Consider again the problem of a car traveling along a banked turn. Sometimes roads have a "reversed" banking angle. That is, the road is tilted "away" from the center of curvature of the road. If the coefficient of static friction between the tires and the road is μs = 0.4, the radius of curvature is 22 m, and the banking angle is 12°, what is the maximum speed at which a car can safely navigate such a turn?
parallel force due to weight=sin12*cos12*mg
friction: mu*mg*cos12
centripetal+parallel=friction
mv^2/r+mg*sin12*cos12=mu(mg)cos12
solve for v.
Draw a sketch, check that I did it in my head, needs checking
To find the maximum speed at which a car can safely navigate a reverse-banked turn, we need to use the concept of centripetal force and take into account the angle of the banking.
The forces acting on the car in this scenario are the gravitational force (mg) acting vertically downward, the normal force (N) acting perpendicular to the surface of the road, and the static friction force (fs) acting horizontally towards the center of the turn. The normal force can be further split into its horizontal and vertical components (Ncosθ and Nsinθ, respectively), where θ is the banking angle.
In order for the car to safely navigate the turn, the maximum static friction force must be equal to or greater than the centripetal force required to keep the car moving in a curved path. The maximum static friction force can be expressed as μsN, where μs is the coefficient of static friction.
The centripetal force required to keep the car on the curved path is given by the equation Fc = mv^2 / r, where m is the mass of the car, v is its velocity, and r is the radius of curvature.
Setting up the equation for the maximum static friction force and the centripetal force, we have:
μsN = mv^2 / r
Now, let's determine the values for the variables in the equation:
μs = 0.4 (given)
r = 22 m (given)
θ = 12° (given)
To find N, we can use the vertical equilibrium of forces:
mg = Nsinθ
Solving for N:
N = mg / sinθ
Now, substitute the value of N into the equation for the maximum static friction force:
μsN = μsmg / sinθ
Substituting this into the equation for the maximum static friction force:
μsmg / sinθ = mv^2 / r
Simplifying the equation:
v^2 = μsrgsinθ / m
Finally, solve for v:
v = √(μsrgsinθ / m)
Substitute the known values:
v = √(0.4 * 9.8 m/s^2 * 22 m * sin12°) / m
Calculating the expression inside the square root:
v = √(8.0496) / m
The maximum speed at which the car can safely navigate the reverse-banked turn depends on the mass of the car. Therefore, without knowing the mass of the car, we cannot calculate the maximum speed.