How can I weigh part of an object?

Question

I have these little pieces of steel:

I can weigh the whole thing on a scale (the one with lines on it is 14.07g), but I just want to weigh part of it (e.g. the green line to the end i.e. excluding the red X).

The weight is not uniformly distributed along the length, so while I could calculate an estimate based on the position of the green line, it wouldn't be sufficiently accurate.

Is there some nondestructive way to weigh just that specific portion of it?

I am mostly blind and can't read the photo. Is the material from which the object is made of uniform density? If so, you might be able to effectively weigh parts of it by first determining the density (weigh it with a scale, and determine volume by immersion). Then you can submerge only a part, determine its volume, and compute its weight. Not sure there's a good way if the density is not uniform. — Rick Goldstein, Nov 15 '21 at 21:38
How well do you need to measure it? If fairly uniform in thickness, the weight of some part is related to the total weight and relative area. — Jon Custer, Nov 15 '21 at 21:59
@Rick "The weight is not uniformly distributed along the length, so while I could calculate an estimate based on the position of the green line, it wouldn't be sufficiently accurate." In fact, it's not uniformly distributed along any other dimension or over any section. — Jason C, Nov 15 '21 at 22:48
I see the only way possible to do this would be cut on the green line and weigh it... — Jon Custer, Nov 15 '21 at 22:53
Possibly ask this question at the Engineering SE site for better answers on applications. — Steeven, Nov 15 '21 at 22:57
Well... I know that if I can balance it I can tell where a line would be where each side is half the total. Can I somehow do like... a binary search style thing until I get to my desired line? — Jason C, Nov 15 '21 at 22:59
@JasonC You misunderstood what I was asking. I wasn't asking about uniform distribution of weight over any particular dimension. I was talking about density of the material (mass per unit volume). The uniform density assumption is what David White used in the answer below. — Rick Goldstein, Nov 16 '21 at 04:12
You might be interested: https://physics.stackexchange.com/q/70839/ — Allure, Nov 16 '21 at 07:49
@Rick Not perfectly uniform density; these particular parts have significant corrosion. It's close though. — Jason C, Nov 16 '21 at 12:37
@JasonC Every balancing measurement (balancing the object on an 'edge') gives you a plane in which the center of mass is contained. This plane has some thickness depending on the accuracy of your balancing measurement. Three planes are sufficient to uniquely determine a point (or tiny region of space, depending on your accuracy). — , Nov 16 '21 at 12:38
@JasonC You are right, I deleted my faulty idea. Of course you control the position of the added mass, but not the distribution of the mass of the object on either side of the line. — , Nov 16 '21 at 12:47
If you really mean "it's not uniformly distributed along any other dimension or over any section", then you've posed an impossible problem. Why? Because, without cutting into the object, all the information you can detect about the mass distribution is fully described by the center of mass and the inertia tensor (the latter being a 3x3 symmetric matrix). So if you give me an object where the answer is X, I can produce another identical-seeming object, by redistributing the density in a way that doesn't alter the center-of-mass and inertia tensor, for which the answer is not X. — Don Hatch, Nov 16 '21 at 19:29

score 20 · Answer 1 · answered Nov 16 '21 at 00:01

20

Use a balance to get the mass of the whole steel object. Then, fill a 100 ml graduate cylinder exactly to the 50 ml mark. Place the steel in the graduate cylinder, obtain the new reading, and subtract 50 ml from that reading to obtain the volume of the steel part. From that information, calculate the density of the steel as $\rho = \frac{m}{V}$.

Next, tie a string around the right end of the steel part. Make sure that the graduate cylinder again has exactly 50 ml of water in it. Submerge the steel part up to the green line, obtain the new graduate cylinder reading, and calculate the volume of the steel part that was submerged. Using that volume and the previously calculated density, calculate the mass of the submerged portion.

answered Nov 16 '21 at 00:01

David White

12,158

1

This sounds like the closest I can get, good idea. Now, these pieces do have a decent layer of rust on them. Do you think the density difference of rust, and its slight porosity, will be negligible? Looking at the sides, most of it only looks to be a few microns thick but there are one or two spots that look to be a millimeter or two. Pic forthcoming. – Jason C Nov 16 '21 at 01:03
https://i.imgur.com/JVE3Rlh.png – Jason C Nov 16 '21 at 01:17
3

@JasonC, this method will calculate an average density, which will give you the weight of the partially submerged volume, assuming that the partial volume has the same amount of average imperfections as the whole. I don't want my question to sound argumentative, but do you want the weight of the partial object with all of its imperfections, or do you want the weight of the partial volume if it didn't have any imperfections on it? – David White Nov 16 '21 at 03:04
12

