- #1
1MileCrash
- 1,342
- 41
Homework Statement
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
Homework Equations
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
[itex]\int^{4}_{0} 64y(1.3(4-y)) dy[/itex]
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.