How Does Pressure Distribute in a Fluid System?

  • Thread starter Thread starter sgstudent
  • Start date Start date
  • Tags Tags
    Act Pressure
AI Thread Summary
Pressure in a fluid system, represented by the equation P=hρg, acts in all directions, including on surrounding gases. In a barometer, the pressure at the bottom is balanced by the normal force, while the pressure difference is exerted throughout the fluid. A small sensor measuring pressure at various points would indicate that pressure remains consistent at the same depth, regardless of the sensor's size. As the sensor size decreases, pressure values converge, showing no abrupt change at the liquid-air interface. Overall, pressure in fluids is uniform at a given depth and acts equally on all surrounding surfaces.
sgstudent
Messages
726
Reaction score
3

Homework Statement


P=hρg is the pressure difference but where does this pressure act upon? In a standard barometer there is a vacuum on top so 0 Pa, and at the bottom the normal force acts on the glass base. So hρg=Pweight. But theoretically, where would the hρg calculated be acted upon? Or will that pressure just be exerted at all directions?

If so in this example http://postimage.org/image/53k5i2cqz/full/ will the hρg be exerted onto the air as well?


Homework Equations



P=hρg

The Attempt at a Solution


I think so because under forces, we learned that by Newton's third law besides the normal force balancing out weight, there is also a reaction force acting on the other body that applied the normal force. So now in terms of pressure in that scenario the liquid would exert the pressure on the gas below.

However, the main part that confuses me is that the normal force balancing the weight will be countered by weight itself. So there is no net force acting on the liquid. So why should there be a pressure being exerted on it?

Thanks for the help!
 
Last edited by a moderator:
Physics news on Phys.org
Yes pressure acts in all directions in a fluid. Let's suppose you can make a tiny spherical sensor that can measure the force acting at say 1000 points all over its surface.

If the sensor was large you might measure a pressure difference between the top and bottom, but the pressure on each side would be the same besause they are at the same depth in the fluid.

As the diameter of this sensor is reduced towards zero would find that the pressure on all the points approached the same value.

If the sensor is really small then you would find that the pressure just either side of the liquid air surface was the same. It doesn't suddenly change polarity from positive to negative as you go through the surface.

Aside: I'm ignoring any effect due to surface tension
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top