The weight of the meterstick is:
[tex]W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N[/tex]
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
[tex]d_1 = 0.50 m - 0.40 m=0.10 m[/tex]
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
[tex]M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm[/tex]
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
[tex](mg) d_2 = 0.20 Nm[/tex]
from which we find the value of d2:
[tex]d_2 = \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m [/tex]
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
