Tricky Tipping Problem; 11 sided figure

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To determine the tipping angle of a regular 11-sided polygon with 1.32-meter sides, the center of mass must be calculated to find when it exceeds the tipping point. The interior angle of the polygon is 147.27 degrees, leading to a derived tipping angle of 16.4 degrees. The area of the cross-section is estimated at 16.3 m² using a formula found online. A suggestion is made to derive the area formula by dividing the polygon into 11 isosceles triangles and applying trigonometry. The discussion emphasizes understanding the geometric principles behind the calculations.
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Homework Statement


A regular polygon of 11 sides is on an adjustable incline. The polygon has sides on 1.32 meters each. At what angle of the incline will the polygon tip over. Extra: what is the area of a cross section of the polygon.

The Attempt at a Solution


This problem has to do with center of mass. I have to calculate at what angle the center of mass will be over the 90 degree mark.

I found that each interior angle of the 11 sided figure is 147.27 degrees. I then don't know how to explain it but I took half of that angle and ended up with 73.63 degrees. Because the center of mass has to be at a right angle to tip over I then did 90-73.63= 16.4 degrees and I think that this is my answer.

The area of the cross section I think is 16.3 m^2 because I looked up an equation on wikipedia that said the area is equal to 9.356*(side length)^2

Can anyone confirm this.
 
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Right on both. You might try thinking about how to derive the wikipedia formula. It's not that hard. Split the polygon into 11 isosceles triangles with a vertex at the center. The central angle is 360/11. Do trig.
 
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