Fluid Mechanics - Pressure measurements

AI Thread Summary
The discussion revolves around solving a fluid mechanics problem involving a U-tube filled with mercury and water. The user successfully calculated the length of the water column in the right arm as 24.5 cm but struggled with determining the height change of mercury in the left arm due to the water pressure. They sought guidance on equating the pressure changes and applying hydrostatic principles. A helpful response provided equations relating the pressures and mass conservation, leading to a successful resolution of the problem. The user expressed gratitude for the clarification and understanding gained from the explanation.
CSUFStudent
Messages
2
Reaction score
0
I'm having difficulty with one problem and was hoping someone could help me out. Here's the problem:

Mercury is poured into a U-tube as in Figure P14.18a. The left arm of the tube has a cross-sectional area A1 of 107.0 cm2, and the right arm has a cross-sectional area A2 of 4.10 cm2. One hundred grams of water are then poured into the right arm, as in Figure P14.18b.

(a) Determine the length of the water column in the right arm of the U-tube
(b) Given that the density of mercury is 13.6 g/cm3, what distance h does the mercury rise in the left arm?

I've attached the picture of the figure to this post

For part (a), given the radius of A2 and the volume of water, I easily got the length of the water column: 24.5cm

Part (b) is giving me a problem. How do I equate the change in height of the mercury due to the pressure of the water on it? I can determine the pressure of the water (P=F/A) on the mercury and thereby determine the upward pressure on the left arm (A2/A1=F2/F1). Once I get the change in height (dy) of the right arm I can use it to figure out the subsequent change on the left arm (A1/dy1=A2/dy2) but getting to that point is difficult for me.

I would greatly appreciate any suggestions or a point in the right direction.

Thanks!
 

Attachments

  • p15-18.gif
    p15-18.gif
    13.8 KB · Views: 610
Physics news on Phys.org
Open the archive attached, I have modified the yours one a bit. Pay attention to the new references z, P_1, and L. P_1 is the pressure just at the interface water-mercury, and L is the length obtained in a).

Pressure at the interface (Hydrostatics):

P_1=P_{at}+\rho_w g L

that pressure is the same at the same height inside the mercury. So that, we can relate the jump of pressures at the left side:

P_1=P_{at}+\rho_{Hg} g (h+z)

So that:

\rho_w g L=\rho_{Hg} g (h+z) (1)

In addition and due to the mass conservation:

\rho_{Hg} z A_2=\rho_{Hg} h A_1 (2)

With (1) and (2) you can solve for h.
 

Attachments

Thank you very much! The solution you provided worked perfectly. I don't think I would have ever come to the same conclusion on my own but with your explanation I understand the physics involved and how you arrived at the answer. Thanks again! :smile:
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top