Answer to Question #130080 in Mechanics | Relativity for Atom

Question

A uniform metre rule of mass 90g is pivotted at the 40cm mark. If the ruler is equilibrium with an unknown mass M, placed at 10cm mark, and 79g mass at the 70cm mark. Find the mass M

Answer 1

The equilibrium condition

$$\sum M_i=0$$ $$-Mgd+m_1gd_1+m_2gd_2=0$$

Hence, the unknown mass

$$M=\frac{m_1d_1+m_2d_2}{d}\\ =\frac{79\:\rm g\times 30\: cm+90\: g\times 10\: cm}{30\:\rm cm}=109\:\rm g$$

Answer: $109\:\rm g$