Weighing Long Things

[Ophiolite]

I know, I just have this mental image of all that mass being pulled downward all along its length, and wondering how all that pull is detected in the relatively small area of the scale.

As I understand it - and I never went past 1st year university physics and that was half a century ago - its at the scale where we get action (the downward acceleration of the timber) initiating a reaction in the form of "resistance" from the scale. Of course, at a microscopic level there are all sorts of actions and reactions going on that we label stress and strain. (All hail the powerful Hooke!)

American physicists have tried to popularize the slug, but we haven't gotten anywhere.

I don't know about that. I've joined more than one American physicist in slugging beers. I was a convert.

 

[Michael]

The longest guitar I am aware of is a little under 5 feet. If you're worried about something 20 feet long, I want to see a picture of your guitar!

Grand Ole Opry House photos

 

[JackRT]

American physicists have tried to popularize the slug, but we haven't gotten anywhere.

Did you forget about the "poundal"?

 

[essentialsaltes]

Did you forget about the "poundal"?

I would certainly prefer to do so.

 

[JackRT]

I would certainly prefer to do so.

I prefer the Systeme Internationale --- EVERY --- SINGLE --- TIME.

 

[Halbhh]

If you lay a 2x4 of wood on something like a normal bathroom scale, and the wood is say, 20 feet long, will you get an accurate measure of its weight? I'm imagining so much of the mass will be being pulled toward the center of the Earth away from the scale, that some of it doesn't get measured? Or no?

FYI, I'm wanting to weigh my guitars because I want to buy a lighter one. I realize I can weigh myself with and without the guitar and subtract the difference, but I think it might be slightly more accurate if I can just lay it on a scale. Anyway I'm still interested in the physics question though.

Yes, if you balance the boards, you will get a reasonably accurate measure, no matter that they are very long. The radius of the earth is so much vastly larger than 20 feet. (imagine one board end was 1 inch further from the center of the Earth than the average of the board, then the ratio of decreased gravity on that end vs the average on the whole board would be roughly 1/(12*5280*4000miles) ~4*10^-9 or near to the order of 1 part in a hundred million.) \

So, board length is unimportant, but getting the board weight well centered on the scale is usually important: it matters for most scales that you are well centered on the scale, instead of having significantly more weight to one side of the scale platform, as you can verify by trying that with your feet on a bathroom scale.

 

[Dave-W]

So if it extended outside Earth's atmosphere, only the section of the wood subject to Earth's gravity would be weighed accurately?

Actually, the farther away it got from the center of the earth, the lighter each section would be.

 

[Chesterton]

The atmosphere is not some magic barrier to gravity.

Anyway, because technically correct is the best kind of correct, assuming the scale is accurate, then it would always measure the weight accurately. The weight is defined as the force of gravity on an object, which will vary depending on the orientation and elevation and location of the object. The mass of the object, which is an intrinsic property that doesn't change with location, is a different thing entirely. Though in many circumstances, we commonly treat them equivalently, because under most circumstances the difference between mass and weight is a proportional factor that doesn't vary much, the distinction is an important one. American physicists have tried to popularize the slug, but we haven't gotten anywhere.

Actually, the farther away it got from the center of the earth, the lighter each section would be.

The atmosphere is not a magic barrier, but distance from a large mass like the Earth is. I wonder how a scale would measure a hypothetical beam which extended outside the atmosphere. Some of it would be normal, some of it lighter, and at the ends it would be weightless.

 

[Chesterton]

Yes, if you balance the boards, you will get a reasonably accurate measure, no matter that they are very long. The radius of the earth is so much vastly larger than 20 feet. (imagine one board end was 1 inch further from the center of the Earth than the average of the board, then the ratio of decreased gravity on that end vs the average on the whole board would be roughly 1/(12*5280*4000miles) ~4*10^-9 or near to the order of 1 part in a hundred million.)

Are you saying there would be some negligible decrease in weight even on just a 20 foot board?

 

[essentialsaltes]

The atmosphere is not a magic barrier, but distance from a large mass like the Earth is. I wonder how a scale would measure a hypothetical beam which extended outside the atmosphere. Some of it would be normal, some of it lighter, and at the ends it would be weightless.

If you assumed a uniform density, it wouldn't be that difficult a calculation, but would require calculus. However, there would also be effects from buoyancy. Just as you are 'lighter' in the water, things are lighter from displacing air, but if your beam stuck out of the atmosphere, some of it wouldn't be 'buoyed' up.

Are you saying there would be some negligible decrease in weight even on just a 20 foot board?

Yes.

