Circular and Satellite Motion Legacy Problem #14 Guided Solution

Justin is driving his car along a horizontal curve. His speed is 23 meters per second, which is a little bit more than 50 miles per hour. The radius of curvature of the turn is 65 meters. And we're asked, what's the minimum value of coefficient of friction which we require to keep Justin's car on the curve? So the way I'm going to approach this problem is I'm going to center my solution around a free body diagram representing the forces acting up on Justin's car. So I think he's on a level highway. So there's a gravity force down and a normal force straight up. When you make a turn, you don't accelerate up and down. You accelerate horizontally. So these two forces, F-grab and F-norm, will balance each other. I think what else is acting up on Justin's car? One obvious choice is the friction forces. It must be because we're asked to determine the coefficient of friction or the mu value, as in the equation, F-frict equal mu F-norm. So I know there's friction force acting towards the center of the circle. There's no other forces acting up on Justin's car, just gravity normal and then friction towards the center. As mentioned, gravity and normal balance each other out, which means the friction force is the net force. So I would write as my first statement, F-net equal F-friction. Now as I talk through the solution, I won't be doing a whole lot of algebra until the very end. I should say I won't be doing a whole lot of calculator work until the end. So I start with the idea that F-net equal F-friction. And I'm going to try to take this equation and derive an equation for mu from it. So for the F-net side of this equation, I'm just going to write M times A. And for the F-frict side of this equation, I'm going to write mu times F-norm. Now if I think a little harder about that mu times F-norm, I could say F-norm is equal to the gravity force, or Mg. So my F-net equal F-frict equation now becomes Ma equal mu Mg. Now I notice there's an M on each side of the equation, so if I divide through by M, I can cancel the M. Now I have an equation that says A equal mu G. Rearranging to get it in the form of mu equal, I would have mu equal A over G. Now the A value can be calculated, because I know the speed of Justin's car is 23 meters per second, and I know the radius of the turn is 65 meters. So if I say A equal V squared over R and plug these numbers in, I get 8.1385 meters per second per second. That's the A in the equation, mu equal A over G. If I divide this A by 9.8, I get the value of mu, the coefficient of friction. It comes out to be 0.83, and that's right on the edge of what a car can stand in making a turn. That's one reason why they have speed limits for turns, and that's another reason why they often bank turns.