Consider a front-wheel drive car that is speeding up on a circular track of radius $R$. The static friction force has a radial component and a tangential component. At some point the speed will become too high for the car to stay on this circular trajectory. At that point the radial component of the friction force will be $\mu_s N$, where $\mu_s$ is the coefficient of static friction and $N$ the normal force. But since the friction force also has a tangential component, the total friction force will be greater than $\mu_s N$. QED.
$\begingroup$
$\endgroup$
Add a comment
|
4 Answers
$\begingroup$
$\endgroup$
3
At that point the radial component of the friction force will be $\mu_sN$
This is your mistake. As long as the car is not slipping, the radial component of the friction force will be given by $F_r = \frac{mv^2}{R}$. However, the total friction force must be less than or equal to $\mu_sN$ in magnitude, not just the radial component. Since the car is speeding up and the only forces acting on it are gravity, normal and friction, where gravity and normal cancel out, the net force on the car is equal to the force of friction. Thus we have: $$\vec F_f = m\vec a = \mu_sN \hat a$$ From this, we can see that the radial component of friction will instead be given by $\mu_sN\frac{a_r}{a}$ where $a_r$ is the radial component of acceleration and $a$ is the magnitude of the total acceleration. An interesting consequence of this is that the maximum speed possible before you start to slip is lower if you're speeding up than if your speed is constant.
-
$\begingroup$ I get your point, and it is a natural objection to my "proof". However, notice that if we accept that it is the total friction force which is equal to μ N , this has practical implications. For example, suppose you are driving your anti-lock-brakes-equipped car along a straight road that is banked. Then since you are "using up" part of your static friction to avoid sliding down, the car stopping distance should be more than the stopping distance if the road is level. Are you aware of any experimental evidence for such behavior? $\endgroup$ Sep 13, 2020 at 2:16
-
$\begingroup$ @JoseMenendez That is correct that the stopping distance would be higher for a banked road. I don't know if there have been any experimental tests of this behaviour. $\endgroup$– SandejoSep 13, 2020 at 2:35
-
$\begingroup$ Thanks. It's a small effect, though. It goes like the square of the sine of the banking angle. Probably impossible to see experimentally due to the crudeness of the μ N theory. All this is inspired by Example 8.6 in the last edition of Knight's textbook, where the author solved an accelerated circular motion problem by assuming that the radial component of the friction force reaches the maximum μ N value, and obscuring this assumption by using "force of wheels" to name the tangential component of the same force. $\endgroup$ Sep 13, 2020 at 6:38
$\begingroup$
$\endgroup$
Sometimes it is useful to introduce the concept of "sticktion". While the tyres still grip, sticktion can build up beyond the level of friction. Once the sticktion maximum is exceeded, the car breaks away and the retarding force reduces to the level of friction.
answered Sep 12, 2020 at 18:32
$\begingroup$
$\endgroup$
It is possible that the friction coeficient depends on the direction because the tire has grooves, it is not isotropic. Let's say that $\mu_R$ is the radial friction coeficient and $\mu_t$ the tangential one.
If $\mu_R < \mu_t$, after a speed threshold the centripetal force vanishes, and the car slides following the local tangent path.
If $\mu_t < \mu_R$, the tangential speed reaches a maximum and the tires starts to slide. But even so, the speed can still increase, despite of some sliding. When the speed is such that $F_R = m\frac{v^2}{R} = \mu_R N$ => $v = \sqrt{\mu_R gR}$, the centripetal force vanishes and the result it the same as before.
answered Sep 12, 2020 at 18:42
$\begingroup$
$\endgroup$
If the vehicle is both accelerating (or braking) and cornering, the total friction force will be greater than either the lateral friction associated with cornering alone or the longitudinal friction associated with accelerating (or braking) alone. Since the total friction force is shared between the two, the vehicle will slip sooner if both accelerating and cornering at the same time, than if only accelerating or only cornering.
This can be illustrated by using the so called Kamm circle of friction. Refer to the figures below. The Kamm circle assumes that the coefficient of static friction is the same both longitudinal and laterally, as well as assumes the normal load supported by the tire is the same for both accelerating and cornering. These assumptions are discussed later.
The figures are an overhead view of one of the vehicle tires. The vehicle direction of motion is upward. The circle radius represents the maximum possible static friction force, or $u_{s}N$ where $N$ is the load (weight) of the vehicle supported by the tire.
Fig 1 shows the vehicle accelerating only where the longitudinal friction force enabling acceleration, $F_{Lon}$, equals the maximum possible static friction force, i.e., where loss of traction is impending.
Fig 2 shows the vehicle only cornering (to the right) where the lateral friction force, $F_{Lat}$ which equals the centripetal force, equals the maximum possible static friction force, i.e., where skidding is impending.
Fig 3 shows the vehicle both accelerating and cornering. As can be seen, the total friction force reaches the maximum possible static friction force sooner that when only accelerating or cornering.
Bottom line: The vehicle will slip sooner if both accelerating and cornering at the same time, then if only accelerating or cornering.
Regarding the assumptions discussed at the beginning.
Since the tire tread is not necessarily the same in the longitudinal and lateral directions, the coefficients of static friction may be different. For example, if the tire is designed more for cornering then the Kamm circle may be an ellipse, with the major axis in the horizontal direction.
Since when accelerating or cornering the weight of the vehicle may shift among tires and on a different area of the tire, the assumption of the same load $N$ may not be correct.
Hope this helps.
