This solution does not make sense to me (dynamics)

[Pascal1p]

Homework Statement

Homework Equations

F=m*a

The Attempt at a Solution

Don't worry it not a homework question of mine. I was just trying to solve some questions (on my own) that I know I have difficulty with to get better at it. Especially with springs I can get confused, so I thought I would try that.
With this question I got stuck and I know that it was because I set the unstretched length as when the spring went a distance 'd' back, but of course he is only at equilibrium then and not unstretched then.

But still I don't get the answer. They talk about that if y=d then N=0, but to me that does not seem right.
y(equilibrium) + d, is the position he would be at when you press it down measured from the unstretched position?
As I see it they saying, at the position the Block A is when it is being pushed down and released from rest, at that point there is no Fn working on it. Which would be silly.
N=0 should be the point when the string is in unstretched position, since at that moment Block A and Pan B should separate? So N=0 when F(spring)= 0= k*s, so when s=0.

 

[haruspex]

But still I don't get the answer. They talk about that if y=d then N=0, but to me that does not seem right.
y(equilibrium) + d, is the position he would be at when you press it down measured from the unstretched position?

I agree it is not well explained. Worse, it is wrong
If you depress the spring by distance d from equilibrium and release, it will oscillate equally about its equilibrium, so will rise distance d above equilibrium.
The mass will separate from the pan when its acceleration is g down (i.e. no normal force). At the instant this happens, the pan's acceleration must be the same, despite being attached to the spring. Hence, at this instant, there is also no force between the pan and the spring, i.e. the spring is at its unstretched length. But note that in that reasoning we have not specified how far the spring was pushed down initially. In other words, as long as you push it down far enough that spring reached its uncompressed length the block will separate from the pan at that instant.
The answer obtained above is only the minimum value for d; any larger value will also work.

 

[Pascal1p]

I agree it is not well explained. Worse, it is wrong
If you depress the spring by distance d from equilibrium and release, it will oscillate equally about its equilibrium, so will rise distance d above equilibrium.
The mass will separate from the pan when its acceleration is g down (i.e. no normal force). At the instant this happens, the pan's acceleration must be the same, despite being attached to the spring. Hence, at this instant, there is also no force between the pan and the spring, i.e. the spring is at its unstretched length. But note that in that reasoning we have not specified how far the spring was pushed down initially. In other words, as long as you push it down far enough that spring reached its uncompressed length the block will separate from the pan at that instant.
The answer obtained above is only the minimum value for d; any larger value will also work.

That minimum value of d, does it correspond with the answer they give?
I don't get that they do Fspring= k*(Yeq + Y) and then say if Y=d then N=0.
Because like you said N=0 is at unstretched position, so when Fspring= 0
Fspring = k*(Yeq+d) that corresponds to the springforce on the pan, when the pan is pushed down with 'distance' d?

 

[haruspex]

That minimum value of d, does it correspond with the answer they give?
I don't get that they do Fspring= k*(Yeq + Y) and then say if Y=d then N=0.
Because like you said N=0 is at unstretched position, so when Fspring= 0
Fspring = k*(Yeq+d) that corresponds to the springforce on the pan, when the pan is pushed down with 'distance' d?

Yes, the value they get is the amount it has to be depressed in order that it will bounce up to the relaxed height. This is very easy to calculate; they really made it harder than it is. We know the equilibrium position is d=total weight/k below the relaxed position. If depressed by y from equilibrium and released it will oscillate +/-y about equilibrium, so to oscillate up to relaxed position it must be depressed by the same d.