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Chapter 11: Problem 12

A person who weighs \(625\) \(\mathrm{N}\) is riding a 98 -N mountain bike. Suppose that the entire weight of the rider and bike is supported equally by the two tires. If the pressure in each tire is \(7.60 \times 10^{5} \mathrm{Pa}\), what is the area of contact between each tire and the ground?

Short Answer

Expert verified
The contact area is approximately \(4.755 \times 10^{-4} \text{ m}^2\) per tire.

Step by step solution

01

Sum the Weights

First, we find the total weight supported by both tires, which is the sum of the weight of the person and the weight of the bike. We have \(625 \, \text{N}\) for the person and \(98 \, \text{N}\) for the bike. Thus, \[\text{Total Weight} = 625 \, \text{N} + 98 \, \text{N} = 723 \, \text{N}.\]
02

Weight Supported by Each Tire

The total weight is supported equally by both tires. Therefore, the weight supported by each tire is half of the total weight. Thus,\[\text{Weight per Tire} = \frac{723 \, \text{N}}{2} = 361.5 \, \text{N}.\]
03

Use Pressure Formula

We know that pressure \(P\) is equal to force \(F\) divided by area \(A\). Mathematically, \[P = \frac{F}{A}.\] Rearranging the formula to solve for area, we have \[A = \frac{F}{P}.\]
04

Plug in the Known Values

Using the weight supported by one tire as the force and the given pressure in the tire, we calculate the area of contact:\[A = \frac{361.5 \, \text{N}}{7.60 \times 10^5 \, \text{Pa}}.\]
05

Calculate the Contact Area

Perform the division to find the area:\[A = \frac{361.5}{7.60 \times 10^5} = 4.755 \times 10^{-4} \, \text{m}^2.\] Thus, the contact area between each tire and the ground is approximately \(4.755 \times 10^{-4} \text{ m}^2\).

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Key Concepts

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

Weight Distribution
Weight distribution is an important concept when dealing with objects placed on surfaces. It refers to how weight is spread across different points or areas. In our exercise, the weight of a person and a bike is distributed equally between the two tires of a mountain bike.

When we sum up the weights of both the rider and the bike, we get a total weight force. This force is the total downward force exerted on the ground by the bike and rider combined. After calculating the total weight to be 723 N, it is important to understand that this weight is equally borne by both tires. Each tire, therefore, supports half of the total weight, which makes it 361.5 N per tire.

Understanding weight distribution helps us in finding out the force each tire exerts on the ground, which is a critical step in solving pressure-related problems, as it affects the calculation of contact pressure.
Contact Area Calculation
Contact area calculation is about finding the surface area through which a force is transmitted between two bodies. In this exercise, we need to determine the area where each tire touches the ground.

Contact area influences not only pressure but also the stability and friction between surfaces. Generally, larger contact areas distribute forces over a wider region, potentially reducing the pressure exerted on that area.

To determine the contact area of each tire, we use the pressure formula in physics. Given that each tire supports a force of 361.5 N and knowing the pressure inside the tire is 7.60 x 10^5 Pa, we can rearrange the pressure formula to solve for the contact area. Substituting the known values into the equation, we calculate the area as 4.755 x 10^-4 m^2. This value represents the size of the contact area between each bike tire and the ground.
Pressure Formula in Physics
The pressure formula is a fundamental element in physics used to determine the force exerted over an area. Pressure is defined as the force applied per unit area and is expressed by the formula:\[P = \frac{F}{A}\]

Where:
  • \( P \) is the pressure,
  • \( F \) is the force,
  • \( A \) is the area.
In this exercise, we use this formula not to find the pressure (since it was given), but to find the area. To do this, the formula is rearranged:\[A = \frac{F}{P}\]

With the known values of force and pressure, you can plug in to find the contact area of the tire. This inverse relationship means that by knowing the pressure and the force, you can determine how much area is in contact.Understanding and applying the pressure formula is vital for solving many real-world physics problems related to contact between surfaces, fluid dynamics, and more.

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Most popular questions from this chapter

An airplane wing is designed so that the speed of the air across the top of the wing is \(251\) \(\mathrm{m} / \mathrm{s}\) when the speed of the air below the wing is \(225\) \(\mathrm{m} / \mathrm{s}\). The density of the air is \(1.29\) \(\mathrm{kg} / \mathrm{m}^{3} .\) What is the lifting force on a wing of area \(24.0\) \(\mathrm{m}^{2} ?\)

A hollow cubical box is \(0.30\) \(\mathrm{m}\) on an edge. This box is floating in a lake with one-third of its height beneath the surface. The walls of the box have a negligible thickness. Water from a hose is poured into the open top of the box. What is the depth of the water in the box just at the instant that water from the lake begins to pour into the box from the lake?

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 \(0.85 \times 10^{5}\) Pa. The inside of the box is completely evacuated. What is the magnitude of the force required to pull the lid off the box?

A mercury barometer reads \(747.0\) \(\mathrm{mm}\) on the roof of a building and \(760.0\) \(\mathrm{mm}\) on the ground. Assuming a constant value of \(1.29\) \(\mathrm{kg} / \mathrm{m}^{3}\) for the density of air, determine the height of the building.

An 81 -kg person puts on a life jacket, jumps into the water, and floats. The jacket has a volume of \(3.1 \times 10^{-2} \mathrm{m}^{3}\) and is completely submerged under the water. The volume of the person's body that is under water is \(6.2 \times 10^{-2} \mathrm{m}^{3} .\) What is the density of the life jacket?
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