Basics of Fluid Mechanics and Pressure

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The discussion focuses on understanding the forces acting on a liquid molecule at the surface of a liquid in a container. Key points include the roles of atmospheric pressure, pressure from the surrounding liquid, and the weight of the liquid molecule, with clarification that buoyancy and surface tension also play a role, albeit minimally for small particles. The conversation emphasizes the importance of analyzing fluids at the continuum level rather than the molecular level, leading to the conclusion that pressure calculations should consider a column of liquid rather than individual molecules. Additionally, when a container accelerates, the pressure equation adjusts to account for both gravitational and inertial effects, illustrating the complexity of fluid dynamics. The analysis concludes with insights on how the liquid surface forms an angle due to pressure gradients and the need for equilibrium in fluid behavior.
  • #31
##P(z) = ρzg_{eff}## ,where z is measured from the surface of water . In this case ##g_{eff} = g+a ##
 
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  • #32
Tanya Sharma said:
##P(z) = ρzg_{eff}## ,where z is measured from the surface of water . In this case ##g_{eff} = g+a ##
Right. So, if V is the volume of displaced water, what is the buoyant force now?

Chet
 
  • #33
##F_b = Vρg_{eff}## , where ##ρ## is density of water .
 
  • #34
Tanya Sharma said:
##F_b = Vρg_{eff}## , where ##ρ## is density of water .
So, from Newton's second law, if M is the mass of the block, what is the displaced volume of water in the accelerating elevator? How does this compare with the displaced volume if the elevator is not accelerating?

Chet
 
  • #35
Chestermiller said:
So, from Newton's second law, if M is the mass of the block, what is the displaced volume of water in the accelerating elevator? How does this compare with the displaced volume if the elevator is not accelerating?

Chet

When the elevator is not accelerating then ##Mg = V_1ρg## .

When the elevator is accelerating up then ##Mg = V_2ρ(g+a)## .

##V_1 = V_2(1+\frac{a}{g})## or ##V_1>V_2## . The displaced fluid when the elevator is accelerating is less i.e the block sinks less ( moves upwards ) as compared to when the elevator is at rest .

Is this what you are suggesting ?
 
  • #36
Tanya Sharma said:
When the elevator is not accelerating then ##Mg = V_1ρg## .

When the elevator is accelerating up then ##Mg = V_2ρ(g+a)## .

##V_1 = V_2(1+\frac{a}{g})## or ##V_1>V_2## . The displaced fluid when the elevator is accelerating is less i.e the block sinks less ( moves upwards ) as compared to when the elevator is at rest .

Is this what you are suggesting ?
No. Try that force balance again, and this time use a free body diagram on the block.

Chet
 
  • #37
##V_1ρg -Mg = 0##

##V_2ρ(g+a) - Mg - Ma = 0## i.e ##V_1=V_2## ?
 
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  • #38
Tanya Sharma said:
##Mg - V_1ρg = 0##

##V_2ρ(g+a) - Mg - Ma = 0## i.e ##V_1=V_2## ?
Yes. This is what will happen after the block stops bobbing. Initially, there will be a transient, and the block will dip and then bob, but, after the transient bobbing dies out, the displaced volume will be the same as without the acceleration.

Chet
 
  • #39
On second thought, I take back what I said about the dip and bobbing. There should not be a transient even if the acceleration is applied suddenly. Sorry for the confusion.
 
  • #40
If we write Newton's 2nd law for the block in the ground frame (not from the accelerated frame) .

##F_{b1} - Mg = 0##

##F_{b2} - Mg = Ma## i.e ##F_{b2} =M(g+a)##

##F_{b2} > F_{b1}## . Since the buoyant force is more in accelerating elevator case , fluid displaced should be more . This is the same reasoning I gave in post #29 .

What is the flaw in this reasoning ?
 
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  • #41
Tanya Sharma said:
If we write Newton's 2nd law for the block in the ground frame (not from the accelerated frame) .

##F_{b1} - Mg = 0##

##F_{b2} - Mg = Ma## i.e ##F_{b2} =M(g+a)##

##F_{b2} > F_{b1}## . Since the buoyant force is more in accelerating elevator case , fluid displaced should be more . This is the same reeasoning I gave in post #29 .

What is the flaw in this reasoning ?
If the water is considered incompressible, then the buoyant force imposed by the water will respond instantly to the acceleration. Your second equation here confirms the analysis in post # 37 showing that the displacement will be the same.

Chet
 
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  • #42
it is F_{b2}>F_{b1} but because the fluid weights more now (it is like the gravitational acceleration g is increased by a) not because there is more volume of fluid displaced.
 
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  • #43
Delta² said:
it is F_{b2}>F_{b1} but because the fluid weights more now (it is like the gravitational acceleration g is increased by a) not because there is more volume of fluid displaced.

You are right .
 
  • #44
Chet

There is an ice cube floating in a glass of water . What happens to the water level if

1) ice cube has a cavity with air inside(the cavity).

2) ice cube has a cavity with water inside(the cavity) .

I think that the water level goes down in the first case and remains same in the second case .

Do you agree ?
 
  • #45
Tanya Sharma said:
Chet

There is an ice cube floating in a glass of water . What happens to the water level if

1) ice cube has a cavity with air inside(the cavity).

2) ice cube has a cavity with water inside(the cavity) .

I think that the water level goes down in the first case and remains same in the second case .

Do you agree ?
I don't quite understand the question. The base case of comparison is a solid ice cube?

Chet
 
  • #46
No . The comparison is with the level of water present in the glass before the ice melts .Sorry for being unclear .

What happens to the original water level if

1) an ice cube having a cavity with air inside(the cavity) , floating on the water melts.

2) ice cube having a cavity with water inside(the cavity) , floating on the water melts .

I think that the water level goes down in the first case and remains same in the second case .

Do you agree ?
 
  • #47
Tanya Sharma said:
No . The comparison is with the level of water present in the glass before the ice melts .Sorry for being unclear .

What happens to the original water level if

1) an ice cube having a cavity with air inside(the cavity) , floating on the water melts.

2) ice cube having a cavity with water inside(the cavity) , floating on the water melts .

I think that the water level goes down in the first case and remains same in the second case .

Do you agree ?
Off hand, I find it rather tricky to reason out. Why don't you just model it, and let the math do the work for you? I would do either a cubical cavity or a spherical cavity.

Chet
 
  • #48
I have done the maths(not much involved) . Just wanted to reassure myself by confirming the result with you :)
 
  • #49
Tanya Sharma said:
I have done the maths(not much involved) . Just wanted to reassure myself by confirming the result with you :)
I confirm your result for item 2. For item 1, it depends on whether you take into account the slight density of the air. If you include the density of the air, then the level goes down slightly, but, if you neglect the density of the air, then I get no change.

Chet
 
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  • #50
I vent done the math but in the first case you assume that after the ice cube melts, the air is released to the atmosphere or dissolved into the water?
 
  • #51
Delta² said:
I vent done the math but in the first case you assume that after the ice cube melts, the air is released to the atmosphere or dissolved into the water?
Please tell us your thoughts on this.

Chet
 

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