### Video Transcript

A car that has a mass of 900.0
kilograms drives along a circular unbanked curve at a constant speed of 25.000
meters per second. The radius of the curve is 500.0
meters. What magnitude force acts on the
car to maintain its speed and direction as it follows the curve? What is the minimum possible value
for the coefficient of static friction between the carโs tires and the road
surface?

We can label the force we want to
solve for capital ๐น and the minimum possible coefficient of static friction weโll
name ๐ sub ๐ . We can start on our solution by
drawing a diagram of the situation. Our car in this example, with a
mass of 900.0 kilograms, moves on a circular section of road with a radius of
curvature 500.0 meters. While it drives on this section, it
maintains a steady speed weโve called ๐ฃ 25.000 meters per second.

Knowing all this, we wanna solve
first for the magnitude of the force acting on the car that keeps it moving in a
circular path on this unbanked road. A force that keeps an object moving
in a circular path is a centripetal force. And itโs equal to the mass of the
object times its speed squared over the radius of the circle it moves in. In our case, weโve been given the
radius ๐, the speed ๐ฃ, and the mass of the car ๐. So, weโre ready to plug in and
solve for ๐น.

When we enter this expression on
our calculator, we find that ๐น is 1125 newtons. Thatโs the center-seeking force
needed to keep this car on a circular path. If we drew a picture of our car
from behind while it was turning on this circular path, the direction of the force
๐น would be to the left, pointing it towards what from this perspective is the
center of the circle. The physical mechanism that creates
this force ๐น is the friction force on the tires of the car from the road.

Wanting to solve for the
coefficient of static friction between these two materials, we can recall that the
force of friction on an object is equal to the coefficient of friction it
experiences multiplied by the normal force acting on it. In our case, we can write that the
force of friction on the car is equal to the coefficient of static friction because
the wheels donโt slide on the roadโs surface multiplied by the mass of the car times
the acceleration due to gravity ๐. Where ๐ weโll treat as exactly 9.8
meters per second squared.

This frictional force of the roadโs
surface on the tires is the physical mechanism by which the centripetal force ๐น sub
๐ is exerted on the car. So, we can say that the coefficient
of static friction times ๐๐ is equal to ๐๐ฃ squared over ๐. And we see that the mass of the car
cancels out from this expression. If we then divide both sides of
this equation by ๐, we see that ๐ sub ๐ is equal to ๐ฃ squared over ๐ times
๐.

When we plug in for the given
values of ๐ฃ, ๐, and ๐, we find that, to four significant figures, ๐ sub ๐ is
0.1276. Thatโs the minimum value that the
coefficient of static friction can be between the roadโs surface and the tires in
order for this magnitude centripetal force to be exerted on the car.