4. Finding yR and xR

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. So, FR is actually nothing but the weight of the overlying fluid.

In fluid statics, the resultant force (FR) acting on a surface submerged in fluid is found by integrating the pressure across that surface. The pressure (p) varies with depth and is given by the formula p = ρgh, where ρ is the fluid density, g is the acceleration due to gravity, and h is the height of the fluid column above the surface. If we assume p is uniform over the area A, we can simplify this integral to F = pA. This resultant force originates from the weight of the fluid above the surface acting downwards.

Think of this as a column of water in a swimming pool. The deeper you are in the pool, the heavier the water above you, which pushes down. If you were to measure the force on the bottom of the pool, you'd notice it corresponds to the weight of all the water directly above that point.

So, we also need to find out X R. What is the logical way? We will do the summation of moment around y axis now. For X R we need to do the moment calculation around y axis so around this axis. How? See, X R into F R. Where is X R? This is X R.

To calculate the x coordinate of the center of pressure (xR), we perform a moment analysis around the y-axis. The resultant force (FR) is multiplied by the x-coordinate of the center of pressure (xR) and this must equal the integral of x multiplied by the differential force (dF) across the area. Thus, we establish the relationship that xR can be found as an integral divided by the total area A. This concept is built upon understanding how distributed forces act on a surface and how they balance out at a point of interest.

Imagine balancing a seesaw. The position where the seesaw balances represents the center of pressure. Each side of the seesaw exerts a force, and depending on how far you sit from the center, the seesaw either tips or stays level. This is similar to how the different pressures vary at depths lead to a resultant force acting not exactly at the centroid, but shifted based on the distribution of pressures.

For finding y R the moment equilibrium should be done around x axis. So, the way we do it y R, F R. So, c y R into F R, this will be equal to integral y dF.

To determine yR, the y-coordinate of the center of pressure, the moment equilibrium around the x-axis is used. The equation setup involves the resultant force (FR) multiplied by the distance from the x-axis to the center of pressure (yR) being equal to the integral of each little force (dF) times its distance from the x-axis (y). This leads to a solution using the second moment of inertia about the x-axis and the area.

Consider a large swing with kids sitting at varying distances from the pivot point. If you want to determine the exact point where the swing will balance, you need to consider not just how many kids there are, but their positions as well. The moments created by each kid need to sum up to zero for the swing to be stable, just like we balance moments in our calculations.

So, this is an important result to note down, F R is equal to gamma A h c. This equation of F R suggests that the magnitude of the resultant force is equal to the pressure at the centroid multiplied by total area.

This highlights how the pressure at the centroid of the submerged area plays a critical role in determining the resultant force (FR). By establishing the equation FR = γAh_c, where γ is the specific weight of the fluid, A is the area, and h_c is the vertical distance from the fluid surface to the centroid, we comprehend how forces act on an area below the fluid's surface. Understanding where the centroid is and how deep it is submerged informs how we measure resultant forces in different orientations.

Picture a boat floating on water. The deeper it sits (submerged more), the more water weight it must displace, leading to a greater upward buoyant force. This parallels how the depth of the centroid affects the force acting on underwater surfaces.

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