Question

A metre rule is balanced on a knife edge at it's centre of gravity G . A mass of 40g is placed at a distance of 30cm from G and another mass x g placed at 10cm from G on the other side kept the ruler in equilibrium. What is the value of x
In

Answer 1

To keep lever in equilibrium, the moments of torque must be balanced. Moments are $M = F \cdot l$, where F is applied force, l - distance between the force and the fulcrum. $F = mg$.

$M_1=M_2 \Rightarrow m_1gl_1 = x g l_2$

$m_1l_1 = xl_2$

$x = \frac{m_1l_1}{l_2}= \frac{40*30}{10} = 120$ g.