Experts are full of valuable knowledge and are ready to help with any question. Credentials confirmed by a Fortune 500 verification firm.

Get a Professional Answer

Via email, text message, or notification as you wait on our site. Ask follow up questions if you need to.

100% Satisfaction Guarantee

Rate the answer you receive.

Ask CarDoc Your Own Question

CarDoc, Independent European Shop owner

Category: Car

Satisfied Customers: 6457

Experience: 20 years of experience working with European cars. Independent European shop owner and technician.

14582033

Type Your Car Question Here...

CarDoc is online now

Fargason: Why is the centripetal force of a car going around

This answer was rated:

★★★★★

Why is the centripetal force of a car going around a curve equal to the static frictional force? I got this from my physics class, but I don't understand the actual physical meaning. If there is a static frictional force pointing towards the center of the curve, that has to mean that there is a force pushing in the opposite direction to the curve that the static frictional force is opposing, but I don't know what that is.

When you say that the tires loose adhesion at 0.8-0.9 grams, (Is it grams?) where are you getting numbers? Do they relate to vectors pointing out from the center of the circle? Is that how people find the coefficients for static and kinetic friction?

Not grams. G=force of gravity. 1 g lateral, would be a force vector pointing from the apex of the circle outward, perpendicular to the force of gravity.

The numbers are going to be different for every car and every tire. We'll take 2 extremes. 1975 Buick Electra, on period-correct tires, can generate a maximum of approximately 0.60g lateral 2012 Renault Formula 1 car can generate about 4-5 g lateral.