# Mechanics of a Large Chest in Space

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1. Jun 7, 2013

### plazprestige

At the moment, I am reading the book Relativity, written by Albert Einstein regarding his special and general theory of relativity. In his introduction to general relativity, he poses a thought experiment:

Suppose we have a large chest in space far away from any appreciable mass (stars, planets, etc.). Inside of this chest, we have an man (it's a large chest) standing on the floor of the chest with strings holding him to the chest floor. Furthermore, he is holding a rock in his hand.

Now let's suppose there is a hook on the outside of the lid of the chest that allows some immaterial being to pull the chest "upward" relative to its current plane with a constant force F.

The book describes how if he let go of the rock, it would fall to the floor. Why would it not fall upward? Here is the direct quote from the book:

Can someone please explain this bolded excerpt? Why is this true?

2. Jun 7, 2013

### plazprestige

Ok, so I understand why the rock will approach the floor now, but I do not understand how the person is under similar conditions as if he were on earth.

3. Jun 7, 2013

### WannabeNewton

As soon as the man releases the body, it is no longer being pushed up by the ground of the accelerating chest to travel along with the chest so it will remain in place relative to the outside observer at rest relative to the chest (pretend now that the chest is transparent), who will see the ground of the chest catch up with the body (at which point it will travel along with the chest again). Relative to the man in the chest, who is traveling along with the chest, the body accelerates down the floor of the chest.

Now, for a different take on things, go to the frame of the chest being pulled up. Einstein's equivalence principle says that the man in the chest can make no distinction between accelerating upwards with uniform force of magnitude $F$ and standing in a uniform gravitational field pointing down, with magnitude $F$. This means if he releases the body he held in his hand, he will simply see it fall down in this apparent gravitational field, much like you will see a rock fall to the ground when you release it whilst standing on the surface of the Earth.

4. Jun 7, 2013

### Staff: Mentor

Assume that the chest is being accelerated at a rate of 10 meters per second per second, which is pretty much the acceleration of a free-falling object near the surface of the earth.

Now the man inside the chest will observe two things:
1) When he lets go of the rock, it will accelerate towards the floor as if the chest were at rest on the surface of the earth. In fact, the floor of the chest is accelerating towards the rock, but from inside the chest there's no way of knowing this, nor distinguishing it from the case of gravity pulling the rock towards a stationary floor.

2) If the man stands on a spring scale, it will read the same weight as if the chest were at rest on the surface of the earth. In fact, the accelerating floor of the chest is pushing against the scale which is turn pushing on the man to accelerate him along with the chest, but again there's no way of knowing this from inside the chest, nor distinguishing it from teh case in which gravity is pulling the man down onto the scale.

Google around for "Einstein equivalence principle" and see what you find.

Last edited: Jun 8, 2013
5. Jun 7, 2013

### tiny-tim

hi plazprestige!
can you think of any differences?

(eg, you've decided you accept that letting go of the rock doesn't show any difference!)

6. May 4, 2016

### PT1

I know it has been a long time since the last comment here. I recently read Relativity by Einstein. While reading about the chest I did notice a possible difference that should be noticed. It is the reason I searched out this thread. Could someone clear this up for me? The situation has a "being" applying a force that is accelerating all three, the chest, the person and the rock which is in the hands of the person, so it feels to the person just as though he/she is standing on the earth. As soon as the rock is dropped, that same force would suddenly be acting on a smaller mass (minus the rock's mass) resulting in a sudden burst in acceleration to the chest and person, which might feel to the person like an elevator starting to rise. When the rock meets the floor the person might feel as though the elevator stopped. Where might I have gone astray?

7. May 4, 2016

### SammyS

Staff Emeritus
(Yes, an old thread. For a moment, I thought tiny-tim was back !)

You are correct about the chest & human system having increased acceleration while the rock falls. The magnitude of this increase depends on the relative mass of the rock, compared with the mass of chest and human

If the problem had read that the external being produced a constant acceleration as Nugatory suggested, rather than a constant force, then the passenger would feel much like she would in a stationary elevator on Earth.

Last edited: May 5, 2016
8. May 5, 2016

### PT1

The above quote by plazprestige from Albert Einstein's Relativity continues;
'The observer will further convince himself that the acceleration of the body towards the floor of the chest is always of the same magnitude, whatever kind of body he may happen to use the experiment.'
If the person and the chest experiences a different amount of acceleration depending on, as you said, 'the relative size of the rock' , then the rock would appear to accelerate towards the floor with a rate that would change depending on its mass compared to that of the person plus the chest also, contrary to what Einstein is suggesting. Am I also correct here?

9. May 5, 2016

### SammyS

Staff Emeritus

I will suppose that using a constant force rather than a constant acceleration was an over-sight by Einstein.

10. May 5, 2016

### OldYat47

Dropping the rock has no effect on the rate of acceleration as the thought experiment is proposed. The acceleration remains constant. And anyway, if the rock were relatively light, say 3 kg, the difference in acceleration would be tiny compared to the mass of the man and the huge chest. Certainly undetectable to the man.

11. May 5, 2016

### SammyS

Staff Emeritus
Do you have details supporting your conclusion?

12. May 5, 2016

### OldYat47

On the ability of the "unknown being" to maintain constant acceleration? No, other than familiarity with similar thought experiments.

