Coefficient of Static Friction Physics Problem?

Question

A curve with a 120-m radius on a level road is banked at the correct angle for a speed of 20 m/s.

If an automobile rounds this curve at 30 m/s, what is the minimum coefficient of static friction between tires and road needed to prevent skidding?

Answer 1

For a banked road the road must be banked so the
Normal force=Vector sum of centripetal force and weight.
Or

N=sqrt(mV^3/R + mg)
Bank angle = arcTan( F(c)/mg)=
Bank angle = arcTan( mV^2/ (Rmg))=
Bank angle =arcTan(V^2/ (Rg))=
B-Angle=18.8 degrees
cos(bank angle)=mg/N

The centripetal force must be equal the force of friction.

Fc=mV^2/R
f= umg cos(bank angle)

u cos(bank) =V^2/(R g cos(bank) )
u =(30m/s)^2/(120 x 9.81 x cos(18.8))
u=0.808

I hope it helps

Answer 2

The rotating passenger can do so provided centripetal force equal to (m v^2)/r or (m w^2)r, where m is mass of passenger, v the speed , w angular velocity and r the radius of the circle in which passenger rotates acts on them. What can provide this, only the wall. How it can do it, if the rotating passenger reacts on it in opposite direction (centrifugal) In fact passenger has an inertial tendency to move tangential but the cylindrical wall does not allow it. in this attempt the wall pushes and the passenger pushes the wall constituting the action and reaction pair of Newton's third law. So Normal reaction, N = m*(0.7*2*pi)^2 * 4.2, So maximum force, f of limiting friction against any tendency of movement is given by f = mu*m*(0.7*2*pi)^2 * 4.2 = 34*m*mu N. In order that this maximum force equal the downward pulling force due to earth mu must be at least equal to the value given by the equation f= 341*m*mu = mg or the minimum value of mu will be m = g/341 = 9.81/341 = 0.03 If mu is more it does not matter the frictional force will be adjusted to be equal to mg if it is less there will be tendency of the passenger to be pushed towards the floor where his feet will touch and he would be likely to fall.