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Chapter 15: Problem 24

Suppose you pour water into a container until it reaches a depth of \(12 \mathrm{cm}\). Next, you carefully pour in a \(7.2-\mathrm{cm}\) thickness of olive oil so that it floats on top of the water. What is the pressure at the bottom of the container?

Short Answer

Expert verified
The total pressure at the bottom is approximately 2044 Pa.

Step by step solution

01

Understand the Problem

We are tasked with finding the pressure at the bottom of a container with two liquids - water and olive oil. First, identify the depths: 12 cm of water and 7.2 cm of olive oil. The pressure at the bottom is due to the weight of both the water and the olive oil.
02

Convert Measurements to Meters

Convert the depths from centimeters to meters, since pressure calculations require SI units. For water, 12 cm is equivalent to 0.12 meters, and for olive oil, 7.2 cm is equivalent to 0.072 meters.
03

Calculate the Pressure Due to Water

Use the formula for pressure due to a liquid: \( P = \rho g h \), where \( \rho \) is the density, \( g \) is acceleration due to gravity, and \( h \) is the height of the liquid column. The density of water is approximately 1000 kg/m鲁 and \( g \approx 9.8 \ m/s虏 \). Thus, the pressure from the water is \( P_{\text{water}} = 1000 \times 9.8 \times 0.12 \).
04

Calculate the Pressure Due to Olive Oil

Similarly, calculate the pressure from olive oil using its density (approximately 920 kg/m鲁). Use the formula \( P = \rho g h \) with \( h = 0.072 \ \text{m} \). Thus, \( P_{\text{olive\ oil}} = 920 \times 9.8 \times 0.072 \).
05

Sum the Pressures to Find Total Pressure at the Bottom

Add the pressures from the water and the olive oil to get the total pressure at the bottom: \( P_{\text{total}} = P_{\text{water}} + P_{\text{olive\ oil}} \). Plug in the values from steps 3 and 4 and compute the total pressure.

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Key Concepts

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

Hydrostatic Pressure
Hydrostatic pressure is the pressure exerted by a fluid at rest due to the gravitational force acting on it. It is experienced in all directions at a given depth within the fluid. This type of pressure increases as you go deeper into the fluid. At the very bottom, it is the highest because it accounts for the weight of the fluid above. Hydrostatic pressure is important for calculating the forces experienced by objects submerged in a fluid, or on surfaces in contact with liquid, like in a container filled with water and olive oil.
  • The formula to calculate hydrostatic pressure is given by: \[ P = \rho g h \] where:
    • \( P \) is the hydrostatic pressure.
    • \( \rho \) is the fluid鈥檚 density.
    • \( g \) is the acceleration due to gravity (approximately \( 9.8 \ m/s^2 \)).
    • \( h \) is the height of the fluid column.
This formula helps us understand how pressure builds up in layers of liquids like our water and olive oil example.
Density of Liquids
The density of a liquid is crucial in determining the hydrostatic pressure exerted at a given depth. Density is defined as the mass of the liquid per unit volume, often given in kg/m鲁. Liquids with higher densities exert more pressure at the same depth compared to those with lower densities.
  • For instance, the density of water is approximately 1000 kg/m鲁, which is higher than many other common liquids.
  • On the other hand, olive oil has a density of about 920 kg/m鲁, which is lower than water.
These densities are used in calculating the pressure contributed by each liquid in the container. As olive oil is less dense, it floats on water rather than mixing.
Pressure Calculation
Pressure calculation in fluids involves determining the force exerted by a fluid on a surface. We use the depth of the liquid and its density to calculate the pressure at various points within the fluid. When dealing with multiple liquid layers, we calculate the pressure contribution of each layer separately and then sum them.
  • The contribution from one liquid layer assumes the formula: \[ P = \rho g h \]
  • Thus for water and olive oil, you'd calculate it separately using their respective depths and densities:
    • Water contributes \( P_{water} = 1000 \times 9.8 \times 0.12 \) N/m虏
    • Olive oil contributes \( P_{olive\ oil} = 920 \times 9.8 \times 0.072 \) N/m虏
Finally, these pressures are added together to retrieve the total pressure at the base of the container, encompassing all the liquid layers.
Liquid Layers
The concept of liquid layers is fundamental when multiple types of liquids are poured together. Due to differences in densities, liquids may form separate layers rather than mixing immediately.
  • Light liquids, such as olive oil, tend to float on denser ones like water.
  • The pressure at any point within the system is the sum of pressures due to each liquid layer above that point. The lower layers contribute more to the pressure due to their weight.
This stacking of liquid layers is quite common in both natural and man-made environments, influencing how we calculate pressures in scenarios such as this exercise.

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Most popular questions from this chapter

IP A pan half-filled with water is placed in the back of an SUV. (a) When the SUV is driving on the freeway with a constant velocity, is the surface of the water in the pan level, tilted forward, or tilted backward? Explain. (b) Suppose the SUV accelerates in the forward direction with a constant acceleration \(a\). Is the surface of the water tilted forward, or tilted backward? Explain. (c) Show that the angle of tilt, \(\theta,\) in part (b) has a magnitude given by tan \(\theta=a / g,\) where \(g\) is the acceleration of gravity.

A hot-air balloon plus cargo has a mass of \(1890 \mathrm{kg}\) and a volume of \(11,430 \mathrm{m}^{3} .\) The balloon is floating at a constant height of \(6.25 \mathrm{m}\) above the ground. What is the density of the hot air in the balloon?

A geode is a hollow rock with a solid shell and an airfilled interior. Suppose a particular geode weighs twice as much in air as it does when completely submerged in water. If the density of the solid part of the geode is \(2500 \mathrm{kg} / \mathrm{m}^{3}\), what fraction of the geode's volume is hollow?

As a storm front moves in, you notice that the column of mercury in a barometer rises to only \(736 \mathrm{mm}\). (a) What is the air pressure? (b) If the mercury in this barometer is replaced with water, to what height does the column of water rise? Assume the same air pressure found in part (a).

A water tank springs a leak. Find the speed of water emerging from the hole if the leak is \(2.7 \mathrm{m}\) below the surface of the water, which is open to the atmosphere.
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