Problem 26 - College Physics

Question

The gauge pressure in both tires of a bicycle is \(690 \mathrm{kPa}\). If the bicycle and the rider have a combined mass of \(90.0 \mathrm{~kg}\), what is the area of contact of each tire with the ground? (Assume that each tire supports half the total weight of the bicycle.)

Step-by-Step Solution

Answer

The area of contact of each tire with the ground is approximately 0.00064 m².

Step 1: Calculate Total Weight

First, calculate the total weight supported by both tires. Use the formula: \[ W = m \cdot g \]where \( m = 90.0 \ kg\) is the combined mass of the bicycle and rider, and \( g = 9.81 \ m/s^2 \) is the acceleration due to gravity. Substituting the values, \[ W = 90.0 \ kg \cdot 9.81 \ m/s^2 = 882.9 \ N \] Thus, the total weight is 882.9 N.

Step 2: Determine Weight Supported by Each Tire

Each tire supports half of the total weight because the bicycle has two tires. Therefore, the weight supported by each tire is:\[ F_{ ext{tire}} = \frac{W}{2} = \frac{882.9 \ N}{2} = 441.45 \ N \]So, each tire supports a force of 441.45 N.

Step 3: Use Pressure Formula to Find Area

The gauge pressure in each tire, \( P = 690 \ kPa \), is given in kilopascals, which we convert to pascals (since 1 kPa = 1000 Pa):\[ P = 690 \times 10^3 \ Pa \]We use the pressure formula:\[ P = \frac{F}{A} \]where \( F = 441.45 \ N \) and \( A \) is the area of contact. Rearranging the formula for area:\[ A = \frac{F}{P} \]Substitute the values:\[ A = \frac{441.45 \ N}{690 \times 10^3 \ Pa} \approx 0.00064 \ m^2 \]Therefore, the area of contact of each tire with the ground is approximately 0.00064 m².

Key Concepts

Gauge Pressure

Gauge pressure is a term used in physics to describe the pressure of a system above atmospheric pressure. It does not include the atmospheric pressure itself. In real-world applications, understanding gauge pressure is vital because several devices, like car and bicycle tires, measure air pressure using this method.
This type of pressure helps indicate how much more pressure is found in a tire compared to the surrounding air.
It's measured in units like Pascals (Pa) or kilopascals (kPa), which are units of force per unit area.

• Atmospheric pressure averages around 101.3 kPa.
• When the gauge pressure reads 690 kPa, it implies the tire pressure is 690 kPa more than the atmospheric pressure.

Understanding gauge pressure is essential when calculating contact areas in scenarios where the force exerted by an object is spread over a surface, such as a bicycle tire on the road.

Contact Area Calculation

The contact area is a fundamental concept in physics, especially when dealing with forces and pressures. It describes the surface area through which an object contacts another surface. This becomes crucial when determining how pressure affects the object in contact.
To calculate the area of contact of a bicycle tire with the ground, one must apply the formula for pressure:

• Pressure (P) is defined as the force (F) per unit area (A).

The formula can be rearranged to solve for area: \[ A = \frac{F}{P} \]In our previous calculation, knowing that:

• Force on each tire is \(441.45 \, N\)
• Gauge pressure is \(690 \, kPa\), which we convert to pascals: \(690,000 \, Pa\)

By substituting these values into the area formula, we find the contact area for each tire. This concept is critical when considering how pressure impacts wear and tear, or the efficiency of motion in wheels.

Bicycle Wheel Physics

Bicycle wheel physics revolves around principles of force, pressure, and contact areas (like gauge pressure discussions).
Understanding these principles is vital for not only calculating the contact area when tire pressure is known but also for understanding the dynamics of motion and stability.

• The weight of a bicycle and its rider is supported by the two wheels.
• Each wheel sustains half of the total weight, promoting stability.

With a combined mass, this weight is calculated as force due to gravity, which is essential in finding the force each tire exerts on the ground.
By comprehending forces and gauge pressures, one can predict how changes in tire pressure might affect traction or the ease of cycling. This physics knowledge helps in optimizing tire pressures for both performance and comfort.