# Dropping a weight onto a spring scale

Say I drop a 5kg weight from a height of 1 meters onto a spring scale like many people have in their bathrooms. On impact the scale will show a higher weight than 5kg.

Question: Which quantities factor into the maximum weight shown on the scale and is there a way to calculate the properties of the spring inside the scale based on this information?

Edit: this is not homework, just something I wondered about when brushing my teeth this morning...

Proposed answer: the stiffness of the spring has influence on the value momentarily displayed; the stiffer, the higher. For unusually soft spring (soft enough that the scale goes down about $5$cm or more when an adult steps on it), the following analysis might allows an estimate of the stiffness of the spring. But with a normal mechanical bathroom scale, the method can't be used; we won't have time to make the reading, it would far off-scale or/and useless, and the experiment is likely to damage the scale unless it, or the mass, is soft or/and elastic. The reading truly depends a lot on the mass of the moving parts of the scale, and what happens to the energy on impact: that could be lost in permanent damage to the scale's surface, or of the dropped mass; or it could be stored as a deformation of the mass or scale's surface, rather than as a deformation of the scale's spring; in which cases the reading is of little use to estimate the stiffness of the spring.

This hopefully improves on an approximation made in the other answer, while using the same hypothesis (doubtful in practice) that

1. the scale has no damping or friction
2. no energy is lost when the mass impacts the scale
3. the scale's moving mechanism has negligible mass compared to the mass dropped

I use the same notation except for the maximum reading of the scale that I rename $R$ (rather than $m$, which is confusing since $m\gg M$ in practice).

• $M=5$kg   mass of the object
• $R$   highest reading displayed by the scale (in kg)
• $x$   corresponding maximum displacement of the scale since contact (in m)
• $h=1$m   dropping height from surface of scale before contact
• $g=9.81$N/kg   gravity of Earth (both local and assumed by scale manufacturer)
• $k$   stiffness of the spring (in N/m)

The scale is such that $$R\;g\;=\;k\;x$$ When the scale displays $R$, all the energy of the mass falling from height $h$ then down by $x$ is stored in the spring, thus $$M\;g\;(h+x)\;=\;{1\over2}\;k\;x^2$$

Eliminating $x$, we get $$2\;M\;(k\;h+R\;g)\;=\;R^2\;g$$

When we plug $h=0$ it follows that $R=2M$ (there is temporary overshoot; that's normal, and in practice the scale will stabilize between its initial reading of $0$ and its maximum reading of $2M$, to the average of that, $M$, as expected).

The stiffness of the spring is thus $$k\;=\;{R\;g\;(R-2\;M)\over2\;M\;h}$$

Because the hypothesis made are so unrealistic, we should take whatever result with a lot of caution, and cross-check; perhaps by reducing $h$, or better by attempting to measure how much the scale goes down under some weight (that would be hard to measure, but most of the error is from that measurement, thus reliably bounded). On top of that, a relative error on $R$ is bound to cause a worse than twice larger relative error on $k$, getting much worse when $R$ is less than a few times $M$.

Solving the equation for $R>0$ we get $$R\;=\;M+\sqrt{M^2+{2\;M\;k\;h\over g}}$$

If we plug $k\;=\;98100$N/m (that is, the scale goes down $1$cm for the weight of a mass of $100$kg) we get a reading of $R=321$kg. My former mechanical bathroom scale did not have that reading (and I'm confident it had a stiffer spring, leading to even larger reading $R$); this confirms that the method can't be used in practice, unless the spring is unusually soft: if we make it $k\;=\;9810$N/m, we get $R=105$kg.

With these later soft spring parameters, the approximation made in the other answer gives $k$ in excess by $+10\%$ (it is hard to tell if that matters compared to other sources of error). Stiffest spring, higher $h$, or lower $M$, make that approximation closer to the theoretical value we obtain in the present answer.

Update: another reason the method can't be used in practice is that, except for unusually soft springs, we won't have enough time to read $R$, since the spring will remain compressed so little time (only a small fraction of the drop time, and that fraction reduces with stiffest springs). Further, the hypothesis made imply that the spring will push the mass up in a rebound throwing it back to height $h$; but in reality the rebound of the mass will be much lower, with a lot of the corresponding energy absorbed by the scale mechanism and surface, and the mass itself, when we have posit there is no such loss.

• @Rations: $R\;g\;=\;k\;x$ is only about the scale; at any moment, the scale's reading $R$ is proportional to the displacement of the scale $x$, and the combination of the scale's spring and mechanism is such that this equation holds, so that the scale's reading (at equilibrium) gives the mass of what's on it. $g$ is the gravity of earth assumed by the scale's manufacturer (or calibration). – fgrieu May 9 '15 at 16:29
• @Rations: When you gently put a mass $M$ just above the scale, and drop it from there, the scale initially oscillates between $0$ and $2M$. It is only after damping (and some loss of energy) that you get a reading of $M$. – fgrieu May 10 '15 at 15:40

If you suppose that the scale works like a spring, which seems reasonable, then during standard use, the displacement $x$ of the scale is proportional to the mass $m$. The equilibrium relation is $$mg=kx,\tag{1}$$ where $k$ is the stiffness of the spring.

Assuming that, when you dropped a mass $M$ from a height $h$, all kinetic energy (which is equal to $Mgh$ because it's been converted from potential energy) is converted into elastic energy, we have the relation $$Mgh=\frac12kx^2.\tag{2}$$ The relation between the measured mass $m$ and the real mass $M$ is therefore $$M=\frac{1}{2}\frac{gm^2}{kh}.$$ Under all these assumptions, we find that the stiffness of the spring is $k=\frac12\frac gh\frac{m^2}{M}$.

• If we make $h=0$, the displayed mass$$m=\sqrt{2\;k\;h\;M\over g}$$ is $m=0$, when it should be $m=M$ (what is displayed when we delicately put the weight on the scale, as we should). Also this solution does not account for the statement's fact that the display $m$ is more than $M$ for all $h>0$. So this can't be exactly right. Perhaps it is a valid approximation in some unstated domain, like $$h\gg{M\;g\over k}$$PS: what a pain that the notation makes $m>M$. – fgrieu Apr 27 '15 at 13:04
• @fgrieu. You are right, I have made two approximations. 1. I have neglected the displacement of the scale in the expression of the total energy: $x\ll h$. 2. I have neglected the damping system in the scale. If I have time, I will add the solution including these two extra parameters. – Tom-Tom Apr 27 '15 at 13:13
• Correction: if we make $h=0$, we should obtain $m=2M$. – fgrieu Apr 27 '15 at 15:42