⦁ The coefficient of Statin Friction between tires or a car and a dry road is 0.62. The mass of car is 1500 kg. What maximum braking force is obtainable (a) on a level road

and (b) on an 8 6° downgrade?

To determine the maximum braking force that can be achieved by a car on different road conditions, we need to consider the coefficient of static friction, the mass of the car, and the effects of gravity on an incline. Let's break this down step by step.

Understanding the Basics

The maximum braking force can be calculated using the formula:

F_max = μ * N

Where:

• F_max is the maximum braking force.
• μ is the coefficient of static friction (0.62 in this case).
• N is the normal force, which is affected by the weight of the car and any incline.

Calculating on a Level Road

On a level road, the normal force (N) is simply the weight of the car, which can be calculated using:

N = m * g

Here, m is the mass of the car (1500 kg) and g is the acceleration due to gravity (approximately 9.81 m/s²).

Calculating the normal force:

N = 1500 kg * 9.81 m/s² = 14715 N

Now, substituting this value into the maximum braking force formula:

F_max = 0.62 * 14715 N = 9145.3 N

Considering an 8° Downward Slope

When the car is on a downgrade, the normal force is reduced due to the component of gravitational force acting parallel to the slope. The normal force can be calculated as:

N = m * g * cos(θ)

Where θ is the angle of the slope (8°). First, we need to find the cosine of 8°:

cos(8°) ≈ 0.9903

Now, we can calculate the normal force on the slope:

N = 1500 kg * 9.81 m/s² * 0.9903 ≈ 14500.5 N

Next, we substitute this value into the maximum braking force formula:

F_max = 0.62 * 14500.5 N ≈ 8991.3 N

Summary of Results

To summarize:

• Maximum braking force on a level road: 9145.3 N
• Maximum braking force on an 8° downgrade: 8991.3 N

This analysis shows how the incline affects the braking force due to changes in the normal force. Understanding these principles is crucial for safe driving and vehicle dynamics.