A car of mass M = 1200kg traveling at 50.0km/hourenters a banked turn covered wi

Question

A car of mass M = 1200kg traveling at 50.0km/hourenters a banked turn covered with ice. The road is banked at an angle ?, and there is no friction between the road and the car's tires. (Figure 1) . Use g = 9.80m/s2 throughout this problem.

Now, suppose that the curve is level (?=0) and that the ice has melted, so that there is a coefficient of static friction ? between the road and the car's tires.(Figure 2) What is ?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 50.0km/hour and that the radius of the curve is given by the value you found for r - 54.1m in Part A.

Explanation / Answer

You want the centripetal acceleration up the slope to be matched by the gravitational down the slope.
Gravity: Ag = g*sin(T)
Centripetal: Ac = (v^2/r)*cos(t)

g*sin(T) = (v^2/r)*cos(t)
r = v^2/[g*tan(T)] = 54 m

Friction has to balance the centripetal force. And friction is just the normal force times the coefficient of static friction. And normal force is just the weight of the car.
Friction = N*mu = m*g*mu
1200*mu = m*(v^2/r)
g*mu = v^2/r
mu = v^2/(r*g)
mu = 0.364

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