What is the force of the track on the car?

[ScullyX51]

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= -M_sN=-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?

 

[PhanthomJay]

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.

 

[thomate1]

Your approach is correct, but you made a mistake in your calculation for the i component. The correct equation for the i component should be F=mv^2/r, where m is the mass of the car and v is its velocity. Plugging in the given values, we get:

F=(500 kg)(8.0 m/s)^2 / 15 m = 2133.33 N = 2.13 kN

This is the force of the track on the car at the top of the hill. To find the normal force, we need to subtract the weight of the car from this force, since the normal force is the force that balances the weight. The weight of the car is mg, which is (500 kg)(9.8 m/s^2) = 4900 N. So the normal force is:

N = F - mg = 2133.33 N - 4900 N = -2766.67 N = -2.77 kN

Since the normal force is pointing upwards, we take the absolute value to get the magnitude, which is 2.77 kN. This is very close to the given answer of 2.8 kN.

Therefore, the force of the track on the car at the top of the hill is 2.77 kN upwards.