Question

Гоночный автомобиль едет по треку, имеющему на повороте радиус R - 50 м, со скоростью 28 м/с. Определите, не вылетит ли с трека автомобиль. Коэффициент трения колёс автомобиля о дорогу равен 0,4. Сделайте рисунок, расставьте все силы, действующие на автомобиль.

Answer 1

Ответ:

Автомобиль некоторое время будет двигаться по инерции – нужно узнать, будет ли сила инерции больше центростремительной силы, которую представляет сила трения в данном случае – ведь она не даёт нам "соскальзывать".

Пусть масса тела равна m, а сила инерции – F;

Должно выполняться равенство: F < Fтр

F = mv² / R – как центробежная сила

F = 15,68m (СИ)

Fтр = μmg – закон Кулона-Амонтона

Fтр = 3,924m (СИ)

Как видим, F > Fтр, а значит автомобиль соскользнёт с трека.

Fтр будет направлено к центру окружности, по которому поворачивается тело, а F – противоположно ей. Причём F больше Fтр в 4 раза.

Объяснение:

Answer 2

Problem Analysis

We are given the following information: - The racing car is moving on a track with a radius of R = 50 m. - The car is moving with a speed of 28 m/s. - The coefficient of friction between the car's wheels and the road is 0.4.

We need to determine whether the car will stay on the track or not and draw a diagram showing all the forces acting on the car.

Solution

To determine whether the car will stay on the track, we need to analyze the forces acting on the car at the turn.

Forces Acting on the Car

1. Centripetal Force (Fc): This force is responsible for keeping the car moving in a circular path. It is directed towards the center of the circle and is given by the equation:

Fc = (m * v^2) / R

Where: - m is the mass of the car. - v is the velocity of the car. - R is the radius of the turn.

2. Frictional Force (Ff): This force acts in the opposite direction to the car's motion and is responsible for providing the necessary centripetal force. It can be calculated using the equation:

Ff = μ * N

Where: - μ is the coefficient of friction. - N is the normal force acting on the car.

3. Weight (W): This force acts vertically downwards and is given by:

W = m * g

Where: - g is the acceleration due to gravity.

Analysis

To determine whether the car will stay on the track, we need to compare the centripetal force (Fc) with the frictional force (Ff). If the frictional force is greater than or equal to the centripetal force, the car will stay on the track. Otherwise, it will slide off.

Let's calculate the forces and compare them.

Calculation

Given: - Radius of the turn (R) = 50 m - Velocity of the car (v) = 28 m/s - Coefficient of friction (μ) = 0.4

We need to calculate the centripetal force (Fc) and the frictional force (Ff).

1. Centripetal Force (Fc): Fc = (m * v^2) / R 2. Frictional Force (Ff): Ff = μ * N

To calculate the normal force (N), we can use the equation: N = m * g

Substituting the value of N in the equation for Ff, we get: Ff = μ * (m * g)

Since the weight (W) is equal to the mass (m) multiplied by the acceleration due to gravity (g), we can rewrite the equation as: Ff = μ * W

Substituting the value of W, we get: Ff = μ * (m * g)

Now, let's calculate the forces.

Calculation Results

1. Centripetal Force (Fc): Fc = (m * v^2) / R Fc = (m * (28 m/s)^2) / 50 m

2. Frictional Force (Ff): Ff = μ * (m * g) Ff = 0.4 * (m * 9.8 m/s^2)

Conclusion

To determine whether the car will stay on the track, we need to compare the centripetal force (Fc) with the frictional force (Ff). If the frictional force is greater than or equal to the centripetal force, the car will stay on the track. Otherwise, it will slide off.

To draw a diagram showing all the forces acting on the car, we need to consider the centripetal force (Fc), the frictional force (Ff), and the weight (W).

Please note that the exact values of the forces cannot be determined without knowing the mass of the car. However, we can compare the forces and determine whether the car will stay on the track based on their relative magnitudes.

Diagram

Here is a diagram showing the forces acting on the car:

          Fc
          ↑
          │
          │
          │
          │
Ff ←─────┼─────→ Ff
          │
          │
          │
          │
          ↓
          W

Please note that the direction of the forces may vary depending on the specific orientation of the turn. The diagram above represents a general case.

Based on the comparison of the forces, we can determine whether the car will stay on the track or not.