RBD BYU 15

Resolving forces on the top sphere ($m$):

• Vertical: \( N_1 \sin\theta = mg \)
• Horizontal: \( N_1 \cos\theta = N \)

Dividing the equations: $$ \tan\theta = \frac{mg}{N} \implies N = \frac{mg}{\tan\theta} = mg \cot\theta $$

The normal reactions from the walls on the two spheres create a couple that tries to rotate the system. This is balanced by the restoring torque from the weight.

Torque Equation : $$ N \times (2r \sin\theta) = Mg R $$ Substituting \( N = mg \cot\theta \): $$ (mg \cot\theta) (2r \sin\theta) = Mg R $$ $$ 2mg r (\frac{\cos\theta}{\sin\theta}) \sin\theta = Mg R $$ $$ 2mg r \cos\theta = Mg R $$

Rearranging to find the mass ratio:

$$ \frac{m}{M} = \frac{R}{2r \cos\theta} $$

Using the geometry where \( \cos\theta \) relates to the radii and wall width, this gives the limiting condition for equilibrium.