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Chapter 2: Problem 14

According to Archimedes' principle, the mass of a floating object equals the mass of the fluid displaced by the object. Use this principle to solve the following problems. (a) A wooden cylinder 30.0 cm high floats vertically in a tub of water (density \(=1.00 \mathrm{g} / \mathrm{cm}^{3}\) ). The top of the cylinder is \(13.5 \mathrm{cm}\) above the surface of the liquid. What is the density of the wood? (b) The same cylinder floats vertically in a liquid of unknown density. The top of the cylinder is \(18.9 \mathrm{cm}\) above the surface of the liquid. What is the liquid density? (c) Explain why knowing the length and width of the wooden objects is unnecessary in solving Parts (a) and (b).

Short Answer

Expert verified
The density of wood is 0.55 g/cm鲁, the density of the unknown liquid is approximately 1.49 g/cm鲁, and the length and width of the wooden objects aren't necessary to solve parts (a) and (b) because only the proportion of height that is submerged is relevant in these calculations.

Step by step solution

01

Calculating the Density of the Wood

First, let's use the volume percent submerged to get the density of the wood according to the following formula: \(\text{Density of object} = \text{Density of fluid} * \frac{\text{Volume submerged}}{\text{Total volume}}\). Here, the density of fluid is the density of water, 1.00 g/cm鲁. The volume percent submerged is equal to the submerged height over the total height, (30.0 cm - 13.5 cm) / 30.0 cm = 0.55 . Substituting these values we get: \(\text{Density of wood} = 1 g/cm鲁 * 0.55 = 0.55 g/cm鲁\) .
02

Calculating the Density of the Unknown Fluid

Next, let's calculate the unknown liquid density where the wooden cylinder is floating, following the same process. Now in this case, the density of the fluid is unknown. However, we know the density of the object (wood) which we calculated to be 0.55 g/cm鲁. The volume percent submerged is (30.0 cm - 18.9 cm) / 30.0 cm = 0.37. Substituting these values into the formula transforms it to: Density of fluid = Density of object / Percent volume submerged = 0.55 g/cm鲁 / 0.37 = 1.486 g/cm鲁. Therefore, the density of the unknown liquid is approximately 1.49 g/cm鲁 when rounded off to two decimal places.
03

Reasoning Regarding Length and Width of the Cylinder

Finally, let's explain why knowing the length and width of the wooden objects is unnecessary to solve parts (a) and (b). Since the densities (whether of the wood or the liquid) are defined as the mass divided by the volume, both height and cross-sectional area of the cylinder would be in the numerator and the denominator of the fraction respectively. Thus, these factors would cancel out, leaving only the vertical height at play in determining volume submerged. Hence, the length and width of the cylinder don't matter in this case, only the proportion of the height that is submerged is important.

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Key Concepts

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

Density Calculations
Density is a key concept in understanding why objects float or sink. It is defined as mass per unit volume and is represented by the formula \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). This means if you have two objects of the same size, the one with the higher mass will have higher density.

For the exercise, calculating the density requires comparing how much of the wooden cylinder is submerged in water, which tells us how dense the wood is compared to water. Water has a density of 1.00 g/cm鲁. If an object has a density less than 1.00 g/cm鲁, it will float, and more of the object will stick out of the water.

In this case, the density of the wooden cylinder is calculated using the proportion of its height that is submerged. Only 55% of the wooden cylinder鈥檚 height is submerged, indicating its density is lower than that of water and is found to be 0.55 g/cm鲁 by the formula: \( \text{Density of wood} = \text{Density of water} \times \text{Volume percent submerged} \).
Buoyancy
Buoyancy explains why objects float and it is based on Archimedes' Principle, which states that a floating object displaces a amount of fluid equal to its own mass. Buoyancy is the force that pushes objects up in a fluid, countering gravity's pull downwards.

A balance exists between the weight of the object and the weight of the fluid displaced; this balance is what enables floating. If the weight of the object is less than or equal to the weight of the fluid it displaces, it floats.

In the exercise, when the wooden cylinder was placed in water and then in another liquid, its different submersion levels indicated different buoyant forces due to different densities of the liquids. The height of the cylinder submerged in water was used to determine its density and subsequently, using this density, the unknown liquid's density was determined as well.
Fluid Mechanics
Fluid mechanics is the study of fluids and how they behave when at rest and when in motion. It combines principles from physics and engineering to explore how fluids exert forces and how they flow.

This field of study includes understanding concepts like pressure, buoyant force, and density dynamics. Archimedes' Principle is a fundamental aspect of fluid mechanics, as it helps explain how objects interact with fluids around them.

In practical terms, for this exercise, fluid mechanics helps us understand how a wooden cylinder floats on water and how it affects its positioning when placed into a liquid with unknown density. By grasping the essence of fluid mechanics, one can predict, solve, and understand fluid-related problems much effectively even without knowing the exact dimensions of the objects, as only relative submerged volume matters.

