Consider again the problem of a car traveling along a banked turn. Sometimes roads have a "reversed" banking angle. That is, the road is tilted "away" from the center of curvature of the road. If the coefficient of static friction between the tires and the road is μs = 0.4, the radius of curvature is 12 m, and the banking angle is 13°, what is the maximum speed at which a car can safely navigate such a turn?

Incorrect: Your answer is incorrect.
m/s

Question

Consider again the problem of a car traveling along a banked turn. Sometimes roads have a "reversed" banking angle. That is, the road is tilted "away" from the center of curvature of the road. If the coefficient of static friction between the tires and the road is μs = 0.4, the radius of curvature is 12 m, and the banking angle is 13°, what is the maximum speed at which a car can safely navigate such a turn?
  
Incorrect: Your answer is incorrect.
 m/s

roadjester · Answer

$\huge \mu_smg=\dfrac {mv^2} r cos\theta$
solve for v.

anonymous · Answer

@vsandrasue you got some hard questions. i didnt get this one

anonymous · Answer

rearrange the equation and solve for v like @roadjester said

vincent-lyon.fr · Answer

This equation
$\large \mu_smg=\dfrac {mv^2} r cos\theta$

does not apply to the case studied here.

In the condition of the motion, acceleration is

$a=g\Large \frac{\mu \cos \theta-\sin\theta}{\mu \sin \theta+\cos\theta}$