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Suppose that we are observing a car driving around an inclined speedway (in circular motion). For example, imagine the Indianapolis 500 Speedway: enter image description here

Now, we can assume that the cars are driving around a circle whose center is above the ground level - therefore, the net force acting on the car must work in accordance with the direction of the radius of this circle. These are all (in my opinion) forces acting on the car:

  1. The force of gravity acting downwards: one component acting towards the asphalt, the other one acting down the slope.
  2. The reaction force normal to the asphalt (opposite to one of the components of the force of gravity)
  3. The force of friction due to turning - I am not really sore which direction it actually acts - I would say that this force is radial.
  4. The force of friction due to the slope

    The fourth force is giving me hard time. I would normally say that this force acts up the slope because it opposes one of the components of the force of gravity - acting down the slope. However, with this assumptions these forces would not add to the desired net force, therefore, there must be something wrong in my assumptions. Could you help me to work it out?
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Friction is always along the surface, and there are not two frictional forces here-- just one. The friction here is called "static friction" (believe it or not, even though the wheels are turning the point in contact with the road does not slide along the road), which is a tricky force because it is whatever it needs to be to keep the car from sliding. So in problems like this, find the static friction force last-- it's whatever it needs to be, but it has to point along the surface. (And you may know this, but the normal force does not balance the gravity component perpendicular to the road. You have to make sure all the forces add up to the necessary centripetal force, and that latter force has components both along and perpendicular to the road.)

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  • $\begingroup$ Thank you for your answer. Actually, I wrote "balances" but did not mean that, what I mean was that it "opposes" this force. So, the solution to my problem - the 4th force is actually pointing down the slope? $\endgroup$ – Aemilius Mar 16 '18 at 14:38
  • $\begingroup$ Yes. Basically, static friction is doing the same thing that the normal force is-- they both simply keep the object from either sliding along the surface or moving up or down from the surface. So in that sense, the two are not really different forces at all, and could easily be combined into one single force (so then there are only 2 forces here) that is whatever is needed to make sure the acceleration is purely radial and equal to the necessary centripetal acceleration. Its component along the road gets called static friction, and perpendicular gets called the normal force. $\endgroup$ – Ken G Mar 16 '18 at 14:41
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Frictional force will always form when a body tends to move in the opposite direction. For example if I’m pushing a block on the ground towards the north direction the frictional force will act towards the south. This is regardless of the material or gravitational strength. You can see this effect even on the moon. Now if the car is turning along a plane circular surface, the car will experience a outward force due to inertia. But because every surface has a non zero frictional coefficient the force would act opposite to direction of a motion( ignoring the motion due to the wheels for now). Thus the frictional force would be inward towards the center of curvature. In your analogy of the race cars on a inclined surface the centripetal force will act outward and at an angle equal to the angle of inclination of the road. Now due to friction the car will experience a force which will act exactly opposite and downward at the same angle regardless of surface. This is the exact reason why drivers brake when approaching a curve because if the speed is too much the car will oversteer because the maximum frictional force is limited to the weight of the car and the frictional coefficient of the surface.

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