Calculating Center of Gravity Position for Hanging Rectangular Shape with String Attachment

I have been having difficulty in solving this problem set for the past few weeks. I had several attempts and I have attached the one I thinks feels correct but will love to be corrected. Thanks!

This is my attempt!

One step on the way is to find the position of the midpoint of the higher of the two 20 cm sides. From there, you can get to the center of gravity.

Jan Ragnar skrev:

Just wonder, how do you got the equation "y = 2 * 20cos(α) - 5√5cos(α + θ)"?

I think 1.5*20*cos(alpha) + 5*sin(alpha) is simpler.

"Rectangular disc" är ett underligt uttryck.

Thank you so much that helped a lot!

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