Question

A 6 ft cylindrical tank with a radius of \(3 \mathrm{ft}\) is filled with water, which has a weight density of \(62.4 \mathrm{lb} / \mathrm{ft}^{3}\). The water is to be pumped to a point \(2 \mathrm{ft}\) above the top of the tank. (a) How much work is performed in pumping all the water from the tank? (b) How much work is performed in pumping \(3 \mathrm{ft}\) of water from the tank? (c) At what point is \(1 / 2\) of the total work done?

Short answer

(a) Work is 20217.6\pi ft-lb; (b) 7577.6\pi ft-lb; (c) Half work at 2.485 ft.

Step-by-step solution

Understand the Problem

The task involves finding the work needed to pump water out of a cylindrical tank. We need the total work to pump all the water, the work to pump 3 ft of water, and find the point for half the total work.

Identify the Work Formula

Work done is calculated using the formula: \[ \text{Work} = \int_{a}^{b} F(x) \, dx \]Where \(F(x)\) represents the force needed to move a small slice of water.

Calculate the Volume of the Tank

The volume \(V\) of a cylinder is computed via the formula: \[ V = \pi r^2 h \]Given that \(r = 3 \) ft and \(h = 6\) ft, the volume is \[ V = \pi (3)^2 (6) = 54\pi \, \text{ft}^3 \].

Calculating Force as a Function of Position

Consider a small slice of water at depth \(x\) with thickness \(\Delta x\). The volume of this slice is \[ V_{\text{slice}} = \pi \times 3^2 \times \Delta x = 9\pi \Delta x \]The weight of this slice is \[ F(x) = 62.4 \times 9\pi \Delta x = 561.6\pi \Delta x \] lb per unit length. The slice is lifted \(x + 2\) feet in height.

Integrate to Find Total Work

The total work needed to pump all the water out is given by integrating the force over distance:\[ W_{\text{total}} = \int_{0}^{6} (561.6\pi (x+2)) \, dx \]Calculating, we get:\[ W_{\text{total}} = 561.6\pi \left[ \frac{x^2}{2} + 2x \right]_{0}^{6} = 561.6\pi \left( \frac{6^2}{2} + 2 \times 6 \right) = 561.6\pi \times 36 = 20217.6\pi \; \text{ft-lb} \]

Calculate Work for 3 ft of Water

To find the work needed to pump 3 feet of water, integrate from 0 to 3:\[ W_{3\text{ft}} = \int_{0}^{3} 561.6\pi (x+2) \, dx \]Calculating, we get:\[ W_{3\text{ft}} = 561.6\pi \left[ \frac{x^2}{2} + 2x \right]_{0}^{3} = 561.6\pi \left( \frac{3^2}{2} + 2 \times 3 \right) = 561.6\pi \times 13.5 = 7577.6\pi \; \text{ft-lb} \]

Determine Point of Half Total Work

To find the point where half the total work is done, we need to solve:\[ \frac{1}{2} \times 20217.6\pi = 561.6\pi \int_{0}^{c}(x+2) \, dx \]Solving:\[ 10108.8\pi = 561.6\pi \left[ \frac{x^2}{2} + 2x \right]_{0}^{c} \]\[ \frac{x^2}{2} + 2x = 18 \Rightarrow x^2 + 4x - 36 = 0 \]Solving the quadratic, we get:\[ x = 2.485 \text{ ft} \]

Key concepts

Cylindrical Tanks

A cylindrical tank is a three-dimensional geometric shape with a circular base and a given height. When dealing with such tanks in physics problems, knowing their dimensions is crucial. In this example, we have a cylindrical tank with a radius of 3 feet and a height of 6 feet.
This tank is filled with water, and our objective is to analyze the work necessary to pump this water out. To comprehend any problem related to work done on cylindrical tanks, one needs to consider:

• The tank's dimensions, which are essential to calculate the volume and thus the mass of the liquid contained within.
• The gravitational forces acting on the liquid, impacting how much effort or work is required to move it.

Understanding these parameters in cylindrical tanks helps us set up equations that express exactly how energy must be transferred to achieve the movement of the liquid within and out of the tank.

Integration in Calculus

Integration is a fundamental concept in calculus used to find areas, volumes, and in our case, the total work done in a system. When solving problems involving the work required to move liquids in a tank, integration plays a key role.The formula \[\text{Work} = \int_{a}^{b} F(x) \, dx\]helps us determine the work done by integrating the force function over the distance of movement. The force function \(F(x)\) in our scenario represents the weight of a thin slice of water at a certain depth.Key points when applying integration in such problems are:

• Identify the variable of integration, usually representing depth or height in relation to the tank.
• Define the limits of integration, such as the bottom to the top of the tank, or any specific section.
• Calculate the integral, which effectively aggregates the small amounts of work needed to shift each infinitesimal slice of liquid from its position to the destination height.

Thus, integration is fundamental in calculating total work by summing up the contributions of each slice of water.

Weight Density of Water

Weight density is a property of any material, including water, that tells us how much weight there is per unit volume. In this problem, the weight density of water is given as 62.4 lb/ft³. This value is essential as it influences the force calculations necessary to determine the work required to pump water out of the tank. In our exercise:

• We consider the tiny slice of water at a certain depth. Its weight is calculated by multiplying its volume by the weight density.
• This weight exerts a force that must be overcome to lift the water, fundamentally tied to the principles of work.
• The weight density allows us to convert volume-based measurements into force, which is necessary for integration and calculating total work done.

This concept shows up frequently in physics problems involving fluids, as fluids have uniform weight density, making calculations systematic and reliable.