Well, my first thought is that of road engineers: Banked curves beg speeders to speed. One cannot on a public highway encourage racing.
Now, the question.
Friction down the incline= mg*.8*cosTheta
so gravity down the incline cant be greater than that.
(1) mgSinTheta<mg*.8*cosTheta
or TanTheta<.8
Now at 60km/hr (16.7m/s) centripetal force will have two components: up the plane (mv^2/r * CosTheta) and normal to the plane (mv^2/r*sinTheta). We still have the up the plane and normal components of the weight of the car.
So, the force up the plane < retarding friction force.
mv^2/r*CosTheta -mgSinTheta < .8(mgCosTheta+mv^2/r * SinTheta)
Divide both sides by cosTheta
mv^2/r -mg TanTheta<.8mg+.8mv^2/r tanTheta
solve for tan Theta in (1)put that into the equation and solve for r.
As a civil engineering intern during one of your summer in college, you are asked to design a curved section of roadway that meets the following conditions: when ice is on the road, and the ceefficient of static friction between the road and rubber is 0.080, a car at rest must not slide into the ditch and a car traveling less than 60km/h must not skid to the outside of the curve.Neglect the effects of air drag and rolling friction.What is the minimum radius of curvature of the curve and at what angle should the road be banked?
1 answer
1 answer