A Spring Supporting a Box Containing a Block

[cj]

A spring of stiffness k supports a box
of mass M in which is placed a block of
mass m.

A) If the system is pulled downward
a distance d from the equilibrium position
and then released, find the force of reaction
between the block and the bottom of the
box as a funtion of time.

B) For what value of d will the block just
begin to leave the bottom of the box at the
top of the vertical oscillations? Neglect any
air resistance.
-----
I'm assuming undamped SHM:
(M+m)\frac{d^2x}{dt^2} + kx = 0 ??

Does the reaction force between the
Block and the bottom of the Box simply
equal the mass of Block times the
acceleration of gravity + acceleration of
the Box?
F_{reaction} = m_{box} \cdot (g + \frac{d^2x}{dt^2}) ??

Is there any more development I could do
to better represent the force of reaction
between the block and the bottom of the
box as a funtion of time?

For Part B -- under what conditions would the Block
leave the bottom of the Box? I'm perplexed here
because it seems that the acceleration of the
Block will always be the same as that of the
Box, but experienced-based intuition tells
me that, yes, the Block could at some point lose
contact with the Box??

 

[HallsofIvy]

Notice that the acceleration of the box, block unit depends upon their total mass but the acceleration of the block alone depends on the mass of the block alone.
As long as the block is in contact with the box, they have the same acceleration. The difference between the force necessary to give the block-box a given acceleration and the force necessary to give the block alone that acceleration is the force of the box on the block.

What happens when the spring reaches its greatest length (i.e. its amplitude)? The spring is attached to the box and will start to pull it back but it is not attached to the block which has its own momentum.

 

[Doc Al]

Just a couple of points to add to what HallsofIvy said.

Does the reaction force between the
Block and the bottom of the Box simply
equal the mass of Block times the
acceleration of gravity + acceleration of
the Box?
F_{reaction} = m_{box} \cdot (g + \frac{d^2x}{dt^2}) ??

No. Three forces act on the box: its weight, the spring, the block-box contact force (the "reaction force").

On the block, only two forces act: its weight and the block-box reaction force. So...
F_{reaction} = m_{block} \cdot (g + a)
where a is the acceleration of the block.

For Part B -- under what conditions would the Block
leave the bottom of the Box? I'm perplexed here
because it seems that the acceleration of the
Block will always be the same as that of the
Box, but experienced-based intuition tells
me that, yes, the Block could at some point lose
contact with the Box??

Note that the box is constrained by the spring. When it rises to a certain point, it will accelerate downward greater than g. Of course, the block is not so constrained: at that point the block-box force will equal zero and they will separate.

 

[TravisEE]

I am trying to solve this problem as well, but I don't understand the block-box contact force that you say acts on the box. Can you further define this force?

 

[John Creighto]

I am trying to solve this problem as well, but I don't understand the block-box contact force that you say acts on the box. Can you further define this force?

If the downward acceleration of the box is greater then g then the block will get thrown from the box.

 

[TravisEE]

The force of reaction between the box and the block would simply be x(t) = d*cos{[k/(M+m)]^(1/2)*t}, and for the block to leave the bottom of the box: d > (m*g)/k. Am I thinking correctly?

 

[John Creighto]

Notice that the acceleration of the box, block unit depends upon their total mass but the acceleration of the block alone depends on the mass of the block alone.

It seems obvious that that should be the case but it would still be interesting to write the problem out as a system of equations. The momentum of the small block would reduce the force which it applies on the larger block during negative acceleration.