# How does a box have a net force of zero when it is on the ground and has an oscillating spring in it?

I was working through David Morin's workbook and I came across this question. Imagine that we have a box with a massless, oscillating spring attached to the ceiling of the box. There is also a ball at the other end of the vertical spring. The box is on (but not attached to) a scale on the ground. The question is to figure out where the ball is when the scale reads the highest (equivalently meaning where is the ball when the box exerts the largest force on the scale). I couldn't figure it out, so I read David Morin's solution.

At the bottom of the motion, the upward force from the spring on the mass is maximum (because the spring is stretched maximally there), which means that the downward force from the spring on the box is maximum (because the spring exerts equal and opposite forces at its ends). This in turn means that the upward force from the scale on the box is maximum (because the net force on the box is always zero, because it isn’t accelerating). And this force is the reading on the scale.

This is where I get confused. I know that if the net force on the box is zero, I can follow Morin's reasoning and therefore how the box will read the highest when the ball is at the bottom of its motion. But, how does one know the net force on the box is zero? For example, how do you know it won't "jump up" during its motion (I haven't tested this, but intuition tells me that if you have a light box, the box will "jump" when the spring is at the top of its motion, meaning the net force on the box can't be zero).

I think it has to do with the assumption that the box is heavy and strong, but I can't understand how this implies that the box must stay in place. Furthermore, even if the box is heavy and strong, I still don't see how the box staying in place is the only possible motion. It seems to me as if it is possible to conserve momentum and energy in the Earth-box-spring system if the box "jumps up" as long as the oscillating ball moves in a specific way (however, I also suspect that this is the reasoning which causing me to be confused).

To restate the question, why does the net force on the box have to be zero and why is it the only possible motion which conserves energy and momentum of the spring, box, and ball system?

• Even if it does jump up, that will happen when the spring is exerting its minimum downward force on the box, not its maximum. Commented Apr 25, 2020 at 19:53
• @ThePhoton I simply said that because it provided a (perhaps false) counterexample to the fact that the net force on the box is always zero. If it jumps up, obviously there must be an net acceleration and hence a net force on the box. Commented Apr 25, 2020 at 19:59

As long as the box rests on the ground (or the scale) you know the net force is zero because the acceleration is zero.

If the box jumps, then the net force is obviously not zero. But this would happen when the downward force on the box is minimum (it would have to actually be negative), so it doesn't affect the answer to the question you were asked about the maximum downward force.

If the scale were resting, say, on a table and the table collapsed, again that would indicate the net force not being zero. This could affect the answer to your question in the real world, but for classroom exercises (outside of civil engineering courses) we're usually allowed to assume our objects rest on an unyielding solid surface.

In any case, the first step to predicting whether the table might collapse would be to calculate the maximum downward force assuming the table doesn't collapse, and then compare that to the strength of the table. If the force exceeds the yield strength of the table, then you'd have a much more complicated problem to predict how the whole system evolves.

• But how do you know for sure the box rests on the ground? How do you know the acceleration must be zero? Commented Apr 26, 2020 at 2:53
• Put a box on the ground. Does it fall into the ground, or does it rest on the ground? By observation we find that resting on the ground is a more common occurrence than falling into the ground. Commented Apr 26, 2020 at 2:56
• If you want to make the problem more difficult, imagine the box is floating on water. Now it will certainly bob up and down as the mass oscillates on the spring. You could certainly explore how to model this system. But it won't be something you can do with a few simple equations. So, like I said in my answer, usually for classroom exercises we're allowed to assume the ground is an unyielding solid surface. Commented Apr 26, 2020 at 2:58
• Does this imply that no matter what the internal workings of the box are, the box must always have a net acceleration of zero? Commented Apr 26, 2020 at 3:11
• If the ground it's resting on doesn't fail in some way, then we know it can't have a positive downward acceleration. As you pointed out, it might jump upward (negative downward acceleration). But again, you were asked about the situation of maximum downward force from the spring, so the jumping upward scenario, while possibly interesting in another context, isn't relevant to the problem you were trying to solve. Commented Apr 26, 2020 at 3:14

It is specified that the box is at rest on the balance!

...The box is on (but not attached to) a scale on the ground...

This puts the box on a scale which has previously been at rest on the ground. Now there are changes in Kinetic Energy during the process of putting the box on the ground, after which it rests.

...(because the net force on the box is always zero, because it isn’t accelerating)...

Now while pulling the spring downwards by some unspecified force and keeping it held there, it would result in a net force on the box downwards which is accounted for by the scale under it. Just like how putting weights over something on the ground increases its weight, but is counteracted by the normal force. Thus, for as long as you're holding down the spring with some force, the body is at rest.

The explanation by David Morin doesn't account for the change when we release it, because it obviously is accelerating and so isn't at rest, BUT when it's on the ground and the spring is held in a position, it is AT REST.

• I'm not too concerned about what is happening before the oscillation. I just want to know why specifically the normal force is zero when you let go of the spring. Commented Apr 26, 2020 at 3:48
• It isn't. Normally, when you hold it, you are obviously adding on extra force, and when you release it, that force needs to manifest in some other form, and so it is manifested as the force that causes a slight weightlessness and an alternate additional weight downward, such that the scale gets back to the value of the weight of the box eventually. It is not at rest when you let go of the spring, it faces a slight oscillation due to the constant shift between upward and downward forces due to the oscillation of the spring.
– stwq
Commented Apr 26, 2020 at 4:20
• It is only at rest for every position of the spring
– stwq
Commented Apr 26, 2020 at 4:23