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Section7.6Fluid Forces¶ permalink

In the unfortunate situation of a car driving into a body of water, the conventional wisdom is that the water pressure on the doors will quickly be so great that they will be effectively unopenable. (Survival techniques suggest immediately opening the door, rolling down or breaking the window, or waiting until the water fills up the interior at which point the pressure is equalized and the door will open. See Mythbusters episode #72 to watch Adam Savage test these options.)

How can this be true? How much force does it take to open the door of a submerged car? In this section we will find the answer to this question by examining the forces exerted by fluids.

We start with pressure, which is related to force by the following equations: \begin{equation*} \text{ Pressure } = \frac{\text{ Force } }{\text{ Area } } \Leftrightarrow \text{ Force } = \text{ Pressure } \times\text{ Area } . \end{equation*}

In the context of fluids, we have the following definition.

Definition7.6.1Fluid Pressure

Let $w$ be the weight–density of a fluid. The pressure $p$ exerted on an object at depth $d$ in the fluid is $p = w\cdot d\text{.}$

We use this definition to find the force exerted on a horizontal sheet by considering the sheet's area.

Example7.6.2Computing fluid force
1. A cylindrical storage tank has a radius of 2 ft and holds 10 ft of a fluid with a weight–density of 50 lb/ft$^3\text{.}$ (See Figure 7.6.3.) What is the force exerted on the base of the cylinder by the fluid?

2. A rectangular tank whose base is a 5 ft square has a circular hatch at the bottom with a radius of 2 ft. The tank holds 10 ft of a fluid with a weight–density of 50 lb/ft$^3\text{.}$ (See Figure 7.6.4.) What is the force exerted on the hatch by the fluid?

Solution

The previous example demonstrates that computing the force exerted on a horizontally oriented plate is relatively easy to compute. What about a vertically oriented plate? For instance, suppose we have a circular porthole located on the side of a submarine. How do we compute the fluid force exerted on it?

Pascal's Principle states that the pressure exerted by a fluid at a depth is equal in all directions. Thus the pressure on any portion of a plate that is 1 ft below the surface of water is the same no matter how the plate is oriented. (Thus a hollow cube submerged at a great depth will not simply be “crushed” from above, but the sides will also crumple in. The fluid will exert force on all sides of the cube.)

So consider a vertically oriented plate as shown in Figure 7.6.5 submerged in a fluid with weight–density $w\text{.}$ What is the total fluid force exerted on this plate? We find this force by first approximating the force on small horizontal strips.

Let the top of the plate be at depth $b$ and let the bottom be at depth $a\text{.}$ (For now we assume that surface of the fluid is at depth 0, so if the bottom of the plate is 3 ft under the surface, we have $a=-3\text{.}$ We will come back to this later.) We partition the interval $[a,b]$ into $n$ subintervals \begin{equation*} a = y_1 \lt y_2 \lt \cdots \lt y_{n+1} = b, \end{equation*} with the $i^\text{ th }$ subinterval having length $\Delta y_i\text{.}$ The force $F_i$ exerted on the plate in the $i^\text{ th }$ subinterval is $F_i = \text{ Pressure } \times \text{ Area }\text{.}$

The pressure is depth $\times w\text{.}$ We approximate the depth of this thin strip by choosing any value $d_i$ in $[y_i,y_{i+1}]\text{;}$ the depth is approximately $-d_i\text{.}$ (Our convention has $d_i$ being a negative number, so $-d_i$ is positive.) For convenience, we let $d_i$ be an endpoint of the subinterval; we let $d_i = y_i\text{.}$

The area of the thin strip is approximately length × width. The width is $\Delta y_i\text{.}$ The length is a function of some $y$-value $c_i$ in the $i^\text{ th }$ subinterval. We state the length is $\ell(c_i)\text{.}$ Thus \begin{align*} F_i \amp = \text{ Pressure } \times \text{ Area }\\ \amp = -y_i\cdot w \times \ell(c_i)\cdot\Delta y_i. \end{align*}

To approximate the total force, we add up the approximate forces on each of the $n$ thin strips: \begin{equation*} F = \sum_{i=1}^n F_i \approx \sum_{i=1}^n -w\cdot y_i\cdot\ell(c_i)\cdot\Delta y_i. \end{equation*}

This is, of course, another Riemann Sum. We can find the exact force by taking a limit as the subinterval lengths go to 0; we evaluate this limit with a definite integral.

Key Idea7.6.6Fluid Force on a Vertically Oriented Plate

Let a vertically oriented plate be submerged in a fluid with weight–density $w$ where the top of the plate is at $y=b$ and the bottom is at $y=a\text{.}$ Let $\ell(y)$ be the length of the plate at $y\text{.}$

1. If $y=0$ corresponds to the surface of the fluid, then the force exerted on the plate by the fluid is \begin{equation*} F=\int_a^b w\cdot(-y)\cdot\ell(y)\ dy. \end{equation*}

2. In general, let $d(y)$ represent the distance between the surface of the fluid and the plate at $y\text{.}$ Then the force exerted on the plate by the fluid is \begin{equation*} F=\int_a^b w\cdot d(y)\cdot\ell(y)\ dy. \end{equation*}

Example7.6.7Finding fluid force

Consider a thin plate in the shape of an isosceles triangle as shown in Figure 7.6.8 submerged in water with a weight–density of 62.4 lb/ft$^3\text{.}$ If the bottom of the plate is 10 ft below the surface of the water, what is the total fluid force exerted on this plate?

Solution
Example7.6.11Finding fluid force

Find the total fluid force on a car door submerged up to the bottom of its window in water, where the car door is a rectangle 40'' long and 27'' high (based on the dimensions of a 2005 Fiat Grande Punto.)

Solution
Example7.6.13Finding fluid force

An underwater observation tower is being built with circular viewing portholes enabling visitors to see underwater life. Each vertically oriented porthole is to have a 3 ft diameter whose center is to be located 50 ft underwater. Find the total fluid force exerted on each porthole. Also, compute the fluid force on a horizontally oriented porthole that is under 50 ft of water.

Solution

We end this chapter with a reminder of the true skills meant to be developed here. We are not truly concerned with an ability to find fluid forces or the volumes of solids of revolution. Work done by a variable force is important, though measuring the work done in pulling a rope up a cliff is probably not.

What we are actually concerned with is the ability to solve certain problems by first approximating the solution, then refining the approximation, then recognizing if/when this refining process results in a definite integral through a limit. Knowing the formulas found inside the special boxes within this chapter is beneficial as it helps solve problems found in the exercises, and other mathematical skills are strengthened by properly applying these formulas. However, more importantly, understand how each of these formulas was constructed. Each is the result of a summation of approximations; each summation was a Riemann sum, allowing us to take a limit and find the exact answer through a definite integral.

The next chapter addresses an entirely different topic: sequences and series. In short, a sequence is a list of numbers, where a series is the summation of a list of numbers. These seemingly–simple ideas lead to very powerful mathematics.

Subsection7.6.1Exercises

Terms and Concepts

In the following exercises, find the fluid force exerted on the given plate, submerged in water with a weight density of 62.4 lb/ft$^3\text{.}$