Question

CLASSWORK: A uniform metre rule of mass 90g is pivoted at the 40cm mark. If the metre rule is in equilibrium with an unknown mass, m, placed at the 10cm mark and a 72g mass at the 70cm mark, calculate the value of m.

Answer 1

Position the ruler so that 0 is on the left, 100 cm is on the right.

Mass of the ruler that hangs on the left and right respectively:

$$m_l=m\cdot40/100=0.4m=36\text{ g},\\ m_r=m\cdot(100-40)/100=0.6m_r=54\text{ g}.$$

Of course, these masses create torque of the ruler, and the lever is a half of the sides around the pivot:

$$\tau_l=0.2m_lg=70.56\text{ Nm},\\ \tau_r=0.3m_rg=158.76\text{ Nm}.$$

Torque created by 72 g mass:

$$\tau_1=0.3\cdot72\cdot g=211.68\text{ Nm}.$$

Torque by the undefined mass:

$$\tau=0.3\cdot gm=2.94m.$$

The ruler is in equilibrium, therefore, torques on the left equal torques on the right:

$$70.56+2.94m=158.76+211.68,\\ m=100.6\text{ g}.$$