Timeline for Why should water fall down from an upside down glass of water?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| May 28, 2021 at 17:51 | comment | added | Bill Watts | Actually the variation with depth is for the liquid sitting on the bottom where the pressure is determined by the weight of the fluid above it. In the case where the fluid is at the top, it will be in free fall at first, and the pressure will be atmospheric all the way around it. As it accelerates, air drag will become significant, causing a pressure difference between top and bottom, but the liquid will still fall. | |
| May 28, 2021 at 8:36 | comment | added | FullBridge | But then the drop in pressure associated with some height $\Delta h$ would be exactly equal to the amount required to hold the water above it and maintain equilibrium. | |
| May 28, 2021 at 6:42 | comment | added | FullBridge | I didn't assume zero pressure at the top, in that case $F_g$ would be $0$. I think you're saying that the pressure at the top should be atmospheric pressure, but why should that be the case? Why can't it be $\frac{P_aA-\rho Vg}{A}$? | |
| May 28, 2021 at 6:19 | comment | added | Bill Watts | How the glass is supported is irrelevant. I am assuming the glass is stationary the way your case is stated. Think about it. Why would you think there is zero pressure at the top of the liquid? Consider a small particle of liquid in the liquid volume somewhere. Pressure acts on it in every direction and the net effect is zero force. The only net force on that particle is gravity. | |
| May 28, 2021 at 6:05 | comment | added | FullBridge | But that pressure would be exerted on the glass right? What if the glass is fixed through some rigid support? | |
| May 28, 2021 at 5:47 | history | answered | Bill Watts | CC BY-SA 4.0 |