How do you balance this beam?

Question

a mass of 3 kg is placed at the 90 cm mark of a .5 kg meter stick. the beam is 100 cm long and the pivot is at the 30 cm mark.

how much mass should be placed at the 20 cm mark to balance the beam?

Answer 1

Let:

P is the weight of the mass to balance the beam.

Now take moments at the pivot. The summation of moments at this point is zero. So;

P(30 -20) = 0.5(50 -30) + 3(90 - 30)

10P = 10 + 180 = 190

P = 19 kg

Answer 2

-10m + 20 x 0.5 + 60 x 3 = 0

m = 19 kg

Answer 3

Get the mass of the two sides of the beam (ie. 30% and 70% of 0.5kg, respectively). These masses act as POINT masses at their centre of mass - ie. for the 30cm side it's as if there is 0.15kg at the 15cm mark, and 0.35kg at the 35cm mark for the 70cm side. Then solve for the unknown mass such that the torques balance.

torque = distance * force

Therefore distance * force on 30cm side should equal dist * force on 70 cm side.