A 0.106 kg meterstick is supported at its 31.1 cm mark by a string attached to the ceiling. A 0.602 kg mass hangs vertically from the 4.71 cm mark. A mass is attached somewhereon the meterstick to keep it horizontal and in both rotational and translational equilibrium. The force applied by the string attaching the meterstick to the ceiling is 23.9 N.
The acceleration of gravity is 9.81 m/s2 . the unknown mass is1.73 kg

b) Find the point where the mass attaches to
the stick. Answer in units of m.

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.