[Phil]

Imagine a postage scale - like this: Scale. It is sitting on a hard level surface.

Say the scale itself weighs 8 ounces (but its capacity is 20 lbs). If I set a second scale in the basket of the 1st scale, the reading of the lower scale would be 8 ounces.

If I place a 3rd scale on top of the 2nd, presumably the 1st scale measures 16 oz and the 2nd scale measures 8 oz, however...

- Picture a stack of many of scales - say, 20. It seems at some point (maybe beginning with the 3rd scale) that some of the force that is exerted downward is going to be absorbed into the 'spring' that is created by the vertical stack of scales. While the bottom scale should stay accurate and measure 10 lbs, the intermediate scales may not be accurate as to the true weight of the scales above it. For instance it wouldn't surprise me at all if the 19th scale does not measure 8 oz.

Well?

[RMB1]

> Well?
Well, I think you are wrong.

Imagine 17 scales in a box (of negligible mass), being weighed on another scale.

The bottom scale measures the weight of the 17 scales.

Now add another scale to the box, the bottom scale now measures the weight of the 18 scales, the last scale has contribute its full "weight" to the reading of the scale.

Now if we rearrange the top 18 scales so they are each weighing each other in a pile (with the first one sitting on the bottom scale) this does not change their mass, so it does not change the weight shown on the bottom scale.

In other words, the interal forces in the collection of the top 18 scales (e.g. tension in the balance springs) does not affect the external force (weight) required to support the 18 scales as a totality.

[Phil]

RMB1 said:

Well, I think you are wrong.

Imagine 17 scales in a box (of negligible mass), being weighed on another scale.

The bottom scale measures the weight of the 17 scales.

Now add another scale to the box, the bottom scale now measures the weight of the 18 scales, the last scale has contribute its full "weight" to the reading of the scale.

I don't disagree.

Quote

Now if we rearrange the top 18 scales so they are each weighing each other in a pile (with the first one sitting on the bottom scale) this does not change their mass, so it does not change the weight shown on the bottom scale.

I don't disagree either.

Read it again. I am more interested in the readings of the intermediate scales, not the bottom one.

[helene_t]

There is no such thing as absorption of force (in this context at least). The intermediate scales will simply show the combined weight of all the scales that are on top of them.

[RMB1]

Phil, on Aug 4 2010, 04:31 PM, said:

For instance it wouldn't surprise me at all if the 19th scale does not measure 8 oz.

Sorry.

I understood "measure" as "weigh" (as shown on the scales beneath).
You mean "measure" as "indicate" (for the weight of the those above).

[hrothgar]

Phil, on Aug 4 2010, 06:31 PM, said:

Imagine a postage scale - like this: Scale. It is sitting on a hard level surface.

Say the scale itself weighs 8 ounces (but its capacity is 20 lbs). If I set a second scale in the basket of the 1st scale, the reading of the lower scale would be 8 ounces.

If I place a 3rd scale on top of the 2nd, presumably the 1st scale measures 16 oz and the 2nd scale measures 8 oz, however...

- Picture a stack of many of scales - say, 20. It seems at some point (maybe beginning with the 3rd scale) that some of the force that is exerted downward is going to be absorbed into the 'spring' that is created by the vertical stack of scales. While the bottom scale should stay accurate and measure 10 lbs, the intermediate scales may not be accurate as to the true weight of the scales above it. For instance it wouldn't surprise me at all if the 19th scale does not measure 8 oz.

Well?

Hi Phil

The following example might make things easier

Rather than trying to conceptualize a stack of scales, instead compare the following

Case 1: A mechanical scale sitting on a solid, flat surface
Case 2: An identical scale sitting on a cushion

As I understand matters, these two scales will often differ in their estimates of weight

From the sounds of things, you are suggesting that the stack of scales is equivalent to a cushion...

<<Just checked with a friend of mine who stated that the cushion doesn't matter... As soon as the system converges to a steady state, the two scales will show the same weight regardless of whether they are sitting on an elastic surface or an inelastic surface>>

[shyams]

Phil, on Aug 4 2010, 04:31 PM, said:

For instance it wouldn't surprise me at all if the 19th scale does not measure 8 oz.

Though this sounds intuitively right, I think it is due to spring movements.

If we tried this in real life, some of the springs would be in a state of motion (vibration). When springs vibrate, a part of the stack is accelerating / decelerating. This adds to or counteracts the earth's gravitational force.

In fact, when such a stack has springs that are vibrating outwards (i.e. the stack is expanding), the topmost scale would also register a weight reading (despite it being empty)

If we achieved a stable system where none of the springs vibrate, then the laws of physics would require that the every one of the scales would show the correct calculated weights

[gwnn]

Agree with helene_t and shyams.

I don't claim to be a Physics expert despite my coming Master's thesis in Physics, especially not in Newtonian mechanics, but there really is nothing else to all this, unless you want to include air pressure or Coriolis forces, but you really shouldn't, unless your scales have significant dimensions in comparison to the height of the atmosphere.

[mikeh]

I used to think it was turtles....turtles all the way down. Now I have to allow for the possibility that it is spring-operated scales all the way down....complicated, it appears, by the theory of relativity.

[gwnn]

hrothgar, on Aug 4 2010, 04:19 PM, said:

<<Just checked with a friend of mine who stated that the cushion doesn't matter... As soon as the system converges to a steady state, the two scales will show the same weight regardless of whether they are sitting on an elastic surface or an inelastic surface>>

Agree with your friend.

[barmar]

hrothgar, on Aug 4 2010, 12:19 PM, said:

<<Just checked with a friend of mine who stated that the cushion doesn't matter... As soon as the system converges to a steady state, the two scales will show the same weight regardless of whether they are sitting on an elastic surface or an inelastic surface>>

So all the people who think it makes a difference whether they use their bathroom scale on a hard tile surface or a plush carpet are wrong.