1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Direction of pressure on a multiple liquid manometer

  1. Oct 19, 2015 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations
    3. The attempt at a solution
    I already have the solution.

    My question is simple. I can't understand how I define the direction of pressure in each "changing" point. I also have the solution if needed.
  2. jcsd
  3. Oct 19, 2015 #2


    User Avatar
    Science Advisor

    Pressure doesn't have direction.
  4. Oct 19, 2015 #3
    The pressure is continuous across the interface at each changing point. Is that what you meant?

  5. Oct 19, 2015 #4
    Okay, although another lecturer pointed out that pressure should be a vector, I shouldn't mention the word direction - you are right.
    If I rephrase it to "how to find out if pressure is positive or negative"? Does it make sense now?

    Here's the solution. I can't understand how I define where I put + and where I put -
    υδ stand for water, ελ for oil and υδρ for mercury. ρ is density
  6. Oct 19, 2015 #5
    Pascal's Law explicitly mentions that " pressure exerted anywhere in a confined incompressible fluid is transmitted equally in all directions throughout the fluid such that the pressure variations (initial differences) remain the same". So, even if pressure did have a direction (it doesn't, it is a dot product of Force and Area on which the force acts) it wouldn't matter if it was positive or negative.
  7. Oct 19, 2015 #6
    If you are moving downward to get the pressure change, you use a plus sign, and if you are moving upward to get the pressure change, you use a minus sign. This is because pressure increases as depth increases, and decreases as depth decreases.
  8. Oct 19, 2015 #7
    Pressure does have direction in the sense that it acts in the direction perpendicular to every surface upon which it acts. This is because, in reality, pressure is an isotropic second order tensor which, when dotted with a normal to the surface, yields a vector (force per unit area) normal to the surface.

  9. Oct 19, 2015 #8
    This conversation is off-topic. As a student, I don't have the knowledge to support or disprove my lecturer's opinion.

    Can you explain again? I don't see what "moving" upwards or downwards means. Does it have to do with the heights?

    I remember the professor saying we should imagine how the fluid would move if the others weren't there - in case this helps you help me.
  10. Oct 19, 2015 #9
    I want you to go back to the figure, and label some more heights (distances) in the figure:

    Label h4 the distance from the bottom of the h2 section to the point where the tube is joined to be big tank.

    Label h5 the distance from the top of h1 to the top of the U tube section containing the oil.

    Label h6 the distance from the bottom of h3 to the bottom of the U tube section containing the mercury.

    I also want you to label some more pressures on the figure.

    Label p3 the pressure at the junction between the tube and the tank.

    Label p4 the pressure in the tube at the boundary between the water and the oil.

    Label p5 the pressure in the oil at the very top of the U.

    Label p6 the pressure in the tube at the boundary between the oil and the mercury.

    Label p7 the pressure at the bottom of the U containing the mercury.

    Get back with me when you get done with this, and I will help you to continue.

  11. Oct 20, 2015 #10
    Here it is. Sorry but I don't have time to make it look more "neat".
    Now, I'm afraid you'll explain it a way I possibly won't understand because it may contain information I don't know how to work with. But I will gladly read it carefully, so go on.
  12. Oct 20, 2015 #11
    OK. But before we begin, please repeat after me: I think I can, I think I can, I think I can,........

    I'm going to do part of the problem, show you what to do, and then I'm going to let you continue.

    I know that, when I'm in a swimming pool, as I swim down to the bottom, the water pressure increases, and, as I swim back up to the top, the water pressure decreases. So, going from a higher point to a lower point, I get an increase in pressure, and going from a lower point to a higher point, I get a decrease in pressure.

    Now, I'm going to start out at point 1, where the pressure is p1. At this location the pressure in the water just near the surface is equal to the air pressure in the head space of the tank. Now, I want to go down to point 3 where the pressure is p3. I know that point 3 is below point 1; the distance below is (h1+h2+h4). Since I have to go downward to get from point 1 to point 3, the pressure at point 3 is:$$p_3=p_1+ρ_{water}g(h_1+h_2+h_4)$$

    Next I want to get p4. I know that point 4 is above point 3; the distance above is ##(h_2+h_4)##. Since I have to move upward to get from point 3 to point 4, the pressure at point 4 is:$$p_4=p_3-ρ_{water}g(h_2+h_4)$$
    Now I have the pressure at the interface between the water and the oil.

    Next, I'm going to move upward through the oil to get the pressure at point 5. I know that point 5 is above point 4; the distance above is ##(h_1+h_5)##. Since I have to move upward to get from point 4 to point 5, the pressure at point 5 is:

    Next, I want to get the pressure at point 6, which represents the interface between the oil and the mercury. Is point 6 above or below point 5? What is the distance between point 5 and point 6? Since I have to move downward to get from point 5 to point 6, is the pressure at point 6 higher or lower than the pressure at point 5? In terms of the pressure at point 5, the density of the oil, and the distance between points 5 and 6, what is the pressure at point 6?
  13. Oct 22, 2015 #12
    We move downward and the distance is (h1+h2+h5). Pressure at point 6 must be p6=p5 + ρoilg(h1+h2+h5).

    Is it safe to assume I can do the same for each one of the 3 interface points? See if I'm moving upwards or downwards and use the corresponding sign?
  14. Oct 22, 2015 #13
    I think you've got it. The last step is to add up all your equations together and see what you get. This will eliminate all the pressures except p1 and p2.

  15. Oct 23, 2015 #14
    Thanks a lot for helping me. I'll come back in case I face another problem :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted