Fluid Mechanics: Topic 4.1 - Hydrostatic force on a plane surface

Summary

This lesson explores the derivation of the hydrostatic force equation on a plane surface, essential for assessing the structural integrity of a water-filled fish tank. It explains calculating the resultant force on the bottom and side walls, considering atmospheric pressure and water depth. The video also introduces the concept of the centroid and center of pressure, illustrating how to integrate these to find the resultant force on an arbitrarily shaped and oriented wall, crucial for fluid mechanics problem-solving.

Takeaways

• 📚 The lesson aims to derive the equation for calculating the hydrostatic force on a plane surface submerged in a fluid.
• 🏞️ A rectangular fish tank filled with water to a depth 'd' is used as an example to explain the concept of hydrostatic force.
• 🌡️ The atmospheric pressure 'Pa' surrounds the tank and contributes to the force exerted on the tank walls.
• 🔽 The bottom wall of the tank experiences a resultant force due to the difference in pressure between the inside and outside of the tank.
• 📉 The atmospheric pressure terms cancel out in the calculation of the resultant force on the bottom wall, leaving a force proportional to the specific weight of the water 'gamma' and the depth 'd'.
• 🔼 The side walls of the tank experience varying pressures due to the linear increase in pressure with depth inside the tank.
• 📏 A coordinate system is established to analyze the pressure forces on an arbitrary section of the side wall at an angle 'theta' to the free surface.
• 📐 The net force due to fluid pressure at a small area 'dA' on the side wall is calculated using trigonometry, considering the depth 'h' as 'y sin(theta)'.
• ∫ The resultant force on the entire wall is found by integrating the net force over all points of the wall, leading to an expression involving the centroid of the wall.
• 📌 The center of pressure 'CP' is identified as the point where the resultant force 'FR' acts, equating to the pressure at the centroid 'PC' times the area.
• 🔧 The equations derived are applicable for any flat surface where the liquid's free surface and the tank's side are exposed to the same pressure, such as atmospheric pressure.
• 🛠️ If the tank is pressurized to a gage pressure 'P0', the resultant force equation must be adjusted to account for this additional pressure.

Q & A

What is the purpose of deriving the equation for calculating the magnitude of the hydrostatic force on a plane surface?

The purpose is to determine whether the fish tank walls can support the water pressure without breaking, which is essential for structural integrity and safety.

Why is it necessary to know the resultant force along the bottom wall of a fish tank?

Knowing the resultant force along the bottom wall helps in understanding the total force exerted on the walls, which is crucial for assessing the tank's ability to withstand the water's pressure.

What is the atmospheric pressure denoted by in the script?

The atmospheric pressure is denoted by 'Pa' in the script.

How does the pressure force on the underside of the bottom wall of the tank relate to the area of the bottom wall?

The pressure force on the underside of the bottom wall is equal to the atmospheric pressure 'Pa' times the area of the bottom wall, and it points upward.

What is the formula for the resultant force acting along the bottom wall of the tank?

The resultant force is calculated by subtracting the downward force (Pa + gamma * d) times the area from the upward force Pa times the area, resulting in -gamma * d * (area of the bottom wall).

What is the significance of the atmospheric pressure terms canceling out in the calculation of the resultant force on the bottom wall?

The cancellation simplifies the equation, leaving only the component of the force due to the water's weight, which is gamma times d times the area, pointing downward.

How does the pressure along the interior of the side walls of the tank change with depth?

The pressure along the interior of the side walls increases linearly with depth, being the atmospheric pressure plus the specific weight of the water times the depth at any point.

What is the role of the centroid in calculating the resultant force on a wall section?

The centroid is the geometric center of the wall and is used to determine the location of the center of pressure, which is where the resultant force acts, simplifying the calculation of the force on the wall section.

What is the expression for the resultant force FR on the entire wall in terms of the centroid?

The resultant force FR is equal to gamma times sin(theta) times yC (the y-coordinate of the centroid) times the area A of the wall.

How can the resultant force equation be rewritten using the pressure at the centroid?

The resultant force equation can be rewritten as the pressure at the centroid PC times the area A, where PC is gamma hC, the pressure at the centroid's depth.

What modification would be needed to the resultant force equation if the tank is pressurized to a gage pressure of P0?

The equation would be modified to (gamma sin(theta) yC plus P0) times the area A, which can also be expressed as (gamma hC plus P0) times A.

Outlines

00:00

📚 Hydrostatic Force Calculation on a Plane Surface

This paragraph introduces the concept of calculating the hydrostatic force exerted on a plane surface, specifically the bottom wall of a rectangular fish tank filled with water. It explains the need to determine the total force exerted on the tank walls to assess their structural integrity. The discussion begins with the resultant force on the bottom wall, considering both the atmospheric pressure and the pressure within the water. The resultant force is calculated by taking the difference between the upward force from the atmosphere and the downward force from the water's weight. The paragraph also touches on the complexities of calculating the force on the side walls due to the varying pressure with depth, setting the stage for a detailed analysis in subsequent sections.

