1.1. Basics of fluid mechanics-I(Cont.)

Okay. Welcome back. So, this is the again starting with the last slide from the last lecture and we figured out what was the average depth of the block and that came out to be, you know, the average depth was 3 blocks here, what where does that average occur? It was at block number 5 this average and we found out the resultant and the resultant came this. Okay.

In the last lecture, we discussed the concept of average depth related to fluid mechanics, specifically focusing on a scenario involving blocks of fluid. We identified that the average depth of the fluid was found to be 3 blocks deep at a certain position, specifically block number 5. This understanding of average depth is crucial as it relates to how fluid behaves under certain conditions.

Think of a swimming pool divided into sections like blocks. If we fill the pool to different levels at different sections, the average depth would be like finding a middle ground to describe the overall height of water, which affects how things float or move in the water.

So, now we proceed with one of the real life examples of moment. So, the question is, an elliptical gate covers the end of a pipe 4 meters in diameter. If the gate is hinged at the top here, what normal force F applied at the bottom of the gate is required to open the gate when the water is 8 meters deep above the top of the pipe, okay, so this, and the pipe is open to atmosphere, okay, on the other side, here. Neglect the weight of the gate, okay.

This example illustrates how physical forces interact in fluid mechanics. The scenario describes an elliptical gate at the end of a pipe. When water is 8 meters deep above the gate, the pressure exerted not only pushes down on the gate but also creates a moment about the top hinge. To find the force required to open the gate, one must consider the pressure at various points and how this translates into a force/torque around the hinge point.

Imagine trying to lift a heavy lid from a pot filled with water. The deeper the water level, the harder it becomes to lift the lid. You can use an object (like a stick) to push the lid from below to help open it, symbolizing how the normal force F acts at the base of the gate.

So, I have kept this figure on the right side. So, that you are able to follow, what we actually we are going to do. So, what is the pressure datum? This is the atmosphere, okay. So, the resultant force is going to be the pressure at the centroid into area that we have seen in the derivation in last lecture. What is the area? Area of the ellipse that we have seen is pi into a b, okay.

To calculate the resultant force acting on the elliptical gate, we begin by determining the pressure at the centroid of the elliptical shape influenced by the water above it. The force exerted by the fluid is calculated using the formula: Resultant Force = Pressure at Centroid × Area. The area of the elliptical shape is found using the formula πab, where 'a' and 'b' are the semi-major and semi-minor axes of the ellipse.

Consider measuring the force on a large trampoline pushed down by people jumping on it. The weight of the jumpers causes a pressure-related force that spreads across the trampoline's surface, much like how water exerts pressure on the surface of a submerged gate.

So, we need to find out y R and x R, y R is given by y c, correct. So, I x c / A from the set of the areas that we had seen in the shown in the last lecture is given as a square / 4. If you are confused, you can go and see it in the last lecture notes.

Once the resultant force has been calculated, the next step is to determine where this force acts. The vertical location (yR) can be found using the formula yR = yc + (IxC/A), where yc is the height of the centroid, IxC is the moment of inertia about the centroid, and A is the area. This helps in understanding how far above the hinge line the resultant force acts.

Think about a seesaw. If you sit at a certain distance from the pivot point, your weight creates a moment that determines how easy it is for someone else to lift the other end. In our case, the location of the resultant force helps define how difficult it will be to open the gate.

Now, the most important part is force that is required to open the gate. So, what all do we know? See, y R is 0.125, right. So, this will become 2.5 + 0.125 that will be 2.625, correct. And this is 2.5 meters, this is 2 and this is total length l total.

After finding yR, we need to calculate the force required to open the gate. The total length (l total) plays a crucial role here. By taking moments about the hinge and relating the force at the bottom of the gate to the resultant force and lengths, we can derive the required force. Moments are all about the balance of forces around a pivot.

Imagine opening a heavy door. If you push it far from the hinges, it’s easier to move than if you push it right at the hinge. Using this concept helps understand how the distance from the hinge affects the force required to open the gate.

So, we also need to, you know, to wind up the, you know, the forces on the plane surfaces. Some of the important results, the average magnitude of the pressure force is the pressure at the centroid.

In fluid mechanics, it's important to consider both vertical and horizontal forces acting on surfaces. The average pressure force acts at the centroid, with horizontal components determined by the pressure at the centroid multiplied by the area, and vertical components related to the weight of the liquid above a surface.

Think of a large balloon under water. The water exerts pressure on all sides. The upward force acting as buoyancy, which can lift the balloon, depends on how much water is above it—a similar concept to how we can calculate forces on structures underwater.

So, now we need to calculate the pressure on the curved surface. So, this surface here, you see this, okay. So, how? We need to know the horizontal component, we need to find out the vertical component because it is curved, right.

When calculating pressure on a curved surface, we must consider both the horizontal and vertical components of force due to the gradient of pressure that changes with depth. The vertical component is particularly important as it corresponds to the weight of the liquid column above the curved surface affecting the pressure exerted.

Visualize a curved waterslide. As you go further down, the pressure from the water above increases. Understanding how this pressure accumulates helps explain both the forces acting on the slide and the experience of sliding down!

To sum the vertical component of the pressure force on a curved surface is equal to the weight of the liquid vertically above the curved surface and extending up to the surface where the pressure is equal to the reference pressure.

In summary, the vertical force acting on a curved surface relates to the weight of the liquid above it, extending to the point where the pressure equals the reference pressure, typically the atmospheric pressure at the surface of the fluid. This principle helps us assess forces on structures like dams or underwater pipes.

Imagine pressing down on a sponge soaked in water. The weight of the water 'sitting' on the sponge creates a downward force—but that force is balanced by the upward force from the sponge itself as it tries to push back against the pressure in the water.

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