Fluids & Solids/ Force (EK)

[SaintJude]

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Q. NO. 98:

View attachment 18582

A. There is no difference
B. 3000 N
C. 101, 300 N.
D. 104, 300 N.

I understand how you get the value P= F/A, but why would you conceptually think that the pressure are not the same?I thought pressure only depends on density and depth...

 

[SaintJude]

Wrong edit. Ignore post

 

[aSagacious]

Pressure is defined as a force distributed over a given area, or algebraically...

P = F/A

You are given that A1 = A2 and that P1 = 101,300 and P2 = 0. Plug and chug.

 

[SaintJude]

I understand how you get the value P= F/A, but why would you conceptually think that the pressure are not the same?

Ok, Is this because the pressure in a vacuum is 0? But then for the "Plug no.2", why can one disregard the atmospheric pressure above the plug?

 

[milski]

Before you start pulling, there are two force acting on each disk - one from above, one from below. The ones from above are the same, since the pressure on each disk is the same. The one below is 0 for the plug with vacuum and whatever the calculated force happens to be for atmospheric pressure on disk with such area. The difference between these two forces is how hard you need to pull.

You can do either P1*A-P2*A or (P1-P2)*A and treat it either as force difference or pressure difference. The result will be the same.

 

[MedPR]

I thought vacuum = 0 pressure. So why is the 0 pressure area under plug #2 pulling?

 

[milski]

I thought vacuum = 0 pressure. So why is the 0 pressure area under plug #2 pulling?

Yes, that's correct. The vacuum is not pulling but it's not pushing either. The atmospheric pressure under plug #1 is pushing up though.

 

[MedPR]

Yes, that's correct. The vacuum is not pulling but it's not pushing either. The atmospheric pressure under plug #1 is pushing up though.

Oh. Well then.. 🙂

 

[milski]

I understand how you get the value P= F/A, but why would you conceptually think that the pressure are not the same?

Ok, Is this because the pressure in a vacuum is 0? But then for the "Plug no.2", why can one disregard the atmospheric pressure above the plug?

You cannot disregard it.

The 'net pressure' on plug #1: Patm+Pwater-Patm = Pwater
The 'net pressure' on plug #2: Patm+Pwater

P2-P1=Patm+Pwater-Pwater=Patm

F=PA=Patm*1=101 300 N

 

[MrNeuro]

You cannot disregard it.

The 'net pressure' on plug #1: Patm+Pwater-Patm = Pwater
The 'net pressure' on plug #2: Patm+Pwater

P2-P1=Patm+Pwater-Pwater=Patm

F=PA=Patm*1=101 300 N

huh can you walk me through plug 1 and plug 2 i know that P1-P2 = pgh

i get that the pressure is Ptotal = Patm + pgh but where does your -Patm for plug 1 come from and how and/or do you apply the fact that there is a vacuum for plug 2?

 

[MedPR]

Yea, after a second look, I'm not seeing how Patm is pushing on the bottom of plug 1. If Plug1 is at the bottom of the pool, there is no atm below it, just the ground.

 

[milski]

huh can you walk me through plug 1 and plug 2 i know that P1-P2 = pgh

i get that the pressure is Ptotal = Patm + pgh but where does your -Patm for plug 1 come from and how and/or do you apply the fact that there is a vacuum for plug 2?

That's the pressure on top of the plug. On the other side of the plug you have atmospheric pressure. Since the force from it is pointed up, you can subtract it from the pressure on top of the plug, which you calculated correctly.

The plug with the vacuum has no pressure below it so there is not any force pushing from below.

 

[milski]

Yea, after a second look, I'm not seeing how Patm is pushing on the bottom of plug 1. If Plug1 is at the bottom of the pool, there is no atm below it, just the ground.

It plugs a drain, it's not on the ground.

If it was on the ground, you have two options. Either you ensure an ideal seal between the plug and the ground. In that case it acts exactly as the one with the vacuum below. Or you have imperfections in the seal and the pressures above and below it are the same. The latter is the realistic case.

 

[MrNeuro]

It plugs a drain, it's not on the ground.

If it was on the ground, you have two options. Either you ensure an ideal seal between the plug and the ground. In that case it acts exactly as the one with the vacuum below. Or you have imperfections in the seal and the pressures above and below it are the same. The latter is the realistic case.

suppose water had seeped into the plug and then assume somehow miraculously the plug reseals....the pressure in the plug is now lower (right?) is there any way you can calculate the pressure inside the seal given the volume and height of the region beneath the seal???

and thanks for that clarification

 

[milski]

suppose water had seeped into the plug and then assume somehow miraculously the plug reseals....the pressure in the plug is now lower (right?) is there any way you can calculate the pressure inside the seal given the volume and height of the region beneath the seal???

and thanks for that clarification

Are we talking about the plug the vacuum below it? Water is incompressible, the vacuum should stay about as good as it was before the water seeped in. The pressure on top will be slightly lower then before since the total depth will be slightly lower. The difference between the two plugs will remain the same, since it depends only on the atmospheric pressure.

 

[MrNeuro]

Are we talking about the plug the vacuum below it? Water is incompressible, the vacuum should stay about as good as it was before the water seeped in. The pressure on top will be slightly lower then before since the total depth will be slightly lower. The difference between the two plugs will remain the same, since it depends only on the atmospheric pressure.

nvm that was a terrible question i totally forgot what the question was please ignore lol

 

[SaintJude]

One final question:

What if this question had asked "Which plug would require more work to pull out?" Interestingly, the plug over the vacuum would take more work, right?

Since the net pressure of Plug 2 is actually higher than 1 and F= P/A , then the force required for pressure 2 would be greater.

What do you guys think?

 

[milski]

Yes, that would be sort of correct. It's hard to define exactly how much work required to pull the plug since that includes a displacement. How far do you need to move the plug to consider it removed? Are they equally thick? But if you assume that everything else is the same for the two situations except the pressures, yes the work required is more for the second one.