fgrieu
Reputation
205
Next privilege 250 Rep.
 Feb 4 awarded Yearling Jan 13 awarded Famous Question Nov 21 awarded Popular Question 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 $m>M$.