1. A uniform meter stick is pivoted at its center. The meter stick has a M1 = 103 g mass suspended at the D1 = 28 cm position. Which is measured from the left end. At what position should a M2 = 74 g mass be suspended to put the system in equilibrium? Give your answer D2 in cm from the left end of the meter stick. 2. Because D2 is measured from the left D2= D2+.5m for the final answer. Xcm= (m1D1+m2D2)/m1+m2 3. (.103kg*.28m+ .074kg*D2)/ (.103kg+.074kg) I get d2=.389m +.5m= .889 m which is wrong. How am I getting the math wrong or am I doing the problem incorrectly.