# How can you make a solid box jump up with no external forces?

Let's say I have a box lying on a flat surface.

Inside the box, I make a heavy mass hang from a spring attached to the top of the box. If we pull the mass all the way down and let it go, I believe it could have enough momentum to hit the top of the box and drag it up in the air a bit.

I would like to make it more than once, so I'm thinking of putting a motor and worm gear or something, to be able to pull the spring down, lock it, then release, repeat (with a microcontroller)

I'm wondering about 3 things:

1. If we don't pull the spring hard/far enough, when we release it it wouldn't provide much force. But if we stretch it too far, or keep it under tension for too long without firing, it would deform and then not work anymore, am I right ?

2. What would be the formula to calculate the relationship between the movable mass's weight vs the whole box's weight, the spring's pulled length or tension and the net upward force ?

3. Is there a more efficient way than this approach, not requiring firing rockets ? :P

Thanks !

• Same way a human can jump with no external forces while trapped inside said box :-) . (Yes, this is feasible) Sep 29 '21 at 14:50

## Question 1:

If we don't pull the spring hard/far enough, when we release it it wouldn't provide much force. But if we stretch it too far, or keep it under tension for too long without firing, it would deform and then not work anymore, am I right?

You are right. The effect you are describing is called creep. most materials suffer from it. However, in this particular problem, I don't expect it to be relevant, or easy to model.

## Question 2:

There are two ways to make an estimation.

• a rough estimation with energies
• an estimation with differential equations of motion

both of them have problems.

### Q 2.1: Energies

IF $$\Delta s$$ is the extension of the spring, then the energy of the pulled string will be equal to:

$$\frac{1}{2}k \Delta s^2$$

Assuming that at the highest point $$m_1, m_2$$ don't have a velocity (this is not the real case), then

$$(m_1+m_2) \cdot g \cdot h =\frac{1}{2}k \Delta s^2$$

therefore: $$h =\frac{k \Delta s^2}{2 (m_1+m_2) \cdot g }$$

NOTE: this is a rough estimation.

### Q 2.2 Differential equations of motion

Assuming the following coordinate system

where:

• $$F_s$$ is the spring force, and its equal to $$F_s = k(y_2-y_1)$$

the following system of Differential equations can be written (there are a few asterisks -see below):

$$\begin{cases}m_1 \ddot{y}_1 =F_s -m_1\cdot g \\ m_2 \ddot{y}_2 =-F_s - m_2\cdot g +N_2 \end{cases}$$

where $$N_2$$ is the reaction from the ground. This is one of the asterisks, in that the $$N_2$$ and $$y_2$$, both of them need to be positive non zero, however only one of them can be (if $$y_2$$ is greater than 0 then the box is not in contact so there is no reaction from the ground). more precisely:

$$N_2 =\begin{cases}0 & y_2>0 \\m_2\cdot g+ k(y_2-y_1) & y_2=0 \\\end{cases}$$

Analytically, I can't think of a neat way to solve this, however you can solve this numerically.

Additionally for the numerical scheme, to simplify things you can assume that if in a timestep $$i$$, $$y_{2,i}<0$$. then in the next timestep $$y_{2,i+1} = -y_{2,i}$$ (this indicates that there is a bounce).

### Question 3:

There are too many ways, your question there needs to focus.

• I'm not too familiar with the equations, in your 1st formula, k is the Coulomb constant and Δs is a unit of time or distance ? And in your equation about N2, we don't need to include the mass of m1 ? I will study for the rest :) Sep 29 '21 at 11:24
• @MB101874 $\Delta s$ is the initial displacement of the spring. ITs measured in m or mm. Regarding the later no. $N_2$ will only depend on the force of the spring. If the spring is extended (down) then $N_2$ is greater. If the spring is compressed, then $N_2$ is less. Sep 29 '21 at 11:28

Depending on how high you want it to go, you may not even need a spring, just a solenoid with sufficient travel and weight to it.

