Static friction on a car on a circle

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When a car moves on a sloped circular track, static friction plays a crucial role in maintaining its circular motion. The direction of static friction can either be towards the center or away from it, depending on the car's speed, the radius of the circle, and the slope's angle. For a car traveling at 20 km/h on a track with a radius of 113 meters and an 8-degree slope, the forces must be analyzed to determine the correct direction of static friction. The resultant force must be horizontal and directed towards the center, equating to the necessary centripetal force. Properly setting up the equations for the forces involved will clarify the direction of static friction needed to keep the car on the circular path.
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When a car is moving on a circular track which is sloped away from the centre of the track, is the static friction keep that car on the circle (constant speed) towards the outside of the circle, opposite the acceleration?

I'm having trouble seeing this because to me, the car would want to fall out of the circle, because of the slope. Which would mean it would need the static friction to be towards the centre.
 
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The static friction can act in both directions, towards the centre and away from it. To move along the circle with speed v, the car needs proper centripetal force, which is the resultant of gravity, normal force and static friction between the road and tyres.

ehild
 
OK but in this case which direction would the static friction be? I just need to know to draw the free body diagram.
 
It depends on the speed of the car, the radius of circle and the angle of the slope.
 
The speed is 20 km/h, the radius is 113m and the angle of the slope is 8 degres
 
Draw the free body diagram. You have two unknown forces: the normal force and the static friction. The normal force is perpendicular to the road, the friction is parallel, but you do not know if it points inward or outward. The resultant has to be horizontal, pointing towards the centre and equal to mv2/R. Write the equations for the force components, horizontal and vertical, assuming a direction for the friction. There are two equations with two unknowns. If you get negative result for the friction it points to the other direction you assumed.

ehild
 
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