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?
I always think problems like this should be easy for me, but they rarely are. I think the key is to do a good force diagram, and then start writing equations for things that should balance:
Forces in the Y direction balance
Forces in the direction of the Rod balance
Torques around the pivot point where the rod meets the edge balance.
FE: Force exerted by edge of bowl on rod
FI: Force exerted by interior of bowl
3r: Length of rod (r is radius of sphere)
L: length of rod that extends beyond the edge of the bowl
t: short for theta, angle between rod and horizontal
P: point where end of rod meets bowl
E: point where rod meets edge of bowl
m: mass of rod
Note that angle from far edge of bowl to center of sphere to point P is 2*t
Eqn 1: mg = FE cos(t) + FI sin(2t)
balance forces in Y direction
Eqn 2: mg sin(t) = FI cos(t)
balance forces in Rod direction
Eqn 3: mg (L/3r) cos(t) (L/2) + FI sin(t) (3r-L) = mg ((3r-L)/3r) cos(t) (3r-L)/2
balance torques around point E; force * lever arm length
That's what I have so far.
Unknowns: t, L, FI, FE but only 3 equations
There is also equilibrium at point P keeping the rod from sliding up or down the bowl, I think that's where the 4th equation is.
Posted by Larry
on 2004-12-31 17:33:26