0
$\begingroup$

Suppose we are riding on a vehicle which is moving on a flat road and the road is rough. When taking a turn,centrifugal force applies on us from the frame of car and frictional force tries to obstruct that. It is said that when centrifugal force becomes more than frictional force,then it will skid. But..

Since we are in the frame of car,we are in equilibrium. In that case net force on us must be $0$ no matter what. So if frictional force exists,then it should always be equal to centrifugal force. So the question of maximum frictional force shouldn't even rise.

I know i am wrong,but i am interested to know where i am going wrong since i am new to pseudo forces.

Improve this question
$\endgroup$
2
  • $\begingroup$ Do you understand the problem in the inertial frame, where the friction provides an unbalanced centripetal force and changes the car's velocity? $\endgroup$
    – rob ♦
    Commented Sep 28, 2021 at 18:48
  • $\begingroup$ No,please enlighten me on that $\endgroup$
    – madness
    Commented Sep 28, 2021 at 19:50

3 Answers 3

Reset to default
3
$\begingroup$

The “centrifugal force” is not an interaction between two objects. For that matter, neither is the “centripetal force.” “Centrifugal” and “centripetal” are directions. The “-fugal” means “fleeing” (related to “fugitive”); the “-petal” means “pointing towards.”

If you’re new to this business (as you say in a comment), don’t leap to the accelerated reference frame attached to the turning vehicle. Accelerated reference frames are complicated. First analyze the problem from the inertial reference frame of the road. The vehicle travels down the straight section of the road at constant velocity. When the vehicle gets to the curve in the road, it’ll still move in a straight line (to the horror of everyone inside) unless some force changes its velocity. That force must have some net component in the centripetal, center-pointing, direction. It might be friction between the road and the tires. It might be that the road is “banked” so that the normal force has a centripetal component. It might be that both of these fail and the vehicle is pushed centerward by the guardrail on the outside of the curve (to the horror of everyone inside).

If the force needed to steer the car along the circle is greater than the maximum static friction between the tires and the road, the tire-to-road interaction will switch from static/rolling friction to sliding. The car may still follow a curved path — but if the radius of curvature is different from the radius of curvature for the road, then the car is going to end up in the ditch.

When you write

Since we are in the frame of car, we are in equilibrium. In that case net force on us must be 0 no matter what.

you are assuming that this reference frame exists, that the car occupies it, and that you can ask questions about it. The idea that “we must be in equilibrium” is a common beginner mistake. There is no such rule. For your velocity to change, you must be in disequilibrium, with a nonzero net force.

The illusory (but, in limited circumstances, computationally helpful) “centrifugal force” arises when you look at the car within an accelerating, non-inertial reference frame. Suppose that your car is parked. You balance your hot coffee cup on the steering wheel and (to the horror of everyone present) stomp on the gas pedal, lurching the car forward. Was the coffee thrown backwards into your lap by a “front-fugal” force? You could say so if you liked; but it’s more parsimonious to say that the coffee was trying to stay in the same place, while the car moved out from underneath it.

Improve this answer
$\endgroup$
4
  • $\begingroup$ What does "parsimonious" mean in this case? (Your answer, especially the last paragraph is quite an interesting one -- upvoted.) $\endgroup$
    – Shane
    Commented Sep 29, 2021 at 3:32
  • 1
    $\begingroup$ “parsimonious (adj): using a minimal number of assumptions, steps, or conjectures.” Why build both the accelerating reference frame and invent a front-fugal force to throw the coffee cup, when you can use an inertial frame where Newton’s laws work and say that only the car moved? $\endgroup$
    – rob ♦
    Commented Sep 29, 2021 at 5:22
  • $\begingroup$ Thanks a lot!Now if we assume the car is going with a tangential acceleration $a_t$ and we are asked about the skiding of car,then at the turning point there is tangential as well as centripetal force,now how can we say that frictional force is the net force of these two?I have been disturbed by this concept for the past few days,why it must be the resultant? $\endgroup$
    – madness
    Commented Sep 29, 2021 at 14:49
  • $\begingroup$ “Centripetal” and “tangential” are both directions; by definition they are perpendicular. The friction force, a real interaction, may have nonzero projection onto both the centripetal and tangential directions. Compare with frictionless-inclined-plane problem, where the vertical weight force must be projected parallel and perpendicular to the direction of motion. $\endgroup$
    – rob ♦
    Commented Sep 29, 2021 at 20:53
1
$\begingroup$

So if frictional force exists, then it should always be equal to centrifugal force

Only up until the maximum possible static friction force between the tires and the road is reached. Between the tires and the road that force is

$$F_{f-max}=\mu_{s}N$$

Where $N$ is the normal (perpendicular) force between the tire and the road (i.e., the portion of the weight of the vehicle supported by the wheel), and $\mu_{s}$ is the coefficient of static friction between the tire and the road.

The centrifugal force is a pseudo force that exists only in the non-inertial (accelerating) frame of the car to explain the effect of the inertia of the vehicle and its occupants. It is the same magnitude as the centripetal force with the opposite sign. The actual force responsible for turning the car (and its occupants) is the centripetal force in the inertial frame of the road which acts towards the center of the turn. That force is provided by static friction between the tires and the road (and static friction between the occupants and their seats to keep them from sliding).

For a car of mass $m$ moving at constant speed of magnitude $v$ and circular motion with radius $r$ the centripetal force $F_{C}$ is given by

$$F_{C}=\frac {mv^2}{r}$$

The static friction force, $F_f$, balances the centripetal force $F_C$, up until the maximum possible static friction force is reached, or until $F_{f-max}=F_{C}$.

Impending skidding thus occurs when

$$\frac {mv^2}{r}=\mu_{s}N$$

Hope this helps.

Improve this answer
$\endgroup$
1
  • $\begingroup$ But see even if car skids away,from our point of view in car frame,we are still at rest,then centrifugal force acts on us but net force should be $0$,isn't it contradictory? $\endgroup$
    – madness
    Commented Sep 28, 2021 at 22:36
-2
$\begingroup$

If centrifugal force is larger than the maximum frictional force, then you will move outward. To stay in the car, the door will provide the extra force you need to have net force equals 0. The forces you experience in the car frame include centrifugal force, frictional force, and reaction force from the door.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.