What is the Largest Ratio R/r for Equilibrium in a Frictionless Spherical Bowl?

[ronaldor9]

Homework Statement

Three identical spheres of radius r are at rest at the bottom of a spherical bowl of radius R. If a fourth sphere is placed on top, what is the largest ratio R/r for equilibrium if there is no friction?

Homework Equations

Obviously F_net=0 since the sphere are in equilibrium, however I'm not sure how to start. Could somebody tell me how to begin?

The Attempt at a Solution

 

[magwas]

I guess by "equilibrum" they mean that the 4th sphere is supported by the three other.
There are three forces (Fa1,Fa2,Fa3) where the three sphere touches the big sphere.
There are three other forces (Fb1,Fb2,Fb3) where the three sphere support the 4th.
There are also forces between the three spheres (Fc12,Fc23,Fc13).
The vertical component of forces Fbn should be enough to support the 4th ball.
Each spheres have a weight which can be expressed by its force F0.

 

[ronaldor9]

Yeah that's as far as I was able to get. I'm having troubles decomposing each of those vectors into x- and y-components.

 

[pongo38]

It isn't just that sum forces is zero. What other sum must be zero for equilibrium? I have solved this problem using accurate engineering drawing to help me understand it, and can now see the solution can be obtained also by other means, including vectors if you must. Could this be about vector products as well as sums?

 

[jtebb]

The answer is R/r = 3

The stack sets up a tetrahedron between the centers of mass of each of the 4 balls. The weight of upper ball is split between each of the 3 supporting balls along the edges of the tetrahedron. If you traveled down one edge of the tetrahedron, you'd reach the center of the supporting ball after 2r and would need to travel another r to reach the required contact point on the bowl. Why is it the required contact point? Because the bowl has to touch at least as high as this point on the support ball (although it could touch above this point, but that would be a smaller bowl) in order to counteract and cancel the component of the weight (1/3 of mg) that is held by this third of the tetrahedron. This cancellation is required for equilibrium. Since the normal from the bowl has to be NORMAL to the bowl, it must be orthogonal to the bowl surface and hence point back toward the center of the sphere that contains the bowl. That means that the large sphere has its center at the top of the tetrahedron (the centre of mass of the upper sphere) and must have a radius of 3r. So R=3r or R/r=3.

Anyone buying this?