How Does Banking Angle Affect Centripetal Force on a Curve?

[lzh]

QUESTION:A car travels at a speed of 16 m/s around a
curve of radius 21 m.
The acceleration of gravity is 9.8 m/s^2 :
http://img262.imageshack.us/img262/3067/bankedcurvera4.jpg
What is the net centripetal force needed to
keep the car from skidding sideways? Answer
in units of N.
if there was no friction, what centripedtal force could be provided just by the banking of the road?
ANSWER:
http://img523.imageshack.us/img523/4331/bankrz7.jpg
i used 2 eqns for this by dividing all the forces into separate components.
y-direct:
mg+fcos72=ncos18
x-direct:
mv^2/r=nsin18+fsin72
i used those and found
n=32718.5452
f=21413.6627
coeff of frict=.65448
net centripetal=30476.19
i found all of the above for the other parts of the question, and they are all correct. But the question i presented above confused me. If no friction is present, the only force contributing to centripetal force would be the normal force. so that'd be nsin18, but it wasn't right.

what am i thinking wrong here?

 

[neutrino]

mg+fcos18=ncos18

Redo this part.

 

[lzh]

oh srry about that, i wrote the eqn down wrong, but i did it the right way on paper. its been changed

 

[husky88]

If no friction is present, the only force contributing to centripetal force would be the normal force. so that'd be nsin18, but it wasn't right.

Yep, that's correct. Maybe there are some other errors in your previous results.

 

[lzh]

but i know for a fact that my normal force is correct...

 

[PhanthomJay]

but i know for a fact that my normal force is correct...

Recalculate the normal force without the friction.

 

[lzh]

ok, ty, i got it