- #106
erobz
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bob012345 said:One thing, the water leaking from the bucket should be less than normal maybe even zero when the bucket is accelerating downward. Does you equation physically allow for that?
I got this when I derived it (there is a possibility I got the sign convention mixed up), but yeah.
$$ P = - \rho \left( \frac{d^2x}{dt^2} - g\right) z $$
That result is substituted into this:
$$ \frac{dM}{dt} = - \rho A_j \sqrt{ \frac{2 P}{ \rho \left( 1 - \tau^2 \right) }} $$
So if ## \frac{d^2x}{dt^2} \rightarrow g ##, then ## P \to 0 ## and ## \frac{d M(t)}{dt} \to 0 ##. The bucket stops leaking.