Question

2. A 0.15kg meterstick is supported at the 50cm mark. A mass of 0.5kg is attached at the 80cm mark.

a. How much mass should be attached to the 40cm mark to keep the meterstick horizontal?

b. Determine the supporting force from the fulcrum on the meterstick.

Answer 1

(a) the equilibrium condition

$$m_1gd_1=m_2gd_2$$

Hence

$$m_2=m_1\frac{d_1}{d_2}=\rm 0.5\: kg\frac{30\: cm}{10\: cm}=1.5\: kg$$

(b) the supporting force from the fulcrum on the meterstick

$$R=m_1g+m_2g=\rm (0.5+1.5)*9.8=19.6\: N$$