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 10 m, and the banking angle is 14°, what is the maximum speed at which a car can safely navigate such a turn?
Can someone explain this one?
I recommend you draw a picture.
The force down the hill is mv^2/r cosTheta+mgsinTheta
That has to be equal to fricion: mgcosTheta*mu
set them equal, solve for v.
check my thinking, I did it in my head, so there may be an error.
To find the maximum speed at which a car can safely navigate a banked turn with a reversed banking angle, we need to consider the forces acting on the car and the conditions for safe driving.
First, let's visualize the situation. In a banked turn, the road is tilted sideways, either towards or away from the center of curvature. Reversed banking angle means that the road is tilted away from the center of curvature.
In this scenario, three forces are acting on the car: the gravitational force (mg) acting downwards, the normal force (N) exerted by the road perpendicular to the road surface, and the frictional force (f) acting in the horizontal direction.
To determine the maximum speed, we need to consider the critical point at which the frictional force is at its maximum while still maintaining static equilibrium, meaning the car does not slip. At this point, the static friction is at its maximum value (μsN), where μs is the coefficient of static friction.
To find the maximum speed, we can equate the gravitational force component parallel to the inclined road (mg sin θ) with the frictional force:
mg sin θ = μsN
Next, we need to relate the normal force (N) to the various forces acting on the car. Considering the components of the normal force:
N = mg cos θ
Now we can substitute the expression for N in terms of mg cos θ:
mg sin θ = μs(mg cos θ)
Simplifying the equation:
tan θ = μs
Now we can substitute the given values into the equation and solve for the angle θ:
μs = 0.4
tan θ = 0.4
Using the inverse tangent function (tan^(-1)), we can find the value of θ:
θ = tan^(-1)(0.4)
θ ≈ 21.8°
The value of the angle θ tells us the angle at which the road is tilted away from the center of curvature.
Finally, we can use the radius of curvature (10 m), the banking angle (14°) and the angle of tilt (21.8°) to find the maximum speed at which the car can safely navigate the turn. We'll use the formula:
v = sqrt(μs * g * r * tan(θ))
Substituting the known values:
v = sqrt(0.4 * 9.8 m/s^2 * 10 m * tan(21.8°))
Calculating this expression will give us the maximum speed at which the car can safely navigate the turn.