Question

A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?

Answer 1

The weight of the meterstick is:

`W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N`
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance

`d_1 = 0.50 m - 0.40 m=0.10 m`
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:

`M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm`

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:

`(mg) d_2 = 0.20 Nm`
from which we find the value of d2:

`d_2 = (0.20 Nm)/(mg)= (0.20 Nm)/((0.5 kg)(9.81 m/s^2))=0.04 m`

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.