What is the derivation for the equation for liquid pressure at depth?

In summary, the pressure of a liquid of constant density at a depth 'h' can be calculated using the equation P = hρg. This applies to all vertical columns with constant cross-section (prisms) regardless of the shape of the container. The weight of the liquid is supported by the walls of the container, not just a single point, and the pressure force on the bottom surface is not equal to the weight of the liquid unless the container is a vertical prism. This equation is valid for all cases with constant density and gravitational acceleration.
  • #1
Bipolarity
776
2
The pressure of a liquid of constant density at depth 'h' is given by the equation:

[tex] P = hρg [/tex]

I am trying to understand the derivation of this equation. I understand it perfectly for liquid columns with rectangular liquid columns but I can't see why it works if the shape of the fluid column is irregular.

[tex] P = \frac{F}{A} = \frac{mg}{\frac{V}{h}} = hρg [/tex]

The problem is that [itex] \frac{V}{h} = A [/itex] only if it is rectangular shaped, right?
So why should we allowed to generalize it for all fluid columns?

BiP
 
Physics news on Phys.org
  • #2
No. For example, the area of a circle with radius r multiplied by a height h gives the volume of a cylinder with radius r and height h.
 
  • #3
ecneicS said:
No. For example, the area of a circle with radius r multiplied by a height h gives the volume of a cylinder with radius r and height h.

Ok but what if the area of the cross-section changed with the height? Then would it still apply?

BiP
 
  • #4
Bipolarity said:
Ok but what if the area of the cross-section changed with the height? Then would it still apply?
Yes, it always applies. Suppose you had liquid in a container of irregular shape and you inserted a tube vertically to some depth. If the pressure at the base of the tube resulting from the liquid in the tube were different from that elsewhere at the same level then there would be a flow into or out of the tube. The level at the top of the tube would now be different from that in the rest of the container.
 
  • #5
Bipolarity said:
[tex] P = \frac{F}{A} = \frac{mg}{\frac{V}{h}} = hρg [/tex]

The problem is that [itex] \frac{V}{h} = A [/itex] only if it is rectangular shaped, right?
So why should we allowed to generalize it for all fluid columns?

BiP

The last equation holds for any vertical column with constant cross-section (prism). If the liquid has any other shape, you can still derive the same equation for imaginary vertical column in the liquid. "vertical" means directed in direction of gravitational acceleration. This way the pressure forces of surrounding liquid on the sides of the column don't affect the force balance in vertical direction (force is a vector and F in the first equation is its vertical component).
 
  • #6
It is important to remember that force is a vector and that pressure force in liquid is perpendicular to the surface. If the liquid container does not have the shape of a vertical prism, then the force on the bottom surface is not equal to the weight of the liquid, because weight of the liquid is also supported by the walls of the container.

Before I understood that I had troubles understanding the liquid pressure on the bottom of a V shaped vessel: I thought it should be infinite, because "the weight of the water is supported by a single point (zero area)". This assumption is wrong, because V shaped vessel has inclined sides and pressure force on them will have a nonzero vertical component.

Only in case of a vertical column we can assume that the pressure force on the bottom equals the weight of the liquid. Of course we can always define an imaginary vertical column in any liquid container. So the formula

[tex] P = \frac{F}{A} = \frac{mg}{\frac{V}{h}} = hρg [/tex]

is valid for all cases where the density and gravitational acceleration are constant.
 

1. What is liquid pressure at depth?

Liquid pressure at depth refers to the force exerted by a liquid at a certain depth, due to the weight of the liquid above it. It is caused by the combined effects of gravity and the density of the liquid.

2. How does the depth of a liquid affect its pressure?

The deeper the liquid, the higher the pressure. This is because as the depth increases, the weight of the liquid above increases, resulting in a greater force being exerted on the bottom of the container or object at that depth.

3. What is the formula for calculating liquid pressure at depth?

The formula for calculating liquid pressure at depth is P = ρgh, where P is the pressure, ρ is the density of the liquid, g is the acceleration due to gravity, and h is the depth of the liquid. This formula is known as the hydrostatic equation.

4. How is liquid pressure at depth measured?

Liquid pressure at depth can be measured using a device called a pressure gauge, which is designed to measure the force exerted by the liquid on a specific area.

5. What are some real-life applications of understanding liquid pressure at depth?

Understanding liquid pressure at depth is important in a variety of fields, including oceanography, engineering, and scuba diving. It can also be used to determine the stability of structures such as dams and to calculate the buoyant force on objects submerged in a liquid.

Similar threads

  • Advanced Physics Homework Help
Replies
5
Views
959
Replies
13
Views
737
  • Engineering and Comp Sci Homework Help
Replies
3
Views
884
Replies
5
Views
1K
  • Mechanical Engineering
Replies
4
Views
985
  • Mechanics
Replies
15
Views
4K
Replies
68
Views
3K
Replies
33
Views
4K
Back
Top