An 825-kg race car can drive around an unbanked turn at a maximum speed of 59 m/s without slipping. The turn has a radius of curvature of 135 m. Air flowing over the car's wing exerts a downward-pointing force (called the downforce) of 2700 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?
centripetal force=m*v^2/r
normal force=friction=(mg+2700)mu
set normal force = centripetal force
solve for mu.
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 when it is driving around the unbanked turn.
a) The maximum speed without slipping occurs when the force of static friction between the tires and the track is equal to the maximum force that can be provided by static friction.
1. Identify the forces acting on the car:
- The force of static friction, \(f_{\text{static}}\), between the tires and the track.
- The normal force, \(N\), exerted by the track on the car.
- The weight of the car, \(W\), which is equal to the mass of the car multiplied by the acceleration due to gravity, \(mg\).
2. Determine the maximum force of static friction:
- The maximum force of static friction is equal to the coefficient of static friction, \(\mu_{\text{static}}\), multiplied by the normal force: \(f_{\text{max}} = \mu_{\text{static}}N\).
3. Write the equation for horizontal forces:
- In the horizontal direction, the net force is equal to \(f_{\text{static}}\): \(f_{\text{static}} = \frac{mv^2}{r}\), where \(m\) is the mass of the car and \(r\) is the radius of curvature of the turn.
4. Write the equation for vertical forces:
- In the vertical direction, the weight of the car is equal to the sum of the normal force and the downward force due to the downforce: \(W = N + F_{\text{downforce}}\).
5. Substitute the maximum force of static friction into the equation for horizontal forces:
- \(\frac{mv^2}{r} = \mu_{\text{static}}N\).
6. Solve for the coefficient of static friction, \(\mu_{\text{static}}\):
- Rearranging the equation: \(\mu_{\text{static}} = \frac{mv^2}{Nr}\).
Now, let's calculate the coefficient of static friction.
Given:
- Mass of the car, \(m = 825 \, \text{kg}\).
- Maximum speed, \(v = 59 \, \text{m/s}\).
- Radius of curvature, \(r = 135 \, \text{m}\).
- Downforce, \(F_{\text{downforce}} = 2700 \, \text{N}\).
First, we need to find the normal force, \(N\):
The weight of the car can be calculated using the formula: \(W = mg\), where \(g\) is the acceleration due to gravity.
Considering that the acceleration due to gravity is approximately \(9.8 \, \text{m/s}^2\), we have:
\(W = (825 \, \text{kg})(9.8 \, \text{m/s}^2)\).
Then, we can write the equation for vertical forces:
\(W = N + F_{\text{downforce}}\).
Rearranging the equation to find \(N\):
\(N = W - F_{\text{downforce}}\).
Next, we can substitute the known values into the equation for the coefficient of static friction:
\(\mu_{\text{static}} = \frac{(825 \, \text{kg})(59 \, \text{m/s})^2}{(W - F_{\text{downforce}})(135 \, \text{m})}\).
Now, we can calculate the coefficient of static friction.
b) To find the maximum speed without the downforce, we need to calculate the coefficient of static friction in the absence of the downward-pointing force.
The steps for finding the maximum speed are the same as in part a), except we assume there is no downward force (i.e., \(F_{\text{downforce}}\) is equal to zero).
So the equation for the coefficient of static friction becomes:
\(\mu_{\text{static}} = \frac{(825 \, \text{kg})(v^2)}{(W)(135 \, \text{m})}\).
Now, we can calculate the maximum speed without the downforce.