A 2000-kg automobile with four-wheel drive and weight distributed equally on the four 60-cm diameter wheels can accelerate from 0 to 27 m/s (60 MPH) in 6 seconds.

a. Calculate the minimum coefficient of static friction between the tires and the ground. (10 pts.)

b. What torque is required on the wheels to achieve this acceleration? (10 pts.)

c. Repeat the calculation in part (a) assuming the automobile has two-wheel drive.

4*2000*g*mu=2000g*(27-0)/6

mu=27/4*6

b. torque=force*.3= 500g*mu*.3

c. change the 4 to 2.

a. To calculate the minimum coefficient of static friction between the tires and the ground, we need to consider the forces involved in the acceleration of the automobile.

First, let's calculate the force required to accelerate the automobile. We can use Newton's second law of motion, which states that force (F) equals mass (m) multiplied by acceleration (a).

Given:
Mass of the automobile (m) = 2000 kg
Acceleration (a) = (27 m/s) / 6 s = 4.5 m/s^2

F = m * a
F = 2000 kg * 4.5 m/s^2
F = 9000 N

Now, let's analyze the forces acting on the automobile during acceleration. The only horizontal force that can provide the required forward acceleration is the frictional force between the tires and the ground. This frictional force can be calculated using the equation:

Frictional force (F_f) = coefficient of static friction (μ) * normal force (N)

For an automobile with weight distributed equally on the four wheels, each wheel carries 25% of the total weight. Therefore, the normal force acting on each wheel is:

Normal force (N) = Weight of the automobile / 4
= (2000 kg * 9.8 m/s^2) / 4
= 4900 N

Now, substituting the known values into the equation for frictional force:

9000 N = μ * 4900 N

Simplifying the equation gives:
μ = 9000 N / 4900 N
μ ≈ 1.84

Therefore, the minimum coefficient of static friction between the tires and the ground is approximately 1.84.

b. To calculate the torque required on the wheels to achieve this acceleration, we need to consider the rotational dynamics of the wheels.

The torque (τ) required to produce linear acceleration can be calculated using the equation:

Torque (τ) = Moment of Inertia (I) * Angular Acceleration (α)

The moment of inertia (I) of a solid cylinder (wheel) rotating about its central axis is given by:

Moment of Inertia (I) = (1/2) * Mass of the wheel (m_wheel) * Radius of the wheel (r_wheel)^2

Given:
Mass of the wheel (m_wheel) = (25% of 2000 kg) = 500 kg
Radius of the wheel (r_wheel) = (60 cm) / 2 = 0.3 m
Angular Acceleration (α) = acceleration (a) / Radius of the wheel (r_wheel)

α = 4.5 m/s^2 / 0.3 m
α = 15 rad/s^2

Now, substituting the known values into the equation for torque:

τ = (1/2) * 500 kg * (0.3 m)^2 * 15 rad/s^2

Simplifying the equation gives:
τ ≈ 1,012.5 N.m

Therefore, the torque required on the wheels to achieve this acceleration is approximately 1,012.5 N.m.

c. If the automobile has two-wheel drive instead of four-wheel drive, the weight distribution changes. In this case, it is generally assumed that each powered wheel carries 40% of the total weight.

The normal force acting on each wheel will be:
Normal force (N) = Weight of the automobile / 2
= (2000 kg * 9.8 m/s^2) / 2
= 9800 N

Now, substituting the known values into the equation for frictional force:

9000 N = μ * 9800 N

Simplifying the equation gives:
μ = 9000 N / 9800 N
μ ≈ 0.92

Therefore, the minimum coefficient of static friction between the tires and the ground for a two-wheel drive automobile is approximately 0.92.