What is mu_min given radius 54m?

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To determine mu_min for a car of mass 1400 kg traveling at 50.0 km/h on a level curve with a radius of 54 m, the centripetal force must equal the frictional force. The equation mu*m*g = m*v^2/r can be simplified to find mu_min. Given the values, the calculation involves substituting the known parameters into the equation. The discussion emphasizes the importance of understanding the relationship between friction, mass, and centripetal force in this context. The user seeks assistance in completing the calculation for mu_min.
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What is mu_min given radius 54m?

A)A car of mass M = 1400 kg traveling at 50.0 km/hour enters a banked turn covered with ice. The road is banked at an angle theta, and there is no friction between the road and the car's tires. http://session.masteringphysics.com/problemAsset/1011163/29/MLD_cm_7_a.jpg

B)Now, suppose that the curve is level (theta = 0) and that the ice has melted, so that there is a coefficient of static friction mu between the road and the car's tires. (Part B figure) What is mu_min, the minimum value of the coefficient of static friction between the tires and the road required to prevent the car from slipping? Assume that the car's speed is still 50.0 km/hour and that the radius of the curve is given by the value you found for r in Part A.
http://session.masteringphysics.com/problemAsset/1011163/29/MLD_cm_7_b.jpgI got A) r= 54.0 m because of r= v^2/[g*tan(theta)] given theta = 20

I'm not sure what I need to do to get mu_min. Can someone please help me?
 
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You know the velocity, radius of the path, and mass of the car. The road applies a force to the car causing it to accelerate in a circular path. The force has magnitude,

mu*m*g

and equals the centripetal force, m*v^2/r
 
thank you so much Spinnor!
 
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