A 0165kg 592cmlong uniform bar has a small 0075kg mass glued

A 0.165-kg, 59.2-cm-long uniform bar has a small 0.075-kg mass glued to its left end and a small 0.150-kg mass glued to the other end. You want to balance this system horizontally on a fulcrum placed just under its center of gravity.

How far from the left end should the fulcrum be placed?

Solution

xcm = (m1x1 + m2x2 + m3x3) / ( m1 + m2 + m3)

xcm = [ ( 0.075 x 0 ) + ( 0.165 x 59.2 /2 ) + ( 0.150 x 59.2) ] / ( 0.075 + 0.165 + 0.150)

= 13.764 / 0.39

= 35.39 cm from left end.