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Banked Tracks For Turning

I'm unsure how to approach this question. It seems to just be a simple use of a = V^2 / r but all of my answers have been wrong.

I've taken the acceleration inwards as 9.81 (due to the question stating that the force of friction inwards is equal to the weight of the car) + sin(20) * 9.81 (due to gravity acting along the slope) and the radius of the track being 44 as stated in the question. Putting this into the formula gets me a velocity of 24.068 which is wrong.

Any advice would be appreciated.

Banked Tracks for Turning 17.4 Isaac Physics
I just figured it out and its hard than I thought it would be.

The trick is that you need to find the centripetal force. This is the total force acting directly towards the centre (not down the slope).

To do this, you need to find the component of the frictional force that is acting centripetally. The frictional force does give a centripetal force equal to the weight because its not acting exactly towards the middle.

Also, the mg sin(20) term that you added to the centripetal force is incorrect. This would be correct in a situation with no friction, but adding friction makes it more complicated. The friction is acting down the slope, so you need to consider the normal force. The vertical component of the normal force will cancel out with the vertical component of the friction, and the weight. You then need to find the component of the normal force which is acting centripetally.

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