1.4.1. Square Gate Problem

Let us start to solve the problems, these very easy problems that there is a square gate of dimensions of 1.5 meter into 1.5 meters. One of the vertical sides of a fully filled water tank has one side on the free surface. It is hinged on the lower horizontal sides. Here it is hinged and is held in position by force applied on the vertical central line at a depth of 0.75 meter below the free surface.

In this problem, we have a square gate measuring 1.5 meters by 1.5 meters attached to a water tank. The water tank is filled to the top, creating pressure against the gate. The gate is fixed at the bottom and can rotate around this hinge. A force is applied to keep the gate in place, and this force is applied at the center of the gate, which is located 0.75 meters below the water's free surface.

Imagine a large door (the gate) that swings open at the bottom (the hinge) and is pushed down by water pressure. The moment you fill a pool with water, the pressure against the sides of the pool increases, causing them to push outward, similar to how the water pushes against this gate.

So this is 1.5 meters and 0.75 meters below means 0.75 meter from the bottom, from the hinge. This is what 0.75 meters, total is 1.5 meters. So what could be the magnitude of this force? What could be this force part. So this is the force. You can easily solve these problems. The problem is that we need to know hydrostatic pressure distributions. Then we can compute the force due to the hydrostatic distribution.

To find the magnitude of the force needed to keep the gate in place, we first need to understand how the hydrostatic pressure acts on the gate. Hydrostatic pressure increases with depth. At 0.75 meters below the water surface, we can calculate the hydrostatic pressure using the formula: Pressure = density × gravity × height.

Think of how much effort it takes to keep a heavy door closed when someone is pushing against it. The deeper you go into the water, the harder this push gets—even more so if you try to keep the door closed while the water tries to push it open.

Once I know what is the force is acting because the fluid is at rest and where it acts the force that the locations. If I know the force, the center of pressures or the force where is acting it, if I know that to the force magnitudes or the at the locations where the force acts then I can take a moment at the hinge locations to compute what will be the force component.

Next, we calculate the force acting on the gate due to the hydrostatic pressure. The force acts at a specific point known as the center of pressure. To maintain equilibrium, we then take moments about the hinge, applying the principle that the total moments about the hinge must equal zero. This lets us solve for the unknown force required to keep the gate closed.

Imagine trying to balance a seesaw. If one side is heavier, you need to apply a force on the lighter side to keep it level. Here, the weight of water pushes against the gate, and we must counterbalance it by estimating where the force acts to keep everything in balance.

Net Force exerted by fluid = (½) x 1.5 ρg x 1.5 x 1.5 F = ρ Taking Moment about hinge (1.5/3) x F = F x 0.75 F = 11.036 KN. As you know in a triangular pressure distribution diagrams the resultant force acts at a one third distance from the bottoms or two-third distance from the free surface.

In summary, we calculated the net forces acting on the gate using hydrostatic pressure principles. The resultant force from the triangular pressure distribution acts at one-third the height of the gate from the bottom. By applying this to the moment about the hinge, we find the force required not just to support the gate but also hold it in place against the water pressure.

Consider a large spoon you are trying to hold underwater. The deeper you push, the more effort you need to keep it steady. Similarly, we calculated that to keep our gate in position, a specific force equivalent to about 11 kN is necessary to counteract the water's push.

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