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13.7: Sample problems and solutions
https://phys.libretexts.org/Courses/Berea_College/Introductory_Physics%3A_Berea_College/13%3A_Simple_Harmonic_Motion/13.07%3A_Sample_problems_and_solutions
At equilibrium, the force due to gravity on one of the small spheres is therefore: $\begin{aligned} F_g=\frac{GMm}{d^2}\end{aligned}$ Assuming that, at equilibrium, the force vector is perpendicular...At equilibrium, the force due to gravity on one of the small spheres is therefore: $\begin{aligned} F_g=\frac{GMm}{d^2}\end{aligned}$ Assuming that, at equilibrium, the force vector is perpendicular to the rod, the torque from one of the large spheres is just the force multiplied by the distance to the axis of rotation.

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