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Chapter 13: Problem 3

Find the pressure due to the fluid at a depth of \(76 \mathrm{~cm}\) in still \((a)\) water \(\left(\rho_{w}=1.00 \mathrm{~g} / \mathrm{cm}^{3}\right)\) and (b) mercury ( \(\left.\rho=13.6 \mathrm{~g} / \mathrm{cm}^{3}\right)\). (a) \(P=\rho_{w} g h=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.76 \mathrm{~m})=7450 \mathrm{~N} / \mathrm{m}^{2}=7.5 \mathrm{kPa}\) (b) \(P=\rho g h=\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.76 \mathrm{~m})=1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2} \approx 1.0 \mathrm{~atm}\)

Short Answer

Expert verified
The pressure at 76 cm in water is 7.5 kPa, and in mercury, it is approximately 1.0 atm.

Step by step solution

01

Convert Depth to Meters

First, convert the depth from centimeters to meters for both water and mercury calculations. Given depth is 76 cm, convert it to meters by dividing by 100: 76 cm = 0.76 m.
02

Calculate Pressure in Water

To find the pressure in water, use the formula: \[ P_w = \rho_w \cdot g \cdot h \]where \(\rho_w = 1000 \text{ kg/m}^3\), the density of water, \(g = 9.81 \text{ m/s}^2\), the acceleration due to gravity, and \(h = 0.76\text{ m}\), the depth.Substitute the values into the formula:\[ P_w = 1000 \times 9.81 \times 0.76 \]This simplifies to:\[ P_w = 7455.6 \text{ N/m}^2 \approx 7.5 \text{ kPa} \]
03

Calculate Pressure in Mercury

For mercury, use the same formula, but with mercury鈥檚 density:\[ P_m = \rho_m \cdot g \cdot h \]where \(\rho_m = 13600 \text{ kg/m}^3\), the density of mercury, \(g = 9.81 \text{ m/s}^2\), and \(h = 0.76 \text{ m}\).Substitute the values into the formula:\[ P_m = 13600 \times 9.81 \times 0.76 \]This simplifies to:\[ P_m = 101188.8 \text{ N/m}^2 \approx 1.01 \times 10^5 \text{ N/m}^2 \approx 1.0 \text{ atm} \]
04

Summarize the Results

From the calculations: - The pressure at 76 cm depth in water is approximately 7.5 kPa.- The pressure at the same depth in mercury is approximately 1.0 atm which equals \(1.01 \times 10^5 \text{ N/m}^2\).

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Key Concepts

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

Density
Density is a fundamental concept when dealing with fluid pressure. It refers to the mass of a substance per unit volume, often denoted by the symbol \( \rho \). In this particular exercise, we are dealing with two different substances: water and mercury.
  • Water has a density of \( 1.00 \text{ g/cm}^3 \), which is equivalent to \( 1000 \text{ kg/m}^3 \) when converted to SI units. This lower density compared to mercury results in lower pressure at the same depth.
  • Mercury, on the other hand, has a density of \( 13.6 \text{ g/cm}^3 \), equal to \( 13600 \text{ kg/m}^3 \) in SI units. This higher density means that mercury exerts more pressure than water at the same depth.
Density directly affects how much pressure a fluid can exert at a given depth. The denser the fluid, the greater the weight above, and therefore the greater the pressure exerted at the surface beneath it. Understanding density is crucial when calculating fluid pressure.
Depth Conversion
Converting units is often essential in physics problems, especially when using formulas that require consistent measurements, such as pressure calculation. In this exercise, the depth is initially given in centimeters (76 cm) and must be converted into meters to use in the pressure calculation formula.
  • The conversion is straightforward; divide the centimeter value by 100 to convert it to meters: \( 76 \text{ cm} = 0.76 \text{ m} \).
  • This conversion ensures that all units are compatible with the other measurements in the formula, like the density expressed in \( \text{kg/m}^3 \) and the gravitational acceleration in \( \text{m/s}^2 \).
Converting between units is an essential skill in physics education, allowing students to correctly execute calculations and comprehend the relationships between different physical quantities.
Physics Education
Physics education focuses on understanding the fundamental principles that govern the natural world, such as fluid pressure. Educational exercises like this help students apply theoretical knowledge to solve practical problems.
  • These problems teach students to identify relevant physical quantities, like density and depth, and how they interact to determine fluid pressure.
  • By completing exercises that require conversion between units and application of consistent units within formulas, students learn to accurately perform calculations involving fluid pressures.
The step-by-step approach helps students build confidence and mastery in physics, encouraging them to explore more complex problems and deepen their understanding of the physical world.
Pressure Calculation
Pressure calculation in fluids involves using the principle that pressure at a depth in a fluid is given by the formula \( P = \rho \cdot g \cdot h \), where \( \rho \) is the density, \( g \) is the gravitational acceleration, and \( h \) is the depth.
  • In the exercise, for water with a density of \( 1000 \text{ kg/m}^3 \), calculate pressure using: \( P_w = 1000 \times 9.81 \times 0.76 = 7455.6 \text{ N/m}^2 \), approximately \( 7.5 \text{ kPa} \).
  • For mercury, with a density of \( 13600 \text{ kg/m}^3 \), the pressure becomes: \( P_m = 13600 \times 9.81 \times 0.76 = 101188.8 \text{ N/m}^2 \), roughly \( 1.0 \text{ atm} \).
This exercise shows how the higher density of mercury results in greater pressure at the same depth compared to water, illustrating the formula's interaction with different elements crucial for pressure assessment.

