Answer:
Explanation:
In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.
We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.
The weigth depends on the size and specific gravity.
W = 1/2 * b * h * L * SG
Then
Teq = 1/2 * b * h * L * SG * b / 2
Teq = 1/4 * b^2 * h * L * SG
The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:
![T1 = \int\limits^h_0 {p(y) * sin(30) * L * (h-y)} \, dy](https://tex.z-dn.net/?f=T1%20%3D%20%5Cint%5Climits%5Eh_0%20%7Bp%28y%29%20%2A%20sin%2830%29%20%2A%20L%20%2A%20%28h-y%29%7D%20%5C%2C%20dy)
The term sin(30) is because of the slope of the wall
The pressure of water is:
p(y) = SGw * (h - y)
Then:
![T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy](https://tex.z-dn.net/?f=T1%20%3D%20%5Cint%5Climits%5Eh_0%20%7BSGw%20%2A%20%28h-y%29%20%2A%20sin%2830%29%20%2A%20L%20%2A%20%28h-y%29%7D%20%5C%2C%20dy)
![T1 = \int\limits^h_0 {SGw * sin(30) * L * (h-y)^2} \, dy](https://tex.z-dn.net/?f=T1%20%3D%20%5Cint%5Climits%5Eh_0%20%7BSGw%20%2A%20sin%2830%29%20%2A%20L%20%2A%20%28h-y%29%5E2%7D%20%5C%2C%20dy)
![T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy](https://tex.z-dn.net/?f=T1%20%3D%20SGw%20%2A%20sin%2830%29%20%2A%20L%20%2A%20%5Cint%5Climits%5Eh_0%20%7B%28h-y%29%5E2%7D%20%5C%2C%20dy)
![T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy](https://tex.z-dn.net/?f=T1%20%3D%20SGw%20%2A%20sin%2830%29%20%2A%20L%20%2A%20%5Cint%5Climits%5Eh_0%20%7B%28h-y%29%5E2%7D%20%5C%2C%20dy)
![T1 = SGw * sin(30) * L * \int\limits^h_0 {h^2 - 2*h*y + y^2} \, dy](https://tex.z-dn.net/?f=T1%20%3D%20SGw%20%2A%20sin%2830%29%20%2A%20L%20%2A%20%5Cint%5Climits%5Eh_0%20%7Bh%5E2%20-%202%2Ah%2Ay%20%2B%20y%5E2%7D%20%5C%2C%20dy)
T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)
T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)
T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)
T1 = 1/3 * SGw * sin(30) * L * h^3
To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)
1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3
In an equilateral triangle h = b * cos(30)
1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3
SG > SGw * 4/3* sin(30) * (cos(30))^2
SG > 1/2 * SGw
For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.
This is avergae specific gravity, including holes.