What is the force of the track on the car?

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The discussion centers on calculating the force of the track on a roller-coaster car at the top of a hill. The car has a mass of 500 kg and a speed of 8.0 m/s while at the top of a hill with a radius of 15 m. The user initially calculated the normal force (N) as equal to the weight (mg), resulting in a value of 4.9 kN, which was incorrect. The correct approach involves recognizing that the centripetal acceleration acts downward, requiring adjustments to the force balance equation. The correct normal force at the top of the hill is determined to be 2.8 kN.
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Homework Statement


A roller-coaster has a mass of 500 kg when fully loaded with passengers. The car passes over a hill of radius 15 m. When at the top of the hill, the car has a speed of 8.0 m/s. WHat is the force of the track on the car at the top of the hill?


Homework Equations


F=ma


The Attempt at a Solution


I drew an FBD: wtih mg pointing down, N pointing up and static friction pointing to the left. ( I think this may be wrong, but I figured it may need static friction to stay on the track).
Then when broken into i and j components I got:
i= -MsN=-MV2/r ...(because the acceleration is centripical)
then for j:
I have N-mg=0
N=mg
and since it was asking for the force of the track on the car, I thought this was the normal force, so I just plugged in the values for m ang, and got 4.9 kN up. This is wrong, it is saying the correct answer is 2.8 kN up. Any suggestions on where I went wrong?
 
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The car is at the top of the hill; you seem to have your i and j mixed up. The centripetal accelertion is vertically down. Ignore friction.
 

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