What Speed Range Keeps a Car from Skidding on a Banked Curve?

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The discussion revolves around determining the speed range at which a car can navigate a banked curve without skidding, given a specific radius and coefficient of friction. A 950 kg car on a 30m radius curve can travel at 40 km/h on icy conditions, but the focus is on calculating speeds with a coefficient of static friction of 0.3. Participants emphasize the importance of understanding forces acting on the car, including friction and normal force, and suggest drawing free body diagrams to visualize these forces. The conversation includes guidance on how to derive the slope of the curve and apply relevant physics equations. Overall, the thread highlights the need for a clear grasp of the underlying physics concepts to solve the problem effectively.
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curve banking problem?

Homework Statement



a curve of radius 30m is banked so that a 950kg car traveling 40 km/h can round it even if the road is so icy that the coefficient of static friction is approximately zero. Find the range of speeds at which a car can travel around this curve without skidding if the coefficient of static friction between the road and the tires is 0.3

Homework Equations



This part i really need help in pls may i ask for any equations to use i can't even understand the problem. I can't interpret the free body diagram pls give me a formula ^^

The Attempt at a Solution



none i tried drawing a slope with an object but i don't know how to get the angle of the slope...
 
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What class are you taking? What equations are introduced in the chapter that this question is from? You need to show more effort before we can help you much.
 
think about this
when the car has maximum speed, friction is down the slope and when speed is minimum, friction is up the slope
 
First ignore the friction and calculate the slope of he curve from the icy case. Draw a diagram to show the forces on the car.
 
How do i calculate the slope pls.. I'm taking physics natural science and the last formula our pro gave us was the coefficient of friction times normal force is equal to force of friction. so i don't know how to get the slope of the problem i don't even know how to derive where the slope came from sorry
 
thank you for the diagram it has been very helpful thank you very much i at least have an idea now on how to sole ^^
 
Extending it to the case of friction just involves adding the extra foces, remember that friction will act parrallel to the road in a direction to stop the car slipping.
As you said above, friction force is the normal force * coefficent of friction.
 
mgb_phys said:
friction will act parrallel to the road in a direction to stop the car slipping.

I guess frictional force considered in this question is the force acting perpendicular to the road which avoids the car slippin towards or away from centre of the circular track(or any kind of track, maybe call it instantaneous center) at lowest and highest speed at the curve respectively.
the frictional force mgb_phys is considering is the force which gets the car going
 
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