05:00

🔍 Integrating Fluid Pressure to Determine Resultant Force on Side Walls

The second paragraph delves into the process of calculating the resultant force of fluid pressure on the side walls of a tank. It outlines the method of integrating the net force over the entire wall surface, taking into account the varying pressures both inside and outside the tank. The paragraph introduces the concept of the centroid and center of pressure, explaining their roles in determining the resultant force's magnitude and direction. The summary of the resultant force equation is presented, showing how it can be simplified using the depth of the centroid from the free surface. The paragraph concludes by highlighting the importance of understanding the difference between the vertical distance (hC) and the distance along the wall's orientation (yC), and how these measurements affect the calculation of the resultant force, especially in the context of a pressurized tank.

Keywords

💡Hydrostatic force

Hydrostatic force refers to the pressure exerted by a fluid at rest on an object submerged in it. In the context of the video, it is crucial for understanding how much force the walls of a fish tank must withstand due to the weight of the water. The script explains calculating this force on a plane surface, which is essential for determining if the tank can support the water without breaking.

💡Fish tank

A fish tank in this video serves as the primary example for explaining hydrostatic force. It is a rectangular container filled with water to a certain depth, and the script uses it to illustrate the concept of hydrostatic pressure and resultant force acting on its walls.

💡Atmospheric pressure (Pa)

Atmospheric pressure, denoted as Pa in the script, is the pressure exerted by the atmosphere at any given point. It plays a significant role in calculating the total force on the bottom of the fish tank, where the atmospheric pressure is added to the pressure due to the water's weight to determine the resultant force.

💡Specific weight (gamma)

Specific weight, symbolized by gamma in the script, is the weight of a fluid per unit volume. It is essential in calculating the hydrostatic force exerted by the water on the tank's walls, especially when determining the pressure at various depths within the tank.

💡Resultant force

Resultant force is the single force that can replace several forces acting in different directions while having the same effect. The script discusses calculating the resultant force along the bottom and side walls of the fish tank, which helps in understanding the overall force exerted by the water.

💡Centroid

The centroid is the geometric center of an object, and in the script, it is used to find the center of pressure on a section of the tank wall. The centroid's depth (hC) from the free surface is used to calculate the resultant force acting on the wall, illustrating a method to simplify complex force calculations.

💡Center of pressure (CP)

The center of pressure is the point where the resultant force due to fluid pressure acts on a submerged surface. In the script, it is used to determine the location (xR, yR) where the equivalent force and moment act on the wall, simplifying the analysis of forces on the tank's walls.

💡Coordinate system

A coordinate system is used in the script to establish a reference frame for analyzing the forces on the tank's walls. It is set up with x and y coordinates to help examine the pressure forces at different points on the wall and to integrate these forces to find the resultant force.

💡Integrate

Integration is a mathematical process used in the script to sum up the small pressure forces (dFnet) acting on all points of the wall to find the resultant force. It is a key step in calculating the total hydrostatic force exerted by the water on the tank's walls.

💡Gage pressure (P0)

Gage pressure, denoted as P0 in the script, is the pressure relative to atmospheric pressure. If the tank is pressurized, the resultant force equation must be adjusted to account for this additional pressure, demonstrating the flexibility required in fluid mechanics calculations.

💡Free surface

The free surface is the upper surface of a liquid that is not subjected to any external pressure other than the atmosphere's. In the script, it is the reference level from which depths are measured, and the pressure at the free surface is considered equal to atmospheric pressure.

Highlights

Deriving the equation for calculating hydrostatic force on a plane surface.

The necessity to determine the total force exerted on fish tank walls to assess structural integrity.

Resultant force on the bottom wall is influenced by atmospheric pressure and water depth.

Atmospheric pressure and water depth create a downward force on the tank's bottom wall.

Resultant force calculation simplifies by canceling atmospheric pressure terms.

The challenge of calculating resultant force on side walls due to varying pressure.

Exterior and interior pressure forces on side walls and their impact on resultant force.

Integration of fluid pressure along the wall to find the resultant force.

Use of trigonometry to express depth h in terms of y and the wall's angle theta.

The concept of centroid and its role in calculating resultant force on a wall section.

The center of pressure and its equivalence to the resultant force of the fluid pressure field.

Expression for resultant force FR in terms of gamma, sin(theta), yC, and area A.

Rewriting the resultant force equation using the pressure at the centroid PC.

Equations' applicability to flat surfaces exposed to the same pressure.

Understanding the difference between vertical distance hC and distance yC along the wall's orientation.

Adjustment of resultant force equation for pressurized tanks with gage pressure P0.

Transcripts

In this lesson, we will derive the equation for calculating the magnitude of the hydrostatic

force on a plane surface.

Here we have a rectangular fish tank filled with water to a depth d.

