Is water pressure on an inclined wall less than the weight of the water?

[Mike_bb]

Hello.

I have question about water pressure in a vessel.
Is water pressure on an inclined wall less than weight of water that acting on inclined wall or picture isn't correct? $F$ is pressure on the picture, $P$ is weight.

Thanks.

 

[Baluncore]

Hydrostatic water pressure is a function of depth.
The hydrostatic pressure at a point has no specific direction.
The forces on the wall of a container are normal to the wall.
https://en.wikipedia.org/wiki/Hydrostatics#Hydrostatic_force_on_submerged_surfaces

Is water pressure on an inclined wall less than weight of water that acting on inclined wall? F is pressure on the picture, P is weight.

Your diagram is confused, with F and P reversed.
Weight is a force, F.
Hydrostatic pressure has no direction, it increases downwards.

 

[Mike_bb]

Hydrostatic water pressure is a function of depth.
The hydrostatic pressure at a point has no specific direction.
The forces on the wall of a container are normal to the wall.
https://en.wikipedia.org/wiki/Hydrostatics#Hydrostatic_force_on_submerged_surfaces


Your diagram is confused, with F and P reversed.
Weight is a force, F.
Hydrostatic pressure has no direction, it increases downwards.

How is it possible that weight is less than force on the wall ?

 

[Baluncore]

How is it possible that weight is less than force on the wall ?

Define what you mean by "weight".

 

[Mike_bb]

Define what you mean by "weight".

$P=mg$

 

[Baluncore]

So what is the weight of water, 1 metre below the surface of the Pacific Ocean?

 

[Mike_bb]

So what is the weight of water, 1 metre below the surface of the Pacific Ocean?

Ok. In my first post I meant that there are two forces: force acting on the wall $F$ and weight $P$ of part of water. Could you explain what's wrong on my picture?

 

[Baluncore]

Ok. In my first post I meant that there are two forces: force acting on the wall F and weight P of part of water. Could you explain what's wrong on my picture?

Weight does not come into the problem, except through an earlier density calculation.
The symbol P should be the hydrostatic pressure.
Hydrostatic pressure; P = depth * density * g ; in units of pascal, Pa = N/m².
That pressure is applied, normal to an area, A, of the wall, resulting in two forces.
One horizontal; F_H = P * A * Cos( θ ) ; newtons.
The other vertical; F_V = P * A * Sin( θ ) ; newtons.
Where θ is the slope of the wall, horizontal = 0°, vertical = 90°.

 

[Mike_bb]

Weight does not come into the problem, except through an earlier density calculation.
The symbol P should be the hydrostatic pressure.
Hydrostatic pressure; P = depth * density * g ; in units of pascal, Pa = N/m².
That pressure is applied, normal to an area, A, of the wall, resulting in two forces.
One horizontal; F_H = P * A * Cos( θ ) ; newtons.
The other vertical; F_V = P * A * Sin( θ ) ; newtons.
Where θ is the slope of the wall, horizontal = 0°, vertical = 90°.

The other vertical; FV = P * A * Sin( θ )

Is this weight of water?

 

[Baluncore]

Is this weight of water?

No.
Weight requires you refer to a specified volume of water that has a mass.
You need to get the word weight out of your analysis.

 

[Mike_bb]

No.
Weight requires you refer to a specified volume of water that has a mass.
You need to get the word weight out of your analysis.

Water has mass M. Is this picture correct?

 

[Mike_bb]

No.
Weight requires you refer to a specified volume of water that has a mass.
You need to get the word weight out of your analysis.

Weight always exists.

 

[russ_watters]

Weight always exists.

The reason weight doesn't really apply here is that pressure by definition acts at a point. A point doesn't have a volume so it doesn't have a weight. That's why the hydrostatic pressure equation has density, not weight.

Now, if you want you can calculate the total force (not pressure) based on the weight of the water, but that's a different and more difficult calculation (integrating the pressure over the surface it is being applied to).

 

[Mike_bb]

The reason weight doesn't really apply here is that pressure by definition acts at a point. A point doesn't have a volume so it doesn't have a weight. That's why the hydrostatic pressure equation has density, not weight.

Now, if you want you can calculate the total force (not pressure) based on the weight of the water, but that's a different and more difficult calculation (integrating the pressure over the surface it is being applied to).

I have two forces: $F$ is acting on the wall. $W$ is weight of water with mass M. Could you explain which of these forces is vertical component of another force?

 

[russ_watters]

I have two forces: $F$ is acting on the wall. $W$ is weight of water with mass M. Could you explain which of these forces is vertical component of another force?

Neither. Weight always acts vertically, period. And since you defined it this way, F is the net force acting on the wall. And in this case they aren't really related.

I feel like you're issue here is that because you are so focused on weight you are actually completely ignoring hydrostatic pressure. Maybe if you first tried a vertical wall you'd see that weight is completely irrelevant for finding the force/pressure on the wall.

 

[PeroK]

Neither. Weight always acts vertically, period. And since you defined it this way, F is the net force acting on the wall. And in this case they aren't really related.

I feel like you're issue here is that because you are so focused on Weight you are actually completely ignoring hydrostatic pressure. Maybe if you first tried a vertical wall you'd see that Weight is completely irrelevant for finding the force/pressure on the wall.