 

[Sophrosyne]

One way to weigh really long items is to use 2 scales, one on each end and add the 2 results together. If you don't have a long item balanced it could throw some scales off as some aren't designed to weigh things properly when the force (weight) is applied at a large enough angle this is why putting the balance in the middle is important. Scales that weigh things suspended above ground probably are less affected than those who sit on the ground (like bathroom scales.
One other things I've found is some bathroom scales if you weigh yourself and pick up an object weighing a half pound it will give the same result. You literally have to weigh something about 3-5 pounds more or less than what is showing to get out of the "range" the scale doesn't give the same result. I often grab the door frame and lift up and let off when I get about 10lbs less the scale will show that amount even when you let go and then step off the scale and you will get a new result sometimes a lb or two different.

 

[Chesterton]

If you assumed a uniform density, it wouldn't be that difficult a calculation, but would require calculus. However, there would also be effects from buoyancy. Just as you are 'lighter' in the water, things are lighter from displacing air, but if your beam stuck out of the atmosphere, some of it wouldn't be 'buoyed' up.

Hmm, I'm not sure I can agree that one could calculate the weight of a partially weightless object.

Yes.

So my intuition wasn't that crazy after all I guess.

 

[Halbhh]

Are you saying there would be some negligible decrease in weight even on just a 20 foot board?

Actually that 1" order of magnitude illustrative difference in board end distance to Earth's center is itself off by orders of magnitude, since the effective curvature of the earth in just 10 feet is much tinier than an inch. (it's like 8" in a mile) But, quickly, even if it was for example as large as 1 part in 10^8 or 10^9 gravity (it's less!), then even that would be a lot less of a force than just a random subtle puff of air hitting the board, like turbulence from an air vent or an open window. So, it'd require great sensitivity and isolation from moving air to even try to measure. Can any kind of scale that could be purchased, even a scientific instrument type available for sale, be as sensitive as 1 part in 10^9? I don't know, but it would surprise me if one can do that.

 

[essentialsaltes]

Hmm, I'm not sure I can agree that one could calculate the weight of a partially weightless object.

I don't know what a 'partially weightless' object is, but (ignoring buoyancy and other effects) I think the answer is

weight = GmM/[R(R+L)]

where
G = Newton constant
m = mass of board
M = mass of earth
R = radius of earth
L = length of board

When L is negligible compared to R, this reduces to GmM/R^2, which is correct.

 

[Chesterton]

I don't know what a 'partially weightless' object is, but (ignoring buoyancy and other effects) I think the answer is

weight = GmM/[R(R+L)]

where
G = Newton constant
m = mass of board
M = mass of earth
R = radius of earth
L = length of board

When L is negligible compared to R, this reduces to GmM/R^2, which is correct.

But at the ends of the board, out in space, M is not a factor.

 

[Halbhh]

But at the ends of the board, out in space, M is not a factor.

dude you got a really long board in mind then, a lot more than 20 feet? even Jupiter shares a gravity tug with Earth.

let's imagine we have a board a lot longer than the earth is wide

let's imagine a board so long that roughly the far end of the board distance from the center of the earth we could just approximate to be 1/2 the board length.

So, how long a board until the earth's gravity on the far end would be say just 1/100th as much as here on the surface of earth at the center of the board?

This is an easy calculation, since gravity falls off as the square of distance, and the earth has a radius of 3,960 miles. You just need then a distance of 10 times (square root of 100) that radius to get to 1/100th the gravitational force, or simply for the far end of the board to be 39,600 miles from the center of the earth, and so the board being twice that length would have a length of about 80,000 miles. Of course it would break already, lol, but let's pretend it has an unbelievable strength. Dude the board is a hazard to satellites!

Imagine the torque on the board clamps though from the lunar tidal forces. oh my

 

[Chesterton]

Of course it would break already, lol, but let's pretend it has an unbelievable strength. Dude the board is a hazard to satellites!

Imagine the torque on the board clamps though from the lunar tidal forces during the lunar month. oh my

Well hey, people are talking about building those space elevators using cables which may be around 25,000 miles tall/long.

 

[sesquiterpene]

So, it'd require great sensitivity and isolation from moving air to even try to measure. Can any kind of scale that could be purchased, even a scientific instrument type available for sale, be as sensitive as 1 part in 10^9? I don't know, but it would surprise me if one can do that.

The range for the mocrobalances I've used is 52g readable to 1 microgram, so it's approaching 1 part in 10^8 for commercially available scales.
edit:they all have enclosures to protect from air currents, and might take 15 seconds to stabilize once you close them

 

[essentialsaltes]

But at the ends of the board, out in space, M is not a factor.

Like the energizer bunny, Gravity keeps going and going.

 

[Francis Drake]

Stand the 4x2 on end and let go for a moment as you take the reading. It will take a few seconds before it starts to tip.

I have also weighed long objects by weighing each end.
If you put one end of the wood on a brick, and the other end on the scales, each end will bear 50% of the load, so just double it. You can check by swapping the scaled from one end to the other.

Its the same as using 2 or even 3 sets of scales on a large object. Add the scale weights together to get the total.