# Fluid force on a dam just trying to reason with the problem

1. Dec 5, 2011

### 1MileCrash

1. The problem statement, all variables and given/known data

A dam is shaped like a right triangle, 4 height and 6 across.

The right triangle is oriented so that the right angle is at the TOP RIGHT. (we are looking at the face of the dam)

Density is a weight density, and is 64 lb/cubic foot

2. Relevant equations

3. The attempt at a solution

OK, to work this out, am I allowed to just flip the dam over? If the pressure increases with depth, it decreases with height, and if I flip it over, it increases with height on the plane.

So I flipped it over.

Now, the area of a rectangular slice of the triangle (which must be horizontal slices so that tiny slices have constant pressure) have an area of delta y X width.

To find the width, I invoked the wrath of similar triangles. Since width is always 3/2 of height, and the height of a "triangle" made from the larger one is given by 4-y (because partitioning a triangle at y=4 would be nothing, while one at y=0 is the whole thing) so the width of the triangle is (3/2)(4-y)delta y

Since I flipped the dam, pressure increases with depth according to p = 64y

And since force = area X pressure,

64y(3/2)(4-y)delta y is the force on any slice of the dam.

Integrating over the full height of the dam, from 0 to 4, should give me total force.

Is

$\int^{4}_{0} 64y(1.3(4-y)) dy$

The total pressure on the dam? Do I roughly know what I'm doing?

Just seeing if I intuitively grasp it before actually reviewing it for final. Always helps me.

2. Dec 5, 2011

### 1MileCrash

For some reason the edit button is missing... integral should have 1.5 in it, not 1.3.