This is what I was suggesting in my original comment to the OP. For this to work, the density of the material must be assumed to be uniform to within the accuracy required. – Rick Goldstein Nov 16 '21 at 04:06
1

@David With all its imperfections. That is, if I were to cut the object at the line. – Jason C Nov 16 '21 at 12:51
My biggest concern (possibly unfounded) with this method is accurately reading the volume in the graduated cylinder. I'll do some math to figure out volume tolerance after I get some coffee. PS ”cut the object at the line" = "cut the object on the same plane as the surface of the water when partially submerged", to be clear. – Jason C Nov 16 '21 at 13:33
@ Jason C, so do you know the answer and was the last comment it? The comments and question seemed to give the impression that the density wasn't constant but now it seems as though you did mean constant density, even though the mass isn't uniformly distributed... – John Hunter Nov 16 '21 at 20:47

score 11 · Answer 2 · edited Nov 16 '21 at 14:03

11

Here is a method that you could try.

For an object that can be modelled as two pieces of mass $m_1$ and $m_2$, with a COM of each in an unknown position due to a possible variable density, in principle the values could be found like this.

The blue numbers and $F_1$ are known, (numbers made up as an example), but the red numbers are unknown. Below $g$ and $\cos\theta$ terms are omitted as they cancel.

The object is hung from thin cotton or tape at P and the other end rested at a slight angle $\theta$ on the green scale, producing a reading $F_1$

Doing moments (sum of torques = 0) around P, gets our first equation

$$m_1(3+r_1)+m_2(3-r_2)=15.5 \times 10 \tag1$$

swapping the object around and hanging from the other end would give

$$m_1(7-r_1)+m_2(7+r_2)=195 \tag2$$

where the scale $F_2$ would now read 19.5

It's also known that $$m_1+m_2 = 35\tag3$$

However there are 4 red unknowns, another equation is needed.

This time balance the object on a knife edge at $Q$. It would be hard to get an exact balance, so hang a small mass (of value $5$ in this example, not shown) at P, using tape to ensure it's right at the end. The reading on the scale now, by coincidence, is also $F_3 = 5$,

Moments around $Q$ gives a fourth equation

$$5\times 7 + 5 \times 3 + m_2r_2 = m_1r_1\tag4$$

In principle these equations can be solved to recover the unknown red values.

edited Nov 16 '21 at 14:03

Jason C

1,004

answered Nov 16 '21 at 07:42

John Hunter

13,700

1

At a glance, this looks promising. Will digest then try it after I get some coffee. – Jason C Nov 16 '21 at 13:10
@ Jason C Yes, it looked promising, but before you spend too much time trying it, there may be a problem. It needs checking whether all four equations are independent, they may not be, so best to wait a while to see if something can be done to improve it. – John Hunter Nov 16 '21 at 13:22
@Jason C Thanks for the edit, at the moment all that comes to mind is either assume the red part has uniform density hence $r_2$ could be found, but the other part can still have variable density, or try something to do with moment of inertia, spinning it, to get one more independent equation, but that's probably difficult in practice. – John Hunter Nov 16 '21 at 14:11
1

The equations will not be independent, you can only do moments once. I used to worry when doing 'solve by moments' problems which point to take as the axis. Eventually through repeating a lot of problems, I started to believe what I'd been told, that it didn't matter. The only difference choice of axis can make is whether some forces produce zero torque (by acting through the axis), simplifying the working. – Neil_UK Nov 16 '21 at 16:21
(I didn't forget; I just have to wait until after work today. Really appreciate your answer! I'll also add my brain to the independent equation mission. This comment will self destruct this evening.) – Jason C Nov 16 '21 at 16:33
wouldn't that be just three unknowns since we can easily get the total mass? (so e.g. $m_2 = m_{tot} - m_1$) – ilkkachu Nov 16 '21 at 17:32
@ Neil_UK moments twice still seems to give independent results – John Hunter Nov 16 '21 at 18:42
@ Jason C Why do you want to know the mass of part of the object? If it's not a silly question?! – John Hunter Nov 16 '21 at 18:42
1

This won't work. For a rigid body, no matter how many measurements you make, all you can obtain is a center of mass and the inertia moments. This will be useless in telling you the exact mass below a given line. At best, you could get bounds on the mass on either side. – Jeffrey Nov 16 '21 at 20:33
@Jeffrey It seems as though the density may have been constant after all, see other comments - and that makes the problem a lot easier – John Hunter Nov 16 '21 at 20:49

How can I weigh part of an object?

2 Answers2