Answer to Question #114887 in Physics for Lawson Victor

Question

A uniform metre rule is balanced on a fulcrum placed at the 40cm mark by suspending a mass of 90g at the 5cm mark. Calculate the mass of the metre rule

Answer 1

The free part of the ruler is 60 cm long. The center of gravity (CoG) is, therefore, at the middle between 100 and 40, or 70 cm. This lever is 30 cm. The mass of this end is 0.6m.

The end with the mass is 40 cm long. Its CoG is at 20 cm mark. The mass of this end is 04m. Additionally, there is 90 g with a 35 cm lever. Equate torques on the free end and the end with the mass:

$$0.6mg\cdot30=0.4mg\cdot20+9\cdot g\cdot35,\\ \space\\ m=\frac{9\cdot35}{0.6\cdot30-0.4\cdot20}=315\text{ g}.$$