down force = F = 870*9.81 + 11000
mu F = m v^2/R
mu (870*9.81 + 11000) = 870 (41)^2/180
for part b, same mu
mu(870*9.81) = 870 v^2/180
v^2 = 180*9.81*mu
mu F = m v^2/R
mu (870*9.81 + 11000) = 870 (41)^2/180
for part b, same mu
mu(870*9.81) = 870 v^2/180
v^2 = 180*9.81*mu
part b: 27.56
(a) The centripetal force that keeps the car moving in a circle is provided by the friction between the tires and the track. The maximum static friction force that can be exerted between two surfaces is given by the equation:
Fs_max = μs * N
Where Fs_max is the maximum static friction force, μs is the coefficient of static friction, and N is the normal force.
In this case, the downward force exerted by the car's weight is balanced by the upward normal force:
N = mg
Where m is the mass of the car and g is the acceleration due to gravity.
Given:
Mass of the car, m = 870 kg
Downforce, Fd = 11000 N
Acceleration due to gravity, g = 9.8 m/s^2
First, let's find the normal force:
N = mg = 870 kg * 9.8 m/s^2 = 8526 N
Next, let's calculate the frictional force:
Fs_max = μs * N
We know that the maximum speed without slipping is 41 m/s, so we can set up an equation using the centripetal force:
Fs_max = m * (v^2 / r)
Where v is the velocity and r is the radius of the turn.
Substituting the values:
Fs_max = 870 kg * (41 m/s)^2 / 180 m
Now we can equate the two equations for Fs_max:
μs * N = 870 kg * (41 m/s)^2 / 180 m
Rearranging, we can find the coefficient of static friction:
μs = (870 kg * (41 m/s)^2 / 180 m) / (8526 N)
Calculating this expression gives us the value of the coefficient of static friction between the track and the car's tires.