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I understand that when a car is turning, it is using static friction to do so.

The wheel is turned where the perpendicular component of the wheel is opposing the direction of motion, creating a force that turns it.

Doesn't centripetal force have to be less than or equal to static friction, in terms of magnitude?

And can someone explain the centripetal force equation and how it relates to the force applied and static friction equation?

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  • $\begingroup$ The question isn't very understandable. By turn, you mean a drift turn that seemingly looks like the vehicle is turning about an axis, or the vehicle turns around an arc. And you should explain better how the static friction (meaning motionless friction) affect a moving car if it's the latter case. $\endgroup$ – TechDroid Apr 17 at 6:06
  • $\begingroup$ I'm asking how a car turns with friction if it's moving around a circle with constant velocity, and is the maximum value of the centripetal force the same as the maximum value of static friction. $\endgroup$ – B V Apr 17 at 9:09
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When a body moving at velocity $v$ maintains a circular path around a point, we say the centrifugal force on the body is equal to the centripetal force. In the case of a car going around a circular bend, the centripetal force negating the centrifugal force due to $v$ is provided by the vehicle wheels in the form of friction. For a vehicle to not drift off track, the velocity $v$ (hence centrifugal force) at which it goes through the bend must be equal or less than the maximum static friction the traction of the four wheels can have on that road.

Therefore in this case, the centripetal force is the static friction. Needless to say, the force of static friction (centripetal force) is equal to the centrifugal force. $$ \mu_s N \geq \frac {mv^2}{r}$$

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  • $\begingroup$ How is the force distributed across the different wheels? The front wheels don't have the same angle when steering, they point towards a center of a circle using Ackerman steering. How would the centripetal force vary in each wheel? $\endgroup$ – B V Apr 17 at 19:08
  • $\begingroup$ From my understanding, the total friction force in the radial direction on all four wheels must equal the centrifugal force on the car. It doesn't matter which direction the wheel is pointing, the static friction of a wheel should be the same. $\endgroup$ – TechDroid Apr 17 at 19:39
  • $\begingroup$ I'm guessing what I'm trying to ask is, is there a way to calculate the specific centripetal force in each wheel given the angle the wheel is turned, and the friction coefficient + weight of the car? I understand that all the wheels must add up to mv^2/r, though $\endgroup$ – B V Apr 17 at 21:08
  • $\begingroup$ And are you sure the static friction is the same no matter what direction the wheel is facing? I doubt this is the case. $\endgroup$ – B V Apr 17 at 22:47
  • $\begingroup$ Considering "static" friction in this case, I believe it's a force tending towards the center of the circular path and perpendicular to the velocity vector of the vehicle. Since coefficient of friction isn't dependent on the orientation of contact of the material with the surface (road), then the friction force is the same regardless of the alignment of the wheels. To look at it from a different angle, the surface of the wheel touching the road is the same regardless of orientation also. $\frac 14 \frac {mv^2}{r}$ should cut it for each wheel. $\endgroup$ – TechDroid Apr 17 at 23:31

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