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?
(In reply to re(2): Gosh
Hugo wrote: "Larry, I am not sure , but shouldn't the first equation be
Eqn 1: mg = FE cos(t) + FI sin(t) ?"
A very good question, and I'm not sure which way it should be. First let me explain why I had equation 1 as:
"Eqn 1: mg = FE cos(t) + FI sin(2t)
balance forces in Y direction"
The big question is: what is the direction of the force FI, the force between the interior of the bowl and the lowest point of the rod. If, in fact, this force acts perpendicular to the surface of the bowl, then force FI points from the point of contact (Point P) towards the center of the sphere.
Picture the rod starting horizontal, touching the left bowl edge and extending beyond the bowl to the right. As the left edge of the rod slides down the bowl, the rod makes angle "t" with the horizontal, but the left tip of the rod traces out an arc of "2t" as measured from the center of the sphere.
So if the force FI acts normal to the bowl surface, the component in the Y direction is FI sin(2t).
But if force FI acts along the rod longitudinally, then force FI is acting in the direction of angle t, so the vertical component would then be FI sin(t).
I'm not sure which way it should be. I vaguely remember something from freshman Physics that all forces are normal forces, except for friction. Another thought I've had is that if the answer I am getting is so far off of the intuitive answer of "about 30 degrees", either it's wrong, or maybe it's because our mathematical bowl is frictionless.
Posted by Larry
on 2005-01-03 01:05:03