Normal force, car on curved road

[EV33]

Homework Statement

A curve of radius 139 m is banked at an angle of 11°. An 866-kg car negotiates the curve at 89 km/h without skidding. Neglect the effects of air drag and rolling friction. Find the following.
(a) the normal force exerted by the pavement on the tires

Homework Equations

there is no set equation for Normal force

The Attempt at a Solution

I would assume that the normal force would be mgcos(theta) after making a free body diargram because mg is straig down, and the normal is pointed perpindicular to the 11 degree road, so to find fn I would say cos(theta)= fn/mg, and then mgcos(theta)is equalto fn, but the answer is (mg)/cos(theta). This equation would suggest that the normal force is greater than the weight force which makes no sense no me.

Why is mg/cos(theta) correct?

 

[rl.bhat]

Normal force fn is perpendicular to the banked pavement. Its vertical component balances the weight of the car and horizontal component provides the required centripetal force to keep the car on the track.

 

[EV33]

Why is this situation anydifferent from this one?

http://www.physicsclassroom.com/Class/vectors/U3l3e.cfm

These are so similar in set up I don't see how they would have different equations.

 

[rl.bhat]

In the frictionless inclined plane, body slides downward due to the component mg along the inclined plane. i.e. mg*sinθ. And the normal farce is mg*cosθ.
But on the frictionless banked road, the car is not sliding along the slope but moving in a curves path. So the mg*cosθ is not helpful here. The centripetal force is provided by the component of the normal reaction. Other component of normal reaction is balanced by the weight of the car.