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bio website Grieu
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visits member for 3 years, 7 months
seen May 27 at 16:59

I'm an engineer with experience in applied cryptography, in particular in Smart Card systems.

Feel free to edit my answers to fix obvious errors, or improve them on expression (I'm not a native English speaker).


May
10
comment Dropping a weight onto a spring scale
@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 $M$.
May
9
comment Dropping a weight onto a spring scale
@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).
Apr
29
revised Dropping a weight onto a spring scale
Revise kind of scale perhaps allowing the method to be used
Apr
29
revised Dropping a weight onto a spring scale
We won't have time to make the reading, and it would far off-scale or/and useless
Apr
29
revised Dropping a weight onto a spring scale
Actually answer
Apr
28
revised Dropping a weight onto a spring scale
Polish comment on relative error on k
Apr
28
revised Dropping a weight onto a spring scale
Discuss relative error
Apr
28
revised Dropping a weight onto a spring scale
Discuss the case of a spring soft enough that the method can work
Apr
27
revised Dropping a weight onto a spring scale
The method is doomed
Apr
27
revised Dropping a weight onto a spring scale
Mention we should not trust the result
Apr
27
revised Dropping a weight onto a spring scale
Explain the case h=0
Apr
27
revised Dropping a weight onto a spring scale
Give hypothesis
Apr
27
revised Dropping a weight onto a spring scale
Expand with numeric values and comparison to other answer giving an approximation
Apr
27
awarded  Teacher
Apr
27
answered Dropping a weight onto a spring scale
Apr
27
comment Dropping a weight onto a spring scale
Correction: if we make $h=0$, we should obtain $m=2M$.
Apr
27
comment Dropping a weight onto a spring scale
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 $m>M$.
Dec
13
comment Will a bullet dropped and a bullet fired from a gun horizontally REALLY hit the ground at the same time when air drag is taken into account?
Problem with this line of thought is that from an experimental standpoint, and when the bullet starts from an height from ground commensurate to a man's size, the effect of $F_y$ on the time it takes for the bullet to reach ground is negligible in both the case of the bullet dropped, and the case of the bullet fired. In the first case, the overwhelming factor is the drop height. In the second case, that's the (practically unavoidable) vertical component of the bullet's initial velocity, and the height.
Dec
10
awarded  Autobiographer
Nov
16
awarded  Announcer