Tanya Sharma
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##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?Tanya Sharma said:##P(z) = ρzg_{eff}## ,where z is measured from the surface of water . In this case ##g_{eff} = g+a ##
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?Tanya Sharma said:##F_b = Vρg_{eff}## , where ##ρ## is density of water .
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
No. Try that force balance again, and this time use a free body diagram on the block.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 ?
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.Tanya Sharma said:##Mg - V_1ρg = 0##
##V_2ρ(g+a) - Mg - Ma = 0## i.e ##V_1=V_2## ?
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.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 ?
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.
I don't quite understand the question. The base case of comparison is a solid ice cube?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 ?
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.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 ?
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.Tanya Sharma said:I have done the maths(not much involved) . Just wanted to reassure myself by confirming the result with you :)
Please tell us your thoughts on this.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?