Question

Fish Tank A rectangular freshwater fish tank with base \(2 \times 4 \mathrm{ft}\) and height 2 \(\mathrm{ft}\) (interior dimensions) is filled to within 2 in. of the top. (a) Find the fluid force against each end of the tank. (b) Suppose the tank is sealed and stood on end (without spilling) so that one of the square ends is the base. What does that do to the fluid forces on the rectangular sides?

Short answer

The fluid force against each end of the tank is 9831.99 lb. When the tank is placed on its end, the maximum fluid force on the rectangular sides increases despite the surface area decreasing.

Step-by-step solution

Find Fluid Force for Each End

The force of the fluid against each end is given by the pressure of the fluid over the surface area. \( F = PA \), where \( P = Dgh \) is the pressure, \( D \) is the density of the fluid (freshwater in this case), \( g \) is the gravitational constant, and \( h \) is the height of the fluid. The surface area of each end is given by \( L \times W \), where \( L \) and \( W \) are the dimensions of the end of the tank. Here, the tank is filled to within 2 inches, or \( h = 2ft - 2/12ft = 22/12 ft \), the density D of freshwater is approximately 62.4 lb/ft³, the area of each end is \( L \times W = 2ft \times 2ft = 4 ft²\). The force exerted by the fluid against each end is : \( F = DghA = 62.4 lb/ft³ \times 9.81 m/s² \times 22/12ft \times 4ft² \)

Conversion of Units

Convert the force of fluid from lb/ft³·m/s²·ft·ft² to lb, using the conversion factor 1lb·ft/s² = 1 lb·ft/s², thus force \( F = 62.4 lb/ft³ \times 9.81 m/s² \times 22/12ft \times 4ft² / 1lb·ft/s² = 9831.99 lb\)

Find Fluid Forces on Rectangular Sides After Position Change

Now suppose the tank is sealed and stood on end with one of the square ends as its base. Since the pressure depends on the depth of the fluid, the pressure does not remain constant along each rectangular side. Due to the increased depth, despite the area being reduced, the maximum pressure (thus force) on the rectangular sides increases.

Key concepts

Pressure Calculation

Understanding pressure calculation is vital when solving fluid force problems in calculus. Pressure, denoted by the symbol 'P', can be thought of as the force applied per unit area on a surface. In fluids, this force is exerted by the fluid on container walls or any object immersed in it. The formula used to calculate the pressure exerted by a fluid is given by
\( P = Dgh \),
where 'D' represents the density of the fluid, 'g' is the acceleration due to gravity (9.81 m/s² on Earth's surface), and 'h' is the height of the fluid column above the point of interest.
When dealing with freshwater, as in the textbook exercise, the density 'D' is approximately 62.4 lb/ft³. The gravitational constant 'g' remains consistent, and the height 'h' varies depending on the depth of the water. Calculating pressure is fundamental because it allows us to determine the force exerted by the fluid at different depths leading to the understanding of how this force impacts objects or surfaces submerged in the fluid.

Surface Area and Force

In fluid dynamics problems, it's crucial to relate the concept of surface area to force. The force 'F' exerted by a fluid against a surface is the product of the pressure 'P' and the area 'A' of the surface. Mathematically, it is expressed as
\( F = PA \).
For a rectangular surface such as the tank ends in the exercise, the surface area can be found by multiplying its length 'L' by its width 'W'. So, if each end of the tank measures 2 feet by 2 feet, the area 'A' would be \( 4 ft² \).
Combining this with the calculated pressure, one can find the total force. In our exercise, the height of the fluid 'h' changes when the tank is filled, adjusting the pressure and thus altering the force on the tank ends. Learning how to calculate the fluid force correctly is important for not only academic settings but also for real-world applications, such as designing tanks and dams.

Fluid Dynamics in Calculus

The field of fluid dynamics in calculus often involves integrating to find the force exerted by fluids over a surface. When a tank is reoriented, the distribution of force changes due to the fluid pressure being dependent on depth. The deeper the fluid, the greater the pressure and therefore the force exerted. In our exercise, when the tank is stood on one end, the fluid depth increases for the now-vertical sides, leading to a variable pressure distribution that is highest at the bottom and decreases toward the top. This is because the pressure is a function of depth 'h', and this depth changes along the height of the tank.
Calculating the force on these new vertical sides requires integrating the pressure function over the surface area, taking into account that the pressure at each height 'y' is \( Dg(y) \), with 'y' varying from the top of the fluid to the bottom. Understanding these concepts in fluid dynamics serves as a foundation for complex applications in engineering and environmental science, where such calculations are routinely carried out to ensure the stability and security of structures interacting with fluids.