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Chapter 8: Problem 8

The volume \(V\) of liquid in a hollow horizontal cylinder of radius \(r\) and length \(L\) is related to the depth of the liquid \(h\) by \\[V=\left[r^{2} \cos ^{-1}\left(\frac{r-h}{r}\right)-(r-h) \sqrt{2 r h-h^{2}}\right] L\\] Determine \(h\) given \(r=2 \mathrm{m}, L=5 \mathrm{m},\) and \(V=8.5 \mathrm{m}^{3}\). Note that if you are using a programming language or software tool that is not rich in trigonometric functions, the arc cosine can be computed with \\[\cos ^{-1} x=\frac{\pi}{2}-\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)\\]

Short Answer

Expert verified
The depth of the liquid in the horizontal cylinder, \(h\), is approximately \(1.082\) meters.

Step by step solution

01

Plug in given values into the equation

Plug in the given values for \(r\), \(L\), and \(V\) into the given equation and isolate \(h\). \[8.5 = \left[ 2^2 \cos^{-1}\left(\frac{2-h}{2}\right) - (2-h)\sqrt{4h-h^2}\right] \cdot 5\]
02

Simplifying and isolate h

First, simplify the equation by dividing by \(5\) and plugging in the given values of \(r\): \[1.7 = 4 \cos^{-1}\left(\frac{2 - h}{2}\right) - (2 - h)\sqrt{4h - h^2}\] Now we can use the alternate arc cosine formula to further simplify the equation: \[1.7 = 4\left[\frac{\pi}{2} - \tan^{-1}\left(\frac{\frac{2 - h}{2}}{\sqrt{1 - \left(\frac{2 - h}{2}\right)^2}}\right)\right] - (2 - h)\sqrt{4h - h^2}\]
03

Solve for h

To solve for \(h\), place the equation into a numerical root-finding algorithm or graphing tool. This is because finding an algebraic solution by hand for this equation is practically impossible. For example, using a graphing tool, plot and find the intersection point of: \[y = 4\left[\frac{\pi}{2} - \tan^{-1}\left(\frac{\frac{2 - h}{2}}{\sqrt{1 - \left(\frac{2 - h}{2}\right)^2}}\right)\right] - (2 - h)\sqrt{4h - h^2}\] and \[y = 1.7\] The value for h at the intersection point will be the depth of the liquid. By plugging the equation into a numerical solver or using a graphing tool, we get \(h \approx 1.082\) meters as the depth of the liquid.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Root-Finding Algorithms
Root-finding algorithms are numerical methods used to determine the root or solution of an equation, where the equation equals zero. In this context, solving for the depth of the liquid, represented by \( h \), in the equation given involves finding the value of \( h \) that satisfies the equality. The equation is complex and cannot be easily rearranged to solve for \( h \) algebraically. Thus, we employ root-finding algorithms, such as the bisection method, Newton's method, or secant method.
  • The **Bisection Method** halves the interval repeatedly and chooses subintervals where the sign of the function changes, indicating a root exists within that interval.
  • **Newton's Method** uses derivatives to approximate the root, providing fast convergence, but it requires the derivative of the function, which could be challenging for complex functions.
  • The **Secant Method** is similar to Newton's but uses secant lines instead of tangent lines. It doesn't require the derivative, making it simpler to implement for complex equations.
By using these methods, as well as computational tools or graphing utilities, we can accurately and efficiently find the correct value of \( h \) that makes the original volume equation true.
Trigonometric Functions
Trigonometric functions and their inverses are essential tools in mathematics, particularly when dealing with geometric and periodic phenomena. In this problem, the volume of liquid in a cylinder is expressed through a trigonometric model that includes the inverse cosine, or arc cosine, function.
  • The model uses **inverse cosine** (\( \cos^{-1}(x) \)), which is the angle whose cosine is \( x \). This is crucial for turning geometric conditions into solvable equations.
  • Since some computational environments lack direct trigonometric capabilities, we use the alternate formula: \(\cos^{-1}(x) = \frac{\pi}{2} - \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\). This formula is instrumental in simplifying the equation to a form that can be more easily solved using numerical methods.
Understanding how to manipulate these functions is pivotal in dealing with complex problems that involve trigonometric behavior, enabling the student to transform the equation into a form that's amenable to numerical or computational solutions.
Mathematical Modeling
Mathematical modeling involves creating mathematical expressions that accurately represent real-world scenarios. In this exercise, the model helps find the depth of liquid \( h \) in a horizontal cylindrical tank. This model accounts for the geometric shape of the cylinder and the physical constraint imposed by liquid depth.
  • The equation given is a volumetric model representing how the depth of liquid correlates with volume, radius, and length of the cylinder.
  • Such models are crucial in engineering and physics, allowing professionals to predict and analyze scenarios like filling times, capacity calculations, and handling of fluids.
  • The complexities often arise with the interaction of geometry (cylinder shape) and depth measurements, which require trigonometric considerations (angles and arcs).
Through mathematical modeling, students learn to translate physical problems into equations that can be tackled using mathematical tools, enabling better analysis and understanding of the world around us.

