Uniform Meterstick in Static Rotational Equilibrium

A uniform meter stick in static rotational equilibrium when a mass m₁ is suspended from the x₁ mark, a mass m₂ is suspended from the x₂ mark,
and the support stand is placed at the xₛ mark. What is the mass of the meterstick, mₘ?

Hint:
F₁ + F₂ + Fₘ + Fₛ = 0
𝛕₁ + 𝛕₂ + 𝛕ₘ + 𝛕ₛ = 0

m₁ g + m₂ g + mₘ g - N = 0
x₁ m₁ g + x₂ m₂ g + mₘ xₘ g - N xₛ = 0
xₘ is the position of the center of mass for the uniform meter stick.
xₘ = 50cm