Finding the frinction coefficient for a car on a banked turn

  • Thread starter Elmnt
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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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
13
1
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.
 
Last edited:
  • #4
13
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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?
 
Last edited:

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