Finding the frinction coefficient for a car on a banked turn

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To determine the minimum coefficient of friction for a car negotiating a banked turn at 35 km/h on a curve designed for 60 km/h, the forces acting on the car must be analyzed using free body diagrams. The normal force (N) acts perpendicular to the surface, while gravitational force (mg) acts downward, and friction provides the necessary centripetal force. The equations involve balancing the forces, where N sin(theta) and friction cos(theta) contribute to the net force required for centripetal acceleration. The frictional force can be expressed as mu times N, leading to the need to solve for the coefficient of friction (mu). Clarification on the net force equation is also sought, indicating a potential misunderstanding in the force balance.
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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?


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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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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.
 
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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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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