A car (m = 2000 kg) is going around an unbanked curve at the recommended speed of 11m/s (24.6 MPH). (a) If the radius of the curvature of the path is 25m and the coefficient of static friction between the rubber tires and the road is µs = 0.70, does the car skid as it goes around the curve? (b) What will happen if the driver ignores the highway speed limit sign and travels at 18 m/s (40.3 MPH) ? (c) What speed is safe for traveling around the curve if the road surface is wet from a recent rainstorm and the coefficient of static friction between the wet roud and the rubber tires is µs = 0.50?

a) car does not skid, b) car skids, c) v = 11.07 m / s

Explanation:

a) When the car around in a curve all force must be exerted by friction, write Newton's second Law

Y axis (vertical)

N - W = 0

N = W = mg

X axis (radial

F = m a

The acceleration is centripetal

a = v² / r

fr = μ N

Let's calculate the maximum friction force

fr = μ m g

fr = 0.70 2000 9.8

fr = 13720 N

Let's calculate the force necessary to take the curve

F = m v² / r

F = 2000 11²/25

F = 9680 N

When examining these two values we see that the maximum value of the friction force is greater than the force to stay in the curve, for which the car does not skid

b) The speed of the driver is v = 18m / s, let's calculate the force to stay in the curve

F = 2000 18²/25

F = 25920 N

This force is greater than the maximum friction force, so it is a skating car

c) The friction coefficient decreases to μ = 0.5

fr = m a

μ mg = m v² / r

v = √μ g r

v = √ (0.50 9.8 25)

v = 11.07 m / s

This is the maximum speed