Question

A horizontal meter stick supported at the 50 cm - cm mark, has a mass of 0.20 kg hanging from it at the 20 - cm mark and a 0.30 kg mass hanging from it at the 70 - cm mark. Determine the position on the meter stick at which one would hang a third mass of 0.40 kg to keep the meter stick balanced.

a. 55 cm
b. 70 cm
c. 40 cm
d. 50 cm
e. 65 cm

Answer 1

Make a drawing. Then consider that for the meter stick to be balanced it has to have zero torque around its support. Torque is given by:\[T=Fd=mgd\]where T is torque; F is force; d is distance; m is mass; and g is the acceleration of gravity. Torque that turns the meter stick counterclockwise is positive while torque that turns the meter stick clockwise is negative. All you have to do is solve this equation then:\[T _{mass 1}+T _{mass2}+T _{mass3}=0\]Hint: distance should be measured from the meter stick's support (i.e. from 50 cm).

Answer 2

Ok thanks. I got 50 cm but I wasn't sure if I did it right

Answer 3

That's the right answer. The system is already has zero torque on it before the third mass is added. Therefore, to maintain that zero torque, the mass has to be added at the meter stick's support, i.e. at 50cm.