1.4. Buoyant Force

So, this is a fluid statics 2, I mean, I call it 2, because we are going to do the surface forces and body forces therefore we need to know, what are we going to study in fluid statics 2. So, in statics 2 its static surface forces, we are going to see force on plane areas. We are going to see force on curved surfaces, like this. This is plane areas, this is plane, this is curved surface and we also will see the buoyant force, a very small detail of it, but I think this is very necessary.

In this section, we are introduced to the concept of buoyant force within the context of fluid statics. The instructor emphasizes the importance of understanding surface forces (forces acting on surfaces of various shapes) and how they relate to body forces (forces acting within the fluid). Buoyant force, a critical aspect of fluid mechanics, arises due to the pressure difference acting on different points in a fluid and will be discussed in the context of plane and curved surfaces.

Imagine a balloon filled with water. When you push it underwater, you can feel the pressure from the water trying to push it back up. This sensation is similar to how buoyant force works—it's the upward push you feel due to the water's pressure.

So, we have to see what are the forces on plane areas, that is horizontal surface. So, if you see, this is a figure that shows, you know, a horizontal surface a depth h, okay. So, this h is the vertical distance to free surface and this what is the P here, okay. And what is the resultant force at the bottom, okay, and P we are assuming 500 kilo Pascal's, okay, that we are going to see.

This chunk discusses how pressure acts on horizontal surfaces submerged in a fluid. The key concept introduced is the depth (h), which influences the pressure exerted on the surface. The pressure at this depth can be quantified using a specific value (in this case, 500 kPa), laying the groundwork for further calculations regarding resultant forces acting at the bottom of a fluid container.

Think of a swimming pool. The deeper you go, the more water is above you. This increases the pressure you feel at deeper levels. If you dive to the bottom of a pool, the pressure on your ears demonstrates how pressure increases with depth.

So, the force resultant force is going to be the integration of pressure into area, So, p is constant so it comes out and that becomes pressure into area pA, where p is rho gh, okay, this is the gauge pressure. So, F = ∫pdA = p∫dA = pA.

In this section, we learn how to calculate the resultant force acting on a submerged surface by integrating pressure over the area of that surface. Since pressure (p) can be expressed as ρgh (where ρ is the fluid density, g is gravitational acceleration, and h is the depth), we can simplify the computation by recognizing that p is constant when considering a specific depth. Thus, the resultant force (F) is essentially the product of pressure and the area (A) of the surface.

Imagine you have a flat board resting at the bottom of a pool. If you were to calculate the total pressure acting on that board, you'd think of how deep the board is and the size of the board itself. The pressure multiplied by the board's area gives you the total force pushing up on it!

F passes through the centroid of the area. This is an important information for you. And therefore, the change in pressure can be equated to rho into a, okay, or we can write in x ∂p/∂x = ρa = 0.

This chunk elucidates additional important characteristics of resultant force. It's stated that the resultant force acts through the centroid of the submerged area, a crucial insight when analyzing forces on surfaces. This also emphasizes that, for forces acting in the horizontal direction (x direction), there is no change in pressure occurring.

Think of balancing a pencil on your finger. The point at which your finger can hold the pencil with the least effort is right at its center — similar to how the resultant force acts at the centroid of a submerged surface in fluid.

Another important thing is, we have to learn and revise again, what are the forces on the plane areas or the inclined surface. So, this has to be taken in a little bit of more detail. What will be the direction of the force? Always perpendicular, normal to the plane, right.

In this part, the focus shifts to inclined surfaces. The instructor highlights that, similar to horizontal surfaces, the resultant force still acts perpendicular to the surface of the area. Importantly, the pressure is no longer constant as it varies with depth along the inclined surface. This means we need to integrate the pressure over the area to find the resultant force precisely.

Think of a water slide. If the slide is angled and you’re sitting on it, the force (due to water) will always push straight out perpendicular to the slide. Just like how water pressure acts differently on different points on an inclined surface.

What will be the magnitude of the force? We have to integrate the pressure over the entire area. The moment of the resultant force must be equal to the moment of the distributed pressure force.

This chunk explains the process involved in determining not just the magnitude of the resultant force but also its line of action. To accurately describe the resultant force on an inclined surface, we must consider the moments caused by the distributed pressure over the area. This entails a balance, which allows us to find precise coordinates for where the resultant force acts.

Consider trying to balance a seesaw with varying weights on either side. Understanding how the weight creates moments about the center allows you to predict how to position yourself for balance — similar to how determining moments on surfaces helps find the resultant force in fluids.

The point through which the resultant force acts is called the center of pressure. This is you might have heard in your fluid mechanics class what calculate the center of pressure. So, this is what the center of pressure is a quick revision for you again.

In this conclusion, the center of pressure, the point of application of the resultant force, is highlighted as a critical concept. Unlike the centroid, the center of pressure will generally not coincide with it because pressure increases with depth, affecting where the force effectively acts.

Think of pushing a shopping cart. If all the weight is distributed evenly, it’s easy to push from the center. However, if one side is heavier, you'd have to push from a different spot to maintain balance — much like how the center of pressure is affected by the distribution of weight in a fluid.

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