In order to determine whether the fish tank walls can support that much water without

breaking, we would need to know the total force exerted on the walls.

First we will determine the resultant force along the bottom wall.

The tank is surrounded on all sides by the atmosphere, which has a pressure Pa.

On the underside of the bottom wall, which is exposed to the atmosphere, the pressure

force is Pa times the area of bottom wall, and this force points upward.

Inside the tank, the pressure force increases as one descends through the water, reaching

a maximum at the bottom of the tank.

The absolute pressure on the upper part of bottom wall at depth d is the atmospheric

pressure plus the specific weight of the water times the depth.

This creates a downward force on the wall equal to the Pa plus gamma d, times the area

of the bottom wall.

If we examine just the fluid pressure forces acting along the bottom wall, the resultant

force is Pa times the area of the bottom wall, minus the quantity Pa plus gamma d, times

the area of the bottom wall.

The atmospheric pressure terms cancel out and we are left with negative gamma times

d times the area.

This means the resultant force points downward with a magnitude of gamma times d times the

area.

Finding the resultant pressure force along the bottom wall was relatively easy because

the pressure is constant along the entire wall.

This is not the case for the side walls.

The pressure along the exterior of the side walls is Pa and the related pressure force

is Pa times the area of side wall.

However, on the interior of the side walls, the pressure increases linearly with depth.

At a depth h, the pressure is the atmospheric pressure plus gamma h.

Multiplying by the area of the side walls gives the related pressure force.

We want to determine the magnitude of the resultant force caused by the fluid pressure

on the side walls.

Here is a different tank.

This tank is open to the atmosphere and partially filled with a liquid of specific weight gamma.

We are going to examine a section of the flat wall, highlighted in pink, that has an arbitrary

shape and which is oriented at an arbitrary angle theta relative to the free surface.

We set up a coordinate system where the x-coordinate comes out of the screen and is oriented along

the free surface.

The y-coordinate is oriented along the section of wall we wish to examine.

We will need to examine some important points on the side wall, so rotate the surface 90

degrees about the y-axis to get a better view.

Point (x,y) is the location of some arbitrary small area dA on the wall.

dA has an area dx times dy, and is at depth h from the free surface.

Point C is the centroid of the wall and is located at coordinates (xC, yC).

The depth of the centroid from the free surface is hC.

The centroid is the geometric center of the wall, and the centroid and center of mass

are the same if the density of the wall is constant.

Point CP is the center of pressure, which is where the resultant force FR acts.

The coordinates of the center of pressure are labeled (xR, yR).

The resultant force and center of pressure location produce an equivalent force and moment

on the wall as the original fluid pressure field.

In order to find the resultant force over the entire highlighted surface, we will start

by examining the force at the location of dA.

The small pressure force on the exterior of the wall at area dA, labeled dFext, is the

atmospheric pressure Pa times the area dA.

The small pressure force on the interior of the wall at area dA, which is labeled dFin,

is the pressure at dA times the area dA.

The pressure at that depth is the atmospheric pressure plus gamma h.

Using trigonometry, h can be rewritten as y times sin(theta).

The net force due to fluid pressure at dA is dFinterior minus dFexterior.

After plugging in the expressions for the two forces, and eliminating Pa times dA, we

obtain gamma times y sin(theta) times dA.

To find the resultant force of the fluid pressure on the entire wall, we integrate dFnet along

all points of the wall.

Plug in the expression for dFnet, and pull gamma sin(theta) out of the integral because

they are constant.

The integral of y dA can be simplified if we recall the definition of the centroid.

The y-coordinate of the centroid, yC, is the integral of y dA, integrated over the entire

surface, divided by the entire surface area A.

We now have an expression for the resultant force FR.

FR is equal to gamma times sin(theta) times yC times the area.

yC times sin(theta) is equal to the depth of the centroid, hC, so we can rewrite the

resultant force equation as gamma times hC times A.

This form of the equation is often more convenient to use when solving fluid mechanics problems.

Gamma hC is the pressure at the centroid, PC, so we can further rewrite the resultant

force equation as the pressure at the centroid PC times the area.

These equations are valid for any flat surface in which the free surface of the liquid and

the side of the tank are exposed to the same pressure.

In this case, they are both exposed to the atmospheric pressure.

We are free to use either yC or hC to find the resultant force and it is important to

understand the difference between the two quantities.

hC is the vertical distance from the free surface to the centroid.

yC is the distance from the free surface to the centroid as measured along the orientation

of the wall.

If the wall is oriented at an angle less than 90 degrees relative to the free surface, hC

will be shorter than yC.

If the wall is vertical and theta is 90 degrees, hC and yC will be the same distance.

In both cases, the distance yR will be greater than yC.

Although not proven in this video, if the tank is pressurized to a gage pressure of

P0, the equation for the resultant force would need to be modified to (gamma sin(theta) yC

plus P0) times the area A, which can be rewritten as (gamma hC plus P0) times A.