On the minor effect of dropping the rock? Sure. Let's assume the box is made of 1/2" yellow pine plywood, 4 feet square by seven feet high. The box would have a mass of about 8.2 kg. Let's assume the man's mass 8.2 kg, the same as the weight of the box. So the total is about 16.4 kg. I did make an error in saying the rock was 3 kg. I should have said 3N, 0.3 kg. So the difference in acceleration would be about 0.018 X 10 m/s^2, about 0.18 m/s^2 difference.

13. May 5, 2016

### jbriggs444

Let us review the proposed thought experiment...
That's not constant acceleration. That's constant force. It will result in constant acceleration only so long as the chest and its contents remain rigidly in place relative to one another.

Last edited: May 5, 2016
14. May 5, 2016

### PT1

Thanks for the comments OldYat47. That the force remains constant was prefaced by Einstein in the quote at the top of the page. I was basing my conclusions only on the details given by Einstein. I should have indicated that one should consider a rock of sufficient mass, say maybe 20 to 25% of the total mass so that the same force would then be working on only 70 to 80% of the original load. If a 91.8 kg man in an 8.2 kg box held a 25 kg rock (total 125 kg) then dropped it, the force needed to accelerate 125 kg at 1g would now be pulling on only 100 kg. And the difference in acceleration would be undetectable?

15. May 5, 2016

### OldYat47

Very true, constant force. My mistake. And as the mass of the rock increases relative to the total mass the acceleration would change to a larger degree.

Be careful with your units. A man with a mass of 91.8 kg would weigh 900 N. I don't think that's what you meant.

It can be confusing since kilograms are generally used as referring to weight, not mass. It's sloppy, but that's the way it is. When this gets debated I always suggest a simple test with an electronic bathroom scale that can be set to pounds or "kilograms" Put a 5 gallon bucket on the scale and add water until it weighs 32.2 pounds. That's one slug mass (sometimes used as one pound mass or lbm). If we reset the scale to kilograms what should the readout say? We know that one slug mass is the same as 14.6 kilograms mass. 14.6 kilograms mass on earth weighs about 143 Newtons. But the scale actually reads out 14.6 "kilograms" (actually Newtons) or about 1.5 kg mass. Like I said, sloppy, but that's the way it is.

16. May 5, 2016

### PT1

Thanks. I won't post excuses but will attempt to clean it up in future.
Thanks. Rather than post excuses, I'll try to keep it cleaner in future.

17. May 5, 2016

### Staff: Mentor

An object whose weight here on earth has a mass of one slug, but one pound-mass is nowhere close to a slug (see https://en.wikipedia.org/wiki/Pound_(mass)).
No, not newtons.
No, that's not the way it is.
If you have an object whose mass is one slug, its weight is about 32.2 lb, as you said. If the scale were switched to displaying kg, it would say 14.6 kg.
1.5 kg $\ne$ 14.6kg, under any circumstances. If the scale's readout was in units of newtons, it would say the 143 newtons that you had earlier. (I'm trusting that your arithmetic was accurate.)

Physicists distinguish between units of mass (slug in the English system and kilogram in the SI "metric" system) and units of force (pounds in the English system and newtons in the SI system). People less particular convert from pounds to kilograms and back, and there is usually no problem because under the more-or-less constant gravitational field on earth, there's no harm in doing so.

However, if you were to take a 1 kg bar of steel to, say, the moon, its weight would be considerably less than the 2.2 lb it would weigh here on earth. The mass would be the same, but with a lower gravitational force, it's weight would be much less.

18. May 5, 2016

### OldYat47

http://www.physics.ucla.edu/k-6connection/Mass,w,d.htm

Please read the linked article. A slug (mass) is defined as a mass that weighs 32.2 pounds (force, since force = mass times acceleration) in Earth's gravitational field at the surface of Earth. The conversion from slug (mass) to kilogram (mass) is one slug (mass) equals 14.59 kilograms (mass). So a scale reading of 32.2 pounds means that a mass of one slug is on the scale. The equal number of kilograms (mass) on a scale would read 143.13 kilograms weight (actually Newtons, but that's how scales are commonly labeled, which is not strictly correct), (9.81 m/s^2 X 14.59 kilograms mass). But if you switch from pounds to kilograms on the scale you don't get 143.13 kilograms weight (or Newtons), the readout will say 14.64 kilograms (weight, actually Newtons)

The Wikipedia article is poorly written. The folks in the physics department at UCLA did a much better job.

19. May 5, 2016

### jbriggs444

14.59 kilograms mass on a scale on the earth weighs approximately 14.59 kilograms-force. This is 143.13 Newtons.

20. May 13, 2016

### OldYat47

First, ignore what I posted before. I was decidedly unwell.

Now that I'm thinking clearly again, let me propose this: If we use a scale that reads out in pounds weight and put a 14.59 kilogram mass on it the readout will indicate 32.2 pounds. If we put the same 14.59 kg mass on a scale calibrated in Newtons the readout will indicate 143.13 Newtons. And if we take the same 14.59 kilograms mass and put it on a scale calibrated in kilograms it will read 14.59 kilograms. On a scale calibrated in slugs mass that same 14.59 kg mass will read 1 slug.

What do you think? Are we closer to agreement?