Question

You are driving a classic 1954 Nash Ambassador with a friend who is sitting to your right on the passenger side of the front seat. The Ambassador has flat bench seats. You would like to be closer to your friend and decide to use physics to achieve your romantic goal by making a quick turn.If the coefficient of static friction between your friend and the car seat is 0.25, and you keep driving at a constant speed of 30 m/s , what is the maximum radius you could make your turn and still have your friend slide your way?

Answer 1

Answer:

If I drive my car on a circular path with a constant velocity, the car and the passengers inside would be applied a centripetal acceleration.

$a = \frac{v^2}{r}$

By Newton's Second Law, the centripetal force can be derived.

$F = ma = \frac{mv^2}{r}$

The passenger would slide towards me if the centripetal force is greater than the static friction force.

$F_s = \mu_s mg = 0.25mg$

$F > F_s\\\frac{m30^2}{r} > 0.25m(9.8)\\r_{max} < 367.3 ~m$

If the radius of the curve is greater than this, then the car would slide off the road.

Explanation:

The one thing that has always confused students is that there is no such thing as centrifugal force. When you take a turn in the car, you slide out because you tend to keep your position while the car below you is changing its position. This is simply Newton's First Law of Inertia. Objects tend to keep their initial position.

The aforementioned centripetal force is the force resulted by a constant circular motion. These two forces should not be confused.