Pseudoforces as seen by observer in a non-inertial frame

Question

Suppose, we have a car, that is accelerating with respect to an inertial frame, with an acceleration $a_c$. So, observers in the inertial frame can claim that there is a force $F_c=ma_c$ acting on the car.

Now suppose, we attach a non-inertial frame of reference to the car. Hence, this reference frame is also accelerating at $a_c$ relative to the inertial frame.

Using transformations and considering the inertial and non-inertial frames to be $A$ and $B$, one can easily show that :

$$ma_A=ma_B+ma_{AB}$$ Here $a_{AB}$ is the relative acceleration of non inertial frame $B$ with respect to $A$.

However, this means : $$ma_B=ma_A-ma_{AB}$$

In our example, the acceleration of the car in inertial frame is $a_c$ and the acceleration of the non-inertial frame attached to the car, is also $a_c$ by definition. Hence, we can write : $$ma_B=ma_A-ma_{AB}=ma_c-ma_c=0$$

Thus, we have proved the obvious that in a reference frame attached to the car, the car seems to be at rest i.e. there is no forces acting on the car.

My problem is that I'm unable to interpret the above properly. If there are no forces acting on the car in the non-inertial frame, as the net acceleration is zero, how can people sitting in the car experience a backward pseudoforce. I know the origins of the pseudoforce due to inertia and so on, but I don't see mathematically how to interpret the above.

In general, the force on any object in it's own non-inertial frame comes out to be $0$. Then how can we say that observers in non-inertial frame experience pseudoforces. In our above equation, the term $ma_{AB}$ represents the pseudoforce. However, isn't this total cancelled out by the acceleration in the inertial frame term $ma_{A}$.

Hence the pseudoforce that observers inside a car feel, when the car accelerates forward, should be cancelled out by the forward acceleration of the car itself. Hence the total force in the non-inertial frame is supposed to be $0$, and yet, there is a distinct pseudoforce acting on the observers.

Answer 1

You assume that the force which acts on the car is the same one that acts on the passengers, which is not correct since the car and the passengers are two different objects.

What you actually feel when the car accelerates forward is the chair pushing you forwards and you pushing the chair backwards. When there is no contact with the chair your body tends to remain in equilibrium (remains at the same velocity). As soon as your body makes contact with the chair, the chair acts with forward force on the body which you can feel.

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Consider a different example which is a bit more intuitive - a passenger on roller skates in an empty wagon that accelerates at $a$ with respect to earth. The roller skates are introduced to cancel the friction force as much as possible.

The passenger and the wagon are two different objects, which means that force acting on the wagon is not the same force acting on the passenger. When there is a net force applied to the wagon, the wagon starts accelerating relative to a stationary (inertial) reference frame $A$. Since no force is applied to the passenger, the passenger remains in equilibrium as seen from $A$, which follows directly from the first Newton's law.

In the wagon (non-inertial) reference frame $B$, which moves together with the wagon in the same direction, the passenger appears to be accelerating at $-a$ (going backwards) although no force has been applied to them. This clearly violates the first Newton's law which says

An object acted on by no net external force remains in equilibrium, i.e. has a constant velocity (which my be zero) and zero acceleration.

This is why we must introduce a force that acts on the passenger in order to make the reference frame $B$ satisfy the first and second Newton's laws. Since this force actually does not exist, it is called a pseudo-force or fictitious force.

To describe this with equations

$$a_{P/A} = a_{P/B} + a_{B/A}$$

where $a_{P/A}$ is the passenger acceleration in $A$, $a_{P/B}$ is the passenger acceleration in $B$, and $a_{B/A}$ is the wagon acceleration in $A$. Since no external force acts on the passenger $a_{P/A} = 0$, the above equation becomes

$$a_{P/B} = -a_{B/A}$$

Suddenly, the passenger has some acceleration although there is no external force acting on them. Introduce a fictitious force to make the first and second Newton's laws work in $B$ and problem solved. However, this clearly violates the third Newton's law since the fictitious force has no reaction pair, i.e. there is no force that acts from the passenger to the wagon. This is why non-inertial reference frames are not suitable for the Newton's laws.

Answer 2

Key here is that how you "feel force" is not determined by the total force acting on you. You can "feel force" even when total force is zero. A simple example is pushing a wall with your hand. If the wall is not moving anywhere, its and your hand's acceleration is zero and thus total force is zero. Yet you feel the push in your hand. This is because your hand is indeed acting a force on the wall but the wall is acting a normal force of similar magnitude but opposite direction on your hand thus making total force zero.

In your car example you are correct that in the non-inertial frame attached to the car the total force acting on a passenger in zero. The feeling of force is due to the person being pushed toward the seat and the seat acting a normal force toward the passenger. As in the example of pushing wall with a hand, the person "feels force" even though total force (real forces plus pseudo forces) acting on the person in the non-inertial frame is zero.