Understanding an equation in a dynamics spring problem

[Bunny-chan]

Homework Statement

A block of mass M = 0.5kg, attached to a spring of elastic constant k = 3N/m on a vertical wall, slides without friction through an horizontal air table. A disk of mass m = 0.05kg is placed on the block, whose surface has a coefficient of static friction \mu _e = 0.8. What is the maximum oscillation amplitude of the block so the disk won't slide off of it?

Homework Equations

\vec F = m\vec a \\ \vec F = -kx \\ F_{\mu e}^{max} = \mu _emg

The Attempt at a Solution

I have already solved this problem, and then I've checked part of a solution on the web containing an insight on the theory behind it:

I don't really understand the second part where the author talks about the non-inertial frame reference, and about the forces involving it, and ends up with the equation

Even though I was able to solve the exercise, I reached the last equation through a more direct way, without thinking too much about it, so this explanation made me a bit confused. Am I being clear enough?

I hope someone can help!

 

[haruspex]

this explanation made me a bit confused.

There are cases where noninertial frames make life simpler, but this is not one of them. The sign of a is irrelevant since reversing it just leads to the other extreme of x, so it matters not whether you write a=μ_eg or a=-μ_eg. This same equation arises whether you think in terms of an inertial frame or a noninertial one.

 

[Bunny-chan]

There are cases where noninertial frames make life simpler, but this is not one of them. The sign of a is irrelevant since reversing it just leads to the other extreme of x, so it matters not whether you write a=μ_eg or a=-μ_eg. This same equation arises whether you think in terms of an inertial frame or a noninertial one.

Hmm. I think I get it. So this is why in these equations we don't really take into account the negative signs? Because we're just interested in the value?

 

[haruspex]

Hmm. I think I get it. So this is why in these equations we don't really take into account the negative signs? Because we're just interested in the value?

No, I'm saying that both signs are correct for maximum displacement; but the question asks for amplitude, which is by definition the magnitude of the maximum displacement, so non-negative.

 

[Bunny-chan]

No, I'm saying that both signs are correct for maximum displacement; but the question asks for amplitude, which is by definition the magnitude of the maximum displacement, so non-negative.

OK. I see. Thank you.