What does it mean for there to be uniform pressure?

AI Thread Summary
Uniform pressure in a fluid means that the pressure is consistent throughout the fluid, regardless of height or other factors. This concept simplifies calculations by assuming that the pressure does not vary across the surface of the closed hemispherical shell. The discussion highlights that while typically pressure increases with depth due to the weight of the fluid above, the uniform pressure assumption negates this variable. It is suggested that this scenario may apply in conditions like space, where gravity is negligible, or with gases where pressure variation is minimal. Overall, the uniform pressure assumption is intended to streamline the problem-solving process.
eprparadox
Messages
133
Reaction score
2

Homework Statement



A closed hemispherical shell of radius R is filled with fluid at uniform pressure p. The net force of the fluid on the curved portion of the shell is given by:

Homework Equations


The Attempt at a Solution


[/B]
I'm not even posting the answers because I'm not looking to get an answer.

I don't understand what it means for the fluid to be "at uniform pressure p".

At any height beneath the surface of this hemispherical shell, shouldn't the pressure be the sum of the weight of the water above it plus atmospheric pressure?

What does the uniform pressure, ## p ## mean?
 
Physics news on Phys.org
I would guess that the object is in space with no significant gravity and the fluid is stationary so there are no dynamic fluid forces.
Just assume the pressure is the same at all points in the fluid.
 
  • Like
Likes CWatters and jbriggs444
I would assume the same
 
eprparadox said:
weight of the water
It does not specify water. A fluid can be a gas, so the variation in pressure may be negligible.
 
It is likely to make the problem a lot easier :-)
 
It most likely means that fluid pressure is not variable across the shell, at all points of the shells surface the pressure is not polarized to anyone particular sector of it .

It is probably to make the question easier
 
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