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Here's an example illustration:

a busy cat

The way I understand it, when the car starts turning, the centrifugal force at first does not exceed the force of friction, which allows the car to stir. But at some point, centrifugal force becomes larger than force of friction, and the car breaks into a slide. But if centrifugal force is fictional, then which force acts against the force of friction in this case? Could you explain from the perspective of forces and balance between them what's really going on in this example?

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  • $\begingroup$ Inertia. The car "wants" to keep going straight and friction can not hold her back (making her rotate) strong enough $\endgroup$ – JalfredP Dec 11 '18 at 23:17
  • $\begingroup$ @JalfredP, well, that makes sense, except that it doesn't fit into the statement "to break static force of friction, one has to apply force stronger than it" and "for every force, there has to be a force opposing it". So, unless I misunderstand something completely, there must me some force that acts stronger than force of friction and allows a car to break into sliding. Like "inertial force", but it's also fictitious as far as I know... $\endgroup$ – A.V. Arno Dec 11 '18 at 23:46
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    $\begingroup$ It is not quite correct to say that "to break the static force of friction one has to apply a force stronger than it." It's better to think that "the force of static friction will apply whatever force necessary to keep the object from sliding, up until a maximum strength." From the point of view of outside the car the car turns its wheels. If there were no static friction the wheels would start to slide, but the surface opposes that with static friction. If the car is going too fast, the static friction can't handle that and it starts to slide. No thinking about "applying forces" at all. $\endgroup$ – Luke Pritchett Dec 12 '18 at 0:30
  • $\begingroup$ Related: physics.stackexchange.com/q/109500/2451 and links therein. $\endgroup$ – Qmechanic Dec 12 '18 at 18:37
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Let us say, for simplicity, that the car is moving by inertia (no force acting on it). Well, not totally, because it is inevitable that, while it moves in a straight line, friction with the ground is decelerating it (i.e. reducing its speed). Then you turn the wheels leftward, at which moment friction starts taking another role, it starts changing the car’s direction, which is another form of acceleration. This creates a typical Newton 3rd law pair of forces: the surface of the ground pushes on the car to change its direction, the car pushes on the surface with a force of equal magnitude and opposite direction. But note that these forces are acting on different bodies. To determine what happens to the car, you have to focus on forces acting on the car. It is subject to gravity pushing downward and normal force pushing upward, both of same magnitude but opposite direction, so they cancel out (this is not a Newton pair, because these forces do act on the same object; if they happen to have same magnitude, it is because normal force, when the ground has enough cohesion, is a sort of self-adjusting mechanism that develops as much strength as needed to avoid penetration). So we are left with a single net force acting on the car, which is the friction exerted by the ground against the turned wheels. That is why, as you have shown in your picture, the car is changing its direction. What is your problem though? The car is not turning as much as you would like and so it is sliding out of the road. Which factors cause this effect? Logically, friction is not good enough, there is not enough rugosity; if friction were enormous, that would be the end of it: imagine that, instead of crashing against the small rugosities of the road, the car crashed against big stone balls… The other factor is the car’s velocity: if it were very small, then it would be easier for the ground to create a trap for it…

This formula calculates the maximum velocity that the car can have, given the friction exerted by the road, if you want to keep it moving within the curvature of a given radius described by the road:

$$m\frac{{{v^2}}}{r} = \mu {F_N} = \mu mg \to \frac{{{v^2}}}{r} = \mu g \to {v_{\max }} = \sqrt {\mu gr} % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBamaala % aabaGaamODamaaCaaaleqabaGaaGOmaaaaaOqaaiaadkhaaaGaeyyp % a0JaeqiVd0MaamOramaaBaaaleaacaWGobaabeaakiabg2da9iabeY % 7aTjaad2gacaWGNbGaeyOKH46aaSaaaeaacaWG2bWaaWbaaSqabeaa % caaIYaaaaaGcbaGaamOCaaaacqGH9aqpcqaH8oqBcaWGNbGaeyOKH4 % QaamODamaaBaaaleaaciGGTbGaaiyyaiaacIhaaeqaaOGaeyypa0Za % aOaaaeaacqaH8oqBcaWGNbGaamOCaaWcbeaaaaa!563D! $$

You can read the formula as stating that friction (which is a coefficient of normal force, which in turn is equal to the car’s weight) is strong enough to act as centripetal force, which is the one given by the expression with m and v and r. But please note that the latter expression is not saying that that the car is being pushed out of the curvature by any force dependent on its mass or its velocity or the radius of curvature. The car just tends to keep its original direction by sheer inertia. The expression is not there because of something that the car does. It is there because we assume that the car is turning around and if this happens, it is because there is enough friction, it just happens that we calculate the value of such friction force by looking at the challenge it was facing (mass and v being two factors that make the challenge harder, radius being another that makes things easier).

Therefore, as you can see, no reference to centrifugal force is really needed to explain what happens.

