91Ó°ÊÓ

Pressure calculation is a common task in physics, especially when dealing with forces distributed over a specific area. In this problem, we are interested in how pressure relates to the weight of a person and their mountain bike, which is transmitted through the tires to the ground.

The concept of pressure is defined as the force exerted per unit area. Mathematically, this is expressed as:

• \( P = \frac{F}{A} \)

where \( P \) is the pressure, \( F \) is the force (or total weight here), and \( A \) is the area of contact.

To solve the exercise, we took the combined weight of the rider and bike, which is supported by both tires equally. We calculated the pressure exerted through the contact area of each tire to find out how much area the tires need to support the given pressure. Understanding this relationship helps in assessing the tire's functionality and ensuring safety through proper pressure management.

The contact area in this context refers to the portion of the tire that makes direct contact with the ground. This is crucial because it determines how the weight of the rider and the mountain bike is distributed over the surface they are on. Understanding this concept helps us manage and predict wear and tear on tires, efficiency, and safety of the ride.

To find the contact area of each tire, we rearrange the pressure formula to solve for the area:

• \( A = \frac{F}{P} \)

In this specific problem, knowing the force on each tire (half the total weight) and the pressure within the tire walls, we calculated that the contact area is approximately \( 4.76 \times 10^{-4} \) square meters per tire.

By understanding how to find the contact area, we gain insight into material limits and optimal pressure for performance and safety.

Newton's laws of motion are foundational principles in physics that help us understand and describe how objects move and interact. They lay the groundwork for most calculations involving forces and motion.

In the context of this exercise, the balancing of weight across the two tires implicitly involves Newton's laws, particularly the second law, which states:

• "The acceleration of an object depends on the net force acting upon the object and the mass of the object."
• Expressed as \( F = ma \), though in this static situation, we're primarily dealing with forces.

Given that the system isn’t accelerating (since the bike is stationary), Newton's first law also subtly plays a role: an object at rest will stay at rest unless acted upon by a net force. The bike and rider's weight pushed down while the equal force from the ground pushed up, providing equilibrium.

Newton's third law, "For every action, there's an equal and opposite reaction," is evident in how the ground pushes back against the tire with an equal force to support the rider and bike. Understanding these laws helps us handle calculations involving contact forces effectively.