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Chapter 11: Problem 11

An airtight box has a removable lid of area \(1.3 \times 10^{-2} \mathrm{m}^{2}\) and negligible weight. The box is taken up a mountain where the air pressure outside the box is \(0.85 \times 10^{5} \mathrm{Pa}\) . The inside of the box is completely evacuated. What is the magnitude of the force required to pull the lid off the box?

Short Answer

Expert verified
The force required is \(1105 \mathrm{N}\).

Step by step solution

01

Understand the Problem

We have an airtight box with a removable lid. The box is taken to a mountain where the air pressure outside is known, and the inside is completely evacuated (meaning the pressure inside is zero). We need to find the force required to lift the lid off the box.
02

Identify Given Values

We are given the area of the lid as \(A = 1.3 \times 10^{-2} \mathrm{m}^{2}\) and the outside pressure \(P_{\text{outside}} = 0.85 \times 10^{5} \mathrm{Pa}\). The pressure inside the box is \(P_{\text{inside}} = 0 \mathrm{Pa}\) because it is evacuated.
03

Calculate Pressure Difference

The force required to remove the lid is due to the pressure difference between the inside and outside of the box. Calculate this pressure difference: \( \Delta P = P_{\text{outside}} - P_{\text{inside}} = 0.85 \times 10^{5} \mathrm{Pa} - 0 \mathrm{Pa} = 0.85 \times 10^{5} \mathrm{Pa} \).
04

Apply the Force Equation

The force required to remove the lid due to the pressure difference is calculated using the equation: \(F = \Delta P \times A\). Substitute the known values: \( F = 0.85 \times 10^{5} \mathrm{Pa} \times 1.3 \times 10^{-2} \mathrm{m}^{2} \).
05

Perform the Calculation

Multiply the pressure difference by the area of the lid to find the force: \(F = 0.85 \times 10^{5} \times 1.3 \times 10^{-2} = 1.105 \times 10^{3} \mathrm{N}\).
06

State the Result

The magnitude of the force required to pull the lid off the box is \(1105 \mathrm{N}\).

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Key Concepts

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

Air Pressure
Air pressure is the force exerted by the weight of air in the Earth's atmosphere. This force acts on every surface, whether it's a plane, a mountain top, or an airtight box on that mountain.
On the mountain top, where the exercise is set, the air pressure is different than at sea level. Usually, at higher altitudes, air pressure decreases because there is less air above the surface to exert pressure.
  • At sea level, standard air pressure is often around \(1.01 \/\, 10^5 \/\, \text{Pa}\).
  • In the exercise, the air pressure is \(0.85 \/\, 10^5 \/\, \text{Pa}\).
Air pressure is crucial for understanding why objects experience forces when placed in environments with varying pressures, especially when the inside of a container is evacuated, creating a vacuum.
Pressure Difference
Pressure difference, as its name implies, refers to the difference in pressure between two points. In our exercise, one point is the inside of the box and the other is the outside of the box on the mountain.

When the box is evacuated, the pressure inside becomes \(0 \/\, \text{Pa}\), meaning it has no air to exert any force. Conversely, the pressure outside remains at \(0.85 \/\, 10^5 \/\, \text{Pa}\). The difference in pressure is what creates a net force that tries to keep the lid sealed on the box.
  • Pressure difference \( \Delta P = P_{\text{outside}} - P_{\text{inside}} = 0.85 \/\, 10^5 \/\, \text{Pa}\).
This pressure difference is fundamental to determining the force required to remove the lid because it creates an unequal force distribution across the area of the lid.
Force Calculation
Force calculation involves determining the force experienced by an object as a result of pressure differences and areas.
The fundamental equation used is \( F = \Delta P \times A \), where \( F \) is the force, \( \Delta P \) is the pressure difference, and \( A \) is the area over which this pressure acts.
  • In the exercise, we have an area of \( A = 1.3 \/\, 10^{-2} \/\, \text{m}^2 \).
  • The calculated force is \( F = 0.85 \/\, 10^5 \/\, \text{Pa} \times 1.3 \/\, 10^{-2} \/\, \text{m}^2 = 1.105 \/\, 10^3 \/\, \text{N} \).
This force is what would be felt if attempting to lift the lid. It shows how pressure and area interact to create a tangible force that can be measured and counteracted, showing the real-world applications of these calculations.

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Most popular questions from this chapter

A paperweight, when weighed in air, has a weight of \(W=6.9 \mathrm{N}\) . When completely immersed in water, however, it has a weight of \(W_{\text { in water }}=4.3 \mathrm{N}\) . Find the volume of the paperweight.

A water line with an internal radius of \(6.5 \times 10^{-3} \mathrm{m}\) is connected to a shower head that has 12 holes. The speed of the water in the line is 1.2 \(\mathrm{m} / \mathrm{s}\) . (a) What is the volume flow rate in the line? (b) At what speed does the water leave one of the holes (effective hole radius \(=4.6 \times 10^{-4} \mathrm{m}\) ) in the head?

Mercury is poured into a tall glass. Ethyl alcohol (which does not mix with mercury) is then poured on top of the mercury until the height of the ethyl alcohol itself is 110 cm. The air pressure at the top of the ethyl alcohol is one atmosphere. What is the absolute pressure at a point that is 7.10 cm below the ethyl alcohol鈥搈ercury interface?

Three fire hoses are connected to a fire hydrant. Each hose has a radius of 0.020 m. Water enters the hydrant through an underground pipe of radius 0.080 m. In this pipe the water has a speed of 3.0 m/s. (a) How many kilograms of water are poured onto a fire in one hour by all three hoses? (b) Find the water speed in each hose.

One of the concrete pillars that support a house is 2.2 \(\mathrm{m}\) tall and has a radius of 0.50 \(\mathrm{m}\) . The density of concrete is about \(2.2 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\) . Find the weight of this pillar in pounds \((1 \mathrm{N}=0.2248 \mathrm{b})\) .
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