Question

A wooden bench 100cm long and a mass of 95g can be balanced from a knife edge when 5g mass is hung 10cm from one end,how far is the knife edge from the center of the edge

Answer 1

The bench is in equilibrium if

$$\tau_1 = \tau_2$$

where $\tau_1, \tau_2$ are the magnitude of the torques that rotate the bench clockwise and counterclockwise about the knife edge respectively.

By definition:

$$\tau_1 = m_1gl_1\\ \tau_2 = m_2gl_2$$

where $m_1 = 5g, m_2 = 95g, l_1 = 50cm-5cm-l_2 = 45cm-l_2$ is the distance of the 5g mass from the knife edge, $l_2$ is the distance of the knife edge from the center of the bench, and $g$ is the gravitational acceleration.

Thus, obtain:

$$5\cdot g\cdot (45-l_2) = 95\cdot g\cdot l_2\\ 5\cdot (45-l_2) = 95\cdot g\cdot l_2\\ 225 - 5l_2 = 95l_2\\ l_2 = 2.5cm$$

Answer. 2.5 cm.