When does the plate of glass loosen from a glass cylinder in water?

AI Thread Summary
The discussion revolves around determining the height of ethanol needed to loosen a glass plate at the bottom of a glass cylinder submerged in water. The key concept is hydrostatic pressure, where the pressure exerted by the water must equal the pressure exerted by the ethanol for the plate to loosen. The densities of water and ethanol are crucial in this calculation, as they influence the pressure at different heights. Participants emphasize understanding the forces acting on the glass plate, particularly the role of hydrostatic pressure. The initial confusion about the problem's details was clarified, leading to a better grasp of the forces involved.
justduy
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Homework Statement


A thin glass plate is pressed at the end of a glass cylinder. The glass cylinder with the glass plate at the bottom is dipped in water so that the plate is 25 cm under the water surface (The glass cylinder i open, meaning only the bottom part is closed, due to the glass plate). They we proceed with filling this cylinder with Ethanol, and the question is: how much Ethanol must we fill until the plate loosens. It asks for the height of Ethanol, meaning how high must we fill it.

(I realize that this is not the easiest way to describe the problem, but being that my English is not that good, this is the best I can do. Sorry in advance)

Density of Ethanol: 0.79 * 10^3 kg/m^3
of Water: 1.00 * 10^3 kg/m^3

The Attempt at a Solution


I don't really know, I looked at the proposed solution, but I understand nothing. The book never said anything else about this. The proposed solution said that the plates loosens when the pressure of water = pressure of ethanol, basically: D(water) * g * h (25 cm) = D(ethanol) * g * h (this is what we are supposed to find).

I don't understand why...
 
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Ignoring the model solution: just thinking about it for a bit - what are the forces on the glass plate?
 
Simon Bridge said:
Ignoring the model solution: just thinking about it for a bit - what are the forces on the glass plate?
It says nothing about what kinds of forces that are in play. The only thing we get to know is that: the plate of glass is pushed at the end of the cylinder of glass which is then put under water.
 
justduy said:
It says nothing about what kinds of forces that are in play. The only thing we get to know is that: the plate of glass is pushed at the end of the cylinder of glass which is then put under water.

Think of hydrostatic pressure ...

ehild
 
It says nothing about what kinds of forces that are in play. The only thing we get to know is that: the plate of glass is pushed at the end of the cylinder of glass which is then put under water.
... if the glass end was not under water, it would just fall off.
Why? What is it about being under water that holds the glass plate in place?
(As ehild points out: think - hydrostatic pressure... you'll see you are told a lot about the forces on the plate.)

This question does not come just out of the blue: you have been doing some coursework as well. What is the subject you are learning in class about now?
 
Simon Bridge said:
... if the glass end was not under water, it would just fall off.
Why? What is it about being under water that holds the glass plate in place?
(As ehild points out: think - hydrostatic pressure... you'll see you are told a lot about the forces on the plate.)

This question does not come just out of the blue: you have been doing some coursework as well. What is the subject you are learning in class about now?

I understood it now. I mistook some information given by the book. I wasn't able to see the difference on p and greek letter ro. I understand now that there is pressure or force affecting the whole glass cylinder, both from the water on the glass cylinder and the ethanol on the glass cylinder.
 
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Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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