Equilibrium. The knot is at rest, so the net force on it is zero. Three forces pull on it: the tension T₁ in the left cord, the tension T₂ in the right cord, and the weight W₃ of the centre mass. The tension in a cord over a smooth pulley is the weight of the mass hanging from its end, so T₁ = m₁g and T₂ = m₂g.

Components. Let α and β be the angles that the left and right cords make with the vertical. Each force can be split into a horizontal and a vertical component, and each set must add to zero:

horizontal: T₁ sinα = T₂ sinβ
vertical: T₁ cosα + T₂ cosβ = W₃

Triangle of forces. If three forces keep a point in equilibrium, they can be drawn head to tail as a closed triangle. Here the sides of the triangle are proportional to m₁, m₂ and m₃, so you can find the angles with the cosine rule:

cosα = (m₁² + m₃² − m₂²) / (2 m₁ m₃)
cosβ = (m₂² + m₃² − m₁²) / (2 m₂ m₃)
cosθ = (m₃² − m₁² − m₂²) / (2 m₁ m₂), θ = α + β

(You can use the sine rule too. Since only the ratios of the masses matter, you can use grams and leave out g.)

What can balance? Each mass must be smaller than the other two added together (the sides of a triangle). With the pulleys at the same height both cords must also pull upwards, so α and β must be less than 90°, which means that m₁² < m₂² + m₃² and m₂² < m₁² + m₃². If you try masses that cannot balance, the page tells you why and leaves the masses as they were.

Try: equal side masses and a heavier centre mass (the angle gets smaller as the centre mass gets bigger); an unequal pair (the heavier side mass is on the cord nearer to the vertical); 3, 4 and 5 units of mass (the angle between the two cords is 90°).

The picture is zoomed so that the knot always fits, so the pulleys move together or apart, and the masses are not drawn at the right height.