What is a fake force? A fake force is one of those forces that introductory texts tell you aren’t real – like centrifugal force. They aren’t real in the sense that they are due to one of the fundamental interactions. Basically, introductory texts (and even blogs like this one – not a bad summary of real vs. fake forces) attack the centrifugal force. This is because it is so common for students to want to use these faux forces in the wrong way. Better to just not use them at all.

Anyway, there are times when faux forces are awesome. Just to be clear, a faux force is needed to use normal newtonian mechanics when the reference frame is accelerating. To show this, let me look at the following problem. I will solve it without and then with fake forces.

Suppose I am in an elevator that is accelerating upwards and I throw a ball up. How high will it go in the elevator (in the reference frame of the elevator)?

Note: v_{0} is the velocity thrown with respect to the elevator. Let me assume that at t = 0 sec, the elevator is not moving (but accelerating) and at a position y = 0 meters. I hope you can convince yourself that if I solve this with this assumption, it would still work if you chose something else for the initial velocities (if you are not convinced, that will be a homework assignment for you). So, here is what I have: (note this is a 1-d problem, so all velocities and accelerations are y-components and thus not vectors)

- mass of ball = m
- initial velocity of the ball with respect to the elevator = v
_{0} - initial position of the ball with respect to the elevator = 0 meters
- initial position of the elevator with respect to the Earth = 0 meters
- acceleration of the elevator with respect to the Earth = a
- initial velocity of elevator (wrtE) = 0 m/s
- initial position of the elevator (wrtE) = 0 m

So, what is the plan? The plan is to deal with this stuff from the perspective of a person on the Earth, outside the elevator because this is an inertial frame where Newton’s laws are supposed to work. The position of the elevator with time will be:

Remember, I already said the initial velocity was 0 m/s and the initial position was 0 meters. This is really just the kinematic equation (which I previously talked about). Now, here is the position of the ball (with respect to the Earth). After the ball is thrown, there is ONLY the gravitational force on the ball. It doesn’t matter that it is in an elevator.

Again, the ball starts at y=0 meters and g is the local gravitational constant = 9.8 N/kg. Now what? Well, in the elevator, the ball appears to be at a height that is the difference between the y position of the ball and the y-position of the elevator:

Clearly, I can combine a couple of terms here and get:

Maybe you see something cool here, but I will talk about that in a second. How about I just look at a sketch of the position of the ball and the elevator:

I made that graph with GraphSketch.com – pretty awesome, check it out. Notice that the quantity I am interested in in the difference between the red and blue lines. I could just plot that, but I want a general solution for the maximum of this value. Whenever you think of max (or min) you should think of a max-min problem (that is unless you have never had calculus). So, the following involves some calculus (Warning).

If I have some function, it could go up and down. How could I find when it is at a maximum or minimum? Here is a sketch:

When the function is at a maximum or minimum (other than as t gets very large or small) then the slope of the function has to be zero as shown by the dotted line. So, to find a max or min point I can just find the slope and see where that is zero. The slope of a function is it’s derivative. I will take the derivative of h(t) with respect to time:

So, the slope changes with time. I can find at what time the slope is zero.

This is the time the function is either a max or min (technically a local max or min). Notice that there is only one. If the slope had been a 2nd order polynomial, it could have had two values for when the slope was zero. If I plug this time for the max or min back into the height function, I will get the max or min height. Hopefully, I should be able to tell if it is a max or min when I plug it in.

Check number 1 – does it have the right units? It would be (m/s)^{2} over m/s^{2} which is meters – that is good. Is it positive? Yes, it should be. Does it give the correct answer for the case where a = 0 m/s^{2}? Yes (I will show this later). Ok. Great.

Now, let me do this using a fake force.

I think you can imagine what this fake force will look like, but let me derive it. Suppose my frame is accelerating with respect to an inertial (non-accelerating frame). Then I can measure the position of a ball two ways y and y’ where I will use y’ as the position of the ball in the inertial frame.

The relationship between y and y’ would be:

Where this assumes that at t = 0 seconds, the two frames are at the same location such that y = y’. Now, I can take the derivative twice (with respect to time). Also, remember that I am only dealing with the y-motion, so these are not vectors.

This says the velocity of the ball in the inertial frame (not accelerating) is the velocity of the ball as measured in the elevator plus at. If I take the derivative again (with respect to time) I get:

My notation is getting awkward. I will use a’ as the acceleration of the ball from the ground. a is the acceleration of the ball from inside the elevator and a_{e} is the acceleration of the elevator with respect to the ground. Now I will solve for a:

Now I will multiply both sides of this expression by the mass of the ball.

First, I replaced ma’ with F_{net}. This is the net of all the real forces. If I want to be in the frame of the accelerating elevator, I have this extra -ma_{e} term. It looks like a force term, so I call it a force – a faux force. (one-d, no vector notation). You could easily re-derive this with vectors and get something like:

So, the point is: if I have an accelerating elevator, I can add this fake force that is the opposite of the acceleration. Note – if the other frame is rotating, things get much more complicated. Now, let me redo the elevator problem from the frame of the accelerating elevator. When the ball is in the air, I have the following free body diagram:

In the y-direction, Newton’s second law looks like:

So, the acceleration of the ball in the frame of the elevator is just like gravity were stronger. So, I can write the kinematic equations (which work for constant acceleration):

This is the position of the ball in the elevator as measured in the elevator. Notice that this is EYE-dentical to the height I derived before. However, now when I take the derivative with respect to time (to find the maximum position) it IS the velocity of the ball in the frame of the elevator. Notice that the highest point is when the velocity is zero.

The ball will be the highest when v = 0 m/s which happens at

Again, this is the exact same time as before. So I get the exact same height.

### Conclusion

I know you were going to say this:

“Why are you making this so hard? Everyone can see that if the elevator is accelerating up, it just just like gravity is stronger. Duh. Also, Einstein said that an accelerating frame is EYE-dentical to gravity anyway.”

This is true, but it is useful to look at it both ways.