Normal force, car on curved road

1. Nov 12, 2009

EV33

1. The problem statement, all variables and given/known data

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

2. Relevant equations

there is no set eqution for Normal force

3. 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?

2. Nov 12, 2009

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.

3. Nov 12, 2009

EV33

4. Nov 12, 2009

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.