Solving for Max Height & Side Lengths of Triangle Inclined at 24 Degrees

[tachu101]

Homework Statement

Given a isosceles triangle with base 1.32 meters and other sides unknown; Given rectangle with base 1.32 meters. These polygons are on an incline of 24 degrees. What are the maximum height of the rectangle and the maximum side lengths of the triangle (not the base) without have the objects tip over.

Homework Equations

The incline angle and the center of mass at 90 degrees is what the question is asking for. So the incline (24 degrees) + (unknown angle) = 90 so right away I know that the angle must be 66 degrees. Then I think that some trig comes into play.

The Attempt at a Solution

If you take half of the rectangle (to make a triangle) you can find the height of the rectangle by doing 1.32tan66=height which would get 2.964 meters. (is this the maximum height)?

The triangle is more complicated. I did the same thing and broke the triangle in half so 1.32 becomes .66 for the base. This then goes into .66tan66= height from base to the center of mass (centroid) which comes out to 1.4823 meters .
Then I think that the length of the base to the centroid and the length from the centroid to the top of the triangle is in the relationship of 1:2. So I tripled the 1.483 to get 4.449 meters as the length of the altitude.
Finally I use the Pythag Therm to get 4.449^2+.66^2= 4.4958 meters (and I think this is the answer)

 

[Doc Al]

Your method looks good. (You may want to recheck your arithmetic, looking for round-off errors.)