# Homework Help: Turning the Corner

1. Nov 2, 2007

### calvinth

A concrete highway curve of radius 80.0 m is banked at a 19.0^\circ angle.

What is the maximum speed with which a 1900 kg rubber-tired car can take this curve without sliding? (Take the static coefficient of friction of rubber on concrete to be 1.0.)

I used the equation v = sq rt (rg*tan(theta)) but that didn't work, so I used
v(max) = sq. rt. (rg*u(coeff. static friction)) and got 28.8 m/s2, but that didn't work either.

If anyone can tell me what I'm missing or went wrong, I'd greatly appreciate it.

2. Nov 4, 2007

### dynamicsolo

I'm afraid you're going have to break down and draw a force diagram for this one. Consider that the car is on a 19ยบ incline tipped so it is leaning toward the center of the turn. The axis it is moving around on the circle is vertical, so the centripetal force we're looking for points horizontally. You have the weight force, the normal force from the road surface, and the static frictional force*, also from the surface, acting on the car. How are these forces related? What will be the horizontal components of these physical forces? The sum of these horizontal components toward or away from the center of the circle will be the centripetal force. This will enable you to solve for a (tangential) speed along the circle.

*I know it sounds weird that a moving car has static friction acting on it; if you've discussed rolling objects in your course, it may have been remarked that static friction is necessary to have rolling, instead of sliding.

Since we are asked for the maximum speed, we have to consider what that means. If there were no friction, going faster would tend to "throw" the car out of the turn, causing it to slide outward off the ramp. So the static friction has to act "downhill" on the banked turn to hold the car on the circle.

(The presence of this friction causes the minimum speed on the ramp to be different. Going too slow would tend to make the car slide inward off the ramp, so the friction must act "uphill" on the banked turn. The reversal of direction of this force leads to a change in the centripetal force and in the required velocity to stay on the circular turn.)