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Most popular questions from this chapter

The general form for a three-dimensional stress field is given by \\[\left[\begin{array}{lll}\sigma_{x x} & \sigma_{x y} & \sigma_{x z} \\ \sigma_{x y} & \sigma_{y y} & \sigma_{y z} \\\\\sigma_{x z} & \sigma_{y z} & \sigma_{z z}\end{array}\right]\\] where the diagonal terms represent tensile or compressive stresses and the off-diagonal terms represent shear stresses. A stress field (in MPa) is given by \\[\left[\begin{array}{ccc}10 & 14 & 25 \\\14 & 7 & 15 \\\25 & 15 & 16\end{array}\right]\\] To solve for the principal stresses, it is necessary to construct the following matrix (again in MPa): \\[\left[\begin{array}{ccc}10-\sigma & 14 & 25 \\\14 & 7-\sigma & 15 \\\25 & 15 & 16-\sigma\end{array}\right]\\] \(\sigma_{1}, \sigma_{2},\) and \(\sigma_{3}\) can be solved from the equation \\[\sigma^{3}-I \sigma^{2}+I I \sigma-I I I=0\\] where \\[\begin{array}{l}I=\sigma_{x x}+\sigma_{y y}+\sigma_{z z} \\\I I=\sigma_{x x} \sigma_{y y}+\sigma_{x x} \sigma_{z z}+\sigma_{y y} \sigma_{zz}-\sigma_{x y}^{2}-\sigma_{x z}^{2}-\sigma_{y z}^{2} \\\I I=\sigma_{x x} \sigma_{y y} \sigma_{z z}-\sigma_{x x} \sigma_{y z}^{2}-\sigma_{y y} \sigma_{x z}^{2}-\sigma_{z z} \sigma_{x y}^{2}+2 \sigma_{x y} \sigma_{x z} \sigma_{y z}\end{array}\\] \(I, I I,\) and \(I I I\) are known as the stress invariants. Find \(\sigma_{1}, \sigma_{2},\) and \(\sigma_{3}\) using a root-finding technique.

In Sec. \(8.4,\) the phase angle \(\phi\) between the forced vibration caused by the rough road and the motion of the car is given by \\[\tan \phi=\frac{2\left(c / c_{c}\right)(\omega / p)}{1-(\omega / p)^{2}}\\] As a mechanical engineer, you would like to know if there are cases where \(\phi=\omega / 3-1 .\) Use the other parameters from the section to set up the equation as a roots problem and solve for \(\omega\).

Figure \(P 8.43\) shows three reservoirs connected by circular pipes. The pipes, which are made of asphalt-dipped cast iron \((\varepsilon=0.0012 \mathrm{m}),\) have the following characteristics: $$\begin{array}{lccc}\text { Pipe } & 1 & 2 & 3 \\\\\text { Length, } \mathrm{m} & 1800 & 500 & 1400 \\\\\text { Diomeler, } \mathrm{m} & 0.4 & 0.25 & 0.2 \\\\\text { Flow, } \mathrm{m}^{3} / \mathrm{s} & 2 & 0.1 & ? \\\\\hline\end{array}$$ If the water surface elevations in Reservoirs \(A\) and \(C\) are 200 and \(172.5 \mathrm{m},\) respectively, determine the elevation in Reservoir \(\mathrm{B}\) and the flows in pipes 1 and 3 . Note that the kinematic viscosity of water is \(1 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}\) and use the Colebrook equation to determine the friction factor (recall Prob. 8.12 ).

You buy a \(\$ 25,000\) piece of equipment for nothing down and \(\$ 5,500\) per year for 6 years. What interest rate are you paying? The formula relating present worth \(P,\) annual payments \(A,\) number of years \(n,\) and interest rate \(i\) is \\[A=P \frac{i(1+i)^{n}}{(1+i)^{n}-1}\\].

\( \mathrm{A}\) total charge \(Q\) is uniformly distributed around a ringshaped conductor with radius \(a\). A charge \(q\) is located at a distance \(x\) from the center of the ring (Fig. \(\mathrm{P} 8.31\) ). The force exerted on the charge by the ring is given by \\[F=\frac{1}{4 \pi e_{0}} \frac{q Q x}{\left(x^{2}+a^{2}\right)^{3 / 2}}\\] where \(e_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \mathrm{m}^{2}\right) .\) Find the distance \(x\) where the force is \(1.25 \mathrm{N}\) if \(q\) and \(Q\) are \(2 \times 10^{-5} \mathrm{C}\) for a ring with a radius of \(0.9 \mathrm{m}\).
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