Question

A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 5.12 g coins stacked over the 24.2 cm mark, the stick is found to balance at the 27.8 cm mark. What is the mass of the meter stick

Answer 1

Answer:

The mass of the meter stick is  $1.66054054054g$

Explanation:

Here since the meter stick is in equilibrium position its net torque and net force should be equal to zero

      Since at the begging the meter stick is balanced at center of the meter stick that means its center of mass should be present at $50.0cm$

Now lets consider the later case where stick is balanced by two 5.12 g coins .

Here torque due to two coins = $t_{c} = 5.12\times(27.8-24.2)\times2\times9.8$

  Torque due to weight of meter stick =  $t_{m}$ =  $m\times(50-27.8)\times9.8$

  where m = mass of the meter stick

    Here $t_{c} = t_{m}$.

Upon equating we will be getting mass of the meter stick =$1.66054054054g$