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Critically Damped Bathroom Scale

  • #1
This is the first time I post anything here, so, if I am doing anything wrong about the rules (it is possible that I didn't correctly understand a topic or two), please let me know. This one is a somewhat classic problem of damping, and I can't understand the basic concepts, so I tried to do my best in solving it, but I don't know how to proceed:

Homework Statement



One wants to build a bathroom scale which platform's deflection is 2.5cm when a man with 91kg is weighing himself. If the motion of the scale is critically damped, determine the spring constant k that has to be used, and the damping constant b. With the 91kg man, what will be the maximum force that the scale will apply under his feet, while the scale returns to its equilibrium position?

(this question was based on the problem 2-42 of the third edition of Symon's Mechanics book)

Homework Equations



x = [x0 + (x0[itex]\gamma[/itex] - v0)t]e^(-[itex]\gamma[/itex]t)

b/2m = [itex]\gamma[/itex]

k = mg/x

Such that:

x = position
x0 = initial position
m = mass of what is being weighed
g = acceleration of gravity (I considered this as 10 m/s² to simplify the algebra)
t = time
k = spring constant
b = damping constant
v0 = initial velocity of the thing that is being weighed, when it is put on the scale

The Attempt at a Solution



I just solved a similar problem, and tried to apply the same ideas:

First, to find the spring constant k, I just used the data given, considering g = 10m/s² so k = 91*10/(2.5*10^-2) = 3.64 * 10^4 Kg/s²

> My first doubt is here: although the order of magnitude of this k makes sense to me, I don't know for sure if I considered the right x...

The problem is to find the [itex]\gamma[/itex], since, to avoid overshooting, I made x0[itex]\gamma[/itex] - v0 = 0 [itex]\rightarrow[/itex] [itex]\gamma[/itex] = v0/x0, but, now, what is this velocity?! Is this the velocity that the scale spring is being compressed? But how can I calculate this? Another point: isn't the x0, intended to be the equilibrium position, equals zero?

[The main difference between this problem and the one that I solved is that the latter was about a mass falling from a certain height to the scale, and now the man is just standing on the scale... So I'm lost about that]

I cordially ask, to whom may help, to explain carefully why anything I did is wrong, and why the correct way is correct, since I am beginning my studies in this topic and I am a little bit "slow" to get some ideas.

Thanks!
 

Answers and Replies

  • #2
rude man
Homework Helper
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First, to find the spring constant k, I just used the data given, considering g = 10m/s² so k = 91*10/(2.5*10^-2) = 3.64 * 10^4 Kg/s²

> My first doubt is here: although the order of magnitude of this k makes sense to me, I don't know for sure if I considered the right x...
that's fine.
The problem is to find the [itex]\gamma[/itex], since, to avoid overshooting, I made x0[itex]\gamma[/itex] - v0 = 0 [itex]\rightarrow[/itex] [itex]\gamma[/itex] = v0/x0, but, now, what is this velocity?! Is this the velocity that the scale spring is being compressed? But how can I calculate this?
Your gamma is not computed from velocity. Velocity does not enter into this computation.
Another point: isn't the x0, intended to be the equilibrium position, equals zero?
yes, the initial position is zero, but it's not the equilibrium position. the equilibrium position is 2.5 cm.

You should approach the problem from basic principles, not by similarity to a problem you've already had.

Start with the basic F = ma equation. How do you write that?
 
Last edited:
  • #3
ehild
Homework Helper
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What does critical damping mean? How are k and b related in case of critical damping?

ehild
 

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