Car shouldn't slid

A free-body diagram for the car is shown at left. Both the normal force, $N$ (blue components) and the friction force, $f$ (red components) have been resolved into horizontal and vertical components. Notice that there are three vectors in the vertical direction, and:

$N \cdot cos(\theta) = mg + f \cdot sin(\theta)$

Using the approximation $f = \mu N$ , where $\mu$ is the coefficient of friction, gives:

$N = \frac{mg}{cos(\theta) - \mu \cdot sin(\theta)}$

If the coefficient of friction is zero, this reduces to the same normal force as we derived for no-friction, which is reassuring. If the coefficient of friction is not zero, notice that the normal force will be larger than it was in the no-friction case. Now, in the horizontal direction:

$F_{net} = N sin(\theta) + f cos(\theta) = N sin(\theta) + \mu \cdot N cos(\theta) = N(sin(\theta) + \mu cos(\theta)) = (\frac{mg}{cos(\theta) - \mu sin(\theta)})(sin(\theta + \mu cos(\theta)) = \boxed{\frac{sin(\theta) + \mu cos(\theta)}{cos(\theta) - \mu sin(\theta)} \cdot mg}.$

Here, the term $f \cdot cos(\theta)$ is friction's contribution to the centripetal force. Also, notice again that if $\mu = 0$ , this result reduces to the same $F_{net}$ as the no-friction case. Now, since the net force provides the centripetal force to turn the car:

$F_{net} = F_{centripetal} \Rightarrow \frac{sin(\theta) + \mu cos(\theta)}{cos(\theta) - \mu sin(\theta)} \cdot mg = \frac{mv^2}{R}$ ;

and solving for $v$ gives:

$v = \sqrt{ (\frac{sin(\theta) + \mu cos(\theta)}{cos(\theta) - \mu sin(\theta)}) \cdot gR} = \sqrt{ (\frac{sin(11) + (0.48) cos(11)}{cos(11) - (0.48) sin(11)})(9.8)(200)} = \boxed{38.2}$ m/s.