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Chapter 11: Problem 33

The atmospheric pressure above a swimming pool changes from 755 to \(765 \mathrm{mm}\) of mercury. The bottom of the pool is a rectangle \((12 \mathrm{m} \times 24 \mathrm{m}) .\) By how much does the force on the bottom of the pool increase?

Short Answer

Expert verified
The force on the bottom of the pool increases by 384,304 N.

Step by step solution

01

Calculate the Pressure Difference in Pascals

First, calculate the change in atmospheric pressure in millimeters of mercury by subtracting the initial pressure from the final pressure: \[\Delta P = 765\, \text{mm} - 755\, \text{mm} = 10\, \text{mm}\] of mercury. To convert this pressure to Pascals (Pa), use the conversion factor \(1\, \text{mm of mercury} = 133.3\, \text{Pa}\). Thus, \[\Delta P = 10\, \text{mm} \times 133.3\, \text{Pa/mm} = 1333\, \text{Pa}.\]
02

Calculate the Area of the Pool's Bottom

The bottom of the swimming pool is a rectangle with dimensions 12 meters by 24 meters. Calculate the area \(A\) using the formula for the area of a rectangle \(A = \text{length} \times \text{width}\). Thus, \[A = 12\, \text{m} \times 24\, \text{m} = 288\, \text{m}^2.\]
03

Calculate the Increase in Force

The force exerted on the bottom of the pool due to the change in atmospheric pressure can be determined using the formula \(F = \Delta P \times A\), where \(F\) is the force, \(\Delta P\) is the change in pressure, and \(A\) is the area. Substitute the known values: \[F = 1333\, \text{Pa} \times 288\, \text{m}^2 = 384,304\, \text{N}.\] Thus, the increase in the force on the bottom due to the changed pressure is 384,304 Newtons.

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Key Concepts

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

Understanding Pascal and Pressure Conversion
Atmospheric pressure is often measured in millimeters of mercury, but it's crucial to understand that this is not the only unit of pressure. In the International System of Units (SI), pressure is measured in Pascals (Pa). One Pascal is equivalent to one Newton per square meter (N/m^2). This means it represents a small force applied over a large area.

To solve problems involving pressure, you might need to convert between units. For example, atmospheric pressure often appears in millimeters of mercury. To convert to Pascals, use the conversion factor:
  • 1 mm of mercury (mmHg) = 133.3 Pa.
By multiplying the change in pressure measured in mmHg by this factor, you convert the value into Pascals, which is more suitable for calculations involving force and area.
Breaking Down Area Calculation
When dealing with objects like a swimming pool, understanding the area is key. The pool鈥檚 surface provides the contact area for pressure to act upon. The bottom of a rectangle-shaped swimming pool can be calculated using the formula:
  • Area (A) = Length 脳 Width.
In our problem, the pool's dimensions are 12 meters by 24 meters. Plug these values into the formula:
  • Area = 12 m 脳 24 m = 288 m虏.
Having the area in square meters aligns with the SI Unit standards and directly interfaces with forces measured in Pascals (Pa). This simplifies further calculations, such as determining force.
Calculating Force Based on Pressure and Area
Force is the push or pull exerted on an object, and in our problem, it's determined by pressure acting over an area. When you have the pressure increase and the area on which it exerts, you can find the resulting force using:
  • Force (F) = Pressure Change (螖P) 脳 Area (A).
Using previously calculated values, where 螖P = 1333 Pascals and A = 288 m虏, substitute to find the force:
  • Force = 1333 Pa 脳 288 m虏 = 384,304 Newtons (N).
This means that due to the increase in atmospheric pressure, the force on the pool bottom increases significantly, highlighting how environmental changes can impact submerged and surfaced structures through pressure variations.

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Most popular questions from this chapter

A blood transfusion is being set up in an emergency room for an accident victim. Blood has a density of \(1060 \mathrm{kg} / \mathrm{m}^{3}\) and a viscosity of \(4.0 \times 10^{-3} \mathrm{Pa} \cdot \mathrm{s} .\) The needle being used has a length of \(3.0 \mathrm{cm}\) and an inner radius of \(0.25 \mathrm{mm} .\) The doctor wishes to use a volume flow rate through the needle of \(4.5 \times 10^{-8} \mathrm{m}^{3} / \mathrm{s} .\) What is the distance \(h\) above the victim's arm where the level of the blood in the transfusion bottle should be located? As an approximation, assume that the level of the blood in the transfusion bottle and the point where the needle enters the vein in the arm have the same pressure of one atmosphere. (In reality, the pressure in the vein is slightly above atmospheric pressure.)

A liquid is flowing through a horizontal pipe whose radius is \(0.0200 \mathrm{m}\). The pipe bends straight upward through a height of \(10.0 \mathrm{m}\) and joins another horizontal pipe whose radius is \(0.0400 \mathrm{m} .\) What volume flow rate will keep the pressures in the two horizontal pipes the same?

Two hoses are connected to the same outlet using a Y-connector, as the drawing shows. The hoses \(A\) and \(B\) have the same length, but hose \(B\) has the larger radius. Each is open to the atmosphere at the end where the water exits. Water flows through both hoses as a viscous fluid, and Poiseuille's \(\operatorname{law}\left[Q=\pi R^{4}\left(P_{2}-P_{1}\right) /(8 \eta L)\right]\) applies to each. In this law, \(P_{2}\) is the pressure upstream, \(P_{1}\) is the pressure downstream, and \(Q\) is the volume flow rate. The ratio of the radius of hose \(\mathrm{B}\) to the radius of hose \(\mathrm{A}\) is \(R_{\mathrm{B}} / R_{\mathrm{A}}=1.50 .\) Find the ratio of the speed of the water in hose \(B\) to the speed in hose \(A\).

A water bed for sale has dimensions of \(1.83 \mathrm{m} \times 2.13 \mathrm{m} \times 0.229 \mathrm{m}\) The floor of the bedroom will tolerate an additional weight of no more than \(6660 \mathrm{N}\). Find the weight of the water in the bed and determine whether the bed should be purchased.

A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is \(1.80 \times 10^{5}\) Pa. Assuming that the top and bottom surfaces of the cap each have an area of \(4.10 \times 10^{-4} \mathrm{m}^{2}\), obtain the magnitude of the force that the screw thread exerts on the cap in order to keep it on the bottle. The air pressure outside the bottle is one atmosphere.
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