To solve this, we'll use calculus to compute the work required to pump the liquid to the rim of the cone.

Step 1: Define the Work Integral

The general formula for work required to lift a volume of liquid is:

$dW = \text{(weight density)} \times \text{(volume slice)} \times \text{(lifting distance)}$

The total work is obtained by integrating over the height of the liquid.

$W = \int_{a}^{b} \text{(weight density)} \times \text{(cross-sectional area)} \times \text{(lifting distance)} \, dy$

where:

• The weight density of the liquid is $51.8$ lb/ft³.
• The radius of the conical tank is $9$ ft, and the height is $36$ ft.
• The liquid is filled up to a height of $33$ ft.
• The volume of a thin slice at height $y$ is given by the area of the cross-section times the thickness $dy$.
• The lifting distance for a slice at height $y$ is $36 - y$ (since the rim is at $y = 36$).

Step 2: Express the Radius as a Function of Height

The cone follows a linear relation between its height and radius. The equation of the cone's radius at any height $y$ can be determined using similar triangles:

$r = \frac{9}{36} y = \frac{y}{4}$

Thus, the cross-sectional area (circle) at height $y$ is:

$A(y) = \pi r^2 = \pi \left(\frac{y}{4}\right)^2 = \frac{\pi y^2}{16}$

Step 3: Set Up the Integral

The volume of a thin slice is:

$dV = A(y) dy = \frac{\pi y^2}{16} dy$

The weight of the slice is:

$\text{Weight} = 51.8 \times dV = 51.8 \times \frac{\pi y^2}{16} dy$

The work to lift this slice to the top is:

$dW = 51.8 \times \frac{\pi y^2}{16} (36 - y) dy$

Thus, the total work integral is:

$W = \int_{0}^{33} 51.8 \times \frac{\pi y^2}{16} (36 - y) dy$

Let's compute this integral. sympy as sp

Define variables

y = sp.Symbol('y')

Given constants

density = 51.8 # lb/ft³ radius_ratio = 1/4 # r = y/4 pi = sp.pi

Cross-sectional area

A_y = pi * (y/4)**2

Weight of a thin slice

dV = A_y weight = density * dV

Lifting distance

lifting_distance = 36 - y

Work integral

W_integral = weight * lifting_distance

Compute the integral from y = 0 to y = 33

W_total = sp.integrate(W_integral, (y, 0, 33))

Evaluate the result numerically

W_total_value = W_total.evalf()

W_total_valueThe amount of work required to pump the liquid to the rim of the tank is 1,370,669 ft-lb (rounded to the nearest whole number).

Would you like a detailed breakdown of the calculations or explanation of any step?

Related Questions:

1. How would the work change if the liquid had a different weight density?
2. What if the cone were only partially filled, say to a height of 20 ft?
3. How does the work compare if the cone were upside down?
4. How much work would it take to pump the entire cone full of liquid?
5. How would the integral setup change if the liquid were pumped to a higher level?

Tip:

When solving work problems in physics and calculus, always check if similar triangles can be used to simplify the radius or cross-sectional area formula.