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Chapter 11: Problem 12

A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is \(1.80 \times 10^{5} \mathrm{~Pa}\). 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.

Short Answer

Expert verified
The force exerted by the screw thread on the cap is 32.39 N.

Step by step solution

01

Understand the Forces

The bottle cap is affected by two pressures: the internal pressure (\(P_{ ext{internal}} = 1.80 \times 10^{5} \ ext{Pa}\)) and the external atmospheric pressure (\(P_{ ext{external}} = 1.01 \times 10^{5} \ ext{Pa}\)). These pressures exert forces on the cap. The difference in these forces due to pressure difference will define the net force that acts on the cap.
02

Calculate Pressure Difference

To find the pressure difference between the inside and outside of the bottle:\[\Delta P = P_{ ext{internal}} - P_{ ext{external}}\]Substitute the given values:\[\Delta P = 1.80 \times 10^{5} \ ext{Pa} - 1.01 \times 10^{5} \ ext{Pa} = 0.79 \times 10^{5} \ ext{Pa}\]
03

Calculate the Force Exerted by the Pressure Difference

The force exerted by the pressure difference on one side of the cap is found using:\[F = \Delta P \cdot A\]where \(A = 4.10 \times 10^{-4} \ ext{m}^2\) is the area of the cap. Plug in the values to solve for the force:\[F = 0.79 \times 10^{5} \ ext{Pa} \times 4.10 \times 10^{-4} \ ext{m}^2 = 32.39 \ ext{N}\]
04

Consider Equal and Opposite Forces

The thread exerts a force to counteract this pressure difference. Therefore, the magnitude of the force exerted by the screw thread to keep the cap in place must be equal to 32.39 N to balance the forces.

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Key Concepts

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

Force Calculation
Understanding how forces are calculated is essential for grasping physics problems. In this scenario, the force we need to calculate is a mechanical force exerted by the pressure difference between the carbon dioxide inside the bottle and the air outside. Force can be simply defined using the equation:
\[ F = ext{Pressure} imes ext{Area} \]
Pressure is a measure of how much force is applied over a given area. The standard unit is Pascal (Pa), which is equal to one Newton per square meter. In any scenario involving pressure, identifying the areas and pressures at play is crucial. Here, the cap's area is given as constant along with the internal and external pressures. With these values, you can determine the force exerted by a specific pressure over a defined area.
Pressure Difference
Pressure difference is a cornerstone concept in understanding how gases and fluids influence each other through force. In the case of a sealed soda bottle, there are two primary pressures: internal pressure from the gas and external atmospheric pressure. The pressure difference determines the net force experienced by the bottle cap.
  • Internal Pressure: This is the pressure from the carbon dioxide inside the bottle, which in this problem is given as \(1.80 \times 10^5 \text{ Pa}\).

  • External Pressure: This is the atmospheric pressure acting on the outside, equal to \(1.01 \times 10^5 \text{ Pa}\).
To find the effective force, we first need to calculate the pressure difference (\(\Delta P\)). This is done by subtracting the external pressure from the internal pressure, resulting in a difference of \(0.79 \times 10^5 \text{ Pa}\). This difference is crucial, as it will directly inform the force calculation used to determine how the cap behaves under these conditions.
Screw Cap Mechanics
The mechanics of a screw cap are fascinating, as they involve the interplay of forces to keep the cap securely on the bottle. In this exercise, the thread of the screw cap counteracts the pressure exerted by the internal gas pressure difference. How does this work?
A thread essentially applies a force that opposes the force produced by the pressure difference. This needs to match exactly to prevent the cap from popping off or loosening. According to Newton's third law, for every action, there is an equal and opposite reaction. Thus, the screw thread forces must equal the force exerted by the pressure difference.
The calculated force using the pressure difference and the cap area, as determined previously, needs to be balanced by the screw cap's mechanics with a force of \(32.39 \text{ N}\). This ensures the cap remains in place despite the internal pressure exertion from the soda's carbon dioxide. Understanding this balance is crucial for designing caps able to withstand various internal pressures.

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Most popular questions from this chapter

Poiseuille's law remains valid as long as the fluid flow is laminar. For sufficiently high speed, however, the flow becomes turbulent, even if the fluid is moving through a smooth pipe with no restrictions. It is found experimentally that the flow is laminar as long as the Reynolds number Re is less than about 2000: \(\mathrm{Re}=2 \bar{v} \rho R / \eta .\) Here \(\bar{v}, \rho,\) and \(\eta\) are, respectively, the average speed, density, and viscosity of the fluid, and \(R\) is the radius of the pipe. Calculate the highest average speed that blood \(\left(\rho=1060 \mathrm{~kg} / \mathrm{m}^{3}, \eta=4.0 \times 10^{-3} \mathrm{~Pa} \cdot \mathrm{s}\right)\) could have and still remain in laminar flow when it flows through the aorta \(\left(R=8.0 \times 10^{-3} \mathrm{~m}\right)\).

In the process of changing a flat tire, a motorist uses a hydraulic jack. She begins by applying a force of \(45 \mathrm{~N}\) to the input piston, which has a radius \(r_{1}\). As a result, the output plunger, which has a radius \(r_{2}\), applies a force to the car. The ratio \(r_{2} / r_{1}\) has a value of 8.3. Ignore the height difference between the input piston and output plunger and determine the force that the output plunger applies to the car.

Interactive Solution \(\underline{11.73}\) at illustrates a model for solving this problem. A pressure difference of \(1.8 \times 10^{3} \mathrm{~Pa}\) is needed to drive water \(\left(\eta=1.0 \times 10^{-3} \mathrm{~Pa} \cdot \mathrm{s}\right)\) through a pipe whose radius is \(5.1 \times 10^{-3} \mathrm{~m} .\) The volume flow rate of the water is \(2.8 \times 10^{-4} \mathrm{~m}^{3} / \mathrm{s} .\) What is the length of the pipe?

The human lungs can function satisfactorily up to a limit where the pressure difference between the outside and inside of the lungs is one-twentieth of an atmosphere. If a diver uses a snorkel for breathing, how far below the water can she swim? Assume the diver is in salt water whose density is \(1025 \mathrm{~kg} / \mathrm{m}^{3}\)

Neutron stars consist only of neutrons and have unbelievably high densities. A typical mass and radius for a neutron star might be \(2.7 \times 10^{28} \mathrm{~kg}\) and \(1.2 \times 10^{3} \mathrm{~m}\). (a) Find the density of such a star. (b) If a dime \(\left(V=2.0 \times 10^{-7} \mathrm{~m}^{3}\right)\) were made from this material, how much would it weigh (in pounds)?
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