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A person who weighs 625 \(\mathrm{N}\) is riding a \(98-\mathrm{N}\) 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 between each tire and the ground is approximately \( 4.76 \times 10^{-4} \text{ m}^2 \).

Step by step solution

01

Understand the Problem

The problem involves finding the area of contact between each tire and the ground under specific conditions. It provides the total weight of the rider and the bike and the pressure inside the tires.
02

Calculate Total Weight

First, compute the total weight supported by the two tires. The total weight is the sum of the rider's weight and the bike's weight. In this case:\[ W = 625 \text{ N} + 98 \text{ N} = 723 \text{ N} \]
03

Determine Weight on Each Tire

Since the entire weight is supported equally by the two tires, each tire supports half of the total weight. Therefore, the weight on each tire is:\[ \text{Weight on each tire} = \frac{723 \text{ N}}{2} = 361.5 \text{ N} \]
04

Use Pressure Formula to Find Area

The pressure on each tire can be expressed as the force per unit area, where force is the weight on each tire. Use the formula for pressure \( P = \frac{F}{A} \), where \( F \) is the force (weight on each tire) and \( A \) is the area. Rearrange to solve for contact area \( A \):\[ A = \frac{F}{P} = \frac{361.5 \text{ N}}{7.60 \times 10^{5} \text{ Pa}} \]
05

Compute the Contact Area

Substitute the known values into the equation to obtain:\[ A = \frac{361.5}{7.60 \times 10^{5}} \approx 4.76 \times 10^{-4} \text{ m}^2 \]

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Key Concepts

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

Contact Area Calculation
In physics, understanding how to calculate the contact area is crucial for solving problems involving pressure and force. The contact area is where two surfaces meet, and it plays a significant role in determining the pressure exerted on a surface. To compute the contact area, you need to know the force acting on the surface and the pressure exerted by the surface.
The formula to calculate the contact area (\( A \)) is derived from the pressure formula:
  • Pressure is defined as the force per unit area: \( P = \frac{F}{A} \).
  • By rearranging this equation, we can solve for the area: \( A = \frac{F}{P} \).
This calculation is essential when working with problems involving objects supported by surfaces, like tires supporting a bicycle and rider. Calculating the contact area helps determine the distribution of force across the surface in contact with the ground.
Force Distribution
Force distribution explains how a force is spread over an area or different parts of a system. When dealing with problems concerning pressure, it's important to understand how force is distributed.
For instance, in our exercise, the total weight of the rider and the bike (723 N) is equally distributed between two tires. This means each tire supports half of the total weight:
  • Weight on each tire = \( \frac{723 \text{ N}}{2} = 361.5 \text{ N} \).
By distributing the force evenly, you ensure that each tire exerts an equal amount of pressure on the ground, which influences the contact area and helps maintain balance and stability.
Pressure Formula
The pressure formula is a fundamental concept in physics that describes the relationship between force and area. Pressure is defined as the force acting perpendicular to a surface, divided by the area over which the force is distributed.
The formula is expressed as:
  • \( P = \frac{F}{A} \)
Where \( P \) is the pressure, \( F \) is the force, and \( A \) is the area.Pressure units are typically in pascals (Pa), where 1 Pa equals 1 Newton per square meter. Understanding how to rearrange this formula allows you to calculate not only the contact area, but also solve for force or pressure, depending on the given variables in a problem. It's this rearrangement that we used to find contact area: \( A = \frac{F}{P} \). This highlights how linked the concepts of force, area, and pressure are.
Problem-Solving Steps in Physics
Effective problem-solving in physics often involves a systematic approach. Understanding this process can aid in tackling various physics scenarios, like the one in our problem.
Here's a step-by-step approach:
  • **Step 1: Comprehend the Problem** - Understand what is being asked. Identify the known and unknown quantities.
  • **Step 2: Analyze the Problem** - Break it down into smaller, more manageable parts.
  • **Step 3: Use the Right Formulas** - Apply relevant physics principles and formulas, like the pressure formula.
  • **Step 4: Solve for the Unknown** - Rearrange equations if necessary to solve for the desired variable.
  • **Step 5: Review** - Check your solution for consistency and accuracy.
Applying these steps makes it easier to navigate the complexities of physics problems and offers a structured way to arrive at a solution.

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Most popular questions from this chapter

A small crack occurs at the base of a \(15.0-\mathrm{m}\) -high dam. The effective crack area through which water leaves is \(1.30 \times 10^{-3} \mathrm{m}^{2}\) (a) Ignoring viscous losses, what is the speed of water flowing through the crack? (b) How many cubic meters of water per second leave the dam?

A hydrometer is a device used to measure the density of a liquid. It is a cylindrical tube weighted at one end, so that it floats with the heavier end downward. The tube is contained inside a large 鈥渕edicine dropper,鈥 into which the liquid is drawn using the squeeze bulb (see the drawing). For use with your car, marks are put on the tube so that the level at which it floats indicates whether the liquid is battery acid (more dense) or antifreeze (less dense). The hydrometer has a weight of \(W=5.88 \times 10^{-2} \mathrm{N}\) and a cross-sectional area of tw \(A=7.85 \times 10^{-5} \mathrm{m}^{2} .\) How far from the bottom of the tube should the mark be put that denotes (a) battery acid \(\left(\rho=1280 \mathrm{kg} / \mathrm{m}^{3}\right)\) and (b) antifreeze \(\left(\rho=1073 \mathrm{kg} / \mathrm{m}^{3}\right) ?\)

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 arew 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.

The atmospheric pressure above a swimming pool changes from 755 to 765 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?

A \(0.10-\mathrm{m} \times 0.20-\mathrm{m} \times 0.30-\mathrm{m}\) block is suspended from a wire and is completely under water. What buoyant force acts on the block?

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