Masses hung from meter stick, fulcrum equilibrium....?

Question

Two masses are hung from a meterstick, which has a mass of 0.248 kg. The first mass is 0.278 kg and it is suspended from the 22.9 cm position and second mass is 1.773 kg and it is suspended from the 75.6 cm position. At what position along the meterstick should you place a fulcrum so that the entire system (meterstick plus the two masses) is balanced?

Answer 1

For the stick to be balanced all torques about the fulcrum must be balanced. Set a variable x, which is the position of the fulcrum
Sum torques about the fulcrum and set equal to zero (g divides out so I won't bother with it)
Assume the fulcrum will be between the 22.9 and 75.6 cm positions
The .278 kg:
-0.278*(x-22.9)
The 1.773
1.773*(75.6-x)
The center of mass of the stick. I will assume that the fulcrum is to the right of the center of mass. The sign will prove me right or wrong
-.248*(x-50)

summing
x*(.278+.248+1.773)=
22.9*0.278+50*0.248+1.773*75.6
the fulcrum is located at 66.46 cm

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