Why Is My Calculation of Maximum Car Velocity on a Banked Curve Incorrect?

[stupidpig]

Hi, I've been working on this problem for a while and I keep on getting same answer! Can someone please tell me what I'm doing wrong.

Problem:
A circular curve is banked so that a car traveling with uniform speed rounding the curve usually relies on friction to keep it from slipping to this left or right.
What is the maximum velocity the car can maintain in order that the car does not move up the plane.(Answer in km/hr).

Radius = 56.4m
Mass of car = 2.3kg
Angle = 34 degree
Coefficient of kinetic friction = 0.41

My work:
N = mgcos(34) = 18.68
Fp = mgsin(34) = 12.6
Fr = (0.41)N = 7.66
Fc = centripetal force = mv^2/r

so here's my final equation to get v:
mv^2/r-Fr = Fp
(2.3)(v^2)/(56.4)-(7.66) = 12.6
v = 22.28m/s = 82.08km/hr

82.08km/hr is so unrealistic for 2.8kg car to bank such a turn.
Heck, even my puny vw golf can't even do it at 82.08km/hr

I must be doing something wrong!

Help me please.
Thanks

 

[Clausius2]

Problem:

Radius = 56.4m= R
Mass of car = 2.3kg= M
Angle = 34 degree= $$\alpha$$
Coefficient of kinetic friction = 0.41=$$\mu$$

Given the notation above, the force of friction must balance the centripetal force and weight:

$$\mu Mgcos \alpha=\frac{Mv^2}{R}cos\alpha+Mgsen\alpha$$

SURPRISE! the problem has no dependence on the mass M!. Is it true in real world?

$$v=\sqrt{gR\frac{\mu+tan\alpha}{1-\mu tan \alpha}}$$

For velocities greater than this, the equilibrium is broken and car would be rejected on the tangent way.

I wish there were some road with $$\mu=1/tan\alpha$$ !