Question

# An automobile moves on a level horizontal road in a circular path of radius 30.5m. The...

An automobile moves on a level horizontal road in a circular path of radius 30.5m. The coefficients of friction between the tires and the road are Ps = 0.50 and Pk = 0.1, respectively. The maximum speed with which this car can round this curve is?

The car is traveling in a circle, assumed at a constant speed, and thus requires an acceleration, inward. That acceleration is centripetal acceleration:
a = v^2/r

What causes this acceleration? Traction...also known as static friction.

What must exist for there to be any type of dry friction? A normal force.

The normal force will do what ever is necessary to prevent the car from sinking in to the asphalt. On an unbanked, flat roadway, the normal force is only responsible for opposing the force of gravity. It is thus:
N = m*g

Relating to maximum deliverable force of static friction:
F = mu_s*m*g

And Newton's 2nd law:
Fnet = F = m*a

Substitute centripetal acceleration and parameters of maximum friction:
mu_s*m*g = m*v^2/r

solve for v (notice m's cancel and result is independent of mass):
v = sqrt(mu_s*r*g)

Data:
mu_s:=0.5; r:=30.5 m; g:=9.8 N/kg;

Result:
v = 12.22 meters/second

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