Finding the frinction coefficient for a car on a banked turn

In summary, a banked circular highway curve with a radius of 202 m is designed for traffic moving at 60 km/h. The minimum coefficient of friction between tires and road needed for cars to negotiate the turn without sliding off the road is dependent on the centripetal force and can be solved using the equation mu = (Nsinθ - Fcosθ)/N or (Mg/cosθ).
  • #1
Elmnt
13
0

Homework Statement



A banked circular highway curve is designed for traffic moving at 60 km/h. The radius of the curve is 202 m. Traffic is moving along the highway at 35 km/h on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to negotiate the turn without sliding off the road?


Homework Equations





The Attempt at a Solution


I am pretty lost with this one. I have tried playing with variables and I have also tried a couple different free body diagrams but haven't gotten anywhere. My free body diagram represents a car on a banked turn with the force due to gravity straight down the normal force perpendicular to the pavement and friction parallel to the pavement pointing into the turn. Any tip or hint to point me in the right direction would be greatly appreciated.
 
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  • #2
Well the frictional force produced by the tires and the road would provide the centripetal force needed for the car to stay on the road.
 
  • #3
You have N perpendicular to the surface, mg straight down and friction acting along the surface.
N sin theta - friction cos theta is the net force that gives you centripetal acceleration (along the horizontal)
Friction is mu times N or mu times mg/cos theta.
So everything is there - just solve for mu.
 
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  • #4
bobaustin said:
You have N perpendicular to the surface, mg straight down and friction acting along the surface.
N sin theta - friction cos theta is the net force that gives you centripetal acceleration (along the horizontal)
Friction is mu times N or mu times mg/cos theta.
So everything is there - just solve for mu.
Wouldn't N sin theta + Friction cos theta = net force?
 
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1. What is the friction coefficient and why is it important for a car on a banked turn?

The friction coefficient is a measure of the amount of friction between two surfaces in contact. For a car on a banked turn, it is important because it determines the maximum speed at which the car can safely navigate the turn without sliding off the track.

2. How is the friction coefficient calculated for a car on a banked turn?

The friction coefficient for a car on a banked turn can be calculated using the equation μ = tanθ, where μ is the friction coefficient and θ is the angle of the banked turn. This equation assumes a level, circular track with a constant radius.

3. What factors can affect the friction coefficient for a car on a banked turn?

The friction coefficient for a car on a banked turn can be affected by several factors, such as the type and condition of the tires, the weight and distribution of the car, and the surface of the track. Other factors, such as temperature and humidity, can also play a role in the friction coefficient.

4. How does the friction coefficient change as the angle of the banked turn increases?

As the angle of the banked turn increases, the friction coefficient for a car on the turn also increases. This is because a steeper banked turn provides more lateral force to counteract the centrifugal force of the car, allowing it to maintain a higher speed without sliding off the track.

5. How can the friction coefficient be adjusted for a car on a banked turn?

The friction coefficient for a car on a banked turn can be adjusted by changing the angle of the banked turn, using different types of tires with varying levels of grip, or adjusting the weight and distribution of the car. Track conditions can also be altered to affect the friction coefficient, such as by applying a layer of rubber or water to the surface.

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