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Max
  • Member for 4 years, 7 months
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Can I use Pascal's principle to destroy any airtight container?
In my original post, I actually made an error in converting from $\text{mm}^2$. I've fixed my post now, but I think you have carried my error over. Also, in my argument, the volume of the pin doesn't matter, because the change in pressure occurs at the instant the pin is pushed. The compressibility of the water dictates only how much resistance it gives to my pushing (although this quantity may increase when the pin is further in). Nonetheless, I think the essence of your argument, that the walls are flexible, correctly addresses my question, but I'd like to wait for corroboration.
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Can I use Pascal's principle to destroy any airtight container?
This confuses me more. I thought water was nearly incompressible. And I can apply 50 N of force to a pin, e.g. by setting a 5 kg brick on it.
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Why can, or can not, a perfectly incompressible fluid exist?
No such infinite forces are known in nature The gravitational force approaches infinity near the center of a black hole; likewise, can't we theorize a substance in which the atomic forces approach infinity under compression?
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Can I use Pascal's principle to destroy any airtight container?
Wow. So it really would break ...? Then how come we can keep mantis shrimps, which can punch with 1500 N of force, in an aquarium? And is it the pressure or the force that causes the break?
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Evaluating sum of torques for different choices of origin when solving equilibrium problem
Thank you. This is what I realized, too. To be clear, this means that the magnitude of the applied force required to keep the ball at rest is completely independent of the coef. of friction, right?
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Evaluating sum of torques for different choices of origin when solving equilibrium problem
I also forgot to account for the normal force from the applied force. I have since figured out the answer to my question, but I'll let it stand in case someone wants to answer (I may write my own too).
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Evaluating sum of torques for different choices of origin when solving equilibrium problem
Typo: should be $ \mu_s mg R \cos \theta$, since it's proportional to the normal force.
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Derivation of gyroscopic precession equation: Why is $d\theta = \frac{dL}{L\sin\theta}$?
The picture shows that $dL$ is proportional to $ds$, i.e. $$dL = \alpha ds$$ and that $L\sin\theta$ is proportional to $r$, i.e. $$L\sin\theta = \beta r$$. I would like to see a mathematical expression showing that both proportions are the same, i.e. that $\alpha = \beta$, so that they cancel out when taking $ds / r$. The constants $\alpha$ and $\beta$ would be the transformations defining your "angular-momentum space." The answers so far have simply assumed that $\alpha = \beta$ in this transformation.