Tricky Tipping Problem; 11 sided figure

  • Thread starter Thread starter tachu101
  • Start date Start date
  • Tags Tags
    Figure
Click For Summary
SUMMARY

The problem involves determining the tipping angle of a regular 11-sided polygon (hendecagon) with sides measuring 1.32 meters each on an incline. The calculated tipping angle is 16.4 degrees, derived from the center of mass analysis, where the interior angle of the polygon is 147.27 degrees. Additionally, the area of the cross-section is calculated to be 16.3 m² using the formula from Wikipedia, which states that the area equals 9.356 times the square of the side length. Confirmation of these calculations is encouraged through further derivation of the area formula.

PREREQUISITES
  • Understanding of center of mass principles
  • Knowledge of polygon geometry, specifically for an 11-sided figure
  • Familiarity with trigonometric functions and calculations
  • Ability to apply area formulas for regular polygons
NEXT STEPS
  • Learn how to derive the area formula for regular polygons
  • Study the properties of isosceles triangles in relation to polygons
  • Explore trigonometric functions relevant to angles and center of mass
  • Investigate the implications of incline angles on stability in physics
USEFUL FOR

Students studying geometry, physics enthusiasts analyzing stability, and educators seeking to explain polygon properties and center of mass concepts.

tachu101
Messages
74
Reaction score
0

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.
 
Physics news on Phys.org
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.
 

Similar threads

Replies
4
Views
3K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
1
Views
2K
Replies
11
Views
2K
Replies
1
Views
2K