# A tank has the shape of an inverted cone

I would like some help with this problem

“A tank has the shape of an inverted circular cone with a radius of 2 feet and a height of 5 feet. If the tank contains gasoline to a death of 1 foot, find the work required to empty the tank through the pipe extending 1 feet above the top of the tank.” Use y as the weight for the gasoline

• Would you like to share what you have tried/thought of so far? Commented May 6, 2020 at 15:19
• I drew the cone acting as the tank with the base radius of 2 feet and the height as 5ft. I started drawing gasoline thats 1 foot in the cone Commented May 6, 2020 at 15:23
• Would you like to edit your question body to include your process? People usually vote to close questions before reading the comments. Commented May 6, 2020 at 15:24
• okie I did it! View the picture Commented May 6, 2020 at 15:33
• I don’t have an idea. new to this! but I want to understand the thought process behind it. This is just the starting point I came up with Commented May 6, 2020 at 15:45

Set the "spike" end on the corn as gravity potential $$0$$. Then the potential energy of the liquid (gasoline or whatever) at this position can be given by the integral $$\int_0^1 \rho_{liquid}g\pi(\frac 2 5 x)^2xdx$$ (be careful about the unit!!)

Then this is completely a physics problem.

• I don’t understand this or any close to physics. I’m sorry Commented May 6, 2020 at 15:59
• @EstherLee $PE=mgh$. $h=x$. $g=g$. $dm=\rho dV=\rho mr^2 dh=\rho m(\frac 2 5 x)^2 dx$. Commented May 6, 2020 at 16:01

A helpful way to solve these types of problems is to set up equations for the initial state of the energy in the system, and the final state of the energy in the system. Since energy must be conserved, the energy in each of these states must be equal. Let us assume that the tip of the cone is at a potential energy of $$0$$. Let $$U_i$$ be the potential energy of the liquid in the initial state and $$U_f$$ be the final potential energy of the liquid.

The initial state of the liquid is at the bottom of the tank and the final state of the liquid is in the pipe the top of the tank. In order to get the liquid to the top, there must be some work done in the initial state (i.e. energy must be supplied for the liquid to move upwards). So, we have the following relationship: \begin{align*} E_i &= E_f\\ U_i + W &= U_f \end{align*}

The final potential energy is relatively easy to solve for, since all of the liquid is in the pipe above the tank. So, the potential energy of the liquid at this point is just: $$U_f = mgh_f$$ where $$m$$ is the mass of the liquid, $$g$$ is the gravitational constant, and $$h_f$$ is the final height of the liquid. We know the values of $$g$$ (since it is a constant) and $$h_f$$ (Since all of the liquid is at the same height, $$h_f$$ = 6 ft). We just need to find the value of $$m$$. Assuming no liquid is spilled, we can solve for the mass of the liquid in the initial state and use that as our value of $$m$$ in the final state. We must know the density, $$\rho$$, of the liquid and from there we can solve for $$m$$: \begin{align*} \rho &= \frac{m}{V}\\ m &= \rho V \end{align*}

Now, we just solve for the volume, $$V$$ in the initial state. This is just the volume of the smaller cone that contains the liquid in its initial state: $$V = \pi r^2 h = \pi [\frac{2}{5}(1 \text{ ft})]^2[1 \text{ ft}]$$

The $$\frac{2}{5}$$ is the ratio of the radius of the full cone to the height of the full cone, which helps us find the radius of smaller cone that contains the liquid in the tank.

From this, we can solve for $$m$$ and thus, $$U_f$$.

Now, we just need to solve for $$U_i$$. Unlike $$U_f$$, we cannot make the simple assumption that all of liquid is at the same potential. So, we cannot just express $$U_f$$ as $$mgh_f$$, since different masses are at different heights. We need to sum all of potential energies, $$dU_f$$, of the small masses, $$dm$$, at their respective heights, $$h$$.

Each $$dm$$ has its own volume, $$dV$$, that we must solve for and then sum (i.e. integrate) all of them together to get $$U_i$$. Just like before: $$dm = \rho dV$$

But, what is $$dV$$? Well, since the volume changes with height, we can express it in terms of $$dh$$: $$dV = \pi r^2dh$$

And just like before, we can express $$r$$ in terms of $$h$$: $$r = \frac{2}{5}h$$

Putting this altogether, we get: $$dV = \pi (\frac{2}{5}h)^2 dh$$

Now, we can go back to our expression for $$dU_f$$: \begin{align*} dU_i &= dm \text{ } g \text{ } h\\ &= \rho dV \text{ } g \text{ } h\\ &= \rho \left[ \pi (\frac{2}{5}h)^2 dh \right] gh \end{align*}

We can now solve for the total potential energy of the initial state by integrating. What are the bounds of the integral? Well, since the liquid in the initial state begins at the tip of the cone, which we have assigned 0 potential energy, the lower bound is 0. The upper bound is the max height of the liquid in the initial state, which is just 1 ft. Therefore, we get: \begin{align*} U_i &= \int_0^1 dU_i\\ &= \int_0^1 \rho g \left[ \pi (\frac{2}{5}h)^2 h \right] dh \end{align*}

After solving for $$U_i$$, you have all you need to solve for the work: $$W = U_f - U_i$$