When you take your 1300-kg car out for a spin, you go around a corner of radius 57.7 m with a speed of 16.5 m/s. The coefficient of static friction between the car and the road is 0.93. Assuming your car doesn't skid, what is the force exerted on it by static friction?

Centripetal force comes from friction:

what is mass*velocity squared divided by radius? That is the force directed inward.

The last data in the problem is not revelant.

6134

To find the force exerted on the car by static friction, we need to apply Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

In this case, the net force is the force of static friction, and the acceleration is the centripetal acceleration. The centripetal acceleration is given by the formula:

a = v^2 / r

where v is the velocity and r is the radius of the corner.

Substituting the given values:

a = (16.5 m/s)^2 / 57.7 m = 4.73 m/s^2

Now, we can calculate the force of static friction using Newton's second law:

F_friction = m * a

Substituting the mass of the car (1300 kg) and the calculated centripetal acceleration (4.73 m/s^2):

F_friction = 1300 kg * 4.73 m/s^2 = 6157 N

Therefore, the force exerted on the car by static friction is 6157 N.