6. Properties of Areas

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 we 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. So, what is the force on the bottom of this tank of water actually, what is the net force on the bottom of this tank?

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 fluid statics, we need to understand the forces acting on plane areas. For a horizontal surface at a certain depth (h), the pressure exerted is influenced by the weight of the fluid above it. Here, P is given a value (500 kPa), indicating the pressure at that depth. The net force (F) acting at the bottom is calculated by integrating the pressure over the area of contact, which simplifies to the product of pressure (p) and area (A). The pressure is generally calculated using the formula p = ρgh (density * gravitational force * height). This relationship shows how the total force at the bottom of a fluid container is directly proportional to both pressure and area.

Think of a swimming pool filled with water. As you go deeper, you can feel the pressure increases against your body. This happens because there is more water above you pushing down. If you were to measure the pressure at different depths with a gauge, you would see it follows the formula for pressure (P = ρgh). Now, if you calculate the total force at the bottom of the pool (like pushing down on your feet), it's determined by how deep the water is (h) and how much space it's taking up (the area of the pool's bottom).

Another important thing is, we have to learn and revise again, what are the forces on the plane areas or 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. So, the force will start acting like this, correct. What will be the magnitude of the force? We have to integrate the pressure over the entire area. Here, the pressure is no longer constant, because it is not it is not at one elevation it is varying see, the h is changing here, here it is different, here it is different, here it is different.

When considering forces on inclined surfaces, it's essential to recognize that these forces act perpendicular (normal) to the surface at all points. Since the pressure depends on depth, it varies along the inclined surface. This means instead of a single pressure value, we must integrate the pressure across the surface area, accounting for the changes in depth (h) at different points along the incline. The integration process calculates the total force acting on the inclined surface by summing all the small forces resulting from varying pressures.

Imagine a sloped waterslide. Water is flowing down the slide at an angle – the pressure felt by someone lying on the slide will not be the same all over their body. At the top part of their back, there's less water above them, and thus lower pressure. As you reach the bottom, the cushion of water above you is heavier, resulting in greater pressure. To feel how much force is acting, you'd have to consider the entire surface area against the water slide, not just one point. This explains why we integrate forces over the entire surface.

So, the question, the biggest question is you have to determine the location direction and magnitude of the resultant force acting on one side of this area due to the liquid in contact with water. If you see the body, okay. I will just erase because this was just to you know, okay, alright. What we say, let the plane in which the surface lies, intersect the free surface. So, this is the free surface here, okay, and let the plane in which the surface lies the body intersect at point O, okay, right. Good.

To determine the resultant force acting on any area submerged in fluid, you'll need to know the force's magnitude, direction, and point of application. This is typically done by understanding how the submerged surface interacts with the fluid above it, represented here as the 'free surface' where the fluid meets the air. From the intersection of the area (let’s say point O), pressures from the surrounding fluid act differently depending on the depth and contribute to the overall resultant force. The focus is on calculating this resultant force both in terms of its direction (normal to the surface) and where it acts, often determined through the centroid of the area submerged.

Visualize a large poster that's partially submerged in a swimming pool. The water creates pressure on the surface of the poster. To find out how much push (force) the water is applying on the poster, you need to look at where the water is deepest (and thus, pushing harder) and how much area of the poster is covered by water. The deeper it is submerged, the more water pressure acts on it. By pinpointing where the center of the push occurs (the centroid), we can accurately assess the total force acting on the poster.

F R is actually nothing but the weight of the overlying fluid, okay. Also F is normal to the surface and towards the surface, if p is positive, okay. 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 direction, - ∂p/∂x = ρa = 0.

The resultant force (F_R) acting on a submerged area corresponds to the weight of the fluid sitting above it; this means it is a direct result of the water’s pressure acting down on the area. This force acts perpendicular to the surface and its line of action will pass through the centroid of the submerged area. It's pertinent to note that changes in pressure are typically assessed in the vertical direction (z), as fluid pressure does not vary horizontally (x direction) under static conditions. Thus, we can express conditions indicating that there’s no acceleration in the horizontal plane.

Think about a balloon submerged in water. The balloon experiences a force pushing it up, equal to the weight of the water above it. This push acts straight up from the centroid of the balloon, which is like the middle of the balloon. If we wanted to check if more water (and thus more weight) was exterting down from above, we could calculate the pressure based on how deep it is underwater. In essence, the balloon's position in the water creates a balance of forces based on the height of the water (pressure) above it.

So, the way we do it y R, F R. So, c y R into F R, so this is y r into F R, this will be equal to integral y dF, right. So what is y dF?. So this is dF and this is y, so all the summation beginning from here, until here, all these small small dF that is there.

To find the coordinates of the center of pressure, we need to sum the moments created by the force acting on the submerged area. The coordinate for the center of pressure in the vertical (y) direction (Y_R) is related to the equilibrium of moments around the x-axis. We can calculate this by equating the moment due to the resultant force (F_R) and the integral of the moment due to differential forces (dF) over the area. The value of Y_R doesn't just depend on simple coordinates but involves integrating the moments of these dF forces that serve to represent small contributions to the overall force applying on the structure.

Consider a seesaw with kids sitting at various distances from the center. Each child represents a differential force (dF) acting on the seesaw due to their weight. If we want to find out the point where it balances (the center of pressure), we calculate the moments created by each child about the center and see where it tips. By summing the effects of each child's weight based on how far they are from the center, we can determine how to balance the seesaw correctly.

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