We are tasked with calculating the work required to pump liquid from a conical container, both to the rim and to a level 5 feet above the rim. The key is to apply principles of fluid mechanics and calculus. The work is given by:

$W = \int_{a}^{b} F(y) \, dy$

where $F(y)$ is the force required to move an elemental volume of liquid from height $y$ to the desired height, and the limits $a$ and $b$ will be determined based on the height of the liquid in the container.

Step 1: Set up variables and geometry of the cone

The cone has:

• Radius $R = 10 \, \text{ft}$,
• Height $H = 40 \, \text{ft}$,
• Liquid filled up to $35 \, \text{ft}$,
• Density $\rho = 50.6 \, \text{lb/ft}^3$.

To calculate the force, we need to express the volume of a thin slice of liquid at height $y$ inside the cone.

At any height $y$, the radius $r(y)$ is proportional to the height by the relationship: $r(y) = \frac{R}{H} y = \frac{10}{40} y = \frac{y}{4}$

Step 2: Volume of a thin slice of liquid

A thin horizontal slice of liquid at height $y$ has a volume given by the differential volume of a cylinder: $dV = \pi r(y)^2 \, dy = \pi \left( \frac{y}{4} \right)^2 \, dy = \pi \frac{y^2}{16} \, dy$

Step 3: Weight of a thin slice of liquid

The weight of this slice is: $dW = \rho \, dV = 50.6 \cdot \pi \frac{y^2}{16} \, dy$

Step 4: Work to pump to the rim

The work to pump a slice from height $y$ to the rim (40 ft) is given by: $dW_{\text{rim}} = \text{weight} \times \text{distance to pump} = 50.6 \cdot \pi \frac{y^2}{16} (40 - y) \, dy$

Thus, the total work required to pump all the liquid to the rim is: $W_{\text{rim}} = \int_0^{35} 50.6 \cdot \pi \frac{y^2}{16} (40 - y) \, dy$

Step 5: Work to pump 5 feet above the rim

To pump the liquid 5 feet above the rim, the distance to pump a slice at height $y$ becomes $45 - y$. Therefore, the work to pump to 5 ft above the rim is: $dW_{\text{above}} = 50.6 \cdot \pi \frac{y^2}{16} (45 - y) \, dy$

The total work required to pump all the liquid 5 feet above the rim is: $W_{\text{above}} = \int_0^{35} 50.6 \cdot \pi \frac{y^2}{16} (45 - y) \, dy$

Step 6: Solving the integrals

Let me calculate these integrals for the total work in both cases.The total work required to:

1. Pump the liquid to the rim is approximately: $621464.19\pi \, \text{ft-lb} \approx 1,952,734.47 \, \text{ft-lb}$

2. Pump the liquid 5 feet above the rim is approximately: $847451.17\pi \, \text{ft-lb} \approx 2,662,046.26 \, \text{ft-lb}$

These are the amounts of work in foot-pounds needed to pump the liquid to the respective levels.

Would you like more details on any part of this solution?

Here are five related questions to deepen your understanding:

1. How does the shape of the container affect the work calculation?
2. What happens to the work if the liquid's density changes?
3. How would the work change if the cone were filled to a different height?
4. Can this method be applied to containers of other shapes (e.g., cylindrical)?
5. How does the gravitational constant factor into work calculations in other units?

Tip: Always ensure units are consistent throughout the calculations to avoid errors.