- #1

JohnnyLaws

- 10

- 0

- Homework Statement
- We have a uniform square of 15 unit of lenght and a triangular hole in it. This triangle is known to be isosceles, and the distance between the center of the square and its edge is 7,5 units, which is equivalent to the height of the triangle. I'll show the image.

- Relevant Equations
- I think I need to use this equation:

(∫xdA)/A

Integral limits and derivation of dA will be approached in the next section.

Essentially, I'm attempting to calculate the center of mass of the square, taking into account that the origin for the center of mass is the bottom left corner of the square, while the origin for the center of mass of the triangle is shifted 7.5 units in the x-direction and remains at y = 0 units relative to the other origin.

Here it is the image of the statement:

As I mentioned in the "relevant equations" section, my approach to solving this exercise involves calculating the difference between the centers of mass of the square and the triangle.

Starting with calculation of center of mass for the square.

Starting with the initial equation for the x-coordinate:

When making small changes in x (dx), we can express dA in terms of dx. In other words, we can write: dA = dx * y.

So, we can rewrite the equation for xc:

Since y in this case is a constant equal to 15, we can simplify the equation to:

Now for yc of the square:

When considering a small change in y (dy), we can express dA in terms of dy. In simpler terms, we can rewrite it as: dA = dy * x

Since x in this case is a constant equal to 15, we can simplify the equation to:

So as we expected for a homgeneous square, the center of mass is in the middle.

Therefore, xc = yc = 7.5.

I'm not certain about what needs to be done for the triangle. I'm unsure how to parametrize this figure. Am I on the right track or is everything I'm thinking incorrect?

I know that's a simple exercise but I can't find anything on internet.

By the way, the solution for this exercise from the center of the square is: xc = 2.07 and yc = 0.

As I mentioned in the "relevant equations" section, my approach to solving this exercise involves calculating the difference between the centers of mass of the square and the triangle.

Starting with calculation of center of mass for the square.

Starting with the initial equation for the x-coordinate:

When making small changes in x (dx), we can express dA in terms of dx. In other words, we can write: dA = dx * y.

So, we can rewrite the equation for xc:

Since y in this case is a constant equal to 15, we can simplify the equation to:

When considering a small change in y (dy), we can express dA in terms of dy. In simpler terms, we can rewrite it as: dA = dy * x

Since x in this case is a constant equal to 15, we can simplify the equation to:

So as we expected for a homgeneous square, the center of mass is in the middle.

Therefore, xc = yc = 7.5.

I'm not certain about what needs to be done for the triangle. I'm unsure how to parametrize this figure. Am I on the right track or is everything I'm thinking incorrect?

I know that's a simple exercise but I can't find anything on internet.

By the way, the solution for this exercise from the center of the square is: xc = 2.07 and yc = 0.