1. Using Definition 7.6.1, we calculate that the pressure exerted on the cylinder's base is \(w\cdot d = 50 \text{ lb/ft } ^3\times 10\text{ ft } = 500\) lb/ft\(^2\text{.}\) The area of the base is \(\pi\cdot 2^2 = 4\pi\) ft\(^2\text{.}\) So the force exerted by the fluid is \begin{equation*} F = 500\times 4\pi = 6283\text{ lb } . \end{equation*} Note that we effectively just computed the weight of the fluid in the tank.

2. The dimensions of the tank in this problem are irrelevant. All we are concerned with are the dimensions of the hatch and the depth of the fluid. Since the dimensions of the hatch are the same as the base of the tank in the previous part of this example, as is the depth, we see that the fluid force is the same. That is, \(F = 6283\) lb. A key concept to understand here is that we are effectively measuring the weight of a 10 ft column of water above the hatch. The size of the tank holding the fluid does not matter.

We approach this problem in two different ways to illustrate the different ways Key Idea 7.6.6 can be implemented. First we will let \(y=0\) represent the surface of the water, then we will consider an alternate convention.

1. We let \(y=0\) represent the surface of the water; therefore the bottom of the plate is at \(y=-10\text{.}\) We center the triangle on the \(y\)-axis as shown in Figure 7.6.9. The depth of the plate at \(y\) is \(-y\) as indicated by the Key Idea. We now consider the length of the plate at \(y\text{.}\) We need to find equations of the left and right edges of the plate. The right hand side is a line that connects the points \((0,-10)\) and \((2,-6)\text{:}\) that line has equation \(x=1/2(y+10)\text{.}\) (Find the equation in the familiar \(y=mx+b\) format and solve for \(x\text{.}\)) Likewise, the left hand side is described by the line \(x=-1/2(y+10)\text{.}\) The total length is the distance between these two lines: \(\ell(y)=1/2(y+10) - (-1/2(y+10)) = y+10\text{.}\)

   The total fluid force is then: \begin{align*} F \amp = \int_{-10}^{-6} 62.4(-y)(y+10)\ dy\\ \amp = 62.4\cdot \frac{176}{3} \approx 3660.8\text{ lb } . \end{align*}

2. Sometimes it seems easier to orient the thin plate nearer the origin. For instance, consider the convention that the bottom of the triangular plate is at \((0,0)\text{,}\) as shown in Figure 7.6.10. The equations of the left and right hand sides are easy to find. They are \(y=2x\) and \(y=-2x\text{,}\) respectively, which we rewrite as \(x= 1/2y\) and \(x=-1/2y\text{.}\) Thus the length function is \(\ell(y) = 1/2y-(-1/2y) = y\text{.}\)

   As the surface of the water is 10 ft above the base of the plate, we have that the surface of the water is at \(y=10\text{.}\) Thus the depth function is the distance between \(y=10\) and \(y\text{;}\) \(d(y) = 10-y\text{.}\) We compute the total fluid force as: \begin{align*} F \amp =\int_0^4 62.4(10-y)(y)\ dy\\ \amp \approx 3660.8\text{ lb } . \end{align*}

The correct answer is, of course, independent of the placement of the plate in the coordinate plane as long as we are consistent.

The car door, as a rectangle, is drawn in Figure 7.6.12. Its length is \(10/3\) ft and its height is 2.25 ft. We adopt the convention that the top of the door is at the surface of the water, both of which are at \(y=0\text{.}\) Using the weight–density of water of 62.4 lb/ft\(^3\text{,}\) we have the total force as \begin{align*} F \amp = \int_{-2.25}^0 62.4(-y)10/3\ dy\\ \amp = \int_{-2.25}^0 -208y\ dy\\ \amp = -104y^2\Big|_{-2.25}^0\\ \amp = 526.5 \text{ lb. } \end{align*}

Most adults would find it very difficult to apply over 500 lb of force to a car door while seated inside, making the door effectively impossible to open. This is counter–intuitive as most assume that the door would be relatively easy to open. The truth is that it is not, hence the survival tips mentioned at the beginning of this section.

We place the center of the porthole at the origin, meaning the surface of the water is at \(y=50\) and the depth function will be \(d(y)=50-y\text{;}\) see Figure 7.6.14

The equation of a circle with a radius of 1.5 is \(x^2+y^2=2.25\text{;}\) solving for \(x\) we have \(x=\pm \sqrt{2.25-y^2}\text{,}\) where the positive square root corresponds to the right side of the circle and the negative square root corresponds to the left side of the circle. Thus the length function at depth \(y\) is \(\ell(y) = 2\sqrt{2.25-y^2}\text{.}\) Integrating on \([-1.5,1.5]\) we have: \begin{align*} F \amp = 62.4\int_{-1.5}^{1.5} 2(50-y)\sqrt{2.25-y^2}\ dy\\ \amp = 62.4\int_{-1.5}^{1.5} \big(100\sqrt{2.25-y^2} - 2y\sqrt{2.25-y^2}\big)\ dy\\ \amp = 6240\int_{-1.5}^{1.5} \big(\sqrt{2.25-y^2}\big)\ dy - 62.4\int_{-1.5}^{1.5} \big(2y\sqrt{2.25-y^2}\big)\ dy. \end{align*}

The second integral above can be evaluated using Substitution. Let \(u=2.25-y^2\) with \(du = -2y\,dy\text{.}\) The new bounds are: \(u(-1.5)=0\) and \(u(1.5)=0\text{;}\) the new integral will integrate from \(u=0\) to \(u=0\text{,}\) hence the integral is 0.

The first integral above finds the area of half a circle of radius 1.5, thus the first integral evaluates to \(6240\cdot\pi\cdot1.5^2/2 = 22,054\text{.}\) Thus the total fluid force on a vertically oriented porthole is \(22,054\) lb.

Finding the force on a horizontally oriented porthole is more straightforward: \begin{equation*} F = \text{ Pressure } \times\text{ Area } = 62.4\cdot50\times \pi\cdot1.5^2 = 22,054\text{ lb } . \end{equation*}

That these two forces are equal is not coincidental; it turns out that the fluid force applied to a vertically oriented circle whose center is at depth \(d\) is the same as force applied to a horizontally oriented circle at depth \(d\text{.}\)