Solving for Equilibrium: Water vs Mercury

[goonking]

Homework Statement

Homework Equations

P₁= P_atm + ρ_watervg
P₂= P_atm + ρ_mer.vg

The Attempt at a Solution

so both sides of the tubes are open to atmosphere pressure and are in equilibrium.
I can get P₁ =P₂

the 1 atm from each side cancels out.
g (gravity) cancels out

I'm left with volume (or distance in this case) of water times density of water = distance of mercury times density of mercury:

(1000 kg/m³ ) (0.12m) = distance of mercury ( 13600 kg/m³)

solving for distance of mercury gives 0.088m or 0.88 cm

is this correct?

 

[Orodruin]

is this correct?

Yes. As long as you interpret the question as how hight the mercury on the other side rises above the mercury water interface. If it is in comparison to the original level the answer will be different.

 

[goonking]

Yes. As long as you interpret the question as how hight the mercury on the other side rises above the mercury water interface. If it is in comparison to the original level the answer will be different.

how would it be in comparison to the original level? what would the answer then be?

 

[Orodruin]

how would it be in comparison to the original level? what would the answer then be?

What do you think? If one end moves down 1 cm, how does the other end move?

 

[goonking]

What do you think? If one end moves down 1 cm, how does the other end move?

up 1 cm.

so if the side with water went down 12 cm, the other side should have went up 12 cm, but that isn't what I got as the answer, I'm wrong here

 

[Orodruin]

And what is the resulting difference in surface levels?

 

[SteamKing]

Homework Statement

View attachment 83575

Homework Equations

P₁= P_atm + ρ_watervg
P₂= P_atm + ρ_mer.vg

The Attempt at a Solution

so both sides of the tubes are open to atmosphere pressure and are in equilibrium.
I can get P₁ =P₂

the 1 atm from each side cancels out.
g (gravity) cancels out

I'm left with volume (or distance in this case) of water times density of water = distance of mercury times density of mercury:

(1000 kg/m³ ) (0.12m) = distance of mercury ( 13600 kg/m³)

solving for distance of mercury gives 0.088m or 0.88 cm

is this correct?

Your calculation should read 0.0088 m or 0.88 cm. Watch those decimals and zeroes.

 

[goonking]

And what is the resulting difference in surface levels?

if the water pushed mercury down 12 cm, and the other side moved up 0.88 cm, then the difference should be 12- 0.88 = 11.12 cm?

 

[Orodruin]

if the water pushed mercury down 12 cm, and the other side moved up 0.88 cm, then the difference should be 12- 0.88 = 11.12 cm?

No, we are talking only about the mercury surfaces now. If one mercury surface goes down by 1 cm, you have said that the other rises by 1 cm - so what is the separation between the two? You have also stated that the separation between the two should be 0.88 cm. What does this mean for how much the mercury rose on the side you did not pour water?

 

[goonking]

No, we are talking only about the mercury surfaces now. If one mercury surface goes down by 1 cm, you have said that the other rises by 1 cm - so what is the separation between the two? You have also stated that the separation between the two should be 0.88 cm. What does this mean for how much the mercury rose on the side you did not pour water?

hmm, if I pour 1 cm of mercury into a tube full of mercury, when it is in equilibrium, i essentially raised both sides of the tubes by 0.5 cm.

so If the water is 12 cm deep, that side that water was added, should increase by 12 / 2 = 6 cm

so the other side should increase by 0.88 / 2 = 0.44 cm

 

[Orodruin]

hmm, if I pour 1 cm of mercury into a tube full of mercury

We are not talking about pouring more mercury in, we are talking about pushing one of the mercury surfaces down.

so If the water is 12 cm deep, that side that water was added, should increase by 12 / 2 = 6 cm

This is wrong.

so the other side should increase by 0.88 / 2 = 0.44 cm

This is correct, but check your reasoning.

 

[goonking]

We are not talking about pouring more mercury in, we are talking about pushing one of the mercury surfaces down.

This is wrong.

This is correct, but check your reasoning.

so the separation between the 2 when one side is pushed down 1 cm, is 2 cm

 

[Orodruin]

so the separation between the 2 when one side is pushed down 1 cm, is 2 cm

Correct, and since the separation was 0.88 cm, one side is pushed down 0.44 cm and the other up by 0.44 cm. The water level rises 12 cm above the surface that is pushed down and therefore is 11.56 cm above the original location of the mercury surface, i.e., 11.12 cm above the mercury surface on the other side.