# Pressure liquid

1. ### batballbat

127
what would the pressure of liquid at a depth be in a container which is slanted?

2. ### batballbat

127
i suspect $$h.d.g.sinAngle$$. correct me if i am wrong

4. ### batballbat

127
in the derivation of the pressure of liquid, the weight is assumed to act perpendicular to the column of liquid. Plz comment

### Staff: Mentor

Gravity pulls straight down, so...

6. ### batballbat

127
of course but i find similar case to the inclined plane. can somebody give a reasoning?

### Staff: Mentor

What do you find similar about an inclined plane? What is YOUR reasoning?

127

### Staff: Mentor

The pressure depends on the depth beneath the surface, which is not slanted. As already stated, the shape of the container--whether slanted or vertical--is irrelevant.

If you want more, give a specific example of what you have in mind with a diagram.

### Staff: Mentor

Sorry, we don't spoonfeed here. If you want to learn/want help, you need to show some effort at trying to figure it out for yourself. Then when you make a wrong turn, we'll nudge you back in the right direction.

11. ### batballbat

127
wouldnt this imply that the liquid would accelerate at g in slanted tubes?

### Staff: Mentor

No. Please describe exactly what you have in mind. Are you talking about hydrostatic pressure? (Which is what I assumed.) Or fluid dynamics?

### Staff: Mentor

g is acceleration due to gravity. If you block that acceleration, the required force is f=mg. Aka, weight.

14. ### DaveC426913

16,129
Since you're talking about pressure-at-depth then you're talking about a container full of fluid (as opposed to, say, an air-filled container with an amount of water placed in it).

Water suspended in water is neutrally buoyant, so its not like some arbitrary mass of water is going to start sliding to the bottom, accelerating under gravity.

15. ### batballbat

127
No. Please describe exactly what you have in mind. Are you talking about hydrostatic pressure? (Which is what I assumed.) Or fluid dynamics?

I am not familiar with the terms but i guess you are asking whether i am talking about stationary fluids or flowing ones. Eg. in an inclined plane, there is a mass a top, even though its weight acts exactly downward, it would rather move along the plane. And the force with with it moves along the plane is lesser according to its slope. Same for liquids. But as dave said while the liquid is continuous and stationary, the force with with a finite upper part of liquid exerts on the lower part will be the same as for the case of liquids in vertical column. This is not clear to me.

Eg. lets take a column of liquid standing upright and pour some water into it. And then slant it a bit. Then the depth of the liquid increases even though it is not continuous on the upper part (i hope this is understood). So as the depth increases although not uniformly, the pressure in one side must increase. Help me out with this.

### Staff: Mentor

Pascal's vases:

Another set:

Notice that no matter how weird the shape, the tops of the liquid surfaces in the different containers are at the same level. The pressure difference in some container from bottom to top does not depend on shape. It depends only the height of the liquid.

17. ### batballbat

127
i havent learned any of hydrodynamics or hydrostatics. So i think this phenomenon is taken as as true. Is there a proof for it?

### Staff: Mentor

The diagram and image (real) I posted in [post=3643843]post #16[/post] are pretty solid evidence. Hydraulic pumps rely on this principle.

### Staff: Mentor

This is getting very frustrating for me because a bunch of people are putting a bunch of effort into helping you learn, but it doesn't seem like you are trying at all. For example, you said you don't know terms like "hydrostatic", but that term is defined in the very first sentence of the link i gave you in the first reply!

"Learning" is not something we can give you: we can tell you where to find the knowledge, but you have to get it into your head.

20. ### sankalpmittal

772
OP , when I was a kid (well now also I am a kid in class 10th ! ) I also used to think the same way as you are currently.

I used to say :
Pressure = (Force x sin θ)/Area

Now I know that I was wrong because - Pressure is "thrust upon area." It is defined this way. Thrust is defined as force acting perpendicularly on a body.

Hence always ,

P=hρg x sin 90o or P = hdg x cos 0o

As sin 90o = cos 0o = 1

So P=hρg

Do you get it now ?