Question

Two weights 0.02 kgf and 0.03 kgf suspended at points situated 0.3 m and 0.1 m, respectively, from the fulcrum at centre of gravity of a metre rod. Where from should a weight of 0.09 kg be suspended in order to make the beam horizontal​

Answer 1

Given:

Two weights 0.02 kgf and 0.03 kgf suspended at points situated 0.3 m and 0.1 m respectively, from the fulcrum at centre of gravity of a metre rod.

To find:

Position at which 0.09 kg weight should be hung so that beam stays horizontal.

Calculation:

The beam will stay horizontal when the net torque by all the forces nullify each other. In other words , the torques shall balance each other.

$\therefore \: \sum( \tau) = 0$

$= > \tau_{1} + \tau_{2} = \tau_{3}$

$= > (0.02 \times 0.3) + (0.03 \times 0.1) = (0.09 \times d)$

$= > 0.006 + 0.003 = 0.09d$

$= > 0.009 = 0.09d$

$= > d = 0.1 \: m$

So the force has to be applied at a distance of 0.1 m from the fulcrum.