Is the Normal Force Component in Superelevated Curves the Centripetal Force?

[fluidistic]

A question about when a car take a superelevated curve. I don't know how to be more precise in English. Take a look at this photo, maybe you'll understand better : http://www.fmciclismo.com/noticias/bmx/imagenes/peralte%20bmx%20copia.jpg
My question is : when you draw the free body diagram, you see that the normal force (N) has a component in direction of the center of the radius of the curve that describe the movement of the car. Does this component is precisely the centripetal force? That would mean that the centripetal acceleration is worth $$\frac{F_c}{m}$$.

 

[Dick]

Yes, that component of the normal force creates the centripetal acceleration. Nice picture, by the way.

 

[fluidistic]

Thanks for your answer. And for the picture, thanks to google .