Moreover, for a rectangular well filled with water, the pressure depends only only the height of the water, not on the width of the well.

 

[Chestermiller]

The force F has a horizontal component, but the weight is purely vertical. The reason for the horizontal component of F is that it has to balance the horizontal component of the water in the center section which pushes outward from the center. The vertical component of F is equal to the weight of the water in the "wedge."

 

[jbriggs444]

Water has mass M. Is this picture correct?
View attachment 359785

I assume that this drawing is intended to depict the total force on the container wall as $F$ and the vertical component of that force as $W$.

We can also safely assume that there is zero (gauge) pressure on the upper surface of the water.

In this case it would be correct to state that $W = Mg$.

However, your understanding of the drawing is flawed. It seems that you (confusingly) expect that $F$ is the normal component of $W$ rather than correctly seeing $W$ as the vertical component of $F$.

If you want to draw a relevant triangle, it would look like this:

The total normal force $F$ is greater than the weight of the supported water $W$.

 

[Mike_bb]

I assume that this drawing is intended to depict the total force on the container wall as $F$ and the vertical component of that force as $W$.

We can also safely assume that there is zero (gauge) pressure on the upper surface of the water.

In this case it would be correct to state that $W = Mg$.

However, your understanding of the drawing is flawed. It seems that you (confusingly) expect that $F$ is the normal component of $W$ rather than correctly seeing $W$ as the vertical component of $F$.

If you want to draw a relevant triangle, it would look like this:

View attachment 359828
The total normal force $F$ is greater than the weight of the supported water $W$.

If we consider brick with mass M on the inclined plane then weight of brick will be directed vertically and it will be more than force that acting on the inclined plane.

 

[russ_watters]

If we consider brick with mass M on the inclined plane then weight of brick will be directed vertically and it will be more than force that acting on the inclined plane.

True, but a brick doesnt have hydrostatic pressure.

 

[Mike_bb]

True, but a brick doesnt have hydrostatic pressure.

Big thanks! Now I understand it.

 

[Mike_bb]

russ_watters,

If we have brick with mass m on the inclined plane and F1 is the force that push brick down perpendicular to the surface by hand then what will the force F2 be equal to? Is $F2$ more than $mg$ (as on the picture)?
I think that $F2=mg*cos A + F1$. Is it true?
Thanks

 

[A.T.]

I think that $F2=mg*cos A + F1$. Is it true?

Yes, if by F2 you mean the normal contact force.

Your diagram shows something else, which doesn't make much sense, with or without F1.

 

[snorkack]

True, but a brick doesnt have hydrostatic pressure.

Depends on the condition of the brick.
When dry, yes, the brick does not have much of hydrostatic pressure and exerts pressure directly down when it is not placed in shear, sidewards stress like arches, flat arches et cetera.
Moisten the brick, as through a leaky roof or too narrow eaves, and the brick will gradually develop hydrostatic pressure till it flows as liquid clay.

 

[jbriggs444]

View attachment 359854

If the brick is accelerating down the slope then it has little to do with a situation involving hydrostatic equilibrium. @russ_watters has made this point in #20 above.

A fluid in equilibrium cannot sustain any shear force. Zero friction at all walls of the container. Not even viscous friction like a moistened unfired clay brick. Or 100 year old pitch.

 

[Mike_bb]

Yes, if by F2 you mean the normal contact force.

Your diagram shows something else, which doesn't make much sense, with or without F1.

Ok,thx. In my diagram $mg$ is less than $F2$. I have a doubt. $mg < F2$?

 

[jbriggs444]

Ok,thx. In my diagram $mg$ is less than $F2$. I have a doubt. $mg < F2$?

You've drawn $F_2$ longer than $mg$. So it seems that you think that $mg \lt F_2$

But what, exactly, is $F_2$?

Is it the contact force of block on plane with the surface of the plane being frictionless?

Is it the normal component of the contact force of block on plane with the surface of the plane being frictionless or otherwise.

Is it the normal component of the sum of $F_1$ and $mg$?

Is it the normal component of $mg$ alone?

If you push down hard enough on the top of the block (thereby increasing both $F_1$ and $F_2$) you can contrive for the normal component of the contact force to exceed $mg$, yes.

 

[A.T.]

Ok,thx. In my diagram $mg$ is less than $F2$. I have a doubt. $mg < F2$?

Who knows? You have never explained what F2 is supposed to be. And the nonsense drawing doesn't help.

 

[Mike_bb]

You've drawn $F_2$ longer than $mg$. So it seems that you think that $mg \lt F_2$

But what, exactly, is $F_2$?

Is it the contact force of block on plane with the surface of the plane being frictionless?

Is it the normal component of the contact force of block on plane with the surface of the plane being frictionless or otherwise.

Is it the normal component of the sum of $F_1$ and $mg$?

Is it the normal component of $mg$ alone?

If you push down hard enough on the top of the block (thereby increasing both $F_1$ and $F_2$) you can contrive for the normal component of the contact force to exceed $mg$, yes.

$mg$ is weight of brick.
$F2$ is pressure force on surface of inclined plane.

 

[Mike_bb]

I solved my problems. Thanks to all!