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I don't understand how the force of friction can face a different direction in each case. To my understanding, without friction there is a net force down the slope. We want the net force to be towards the right (centripetal acceleration). So why isn't the force of friction always up the slope to cancel the y component of the net force?

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You o want the net force to be to the right, but when the component of gravity that is to the right is already larger than that, then you need friction to point the other way. So the free-body diagrams are correct, all that remains is to use the v and r to figure out the centripetal force, and then make friction, gravity, and the normal force add up to it.

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  • $\begingroup$ Don't you mean "when the component of the normal force that is to the right"? Gravity is always vertical and never has a horizontal component. Or do you mean "the component of gravity down the slope"? $\endgroup$ – sammy gerbil Dec 12 '16 at 1:22
  • $\begingroup$ Yes, I meant to the right and down the slope, when that component of gravity is larger than the component of the centripetal force to the right and down the slope, then there will need to be a friction upward and to the left. You could also do it horizontally as you suggest, but you will need to know the normal force, so it will end up the same way. $\endgroup$ – Ken G Dec 12 '16 at 4:30
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When the car is at rest on the banked track, the component of the gravitational force on the car in the direction parallel to the slope is down the slope (in your figures). For the car to be in equilibrium (at rest) on the bank, the frictional force (parallel to the slope) exerted by the track on the car must be up the slope.

When the car starts traveling at a low velocity around the track, the gravitational force component down the slope stays the same, but the frictional force exerted by the track on the car up the slope must decrease to provide a net force down the slope to allow the car to experience centripetal acceleration. As the speed increases, at some point, the frictional force exerted by the track will drop to zero, and the gravitational component down the slope will be sufficient to solely match the mass times the centripetal acceleration. This is called the critical speed. At speeds exceeding the critical speed, the track will have to exert a frictional force down the slope (i.e., in the same direction as the gravitational component) in order to provide the centripetal acceleration.

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Perhaps it is best explained by using some diagrams showing the forces acting on the car being added together to produce a net inward force $\vec F_{\rm net}$ which produces the centripetal acceleration.

enter image description here

Diagram 3 shows the "ideal" banking situation where no frictional force is required.

The constraints are:

  • The gravitational force $\vec F^{\;\rm grav}_{\;\rm EC}$ is fixed in magnitude and direction.
  • The normal reaction force $\vec F^{\;\rm norm}_{\;\rm TC}$ is fixed in direction relative to the gravitational force $\phi$ by the angle of the slope $\theta$.
  • The frictional force $\vec F^{\;\rm s-fric}_{\;\rm TC}$is has to be at right angles to the normal force.
  • The net force $F_{\rm net}$ must be horizontal and towards the centre of the curve.

So you want the car to go round the corner faster than the ideal banking situation.
This requires a larger net inward force and the only way that can be done is to make the magnitude of the normal force larger, remember you cannot change its direction, and add an inward, down the slope, friction force at right angles to the normal force as in diagram 1.

For the car to go slower around the corner diagram 2 shows the way.
Reduce the normal force and add a frictional force up the slope.

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