I have a car on cone-shaped, circular track, and the plane is frictionless incli

Question

I have a car on cone-shaped, circular track, and the plane is frictionless inclined at 45 degrees. The car weighs 11,000 kg and it's driving in a circle at a steady distance from the center of the circular track. How do I calculate the force exerted on the car by the track - PLEASE explain why the normal force here (according to my solution book) is equal to mg/costheta, and not mg*costheta. I am totally confused. Does it have to do with the absence of fiction? Or the fact that the car is not moving down the slope? Help! My exam is tomorrow and I am confused.

Explanation / Answer

Hi, Firstly calm down! The easiest thing about physics is that a complicated problem can easily be calculated using simple diagrams. Now, imagine the car is on the cone shaped circular path. Draw a vertical section of this slope. You would see a vertical line, a horizontal line, and the line joining the two at 45 degrees. Now using trigo, can you calculate the rest of the force of the track on the car? when a cos is not multiplied by and instead divided by, it means your trying to do the reverse of the equation. In your situation its (to be evaluated)*cos = Some Value ==> (to be evaluated) = some value/cos. I hope this would be able to clear up some of your stress on the subject!

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