A turn in the road with a radius of 30m is banked at an angle of 40 degrees. If [itex]\mu[/itex]=.75, what is the minimum and max speed a 1200 kg car can travel without sliding off the turn.
F=ma, Mv^2/R, Fw = mg
The Attempt at a Solution
Alright, so using the information given i made a force diagram. Basically what i did is make the car, then a vector going to the right as Fc, another vector which makes the hypotenuse coming from the car to North-East, as Fn since it is perpendicular to the curved road, and Since Fw is down, upwards it would fit in the triangle to make a triangle with the legs of Fw and Fc, and a hypotenuse of Fn.
Since the velocity isn't given I solved for the velocity by replacing Fc with Mv^2/r.
(tan(40)*m*v^2)/r = mg
Plugging in all the values I got a Velocity of 18.7 m/s, but this is without friction so its better not to use this, and to find the Fn and Ff.
So mg/sin(40)=Fn. Fn works out to be 18295N, so using [itex]\mu[/itex]*Fn = Ff, the Ff is 13722N.
I'm honestly very confused on where to go from here, not even sure if what i've done so far is correct.. If anyone could help it'd be greatly appreciated..