Now imagine that a passenger inside the car feels that he is being expelled out of it. The safety belt retains him, nevertheless. This can be described from an inertial reference frame, like the ground. In this frame, the passenger, again, just moves by inertia. The force that holds it is the same as the one that makes the car turn, the ground’s friction, which is communicated to him through the belt. Then the passenger pushes back on the belt, just like the car pushes on the road. Still we made no reference to centrifugal force.

But we can also describe things from an accelerated reference frame, like the turning car itself. The nature of any reference frame is that it is motionless and it is others who move. So to account for the fact that the passenger feels something pulling him out of the car, in this peculiar (accelerated) frame you must invent a fictitious force, which is called centrifugal force, and is exactly cancelled by the tension exerted by the belt.

Edit: The term "reactive centrifugal force" is sometimes also used meaning the force exerted by the car on pavement or by the passenger on the safety belt. That is a real force. But don't expect to compare it against the friction force and detect which one is the winner, because by Newton 3rd law they are always equal. Give up with the mindset where you see forces as qualities that objects have and bring to the interaction. Force is a measure of the interaction. Let the car crash against a wall and you will find a strong force, let it slide on ice and there will be almost none. If you still want to see things as a battle, you can, but don't choose the Newton 3rd law pair of forces as the fighters. You must choose forces acting on a single body... and being currently active. You are comparing friction on one side and the historical force that set the car in motion on the other side. But the latter force is not operative. That does not mean it is irrelevant: it indirectly shows up in the form of the inertia of the body (m and v), which is what triggers the interaction and as explained in the answer it has the effect of making the challenge harder for the pavement.

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Same problem exists when the car breaks in a straight line, it is simple inertia from Newton's third law of motion: for every action there is an equal and oposite reaction.

So pavement pushes car left (friction), then car pushes pavement right, which is what you perceive as the "fictitious" force.

I would argue that centrifugal force is only fictitious outside the car.

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  • $\begingroup$ "it is simple inertia" This certainly makes sense intuitively, but doesn't fit (at least from a perspective of not so good at physics myself) with the explanation of static force of friction. Because it says that a force must be applied against it to overcome it. Can just inertia be applied against it too? So basically, inertia per se can act as force and do work? $\endgroup$ – A.V. Arno Dec 12 '18 at 0:12
  • $\begingroup$ Adrian, you mixed Newton's laws of motion. Law 1: An object at rest tends to stay at rest and an object in motion tends to stay in motion at constant velocity (law of inertia); Law 2: acceleration requires a net force, usually expressed as $F=ma$; Law 3: For every action, there is an equal and oppositely directed reaction. $\endgroup$ – David White Dec 12 '18 at 1:52
  • $\begingroup$ Fixed "second law" to "third law", thank you. $\endgroup$ – Adrian Constantin Dec 12 '18 at 10:49
  • $\begingroup$ To be correct "inertia" is what keeps the moving object in motion when no force is acting on it. Intuitively I was thinking of it as also the reason for putting up a "reaction" when an "action" is applied on the object. So the car starts sliding when friction can not keep it on path, and you believe there must an external force that is overcoming (cancelling) friction. But if that were the case, there would be no effect on the car from friction any more. $\endgroup$ – Adrian Constantin Dec 12 '18 at 11:11
  • $\begingroup$ Even when the car slides, friction is still there. I believe your question is: why is friction sometimes enough and sometimes not enough for the expected path of the car, so it is like you are asking why do I need a bigger force to make a sharp turn, then I need to make a large turn ? Because acceleration is bigger for the sharp turn (sharp turn means: shorter time, bigger change in speed). And the force needed to accelerate the car is F = ma, or in our case we should say: a = F / m. So to get bigger a, you have to start with bigger F. $\endgroup$ – Adrian Constantin Dec 12 '18 at 11:13
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You question is about the application of Newton's first two laws in an inertial frame of reference and in an accelerated frame of reference.

Assuming that the Earth is an inertial frame of reference then your car will continue to move at a constant velocity unless there is a force acting on it.
So to turn a corner there must be a horizontal force on the car to produce a centripetal acceleration and in your example that force is initially a static frictional force.

Remembering that the static friction force can have any value up to a maximum of $\mu_{\rm static} N = \mu_{\rm static} mg$ then the car (assumed to be a point mass $m$) can turn a corner as long as this inequality is satisfied $\dfrac{mv^2}{r} \le \mu_{\rm static}mg\Rightarrow \dfrac{v^2}{r} \le \mu_{\rm static}g$.

If the car is travelling at a maximum speed for a given corner radius then $\dfrac{v_{\rm maximum}^2}{r} = \mu_{\rm static}g$

Now suppose that the bend become sharper then $r$ decreases and the friction cannot supply enough force to prevent the car slipping relative to the ground.
The car will trying to travel along a less curved path (larger $r$) path and/or at a slower speed with the frictional force now being smaller, $\mu_{\rm dynamic} mg$.
All the above is really a use of Newton's second law $F=ma$.


