Pressure force on curved surfaces: vertical component

[Est120]

Homework Statement:

Find the vertical force

Relevant Equations:

F=Pave*A, Pave=rho*g*hc

hi guys, i can't understand why they calculate F yin this way, the part of floor that is the vertical proyection has less water than the floor in the left so i tought Fy would be less, please can someone explain this concept to me?

 

[Chestermiller]

I don't understand the diagram. Where is the soil, and which surface are you trying to find the pressure force on?

 

[Est120]

the image is ugly but the concept I am still trying to understand is the following: in the curved body of water just down the cilinder why Fy is calculated that way? I am assuming that the force due to pressure on the ground is rho*g*5 and the floor exerts this force in opposite direction on the fluid but , it is still valid for the curved body of water? even if it doesn't has any "weight" of water above? i put 2 images to show that people solve this kind of problems this way but doesn't really explain this concept (as usually)

 

[Chestermiller]

How would you determine the upward force that the water exerts on the cylinder?

 

[Est120]

i would just calculate FR from Fh and Fv then the force of the water on the cilinder is the same in magnitude but in opposite direction pointing towards the center, the problem is that i can't understand why Fy is calculated that way, it doesn't matter that the portion of fluid in the curved body have less water above?

 

[Chestermiller]

i would just calculate FR from Fh and Fv then the force of the water on the cilinder is the same in magnitude but in opposite direction pointing towards the center, the problem is that i can't understand why Fy is calculated that way, it doesn't matter that the portion of fluid in the curved body have less water above?

Suppose, rather than calculating the upward force exerted by the water on the curved quarter of the cylinder, you calculated the downward force exerted by the curved quarter of the cylinder on the limited slab of water directly beneath it. By Newton's 3rd law, this would be equal in magnitude to the upward force of the water on the cylinder. Would this be OK with you?

 

[haruspex]

the part of floor that is the vertical proyection has less water than the floor in the left so i tought Fy would be less

Pressure at a point in a fluid depends on the height of fluid above the point. It is completely independent of the depth below the point.
The pressure at a depth of 10m in the ocean is the same whether just offshore or over the Marianas Trench.

 

[Est120]

that is what I am trying to say, at the left the floor has 5m of fluid above but at the right (the floor that is the vertical proyection of quarter of the cilinder) has only 0.8m of fluid above, I am very confused about this kind of calculations i never get the why in a way that i can understand clearly rather than memorize the concept

 

[haruspex]

that is what I am trying to say, at the left the floor has 5m of fluid above but at the right (the floor that is the vertical proyection of quarter of the cilinder) has only 0.8m of fluid above, I am very confused about this kind of calculations i never get the why in a way that i can understand clearly rather than memorize the concept

At a given horizontal level in the fluid the pressure is the same everywhere; if not, fluid would flow from the higher pressure to the lower.
Don't worry about the fluid below a point on the curved surface, just consider the depth of that point below a point in the fluid where the pressure is known, e.g. at a surface exposed to the atmosphere.

 

[Chestermiller]

that is what I am trying to say, at the left the floor has 5m of fluid above but at the right (the floor that is the vertical proyection of quarter of the cilinder) has only 0.8m of fluid above, I am very confused about this kind of calculations i never get the why in a way that i can understand clearly rather than memorize the concept

You haven't responded to my post #6. I would appreciate a response.

 

[Est120]

all my issue is about Fy, why it has that value?, the force the liquid exerts on the floor at the right is the same

Suppose, rather than calculating the upward force exerted by the water on the curved quarter of the cylinder, you calculated the downward force exerted by the curved quarter of the cylinder on the limited slab of water directly beneath it. By Newton's 3rd law, this would be equal in magnitude to the upward force of the water on the cylinder. Would this be OK with you?

yes, that is ok i can clearly understand that concept my issue was Fy calculated in that manner, thank you for the response

 

[Chestermiller]

all my issue is about Fy, why it has that value?, the force the liquid exerts on the floor at the right is the same

yes, that is ok i can clearly understand that concept my issue was Fy calculated in that manner, thank you for the response

The free body diagram for the "shoe shaped" region of water situated directly underneath the curved surface, and sandwiched in-between the cylinder and the base, is shown in your original figure (in post #1). It shows the vertical force due to the weight of the fluid contained in this region, the upward force from the base, the horizontal force exerted by the fluid on the left, and the horizontal and vertical components of the force from the cylinder surface acting on the fluid. Do you see these forces in the figure? We are going to do a vertical equilibrium force balance on this shoe-shaped region. Is it the upward vertical force from the base that you do not understand?

 

[Lnewqban]

that is what I am trying to say, at the left the floor has 5m of fluid above but at the right (the floor that is the vertical proyection of quarter of the cilinder) has only 0.8m of fluid above, I am very confused about this kind of calculations i never get the why in a way that i can understand clearly rather than memorize the concept

Please, see:
https://en.m.wikipedia.org/wiki/Pascal's_law

Imagine replacing the cylinder with a 45-degree floodgate.
Does that change modify the direction and intensity of the average force trying to open that gate respect to the force acting on the cylinder?

 

[Chestermiller]

Please, see:
https://en.m.wikipedia.org/wiki/Pascal's_law

Imagine replacing the cylinder with a 45-degree floodgate.
Does that change modify the direction and intensity of the average force trying to open that gate respect to the force acting on the cylinder?

This changes the weight of the water in the free body, which changes the net downward force from the overlying surface.

 

[haruspex]

my issue was Fy calculated in that manner

Can you explain more clearly what it is that you do not understand? Since you do not seem to be finding our posts helpful, maybe we do not understand your difficulty.

As I explained in post #9, the pressure at a point in a given fluid depends only on its depth below the free surface, so at all points across the bottom of the triangularish region it is the same. Considered as a single force, it therefore acts at the midpoint.

 

[Est120]

your post helped me a lot!, thanks i had the issue with fy calculated in that manner because i tought that the force acting on the shaped region base was simply the weight of the fluid in that shaped region, e.g in the floor at the left the force acting is rho*g*5*A which is simply the hydrostatic formula for the pressure times the area