Question

A uniform wooden metre rule of mass 90g is pivoted at the 40cm mark.if the rule is in equilibrium with an unknown mass M placed at the 10cm mark and 72g mass at the 70cm mark,determine M

Answer 1

The Principle of Moments gives

$$Mgd=m_1gd_1+m_2gd_2$$

So

$$M=\frac{m_1d_1+m_2d_2}{d}$$

$$=\frac{90\times5+72\times 30}{30}=87\:\rm{g}$$