A car of mass = 1100 traveling at 65.0 enters a banked turn covered with ice. Th

Question

A car of mass = 1100 traveling at 65.0 enters 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 = 9.80 throughout this problem. Now, suppose that the curve. What is , 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 65.0 and that the radius of the curve is given by the value you found for in Part A.( which was r= 91.4m )

Explanation / Answer

speed=65 km/hr = 18.055 m/s

v = sqrt(r*g*tan?)

v^2 =r * g * tan?

18.055^2 = r * 9.8 * tan 20

r = 91.4 m

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Friction has to balance thecentripetal force.

And friction is just the normal force times thecoefficient of static friction. And normal force is just the weight of the car.

Friction = N*mu = m*g*mu

1100 * mu = m*(v^2/r)

g*mu = v^2/r

mu = v^2/(r*g)

= 18.055^2 / (91.4 * 9.8)

mu = 0.364

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