energy-conservationhomework-and-exercisesnewtonian-mechanicsspring
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…
Analyzing the acceleration of the center of mass of the system might be the easiest way to go since we could avoid worrying about internal interactions.
Let's use Newton's second law: $\sum F=N-Mg=Ma_\text{cm}$, where $M$ is the total mass of the hourglass enclosure and sand, $N$ is what you read on the scale (normal force), and $a_\text{cm}$ is the center of mass acceleration. I have written the forces such that upward is positive
The center of mass of the enclosure+sand moves downward during process, but what matters is the acceleration. If the acceleration is upward, $N>Mg$. If it is downward, $N<Mg$. Zero acceleration means $N=Mg$. Thus, if we figure out the direction of the acceleration, we know how the scale reading compares to the gravitational force $Mg$.
The sand that is still in the top and already in the bottom, as well as the enclosure, undergoes no acceleration. Thus, the direction of $a_\text{cm}$ is the same as the direction of $a_\text{falling sand}$ . Let's just focus on a bit of sand as it begins to fall (initial) and then comes to rest at the bottom (final). $v_\text{i, falling}=v_\text{f, falling}=0$, so $a_\text{avg, falling}=0$. Thus, the (average) acceleration of the entire system is zero. The scale reads the weight of the system.
The paragraph above assumed the steady state condition that the OP sought. During this process, the center of mass apparently moves downward at constant velocity. But during the initial "flip" of the hour glass, as well as the final bit where the last grains are dropping, the acceleration must be non-zero to "start" and "stop" this center of mass movement.
If the two devices are sharing the load, they will each show approximately half the weight. But if you hang one below the other, then each feels the full load. Think about it this way: the one nearest the suitcase "feels" the suitcase. It doesn't care whether you are holding it with your hand, attached to a spring, or whether it is dangling from the side of Mount Rushmore.
The top spring is measuring the weight of "the thing below it" - that is the suitcase plus the other spring gage. It doesn't care whether the spring gage is in the suitcase, or attached to the top.
If you were right, be nice about it to your girlfriend... I don't want to cause trouble.
PS - I think your confusion comes about from the difference between force and displacement. When you have two springs in series, then indeed you need to pull twice as far in order to get the same force registered. That is because the spring constant of the combination is half as big. But each device, since it registers half of the total displacement, registers the same force as before.
Best Answer
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 $k=\frac12\frac gh\frac{m^2}{M}$.