The force generated will depend on $$F=ma$$ with the $$m$$ of the thing moving and how fast it starts and stops moving, but this is probably difficult to determine. It would be better to run some tests.

• I would like to upvote this too, but I can't :( I like your solution too Sep 29 '21 at 17:05
• Interesting idea, and worth a try (+1), but given that you'd presumably have to shift the mass of some batteries I'm not sure. Another F=ma to consider is the force the solenoid can exert on a plunger of mass m. You could make the battery move with the plunger, but it wouldn't help much if the fixed force reduced the acceleration. Sep 30 '21 at 10:34
• "I would like to upvote this too, but I can't :(", why can't you?
– r13
Sep 30 '21 at 17:08
• @r13 it seems I didn't have enough reputation points to be able to do so :) Oct 1 '21 at 0:10
• @MB101874 You only need a reputation score of 15 to "vote up", 50 to "comment", and 125 tp "down-vote". See the link. engineering.stackexchange.com/tour
– r13
Oct 1 '21 at 1:03

Just to be clear, there are external forces at work here! When you are extending the spring (or whatever), the box doesn't move downwards due to the restoring force of the ground it's sitting on. If you tried to make the box move while it was floating in zero-gee, conservation of momentum would make that impossible.

• you are right ahah Sep 29 '21 at 17:04

1.) "... if we stretch it too far, or keep it under tension for too long without firing, it would deform and then not work anymore, am I right?

In a general sense, your thoughts are fine, but let's be more specific about the different scenarios:

(1a.) - If the spring is stretched beyond its breaking strength, it breaks and would no longer work.

(1b.) - If the spring is stretched beyond its yield strength, it will have a permanent elongation in it that renders it become ineffective/defective.

(1c.) - If the spring is stretched beyond its ultimate tensile strength, it will continue to elongate without additional force until failure.

(1d.) - If the spring is stretched without the load removed for a long period of time, it will suffer the phenomenon called "creep" - an increase in strain without additional stress, and the time dependant strain is not recoverable (the spring will not return to its original length).

2.) "What would be the formula to calculate the relationship between the movable mass's weight vs the whole box's weight, the spring's pulled length or tension and the net upward force?"

I don't quite understand what is "relationship" you are looking for. Maybe the sketch below can offer some insight.

In the sketch above, it is clear that in a place with the gravitational acceleration \$g =0", as shown in the case (2a), the entire system (box + spring) will "float" in space if no external force is applied; if an external force is applied, the system will "drift" freely up or down, depends on the direction of the applied force.

3.) Is there a more efficient way than this approach, not requiring firing rockets?

Again, I don't quite clear on your approach. Do you mean to stretch the spring all the way down then release it? If so, the sketch below indicates it is possible if the rebound force ($$k\Delta s$$) is greater than the total weight of the system. That is $$\sum W < k\Delta s$$, then the system will bounce upward without external force.

• For #3, I wasn't clear enough; I meant not using springs. RC_23 suggested using a solenoid. Do you have other ideas ? :) Sep 30 '21 at 10:24
• If the system weights 10 N and the rebound of the spring 1s 15 N, there is a net upward force of 5 N, the spring is attached to the box, so where the box will move in this case?
– r13
Sep 30 '21 at 11:10
• You can explore the way to make a basic rocket that can shoot up to the sky by combustion and thrust. You need to find the right type and mix-proportion of the fuel then.
– r13
Sep 30 '21 at 17:04
• I think you would be interested in this article, pay attention to the subjects - "Chemistry" & "Physics" of Fireworks". explainthatstuff.com/howfireworkswork.html#physics
– r13
Sep 30 '21 at 18:20
• I would say the whole system would get the upward force of 5N, so reaching 1.27m if it weighted 1kg, if I understand correctly from link Unfortunately it's a closed box with no opening so I can't use thrust :( Oct 1 '21 at 0:22