To solve this problem we will apply the concepts related to translational and rotational equilibrium of bodies. As a guide we will make a free body diagram of the system that will allow us to make the sum of Forces and Torques easier. Our values are given by
[tex]M = 0.106 kg = \text{Mass of stick at 31.1 cm }[/tex]
[tex]m = 0.602 kg = \text{Mass hangs at 4.71 cm}[/tex]
[tex]T = 23.9 N = \text{The force on the string}[/tex]
[tex]m_1 = \text{Unknown mass }[/tex]
If the system is in translatary equllibrium then
[tex]\sum F = 0[/tex]
[tex]T - mg - Mg-m_1g = 0[/tex]
Rearranging for the Unknown mass we have,
[tex]m_1g= T + mg + Mg[/tex]
[tex]m_1 = \frac{T}{g}+ m + M[/tex]
[tex]m_1 = \frac{23.9}{9.8}+0.106+0.602[/tex]
[tex]m_1 = 3.1467kg[/tex]
Through this mass it is possible to perform summation of moments at the extreme left-hand point with which it will be possible to find the distance of the unknown mass.
[tex]\sum \tau = 0[/tex]
[tex](mg)(4.71cm) + (Mg)(50cm) + (m_1g)(50cm + x) - (T)(31.1 cm) = 0[/tex]
[tex](0.602*9.8)(4.71cm) + (0.602*9.8)(50cm) + (3.1467*9.8)(50cm + x) - (23.9)(31.1 cm) = 0[/tex]
[tex]x = -36.36cm[/tex]
Therefore the point where the mass attaches to the stick is
[tex]d = 50cm-36.36cm = 13.64cm[/tex] from left to right according to the diagram.
