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01SpAcE01
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Member warned about posting with no effort and no template
RUber said:What are you using to determine the width of the triangle? Is that just some constant times the depth, essentially the pressure from the column of water?
Does the column have volume or are you just working with 2D? I don't know what physical constants to use, so I will call it C.
Your triangle is defined by one side which is 125mm high, and 125Cmm wide. The width at any given height h is C(125-h). You could either use calculus or geometry to find out the area contained between h = 0 and h= 10mm.
According to Pascal's Law, hydrostatic pressure is proportional to the depth of the fluid:01SpAcE01 said:Thanks for the response, sorry therre wasn't enough info. This is 3 dimensional. Needing to now how much force 2 litres of water will put on the highlighted wall of 10mm with a full water body height of 125mm.
01SpAcE01 said:What do you mean with integrate over the circular area? It's been too long since I did integration. Let's say the area of that 10 mm wall was 3x10^-4 m^2, could you guide me on calculating the approximate force applied aganst it form the water? -I'd really appreciate the help. I'm trying to manage too much at the moment.
Attached is a simplified model of the part.
RUber said:I think your first diagram may have been sufficient. The question is, what are you going to use as your constant?
If you have 2L = 2000 cm^3 of water, in something 125mm=12.5 cm high, you can assume that the cross-sectional area of the column is 160cm^2.
As SteamKing mentioned, gravity is 9.81 m/sec^2 and the density of water is 1 g/cm^3, so you still return to
P(h) = C (125-h), where C is the pressure per unit height of the column.
You are looking for the pressure from h = 0 to h = 10.
##\int_0^{10} C(125-h) = 125(10) - \frac{10^2}{2}##
Or, with geometry,
The width of the triangle at h= 10 is 115C.
The width of the triangle at h = 0 is 125C.
The rectangle with dimensions 10mm x 115Cmm has area ...
The triangle with height 10mm and base length 10C has area...
Total area = area of rectangle + area of triangle.
Both approaches lead to the same result. So...what is C?
RUber said:For the physical constant, check out what SteamKing posted. It will be based on the volume per unit height of the column, Gravity, and the density of water. You should have all those values.
Well, what are these other people telling you? How do you know they aren't wrong?01SpAcE01 said:How can it be that nobody is clear on this. I've had other people look at this and they are telling me different. Please someone clarify this.
To calculate the resultant force on a segment of submerged wall, you will need to first determine the pressure distribution on the wall. This can be done by multiplying the density of the fluid by the gravitational acceleration and the depth of the fluid at that point. Then, you will need to calculate the horizontal and vertical components of the pressure distribution. Finally, you can use the Pythagorean theorem to find the magnitude of the resultant force and use trigonometric functions to determine its direction.
The resultant force on a submerged wall is affected by several factors, including the density and depth of the fluid, the size and shape of the wall, and the angle at which the wall is positioned relative to the fluid flow. The presence of other objects in the fluid, such as other walls or obstacles, can also impact the resultant force on the wall.
The angle of the submerged wall can significantly affect the resultant force. If the wall is perpendicular to the fluid flow, the resultant force will be at its maximum magnitude. As the angle of the wall decreases, the magnitude of the resultant force will also decrease. If the wall is parallel to the fluid flow, there will be no resultant force acting on it.
Yes, the resultant force on a submerged wall can be negative. This can occur if the direction of the fluid flow is opposite to the direction of the wall, resulting in a negative horizontal component of the pressure distribution. In this case, the resultant force would be pulling the wall towards the fluid, instead of away from it.
No, the resultant force on a submerged wall is not the same at all points along the wall. It varies based on the pressure distribution, which is affected by the factors mentioned earlier. The resultant force will be at its maximum at the point where the wall is perpendicular to the fluid flow and will decrease as the angle of the wall changes. Additionally, the resultant force may be different at different depths of the fluid, depending on the pressure distribution at those points.