How Does Centroid Affect Hydrostatic Pressure Calculation?

[Jbreezy]

Homework Statement

Can you please help me I think I have confused something.We did this in class.
A triangular plate with base = 10 m and height 12 m is placed under water with the base at the top at a depth of 8 meters under water and the apex 20 meters under water. Find the hydrostatic force on one side of the plate.

Homework Equations

OK So I know Pressure = density x gravity x height. So in this case 9800 x 12 = 117600
The area is (10 x 12) / 2 = 60
This is were I'm confused F = P x A = 705600 N Which is what you get if you do the 60 times 117600
But I'm so confused where does centroid come into this problem? I thought it was supposed to be the pressure acting on the centroid So in the first equation part where I had Pressure times gravity time height I think it should be pressure x gravity x (1/3 the height of the triangle) Because all the pressure acts at that one point.

The Attempt at a Solution

 

[HallsofIvy]

Your error is that the pressure of the water at each depth is not the same. Force= pressure times area is true only if the pressure is the same at each point. The pressure on a surface, at depth, d m below the level of the water, is the weight of a column of water above a square of 1 cm by 1 cm d m height (divided by the area of that square so the units are weight per area). One thing we need to know to answer this question is the orientation of the triangle. One thing we need to know about this is the actual geometry of the triangle. You say it has "base 10m and height 12m" but don't say whether this is an isosceles triangle- that can affect the answer.

Assuming it is an isosceles triangle we can set up a coordinate system with origin at the vertex. Since the 10 m base is 12 m above the vertex, the other two vertices are at (-5, 12) and (5 12). Two sides are a line from (0, 0) to (-5, 12) which has equation y= -(12/5)x and y= (12/5)x. At each height, y, then, a line from one side to another extends from x= -(5/12)y to x= (5/12)y, a length of (5/12)y- (-(5/12)y)=(10/12)y= (5/6)y. The height from y to the top of the triangle, because y is measured from the vertex, is 12- y and, because that base is 8m below the top of the water, the depth, at each y, is 20- y. So the weight of water on a one cm by one cm square is the density of water times 20- y. Integrate that as y goes from 0 to 12.

 

[Jbreezy]

No I'm trying to do it using centroids. How do I do it using that ? I need to know how to do it with centroids. I wrote the problem exactly as is. I"m not sure what type to triangle it is sorry. See my issue is ow to incorporate the centroid into this. I know that the pressure at each depth is not the same. Which is why I want to use the centroid of the triangle seeing that it is symmetric. ... I have a note from class that says F = P time A times Centroid. But when I went back over the example I don't see how they incorporated that centroid into the answer. I feel like all they did was force times area times pressure. Which is maybe what you are referring to as my mistake of saying the pressure acted the same everywhere? I'm not sure.

 

[Jbreezy]

Why can't it be Force = (density x gravity x height of centroid) x area of triangle ?
Where Pressure = (density x gravity x height of centroid ) ? Why can't this work?
So it would be

F = 9800 x (4 for height of centroid ) x ( (10 x 12) / 2 ) = 2352000