Question

A meterstick of negligible mass has a 0.20-kg mass at its 35 cm mark and a 0.40-kg mass at the 75-cm mark Where should the fulcrum be so the meterstick is balanced? (Please be clear and concise. Write legibly. Typing work is strongly suggested.)

Answer 1

About the fulcrum the net torque on the meter stick will be zero

So let say the fulcrum is at distance "x" from the origin

Torque of left mass = torque of right mass

$0.20g*(x - 35) = 0.40g*(75 - x)$

$(x - 35) =2*(75 - x)$

$3x =150 + 35 = 185$

$x = 61.67 cm$