The introduction of fictitious forces is necessitated by the fact that Newton's laws do no work in accelerated frames of reference.
In this case if the frame of reference was the car, ie the car was at rest relative to that frame, there would be a frictional force on the car and yet the car was not accelerating relative to this new frame.
The way out of this so that Newton's laws could still be used is to introduce another force, a "fictitious" force, which in this case was equal in magnitude and opposite in direction to the frictional force, you called it the centrifugal force.
In this way the net force on the car would be zero and Newton's laws would apply.
When the car started to slip relative to the ground the value of the centrifugal force would change and it would then equal the value of the new dynamic frictional force and again the net force on the car would be zero.
When you are going around the corner and not accelerating relative to the car (the frame of reference) you would say that there are two forces acting on you, the force of the car seat and seat belt on you and a centrifugal force ie you have a net force on you of zero which then consistent with Newton's laws.

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"Fictitious" is physics jargon, and doesn't mean the same thing it does in everyday life. It doesn't really mean "made up" it just means it doesn't exist in an inertial frame. Your picture is drawn in an inertial frame: there is no centrifugal force. Friction keeps the car on the road as long as it doesn't try to turn too fast. Once the wheels are turned so far that the lateral force required to keep pushing the car to go further to the left is more than friction can supply, it starts to slide.

Centrifugal force is present when you analyze motion from within a rotating frame of reference (such as a frame fixed to a car that's going around a curve). A passenger inside the car feels a force pushing her toward the right as the car turns toward the left, because for that person, the car is a natural frame of reference, and as it follows a curved path "fictitious" forces arise. From the passenger's point of view, which ignores the world outside and considers all motion relative to the car, it is indeed centrifugal force that pushes the car into the ditch.

Edit following discussion with OP:

The intuition that a force is required to push a cornering car off its intended path suggests that you have the impression that following the curve is the “natural inclination” of the car—else why would a force be required to cause it not to? In reality, the “natural inclination” of the car is to follow a straight path. This inclination is overcome by the force of friction when a car navigates a curve. If the curve is too sharp or the car is too fast, the “natural inclination” wins: the car follows something closer to a straight path than the road does because the maximum possible value of the frictional force between the tires and pavement is not large enough.

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  • $\begingroup$ "lateral force" So, is this the term for it? Because google shows this term only as a characteristic of tires, not as a physical force. $\endgroup$ – A.V. Arno Dec 11 '18 at 23:43
  • $\begingroup$ You can call it "centripetal force". It's the force required to make something change its direction of motion. In this case it's provided by friction between the road and the tires. For a stone whirling around on a string, it's the tension in the string. etc etc $\endgroup$ – Ben51 Dec 11 '18 at 23:44
  • $\begingroup$ now I'm even more confused. Centripetal force pulls car into turning (apparently, centripetal in this case being the force of friction, because it prevents car from continuing going forward and thus makes it turn). But what force acts opposite to it? $\endgroup$ – A.V. Arno Dec 11 '18 at 23:49
  • $\begingroup$ Sounds like you're not so confused. $\endgroup$ – Ben51 Dec 11 '18 at 23:50
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    $\begingroup$ In the inertial frame, that is not a correct description of the situation. There is NO force which "overcomes" friction. There is only one force on the car: friction. That's why it goes in a curve instead of straight. When the driver asks the car to turn so sharply that friction can no longer provide the needed force, the wheels start to slide. $\endgroup$ – Ben51 Dec 11 '18 at 23:53
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A turn requires an inwards radial acceleration $a_r$. Static friction $f_s$ is what causes it:

$$f_s=ma_r$$

There are no further forces in this equation - no "centrifugal force" or anything else.

Static friction has a limit, though:

$$f_s\leq \mu_s n$$

To keep a certain turning radius, a certain radial acceleration $a_r$ is required. If that $a_r$ can't be achieved, then the turn must widen - meaning, then the car will slide outwards to make a bigger "circle", a less sharp turn. And this is exactly what happens: when you turn so sharp that the $a_r$ must be so big, that the static friction can't achieve it (it has to go above it limit to achieve it), then static friction let's go (and kinetic friction takes over), and the car slides.

About fictitious forces

Imagine standing in a bus, not holding the rail.

The bus brakes hard.

You fall forwards.

Was it a force that pushed you forwards? From your eyes it certainly seems like it - because your eyes see the bus as stationary. As if the bus wasn't moving, but you were. So something must have pushed you.

But we know that this is an illusion. Your body was in full speed forward along with the bus. Suddenly the bus braked, but your body still continued forward at that speed. It has inertia. No force involved here (except for the force that stops the bus). It is purely an illusion. If you consider the bus as your frame, then you must invent a non-existing or fictitious force that pushes you forward - otherwise Newton's laws wouldn't hold true.

The same is the case when something moves in circular motion (a turn is a small part of a circular path). You feel squeezed sideways when sitting in a turning car, as if some "centrifugal force" pushes you towards the door. But in reality, your body just tries to continue moving straight, and the car moves sideways and pulls you along. The "centrifugal force" doesn't really exist.

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