Given a solid bowl (an idealized hemisphere, the opening facing straight up), let a slender rod rest in the bowl supported at two points: a point on the bowl's edge, and some point on the bowl's interior surface.

The bowl's edge exerts a force on the rod which is perpendicular to the rod's length.

The bowl's interior exerts, on the rod's end, a force which is perpendicular to the bowl's surface (at the point where they meet).

If the rod's length is three times the radius of the bowl, what is the angle between the rod, and the plane of the bowl's edge?

Gosh, Larry. While I am nonetheless very impressed by your solution and will not argue with the accuracy of Excel, it really bothers me that a rod can behave that way.

Sketching a cross-section of the hemisphere and assuming the radius perpendicular to the horizontal plane as an equilibrium reference line, when a third of the portion of the rod within the hemisphere intersects with the perpendicular, the rod is at 30 degrees.

I shall simply need to obtain a reasonably perfect hemisphere and uniform rod of the required length and actually test your solution. After reading your solution, I rummaged my kitchen to no avail. You'd think any good housewife would keep those things handy. Hugs and congratulations. -Cee-

Edited on January 1, 2005, 5:47 pm

Posted by CeeAnne on 2005-01-01 17:44:10