Since it is increasingly difficult to get a stimulating response going on this site, I thought I might offer a simple high school physics question I have been pondering for a while. Although it is totally irrelevant to politics, economics, history or religion, I would like some nerdy thoughts on this.
If I were to obtain four identical sets of bathroom scales and pile them on top of each other with the bottom one on a hard floor, and then I stand on the top one, what would the scales read?
Each scale operates on Hooke's Law with the reading being directly proportional to the amount of compression of a spring. My initial reaction was that the scales would operate independently once the system stabilised. If so, they would all read my weight plus the weight of any scales above them, but there is another school of thought:
Since the springs in each scale are in series with the others then each would compress by only one quarter of the distance that it would if used on its own. So the reading would be one quarter of my weight plus a proportional correction for the weight of any scales above them.
Add a Comment (25)
Email This
Message Author
Statistics
My solution
I've got to go with the second school of thought.
If all four scales read the full weight (option 1), that would imply all four springs were compressed with the force that would compress a single spring the same distance. But the springs also exert force on each other, and those forces add.
To be in equilibrium the sum of the forces exerted by the springs has to equal the force compressing them. So each one gets 1/4 of the weight.
The First One Is Correct
The key to this problem, like any problem, is ensuring that all of the forces are accounted for correctly. When you stand on a scale, you experience two forces: a downward force (gravity, mg) and an upward force (the spring in the scale, kx). However, every other scale, but the bottom one, experience 3 forces: their weight and both downward and upward forces from the scales above and below them. The bottom scale is immobile, so only its compression is relevant. Additionally, since we are looking at an equilibrium problem, the total forces acting on each scale must be zero. The equations to solve are then:
0 = k d1 - Myou g
0 = k (d2 - d1) - Mscale g
0 = k (d3 - d2) - Mscale g
0 = k (d4 - d3) - Mscale g
where di is the compression that scale i experiences counting from the top down. When you solve for the di's you find that
di = g [ Myou + (i - 1) Mscale ] /k
which a correctly calibrated scale would read the term in parentheses directly. So, each additional scale will read the mass of the all the ones above it and you.
So, what's next?
Evil Scales
No matter what the scales eventually read ... it's probably still disappointing to the person standing on them.
k.
Adding 4 springs together, of stiffness K, to form a longer spring gives you a new spring of new stiffness Knew which is
1/Knew = 1/k + 1/k + 1/k + 1/k
Knew = k/4
= > Applying the same force F to the new spring will make it shrink 4 times more i.e. each spring will shrink by the same amount, as if it's on it's own, and read the original weight. => one is correct.
My thoughts on this problem..
Well, I don't have degree in anything. But I believe that, if I understood your problem correctly, the top[1st] scale would read your true weight, the 2nd would read your true weight plus the weight of the top scale. And so on and so fourth. So the 4th scale would read all of your previous scales' weights and your true weight. The scale would act independently. The only force acting on the objects is the gravitational force pointing to the center of the Earth. So in essence I agree with your first answer.
Conclusion
I think we have to give this one to option one. Perhaps the exercise serves as an example of where intuition cannot be trusted and people with scientific training can think more clearly.
However, I think my problem may have been caused by my education. Although my initial gut reaction was option one, being an electronics engineer, I began thinking of it in terms an equivalent circuit having resistors (springs) in series reducing the current - totally wrong. I comfort myself by noting that Scott's intuitive guess was also off the mark.
We have the mathematical answer so there is no need for an experiment. However, in the spirit of Archimedes, does anyone have a couple of bathroom scales? If so, a report please.
Another thought experiment
Let's imagine that you place 4 springs on top of each other, separated by a piece of paper. Is the pressure on each layer of paper the same? Probably, assuming that the springs are very light in weight. So you remove the paper, now you have four springs, each resting on the other. Is the compression on each spring the same? Probably. Now you take out your welding torch and weld each spring to the others. Is each quarter of the big spring under the same compression? Probably. So the solution in my opinion is that you would have 1/4 weight on each spring.
RSS
Scales by NomadSoul :: NR6 :: Show
Although I am certainly a nerd, I can't contribute much of use to this because my nerdiness is not of the mathematical kind. However, I will leave you with some thoughts that occurred to me. Feel free to disregard them... (I know I probably should have)...
If you layer the scales on top of each other in an overlapping fashion, do the scales look like scales? And if you stack them and the stack got too high, would you have to scale the scales to get to the top? And just what scale are these scales? Are train scales subject to the same laws as bathroom scales--that is, can they be measured on the same scale? If you tweak the springs of these scales, can you play scales on them?
Forgive the lightness of my tone, but there are just so many possibilities to be weighed. Puts a spring in my step!
Ah, it is early, and I am still sober!