{432}dy $

Mass of each slice: $ M = \delta V = \frac{1500kg}{3m^3} *\frac{121\pi y^2}{432}dy = \frac{15125\pi y^2}{36}dy$

Force of each slice: $ F = Mg = 9.8*M = 9.8\frac{15125\pi y^2}{36}dy = \frac{148225\pi y^2}{36}dy$

$ \int_{6}^{12} \frac{148225\pi y^2}{36}(12-y) dy = 4891425\pi $

I've tried all sorts of different stuff. This was my latest method.

$\endgroup$
1
$\begingroup$

$$\text{Force = area} \cdot \text{density} \cdot \text{depth}$$

Think of the top of the cone as $y=0$ and the bottom as $y=12$. The vertex of the cone is $(0,12)$ and the point at the far right edge of the top of the cone is $(11,0)$. Using these points, you can find the equation of the line that is the edge of the cone \begin{align*} m &= -\frac{12}{11},\\ y &= -\frac{12}{11}x + 12. \end{align*} Convert it to $x$ because we will be integrating with respect to $y$ $$x = -\frac{11}{12y} + 11.$$ This is also the radius, $r$.

We will integrate using cylinders of face area $\pi r^2$ and depth $dy$. Density is $1500$, as given.

Because work is $\text{force}\cdot \text{distance}$, we must take into account the distance the cocoa must be lifted. Since we pump to the top of the container, $y=0$, this is simply our current depth, $y$.

We will take the integral, but only for the levels that we actually have cocoa to pump at.

Therefore work equals the integral $$\int_6^{12} \pi\left(-\frac{11}{12}y + 12\right)^2 1500 y\, dy$$

This comes out to $712748.833$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.