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Most popular questions from this chapter

During the early part of the 20 th century, sulfanilamide (an antibacterial drug) was only administered by injection or in a solid pill. In \(1937,\) a pharmaceutical company decided to market a liquid formulation of the drug. Since sulfanilamide was known to be highly insoluble in water and other common pharmaceutical solvents, a number of alternative solvents were tested and the drug was found to be soluble in diethylene glycol (DEG). After satisfactory results were obtained in tests of flavor, appearance, and fragrance, 240 gallons of sulfanilamide in DEG were manufactured and marketed as Elixir Sulfanilamide. After a number of deaths were determined to have been caused by the formulation, the Food and Drug Administration (FDA) mounted a campaign to recall the drug and recovered about 232 gallons. By this time, 107 people had died. The incident led to passage of the 1938 Federal Food, Drug, and cosmetic Act that significantly tightened FDA safety requirements. Not all of the quantities needed in solving the following problems can be found in the text. Give sources of such information and list all assumptions. (a) The dosage instructions for the elixir were to "take 2 to 3 teaspoons in water every four hours." Assume each teaspoon was pure DEG, and estimate the volume (mL) of DEG a patient would have consumed in a day. (b) The lethal oral dose of diethylene glycol has been estimated to be 1.4 mL DEG/kg body mass. Determine the maximum patient mass \(\left(1 \mathrm{b}_{\mathrm{m}}\right)\) for which the daily dose estimated in Part (a) would be fatal. If you need values of quantities you cannot find in this text, use the Internet. Suggest three reasons why that dose could be dangerous to a patient whose mass is well above the calculated value. (c) Estimate how many people would have been poisoned if the total production of the drug had been consumed. (d) List steps the company should have taken that would have prevented this tragedy.

Use a spreadsheet program to fit a straight line \((y=a x+b),\) to tabulated \((x, y)\) data. Your program should evaluate the slope \(a\) and intercept \(b\) of the best fit to the data, and then calculate values of \(y\) using the estimated \(a\) and \(b\) for each tabulated value of \(x\). Calculate the average deviation (residual) of the estimated \(y\) from the calculated value, and comment upon the quality of the fit to the data. Test your program by fitting a line to the data in the following table: $$\begin{array}{|c|c|c|c|c|c|}\hline x & 1.0 & 1.5 & 2.0 & 2.5 & 3.0 \\\\\hline y & 2.35 & 5.53 & 8.92 & 12.15 & 15.38 \\\\\hline\end{array}$$

A process instrument reading, \(Z\) (volts), is thought to be related to a process stream flow rate \(\dot{V}(\mathrm{L} / \mathrm{s})\) and pressure \(P(\mathrm{kPa})\) by the following expression: $$Z=a \dot{V}^{b} P^{c}$$ Process data have been obtained in two sets of runs- -one with \(\dot{V}\) held constant, the other with \(P\) held constant. The data are as follows: $$\begin{array}{|c|c|c|c|c|c|c|c|}\hline \text { Point } & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\\\hline \dot{V}(\mathrm{L} / \mathrm{s}) & 0.65 & 1.02 & 1.75 & 3.43 & 1.02 & 1.02 & 1.02 \\\\\hline P(\mathrm{kPa}) & 11.2 & 11.2 & 11.2 & 11.2 & 9.1 & 7.6 & 5.4 \\\\\hline Z(\text { volts }) & 2.27 & 2.58 & 3.72 & 5.21 & 3.50 & 4.19 & 5.89 \\\\\hline\end{array}$$ (a) Suppose you had only performed runs \(2,3,\) and \(5 .\) Calculate \(a, b,\) and \(c\) algebraically from the data for these three runs. (b) Now use a graphical method and all the data to calculate \(a, b,\) and \(c .\) Comment on why you would have more confidence in this result than in that of Part (a). (Hint: You will need at least two plots.)

The temperature in a process unit is controlled by passing cooling water at a measured rate through a jacket that encloses the unit. The exact relationship between the unit temperature \(T\left(^{\circ} \mathrm{C}\right)\) and the water flow rate \(\phi(\mathrm{L} / \mathrm{s})\) is extremely complex, and it is desired to derive a simple empirical formula to approximate this relationship over a limited range of flow rates and temperatures. Data are taken for \(T\) versus \(\phi\). Plots of \(T\) versus \(\phi\) on rectangular and semilog coordinates are distinctly curved (ruling out \(T=a \phi+b\) and \(T=a e^{b \phi}\) as possible empirical functions), but a log plot appears as follows: A line drawn through the data goes through the points \(\left(\phi_{1}=25 \mathrm{L} / \mathrm{s}, T_{1}=210^{\circ} \mathrm{C}\right)\) and \(\left(\phi_{2}=40 \mathrm{L} / \mathrm{s},\right.\) \(\left.T_{2}=120^{\circ} \mathrm{C}\right)\). (a) What is the empirical relationship between \(\phi\) and \(T ?\) (b) Using your derived equation, estimate the cooling water flow rates needed to maintain the process unit temperature at \(85^{\circ} \mathrm{C}, 175^{\circ} \mathrm{C},\) and \(290^{\circ} \mathrm{C}\). (c) In which of the three estimates in Part (b) would you have the most confidence and in which would you have the least confidence? Explain your reasoning.

The Reynolds number is a dimensionless group defined for a fluid flowing in a pipe as \(R e=D u \rho / \mu\) where \(D\) is pipe diameter, \(u\) is fluid velocity, \(\rho\) is fluid density, and \(\mu\) is fluid viscosity. When the value of the Reynolds number is less than about \(2100,\) the flow is laminar- -that is, the fluid flows in smooth streamlines. For Reynolds numbers above \(2100,\) the flow is turbulent, characterized by a great deal of agitation. Liquid methyl ethyl ketone (MEK) flows through a pipe with an inner diameter of 2.067 inches at an average velocity of \(0.48 \mathrm{ft} / \mathrm{s}\). At the fluid temperature of \(20^{\circ} \mathrm{C}\) the density of liquid \(\mathrm{MEK}\) is \(0.805 \mathrm{g} / \mathrm{cm}^{3}\) and the viscosity is 0.43 centipoise \(\left[1 \mathrm{cP}=1.00 \times 10^{-3} \mathrm{kg} /(\mathrm{m} \cdot \mathrm{s})\right]\). Without using a calculator, determine whether the flow is laminar or turbulent. Show your calculations.
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