A 950-kg race car can drive around an unbanked turn at a maximum speed of 46 m/s without slipping. The turn has a radius of 120 m. Air flowing over the car's wing exerts a downward-pointing force (called the downforce) of 12000 N on the car. (a) What is the coefficient of static friction between the track and the car's tires? (b) What would be the maximum speed if no downforce acted on the car?

a) μ = Fc/FN (i)

FN =(950x9.8)+12000= 21310
Fc =(950)(46)^2/120= 16751
Putting values in (i) μ=0.7860 (no unit)

b) v= sqrt(rμg)= sqrt[(120)(0.7860)(9.8)]
v= 30.40 m/s^2

To solve these questions, we need to use the centripetal force equation and relate it to the forces acting on the car.

(a) Let's start by calculating the net force on the car for the given maximum speed. The net force is directed towards the center of the turn and is given by the equation:

F_net = m * v^2 / r

where F_net is the net force, m is the mass of the car, v is the velocity, and r is the radius of the turn.

Plugging in the given values:

F_net = 950 kg * (46 m/s)^2 / 120 m
= 16158.33 N

The net force is equal to the sum of the force of static friction and the downward-pointing force (downforce) exerted by the air on the car's wing:

F_net = F_friction + F_downforce

F_friction = F_net - F_downforce
= 16158.33 N - 12000 N
= 4158.33 N

The force of friction is given by the equation:

F_friction = μ * F_normal

where μ is the coefficient of static friction and F_normal is the normal force.

Since the car is not slipping, the normal force is equal to the gravitational force acting on the car:

F_normal = m * g

where g is the acceleration due to gravity, approximately 9.8 m/s^2.

F_friction = μ * m * g

Now we can solve for the coefficient of static friction:

μ = F_friction / (m * g)
= 4158.33 N / (950 kg * 9.8 m/s^2)
≈ 0.44

Therefore, the coefficient of static friction between the track and the car's tires is approximately 0.44.

(b) To find the maximum speed without any downforce acting on the car, we will use the same equation:

F_net = m * v^2 / r

This time, the downward-pointing force (downforce) is not acting on the car:

F_net = F_friction

F_friction = μ * F_normal

F_friction = μ * m * g

Setting F_friction equal to the net force:

μ * m * g = m * v^2 / r

Now we can solve for the maximum speed:

v = √(μ * g * r)

Plugging in the given values:

v = √(0.44 * 9.8 m/s^2 * 120 m)
= √(509.76)
≈ 22.6 m/s

Therefore, the maximum speed without any downforce acting on the car would be approximately 22.6 m/s.

To find the coefficient of static friction between the track and the car's tires, we need to consider the forces acting on the car while it is turning at maximum speed without slipping.

Let's start by considering the forces acting on the car:

1. The car's weight (W): This is given by W = mg, where m is the mass of the car (950 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2). So, W = 950 kg * 9.8 m/s^2 = 9310 N.

2. The normal force (N): According to Newton's third law, the normal force is equal in magnitude and opposite in direction to the downward force acting on the car (downforce). So, N = 12000 N.

3. The centripetal force (Fc): This is the force that keeps the car moving in a circular path. It is given by Fc = (m * v^2) / r, where m is the mass of the car, v is the velocity of the car, and r is the radius of the turn. So, Fc = (950 kg * (46 m/s)^2) / 120 m = 169591.67 N.

Now, let's analyze the forces in the horizontal direction:

The only force acting in the horizontal direction is the force of static friction (fs). The maximum value of fs can be expressed as fs = μs * N, where μs is the coefficient of static friction.

At maximum speed without slipping, the force of static friction must provide the centripetal force:

fs = Fc = (950 kg * (46 m/s)^2) / 120 m = 169591.67 N.

Now, we can determine the coefficient of static friction (μs):

μs * N = 169591.67 N,
μs * 12000 N = 169591.67 N,

Dividing both sides of the equation by 12000 N:

μs = 169591.67 N / 12000 N ≈ 14.13.

Therefore, the coefficient of static friction between the track and the car's tires is approximately 14.13.

Now, let's find the maximum speed if no downforce acted on the car.

Without the downforce, the normal force (N) would only be equal to the weight (W) of the car, which is 9310 N. Using the same equation as before to find the maximum speed without slipping:

fs = Fc,
μs * N = (m * v^2) / r.

Substituting the values:

μs * 9310 N = (950 kg * v^2) / 120 m.

Now, we can solve for v:

v^2 = (μs * 9310 N * 120 m) / 950 kg,
v^2 = 14.13 * 9310 N * 120 m / 950 kg.

Taking the square root of both sides:

v ≈ √(14.13 * 9310 N * 120 m / 950 kg).

Calculating this expression will give you the maximum speed without downforce acting on the car.