/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 A right circular cone of base ra... [FREE SOLUTION] | 91Ó°ÊÓ

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A right circular cone of base radius \(R\), height \(H\), and known density \(\rho_{s}\) floats base down in a liquid of unknown density \(\rho_{f}\). A height \(h\) of the cone is above the liquid surface. Derive a formula for \(\rho_{f}\) in terms of \(\rho_{s}, R,\) and \(h / H,\) simplifying it algebraically to the greatest possible extent. [Recall Archimedes' principle, stated in the preceding problem, and note that the volume of a cone equals (base area)(height)/3.]

Short Answer

Expert verified
The required equation is \(\rho_{f} = \rho_{s}\frac{h}{H}\)

Step by step solution

01

Define volume of full cone and submerged part

Firstly, we will define the volume of the full cone as \(V_{f}=\frac{1}{3}\pi R^{2}H\) based on the geometry. The volume of the submerged part of the cone is \(V_{h} = \frac{1}{3}\pi R^{2}(H-h)\) where \(H-h\) is the submerged height of the cone.
02

Apply the Archimedes' principle

Based on Archimedes' principle, we know the weight of the displaced fluid equals the weight of the floating body. Let's set these equal to each other: \(V_{h}\rho_{f}g = V_{f}\rho_{s}g\). The gravity \(g\) will cancel out from both sides.
03

Define relation between radius, height and the ratio

Since the cone is similar in shape whether we look at the full cone or only the submerged part, we can derive a relation between heights and radii of the full cone and submerged part. By similar triangles, we have \(r/R = h/H\) or \(r = Rh/H\). Substituting \(r^2\) into the formula for the volume of the submerged part of the cone, we get \(V_{h} = \frac{1}{3}\pi (Rh/H)^2 (H-h)\)
04

Solve for density of the fluid

Substituting the expressions for \(V_{h}\) and \(V_{f}\) into the Archimedes' principle equation, we solve for the density of the fluid. After simplifying, we will get \(\rho_{f} = \rho_{s}\frac{h}{H}\).

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Key Concepts

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

Archimedes' Principle
Archimedes' principle is a foundational concept in fluid mechanics that enables us to understand how objects behave when immersed in a fluid. The principle states that the upward buoyant force exerted on a body immersed in a fluid is equal to the weight of the fluid that the body displaces. This principle is succinctly captured in the formula:
\[\begin{equation}F_{b} = V_{\text{displaced}} \times \rho_{\text{fluid}} \times g\text{,}\end{equation}\]where
  • \(F_{b}\) is the buoyant force,
  • \(V_{\text{displaced}}\) is the volume of the fluid displaced by the object,
  • \(\rho_{\text{fluid}}\) is the density of the fluid, and
  • \(g\) is the acceleration due to gravity.
Because of this principle, an object will float if the buoyant force is equal to or greater than the weight of the object. If the weight of the object is less than the displaced fluid's weight, the object will float partially submerged until this equilibrium is met, a scenario we often encounter in textbook exercises involving buoyancy.
Volume of a Cone
The volume of a cone is a straightforward yet essential geometric calculation, especially when dealing with problems related to buoyancy and displacement of fluids. The formula to calculate the volume of a right circular cone is given by:
\[\begin{equation}V = \frac{1}{3}\text{Ï€}R^2H\text{,}\end{equation}\]where
  • \(R\) is the radius of the base, and
  • \(H\) is the height of the cone.
Understanding how to calculate the volume of a cone is not only crucial for solving problems involving solid objects but also for determining the volume of fluid displaced when such objects are submerged. When calculating the density of the floating material in a fluid, knowing the volume of the submerged part is a key step, as demonstrated in the given exercise.
Buoyancy
Buoyancy is the force that keeps objects afloat in a fluid. The buoyant force is always directed upwards because it counters the gravitational pull, and its magnitude is equal to the weight of the fluid displaced by the submerged part of the object. For objects floating at equilibrium in a fluid, as described in our textbook exercise, the buoyant force exactly equals the weight of the object. If you've ever held a balloon underwater and felt it push up against your hand, you've experienced buoyancy first-hand. This upward force has a profound impact on everything from ship design to the behavior of submarines and even the principles behind hot air balloons.
Density of Materials
Density, symbolized by \(\rho\), is a property of materials that represents the mass per unit volume. It is calculated using the formula:
\[\begin{equation}\rho = \frac{m}{V}\text{,}\end{equation}\]where
  • \(m\) is the mass of the material, and
  • \(V\) is its volume.
This concept becomes particularly important when we assess whether an object will float or sink in a fluid. If the density of the object is less than that of the fluid, it will float, as is the case in our example with a cone floating in a liquid. The density of the submerged part of the object in relation to the fluid it displaces is crucial in calculating the buoyant force and, by extension, the overall floating condition according to Archimedes' principle.
Geometric Similarity
Geometric similarity plays a role in problems where objects have the same shape but different sizes, like the cones in our exercise. Two geometric figures are similar if their corresponding angles are equal, and the lengths of their corresponding sides are proportional. This concept is invaluable in problems where we calculate volumes or surface areas of submerged objects. In the case of the cone from the textbook exercise, the geometric similarity of the full cone and the portion submerged in the fluid allows us to relate their dimensions, which leads to a simplification when calculating the volume of the submerged part—a calculation that is pivotal for using Archimedes' principle accurately.

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Most popular questions from this chapter

In modeling the effect of an impurity on crystal growth, the following equation was derived: \(\frac{G-G_{\mathrm{L}}}{G_{0}-G}=\frac{1}{K_{\mathrm{L}} C^{m}}\) where \(C\) is impurity concentration, \(G_{\mathrm{L}}\) is a limiting growth rate, \(G_{0}\) is the growth rate of the crystal with no impurity present, and \(K_{\mathrm{L}}\) and \(m\) are model parameters. In a particular experiment, \(G_{0}=3.00 \times 10^{-3} \mathrm{mm} / \mathrm{min},\) and \(G_{\mathrm{L}}=1.80 \times 10^{-3} \mathrm{mm} / \mathrm{min} .\) Growth rates are measured for several impurity concentrations \(C\) (parts per million, or ppm), with the following results: $$\begin{array}{|c|c|c|c|c|c|}\hline C(\mathrm{ppm}) & 50.0 & 75.0 & 100.0 & 125.0 & 150.0 \\\\\hline G(\mathrm{mm} / \mathrm{min}) \times 10^{3} & 2.50 & 2.20 & 2.04 & 1.95 & 1.90 \\\\\hline\end{array}$$ (For example, when \(\left.C=50.0 \mathrm{ppm}, G=2.50 \times 10^{-3} \mathrm{mm} / \mathrm{min}\right)\). (a) Determine \(K_{\mathrm{L}}\) and \(m,\) giving both numerical values and units. (b) A solution is fed to a crystallizer in which the impurity concentration is 475 ppm. Estimate the expected crystal growth rate in (mm/min). Then state why you would be extremely skeptical about this result.

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