Does Pressure Distribute Evenly in Water Columns?

AI Thread Summary
Pressure in a water column is uniform at a given depth, meaning it is distributed evenly across the entire area. When calculating pressure for a specific area, such as a 1-inch diameter, the relevant force is the total force acting on that area, which is derived from the overall column. The relationship between force, pressure, and area is defined by the equation F = P * A. To find the pressure for a specific area, one must divide the total force by the area of interest. Thus, pressure is not dependent on the entire column but rather on the force applied to the specific area being evaluated.
Idea04
Messages
194
Reaction score
1
If you applied a force to a 1 foot diameter column of water and wanted to determine the pressure for only 1 inch diameter area of water would you only take the force directly above that 1 inch diameter. Or would the force for 1 inch diameter be the same as the force from the whole area of water.
I think the pressure is evenly distributed and it will all be the same force for every square inch but my calculations show otherwise.
 
Physics news on Phys.org
The pressure at the bottom of a static column of water is uniform. More area means more force, because force is pressure multiplied by area:

F = P * A
 
Force can be applied to a 1-foot diameter column of water,
but it is peculiar to ask what the Pressure is for a 1-inch diameter Area,
because Pressure is the Force through a 1-square-inch Area.

Probably the question wanted you to compute the Pressure for the surface,
which would be the Force divided by the Area (ie, about F / 110sq.in).
 
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