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Most popular questions from this chapter

A mass (or load) acting downward on a piston confines a fluid of density \(\rho\) in a closed container, as shown in Fig. 13-2. The combined weight of the piston and load on the right is \(200 \mathrm{~N}\), and the cross-sectional area of the piston is \(A=8.0 \mathrm{~cm}^{2}\). Find the total pressure at point- \(B\) if the fluid is mercury and \(h=25 \mathrm{~cm}\left(\rho_{\mathrm{Hg}}=13600 \mathrm{~kg} / \mathrm{m}^{3}\right) .\) What would an ordinary pressure gauge read at \(B ?\) Recall what Pascal's principle tells us about the pressure applied to the fluid by the piston and atmosphere: This added pressure is applied at all points within the fluid. Therefore, the total pressure at \(B\) is composed of three parts: Pressure of the atmosphere \(=1.0 \times 10^{5} \mathrm{~Pa}\) Pressure due to the piston and load \(=\frac{F_{W}}{A}=\frac{200 \mathrm{~N}}{8.0 \times 10^{-4} \mathrm{~m}^{2}}=2.5 \times 10^{5} \mathrm{~Pa}\) Pressure due to the height \(h\) of fluid \(=h p g=0.33 \times 10^{5} \mathrm{~Pa}\) In this case, the pressure of the fluid itself is relatively small. We have Total pressure at \(B=3.8 \times 10^{5} \mathrm{~Pa}\) The gauge pressure does not include atmospheric pressure. Therefore, Gauge pressure at \(B=2.8 \times 10^{5} \mathrm{~Pa}\)

A foam plastic \(\left(\rho_{p}=0.58 \mathrm{~g} / \mathrm{cm}^{3}\right)\) is to be used as a life preserver. What volume of plastic must be used if it is to keep 20 percent (by volume) of an 80 -kg man above water in a lake? The average density of the man is \(1.04 \mathrm{~g} / \mathrm{cm}^{3}\). Keep in mind that a density of \(1 \mathrm{~g} / \mathrm{cm}^{3}\) equals \(1000 \mathrm{~kg} / \mathrm{m}^{3}\). At equilibrium $$\begin{array}{l} F_{B} \text { on man }+F B \text { on plastic }=\text { Weight of man }+\text { Weight of plastic } \\ \quad\left(\rho_{w}\right)\left(0.80 V_{m}\right) g+\rho_{w} V_{p} g=\rho_{m} V_{m} g+\rho_{p} V_{p} g \end{array}$$ or $$\left(\rho_{w}-\rho_{p}\right) V_{p}=\left(\rho_{m}-0.80 \rho_{w}\right) V_{m}$$ where subscripts \(m, w\), and \(p\) refer to man, water, and plastic, respectively. But \(\rho_{m} V_{m}=80 \mathrm{~kg}\) and so \(V_{m}=(80 / 1040) \mathrm{m}^{3}\). Substitution gives $$\left[(1000-580) \mathrm{kg} / \mathrm{m}^{3}\right] V_{p}=\left[(1040-800) \mathrm{kg} / \mathrm{m}^{3}\right]\left[(80 / 1040) \mathrm{m}^{3}\right]$$ from which \(V_{p}=0.044 \mathrm{~m}^{3}\).

Atmospheric pressure is about \(1.0 \times 10^{5} \mathrm{~Pa}\). How large a force does the still air in a room exert on the inside of a window pane that is \(40 \mathrm{~cm} \times 80 \mathrm{~cm}\) ? The atmosphere exerts a force normal to any surface placed in it. Consequently, the force on the window pane is perpendicular to the pane and is given by $$F=P A=\left(1.0 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(0.40 \times 0.80 \mathrm{~m}^{2}\right)=3.2 \times 10^{4} \mathrm{~N}$$ Of course, a nearly equal force due to the atmosphere on the outside keeps the window from breaking.

A glass tube is bent into the form of a U. A \(50.0-\mathrm{cm}\) height of olive oil in one arm is found to balance \(46.0 \mathrm{~cm}\) of water in the other. What is the density of the olive oil?

On a day when the pressure of the atmosphere is \(1.000 \times 10^{5} \mathrm{~Pa}\), a chemist distills a liquid under slightly reduced pressure. The pressure within the distillation chamber is read by an oil-filled manometer (density of oil \(=0.78 \mathrm{~g} / \mathrm{cm}^{3}\) ). The difference in heights on the two sides of the manometer is \(27 \mathrm{~cm}\). What is the pressure in the distillation chamber?
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