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.
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?
1 answer
1 answer