(This is curiosity, not a practical question. It was inspired by standing still for a digital scale, and considering whether it would be possible to make a scale that could handle being jumped on, juggled on, etc.)

Setup: An impermeable rigid box with arbitrary contents on top of (fixed to) an ideal scale, in a known uniform gravitational field and no electromagnetic field. Assume classical mechanics (i.e. no relativistic effects or uncertainty).

Question: How long must data (i.e. the force on the scale) be collected in order to know the mass of the box and contents, given that the contents of the box are doing their best to make it difficult (which I think can only consist of accelerating masses vertically)?

I think that it is not possible to get an exact answer after a finite time; for example, consider if the box contains a mass being accelerated downward arbitrarily slowly — thus reducing the total weight by a small fraction, until it touches bottom. Is there an upper bound on the error, given a time interval and some property of the collected data (e.g. the maximum or mean force)?


The box is subjected to two forces, gravity and the normal force from the scale, so $F_N(t) - mg = ma(t)$. Presumably you can't directly measure the instantaneous acceleration of the box's center of mass, so you can't just plug in $a(t)$ and $F_N(t)$ at a given time and calculate $m$ that way. What you can do is take the time average of the force,

$$\langle F_N(t)\rangle_T = \frac{1}{T}\int_0^T F_N(t)\mathrm{d}t = \frac{m}{T}\int_0^T [a(t) + g]\mathrm{d}t = m\biggl(\frac{v(T) - v(0)}{T} + g\biggr)$$

As your measurement time $T$ increases, the contribution of the fractional term becomes less and less. So as long as you can establish some upper bound $\bar a_\text{max} < g$ such that $\bigl|\frac{v(T) - v(0)}{T}\bigr| \le \bar a_\text{max}$, you can constrain the mass to lie within

$$\frac{\langle F_N(t)\rangle_T}{g + \bar{a}_\text{max}} \le m \le \frac{\langle F_N(t)\rangle_T}{g - \bar{a}_\text{max}}$$

If nothing else, you can get your upper bound from relativity, $\bar{a}_\text{max} = \frac{2c}{T}$, although that's pretty impractical because you'd have to measure for two years just to get $\bar a_\text{max} < g$.

More realistically, the object is probably constrained to some limited distance $d$ because it's attached to the scale. That means it can't spend longer than $\frac{d}{v}$ at any given velocity $v$. Assuming your initial and final data points are not correlated to the object's motion, you're more likely to choose times when it is moving slowly than when it's moving quickly, although I'm not sure how best to quantify that.

  • 1
    $\begingroup$ This is definitely in the right direction; the need to choose $\bar{a}_{max}$ is the hole. There is a $d$: the height of the box — that's why the box exists. $\endgroup$ – Kevin Reid Jan 15 '12 at 3:06

It all comes down to where the centre of mass of the objects inside the box are. Most scales only measure accurately if the centre of mass is right over the sensors. But if the motion is random, you can probably build a statistical distribution for the centre of mass and calculate an approximate value. Practically, you can never get an exact value anyway so this isn't that big a deal if you wait a while. Obviously, the longer you wait, the more accurate your mass is as you reduce the error in the distribution.

  • $\begingroup$ I'm assuming an ideal scale, so the center of mass isn't a problem, and I want an upper bound on the error more specific than ‘it will improve over time’. $\endgroup$ – Kevin Reid Jan 15 '12 at 1:53
  • $\begingroup$ If the center of mass isn't a problem then the measurement doesn't change in time. Unless your masses are jumping up and down? In which case this is more of a statistics problem than a physics one. You can never get an exact answer from a measurement but what you normally do is choose how much percentage error you are ok with and calculate the time required to get to that by assuming a particular statistical distribution, like say, a normal distribution. Sorry I couldn't be more help. $\endgroup$ – Durand Jan 15 '12 at 2:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.